Essential Biochemistry, 5th Edition [5 ed.] 9781119712855, 9781119713203, 9781119718987

Table of contents :
Cover
Amino Acid Structures and Abbreviations
Title Page
Copyright
Brief Contents
Contents
Preface
Part 1: Foundations
Chapter 1: The Chemical Basis of Life
1.1 What Is Biochemistry?
1.2 Biological Molecules
Cells contain four major types of biomolecules
There are three major kinds of biological polymers
Box 1.A Units Used in Biochemistry
1.3 Energy and Metabolism
Enthalpy and entropy are components of free energy
ÄG is less than zero for a spontaneous process
Life is thermodynamically possible
1.4 The Origin of Cells
Prebiotic evolution led to cells
Box 1.B How Does Evolution Work
Eukaryotes are more complex than prokaryotes
The human body includes microorganisms
Chapter 2: Aqueous Chemistry
2.1 Water Molecules and Hydrogen Bonds
Hydrogen bonds are one type of electrostatic force
Water dissolves many compounds
Box 2.A Why Do Some Drugs Contain Fluorine
2.2 The Hydrophobic Effect
Amphiphilic molecules experience both hydrophilic interactions and the hydrophobic effect
The hydrophobic core of a lipid bilayer is a barrier to diffusion
Box 2.B Sweat, Exercise, and Sports Drinks
2.3 Acid–Base Chemistry
[H+] and [OH–] are inversely related
The pH of a solution can be altered
Box 2.C Atmospheric CO2 and Ocean Acidification
A pK value describes an acid’s tendency to ionize
The pH of a solution of acid is related to the pK
2.4 Tools and Techniques: Buffers
2.5 Clinical Connection: Acid–Base Balance in Humans
Part 2: Molecular Structure and Function
Chapter 3: Nucleic Acid Structure and Function
3.1 Nucleotides
Nucleic acids are polymers of nucleotides
Some nucleotides have other functions
3.2 Nucleic Acid Structure
DNA is a double helix
RNA is single-stranded
Nucleic acids can be denatured and renatured
3.3 The Central Dogma
Box 3.A Replication, Mitosis, Meiosis, and Mendel’s Laws
DNA must be decoded
A mutated gene can cause disease
Genes can be altered
Box 3.B Genetically Modified Organisms
3.4 Genomics
The exact number of human genes is not known
Genome size varies
Genomics has practical applications
Box 3.C Viruses
Chapter 4: Protein Structure
4.1 Amino Acids, the Building Blocks of Proteins
The 20 amino acids have different chemical properties
Box 4.A Does Chirality Matter?
Box 4.B Monosodium Glutamate
Peptide bonds link amino acids in proteins
The amino acid sequence is the first level of protein structure
4.2 Secondary Structure: The Conformation of the Peptide Group
The á helix exhibits a twisted backbone conformation
The â sheet contains multiple polypeptide strands
Proteins also contain irregular secondary structure
4.3 Tertiary Structure and Protein Stability
Proteins can be described in different ways
Globular proteins have a hydrophobic core
Protein structures are stabilized mainly by the hydrophobic effect
Box 4.C Thioester Bonds as Spring-Loaded Traps
Protein folding is a dynamic process
Box 4.D Baking and Gluten Denaturation
Disorder is a feature of many proteins
Protein functions may depend on disordered regions
4.4 Quaternary Structure
4.5 Clinical Connection: Protein Misfolding and Disease
4.6 Tools and Techniques: Analyzing Protein Structure
Chromatography takes advantage of a polypeptide’s unique properties
Mass spectrometry reveals amino acid sequences
Box 4.E Mass Spectrometry Applications
Protein structures are determined by NMR spectroscopy, X-ray crystallography, and cryo-electron microscopy
Chapter 5: Protein Function
5.1 Myoglobin and Hemoglobin: Oxygen-Binding Proteins
Oxygen binding to myoglobin depends on the oxygen concentration
Myoglobin and hemoglobin are related by evolution
Oxygen binds cooperatively to hemoglobin
A conformational shift explains hemoglobin’s cooperative behavior
Box 5.A Carbon Monoxide Poisoning
H+ ions and bisphosphoglycerate regulate oxygen binding to hemoglobin in vivo
5.2 Clinical Connection: Hemoglobin Variants
5.3 Structural Proteins
Actin filaments are most abundant
Actin filaments continuously extend and retract
Tubulin forms hollow microtubules
Keratin is an intermediate filament
Collagen is a triple helix
Box 5.B Vitamin C Deficiency Causes Scurvy
Collagen molecules are covalently cross-linked
Box 5.C Bone and Collagen Defects
5.4 Motor Proteins
Myosin has two heads and a long tail
Myosin operates through a lever mechanism
Kinesin is a microtubule-associated motor protein
Box 5.D Myosin Mutations and Deafness
Kinesin is a processive motor
5.5 Antibodies
Immunoglobulin G includes two antigen-binding sites
B lymphocytes produce diverse antibodies
Researchers take advantage of antibodies’ affinity and specificity
Chapter 6: How Enzymes Work
6.1 What Is an Enzyme?
Enzymes are usually named after the reaction they catalyze
6.2 Chemical Catalytic Mechanisms
A catalyst provides a reaction pathway with a lower activation energy barrier
Enzymes use chemical catalytic mechanisms
Box 6.A Depicting Reaction Mechanisms
The catalytic triad of chymotrypsin promotes peptide bond hydrolysis
6.3 Unique Properties of Enzyme Catalysts
Enzymes stabilize the transition state
Efficient catalysis depends on proximity and orientation effects
The active-site microenvironment promotes catalysis
6.4 Chymotrypsin in Context
Not all serine proteases are related by evolution
Enzymes with similar mechanisms exhibit different substrate specificity
Chymotrypsin is activated by proteolysis
Protease inhibitors limit protease activity
6.5 Clinical Connection: Blood Coagulation
Chapter 7: Enzyme Kinetics and Inhibition
7.1 Introduction to Enzyme Kinetics
7.2 Derivation and Meaning of the Michaelis–Menten Equation
Rate equations describe chemical processes
The Michaelis–Menten equation is a rate equation for an enzyme-catalyzed reaction
KM is the substrate concentration at which velocity is half-maximal
The catalytic constant describes how quickly an enzyme can act
kcat/KM indicates catalytic efficiency
KM and Vmax are experimentally determined
Not all enzymes fit the simple Michaelis–Menten model
7.3 Enzyme Inhibition
Some inhibitors act irreversibly
Competitive inhibition is the most common form of reversible enzyme inhibition
Transition state analogs inhibit enzymes
Other types of inhibitors affect Vmax
Box 7.A Inhibitors of HIV Protease
Allosteric enzyme regulation includes inhibition and activation
Several factors may influence enzyme activity
7.4 Clinical Connection: Drug Development
Chapter 8: Lipids and Membranes
8.1 Lipids
Fatty acids contain long hydrocarbon chains
Box 8.A Omega-3 Fatty Acids
Some lipids contain polar head groups
Lipids perform a variety of physiological functions
Box 8.B The Lipid Vitamins A, D, E, and K
8.2 The Lipid Bilayer
The bilayer is a fluid structure
Natural bilayers are asymmetric
8.3 Membrane Proteins
Integral membrane proteins span the bilayer
An á helix can cross the bilayer
A transmembrane â sheet forms a barrel
Lipid-linked proteins are anchored in the membrane
8.4 The Fluid Mosaic Model
Membrane proteins have a fixed orientation
Lipid asymmetry is maintained by enzymes
Chapter 9: Membrane Transport
9.1 The Thermodynamics of Membrane Transport
Ion movements alter membrane potential
Membrane proteins mediate transmembrane ion movement
9.2 Passive Transport
Porins are â barrel proteins
Ion channels are highly selective
Gated channels undergo conformational changes
Box 9.A Pores Can Kill
Aquaporins are water-specific pores
Some transport proteins alternate between conformations
9.3 Active Transport
The Na,K-ATPase changes conformation as it pumps ions across the membrane
ABC transporters mediate drug resistance
Secondary active transport exploits existing gradients
9.4 Membrane Fusion
SNAREs link vesicle and plasma membranes
Box 9.B Antidepressants Block Serotonin Transport
Endocytosis is the reverse of exocytosis
Autophagosomes enclose cell materials for degradation
Box 9.C Exosomes
Chapter 10: Signaling
10.1 General Features of Signaling Pathways
A ligand binds to a receptor with a characteristic affinity
Most signaling occurs through two types of receptors
The effects of signaling are limited
10.2 G Protein Signaling Pathways
G protein–coupled receptors include seven transmembrane helices
The receptor activates a G protein
The second messenger cyclic AMP activates protein kinase A
Arrestin competes with G proteins
Signaling pathways must be switched off
The phosphoinositide signaling pathway generates two second messengers
Many sensory receptors are GPCRs
Box 10.A Opioids
10.3 Receptor Tyrosine Kinases
The insulin receptor dimer changes conformation
The receptor undergoes autophosphorylation
Box 10.B Cell Signaling and Cancer
10.4 Lipid Hormone Signaling
Eicosanoids are short-range signals
Box 10.C Inhibitors of Cyclooxygenase
Chapter 11: Carbohydrates
11.1 Monosaccharides
Most carbohydrates are chiral compounds
Cyclization generates á and â anomers
Monosaccharides can be derivatized in many different ways
Box 11.A The Maillard Reaction
11.2 Polysaccharides
Lactose and sucrose are the most common disaccharides
Starch and glycogen are fuel-storage molecules
Cellulose and chitin provide structural support
Box 11.B Cellulosic Biofuel
Bacterial polysaccharides form a biofilm
11.3 Glycoproteins
Oligosaccharides are N-linked or O-linked
Oligosaccharide groups are biological markers
Box 11.C The ABO Blood Group System
Proteoglycans contain long glycosaminoglycan chains
Bacterial cell walls are made of peptidoglycan
Part 3: Metabolism
Chapter 12: Metabolism and Bioenergetics
12.1 Food and Fuel
Cells take up the products of digestion
Monomers are stored as polymers
Fuels are mobilized as needed
12.2 Metabolic Pathways
Some major metabolic pathways share a few common intermediates
Many metabolic pathways include oxidation–reduction reactions
Metabolic pathways are complex
Human metabolism depends on vitamins
Box 12.A The Transcriptome, the Proteome, and the Metabolome
Box 12.B Iron Metabolism
12.3 Free Energy Changes in Metabolic Reactions
The free energy change depends on reactant concentrations
Unfavorable reactions are coupled to favorable reactions
Energy can take different forms
Regulation occurs at the steps with the largest free energy changes
Chapter 13: Glucose Metabolism
13.1 Glycolysis
Energy is invested at the start of glycolysis
ATP is generated near the end of glycolysis
Box 13.A Catabolism of Other Sugars
Some cells convert pyruvate to lactate or ethanol
Box 13.B Alcohol Metabolism
Pyruvate is the precursor of other molecules
13.2 Gluconeogenesis
Four gluconeogenic enzymes plus some glycolytic enzymes convert pyruvate to glucose
Gluconeogenesis is regulated at the fructose bisphosphatase step
13.3 Glycogen Synthesis and Degradation
Glycogen synthesis consumes the energy of UTP
Glycogen phosphorylase catalyzes glycogenolysis
13.4 The Pentose Phosphate Pathway
The oxidative reactions of the pentose phosphate pathway produce NADPH
Isomerization and interconversion reactions generate a variety of monosaccharides
A summary of glucose metabolism
13.5 Clinical Connection: Disorders of Carbohydrate Metabolism
Glycogen storage diseases affect liver and muscle
Chapter 14: The Citric Acid Cycle
14.1 The Pyruvate Dehydrogenase Reaction
The pyruvate dehydrogenase complex contains multiple copies of three different enzymes
Pyruvate dehydrogenase converts pyruvate to acetyl-CoA
14.2 The Eight Reactions of the Citric Acid Cycle
1. Citrate synthase adds an acetyl group to oxaloacetate
2. Aconitase isomerizes citrate to isocitrate
3. Isocitrate dehydrogenase releases the first CO2
4. á-Ketoglutarate dehydrogenase releases the second CO2
5. Succinyl-CoA synthetase catalyzes substrate-level phosphorylation
6. Succinate dehydrogenase generates ubiquinol
7. Fumarase catalyzes a hydration reaction
8. Malate dehydrogenase regenerates oxaloacetate
14.3 Thermodynamics of the Citric Acid Cycle
The citric acid cycle is an energy-generating catalytic cycle
The citric acid cycle is regulated at three steps
Box 14.A Mutations in Citric Acid Cycle Enzymes
The citric acid cycle probably evolved as a synthetic pathway
14.4 Anabolic and Catabolic Functions of the Citric Acid Cycle
Citric acid cycle intermediates are precursors of other molecules
Anaplerotic reactions replenish citric acid cycle intermediates
Box 14.B The Glyoxylate Pathway
Chapter 15: Oxidative Phosphorylation
15.1 The Thermodynamics of Oxidation–Reduction Reactions
Reduction potential indicates a substance’s tendency to accept electrons
The free energy change can be calculated from the change in reduction potential
15.2 Mitochondrial Electron Transport
Mitochondrial membranes define two compartments
Complex I transfers electrons from NADH to ubiquinone
Other oxidation reactions contribute to the ubiquinol pool
Complex III transfers electrons from ubiquinol to cytochrome c
Complex IV oxidizes cytochrome c and reduces O2
Respiratory complexes associate with each other
Box 15.A Reactive Oxygen Species
15.3 Chemiosmosis
Chemiosmosis links electron transport and oxidative phosphorylation
The proton gradient is an electrochemical gradient
15.4 ATP Synthase
Proton translocation rotates the c ring of ATP synthase
The binding change mechanism explains how ATP is made
The P:O ratio describes the stoichiometry of oxidative phosphorylation
Box 15.B Uncoupling Agents Prevent ATP Synthesis
The rate of oxidative phosphorylation reflects the need for ATP
Box 15.C Powering Human Muscles
Chapter 16: Photosynthesis
16.1 Chloroplasts and Solar Energy
Pigments absorb light of different wavelengths
Light-harvesting complexes transfer energy to the reaction center
16.2 The Light Reactions
Photosystem II is a light-activated oxidation–reduction enzyme
The oxygen-evolving complex of Photosystem II oxidizes water
Cytochrome b6f links Photosystems I and II
A second photooxidation occurs at Photosystem I
Chemiosmosis provides the free energy for ATP synthesis
16.3 Carbon Fixation
Rubisco catalyzes CO2 fixation
The Calvin cycle rearranges sugar molecules
Box 16.A The C4 Pathway
The availability of light regulates carbon fixation
Calvin cycle products are used to synthesize sucrose and starch
Chapter 17: Lipid Metabolism
17.1 Lipid Transport
17.2 Fatty Acid Oxidation
Fatty acids are activated before they are degraded
Each round of â oxidation has four reactions
Degradation of unsaturated fatty acids requires isomerization and reduction
Oxidation of odd-chain fatty acids yields propionyl-CoA
Some fatty acid oxidation occurs in peroxisomes
17.3 Fatty Acid Synthesis
Acetyl-CoA carboxylase catalyzes the first step of fatty acid synthesis
Fatty acid synthase catalyzes seven reactions
Other enzymes elongate and desaturate newly synthesized fatty acids
Box 17.A Fats, Diet, and Heart Disease
Fatty acid synthesis can be activated and inhibited
Box 17.B Inhibitors of Fatty Acid Synthesis
Acetyl-CoA can be converted to ketone bodies
17.4 Synthesis of Other Lipids
Triacylglycerols and phospholipids are built from acyl-CoA groups
Cholesterol synthesis begins with acetyl-CoA
A summary of lipid metabolism
Chapter 18: Nitrogen Metabolism
18.1 Nitrogen Fixation and Assimilation
Nitrogenase converts N2 to NH3
Ammonia is assimilated by glutamine synthetase and glutamate synthase
Transamination moves amino groups between compounds
Box 18.A Transaminases in the Clinic
18.2 Amino Acid Biosynthesis
Several amino acids are easily synthesized from common metabolites
Amino acids with sulfur, branched chains, or aromatic groups are more difficult to synthesize
Box 18.B Homocysteine, Methionine, and One-Carbon Chemistry
Box 18.C Glyphosate, the Most Popular Herbicide
Amino acids are the precursors of some signaling molecules
Box 18.D Nitric Oxide
18.3 Amino Acid Catabolism
Amino acids are glucogenic, ketogenic, or both
Box 18.E Diseases of Amino Acid Metabolism
18.4 Nitrogen Disposal: The Urea Cycle
Glutamate supplies nitrogen to the urea cycle
The urea cycle consists of four reactions
18.5 Nucleotide Metabolism
Purine nucleotide synthesis yields IMP and then AMP and GMP
Pyrimidine nucleotide synthesis yields UTP and CTP
Ribonucleotide reductase converts ribonucleotides to deoxyribonucleotides
Thymidine nucleotides are produced by methylation
Nucleotide degradation produces urate or amino acids
Chapter 19: Regulation of Mammalian Fuel Metabolism
19.1 Integration of Fuel Metabolism
Organs are specialized for different functions
Metabolites travel between organs
Box 19.A The Intestinal Microbiota Contribute to Metabolism
19.2 Hormonal Control of Fuel Metabolism
Insulin is released in response to glucose
Insulin promotes fuel use and storage
mTOR responds to insulin signaling
Glucagon and epinephrine trigger fuel mobilization
Additional hormones influence fuel metabolism
AMP-dependent protein kinase acts as a fuel sensor
Fuel metabolism is also controlled by redox balance and oxygen
19.3 Disorders of Fuel Metabolism
The body generates glucose and ketone bodies during starvation
Box 19.B Marasmus and Kwashiorkor
Obesity has multiple causes
Diabetes is characterized by hyperglycemia
Obesity, diabetes, and cardiovascular disease are linked
19.4 Clinical Connection: Cancer Metabolism
Aerobic glycolysis supports biosynthesis
Cancer cells consume large amounts of glutamine
Part 4: Genetic Information
Chapter 20: DNA Replication and Repair
20.1 The DNA Replication Machinery
Replication occurs in factories
Helicases convert double-stranded DNA to single-stranded DNA
DNA polymerase faces two problems
DNA polymerases share a common structure and mechanism
DNA polymerase proofreads newly synthesized DNA
An RNase and a ligase are required to complete the lagging strand
20.2 Telomeres
Telomerase extends chromosomes
Box 20.A HIV Reverse Transcriptase
Is telomerase activity linked to cell immortality
20.3 DNA Damage and Repair
DNA damage is unavoidable
Repair enzymes restore some types of damaged DNA
Base excision repair corrects the most frequent DNA lesions
Nucleotide excision repair targets the second most common form of DNA damage
Double-strand breaks can be repaired by joining the ends
Recombination also restores broken DNA molecules
Box 20.B Gene Editing with CRISPR
20.4 Clinical Connection: Cancer as a Genetic Disease
Tumor growth depends on multiple events
DNA repair pathways are closely linked to cancer
20.5 DNA Packaging
DNA is negatively supercoiled
Topoisomerases alter DNA supercoiling
Eukaryotic DNA is packaged in nucleosomes
20.6 Tools and Techniques: Manipulating DNA
Cutting and pasting generates recombinant
DNA
The polymerase chain reaction amplifies DNA
DNA sequencing uses DNA polymerase to make a complementary strand
Chapter 21: Transcription and RNA
21.1 Initiating Transcription
What is a gene?
DNA packaging affects transcription
DNA also undergoes covalent modification
Transcription begins at promoters
Transcription factors recognize eukaryotic promoters
Mediator integrates multiple regulatory signals
Box 21.A DNA-Binding Proteins
Prokaryotic operons allow coordinated gene expression
21.2 RNA Polymerase
RNA polymerases have a common structure and mechanism
Box 21.B RNA-Dependent RNA Polymerase
RNA polymerase is a processive enzyme
Transcription elongation requires changes in RNA polymerase
Transcription is terminated in several ways
21.3 RNA Processing
Eukaryotic mRNAs receive a 5' cap and a 3' poly(A) tail
Splicing removes introns from eukaryotic RNA
mRNA turnover and RNA interference limit gene expression
Box 21.C The Nuclear Pore Complex
rRNA and tRNA processing includes the addition, deletion, and modification of nucleotides
RNAs have extensive secondary structure
Chapter 22: Protein Synthesis
22.1 tRNA and the Genetic Code
The genetic code is redundant
tRNAs have a common structure
tRNA aminoacylation consumes ATP
Editing increases the accuracy of aminoacylation
tRNA anticodons pair with mRNA codons
Box 22.A The Genetic Code Expanded
22.2 Ribosome Structure
The ribosome is mostly RNA
Three tRNAs and one mRNA bind to the ribosome
22.3 Translation
Initiation requires an initiator tRNA
The appropriate tRNAs are delivered to the ribosome during elongation
The peptidyl transferase active site catalyzes peptide bond formation
Box 22.B Antibiotic Inhibitors of Protein Synthesis
Release factors mediate translation termination
Translation is efficient and dynamic
22.4 Post-Translational
Events
Chaperones promote protein folding
The signal recognition particle targets some proteins for membrane translocation
Many proteins undergo covalent modification
Glossary
Odd-Numbered Solutions
Index
Nucleic Acid Bases, Nucleosides, and Nucleotides
Common Functional Groups and Linkages in Biochemistry and Useful Constants and Key Equations
EULA

Citation preview

Amino Acid Structures and Abbreviations Hydrophobic amino acids COO– CH3

COO– H

C

CH3

H

NH+ 3

H

C

CH2

NH+ 3

CH CH3

H

CH

H

CH

C

Leucine (Leu, L)

H

CH2

CH3

H

C

CH2

CH2

CH2

N H

Tryptophan (Trp, W)

COO– CH2 H C CH2 H2N+ CH2

CH3

S

NH+ 3

Proline (Pro, P)

Methionine (Met, M)

Isoleucine (Ile, I)

C

NH+ 3

COO–

NH+ 3

CH3

CH2

Phenylalanine (Phe, F)

COO– CH3

CH3

C

COO–

NH+ 3

Valine (Val, V)

Alanine (Ala, A)

COO–

C NH+ 3

COO–

Polar amino acids COO– H

C

CH2

OH

H

NH+ 3

H

C

CH2

OH

H

NH+ 3

NH2

H

NH+ 3

C

CH2

OH

H

COO–

O CH2

C

NH2 H

C

CH2

NH+ 3

CH2

SH

Cysteine (Cys, C)

COO–

N

H

N H

C

H

NH+ 3

Glycine (Gly, G)

Histidine (His, H)

Glutamine (Gln, Q)

C

NH+ 3

Tyrosine (Tyr, Y)

NH+ 3

Asparagine (Asn, N)

CH2

NH+ 3

COO–

O C

C

Threonine (Thr, T)

Serine (Ser, S)

COO–

CH

C

COO–

COO–

COO– CH3

Charged amino acids COO– H

C

CH2

COO–

O C

–

O

NH+ 3

Aspartate (Asp, D)

H

C

CH2

CH2

C

NH+ 3

Glutamate (Glu, E)

COO–

COO–

O O

–

H

C

CH2

CH2

CH2

NH+ 3

CH2

NH+ 3

H

C

CH2

NH2 CH2

CH2

NH

NH+ 3

Lysine (Lys, K)

Arginine (Arg, R)

C

NH+ 2

Essential Biochemistry Fifth Edition

CH A RLOT TE W. PR AT T Seattle Pacific University

K ATHLEEN CORNELY Providence College

SENIOR DESIGNER Thomas Nery SENIOR MANAGER, COURSE DEVELOPMENT AND PRODUCTION Svetlana Barskaya COURSE CONTENT DEVELOPERS  Corrina Santos and Andrew Moore SENIOR COURSE PRODUCTION OPERATIONS SPECIALIST Patricia Gutierrez ASSOCIATE PUBLISHER Sladjana Bruno SENIOR EDITOR Jennifer Yee EDITORIAL ASSISTANT Samantha Hart MARKETING MANAGER Michael Olsen COVER IMAGE © shunyufan/E+/Getty Images This book was typeset in 9.5/12 STIX Two Text Regular at Lumina Datamatics Ltd. Copyright © 2021 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. ISBN 9781119712855 Library of Congress Cataloging-in-Publication Data Names: Pratt, Charlotte W., author. | Cornely, Kathleen, author. Title: Essential biochemistry / Charlotte W. Pratt, Seattle Pacific University, Kathleen Cornely, Providence College. Description: Fifth edition. | Hoboken : Wiley, 2021. | Includes index. Identifiers: LCCN 2020055734 (print) | LCCN 2020055735 (ebook) | ISBN 9781119713203 (paperback) | ISBN 9781119718987 (adobe pdf) | ISBN 9781119712855 (epub) Subjects: LCSH: Biochemistry. Classification: LCC QD415 .P67 2021 (print) | LCC QD415 (ebook) | DDC 572--dc23 LC record available at https://lccn.loc.gov/2020055734 LC ebook record available at https://lccn.loc.gov/2020055735 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

About the Authors C H A R LO T T E P R AT T received a B.S. in biology from the University of Notre Dame and a Ph.D. in biochemistry from Duke University. She is a protein chemist who has conducted research in blood coagulation and inflammation at the University of North Carolina at Chapel Hill. She is currently Associate Professor in the Biology Department at Seattle ­Pacific ­University. Her interests include molecular evolution, enzyme action, and the relationship between metabolic processes and disease. She has written numerous research and review ­articles, has worked as a textbook editor, and is a co-author, with Donald Voet and Judith G. Voet, of Fundamentals of Biochemistry, published by John Wiley & Sons, Inc. K AT H LE E N CO R N E LY holds a B.S. in chemistry from Bowling Green (Ohio) State University, an M.S. in biochemistry from Indiana University, and a Ph.D. in nutritional b ­ iochemistry from Cornell University. She is currently the Robert H. Walsh ’39 Endowed Professor in Chemistry and Biochemistry at Providence College, where she has focused on expanding the use of case studies and guided inquiry across a broad spectrum of classes. Her interest in active pedagogy has led to her involvement in national programs including Project ­Kaleidoscope, the POGIL Project, and the Howard Hughes Medical Institute SEA PHAGES program, which has also fueled her current experimental research in phage genomics. She has been a member of the editorial board of Biochemistry and Molecular Biology Education and has served for several years as coordinator of the undergraduate poster competition at the annual meeting of the American Society for Biochemistry and Molecular Biology.

v

Brief Contents PREFACE

Part 1

xiv

Foundations

Part 3

Metabolism

12

Metabolism and Bioenergetics 337

1

The Chemical Basis of Life 1

13

Glucose Metabolism 366

2

Aqueous Chemistry 27

14

The Citric Acid Cycle 403

15 Oxidative Phosphorylation 428 Part 2

Molecular Structure and Function

3

Nucleic Acid Structure and Function 57

4

Protein Structure 86

5

Protein Function 125

6

How Enzymes Work 167

7

Enzyme Kinetics and Inhibition 198

8

Lipids and Membranes 234

9

Membrane Transport 258

10

Signaling 287

11

Carbohydrates 315

vi

16 Photosynthesis 458 17

Lipid Metabolism 483

18

Nitrogen Metabolism 518

19 Regulation of Mammalian Fuel Metabolism 555 Part 4

Genetic Information

20 DNA Replication and Repair 582 21

Transcription and RNA 627

22

Protein Synthesis 663

Contents PREFACE

Part 1

xiv

Part 2

Foundations

1 The Chemical Basis of Life

1

1.1 1.2

What Is Biochemistry? 1 Biological Molecules 3 Cells contain four major types of biomolecules 3 There are three major kinds of biological polymers 6 Box 1.A Units Used in Biochemistry 7 1.3 Energy and Metabolism 10 Enthalpy and entropy are components of free energy 11 ∆G is less than zero for a spontaneous process 12 Life is thermodynamically possible 12 1.4 The Origin of Cells 14 Prebiotic evolution led to cells 15 Box 1.B How Does Evolution Work? 17 Eukaryotes are more complex than prokaryotes 17 The human body includes microorganisms 19

2 Aqueous Chemistry 2.1

27

Water Molecules and Hydrogen Bonds 27 Hydrogen bonds are one type of electrostatic force 29 Water dissolves many compounds 31 Box 2.A Why Do Some Drugs Contain Fluorine? 31 2.2 The Hydrophobic Effect 33 Amphiphilic molecules experience both hydrophilic interactions and the hydrophobic effect 35 The hydrophobic core of a lipid bilayer is a barrier to diffusion 35 Box 2.B Sweat, Exercise, and Sports Drinks 36 2.3 Acid–Base Chemistry 37 [H+] and [OH–] are inversely related 38 The pH of a solution can be altered 39 Box 2.C Atmospheric CO2 and Ocean Acidification 39 A pK value describes an acid’s tendency to ionize 40 The pH of a solution of acid is related to the pK 41 2.4 Tools and Techniques: Buffers 44 2.5 Clinical Connection: Acid–Base Balance in Humans 46

Molecular Structure and Function

3 Nucleic Acid Structure and Function

57

3.1

Nucleotides 57 Nucleic acids are polymers of nucleotides 58 Some nucleotides have other functions 60 3.2 Nucleic Acid Structure 61 DNA is a double helix 62 RNA is single-stranded 64 Nucleic acids can be denatured and renatured 64 3.3 The Central Dogma 67 Box 3.A Replication, Mitosis, Meiosis, and Mendel’s Laws 67 DNA must be decoded 70 A mutated gene can cause disease 71 Genes can be altered 72 Box 3.B Genetically Modified Organisms 73 3.4 Genomics 74 The exact number of human genes is not known 75 Genome size varies 75 Genomics has practical applications 77 Box 3.C Viruses 78

4 Protein Structure

86

4.1

Amino Acids, the Building Blocks of Proteins 86 The 20 amino acids have different chemical properties 88 Box 4.A Does Chirality Matter? 89 Box 4.B Monosodium Glutamate 91 Peptide bonds link amino acids in proteins 91 The amino acid sequence is the first level of protein structure 94 4.2 Secondary Structure: The Conformation of the Peptide Group 95 The α helix exhibits a twisted backbone conformation 96 The β sheet contains multiple polypeptide strands 96 Proteins also contain irregular secondary structure 98

vii

viii  CON TENTS

4.3

Tertiary Structure and Protein Stability 99 Proteins can be described in different ways 99 Globular proteins have a hydrophobic core 100 Protein structures are stabilized mainly by the hydrophobic effect 101 Box 4.C Thioester Bonds as Spring-Loaded Traps 103 Protein folding is a dynamic process 103 Box 4.D Baking and Gluten Denaturation 104 Disorder is a feature of many proteins 105 Protein functions may depend on disordered regions 106 4.4 Quaternary Structure 107 4.5 Clinical Connection: Protein Misfolding and Disease 109 4.6 Tools and Techniques: Analyzing Protein Structure 111 Chromatography takes advantage of a polypeptide’s unique properties 111 Mass spectrometry reveals amino acid sequences 114 Box 4.E Mass Spectrometry Applications 116 Protein structures are determined by NMR spectroscopy, X-ray crystallography, and cryo-electron microscopy 116

5 Protein Function 5.1

125

Myoglobin and Hemoglobin: Oxygen-Binding Proteins 126 Oxygen binding to myoglobin depends on the oxygen concentration 127 Myoglobin and hemoglobin are related by evolution 128 Oxygen binds cooperatively to hemoglobin 129 A conformational shift explains hemoglobin’s cooperative behavior 130 Box 5.A Carbon Monoxide Poisoning 130 H+ ions and bisphosphoglycerate regulate oxygen binding to hemoglobin in vivo 132 5.2 Clinical Connection: Hemoglobin Variants 134 5.3 Structural Proteins 136 Actin filaments are most abundant 137 Actin filaments continuously extend and retract 138 Tubulin forms hollow microtubules 139 Keratin is an intermediate filament 142 Collagen is a triple helix 144 Box 5.B Vitamin C Deficiency Causes Scurvy 144 Collagen molecules are covalently cross-linked 145 Box 5.C Bone and Collagen Defects 147 5.4 Motor Proteins 148 Myosin has two heads and a long tail 148 Myosin operates through a lever mechanism 150

Kinesin is a microtubule-associated motor protein 151 Box 5.D Myosin Mutations and Deafness 151 Kinesin is a processive motor 152 5.5 Antibodies 154 Immunoglobulin G includes two antigen-binding sites 154 B lymphocytes produce diverse antibodies 156 Researchers take advantage of antibodies’ affinity and specificity 157

6 How Enzymes Work

167

6.1

What Is an Enzyme? 167 Enzymes are usually named after the reaction they catalyze 170 6.2 Chemical Catalytic Mechanisms 171 A catalyst provides a reaction pathway with a lower activation energy barrier 173 Enzymes use chemical catalytic mechanisms 173 Box 6.A Depicting Reaction Mechanisms 175 The catalytic triad of chymotrypsin promotes peptide bond hydrolysis 177 6.3 Unique Properties of Enzyme Catalysts 180 Enzymes stabilize the transition state 180 Efficient catalysis depends on proximity and orientation effects 181 The active-site microenvironment promotes catalysis 182 6.4 Chymotrypsin in Context 183 Not all serine proteases are related by evolution 183 Enzymes with similar mechanisms exhibit different substrate specificity 184 Chymotrypsin is activated by proteolysis 185 Protease inhibitors limit protease activity 186 6.5 Clinical Connection: Blood Coagulation 187

7 Enzyme Kinetics and Inhibition 7.1 7.2

198

Introduction to Enzyme Kinetics 198 Derivation and Meaning of the Michaelis–Menten Equation 201 Rate equations describe chemical processes 201 The Michaelis–Menten equation is a rate equation for an enzyme-catalyzed reaction 202 KM is the substrate concentration at which velocity is half-maximal 204 The catalytic constant describes how quickly an enzyme can act 204 kcat/KM indicates catalytic efficiency 205

CO NTENTS   ix

KM and Vmax are experimentally determined 205 Not all enzymes fit the simple Michaelis–Menten model 207 7.3 Enzyme Inhibition 209 Some inhibitors act irreversibly 209 Competitive inhibition is the most common form of reversible enzyme inhibition 210 Transition state analogs inhibit enzymes 212 Other types of inhibitors affect Vmax 213 Box 7.A Inhibitors of HIV Protease 214 Allosteric enzyme regulation includes inhibition and activation 216 Several factors may influence enzyme activity 219 7.4 Clinical Connection: Drug Development 219

8 Lipids and Membranes

234

8.1

Lipids 234 Fatty acids contain long hydrocarbon chains 235 Box 8.A Omega-3 Fatty Acids 236 Some lipids contain polar head groups 237 Lipids perform a variety of physiological functions 239 Box 8.B The Lipid Vitamins A, D, E, and K 240 8.2 The Lipid Bilayer 241 The bilayer is a fluid structure 242 Natural bilayers are asymmetric 243 8.3 Membrane Proteins 244 Integral membrane proteins span the bilayer 245 An α helix can cross the bilayer 245 A transmembrane β sheet forms a barrel 246 Lipid-linked proteins are anchored in the membrane 246 8.4 The Fluid Mosaic Model 248 Membrane proteins have a fixed orientation 249 Lipid asymmetry is maintained by enzymes 250

9 Membrane Transport 9.1

258

The Thermodynamics of Membrane Transport 258 Ion movements alter membrane potential 259 Membrane proteins mediate transmembrane ion movement 260 9.2 Passive Transport 263 Porins are β barrel proteins 263 Ion channels are highly selective 264 Gated channels undergo conformational changes 265 Box 9.A Pores Can Kill 265 Aquaporins are water-specific pores 266

Some transport proteins alternate between conformations 268 9.3 Active Transport 269 The Na,K-ATPase changes conformation as it pumps ions across the membrane 269 ABC transporters mediate drug resistance 271 Secondary active transport exploits existing gradients 271 9.4 Membrane Fusion 272 SNAREs link vesicle and plasma membranes 273 Box 9.B Antidepressants Block Serotonin Transport 275 Endocytosis is the reverse of exocytosis 276 Autophagosomes enclose cell materials for degradation 277 Box 9.C Exosomes 278

10 Signaling

287

10.1 General Features of Signaling Pathways 287 A ligand binds to a receptor with a characteristic affinity 288 Most signaling occurs through two types of receptors 289 The effects of signaling are limited 290 10.2 G Protein Signaling Pathways 291 G protein–coupled receptors include seven transmembrane helices 292 The receptor activates a G protein 293 The second messenger cyclic AMP activates protein kinase A 294 Arrestin competes with G proteins 296 Signaling pathways must be switched off 296 The phosphoinositide signaling pathway generates two second messengers 297 Many sensory receptors are GPCRs 298 Box 10.A Opioids 299 10.3 Receptor Tyrosine Kinases 300 The insulin receptor dimer changes conformation 300 The receptor undergoes autophosphorylation 302 Box 10.B Cell Signaling and Cancer 303 10.4 Lipid Hormone Signaling 303 Eicosanoids are short-range signals 305 Box 10.C Inhibitors of Cyclooxygenase 306

11 Carbohydrates

315

11.1 Monosaccharides 315 Most carbohydrates are chiral compounds 316 Cyclization generates α and β anomers 317 Monosaccharides can be derivatized in many different ways 318

x  CON TEN TS

Box 11.A The Maillard Reaction 319 11.2 Polysaccharides 320 Lactose and sucrose are the most common disaccharides 321 Starch and glycogen are fuel-storage molecules 321 Cellulose and chitin provide structural support 322 Box 11.B Cellulosic Biofuel 323 Bacterial polysaccharides form a biofilm 324 11.3 Glycoproteins 325 Oligosaccharides are N-linked or O-linked 325 Oligosaccharide groups are biological markers 326 Box 11.C The ABO Blood Group System 327 Proteoglycans contain long glycosaminoglycan chains 327 Bacterial cell walls are made of peptidoglycan 328 Part 3

Metabolism

12 Metabolism and Bioenergetics

337

12.1 Food and Fuel 337 Cells take up the products of digestion 338 Monomers are stored as polymers 339 Fuels are mobilized as needed 340 12.2 Metabolic Pathways 343 Some major metabolic pathways share a few common intermediates 343 Many metabolic pathways include oxidation–reduction reactions 344 Metabolic pathways are complex 346 Human metabolism depends on vitamins 347 Box 12.A The Transcriptome, the Proteome, and the Metabolome 348 Box 12.B Iron Metabolism 351 12.3 Free Energy Changes in Metabolic Reactions 352 The free energy change depends on reactant concentrations 352 Unfavorable reactions are coupled to favorable reactions 354 Energy can take different forms 356 Regulation occurs at the steps with the largest free energy changes 357

13 Glucose Metabolism

366

13.1 Glycolysis 366 Energy is invested at the start of glycolysis 367 ATP is generated near the end of glycolysis 373 Box 13.A Catabolism of Other Sugars 378 Some cells convert pyruvate to lactate or ethanol 379

Box 13.B Alcohol Metabolism 380 Pyruvate is the precursor of other molecules 381 13.2 Gluconeogenesis 383 Four gluconeogenic enzymes plus some glycolytic enzymes convert pyruvate to glucose 383 Gluconeogenesis is regulated at the fructose bisphosphatase step 385 13.3 Glycogen Synthesis and Degradation 386 Glycogen synthesis consumes the energy of UTP 386 Glycogen phosphorylase catalyzes glycogenolysis 388 13.4 The Pentose Phosphate Pathway 389 The oxidative reactions of the pentose phosphate pathway produce NADPH 389 Isomerization and interconversion reactions generate a variety of monosaccharides 390 A summary of glucose metabolism 392 13.5 Clinical Connection: Disorders of Carbohydrate Metabolism 393 Glycogen storage diseases affect liver and muscle 394

14 The Citric Acid Cycle

403

14.1 The Pyruvate Dehydrogenase Reaction 403 The pyruvate dehydrogenase complex contains multiple copies of three different enzymes 404 Pyruvate dehydrogenase converts pyruvate to acetyl-CoA 404 14.2 The Eight Reactions of the Citric Acid Cycle 406 1. Citrate synthase adds an acetyl group to oxaloacetate 407 2. Aconitase isomerizes citrate to isocitrate 409 3. Isocitrate dehydrogenase releases the first CO2 410 4. α-Ketoglutarate dehydrogenase releases the second CO2 410 5. Succinyl-CoA synthetase catalyzes substrate-level phosphorylation 411 6. Succinate dehydrogenase generates ubiquinol 412 7. Fumarase catalyzes a hydration reaction 412 8. Malate dehydrogenase regenerates oxaloacetate 412 14.3 Thermodynamics of the Citric Acid Cycle 413 The citric acid cycle is an energy-generating catalytic cycle 413 The citric acid cycle is regulated at three steps 414 Box 14.A Mutations in Citric Acid Cycle Enzymes 415 The citric acid cycle probably evolved as a synthetic pathway 415

CO NTENTS   xi

14.4 Anabolic and Catabolic Functions of the Citric Acid Cycle 416 Citric acid cycle intermediates are precursors of other molecules 416 Anaplerotic reactions replenish citric acid cycle intermediates 418 Box 14.B The Glyoxylate Pathway 419

15 Oxidative Phosphorylation

428

15.1 The Thermodynamics of Oxidation–Reduction Reactions 428 Reduction potential indicates a substance’s tendency to accept electrons 429 The free energy change can be calculated from the change in reduction potential 431 15.2 Mitochondrial Electron Transport 432 Mitochondrial membranes define two compartments 433 Complex I transfers electrons from NADH to ubiquinone 434 Other oxidation reactions contribute to the ubiquinol pool 436 Complex III transfers electrons from ubiquinol to cytochrome c 437 Complex IV oxidizes cytochrome c and reduces O2 439 Respiratory complexes associate with each other 441 Box 15.A Reactive Oxygen Species 442 15.3 Chemiosmosis 443 Chemiosmosis links electron transport and oxidative phosphorylation 443 The proton gradient is an electrochemical gradient 443 15.4 ATP Synthase 445 Proton translocation rotates the c ring of ATP synthase 445 The binding change mechanism explains how ATP is made 447 The P:O ratio describes the stoichiometry of oxidative phosphorylation 447 Box 15.B Uncoupling Agents Prevent ATP Synthesis 448 The rate of oxidative phosphorylation reflects the need for ATP 448 Box 15.C Powering Human Muscles 449

16 Photosynthesis

458

16.1 Chloroplasts and Solar Energy 458 Pigments absorb light of different wavelengths 459 Light-harvesting complexes transfer energy to the reaction center 461

16.2 The Light Reactions 463 Photosystem II is a light-activated oxidation–reduction enzyme 463 The oxygen-evolving complex of Photosystem II oxidizes water 464 Cytochrome b6f links Photosystems I and II 466 A second photooxidation occurs at Photosystem I 467 Chemiosmosis provides the free energy for ATP synthesis 469 16.3 Carbon Fixation 471 Rubisco catalyzes CO2 fixation 471 The Calvin cycle rearranges sugar molecules 472 Box 16.A The C4 Pathway 473 The availability of light regulates carbon fixation 475 Calvin cycle products are used to synthesize sucrose and starch 476

17 Lipid Metabolism

483

17.1 Lipid Transport 483 17.2 Fatty Acid Oxidation 486 Fatty acids are activated before they are degraded 487 Each round of β oxidation has four reactions 488 Degradation of unsaturated fatty acids requires isomerization and reduction 491 Oxidation of odd-chain fatty acids yields propionyl-CoA 492 Some fatty acid oxidation occurs in peroxisomes 494 17.3 Fatty Acid Synthesis 495 Acetyl-CoA carboxylase catalyzes the first step of fatty acid synthesis 496 Fatty acid synthase catalyzes seven reactions 497 Other enzymes elongate and desaturate newly synthesized fatty acids 500 Box 17.A Fats, Diet, and Heart Disease 500 Fatty acid synthesis can be activated and inhibited 501 Box 17.B Inhibitors of Fatty Acid Synthesis 502 Acetyl-CoA can be converted to ketone bodies 503 17.4 Synthesis of Other Lipids 505 Triacylglycerols and phospholipids are built from acyl-CoA groups 505 Cholesterol synthesis begins with acetyl-CoA 508 A summary of lipid metabolism 510

18 Nitrogen Metabolism

518

18.1 Nitrogen Fixation and Assimilation 518 Nitrogenase converts N2 to NH3 519

xii  CON TENTS

Ammonia is assimilated by glutamine synthetase and glutamate synthase 519 Transamination moves amino groups between compounds 521 Box 18.A Transaminases in the Clinic 523 18.2 Amino Acid Biosynthesis 523 Several amino acids are easily synthesized from common metabolites 524 Amino acids with sulfur, branched chains, or aromatic groups are more difficult to synthesize 526 Box 18.B Homocysteine, Methionine, and One-Carbon Chemistry 527 Box 18.C Glyphosate, the Most Popular Herbicide 528 Amino acids are the precursors of some signaling molecules 530 Box 18.D Nitric Oxide 531 18.3 Amino Acid Catabolism 532 Amino acids are glucogenic, ketogenic, or both 532 Box 18.E Diseases of Amino Acid Metabolism 535 18.4 Nitrogen Disposal: The Urea Cycle 536 Glutamate supplies nitrogen to the urea cycle 537 The urea cycle consists of four reactions 538 18.5 Nucleotide Metabolism 540 Purine nucleotide synthesis yields IMP and then AMP and GMP 541 Pyrimidine nucleotide synthesis yields UTP and CTP 542 Ribonucleotide reductase converts ribonucleotides to deoxyribonucleotides 543 Thymidine nucleotides are produced by methylation 544 Nucleotide degradation produces urate or amino acids 545

19 Regulation of Mammalian Fuel Metabolism

555

19.1 Integration of Fuel Metabolism 555 Organs are specialized for different functions 556 Metabolites travel between organs 557 Box 19.A The Intestinal Microbiota Contribute to Metabolism 558 19.2 Hormonal Control of Fuel Metabolism 560 Insulin is released in response to glucose 560 Insulin promotes fuel use and storage 561 mTOR responds to insulin signaling 563 Glucagon and epinephrine trigger fuel mobilization 564 Additional hormones influence fuel metabolism 565 AMP-dependent protein kinase acts as a fuel sensor 566

Fuel metabolism is also controlled by redox balance and oxygen 566 19.3 Disorders of Fuel Metabolism 568 The body generates glucose and ketone bodies during starvation 568 Box 19.B Marasmus and Kwashiorkor 568 Obesity has multiple causes 569 Diabetes is characterized by hyperglycemia 570 Obesity, diabetes, and cardiovascular disease are linked 572 19.4 Clinical Connection: Cancer Metabolism 573 Aerobic glycolysis supports biosynthesis 573 Cancer cells consume large amounts of glutamine 574

Part 4

Genetic Information

20 DNA Replication and Repair

582

20.1 The DNA Replication Machinery 582 Replication occurs in factories 583 Helicases convert double-stranded DNA to single-stranded DNA 584 DNA polymerase faces two problems 585 DNA polymerases share a common structure and mechanism 587 DNA polymerase proofreads newly synthesized DNA 589 An RNase and a ligase are required to complete the lagging strand 590 20.2 Telomeres 593 Telomerase extends chromosomes 594 Box 20.A HIV Reverse Transcriptase 595 Is telomerase activity linked to cell immortality? 596 20.3 DNA Damage and Repair 596 DNA damage is unavoidable 597 Repair enzymes restore some types of damaged DNA 598 Base excision repair corrects the most frequent DNA lesions 598 Nucleotide excision repair targets the second most common form of DNA damage 599 Double-strand breaks can be repaired by joining the ends 601 Recombination also restores broken DNA molecules 601 Box 20.B Gene Editing with CRISPR 602 20.4 Clinical Connection: Cancer as a Genetic Disease 604 Tumor growth depends on multiple events 605

CO NTENTS   xiii

DNA repair pathways are closely linked to cancer 605 20.5 DNA Packaging 607 DNA is negatively supercoiled 607 Topoisomerases alter DNA supercoiling 608 Eukaryotic DNA is packaged in nucleosomes 610 20.6 Tools and Techniques: Manipulating DNA 611 Cutting and pasting generates recombinant DNA 612 The polymerase chain reaction amplifies DNA 614 DNA sequencing uses DNA polymerase to make a complementary strand 615

21 Transcription and RNA

627

21.1 Initiating Transcription 627 What is a gene? 628 DNA packaging affects transcription 628 DNA also undergoes covalent modification 631 Transcription begins at promoters 631 Transcription factors recognize eukaryotic promoters 633 Mediator integrates multiple regulatory signals 634 Box 21.A DNA-Binding Proteins 635 Prokaryotic operons allow coordinated gene expression 636 21.2 RNA Polymerase 638 RNA polymerases have a common structure and mechanism 639 Box 21.B RNA-Dependent RNA Polymerase 640 RNA polymerase is a processive enzyme 641 Transcription elongation requires changes in RNA polymerase 642 Transcription is terminated in several ways 644 21.3 RNA Processing 645 Eukaryotic mRNAs receive a 5′ cap and a 3′ poly(A) tail 645 Splicing removes introns from eukaryotic RNA 646 mRNA turnover and RNA interference limit gene expression 649 Box 21.C The Nuclear Pore Complex 649

rRNA and tRNA processing includes the addition, deletion, and modification of nucleotides 652 RNAs have extensive secondary structure 653

22 Protein Synthesis

663

22.1 tRNA and the Genetic Code 663 The genetic code is redundant 664 tRNAs have a common structure 665 tRNA aminoacylation consumes ATP 666 Editing increases the accuracy of aminoacylation 667 tRNA anticodons pair with mRNA codons 668 Box 22.A The Genetic Code Expanded 669 22.2 Ribosome Structure 669 The ribosome is mostly RNA 670 Three tRNAs and one mRNA bind to the ribosome 671 22.3 Translation 673 Initiation requires an initiator tRNA 673 The appropriate tRNAs are delivered to the ribosome during elongation 675 The peptidyl transferase active site catalyzes peptide bond formation 677 Box 22.B Antibiotic Inhibitors of Protein Synthesis 679 Release factors mediate translation termination 680 Translation is efficient and dynamic 681 22.4 Post-Translational Events 683 Chaperones promote protein folding 684 The signal recognition particle targets some proteins for membrane translocation 685 Many proteins undergo covalent modification 687 GLOSSARY

G-1

ODD-NUMBERED SOLUTIONS INDEX

I-1

S-1

Preface Our goal in writing Essential Biochemistry has been to provide students with a succinct guide to modern biochemistry that includes plenty of chemistry, biological context, and problemsolving opportunities. We want our book to be readable and practical, presenting the basic concepts of molecular structure and function, metabolism, and molecular biology along with some insights into the biochemistry of health and disease. This fifth edition of Essential Biochemistry is our most ambitious revision—almost every page has at least minor changes. We looked for better ways to focus on the chemistry behind the biology, and we added some new topics to help students make connections between biochemistry and other subjects. We worked hard to integrate the updates so that students would find this book—like the previous editions—clear, concise, and manageable. We kept the organization mostly the same, and the chapters include all the helpful pedagogical features you’re used to. And because we believe that students learn by doing, at the end of each chapter you’ll find even more problems for students to test their understanding. Of the 1960 problems— averaging about 90 per chapter—20% are new to this edition. Among the major differences in the new edition is a streamlined Chapter 3, now called Nucleic Acid Structure and Function, to provide a solid foundation for understanding protein structure and function. New material also helps students relate biochemistry to previously encountered biological concepts such as chromosome division and the laws of inheritance. Material related to manipulating DNA has been moved to Chapter 20 (DNA Replication and Repair), which provides the appropriate context for appreciating techniques for copying, pasting, and sequencing DNA. In this edition, Chapter 4 includes an updated section on protein folding and stability, with information about metamorphic proteins and intrinsically disordered proteins. We added a new section on antibody structure and function to Chapter 5 to present key features of immunoglobulins and some current applications. We used a new approach to define uncompetitive and noncompetitive enzyme inhibition, which should help students better visualize these situations (Chapter 7). Additional diagrams now present all the intermediates of the pentose phosphate pathway (Chapter 13) and the Calvin cycle (Chapter 16). Other notable changes are 10 new boxes covering topics such as virus structures and viral replication, thioesters, opioids, the Maillard reaction, iron metabolism, and the nuclear pore complex. Significant updates in the text itself address practical application of genomics (Chapter 3), autophagy (Chapter 9), the structures and mechanisms of sensory and other signaling proteins (Chapter 10), supermolecular complexes in respiration and photosynthesis (Chapters 15 and 16), the regulation of metabolism by sterols and oxygen (Chapters 17 and 19), and next-generation DNA sequencing (Chapter 20). xiv

In addition to many small adjustments to improve clarity and readability, we have refreshed some diagrams and have moved information from figure captions to the text so that students won’t overlook key details. Finally, the list of selected readings for each chapter has been revised with the goal of providing current and accessible reviews that students can follow. We hope you’ll find the fifth edition a valuable resource for your students. As always, we welcome your feedback and suggestions, Charlotte Pratt Kathleen Cornely

Pedagogical Elements We hope that students will not simply memorize what they read but will use the textbook as a guide and learn to think and explore on their own. To that end, we have built into the chapters a variety of features to facilitate learning outside the lecture hall. •  Do You Remember review questions at the start of each chapter help students tie new topics to previously encountered material. •  Learning Objectives for each section are based on verbs, giving students an indication of what they need to be able to do, not just know. •  Before Going On study hints at the end of each section suggest activities that support learning. •  Questions following selected tables and figures prompt students to inspect information more closely and make comparisons. •  Key sentences summarizing main points are printed in italics to help students focus and review. •  Metabolism overview figures introduced in Chapter 12 and revisited in subsequent chapters help students place individual metabolic pathways into a broader context. •  Chapter Summaries, organized by section headings, highlight the most important concepts in each section. •  Key terms are in boldface and are listed at the end of the chapter, with complete definitions in the Glossary. •  Tools and Techniques sections appear at the end of Chapters 2, 4, and 20 to showcase practical aspects of biochemistry and provide an overview of experimental techniques that students will encounter in their reading or laboratory experience. •  Clinical Connection sections in Chapters 2, 4, 5, 6, 7, 13, 19, and 20 provide more in-depth exploration of the biochemical basis of diseases.

P reface   xv

•  Selected Readings are listed at the end of each chapter, with short descriptions, for students to consult as sources of additional information and context. •  Problems for each chapter—20% more than in the last edition—are grouped by section and are offered in pairs, with the answers to odd-numbered problems provided in an appendix.

Organization We have chosen to focus on aspects of biochemistry that tend to receive little coverage in other courses or present a ­ challenge  to many students. Thus, in this textbook, we devote proportionately more space to topics such as acid–base chemistry, enzyme mechanisms, enzyme kinetics, oxidation–­ reduction reactions, oxidative phosphorylation, photosynthesis, and the enzymology of DNA replication, transcription, and translation. At the same time, we appreciate that students can become overwhelmed with information. To counteract this tendency, we have intentionally left out some details, particularly in the chapters on metabolic pathways, in order to emphasize some general themes, such as the stepwise nature of pathways, their evolution, and their regulation. The 22 chapters of Essential Biochemistry are relatively short, so that students can spend less time reading and more time extending their learning through active problem-solving. Most of the problems require some analysis rather than simple recall of facts. Many problems based on research data provide students a glimpse of the “real world” of science and medicine. Although each chapter of Essential Biochemistry, Fifth Edition is designed to be self-contained so that it can be covered at any point in the syllabus, the 22 chapters are organized into four parts that span the major themes of biochemistry, including some chemistry background, structure–function relationships, the transformation of matter and energy, and how genetic information is stored and made accessible. Part 1 of the textbook includes an introductory chapter and a chapter on water. Students with extensive exposure to chemistry can use this material for review. For students with little previous experience, these two chapters provide the chemistry background they will need to appreciate the molecular structures and metabolic reactions they will encounter later. Part 2 begins with a chapter on the genetic basis of macromolecular structure and function (Chapter 3, Nucleic Acid Structure and Function). This is followed by chapters on protein structure (Chapter 4) and protein function (Chapter 5), with coverage of myoglobin and hemoglobin, cytoskeletal and motor proteins, and antibodies. An explanation of how enzymes work (Chapter 6) precedes a discussion of enzyme kinetics (Chapter 7), an arrangement that allows students to grasp the importance of enzymes and to focus on the chemistry

of enzyme-catalyzed reactions before delving into the more quantitative aspects of enzyme kinetics. A chapter on lipid chemistry (Chapter 8, Lipids and Membranes) is followed by two chapters that discuss critical biological functions of membranes (Chapter 9, Membrane Transport, and Chapter 10, Signaling). The section ends with a chapter on carbohydrate chemistry (Chapter 11), completing the survey of molecular structure and function. Part 3 begins with an introduction to metabolism that provides an overview of fuel acquisition, storage, and mobilization as well as the thermodynamics of metabolic reactions (Chapter 12). This is followed, in traditional fashion, by chapters on glucose and glycogen metabolism (Chapter 13); the citric acid cycle (Chapter 14); electron transport and oxida­tive phosphorylation (Chapter 15); the light and dark reactions of photosynthesis (Chapter 16); lipid catabolism and biosynthesis (Chapter 17); and pathways involving nitrogen-containing compounds, including the synthesis and degradation of amino acids, the synthesis and degradation of nucleotides, and the nitrogen cycle (Chapter 18). The final chapter of Part 2 ­explores the integration of mammalian metabolism, with extensive discussions of hormonal control of metabolic pathways, disorders of fuel metabolism, and cancer (Chapter 19). Part 4, the management of genetic information, ­includes three chapters, covering DNA replication and repair (Chapter 20), transcription (Chapter 21), and protein ­ synthesis (Chapter 22). Because these topics are typically also covered in other courses, Chapters 20–22 emphasize the relevant biochemical details, such as topoisomerase ­action, nucleosome structure, mechanisms of polymerases and other enzymes, structures of accessory proteins, proofreading strategies, and ­chaperone-assisted protein folding.

The WileyPLUS Advantage WileyPLUS is a research-based online environment for effective teaching and learning. The customization features, quality question banks, interactive eTextbook, and analytical tools allow you to quickly create a customized course that tracks student learning trends. Your students can stay engaged and on track with the use of intuitive tools like the syncing cal­ endar and the student mobile app. All Wiley educational products and services are born accessible, designed for users of all abilities. Guided Tours cover the major topics of the course. These multi-part tutorials explain biochemistry in time and space. Interactive questions at the end of each tour reinforce learning. Assignable End-of-Chapter Questions, over 20 per chapter, can be assigned to students through WileyPLUS. Twenty-four Sample Calculation Videos walk students through each step of the sample calculations.

xvi  Preface

Brief Bioinformatics Exercises crafted by Rakesh Mogul at California State Polytechnic University, Pomona, provide detailed instructions for novices to access and use bioinformatics databases and software tools. Each of the 57 exercises includes multiple-choice questions to help students gauge their success in learning from these resources. Do You Remember Practice Quizzes help students prepare for new material by reinforcing relevant topics from previous chapters. Concept Check Questions for each section allow students to test their knowledge. Discussion Questions are thought-provoking questions that serve as a point of departure for student discussion and engagement with content. Twenty-three Animated Process Diagrams bring multi-step figures to life. Bioinformatics Projects, written by Paul Craig at Rochester Institute of Technology, provide guidance for 12 extended explorations of online databases, with questions, many openended, for students to learn on their own.

Additional Instructor Resources in WileyPLUS •  PowerPoint Art Slides. •  Exercise Questions with immediate descriptive feedback by Rachel Milner, University of Alberta; Adrienne Wright, University of Alberta; and Mary Peek, Georgia Institute of Technology. •  Test Bank Questions by Scott Lefler, Arizona State University and Anne Grippo, Arkansas State University. •  Practice and Pre-Lecture Questions by Steven Vik, Southern Methodist University, and Mary Peek, Georgia Institute of Technology. •  PowerPoint Lecture Slides by Mary Peek, Georgia Institute of Technology. •  Personal Response System (“Clicker”) Questions by Gail Grabner, University of Texas at Austin, and Mary Peek, Georgia Institute of Technology.

Acknowledgments Unless otherwise noted, molecular graphics images were created using data from the Protein Data Bank (www.rcsb.org) with the PyMol Molecular Graphics System Version 2.1, Schrödinger LLC, originally developed by Warren DeLano, or with the Swiss-Pdb Viewer [Guex, N., Peitsch, M.C., SWISS-MODEL and the Swiss-Pdb Viewer: an environment for comparative protein modeling, Electrophoresis 18, ­2714–2723 (1997). We would like to thank everyone who helped develop Essential Biochemistry, Fifth Edition, including Associate Publisher Sladjana Bruno, Senior Editor Jennifer Yee, Senior Course Production Operations Specialist Patricia Gutierrez, Course Content Developers Corrina Santos and Andrew Moore, Senior Designer Thomas Nery, Editorial Assistant Samantha Hart, and the Lumina Datamatics team. We also thank all the reviewers who provided essential feedback on manuscript and media, corrected errors, and made valuable suggestions for improvements that have been so important in the writing and development of Essential Biochemistry, Fifth Edition

Fifth Edition Reviewers: Arkansas Cindy L. White, Harding University California M. Nidanie Henderson-Stull, Soka University of America Rakesh Mogul, California State Polytechnic University, Pomona Florida Vijaya Narayanan, Florida International University Massachusetts Emily Westover, Brandeis University  Michigan Allison C. Lamanna, University of Michigan

Nebraska Jing Zhang, University of Nebraska, Lincoln North Dakota Dennis R. Viernes, University of Mary New York Youngjoo Kim, SUNY College at Old Westbury Texas Autumn L. Sutherlin, Abilene Christian University Virginia Michael Klemba, Virginia Tech Canada Isabelle Barrette-Ng, University of Windsor Sian T. Patterson, University of Toronto

Previous Edition Reviewers: Arkansas Anne Grippo, Arkansas State University Arizona Allan Bieber, Arizona State University Matthew Gage, Northern Arizona University Scott Lefler, Arizona State University, Tempe Allan Scruggs, Arizona State University, Tempe Richard Posner, Northern Arizona State University California Elaine Carter, Los Angeles City College Daniel Edwards, California State University, Chico Gregg Jongeward, University of the Pacific Pavan Kadandale, University of California, Irvine Paul Larsen, University of California, Riverside

P reface   xvii Rakesh Mogul, California State Polytechnic University, Pomona Brian Sato, University of California, Irvine Colorado Paul Azari, Colorado State University Andrew Bonham, Metropolitan State University of Denver Johannes Rudolph, University of Colorado Connecticut Matthew Fisher, Saint Vincent’s College Florida David Brown, Florida Gulf Coast University Georgia Chavonda Mills, Georgia College Mary E. Peek, Georgia Tech University Rich Singiser, Clayton State University Hawaii Jon-Paul Bingham, University of Hawaii-Manoa, College of Tropical Agriculture and Human Resources Illinois Lisa Wen, Western Illinois University Gary Roby, College of DuPage Jon Friesen, Illinois State University Constance Jeffrey, University of Illinois, Chicago Stanley Lo, Northwestern University Kristi McQuade, Bradley University Indiana Brenda Blacklock, Indiana University-Purdue University Indianapolis Todd Hrubey, Butler University Christine Hrycyna, Purdue University Mohammad Qasim, Indiana University-Purdue University Iowa Don Heck, Iowa State University Kansas Peter Gegenheimer, The University of Kansas Ramaswamy Krishnamoorthi, Kansas State University Louisiana James Moroney, Louisiana State University Jeffrey Temple, Southeastern Louisiana University Maine Robert Gundersen, University of Maine, Orono Massachusetts Jeffry Nichols, Worcester State University Michigan Marilee Benore, University of Michigan Kim Colvert, Ferris State University Kathleen Foley, Michigan State University Deborah Heyl-Clegg, Eastern Michigan University Melvin Schindler, Michigan State University Jon Stoltzfus, Michigan State University Mark Thomson, Ferris State University Minnesota Sandra Olmsted, Augsburg College Tammy Stobb, St. Cloud State University Mississippi Jeffrey Evans, University of Southern Mississippi James R. Heitz, Mississippi State University Arthur Chu, Delta State University Missouri Karen Bame, University of Missouri, Kansas City Nuran Ercal, Missouri University of Science & Technology

Nebraska Jodi Kreiling, University of Nebraska, Omaha Madhavan Soundararajan, University of Nebraska Russell Rasmussen, Wayne State College New Jersey Yufeng Wei, Seton Hall University Bryan Spiegelberg, Rider University New Mexico Beulah Woodfin, University of New Mexico New York Wendy Pogozelski, SUNY Geneseo Susan Rotenberg, Queens College of CUNY Sergio Abreu, Fordham University Ohio Edward Merino, University of Cincinnati Heeyoung Tai, Miami University Lai-Chu Wu, The Ohio State University Oklahoma Charles Crittell, East Central University Oregon Jeannine Chan, Pacific University Steven Sylvester, Oregon State University Pennsylvania Mahrukh Azam, West Chester University of Pennsylvania Jeffrey Brodsky, University of Pittsburgh David Edwards, University of Pittsburgh School of Pharmacy Robin Ertl, Marywood University Amy Hark, Muhlenberg College Justin Huffman, Pennsylvania State University, Altoona Michael Sypes, Pennsylvania State University Sandra Turchi-Dooley, Millersville University Laura Zapanta, University of Pittsburgh Rhode Island Lenore Martin, University of Rhode Island Erica Oduaran, Roger Williams University South Carolina Carolyn S. Brown, Clemson University Weiguo Cao, Clemson University Ramin Radfar, Wofford College Paul Richardson, Coastal Carolina University Kerry Smith, Clemson University Tennessee Meagan Mann, Austin Peay State University Texas Johannes Bauer, Southern Methodist University David W. Eldridge, Baylor University Edward Funkhouser, Texas A&M University Gail Grabner, University of Texas, Austin Barrie Kitto, University of Texas at Austin Marcos Oliveira, Feik School of Pharmacy, University of the Incarnate Word Richard Sheardy, Texas Woman’s University Linette Watkins, Southwest Texas State University Utah Craig Thulin, Utah Valley University Wisconsin Sandy Grunwald, University of Wisconsin La Crosse Canada Isabelle Barrette-Ng, University of Windsor

CHAPTER 1

This first chapter offers a preview of the study of biochemis­ try, broken down into three sections that reflect how topics in this book are organized. First come brief descriptions of the four major types of small biological molecules and their polymeric forms. Next is a summary of the thermodynamics that apply to metabolic reactions. Finally, there is a discus­ sion of the origin of self-replicating life-forms and their evo­ lution into modern cells. These short discussions introduce some of the key players and major themes of biochemistry and provide a foundation for the topics that will be encoun­ tered in subsequent chapters.

DSP/deepseaphotography

The Chemical Basis of Life

The chemical reactions of living systems take place across a wide range of conditions. Although many microbial species can tolerate extreme heat, multicellular organisms require much more temper­ ate habitats. One exception is Alvinella pompejana, the Pompeii worm, which lives near deep-sea hydrothermal vents and thrives at 42°C (107°F). Hair-like colonies of symbiotic bacteria may help insulate its body.

1.1 What Is Biochemistry? LEARNING OBJECTIVE Recognize the main themes of biochemistry. Biochemistry is the scientific discipline that seeks to explain life at the molecular level. It uses the tools and terminology of chemistry to describe the various attributes of living organ­ isms. Biochemistry offers answers to such fundamental questions as “What are we made of?” and “How do we work?” Biochemistry is also a practical science: It generates powerful techniques that underlie advances in other fields, such as genetics, cell biology, and immu­ nology; it offers insights into the treatment of diseases such as cancer and diabetes; and it improves the efficiency of industries such as wastewater treatment, food production, and drug manufacturing. Some aspects of biochemistry can be approached by studying individual molecules iso­ lated from cells. A thorough understanding of each molecule’s physical structure and chem­ ical reactivity helps lead to an understanding of how molecules cooperate and combine to form larger functional units and, ultimately, the intact organism (Fig. 1.1). However, just as a clock completely disassembled no longer resembles a clock, information about a multi­ tude of biological molecules does not necessarily reveal how an organism lives. Biochemists therefore investigate how organisms behave under different conditions or when a particular molecule is modified or absent. In addition, they collect vast amounts of information about molecular structures and functions—information that is stored and analyzed by computer, a field of study known as bioinformatics. A biochemist’s laboratory is as likely to hold racks of test tubes as flasks of bacteria or computers. Chapters 3 through 22 of this book are divided into three groups that roughly correspond to three major themes of biochemistry:

1

2

C hA pTER 1

The Chemical Basis of Life Organism

Organ

Cell

Organelle

Liver Hepatocyte

Mitochondrion Human

Citrate synthase

DNA

Citrate

Ubiquinone

Photodisc/Getty Images

Levels of organization in a living organism. Biochemistry focuses on the structures and functions of molecules. Interac­ tions between molecules give rise to higher­order structures (for example, organelles), which may themselves be components of larger entities, leading ultimately to the entire organism. FIGURE 1.1

Molecules

1. Living organisms are made of macromolecules. Some molecules are responsible for the physical shapes of cells. Others carry out various activities in the cell. (For convenience, we often use cell interchangeably with organism since the simplest living entity is a single cell.) In all cases, the structure of a molecule is intimately linked to its function. Understanding a molecule’s structural characteristics is therefore an important key to understanding its functional significance. 2. Organisms acquire, transform, store, and use energy. The ability of a cell to carry out metabolic reactions—to synthesize its constituents and to move, grow, and reproduce— requires the input of energy. A cell must extract this energy from the environment and spend it or store it in a manageable form. 3. Biological information is transmitted from generation to generation. Modern human beings look much like they did 100,000 years ago. Certain bacteria have persisted for millions, if not billions, of years. In all organisms, the genetic information that specifies a cell’s structural composition and functional capacity must be safely maintained and transmitted each time the cell divides. Several other themes run throughout biochemistry and we will highlight these where appropriate. 4. Cells maintain a state of homeostasis. Even within its own lifetime, a cell may dramati­ cally alter its shape or metabolic activities, but it does so within certain limits. In order to remain in a steady, nonequilibrium state—homeostasis—the cell must recognize changing internal and external conditions and regulate its activities. 5. Organisms evolve. Over long periods of time, the genetic composition of a population of organisms changes. Examining the molecular makeup of living organisms allows biochemists to identify the genetic features that distinguish groups of organisms and to trace their evolutionary history. 6. Diseases can be explained at the biochemical level. Identifying the molecular defects that underlie human diseases, or investigating the pathways that allow one organism to infect another, is the first step in diagnosing, treating, preventing, or curing a host of ailments.

1.2  Biological Molecules  3

1.2 Biological Molecules LEARNING OBJECTIVES Identify the major classes of biological molecules. • List the elements found in biological molecules. • Draw and name the common functional groups in biological molecules. • Draw and name the common linkages in biological molecules. • D  istinguish the main structural features of amino acids, carbohydrates, nucleotides, and lipids. • I dentify the monomers and linkages in polypeptides, polysaccharides, and nucleic acids. • Summarize the biological functions of the major classes of biological molecules.

Even the simplest organisms contain a staggering number of different molecules, yet this number represents only an infinitesimal portion of all the molecules that are chemically possible. For one thing, only a small subset of the known elements are found in living systems (Fig. 1.2). The most abundant of these are C, N, O, and H, followed by Ca, P, K, S, Cl, Na, and Mg. Certain trace elements are also present in very small quantities. Virtually all the molecules in a living organism contain carbon, so biochemistry can be considered to be a branch of organic chemistry. In addition, nearly all biological molecules are constructed from H, N, O, P, and S. Most of these molecules belong to one of a few struc­ tural classes, which are described below. Similarly, the chemical reactivity of biomolecules is limited relative to the reactivity of all chemical compounds. A few of the functional groups and intramolecular linkages that are common in biochemistry are listed in Table 1.1. Familiarity with these functional groups is essential for understanding the behavior of the different types of biological molecules we will encounter throughout this book.

Cells contain four major types of biomolecules Most of the cell’s small molecules can be divided into four classes. Although each class contains many members, they are united under a single structural or functional definition. ­Identifying a particular molecule’s class may help predict its chemical properties and possibly its role in the cell.

1 H 11 Na 19 K

12 Mg 20 Ca

23 V

24 25 26 27 28 29 30 Cr Mn Fe Co Ni Cu Zn 42 48 Mo Cd 74 W

5 B 13 Al

6 C 14 Si

7 N 15 P 33 As

8 O 16 S 34 Se

9 F 17 Cl 35 Br 53 I

  FIGURE 1.2   Elements found in biological systems.  The most abundant elements are most darkly shaded; trace elements are most lightly shaded. Not every organism contains every trace element. Biological molecules primarily contain H, C, N, O, P, and S.

4  C h apter 1  The Chemical Basis of Life

TA BLE 1 .1   Common Functional Groups and Linkages in Biochemistry

Structure a

Compound name Amine b

RNH2 or R2NH or R3N or

Functional group

RNH 3 R2NH 2 R3NH

N



or

(amino group)

N

Alcohol

ROH

OH (hydroxyl group)

Thiol

RSH

SH (sulfhydryl group)

Ether

ROR

O

O Aldehyde

R

C

O H

C

O Ketone

R

C

C

R

OH or

C O

OH (carboxyl group) or

R

C

O

C

O (carboxylate group)

C

R

C

(acyl group)

C

(carbonyl group), R

C O

O OR

C

O

(ester linkage)

O

(thioester linkage)

C

N

(amido group)

C

N

S

S Thioester

O

R

R

(acyl group)

C

(carbonyl group), R

O

O Ester

O

O

O Carboxylic acid b (Carboxylate)

(ether linkage)

C

OR

O

Amide

Imine b

R

C O

NH2

R

C O

NHR

R

C

NR2

R R

NH or NR or

O

R R

NH 2 NHR

O R

O

P

R

O

P

OH or

O O

O



P

O

O

P

O O

R

O

P O

O

H

(imino group)

(phosphoester linkage)

OH

O

OH or

P

P

OH or

O

P

O O



O

O

P OH

O (phosphoryl group)

O O

OH O

OH O

Diphosphoric acid ester b, d

P

OH

O R



N

O

OH O

Phosphoric acid ester b, c

C

or

P OH O

O O O P

P

O

OH OH or

OH

(phosphoanhydride linkage)

O P O

O O

P O

(diphosphoryl group, pyrophosphoryl group) a 

R represents any carbon-containing group. In a molecule with more than one R group, the groups may be the same or different.

b 

Under physiological conditions, these groups are ionized and hence bear a positive or negative charge.

c 

2− When R = H, the molecule is inorganic phosphate (abbreviated Pi ), usually ​​H2PO​  − 4​  ​​ or ​​HPO​  4​  ​​.

d 

When R = H, the molecule is pyrophosphate (abbreviated PPi ).

Question  Cover the structure column and draw the structure for each compound listed on the left. Do the same for each functional group.

O

1.2  Biological Molecules  5

COO– H C

CH3

NH+ 3

a. A structural formula includes all the atoms and the major bonds. Some bonds, such as the C—O and N—H bonds, are implied. The central carbon has tetrahedral geometry. The horizontal bonds extend slightly above the plane of the page and the vertical bonds extend slightly behind it.

b. A ball-and-stick representation more accurately reveals the arrangement of atoms in space, although it does not show their relative sizes or electrical charges. The atoms are color-coded by convention: C gray, N blue, O red, and H white.

c. The space-filling model best represents the actual shape of the molecule but may obscure some of its atoms and linkages. Each atom is shown as a sphere whose radius (the van der Waals radius) corresponds to the distance of closest approach by another atom.

  FIGURE 1.3   Representations of alanine.  a. Structural formula,  b. ball-and-stick model, and

c. space-filling model.

1. Amino Acids 

Among the simplest compounds are the amino acids, so named because they contain an amino group (NH2) and a carboxylic acid group (COOH). Under − physiological conditions, these groups are actually ionized to ​NH​  + 3​  ​​ and COO . The com­ mon amino acid alanine—like other small molecules—can be depicted in different ways, for example, by a structural formula, a ball-and-stick model, or a space-filling model (Fig. 1.3). Other amino acids resemble alanine in basic structure, but instead of a methyl group (CH3), they have another group—called a side chain or R group—that may also contain N, O, or S; for example,

COO– H C

CH2

COO–

O

H C

C NH2

NH+ 3

CH2

SH

NH+ 3

Asparagine

Cysteine

2. Carbohydrates 

Simple carbohydrates (also called monosaccharides or just sug­ ars) have the formula (CH2O)n, where n is ≥ 3. Glucose, a monosaccharide with six carbon atoms, has the formula C6H12O6. It is sometimes convenient to draw it as a ladder-like chain (left); however, glucose forms a cyclic structure in solution (right):

O

H C

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OH

H HO Glucose

CH2OH O H OH H H

H OH

OH

In the representation of the cyclic structure, the darker bonds project in front of the page and the lighter bonds project behind it. In many monosaccharides, one or more hydroxyl groups are replaced by other groups, but the ring structure and multiple OH groups of these mole­ cules allow them to be easily recognized as carbohydrates.

6  C h apter 1  The Chemical Basis of Life

3. Nucleotides 

A five-carbon sugar, a nitrogen-containing ring, and one or more phosphate groups are the components of nucleotides. For example, adenosine triphosphate (ATP) contains the nitrogenous group adenine linked to the monosaccharide ribose, to which a triphosphate group is also attached:

NH2

O −

O

O O

P O−

P

N

N

Triphosphate

O O

O−

O

P O−

H

N

N O

CH2 H

H

OH

OH

Adenosine triphosphate (ATP)

Adenine

Ribose

H

The most common nucleotides are mono-, di-, and triphosphates containing the nitrogenous ring compounds (or “bases”) adenine, cytosine, guanine, thymine, or uracil (abbreviated A, C, G, T, and U).

4. Lipids 

The fourth major group of biomolecules consists of the lipids. These com­ pounds cannot be described by a single structural formula since they are a diverse collection of mole­cules. However, they all tend to be poorly soluble in water because the bulk of their structure is hydrocarbon-like. For example, palmitic acid consists of a highly insoluble chain of 15 carbons attached to a carboxylic acid group, which is ionized under physiological condi­ tions. The anionic lipid is therefore called palmitate.

O H3C

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

C

O−

CH2

Palmitate

Cholesterol, although it differs significantly in structure from palmitate, is also poorly soluble in water because of its hydrocarbon-like composition.

CH3 CH CH3

CH2

CH2

CH2

CH3 CH CH3

CH3 HO

Cholesterol

Cells also contain a few other small molecules that cannot be easily classified into the groups above or that are constructed from molecules belonging to more than one group.

There are three major kinds of biological polymers In addition to small molecules consisting of relatively few atoms, organisms contain macro­ molecules that may consist of thousands of atoms. Such huge molecules are not synthesized in one piece but are built from smaller units. This is a universal feature of nature: A few kinds

1.2  Biological Molecules  7

Box 1.A Units Used in Biochemistry Biochemists follow certain conventions when quantifying objects on a molecular scale. For example, the mass of a molecule can be expressed in atomic mass units; however, the masses of biolog­ical molecules—especially very large ones—are typically given with­ out units. Here it is understood that the mass is expressed relative to one-twelfth the mass of an atom of the common carbon isotope 12 C (12.011 atomic mass units). Occasionally, units of daltons (D) are used (1 dalton = 1 atomic mass unit), often with the prefix kilo, k (kD). This is useful for macromolecules such as proteins, many of which have masses in the range from 20,000 (20 kD) to over 1,000,000 (1000 kD). The standard metric prefixes are also necessary for express­ ing the minute concentrations of biomolecules in living cells. Con­ centrations are usually given as moles per liter (mol · L−1 or M), with the appropriate prefix, such as m, µ, or n:

mega (M)

106

10−9

10

pico (p)

10−12

milli (m)

10−3

femto (f)

10−15

−6

10

For example, the concentration of the sugar glucose in human blood is about 5 mM, but many intracellular molecules are pres­ ent at concentrations of µM or less. Distances are customarily expressed in angstroms, Å (1 Å = 10−10 m) or in nanometers, nm (1 nm = 10−9 m). For example, the distance between the centers of carbon atoms in a CC bond is about 1.5 Å and the diameter of a DNA molecule is about 20 Å. Question  The diameter of a typical spherical bacterial cell is about 1 µm. What is the cell’s volume?

of building blocks can be combined in different ways to produce a wide variety of larger structures. This is advantageous for a cell, which can get by with a limited array of raw materi­ als. In addition, the very act of chemically linking individual units (monomers) into longer strings (polymers) is a way of encoding information (the sequence of the monomeric units) in a stable form. Biochemists use certain units of measure to describe both large and small molecules (Box 1.A). Amino acids, monosaccharides, and nucleotides each form polymeric structures with widely varying properties. In most cases, the individual monomers become covalently linked in head-to-tail fashion:

Residue

Monomers

nano (n)

kilo (k) micro ( µ)

3

Polymer

The linkage between monomeric units is characteristic of each type of polymer. The mono­ mers are called residues after they have been incorporated into the polymer. Strictly speak­ ing, lipids do not form polymers, although they do tend to aggregate to form larger structures such as cell membranes. Most of the mass of a cell consists of polymers, with proteins accounting for the greatest share (Fig. 1.4). Inorganic ions Small molecules Nucleic acids Carbohydrates Proteins Lipids

  FIGURE 1.4   Mass of a mammalian cell.  Proteins and lipids account for about 75% of the dry mass of a typical mammalian cell.

8  C h apter 1  The Chemical Basis of Life

1. Proteins 

Polymers of amino acids are called polypeptides or proteins. Twenty dif­ ferent amino acids serve as building blocks for proteins, which may contain many hundreds of amino acid residues. The amino acid residues are linked to each other by amide bonds called peptide bonds. A peptide bond (arrow) links the two residues in a dipeptide (the side chains of the amino acids are represented by R1 and R2).

+

H3N

R1 O

R2

C

C

N

H a.

b.   FIGURE 1.5   Structure of human endothelin.  The 21 amino acid residues of this poly­ peptide, shaded from blue to red, form a compact structure. In a, each amino acid residue is represented by a sphere. The balland-stick model b shows all the atoms except hydrogen.

O C

C

O–

H H

Because the side chains of the 20 amino acids have different sizes, shapes, and chemical prop­ erties, the exact conformation (three-dimensional shape) of the polypeptide chain depends on its amino acid composition and sequence. For example, the small polypeptide endothelin, with 21 residues, assumes a compact shape in which the polymer bends and folds to accom­ modate the functional groups of its amino acid residues (Fig. 1.5). The 20 different amino acids can be combined in almost any order and in almost any pro­ portion to produce myriad polypeptides, all of which have unique three-dimensional shapes. This property makes proteins as a class the most structurally variable and therefore the most functionally versatile of all the biopolymers. Accordingly, proteins perform a wide variety of tasks in the cell, such as mediating chemical reactions and providing structural support.

2. Nucleic Acids  Polymers of nucleotides are termed polynucleotides or nucleic acids, better known as DNA and RNA. Unlike polypeptides, with 20 different amino acids available for polymerization, each nucleic acid is made from just four different nucleotides. For exam­ple, the residues in RNA contain the bases adenine, cytosine, guanine, and uracil, whereas the residues in DNA contain adenine, cytosine, guanine, and thymine. Polymeriza­ tion involves the phosphate and sugar groups of the nucleotides, which become linked by phosphodiester bonds. O– –

O

P

O

O

CH2 O H

Phosphodiester bond

Base

H

H

O

H

–

O P

H

O

O

CH2 O H

Base

H

H

OH

H

H

In part because nucleotides are much less variable in structure and chemistry than amino acids, nucleic acids tend to have more regular structures than proteins. This is in keeping with their primary role as carriers of genetic information, which is contained in their sequence of

1.2  Biological Molecules  9

nucleotide residues rather than in their three-dimensional shape (Fig. 1.6). Nevertheless, many nucleic acids do bend and fold into compact globular shapes, as proteins do.

CGUACG a.

3. Polysaccharides  Polysaccharides usually contain only one or a few different types of monosaccharide residues, so even though a cell may synthesize dozens of differ­ ent kinds of monosaccharides, most of its polysaccharides are homogeneous polymers. This tends to limit their potential for carrying genetic information in the sequence of their resi­ dues (as nu­cleic acids do) or for adopting a large variety of shapes and mediating chemical reactions (as proteins do). On the other hand, polysaccharides perform essential cell functions by serving as fuel-storage molecules and by providing structural support. For example, plants link the monosaccharide glucose, which is a fuel for virtually all cells, into the polysaccharide starch for long-term storage. The glucose residues are linked by glycosidic bonds (the bond is shown in red in this disaccharide):

H HO

CH2OH O H OH H H

OH

H

H O

CH2OH O H OH H H

H OH

OH

Glucose monomers are also the building blocks for cellulose, the extended polymer that helps make plant cell walls rigid (Fig. 1.7). The starch and cellulose polymers differ in the arrange­ ment of the glycosidic bonds between glucose residues. The brief descriptions of biological polymers given above are generalizations, meant to convey some appreciation for the possible structures and functions of these macromolecules. Exceptions to the generalizations abound. For example, some small polysaccharides encode information that allows cells bearing the molecules on their surfaces to recognize each other. Likewise, some nucleic acids perform structural roles, for example, by serving as scaffolding in ribosomes, the small machines where protein synthesis takes place. Under certain condi­ tions, proteins are called on as fuel-storage molecules. A summary of the major and minor functions of proteins, polysaccharides, and nucleic acids is presented in Table 1.2.

Glucose

Starch

Cellulose

  FIGURE 1.7   Glucose and its polymers.  Both starch and cellulose are polysaccharides containing glucose residues. They differ in the type of chemical linkage between the monosaccharide units. Starch molecules have a loose helical conformation, whereas cellulose molecules are extended and relatively stiff.

b.   FIGURE 1.6   Structure of a nucleic acid.  a. Sequence of nucleotide residues, using one-letter abbreviations.  b. Ball-and-stick model of the polynucleotide, show­ ing all atoms except hydrogen (this structure is a six-residue segment of RNA).

10  C ha pter 1   The Chemical Basis of Life

TAB L E 1. 2   Functions of Biopolymers

Encode information

Carry out metabolic reactions

Proteins

—

Nucleic acids

✔

Polysaccharides

✓

Biopolymer

Store energy

Support cellular structures

✔

✓

✔

✓

—

✓

—

✔

✔

✔ major function ✓ minor function

Before Going On • List the six most abundant elements in biological molecules. • Name the common functional groups and linkages shown in Table 1.1. • Give the structural or functional definitions for amino acids, monosaccharides, nucleotides, and lipids. • Describe the advantage of building a polymer from monomers. • Give the structural definitions and major functions of proteins, polysaccharides, and nucleic acids. • Name the linkage in each type of polymer. • List the major functions of proteins, polysaccharides, and nucleic acids.

1.3 Energy and Metabolism LEARNING OBJECTIVES Explain how enthalpy, entropy, and free energy apply to biological systems. • Define enthalpy, entropy, and free energy. • Write the equation that links changes in enthalpy, entropy, and free energy. • Relate changes in enthalpy and entropy to the spontaneity of a process. • Describe the energy flow that makes living systems thermodynamically possible.

Assembling small molecules into polymeric macromolecules requires energy. And unless the monomeric units are readily available, a cell must synthesize the monomers, which also requires energy. In fact, cells require energy for all the functions of living, growing, and reproducing. It is useful to describe the energy in biological systems using the terminology of thermody­ namics (the study of heat and power). An organism, like any chemical system, is subject to the laws of thermodynamics. According to the first law of thermodynamics, energy cannot be cre­ ated or destroyed. However, it can be transformed. For example, the energy of a river flowing over a dam can be harnessed as electricity, which can then be used to produce heat or perform mechanical work. Cells can be considered to be very small machines that use chemical energy to drive metabolic reactions, which may also produce heat or carry out mechanical work.

1.3  Energy and Metabolism  11

Enthalpy and entropy are components of free energy The energy relevant to biochemical systems is called the Gibbs free energy (after the scien­ tist who defined it) or just free energy. It is abbreviated G and has units of joules per mol (J · mol−1). Free energy has two components: enthalpy and entropy. Enthalpy (abbreviated H, with units of J · mol−1) is taken to be equivalent to the heat content of the system. Entropy (abbreviated S, with units of J · K−1 · mol−1) is a measure of how the energy is dispersed within that system. Entropy can therefore be considered to be a measure of the system’s disorder or randomness, because the more ways a system’s components can be arranged, the more dis­ persed its energy. For example, consider a pool table at the start of a game when all 15 balls are arranged in one neat triangle (a state of high order or low entropy). After play has begun, the balls are scattered across the table, which is now in a state of disorder and high entropy (Fig. 1.8). Free energy, enthalpy, and entropy are related by the equation

G = H − TS

(1.1)

where T represents temperature in Kelvin (equivalent to degrees Celsius plus 273). Tempera­ ture is a coefficient of the entropy term because entropy varies with temperature; the entropy of a substance increases when it is warmed because more thermal energy has been dispersed within it. The enthalpy of a chemical system can be measured, although with some difficulty, but it is next to impossible to measure a system’s entropy because this would require counting all the possible arrangements of its components or all the ways its energy could be spread out among them. Therefore, it is more practical to deal with changes in these quantities (change is indicated by the Greek letter delta, ∆) so that ∆G = ∆H − T∆S

(1.2)

Biochemists can measure how the free energy, enthalpy, and entropy of a system differ before and after a chemical reaction. For example, exothermic reactions are accompanied by the release of heat to the surroundings (Hfinal − Hinitial = ∆H < 0), whereas endothermic ­reactions absorb heat from the surroundings (∆H > 0). Similarly, the entropy change, Sfinal − S­initial = ∆S, can be positive or negative. When ∆H and ∆S for a process are known, Equation 1.2 can be used to calculate the value of ∆G at a given temperature (see Sample Calculation 1.1).

a.

b.

  FIGURE 1.8   Illustration of entropy.  Entropy is a measure of the dispersal of energy in a system, so it reflects the system’s randomness or disorder.  a. Entropy is low when all the balls are arranged in a single area of the pool table.  b. Entropy is high after the balls have been scattered, because there are now a large number of different possible arrangements of the balls on the table.

Question  Compare the entropy of a ball of yarn before and after a cat has played with it.

12  C ha pter 1   The Chemical Basis of Life

SA MP L E CALCU LAT I O N 1 . 1 Problem  Use the information below to calculate the change in enthalpy and the change in entropy for the reaction A → B. Enthalpy (kJ · mol−1)

Entropy (J · K−1 · mol−1)

A

60

22

B

75

97

SEE SAMPLE CALCULATION VIDEOS

Solution ∆H = HB − HA ∆S = SB − SA

= 75 kJ · mol−1 − 60 kJ · mol−1

= 97 J · K−1 · mol−1

−1



= 15 kJ · mol − 22 J · K−1 · mol−1



= 15,000 J · mol−1

= 75 J · K−1 · mol−1

∆G is less than zero for a spontaneous process A china cup dropped from a great height will break, but the pieces will never reassemble themselves to restore the cup. The thermodynamic explanation is that the broken pieces have less free energy than the intact cup. In order for a process to occur, the overall change in free energy (∆G) must be negative. For a chemical reaction, this means that the free energy of the products must be less than the free energy of the reactants: ∆G = Gproducts − Greactants < 0

(1.3)

When ∆G is less than zero, the reaction is said to be spontaneous or exergonic. A nonspontaneous or endergonic reaction has a free energy change greater than zero; in this case, the reverse reaction is spontaneous: A → B

B→A

∆G > 0

∆G < 0

Nonspontaneous Spontaneous Note that thermodynamic spontaneity does not indicate how fast a reaction occurs, only whether it will occur as written. (The rate of a reaction depends on other factors, such as the concentra­ tions of the reacting molecules, the temperature, and the presence of a catalyst.) When a reac­ tion, such as A → B, is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so there is no net change in the system. In this situation, ∆G = 0. A quick examination of Equation 1.2 reveals that a reaction that occurs with a decrease in enthalpy and an increase in entropy is spontaneous at all temperatures because ∆G is always less than zero. These results are consistent with everyday experience. For example, heat moves spontaneously from a hot object to a cool object, and items that are neatly arranged tend to become disordered, never the other way around. (This is a manifestation of the second law of thermodynamics, which states that energy tends to spread out.) Accordingly, reactions in which the enthalpy increases and entropy decreases do not occur. If enthalpy and entropy both increase or both decrease during a reaction, the value of ∆G then depends on the tem­ perature, which governs whether the T∆S term of Equation 1.2 is greater than or less than the ∆H term. This means that a large increase in entropy can offset an unfavorable (positive) change in enthalpy. Conversely, the release of a large amount of heat (∆H < 0) during a reac­ tion can offset an unfavorable decrease in entropy (see Sample Calculation 1.2).

Life is thermodynamically possible In order to exist, life must be thermodynamically spontaneous. Does this hold at the molec­ ular level? When analyzed in a test tube (in vitro, literally “in glass”), many of a cell’s met­ abolic reactions have free energy changes that are less than zero, but some reactions do not. Nevertheless, the nonspontaneous reactions are able to proceed in vivo (in a living organism) because they occur in concert with other reactions that are thermodynamically favorable. Consider two reactions in vitro, one nonspontaneous (∆G > 0) and one spontaneous (∆G < 0): A → B   ∆G = +15 kJ · mol−1 (nonspontaneous) B → C  ∆G = −20 kJ · mol−1 (spontaneous)

1.3  Energy and Metabolism  13

SA MP L E CA LCULAT I ON 1. 2 Problem  Use the information given in Sample Calculation 1.1 to determine whether the reaction A → B is spontaneous at 25°C.



= 15,000 − 22,400 J · mol−1



= −7400 J · mol−1

Solution  Substitute the values for ∆H and ∆S, calculated in Sample Calculation 1.1, into Equation 1.2. To express the tem­ perature in Kelvin, add 273 to the temperature in degrees Celsius: 273 + 25 = 298 K.



= −7.4 kJ · mol−1

∆G = ∆H − T∆S

Because ∆G is less than zero, the reaction is spontaneous. Even though the change in enthalpy is unfavorable, the large increase in entropy makes ∆G favorable.

= 15,000 J · mol−1 − 298 K (75 J · K−1 · mol−1)

When the reactions are combined, their ∆G values are added, so the overall process has a negative change in free energy: A + B → B + C   ∆G = (15 kJ · mol−1) + (−20 kJ · mol−1)       A → C  ∆G = −5 kJ · mol−1 This phenomenon is shown graphically in Figure 1.9. In effect, the unfavorable “uphill” reaction A → B is pulled along by the more favorable “downhill” reaction B → C. Cells couple unfavorable metabolic processes with favorable ones so that the net change in free energy is negative. Note that it is permissible to add ∆G values because the free energy, G, depends only on the initial and final states of the system, without regard to the specific chem­ ical or mechanical work that occurred in going from one state to the other. Most macroscopic life on earth today is sustained by the energy of the sun (this was not always the case, nor is it true of all organisms). In photosynthetic organisms, such as green plants, light energy excites certain molecules so that their subsequent chemi­ cal reactions occur with a net negative change in free energy. These thermodynamically favorable (spontaneous) reactions are coupled to the unfavorable synthesis of monosac­ charides from atmospheric CO2 (Fig. 1.10). In this process, the carbon is reduced. Reduc­ tion, the gain of electrons, is accomplished by the addition of hydrogen or the removal of oxygen (the oxidation states of carbon are reviewed in Table 1.3). The plant—or an ani­ mal that eats the plant—can then break down the monosaccharide to use it as a fuel to power other metabolic activities. In the process, the carbon is oxidized—it loses elec­ trons through the addition of oxygen or the removal of hydrogen—and ultimately becomes CO2. The oxidation of carbon is thermodynamically favorable, so it can be cou­ pled to energy-requiring processes such as the synthesis of building blocks and their polymerization to form macromolecules.

B

Free energy (G)

B

A C

Reaction coordinate   FIGURE 1.9   Free energy changes in coupled reactions.  A nonspontaneous reaction, such as A → B, which has a positive value of ∆G, can be coupled to another reaction, B → C, which has a negative value of ∆G and is therefore spontaneous. The reactions are coupled because the product of the first reaction, B, is a reactant for the second reaction.

Question  Which reaction occurs spontaneously in reverse: C → B, B → A, or C → A?

SEE SAMPLE CALCULATION VIDEOS

14  C ha pter 1   The Chemical Basis of Life

Free energy

Light energy

O

C

O

Reduction (unfavorable)

H C

OH

Carbon of a monosaccharide

Oxidation (favorable)

O

C

O

  FIGURE 1.10   Reduction and reoxidation of carbon compounds.  The sun pro­

TA BLE 1 .3   Oxidation States of Carbon

Compounda

Formula

Carbon dioxide Carbon dioxide most oxidizedmost oxidized (least reduced(least ) reduced )

O

Acetic acid Acetic acid

H

C H C H

Carbon monoxide Carbon monoxide

C

O

Formic acid Formic acid

H

C

Acetone

H H C H

Acetone

H

AcetaldehydeAcetaldehyde

H

C H

OO

Virtually all metabolic processes occur with the aid of catalysts called enzymes, most of which are proteins (a catalyst greatly increases the rate of O a reaction without itself undergoing any net change). For example, specific CH C C enzymes catalyze the formation of peptide, phosphodiester, and glycosidic link­ OH OH ages during polymer synthesis. Other enzymes catalyze cleavage of these bonds H to break the polymers into their monomeric units. C O A living organism—with its high level of organization of atoms, molecules, O O and larger structures—represents a state of low entropy relative to its surround­ H C ings. Yet the organism can maintain this thermodynamically unfavorable state OH OH as long as it continually obtains free energy from its food. Thus, living organ­ O isms do indeed obey the laws of thermodynamics. When the organism ceases to O H H H obtain a source of free energy from its surroundings or exhausts its stored food, C H C C HC C H the chemical reactions in its cells reach equilibrium (∆G = 0), which results H H H in death. H O O CH C C Before Going On H H H

C

Ethanol

Ethanol

H H

O

CH HC

C

H

H

H

H

C

C CH OH

C

H

H

H

H

HH

H Ethane

Ethane

H

C

a

H

• Show how increasing temperature affects ∆G when ∆H and ∆S are constant.

H OH

• Explain how thermodynamically unfavorable reactions proceed in vivo. • Explain why an organism must have a steady supply of food. • Describe the cycle of carbon reduction and oxidation in photosynthesis and in the breakdown of a compound such as a monosaccharide.

H C

C

HH

The Origin of Cells

H 1.4

H

H

H

H

C

CH HC

C

LE A RNING OBJECTIVES H

H

H

H

H

Summarize the evolutionary history of cells.

H

H

Methane Methane least oxidizedleast oxidized (most reduced(most ) reduced )

O

C

C

Ethene

C H

H

Ethene

• Make up values for ∆H and ∆S to generate ∆G values corresponding to a spontaneous and a nonspontaneous reaction.

H

H

Acetylene Acetylene

O

OH

H

Formaldehyde Formaldehyde

C

vides the free energy to convert CO2 to reduced compounds such as monosaccharides. The reoxidation of these compounds to CO2 is thermodynamically spontaneous, so free energy can be made available for other metabolic processes. Note that free energy is not actually a substance that is physically released from a molecule.

C

HH

H

Compounds are listed in order of decreasing ­oxidation state of the red carbon atom.

C H

H

• List the events that must have occurred during prebiotic evolution. • Name the three domains of life. • Distinguish prokaryotic and eukaryotic cells. • Summarize the importance of the human microbiota.

Every living cell originates from the division of a parental cell. Thus, the ability to replicate (make a replica or copy of itself) is one of the universal characteristics of living organisms. In order to leave descendants that closely resemble itself, a cell must contain a set of instructions—and the means for carrying them out—that can be transmitted from generation to generation. Over time, the instructions change gradually, so that species also change, or evolve. By carefully examining an organism’s genetic information and the cellular machinery that supports it, biochemists can draw some conclusions about the organ­ ism’s relationship to more ancient life-forms. The history of evolution is therefore contained not just within the fossil record but also in the molecular makeup of all living cells. For exam­ ple, nucleic acids participate in the storage and transmission of genetic information in all organisms, and the oxidation of glu­ cose is an almost universal means for generating metabolic free energy. Consequently, DNA, RNA, and glucose must have been present in the ancestor of all cells.

SPL/Science Source

1.4  The Origin of Cells  15

  FIGURE 1.11  A hydrothermal vent.  Life may have originated

at these “black smokers,” where high temperatures, H2S, and metal sulfides might have stimulated the formation of biological molecules.

Prebiotic evolution led to cells Experimental evidence strongly supports the hypothesis that early in the earth’s history, small biological molecules such as amino acids and monosaccharides—and even the more elabo­ rate nucleotides—arose spontaneously from inorganic (prebiotic) materials like H2O, CH4, and HCN. Such reactions, which need an energy source, might have occurred near hydrother­ mal vents, where water as hot as 350°C emerges from the sea floor (Fig. 1.11), or in terrestrial ponds exposed to lighting and ultraviolet radiation. A less-popular hypothesis is that organic molecules were delivered to the earth by meteorites, although the origin of those molecules also needs an explanation. Researchers have recovered amino acids and other biological mol­ ecules from hydrothermal vents and from laboratory experiments designed to mimic condi­ tions on the early earth. Even with some uncertainty about the exact temperatures, reactant concentrations, and the involvement of catalysts like iron or nickel, organic molecules could have formed through hypothetical processes such as these:

H O

O 3H

H

C H

C

O C H

H C H

H O

H

5H C

H C H + H C

N +

C

N

N

C

H

Glyceraldehyde (carbohydrate)

O

N

N

H C C

N C N H

Adenine (nucleotide base)

H

O

H

O H

O

C

H C

N

H H

H

Glycine (amino acid)

Regardless of how they formed, the first biological building blocks must have accu­ mulated, acquired reactive phosphate or thioester groups, and reached concentrations that allowed the formation of polymers. Polymerization might have been stimulated when the organic molecules—often bearing anionic (negatively charged) groups—aligned themselves on a cationic (positively charged) mineral surface:

H

16  C ha pter 1   The Chemical Basis of Life − Monomers

Polymer

−

− − +

+

+

+

−

−

−

−

+

+

+

+

− − − − +

+

+

+

Clay

In fact, in the laboratory, common clay promotes the polymerization of nucleotides into RNA. Primitive polymers would have had to gain the capacity for self-replication. Otherwise, no mat­ ter how stable or chemically versatile, such molecules would never have given rise to any­ thing larger or more complicated: The probability of assembling a fully functional cell from a solution of thousands of separate small molecules is practically nil. Because RNA in modern cells represents a form of genetic information and participates in all aspects of expressing that information, it may be similar to the first self-replicating biopolymer. It might have made a copy of itself by first making a complement, a sort of mirror image, that could then make a complement of itself, which would be identical to the original molecule (Fig. 1.12). A replicating molecule’s chances of increasing in number depend on natural selection, the phenomenon whereby the entities best suited to the prevailing conditions are the likeliest to survive and multiply (Box 1.B). This would have favored a replicator that was chemically stable and had a ready supply of building blocks and free energy for making copies of itself. Accordingly, it would have been advanta­geous to become enclosed in some sort of membrane that could prevent valuable small molecules from diffusing away. Natural selection would also have favored replicating systems that developed the means for synthesizing their own building blocks and for more efficiently harnessing sources of free energy. Fossil evidence for microscopic life dates to about 3.5 billion years ago, about a billion years after the earth formed from interstellar dust. The first cells were probably able to “fix” CO2—that is, convert it to reduced organic ­c ompounds—using the free energy released in the oxidation of readily available inor­ ganic compounds such as H2S or Fe2+. Vestiges of these processes can be seen in modern metabolic reactions that involve sulfur and iron. Later, photosynthetic organisms similar to present-day cyanobacteria (also called bluegreen algae) used the sun’s energy to fix CO2: CO2 + H2O → (CH2O) + O2 The concomitant oxidation of H 2 O to O 2 dramatically increased the concentra­ tion of atmospheric O 2, about 2.5 billion years ago, and made it possible for aerobic 1. The polyA molecule serves as a template for the synthesis of a polymer containing uracil nucleotides, U, which are complementary to adenine nucleotides (in modern RNA, A pairs with U). U A A A A

A Original

U

3. The polyU molecule serves as a template for the synthesis of a new complementary polyA chain.

2. The two polymer chains separate.

U

U U

A

A A A A A A

U U U U U

A A A A A

+

U U U U U

A

4. The chains again separate and the polyU polymer is discarded, leaving the original polyA molecule and its exact copy.

A

A

A A A A A

Original Complement

  FIGURE 1.12  Possible mechanism for the self-replication of a primitive RNA molecule. 

For simplicity, the RNA molecule is shown as a polymer of adenine nucleotides, A. Question  Draw a diagram showing how polyU would be replicated.

+

U U U U U

A A A A A

U U U U U

A A A A A

+

A A A A A

Original Copy

1.4  The Origin of Cells  17

Box 1.B How Does Evolution Work? Documenting evolutionary change is relatively straightforward, but the mechanisms whereby evolution occurs are prone to misunder­ standing. Populations change over time, and new species arise as a result of natural selection. Selection operates on individuals, but its effects can be seen in a population only over a period of time. Most populations are collections of individuals that share an overall genetic makeup but also exhibit small variations due to random alter­ ations (mutations) in their genetic material as it is passed from parent to offspring. In general, the survival of an individual depends on how well suited it is to the particular conditions under which it lives. Individuals whose genetic makeup grants them the greatest rate of survival have more opportunities to leave offspring with the same genetic makeup. Consequently, their characteristics become widespread in a population and, over time, the popula­ tion appears to adapt to its environment. A species that is well suited to its environment tends to persist; a poorly adapted species fails to reproduce and therefore dies out. Because evolution is the result of random variations and changing probabilities for successful reproduction, it is inherently random and unpredictable. Furthermore, natural selection acts on the raw materials at hand. It cannot create something out of nothing but must operate in increments. For example, the insect wing did not suddenly appear in the offspring of a wingless par­ ent but most likely developed bit by bit, over many generations, by modification of a heat-exchange appendage. Each step of the

wing’s development would have been subject to natural selection, eventually making an individual that bore the appendage more likely to survive, perhaps by being able to first glide and then actu­ ally fly in pursuit of food or to evade predators. Although we tend to think of evolution as an impercepti­ bly slow process, occurring on a geological time scale, it can be observed. For example, under optimal conditions, the bacterium Escherichia coli requires only about 20 minutes to produce a new generation. In the laboratory, a culture of E. coli cells can prog­ ress through about 2500 generations in a year (in contrast, 2500 human generations would require about 60,000 years). Hence, it is possible to subject a population of cultured bacterial cells to some “artificial” selection—for example, by making an essential nutrient scarce—and observe how the genetic composition of the population changes over time. Experiments such as these have revealed that even after an initial period of rapid adaptation to the new conditions, the popu­ lation continues to change, suggesting that evolution is an ongo­ ing process with no fixed end point. In addition, so-called neutral mutations, which do not significantly affect an organism’s fitness, randomly accumulate. For this reason, it is impossible to explain all genetic changes in terms of adaptations to specific conditions. Question  Why can’t acquired (rather than genetic) characteristics serve as the raw material for evolution?

(oxygen-using) organisms to take advantage of this powerful oxidizing agent. The anaerobic origins of life are still visible in the most basic metabolic reactions of modern organ­ isms; these reactions proceed in the absence of oxygen. Now that the earth’s atmosphere contains about 20% oxygen, anaerobic organisms have not disappeared, but they have been restricted to microenvironments where O2 is scarce, such as the digestive systems of ani­ mals or underwater sediments.

Eukaryotes are more complex than prokaryotes The earth’s present-day life-forms are of two types, which are distinguished by their cellular architecture: 1. Prokaryotes are small unicellular organisms that lack a discrete nucleus and usually contain no internal membrane systems. This group comprises two subgroups that are remarkably different metabolically, although they are similar in appearance: the eubacteria (usually just called bacteria), exemplified by E. coli, and the archaea (or archaebacteria), best known as organisms that inhabit extreme environments, although they are actually found almost everywhere (Fig. 1.13). 2. Eukaryotic cells are usually larger than prokaryotic cells and contain a nucleus and other membrane-bounded cellular compartments (such as mitochondria, chloroplasts, and endoplasmic reticulum). Eukaryotes may be unicellular or multicellular. This group (also called the eukarya) includes microscopic organisms as well as familiar macroscopic plants and animals (Fig. 1.14). By analyzing the sequences of nucleotides in certain genes that are present in all species, it is possible to construct a diagram that indicates how the bacteria, archaea, and eukarya are related. The number of sequence differences between two groups of organisms indicates how long ago they diverged from a common ancestor. Species with similar sequences have a longer shared evolutionary history than species with dissimilar sequences. This sort of analysis has produced the evolutionary tree shown in Figure 1.15.

Dr. David J. Patterson/Science Source

E. Gray/Science Source

18  C ha pter 1   The Chemical Basis of Life

  FIGURE 1.14   A eukaryotic cell.  The paramecium, a one-

celled organism, contains a nucleus and other membrane-bounded compartments.

  FIGURE 1.13  Prokaryotic cells. 

These single-celled Escherichia coli bacteria lack a nucleus and internal membrane systems.

Question  Describe the visible differences between prokaryotic and eukaryotic cells (Figures 1.13 and 1.14).

  FIGURE 1.15  Evolutionary tree based on nucleotide sequences.  This diagram reveals that the ancestors of archaea and bacteria separated before the eukarya emerged from an archaea-like ancestor. Note that the closely spaced fungi, plants, and animals are actually more similar to each other than are many groups of prokaryotes.

Prokaryotes

Eukaryotes Animals

Bacteria

Eukarya Fungi Plants

Archaea

Eukaryotic cells are likely to have evolved from a mixed population of prokaryotic cells about 1.8 billion years ago. Over many generations of living in close proximity and sharing each other’s metabolic products, some of the bacterial cells became stably incorporated inside archaeal cells, which accounts for the mosaic character of modern eukaryotic cells (Fig. 1.16). The descendants of the once free-living bacteria became mitochondria, which carry out most of the eukaryote’s oxidative metabolism, or chloroplasts, which carry out photosynthesis in plants and closely resemble the photosynthetic cyanobacteria. In fact, both mitochondria and chloroplasts contain their own genetic material and protein-synthesizing machinery and can grow and divide independently of the rest of the cell. Remnants of the archaeal ancestor of eukaryotes are found in the eukaryotic cytoskeleton and DNA-replication enzymes. Nucleus

DNA

Organelles

  FIGURE 1.16  Possible origin of eukaryotic cells.  The close

association of different kinds of free-living cells gradually led to the modern eukaryotic cell, which appears to be a mosaic of bacterial and archaeal features and contains organelles that resemble whole bacterial cells.

1.4  The Origin of Cells  19

Science Source/Science Source

The evolution of eukaryotes included the development of extensive intracellular membranes, many of which enclose discrete compartments, or Plasma organelles, with specialized functions, such as lysosomes (for the degradation membrane of macromolecules), p ­ eroxisomes (for some oxidative reactions), and vacuoles (for storage). Like mitochondria and chloroplasts, the nucleus is surrounded Endoplasmic reticulum by a double membrane; the outer nuclear membrane bends and folds exten­ sively to form the endoplasmic reticulum. Some key features of eukaryotic Mitochondrion cell structure are shown in Figure 1.17. A few types of prokaryotic cells also Nucleus contain membrane-enclosed compartments devoted to storage or to chemical processes that could potentially damage the rest of the cell, but these cells lack a nucleus and the variety of organelles that characterize eukaryotic cells. Many types of prokaryotic and eukaryotic cells live in colonies, which may increase metabolic efficiency through the cooperation of individual cells. However, a true multicellular lifestyle, which entails the division of labor and specialization among different types of cells, occurs only in eukaryotes. Multicellular organisms first appeared in the fossil record about 600 million years ago.   FIGURE 1.17  Eukaryotic cell structure.  Key The earth currently sustains about 10 million different species components of an animal cell are labeled in this (although estimates vary widely). Perhaps some 500 million species have electron micrograph. A typical plant cell also contains chloroplasts and is surrounded by a cell wall. appeared and vanished over the course of evolutionary history. It is unlikely that the earth harbors more than a few mammals that have yet to be discovered, but new microbial species are routinely described. Although the number of known prokaryotes (about 10,000) is much less than the number of known eukaryotes (for example, there are about 900,000 known species of insect), p ­ rokaryotic metabolic lifestyles are amazingly varied.

The human body includes microorganisms The human body consists of an estimated 10 trillion (1013) cells, and there may be as many as 40 to 100 trillion microorganisms living in and on the body, including bacteria, archaea, and fungi (viruses are also present but are molecular parasites rather than true cells). The “for­ eign” cells, mostly living in the intestine, form an integrated community called the microbiota. Very few of these cells are pathogens that cause disease; in fact, a diverse and stable microbiota actually helps to prevent the growth of harmful species. Analysis of microbial DNA has allowed characterization of the microbiome, the genetic information contributed by the organisms that make up the microbiota. The microbiome reveals that the human species hosts several thousand different species of microorganisms. Many have not yet been identified, and relatively few have been cultured. However, an indi­ vidual person typically harbors only a few hundred species of microorganisms. Establishing this community of cells begins at birth and is mostly complete after about a year. The mix of species remains fairly constant throughout the person’s lifetime, although the proportions may fluctuate somewhat. The microbiota can vary markedly among individuals, even in the same household. However, the overall metabolic capabilities of the microbial community seem to matter more than which species are actually present. Contrary to expectations, there is no core microbiota that is common to all individuals. The gut microbiota in particular plays a large role in digestion, providing nutrients to the host and regulating metabolic functions. However, byproducts of microbial metabolism can act as hormones and neurotransmitters with possible links to anxiety and depression. In turn, antibiotics, other types of drugs, and even some personal care products can alter the commu­ nity of microorganisms and the chemicals they produce. All the while, the human immune system must ignore—or “tolerate”—the microbiota. If this balance is disrupted, the immune system may react to the microorganisms, generating inflammatory responses that can lead to diabetes or inflammatory bowel disease. Untangling these complicated relationships is the goal of the Integrated Human Microbiome Project (https://hmpdacc.org/ihmp/), which orga­ nizes data describing the species that inhabit the human body and contribute to human health and disease.

20  C ha pter 1   The Chemical Basis of Life

Before Going On • Describe how simple prebiotic compounds could give rise to biological monomers and polymers. • Explain why anaerobic organisms arose before aerobic organisms. • Describe the differences between prokaryotes and eukaryotes. • Explain why eukaryotic cells appear to be mosaics. • List some functions of the human microbiota.

Summary 1.2  Biological Molecules •  The most abundant elements in biological molecules are H, C, N, O, P, and S, but a variety of other elements are also present in living systems. •  The major classes of small molecules in cells are amino acids, monosaccharides, nucleotides, and lipids. The major types of bio­ logical polymers are proteins, nucleic acids, and polysaccharides.

1.3  Energy and Metabolism

•  Life is thermodynamically possible because unfavorable ender­ gonic processes are coupled to favorable exergonic processes.

1.4  The Origin of Cells •  The earliest cells may have evolved in concentrated solutions of molecules or near hydrothermal vents. •  Eukaryotic cells contain membrane-bounded organelles. Prokar­ yotic cells, which are smaller and simpler, include the bacteria and the archaea.

•  Free energy has two components: enthalpy (heat content) and entropy (disorder). Free energy decreases in a spontaneous process.

Key Terms bioinformatics homeostasis trace element amino acid carbohydrate monosaccharide nucleotide lipid monomer polymer residue polypeptide protein peptide bond conformation polynucleotide nucleic acid

phosphodiester bond polysaccharide glycosidic bond free energy (G) enthalpy (H) entropy (S) exothermic reaction endothermic reaction ∆G spontaneous process exergonic reaction nonspontaneous process endergonic reaction in vitro in vivo reduction oxidation

enzyme replication evolution complement natural selection aerobic anaerobic prokaryote bacteria archaea eukaryote eukarya organelle microbiota microbiome

Bioinformatics Brief Bioinformatics Exercises 1.1  The Periodic Table of the Elements and Domains of Life

1.2  Organic Functional Groups and the Three-Dimensional Struc­ ture of Vitamin C

Problems  21

Problems 1.2  Biological Molecules

H3C

CH3

CH3

CH3 CH2OH

1.  Use Table 1.1 to assign the appropriate compound name to each molecule.

Retinol

O a. H3C

(CH2)14

b. H3C

CH2

C

OH

H3C

CH3 O

CH3

CH3

C

NH2

H

Retinal

O c. H3C

C

d. H3C

CH2

O

CH2

CH3

H3C

OH

H3C

O

CH3

OH b. H3C

O

O

P

H3C

OH

CH3

N

SH

CH2

O d. H3C

C

H3C

H2 C

CH3

C H2

H C

O C H

C

6.  The structures of several molecules are shown below. Use Table 1.1 to identify the functional groups in each structure.

O

H

HO

O

OH

O

C OH

b.

H

N

O2N

H3C d.

H3C H2 C

CH3 H3C

C H

CH3

S

N

O

Nicotinic acid (niacin)

H3CO

CH3 (CH2

H3CO

CH3

CH3 CH

C

CH2)10H

O Coenzyme Q

O CH2

OH

O

O C H2

C

CH2OH

Vitamin C

c.

CH3

O

3.  Use Table 1.1 to assign the appropriate compound name to each molecule. a.

OH

5.  Investigators synthesized a series of compounds that showed promise as drugs for the treatment of Alzheimer’s disease. The struc­ ture of one of the compounds is shown below. Use Table 1.1 to iden­ tify the functional groups in this compound.

O c.

C Retinoic acid

2.  Use Table 1.1 to assign the appropriate compound name to each molecule. a.

CH3 O

CH3

CH3

P O−

O O

P O−

O−

4.  The structures of three forms of vitamin A are shown. Use Table  1.1 to assign the appropriate compound name to each molecule.

7.  Coenzyme A is an important carrier of acetyl groups in metabo­ lism. Its structure is shown in Figure 3.2. Use Table 1.1 to identify the functional groups in this molecule. 8.  If an acetyl group is attached to the sulfhydryl group of coenzyme A (see Solution 1.7), what new functional group is formed? 9.  Name the four types of small biological molecules. Which three are capable of forming polymeric structures? What are the names of the polymeric structures that are formed?

22  C ha pter 1   The Chemical Basis of Life 10.  To which of the four classes of biomolecules do the following compounds belong? a.

CH2OH O H OH H

H HO

OCH3 N HO

OH

H

C

CH3

O

O NH2

O

P

O

O− H

c. HS

CH2

O

CH2

CH2

H

O

H

NH2

N

H

O−

12.  The compound shown below was isolated from the starfish Anthenea aspera. What two types of biomolecules are found in this compound?

N

−O

N

H

P

N

O−

b.

O

O

CH2

H

NH

H

d.

H

H

OH

OH

CH

COO−

N

H

O

O

OH

HO OH

NH+ 3

HO

d. d.

C

O

11.  To which of the four classes of biomolecules do the following compounds belong? a.

O O−

b.

O +

H3N

CH

OH OCH3

13.  The nutritive quality of food can be analyzed by measuring the amounts of the chemical elements it contains. Most foods are mix­ tures of the three major types of molecules:  a.  fats (lipids),  b. car­ bohydrates, and  c.  proteins. What elements are present in each of these types of molecules?

O R

NH

O

C

O−

CH2

14.  A compound present in many foods has the formula C44H86O8NP. To which class of molecules does this compound belong? Explain your answer. 15.  A healthy diet must include some protein. Assuming you had a way to measure the amount of each element in a sample of food, which element would you measure in order to tell whether the food contained protein? 16.  The structures of three compounds are shown below. Based on your answer to Problem 15, which of the three compounds would you add to a food sample so that it would appear to contain more protein? Which of the three compounds would already be present in a food sample that actually did contain protein? Explain.

Se

O

H c.

HO H

CH2OH O H OH H H

OH

H

O C H OH

H

C

H2N

N

3N

CH CH2

NH2

C

O

O− A

B

C

CH2

N

N

OH

2− CH2OPO3

+H

NH2

C

O−

Problems  23 17.  The structure of the compound urea is shown. Urea is a waste product of metabolism excreted by the kidneys into the urine. Why do doctors tell patients with kidney damage that they should con­ sume a low-protein diet?

O−

b.

H3C O

+

H3N

C Urea

O

19.  Consult Table 4.1 for the structures of the amino acids serine and lysine. What functional group does serine have that lysine does not? 20.  Many amino acid side chains are modified after translation. Use the structures of serine and lysine from Problem 19 to draw the structures of the following modified amino acids:  a. phosphoserine, b.  hydroxylysine, and  c. acetyllysine. 21.  The “straight-chain” structure of glucose is shown in Section 1.2. What functional groups are present in the glucose molecule?

CH3

c.

H3C

−O

O

P −O

O

5′

CH2

4′

H

C

O

HO

C

H

H

C

OH

H

C

OH

H

H

3′

2′

O

P −O

4′

4

H

3

H

2

C

CH

H N

CH2OH O H OH H

5

H

3

2

H

OH

Galactose

OH

Glucose Lactose

H

O CH

C

OCH3

O

Aspartame 27.  Compare the solubilities in water of alanine, glucose, palmitate, and cholesterol, and explain your reasoning.

NO2

6

H

H

CH2

Cytosine

1(β)O 4

2′

OH

N H

25.  What types of linkages are found in the following compounds?

HO

3′

28.  Cell membranes are largely hydrophobic structures. Which compound will pass through a membrane more easily, glucose or 2,4-dinitro­phenol? Explain.

24.  What are the structural components of the biological molecules called nucleotides?

CH2OH 5 O H OH H

H

O

O−

N

6

H

N

OH

C

NH2

Uracil

O

CH2

CH2

23.  The structures of the nitrogenous bases uracil and cytosine are shown below. How do their functional groups differ?

O

N

5′

H

+H N 3

Fructose

HN

NH2

H

H

O O

O

N

O

CH2OH

O

NH

O

26.  The structure of the artificial sweetener aspartame is shown below. What type of linkage is found in this compound?

CH2OH

a.

O

O

22.  Consider the monosaccharide fructose.  a.  How does its molec­ ular formula differ from that of glucose?  b.  How does its structure differ from the structure of glucose?

N H

O−

N H

NH2

18.  The structures of the amino acids asparagine (Asn) and cysteine (Cys) are shown in Section 1.2. What functional group does Asn have that Cys does not? What functional group does Cys have that Asn does not?

O

O

O

H N

N H

O H2N

CH3

OH 1(β)

H

NO2 2,4-Dinitrophenol 29.  What polymeric molecule forms a more regular structure, DNA or protein? Explain this observation in terms of the cellular roles of the two different molecules. 30.  What are the two major biological roles of polysaccharides? 31.  Pancreatic amylase digests the glycosidic bonds that link glucose residues together in starch. Would you expect this enzyme to digest the glycosidic bonds in cellulose as well? Explain why or why not. 32.  The complete digestion of starch in mammals yields 4 kilocalo­ ries per gram (see Problem 31). What is the energy yield for cellulose?

24  C ha pter 1   The Chemical Basis of Life

1.3  Energy and Metabolism 33.  What is the sign of the entropy change for each of the following processes?  a.  Water freezes.  b.  Water evaporates.  c.  Dry ice sub­ limes.  d.  Sodium chloride dissolves in water.  e.  Several different types of lipid molecules assemble to form a membrane. 34.  Does entropy increase or decrease in the following reactions in aqueous solution?

COO−

a. a. COO−

O + CO2(g)

C

C

O

CH2

CH3

C

H

O + H+

C

CH3

O + CO2(g)

CH3

35.  Which has the greater entropy, a polymeric molecule or a mix­ ture of its constituent monomers? 36.  How does the entropy change when glucose undergoes combustion? C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 37.  A soccer coach keeps a couple of instant cold packs in her bag in case one of her players suffers a muscle injury. Instant cold packs are composed of a plastic bag containing a smaller water bag and solid ammonium nitrate. In order to activate the cold pack, the bag is kneaded until the smaller water bag breaks, which allows the released water to dissolve the ammonium nitrate. The equation for the dissolution of ammonium nitrate in water is shown below. How does the cold pack work? H2O

NH4NO3 (s)

− ​​NH​  + 4​  ​​  (aq) + ​​NO​  3​  ​​  (aq)

ΔH = 26.4 kJ · mol−1 38.  Campers carry hot packs with them, especially when camping during the winter months or at high altitudes. The design is simi­ lar to that described in Problem 37, except that calcium chloride is used in place of the ammonium nitrate. The equation for the disso­ lution of calcium chloride in water is shown below. How does the hot pack work? H2O

CaCl2 (s)

43.  For a given reaction, the value of ΔH is 15 kJ · mol−1 and the value of ΔS is 51 J · K−1 · mol−1. Above what temperature will this reaction be spontaneous? 44.  Which of the following processes are spontaneous?  a.  A reac­ tion that occurs with any size decrease in enthalpy and any size increase in entropy.  b.  A reaction that occurs with a small increase in enthalpy and a large increase in entropy.  c.  A reaction that occurs with a large decrease in enthalpy and a small decrease in entropy.  d.  A reaction that occurs with any size increase in enthalpy and any size decrease in entropy. 45.  Given that the process described in Problem 37 is spontaneous, what are the signs of ΔS and ΔG? Confirm your answer by using the enthalpy change provided to calculate the sign and magnitude of ΔS. Assume a temperature of 25°C.

COO− b. COO−

42.  Is the reaction described in Problem 41 favorable at  a.  4°C and b. 37°C?

Ca2+(aq) + 2 Cl−(aq)  ΔH = −81 kJ · mol−1

39.  Urea (NH2CONH2) dissolves readily in water; i.e., this is a spon­ taneous process. A beaker containing the dissolved compound is cold to the touch. What conclusions can you make about the sign of the a.  enthalpy change and  b.  entropy change for this process? 40.  Would you expect the reaction shown in Problem 36 to be ender­ gonic or exergonic? Explain. 41.  Calculate ∆H and ∆S, as described in Sample Calculation 1.1, for the reaction in which reactant A is converted to product B.

46.  Given that the process described in Problem 38 is spontaneous, what are the signs of ΔS and ΔG? Confirm your answer by using the enthalpy change provided to calculate the sign and magnitude of ΔS. Assume a temperature of 25°C. 47.  The hydrolysis of pyrophosphate at 25°C is spontaneous. The enthalpy change for this reaction is −14.3 kJ · mol−1. What is the sign and the magnitude of ΔS for this reaction? 48.  Phosphoenolpyruvate donates a phosphate group to ADP to produce pyruvate and ATP. The ∆G value for this reaction at 25°C is −63 kJ · mol−1 and the value of ΔS is 190 J · K−1 · mol−1. What is the value of ∆H? Is heat absorbed from or released to the surroundings? 49.  A monoclonal antibody binds to the protein cytochrome c. The ΔH value for binding at 25°C is −87.9 kJ · mol−1 and the ΔS is −118 J · K−1 · mol−1.  a.  Does entropy increase or decrease when the anti­ body binds to the protein?  b.  Calculate ΔG for the formation of the antibody–protein complex. Does the complex form spontaneously? c.  The ΔG value for the binding of a second monoclonal antibody to cytochrome c is −58.2 kJ · mol−1. Which antibody binds more readily to the protein? 50.  Phosphofructokinase catalyzes the transfer of a phosphate group (from ATP) to fructose-6-phosphate to produce fructose1,6-bisphosphate. The ΔH value for this reaction is −9.5 kJ · mol−1 and the ΔG is −17.2 kJ · mol−1 at 37°C.  a.  Is heat absorbed from or released to the surroundings?  b.  What is the value of ΔS for the reaction? Does this reaction proceed with an increase or decrease in entropy?  c.  Which component makes a greater contribution to the free energy change: the ΔH or ΔS value? Comment on the significance of this observation. 51.  Glucose can be converted to glucose-6-phosphate: glucose + phosphate → glucose-6-phosphate + H2O ΔG is 13.8 kJ · mol−1 a.  Is this reaction favorable? Explain. b.  Suppose the synthesis of glucose-6-phosphate is coupled with the hydrolysis of ATP. Write the overall equation for the coupled process and calculate the ΔG for the coupled reaction. Is the conversion of glucose to glucose-6-phosphate favorable under these conditions? Explain. ATP + H2O → ADP + phosphate   ∆G = −30.5 kJ · mol−1

H (kJ · mol−1)

S (J · K−1 · mol−1)

A

54

22

52.  Glyceraldehyde-3-phosphate (GAP) is converted to 1,3-bis­ phosphoglycerate (1,3BPG) as shown.

B

60

43

GAP + Pi + NAD+ → 1,3BPG + NADH

ΔG = 6.7 kJ · mol−1

Problems  25 a.  Is this reaction spontaneous? b.  The reaction shown above is coupled to the following reaction in which 1,3BPG is converted to 3-phosphoglycerate (3PG): 1,3BPG + ADP → 3PG + ATP

ΔG = −18.8 kJ · mol−1

Write the equation for the overall conversion of GAP to 3PG. Is the coupled reaction favorable? 53.  Place these molecules in order from the most oxidized to the most reduced.

C

60.  Which yields more free energy when completely oxidized, stea­ rate or α-linolenate?

H3C

H

O H

59.  In some cells, lipids such as palmitate (shown in Section 1.2), rather than monosaccharides, serve as the primary metabolic fuel. a. Consider the oxidation state of palmitate’s carbon atoms and explain how it fits into a scheme such as the one shown in Figure 1.10.  b.  On a per-carbon basis, which would make more energy available for metabolic reactions: palmitate or glucose?

OH

C

H

O

H

C

H3C

B

b.  An animal eats the plant and breaks down the monosaccharide in order to obtain energy for cellular processes. 55.  Given the following reactions, tell whether the reactant is being oxidized or reduced. Reactions may not be balanced.

O (CH2)14

O O−

C

C

S

CoA

CH2

CH2 OH

C

COO−

O

COO−

CH

CH2

CH

CH

COO−

COO−

OH

+H

3N

CH

C

O−

S

C

CH

62.  An enzyme binds to a substrate, forming an enzyme–substrate complex. Explain how this process obeys the laws of thermodynamics.

1.4  The Origin of Cells 63.  Why is molecular information so important for classifying and tracing the evolutionary relatedness of bacterial species but less important for vertebrate species?



Species A

T C G T C G A G T C



Species B

T G G A C T A G C C



Species C

T G G A C C A G C C

O + H2

2

+H

3N

CH

C

NH+ 3

H7

O−

CH2

CH2 −O

61.  A protein folds from a random coil into a highly ordered struc­ ture. How does this process, which proceeds with a loss of entropy, obey the laws of thermodynamics?

H15

CH2 S

COO−

66.  A portion of the evolutionary tree for a flu virus is shown here. Different strains are identified by an H followed by a num­ ber.  a.  Identify two pairs of closely related flu strains.  b. Which strain(s) is(are) most closely related to strain H3?

O

d.

(CH2)6

65.  Draw a simple evolutionary tree that shows the relationships between species A, B, and C based on the DNA sequences given here.

COO−

c. COO−

CHCH2)3

64.  The first theories to explain the similarities between bacteria and mitochondria or chloroplasts suggested that an early eukaryotic cell engulfed but failed to fully digest a free-living prokaryotic cell. Why is such an event unlikely to account for the origin of mitochondria or chloroplasts?

COO−

b. COO−

CH

8 CH3

(CH

α-Linolenate

a.  Monosaccharides are synthesized from carbon dioxide by plants during photosynthesis.

CH3

CH2

C

54.  Identify the process described in the following statements as an oxidation or reduction process.

a.

COO−

Stearate

O

H A

(CH2)16

H10 H3

SH

O 56.  For each of the reactions in Problem 55, tell whether an oxidizing agent or a reducing agent is needed to accomplish the reaction. 57.  Rank the forms of vitamin A (see Problem 4) in order from most oxidized to most reduced. 58.  The reaction shown in Problem 52 requires the coenzyme NAD+, which is reduced to NADH during the reaction. Which is more oxi­ dized, GAP or 1,3BPG?

H4 H14

67.  Propose an explanation why taking antibiotics sometimes leads to illness caused by the intestinal bacterium Clostridium difficile. 68.  Clinicians know that a drug may be ineffective as an antibiotic when tested with pure cultures of bacteria in the laboratory. Yet when the drug is orally administered to a patient, it has the desired antibac­ terial effect against the same bacteria. Explain.

26  C ha pter 1   The Chemical Basis of Life

Selected Readings Archibald, J.M., Endosymbiosis and eukaryotic cell evolution, Current Biol. 25, R911–R921, doi: 10.1016/j.cub.2015.07.055 (2015). [An extensive review of hypotheses about the origins of mitochondria and chloroplasts in eukaryotic cells.] Baum, B. and Baum, D.A., The merger that made us, BMC Biology 18, 72, doi: 10.1186/s12915-020-00806-3 (2020). [Briefly summarizes the outside-in and inside-out models for the fusion of prokaryotic cells that generated the first eukaryotes.] Cullen, C.M., Aneja, K.K., Beyhan, S., Cho, C.E., Woloszynek, S., Con­ vertino, M., McCoy, S.J., Zhang, Y., Anderson, M.Z., ­Alvarez-Ponce, D., Smirnova, E., Karstens, L., Dorrestein, P.C., Li, H., Sen Gupta, A., Cheung, K., Powers, J.G., Zhao, Z., and Rosen, G.L., Emerging pri­­ orities for microbiome research, Front. Microbiol. 11, doi: 10.3389/ fmicb.2020.00136 (2020). [Summarizes current understanding and outlines some approaches to studying microbiota.]

Koonin, E.V., Splendor and misery of adaptation, or the importance of neutral null for understanding evolution, BMC Biology 14, 114, doi: 10.1186/s12915-016-0338-2 (2016). [Includes a caution about assigning an explanatory story to every genetic change.] Pagel, M., Natural selection 150 years on, Nature 457, 808–811 (2009). [A short summary of the key points of evolution by natural selec­ tion, including the branching pattern of descent and the process of speciation.] Szostak, J.W., The narrow road to the deep past: In search of the chemistry of the origin of life, Agnew Chem. Int. Ed. Engl. 56, 11037–11043, doi: 10.1002/anie.201704048 (2017). [A review of pre­ biotic evolution and the origin of cells.]

Chapter 1 Credits Figure 1.4 Adapted from Palm, W. and Thompson, C.B., Nature 546, 234–242 (2017). Figure 1.5 Image based on 1EDN. Janes, R.W., Peapus, D.H., Wallace, B.A., The crystal structure of human endothelin, Nat. Struct. Biol. 1, 311–319 (1994). Figure 1.6 Image based on Nucleic Acids Database ARF0108. Biswas, R., Mitra, S.N., Sundaralingam, M., 1.76 A structure of a pyrimidine

start alternating A-RNA hexamer r(CGUAC)dG, Acta Crystallogr., Sect. D 54, 570–576 (1998). Figure 1.15 Adapted from Wheelis, M.L., Kandler, O., and Woese, C.R., Proc. Natl. Acad. Sci. USA 89, 2930–2934 (1992).

Aqueous Chemistry DO YOU REMEMBER? • Organisms maintain a state of homeostasis (Section 1.1).

HHelene/Shutterstock.com

CHAPTER 2

• Biological molecules are composed of a subset of all possible elements and functional groups (Section 1.2). • The free energy of a system is determined by its enthalpy and entropy (Section 1.3).

Wood ants defend themselves by squirting enemies with formic acid from a gland on their abdomen. They also use formic acid to disinfect their nests. One species, Formica paralugubris, takes this antimicrobial defense even further. When these ants drag particles of tree resin—which is already naturally antimicrobial—back to their nest, they intentionally coat it with formic acid to further enhance its antifungal activity.

Water is a fundamental requirement for life, so it is important to understand the ­structural and chemical properties of water. Not only are most biological molecules surrounded by water, but their molecular structure is in part governed by how their component groups interact with water. In addition, water plays a role in how these molecules assemble to form larger structures or undergo chemical transformation. In fact, water itself— or its H+ and OH− ­constituents—participates directly in many biochemical processes. Therefore, an examination of water is a logical prelude to exploring the structures and functions of biomolecules in the following chapters.

2.1 Water Molecules and Hydrogen Bonds LEARNING OBJECTIVES Explain water’s properties in term of its ability to form hydrogen bonds. • Describe the electronic structure of a water molecule. • Identify hydrogen bond donor and acceptor groups. • List the other types of weak noncovalent forces that affect biological molecules. • Describe how water interacts with polar and charged solutes.

27

28  C ha pter 2   Aqueous Chemistry

What is the nature of the substance that accounts for about 70% of the mass of most organisms? The human body, for example, is about 60% water by weight, most of it in the extracellular fluid (the fluid surrounding cells) and inside cells:

Intracellular water (40%)

Non-water (40%)

H

Extracellular water (15%)

O

H   FIGURE 2.1  Electronic structure of the water molecule.  Four electron orbitals, in an approximately tetrahedral arrangement, surround the central oxygen. Two orbitals participate in bonding to hydrogen (gray), and two contain unshared electron pairs.

Water in circulatory system (5%)

In an individual H2O molecule, the central oxygen atom forms covalent bonds with two hydrogen atoms, leaving two unshared pairs of electrons. The molecule therefore has approximately tetrahedral geometry, with the oxygen atom at the center of the tetrahedron, the hydrogen atoms at two of the four corners, and electrons at the other two corners (Fig. 2.1). As a result of this electronic arrangement, the water molecule is polar; that is, it has an uneven distribution of charge. The oxygen atom bears a partial negative charge (indicated by the symbol δ−), and each hydrogen atom bears a partial positive charge (indicated by the symbol δ+): δ+

δ+

δ−

This polarity is the key to many of water’s unique physical properties. Neighboring water molecules tend to orient themselves so that each partially positive hydrogen is aligned with a partially negative oxygen:

δ− δ+

  FIGURE 2.2   Structure of ice.  Each water molecule acts as a donor for two hydrogen bonds and an acceptor for two hydrogen bonds, thereby interacting with four other water molecules in the crystal. (Only two layers of water molecules are shown here.)

Question  Identify a hydrogen bond donor and acceptor in this structure.

This interaction, shaded yellow here, is known as a hydrogen bond. This weak electrostatic attraction between oppositely charged particles actually has some covalent character. In addition, the bond has directionality, or a preferred orientation. Each water molecule can potentially participate in four ­hydrogen bonds, since it has two hydrogen atoms to “donate” to a hydrogen bond and two pairs of unshared electrons that can “accept” ah ­ ydrogen bond. In ice, a crystalline form of water, each water molecule does indeed form hydrogen bonds with four other water ­molecules (Fig. 2.2). This regular, lattice-like structure breaks down when the ice melts. In liquid water, each molecule can potentially form hydrogen bonds with up to four other water molecules, but each bond has a lifetime of only about 10−12 s. As a result, the structure of water is

2.1  Water Molecules and Hydrogen Bonds  29

Because of its ability to form hydrogen bonds, water is highly cohesive. This accounts for its high surface tension, which allows certain insects to walk on water (Fig. 2.3). The cohesiveness of water molecules also explains why water remains a liquid, whereas molecules of similar size, such as CH4 and H2S, are gases at room temperature (25°C). At the same time, water is less dense than other liquids because hydrogen bonding demands that individual molecules not just approach each other but interact with a certain orientation. This geometrical constraint also explains why ice floats; for other materials, the solid is denser than the liquid.

Hydrogen bonds are one type of electrostatic force Powerful covalent bonds define basic molecular constitutions, but much weaker noncovalent bonds, including hydrogen bonds, govern the final three-dimensional shapes of molecules and how they interact with each other. For example, about 460 kJ · mol−1 (110 kcal · mol−1) of energy is required to break a covalent O—H bond. However, a hydrogen bond in water has a strength of only about 20 kJ · mol−1 (4.8 kcal · mol−1). Other noncovalent interactions are weaker still. Hydrogen bonds, with a length of about 1.8 Å, are longer and hence weaker than a covalent O—H bond (which is about 1 Å long). However, a completely noninteracting O and H would approach no closer than about 2.7 Å, which is the sum of their van der Waals radii (the van der Waals radius of an isolated atom is the distance from its nucleus to its effective electronic surface).

O

Covalent bond

H 1Å

O

H

Hydrogen bond

1.8 Å

O

H

No bond

2.7 Å

Hydrogen bonds usually involve N—H and O—H groups as hydrogen donors and the electronegative N and O and occasionally S atoms as hydrogen acceptors (electronegativity is a measure of an atom’s affinity for electrons; Table 2.1). Water, therefore, can form hydrogen

Hermann Eisenbeiss/ Science Source

continually flickering as water molecules rotate, bend, and reorient themselves. Theoretical calculations and spectroscopic data suggest that water molecules participate in only two strong hydrogen bonds, one as a donor and one as an acceptor, generating transient hydrogen-bonded clusters such as the prism structure shown here:

  FIGURE 2.3   A water strider supported by the surface tension of water.

30  C ha pter 2   Aqueous Chemistry

TA B L E 2. 1   Electronegativities of Some Elements

Element

Electronegativity

H

2.20

C

2.55

S

2.58

N

3.04

O

3.44

F

3.98

bonds not just with other water molecules but with a wide variety of other compounds that bear N- and O-containing functional groups:

H

H O

H

H

O

Water–alcohol

R O

R

H

H

N

Water–amine

R

Likewise, these functional groups can form hydrogen bonds among themselves. For example, the complementarity of bases in DNA and RNA is determined by their ability to form hydrogen bonds with each other. Here, three N—H groups are hydrogen bond donors, and N and O atoms are acceptors:

Guanine

Cytosine

Among the noncovalent interactions that occur in biological molecules are electrostatic interactions between charged groups such as carboxylate (—COO−) and amino (—​​NH​ + 3​  ​​) groups. These ionic interactions are intermediate in strength to covalent bonds and hydrogen bonds (Fig. 2.4). Other electrostatic interactions occur between particles that are polar but not actually charged, for example, two carbonyl groups:

400

Bond strength (kJ . mol−1)

350 300 250 200 150

C

100 50 0

Covalent bond

Ionic Hydrogen interaction bond

van der Waals interaction

  FIGURE 2.4   Relative strengths of bonds in ­biological molecules.

δ–

O

δ+

C

O

These forces, called van der Waals interactions, are usually weaker than hydrogen bonds. The interaction between two strongly polar groups is known as a dipole–dipole interaction and has a strength of about 9 kJ · mol−1. Very weak van der Waals interactions, called London dispersion forces, occur between nonpolar molecules as a result of small fluctuations in their distribution of electrons that create a temporary separation of charge. Nonpolar groups such as methyl

2.1  Water Molecules and Hydrogen Bonds  31

groups can therefore experience a small attractive force, in this case about 0.3 kJ · mol−1:

H C

δ+

H

H

δ–

H

H

C

δ+

Universal History Archive/Getty Images

δ–

H

Not surprisingly, these forces act only when the groups are very close, and their strength quickly falls off as the groups draw apart. If the groups approach too closely, however, their van der Waals radii collide and a strong repulsive force overcomes the attractive force. Although hydrogen bonds and van der Waals interactions are individually weak, biological molecules usually contain multiple groups capable of participating in these intermolecular interactions, so their cumulative effect can be significant (Fig. 2.5). These weak forces also determine how biological moleucles can “recognize” or bind noncovalently to each other. Drug molecules are typically designed to optimize the weak interactions that govern their therapeutic activity (Box 2.A).

  FIGURE 2.5   The cumulative effect of small forces.  Just as the fictional giant Gulliver was restrained by many small tethers at the hands of the tiny Lilliputians, the structures of macromolecules are constrained by the effects of many weak noncovalent interactions.

Water dissolves many compounds Unlike most other solvent molecules, water molecules are able to form hydrogen bonds and participate in other electrostatic interactions with a wide variety of compounds. Water has a relatively high dielectric constant, which is a measure of a solvent’s ability to diminish the electrostatic attractions between dissolved ions (Table 2.2). The higher the dielectric constant of the solvent, the less able the ions are to associate with each other. The polar water

Box 2.A Why Do Some Drugs Contain Fluorine? As mentioned in Section 1.2, the most abundant elements in biological molecules are H, C, N, O, P, and S. Fluorine, the 13th most abundant element on earth, appears in animal bodies mainly in bones and teeth. Fluoride ions occasionally take the place of hydroxyl groups in crystals of hydroxyapatite, Ca10(PO4)6(OH)2, the mineral that accounts for most of the mass of bones and teeth. Although fluoride’s role in bone structure is controversial, it strengthens dental enamel and makes teeth more resistant to the demineralization that occurs during the formation of dental caries (cavities). For this reason, small amounts of fluoride are often added to toothpastes and to municipal water supplies, particularly in the U.S. Despite its participation in tooth structure, fluorine almost never appears in naturally occurring organic compounds. Why, then, do about one-quarter of all drug molecules, including the widely prescribed Prozac (fluoxetine, an antidepressant; Box 9.B), fluorouracil (an anticancer agent; Section 7.3), and Ciprofloxacin (an antibacterial agent; Section 20.5), contain F? In designing an effective drug, pharmaceutical scientists often intentionally introduce F in order to alter the drug’s chemical or biological properties without significantly altering its shape. The small fluorine can take the place of hydrogen in a ­chemical structure, but with its high electronegativity (see Table 2.1),

F behaves much more like O than H. Consequently, transforming a relatively inert C—H group into an electron-­withdrawing C—F group can decrease the basicity of nearby amino groups (see Section 2.3). Fewer positive charges in a drug allow it to more easily pass through membranes to enter cells and exert its biological effect.

O

H N

CH3

F3C Prozac (Fluoxetine) In addition, the polar C—F bond can participate in hydrogen bonding (C—F · · · H—C) or other dipole–dipole interactions (such as C—F · · · CO), potentially augmenting the inter­ molecular attraction between a drug and its target molecule in the body. Better binding usually means that the drug will be effective at lower concentrations and will have fewer side effects. Question  Identify the hydrogen-bonding groups in Prozac.

32  C ha pter 2   Aqueous Chemistry

TAB L E 2. 2   Dielectric Constants for Some Solvents at Room Temperature

Solvent

Dielectric constant

Formamide (HCONH2)

109

Water

80

Methanol (CH3OH)

33

Ethanol (CH3CH2OH)

25

1-Propanol (CH3CH2CH2OH)

20

1-Butanol (CH3CH2CH2CH2OH)

18

Benzene (C6H6)

2

Question  Compare the hydrogen-bonding ability of these solvents.

molecules surround ions (for example, the Na+ and Cl− ions from the salt NaCl) by aligning their partial charges with the oppositely charged ions.

Wolfgang Baumeister, Max Planck Institute for Biochemistry.

Na+

Cl–

Because the interactions between the polar water molecules and the ions are stronger than the attractive forces between the Na+ and Cl− ions, the salt dissolves (the dissolved particle is called a solute). Each solute ion surrounded by water molecules is said to be solvated (or hydrated, to indicate that the solvent is water). Biological molecules that bear polar or ionic functional groups are also readily solubilized, in this case because the groups can form hydrogen bonds with the solvent water mol­ecules. Glucose, for example, with its six hydrogen-bonding oxygens, is highly soluble in water:

  FIGURE 2.6   Portion of a Dictyoste-

lium cell visualized by cryoelectron tomography.  In this technique, the cells are rapidly frozen so that they retain their fine structure, and two-dimensional electron micrographs taken from different angles are merged to re-create a three-­ dimensional image. The red structures are filaments of the protein actin, ribosomes and other macromolecular complexes are colored green, and membranes are blue. Small molecules (not visible) fill the spaces between these larger cell components.

Note that when we describe the behavior of a single molecule, such as glucose in this example, we are really describing the average behavior of a huge number of molecules. (Most biochemical techniques cannot assess the activity of an individual molecule.) The concentration of glucose in human blood is about 5 mM. In a solution of 5 mM ­glucose in water, there are about 10,000 water molecules for every glucose molecule (the water molecules are present at a concentration of about 55.5 M). However, biological molecules are never found alone in such dilute conditions in vivo, because a large number of small molecules, large polymers, and macromolecular aggregates collectively form a solution that is more like a hearty stew than a thin, watery soup (Fig. 2.6). Inside a cell, the spaces between molecules may be only a few Å wide, enough room for only two water molecules to fit. This allows solute molecules, each with a coating of properly oriented water molecules, to slide past each other. This thin coating, or shell, of water may be enough to keep molecules from coming into

2.2  The Hydrophobic Effect  33

van der Waals contact (van der Waals interactions are weak but attractive), thereby helping maintain the cell’s contents in a crowded but fluid state. Consequently, although there are a lot of water molecules in biological systems, those molecules may not be free to behave like the water molecules in bulk solution. In fact, the water molecules that associate closely with biological molecules, especially through hydrogen bonding, can be considered to be part of the macromolecules’ structure. For example, every gram of DNA has about 0.6 g of tightly bound water molecules that affect its overall stability.

Before Going On • Explain why a water molecule is polar. • Draw three hydrogen-bonded water molecules. • Describe the structure of liquid water. • Compare the strengths of covalent bonds, hydrogen bonds, ionic interactions, and van der Waals interactions. • Describe what happens when an ionic substance dissolves in water. • Explain why water is a more effective solvent than ammonia or methanol.

2.2 The Hydrophobic Effect LEARNING OBJECTIVES Relate the solubility of substances to the hydrophobic effect. • Explain the hydrophobic effect in terms of water’s entropy. • Predict the water solubility of hydrophobic and hydrophilic substances. • Describe how amphiphilic substances behave in water. • Explain why a lipid bilayer is a barrier to diffusion. Glucose and other readily hydrated substances are said to be hydrophilic (water-loving). In contrast, a compound such as dodecane (a C12 alkane),

which lacks polar groups, is relatively insoluble in water and is said to be hydrophobic (water-fearing). Although pure hydrocarbons are rare in biological systems, many biological molecules contain hydrocarbon-like portions that are insoluble in water. When a nonpolar substance such as vegetable oil (which consists of hydrocarbon-like molecules) is added to water, it does not dissolve but forms a separate phase. In order for the water and oil to mix, free energy must be added to the system (for example, by stirring vigorously or applying heat). Why is it thermodynamically unfavorable to dissolve a hydrophobic substance in water? One possibility is that enthalpy is required to break the hydrogen bonds among solvent water molecules in order to create a “hole” into which a nonpolar molecule can fit. Experimental measurements, however, show that the free energy barrier (ΔG) to the solvation process depends much more on the entropy term (ΔS) than on the enthalpy term (ΔH; recall from Chapter 1 that ΔG = ΔH − TΔS; Equation 1.2). This is because when a hydrophobic molecule is hydrated, it becomes surrounded by a layer of water molecules that cannot participate in normal hydrogen bonding with each other but instead must align themselves so that their polar ends are not oriented toward the nonpolar solute. This constraint on the structure of water represents a loss of entropy in the system, because now the highly

34

C hA pTER 2

Aqueous Chemistry Nonpolar molecule

Water

Layer of constrained water molecules

FIGURE 2.7 Hydration of a nonpolar molecule. When a nonpolar substance (green) is added to water, the system loses entropy because the water molecules surrounding the nonpolar solute (orange) lose their freedom to form hydrogen bonds. The loss of entropy is a property of the entire system, not just the water molecules nearest the solute, because these molecules are continually changing places with water molecules from the rest of the solution. The loss of entropy presents a thermodynamic barrier to the hydration of a nonpolar solute.

mobile water molecules have lost some of their freedom to rapidly form, break, and re-form hydrogen bonds with other water molecules (Fig. 2.7). Note that the loss of entropy is not due to the formation of a frozen “cage” of water molecules around the nonpolar solute, as commonly pictured, because in liquid water the solvent molecules are in constant motion. When a large number of nonpolar molecules are introduced into a sample of water, they do not disperse and become individually hydrated, each surrounded by a layer of water molecules. Instead, the nonpolar molecules are forced together, away from contact with water molecules. (This explains why small oil droplets coalesce into one large oily phase.) Although the entropy of the nonpolar molecules is thereby reduced, this thermodynamically unfavorable event is more than offset by the increase in the entropy of the surrounding water molecules, which regain their ability to interact freely with other water molecules (Fig. 2.8). The exclusion of nonpolar substances from an aqueous solution is known as the hydrophobic effect. It is a powerful force in biochemical systems, even though it is not a bond or an attractive interaction in the conventional sense. The nonpolar molecules do not experience any additional attractive force among themselves; they aggregate only because they are driven out of the aqueous phase by the unfavorable entropy cost of individually hydrating them. The hydrophobic effect governs the structures and functions of many biological molecules.

a.

b.

FIGURE 2.8 Aggregation of nonpolar molecules in water. a. The individual hydration of dispersed nonpolar molecules (green) decreases the entropy of the system because the hydrating water molecules (orange) are not as free to form hydrogen bonds. b. Aggregation of the nonpolar molecules increases the entropy of the system, since the number of water molecules required to hydrate the aggregated solutes is less than the number of water molecules required to hydrate the dispersed solute molecules. This increase in entropy accounts for the spontaneous aggregation of nonpolar substances in water.

Question Explain why it is incorrect to describe the behavior shown in part b in terms of “hydrophobic bonds.”

2.2  The Hydrophobic Effect  35

For example, each polypeptide chain of a protein folds into a globular mass so that its hydrophobic groups are in the interior, away from the solvent, and its polar groups are on the exterior, where they can interact with water. Similarly, the structure of the lipid membrane that surrounds all cells is maintained by the hydrophobic effect acting on the lipids.

Polar head group Nonpolar tail

Amphiphilic molecules experience both hydrophilic interactions and the hydrophobic effect Consider a molecule such as the fatty acid palmitate:

The hydrocarbon “tail” of the molecule (on the right) is nonpolar, while its carboxylate “head” (on the left) is strongly polar. Molecules such as this one, which have both hydrophobic and hydrophilic portions, are said to be amphiphilic or amphipathic. What happens when amphiphilic molecules are added to water? In general, the polar groups of amphiphiles orient themselves toward the solvent molecules and are therefore hydrated, while the nonpolar groups tend to aggregate due to the hydrophobic effect. As a result, the amphiphiles may form a spherical micelle, a particle with a solvated surface and a hydrophobic core (Fig. 2.9). Depending in part on the relative sizes of the hydrophilic and hydrophobic portions of the amphiphiles, the molecules may form a sheet rather than a spherical micelle. The amphiphilic lipids that provide the structural basis of biological membranes form two-layered sheets called bilayers, in which a hydrophobic layer is sandwiched between hydrated polar surfaces (Fig. 2.10). The structures of biological membranes are discussed in more detail in Chapter 8. The formation of micelles or bilayers is thermodynamically favored because the polar head groups can interact with solvent water molecules and the nonpolar tails are sequestered from the solvent.

  FIGURE 2.9  Cross-section of a micelle formed by amphiphilic molecules.  The hydrophobic tails of the molecules aggregate, out of contact with water, due to the hydrophobic effect. The polar head groups are exposed to and can interact with the solvent water molecules.

Polar head group Nonpolar tails

The hydrophobic core of a lipid bilayer is a barrier to diffusion To eliminate its solvent-exposed edges, a lipid bilayer tends to close up to form a vesicle, shown cut in half:

Many of the subcellular compartments (organelles) in eukaryotic cells have a similar structure. When the vesicle forms, it traps a volume of the aqueous solution. Polar solutes in the enclosed compartment tend to remain there because they cannot easily pass through the hydrophobic interior of the bilayer. The energetic cost of transferring a hydrated polar group through the nonpolar lipid tails is too great. (In contrast, small nonpolar molecules such as O2 can pass through the bilayer relatively easily.)

  FIGURE 2.10   A lipid bilayer.  The amphiphilic lipid molecules form two layers so that their polar head groups are exposed to the solvent while their hydrophobic tails are sequestered in the interior of the bilayer, away from water. The likelihood of amphiphilic molecules forming a bilayer rather than a micelle depends in part on the sizes and nature of the hydrophobic and hydrophilic groups. One-tailed lipids tend to form micelles (see Fig. 2.9), and two-tailed lipids tend to form bilayers.

Question  Indicate where a sodium ion and a benzene molecule would be located.

36  C ha pter 2   Aqueous Chemistry

INTRACELLULAR

EXTRACELLULAR

b.

  FIGURE 2.11   A bilayer ­prevents

the diffusion of polar substances.  a. Solutes spontaneously diffuse from a region of high concentration to a region of low concentration.  b. A lipid bilayer, which presents a thermodynamic barrier to the passage of polar substances, prevents the diffusion of polar substances out of the inner compartment (it also prevents the inward diffusion of polar substances from the external solution).

160 Concentration (mM)

a.

Concentration (mM)

160 120 80 40 0

Na+

K+

Cl−

120 80 40 0

Na+

K+

Cl−

  FIGURE 2.12   Ionic composition of intracellular and ­extracellular

fluid.  Human cells contain much higher concentrations of potassium than of sodium or chloride; the opposite is true of the fluid outside the cell. The cell membrane helps maintain the concentration differences.

Normally, substances that are present at high concentrations tend to diffuse to regions of lower concentration. (This movement “down” a concentration gradient is a spontaneous process driven by the increase in entropy of the solute molecules.) A barrier such as a bilayer can prevent this diffusion (Fig. 2.11). This helps explain why cells, which are universally enclosed by a membrane, can maintain their specific concentrations of ions, small molecules, and biopolymers even when the external concentrations of these substances are quite different (Fig. 2.12). The solute composition of intracellular compartments and other biological fluids is carefully regulated. Not surprisingly, organisms spend a considerable amount of metabolic energy to maintain the proper concentrations of water and salts, and losses of one or the other must be compensated (Box 2.B).

Box 2.B Sweat, Exercise, and Sports Drinks Animals, including humans, generate heat, even at rest, due to their metabolic activity. Some of this heat is lost to the environment by radiation, convection, conduction, and—in terrestrial animals—the vaporization of water. Evaporation has a significant cooling effect because about 2.5 kJ of heat is given up for every gram (mL) of water lost. In humans and certain other animals, an increase in skin temperature triggers the activity of sweat glands, which secrete a solution containing (in humans) about 50 mM Na+, 5 mM K+, and 45 mM Cl−. The body is cooled as the sweat evaporates from its surface. The evaporation of water accounts for a small portion of a resting body’s heat loss, but sweating is the main mechanism for dissipating heat generated when the body is highly active. During vigorous exercise or exertion at high ambient temperatures, the body may experience a fluid loss of up to 2 L per hour. Athletic training not only improves the performance of the muscles and cardiopulmonary system, it also increases the capacity for sweating so that the athlete begins to sweat at a lower skin temperature and loses less salt in the secretions of the sweat glands. However, regardless of training, a fluid loss representing more than 2% of the body’s weight may impair cardiovascular function. In fact,

“heat exhaustion” in humans is usually due to dehydration rather than an actual increase in body temperature. Numerous studies have concluded that athletes seldom drink enough before or during exercise. Ideally, fluid intake should match the losses due to sweat and the rate of intake should keep pace with the rate of sweating. So what should the conscientious athlete drink? For activities lasting less than about 90 minutes, especially when periods of high intensity alternate with brief periods of rest, water alone is sufficient. Commercial sports drinks containing carbohydrates can replace the water lost as sweat and also provide a source of energy. However, this carbohydrate boost may be an advantage only during prolonged sustained activity, such as during a marathon, when the body’s own carbohydrate stores are depleted. A marathon runner or a manual laborer in the hot sun might benefit from the salt found in sports drinks, but most athletes do not need the supplemental salt (although it does make the carbohydrate solution more palatable). A normal diet usually contains enough Na+ and Cl− to offset the losses in sweat. Question  Compare the ion concentrations of sweat and extracellular fluid.

2.3  Acid–Base Chemistry  37

Before Going On • Describe the changes in entropy that occur when nonpolar substances are added to water. • Explain how you can distinguish hydrophobic and hydrophilic substances. • Explain why polar molecules dissolve more easily than nonpolar substances in water. • Explain how a molecule can be both hydrophilic and hydrophobic. Give an example. • Explain why a lipid bilayer is a barrier to the diffusion of polar molecules.

2.3 Acid–Base Chemistry LEARNING OBJECTIVES Determine the effect of acids and bases on a solution’s pH. • Recognize the relationship between the concentrations of H+ and OH−. • Predict how the pH changes when acid or base is added to water. • Relate an acid’s pK value to its tendency to ionize. • Perform calculations using the Henderson–Hasselbalch equation. • Predict the ionization states of acid–base groups at a given pH.

Water is not merely an inert medium for biochemical processes; it is an active participant. Its chemical reactivity in biological systems is in part a result of its ability to ionize. This can be expressed in terms of a chemical equilibrium: ​ ​H​ 2​O ⇌ ​H​ +​+ ​OH​ −​ The products of water’s dissociation are a hydrogen ion or proton (H+) and a hydroxide ion (OH−). Aqueous solutions do not actually contain lone protons. Instead, the H+ can be visualized as combining with a water molecule to produce a hydronium ion (H3O+):

H

+

O

H

H However, the H+ is somewhat delocalized, so it probably exists as part of a larger, fleeting structure such as

H

H H

H O H

+

H

O

or

H

O

O H

H

+

O

H

H

H H

O

H

Because a proton does not remain associated with a single water molecule, it appears to be relayed through a hydrogen-bonded network of water molecules (Fig. 2.13). This rapid proton jumping means that the effective mobility of H+ in water is much greater than the mobility of other ions that must physically diffuse among water molecules. Consequently, acid–base reactions are among the fastest biochemical reactions.

SEE GUIDED TOUR Water and pH

38  C ha pter 2   Aqueous Chemistry

+

  FIGURE 2.13   Proton jumping.  A proton associated with one water molecule (as a hydronium ion, at left) appears to jump rapidly through a network of ­hydrogen-bonded water molecules.

[H+] and [OH–] are inversely related Pure water exhibits only a slight tendency to ionize, so the resulting concentrations of H+ and OH− are actually quite small. According to the law of mass action, the ionization of water can be described by a dissociation constant, K, which is equivalent to the concentrations of the reaction products divided by the concentration of un-ionized water: [​H​ +​][​OH​ −​] ​ K = ​ __________    ​​  [​H​ 2​O]

(2.1)

The square brackets represent the molar concentrations of the indicated species. Because the concentration of H2O (55.5 M) is so much greater than [H+] or [OH−], it is considered to be constant, and the dissociation of water is described by Kw, the ionization constant of water: ​ ​Kw​  ​= K [​H​ 2​O] = [​H​ +​][​OH​ −​]​

(2.2)

Kw is 10−14 at 25°C. In a sample of pure water, [H+] = [OH−], so [H+] and [OH−] must both be equal to 10−7 M: ​ ​Kw ​  ​= ​10​ −14​= [​H​ +​][​OH​ −​] = (​10​​  −7​ M) (​10​ −7​M)​

(2.3)

Since the product of [H+] and [OH−] in any solution must be equal to 10−14, a hydrogen ion concentration greater than 10−7 M is balanced by a hydroxide ion concentration less than 10−7 M (Fig. 2.14). A solution in which [H+] = [OH−] = 10−7 M is said to be neutral; a solution with [H+] > 10−7 M ([OH−] < 10−7 M) is acidic, and a solution with [H+] < 10−7 M ([OH−] > 10−7 M) is basic. To more easily describe such solutions, the hydrogen ion concentration is expressed as a pH: ​ pH = −log [​H​ +​]​ 1

[H+] (M)

1

10−2

10−2

10−4

10−4

10−6

10−6

10−8

10−8

10−10

10−10

10−12

10−12

10−14

10−14

  FIGURE 2.14   Relationship between [H+] and [OH–]. 

[OH−] (M)

The product of [H+] and [OH−] is Kw, which is equal to 10−14. Consequently, when [H+] is greater than 10−7 M, [OH−] is less than 10−7 M, and vice versa.

(2.4)

2.3  Acid–Base Chemistry  39 pH

[H+] (M)

14

10−14

pH

13

10−13

Pancreatic juice

7.8–8.0

12

10−12

Blood

7.4

Saliva

6.4–7.0

11

10−11

Urine

5.0–8.0

10

10−10

Gastric juice

1.5–3.0

9

10−9

8

10−8

7

10−7

6

10−6

5

10−5

4

10−4

3

10−3

2

10−2

1

10−1

0

1

TAB L E 2. 3   pH Values of Some Biological Fluids

Fluid

Basic

Accordingly, a neutral solution has a pH of 7, an acidic solution has a pH < 7, and a basic solution has a pH > 7 (Fig. 2.15). Note that because the pH scale is logarithmic, a difference of one pH unit is equivalent to a 10-fold difference in [H+]. The so-called physiological pH, the normal pH of human blood, is a near-neutral 7.4. The pH values of some other body fluids are listed in Table 2.3. The pH of the environment is also a concern (Box 2.C).

Neutral

Acidic

The pH of a solution can be altered The pH of a sample of water can be changed by adding a substance that affects the existing balance between [H+] and [OH−]. Adding an acid increases the concentration of [H+] and decreases the pH; adding a base has the opposite effect. Biochemists define an acid as a substance that can donate a proton and a base as a substance that can accept a proton. For example, adding hydrochloric acid (HCl) to a sample of water increases the hydrogen ion concentration ([H+] or [H3O+]) because the HCl donates a proton to water: ​ HCl + ​H​ 2​O → ​H​ 3​O​ +​+ ​Cl​ −​​ Note that in this reaction, H2O acts as a base that accepts a proton from the added acid. Similarly, adding the base sodium hydroxide (NaOH) increases the pH (decreases [H+]) by introducing hydroxide ions that can recombine with existing hydrogen ions: ​ NaOH + ​H​ 3​O​ +​→ ​Na​ +​+ 2 ​H​ 2​O​

  FIGURE 2.15    ­ Relationship between pH and [H+].  Because pH is equal to –log [H+], the greater the [H+], the lower the pH. A solution with a pH of 7 is neutral, a solution with a pH < 7 is acidic, and a solution with a pH > 7 is basic.

Question  What is the difference in H+ concentration between a solution at pH 4 and a solution at pH 8?

Box 2.C Atmospheric CO2 and Ocean Acidification The human-generated increase in atmospheric carbon dioxide that is contributing to global warming is also impacting the chemistry of the world’s oceans. Atmospheric CO2 dissolves in water and reacts with it to generate carbonic acid. The acid immediately dissociates to form protons (H+) and bicarbonate ​(​HCO​− 3  ​ ​)​: ​​CO​  2​​ + ​H​  2​​O ⇌ ​H​  2​​​CO​  3​​ ⇌ ​H​​  +​ + ​HCO​ − 3​  ​​ The addition of hydrogen ions from CO2-derived carbonic acid therefore leads to a decrease in the pH. Currently, the earth’s oceans are slightly basic, with a pH of approximately 8.0. It has been estimated that over the next 100 years, the ocean pH will drop to about 7.8. Although the oceans act as a CO2 “sink” that helps mitigate the increase in atmospheric CO2, the increase in acidity in the marine environment represents an enormous challenge to organisms that must adapt to the new conditions. Many marine organisms, including mollusks, many corals, and some plankton, use dissolved carbonate ions ​(​CO​3 2− ​  ​)​to construct protective shells of calcium carbonate (CaCO3). However, carbonate ions can combine with H+ to form bicarbonate:

+ − ​​CO​ 2− 3​  ​ + ​H​​  ​ ⇌ ​HCO​ 3​  ​​

Consequently, the increase in ocean acidity could decrease the availability of carbonate and thereby slow the growth of shell-building organisms. This not only would affect the availability of shellfish for human consumption but also would impact huge numbers of unicellular organisms at the base of the marine food chain. It is possible that acidification of the oceans could also dissolve existing calcium carbonate–based materials, such as coral reefs: 2+ ​​CaCO​  3​​ + ​H​​  +​ ⇌ ​HCO​ − 3​  ​ + ​Ca​​  ​​

This could have disastrous consequences for these speciesrich ecosystems. Question  Paradoxically, some marine organisms appear to benefit from increased atmospheric CO2. Write an equation that describes how increased bicarbonate concentrations in seawater could promote shell growth.

40  C ha pter 2   Aqueous Chemistry

In this reaction, H3O+ is the acid that donates a proton to the added base. Note that acids and bases must operate in pairs: An acid can function as an acid (proton donor) only if a base (proton acceptor) is present, and vice versa. Water molecules can serve as both acid and base. The final pH of the solution depends on how much H + (for example, from HCl) has been introduced or how much H+ has been removed from the solution by its reaction with a base (for example, the OH− ion of NaOH). Substances such as HCl and NaOH are known as “strong” acids and bases because they ionize completely in water. The Na+ and Cl− ions are called spectator ions and do not affect the pH. Keep in mind that pH is a property of a solution, not a characteristic of any individual reactant in the mixture. Calculating the pH of a solution of strong acid or base is straightforward (see Sample Calculation 2.1).

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SA MP L E CALCU LAT I ON 2 . 1 Problem  Calculate the pH of 1 L of water to which is added a. 10 mL of 5.0 M HCl or b. 10 mL of 5.0 M NaOH. Solution

(0.01 L)(5.0 M)    a.  The final concentration of HCl is _____________ ​​   ​  = 0.050 M​ (1.01 L) Since HCl dissociates completely, the added [H+] is equal to [HCl], or 0.050 M (the existing hydrogen ion concentration, 10−7 M, can be ignored because it is much smaller).

pH = −log [​H​ +​]



= −log 0.050



= 1.3

b. The final concentration of NaOH is 0.050 M. Since NaOH dissociates completely, the added [OH−] is 0.050 M. Use Equation 2.2 to calculate [H+]. ​ ​Kw​  ​= ​10​ −14​= [​H​ +​][​OH​ −​] [​H​ +​] = ​10​ −14​/[​OH​ −​] = ​10​ −14​/(0.050 M)

= 2.0 × ​10​ −13​M pH = −log [​H​ +​]



= −log (2.0 × ​10​ −13​)



= 12.7​

A pK value describes an acid’s tendency to ionize Most biologically relevant acids and bases, unlike HCl and NaOH, do not dissociate completely when added to water. In other words, proton transfer to or from water is not complete. Therefore, the final concentrations of the acidic and basic species (including water itself) must be expressed in terms of an equilibrium. For example, acetic acid partially ionizes, or donates only some of its protons to water: ​ ​CH​ 3​COOH + ​H​ 2​O ⇌ ​CH​ 3​COO​ −​+ ​H​ 3​O​ +​ The equilibrium constant for this reaction takes the form [​CH​ 3​COO​ −​][​H​ 3​O​ +​]     ​​ ​ K = _________________ ​     [​CH​ 3​COOH][​H​ 2​O]

(2.5)

Because the concentration of H2O is much higher than the other concentrations, it is considered constant and is incorporated into the value of K, which is then formally known as Ka, the acid dissociation constant: [​CH​ 3​COO​ −​][​H​ +​] ​        ​​ ​ ​Ka​  ​= K [​H​ 2​O] = _______________ [​CH​ 3​COOH]

(2.6)

The acid dissociation constant for acetic acid is 1.74 × 10−5. The larger the value of Ka, the more likely the acid is to ionize; that is, the greater its tendency to donate a proton to water. The smaller the value of Ka, the less likely the compound is to donate a proton. Acid dissociation constants, like hydrogen ion concentrations, are often very small numbers. Therefore, it is convenient to transform the Ka to a pK value as follows: ​ pK = −log ​Ka​  ​​

(2.7)

The term pKa is also used, but for simplicity we will use pK. For acetic acid, ​ pK = −log (1.74 × ​10​ −5​) = 4.76​

(2.8)

2.3  Acid–Base Chemistry  41

The larger an acid’s Ka, the smaller its pK and the greater its “strength” as an acid. In a solution of an acid with a low pK value, most of the molecules are in the unprotonated form. For an acid with a high pK value, most of the molecules remain protonated. Consider an acid such as the ammonium ion, ​NH​+ 4  ​ ​: + ​NH​+ 4  ​ ​⇌ ​NH​ 3​+ ​H​  ​

Its Ka is 5.62 × 10−10, which corresponds to a pK of 9.25. This indicates that the ammonium ion is a relatively weak acid, a compound that tends not to donate a proton. On the other hand, ammonia (NH3), which is the conjugate base of the acid ​​NH​ 4+​  ​​, readily accepts a ­proton. The pK values of some compounds are listed in Table 2.4. A polyprotic acid, a compound with more than one acidic hydrogen, has a pK value for each dissociation (called pK1, pK2, etc.). The first proton dissociates with the lowest pK value. Subsequent protons are less likely to dissociate and so have higher pK values.

The pH of a solution of acid is related to the pK When an acid (represented as the proton donor HA) is added to water, the final hydrogen ion concentration of the solution depends on the acid’s tendency to ionize: ​HA ⇌ ​A​ −​+ ​H​ +​

TA B L E 2.4  pK Values of Some Acids

Formula a

Name

pK

Trifluoroacetic acid

CF3COOH

Phosphoric acid

H3PO4

2.15 b

Formic acid

HCOOH

3.75

Succinic acid

HOOCCH2CH2COOH

4.21 b

CH3COOH

4.76

Acetic acid

0.18

Succinate

HOOCCH2CH2COO−

Thiophenol

C6H5SH

5.64 c 6.60

Phosphate

​H​ 2​PO​− 4  ​ ​

6.82 c

N-(2-acetamido)-2-amino  ethanesulfonic acid (ACES)

​H​ 2​NCOCH​ 2​N ​H ​  ​ 2​CH​ 2​CH​ 2​SO​− 3  ​ ​

+

6.90

+

NH 7.00

Imidazolium ion N H p-Nitrophenol

HO

N-2-hydroxyethylpiperazine  Nʹ-2-ethanesulfonic acid (HEPES) Glycinamide

NO2

+

HOCH2CH2HN +

H3NCH2CONH2

7.24

NCH2CH2SO–3

7.55 8.20

Tris(hydroxymethyl)  aminomethane (Tris)

​​(​HOCH​ 2​)​  3​​C​N ​H ​  ​ 3​

8.30

Boric acid

H3BO3

9.24

+

Ammonium ion

​​ H​ + N 4​  ​​

Phenol

C6H5OH

Methylammonium ion Phosphate

​CH​ 3​N​H+ 3​  ​ ​

​H​PO​2− 4  ​  ​

a 

The acidic hydrogen is highlighted in red; b pK1; c pK2; d pK3.

Question  Determine the charge of the conjugate base of each acid.

9.25 9.90 10.60 12.38 d

42  C ha pter 2   Aqueous Chemistry

In other words, the final pH depends on the equilibrium between HA and A−, [​A​ −​][​H​ +​] ​     ​​  ​ ​Ka​  ​= ________ [HA]

(2.9)

so that [HA] ​[​H​ +​] = ​Ka​  ​  ​ _______  ​​  [​A​ −​]

(2.10)

We can express [H+] as a pH, and Ka as a pK, which yields [HA] ​ −log [​H​ +​] = −log ​Ka​  ​− log _____ ​  −  ​​  [​A​  ​]

(2.11)

or [​A​ −​] ​ pH = pK + log _____ ​     ​​  [HA]

SEE SAMPLE CALCULATION VIDEOS

(2.12)

Equation 2.12 is known as the Henderson–Hasselbalch equation. It relates the pH of a solution to the pK of an acid and the concentration of the acid (HA) and its conjugate base (A−). This equation makes it possible to perform practical calculations to predict the pH of a solution (see Sample Calculation 2.2) or the concentrations of an acid and its conjugate base at a given pH (see Sample Calculation 2.3).

SA MP L E CALCU LAT I ON 2 . 2 Problem  Calculate the pH of a 1-L solution to which has been added 6.0 mL of 1.5 M acetic acid and 5.0 mL of 0.4 M sodium acetate. Solution  First, calculate the final concentrations of acetic acid (HA) and acetate (A−). The final volume of the solution is 1 L + 6 mL + 5 mL = 1.011 L. (0.006 L) (1.5 M) ​[HA] = ______________ ​      ​  = 0.0089 M 1.011 L (0.005 L) (0.4 M)    ​   ​  = 0.0020 M​ [​A​ −​] = ______________ 1.011 L

Next, substitute these values into the Henderson–Hasselbalch equation using the pK for acetic acid given in Table 2.4: [​A​ −​]  ​  ​ pH = pK + log ​ _____  [HA] pH = 4.76 + log ______ ​  0.0020 ​  0.0089 = 4.76 − 0.65

= 4.11​

SA MP L E CALCU LAT I ON 2 . 3 Problem  Calculate the concentration of formate in a 10-mM solution of formic acid at pH 4.15. Solution  The solution of formic acid contains both the acid species (formic acid) and its conjugate base (formate). Use the Henderson–Hasselbalch equation to determine the ratio of formate (A−) to formic acid (HA) at pH 4.15, using the pK value given in Table 2.4. −

[​A​  ​] ​ pH = pK + log _____ ​     ​  [HA]

[​A​ −​] log ​ _____   ​ = pH − pK = 4.15 − 3.75 = 0.40 [HA] [​A​ −​] ​ _____   ​ = 2.51 or [​A​ −​] = 2.51[HA]​ [HA]



Since the total concentration of formate and formic acid is 0.01 M, [A−] + [HA] = 0.01 M, and [HA] = 0.01 M − [A−]. Therefore, ​[​A​ −​] = 2.51[HA] [​A​ −​] = 2.51(0.01 M − [​A​ −​]) [​A​ −​] = 0.0251 M − 2.51[​A​ −​]

3.51[​A​ −​] = 0.0251 M

[​A​ −​] = 0.0072 M or 7.2 mM​

2.3  Acid–Base Chemistry  43

The Henderson–Hasselbalch equation indicates that when the pH of a solution of acid is equal to the pK of that acid, then the acid is half dissociated; that is, exactly half of the molecules are in the protonated HA form and half are in the unprotonated A− form. You can prove to yourself that when [A−] = [HA], the log term of the Henderson–Hasselbalch equation becomes zero (log 1 = 0) and pH = pK. When the pH is far below the pK, the acid exists mostly in the HA form; when the pH is far above the pK, the acid exists mostly in the A− form. Note that A− signifies a deprotonated acid; if the acid (HA) bears a positive charge to begin with, the dissociation of a proton yields a neutral species, still designated A−. Knowing the ionization state of an acidic substance at a given pH can be critical. For example, a drug that has no net charge at pH 7.4 may readily enter cells, whereas a drug that bears a net positive or negative charge at that pH may remain in the bloodstream and be therapeutically useless (see Sample Calculation 2.4).

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SA MP L E CA LCULAT I O N 2. 4 2− At pH 6.82, ​[​H​ 2​PO​− 4  ​ ​] = [​HPO​4  ​  ​]​.

Problem  Determine which molecular species of phosphoric acid predominates at pH values of a. 1.5, b. 4, c. 9, and d. 13.

Between pH 6.82 and 12.38, the ​HPO​2− 4  ​  ​species predominates.

Solution  From the pK values in Table 2.4, we know that:

3− At pH 12.38, ​[​HPO​2− 4  ​  ​] = [​PO​4  ​  ​]​.

Below pH 2.15, the fully protonated H3PO4 species predominates.

Above pH 12.38, the fully deprotonated ​PO​3− 4  ​  ​species predominates.

At pH 2.15, ​[​H​ 3​PO​ 4​] =

[​H​ 2​PO​− 4  ​ ​]​.

Between pH 2.15 and 6.82, the

​ ​ 2​PO​− H 4  ​ ​species

Therefore, the predominant species at the indicated pH values are 2− 3− a. H3PO4, b. ​H​ 2​PO​− 4  ​ ​, c. ​HPO​4  ​  ​, and d. ​PO​4  ​  ​.

predominates.

The pH-dependent ionization of biological molecules is also key to understanding their structures and functions. Under physiological conditions, some of the functional groups on biological molecules act as acids and bases. Their ionization states depend on their respective pK values and on the pH ([H+]) of their environment. For example, at physiological pH, a polypeptide bears multiple ionic charges because its carboxylic acid (—COOH) groups are ionized to carboxylate (—COO−) groups and its amino (—NH2) groups are protonated (​ —​NH​3+  ​ ​)​. This is because the pK values for the carboxylic acid groups are about 4, and the pK values for the amino groups are above 10. Consequently, below pH 4, both the carboxylic acid and amino groups are mostly protonated; above pH 10, both groups are mostly deprotonated. pH < 4

4 < pH < 10

pH > 10

—COOH

— COO−

— COO−

—​​NH​ + 3​  ​​

—​​NH​ + 3​  ​​

—NH2

Note that a compound containing a —COO− group is sometimes still called an “acid,” even though it has already given up its proton. Similarly, a “basic” compound may already have accepted a proton. In addition to amino and carboxyl groups, pH-sensitive phosphoryl and imino groups (see Table 1.1) influence the chemistry of biological molecules. For example, phosphoryl groups, with pK values near neutral, tend to exist as mixtures of ionic forms under cellular conditions:

O R

O

P

O O

–

+

+ H

R

O

O–

P

O–

OH

Likewise, imino groups are sensitive to the ambient pH. An imino group, also known as a Schiff base (except when R = H), has a pK value near neutral, so it often exists as a mixture of acidic and basic forms:

R

R C

R

N

R + H+

Imine (Schiff base)

C R

+

H

N R

Iminium ion (protonated Schiff base)

44  C ha pter 2   Aqueous Chemistry

Under physiological conditions, amido, hydroxyl, and sulfhydryl groups (see Table 1.1) are only rarely found in ionized forms in biological molecules.

Before Going On • Draw the structure of a water molecule, including all its electrons, before and after ionization. • Map the fluids in Table 2.3 onto the scale in Figure. 2.15. • For each fluid in Table 2.3, determine whether the addition of acid or base would bring the pH to neutral. • Rearrange the Henderson–Hasselbalch equation to isolate each term. • Predict the net charge of each molecule in Table 2.4 at pH 7.0. • Draw the ionic forms of phosphoryl and imino groups that would predominate at very low and very high pH.

2.4 Tools and Techniques: Buffers LEARNING OBJECTIVES Describe how buffer solutions resist changes in pH. • Recognize the acidic and basic species in a buffer solution. • Use the Henderson–Hasselbalch equation to devise a recipe for a buffer solution. • Determine the useful pH range of a buffer solution.

When a strong acid such as HCl is added to pure water, all the added acid contributes directly to a decrease in pH. However, when HCl is added to a solution containing a weak acid (HA) in equilibrium with its conjugate base (A−), the pH does not change so dramatically, because some of the added protons combine with the conjugate base to re-form the acid and therefore do not contribute to an increase in [H+]. ​ HCl → ​H​ +​+ ​Cl​ −​ large increase in [​H​ +​]

HCl + ​A​ −​→ HA + H+ + ​Cl​ −​ small increase in [​H​ +​]​

Conversely, when a strong base (such as NaOH) is added to the solution of weak acid/conjugate base, some of the added hydroxide ions accept protons from the acid to form H2O and therefore do not contribute to a decrease in [H+]. ​ NaOH → ​Na​ +​+ ​OH​ −​ large decrease in [​H​ +​]

NaOH + HA → ​Na​ +​+ ​A​ −​+ OH− + ​H​ 2​O small decrease in [​H​ +​]​

The weak acid/conjugate base system (HA/A−) acts as a buffer against the added acid or base by preventing the dramatic changes in pH that would otherwise occur. The buffering activity of a weak acid, such as acetic acid, can be traced by titrating the acid with a strong base (Fig. 2.16). At the start of the titration, all the acid is present in its protonated (HA) form. As base (for example, NaOH) is added, protons begin to dissociate from the acid, producing A−. The continued addition of base eventually causes all the protons to dissociate, leaving all the acid in its conjugate base (A−) form. At the midpoint of the titration, exactly half the protons have dissociated, so [HA] = [A−] and pH = pK (Equation 2.12). The broad, flat shape of the titration curve shown in Figure 2.16 indicates that the pH does not change drastically with added acid or base when the pH is near the pK. The effective buffering capacity of an acid is generally taken to be within one pH unit of its pK. For acetic acid (pK = 4.76), this would be pH 3.76–5.76.

2.4  Tools and Techniques: Buffers  45

8 End point 7

Midpoint [CH3COOH] = [CH3COO−]

6 pK Effective buffering range

5 pH

4 [CH3COOH] < [CH3COO−]

[CH3COOH] > [CH3COO−]

3 2

Start point

1

0

0.5

1

H+ ions dissociated   FIGURE 2.16     Titration of acetic acid.  At the start point (before base is added),

the acid is present mainly in its CH3COOH form. As small amounts of base are added, protons dissociate until, at the midpoint of the titration (where pH = pK), [CH3COOH] = [CH3COO−]. The addition of more base causes more protons to dissociate until nearly all the acid is in the CH3COO− form (the end point). The shaded area indicates the effective buffering range of acetic acid. Within one pH unit of the pK, additions of acid or base do not greatly perturb the pH of the solution. Question  Sketch the titration curve for ammonia.

Biochemists nearly always perform experiments in buffered solutions in order to maintain a constant pH when acidic or basic substances are added or when chemical reactions produce or consume protons. Without buffering, fluctuations in pH would alter the ionization state of the molecules under study, which might then behave differently. Before biochemists appreciated the importance of pH and buffering, experimental results were often poorly reproducible, even within the same laboratory. A buffer solution is typically prepared from a weak acid and the salt of its conjugate base (see Sample Calculation 2.5). The two are mixed together in the appropriate ratio, according to the Henderson–Hasselbalch equation, and the final pH is adjusted if necessary by adding a small amount of concentrated HCl or NaOH. In addition to choosing a buffering compound with a pK value near the desired pH, a biochemist must consider other factors, including the compound’s solubility, stability, toxicity to cells, reactivity with other molecules, and cost.

SEE SAMPLE CALCULATION VIDEOS

SA MP L E CA LCULAT IO N 2. 5 Problem  How many mL of a 2.0-M solution of boric acid must be added to 600 mL of a solution of 10 mM sodium borate in order for the pH to be 9.45?

Substitute the known pK (from Table 2.4) and the desired pH:

Solution  Rearrange the Henderson–Hasselbalch equation to isolate the [A−]/[HA] term:

The starting solution contains (0.6 L)(0.01 mol · L−1) = 0.006 moles of borate (A−). The amount of boric acid (HA) needed is 0.006 mol/1.62 = 0.0037 mol. Since the stock boric acid is 2.0 M, the volume of boric acid to be added is (0.0037 mol)/(2.0 mol · L−1) = 0.0019 L or 1.9 mL.

[​A​ −​] ​ pH = pK + log ​ _____   ​  [HA] − [​A​  ​]  ​ = pH − pK log ​ _____  [HA] [​A​ −​]  ​ = ​10​​  (pH−pK)​​ ​ _____  [HA]

[​A​ −​] ​​ _____   ​ = ​10​​  (9.45−9.24)​ = ​10​ 0.21​= 1.62​ [HA]

46  C ha pter 2   Aqueous Chemistry

One commonly used laboratory buffer system that mimics physiological conditions contains a mixture of dissolved NaH2PO4 and Na2HPO4 for a total phosphate concentration of 10 mM. The Na+ ions are spectator ions and are usually not significant because the buffer solution usually also contains about 150 mM NaCl (see Fig. 2.12). In this “phosphate-buffered saline,” the equilibrium between the two species of phosphate ions can “soak up” added acid 2− (producing more ​H​ 2​PO​− 4  ​ ​or added base (producing more ​HPO​4  ​  ​). pK = 6.82 H2PO–4

H+ + HPO2– 4

Acid added

H2PO–4

H+ + HPO2– 4

Base added

H2PO–4

H+ + HPO2– 4

This phenomenon illustrates Le Châtelier’s principle, which states that a change in concentration of one reactant will shift the concentrations of other reactants in order to restore equilibrium. In the human body, the major buffering systems involve bicarbonate, phosphate, and other ions.

Before Going On • Write an equation that describes the equilibrium in a solution containing an acetate/acetic acid buffer and describe what happens when HCl or NaOH is added. • Identify the buffering range of each acid in Table 2.4.

2.5 Clinical Connection: Acid–Base Balance in Humans LEARNING OBJECTIVES Explain how the human body maintains a constant pH. • Write the equations that describe operation of the bicarbonate buffer system in the human body. • Describe the roles of the lungs and kidneys in maintaining blood pH homeostasis. • Summarize the causes and treatments of acidosis and alkalosis.

The cells of the human body typically maintain an internal pH of 6.9–7.4. The body does not normally have to defend itself against strong inorganic acids, but many metabolic processes generate acids, which must be buffered so that they do not cause the pH of blood to drop below its normal value of 7.4. The functional groups of proteins and phosphate groups can serve as biological buffers; however, the most important buffering system involves CO2 (itself a product of metabolism) in the blood plasma (plasma is the fluid component of blood). CO2 reacts with water to form carbonic acid, H2CO3: ​ ​CO​ 2​+ ​H​ 2​O ⇌ ​H​ 2​CO​ 3​ This freely reversible reaction is accelerated in vivo by the enzyme carbonic anhydrase, which is present in most tissues and is particularly abundant in red blood cells. Carbonic acid i­ onizes to bicarbonate, ​​HCO​ − 3​  ​​(see Box 2.C): ​ ​H​ 2​CO​ 3​⇌ ​H​ +​+ ​HCO​− 3  ​ ​

2.5  Clinical Connection: Acid–Base Balance in Humans  47 Normal conditions

H+ + HCO–3

H2CO3

H2O + CO2

H2CO3 H2CO3 H2CO3

H2O + CO2 H2O + CO2 H2O + CO2

H2CO3 H2CO3 H2CO3

H2O + CO2 H2O + CO2 H2O + CO2

Excess acid

H + HCO–3 H+ + HCO–3 H+ + HCO–3 Insufficient acid

H+ + HCO–3 H+ + HCO–3 H + HCO–3

  FIGURE 2.17   The bicarbonate buffer system.  Elimination or retention of CO2 can shift the ­equilibrium in order to promote or prevent the loss of H+ from the body.

so that the overall reaction is ​ ​CO​ 2​+ ​H​ 2​O ⇌ ​H​ +​+ ​HCO​− 3  ​ ​ 2− The pK for this process is 6.1 (the ionization of ​​HCO​ − 3​  ​​ to ​​CO​ 3​  ​​occurs with a pK of 10.3 and is therefore not significant at physiological pH). Although a pK of 6.1 appears to be just outside the range of a useful physiological buffer (which would be within one pH unit of 7.4), the effectiveness of the bicarbonate buffer system is augmented by the fact that excess hydrogen ions can not only be buffered but can also be eliminated from the body. This is possible because after the H+ combines with ​​HCO​ 3−​  ​​ to re-form H2CO3, which rapidly equilibrates with CO2 + H2O, some of the CO2 can be given off as a gas in the lungs. If the body needs to retain more H+ to maintain a constant pH, breathing can be adjusted so that less gaseous CO2 is lost during exhalation (Fig. 2.17). Changes in pulmonary function can adequately adjust blood pH on the order of minutes to hours; however, ­longer-term adjustments of hours to days are made by the Ion exchanger kidneys, which use a variety of mechanisms to excrete or retain H+, bicarbonate, and other ions. In fact, the kidNa+ Na+ neys play a major role in the buffering of metabolic acids. ­Normal metabolic activity generates acids as the result of the degradation of amino acids, the incomplete oxida1 HCO–3 H+ H+ HCO–3 tion of glucose and fatty acids, and the ingestion of acidic groups in the form of phosphoproteins and phospholipids. 2 3 The ​​HCO​ 3−​  ​​required to buffer these acids is initially filtered out of the bloodstream in the kidneys, but the kidneys actively reclaim most of this bicarbonate before it is lost in CO2 CO2 the urine (Fig. 2.18). In addition to reabsorbing filtered ​HCO​3−  ​ ​, the kidneys also generate additional ​​HCO​ 3−​  ​​to offset losses due to the FILTRATE KIDNEY CELL buffering of metabolic acids and the exhalation of CO2. Metabolic activity in the kidney cells produces CO2, which is converted to ​H​ +​ + ​HCO​3 −​ ​. The cells actively secrete the   FIGURE 2.18   Bicarbonate reabsorption.  H+ leaves the kidney H+, which is lost via the urine, accounting for the mildly cells in exchange for Na+ (step 1). The expelled H+ combines with ​​ acidic pH of normal urine. The bicarbonate remaining in HCO​ − 3​  ​​in the filtrate, forming CO2 (step 2). Because it is nonpolar, the − the cell is returned to the bloodstream in exchange for Cl CO2 can diffuse into the kidney cell, where it is converted back to (Fig. 2.19). H+ + ​​HCO​ − 3​  ​​ (step 3).

48  C ha pter 2   Aqueous Chemistry

H+ pump

Ion exchanger

Cl– H+

1

HCO–3

H+

Cl– 2

HCO–3

CO2 FILTRATE

KIDNEY CELL

BLOOD

  FIGURE 2.19   Bicarbonate production.  Carbonic acid–derived protons are pumped out of the kidney cell (step 1), and the bicarbonate remaining in the cell is returned to the bloodstream in exchange for Cl− (step 2).

Some ​​HCO​ − 3​  ​​also accumulates as a result of the metabolism of the amino acid glutamine in the kidneys.

COO H

C

CH2

O CH2

NH2



NH3

C

Glutamine

The two amino groups are removed as ammonia (NH3), which is ultimately excreted in the urine. Because the ammonium ion ​(​NH​+ 4  ​ ​)​has a pK value of 9.25, nearly all the ammonia molecules become protonated at physiological pH. The consumption of protons from carbonic acid (ultimately, CO2) leaves an excess of ​HCO​− 3  ​ ​. Certain medical conditions can disrupt normal acid–base balance, leading to acidosis (blood pH less than 7.35) or alkalosis (blood pH greater than 7.45). The activities of the lungs and kidneys, as well as other organs, may contribute to the imbalance or respond to help correct the imbalance. The most common disorder of acid–base chemistry is metabolic acidosis, which is caused by the accumulation of the acidic products of metabolism and can develop during shock, starvation, severe diarrhea (in which bicarbonate-rich digestive fluid is lost), ­certain genetic diseases, and renal failure (in which damaged kidneys eliminate too little acid). ­Metabolic acidosis can interfere with cardiac function and oxygen delivery and contribute to central nervous system depression. Despite its many different causes, one common symptom of metabolic acidosis is rapid, deep breathing. The increased ventilation helps compensate for the acidosis by “blowing off” more acid in the form of CO2 derived from H2CO3. However, this mechanism also impairs O2 uptake by the lungs. Metabolic acidosis can be treated by administering sodium bicarbonate (NaHCO 3). In chronic metabolic acidosis, the mineral component of bone serves as a buffer, which leads to the loss of calcium, magnesium, and phosphate, and ultimately to osteoporosis and fractures. Metabolic alkalosis, a less common condition, can result from prolonged vomiting, which represents a loss of HCl in gastric fluid. This specific disorder can be treated by infusing NaCl. Metabolic alkalosis is also caused by overproduction of mineralocorticoids (hormones produced by the adrenal glands), which leads to abnormally high levels of H+ excretion and Na+ retention. In this case, the disorder does not respond to saline infusion. Individuals with metabolic alkalosis may experience apnea (cessation of breathing) for 15 s or more and cyanosis (blue coloration) due to inadequate oxygen uptake. Both these symptoms reflect the body’s effort to compensate for the high blood pH by minimizing the loss of CO2 from the lungs. What happens when abnormal lung function is the cause of an acid–base imbalance? Respiratory acidosis can result from impaired pulmonary function caused by airway blockage, asthma (constriction of the airways), and emphysema (loss of alveolar tissue). In all

Key Terms  49

cases, the kidneys respond by adjusting their activity, mainly by increasing the synthesis of the enzymes that break down glutamine in order to produce NH3. Excretion of ​​NH​ 4+​  ​​ helps correct the acidosis. However, renal compensation of respiratory acidosis takes hours to days (the time course for adjusting enzyme levels), so this condition is best treated by restoring pulmonary function through bronchodilation, supplemental oxygen, or mechanical ventilation (assisted breathing). Some of the diseases that impair lung function (for example, asthma) can also contribute to respiratory alkalosis, but this relatively rare condition is more often caused by hyperventilation brought on by fear or anxiety. Unlike the other acid–base disorders, this form of respiratory alkalosis is seldom life-threatening.

Before Going On • Review the equations that describe the interconversion of carbon dioxide and bicarbonate. • Compare the metabolic adjustments made by the lungs and kidneys during acid–base homeostasis. • Discuss how impaired lung or kidney function could lead to acidosis or alkalosis.

Summary 2.1  Water Molecules and Hydrogen Bonds •  Water molecules are polar; they form hydrogen bonds with each other and with other polar molecules bearing hydrogen bond donor or acceptor groups. •  The electrostatic forces acting on biological molecules also include ionic interactions and van der Waals interactions. •  Water dissolves polar and ionic substances.

value. The pH of a solution can be altered by adding an acid (which donates protons) or a base (which accepts protons). •  The tendency for a proton to dissociate from an acid is expressed as a pK value. •  The Henderson–Hasselbalch equation relates the pH of a solution of a weak acid and its conjugate base to the pK and the concentrations of the acid and base.

2.4  Tools and Techniques: Buffers

2.2  The Hydrophobic Effect •  Nonpolar (hydrophobic) substances tend to aggregate rather than disperse in water in order to minimize the decrease in entropy that would be required for water molecules to surround each nonpolar molecule. This is the hydrophobic effect. •  Amphiphilic molecules, which contain both polar and nonpolar groups, may aggregate to form micelles or bilayers.

2.3  Acid–Base Chemistry •  The dissociation of water produces hydroxide ions (OH−) and protons (H+) whose concentration can be expressed as a pH

•  A buffered solution, which contains an acid and its conjugate base, resists changes in pH when more acid or base is added.

2.5  Clinical Connection: Acid–Base Balance in Humans •  The body uses the bicarbonate buffer system to maintain a constant internal pH. Homeostatic adjustments are made by the lungs, where CO2 is released, and by the kidneys, which excrete H+ and ammonia.

Key Terms polarity hydrogen bond ionic interaction van der Waals radius electronegativity

van der Waals interaction dipole–dipole interaction London dispersion forces dielectric constant solute

solvation hydration hydrophilic hydrophobic hydrophobic effect

50  C ha pter 2   Aqueous Chemistry amphiphilic amphipathic micelle bilayer vesicle hydronium ion proton jumping ionization constant of water (Kw) neutral solution

acidic solution basic solution pH acid base acid dissociation constant (Ka) pK conjugate base polyprotic acid

Henderson–Hasselbalch equation buffer Le Châtelier’s principle acidosis alkalosis metabolic acidosis metabolic alkalosis respiratory acidosis respiratory alkalosis

Bioinformatics Brief Bioinformatics Exercises 2.1  Structure and Solubility 2.2  Amino Acids, Ionization, and pK Values

Problems 2.1  Water Molecules and Hydrogen Bonds 1.  The H—C—H bond angle in the perfectly tetrahedral CH 4 ­ olecule is 109°. Explain why the H—O—H bond angle in water is m only about 104.5°. 2.  Each CO bond in CO2 is polar, yet the whole molecule is nonpolar. Explain.

6.  The enzyme dihydrofolate reductase is required for DNA synthesis and, as such, is an attractive drug target in cancer therapy. The drug methotrexate (MTX) competes with the substrate DHF for binding to the enzyme. Identify the hydrogen bond acceptor and donor groups in MTX and DHF. Use an arrow to point toward each acceptor and away from each donor.

H

3.  Which compound has a higher boiling point, H2O or H2S? Explain. 4.  Consider the following molecules and their melting points listed below. How can you account for the differences in melting points among these molecules of similar size? Molecular weight (g · mol–1)

Melting point (°C)

18.0 17.0 16.0

0 –77 –182

Water, H2O Ammonia, NH3 Methane, CH4

5.  Identify the hydrogen bond acceptor and donor groups in the following molecules. Use an arrow to point toward each acceptor and away from each donor.

CH2 O

O +H

3N

CH

C

N

O

Aspartame

S O

Sulfanilamide

N

N+

H

H

N

H MTX

NH2

O

H

O

O N H

N

Uric acid

H

O

N

R

N

N N

N H

DHF

7.  Identify the hydrogen bonding patterns that are identical in the two molecules shown in Problem 6. (Investigators carried out this exercise in order to determine how the DHF substrate and the MTX inhibitor bind to the enzyme.) 8.  An exhaustive study with over 6000 fluorinated compounds was carried out in order to determine the ability of these compounds to participate in hydrogen bonding (see Box 2.A). The structure of one of these compounds is shown below. Identify the hydrogen bond acceptor and donor groups. Use an arrow to point toward each acceptor and away from each donor.

N N

H

O

H

N

N

N H

CH3

H

H

R H

O

O H2N

C

N

N

N H

H H3C

F

H

CH2 COO–

CH

N

9.  In 2007, pets consuming a brand of pet food containing melamine (Compound B in Problem 1.16) and cyanuric acid (both nontoxic when consumed alone) suffered from renal failure when the two compounds combined to form crystalline melamine cyanurate in the kidney. Draw the structure of melamine cyanurate, which forms when melamine and cyanuric acid (see below) form hydrogen bonds with each other.

H3C

C

CH3

O H3C

H

H H

N

N

N

H

H

N

O

H

H

Melamine

CH2

CH3

16.  What intermolecular interactions are likely to form between the two amino acid side chains shown in the protein diagrams below?

O

N

NH2

d.  CsCl

NH

HN

N

N

c.  H C 3

O

Problems  51

C

a. 

Cyanuric acid

10.  A recent study showed that the structure of melamine cyanurate (see Solution 9) could be disrupted by adding a compound called DHI, shown below. DHI disrupts the hydrogen bonds in melamine cyanurate and forms new ones with cyanuric acid (CYA, see Problem 9), releasing melamine. Draw the interactions formed between DHI and cyanuric acid.

COOH

O

HO

C CH2

Asn

b. 

NH2

NH+ 3 (CH2)4

DHI

Ser

–

OOC Asp

CH2

CH2

Lys

HO

N

OH

11.  Examine Table 2.1. What is the relationship between an atom’s electronegativity and its ability to participate in hydrogen bonding?

c. 

12.  Do intermolecular hydrogen bonds form in the compounds below? Draw the hydrogen bonds where appropriate.

CH3

O A

H3C

B

N

C N

  17.  The solubilities of several alcohols in water are shown below (their structures are shown in Table 2.2). Note that propanol is miscible in water (i.e., any amount will dissolve). However, as the size of the alcohol increases, solubility decreases. Explain.

H3C

H CH2

OH

D

H3C

CH2

Cl and H O H

H3C

CH2

O

CH2

CH3 and H

N H

H

13.  Occasionally, when a carbon atom is polarized by an adjacent electronegative atom, a C—H group can form a weak hydrogen bond with a carbonyl group. Draw a structure to represent this situation and add δ+ and δ– symbols where appropriate. 14.  Two hydrogen bonds form between the DNA bases adenine and thymine. Draw these bonds as dashed lines.

CH3

H

N N

N N

N Adenine

H

N

O N H

O

Thymine

15.  What are the most important intermolecular interactions in the following substances?

O a.  H3C

C

CH3

O b.  H3C  

H3C CsCl

C

NH2

CH2

Phe

Ala

H

C

E

CH2

CH3

Alcohol

Solubility in water (mmol · 100 g–1)

Propanol Butanol Pentanol

Miscible 0.11 0.05

18.  Is the solubility trend for the alcohols in Table 2.2 related to their dielectric constants? 19.  Water is unusual in that its solid form is less dense than its liquid form. This means that when a pond freezes in the winter, ice is found as a layer on top of the pond, not on the bottom. What are the biological advantages of this? 20.  Ice skaters skate not on the solid ice surface itself but on a thin film of water that forms between the skater’s blade and the ice. What unique property of water makes ice skating possible? 21.  Ammonium sulfate, (NH4)2SO4, is a water-soluble salt. Draw the structures of the hydrated ions that form when ammonium sulfate dissolves in water. 22.  Which of the four primary alcohols listed in Table 2.2 would be the best solvent for ammonium ions? What can you conclude about the polarity of these solvents, which all contain an —OH group that can form hydrogen bonds? 23.  The amino acid glycine is sometimes drawn as structure A below. However, the structure of glycine is more accurately represented by structure B. Glycine has the following properties: white crystalline solid, high melting point, and water solubility. Why does

)15

52  C ha pter 2   Aqueous Chemistry structure B more accurately represent the structure of glycine than structure A?

H2N

H3N

Structure A

CH2

CH3

COO–

Structure B

H3C

A

B

25.  Explain why water forms nearly spherical droplets on the surface of a freshly waxed car. Why doesn’t water bead on a clean windshield?

C

26.  A paper clip floats on the surface of water contained in a beaker. What happens if a drop of soap solution is added to the water?

CH3

Hexadecyltrimethylammonium

N+

CH3

OH CH3

CH3

CH3

HO O– O

CH3

Cholate

HO

HC

OH CH3

CH3

CH3

CH3

OH

H

(CH2)9

P

CH3

CH3 Dimethyldecylphosphine oxide

H

O P

CH3

CH3

HO

CH2OH O H OH H H

phosphine oxide

HO

CH3

O O

O

C C

(CH2)11 (CH2)11

HO

CH2

O

H3C

CH

COO−

CH3

CH3

CH3

OH O–

33.  The compound Cholate bis-(2-ethylhexyl)sulfosuccinate (abbreviated AOT) is capable of forming “reverse” micelles in the hydrocarbon solvent OH isooctane (2,2,4-trimethylpentane). Scientists have investigated the use of reverse micelles for extracting water-soluble proteins. A two-phase system is formed: the hydrocarbon phase containing the reverse micelles and the water phase containing the protein. After a certain period of time, the protein is transferred to the reverse micelle.  a.  Draw the structure of the reverse micelle that AOT would form in isooctane.  b.  Where would the protein be located in the reverse micelle?

CH2CH3

29.  The structures of dimethyldecylphosphine oxide and n-octyl gluCH2OH coside, both nonionic detergents, are shown here. Identify the polar and the nonpolar regions of these amphipathic molecules. O

O

CH2COO−

32.   Which compounds in Problem 31 are capable of forming micelles? Which are capable of forming bilayers? O

28.  Identify the regions of the detergents shown in Problem 27 that are able to form favorable interactions with water. Identify these interactions.

H3C

H3C

CH3 Cl− N CH3

CH2

E

N+

(CH2)15

(CH2)11

D

27.  The structures of hexadecyltrimethylammonium (a cationic detergent) and cholate (an anionic detergent) are shown here. Identify the polar and the nonpolar regions of these amphipathic molecules.

H3C

H3C

CH3

2.2  The Hydrophobic Effect

CH3

N+

(CH2)11

CH3

24. a.  Water has a surface tension that is nearly three times greater than that of ethanol. Explain.  b.  The surface tension of water decreases with increasing temperature. Explain.

methylammonium

)9

+

COOH

CH2

31.  Consider the structures of the molecules below. Are these molecules polar, nonpolar, or amphiphilic?

O

H3C

(CH2)3

CH

CH2

O

C

CH2 O

H3C

(CH2)3

CH

CH2

O

C

CH

CH2CH3

O

S

O

O

Bis-(2-ethylhexyl)sulfosuccinate (AOT)

(CH2)7 CH O household 3 are amphiphilic substances, often the salts 34.  Many soaps H of long-chain fatty acids, that form water-soluble micelles. An examOH H ple is sodium dodecyl sulfate (SDS), an anionic detergent.  a. IdenH

tify the polar and nonpolar portions of the SDS molecule.  b. Draw the OHstructure of the micelle formed by SDS.  c.  Explain how the SDS micelles “wash away” water-insoluble substances such as cooking n-Octylglucoside grease.

H

O

O

(CH2)7 CH3

H

OH

n-Octylglucoside

30.  Identify the regions of the detergents shown in Problem 29 that are able to form favorable interactions with water. Identify these interactions.

H3C

(CH2)11

O

S

O− Na+

O

Sodium dodecyl sulfate (SDS) 35.  Just as a dissolved substance tends to move spontaneously down its concentration gradient, water also tends to move from an area of high concentration (low solute concentration) to an area of low concentration (high solute concentration), a process known as osmosis.  a.  Explain why a lipid bilayer would be a barrier to osmosis.  b.  Why are isolated human cells placed in a solution that

Problems  53 typically contains about 150 mM NaCl? What would happen if the cells were placed in pure water? 36.  Fresh water is obtained from seawater in a desalination process using reverse osmosis. In reverse osmosis, seawater is forced through a membrane; the salt remains on one side of the membrane and fresh water is produced on the other side. In what ways does reverse osmosis differ from osmosis described in Problem 35? 37.  Which of the substances in Problem 15 would be able to easily cross a bilayer? Which substances could not? Explain your answer. 38.  Drug delivery is a challenging problem because the drug molecules must be sufficiently water-soluble to dissolve in the blood, but also sufficiently nonpolar to pass through the cell membrane for delivery. A medicinal chemist proposes to encapsulate a water-­ soluble drug into a vesicle (see Section 2.2). How does this strategy facilitate the delivery of the drug to its target cell? 39.  A specialized protein pump in the red blood cell membrane exports Na + ions and imports K + ions in order to maintain the sodium and potassium ion concentrations shown in Figure 2.12. Does the movement of these ions occur spontaneously or is this an energy-requiring process? Explain. 40.  Estimate the amount of Na+ lost in sweat during 15 minutes of vigorous exercise. What is the mass of potato chips (200 mg Na+ per ounce) you would have to consume in order to replace the lost sodium? 41.  The bacterium E. coli can adapt to changes in the solute concentration of its growth medium. The cell consists of a cytoplasmic compartment bounded by a cell membrane surrounded by a porous cell wall; both the membrane and the cell wall allow the passage of water and ions. Under nongrowing conditions, only cytoplasmic water content is regulated. What happens to the cytoplasmic volume if E. coli is grown in a growth medium with  a.  a high salt concentration or  b.  a low salt concentration? E. coli

Periplasmic space

Cytoplasm

Cytoplasmic membrane

48.  Fill in the blanks of the following table:

______ ______ neutral ______ ______

______ 7.42 ______ ______ ______

1.3 × 10–8 ______ ______ ______ 7.9 × 10–3

______ ______ ______ 6.3 × 10–8 ______

51.  Give the conjugate base of the following acids:  a. ​​HC​ 2​​​O​ 4−​  ​​;  b. ​HS​O− ​H​ 2​P​O− HC​O− HAs​O2− HP​O2− H​O− 3​  ​  ​​;  c. ​ 4​  ​  ​​;  d. ​ 3​  ​  ​​;  e. ​ 4​  ​  ​​;  f. ​ 4​  ​  ​​;  g. ​ 2​  ​  ​​. 52.  Give the conjugate acid of the species listed in Problem 51. 53.  Calculate the pH of 500 mL of water to which  a.  20 mL of 1.0 M HNO3 or  b.  15 mL of 1.0 M KOH has been added. 54.  Calculate the pH of 1.0 L of water to which  a.  1.5 mL of 3.0 M HCl or  b.  1.5 mL of 3.0 M NaOH has been added. 55.  Calculate the pH of 500 mL of water to which  a.  10 mL of 0.50 M HBr or  b.  25 mL of 0.25 M NaOH has been added. 56.  Calculate the pH of 1.0 L of water to which  a.  15 mL of 2.0 M HI or  b.  25 mL of 2.0 M Ba(OH)2 has been added. 57.  Identify the acidic hydrogens in the following compounds, using Table 2.4 as a guide:

CH2 C

COOH

COOH

CH2

+

N H2

COOH

Piperidine

Citric acid

O +

H3N

CH

C

OH

HO

O

O

C

C

CH2

Oxalic acid

CH2

O

CH2 +

O

NH3

O

N H

Barbituric acid

Lysine

44.  Like all equilibrium constants, the value of Kw is temperature dependent. Kw is 1.47 × 10–14 at 30°C. What is “neutral” pH at this temperature?

O O

OH

NH

CH2

43.  Compare the concentrations of H2O and H+ in a sample of pure water at pH 7.0 at 25°C.

47.  Draw a diagram similar to Figure 2.13 showing how a hydroxide ion appears to jump through an aqueous solution.

[OH–] (M)

50.  The pH of urine has been found to be correlated with diet. Acidic urine results when meat and dairy products are consumed, because the oxidation of sulfur-containing amino acids in the proteins produces protons. Consumption of fruits and vegetables leads to alkaline urine, because these foods contain plentiful amounts of potassium and magnesium carbonate salts. Why would the presence of these salts result in alkaline urine?

2.3  Acid–Base Chemistry

46.  What is the pH of a solution of 1.0 × 10–9 M NaOH?

[H+] (M)

49.  Several hours after a meal, partially digested food leaves the stomach and enters the small intestine, where pancreatic juice is added. How does the pH of the partially digested mixture change as it passes from the stomach into the intestine (see Table 2.3)?

Cell wall

45.  What is the pH of a solution of 1.0 × 10–9 M HCl?

pH

Pancreatic juice Blood Saliva Urine Gastric juice

HO

42.  Under growth conditions, E. coli regulates cytoplasmic K + content (in addition to water; see Problem 41) in order to prevent large changes in cell volume. How might E. coli regulate both the cytoplasmic concentrations of K+ and water when grown in a lowsalt medium? What happens when E. coli is grown in a high-salt medium?

Acid, base, or neutral?

+

NH

CH2

CH2

S O

O−

4-Morphine ethanesulfonic acid (MES)

54  C ha pter 2   Aqueous Chemistry 58.  Rank the following according to their strength as acids:

A B C D E

Acid

Ka

pK

Citrate Succinic acid Succinate Formic acid Citric acid

_________ 6.17 × 10−5 2.29 × 10−6 1.78 × 10−4 _________

4.76 ______ ______ ______ 3.13

71.  The pK of CH3CH2N​​H3​ +​  ​​is 10.7. Would the pK of FCH2CH2N​​H3​ +​  ​​ be higher or lower? 72.  The structure of pyruvic acid is shown below.  a.  Draw the structure of pyruvate.  b.  Using what you have learned about acidic functional groups, which form of this compound is likely to predominate in the cell at pH 7.4? Explain.

CH3 C

59.  Imidazole, shown here in its unprotonated form, has a pK value near 7.0 (see Table 2.4). Draw the structure of imidazole that predominates at the pH of blood.

N Imidazole 60.  The blood of individuals with uncontrolled diabetes contains ketone bodies, such as the one shown here. Identify the acidic functional group(s) in acetoacetic acid and draw the structure of the molecule as it would appear at the pH of blood.

H3C

C

O CH2

COOH

Pyruvic acid

2.4  Tools and Techniques: Buffers

NH

O

O

OH

Acetoacetic acid 61.  Calculate the pH of a 500-mL solution to which has been added 10 mL of 50 mM boric acid and 20 mL of 20 mM sodium borate. 62.  Calculate the pH of a 500-mL solution to which has been added 20 mL of 100 mM glycinamide hydrochloride (Cl‒ +H3NCH3CONH2) and 35 mL of 50 mM glycinamide (H2NCH3CONH2). 63.  A solution is made by mixing 50 mL of a solution of 2.0 M K2HPO4 and 25 mL of a solution of 2.0 M KH 2PO4. The solution is diluted to a final volume of 200 mL. What is the final pH of the solution? 64.  A solution is made by mixing 25 mL of a solution of 2.0 M K2HPO4 and 50 mL of a solution of 2.0 M KH 2PO4. The solution is diluted to a final volume of 200 mL. What is the final pH of the solution? 65.  Many insect larvae, like the ant shown at the start of the chapter, secrete acid for defense and to facilitate feeding on plant leaves. The fluid secreted by the Theroa zethus caterpillar contains approximately 6.53 M formic acid. Calculate the pH of this solution, assuming that only 1% of the acid is in the dissociated (A–) form. 66.  The fluid secreted by the Theroa zethus caterpillar (see Problem 65) also contains approximately 50 mM butyric acid (pK = 4.82). If the secretion contained only the butyric acid and 90% was in the undissociated (HA) form, what would be the solution’s pH? 67.  Calculate the concentration of acetate in a 50 mM solution of acetic acid buffer at pH 5.0. 68.  Calculate the concentration of HP​​O​ 42–​  ​​in a 100 mM solution of phosphate at pH 7.50. 69.  The pK values for succinic acid and succinate are provided in Table 2.4. Draw the structure of the species that predominates at pH 2, 5, and 7. 70.  The amino acid glycine (H2N—CH2—COOH) has pK values of 2.35 and 9.78. Indicate the structure and net charge of the molecular species that predominates at pH 2, 7, and 10.

73.  Which would be a more effective buffer at pH 8.0 (see Table 2.4)?  a.  10 mM HEPES buffer or 10 mM glycinamide buffer;  b.  10 mM Tris buffer or 20 mM Tris buffer;  c.  10 mM boric acid or 10 mM sodium borate. 74.  Which would be a more effective buffer at pH 5.0 (see Table 2.4)?  a.  10 mM acetic acid buffer or 10 mM HEPES buffer;  b.  10 mM acetic acid buffer or 20 mM acetic acid buffer;  c.  10 mM acetic acid or 10 mM sodium acetate. 75.  Phosphoric acid, H3PO4, has three ionizable protons.  a. Sketch the titration curve. Indicate the pK values and the species that predominate in each area of the curve.  b.  Write the equations for the dissociation of the three ionizable protons.  c.  Which two phosphate species are present in the blood at pH 7.4?  d.  Which two phosphate species would be used to prepare a buffer solution of pH 11? 76.  The structure of acetylsalicylic acid (aspirin) is shown. Is aspirin more likely to be absorbed (pass through a lipid membrane) in the stomach (pH ∼2) or in the small intestine (pH ∼8)? Explain.

O C

pK = 2.97

OH O

C

CH3

O Acetylsalicylic acid (aspirin) 77.  Calculate the ratio of imidazole to the imidazolium ion in a solution at pH 7.4. 78.  What volume (in mL) of a 2.0 M solution of imidazolium ­chloride must be added to a 500-mL solution of 10 mM imidazole in order to achieve a pH of 6.5? 79.  Calculate the ratio of methylamine to the methylammonium ion in a solution at pH 10.5. 80.  Methylamine is sold by a manufacturer as a 40% solution by weight. The molar mass of methylamine is 31.1 g ⋅ mol–1. What volume of this solution would need to be added to 1.0 L of a 50-mM solution of methylammonium bromide in order to achieve a pH of 10.5? 81.  What is the volume (in mL) of glacial acetic acid (17.4 M) that would need to be added to 500 mL of a solution of 0.20 M sodium acetate in order to achieve a pH of 5.0? 82.  What is the mass of NaOH that would need to be added to 500 mL of a solution of 0.20 M acetic acid in order to achieve a pH of 5.0? 83.  An experiment requires the buffer HEPES, pH = 8.0 (see Table 2.4).  a.  Write an equation for the dissociation of HEPES in water. Identify the weak acid and the conjugate base.  b.  What is the

Problems  55 effective buffering range for HEPES?  c.  The buffer will be prepared by making 1.0 L of a 0.10 M solution of HEPES. Hydrochloric acid will be added to achieve the desired pH. Describe how you will make 1.0 L of 0.10 M HEPES (A−). (HEPES is supplied by the chemical company as a sodium salt with a molar mass of 260.3 g ⋅ mol−1.)  d. What is the volume (in mL) of 6.0 M HCl that must be added to the 0.10 M HEPES to achieve the desired pH of 8.0? Describe how you will make the buffer. 84.  An alternative way of preparing the buffer described in Problem 83 involves preparing solutions of the sodium salt form of HEPES (A−) and the weak acid form of HEPES (HA) and then combining the solutions to obtain a buffer with the desired pH and concentration.  a.  Describe how you would prepare 250 mL of 1.0-M stock solutions of the A− and HA forms of HEPES (supplied as solids by the manufacturer with molar masses of 260.3 g ⋅ mol−1 and 238.3 g ⋅ mol−1, respectively).  b.  Calculate the volume of each stock solution needed to prepare 1.0 L of the 0.10 M HEPES buffer described in Problem 83 and describe how you would prepare the buffer. 85.   One liter of a 0.10 M Tris buffer (see Table 2.4) is prepared and adjusted to a pH of 8.2.

CH2OH HOH2C

C

NH2

CH2OH

Tris(hydroxymethyl)aminomethane a.  Write the equation for the dissociation of Tris in water. Identify the weak acid and the conjugate base.  b.  What is the effective buffering range for Tris?  c.  What are the concentrations of the conjugate acid and weak base at pH 8.2?  d.  What is the ratio of conjugate base to weak acid if 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer? What is the new pH? Has the buffer functioned effectively? Compare the pH change to that of Problem 54a in which the same amount of acid was added to the same volume of pure water.  e.  What is the ratio of conjugate base to weak acid if 1.5 mL of 3.0 M NaOH is added to 1.0 L of the buffer? What is the new pH? Has the buffer functioned effectively? Compare the pH change to that of Problem 54b in which the same amount of base was added to the same volume of pure water. 86.  One liter of a 0.1 M Tris buffer (see Table 2.4) is prepared and adjusted to a pH of 2.0.  a.  What are the concentrations of the conjugate base and weak acid at this pH?  b.  What is the pH when 1.5 mL of 3.0 M HCl is added to 1.0 L of the buffer? Has the buffer functioned effectively? Explain.  c.  What is the pH when 1.5 mL of 3.0 M NaOH is added to 1.0 L of the buffer? Has the buffer functioned effectively? Explain. 87.  An experiment requires an imidazole buffer at pH 7.0 (see Problem 77). The buffer is prepared by making a solution of imidazole, then adding hydrochloric acid to achieve the desired pH.  a. Describe how to prepare 1.0 L of a 0.10-M solution of imidazole (supplied by the manufacturer as a solid with a molar mass of 68.1 g ⋅ mol−1).  b.  What is the volume (in mL) of 6.0 M HCl that must be added to this solution to achieve the desired pH? Describe how you will make the buffer. 88.  An alternative way of preparing the buffer described in Problem 87 is described in Problem 84 and involves preparing solutions of imidazole (A−) and imidazolium (HA; supplied by the manufacturer as a hydrobromide salt with a molar mass of 149 g ⋅ mol−1) and then combining the solutions to obtain a buffer with the desired pH and concentration.  a.  Describe how you would prepare 100 mL

of 1.0-M stock solutions of imidazole and imidazolium hydrobromide.  b.  Calculate the volume of each stock solution needed to prepare 1.0 L of the 0.10 M imidazole buffer described in Problem 87 and describe how you would prepare the buffer.

2.5  Clinical Connection: Acid–Base Balance in Humans 89.  The pH of blood is maintained within a narrow range (7.35–7.45). Carbonic acid, H2CO3, participates in blood buffering.  a.  Write the equations for the dissociation of the two ionizable protons.  b. The pK for the first ionizable proton is 6.1; the pK for the second ionizable proton is 10.3. Use this information to identify the weak acid and the conjugate base present in the blood.  c.  Calculate the concentration of carbonic acid in a sample of blood with a bicarbonate concentration of 24 mM and a pH of 7.4. 90.  Impaired pulmonary function can contribute to respiratory acidosis. Using the appropriate equations, explain how the failure to eliminate sufficient CO2 through the lungs leads to acidosis. 91.  Metabolic acidosis often occurs in patients with impaired circulation from cardiac arrest. Mechanical hyperventilation can be used to alleviate acidosis. Explain why this strategy is effective. 92.  Mechanical hyperventilation (a standard treatment for acidosis as described in Problem 91) cannot be used in patients with impaired pulmonary function because these patients often have acute lung injury (ALI).  a.  Sodium bicarbonate is an effective treatment for metabolic acidosis. Would this be an acceptable treatment for ALI patients?  b.  A group of physicians at San Francisco General Hospital advocates using Tris (see Problem 85) to treat metabolic acidosis in these patients. Explain why this procedure is effective and why it is an acceptable treatment for ALI patients. 93.  An individual who develops alkalosis by hyperventilating is encouraged to breathe into a paper bag for several minutes. Why does this treatment correct the alkalosis? 94.  A patient who took a large overdose of aspirin is brought into the emergency room for treatment. She suffers from respiratory alkalosis and the pH of her blood is 7.5. Determine the ratio of H ​ C​O− 3​  ​  ​​ to H2CO3 in the patient’s blood at this pH. How does this compare to the ratio of ​HC​O3​ −​  ​​ to H2CO3 in normal blood? Can the H2CO3/​HC​O3​ −​  ​​ system work effectively as a buffer in this patient under these conditions? 95.  Metabolic acidosis is a general term that describes a number of disorders in metabolism in the body that result in a lowering of the blood pH from 7.4 to 7.35 or below. The kidney plays a vital role in regulating blood pH. The kidney can either excrete or reabsorb various ions, including phosphate, H2P​​O​ 4−​  ​​; ammonium, N​​H​ 4+​  ​​; or bicarbonate, HC​​O3​ −​  ​​. Which ions are excreted and which ions are reabsorbed in metabolic acidosis? Explain, using relevant chemical equations. 96.  Metabolic alkalosis occurs when the blood pH rises to 7.45 or greater. Which ions are excreted and which ions are reabsorbed in metabolic alkalosis (see Problem 95)? 97.  Kidney cells excrete H+ and reabsorb ​HCO​− 3  ​ ​from the filtrate. The movement of each of these ions is thermodynamically unfavorable. However, the movement of each becomes possible when it is coupled to another, thermodynamically favorable ion transport process. Explain how the movement of other ions drives the transport of H+ and ​HCO​− 3  ​ ​. 98.  In uncontrolled diabetes, the body converts fats to ketone bodies. One of these ketone bodies is acetoacetic acid (see Problem 60), which accumulates in the bloodstream. Do ketone bodies contribute

56  C ha pter 2   Aqueous Chemistry to acidosis or alkalosis? How might the body compensate for this acid–base imbalance? 99.  Kidney cells have a carbonic anhydrase on their external surface as well as an intracellular carbonic anhydrase. What are the functions of these two enzymes?

100.  Explain why the lungs can rapidly compensate for metabolic acidosis, whereas the kidneys are slow to compensate for respiratory acidosis.

Selected Readings Ball, P., Water, water, everywhere? Nature 427, 19–20 (2004). [A brief discussion of water as the matrix of life.] Garde, S. and Patel, A.J., Unraveling the hydrophobic effect, one molecule at a time, Proc. Natl. Acad. Sci. 108, 16491–16492, doi: 10.1073/pnas.1113256108 (2011). [A brief commentary that summarizes some key features of the hydrophobic effect.] Privalov, P.L. and Crane-Robinson, C., Role of water in the formation of macromolecular structures, Eur. Biophys. J. 46, 203–224, doi: 10.1007/s00249-016-1161-y (2017). [An extensive review that includes descriptions of the contribution of water to the structures and binding properties of proteins and DNA.]

Richardson, J.O., Pérez, C., Lobsiger, S., Reid, A.A., Temelso, B., Shields, G.C., Kisiel, Z., Wales, D.J., Pate, B.H., and Althorpe, S.C., Concerted hydrogen-bond breaking by quantum tunneling in the water hexamer prism, Science 351, 1310–1313, doi: 10.1126/science. aae0012 (2016). [Describes the structure and dynamics of a hexamer of water molecules.] Seifter, J.L. and Chang, H.-Y., Extracellular acid–base balance and ion transport between body fluid compartments, Physiology 32, 367–379, doi: 10.1152/physiol.00007 (2017). [Focuses on plasma membrane transporters involved in maintaining intra- and extracellular pH.]

CHAPTER 3

Nucleic Acid Structure and Function DO YOU REMEMBER? • Cells contain four major types of biological molecules and three major types of polymers (Section 1.2). • Modern prokaryotic and eukaryotic cells apparently evolved from simpler nonliving systems (Section 1.4). • Noncovalent forces, including hydrogen bonds, ionic interactions, and van der Waals forces, act on biological molecules (Section 2.1).

The inherited characteristics of every organism are ­determined by its DNA, but the amount of DNA varies considerably. Each cell of the pine tree that produced this cone contains about 20 billion base pairs of DNA, about seven times more DNA than in a human cell.

All the structural components of a cell and the machinery that carries out the cell’s activities are ultimately specified by the cell’s genetic material—DNA. Therefore, before examining other types of biological molecules and their metabolic transformations, we must consider the nature of DNA, including its chemical structure and how its biological information is organized and expressed.

3.1 Nucleotides LEARNING OBJECTIVES Recognize the structures of nucleotides. • Identify the base, sugar, and phosphate groups of nucleotides. • Recognize nucleotide derivatives.

Gregor Mendel was certainly not the first to notice that an organism’s characteristics (for example, flower color or seed shape in pea plants) were passed to its progeny, but in 1865 he was the first to describe their predictable patterns of inheritance. By 1903, Mendel’s inherited factors (now called genes) were recognized as belonging to chromosomes (a word that means “colored bodies”), which are visible by light microscopy (Fig. 3.1). Eventually, chromosomes were shown to be composed of proteins, which had first been described in 1838 by Gerardus Johannes Mulder, and nucleic acids, which had been discovered in 1869 by Friedrich Miescher. Proteins, with their 20 different types of amino acids and great diversity in size and shape, were the obvious candidates to be carriers of genetic information in chromosomes. Nucleic acids, in contrast, seemed uninteresting and contained only four different types of structural units, called nucleotides. In DNA (deoxyribonucleic

57

58  C ha pter 3   Nucleic Acid Structure and Function

acid), these components—abbreviated A, C, G, and T—were thought to occur as simple repeating tetranucleotides, for example, —ACGT-ACGT-ACGT-ACGT—

Dr. P. Boyer/Science Source

In 1950, when Erwin Chargaff showed that the nucleotides in DNA were not all present in equal numbers and that the nucleotide composition varied among species, it became apparent that DNA might be complex enough to be the genetic material after all. Several other lines of research also pointed to the importance of DNA, and the race was on to decipher its molecular structure.

Nucleic acids are polymers of nucleotides Each nucleotide of DNA includes a nitrogen-containing base. The bases adenine (A) and guanine (G) are known as purines because they resemble the organic compound purine:

NH2

  FIGURE 3.1   Human chromosomes

from amniocentesis.  In this image, the chromosomes have been stained with fluorescent dyes.

N

N

N

N Adenine

O

H

N

N

H

Guanine

6

2

N

N

H2N

N1

3

N 7

5 4

9

N

H

8

N

Purine

H

The bases cytosine (C) and thymine (T) are known as pyrimidines because they resemble the organic compound pyrimidine:

NH2 N O

O

H

N3 2

O

N

CH3

N N

H

1

5 6

N

H

Cytosine

4

Thymine

Pyrimidine

Ribonucleic acid (RNA) contains the pyrimidine uracil (U) rather than thymine:

H O

O

N N H

Uracil

so that DNA contains the bases A, C, G, and T, whereas RNA contains A, C, G, and U. The purines and pyrimidines are known as bases because they can participate in acid–base reactions. However, they donate or accept protons only at extremely low or high pH, so this behavior is not relevant to their function inside cells. Linking atom N9 in a purine or atom N1 in a pyrimidine to a five-carbon sugar forms a nucleoside. In DNA, the sugar is 2ʹ-deoxyribose; in RNA, the sugar is ribose (the sugar atoms are numbered with primes to distinguish them from the atoms of the attached bases). 5′

HOCH2 O Ribose

4′

H

H

3′

OH

5′

HOCH2 O

Base H 1′

2′

H

OH

2′-Deoxyribose

4′

H

H

3′

OH

Base H 1′

2′

H

H

A nucleotide is a nucleoside to which one or more phosphate groups are linked, usually at C5ʹ of the sugar. Depending on whether there are one, two, or three phosphate groups, the

3.1 Nucleotides  59

nucleotide is known as a nucleoside monophosphate, nucleoside diphosphate, or nucleoside triphosphate and is represented by a three-letter abbreviation, for example,

NH2 N

N O −

O

P

N

N O

H2C

O−

H

−

O

O H

H

O

O P

O

H

H

O

O −

O H

N

N

N

H2C

O

O−

O−

H

N

N

H2N

P

NH2

O

H

O

H

P

O

O−

OH OH

OH OH

Guanosine diphosphate (GDP)

Adenosine monophosphate (AMP)

P O−

H

N3 O

2

4 1

5 6

CH3

N

5-Methylcytosine residue

CH3

N N1 2

6 3

N

5 4

N 7

9

8

N

N 6-Methyladenine residue

TA B L E 3.1   Nucleic Acid Bases, Nucleosides, and Nucleotides

Base

Nucleoside a

Nucleotides a

Adenine (A)

Adenosine

Adenylate; adenosine monophosphate (AMP)   adenosine diphosphate (ADP)   adenosine triphosphate (ATP)

Cytosine (C)

Cytidine

Cytidylate; cytidine monophosphate (CMP)   cytidine diphosphate (CDP)   cytidine triphosphate (CTP)

Guanine (G)

Guanosine

Guanylate; guanosine monophosphate (GMP)   guanosine diphosphate (GDP)   guanosine triphosphate (GTP)

Thymine (T)

b

Thymidine

Thymidylate; thymidine monophosphate (TMP)   thymidine diphosphate (TDP)   thymidine triphosphate (TTP)

Uracil (U)

c

Uridine

Uridylate; uridine monophosphate (UMP)   uridine diphosphate (UDP)   uridine triphosphate (UTP)

a 

Nucleosides and nucleotides containing 2ʹ-deoxyribose rather than ribose may be called deoxynucleosides and deoxynucleotides. The nucleotide abbreviation is then preceded by “d.” b 

Thymine is found in DNA but not in RNA.

c 

Uracil is found in RNA but not in DNA.

P O−

O

H2C H

Cytidine triphosphate (CTP)

Deoxynucleotides are named in a similar fashion, and their abbreviations are preceded by “d.” The deoxy counterparts of the compounds shown above would therefore be deoxy­ adenosine monophosphate (dAMP), deoxyguanosine diphosphate (dGDP), and deoxycytidine triphosphate (dCTP). The names and abbreviations of the common bases, nucleosides, and nucleotides are summarized in Table 3.1. After they have been incorporated into DNA or RNA, the bases of some nucleotides may be chemically modified. This occurs extensively in some classes of RNA (described more fully in Section 21.3). The most common DNA modification is methylation, generating residues containing 5-methylcytosine and N 6-methyladenine.

NH2

O

N

O

O

O H

H

H

OH OH

60  C ha pter 3   Nucleic Acid Structure and Function

Some nucleotides have other functions In addition to serving as the building blocks for DNA and RNA, nucleotides perform a variety of functions in the cell. They are involved in energy transduction, intracellular signaling, and regulation of enzyme activity. Some nucleotide derivatives are essential players in the metabolic pathways that synthesize biomolecules or degrade them in order to “capture” free energy. For example, co­enzyme A (CoA; Fig. 3.2a) is a carrier of other molecules during their synthesis and degradation. Two nucleotides are linked in the compounds nicotinamide adenine dinucleotide (NAD; Fig. 3.2b) and flavin adenine dinucleotide (FAD; Fig. 3.2c), which undergo reversible oxidation and reduction during a number of metabolic reactions. Interestingly, a portion of the structures of each of these molecules is derived from a vitamin, a compound that must be obtained from the diet. HN CH2 C

CH2

SH

  FIGURE 3.2   Some nucleotide derivatives.  The adenosine group of each of these compounds is shown in red. Note that each also contains a vitamin derivative.

O

CH2

Question  Locate the nitrogenous base(s) and sugar(s) in each structure.

CH2 NH

Pantothenic acid residue

NH2

C

O

HO

C

H

H3C

C

CH3

CH2

N

N O

O

O

P

O

O−

P

O

CH2

O− H

−

O

N

N O

H

H

O

OH

Adenosine

H

C

Nicotinamide +

O−

P

O Ribose

CH2 H

Coenzyme A (CoA) contains a residue of pantothenic acid (pantothenate), also known as vitamin B5. The sulfhydryl group is the site of attachment of other groups.

O

NH2

CH2

Riboflavin

HO

C

H

HO

C

H

HO

C

H

O

O

P

O

O−

P

O

CH2

O− H

N

N O

H

H

OH

OH

CH2 H3C

N

H3C

N

N

O NH

O

Flavin adenine dinucleotide (FAD)

Oxidation and reduction of flavin adenine dinucleotide (FAD) occurs at the riboflavin group (also known as vitamin B2).

c.

H

P

O Adenosine

P O

COO− H

H

H

N

OH

Niacin

NH2

O−

N

N

O

N

N

O

HO

a.

O

NH2

N

O

Coenzyme A (CoA)

O

O− CH2 H

H

HO

N

N O H

Adenosine

H

OH

Nicotinamide adenine dinucleotide (NAD)

The nicotinamide group of nicotinamide adenine dinucleotide (NAD) is a derivative of the vitamin niacin (also called nicotinic acid or vitamin B3; see inset) and undergoes oxidation and reduction. The related compound nicotinamide adenine dinucleotide phosphate (NADP) contains a phosphoryl group at the adenosine C2ʹ position.

b.

3.2  Nucleic Acid Structure  61

Before Going On • Practice drawing the structures of the two purine bases, the three pyrimidine bases, ribose, and deoxyribose. • Sketch the overall structure of a nucleoside and a nucleotide.

3.2 Nucleic Acid Structure LEARNING OBJECTIVES Describe the structure and stabilizing forces in DNA. • Summarize the physical features of the DNA double helix. • Distinguish the structures of RNA and DNA. • Recount the events in nucleic acid denaturation and renaturation.

In a nucleic acid, the linkage between nucleotides is called a phosphodiester bond because a single phosphate group forms ester bonds to both C5ʹ and C3ʹ. During DNA synthesis in a cell, when a nucleoside triphosphate is added to the polynucleotide chain, a diphosphate group is eliminated. Once incorporated into a polynucleotide, the nucleotide is formally known as a nucleotide residue. Nucleotides consecutively linked by phosphodiester bonds form a polymer in which the bases project out from a backbone of repeating sugar–phosphate groups. 5′ end −

O

O− N

O 5′ CH2 4′

H

O

H

H

P −

O

O

Nucleic Acid Structure

Adenine

N

H

2′

O

N

N H

3′

Phosphodiester bond O

SEE GUIDED TOUR

NH2

O

P

O N

5′

CH2 4′

H

O

H

H

P −

O

O

O H3C

5′

CH2 4′

H

Guanine

NH2

N

H

2′

O O

N H

3′

NH

O

H

H

P −

O

O

NH2

H

2′

O

Thymine

O

N H

3′

O

NH

N 5′

CH2 4′

H 3′ end

O

H 3′

OH

N H 2′

H

H

Cytosine

O

62  C ha pter 3   Nucleic Acid Structure and Function

The end of the polymer that bears a phosphate group attached to C5ʹ is known as the 5ʹ end, and the end that bears a free OH group at C3ʹ is the 3ʹ end. By convention, the base sequence in a polynucleotide is read from the 5ʹ end (on the left) to the 3ʹ end (on the right).

DNA is a double helix A DNA molecule consists of two polynucleotide strands linked by hydrogen bonds (hydrogen bonding is discussed in Section 2.1). The structure of this molecule, elucidated by James Watson and Francis Crick in 1953, incorporated Chargaff’s earlier observations about DNA’s base composition. Specifically, Chargaff noted that the amount of A is equal to the amount of T, the amount of C is equal to the amount of G, and the total amount of A + G is equal to the total amount of C + T. Chargaff’s “rules” could be satisfied by a molecule with two polynucleotide strands in which A and C in one strand pair with T and G in the other. Two hydrogen bonds link adenine and thymine, and three hydrogen bonds link guanine and cytosine:

H N Adenine

N

N

N

H

O H

N

CH3

N O

Thymine

N H

N Guanine

H

O

N N

N

N

H

N

N

H

O

Cytosine

N

H 10.85 Å

All the base pairs, which consist of a purine and a pyrimidine, have the same molecular dimensions (about 11 Å wide). Consequently, the sugar–phosphate backbones of the two strands of DNA are separated by a constant distance, regardless of whether the base pair is A:T, G:C, T:A, or C:G. Although DNA can be shown as a ladder-like structure (left), with the two sugar–­ phosphate backbones as the vertical supports and the base pairs as the rungs, the two strands of DNA twist around each other to generate the familiar double helix (right). Sugar–phosphate backbones

A

T

G

C

T

A A

T

C

G

C

G

T

A G

T

T

C A

G

C

A

T A

3.2  Nucleic Acid Structure  63

This conformation allows successive base pairs, which are essentially planar, to stack on top of each other with a center-to-center distance of only 3.4 Å. In fact, Watson and Crick derived this model for DNA not just from Chargaff’s rules but also from Rosalind Franklin’s studies of the diffraction (scattering) of an X-ray beam by a DNA fiber, which suggested a helix with a repeating spacing of 3.4 Å. The major features of the DNA molecule include the following (Fig. 3.3): 1. The two polynucleotide strands are antiparallel; that is, their phosphodiester bonds run in opposite directions. One strand has a 5ʹ → 3ʹ orientation, and the other has a 3ʹ → 5ʹ orientation. 2. The DNA “ladder” is twisted in a right-handed fashion. (If you climbed the DNA helix as if it were a spiral staircase, you would hold the outer railing—the sugar–phosphate backbone—with your right hand.) 3. The diameter of the helix is about 20 Å and it completes a turn about every 10 base pairs, which corresponds to an axial distance of about 34 Å. 4. The twisting of the DNA “ladder” into a helix creates two grooves of unequal width, the major and minor grooves. 5. The sugar–phosphate backbones define the exterior of the helix and are exposed to the solvent. The negatively charged phosphate groups bind Mg2+ cations in vivo, which helps minimize electrostatic repulsion between these groups. 6. The base pairs are located in the center of the helix, approximately perpendicular to the helix axis. 7.  The base pairs stack on top of each other, so the core of the helix is solid (see Fig. 3.3b). Although the planar faces of the base pairs are not accessible to the solvent, their edges are exposed in the major and minor grooves (this allows certain DNA-binding proteins to recognize specific bases).

5′ end 3′ end

Major groove 34 Å (10 bp)

  FIGURE 3.3   Model of DNA.  a. Ball-and-stick model with atoms colored: C gray, O red, N blue, and P gold (H atoms are not shown).  b. Space-filling model with the sugar–phosphate backbone in gray and the bases color-coded: A green, C blue, G yellow, and T red.

Minor groove 3′ end 5′ end 20 Å

a.

b.

Question  How many nucleotides are shown in this double helix?

64  C ha pter 3   Nucleic Acid Structure and Function

  FIGURE 3.4   A transfer RNA molecule.  This 76-nucleotide single-stranded RNA molecule folds back on itself so that base pairs form between complementary segments.

In nature, DNA seldom assumes a perfectly regular conformation because of small sequence-dependent irregularities. For example, base pairs can roll or twist like propeller blades, and the helix may wind more tightly or loosely at certain nucleotide sequences. DNA-binding proteins may take advantage of these small variations to locate their specific binding sites, and they in turn may further distort the DNA helix by causing it to bend or partially unwind. The size of a DNA segment is expressed in units of base pairs (bp) or kilobase pairs (1000 bp, abbreviated kb). Most naturally occurring DNA molecules comprise thousands to millions of base pairs. A short single-stranded polymer of nucleotides is usually called an oligonucleotide (oligo is Greek for “few”). In a cell, nucleotides are polymerized by the action of enzymes known as polymerases. The phosphodiester bonds linking nucleotide residues can be hydrolyzed by the action of nucleases. An ­e xonuclease removes a residue from the end of a polynucleotide chain, whereas an ­endonuclease cleaves at some other point along the chain. Polymerases and nucleases are usually specific for either DNA or RNA. In the absence of these enzymes, the structures of nucleic acids are remarkably stable. The hydrogen bonds between polynucleotide strands, however, are relatively weak and break to allow the strands to separate during replication and transcription, described below.

RNA is single-stranded RNA, which is a single-stranded polynucleotide, has greater conformational freedom than DNA, whose structure is constrained by the requirements of regular base-pairing between its two strands. An RNA strand can fold back on itself so that base pairs form between complementary segments of the same strand. Complemen­tarity refers to the ability of bases to form hydrogen bonds with their standard partners: A is complementary to T and U, and G is complementary to C. Consequently, RNA molecules tend to assume intricate three-dimensional shapes (Fig. 3.4). Unlike DNA, whose regular structure is suited for the long-term storage of genetic information, RNA can assume more active roles in expressing that information. For example, the molecule shown in Figure 3.4, which carries the amino acid phenylalanine, interacts with a number of proteins and other RNA molecules during protein synthesis. The residues of RNA are also capable of base-pairing with a complementary single strand of DNA to produce an RNA–DNA hybrid double helix (Fig. 3.5). A double helix involving RNA is wider and flatter than the standard DNA helix (its diameter is about 26 Å, and it makes one helical turn every 11 residues). In addition, its base pairs are inclined to the helix axis by about 20°. These structural differences relative to the standard DNA helix primarily reflect the presence of the 2ʹ OH groups in RNA. A double-stranded DNA helix can adopt this same helical conformation; it is known as A-DNA. The standard DNA helix shown in Figure 3.3 is known as B-DNA. Other confor­ mations of DNA have been described, and there is evidence that they exist in vivo, at least for certain nucleotide sequences, but their functional significance is not completely understood.

Nucleic acids can be denatured and renatured   FIGURE 3.5   An RNA–DNA

hybrid helix.  In a double helix formed by one strand of RNA (red) and one strand of DNA (blue), the planar base pairs are tilted and the helix does not wind as steeply as in a standard DNA double helix (compare with Fig. 3.3).

The pairing of polynucleotide strands in a double-stranded nucleic acid is possible because bases in each strand form hydrogen bonds with complementary bases in the other strand: A is the complement of T (or U) and G is the complement of C. However, the structural stability of a double helix does not depend significantly on hydrogen bonding between complemen­ tary bases. (If the strands were separated, the bases could still satisfy their hydrogen-bonding requirements by forming hydrogen bonds with solvent water molecules.) Instead, stability depends mostly on stacking interactions, which are a form of van der Waals interaction, between adjacent base pairs. A view down the helix axis shows that stacked base pairs do not overlap

3.2  Nucleic Acid Structure  65

exactly, due to the winding of the helix (­ Fig. 3.6). Although individual stacking interactions are weak, they are additive along the length of a DNA molecule. In addition to stacking interactions, tightly associated water molecules help stabilize the DNA helix. The major groove is wide enough to accommodate several somewhat disordered water molecules, but in the minor groove, water molecules fit in a more rigid ­single file (Fig. 3.7). This water must be displaced when the helix structure is disrupted. Interestingly, strands containing mostly A and T separate more easily than strands containing mostly G and C. This could be due to the slightly greater stacking energy of G:C base pairs compared to A:T base pairs, but it could also reflect the greater constriction of water molecules in AT-rich DNA, which has a narrower minor groove, such that disrupting the helical structure results in a larger increase in the entropy of the released water molecules. Although G:C base pairs have one more hydrogen bond than A:T base pairs, this feature does not account for the easier separation of AT-rich DNA strands. The loss of helical structure in a sample of DNA molecules can be quantified   FIGURE 3.6   Axial view of DNA base pairs.  in the melting temperature (Tm). To determine the melting point of the DNA, A view down the central axis of the DNA helix the temperature is slowly increased. At a sufficiently high temperature, the base shows the overlap of neighboring base pairs (only pairs begin to unstack, hydrogen bonds break, and the two strands begin to sepathe first two nucleotide pairs are highlighted). rate. This process continues as the temperature rises, until the two strands come Question  Locate the base and sugar in the completely apart. The melting, or denaturation, of the DNA can be recorded as nucleotides with the blue bases. Identify a melting curve (Fig. 3.8) by monitoring an increase in the absorbance of ultra­ the bases. violet (260 nm) light (the aromatic bases absorb more light when unstacked). The midpoint of the melting curve (that is, the temperature at which half the DNA has separated into single strands) is the Tm. Table 3.2 lists the GC content and the melting point of the DNA from different species. Since manipulating DNA in the laboratory frequently requires the thermal separation of paired DNA strands, it is sometimes helpful to know the DNA’s GC content. When the temperature is lowered slowly, denatured DNA can renature; that is, the separated strands can re-form a double helix by reestablishing hydrogen bonds between the complementary strands and by restacking the base pairs. The maximum rate of re­naturation occurs

Melting begins

  FIGURE 3.7   A model of DNA with

bound water molecules.  Water molecules (red) in the minor groove form an ordered “spine” along the double helix, whereas water molecules (blue) in the major groove are less ordered.

Relative absorbance at 260 nm

McDermott, M.L. et al. 2017/ACS

1.4

Melting ends

+

1.3

1.2

1.1

1.0

Tm

30

50

70 Temperature (°C)

90

  FIGURE 3.8   A DNA melting curve.  Thermal denaturation (melting, or strand separation) of DNA results in an increase in ultraviolet absorbance relative to the absorbance at 25°C. The melting point, Tm, of the DNA sample is defined as the midpoint of the melting curve.

66  C ha pter 3   Nucleic Acid Structure and Function

TAB L E 3. 2   GC Content and Melting Points of DNA

Source of DNA

GC content (%)

Tm (°C)

Dictyostelium discoideum (fungus)

23.0

 79.5

Clostridium butyricum (bacterium)

37.4

 82.1

Homo sapiens

40.3

 86.5

Streptomyces albus (bacterium)

72.3

100.5

High heat (melting)

Cooling to 20–25°C below Tm (renaturation)

Rapid cooling to temperature much lower than Tm (improper base pairing)

Rewarming to 20–25°C below Tm (renaturation)

  FIGURE 3.9   Renaturation of DNA.  DNA strands that have been melted apart can renature at a temperature of 20–25°C below the Tm. At much lower temperatures, base pairs may form between short complementary segments within and between the single strands. Correct renaturation is possible only if the sample is rewarmed so that the improperly paired strands can separate and reanneal.

at about 20–25°C below the melting temperature. If the DNA is cooled too rapidly, it may not fully renature because base pairs may form randomly between short complementary segments. At low temperatures, the improperly paired segments are frozen in place since they do not have enough thermal energy to melt apart and find their correct complements (Fig. 3.9). The rate of renaturation of denatured DNA depends on the length of the double-stranded molecule: Short segments come together (anneal) faster than longer segments because the bases in each strand must locate their partners along the length of the complementary strand. The ability of short single-stranded nucleic acids (either DNA or RNA) to hybridize with longer polynucleotide chains is the basis for a number of useful laboratory techniques (described in detail in Section 20.6). For example, an oligonucleotide probe that has been tagged with a fluorescent group can be used to detect the presence of a complementary nucleic acid sequence in a complex mixture.

Before Going On • Explain how Chargaff’s rules helped reveal the structure of DNA. • Describe the arrangement of the base pairs and sugar–phosphate backbones in DNA. • List the ways that RNA differs from DNA. • Describe the molecular events in DNA denaturation and renaturation.

3.3 The Central Dogma

3.3

67

Old Old

The Central Dogma

LEARNING OBJECTIVES Summarize the biological roles of DNA and RNA. • Distinguish replication, transcription, and translation. • Decode a nucleotide sequence to an amino acid sequence. • Describe how a mutation can cause a disease. • Summarize the goals and challenges of gene therapy. The complementarity of the two strands of DNA is essential for its function as the storehouse of genetic information, since this information must be replicated (copied) for each new generation. As first suggested by Watson and Crick, the separated strands of DNA direct the synthesis of complementary strands, thereby generating two identical double-stranded molecules (Fig. 3.10). The parental strands are said to act as templates for the assembly of the new strands because their sequence of nucleotides determines the sequence of nucleotides in the new strands. When a cell divides, each daughter cell receives genetic information—sequences of nucleotides—in the form of DNA molecules containing one old strand and one new strand (Box 3.A). A similar phenomenon is responsible for the expression of that genetic information, a process in which nucleotide sequences are used to direct the synthesis of proteins that carry out the cell’s activities. First, a portion of the DNA, a gene, is transcribed to produce a complementary strand of RNA; then the RNA is translated into a polypeptide chain (protein). This paradigm, known as the central dogma of molecular biology, was formulated by Francis Crick. It can be shown schematically as lication rep

transcription DNA

translation RNA

Protein

New

Old New

New

New Old

FIGURE 3.10 DNA replication. The double helix unwinds so that each parental strand can serve as a template for the synthesis of a new complementary strand. The result is two identical double-helical DNA molecules.

Question Label the 5ʹ and 3ʹ ends of each strand.

Although it is tempting to think of a cell’s DNA as its “brain,” the DNA does not issue commands to the rest of the cell. Instead, DNA simply holds genetic information—the instructions for synthesizing proteins.

Box 3.A Replication, Mitosis, Meiosis, and Mendel’s Laws Each chromosome contains one long double-stranded DNA molecule that must be replicated at some point before cell division begins. Prokaryotic cells typically harbor one chromosome in the form of a circular DNA molecule, and replication and cell division are relatively straightforward. However, eukaryotic cells generally have multiple linear chromosomes that must each be replicated

Chromatids separate during cell division

DNA replicates

Centromere

Chromosome

and then allocated equally to the two daughter cells during cell division—a more complicated process. Linking the chemistry of DNA to the biology of cell division demands a closer look at a typical eukaryotic chromosome. The two DNA strands are shown in blue. Replication generates two identical DNA molecules, each with one old (blue) and one new (purple) strand.

Sister chromatids

68  C ha pter 3   Nucleic Acid Structure and Function

Biophoto Associates/ Science Source

The two DNA molecules, called sister chromatids, remain attached at a more-or-less central point known as the centromere. The two chromatids are clearly apparent early in cell division, when the chromosomes are highly condensed.

Each daughter cell receives one of the sister chromatids when the centromere attachment is broken. The process of mitosis describes the allocation of eukaryotic chromosomes during cell division. For a cell with six chromosomes, the process can be diagrammed as follows: DNA replicates

Chromosomes line up

Sister chromatids separate

Cell splits

Two daughter cells, each with 6 chromosomes

Parent cell with 6 chromosomes

The final result is two daughter cells that are genetically identical to the parent cell (in this example, the parent cell and the daughter cells all have six chromosomes). This mode of cell division is standard for unicellular organisms and for the cells within multicellular organisms. In animals that reproduce sexually, virtually all the cells are diploid. They have two sets of chromosomes, one inherited from each parent. However, in order to reproduce, these animals have specialized cells in the ovaries and testes to generate gametes, which are haploid, having just one set of genetic information. At fertilization, two haploid gametes—one from each ­parent—combine to produce a new diploid organism. Meiosis is the variation of mitosis that occurs in the reproductive tissues of animals as well as plants in order to produce haploid cells. For example, diploid human cells with 46 chromosomes undergo meiosis to produce egg or sperm cells with just 23 chromosomes. Before meiosis, the DNA of each chromosome is replicated. Meiosis begins with the pairing of homologous chromosomes, which have the same genes but slightly different DNA sequences. In humans, the two versions of Chromosome 1 form a pair, the two versions of Chromosome 2 form a pair, and so on for a total of 23 pairs. The model cell above, with six chromosomes, forms three pairs at the start of meiosis. Here, the sets of homologous chromosomes are colored blue and red. When the cell splits, the chromosome pairs also split up, so that the two daughter cells are haploid. However, because each chromosome still consists of two sister chromatids, a second division, exactly like mitosis, must then occur to produce four daughter cells, the gametes, containing only one DNA molecule to represent each chromosome.

DNA replicates

Diploid cell with 6 chromosomes

Homologous chromosomes form pairs

Homologous chromosomes separate, and cell splits

Sister chromatids separate, and cells split

Two haploid cells with replicated chromosomes

Four haploid gametes, each with 3 chromosomes

At fertilization, a haploid gamete from one parent fuses with a haploid gamete from a different parent to generate almost unlimited new diploid combinations of genetic information in the next generation. Consequently, meiosis leads to the genetic diversity that is essential for natural selection. Additional diversity is generated early in meiosis, when homologous pairs exchange segments of DNA.

3.3  The Central Dogma  69 Meiosis also explains why parental traits may or may not appear in the offspring. A diploid parent has two sets of homologous chromosomes, so there are two possible versions, or alleles, for each gene. During meiosis, when pairs of homologous chromosomes separate, the two alleles end up in different daughter cells. Thus, only one allele is passed from the parent to an individual in the next generation. For example, in a certain animal, a gene that codes for an enzyme called “B” has two alleles: the B allele codes for a functional version of the enzyme and the b allele codes for a nonfunctional version of the enzyme. A parent bearing two B alleles passes a B allele to each of its offspring, and a parent bearing two b alleles passes on a b allele.

B

B

DNA replicates

B

B

B

B

B

B

Homologous chromosomes separate

B

B

B

B

Sister chromatids separate

B

B

BB parent cell Haploid gametes

When a BB individual mates with a bb individual, all the offspring will receive one B allele and one b allele. Since the B allele codes for a functional enzyme, all the Bb offspring will have that functionality. Gamete from BB parent

Gamete from bb parent

Fertilization

B

b B

b

Bb diploid offspring When a Bb individual mates with another Bb individual, each parent produces gametes containing either a B allele or a b allele.

B

B

b

b

B

b

Meiosis

Meiosis B

b

B

B

b

b

Bb parent 1

Bb parent 2 Parent 1 gametes

Parent 2 gametes

All the possible combinations of gametes during fertilization can be predicted by setting up a simple grid: Parent 1 gametes

Parent 2 gametes

B

b

B

BB

Bb

b

Bb

bb

70  C ha pter 3   Nucleic Acid Structure and Function There is a 1-in-4 (25%) chance that a given offspring individual will have two B alleles, a 2-in-4 (50%) chance that it will have one B and one b allele, and a 1-in-4 (25%) chance that it will have two b alleles. In a large litter, approximately three-quarters of the offspring will have a functional enzyme (they are either BB or Bb), and one-quarter of the offspring will have the nonfunctional enzyme (they are bb). This three-to-one ratio was one of the patterns that Mendel documented in his classic studies of inheritance in pea plants. Although Mendel’s work preceded the discoveries of DNA, mitosis, and meiosis, the laws of inheritance follow solidly from the structure of DNA, its mode of replication, and the way chromosomes are passed to daughter cells.

DNA must be decoded Even in the simplest organisms, DNA is an enormous molecule, and many organisms contain several different DNA molecules as separate chromosomes. An organism’s complete set of genetic information is called its genome. A genome may comprise several hundred to perhaps 35,000 genes. To transcribe a gene, one of the two strands of DNA serves as a template for an enzyme called RNA polymerase to synthesize a complementary strand of RNA. The RNA therefore has the same sequence (except for the substitution of U for T) and the same 5ʹ → 3ʹ orientation as the nontemplate strand of DNA. This strand of DNA is often called the coding strand or sense strand (the template strand is called the noncoding strand).

5′

3′

A

C

DNA

G C T

G

A

A

G G T A G C C A T C

5′ T

DNA

Coding strand (nontemplate)

T

3′

A R NA C U T G A G C UG A T C C G A

C G C T T A C G A G C G A A T G C T

3′

5′

Noncoding strand (template)

The transcribed RNA is known as messenger RNA (mRNA) because it carries the same genetic message as the gene. The mRNA is translated by a ribosome, a cellular particle consisting of protein and ribosomal RNA (rRNA). At the ribosome, small molecules called transfer RNA (tRNA), which carry amino acids, recognize sequential sets of three bases (known as codons) in the mRNA through complementary base-pairing (a tRNA molecule is shown in Fig. 3.4). The ribosome covalently links the amino acids carried by successive tRNAs to form a polypeptide. The protein’s amino acid sequence therefore ultimately depends on the nucleotide sequence of the DNA.

5 DNA 3

C T C A G T G C C G A G T C A C G G

3 5

Transcription

mRNA 5 tRNAs

C U C A G U G C C G A G U C A C G G Leu

Ser

3

Ala Translation

Protein

Leucine

Serine

Alanine

The correspondence between amino acids and mRNA codons is known as the genetic code. There are a total of 64 codons: 3 of these are “stop” signals that terminate translation

3.3  The Central Dogma  71

TA B L E 3.3   The Standard Genetic Codea

First position (5ʹ end)

U

C

A

G

Third position (3ʹ end)

U

UUU Phe

UCU Ser

UAU Tyr

UGU Cys

U

UUC Phe

UCC Ser

UAC Tyr

UGC Cys

C

C

A

G

Second position

UUA Leu

UCA Ser

UAA Stop

UGA Stop

A

UUG Leu

UCG Ser

UAG Stop

UGG Trp

G

CUU Leu

CCU Pro

CAU His

CGU Arg

U

CUC Leu

CCC Pro

CAC His

CGC Arg

C

CUA Leu

CCA Pro

CAA Gln

CGA Arg

A

CUG Leu

CCG Pro

CAG Gln

CGG Arg

G

AUU Ile

ACU Thr

AAU Asn

AGU Ser

U

AUC Ile

ACC Thr

AAC Asn

AGC Ser

C

AUA Ile

ACA Thr

AAA Lys

AGA Arg

A

AUG Met

ACG Thr

AAG Lys

AGG Arg

G

GUU  Val

GCU Ala

GAU Asp

GGU Gly

U

GUC Val

GCC Ala

GAC Asp

GGC Gly

C

GUA Val

GCA Ala

GAA Glu

GGA Gly

A

GUG Val

GCG Ala

GAG Glu

GGG Gly

G

a 

The 20 amino acids are abbreviated; Ala, alanine; Arg, arginine; Asn, asparagine; Asp, aspartate; Cys, cysteine; Gly, glycine; Gln, glutamine; Glu, glutamate; His, histidine; Ile, isoleucine; Leu, leucine; Lys, lysine; Met, methionine; Phe, phenylalanine; Pro, proline; Ser, serine; Thr, threonine; Trp, tryptophan; Tyr, tyrosine; and Val, valine.

Question  How many amino acids would be uniquely specified by a genetic code that consisted of just the first two nucleotides in each codon?

and the remaining 61 represent, with some redundancy, the 20 standard amino acids found in proteins. Table 3.3 shows which codons specify which amino acids. In theory, knowing a gene’s nucleotide sequence should be equivalent to knowing the amino acid sequence of the protein encoded by the gene. However, as we will see, genetic information is often “processed” at several points before the protein reaches its mature form. Keep in mind that the rRNA and tRNA required for protein synthesis, as well as other types of RNA, are also encoded by genes. The “products” of these genes are the result of transcription without translation.

A mutated gene can cause disease Because an organism’s genetic material influences the organism’s entire repertoire of activities, it is vitally important to unravel the sequence of nucleotides in that organism’s DNA. Over the past 60 years, researchers have developed a variety of methods to sequence DNA. These techniques take advantage of naturally occurring DNA polymerase enzymes that construct complementary copies of a given template in a way that allows the identification of each new nucleotide incorporated (Section 20.6 provides more details). Huge amounts of nucleic acid sequence data—representing individual genes as well as entire genomes—are available for analysis. The process of DNA replication is highly accurate, but mistakes do happen. In addition, DNA can be physically and chemically damaged, and not all cellular repair mechanisms are able to perfectly restore it. Consequently, the information in DNA may change over time. A mutation is simply a permanent change in DNA and may take the form of a single-­nucleotide substitution, an insertion or deletion of nucleotides, or a rearrangements of ­chromosomal segments. Some of these sequence variations cause disease. In traditional approaches to understanding human genetic diseases, researchers used the defective protein associated with a particular disease to track down the corresponding gene. (Current methods for discovering disease mutations rely on directly sequencing the DNA.) One classic example is the genetic change responsible for sickle cell disease. In the

72  C ha pter 3   Nucleic Acid Structure and Function

gene for one hemoglobin protein chain, the codon GAG normally codes for the amino acid glutamate (Glu). In the defective gene, the sequence has mutated to GTG, which codes for valine (Val). This amino acid substitution affects the protein’s structure and function. Normal gene · · · ACT CCT GAG GAG AAG · · · Protein   · · · Thr – Pro – Glu – Glu –  Lys · · · Mutated gene · · · ACT  CCT  GTG  GAG  AAG · · · Protein · · · Thr  –  Pro –  Val –  Glu –  Lys · · · As many as 10,000 human disorders are monogenetic diseases, such as sickle cell disease and cystic fibrosis, where a defect in a specific gene explains the molecular basis of the disease. In many cases, a variety of different mutations have been catalogued for a particular disease gene, which explains in part why symptoms of the disease vary between individuals. The database known as OMIM (Online Mendelian Inheritance in Man, omim.org) contains information on thousands of genetic variants, including the clinical features of the resulting disorder and its biochemical basis. The Genetic Testing Registry (www.ncbi.nlm.nih.gov/gtr/) is a database of the diseases that can be detected through analysis of DNA, carried out by either clinical or research laboratories. Unlike monogenetic diseases, polygenic diseases are linked to variations in a number of different genes (many common human disorders such as cardiovascular disease and cancer also have environmental components). Researchers use genome-wide association studies (GWAS) to correlate the locations of sequence differences with particular diseases. For example, as many as 350 genetic variations have been associated with type 2 diabetes, but some variations are quite rare, and probably only 40–60 actually increase a person’s risk of developing the disorder. The risk tied to any particular genetic variant is low, and the entire set of variations cannot entirely explain the heritability of the disease. In any case, DNA sequence variations can be difficult to interpret, as they may affect protein-coding genes as well as noncoding regulatory DNA segments. Several commercial enterprises offer individual genome-sequencing services (at a cost of only a few hundred dollars), but until genetic information can be reliably translated into effective plans to prevent or treat diseases, the practical value of “personal genomics” is somewhat limited. On average, the DNA of any two humans differs at 3 million sites. These ­single-nucleotide polymorphisms (SNPs) are compiled in databases. A person begins life with an average of 60 new genetic changes that were not present in either parent. Of course, not all genetic variations have negative consequences. The vast majority of variations have no discernible effect on an individual’s ability to survive and reproduce. Some variations may be beneficial, depending on the circumstances, and natural selection acts on this inherent genetic variation, favoring the persistence of certain genes in subsequent generations.

Genes can be altered Understanding genetic information and how it is expressed makes genetic engineering possible. Using a variety of laboratory techniques (described more fully in Section 20.6), researchers can extract a gene from one organism and combine it with other DNA to produce recombinant DNA. The new DNA can then be further manipulated in the laboratory or introduced into another organism to direct the synthesis of a gene product. Sometimes a gene is intentionally modified by “site-directed mutagenesis.” The near-­universal interpretation of the genetic code means that almost any host cell can “read” the codons in a foreign gene to construct the appropriate protein. Introducing human genes into cultured bacterial, fungal, or mammalian cells is an economical way to produce certain proteins that are useful as drugs (Table 3.4). The same technology is used to genetically modify whole organisms (Box 3.B). Making intentional changes to an individual human’s genetic makeup is the goal of gene therapy. In traditional gene therapy, an extra gene is delivered to a patient’s cells in order to compensate for a defective gene. The first successful gene therapy trials treated ­children with severe combined immunodeficiency (SCID), a normally fatal condition caused

3.3  The Central Dogma  73

TA B L E 3.4   Some Recombinant Protein Products

Protein

Purpose

Insulin

Treat insulin-dependent diabetes

Growth hormone

Treat certain growth disorders in children

Erythropoietin

Stimulate production of red blood cells; useful in   kidney dialysis

Tissue plasminogen activator

Promote clot lysis following myocardial infarction   or stroke

Colony stimulating factor

Promote white blood cell production after bone   marrow transplant

Introducing a foreign gene into a single host cell alters the genetic makeup of that cell and all its descendants. However, if the cell is part of a multicellular organism, such as an animal or plant, more work is required to generate a transgenic organism whose cells all contain the foreign gene. In mammals, modified DNA can be introduced into fertilized eggs, which are then implanted in a foster mother. Some of the resulting embryos’ cells (possibly including their reproductive cells) will contain the foreign gene. When the animals mature, they must be bred in order to yield offspring whose cells are all transgenic. Transgenic plants are easier to develop: a few cells containing the modified DNA can sometimes be coaxed into developing into an entire plant. Most of the transgenic animals that have been developed are used for research or are still being developed for commercial purposes. Some examples are virus-resistant pigs and chickens, cows that resist bovine tuberculosis, and cows that produce milk with a healthier assortment of fats. Transgenic salmon that grow faster than their wild relatives are already on the market. Mosquitoes that have been engineered for infertility (to prevent the spread of malaria) are being field-tested. Transgenic crops have been commercially available for over 20 years and in many areas are more popular than conventional crops. Approximately 80–90% of the U.S. corn (maize), soybean, and cotton harvests are genetically modified. Some genetically modified produce is destined for animal feed or industrial uses, but transgenic fruits and vegetables have also been developed for direct human consumption. Among the desirable traits that have been engineered into various plant species is protection from insect pests, leading to higher yields and less reliance on pesticides (which are costly and potentially toxic to animals). Plants engineered for resistance to weed-killing herbicides, such as the Brassica species that produce canola oil (see photo), also give higher yields.

Daniel/123RF

Box 3.B Genetically Modified Organisms

Researchers have developed transgenic crops with improved nutritional value, such as a strain of rice containing high levels of iron and vitamin A, and crops that can flourish under suboptimal conditions such as drought, high temperatures, and high soil salinity. Concerns about the safety of foods containing foreign genes have limited their acceptance by consumers in some places. Transgenic organisms also present some biological risks. For example, genes that code for insecticides (such as the bacterial toxin known as Bt, which is intended to kill the insect larvae that would otherwise feast on the plants) can make their way into wild plants that support beneficial insects. Similarly, herbicide-­resistance genes can jump to weed species, making them even more difficult to control. Nevertheless, crops that can more efficiently sustain the world’s growing population, particularly as the global climate changes, are necessary, and it is difficult to image how to develop these plants without taking advantage of the tools of genetic engineers. Question  What molecular components should be analyzed in order to detect the presence of transgenic material in food items?

by a single-gene defect. Bone marrow cells were removed from each patient and cultured in the presence of an engineered virus capable of transferring a normal version of the defective gene. When the modified cells were infused into the patient, they differentiated into functional immune system cells that produced the missing protein. Some of the diseases treated by gene therapy are listed in Table 3.5. Disorders related to blood cells have been attractive targets for gene therapy, since a small number of treated

74  C ha pter 3   Nucleic Acid Structure and Function

TA B L E 3. 5   Some Diseases Treated by Gene Therapy

Disease

Symptoms

Adrenoleukodystrophy

Neurodegeneration

Hemophilia

Excessive bleeding

Junctional epidermolysis bullosa

Severe skin blistering

Leber’s congenital amaurosis

Blindness

Severe combined immunodeficiency

Immunodeficiency

β-Thalassemia

Anemia

stem cells from bone marrow can regenerate a healthy population of blood cells. Introducing engineered genes to other tissues with lower rates of cellular turnover, such as muscle or brain, is more difficult. In any case, using viruses to deliver genes is inherently risky. For one thing, the engineered viruses can behave unpredictably, sometimes triggering a fatal immune response or inserting themselves into the host cell’s chromosomes at random, which may interrupt the functions of other genes and cause cancer. Other challenges involve delivering the therapeutic gene to enough host cells to “correct” the disease, ensuring that the gene’s expression is regulated properly so that the appropriate amount of protein is produced, and making the genetic change permanent. Some of these potential problems can be minimized by newer gene-therapy protocols that use CRISPR gene-editing technology (described in Box 20.B) to fix—rather than just ­supplement—a defective gene. Hundreds of potential therapies are being tested, but as in traditional gene therapy, none are entirely risk-free. Off-target effects (altering gene sequences in addition to the defective gene) and the irreversible nature of gene editing present technical and ethical issues that require careful consideration.

Before Going On • Draw a diagram to illustrate each step of the central dogma. • Practice locating the codons for each of the 20 amino acids. • Explain the relationship between mutations and disease. • Describe the usefulness of identifying genetic variations between individuals. • Explain the value and limitations of genome-wide association studies. • List some positive and negative outcomes of gene therapy.

3.4 Genomics LEARNING OBJECTIVES Identify the types of information provided by genomic analysis. • Explain how genes are identified. • Compare the genomes of different species.

Analyzing one gene at a time provides limited information, but thanks to DNA sequencing technology, entire genomes can be explored and compared. By tallying the nucleotide sequence differences between two similar groups of organisms, researchers can estimate how long the groups have been independently evolving and accumulating changes since splitting from a common

3.4 Genomics  75

ancestor. This sort of analysis allows the construction of branching trees (such as the one in Fig. 1.15) to depict the course of evolution. However, the work is far from finished.

The exact number of human genes is not known For many organisms, including humans, the exact number of genes has not yet been determined, and different methods for identifying genes yield different estimates. For example, a computer can scan a long DNA sequence for an open reading frame (ORF), that is, a stretch of nucleotides that can potentially be transcribed and translated to a polypeptide. The ORF begins with a “start” codon: ATG in the coding strand of DNA, which corresponds to AUG in RNA (see Table 3.3). This codon specifies methionine, the initial residue of all newly synthesized polypeptides. The ORF ends with one of the three “stop” codons: DNA coding sequences of TAA, TAG, or TGA, which correspond to the three mRNA stop codons (see Table 3.3). For example, Open reading frame · · · ACAGACACCATGAAGTCTGCCTACTGC · · · CCTGTGGGGCAAGGTTAA GCTCGCTTT · · · Start codon

Stop codon

An alternative approach is to examine the population of RNA molecules in a single cell or organism. While this kind of survey captures many protein-coding segments of DNA (which correspond to messenger RNA), it also sometimes reveals that a significant portion of an organism’s genome is transcribed but not translated. The importance of all this noncoding RNA (ncRNA) is not fully understood. Genes can also be identified by comparing a DNA sequence with the sequences of known genes in the same species or in other species. Genes with similar functions tend to have similar sequences; such genes are said to be homologous. Even an inexact match can still indicate a protein’s overall functional category, such as enzyme or hormone receptor, although its exact role in an organism may not be obvious. How many genes are in the human genome? The best estimates put the tally at about 21,000 protein-coding genes. The human genome probably also contains a comparable number of noncoding genes as well as thousands of pseudogenes, which are nonfunctional copies of true genes and appear to be evolutionary leftovers. An analysis of 17,000 protein-coding genes in 44 different human tissues has revealed that only about half of these genes are expressed in all locations. Other studies in which protein-coding genes are inactivated, one at a time, suggest that even fewer human genes—roughly 2000 or about 10%—are absolutely essential for most cells. This core set of genes encodes proteins that are produced in relatively large amounts and carry out the most basic cellular activities. When searching for genetic variations that could be tied to diseases, researchers can avoid labor-intensive genome-wide association studies (Section 3.3) and instead focus only on the exome, the set of DNA sequences that are actually used to direct protein synthesis in most cells. The human exome represents only a tiny fraction of the genome but includes at least 180,000 distinct segments of DNA, far more than the number of genes. This is because virtually all protein-coding genes in complex eukaryotes include multiple short expressed sequences (called exons). The exons in each gene are separated by intervening sequences (called introns) that are cut out of the messenger RNA before it is translated. The mRNA splicing process is described in Section 21.3.

Genome size varies The genomes of almost 100,000 different organisms have been sequenced—from the miniscule circular DNAs of parasitic bacteria to the enormous multi-chromosome genomes of plants and mammals. Some organisms whose genomes have been fully sequenced are listed in Table 3.6. The list includes species that are widely used as model organisms for different types of biochemical studies (Fig. 3.11). Note that for plants and animals, which are diploid (see Box 3.A), genomic information usually refers to the haploid state.

76  C ha pter 3   Nucleic Acid Structure and Function

TAB L E 3 . 6   Genome Size and Gene Number

Organism

Genome size (Mb) a

Bacteria

0.58

515

1.85

1,620

4.64

4,240

1.74

1,760

12.16

6,000

119.70

25,300

382.40

21,900

102.00

18,800

137.70

13,000

3279.00

21,000

Mycoplasma genitalium Haemophilus influenzae Escherichia coli Archaea Methanocaldococcus jannaschii Fungi Saccharomyces cerevisiae (yeast) Plants Arabidopsis thaliana Oryza sativa (rice) Animals Caenorhabditis elegans (nematode) Drosophila melanogaster (fruit fly) Homo sapiens

Protein-coding genes b

a 

1 Mb is 1000 kb or 1 million base pairs of DNA.

b 

RNA genes are not included.

Data from NCBI https://www.ncbi.nlm.nih.gov/genome/.

Andrew Syred/Science Source

Caenorhabditis elegans is a small (1 mm) and transparent roundworm. As a multicellular organism, it bears genes not found in unicellular organisms.

Baker’s yeast, Saccharomyces cerevisiae, is one of the simplest eukaryotic organisms, with around 6000 genes.

Dr. Jeremy Burgess/Science Source

Escherichia coli, a normal inhabitant of the mammalian digestive tract, is a metabolically versatile bacterium that tolerates both aerobic and anaerobic conditions.

Sinclair Stammers/Science Source

  FIGURE 3.11   Some model

organisms.

Dr. Kari Lounatmaa/Science Source

Question  What is the relationship between genome size and gene number in prokaryotes? How does this differ in eukaryotes?

The plant kingdom is represented by Arabidopsis thaliana, which has a short generation time and readily takes up foreign DNA.

3.4 Genomics  77 Unique sequences

Protein-coding genes

Interspersed repeats

Tandem repeats

  FIGURE 3.12   Coding and noncoding portions of the human genome.  Approximately 1.4% of the genome codes for proteins. Interspersed repetitive sequences make up 48% of the genome, and highly repetitive tandem repeat sequences account for about 3%.

Question  What percentage of the human genome consists of unique (nonrepeating) DNA sequences?

Not surprisingly, organisms with the simplest lifestyles tend to have the smallest amount of DNA and the fewest genes. For example, M. genitalium (see Table 3.6) is a human pathogen that relies on its host to provide nutrients; this organism contains fewer genes than free-living bacteria such as E. coli. Multicellular organisms generally have even more DNA and more genes, presumably to support the activities of their many specialized cell types. Exceptions abound; for example, the largest known genome belongs to the plant Paris japonica, which has about 50 times more DNA than humans but probably has a similar number of genes. In prokaryotic genomes, all but a few percent of the DNA represents genes for proteins and RNA. The proportion of noncoding DNA generally increases with the complexity of the organism. For example, about 30% of the yeast genome, about half of the Arabidopsis genome, and over 98% of the human genome is noncoding DNA. Although up to 80% of the human genome may actually be transcribed to RNA, the protein-coding segments account for only about 1.4% of the total (Fig. 3.12). In large genomes, much of the noncoding DNA consists of repeating sequences with no known function. Some repeating sequences are the remnants of virus-like transposable elements, short segments of DNA that are copied many times and inserted randomly into the chromosomes. Not all viruses behave in this way, but all viruses can be considered to be molecular parasites (Box 3.C). The human genome contains two types of repetitive DNA sequences. About 48% of human DNA consists of interspersed repetitive sequences, which are blocks of hundreds or thousands of nucleotides scattered throughout the genome. The most numerous of these are present in hundreds of thousands of copies. Tandemly repeated DNA accounts for another 3% of the human genome. This type of DNA consists of short sequences (usually 2 to 10 bp) that are repeated, side by side, many times. The purpose—if any—of all this repetitive DNA is not understood, but it does help to explain why certain very large genomes actually include only a modest number of genes, and it explains why large genomes are able to accumulate sequence variations that have little effect on an individual’s fitness (see Section 1.4).

Genomics has practical applications Genomics, the study of genomes, can sometimes provide a rough snapshot of the metabolic capabilities of a given organism. For example, humans and fruit flies differ significantly in the proportion of genes involved in the immune response and responses to external stimuli

78  C ha pter 3   Nucleic Acid Structure and Function

Box 3.C Viruses Although viruses are sometimes lumped with bacteria as “germs,” the two groups have little in common. Viruses are not cells, they do not carry out metabolic activities, and they cannot replicate on their own. Instead, they must infect a host cell, either a prokaryote or eukaryote, and take advantage of that cell’s resources. Bacterial viruses are sometimes called bacteriophages (literally, bacteria-eaters) because they kill their host cells. Most viruses are much smaller than the cells they infect and have a simple structure: a nucleic acid genome surrounded by a protein capsid (shell), which may be surrounded by an outer membrane envelope with proteins embedded in it. Not all viruses have an outer membrane, and some virus capsids also enclose proteins that assist with infection or viral replication. Surface protein Membrane Capsid Nucleic acid

In all cases, the virus first attaches to a protein or another molecule on the surface of a host cell. This attachment may be quite specific, which limits the range of hosts that the virus can infect. In general, bacteriophages inject their nucleic acids into a bacterial cell, leaving the emptied capsid on the surface, whereas eukaryotic viruses are internalized before the capsid breaks apart and releases the nucleic acids. Viral genomes exhibit huge variety. They may be RNA or DNA, single- or double-stranded, single molecules or multiple chains or even circles. Viral genomes range in size from about 4000 nucleotides in a single RNA chain to 2,500,000 base pairs of ­ double-stranded DNA, with anywhere from 4 to 1000 genes. Not surprisingly, some viruses have a simple and quick life cycle, whereas others follow a much more elaborate sequence of infection, replication, and assembly of new virions (individual virus particles). All viruses require some of the host cell’s enzymes to help replicate their genomes and produce viral proteins. Doublestranded viral DNA genomes are replicated and transcribed in the usual way, but single-stranded viral DNA is usually converted by host enzymes to double-stranded DNA that can guide the synthesis of new viral DNA genomes and viral mRNAs. RNA viruses use a variety of strategies. So-called positive-sense RNA viruses

have single-stranded genomes that can be directly translated into viral proteins, but their replication requires the synthesis of a complementary chain to act as a template for synthesis of more ­single-stranded RNA genomes. The virus known as SARS-CoV-2, the cause of the COVID-19 pandemic, is among the largest RNA viruses, with a positive-sense RNA genome of about 30,000 nucleotides. Four of its genes code for structural proteins, including the large spike protein that gives the virus its crownlike appearance. Several other genes code for proteins that replicate the RNA genome and interfere with the host cell’s antiviral defenses. Negative-sense RNA viruses must first be transcribed, often by a viral RNA-directed RNA polymerase, to complementary RNA that can be translated. The retroviruses use their own reverse transcriptase enzyme to make a DNA copy of their RNA genome, and the DNA may then be integrated into the host cell’s chromosome. The subsequent excision of viral DNA may leave characteristic “footprint” sequences (the source of some repetitive DNA sequences). The excised virus may also bring along a host gene or two; viruses are one way for genes to move around the genome or be transferred between species. While the host cell’s DNA or RNA polymerases and ribosomes are occupied with producing viral components, normal cellular activities may be neglected. In addition, some viruses encode proteins that specifically inhibit the host cell’s normal functions and block antiviral responses, such as the degradation of doublestranded RNA, which almost always represents a replicating virus. New viruses assemble inside the host cell or at its surface as viral proteins and nucleic acids accumulate. The capsid proteins tend to self-assemble, typically forming hollow 20-sided icosa­ hedral shapes or helical tubes.

Icosahedral capsid

Helical capsid

Some bacteriophage genomes are actively threaded into preassembled capsids, whereas other virus capsids assemble around the viral nucleic acids. Additional steps of viral genome replication and protein maturation may occur inside the newly assembled virus before it becomes a complete virion capable of infecting a new host cell. The size of the virion “burst” ranges from perhaps 50 (for certain bacteriophage-infected prokaryotes) to over 50,000 (for some virus-infected eukaryotic cells).

(Fig. 3.13). An unusual number of genes belonging to one category might indicate some unusual biological properties in an organism. This sort of information can be useful for developing drugs to inhibit the growth of a pathogenic organism according to its unique metabolism. For example, as many as 2000 of the 6000 genes in the malaria parasite Plasmodium may code for proteins that are potential drug targets.

3.4 Genomics  79 Stimulus response Immune system

a. Human

b. Drosophila

  FIGURE 3.13   Functional classification of genes.  This diagram groups  a. human genes and 

b. Drosophila genes according to the biochemical function of the gene product. Humans devote a larger proportion of genes to the immune response (3.1% versus 1.0% in Drosophila) and to responses to external stimuli (6.0% versus 3.2% in Drosophila).

Curiously, many prokaryotes and small eukaryotes can acquire genes from other species through horizontal gene transfer. (Vertical gene transfer is the conventional pathway from parent to offspring, described in Box 3.A.) Horizontal gene transfer is one mechanism responsible for the spread of antibiotic-resistance genes among different bacterial species. In general, swapping genes with another organism or picking up stray DNA from the environment is a way to increase genetic diversity and potentially increase fitness. Although bacteria can readily acquire mammalian genes, the opposite does not appear to occur. The human microbiota, however, adds millions of microbial genes to the overall genetic makeup of the human body. Genomic analysis can also provide insights into the functionality of microbial communities in almost any sample, such as a spoonful of soil or a jar of seawater. In an approach known as metagenomics, all the bits of DNA in a sample are sequenced, and the genomes of the component species are reconstructed from the fragments. Although this method does not always reveal the presence of rare species, it is a practical way to assess the overall diversity of the community. In fact, it is often the only way, because most prokaryotes cannot be grown in the laboratory for close study. For larger organisms, DNA barcoding is a technique for identifying the species present in a sample of environmental DNA. In this case, researchers look for the sequence (the “barcode”) of a specific gene that is known to vary between species but not between members of that species. DNA barcoding can be used to detect the presence of endangered animals from traces they leave behind (fur, feces, and so on) without having to capture or harm the animal.

Before Going On • Describe the approaches used to identify genes. • Outline the correlation between gene number and organismal lifestyle. • List some ways in which the human genome differs from a bacterial genome. • List some practical applications of genomics.

80  C ha pter 3   Nucleic Acid Structure and Function

Summary 3.1  Nucleotides •  The genetic material in virtually all organisms consists of DNA, a polymer of nucleotides. A nucleotide contains a purine or pyrimidine base linked to a ribose group (in RNA) or a deoxyribose group (in DNA) that also bears one or more phosphate groups.

3.2  Nucleic Acid Structure •  DNA contains two antiparallel helical strands of nucleotides linked by phosphodiester bonds. Each base pairs with a complementary base in the opposite strand: A with T and G with C. The structure of RNA, which is single-stranded and contains U rather than T, is more variable. •  Nucleic acid structures are stabilized primarily by stacking interactions between bases. The separated strands of DNA can reanneal.

•  The sequence of nucleotides in a segment of DNA can reveal mutations that cause disease. Genetic variations across the genome may contribute to polygenic disease. •  In gene therapy, a normal gene is introduced or a gene is edited in order to cure a genetic disease.

3.4  Genomics •  Even with multiple techniques for identifying genes, the total number of genes in the human genome is not known. •  In general, the size of the genome and the number of ­protein-coding genes increases with organismal complexity. Large genomes may contain repetitive DNA sequences. •  Genomic analysis may reveal the functionality of organisms or the species diversity of communities.

3.3  The Central Dogma •  The central dogma summarizes how the sequence of nucleotides in DNA is transcribed into RNA, which is then translated into protein according to the genetic code.

Key Terms chromosome nucleic acid nucleotide DNA (deoxyribonucleic acid) base purine pyrimidine RNA (ribonucleic acid) nucleoside deoxynucleotide vitamin phosphodiester bond polynucleotide residue 5ʹ end 3ʹ end base pair sugar–phosphate backbone antiparallel major groove minor groove bp kb oligonucleotide polymerase nuclease exonuclease endonuclease complement

A-DNA B-DNA stacking interactions melting temperature (Tm) denaturation renaturation anneal probe replication gene expression sister chromatids centromere mitosis diploid haploid gametes meiosis homologous chromosomes allele gene transcription translation central dogma of molecular biology genome coding strand noncoding strand messenger RNA (mRNA) ribosome ribosomal RNA (rRNA)

transfer RNA (tRNA) codon genetic code mutation monogenetic disease polygenic disease genome-wide association study (GWAS) single-nucleotide polymorphism (SNP) recombinant DNA transgenic organism gene therapy open reading frame (ORF) noncoding RNA (ncRNA) homologous genes pseudogene exome exon intron transposable element virus bacteriophage capsid virion interspersed repetitive DNA tandemly repeated DNA genomics horizontal gene transfer metagenomics DNA barcoding

Problems  81

Bioinformatics Brief Bioinformatics Exercises 3.1  Drawing and Visualizing Nucleotides 3.2  The DNA Double Helix 3.3  Melting Temperature and the GC Content of Duplex DNA

3.4  Analysis of Genomic DNA Bioinformatics Project Databases for the Storage and “Mining” of Genome Sequences

Problems 3.1 Nucleotides

that the triple helix was stabilized by hydrogen bonds between the interior phosphate groups. What are the flaws in this model?

1.  The identification of DNA as the genetic material began with Griffith’s “transformation” experiment conducted in 1928. Griffith worked with Pneumococcus, an encapsulated bacterium that forms smooth colonies when plated on agar and causes death when injected into mice. A mutant Pneumococcus lacking the enzymes needed to synthesize the polysaccharide capsule (required for virulence) forms rough colonies when plated on agar and does not cause death when injected into mice. Griffith found that heat-treated wild-type Pneumo­ coccus did not cause death when injected into the mice because the heat treatment destroyed the polysaccharide capsule. However, if Griffith mixed heat-treated wild-type Pneumococcus and the mutant unencapsulated Pneumococcus together and injected this mixture, the mice died. Even more surprisingly, upon autopsy, Griffith found live, encapsulated Pneumococcus bacteria in the mouse tissue. Griffith concluded that the mutant Pneumococcus had been “transformed” into disease-causing Pneumococcus, but he could not explain how this occurred. Using your current knowledge of how DNA works, explain how the mutant Pneumococcus became transformed.

5.  Classify the following as purines or pyrimidines:  a. hypoxanthine;  b. xanthine; c.  the nitrogenous base shown in Problem 9; and  d.  the compound described in Problem 12.

2.  In 1944, Avery, MacLeod, and McCarty set out to identify the chemical agent capable of transforming mutant unencapsulated Pneumococcus to the deadly encapsulated form (see Problem 1). They isolated a viscous substance with the chemical and physical properties of DNA that was capable of transformation. If proteases (enzymes that degrade proteins) or ribonucleases (enzymes that degrade RNA) were added prior to the experiment, transformation could still occur. What did these treatments tell the investigators about the molecular identity of the transforming factor?

9.  Certain strains of E. coli incorporate the nitrogenous base shown below into nucleotides. For which base is this one a substitute?

3.  In 1952, Alfred Hershey and Martha Chase carried out experiments using bacteriophages, which consist of nucleic acid enclosed by a protein capsid (coat). They first labeled the bacteriophages with the radioactive isotopes 35S and 32P. Because proteins contain sulfur but not phosphorus, and DNA contains phosphorus but not sulfur, each type of molecule was separately labeled. The radiolabeled bacteriophages were allowed to infect the bacteria and then the preparation was treated to separate the empty capsids (ghosts) from the bacterial cells. The ghosts were found to contain most of the 35S label, whereas 30% of the 32P was found in the new bacteriophages produced by the infected cells. What does this experiment reveal about the roles of bacteriophage DNA and protein? 4.  In February 1953 (two months before Watson and Crick published their paper describing DNA as a double helix), Linus Pauling and Robert Corey published a paper in which they proposed that DNA adopts a triple-helical structure. In their model, the three chains were tightly packed together, with the phosphates on the inside of the triple helix and the nitrogenous bases on the outside. They proposed

O

O N N H

N

NH

NH

N H

N

Hypoxanthine

N H

O

Xanthine

6.  Examine the structure of GDP in Section 3.1. Identify the functional groups and linkages in this molecule (see Table 1.1). 7.  Describe the chemical difference between uracil and thymine. 8. a. How do the adenosine groups in FAD and CoA differ (see Fig. 3.2)?  b.  What kind of linkage joins the two nucleotides in NAD and FAD (see Fig. 3.2)?

O Cl

NH N H

O

10.  An E. coli culture is grown in the presence of the base shown in Problem 9. A control culture is grown in the absence of this modified base. Compare the masses of the DNA isolated from E. coli in these two cultures. 11.  The compound 8-chloroadenosine interferes with several cellular processes and inhibits the proliferation of cancer cells. Draw the structure of this compound. 12.  The simple synthesis of the antiviral compound 5-bromo-2′-­ deoxyuridine was recently reported. Draw the structure of this compound. For what base is this compound a substitute? 13.  In some RNA molecules, a cytosine base is first methylated at N3 and then oxidatively deaminated in a process that replaces the amino group at position 4 with a carbonyl group. Draw the modified base. 14.  DNA damage occurs when adenine, guanine, and cytosine undergo oxidative deamination. Draw the structures of the modified bases.

82  C ha pter 3   Nucleic Acid Structure and Function 15.  Investigators recently identified a bacterial enzyme that acts on 5-hydroxyuridine. Draw the structure of this nucleoside. 16.  The compound bredinin has antiviral activity.  a.  Is bredinin a base, a nucleoside, or a nucleotide? b.  One of its functional groups has a pK lower than 7 and donates a proton to another functional group with a pK greater than 7. Using Table 2.4 as a guide to the types of functional groups that can donate and accept protons, draw the structure of ionized bredinin.

27.  The adenine derivative hypoxanthine (see Problem 5) can base pair with cytosine, adenine, and uracil. Draw the structures of these base pairs. 28.  The nucleotide bases can undergo tautomerization, a form of isomerization that involves the rearrangement of hydrogen atoms. For example, guanine can form the enol tautomer shown here, which can pair with thymine. Draw the structure of the resulting base pair, which has three hydrogen bonds.

O N CH2OH

H

N

O

H

H

OH

OH

OH

NH2 OH

H

Bredinin

3.2  Nucleic Acid Structure 17.  Draw a CA (ribo)dinucleotide and label the phosphodiester bond. How would the structure differ if it were DNA? 18.  Draw the structure of the signaling molecule cyclic triadeny­late (a cyclic trinucleotide with three adenosine monophosphate residues). 19.  The dinucleotide cyclic guanosine monophosphate–adenosine monophosphate (cGAMP) is an intracellular signaling molecule involved in antiviral defense. The molecule has two phosphodiester linkages, one between the 2′-OH of GMP and the 5′-phosphate of AMP and the other between the 3′-OH of AMP and the 5′-phosphate of GMP. Draw this dinucleotide. 20.  Another intracellular signaling molecule is cyclic ADP-ribose (cADPR), in which the two phosphate groups at the 5′ position of ADP are linked to a second ribose at its 5′ carbon. This second ribose in turn is covalently linked to N1 of the ADP. Draw this nucleotide. 21.  A diploid organism with a 30,000-kb haploid genome contains 19% T residues. Calculate the number of A, C, G, and T residues in the DNA of each cell in this organism. 22.  Do Chargaff’s rules hold true for RNA? Explain why or why not. 23.  A well-studied bacteriophage has 97,004 base pairs in its double-stranded DNA genome.  a.  There are 24,182 G residues in the genome. Calculate the number of C, A, and T residues.  b. Why does GenBank report a total of 48,502 bases for this bacteriophage genome?

N

N H2N

N

N H

Guanine (enol tautomer) 29.  Explain whether the following statement is true or false: Because a G:C base pair is stabilized by three hydrogen bonds, whereas an A:T base pair is stabilized by only two hydrogen bonds, GC-rich DNA is harder to melt than AT-rich DNA. 30.  Hydrogen bonding does not make a significantly large contribution to the overall stability of the DNA molecule. Explain. 31.  How can the hydrophobic effect (Section 2.2) explain why DNA adopts a helical structure? 32. a.  Would you expect proteins to bind more favorably to the major groove or the minor groove of DNA? Explain.  b. Eukaryotic DNA is packaged with histones, small proteins with a high lysine and arginine content (see Figure 4.2 for the amino acid structures). Why do histones have a high affinity for DNA? 33. a.  What is the Tm of the DNA sample whose melting curve is shown in Figure 3.8?  b.  Draw melting curves that would be obtained from the DNA of Dictyostelium discoideum and Streptomyces albus (see Table 3.2). 34. a.  Construct a plot of Tm (°C) versus % GC content, using the data provided in Table 3.2.  b.  Use the plot to estimate the melting temperature of the DNA from an organism whose genome contains equal amounts of all four nucleotides.  c.  To what temperature would you have to cool the DNA to allow it to reanneal? 35.  What might you find in comparing the GC content of DNA from Thermus aquaticus (in hot springs) or Pyrococcus furiosus (in hydrothermal vents) and DNA from bacteria in a typical backyard pond? 36.  Explain why the melting temperature of a sample of double-helical DNA increases when the Na+ concentration increases.

25.  The SARS-CoV-2 virus was responsible for a global pandemic in 2020. Its genome contains 29,811 nucleotides, with 8903 A residues, 5482 C residues, 5852 G residues and 9574 U residues.  a. Calculate the percentages of A, C, G, and U in the genome.  b.  What type of genome does this virus have (see Problem 23)?

37. a.  You have a short piece of synthetic RNA that you want to use as a probe to identify a gene in a sample of DNA. The RNA probe has a tendency to hybridize with sequences that are only weakly complementary. Should you increase or decrease the temperature to improve your chances of tagging the correct sequence?  b.  You have a short piece of single-stranded DNA that you want to hybridize to another strand of DNA with one mismatched base pair between the two strands. Should you increase or decrease the temperature to improve your chances of annealing the two strands?

26.  Rotavirus infection is the leading cause of childhood diarrhea worldwide. The genome of the rotavirus A strain consists of 2265 A residues, 1037 C residues, 1037 G residues, and 2265 U residues.  a.  Calculate the percentages of each base in the genome.  b. What type of genome does this virus have (see Problem 23)?

38.  The laboratory technique known as fluorescence in situ hybridization (FISH) allows a fluorescent oligonucleotide probe to hybridize with a cell’s chromosomes, which are typically spread on a microscope slide. Explain why the chromosome preparation must be heated before the probe is added to it.

24.  The complete genome of a virus contains 1609 A residues, 1132 C residues, 1180 G residues, and 1578 T residues. What can you conclude about the structure of the viral genome (see Problem 23), given this information?

Problems  83

3.3  The Central Dogma 39.  Discuss the shortcomings of the following definitions for gene:  a.  A gene is the information that determines an inherited characteristic such as flower color.  b.  A gene is a segment of DNA that encodes a protein.  c.  A gene is a segment of DNA that is transcribed in all cells. 40.  The semiconservative nature of DNA replication (as shown in Fig. 3.10) was proposed by Watson and Crick in 1953, but it was not experimentally verified until 1958. Meselson and Stahl grew bacteria on the “heavy” nitrogen isotope 15N, producing DNA that was denser than normal. The food source was then abruptly switched to one containing only 14N. Bacteria were harvested and the DNA isolated by density gradient centrifugation.  a.  What is the density of the DNA of the first-generation daughter DNA molecules? Explain.  b. What is the density of the DNA isolated after two generations? Explain. c.  What results would Meselson and Stahl have obtained had DNA replicated conservatively? 41.  A portion of a gene is shown below. 5 ʹ-ATGATTCGCCTCGGGGCTCCCCAGTCGCTGGT-3ʹ 3ʹ-TACTAAGCGGAGCCCCGAGGGGTCAGCGACCA-5ʹ The mRNA transcribed from this gene has the following sequence: 5ʹ-AUGAUUCGCCUCGGGGCUCCCCAGUCGCUGGU-3ʹ a.  Identify the coding and noncoding strands of the DNA.  b.  Explain why only the coding strands of DNA are commonly published in databanks. 42.  A segment of the coding strand of a gene is shown below. ACACCATGGTGCATCTGACT a.  Write the sequence of the complementary strand that DNA polymerase would make. b.  Write the sequence of the mRNA that RNA polymerase would make from the gene segment. 43.  In the early 1960s, Marshall Nirenberg deciphered the genetic code by designing an experiment in which he synthesized a polynucleotide strand consisting solely of U residues, then added this strand to a test tube containing all of the components needed for protein synthesis.  a.  What polypeptide was produced by this “cell-free” system?  b.  What polypeptides were produced when poly A, poly C, and poly G were added to the cell-free system? 44.  Har Gobind Korana extended Nirenberg’s work (see Problem 43) by synthesizing polynucleotides with precisely defined sequences.  a.  What polypeptide(s) would be produced if a poly-­ GUGUGU··· were added to the cell-free system described in Problem 43?  b.  Do these results help to decipher the identities of the codons involved? 45.  How many different codons are possible in nucleic acids containing four different nucleotides if a codon consists of  a.  a single nucleotide, or consecutive sequences of  b.  two nucleotides,  c. three nucleotides, or d. four nucleotides? Does your answer help explain why codons consist of three nucleotides? 46.  Synthetic biologists at the Scripps Institute expanded the genetic repertoire by adding two new bases into living bacterial cells. The two bases are named d5SICS and dNaM, and they base-pair with one another. How many different codons are possible in nucleic acids containing six different nucleotides if a codon consists of a consecutive sequence of three nucleotides? 47.  An open reading frame (ORF) is a portion of the genome that potentially codes for a protein. A given mRNA sequence potentially has three different reading frames, only one of which is correct (the selection of the correct ORF will be discussed more fully in

Section 22.3). A portion of the gene for a type II human collagen is shown.  a.  What are the amino acid sequences that can potentially be translated from each of the three possible reading frames? b. Collagen’s amino acid sequence consists of repeating triplets in which every third amino acid is glycine. Does this information help you identify the correct reading frame? AGGTCTTCAGGGAATGCCTGGCGAGAGGGGAGCAGC 48.  Shown below is the coding strand of a portion of the gene that codes for the β-globin chain of hemoglobin.  a.  What are the amino acid sequences that could potentially be translated from each of the three reading frames?  b.  Examine your results and decide which of the three reading frames is the most likely to be correct. GTGCATCTGACTCCTGAGGAGAAGTC 49.  A portion of a sequence from human chromosome 22 is shown below. a. What are the amino acid sequences that could potentially be translated from this sequence if it is from the coding strand?  b.  What are the amino acid sequences that could potentially be translated if this sequence is from the noncoding strand? c. If you are told that the sequence is in the middle of a gene, does this help you to identify the correct reading frame? TTCCAATGACTGAGTTCTCTCTCTAGAGAG 50.  Is it possible for the same segment of DNA to encode two different proteins? Explain. 51. a.  One form of the disease adrenoleukodystrophy (ALD) is caused by the substitution of serine for asparagine in the ALD protein. List the possible single-nucleotide alterations in the DNA of the ALD gene that could cause this genetic disease. b. In another form of ALD, a CGA codon is converted to a UGA codon. Explain how this mutation affects the ALD protein. 52.  A mutation occurs when there is a base change in the DNA sequence. Some base changes do not lead to changes in the amino acid sequence of the resulting protein. Explain why. 53.   The disease cystic fibrosis is the result of a mutation in the gene that encodes the cystic fibrosis transmembrane regulator (CFTR), a channel protein that allows chloride to exit the cell. A partial sequence of the CFTR gene is shown below, with the correct reading frame indicated. The most serious form of the disease results from the deletion of three consecutive nucleotides, as shown.  ormal gene   504  505  506  507   508  509  510  511  512 n sequence ···GAA AAT ATC ATC TTT GGT GTT TCC TAT···  utated gene m sequence ···GAA AAT ATC AT– – – T GGT GTT TCC TAT··· a .  What are the sequences of the normal and mutated proteins?  b.  Explain why this form of the disease is referred to as ∆F508. 54.  Another patient with a less severe form of cystic fibrosis (see Problem 53) has a mutation in a different region of the CFTR gene. The patient’s DNA sequence is ···AAT AGA TAC AG··· (the normal sequence of the CFTR gene is ···AAT ATA GAT ACA G···). How has the DNA sequence changed and how does this affect the encoded protein? 55.  Describe the result when DNA containing a normal G:C base pair undergoes replication during which G tautomerizes (see Problem 28) and pairs with T in the new DNA strand. After cell division, what will the DNA look like in the two daughter cells? Assuming that the G remains in its tautomeric form, what will the DNA look like after those cells divide again?

84  C ha pter 3   Nucleic Acid Structure and Function 56.  A mutation was reported in the gene that encodes ApoB, a very large protein that is a component of low-density lipoprotein (LDL). LDL containing the defective protein is unable to bind to its hepatic receptor, resulting in hypercholesterolemia, which predisposes the patient to heart disease at an early age. A single nucleotide substitution changes one residue from arginine to glutamine. A partial ApoB gene sequence is shown below. Select the correct reading frame and give the amino acid sequence for both the normal and the mutated protein. ···CTGGCCGGCTCAATGGAGAGTCC···

62.  A portion of the sequence from the DNA coding strand of the chick ovalbumin gene is shown. Determine the partial amino acid sequence of the encoded protein. CTCAGAGTTCACCATGGGCTCCATCGGTGCAGCAAGCATGGAA(1104 bp)-TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA 63.  A partial sequence of a newly discovered bacteriophage is shown below.  a.  Identify the longest open reading frame (ORF).  b.  Assuming that the ORF has been correctly identified, where is the most likely start site?  ATGGGATGGCTGAGTACAGCACGTTGAATGAGGCGATGGT     CCGCTGGTGATG

57.  The genome of the bacterium Carsonella ruddii contains 159 kb of DNA with 182 ORFs. What can you conclude about the habitat or lifestyle of this bacterium? 58.  In theory, both strands of DNA can code for proteins; that is, genes can be overlapping. Propose an explanation for why overlapping genes are more commonly observed in prokaryotes than in eukaryotes. 59.  For many years, biologists and others have claimed that humans and chimpanzees are 98% identical at the level of DNA. Both the human and chimp genomes, which are roughly the same size, have now been sequenced, and the data reveal approximately 35 million nucleotide differences between the two species. How does this number compare to the original claim? 60.  When genomes of various organisms were sequenced, biologists expected that the DNA content (the C-value) would always be positively correlated with organismal complexity. However, no such correlation has been demonstrated. In fact, some plant and algae genomes are many times the size of the human genome. The C-value paradox is the term that refers to this puzzling lack of correlation between DNA content and organismal complexity. What questions do biologists need to ask as they attempt to solve the paradox? 61.  A portion of the sequence from the DNA coding strand of the myoglobin gene from a species of mollusk is shown below. Identify the correct open reading frame and give the amino acid sequence for the first few encoded amino acids.   ACCACCCCAACTGAAAATGTCTCTCTCTGATGC

64.  The genome of the bacteriophage described in Problem 63 contains 59 kb and 105 ORFs. None of the ORFs code for tRNAs. How does the bacteriophage replicate its DNA and synthesize the structural proteins necessary to replicate itself? 65.  If each person’s genome contains an SNP every 300 nucleotides or so, how many SNPs are in that person’s genome? 66.  Assuming that genes and SNPs are distributed evenly throughout the human genome, estimate how many protein-coding genes are likely to differ between two individuals. 67.  A genome-wide association study was carried out to identify the SNPs located on chromosome 1 that were correlated with an intestinal disease. The locations of three genes on chromosome 1 (between positions 6.3 × 107 and 6.6 × 107) are shown in the figure below. A –log10 P-value of 7 or greater is assumed to be associated with the disease.  a.  Which chromosomal locations show the strongest correlation with the disease?  b.  Which genes contain SNPs associated with the disease and which do not? gene A 14 12 10 8 6 4 2 0

gene B

gene C

−log10 P-value

3.4 Genomics

67,300,000

67,400,000

67,500,000

68.  Use the information in the figure below to determine the chromosomal locations for the SNPs that are most closely associated with a colon disease. In this study, a –log10 P-value of 5 was used as the cutoff. 

22.5 20.0

−log10 P-value

17.5 15.0 12.5 10.0 7.5 5.0 2.5 0 Chr1 Chr13

Chr2 Chr14

Chr3 Chr15

Chr4

Chr5

Chr16

Chr6 Chr17

Chr7 Chr18

Chr8 Chr19

Chr9 Chr20

Chr10 Chr21

Chr11 Chr22

Chr12 Chrx

Chapter 3 Credits  85

Selected Readings Cannon, M.E. and Mohlke, K.L., Deciphering the emerging complexities of molecular mechanisms at GWAS loci, Am. J. Hum. Genet. 103, 637–653, doi: 10.1016/j.ajhg.2018.10.001 (2018). [Reviews the methods and limitations of genome-wide association studies.] High, K.A. and Roncarolo, M.G., Gene therapy, N. Engl. J. Med. 381, 455–464, doi: 10.1056/NEJMra1706910 (2019). [Summarizes the general approach of gene therapy and describes some proven applications.] Lander, E.S., Initial impact of the sequencing of the human genome, Nature 470, 187–197 (2011). [Reviews genome structure and provides insights into its variation and links to human diseases.] McDermott, M.L., Vanselous, H., Corcelli, S.A., and Petersen, P.B., DNA’s chiral spine of hydration, ACS Cent. Sci. 3, 708–714, doi: 10.1021/acscentsci.7b00100 (2017). [Presents spectroscopic evidence of water binding in DNA’s major and minor grooves.]

Rigden, D.J. and Fernández, X.M., The 28th annual Nucleic Acids Research database issue and molecular biology database collection, Nucleic Acids Res. 49. D1–D9. doi: 10.1093/nar/gkaa1216 (2021). [Introduces the annual database issue, which includes a summary of bioinformatics resources and some additions to the list of 1641 molecular biology databases. The entire issue is available at https:// academic.oup.com/nar.] Salzberg, S.L., Open questions: How many genes do we have? BMC Biology 16, 94, doi: 10.1186/s12915-018-0564-x (2018). [A brief summary of differences between sequence databases.] Travers, A. and Muskhelishvili, G., DNA structure and function, FEBS J. 282, 2279–2295, doi: 10.1111/febs.13307 (2015). [Argues that the physicochemical properties of DNA are ideal for information storage and retrieval.]

Chapter 3 Credits Figure 3.4 Image based on 4TRA. Westhof, E., Dumas, P., Moras, D., Restrained refinement of two crystalline forms of yeast aspartic acid and phenylalanine transfer RNA crystals, Acta Crystallogr. A 44, 112–123 (1988). Figure 3.5 Image based on 1FIX. Horton, N.C., Finzel, B.C., The structure of an RNA/DNA hybrid: a substrate of the ribonuclease activity of HIV-1 reverse transcriptase, J. Mol. Biol. 264, 521–533 (1996).

Figure 3.7 Image from ACS Cent. Sci. 2017, 3, 7, 708–714 Publication Date: May 24, 2017. Copyright © 2017 American Chemical Society. Table 3.2 Data from T.A. Brown (ed.), Molecular Biology LabFax, VI., Academic Press (1998), pp. 233–237. Figure 3.13 Data from the Protein Analysis through Evolutionary Relationships classification system, www.pantherdb.org/.

CHAPTER 4

Protein Structure 123RF

DO YOU REMEMBER?

Behind the colored iris lies the lens of the eye, a 1-cm-diameter flattened sphere of cells that contain little besides proteins known as crystallins. Despite their name, these proteins do not actually form crystals, which would scatter light and defeat the purpose of the lens as a tool to focus light. Instead, the crystallins, which are very soluble, form a highly concentrated solution—almost like glass—that allows light to pass through. The conformations of the crystallins must be maintained for many decades to preserve the transparency of the lens.

• Cells contain four major types of biological molecules and three major types of polymers (Section 1.2). • Noncovalent forces, including hydrogen bonds, ionic interactions, and van der Waals forces, act on biological molecules (Section 2.1). • The hydrophobic effect, which is driven by entropy, excludes nonpolar substances from water (Section 2.2). • An acid’s pK value describes its tendency to ionize (Section 2.3). • The biological information encoded by a sequence of DNA is transcribed to RNA and then translated into the amino acid sequence of a protein (Section 3.3).

Proteins account for about half of the solid mass of a typical cell, and they carry out a wide variety of functions. They provide structural stability and motors for movement; they form the molecular machinery for harvesting free energy and using it to carry out other metabolic activities; they participate in the expression of genetic information; and they mediate communication between the cell and its environment. In subsequent chapters, we will describe in more detail these protein-driven phenomena, but for now we will focus on protein structure. We will look first at the amino acid components of proteins. Next comes a discussion of how protein chains fold into three-dimensional shapes. Finally, the Tools and Techniques section of this chapter examines some of the procedures for purifying and sequencing proteins and determining their structures.

4.1 Amino Acids, the Building Blocks of Proteins LEARNING OBJECTIVES Identify the 20 amino acids that occur in proteins. • Locate the functional groups in amino acids. • Classify amino acid side chains as hydrophobic, polar, or charged. • Draw a simple peptide and label its parts. • Determine the net charge of peptides. • Define the four levels of protein structure. 86

4.1  Amino Acids, the Building Blocks of Proteins  87

Proteins come in a huge variety of shapes and sizes (Fig. 4.1) but they are all built in the same way. Each protein consists of one or more polypeptides, which are chains of polymerized amino acids. A cell may contain dozens of different amino acids, but only 20 of these—called the “standard” amino acids—are commonly found in proteins. As ­introduced in Section 1.2, an amino acid is a small molecule containing an amino group (—NH+3 ) and a carboxylate group (—COO−) as well as a side chain of variable structure, called an R group:

COO− H

C

R

NH+ 3 Note that at physiological pH, the carboxyl group is unprotonated and the amino group is protonated, so an isolated amino acid bears both a negative and a positive charge.

DNA polymerase (E. coli Klenow fragment) Synthesizes a new DNA chain using an existing DNA strand as a template (more in Section 20.1)

Plastocyanin (poplar) Shuttles electrons as part of the apparatus for converting light energy to chemical energy (more in Section 16.2)

Insulin (pig) Released from the pancreas to signal the availability of the metabolic fuel glucose (more in Section 19.2)

Maltoporin (E. coli) Permits sugars to cross the bacterial cell membrane (more in Section 9.2)

Maltoporin (E. coli) Permits sugars to cross the bacterial cell membrane (more in Section 9.2)

Phosphoglycerate kinase (yeast) Catalyzes one of the central reactions in metabolism (more in Section 13.1)

  FIGURE 4.1   A gallery of protein structures.  These space-filling models are all shown at approximately the same scale. In proteins that consist of more than one chain of amino acids, the chains are shaded differently.

88  C ha pter 4   Protein Structure

The 20 amino acids have different chemical properties The identities of the R groups distinguish the 20 standard amino acids. The R groups can be classified by their overall chemical characteristics as hydrophobic, polar, or charged, as shown in Figure 4.2, which also includes the one- and three-letter codes for each amino acid. These compounds are formally called α-amino acids because the amino and carboxylate (acid) groups are both attached to a central carbon atom known as the α carbon (abbreviated Cα). Figure 4.2 also includes the one- and three-letter codes for each amino acid. The three-­ letter code is usually the first three letters of the amino acid’s name. The one-letter code is derived as follows: If only one amino acid begins with a particular letter, that letter is used: C = cysteine, H = histidine, I = isoleucine, M = methionine, S = serine, and V = valine. If more than one amino acid begins with a particular letter, the letter is assigned to the most abundant amino acid: A = alanine, G = glycine, L = leucine, P = proline, and T = threonine. Most of the others are phonetically suggestive: D = aspartate (“asparDate”), F = phenylalanine (“Fenylalanine”), N = asparagine (“asparagiNe”), R = arginine (“aRginine”), W = tryptophan (“tWyptophan”), and Y = tyrosine (“tYrosine”). The rest are assigned as follows:

Hydrophobic amino acids Hydrophobic amino acids COO– CH3 COO– Hydrophobic amino acids COO– CH COO– C CH3 COO– CH3 H C NH+ 3 H C +CH3 NH3 NH+ 3 (Ala, A) Alanine H

Alanine (Ala, A) COO–(Ala, A) CH3 Alanine COO–

CH3 C CH2 CH – COO CH3 + H C NH3CH2 CH CH3 H C +CH2 CH NH3 CH3 Leucine (Leu, CH L) NH+ 3 3 H

Leucine (Leu, L) Polar amino Leucine (Leu,acids L) – Polar amino acids COO Polar amino acids – COO H C CH2 OH COO– CH2 OH H NH C + 3 H C +CH2 OH Serine NH3 (Ser, S) NH+ 3 (Ser, S) Serine – COO(Ser, O Serine S)

COO–

COO–

CH3 C – COO CH3 + CH H C CH3 NH3 CH H C + CH3 NH3 + CH NH 3 Valine (Val, V) 3

– COO C CH2 – COO CH2 H NH C + 3 H C +CH2 NH3 NH+ 3 Phenylalanine (Phe, F)

Valine (Val, V) – CH COO Valine (Val, V) 3

Phenylalanine (Phe, F) COO– Phenylalanine (Phe, F)

H

COO–

CH CH3 CH2 C – CH COO + CH 3 CH2 H C NH3 CH CH2 H C + NH3 Isoleucine (Ile, I) + NH3 H

– COO C CH2 – COO CH2 H C NH+ 3 N H C +CH2 H NH3 N NH+ 3 H Tryptophan (Trp, N W) H W) Tryptophan (Trp,

H

CH3 CH3 CH3

Isoleucine (Ile, I) Isoleucine (Ile, I)

H

COO– Tryptophan (Trp, W)

– 2 CH COO C – CH CH COO 2 2 HH N C+ 2 CH 22 2 H C+ CHCH H2N CHCH Proline (Pro, 2 P)2 H2N+ CH2P) Proline (Pro,

COO–

C CH2 CH2 S CH3 COO– CH2 CH2 S CH3 H NH C + 3 H C +CH2 CH2 S CH3 NH3 Methionine (Met, M) NH+ 3 H

H

Methionine (Met, M) Methionine (Met, M)

COO– CH3

Proline (Pro, P)

COO–

– CH CH3 OH C H COO – CH COO 3 OH CH C + H NH 3 CH OH H C + Threonine (Thr, T) NH3 NH+ 3 Threonine (Thr, T)

– H COO C CH2 – COO CH2 H NH C + 3 H C +CH2 Tyrosine (Tyr, Y) NH 3 + NH 3 Tyrosine (Tyr, Y)

COO– (Thr, T) O Threonine

COO–

– H COO C CH2 SH – COO CH2 SH H NH C + 3 H C +CH2 SH Cysteine NH3 (Cys, C) NH+ 3 (Cys, C) Cysteine

OH OH OH

– COO Tyrosine (Tyr, Y) N – C NH2 H COO H COO C CH2 O C CH2 CH2 O C NH2 H COO C CH2 – – – N COO COO O COO O N CH2 CH2 C NH2 H NH H NH C + C + CH2 CH2 C NH2 H NH C + 3 3 3 H N H C +CH2 C NH2 H C +CH2 CH2 C NH2 H C +CH2 N NH3 NH NH3 (His,H 3 Asparagine (Asn, N) Glutamine (Gln, Q) Histidine H) N NH+ NH+ NH+ 3 3 3 H –

Asparagine (Asn, N) Charged amino Asparagine (Asn,acids N) – Charged amino COO O acids Charged – COO–amino O acids C CH2 C O COO– O CH2 C O– H NH C + 3 C D)O– HAspartate C CH2(Asp, NH+ 3 NH+ Aspartate (Asp, D) 3 H

Aspartate (Asp, D)

COO–(Cys, C) Cysteine

–

Glutamine (Gln, Q) Glutamine (Gln, Q)

COO–

Histidine (His, H) Histidine (His, H)

C O C CH2 CH2 O COO– O CH CH O– C H NH C + 2 2 3 – C H Glutamate C +CH2 CH 2 (Glu, E) O NH3 NH+ Glutamate (Glu, E) 3 H

–

Glutamate (Glu, E)

Glycine (Gly, G) Glycine (Gly, G)

COO–

O

COO–

– COO C H – COO H H NH C + 3 H C +H 3 G) GlycineNH (Gly, NH+ 3

H

H

COO–

C CH2 CH2 CH2 COO– H NH C + CH2 CH2 CH2 3 H C +CHLysine CH(Lys, CH 2 2 K)2 NH3 + NH3 Lysine (Lys, K)

Lysine (Lys, K)

COO–

CH2 CH2 CH2

NH+ 3 NH+ 3 NH+ 3

NH2

COO–

C 2NH+ C CH2 CH2 CH2 NH NH 2 – NH COO CH2 CH2 CH2 NH C 2NH+ H NH C + 2 3 CH2 R)NH C NH+ CH2 (Arg, H C +CH2Arginine 2 NH3 + NH3 Arginine (Arg, R) H

Arginine (Arg, R)

  FIGURE 4.2   Structures and abbreviations of the 20 standard amino acids.  The amino acids

can be classified according to the chemical properties of their R groups as hydrophobic, polar, or charged. The side chain (R group) of each amino acid is shaded. Question  Identify the functional groups in each amino acid. Refer to Table 1.1.

4.1  Amino Acids, the Building Blocks of Proteins  89

E = glutamate (near D, aspartate), K = lysine, and Q = glutamine (near N, asparagine). The carbon atoms of amino acids are sometimes assigned Greek letters, beginning with Cα, the carbon to which the R group is attached. Thus, glutamate has a γ-carboxylate group, and lysine has an ɛ-amino group. Nineteen of the 20 standard amino acids are asymmetric, or chiral, molecules. Their chirality, or handedness (from the Greek cheir, “hand”), results from the asymmetry of the alpha carbon. The four different substituents of Cα can be arranged in two ways. For alanine, a small amino acid with a methyl R group, the possibilities are

COO−

COO− H3N+

Cα

H

CH3

H

Cα

NH+ 3

CH3

You can use a simple model-building kit to satisfy yourself that the two structures are not identical. They are nonsuperimposable mirror images, like right and left hands. The amino acids found in proteins all have the form on the left. For historical reasons, these are designated l amino acids (from the Greek levo, “left”). Their mirror images, which rarely occur in proteins, are the d amino acids (from dextro, “right”). Molecules related by mirror symmetry are physically indistinguishable and are usually present in equal amounts in synthetic preparations. However, the two forms behave differently in biological systems (Box 4.A). It is advisable to become familiar with the structures of the standard amino acids, since their side chains ultimately help determine the three-dimensional shape of a protein, its solubility, its ability to interact with other molecules, and its chemical reactivity.

Amino Acids with Hydrophobic Side Chains 

Some amino acids have nonpolar (hydrophobic) side chains that interact very weakly or not at all with water. The aliphatic (hydrocarbonlike) side chains of alanine (Ala), valine (Val), leucine (Leu), isoleucine (Ile), and phenylalanine (Phe) obviously fit into this group. Although the side chains of methionine (Met) and tryptophan (Trp) include atoms with unshared electron pairs, the bulk of their side chains is nonpolar. Proline (Pro) is unique among the amino acids because its aliphatic side chain is also covalently linked to its amino group. Glycine (Gly) is sometimes included with the hydrophobic amino acids. In proteins, the hydrophobic amino acids are almost always located in the interior of the mol­ecule, among other hydrophobic groups, where they do not interact with water. And because they lack reactive functional groups, the hydrophobic side chains do not directly participate in mediating chemical reactions.

Box 4.A Does Chirality Matter? The importance of chirality in biological systems was brought home in the 1960s when pregnant women with morning sickness were given the sedative thalidomide, which was a mixture of right- and left-handed forms. The active form of the drug has the structure shown here.

H N O

O

H

C

O

N

Tragically, its mirror image, which was also present, caused severe birth defects, including abnormally short or absent limbs. Although the mechanisms of action of the two forms of thalidomide are not well understood, different responses to the two forms can be rationalized. An organism’s ability to distinguish chiral molecules results from the handedness of its molecular constituents. For example, proteins contain all l amino acids, and polynucleotides coil in a right-handed helix (see Fig. 3.3). The lessons learned from thalidomide have made drugs more costly to develop and test but have also made them safer.

Thalidomide

O

Question  Which of the 20 amino acids is not chiral?

90  C ha pter 4   Protein Structure

Amino Acids with Polar Side Chains 

The side chains of the polar amino acids can interact with water because they contain hydrogen-bonding groups. Serine (Ser), threonine (Thr), and tyrosine (Tyr) have hydroxyl groups; cysteine (Cys) has a thiol group; and asparagine (Asn) and glutamine (Gln) have amide groups. All these amino acids, along with histidine (His, which bears a polar imidazole ring), can be found on the solvent-exposed surface of a protein, although they also occur in the protein interior, provided that their hydrogen-­ bonding requirements are satisfied by their proximity to other hydrogen bond donor or acceptor groups. Glycine (Gly), whose side chain consists of only an H atom, cannot form hydrogen bonds but is included with the polar amino acids because it is neither hydrophobic nor charged. Depending on the presence of nearby groups that increase their polarity, some of the polar side chains can ionize at physiological pH values. For example, the neutral (basic) form of histidine can accept a proton to form an imidazolium ion (an acid):

COO− HC

N

CH2 N H

NH+ 3 Base

COO−

H+

HC H+

NH+

CH2

NH+ 3 Acid

N H

As we will see, the ability of histidine to act as an acid or a base gives it great versatility in catalyzing chemical reactions. Similarly, the thiol group of cysteine can be deprotonated, yielding a thiolate anion:

COO− HC

COO−

H+

CH2

HC

SH H+

NH+ 3

CH2

S−

NH+ 3

Occasionally, cysteine’s thiol group undergoes oxidation with another thiol group, such as another Cys side chain, to form a disulfide bond:

COO−

COO− HC

CH2

S

S

CH2

CH NH+ 3

NH+ 3 Disulfide bond

In certain situations, the hydroxyl groups of serine, threonine, and tyrosine undergo chemical reactions in which the O—H bond is cleaved.

Amino Acids with Charged Side Chains  Four amino acids have side chains that are virtually always charged under physiological conditions. Aspartate (Asp) and glutamate (Glu), which bear carboxylate groups, are negatively charged. Lysine (Lys) and arginine (Arg) are positively charged. Histidine, described above, can also bear a positive charge. Amino acids with charged side chains are usually located on the protein’s surface, where the ionic groups can be surrounded by water molecules or interact with other polar or ionic substances. Note that the charges of these side chains depend on their ionization state, which is sensitive to the local pH. The amino acids with acidic or basic side chains can also participate in acid–base reactions. Although it is convenient to view amino acids merely as the building blocks of proteins, many amino acids play key roles in regulating physiological processes (Box 4.B).

4.1  Amino Acids, the Building Blocks of Proteins  91

Box 4.B Monosodium Glutamate the form of monosodium glutamate (MSG) is sometimes added to processed foods as a flavor enhancer. For example, a low-salt food item can be made more appealing by adding MSG to it. At one time, it was thought that consuming excess MSG from restaurant meals or processed foods led to symptoms such as muscle tingling, headaches, and drowsiness—all of which could potentially reflect the role of glutamate in the nervous system. However, MSG is also naturally present in many foods, including cheese, tomatoes, and soy sauce. Furthermore, scientific studies have been unable to show a consistent link between MSG intake and neurological symptoms in humans.

A number of amino acids and compounds derived from them function as signaling molecules in the nervous system (we will look at some of these in more detail in Section 18.2). Among the amino acids with signaling activity is glutamate, which most often operates as an excitatory signal and is necessary for learning and memory. Because glutamate is abundant in dietary proteins and because the human body can manufacture it, glutamate deficiency is rare. However, is there any danger in eating too much glutamate? Glutamate binds to receptors on the tongue that register the taste of umami—one of the five human tastes, along with sweet, salty, sour, and bitter. By itself, the umami taste is not particularly pleasing, but when combined with other tastes, it imparts a sense of savoriness and induces salivation. For this reason, glutamate in

Question  Which group of glutamate is ionized to pair with a single sodium ion in MSG?

Peptide bonds link amino acids in proteins The polymerization of amino acids to form a polypeptide chain involves the condensation of the carboxylate group of one amino acid with the amino group of another (a condensation reaction is one in which a water molecule is eliminated): +

H3N

R1 C

H

O + H

C −

O

H

R2

+

N

C

H

H

SEE GUIDED TOUR Protein Structure

O C O−

H2O +

H3N

R2

R1 O C

C

H

N

C

H

H

O C O−

Peptide bond

The resulting amide bond linking the two amino acids is called a peptide bond. The remaining portions of the amino acids are called amino acid residues. In a cell, peptide bond formation is carried out in several steps involving the ribosome and additional RNA and protein factors (Section 22.3). Peptide bonds can be broken, or hydrolyzed, by the action of ­exopeptidases or endopeptidases (enzymes that act from the end or the middle of the chain, respectively). A hydrolysis reaction is the reverse of a condensation reaction. By convention, a chain of amino acid residues linked by peptide bonds is written or drawn so that the residue with a free amino group is on the left (this end of the polypeptide is called the N-terminus) and the residue with a free carboxylate group is on the right (this end is called the C-terminus):

N-terminus

+

H3N

R1

O

CH

C

N

R2

O

CH

C

H Residue 1

N

R3

O

CH

C

H Residue 2

N

R4

O

CH

C

O−

H Residue 3

Residue 4

Note that, except for the two terminal groups, the charged amino and carboxylate groups of each amino acid are eliminated in forming peptide bonds. The electrostatic properties of the polypeptide therefore depend primarily on the identities of the side chains (R groups) that project out from the polypeptide backbone.

C-terminus

92  C ha pter 4   Protein Structure

TAB L E 4. 1  pK Values of Ionizable Groups in Amino Acids

Group a

pK

COOH

C-terminus

3.5

O CH2

Asp

C

OH

3.9

O CH2

Glu

CH2

C

OH

4.1

+

CH2

His

Cys

CH2

N-terminus

NH+ 3

Tyr

CH2

Lys

CH2

NH

6.0

N H SH

8.4 9.0

OH CH2

CH2

10.5

NH+ 3

CH2

10.5

NH2 CH2

Arg a

CH2

CH2

NH

C

NH+ 2

12.5

The ionizable proton is indicated in red.

The pK values of all the charged and ionizable groups in amino acids are given in Table 4.1 (recall from Section 2.3 that a pK value is a measure of a group’s tendency to ionize). Thus, it is possible to calculate the net charge of a protein at a given pH (see Sample Calculation 4.1). At best, this value is only an estimate, since the side chains of polymerized amino acids do not behave as they do in free amino acids. This is because of the electronic effects of the peptide bond and other functional groups that may be brought into proximity when the polypeptide chain folds into a three-dimensional shape. The chemical properties of a side chain’s immediate neighbors, its microenvironment, may alter its polarity, thereby altering its tendency to lose or accept a proton. Nevertheless, the chemical and physical properties of proteins depend on their constituent amino acids, so proteins exhibit different behaviors under given laboratory conditions. These differences can be exploited to purify a protein, that is, to isolate it from a mixture containing other molecules (see Section 4.6). Most proteins contain all 20 amino acids, with some tending to appear more often than others (Fig. 4.3). 10

Percent

8 6 4 2

Pr o Se r Th r Tr p Ty r Va l

Al a Ar g As n As p Cy s G lu G ln G ly H is Ile Le u Ly s M et Ph e

0 Amino acid   FIGURE 4.3   Percent occurrence of amino acids in proteins.

4.1  Amino Acids, the Building Blocks of Proteins  93

SA MP L E CA LCULAT I O N 4. 1 Problem  Estimate the net charge of the polypeptide chain below at physiological pH (7.4) and at pH 5.0.

At pH 5.0, His is likely to be protonated, giving the polypeptide a net charge of +1:

Ala–Arg–Val–His–Asp–Gln Solution  The polypeptide contains the following ionizable groups, whose pK values are listed in Table 4.1: the N-terminus (pK = 9.0), Arg (pK = 12.5), His (pK = 6.0), Asp (pK = 3.9), and the C-terminus (pK = 3.5). At pH 7.4, the groups whose pK values are less than 7.4 are mostly deprotonated and the groups with pK values greater than 7.4 are mostly protonated. The polypeptide therefore has a net charge of 0: Group Charge N-terminus Arg His Asp C-terminus net charge

Group

Charge

N-terminus Arg His Asp C-terminus

+1 +1 +1 –1 –1

net charge

+1

+1 +1  0 –1 –1 0

Most polypeptides contain between 100 and 1000 amino acid residues, although some contain thousands of amino acids (Table 4.2). Very short polypeptides are often called ­oligopeptides (oligo is Greek for “few”) or just peptides. Some examples of peptides are the 9-residue hormones oxytocin (involved in childbirth and social bonding in mammals) and vasopressin (which regulates water homeostasis). Conotoxins are 10- to 30-residue peptides produced by Conus snails to paralyze their prey. Like many polypeptides that are secreted from cells, these peptides include Cys residues that form intramolecular disulfide bonds. Cys–Tyr–Ile–Gln–Asn–Cys–Pro–Leu–Gly Oxytocin

Cys–Tyr–Phe–Gln–Asn–Cys–Pro–Arg–Gly Vasopressin

Glu–Cys–Cys–Asn–Pro–Ala–Cys–Gly–Arg–His–Tyr–Ser–Cys A conotoxin

Since there are 20 different amino acids that can be polymerized to form polypeptides, even peptides of similar size can differ dramatically from each other, depending on their complement of amino acids. The potential for sequence variation is enormous. For a modest-sized polypeptide of 100 residues, there are 20 100 or 1.27 × 10 130 possible amino acid sequences. This number is clearly unattainable in nature, since there are only about 10 79 atoms in the universe, but it illustrates the tremendous structural variability of proteins. Unraveling the amino acid sequence of a protein may be relatively straightforward if its gene has been sequenced (see Section 3.4). In this case, it is just a matter of reading successive sets of three nucleotides in the DNA as a sequence of amino acids in the protein. However, this exercise may not be accurate if the gene’s mRNA is spliced before being translated or if the protein is hydrolyzed or otherwise covalently modified immediately after it is synthesized. Of course, nucleic acid sequencing is of no use if the protein’s gene has not been identified. The alternative is to use a technique such as mass spectrometry to directly determine the protein’s amino acid sequence (Section 4.6).

SEE SAMPLE CALCULATION VIDEOS

94

C hA pTER 4

Protein Structure

TAB L E 4. 2

Composition of Some Proteins

Number of amino acid residues

Protein

Number of polypeptide chains

Molecular mass (D)

Insulin (bovine)

51

2

5733

Rubredoxin (Pyrococcus)

53

1

5878

Myoglobin (human)

153

1

17,053

Phosphorylase kinase (yeast)

416

1

44,552

Hemoglobin (human)

574

4

61,972

Reverse transcriptase (HIV)

986

2

114,097

Nitrite reductase (Alcaligenes)

1029

3

111,027

C-reactive protein (human)

1030

5

115,160

Pyruvate decarboxylase (yeast)

1112

2

121,600

Immunoglobulin (mouse)

1316

4

145,228

Ribulose bisphosphate carboxylase (spinach)

5048

16

567,960

Glutamine synthetase (Salmonella)

5628

12

621,600

Carbamoyl phosphate synthetase (E. coli)

5820

8

637,020

The amino acid sequence is the first level of protein structure The sequence of amino acids in a polypeptide is called the protein’s primary structure. There are as many as four levels of structure in a protein (Fig. 4.4). Under physiological conditions, a polypeptide very seldom assumes a linear extended conformation but instead usually folds up to form a more compact shape, typically consisting of several layers. The conformation of the polypeptide backbone (exclusive of the side chains) is known as secondary structure. The complete three-dimensional conformation of the polypeptide, including its backbone atoms and all its side chains, is the polypeptide’s tertiary structure. In a protein that consists of more than one polypeptide chain, the quaternary structure refers to the

Primary structure The sequence of amino acid residues

Secondary structure The spatial arrangement of the polypeptide backbone

–Lys–Val–Asn–Val–Asp–

FIGURE 4.4

Levels of protein structure in hemoglobin.

Tertiary structure The three-dimensional structure of an entire polypeptide, including all its side chains

Quaternary structure The spatial arrangement of polypeptide chains in a protein with multiple subunits

4.2  Secondary Structure: The Conformation of the Peptide Group  95

spatial arrangement of all the chains. In the following sections we will consider the second, third, and fourth levels of protein structure.

Before Going On • Draw the structures and give the one- and three-letter abbreviations for the 20 standard amino acids. • Divide the 20 amino acids into groups that are hydrophobic, polar, and charged. • Identify the polar amino acids that are sometimes charged. • Draw a tripeptide and identify its peptide bonds, backbone, side chains, N-terminus, C-terminus, and net charge. • Describe the four levels of protein structure.

4.2 Secondary Structure: The Conformation of the Peptide Group LEARNING OBJECTIVES Recognize the common types of regular secondary structure. • Explain the limited flexibility of the peptide chain. • Describe the features of an α helix. • Describe the features of parallel and antiparallel β sheets. • Define irregular secondary structure.

In the peptide bond that links successive amino acids in a polypeptide chain, the electrons are somewhat delocalized so that the peptide bond has two resonance forms:

O−

O C

C

N H

+

N H

Due to this partial (about 40%) double-bond character, there is no rotation around the C—N bond. In a polypeptide backbone, the repeating N—Cα—C units of the amino acid residues can therefore be considered to be a series of planar peptide groups (where each plane contains the atoms involved in the peptide bond):

O

H N

Cα

C

H N H

Cα

C O

N

O Cα

C

N H

Cα

C O

Here the H atom and R group attached to Cα are not shown. The polypeptide backbone can still rotate around the alpha carbon. The degree of rotation is described by torsion angles known as ϕ (for the N—Cα bond) and ψ (for the

96

C hA pTER 4

Protein Structure

Cα—C bond). However, the values of ϕ and ψ are somewhat limited because rotation of the two peptide groups around Cα can bring atoms from neighboring residues too close, as shown here:

ψ ϕ

The atoms are color-coded: C gray, O red, N blue, and H white, and the van der Waals surfaces of the two carbonyl O atoms are shown. As the resonance structures indicate, the groups involved in the peptide bond are strongly polar, with a tendency to form hydrogen bonds. The backbone amino groups are hydrogen bond donors and the carbonyl oxygens are hydrogen bond acceptors. Under physiological conditions, the polypeptide chain tends to fold so that it can satisfy its hydrogen-bonding requirements. At the same time, the polypeptide backbone must adopt a conformation (a secondary structure) that minimizes steric strain. In addition, side chains must be positioned in a way that minimizes their steric interference. To meet these criteria, the polypeptide backbone often assumes a repeating conformation, known as regular secondary structure, such as an α helix or a β sheet.

The α helix exhibits a twisted backbone conformation The α helix was first identified through model-building studies carried out by Linus Pauling. In this type of secondary structure, the polypeptide backbone twists in a right-handed helix (the DNA helix is also right-handed; see Section 3.2 for an explanation). There are 3.6 residues per turn of the helix and, for every turn, the helix rises 5.4 Å along its axis. In the α helix, the carbonyl oxygen of each residue forms a hydrogen bond with the backbone NH group four residues ahead. The backbone hydrogen-bonding tendencies are thereby met, except for the four residues at each end of the helix (Fig. 4.5). The average α helix is about 10 residues long. Like the DNA helix, whose side chains fill the helix interior (see Figure 3.3b), the α helix is solid—the atoms of the polypeptide backbone are in van der Waals contact. However, in the α helix, the side chains extend outward from the helix (Fig. 4.6). FIGURE 4.5 The α helix. In this conformation, the polypeptide backbone twists in a righthanded fashion so that hydrogen bonds (dashed lines) form between CO and N—H groups four residues farther along. Atoms are color-coded: Cα light gray, carbonyl C dark gray, O red, N blue, side chain purple, H white. (Based

on a drawing by Irving Geis.)

Question How many amino acid residues are shown here? How many hydrogen bonds?

The β sheet contains multiple polypeptide strands Pauling, along with Robert Corey, also built models of the β sheet. This type of secondary structure consists of aligned strands of polypeptide whose hydrogen-bonding requirements are met by bonding between neighboring strands. The strands of a β sheet can be arranged in two ways (Fig. 4.7): In a parallel β sheet, neighboring chains run in the same direction; in an antiparallel β sheet, neighboring chains run in opposite directions. Each residue forms two hydrogen bonds with a neighboring strand, so all hydrogen-bonding requirements are met. The first and last strands of the sheet must find other hydrogen-bonding partners for their outside edges. A single β sheet may contain from 2 to more than 12 polypeptide strands, with an average of 6 strands, and each strand has an average length of 6 residues. In a β sheet, the amino acid side chains extend from both faces (Fig. 4.8).

4.2  Secondary Structure: The Conformation of the Peptide Group  97

b.

a.

  FIGURE 4.6  An α helix from myoglobin.  a. Ball-and-stick model

of residues 100–118 of myoglobin.  b. Space-filling model. Backbone atoms are green and side chains are gray.

C

O

H

N

O

C

C

H

N

O

C

O

C

C

C

Parallel

H

C

C O

N H C C C

O

O

C O

N H

N

Antiparallel

  FIGURE 4.7   β sheets.  In a parallel β sheet and an antiparallel β sheet, the polypeptide backbone is extended. In both types of β sheet, hydrogen bonds form between the amino and carbonyl groups of adjacent strands. The H and R attached to Cα are not shown. Note that the strands are not necessarily separate polypeptides but may be segments of a single chain that loops back on itself.

Question  How many amino acid residues are shown in each chain?

C

C C

C

O

H N

C N H

N H

C

N

C

C

C

C

O

O

N

C

C

N H

N H

N H

O

C O

C

H

C C

C

O

N H

H N

N

C

C O

H

N

C

C

C

O

C

N

C C

O

N H

C

N H

C C

O

O N H

C

C

H

C

C

N H

N H

H

O

C

O

C

C

H N C

O

C C

O

N

C

C O

H

N

O

C C

O

C N H

N H

C C

O

C

C

N H

C C

O

N

C C

C

N H

C H

O

C

N H

N C

C

C O

N

N

N

C C

98  C ha pter 4   Protein Structure

a.

b.

  FIGURE 4.8   Side view of two parallel strands of a β sheet.  a. Ball-and-stick model of a β sheet

from carboxypeptidase A.  b. Space-filling model. Backbone atoms are green and side chains (gray) point alternately to each side of the β sheet.

Proteins also contain irregular secondary structure α Helices and β sheets are classified as regular secondary structures, because their component residues exhibit backbone conformations that are the same from one residue to the next. In fact, these elements of secondary structure are easily recognized in the three-­dimensional structures of a huge variety of proteins, regardless of their amino acid composition. Of course, depending on the identities of the side chains and other groups that might be present, α helices and β sheets may be slightly distorted from their ideal conformations. For example, the final turn of some α helices becomes “stretched out” (longer and thinner than the rest of the helix). In every protein, elements of secondary structure (individual α helices or strands in a β sheet) are linked together by polypeptide loops of various sizes. A loop may be a relatively simple hairpin turn, as in the connection of two antiparallel β strands (which are shown below as flat arrows; left), or it may be quite long, especially if it joins successive strands in a parallel β sheet (right):

These connecting loops, as well as other segments of the polypeptide chain, are usually described as irregular secondary structure; that is, the polypeptide does not adopt a defined secondary structure in which successive residues have the same backbone conformation. Most proteins contain a combination of regular and irregular secondary structure. But note that “irregular” does not necessarily mean “disordered.” In many proteins, the peptide backbone adopts a single, unique conformation. However, as described below, some proteins include segments that are truly disordered and can adopt many different conformations.

Before Going On • Draw a polypeptide backbone and indicate which bonds can rotate freely. • Identify all the hydrogen bond donor and acceptor groups in the backbone. • Explain how an α helix and a β sheet satisfy a polypeptide’s hydrogen-bonding needs. • Compare parallel and antiparallel β sheets. • Summarize the difference between regular and irregular secondary structure.

4.3  Tertiary Structure and Protein Stability  99

4.3 Tertiary Structure and Protein Stability LEARNING OBJECTIVES Describe the forces that stabilize protein structure. • Analyze protein structures presented in different styles. • Explain how the hydrophobic effect stabilizes globular protein structures. • Describe the intramolecular interactions that can stabilize protein structures. • Summarize the process of protein folding. • Identify different types of disorder in protein structures. • List some functions of disordered protein regions.

According to the traditional view, the three-dimensional shape of a polypeptide is its tertiary structure, which includes the chain’s regular and irregular secondary structure (that is, the overall folding of its peptide backbone) as well as the spatial arrangement of all its side chains. However, the idea that every atom has a single designated place mainly reflects historical approaches for evaluating protein structures and does not apply to all proteins. The current understanding is that protein structures lie along a spectrum of possibilities from highly ordered, fixed shapes to highly disordered and dynamic shapes. However, all proteins are subject to the same thermodynamic forces, so, for convenience, we will focus first on the principles that govern proteins with unique shapes and then investigate proteins that do not conform as closely to these traditional “rules.”

a.

Proteins can be described in different ways Most of the ~160,000 proteins whose structures are known in atomic detail were studied using X-ray crystallography (Section 4.6), a technique that captures the conformation of a protein fixed in a crystalline array. This was the technique used by John Kendrew to determine the first protein structure—of myoglobin—in 1958. After painstakingly determining the position of every backbone and side-chain group, Kendrew was mildly disappointed to find that, unlike the elegant, symmetric DNA structure that had been published just a few years earlier, the myoglobin structure was irregular and complex (myoglobin structure is examined in detail in Section 5.1). Many early studies of protein structure examined compact globular proteins that formed crystals suitable for analysis. Fibrous proteins, in contrast, are usually highly elongated (examples of these are presented in Section 5.3). The proteins shown in Figure 4.1 are all globular proteins. To make sense of structures like these, researchers use different approaches. For example, the tertiary structure of the well-studied triose phosphate isomerase can be represented as atoms in space-filling form, as lines that connect Cα of each residue, or as ribbons that highlight secondary structures (Fig. 4.9). Different rendering styles convey different types of information, but keep in mind that proteins, like all molecules, are solid objects with no empty space inside. Well-ordered globular proteins can be classified by their common features, such as the arrangement of helices and sheets. For example, four classes are recognized by the CATH system (the name refers to a hierarchy of organizational levels: Class, Architecture, Topology, and Homology). Ordered proteins may contain mostly α structure, mostly β structure, a combination of α and β, or very few regular secondary structural elements. Examples of each class are shown in Figure 4.10.

b.

c.   FIGURE 4.9   Triose phosphate isomerase.  a. Spacefilling model. All atoms (except H) are shown (C gray, O red, N blue).  b. Polypeptide backbone. The trace connects the α carbons of successive amino acid residues. c. Ribbon diagram. The ribbon represents the overall conformation of the backbone.

100  C ha pter 4   Protein Structure

b.

a.

c.

d.

  FIGURE 4.10   Classes of protein structure.  In each protein, α helices are colored red, and β strands are colored yellow.  a. Growth hormone, an all-α protein.  b. β/γ-Crystallin, an all-β protein.  c. Flavodoxin, an α/β protein. d. Tachystatin, a protein with little secondary structure.

  FIGURE 4.11   A two-domain protein.  In this model of glyceraldehyde-3-phosphate dehydrogenase, the small domain is red and the large domain is green.

Question  Which of the proteins in Figure 4.10 consists of two domains?

Globular proteins have a hydrophobic core

  FIGURE 4.12  Hydrophobic and hydrophilic residues in myoglobin.  Hydrophobic side chains belonging to Ala, Ile, Leu, Met, Phe, Pro, Trp, and Val (green) mostly cluster in the protein interior and polar and charged side chains (purple) predominate on the protein surface. Backbone atoms are gray.

Globular protein shapes typically contain at least two layers of a secondary structure. This means that the protein has definite surface and core regions. On the protein’s surface, some backbone and side-chain groups are exposed to the aqueous environment; in the core, these groups are sequestered from water. In other words, the protein comprises a hydrophilic surface and a hydrophobic core. A polypeptide segment that has folded into a single structural unit with a hydrophobic interior is often called a domain. Some small proteins consist of a single domain. Larger proteins may contain several domains, which may be structurally similar or dissimilar (Fig. 4.11). The core of a domain or a small protein is typically rich in a regular secondary structure. This is because the formation of α helices and β sheets, which are internally h ­ ydrogen-bonded, minimizes the hydrophilicity of the polar backbone groups. Irregular secondary structures (loops) are more often found on the solvent-exposed surface of the domain or protein, where the polar backbone groups can form hydrogen bonds with water molecules. The requirement for a hydrophobic core and a hydrophilic surface also places constraints on amino acid sequences. The location of a particular side chain in a protein’s tertiary structure is related to its hydrophobicity: the more hydrophobic the residue, the more likely it is to be located in the protein interior. For example, highly hydrophobic residues such as Phe and Met are almost always buried. In the protein interior, side chains pack together, leaving essentially no empty space or space that could be occupied by a water molecule. (Fig. 4.12). Polar and charged groups that are positioned in the nonpolar interior must be “neutralized” by interacting with other polar or charged groups.

4.3 Tertiary Structure and Protein Stability Folded

101

Unfolded

Unfavorable solvation FIGURE 4.13 The hydrophobic effect in protein folding. In a folded protein, hydrophobic regions (represented by green segments of the polypeptide chain) are sequestered in the protein interior. Unfolding the protein exposes these segments to water. This arrangement is energetically unfavorable, because the presence of the hydrophobic groups interrupts the hydrogen-bonded network of water molecules.

Protein structures are stabilized mainly by the hydrophobic effect Surprisingly, the fully folded conformation of a typical globular protein is only marginally more stable than its unfolded form. The difference in thermodynamic stability amounts to about 0.4 kJ · mol–1 per amino acid, or about 40 kJ · mol–1 for a 100-residue polypeptide. This is equivalent to the amount of free energy required to break just two hydrogen bonds (about 20 kJ · mol–1 each). This quantity seems incredibly small—considering the number of potential interactions among all of a protein’s backbone and side-chain atoms—yet many proteins do fold into a stable three-dimensional arrangement of atoms. The largest force governing protein structure is the hydrophobic effect (introduced in Section 2.2), which causes nonpolar groups to aggregate in order to minimize their contact with water. This partitioning is not due to any strong attractive force between hydrophobic groups. Rather, the hydrophobic effect is driven by the increase in entropy of the solvent water molecules, which would otherwise have to order themselves around each hydrophobic group. As we have seen, hydrophobic side chains are located predominantly in the interior of a protein. This arrangement stabilizes the folded polypeptide backbone, since unfolding it or extending it would expose the hydrophobic side chains to the solvent (Fig. 4.13). In the core of the protein, the hydrophobic groups experience only very weak attractive forces (van der Waals interactions); they stay in place mainly due to the hydrophobic effect. Hydrogen bonding by itself is not a major determinant of protein stability, because in an unfolded protein, polar groups could just as easily form energetically equivalent hydrogen bonds with water molecules. Instead, hydrogen bonding—such as occurs in the formation of α helices or β sheets—may help the protein to fine-tune a folded conformation that is already largely stabilized by the hydrophobic effect. Once folded, many polypeptides appear to maintain their shapes through various other types of interactions, the most common being ion pairs, interactions with zinc ions, and covalent cross-links such as disulfide bonds. Ion pairs are electrostatic interactions between charged groups, which may be amino acid side chains or the N- and C-terminal groups of a polypeptide (Fig. 4.14). Although

Asp

−

+ − + Arg

Lys Glu

a.

b.

FIGURE 4.14 Examples of ion pairs in myoglobin. a. The ɛ-amino group of Lys 77 interacts with the carboxylate group of Glu 18. b. The carboxylate group of Asp 60 interacts with Arg 45. The atoms are colorcoded: C gray, N blue, and O red. Note that these intramolecular interactions occur between side chains that are near each other in the protein’s tertiary structure but are far apart in the primary structure.

Question Identify amino acid side chains that could form two other types of ion pairs.

102  C ha pter 4   Protein Structure

a.

b.

  FIGURE 4.15   Zinc fingers.  a. A zinc finger with one Zn2+

(purple sphere) coordinated by two Cys and two His residues, from Xenopus transcription factor IIIA.  b. A zinc finger with two Zn2+ coordinated by six Cys residues, from the yeast transcription factor GAL4.

  FIGURE 4.16   Disulfide bonds in lysozyme, an extracellular protein.  This 129-residue enzyme from hen egg white contains eight Cys residues (yellow), which form four disulfide bonds that link different sites on the polypeptide backbone.

the resulting interaction is strong, it does not contribute much to protein stability. This is because the favorable free energy of the electrostatic interaction is offset by the loss of entropy when the charged groups become fixed in the ion pair. For a buried ion pair, there is the additional energetic cost of dehydrating the charged groups in order for them to enter the hydrophobic core. Domains known as zinc fingers are common in DNA-binding proteins. These structures consist of 20–60 residues with one or two Zn2+ ions. The Zn2+ ions are tetrahedrally coordinated by the side chains of cysteine and/or histidine and sometimes aspartate or glutamate (Fig. 4.15). Protein domains of this size are too small to assume a stable tertiary structure without a metal ion cross-link. Zinc is an ideal ion for stabilizing proteins: It can interact with ligands (S, N, or O) provided by several amino acids, and it has only one oxidation state (unlike Cu or Fe ions, which readily undergo oxidation–reduction reactions under cellular conditions). Disulfide bonds (a type of covalent bond, shown in Section 4.1) can form within and between polypeptide chains. Experiments show that even when the cysteine residues of certain proteins are chemically blocked, the proteins may still fold and function normally. This suggests that disulfide bonds are not essential for stabilizing these proteins. In fact, disulfides are rare in intracellular proteins, since the cytoplasm is a reducing environment. They are more plentiful in proteins that are secreted to an extracellular (oxidizing) environment (Fig. 4.16). Here, the bonds may help prevent protein unfolding under relatively harsh extracellular conditions. Other covalent cross-links include thioester bonds, which appear in a few proteins (Box 4.C), and isopeptide bonds, in which a carboxyl group on a side chain or at the C-terminus condenses with an amino group on a side chain or at the N-terminus. A Lys–Glu isopeptide bond is shown here:

C H2

H2 C

C H2

H2 C

O N H

C

Lys

C H2 Glu

Isopeptide bond

H2 C

4.3  Tertiary Structure and Protein Stability  103

Box 4.C Thioester Bonds as Spring-Loaded Traps Thioester bonds (Table 1.1) appear in some small biological molecules such as acetyl-CoA (Section 12.3) as well as in some proteins. For example, some enzymes briefly form thioesters during the transformation of reactants to products. The small protein ubiquitin temporarily attaches to a protein called E2 via a thioester linkage before making a covalent cross-link with the lysine side chain of another protein (Section 12.1). Thioesters also appear as internal cross-links in several specialized proteins that snap open and grab on to other molecules. These internal thioester bonds appear to form spontaneously when polypeptide folding brings a Cys side chain close to a Gln side chain.

Polypeptide backbone

CH CH2 S Cys

CH

C O

CH2 CH2 Gln

The buried thioester groups are inaccessible to water and other nucleophiles that readily attack the electron-poor carbon atom. Breaking a thioester bond is a highly favorable reaction; it releases

about the same amount of free energy as removing a phosphate group from ATP. Thioesters are more reactive than conventional esters because the electron delocalization (resonance) involving the large S atom is not as efficient as the C—O resonance that stabilizes oxygen esters. Proteins that contain internal thioesters must undergo a conformational change that exposes the reactive group to hydroxyl or amino groups, which rapidly react to form covalent adducts:

O

O S

C

+ ROH

S

H + R

O

C

In this way, the thioester-containing protein can latch on to something else. For example, the complement proteins C3 and C4, which help the immune system tag and eliminate pathogens (disease-causing organisms), become activated, undergo a conformational change, and quickly react with the amino groups of proteins and the hydroxyl groups of cell-wall carbohydrates on the pathogen’s surface. In C3 and C4, the Cys and Gln residues that form the thioester are separated by only two other amino acids. Consequently, springing the thioester “trap” may also be facilitated by physical strain in the polypeptide when the protein changes conformation. Some pathogens also position proteins with internal thioesters at the end of long surface extensions; when they contact a prospective host cell, the protein conformation changes, allowing the reactive thioester to form a covalent bond with the target, facilitating infection.

Protein folding is a dynamic process How do newly made proteins assume their proper conformation in the crowded environment of the cell? A polypeptide begins to fold as soon as it emerges from the ribosome, so part of the chain may adopt its mature tertiary structure before the entire chain has been synthesized. It is difficult to monitor this process in the cell, so studies of protein folding in vitro usually rely on small globular proteins that have been chemically unfolded (denatured) and then allowed to refold (renature). In the laboratory, proteins can be denatured by adding highly soluble substances such as salts or urea (NH2—CO—NH2). Large amounts of these solutes interfere with the structure of the solvent water, thereby attenuating the hydrophobic effect and causing the proteins to unfold. When the solutes are removed, the proteins renature. Irreversible protein denaturation is a key part of cooking all sorts of foods, including baked goods (Box 4.D). Although denaturation–renaturation experiments are difficult or impossible for large proteins, biochemists assume that a newly synthesized polypeptide follows some sort of folding pathway in order to reach its most stable conformation or native structure. Contrary to expectations, secondary structures such as α helices and β sheets do not form immediately. Instead, the first stages of protein folding can be described as a “hydrophobic collapse,” when nonpolar groups are forced out of contact with water. Once the hydrophobic core of the protein begins to take shape, the hydrogen bonds responsible for regular secondary structure can form, and other structures and side chains jostle themselves into their final positions in the tertiary structure (Fig. 4.17). Different proteins follow different folding pathways, and even a single polypeptide chain may achieve its native structure in slightly different ways. In a cell, the protein folding process may be aided by additional proteins known as molecular chaperones. Some of these associate with the ribosome to assist with the initial

SEE ANIMATED PROCESS DIAGRAM Denaturation and renaturation of RNase A

104  C ha pter 4   Protein Structure

Box 4.D Baking and Gluten Denaturation or by chemical agents such as sodium bicarbonate (baking soda). When the dough is cooked, the gas bubbles expand further. The heat also drives out water (and any ethanol produced by the yeast) and fully denatures the gluten, causing it to stiffen. The result is a spongy and chewy loaf of bread. Cake batter also contains flour but differs from bread dough by the presence of additional fats (lipids), often introduced in the form of butter, vegetable oil, or eggs (although egg whites are largely fat-free, the yolks are not). The hydrophobic lipids bind to nonpolar segments of glutenins and gliadins to inhibit their intermolecular interactions. Moreover, after the flour is added, cake batter is only minimally mixed and not kneaded, so gluten formation is less extensive than in bread dough. Baking denatures the proteins and solidifies the limited gluten network, yielding a fluffy, crumbly cake rather than a chewy loaf.

A familiar example of protein denaturation is the structural transformation of an egg white—from a clear viscous liquid to an opaque solid—when heated. Ovalbumin and other proteins in the egg white unfold during cooking and aggregate, irreversibly in this case. Thermal denaturation also explains what happens to bread dough in the oven. Contrary to popular belief, raw wheat flour does not contain any protein called gluten. Rather, it contains proteins called glutenins and gliadins. Glutenins form extended coiled shapes and are not highly soluble, due to an abundance of uncharged side chains. When added to water to make bread dough and then kneaded, the glutenins stretch into sheets and trap the smaller globular gliadins. This network of proteins is known as gluten. The glutenins make gluten elastic, while the interspersed gliadins prevent it from becoming too solid. During the dough-rising phase, the sheets of gluten capture bubbles of CO2 gas generated by yeast

  FIGURE 4.17   A protein folding pathway.  In this diagram for a hypothetical globular protein, the polypeptide assumes a progressively more ordered conformation.

stages of folding (described in more detail in Section 22.4). Other chaperones escort newly made proteins as they move to other parts of the cell or undergo post-translational processing. Depending on the protein, this might mean removal of some amino acid residues or the covalent attachment of another group such as a lipid, carbohydrate, or phosphate group (Fig. 4.18). The attached groups usually have a discrete biological function and may also help to stabilize the folded conformation of the protein. Metal ions or small organic molecules

O

HOCH2 H

O H3C

(CH2)14

a.

C

S

CH2

O

O H OH H

H HN b.

  FIGURE 4.18   Some covalent modifications of proteins. a. A 16-carbon fatty acid (palmitate, in red) is linked by a thioester bond to a Cys residue.  b. A chain of several

NH

C

CH2 O− −

H C

CH3

O

P O

O

CH2

c.

O

carbohydrate units (here only one sugar residue is shown, in red) is linked to the amide N of an Asn side chain.  c. A phosphoryl group (red) is esterified to a Ser side chain.

4.3 Tertiary Structure and Protein Stability

may also associate with the protein. Some proteins contain several polypeptide chains, which fold individually before assembling. All the information required for a protein to fold is contained in its amino acid sequence. Unfortunately, there are no completely reliable methods for predicting how a polypeptide chain will fold. In fact, it is difficult to determine whether a relatively short amino acid sequence will form an α helix, β sheet, or no particular structure at all. This uncertainty presents a formidable obstacle to assigning three-dimensional structures—and possible functions—to the burgeoning number of proteins identified through genome sequencing (see Section 3.4).

Disorder is a feature of many proteins Traditional views of protein structure—largely built on experiments with small globular proteins—hold that each protein has just one stable, low-energy conformation. However, a few proteins have been captured in different conformations in crystals, and many crystallized proteins include short mobile segments. A few proteins appear to entirely lack any secondary or tertiary structure yet are able to perform distinct functions. Most tellingly, analyses of thousands of sequences obtained from genomic data suggest that many polypeptide chains do not include enough nonpolar residues to fold into a conventional globular shape with a hydrophobic core. In the modern view of protein structure, globular proteins with only one conformation are simply one end of a continuum of the structural options for polypeptide chains. Of course, no protein exists as a completely rigid block. All proteins are inherently flexible, because individual bonds in the polypeptide chain can rotate, bend, and stretch, and secondary and tertiary structures are stabilized by relatively weak noncovalent forces. In addition to the small, unavoidable, and random fluctuations in their conformations, many proteins must undergo other types of structural changes in order to carry out their biological functions. For example, to catalyze a chemical reaction, an enzyme must closely contact the reacting substances while they are being transformed to reaction products. When an extracellular signaling molecule binds to a cell-surface receptor protein, the protein’s conformation changes in a way that triggers additional events inside the cell. Such conformational adjustments might involve repositioning a few side chains or a loop of polypeptide chain, or they might require more dramatic shifts in the overall arrangement of secondary structures or entire domains. The absence of conformational order in proteins can take different forms, with a range of functional implications. For example, some proteins are prone to denaturation, requiring periodic visits to chaperones to restore their functional conformations. So-called chameleonic sequences of up to 20 residues may form an α helix in one protein but a β sheet in another. Although all proteins fold in a way that allows them to proceed from an unstable, highenergy state to a lower-energy state, the energetic landscape may not always resemble a simple funnel (Fig. 4.19). Conventional or monomorphic proteins have a single stable tertiary

Metamorphic protein

Intrinsically disordered protein

Free energy

Monomorphic protein

FIGURE 4.19 Energy diagrams for proteins. Monomorphic proteins have a single low-energy conformation; metamorphic proteins can adopt multiple shapes that have roughly equivalent energy. Intrinsically disordered proteins are not energetically constrained to occupy any particular structural space. These are idealized diagrams, and proteins may include multiple segments with different degrees of disorder.

Question Which sides of this diagram correspond to low entropy and high entropy?

105

106  C ha pter 4   Protein Structure

a.

b.

  FIGURE 4.20   A metamorphic protein with two stable conformations. 

a. One form of the protein lymphotactin (also called XCL1) consists of a threestranded β sheet and an α helix.  b. An alternate form with an all-β structure interconverts rapidly with the first form.

  FIGURE 4.21   A protein with intrinsically disordered regions.  The signaling molecule interleukin-21 includes four ordered helices (light blue) and disordered segments of polypeptide (magenta) at the N-terminus, C-terminus, and an internal loop. The disordered segments can adopt different positions; just one possible protein structure is depicted here.

structure, whereas metamorphic proteins may have two or more possible conformations. Intrinsically disordered proteins have no fixed structure at all. Metamorphic proteins switch easily between energetically equivalent conformations, typically about once per second (Fig. 4.20). Different forms of these “transformer” proteins are in dynamic equilibrium, but the balance can shift in response to changes in cellular pH, oxidation or reduction of disulfide groups, the addition of phosphoryl groups, the presence of ions such as Mg2+, or binding to other molecules. The alternate structures may exhibit different degrees of functionality (for example, an enzyme that is more active or less active in a certain conformation) or they may have entirely different functions (some dual-function proteins are known as “moonlighting” proteins). Many proteins include disordered segments interspersed with more defined globular regions. An estimated 44% of the human proteome contains disordered segments of over 30 residues (the proteome is the set of all proteins made by an organism). These intrinsically disordered regions, which may contain as many as several hundred amino acids, have no fixed secondary or tertiary structure (­ Fig. 4.21). As expected, they are rich in amino acids bearing polar and charged side chains.

Protein functions may depend on disordered regions What can disordered protein segments do? Some function as linkers or spacers, such as the “hinge” regions that connect domains in antibody proteins (Section 5.5). A few disordered proteins actually perform like molecular springs. Disordered segments commonly appear in proteins that participate in the biomineralization of hard structures such as bones and teeth. Here, the flexible proteins wrap around and solubilize calcium phosphate, which prevents the crystals from precipitating or forming large brittle aggregates. However, the functions of most disordered protein regions are related to binding other proteins. Although it might be tempting to picture intrinsically disordered protein regions as long, extended polypeptide segments, the chains may also transiently adopt more compact structures with well-defined secondary structures, as if “trying out” different conformations. One of these conformations may be suitable for interacting with another molecule, and that conformation may then become locked in place when the two molecules bind (in this respect, the proteins resemble metamorphic proteins). In some cases, a disordered region allows a protein to interact with many different potential binding partners, each of which requires a slightly different conformation for a productive binding interaction (Fig. 4.22). Hence, the disorder itself is key to the protein’s function. An intrinsically disordered protein region may retain some flexibility, even after binding another molecule. If the protein functions as scaffolding for assembling a large multiprotein

4.4 Quaternary Structure

complex, it can adjust its conformation to accommodate each new addition. Intrinsic disorder also means that the complex can adopt more than one structural state. Rather than existing in a fully bound or fully free state, one of these “fuzzy complexes” can exhibit multiple behaviors consistent with different degrees of partial binding. Presumably, such options give cells more ways to fine-tune the functions of proteins. Finally, one newly recognized function of disordered proteins is to aggregate in a tangled network that undergoes liquid-liquid phase separation, generating a droplet with a more gel-like consistency than the surrounding fluid. The protein-rich droplet is sometimes called a membraneless organelle (because conventional organelles are surrounded by a lipid bilayer; Section 8.4). The droplet can harbor additional proteins and other cellular components such as RNA. Membraneless organelles, which may occur in the cytosol or the nucleus, appear to function primarily in organizing and concentrating molecules in order to maximize the efficiency of certain cellular activities, such as processing RNA.

FIGURE 4.22 Schematic view of an intrinsically disordered protein interacting with other proteins. In this diagram, the protein at the center can bind only one other protein at a time; some intrinsically disordered proteins can bind multiple molecules simultaneously.

Before Going On • Explain why a globular protein has a hydrophilic surface and a hydrophobic core. • Name some residues that are likely to be located on the protein surface and some residues that are likely to be located in its core. • List the covalent and noncovalent forces that can stabilize protein structures. • Describe what happens as a polypeptide chain folds. • Explain why proteins are not rigid. • Distinguish monomorphic, metamorphic, and intrinsically disordered proteins and propose some possible functions for each type. • Explain why biochemists know more about globular proteins than disordered proteins.

4.4

107

Quaternary Structure

LEARNING OBJECTIVES List the advantages of having a protein with quaternary structure. • Recognize quaternary structure in proteins. • Explain why large proteins almost always have quaternary structure.

Most small proteins consist of a single polypeptide chain, but most proteins with masses greater than 100 kD contain multiple chains. The individual chains, called subunits, may all be identical, in which case the protein is known as a homodimer, homotrimer, homotetramer, and so on (homo- means “the same”). If the chains are not all identical, the prefix hetero- (“different”) is used. The spatial arrangement of these polypeptides is known as the protein’s quaternary structure. The forces that hold subunits together are similar to those that determine the tertiary structures of the individual polypeptides. That is, the hydrophobic effect is mostly responsible for maintaining quaternary structure. Accordingly, the interface (the area of contact) between two subunits is mostly nonpolar, with closely packed side chains. Hydrogen bonds, ion pairs,

108  C ha pter 4   Protein Structure   FIGURE 4.23   Some proteins with quaternary structure.  The alpha carbon backbone of each polypeptide is shown.  a. Nitrite reductase, an enzyme with three identical subunits, from Alcaligenes.  b. E. coli fumarase, a homotetrameric enzyme.  c. Human hemoglobin, a heterotetramer with two α subunits (blue) and two β subunits (red). d. Bacterial methane hydroxylase, whose two halves (right and left in this image) each contain three kinds of subunits.

Question  Which of the proteins shown in Figure 4.1 has quaternary structure?

a.

b.

c.

d.

and disulfide bonds contribute to a lesser extent but help dictate the exact geometry of the interacting subunits. Among the most common quaternary structures in proteins are symmetrical arrangements of two or more identical subunits (Fig. 4.23). Even in proteins with nonidentical subunits, the symmetry is based on groups of subunits. For example, hemoglobin, a heterotetramer with two αβ units, can be considered to be a dimer of dimers (see Fig. 4.23c). Some metamorphic proteins can form different quaternary structures. The subunits apparently dissociate, change their tertiary structure, then reassociate, sometime forming a complex with a different number of subunits. The advantages of a multisubunit protein structure are many. For starters, extremely large proteins can be constructed by the incremental addition of small building blocks that are encoded by a single gene. For example, the capsid that serves as a protective shell for a virus (see Box 3.C) self-assembles from multiple copies of a few proteins. Herpesviruses are an extreme case, where about 3000 individual proteins snap together to form an almost spherical capsid. Modular construction also makes sense for certain structural proteins that— due to their enormous size—cannot be synthesized all in one piece or must be assembled outside the cell. Moreover, the impact of the inevitable errors in transcription and translation can be minimized if the affected polypeptide is small and readily replaced. Finally, the interaction between subunits in a multisubunit protein affords an opportunity for the subunits to influence each other’s behavior or work cooperatively. The result is a way of regulating function that is not possible in single-subunit proteins or in multisubunit proteins whose subunits each operate independently. In Chapter 5, we will examine the cooperative behavior of hemoglobin, which has four interacting oxygen-binding sites.

Before Going On • Explain how to tell whether a protein has quaternary structure. • Explain why large proteins almost always have quaternary structure.

4.5  Clinical Connection: Protein Misfolding and Disease  109

4.5 Clinical Connection: Protein Misfolding and Disease LEARNING OBJECTIVES Identify the common features of amyloid diseases. • List the possible fates of a misfolded protein. • Describe the overall structure of an amyloid fibril.

Newly synthesized polypeptide

Dennis Selkoe and Marcia Podlisny, Harvard University Medical School

In a cell, a protein may occasionally fail to assume its proper tertiary or quaternary structure. Sometimes this is due to a genetic mutation, such as the substitution of one amino acid for another, that makes it difficult or impossible for the protein to fold. Regardless of the cause, the result may be disastrous for the cell. In fact, a variety of human diseases have been linked to the presence of misfolded proteins. Normally, chaperones help a misfolded protein recover its native conformation. If the protein cannot be salvaged in this way, it is usually degraded to its component amino acids (Fig. 4.24). The operation of this quality control system could explain why some mutated proteins, which are synthesized at a normal rate but fold incorrectly, never reach their intended cellular destinations. In some diseases, misfolded proteins aggregate to form long insoluble fibers. Although the fibers can occur throughout the body, their appearance in the brain is a feature of the deadliest disorders, including Alzheimer’s disease, Parkinson’s disease, and the transmissible spongiform encephalopathies. The aggregated proteins—a different type in each disease—are commonly called amyloid deposits (a name originally referring to their starch-like appearance). At one time, it seemed obvious that fibrous deposits within and between neurons (nerve cells) caused neurological abnormalities and eventually cell death. This made sense because neurons are long-lived cells and the damage could accumulate slowly over time (the diseases mentioned above develop over many years). However, neurodegeneration and symptoms such as memory loss seem to begin before protein aggregates are detectable. Furthermore, almost any protein can be coaxed into forming fibers under the right conditions. These observations seem to rule out a straightforward cause-and-effect scenario. Some evidence suggests that the fibers themselves might not cause disease but instead provide a way to safely detoxify and tuck away misbehaving proteins, particularly when the cell’s normal protein-disposal system is overwhelmed. The real mystery may not be what the fibers do but why they form in the first place. Alzheimer’s disease, the most common neurodegenerative disease, is accompanied by both extracellular “plaques” and intracellular “tangles” in brain tissue (Fig. 4.25). The

misfolding aggregation

folding

Misfolded structure degradation

Native structure

refolding

  FIGURE 4.24   The fates of a misfolded protein.

  FIGURE 4.25   Brain tissue from Alzheimer’s disease.  Amyloid deposits (large red areas) and intracellular tangles (smaller dark shapes) form in parts of the brain responsible for memory and cognition.

  FIGURE 4.26   Model of an amyloid-β fibril containing two filaments.  Each 7-nm-­ diameter filament consists of stacked β strands (flat arrows), with each 42-residue amyloid-β peptide corresponding to one “rung” of the filament.

extracellular amyloid material consists primarily of a protein called amyloid-β, which is a 40- or 42-residue fragment of the much larger amyloid precursor protein located in the cell membrane. Normal brain tissue contains some extracellular amyloid-β, but neither its function nor the function of its precursor protein is completely understood. Curiously, amyloid-β exists as monomers, dimers, and larger insoluble aggregates in equilibrium. Each form behaves differently, but the most toxic form appears to be the dimer. A variety of evidence suggests that amyloid-β might interfere with the operation of membrane transport proteins that take up glutamate (which acts as a signaling molecule between neurons), or it might directly affect membrane integrity in the neurons or the cells that support them. At some point, amyloid-β chains, which are relatively hydrophobic, form extended β strands that stack next to each to generate long fibrils (fine fibers; Fig. 4.26). The intracellular protein known as Tau also forms amyloid fibers in Alzheimer’s disease. Normally, Tau is involved in the assembly of microtubules, a component of the cytoskeleton (Section 5.3). On its own, Tau is an intrinsically disordered protein and for this reason may be prone to aggregate. However, Tau fibrils are not merely an alternate protein conformation, because their formation is irreversible. (Once formed, amyloid fibers tend to be resistant to degradation by cellular enzymes.) Like amyloid-β, a segment of the Tau protein adopts a mostly β secondary structure so that strands align perpendicularly to the fibril axis (Fig. 4.27). Tau fibers accompany a number of neurogenerative diseases other than Alzheimer’s disease, including chronic traumatic encephalopathy from repeated head impact injuries sustained by soldiers in combat or by players of high-contact sports. Different disorders feature Tau fibers with slightly different structures, consistent with chemical modification of the protein, such as phosphorylation. The toxicity of Tau or its fibrils is not understood, but as in other amyloid diseases, symptoms may appear before the fibers are detectable. In Parkinson’s disease, cells in one portion of the brain accumulate fragments of a protein known as α-synuclein, which appears to play a role in neurotransmission. α-Synuclein is a small (140-residue) intrinsically disordered protein that seems to adopt relatively compact structures under cellular conditions. Like other amyloid proteins, it shifts to a β ­conformation—possibly in response to some intra- or extracellular event—when it aggregates. The accumulation of α-synuclein deposits (called Lewy bodies) is associated with the death of neurons, leading to the tremor, muscular rigidly, and slow movements that are symptoms of Parkinson’s disease. The transmissible spongiform encephalopathies (TSEs) are a group of rare fatal disorders in which the brain develops a spongy appearance. Once thought to be caused by a virus, the TSEs are actually triggered by an infectious protein called a prion. Interestingly, normal human brain tissue contains the same 253–amino acid protein, named PrPC (C for cellular), which occurs on neural cell membranes and seems to be required for normal brain function. PrPC has mostly α secondary structure. Introduction of the prion version of the protein, which has mostly β structure, causes PrPC to switch to the same β conformation and aggregate. This generates a stable fiber of hydrogen-bonded β strands that resembles other amyloid fibers. Neurodegeneration may eventually result from the loss of functional PrPC as it misfolds, or from the toxic effects of the amyloid fibers.

β2 V306

β3 β1

β8

β7 β4 β6 β5

F378

  FIGURE 4.27   Model of the filament-forming segments of three Tau peptides.  The eight β strands of each 72-residue segment stack next to the β strands of adjacent Tau polypeptides.

Fitzpatrick et al. 2017/Springer Nature

Gremer et al. 2017/AAAS

110  C ha pter 4   Protein Structure

4.6  Tools and Techniques: Analyzing Protein Structure  111

Prion diseases can be transmitted by ingesting foods derived from infected animals. Somehow, the prion protein must be absorbed, without being digested, and transported to the central nervous system. Intriguingly, there is evidence that Parkinson’s disease may be triggered by changes in the intestinal microbiota (see Section 1.4), suggesting that signs of amyloid diseases might appear in the digestive system before the brain is affected. Early diagnosis would be helpful, because by the time neurological symptoms are obvious, there is no effective treatment.

Before Going On • List some diseases that are associated with the presence of amyloid fibers. • Summarize the evidence suggesting that amyloid fibers cause disease and the evidence suggesting that amyloid fibers are a by-product of disease. • Describe the common structural features of the proteins that form fibrils.

4.6 Tools and Techniques: Analyzing Protein Structure LEARNING OBJECTIVES Describe the techniques for purifying and analyzing proteins. • Explain how chromatography can separate molecules on the basis of size, charge, or specific binding behavior. • Determine the isoelectric points of amino acids. • Summarize how the sequence of a polypeptide can be determined by mass spectrometry. • D  escribe how the three-dimensional arrangement of atoms in a protein can be deduced by analyzing nuclear magnetic resonance or by measuring the diffraction of X-rays or electrons.

Like nucleic acids (Section 3.5), proteins can be purified and analyzed in the laboratory. In this section, we examine some commonly used methods for isolating proteins, determining their sequence of amino acids, and visualizing their three-dimensional structures.

Chromatography takes advantage of a polypeptide’s unique properties As described in Section 4.1, a protein’s amino acid sequence determines its overall chemical characteristics, including its size, shape, charge, and ability to interact with other substances. A variety of laboratory techniques have been devised to exploit these features in order to separate proteins from other cellular components. Keep in mind that the exact charge and conformation of individual molecules vary slightly in a population of otherwise identical mol­ecules, because acid–base groups exist in equilibrium between protonated and unprotonated forms and bonded atoms have some rotational freedom. Consequently, laboratory techniques assess the average chemical and physical properties of the population of molecules. One of the most powerful techniques is chromatography. Originally performed with solvents moving across paper, chromatography now typically uses a column packed with a porous matrix (the stationary phase) and a buffered solution (the mobile phase) that

112  C ha pter 4   Protein Structure

Protein mixture

Small proteins

Large proteins

a.

b.   FIGURE 4.28   Size-exclusion chromatography.  a. Small molecules (blue) can enter the spaces inside the porous beads of the stationary phase, while larger molecules (gold) are excluded.  b. When a mixture of proteins (green) is applied to the top of a sizeexclusion column, the large proteins (gold) migrate more quickly than small proteins (blue) through the column and are recovered by collecting the material that flows out of the bottom of the column. In this way, a mixture of proteins can be separated according to size.

percolates through the column. Proteins or other solutes pass through the column at different rates, depending on how they interact with the stationary phase. In size-exclusion chromatography (also called gel filtration chromatography), the stationary phase consists of tiny beads with pores of a characteristic size. If a solution containing proteins of different sizes is applied to the top of the column, the proteins will move through the column as fluid drips out the bottom. Larger proteins will be excluded from the spaces inside the beads and will pass through the column faster than smaller proteins, which will spend time inside the beads. The proteins gradually become separated and can be recovered by collecting the solution that exits the column (Fig. 4.28). A protein’s net charge at a particular pH can be exploited for its purification by ion exchange chromatography. In this technique, the solid phase typically consists of beads CH2 derivatized with positively charged diethylaminoethyl (DEAE) groups or negatively charged DEAE CH2 CH NH+ 2 carboxymethyl (CM) groups:

CH2

DEAE

CH2

CH2

CH2 NH+ CH2

CH3 CM

CH2

COO−

CH3

Negatively charged proteins − will bind tightly to the DEAE groups, while uncharged and CM CH2 COO positively charged proteins pass through the column. The bound proteins can then be dislodged by passing a high-salt solution through the column so that the dissolved ions can compete with the protein molecules for binding to DEAE groups (Fig. 4.29). Alternatively, the pH of the solvent can be decreased so that the bound protein’s anionic groups become protonated, loosening their hold on the DEAE matrix. Similarly, positively charged proteins will bind to CM groups (while uncharged and anionic proteins flow through the column) and can subsequently be dislodged by solutions with a higher salt concentration or a higher pH. The success of ion exchange can be enhanced by knowing something about the protein’s net charge (see Sample Calculation 4.1) or its isoelectric point, pI, the pH at which it carries no net charge. For a molecule with two ionizable groups, the pI lies between the pK values of those two groups:

pI = 1/2 (pK1 + pK2)

(4.1)

CH3 CH3

4.6  Tools and Techniques: Analyzing Protein Structure  113 High salt

Protein mixture

High salt

Bound proteins Uncharged proteins Cationic proteins

  FIGURE 4.29   Ion exchange chromatography.  When a mixture of proteins is applied to the top of a positively charged anion exchange column (e.g., a DEAE matrix), negatively charged proteins bind to the matrix, while uncharged and cationic proteins flow through the column. The desired protein can be dislodged by applying a high-salt solution (whose anions compete with the protein for binding to the DEAE groups).

Calculating the pI of an amino acid is relatively straightforward (Sample Calculation 4.2). However, a protein may contain many ionizable groups, so although its pI can be estimated from its amino acid composition, its pI is more accurately determined experimentally.

SA MP L E CA LCULAT I O N 4. 2 Problem  Estimate the isoelectric point of arginine. Solution  In order for arginine to have no net charge, its α-­ carboxyl group must be unprotonated (negatively charged), its α-amino group must be unprotonated (neutral), and its side chain must be protonated (positively charged). Because protonation

of the α-amino group or deprotonation of the side chain would change the amino acid’s net charge, the pK values of these groups (9.0 and 12.5) should be used with Equation 4.1: pI = 1/2 (9.0 + 12.5) = 10.75

Other binding behaviors can be adapted for chromatographic separations. For example, a small molecule can be immobilized on the chromatographic matrix, and proteins that can specifically bind to that molecule will stick to the column while other substances exit the column without binding. This technique, called affinity chromatography, is a particularly powerful separation method because it takes advantage of a protein’s unique ability to interact with another molecule, rather than one of its general features such as size or charge. High-performance liquid chromatography (HPLC) is the name given to chromatographic separations, often analytical in nature rather than preparative, that are carried out in closed columns under high pressure, with precisely controlled flow rates and automated sample application. Proteins are sometimes analyzed or isolated by electrophoresis, in which charged molecules move through a gel-like ma­trix such as polyacrylamide under the influence of an electric field. In sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE), both the sample and the gel contain the detergent SDS, which binds to proteins to give them a uniform density of negative charges. When the electric field is applied, the proteins all move toward the positive electrode at a rate depending on their size, with smaller proteins migrating faster than larger ones. After staining, the proteins are visible as bands in the gel (Fig. 4.30).

SEE SAMPLE CALCULATION VIDEOS

114  C ha pter 4   Protein Structure

Mass spectrometry reveals amino acid sequences

−

Bio-Rad Laboratories, ©2012

A standard approach to sequencing a protein has several steps: 1. The sample of protein to be sequenced is purified (for example, by chromatographic or other methods) so that it is free of other proteins. 2. If the protein contains more than one kind of polypeptide chain, the chains are separated so that each can be individually sequenced. In some cases, this requires breaking (reducing) disulfide bonds.

+

3. Large polypeptides must be broken into smaller pieces ( [Na+]in. Sodium ions move into the cell, down their concentration gradient, along with glucose molecules via a symport protein that transports Na+ and glucose simultaneously. Glucose thereby becomes more concentrated inside the cell, which it then exits, down its concentration gradient, via a passive uniport GLUT transporter. Energetically favorable movements are indicated by green arrows; energy-requiring movements are indicated by red arrows. Question  Identify the uniport, symport, and antiport proteins in this diagram.

food and then release it into the bloodstream. Other types of secondary active transporters take advantage of the free energy of a proton gradient, transporting solutes across a membrane along with a proton.

Before Going On • Compare the mechanisms of active and passive transporters. • List the ways that the ATP reaction is used to move solutes across membranes. • Recount the steps of the Na,K-ATPase mechanism.

9.4 Membrane Fusion LEARNING OBJECTIVES Describe the process of membrane fusion. • Summarize the events that occur at a nerve–muscle synapse. • Describe the role of SNARES and membrane curvature in vesicle fusion. • Compare exocytosis and endocytosis. • Describe the role of membranes in autophagy.

9.4  Membrane Fusion  273

The final steps in the transmission of a signal from one neuron to the next, or to a gland or muscle cell, culminate in the release of substances known as neurotransmitters from the end of the axon. Common neurotransmitters include amino acids and compounds derived from them. In the case of a synapse linking a motor neuron and its target muscle cell, the neurotransmitter is acetylcholine:

O H3C

C

CH3 O

CH2

CH2

N+ CH3 CH3

Acetylcholine

Acetylcholine is stored in membrane-bounded compartments, called synaptic vesicles, about 40 nm in diameter. When the action potential reaches the axon terminus, voltage-gated Ca2+ channels open and allow the influx of extracellular Ca2+ ions. The increase in the local intracellular Ca2+ concentration, from less than 1 μM to as much as 100 μM, triggers exocytosis of the vesicles (exocytosis is the fusion of the vesicle with the plasma membrane such that the vesicle contents are released into the extracellular space). The acetylcholine diffuses across the synaptic cleft, the narrow space between the axon terminus and the muscle cell, and binds to integral membrane protein receptors on the muscle cell surface. This binding initiates a sequence of events that result in muscle contraction (Fig. 9.19). In general, the response of the cell that receives a neurotransmitter depends on the nature of the neurotransmitter and the cellular proteins that are activated when the neurotransmitter binds to its receptor. We will explore other receptor systems in Chapter 10. The events at the nerve–muscle synapse occur rapidly, within about one millisecond, but the effects of a single action potential are limited. First, the Ca2+ that triggers neurotransmitter release is quickly pumped back out of the cell by the Ca2+-ATPase. Second, acetylcholine in the synaptic cleft is degraded within a few milliseconds by a lipid-linked or soluble acetylcholinesterase, which catalyzes the reaction

O H3C

C

O

CH2

CH2

H+

H2O

+

N(CH3)3

Acetylcholine

acetylcholinesterase

O H3C

C

Acetate

O− + HO

CH2

CH2

+

N(CH3)3

Choline

Other types of neurotransmitters are recycled rather than destroyed. They are transported back into the cell that released them by the action of secondary active transporters (Box 9.B). A neuron may contain hundreds of synaptic vesicles, only a small percentage of which undergo exocytosis at a time, so the cell can release neurotransmitters repeatedly (as often as 50 times per second).

SNAREs link vesicle and plasma membranes Membrane fusion is a multistep process that begins with the targeting of one membrane (for example, the vesicle) to another (for example, the plasma membrane). A number of proteins participate in tethering the two membranes and readying them for fusion. However, many of these proteins may be only accessory factors for the SNAREs, the proteins that physically pair the two membranes and induce them to fuse.

274  C hapter 9   Membrane Transport Ca2+ channel Action potential

SYNAPTIC CLEFT

AXON

MUSCLE CELL

Acetylcholine receptors

Synaptic vesicles Acetylcholine

Ca2+

1. When an action potential reaches the axon terminus, it causes voltage-gated Ca 2+ channels to open.

2. The increase in intracellular Ca 2+ ion concentration triggers the fusion of synaptic vesicles with the plasma membrane so that the neurotransmitter acetylcholine is released into the synaptic cleft.

3. Acetylcholine binding to receptors on the surface of the muscle cell leads to muscle contraction. The signal is short-lived because acetylcholine remaining in the synaptic cleft is rapidly degraded. Muscle contraction

  FIGURE 9.19   Events at the nerve–muscle synapse.

Question  Describe the events that must occur to restore the neuron and muscle cell to their original states.

SNAREs are integral membrane proteins (their name is coined from the term “soluble N-ethylmaleimide-sensitive-factor attachment protein receptor”). Two SNAREs from the plasma membrane and one from the synaptic vesicle form a complex that includes a 120-Å-long coiled-coil structure containing four helices (two of the SNAREs contribute one helix each, and one SNARE contributes two helices). The four helices, each with about

9.4  Membrane Fusion  275

Box 9.B Antidepressants Block Serotonin Transport The neurotransmitter serotonin, a derivative of tryptophan, is released by cells in the central nervous system.

H3C

+

H+ N H

O

NH3

CF3

HO N H

H3C

H+ H N H

Fluoxetine

H

Cl

Serotonin

Serotonin signaling leads to feelings of well-being, suppression of appetite, and wakefulness, among other things. Seven different families of receptor proteins respond to serotonin signals, sometimes in opposing ways, so the pathways by which this neurotransmitter affects mood and behavior have not been completely defined. Unlike acetylcholine, serotonin is not broken down in the synapse, but instead about 90% of it is transported back into the cell that released it and is reused. Because the extracellular concentration of serotonin is relatively low, it reenters cells via symport with Na+, whose extracellular concentration is greater than its intracellular concentration. The rate at which the serotonin transporter takes up the neurotransmitter helps regulate the extent of signaling. Research suggests that genetic variation in the transporter protein may explain an individual’s susceptibility to conditions such as depression and post-traumatic stress disorder, but such correlations are difficult to prove, since the level of expression of the transporter appears to vary between individuals and even within an individual. Drugs known as selective serotonin reuptake inhibitors (SSRIs) block the transporter and thereby enhance serotonin signaling. Some of the most widely prescribed drugs are SSRIs, including fluoxetine (Prozac®) and sertraline (Zoloft®).

Sertraline

Cl

These drugs are used primarily as antidepressants, although they are also prescribed for anxiety disorders and obsessive-compulsive disorder. Despite decades of research, the interactions between the serotonin transporter and the drugs are not completely understood at the molecular level, and some studies indicate that different inhibitors may bind to different sites on the transporter. The results of rigorous clinical tests suggest that SSRIs are most effective at treating severe disorders, whereas in mild cases of depression, the SSRIs are about as effective as a placebo. One of the challenges in assessing the clinical effectiveness of drugs such as fluoxetine and sertraline is that depression is difficult to define biochemically. Furthermore, serotonin signaling pathways are complex, and the body responds to SSRIs with changes in gene expression and adjustments in other signaling pathways so that antidepressive effects may not be apparent for several weeks. The list of SSRI side effects is long and highly variable among individuals, and, disturbingly, includes a small increase in risk of suicidal behavior. Question  What types of food could contribute to serotonin production in the body?

70 residues, line up in parallel fashion (Fig. 9.20). Unlike other coiled-coil proteins such as keratin (Section 5.3), the four-helix bundle is not a perfectly geometric structure but varies irregularly in diameter. The mutual interactions between SNAREs in the vesicle and plasma membranes serve as an addressing system so that the proper membranes fuse with each other. Initially, the individual SNARE proteins are unfolded, and they spontaneously zip up to form the four-­ helix complex. This action necessarily brings the two membranes close together (Fig. 9.21). Although the formation of the SNARE complex is thermodynamically favorable, membrane fusion may require the presence of additional proteins. The rapid rate of acetylcholine release in vivo indicates that at least some synaptic vesicles are already docked at the plasma membrane, awaiting the Ca2+ signal for fusion to proceed. Experiments with pure lipid vesicles demonstrate that SNAREs are not essential for membrane fusion to occur in vitro, but the rate of fusion does depend on the membranes’

  FIGURE 9.20   Structure of the four-helix bundle of the SNARE complex.  The three proteins

(one includes two helices) are in different colors. Portions of the SNAREs that do not form the helix bundle were cleaved off before X-ray crystallography.

276  C hapter 9   Membrane Transport Vesicle

SEE ANIMATED PROCESS DIAGRAM

SNAREs

Model for SNARE-mediated vesicle fusion

SNARE complex

Plasma membrane

  FIGURE 9.21   Model for SNARE-mediated membrane fusion.  Formation of the complex of SNAREs from the vesicle and plasma membranes brings the membranes close together so that they can fuse.

Question  Why does disassembly of the SNARE complex require ATP?

Vesicle

Plasma membrane

lipid compositions. The explanation for this observation is that the lipid bilayers of the fusing membranes undergo rearrangement: The lipids in the contacting leaflets must mix before a pore forms (Fig. 9.22). Certain types of lipids appear to promote the required changes in membrane curvature. In living cells, bilayer shape changes could be facilitated by the tension exerted by the SNARE complex. In addition, membrane lipids may undergo active remodeling. For example, the enzymatic removal of an acyl chain would convert a cylindrical lipid to a cone-shaped lipid. Clustering of such lipids would cause the bilayer to bow outward.

Removal of lipid tail

Removal of head group

Conversely, removing large lipid head groups would cause the bilayer to bow inward.

Endocytosis is the reverse of exocytosis

  FIGURE 9.22   Schematic view of membrane fusion.  For simplicity, the vesicle and plasma membranes are depicted as bilayers.

When neurotransmitter vesicles fuse with the plasma membrane during exocytosis, the plasma membrane gains membrane proteins and lipids. These materials must be removed and recycled in order to maintain the shape of the neuron and to generate new neurotransmitter vesicles for the next round of neurotransmission. One possibility is that soon after the vesicle and plasma membranes fuse, the “fusion pore” (shown in Fig. 9.22) pinches shut and the vesicle re-forms. The empty vesicle can then be refilled with neurotransmitters. Another mechanism involves endocytosis, in which a new vesicle forms by budding inward from the plasma membrane. This process also requires the membrane to pinch off (the reverse of the process shown in Fig. 9.22) but could occur anywhere on the cell surface. Cells are capable of several kinds of endocytosis (Fig. 9.23). Pinocytosis is the formation of small intracellular vesicles that contain extracellular fluid plus whatever small molecules happen to be present. In receptor-mediated endocytosis, the materials brought into the cell must first bind specifically to a protein receptor on the cell surface. This binding event triggers the membrane shape changes that lead to formation of an intracellular vesicle called an endosome. The endosome may subsequently fuse with a lysosome (or lysosomal enzymes may be delivered to the vesicle) so that its contents can be digested by the lysosomal enzymes. Formation of membranous vesicles—either at the cell surface or involving other membrane systems—often involves “coated” vesicles. So-called coat proteins form a lattice or cage on the cytosolic side of the membrane, forcing the membrane to bud into a spherical shape

9.4  Membrane Fusion  277

EXTRACELLULAR SPACE

CYTOSOL

Receptormediated endocytosis

Pinocytosis

Receptor

Coat proteins

  FIGURE 9.23  Endocytosis.  Pinocytosis captures extracellular materials

nonspecifically. In receptor-mediated endocytosis, a molecule binds to a specific receptor, which is then internalized along with its bound ligand. Coat proteins just beneath the membrane interact with the receptors and help mediate the membrane shape changes required to pinch off a portion of the cell membrane and form a vesicle.

  FIGURE 9.24   Clathrin assembly.  Three clathrin subunits are shown in this model derived from electron micrographs. Multiple subunits assemble to form a lattice that surrounds and defines the shape of an intracellular vesicle.

and then maintaining that structure as the vesicle makes its way to another site in the cell. Vesicle transport often involves a motor protein such as kinesin (Section 5.4), which follows microtubule tracks. Vesicle trafficking between the endoplasmic reticulum and Golgi apparatus, and between the Golgi apparatus and the plasma membrane, involves coated vesicles. One well known coat protein is clathrin, whose spindly subunits assemble into a regular geometric structure around a vesicle (Fig. 9.24). Clathrin and other coat proteins must form strong yet flexible networks that change shape as the vesicle matures and separates from its parent membrane. An interesting type of vesicle formation occurs in the production of exosomes (Box 9.C).

Autophagosomes enclose cell materials for degradation Vesicle formation within the cytoplasm is the basis for autophagy, literally, self-eating. Autophagy, which occurs in all eukaryotes, maintains cellular homeostasis by recycling intracellular components that are no longer needed or are potentially harmful. The undesirable items include normal cytoplasmic contents as well as damaged organelles, aggregated proteins, or certain intracellular pathogens. The rate of autophagy increases in response to certain stressors, such as starvation, hypoxia, and viral infection, and poorly regulated autophagy may underlie some disease states. The process of autophagy begins with membrane budding from the endoplasmic reticulum (the main source of a cell’s lipids; Section 8.4). The resulting vesicle, the phagophore, assumes a cuplike shape as it begins to surround material to be degraded (Fig. 9.25). In some instances, the phagophore randomly encloses a portion of cytoplasm, but in other cases, autophagy specifically targets some unwanted cellular component. The phagophore expands as vesicles deliver additional membrane material, until membrane fusion produces the autophagosome, a compartment surrounded by a double membrane that fully encloses the items to be recycled. Finally, the delivery of lysosomal enzymes generates the autolysosome, in which the captured macromolecules are degraded to monomeric units that can be re-used by the cell. Proteins belonging to the ATG (autophagyrelated gene) family help orchestrate autophagosome maturation and fusion with a

278

C hA PTER 9

Membrane Transport

lysosome, as shown by abnormal autophagy in organisms with ATG mutations. However, the factors that select targets for autophagy and regulate the overall process are not completely understood.

Box 9.C Exosomes Exosomes, also known as microvesicles, are small extracellular membrane-enclosed particles. Although they were once believed to function as a sort of microscopic garbage can containing unwanted cellular components, they are now believed to participate in informal cell–cell communication. Virtually all eukaryotic cells release exosomes, which are typically 30 to 100 nm in diameter and contain an assortment of the cell’s proteins, lipids, and nucleic acids. Rather than budding directly from the plasma membrane, exosomes are formed within an intracellular compartment, which then fuses with the plasma membrane to release the exosomes (see diagram below). Some cells produce exosomes constitutively (constantly), but others seem to release them in response to stress or some other stimulation. In animals, exosomes may circulate throughout the body, so this transport system may be a mechanism for sharing

information about what’s going on inside a certain type of cell. Recipient cells appear to take up exosomes and their cargo by receptor-mediated endocytosis or by direct membrane fusion. The RNA molecules delivered by exosomes may be particularly useful in cell–cell communication, because mRNAs indicate which genes are being expressed and small RNAs help regulate gene expression. Cancer cells release exosomes containing components that could make neighboring cells more accommodating of tumor expansion. Exosomes also appear to play a role in coordinating defenses during an immune response. Because exosomes are present in all body fluids, including blood and urine, they may provide diagnostic information that previously required more invasive procedures such as tissue biopsies. Exosomes have also attracted attention as systems for delivering drugs, gene-edited DNA, or materials that promote tissue healing after injury.

CYTOSOL

Exosomes

Damaged organelle

Endoplasmic reticulum

Vesicles

Phagophore

Lysosome containing hydrolytic enzymes

Autophagosome

Autolysosome

FIGURE 9.25 Autophagy. The cuplike phagophore originates from the endoplasmic reticulum and surrounds unneeded cellular material, such as a damaged organelle. Expansion and membrane fusion produce the autophagosome, a compartment with a double membrane. Fusion with a lysosome introduces hydrolytic enzymes, generating an autolysosome.

Question Explain why lysosomal enzymes must be able to degrade a lipid bilayer in addition to macromolecules.

Key Terms  279

Inadequate autophagy is believed to contribute to the accumulation of misfolded proteins that characterize a number of neurodegenerative diseases (Section 4.5). Similarly, a low rate of autophagy may lead to the pathology of metabolic syndrome (Section 19.3); in this so-called lifestyle disease, cells saturated with sugar and fat may not engage in the ­starvation-induced housecleaning that autophagy normally accomplishes. Conversely, high rates of autophagy have been documented in some types of cancer cells, which apparently use autophagy to maximize the supply of small molecules that support rapid cell growth. I­ nhibiting autophagy is therefore an attractive option for treating some forms of cancer, ­provided that normal rates of autophagy can be maintained in noncancerous cells.

Before Going On • Explain how acetylcholine inside a nerve cell reaches a muscle cell. • Relate the structures of SNARES and clathrin to their functions. • Explain why membrane fusion and changes in bilayer curvature are essential for endocytosis, exocytosis, and autophagy.

Summary 9.1  The Thermodynamics of Membrane Transport •  The transmembrane movements of ions generate changes in membrane potential, Δψ, during neuronal signaling. •  The free energy change for the transmembrane movement of a substance depends on the concentrations on each side of the membrane and, if the substance is charged, on the membrane potential.

9.2  Passive Transport

9.3  Active Transport •  Active transporters such as the Na,K-ATPase and ABC transporters require the free energy of ATP to drive the transmembrane movement of substances against their concentration gradients. •  Secondary active transport allows the favorable movement of one substance to drive the unfavorable transport of another substance.

9.4  Membrane Fusion

•  Passive transport proteins such as porins allow the transmembrane movement of substances according to their concentration gradients. Aquaporins mediate the transport of water molecules. •  Ion channels have a selectivity filter that allows passage of one type of ion. Gated channels open and close in response to some other event. •  Membrane proteins such as the passive glucose transporter undergo conformational changes that alternately expose ligand-binding sites on each side of the membrane.

•  During the release of neurotransmitters, intracellular vesicles fuse with the cell membrane. SNARE proteins in the vesicle and target membranes form a four-helix structure that brings the fusing membranes close together. Changes in bilayer curvature are also necessary for fusion to occur. •  In endocytosis, a vesicle buds off from an existing membrane to form an intracellular vesicle. •  In autophagy, a double-membrane organelle forms around unneeded cellular materials and fuses with a lysosome.

Key Terms membrane potential (Δψ) gas constant (R) Z Faraday constant (F ) action potential axon myelin sheath passive transport active transport porin gated channel

complement osmosis aquaporin uniporter symporter antiporter ATPase ABC transporter secondary active transport neurotransmitter synaptic vesicle

exocytosis endocytosis pinocytosis receptor-mediated endocytosis endosome exosome (microvesicle) autophagy phagophore autophagosome autolysosome

280  C hapter 9   Membrane Transport

Bioinformatics Brief Bioinformatics Exercises 9.1  Maltoporin and OmpF

Problems 9.1  The Thermodynamics of Membrane Transport 1.  Calculate the membrane potential at 20°C when a. [Na+]in = 10 mM and [Na+]out = 100 mM and  b. [Na+]in = 40 mM and [Na+]out = 25 mM. +

2.  Calculate the membrane potential at 20°C when a. [K ] in = 155 mM and [K+]out = 4 mM and  b. [K+]in = 410 mM and [K+]out = 12 mM. 3.  Calculate the intracellular concentration of Na+ in the giant squid axon when the extracellular concentration is 440 mM. Assume a membrane potential of −55 mV at 20°C. 4.  Calculate the extracellular concentration of K+ in frog muscle when the intracellular concentration is 124 mM. Assume a membrane potential of 100 mV at 20°C. 5.  The resting membrane potential maintained in most nerve cells is about −70 mV. Use Equation 9.2 to calculate the ratio of [Na+]in/ [Na+]out at the resting potential. 6.  When a nerve is stimulated, the membrane potential increases from −70 mV to +50 mV. Calculate the ratio of [Na+]in/[Na+]out in the depolarized nerve cell. How does this ratio compare to the ratio you calculated in Problem 5 and what is the significance of the change? 7.  Use Equation 9.4 to calculate the free energy change for the movement of Na+ into a cell at the resting potential (described in Problem 5). Assume the temperature is 37°C. Is this process favorable? 8.  Use Equation 9.4 to calculate the free energy change for the movement of Na+ into the depolarized nerve cell described in Problem 6. Assume the temperature is 37°C. How does this compare with the value you calculated in Problem 7 and what is the significance of the difference? 9.  In typical marine organisms, the intracellular concentrations of Na+ and Ca2+ are 10 mM and 0.1 μM, respectively. Extracellular concentrations of Na+ and Ca2+ are 450 mM and 4 mM, respectively. Use Equation 9.4 to calculate the free energy changes at 20°C for the transmembrane movement of these ions. In which direction do the ions move? Assume the membrane potential is −70 mV. 10.  Calculate the free energy changes at 20°C for the transmembrane movement of Na+ and K+ ions using the conditions presented in Figure 9.1. Assume the membrane potential is −70 mV. In which direction do the ions move? 11.  The concentration of Ca2+ in the endoplasmic reticulum (outside) is 1 mM, and the concentration of Ca2+ in the cytosol (inside) is 0.1  μM. Calculate ΔG at 37°C when the membrane potential is a.  −50 mV (cytosol negative) and  b.  +50 mV. In which case is Ca2+ movement more thermodynamically favorable? 12.  The concentration of Ca2+ in the cytosol (inside) is 0.1 μM and the concentration of Ca2+ in the extracellular medium (outside) is 2 mM. Calculate ΔG at 37°C, assuming the membrane potential is −50 mV. In which direction is Ca2+ movement thermodynamically favored?

13.  Calculate the free energy change for the movement of K+ into a cell when the K+ concentration outside is 15 mM and the cytosolic K+ concentration is 50 mM. Assume that T = 20°C and Δψ = −50 mV (inside negative). Is this process spontaneous? 14.  Calculate the free energy cost of moving Na+ across a membrane from a compartment (outside) where [Na+] = 100 mM to a compartment (inside) where [Na+] = 25 mM. Assume that T = 20°C and Δψ = +50 mV. Is this process spontaneous? 15.  Calculate the free energy change for the movement of Cl− from the extracellular medium, where [Cl−] = 120 mM, into the cytosol, where [Cl−] = 5 mM. Assume that T = 37°C and Δψ = −50 mV. Is this process spontaneous? 16.  The concentration of H+ in the lysosomes (outside) is 32 μM, and the concentration of H+ in the cytosol (inside) is 63 nM. Calculate ΔG at 37°C when the membrane potential is −30 mV (cytosol negative). In which direction is H+ movement thermodynamically favorable? 17.  A high fever can interfere with normal neuronal activity. Since temperature is one of the terms in Equation 9.1, which defines membrane potential, a fever could potentially alter a neuron’s resting membrane potential. a.  Calculate the effect of a change in temperature from 98°F to 104°F (37°C to 40°C) on a neuron’s membrane potential. Assume that the normal resting potential is −70 mV and that the distribution of ions does not change.  b.  How else might an elevated temperature affect neuronal activity? 18.  During mitochondrial electron transport (Chapter 15), protons are transported across the inner mitochondrial membrane from inside the mitochondrial matrix to the intermembrane space. The pH of the mitochondrial matrix is 7.78. The pH of the intermembrane space is 6.88. a.  What is the membrane potential under these conditions?  b.  Use Equation 9.4 to determine the free energy change for proton transport at 37°C. 19. a.  Calculate the value of ΔG for the movement of glucose from outside to inside at 20°C when the extracellular concentration is 5 mM and the cytosolic concentration is 0.5 mM.  b.  What is the free energy cost of moving glucose from the outside of the cell (where its concentration is 0.5 mM) to the cytosol (where its concentration is 5 mM) when T = 20°C? 20.   Calculate the value of ΔG for the movement of fructose from the portal vein circulation (where its concentration is 1 mM) to the inside of a liver cell (where its concentration is 0.1 mM) at 37°C. 21.  Glucose absorbed by the epithelial cells lining the intestine leaves these cells and travels to the liver via the portal vein. After a high-­carbohydrate meal, the concentration of glucose in the portal vein can reach 15 mM. a.  What is the ΔG for transport of glucose from portal vein blood to the interior of the liver cell where the concentration of glucose is 0.5 mM?  b.  What is the ΔG for transport under fasting conditions when the blood glucose level falls to 4 mM? Assume the temperature is 37°C.

Problems  281

23.  Rank the rate of transmembrane diffusion of the following compounds:

O H3C

C

O NH2

H3C

CH2

A. Acetamide

CH2

C

NH2

B. Butyramide

O H2N

C C. Urea

NH2

24.  The permeability coefficient indicates a solute’s tendency to move from an aqueous solvent to a nonpolar lipid membrane. The permeability coefficients for the compounds shown in Problem 23 are listed in the table. How do the permeability coefficients assist you in ranking the rate of the transmembrane diffusion of these compounds? Permeability coefficient (cm · s−1) 9 × 10−6 1 × 10−5 1 × 10−7

Acetamide Butyramide Urea

25.  The permeability coefficients (see Problem 24) of glucose and mannitol for both natural and artificial membranes are shown in the table below. a.  Compare the permeability coefficients for the two solutes (structures shown below). Which solute moves more easily through the synthetic bilayer and why?  b.  Compare the permeability coefficients of each solute for the two types of membranes. For which solute is the difference more dramatic and why? Permeability coefficient (cm · s−1) Glucose Mannitol Synthetic lipid bilayer Red blood cell membrane



2.4 × 10−10 2.0 × 10−4

CHO

4.4 × 10−11 5.0 × 10−9

CH2OH

H

C

OH

HO

C

H

HO

C

H

HO

C

H

H

C

OH

H

C

OH

H

C

OH

H

C

OH

CH2OH Glucose

CH2OH Mannitol

26.  Explain why carbon dioxide can cross a cell membrane without the aid of a transport protein.

9.2  Passive Transport 27.  The bacterium Pseudomonas aeruginosa expresses a phosphatespecific porin when phosphate in its growth medium is limiting. Noting that there are three lysines clustered in the surface-exposed amino terminal region of the protein, investigators constructed mutants in which the lysine residues were replaced with glutamate residues. a.  Why did the investigators hypothesize that lysine residues might play an important role in phosphate transport in the

bacterium?  b.  Predict the effect of the Lys → Glu substitution on the transport activity of the porin. 28.  As noted in the text, the OmpF porin in E. coli has a constricted diameter that prevents the passage of large substances. The protein loops in this constricted site contain a D-E-K-A sequence, which makes the porin weakly selective for cations. If you wanted to construct a mutant porin that was highly selective for calcium ions, what changes would you make to the amino acid sequence at the constricted site? 29.  As a result of the activity of the acetylcholine receptor, muscle cells undergo depolarization, but the change in membrane potential is less dramatic and slower than in neurons. a.  The acetylcholine receptor is also a gated ion channel. What triggers the gate to open?  b.  The acetylcholine receptor/ion channel is specific for Na+ ions. Do Na+ ions flow in or out?  c.  How does the Na+ flow through the ion channel change the membrane potential? 30.  Explain why acid-gated channel proteins include a set of Asp or Glu residues in their acid sensors. How would these groups participate in gating? 31.  When a bacterial cell is transferred from a solute-rich environment to pure water, mechanosensitive channels open and allow the efflux (outflow) of cytoplasmic contents. Use osmotic effects to explain how this prevents cell lysis. 32.  Ammonia, like water, was long believed to diffuse across membranes without the aid of a channel protein. Researchers deleted the gene Rhcg from mice and examined the NH3 permeability of their kidney cells. Cells from wild-type mice exhibited an NH3 flux about three times higher than the mutant cells. a.  What do these results suggest about the role of the protein encoded by Rhcg?  b.  In the light of what you know about aquaporin, does this surprise you? 33.  The selectivity filter in the bacterial chloride channel ClC is formed in part by the OH groups of Ser and Tyr side chains and main chain NH groups. a.  Draw a figure showing plausible interactions between these groups and Cl−.  b.  Explain how these groups would allow Cl− but not cations to pass through. 34.  A channel protein specific for calcium ions was found to have a pore lined with six glutamate residues that participate in coordinating the calcium ion. What would happen to the selectivity of the pore if the Glu residues were mutated to Asp residues? 35.  The kinetics of transport through protein transporters can be described using Michaelis–Menten terminology. The transported substance is analogous to the substrate, and the protein transporter is analogous to the enzyme. KM and Vmax values can be determined to describe the binding and transport process where KM is a measure of the affinity of the transported substance for its transporter and Vmax is a measure of the rate of the transport process. a.  A plot of the glucose transport rate versus glucose concentration for the passive glucose transporter of red blood cells is shown below. Explain why the plot yields a hyperbolic curve.  b.  Estimate the Vmax and the KM values for this glucose transporter. 1.0 Glucose flux (mM . cm−2 . s−1 × 106 )

22.  What is the ΔG for transport of glutamate from the outside of the cell, where the concentration is 0.1 mM, to the inside of the cell, where the concentration is 10 mM? Assume the membrane potential is −70 mV. Is this a spontaneous process?

0.5

0

0

2

4

6

[Glucose] mM

8

10

282  C hapter 9   Membrane Transport

38.  If a propyl group is added to the hydroxyl group on C1 of glucose, the modified glucose is unable to bind to its transporter on the extracellular surface. If the propyl group is added on C6 of glucose, the modified glucose is unable to bind to its transporter on the cytosolic surface of the membrane. What do these observations reveal about the mechanism of passive glucose transport? 39.  Glutamate acts as a neurotransmitter in the brain and is reabsorbed and recycled by glial cells associated with the neurons. The glutamate transporter also transports 3 Na+ and 1 H+ along with glutamate, and it transports 1 K+ in the opposite direction. What is the net charge imbalance across the cell membrane for each glutamate transported? 40.  The bacterium Oxalobacter formigenes is found in the intestine, where it plays a role in digesting the oxalic acid found in some fruits and vegetables (spinach is a particularly rich source). The bacterium takes up oxalic acid from the extracellular medium in the form of oxalate. Once inside the cell, the oxalate undergoes decarboxylation to produce formate, which leaves the cell in antiport with the oxalate. a.  Draw a diagram of this process.  b.  What is the net charge imbalance generated by this system?  c.  Why do you suppose the investigators who elucidated this mechanism referred to the process as a “hydrogen pump”?

−OOC

COO−

Oxalate

decarboxylase

H+

H

COO−

Formate

CO2

41.  As discussed in Section 5.1, tissues produce CO2 as a waste product during respiration. The CO2 enters the red blood cell and combines with water to form carbonic acid in a reaction catalyzed by carbonic anhydrase. A red blood cell protein called Band 3 transports​​ HCO​  3−​  ​​ions in exchange for Cl− ions. a.  What role does Band 3 play in transporting CO2 to the lungs where it can be exhaled?  b.  Band 3 is unusual in that it can transport ​​HCO​ 3−​  ​​in exchange for Cl− ions, in either direction, via passive transport. Why is it important that Band 3 operates in both directions? 42.  A group of investigators was interested in the mechanism of lactate transport in Plasmodium vivax, which causes malaria. They carried out experiments in which they isolated parasites and suspended them in a buffer containing [14C]-lactate at two different pH values, as shown in the figure. Propose a mechanism for lactate transport consistent with the data.

[14C]-Lactate distribution ratio (inside:outside)

37.  Experiments with erythrocyte “ghosts” were carried out to learn more about the glucose transporter in red blood cells. “Ghosts” are prepared by lysing red blood cells in a hypotonic medium and washing away the cytoplasmic contents. Suspension in an isotonic buffer allows the ghost membranes to reseal. If ghosts are prepared so that the enzyme trypsin is incorporated into the ghost interior, glucose transport does not occur. However, glucose transport is not affected if trypsin is located in the extracellular ghost medium. What can you conclude about the glucose transporter, given these observations?

15 pH 7.1

pH 5.5

10

5

0

0

1 Time (min)

2

43.  Glucose uptake by cancer cells via the GLUT1 transport protein occurs far more rapidly than in normal cells. The cancer cell metabolizes the glucose to lactic acid, which exits the cell via the MCT4 transport protein. In response, the activity of carbonic anhydrase (see Section 2.5 and Problem 41), whose catalytic domain is on the outside of the cancer cell, increases. The carbonic anhydrase is associated with a bicarbonate transporter, which assists in maintaining the slightly alkaline pH that cancer cells prefer. a.  Why is it advantageous for the cancer cell to up-regulate the activity of carbonic anhydrase?  b.  List several drug targets that have the potential to interfere with this process and kill the cancer cell. 44.  Liver cells use a choline transport protein to import choline from the portal circulation. Choline transport was measured in mouse cells transfected with the gene for the hepatic transporter. The uptake of radioactively labeled choline by the transfected cells was measured at increasing choline concentrations.

CH3 HO

(CH2)2

+

N

CH3

CH3 Choline a.  Use Michaelis–Menten terminology as described in Problem 35 to estimate KM and Vmax from the curve shown.

Uptake velocity (pmol choline . mg−1 . min−1)

36.  The compound 6-O-benzyl-d-galactose competes with glucose for binding to the glucose transporter. a.  Sketch a curve similar to the one shown in Problem 35 that illustrates the kinetics of glucose transport in the presence of this inhibitor.  b.  How would the KM and Vmax values of 6-O-benzyl-d-galactose compare to those estimated for glucose in Problem 35? Explain.

7 6 5 4 3 2 1 0

0

50

100

150

200

[Choline] (μM)

b.  Plasma concentrations of choline range from 10 to 80 μM, although the concentration may be higher in the portal circulation after ingestion of choline. Does the transporter operate effectively at these concentrations?  c.  The choline transport protein is inhibited by low external pH and stimulated by high external pH. What role might protons play in the transport of choline?  d.  Propose a mechanism that explains how tetramethylammonium ions (TEA, see structure below) inhibit choline transport.

Problems  283 48.  Ouabain, an extract of the East African ouabio tree, has been used as an arrow poison. Ouabain binds to an extracellular site on the Na,K-ATPase and prevents hydrolysis of the phosphorylated intermediate. Why is ouabain a lethal poison?

CH3 CH2 +

CH2 N

CH2

CH3

CH2 CH3 Tetraethylammonium (TEA) 45.  Unidirectional glucose transport into the brain was measured in the presence and absence of phlorizin. The velocity of transport was determined at various glucose concentrations and the data are displayed as a Lineweaver–Burk plot. a.  Calculate the KM and the Vmax (see Problem 35) in the presence and absence of phlorizin.  b. What kind of inhibitor is phlorizin? Explain. 8

1/v (μmol−1 . min . g)

7 6

With inhibitor y = 13.8x + 0.65

5

3 No inhibitor y = 5.4x + 0.64

1 −0.2 −0.1

0

0.1

0.2

1/[Glucose] (mM

0.3

0.4

50.  In eukaryotes, ribosomes (approximate mass 4 × 106 D) are synthesized inside the nucleus, which is enclosed by a double membrane. Protein synthesis occurs in the cytosol. a.  Would a glucose transporter-type protein be capable of transporting ribosomes into the cytosol? Would a porin-type transporter be able to do so? Explain why or why not.  b.  Do you think that free energy would be required to move ribosomes from the nucleus to the cytoplasm? Why or why not? 51.  The retina of the eye contains equal amounts of endothelial and pericyte cells. Basement membrane thickening in pericytes occurs during the early stages of diabetic retinopathy, eventually leading to blindness. Glucose uptake was measured in both types of cells in culture in the presence of increasing amounts of sodium. The results are shown in the graph. a.  How does glucose uptake vary in the two cell types?  b.  What information is conveyed by the shapes of the curves?  c.  By what mechanism might the pericytes use sodium ions to assist with glucose import?

4

2

49.  The parasite Trypanosoma brucei, which causes sleeping sickness, uses proline as an energy source during one stage of its life cycle. The properties of its proline-specific transporter were investigated in a series of experiments. It was discovered that l-hydroxyproline (see Section 5.3) is a potent inhibitor of the transporter whereas d-proline is not, and that proline transport is not affected by pH, Na+, K+, or ouabain (see Problem 48). Incubation of the cells with iodoacetate (which decreases intracellular ATP) decreases the rate of proline transport into the parasite. What do these experiments reveal about the nature of the proline transporter in trypanosomes?

0.5

−1)

46.  Bacterial fermentation of fiber-containing foods in the intestine generates butyrate, a short-chain fatty acid. A carboxylate transporter in the intestinal epithelial cells takes up butyrate, which is used by the cells as an energy source. a.  Estimate the KM and Vmax (see Problem 35) for the carboxylate transporter from the graph below.  b. The carboxylate transporter is regulated by the hormone leptin, which is secreted by adipose tissue and acts as an appetite suppressant. What is the effect of leptin on the KM and Vmax values of the transporter? Explain the significance of these values.

Glucose uptake (nmol . mg−1 . h−1)

H3C

1.6 1.4

Endothelial cells

1.2 1.0

Pericytes

0.8 0.6 0.4 0.2

0

100

Concentration of

15 With leptin Without leptin

10

Na+

150

200

ions (mM)

52.  Use the information provided in the figure to estimate the KM and Vmax (see Problem 35) for glucose uptake by the pericyte transporter described in Problem 51. 0.30

5

0

0

2

4 6 [Butyrate] (mM)

8

10

9.3  Active Transport 47.  The Na,K-ATPase first binds sodium ions and then reacts with ATP to form an aspartyl phosphate intermediate. Draw the structure of the phosphorylated Asp residue.

1/v (nmol−1 . mg . hr)

Uptake (nmol . cm−2 . hr−1)

50

0.25 0.20 0.15 0.10 0.05 0 −0.5

0

0.5

1

1/[S] (mM−1)

1.5

2

284  C hapter 9   Membrane Transport 53.  Many ABC transporters are inhibited by vanadate, a phosphate analog. Why is vanadate an effective inhibitor of these transporters?

you expect this activity to be consistent with fluoxetine’s ability to treat the symptoms of depression?

54.  The ABC multidrug transporter LmrA from Lactococcus exports cytotoxic compounds like vinblastine. There are two binding sites for vinblastine on the transporter; one vinblastine binds with an association constant (equivalent to 1/Kd) of 150 nM; the association constant of the second vinblastine is 30 nM. What do these values tell you about the mode of vinblastine binding to LmrA?

66.  Serotonin and other so-called monoamine signaling molecules are degraded in the liver, beginning with a reaction catalyzed by a monoamine oxidase (MAO). Explain why doctors avoid prescribing an MAO inhibitor along with an SSRI.

55.  Kidney cells include two antiport proteins, an H+/Na+ exchanger and a Cl−/​​HCO​  3−​  ​​exchanger (see Fig. 2.18). What is the source of free energy that drives the transmembrane movement of these ions? 56.  Many cells have a mechanism for exporting ammonium ions. Describe how this could occur through secondary active transport. 57.  The PEPT1 transporter aids in digestion by transporting di- and tripeptides into the cells lining the small intestine. There are three components of this system: (a) a symport transporter that ferries di- and tripeptides across the membrane, along with an H+ ion, (b) an Na+/H+ antiporter, and (c) an Na,K-ATPase. Draw a diagram that illustrates how these three transporters work together to transport peptides into the cell. 58.  The H,K-ATPase in gastric parietal cells is responsible in part for the secretion of hydrochloric acid into the stomach to aid in protein digestion. Parietal cells have two membrane domains: the apical membrane, which faces the stomach lumen, and the basolateral membrane, which has access to the circulatory system. There are five components of the secretion system: (a) the H,K-ATPase antiporter, (b) a Cl−/​​HCO​  3−​  ​​antiporter similar to Band 3 in blood cells (see Problem 41), (c) the Na,K-ATPase, (d) a K+/Cl− symporter, and (e) ­cytosolic ­carbonic anhydrase. Draw a diagram that illustrates how these transporters work to secrete HCl into the stomach lumen.

67.  The toxin produced by Clostridium tetani, which causes tetanus, is a protease that cleaves and destroys SNAREs. Explain why this activity would lead to muscle paralysis. 68.  The drug known as Botox is a preparation of botulinum toxin, which is similar to the tetanus toxin (Problem 67). Describe the biochemical basis for its use by plastic surgeons, who inject small amounts of it to alleviate wrinkling in areas such as around the eyes. 69.  Phosphatidylinositol is a membrane glycerophospholipid whose head group includes a monosaccharide (inositol) group. A kinase adds another phosphate group to a phosphorylated phosphatidylinositol. Why might this activity be required during the production of a new vesicle, which forms by budding from an existing membrane?

O– O

O

P O

H2C O

CH O

OH

CH2

H

H

H HO OH H H

R1 R2

O

OH

kinase

H O– P

O–

O

9.4  Membrane Fusion 59.  Myasthenia gravis is an autoimmune disorder characterized by muscle weakness and fatigue. Patients with the disease generate antibodies against the acetylcholine receptor in the postsynaptic cell; this results in a decrease in the number of receptors. The disease can be treated by administering drugs that inhibit acetylcholinesterase. Why is this an effective strategy for treating the disease? 60.  Lambert–Eaton syndrome is another autoimmune muscle disorder (see Problem 59), but in this disease, antibodies against the voltage-gated calcium channels in the presynaptic cell prevent the channels from opening. Patients with this disease suffer from muscle weakness. Explain why. 61.  Like chymotrypsin, acetylcholinesterase is a member of the serine protease family because it contains an active site serine and, like chymotrypsin, it reacts with DIPF (see Section 6.2). Draw the structure of the enzyme’s catalytic residue modified by DIPF. 62.  Parathion and malathion are organophosphorus compounds similar to DIPF (see Problem 61). These compounds are sometimes used as insecticides. Why are these compounds deadly poisons? 63.  The Na,K-ATPase is essential for establishing the ion gradients that make neuronal signaling possible. Explain why the pump is also required to recycle serotonin. 64.  The illicit drug 3,4-methylenedioxymethylamphetamine (MDMA, also known as Ecstasy) decreases the expression of the serotonin transporter in the brain. Explain how this would alter the MDMA user’s mood. 65.  In addition to being an SSRI, fluoxetine (see Box 9.B) may also bind to and block signaling by one type of serotonin receptor. Would

O– O

O H2C O

CH O

R1 R2

O

P CH2

OH H

H

H HO OH H H

O

O O H O– P

P

O–

O–

O–

O 70.  Some studies show that prior to membrane fusion, the proportion of diacylglycerol in the bilayer increases. Explain how the presence of this lipid would aid the fusion process. 71.  During starvation, yeast cells increase the rate of autophagy. Explain why this makes sense. 72.  An early event in autophagy is the phosphorylation of phosphatidylinositol in the endoplasmic reticulum membrane to generate a lipid with an additional phosphate group in its head group ­(phosphatidylinositol-3-phosphate). Explain how this lipid modification could promote the formation of the phagophore. 73.  Draw a graph (similar to the one shown in Problem 35) that compares the kinetics of substance delivery into the cell via pinocytosis and receptor-mediated endocytosis (see Fig. 9.23). 74.  Low-density lipoprotein (LDL, see Problem 8.78 for a description) enters the cell via receptor-mediated endocytosis by binding to a ­specific cell-surface receptor that localizes to an invaginated

Chapter 9 Credits  285 b. LDL internalized (μg)

area on the membrane surface called a clathrin-coated pit (see Fig. 9.23). Once internalized, the LDL particle is degraded; cholesterol is available to the cell for biosynthetic reactions, and the rate-­limiting enzyme that catalyzes the cell’s own synthesis of cholesterol is inactivated. The experimental data used to elucidate this process in a normal person are shown below. JD, a patient with severe familial hypercholesterolemia, inherited defective alleles from each parent: an allele from his mother that prevented production of LDL receptors and an allele from his father that produced defective LDL receptors. Redraw these graphs to show how the normal process would be altered in this patient and explain your reasoning.

0.1

0.5

0

0

20

40

60

80

100

80

100

LDL (μg . mL−1)

LDL bound (μg)

a. 0.2

0.1

0

0

20

40

60

80

100

LDL (μg . mL−1)

Cholesterol synthesis (nmol . h−1)

c. 0.4

0.2

0

0

20

40

60

LDL (μg . mL−1)

Selected Readings de la Ballina, L.R., Munson, M.J., and Simonsen, A., Lipids and ­lipid-binding proteins in selective autophagy. J. Mol. Biol. 432, 135–159, doi: 10.1016/j.jmb.2019.05.051 (2020). [A thorough review of current knowledge of some of the molecules that direct auto­ phagy, with helpful diagrams.] Deng, D. and Yan, N., GLUT, SGLT, and SWEET: Structural and mechanistic investigations of the glucose transporters, Prot. Sci. 25, 546–558, doi: 10.1002/pro.2858 (2015). [Provides a brief overview of the different types of glucose transport proteins and their physiological significance.] Jahn, R. and Fasshauer, D., Molecular machines governing exocy­tosis of synaptic vesicles, Nature 490, 201–207 (2012). [Reviews many of the events involved in nerve cell signaling, including vesicle fusion.] Morth, J.P., Pederson, B.P., Buch-Pederson, M.J., Andersen, J.P., Vilsen, B., Palmgren, M.G., and Nissen, P., A structural overview of

the plasma membrane Na+, K+-ATPase and H+-ATPase ion pumps, Nat. Rev. Mol. Cell Biol. 12, 60–70 (2011). [Summarizes the structures and physiological importance of two active transporters.] Ozu, M., Galizia, L., Acuña, C., and Amodeo, G., Aquaporins: More than functional monomers in a tetrameric arrangement, Cells 7, 209, doi: 10.3390/cells7110209 (2018). [Reviews aquaporin structure and discusses how the flow of water and other substances is regulated.] Roux, B., Ion channels and ion selectivity, Essays Biochem. 61, 201– 209, doi: 10.1042/EBC20160074 (2017). [Discusses the operation of ion channel proteins, focusing mostly on potassium transport.] Zhang, Y., Liu, Y., Liu, H., and Tang, W.H., Exosomes: Biogenesis, biologic function and clinical potential, Cell Biosci. 9, 19, doi: 10.1186/ s13578-019-0282-2 (2019). [Reviews the formation, activities, and practical applications of exosomes.]

Chapter 9 Credits Figure 9.5 Image based on 1OPF. Cowan, S.W., Schirmer, T., Pauptit, R.A., Jansonius, J.N., The structure of OmpF porin in a tetragonal crystal form, Structure 3, 1041–1050 (1995). Figures 9.6 and 9.7 Images based on 1BL8. Doyle, D.A., Cabral, J.M., Pfuetzner, R.A., Kuo, A., Gulbis, J.M., Cohen, S.L., Chait, B.T., Mackinnon, R., Potassium channel (KcsA) from Streptomyces lividans, Science 280, 69–77 (1998).

Figure 9.10 Image based on 1FQY. Murata, K., Mitsuoka, K., Hirai, T., Walz, T., Agre, P., Heymann, J.B., Engel, A., Fujiyoshi, Y., Structural determinants of water permeation through aquaporin-1, Nature 407, 599–605 (2000). Figure 9.13 Image of GLUT3 (left) based on 4ZWC. Deng, D., Sun, P.C., Yan, C.Y., Yan, N., Crystal structure of maltose-bound human GLUT3 in the outward-open conformation at 2.6 Angstrom, Nature

286  C hapter 9   Membrane Transport 526, 391–396 (2015). Image of GLUT1 (right) based on 4PYP. Deng, D., Yan, C.Y., Xu, C., Wu, J.P., Sun, P.C., Hu, M.X., Yan, N., Crystal structure of the human glucose transporter GLUT1, Nature 510, 121–125 (2014). Figure 9.16 Image based on 4HQJ. Nyblom, M., Poulsen, H., Gourdon, P., Reinhard, L., Andersson, M., Lindahl, E., Fedosova, N., Nissen, P., Crystal structure of Na+, K+-ATPase in the Na+-bound state, Science 342, 123–127 (2013). Figure 9.17 Image based on 3G5U. Aller, S.G., Yu, J., Ward, A., Weng, Y., Chittaboina, S., Zhuo, R., Harrell, P.M., Trinh, Y.T., Zhang, Q.,

Urbatsch, I.L., Chang, G., Structure of P-glycoprotein reveals a molecular basis for poly-specific drug binding, Science 323, 1718– 1722 (2009). Figure 9.20 Image based on 1SFC. Sutton, R.B., Brunger, A.T., Neuronal synaptic fusion complex, Nature 395, 347–353 (1998). Figure 9.24 Image based on 1XI4. Fotin, A., Cheng, Y., Sliz, P., Grigorieff, N., Harrison, S.C., Kirchhausen, T., Walz, T., Molecular model for a complete clathrin lattice from electron cryomicroscopy, Nature 432, 573–579 (2004).

CHAPTER 10

DO YOU REMEMBER? • Some proteins can adopt more than one stable conformation (Section 4.3). • Allosteric regulators can inhibit or activate enzymes (Section 7.3). • Cholesterol and other lipids that do not form bilayers have a variety of other functions (Section 8.1). • Integral membrane proteins completely span the bilayer by forming one or more α helices or a β barrel (Section 8.3). • Conformational changes resulting from ATP hydrolysis drive Na+ and K+ transport in the Na,K-­ATPase (Section 9.3).

Shutterbug75/907 images/Pixabay

Signaling

The taste of a piece of fruit—­sweet or bitter—­depends on whether its component molecules can bind to certain receptors on the surface of cells in the taste buds. By themselves, the molecules and the receptors have no effect, but the appropriate combination of signal plus receptor generates a significant intracellular response.

 ll cells, including prokaryotes, must have mechanisms for sensing external conditions A and responding to them. Because the cell membrane creates a barrier between outside and inside, communication typically involves an extracellular molecule binding to a cell-­surface receptor. The receptor then changes its conformation to transmit information to the cell interior. A signaling pathway requires many proteins, from the receptor itself to the intracellular ­proteins that ultimately respond to the signal by changing their behavior. We begin this chapter by describing some characteristics of signaling pathways and then examine some well-­ known systems that involve G proteins and receptor tyrosine kinases.

10.1 General Features of Signaling Pathways LEARNING OBJECTIVES Summarize the properties of a receptor. • Quantify ligand binding in terms of a dissociation constant. • Recount the events in the two main types of signal transduction. • Describe the factors that limit signaling.

Every signaling pathway requires a receptor, most commonly an integral membrane protein, that specifically binds a small molecule called a ligand. A receptor does not merely bind its ligand in the way that hemoglobin binds oxygen; rather, a receptor interacts with its ligand in

287

288  C ha pter 10   Signaling

TA BLE 1 0 .1   Examples of Extracellular Signals

Hormone

Chemical class

Source

Physiological function

Auxin

Amino acid derivative

Most plant tissues

Promotes cell elongation and flowering in plants

Cortisol

Steroid

Adrenal gland

Suppresses inflammation

Epinephrine

Amino acid derivative

Adrenal gland

Prepares the body for action

Erythropoietin

Polypeptide (165 residues)

Kidneys

Stimulates red blood cell production

Growth hormone

Polypeptide (19 residues)

Pituitary gland

Stimulates growth and metabolism

Nitric oxide

Gas

Vascular endothelial cells

Triggers vasodilation

Thromboxane

Eicosanoid

Platelets

Activates platelets and triggers vasoconstriction

such a way that some kind of response occurs inside the cell. This is signal transduction: the signal itself does not enter the cell, but information is transmitted.

A ligand binds to a receptor with a characteristic affinity Extracellular signals in animals and plants can take many forms, including amino acids and their derivatives, peptides, lipids, and other small molecules (Table 10.1), as well as larger molecules such as glycoproteins. Some signal molecules are formally called hormones, which are substances produced in one tissue that affect the functions of other tissues, but many signal molecules go by other names. For example, interleukins are involved in communication between white blood cells (leukocytes, from the Greek leukos, white). Signal molecules may travel around the entire body or reach only a local area; some signals act on the same cells that produce them. Bacteria often produce small molecules for a form of communication known as quorum sensing: The cells can assess their population density before committing to produce a protective biofilm (described in Section 11.2) or release a toxin. Signaling molecules behave much like enzyme substrates: They bind to their receptors with high affinity, reflecting the structural and electronic complementarity between each ligand and its binding site. Receptor–­ligand binding can be written as a reversible reaction, where R represents the receptor and L the ligand: R + L ​⇌​R · L

1.0

Biochemists express the strength of receptor–­ligand binding as a dissociation constant, Kd, which is the reciprocal of the association constant. For this reaction,

[R . L] [R]T 0.5

[​ ​​R]​ [​​​ ​​L]​ ​​ ​​   ​​  ​ Kd​  ​​ = ______ [​ ​​R  · L​]​​ Kd

[L]   FIGURE 10.1   Receptor–­ligand binding.  As the ligand concentration [L] increases, more receptor molecules bind ligand. Consequently, the fraction of receptors that have bound ligand [R · L] approaches 1.0. [R]T is the total concentration of receptors.

Question  Compare this graph to Figures 5.3 and 7.5.

(10.1)

(See Sample Calculation 10.1.) In keeping with other binding phenomena, such as oxygen binding to myoglobin (Section 5.1) or substrate binding to an enzyme (Section 7.2), Kd is the ligand concentration at which the receptor is half-­saturated with ligand; in other words, half the receptor molecules have bound ligand (Fig. 10.1). A ligand that binds to a receptor and elicits a biological effect is known as an agonist. For example, adenosine is the natural agonist of the adenosine receptor. Adenosine signaling in cardiac muscle slows the heart, and adenosine signaling in the brain leads to a decrease in neurotransmitter release, producing a sedative effect.

10.1  General Features of Signaling Pathways  289

SA M P L E CA LCU LAT I O N 10. 1 Problem  A sample of cells has a total receptor concentration of 10 mM. Twenty-­five percent of the receptors are occupied with ligand and the concentration of free ligand is 15 mM. Calculate Kd for the receptor–­ligand interaction. Solution  Because 25% of the receptors are occupied, [R · L] = 2.5 mM and [R] = 7.5 mM. Use Equation 10.1 to calculate Kd:

[​ ​R]​[​L​]​  ​ ​    ​ ​Kd​  ​= ______ [​ R · L]​

(0.0075​)​(​0.015)      = ​ _____________  ​ (0.0025)



= 0.045 M = 45 mM​

NH2

SEE SAMPLE CALCULATION VIDEOS

N

N

N

N HOH2C

Adenosine

O

H

H

H

OH

OH

H

Caffeine is an antagonist of the adenosine receptor because it binds to the receptor but does not trigger a response.

H3C O

O

CH3 N

N N

N

CH3 Caffeine

It functions like a competitive enzyme inhibitor (Section 7.3). As a re­sult, caffeine keeps the heart rate high and produces a sense of wakefulness. Like caffeine, the majority of drugs currently in clinical use act as agonists or antagonists for various receptors involved in regulating such things as blood pressure, reproduction, and inflammation. Ligands typically bind to a receptor with high affinity and high specificity, but because the ligand–­receptor interactions are noncovalent, binding is reversible. Consequently, a cell responds to a signal molecule—­or a drug—­only while that molecule is associated with its receptor.

Most signaling occurs through two types of receptors When an agonist binds to the adenosine receptor, which is a transmembrane protein, the receptor undergoes a conformational change so that it can interact with an intracellular protein called a G protein. Such receptors are therefore called G protein–­coupled receptors (GPCRs). G proteins are named for their ability to bind guanine nucleotides (GTP and GDP). In response to receptor–­ligand binding, the G protein becomes activated and in turn interacts with and thereby activates additional intracellular proteins. Often, one of these is an enzyme that generates a small molecule product that diffuses throughout the cell. These small molecules are called second messengers because they represent the intracellular response to the extracellular, or first, message that binds to the GPCR. A variety of substances

290  C ha pter 10   Signaling a.

b.

Ligand G protein– coupled receptor

Ligand

Enzyme

Receptor tyrosine kinase

P

G protein (inactive)

Substrate G protein (active)

Second messenger Kinase 1 (inactive)

Target proteins

P

ATP +

ADP + P ATP +

Kinase 2 (inactive)

Kinase 1 (active)

ADP +

Kinase 2 (active) P

Changes in metabolic activity and gene expression Changes in metabolic activity and gene expression   FIGURE 10.2   Overview of signal transduction pathways.  Ligand binding to a cell-­surface

receptor causes a signal to be transduced to the cell interior, leading eventually to changes in the cell’s behavior.  a. Ligand binding to a G protein–­coupled receptor triggers the activation of a G protein, which then activates an enzyme that produces a second messenger. Second messenger molecules diffuse away to activate or inhibit the activity of target proteins in the cell.  b. Ligand binding to a receptor tyrosine kinase activates the kinase activity of the receptor so that intracellular proteins become phosphorylated. A series of kinase reactions activates or inhibits target proteins by adding phosphoryl groups to them. Question  Which pathway components are enzymes?

SEE GUIDED TOUR Signal Transduction

serve as second messengers in cells, including nucleotides, nucleotide derivatives, and the polar and nonpolar portions of membrane lipids. The presence of a second messenger can alter the activities of cellular proteins, leading ultimately to changes in metabolic activity and gene expression. These events are summarized in Figure 10.2a. A second type of receptor, also a transmembrane protein, becomes activated as a kinase as a result of ligand binding. A kinase is an enzyme that transfers a phosphoryl group from ATP to another molecule. In this case, the phosphoryl group is condensed with the hydroxyl group of a tyrosine side chain on a target protein, so these receptors are termed receptor tyrosine kinases. In some signal transduction pathways involving receptor tyrosine kinases, the target protein is also a kinase that becomes catalytically active when phosphorylated. The result may be a series of kinase-­activation events that eventually lead to changes in metabolism and gene expression (Fig. 10.2b). Some receptor systems include both G proteins and tyrosine kinases, and others operate by entirely different mechanisms. For example, the acetylcholine receptor on muscle cells (Fig. 9.19) is a ligand-­gated ion channel. When acetylcholine is released into the neuromuscular synapse and binds to the receptor, Na+ ions flow into the muscle cell, causing depolarization that leads to an influx of Ca2+ ions to trigger muscle contraction.

The effects of signaling are limited The multistep nature of signaling pathways and the participation of enzyme catalysts ensure that the signal represented by an extracellular ligand will be amplified as it is transduced inside the cell (see Fig. 10.2). Consequently, a relatively small extracellular signal can have

10.2  G Protein Signaling Pathways  291

a dramatic effect on a cell’s behavior. The cell’s responses to the signal, however, are regulated in various ways. The speed, strength, and duration of a signaling event may depend on the cellular location of the components of the signaling pathway. There is evidence that components for some pathways are preassembled in multiprotein complexes in or near the plasma membrane so that they can be quickly activated when the ligand docks with its receptor. Components that must diffuse long distances to reach their targets, or that move from the cytoplasm to the nucleus, may need more time to trigger cellular responses. Because signaling pathways tend to be branched rather than completely linear, the same intracellular components may participate in more than one signal transduction pathway, so two different extracellular signals could ultimately achieve the same intracellular results. Alternatively, two signals could cancel each other’s effects. The response of a given cell, which expresses many different types of receptors, therefore depends on how various signals are integrated. Similarly, different types of cells may include different intracellular compo­nents and therefore respond to the same ligand in different ways. In a biological system that obeys the law of homeostasis, any process that is turned on must eventually be turned off. Such control applies to signaling pathways. For example, shortly after a G protein has been activated by its associated receptor, it becomes inactive again. The action of kinases is undone by the action of enzymes that remove phosphoryl groups from target proteins. These and other reactions restore the signaling components to their resting state so that they can be ready to respond again when another ligand binds to its receptor. Finally, most people are aware that a strong odor loses its pungency after a few minutes. This occurs because olfactory receptors, like other types of receptors, become desensitized. In other words, the receptors become less able to transmit a signal even when continually exposed to ligand. Desensitization may allow the signaling machinery to reset itself at a certain level of stimulation so that it can better respond to subsequent changes in ligand concentration.

Before Going On • Compare hormone receptors to enzymes and simple binding proteins. • Explain why an extracellular signal needs a second messenger. • Describe how extracellular signals are amplified inside a cell. • Draw simple diagrams of signaling pathways involving G proteins and receptor tyrosine kinases. • Explain why a receptor would need to be turned off or desensitized.

10.2 G Protein Signaling Pathways LEARNING OBJECTIVES Describe signaling via G protein–­coupled receptors. • Summarize the roles of receptors, G proteins, nucleotides, and lipids in signaling. • Describe how a kinase is activated. • List the mechanisms that limit, compete with, and terminate the G protein signaling pathway. • Explain how the same hormone can elicit different responses in different cells and how different hormones can elicit the same response in a cell. • List some examples of G protein–­coupled receptors and their ligands.

292  C ha pter 10   Signaling

Over 800 genes in the human genome encode G protein–­coupled receptors, and these proteins are responsible for transducing the majority of extracellular signals. In this section we will describe some features of these receptors, their associated G proteins, and various second messengers and their intracellular targets.

G protein–­coupled receptors include seven transmembrane helices

  FIGURE 10.3   The β2-adren-

ergic receptor.  The backbone structure of the protein is colored in rainbow order from the N-­terminus (blue) to the C-­terminus (red). A ligand is shown in space-­ filling form in blue. Question  Explain why the receptor for a polar hormone must be a transmembrane protein.

Koehl A et al 2019/Springer Nature

The GPCRs are known as 7-transmembrane (7TM) receptors because they include seven α helices, which are arranged much like those of the membrane protein bacteriorhodopsin (Fig. 8.8). Many G protein–­coupled receptors are palmitoylated at a cysteine residue, so they are also lipid-­linked proteins (Section 8.3). In the GPCR family, the helical segments are more strongly conserved than the loops that join them on the intracellular and extracellular sides of the membrane. The structure of one of these proteins, the β2-adrenergic receptor, is shown in Figure 10.3. The ligand-­binding site of this GPCR is defined by a portion of the helical core of the protein as well as its extracellular loops. Other GPCRs bind signal molecules in this same general location, with the exact size and polarity of the binding pocket depending on the nature of the signal, which is unique for each receptor. Some GPCRs are equipped with an extracellular protein domain that contributes to the binding site for a ligand, particularly if it is a large peptide. Still other receptors have extensive extracellular domains even though they bind relatively small ligands. In some G protein–­coupled receptors, domains that extend beyond the transmembrane portion of the protein mediate the formation of receptor dimers that are essential for signal transduction (Fig. 10.4). In fact, experiments with receptor proteins in living cells or reconstituted in artificial membranes indicate that many GPCRs may form dimers or oligomers, either with the same or different GPCRs. Although the association in some cases may be transient, lasting up to a few seconds, it may affect the kinetics of ligand binding or signal transduction.

  FIGURE 10.4   Model of the glutamate receptor mGlu5.  The functional receptor is a dimer with two 7TM domains embedded in the membrane (at the bottom). The glutamate binding sites are in the “Venus flytrap” domains (at top), which come together after ligand binding.

10.2 G Protein Signaling Pathways

The physiological ligands for the β2-adrenergic receptor are the hormones epinephrine and norepinephrine, which are synthesized by the adrenal glands from the amino acid tyrosine.

CH3 + H2N

+

H3N

CH2 CH

HO

CH2 CH

OH

OH

HO OH

Epinephrine

OH

Norepinephrine

These same substances, sometimes called adrenaline and noradrenaline, also function as neurotransmitters. They are responsible for the fight-or-flight response, which is characterized by fuel mobilization, dilation of blood vessels and bronchi (airways), and increased cardiac action. Antagonists that prevent signaling via the β2-adrenergic receptor, known as β-blockers, are used to treat high blood pressure. How does the receptor transmit the extracellular hormonal signal to the cell interior? Signal transduction depends on conformational changes involving the receptor’s transmembrane helices. Two of the helices shift slightly to accommodate the ligand, which repositions one of the cytoplasmic protein loops. Studies with a variety of different ligands indicate that the receptor can actually adopt a range of conformations, suggesting that the receptor is not merely an on–off switch but can mediate the effects of strong as well as weak agonists.

The receptor activates a G protein Ligand-induced conformational changes in a G protein–coupled receptor open up a pocket on its cytoplasmic side to create a site where another protein—either a G protein or a protein called arrestin—can bind. The G protein is presumably already close to the receptor, since it has covalently attached lipids that keep it tethered to the cytoplasmic leaflet of the plasma membrane. The trimeric G proteins associated with GPCRs consist of three subunits, designated α, β, and γ (Fig. 10.5); other types of G proteins do not have this three-part structure.

a.

b.

FIGURE 10.5 A GPCR–G protein complex. a. Side view. The GPCR is purple, a bound agonist is red, and the G protein is yellow, green, and blue. b. The G protein β subunit (green) has a propeller-like structure. The small γ subunit (yellow) associates tightly with the β subunit. The α subunit (blue) binds a guanine nucleotide (GDP, orange) in a cleft between two domains.

SEE ANIMATED PROCESS DIAGRAM Overview of heterotrimeric G protein signaling

293

294  C ha pter 10   Signaling

In the resting state, GDP is bound to the G protein’s α subunit, but association with the hormone–­receptor complex causes the G protein to release the GDP and bind GTP in its place. The third phosphate group of GTP is not easily accommodated in the αβγ trimer, so the α subunit dissociates from the β and γ subunits, which remain together. Once they dissociate, the α subunit and the βγ dimer are both active; that is, they interact with additional cellular components in the signal transduction pathway. However, since both the α and β subunits include lipid anchors, the G protein subunits do not diffuse far from the receptor that activated them. The signaling activity of the G protein is limited by the intrinsic GTPase activity of the α subunit, which slowly converts the bound GTP to GDP:

αGDPβγ

inactive GDP Pi GTP

αGTP

+

active

βγ

​GTP + ​H​ 2​O → GDP + ​P​ i​

active

  FIGURE 10.6   The G protein cycle.  The αβγ trimer, with GDP bound to the α subunit, is inactive. Ligand binding to a receptor associated with the G protein triggers a conformational change that causes GTP to replace GDP and the α subunit to dissociate from the βγ dimer. Both portions of the G protein are active in the signaling pathway. The GTPase activity of the α subunit returns the G protein to its inactive trimeric state.

Hydrolysis of the GTP allows the α and βγ units to reassociate as an inactive trimer (Fig. 10.6). The cost for the cell to switch a G protein on and then off again is the free energy of the GTP hydrolysis reaction (GTP is energetically equivalent to ATP).

The second messenger cyclic AMP activates protein kinase A

Although humans have over 800 GPCRs, the number of G proteins is more modest: There are 16 genes for the α subunit, 5 for β, and 12 for γ. Together, these genes code for a small set of trimeric G proteins that interact with various targets in the cell and activate or inhibit them. A single receptor may interact with more than one G protein, so the effects of ligand binding are amplified at this point. One of the major targets of activated G proteins is an integral membrane enzyme called adenylate cyclase. When the α subunit of the G protein binds, the enzyme’s catalytic domains convert ATP to a molecule known as cyclic AMP (cAMP). cAMP is a second messenger that can freely diffuse through the cytosol.

NH2 N

N O −

O

P O

O

O O −

O

P O

−

H

P

O

O

H

H

HO

PPi

N

H H2 C

adenylate cyclase

H

N

N

H N O

H2C

−

NH2

H

O

OH

−

O

OH

ATP

N

N O H O

P

H

H

H

OH

Cyclic AMP (cAMP)

Among the targets of cAMP is an enzyme called protein kinase A or PKA. In the absence of cAMP, this kinase is an inactive tetramer of two regulatory (R) and two catalytic (C) subunits (Fig. 10.7). A segment of each R subunit occupies the active site in a C subunit so that the kinase is unable to phosphorylate any substrates. cAMP binding to the regulatory subunits relieves the inhibition, causing the tetramer to release the two active catalytic subunits.

R C C R

inactive

4 cAMP

R R

+

C

active

+

C

active

In a cell, the kinase subunits may become active even without full dissociation of the tetramer. cAMP functions as an allosteric activator of the kinase, and the level of cAMP determines the level of activity of protein kinase A.

10.2  G Protein Signaling Pathways  295

Protein kinase A is known as a Ser/Thr kinase because it transfers a phosphoryl group from ATP to the serine or threonine side chain of a target protein.

CH2

O

PO2− 3

CH

O

R

PO2− 3

CH3 Phospho-Ser

Phospho-Thr

C

The substrates for the reaction bind in a cleft defined by the two lobes of the protein (Fig. 10.8a). Other kinases share this same core structure but often have additional domains that determine their subcellular location or provide additional regulatory functions. C In addition to regulation by cAMP binding to the R subunit, protein kinase A itself is regulated by phosphorylation. The protein’s so-­called activation loop, a segment of polypeptide near the entrance to the active site, includes a phosphorylatable threonine residue. R When the loop is not phosphorylated, the kinase’s active site is blocked. When phosphorylated, the loop swings aside and the kinase’s catalytic activity increases; for some protein kinases, activity increases by several orders of magnitude. This activation effect is not merely a matter of improving substrate access to the active site but also appears to involve   FIGURE 10.7   Inactive protein conformational changes that affect catalysis. For example, the negatively charged phospho-­ kinase A.  In the inactive complex, Thr interacts with a positively charged arginine residue in the active site. Efficient catalysis the two regulatory subunits (R) block requires that this arginine residue and an adjacent aspartate residue be re-­positioned for the active sites of the two catalytic subphosphoryl-­group transfer from ATP to a protein substrate (Fig. 10.8b). units (C). The targets of protein kinase A include enzymes involved in glycogen metabolism (Section 19.2). One result of signaling via the β2-adrenergic receptor, which leads to protein kinase A activation by cAMP, is the phosphorylation and activation of an enzyme called glycogen phosphorylase, which catalyzes the removal of glucose residues, the cell’s primary metabolic fuel, from glycogen, the cell’s glucose-­storage depot. Consequently, a signal such as epinephrine can mobilize the metabolic fuel needed to power the body’s fight-­or-­flight response. The enzymes that phosphorylate the activation loops of protein kinase A and other cell-­ signaling kinases apparently operate when the kinase is first synthesized, so the kinase is already “primed” and needs only to be allosterically activated by the presence of a second messenger. However, this regulatory mechanism begs the question of what activates the kinase that phosphorylates the kinase. As we will see, kinases that act in series are common in biological signaling pathways, and many of these pathways are interconnected, making it difficult to trace simple cause-­and-­effect relationships.

ATP phospho-Thr Arg

Asp substrate

a.   FIGURE 10.8   Protein kinase A.  a. The backbone of the catalytic subunit is light green, with its activation loop dark green. The phospho-­Thr residue (right side) and ATP (left side) are shown in ball-­and-­ stick form. A peptide that mimics a target protein is blue.  b. Close-­up view of the active site region. When

b. the activation loop is phosphorylated, the phospho-­ Thr residue interacts with an Arg residue, and the adjacent Asp residue is positioned near the third phosphate group of ATP and the peptide substrate. Atoms are color-­coded: C gray or green, O red, N blue, and P gold.

296  C ha pter 10   Signaling

Arrestin competes with G proteins EXTRACELLULAR SPACE

Interacting with a G protein is just one option for the GPCR–­ligand complex. An arrestin protein can occupy the same binding site the G protein uses on the cytoplasmic side of the receptor (Fig. 10.9). Because arrestin competes with the G protein, it hinders or stops the receptor’s ability to signal via the G protein pathway (hence the name arrestin). Arrestins recognize the GPCR–­ ligand complex and undergo a conformational change that exposes Arg and Lys side chains that are particularly effective at binding GPCRs that have been phosphorylated at some point in the signaling process. Consequently, CYTOSOL an arrestin can interact with the receptor in two ways, either of which may activate the arrestin. The activated arrestin can interact with additional intracellular components, including an assortment of kinases. In a way, the arrestin serves as a scaffold or bridge between the enzymes and the receptor, thereby expanding the intracellular effects of the initial hormone signal that docked with the receptor. In fact, the arrestin may remain active even after it dissociates from the G protein–­coupled receptor, prolonging signaling while the recep  FIGURE 10.9   A receptor–­arrestin complex.  tor interacts with additional arrestins or G proteins. Rhodopsin, a model G protein–­coupled receptor, Arrestin binding to a G protein–­coupled receptor also promotes the is shown as a purple ribbon, and the intracellular receptor’s internalization by endocytosis (Section 9.4). This diminishes arrestin protein as a pink ribbon. the cell’s ability to receive subsequent hormone signals and accounts for Question  Compare this figure and Figure 10.5a. the early hypothesis that arrestin functioned simply as an off switch for G protein–­coupled receptors. Internalized receptors may retain some signal transduction activity inside the cell. The significance of this intracellular signaling is not known, although GPCRs have been identified on a variety of intracellular membranes, including the nuclear membrane and mitochondrial membrane, well out of reach of extracellular signals.

Signaling pathways must be switched off What happens after a ligand binds to a receptor, a G protein or an arrestin responds, a second messenger is produced, an effector enzyme such as a protein kinase is activated, and target proteins are phosphorylated? To restore the cell to a resting state, any or all of the events of the signal transduction pathway can be blocked or reversed. First, the dissociation of an extracellular ligand from its receptor can halt a signal transduction process, or the receptor can become desensitized. Desensitization of a G protein–­coupled receptor begins with phosphorylation of the ligand-­bound receptor by a GPCR kinase, followed by arrestin binding and endocytosis. Even without arrestin, G protein signaling is limited. We have already seen that the intrinsic GTPase activity of G proteins stops their activity. Moreover, second messengers often have short lifetimes due to their rapid degradation in the cell. For example, cAMP is hydrolyzed by the enzyme cAMP phosphodiesterase:

NH2 N

N H H2 C O −

O

OH P

NH2 H N

N O H O cAMP

H2O phosphodiesterase

H

O −

O

H

P

O

OH

O

H

H N

N O

H2C

−

H

N

N

H

HO AMP

H

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OH

10.2  G Protein Signaling Pathways  297

Caffeine, for example, in addition to being an adenosine receptor antagonist, can diffuse inside cells and inhibit cAMP phosphodiesterase. As a result, the cAMP concentration remains high, the action of protein kinase A is sustained, and stored metabolic fuels are mobilized, readying the body for action rather than sleep. Some of the cell’s G proteins may inhibit rather than activate adenylate cyclase and therefore decrease the cellular cAMP level. Some G proteins activate cAMP phosphodiesterase, with similar effects on cAMP-­dependent processes. A cell’s response to a hormone signal depends in part on which G proteins respond. Because a single type of hormone may stimulate several types of G proteins, the signaling system may be active for only a brief time before it is turned off. The phosphorylations catalyzed by protein kinase A and other kinases can be reversed by the work of protein phosphatases, which catalyze a hydrolytic reaction to remove phosphoryl groups from protein side chains. Like kinases, phosphatases are generally specific for serine or threonine, or tyrosine, although some “dual specificity” phosphatases remove phosphoryl groups from all three side chains. The active-­site pocket of the Tyr phosphatases is deeper than the pocket of Ser/Thr phosphatases in order to accommodate the larger phospho-­ Tyr side chain. Some protein phosphatases are transmembrane proteins; others are entirely intracellular. They tend to have multiple domains or multiple subunits, consistent with their ability to form numerous protein–­protein interactions and participate in complex regulatory networks.

The phosphoinositide signaling pathway generates two second messengers The diversity of G protein–­coupled receptors, together with the diversity of G proteins, creates almost unlimited possibilities for adjusting the levels of second messengers and altering the activities of cellular enzymes. Epinephrine, the hormone that activates the β2-adrenergic receptor, also binds to a receptor known as the α-­a drenergic receptor, which uses the ­phosphoinositide signaling system. The α-­and β-­adrenergic receptors are situated in different tissues and mediate different physiological effects, even though they bind the same hormone. The G protein associated with the α-­adrenergic receptor activates the cellular enzyme phospholipase C, which acts on the membrane lipid phosphatidylinositol bisphosphate. Phosphatidylinositol is a minor component of the plasma membrane (4–5% of the total phospholipids), and the bisphosphorylated form (with a total of three phosphate groups) is rarer still. Phospholipase C converts this lipid to inositol trisphosphate and diacylglycerol.

H

O −

P

O

O CH2

CH

O

O

C

OC

R1

CH2 O

O H

OPO2− 3

OH H OH HO H

H −2

H OPO2− 3

H

Phosphatidylinositol bisphosphate

H2O phospholipase C

O3PO H

OPO2− 3

OH H OH HO H

H OPO2− 3

H

Inositol trisphosphate

R2

The highly polar inositol trisphosphate is a second messenger that can directly activate kinases. It also undergoes further phosphorylation to generate an additional second messenger containing four phosphoryl groups. However, its key activity seems to be triggering the opening of calcium channels in the endoplasmic reticulum membrane, which allows Ca2+ ions to flow into the cytosol. The influx of Ca2+ initiates numerous events in the cell, including the activation of a Ser/Thr kinase known as protein kinase B or Akt.

OH CH2

CH

+ O

O

C

OC

R1

CH2 O

R2

Diacylglycerol

298  C ha pter 10   Signaling

a.

b.   FIGURE 10.10  Calmodulin.  a. Isolated calmodulin has an extended shape. The four bound Ca2+ ions are shown as blue spheres.  b. When bound to a target protein (blue helix), calmodulin’s long central helix unwinds and bends so that the protein can wrap around its target.

In some cases where Ca2+ ions alter an enzyme’s activity, the change is mediated by a Ca2+-binding protein known as calmodulin. This small (148-residue) protein binds two Ca2+ ions in each of its two globular domains, which are separated by a long α helix (Fig. 10.10a). Free calmodulin has an extended shape, but in the presence of Ca 2+ and a target protein, the helix partially unwinds and calmodulin bends in half to grasp the target protein to activate or inhibit it (Fig. 10.10b). The hydrophobic diacylglycerol product of the phospholipase C reaction is also a second messenger. Although it remains in the cell membrane, it can diffuse laterally to activate protein kinase C, which phosphorylates its target proteins at serine or threonine residues. In its resting state, protein kinase C is a cytosolic protein with an activation loop blocking its active site. Noncovalent binding to diacylglycerol docks the enzyme at the membrane surface so that it changes its conformation, repositions the activation loop, and becomes catalytically active. As in protein kinase A (whose sequence is about 40% identical), phosphorylation of a threonine residue in the activation loop, a requirement for catalytic activity, has already occurred. Full activation of some forms of protein kinase C also requires Ca2+, which is presumably available as a result of the activity of the inositol trisphosphate second messenger. Among the targets of protein kinase C are proteins involved in the regulation of gene expression and cell division. Phospholipase C can be activated not only by a G protein–­coupled receptor such as the α-­adrenergic receptor but also by other signaling systems involving receptor tyrosine kinases. This is an example of cross-­talk, the interconnections between signaling pathways that share some intracellular components. The phosphoinositide signaling pathway is regulated in part by the action of lipid phosphatases that remove phosphoryl groups from the phosphatidylinositol bisphosphate substrate that gives rise to the second messengers. Another example of the overlap between signaling pathways involves sphingolipids such as sphingomyelin (Fig. 8.2), which is a normal component of membranes. Ligand binding to certain receptor tyrosine kinases leads to activation of sphingomyelinases that release sphingosine and ceramide (ceramide is sphingomyelin without its phosphocholine head group). Ceramide is a second messenger that activates kinases, phosphatases, and other cellular enzymes. Sphingosine, which undergoes phosphorylation (by a receptor tyrosine kinase–­dependent mechanism) to sphingosine-1-phosphate, is both an intracellular and extracellular signaling molecule. It inhibits phospholipase inside the cell, apparently exits the cell via an ABC transport protein (Section 9.3), and then binds to a G protein–­coupled receptor to trigger additional cellular responses.

Many sensory receptors are GPCRs As estimated one-­t hird of therapeutic drugs—­including antihistamines, antipsychotics, and opioid painkillers—­target G protein–­c oupled receptors and their associated proteins (Box 10.A). GPCRs also respond to light, odors, and tastes. In humans, olfaction, the sense of smell, depends on roughly 400 GPCRs (some mammals have over 1000). Each sensory neuron in the nose expresses just one type of receptor, but a given ligand, called an odorant, can bind to several different receptors, so the perception of an odor depends on the signaling activity in multiple neurons. This combinatorial system allows humans to distinguish hundreds of thousands of different odorants despite a somewhat limited set of receptors. When an odorant binds to an olfactory receptor, its associated G protein activates aden­ ylate cyclase, and the resulting cAMP second messenger opens a nonspecific ion channel, which facilitates the influx of Na+ and Ca2+ ions. This, in turn, triggers the opening of a Ca2+-dependent chloride channel, and the ensuing exit of Cl –­ions further depolarizes the cell. These changes generate the action potential that travels the length of the neuron toward the brain. In animals, gustation, the sense of taste, works a bit differently. First, neurons themselves do not have taste receptors but instead receive input indirectly from epithelial cells located throughout the mouth but mostly in taste buds on the tongue. Of the five taste qualities—­sweet, salty, sour, bitter, and umami—­two are not actually detected by receptors. The perception of sourness (acidity) depends mainly on the ability of protons to affect ion

10.2  G Protein Signaling Pathways  299

Box 10.A Opioids Morphine is arguably the world’s most powerful and most abused drug. Produced by the opium poppy Papaver somniferum, its usefulness lies in its ability to relieve pain without loss of consciousness. Morphine has been used for millennia; it was the major ingredient of laudanum, popular in the eighteenth and nineteenth centuries. The poppy metabolite codeine, which is readily converted to morphine by cytochrome P450 in the liver (Section 7.4), was until recently a common ingredient in cough suppressants. Morphine along with its natural and synthetic derivatives, such as heroin, oxycodone, methadone, and fentanyl, are collectively known as opioids.

CH3 N

HO

O

Morphine

OH

Morphine and other opioids bind to three different types of G protein–­coupled receptors that are distributed in an overlapping fashion in many parts of the brain as well as in other tissues, notably the intestine (which helps explain the gastrointestinal symptoms associated with opioid use). The receptors interface with arrestin and with an inhibitory G protein that diminishes cAMP production within the cell. G protein–­mediated signaling also increases K+ channel function that hyperpolarizes the cell, blocks Ca2+ transport, and inhibits neurotransmitter release. This

set of responses effectively reduces the transmission of pain signals. Opioid binding to the receptor also activates arrestin, which leads to internalization of the receptor and counters the G protein effects. The human body naturally produces about 30 endogenous opioids, including the enkephalins and endorphins, peptides that have morphinelike effects and are presumably essential for regulating normal pain perception, mood, and other functions. Administering exogenous opioids replicates the pain-­relieving and mood-­elevating effects of the endogenous signals but leads to drug dependence because the body stops manufacturing its own opioids. When an opioid drug is discontinued, the individual experiences severe withdrawal symptoms that include very real physical distress, not just a sensation of craving. As with any drug, chronic use leads to tolerance, an adaptation that requires ever-­increasing doses to achieve the same effect. Unfortunately, in addition to blocking pain, opioids suppress the respiratory center in the brain, which controls breathing. Because tolerance to a drug’s pain-­relieving and euphoric effects develops faster than tolerance to respiratory depression, opioid-­addicted individuals are at risk of ingesting a dose that causes them to stop breathing and die. Understanding ligand–­receptor systems suggests some possible solutions to opioid addiction and overdosing. For example, administering an opioid receptor antagonist such as naloxone, which quickly enters the brain and competes with agonists for binding to the opioid receptors, can quickly reverse the effects of an opioid overdose. Newly developed opioid drugs are designed to exploit the subtle conformational differences in the ligand–­ receptor complex in order to bias the system toward the G protein pathway for analgesia (pain relief) rather than the arrestin pathway that favors drug tolerance and increases the risk of respiratory side effects.

channels, opening Na+ channels and closing K+ channels. The cell becomes depolarized, Ca2+ channels open, and neurotransmitters are released. Nearby neurons then convey the signal to the brain. In a similar fashion, saltiness is detected by the opening of sodium channels. Curiously, low concentrations of Na+ are perceived as appetizing, whereas very high concentrations are distasteful. The “hot” substances in peppers, wasabi, and mustard bind to receptors that perceive pain or heat. The G protein–­c oupled receptors that respond to sweet, bitter, and umami tastes, although not as numerous as olfactory receptors, can respond to a few thousand different ligands, called tastants. Ligand binding to these GPCRs initiates the phosphoinositide pathway. Inositol trisphosphate opens Ca2+ channels; this leads to the opening of Na+ channels and cell depolarization. Voltage-­gated Ca2+ channels then open, and the influx of Ca2+ triggers neurotransmitter release that activates a sensory neuron. Sweet substances bind to a dimeric GPCR that resemble the brain’s glutamate receptor (Fig. 10.4), with subunits known as TAS1R2 and TAS1R3 (Fig. 10.11). Umami receptors, which bind substances such as glutamate, aspartate, and the purine nucleotides IMP and GMP, are dimers of TAS1R1 and TAS1R3. Humans also have 25 types of bitter taste receptor subunits, the TAS2Rs, which collectively appear able to recognize a large number of agonists, although with relatively low affinity. Somewhat nonspecific binding could be advantageous for detecting a wide array of potentially toxic substances, which tend to have a bitter taste. Surprisingly, olfactory receptors and taste receptors have been identified in many nonsensory organs in the body, but these GPCRs do not necessarily respond to conventional odorants or tastants or transmit information to the brain. Their function is still mostly a puzzle.

300  C ha pter 10   Signaling

Before Going On • Describe the structure of a G protein–­coupled receptor. • Describe the activity cycle of a G protein. • Draw a diagram to illustrate the steps of signaling via a G protein, from hormone–­ receptor binding to second messenger degradation. Include arrestin. • Draw a diagram to illustrate the phosphoinositide signaling pathway. • For each pathway, identify points where signaling activity can be switched off. Kim S-K et al. 2017/PNAS

• Explain why it is an advantage for receptors and G proteins to be lipid-­linked. • Compare the relative concentrations of extracellular signal molecules and second messengers. • Discuss the advantages and disadvantages of cross-­talk. • Propose an evolutionary explanation for the ability to sense sweet, salty, sour, bitter, and umami substances.   FIGURE 10.11   Model of the TAS1R2–TAS1R3 sweet receptor.  This structure has a more “closed” conformation than the dimeric glutamate receptor shown in Figure 10.4.

Question  Identify the ligand-­ binding and transmembrane domains.

10.3 Receptor Tyrosine Kinases LEARNING OBJECTIVES Describe the receptor tyrosine kinase signaling pathway. • Compare G protein–­coupled receptors and receptor tyrosine kinases. • Distinguish the two mechanisms by which receptor tyrosine kinases activate target proteins. • Explain how kinases and transcription factors mediate cellular responses over different time scales.

A number of hormones and other signaling molecules that regulate cell growth, division, and immune responses bind to cell-­surface glycoproteins that operate as tyrosine kinases. These receptors are typically built from protein subunits containing an extracellular ligand-­binding site, a single membrane-­spanning helix, and an intracellular tyrosine kinase domain. At one time, ligand binding was believed to bring two separate monomeric receptors together to form a catalytically active dimer, but the current understanding is that the receptor subunits are often already associated but inactive. Ligand binding induces conformational changes that bring the receptor subunits closer together, in a scissor-­like rearrangement. The intracellular tyrosine kinase domains, which are far apart in the absence of ligand, are repositioned close enough so that each can phosphorylate the other, a key step in signal transduction. Receptor tyrosine kinases function as homodimers or heterodimers, depending on the type of ligand. Each monomer has a ligand-­binding site, and many structural studies reveal symmetric complexes, typically with two ligands per receptor dimer. However, much remains unknown about the exact sequence of events that link the binding of the first ligand, the binding of the second, changes in quaternary structure, and activation of the tyrosine kinase domains. Like G protein–­coupled receptors, receptor tyrosine kinases are likely to adopt a series of intermediate structures—­including asymmetric complexes—­in the transition from inactive to fully active.

The insulin receptor dimer changes conformation The insulin receptor is one of approximately 60 receptor tyrosine kinases in humans. Cells throughout the body express the receptor, which recognizes insulin, a 51-residue polypeptide hormone that regulates many aspects of fuel metabolism. The receptor is constructed from two long polypeptides that are cleaved after their synthesis, so the mature receptor has

10.3 Receptor Tyrosine Kinases

an (αβ)2 structure in which all four polypeptide segments are held together by disulfide bonds (Fig. 10.12a). Each αβ unit functions as one monomer. The extracellular ligand-binding portion of the insulin receptor has an inverted V shape with multiple structural domains (Fig. 10.12b). Once the first insulin molecule makes contact, the receptor may “explore” additional conformations before the second contact is established. This process requires segments of the α chain to dramatically reorient. When fully bound, the insulin interacts with portions of each αβ monomer. According to some models, the second insulin molecule binds with a lower affinity (an example of negative cooperative behavior; Section 5.1). By this point, the receptor–insulin complex is T-shaped, a conformation that brings the transmembrane helices and their attached tyrosine kinase domains close together (Fig. 10.13). Two more insulin molecules appear to bind cooperatively to two additional low-affinity sites to stabilize the receptor in a fully active conformation. The existence of four discrete insulin-binding sites on one receptor raises the possibility that different signals could be transduced across a range of physiological insulin concentrations. α

α

β

β

Transmembrane helix

Tyrosine kinase domain

a.

b.

FIGURE 10.12 Structure of the insulin receptor. a. Schematic diagram showing the two α and two β subunits joined by disulfide bonds (horizontal lines). b. Model of the extracellular portion of the receptor, with the structural domains shown in the same colors as in part a. The cell surface is at the bottom.

Insulin

FIGURE 10.13 Conformational changes in the insulin receptor. This diagram is based on X-ray and cryo-EM models of the insulin receptor. Two bound insulin molecules are shown. The structural domains are colored as in Figure 10.12a.

301

302  C ha pter 10   Signaling

The receptor undergoes autophosphorylation Inside the cell, the two tyrosine kinase domains phosphorylate each other, using ATP as a source of the phosphoryl groups. Because the receptor subunits appear to phosphorylate themselves, this process is termed autophosphorylation. Each tyrosine kinase domain has the typical kinase core structure, including an activation loop that lies across the active site to prevent substrate binding. In the insulin receptor, phosphorylation of three tyrosine residues in the kinase activation loop causes the loop to swing aside to better expose the active site (Fig. 10.14). This conformational change allows the enzyme to interact with additional protein substrates and to transfer a phosphoryl group from ATP to a tyrosine side chain on these target proteins. Not all receptor tyrosine kinases phosphorylate other proteins; some activate their intracellular target by other mechanisms. For example, to initiate the responses that promote cell growth and division, many growth factor receptors phosphorylate various intracellular target proteins. They also switch on other pathways that involve small monomeric G proteins (which are also GTPases) such as Ras. The tyrosine kinase domain of the receptor does not directly interact with Ras but instead relies on one or more adapter proteins that form   FIGURE 10.14   Activation of the insulin a bridge between Ras and the phospho-­Tyr residues of the receptor (Fig. 10.15). receptor tyrosine kinase.  The backbone These proteins also stimulate Ras to release GDP and bind GTP. structure of the inactive tyrosine kinase domain Like other G proteins, Ras is active as long as it has GTP bound to it. The of the insulin receptor is shown in light blue, with Ras · GTP complex allosterically activates a Ser/Thr kinase, which becomes the activation loop in dark blue. The structure active and phosphorylates another kinase, activating it, and so on. Cascades of of the active tyrosine kinase domain is shown in several kinases can therefore amplify the initial growth factor signal. light green, with the activation loop in dark green. The ultimate targets of Ras-­dependent signaling cascades are nuclear The phospho-­Tyr side chains are shown in stick proteins, which, when phosphorylated, bind to specific sequences of DNA to form with atoms color coded: C green, O red, and P orange. induce (turn on) or repress (turn off) gene expression. The altered activity of these transcription factors means that the original hormonal signal not only Question  How would a phosphatase affect alters the activities of cellular enzymes on a short time scale (seconds to minthe activity of the insulin receptor? utes) via phosphorylation but also affects protein synthesis, a process that may require several hours or more. Ras signaling activity is shut down by the action of proteins that enhance the GTPase activity of Ras so that it returns to its inactive GDP-­bound form. In addition, phosphatases reverse the effects of the various kinases. The hormone receptor itself may eventually be inactivated by phosphorylation or dephosphorylation or removed from the cell surface by endocytosis. Like the other signaling pathways we have examined, the receptor tyrosine kinase

Receptor tyrosine kinase

Ligand binding to the receptor triggers autophosphorylation of the tyrosine kinase domains.

Adapter proteins bridge the receptor and Ras and induce Ras to release GDP and bind GTP.

The active Ras can now initiate a kinase cascade.

Ligand

GDP

P

P

Ras (inactive)

GDP

P

P Adapter proteins Kinase (inactive)

GTP Ras (active)

P

P

ATP +

GTP

ADP +

Kinase (active)

P

  FIGURE 10.15   The Ras pathway.  Ras links receptor–­hormone binding to an intracellular

kinase cascade.

Question  Describe the role of protein conformational changes in each step of the pathway.

10.4  Lipid Hormone Signaling  303

Box 10.B Cell Signaling and Cancer The progress of a cell through the cell cycle, from DNA replication through the phases of mitosis, depends on the orderly activity of signaling pathways. Cancer, which is the uncontrolled growth of cells, can result from a variety of factors, including overactivation of the signaling pathways that stimulate cell growth. In fact, the majority of cancers include mutations in the genes for proteins that participate in signaling via Ras and the phosphoinositide pathways. These altered genes are termed oncogenes, from the Greek onkos, meaning “tumor.” Oncogenes were first discovered in certain cancer-­ causing vi­ruses. The viruses presumably picked up the normal gene from a host cell, then mutated. In some cases, an oncogene encodes a growth factor receptor that has lost its ligand-­b inding domain but retains its tyrosine kinase domain. As a result, the kinase is constitutively (constantly) active, promoting cell growth and division even in the absence of the growth factor. Some RAS oncogenes generate mutant forms of Ras that hydrolyze GTP extremely slowly, thus maintaining the signaling pathway in the “on” state. Note that oncogenic mutations can strengthen an activating event or weaken an inhibitory event; in either case, the outcome is excessive signaling activity. The importance of various kinases in triggering or sustaining tumor growth has made these enzymes attractive targets for anti­ cancer drugs. Some forms of leukemia (a cancer of white blood cells) are triggered by a chromosomal rearrangement that generates a kinase with constitutive signaling activity, called Bcr-­Abl.

The drug imatinib (Gleevec®) specifically inhibits this kinase without affecting any of the cell’s numerous other kinases. The result is an effective anticancer treatment with few side effects.

N

H

H

N

N

NH C 3

N

N C

N

CH3

O

Gleevec® (imatinib)

The engineered antibody known as trastuzumab (Herceptin®) binds as an antagonist to a growth factor receptor that is overexpressed in many breast cancers. Other antibody-­based drugs target similar receptors in other types of cancers. Clearly, understanding the operation of growth-­signaling pathways—­both normal and mutated—­is essential for the ongoing development of effective anticancer treatments. Question  Using Figures 10.2 and 10.15 as a guide, identify other types of signaling proteins that are potential anticancer drug targets.

pathways are not linear and they are capable of cross-­talk. For example, some receptor tyrosine kinases directly or indirectly (via Ras) activate the kinase that phosphorylates phosphatidylinositol lipids, thereby promoting signaling through the phosphoinositide pathway. Abnormalities in these signaling pathways can promote tumor growth (Box 10.B).

Before Going On • Draw a diagram to illustrate receptor tyrosine kinase signaling, including tyrosine phosphorylation and Ras activation. • Explain how the free energy of ATP and GTP hydrolysis is used to turn on cellular responses to extracellular signals. • Compare kinases and transcription factors in terms of the time required for their effects. • Compare phosphorylation and proteolysis (Section 6.4) as mechanisms for activating an enzyme.

10.4 Lipid Hormone Signaling LEARNING OBJECTIVES Compare lipid signaling to other signal transduction pathways. • Recognize lipid hormones. • Describe how lipid hormones regulate gene expression. • Explain how eicosanoids differ from other signaling molecules.

304  C ha pter 10   Signaling

Some hormones do not need to bind to cell-­surface receptors because they are lipids and can cross the membrane to interact with intracellular recepO− tors. For example, retinoic acid and the thyroid hormones thyroxine (T4) and triiodothyronine (T3) belong to this class of hormones (Fig. 10.16). Retinoic acid (retinoate), a compound that regulates cell growth and CH3 differentiation, particularly in the immune system, is synthesized from Retinoate retinol, a derivative of β-­c arotene (Box 8.B). The thyroid hormones, I O which generally stimulate metabolism, are derived from a large precurHO I sor protein called thyroglobulin: Tyrosine side chains are enzymatically O− iodinated, then two of these residues undergo condensation, and the NH+ 3 hormones are liberated from thyroglobulin by proteolysis. I O The 27-carbon cholesterol, introduced in Section 8.1, is the precursor I of a large number of hormones that regulate metabolism, salt and water Thyroxine (T4) balance, and reproductive functions. Androgens (which are primarily male hormones) have 19 carbons, and estrogens (which are primarily I O female hormones) contain 18 carbons. Cortisol, a C21 glucocorticoid horHO I mone, affects the metabolic activities of a wide variety of tissues. Retinoic − O acid, thyroid hormones, and steroids are all hydrophobic molecules that are NH+ 3 carried in the bloodstream either by specific carrier proteins or by albumin, O a sort of all-­purpose binding protein. I The receptors to which the lipid hormones bind are located inside Triiodothyronine (T3) the appropriate target cell, either in the cytoplasm or the nucleus. Ligand binding often—­but not always—­causes the receptors to form dimers. Each OH receptor subunit is constructed from several modules, which include a O ligand-­binding domain and a DNA-­binding domain. The ligand-­binding CH3 domains are as varied as their hormone ligands, but the DNA-­binding HO OH domains exhibit a common structure that includes two zinc fingers, which H3C are cross-­links formed by the interaction of four cysteine side chains with a Zn2+ ion (Section 4.3). In the absence of a ligand, the receptor cannot bind to DNA. Following ligand binding and dimerization, the receptor moves O Cortisol to the nucleus (if it is not already there) and binds to specific DNA sequences called hormone response elements. Although the nucle  FIGURE 10.16   Some lipid hormones. otide sequences of the hormone response elements vary for each receptor–­l igand complex, they are all composed of two identical 6-bp sequences separated by a few base pairs. Simulta­n­eous binding of the two hormone response element sequences explains why many of the lipid hormone receptors are dimers (Fig. 10.17). The receptors function as transcription factors so that the genes near the hormone response elements may experience higher or lower levels of expression. For example, glucocorticoids such as cortisol stimulate the production of phosphatases, which dampen the stimulating effects of kinases. This property makes cortisol and its derivatives useful as drugs to treat conditions such as chronic inflammation or asthma. However, because so many tissues respond to glucocorticoids, the side effects of these drugs can be significant and tend to limit their long-­term use.   FIGURE 10.17   The glucocorticoid receptor–­DNA complex.  The changes in gene expression triggered by steroids and The two DNA-­binding zinc finger domains of the glucocorticoid other lipid hormones require many hours to take effect. Howreceptor are blue and green. The Zn2+ ions are shown as gray ever, cellular responses to some lipid hormones are evident spheres. The hormone response element sequences of the DNA within seconds or minutes, indicating that the hormones also (bottom) are colored red. Two protein helices make sequence-­ specific contacts with nucleotides. participate in signaling pathways with shorter time courses, such as those centered on G proteins and/or kinases. In these Question  What is the surface charge of the receptor’s DNA-­ instances, the receptors must be located on the cell surface. binding domain?

H3C

CH3

CH3

CH3

O

10.4  Lipid Hormone Signaling  305

Eicosanoids are short-­range signals

COO−

Many of the hormones discussed in this chapter are synthesized and stockpiled to some extent before they Arachidonate are released, but some lipid hormones are synthesized as a response to other signaling events (sphingosinecyclooxygenase 1-­phosphate is one example; Section 10.2). The lipid hormones called eicosanoids are produced when the enzyme O COO− phospholipase A2 is activated by phosphorylation and by the presence of Ca2+. One substrate of the phospholipase O is the membrane lipid phosphatidylinositol. In this lipid, OH cleavage of the acyl chain attached to the second glycerol Prostaglandin H2 carbon often releases arachidonate, a C20 fatty acid (the term eicosanoid comes from the Greek eikosi, meaning “twenty”). Arachidonate, a polyunsaturated fatty acid with COO− other biologically active lipids four double bonds, is further modified by the action of OH enzymes that catalyze cyclization and oxidation reactions (Fig. 10.18). A wide variety of eicosanoids can be produced in a tissue-­dependent fashion, and their functions are similarly varied. Eicosanoids regulate such HO O things as blood pressure, blood coagulation, inflammaOH tion, pain, and fever. The eicosanoid thromboxane, for Thromboxane example, helps activate platelets (cell fragments that participate in blood coagulation) and induces vasoconstriction. Other eicosanoids have the opposite effects:   FIGURE 10.18   Arachidonate conversion to eicosanoid signal They prevent platelet activation and promote vasodila- molecules.  The first step is catalyzed by cyclooxygenase. Only two tion. The use of aspirin as a “blood thinner” stems from of the many dozens of eicosanoids are shown here. its ability to inhibit the enzyme that initiates the conversion of arachidonate to thromboxane (see Fig. 10.18). A number of other drugs interfere with the production of eicosanoids by blocking the same enzymatic step (Box 10.C). The receptors for eicosanoids are G protein–­c oupled receptors that trigger cAMP-­ dependent and phosphoinositide-­d ependent responses. However, eicosanoids are degraded relatively quickly. This instability, along with their hydrophobicity, means that their effects are relatively limited in time and space. Eicosanoids tend to elicit responses only in the cells that produce them and in nearby cells. In contrast, many other hormones travel throughout the body, eliciting effects in any tissue that exhibits the appropriate receptors. For this reason, eicosanoids are sometimes called local mediators rather than hormones. Similar signaling molecules operate in plants. For example, the lipid hormone jasmonate is synthesized as a result of localized herbivore damage and quickly spreads to other parts of the plant to trigger defensive responses.

O

O

CH3

Jasmonate

O

−

Volatile derivatives of jasmonate can travel through the air to spread the alarm to neighboring plants.

306  C ha pter 10   Signaling

Box 10.C Inhibitors of Cyclooxygenase The bark of the willow Salix alba has been used since ancient times to relieve pain and fever. The active ingredient is acetyl­ salicylate, or aspirin.

O C

O−

O

C

CH3

O

Acetylsalicylate (aspirin) Aspirin was first prepared in 1853, but it was not used clinically for another 50 years or so. Effective promotion of aspirin by the Bayer chemical company at the start of the twentieth century marked the beginning of the modern pharmaceutical industry. Despite its popularity, aspirin’s mode of action was not discovered until 1971. It inhibits the production of prostaglandins (which induce pain and fever, among other things) by inhibiting the activity of cyclooxygenase (also known as COX), the enzyme that acts on arachidonate (see Fig. 10.18). COX inhibition results from acetylation of a serine residue located near the active site in a cavity that accommodates the arachidonate substrate. Other pain-­relieving substances such as ibuprofen also bind to COX to prevent the synthesis of prostaglandins, although these drugs do not acetylate the enzyme.

CH

CH

CH2

O HO

NH

C

CH3

Acetaminophen

CH3

H3C

One shortcoming of aspirin is that it inhibits more than one COX isozyme. COX-1 is a constitutively expressed enzyme that is responsible for generating various eicosanoids, including those that maintain the stomach’s protective layer of mucus. COX-2 expression increases during tissue injury or infection and generates eicosanoids involved in inflammation. Long-­term aspirin use suppresses the activity of both isozymes, which can lead to side effects such as gastric ulcers. Rational drug design (Section 7.4), based on the slightly different structures of COX-1 and COX-2, led to the development of drugs that bind only to the active site of COX-2 because they are too large to fit into the COX-1 active site. The drugs therefore can selectively block the production of pro-­inflammatory eicosanoids without damaging gastric tissue. Unfortunately, the side effects of the drugs include an increased risk of heart attacks, through a mechanism that is not fully understood. As a result, only one drug—­celecoxib (Celebrex®)—is currently in use. If nothing else, this story illustrates the complexity of biological signaling pathways and the difficulty of understanding how to manipulate them for therapeutic reasons. A third COX isozyme, COX-3, is expressed at high levels in the central nervous system. It is the target of the widely used drug acetaminophen (Section 7.4), which reduces pain and fever and does not appear to incur the side effects of the COX-2–specific inhibitors.

COO− Question  Which of the drugs shown here is chiral (see Section 4.1), with two different configurations?

H3C Ibuprofen

Before Going On • List some types of lipid hormones and their physiological effects. • Explain why lipid hormones have intracellular receptors. • Compare the timing of the responses to steroid hormones and eicosanoids. • Draw a model of a cell and add shapes to represent all the types of receptors, target enzymes, and other signal-­transduction machinery mentioned in the chapter, placing each component in the membrane, cytosol, or nucleus as appropriate. • Make a list of the drugs mentioned in this chapter and indicate how they interfere with signaling.

Bioinformatics  307

Summary 10.1  General Features of Signaling Pathways •  Agonist or antagonist binding to a receptor can be quantified by a dissociation constant. •  G protein–­coupled receptors and receptor tyrosine kinases are the most common types of receptors. •  While signaling systems amplify extracellular signals, they are also regulated so that signaling can be turned off, and the receptor may become desensitized.

10.2  G Protein Signaling Pathways •  A ligand such as epinephrine binds to a G protein–­c oupled receptor. A G protein responds to the receptor–­ligand complex by releasing GDP, binding GTP, and splitting into an α subunit and a βγ dimer. Alternatively, arrestin may bind to the complex and activate additional proteins. •  The α subunit of the G protein activates adenylate cyclase, which converts ATP to cAMP. cAMP is a second messenger that triggers a conformational change in protein kinase A that repositions its activation loop to achieve full catalytic activity. •  cAMP-­dependent signaling activity is limited by the reduction of second messenger production through the GTPase activity of G proteins and the action of phosphodiesterases and by the activity of phosphatases that reverse the effects of protein kinase A. Ligand dissociation and receptor desensitization through phosphorylation and arrestin binding also limit signaling via G protein–­coupled receptors.

•  G protein–­coupled receptors that lead to activation of phospho­ lipase C generate inositol trisphosphate and diacylglycerol ­second messengers, which activate protein kinase B and protein kinase C, respectively. •  Signaling pathways originating with different G protein–­coupled receptors and receptor tyrosine kinases overlap through activation or inhibition of the same intracellular components, such as kinases, phosphatases, and phospholipases. •  Odorant binding to olfactory receptors leads to cAMP production that triggers cell depolarization. Tastants activate the phosphoinositide pathway that leads to neurotransmitter release.

10.3  Receptor Tyrosine Kinases •  Ligand binding to a receptor tyrosine kinase causes conformational changes in the receptor dimer that bring the cytoplasmic tyrosine kinase domains close enough to phosphorylate each other. •  In addition to acting as kinases, the receptor tyrosine kinases initiate other kinase cascades by activating the small monomeric G protein Ras.

10.4  Lipid Hormone Signaling •  Steroids and other lipid hormones bind primarily to intracellular receptors that dimerize and bind to hormone response elements on DNA to induce or repress the expression of nearby genes. •  Eicosanoids, which are synthesized from membrane lipids, function as signals over short ranges and for a limited time.

Key Terms receptor ligand signal transduction hormone quorum sensing dissociation constant (Kd) agonist antagonist G protein GPCR

second messenger kinase receptor tyrosine kinase desensitization cAMP phosphatase phosphoinositide signaling system cross-­talk olfaction odorant

opiod gustation tastant autophosphorylation transcription factor oncogene hormone response element eicosanoid

Bioinformatics Brief Bioinformatics Exercises 10.1.  G Protein–­Coupled Receptors and Receptor Tyrosine Kinases

10.2.  Biosignaling and the KEGG Database

308  C ha pter 10   Signaling

Problems 10.1  General Features of Signaling Pathways 1.  Which of the signaling molecules listed in Table 10.1 would not require a cell-surface receptor? 2.  Animals “taste” the carbonation (fizziness) of sodas when the dissolved CO2 reacts with water to generate bicarbonate and protons in a reaction catalyzed by carbonic anhydrase expressed on the surface of cells in the tongue. Does this enzyme fit the definition of a CO 2 receptor? 3.  A sample of cells has a total receptor concentration of 25 mM. Ninety percent of the receptors have bound ligand and the concentration of free ligand is 125 µM. What is the Kd for the receptor–ligand interaction? 4.  A sample of cells has a total receptor concentration of 50 mM. Fifty percent of the receptors have bound ligand and the concentration of free ligand is 5 mM.  a.  What is the Kd for the receptor–ligand interaction?  b.  What is the relationship between Kd and [L]? 5.  The Kd for a receptor–ligand interaction is 3 mM. When the concentration of free ligand is 18 mM and the concentration of free receptor is 5 mM, what is the concentration of receptor that is occupied by ligand? 6.  The total concentration of receptors in a sample is 20 mM. The concentration of free ligand is 5 mM and the Kd is 10 mM. Calculate the percentage of receptors that are occupied by ligand. 7.  The total concentration of receptors in a sample is 10 mM. The concentration of free ligand is 2.5 mM and the Kd is 1.5 mM. Calculate the percentage of receptors that are occupied by ligand. 8.  The total concentration of receptors in a sample is 10 mM. The concentration of free ligand is 2.5 mM and the Kd is 0.3 mM. Calculate the percentage of receptors that are occupied by ligand. Compare the answer to this problem with the answer you obtained in Problem 7 and explain the difference. 9.  Use the plot below to estimate a value for Kd.

0.6

12.  The [R · L]:[R]T ratio gives the fraction of receptors that have bound ligand. Use the expression you derived in Problem 11 to express [R · L] as a fraction of [R]T for the following situations:  a.  Kd = 5[L],  b.  Kd = [L], and  c. 5Kd = [L]. 13.  If a cell has 1000 surface receptors for erythropoietin, and if only 10% of those receptors need to bind the ligand to achieve a maximal response, what ligand concentration is required to achieve a maximal response? Use the equation you derived in Problem 11. The Kd for erythropoietin is 1.0 × 10–10 M. 14.  Suppose the number of surface receptors on the cell described in Problem 13 decreases to 150. What ligand concentration is required to achieve a maximal response? 15.  ADP binds to platelets in order to initiate the activation process. Two binding sites were identified on platelets, one with a Kd of 0.35 µM and one with a Kd of 7.9 µM.  a.  Which site is a low-­affinity binding site and which is a high-affinity binding site?  b. The ADP concentration required to activate a platelet is in the range of 0.1–0.5 µM. Which receptor will be more effective at activating the platelet?  c.  Two ADP agonists were also found to bind to platelets: 2-methythio-ADP bound with a Kd of 7 µM and 2-(3-aminopropylthio)ADP bound with a Kd of 200 µM. Can these agonists effectively compete with ADP for binding to platelets?  d.  In the study, 160,000 high-affinity binding sites were identified on each platelet. What is the concentration of ADP required to achieve 85% binding to the high-affinity site? Use the equation you derived in Problem 11. 16.  The peptide hormone glucagon, released during fasting, binds to a high-affinity receptor with a Kd value of 0.8 nM.  a.  What percentage of glucagon receptors are occupied when the blood glucagon concentration is 70 pM after one day of fasting? Use the equation you derived in Problem 11.  b.  A glucagon derivative missing the N-terminal histidine residue has a Kd value of 50 nM. What does this reveal about the structural requirements for glucagon binding to its receptor? 17.  Like the Michaelis–Menten equation, the equation derived in Solution 11 can be converted to an equation for a straight line. A ­double-reciprocal plot for a ligand binding to its receptor is shown below. Use the information in the plot to estimate a value for Kd.

0.2

0.1

0.2

0.3 0.4 [L] (mM)

0.5

0.6

10

0.7

1/[R . L] (μM−1)

[R. L]/[R]T

1.0

value of [R], like [R · L], is difficult to evaluate, but various experimental techniques can be used to determine [R]T, the total number of receptors. [R]T is the sum of [R] and [R · L]. Using this information, begin with Equation 10.1 and derive an expression for the [R · L]/ [R]T ratio. (Note that your derived expression will be similar to the Michaelis–Menten equation and that Equations 7.9 through 7.17 may give you an idea of how to proceed.)

10. a.  In an experiment, the ligand adenosine is added to heart cells in culture. The number of receptors with ligand bound is measured and the data yield a curve like the one shown in Figure 10.1. What would the results look like if the experiment were repeated in the presence of caffeine?  b.  How does the apparent Kd in the presence of caffeine compare to the Kd in its absence? 11.  Kd is defined in Equation 10.1, which shows the relationship between the free receptor concentration [R], the ligand concentration [L], and the concentration of receptor–ligand complexes [R · L]. The

8 6 4 2

−4

−3

−2

−1 0 1 1/[L] (μM−1)

2

3

4

Problems  309 18.  A double-reciprocal plot (see Problem 17) for a ligand binding to its receptor is shown below. Use the information in the plot to calculate a value for Kd.

24.  In the liver, glucagon and epinephrine bind to different members of the G protein–coupled receptor family, yet binding of each of these ligands results in the same response—glycogen breakdown. How is it possible that two different ligands can trigger the same cellular response?

0.10

1/[R . L] (μM−1)

0.08 y = 0.1154x + 0.0077 0.06

25.  Many receptors become desensitized in the presence of high concentrations of signaling ligand. This can occur in a variety of ways, such as removal of the receptors from the cell surface by endocytosis. Why is this an effective desensitization strategy?

0.04 0.02

−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

1/[L] (μM−1)

7 6

y = −0.067x + 6.890

A active

Binactive

Bactive

Cinactive

Cactive

Dinactive

5 [R . L]/[L]

26.  What is the advantage of activating D using the strategy shown in the figure? Why is this strategy more effective than a simple activation of D that occurs in one step?

A inactive

19.  A Scatchard plot is another method of representing ligand binding data using a straight line (see Problem 17). In a Scatchard plot, [R · L]/[L] is plotted versus [R · L]. The slope is equal to –1/Kd. Use the Scatchard plot provided to calculate a value for Kd for calmodulin binding to calcineurin.

4

Dactive

10.2  G Protein Signaling Pathways

3 2 1 20

40 60 [R . L] (nM)

80

100

20.  A nucleoside derivative binds to lymphocytes and elicits a wide variety of cellular responses. The binding of the nucleoside to its receptor was analyzed using a Scatchard plot (see Problem 19). Use the plot provided to calculate the Kd for nucleoside–receptor binding. 40

0 0.1

28.  The cell contains many types of guanine nucleotide exchange factors (GEFs) that regulate the activity of monomeric GTPases by promoting the exchange of GDP for GTP. Explain how G protein– coupled receptors also function as GEFs. 29.  Naturally occurring mutations in the genes that code for GPCRs have provided insights into GPCR function and the diseases that often result from these mutations. List the types of mutations in a GPCR that would result in loss of the receptor’s function.

31.  Some G protein–linked receptors are associated with a protein called RGS (regulator of G protein signaling). RGS stimulates the GTPase activity of the G protein associated with the receptor. What effect does RGS have on the signaling process?

20

10

27.  As described in the text, G protein–coupled receptors are often palmitoylated at a Cys residue.  a.  Draw the structure of the palmitate (16:0) residue covalently linked to a Cys residue.  b.  What is the role of this fatty acyl chain?  c.  What is the expected result if the Cys residue is mutated to a Gly?

30.  Mutations in a GPCR gene (see Problem 29) occasionally result in increased receptor activity. List the types of mutations in a GPCR that would result in gain of function of the receptor.

30 [L.R]/[L]

23.  Epinephrine can bind to several different types of G protein– linked receptors. Each of these receptors triggers a different cellular response. Explain how this is possible.

y = −124x + 45

0.15

0.2

0.25

[R . L] (nM)

32.  An Asp residue in the third transmembrane helix of the epinephrine receptor (a GPCR) interacts with the ligand epinephrine.  a.  What type of interaction is likely to form between the receptor and epinephrine?  b.  How might an Asp → Glu mutation affect ligand binding?  c.  How might an Asp → Asn mutation affect ligand binding?

21.  Why might it be difficult to purify cell-surface receptors using the techniques described in Section 4.6?

33.  How do epinephrine and norepinephrine differ from tyrosine, their parent amino acid?

22.  Affinity chromatography is often used as a technique to purify cell-surface receptors. Describe the steps you would take to purify a cell-surface receptor using this technique.

34.  β2-Adrenergic receptor antagonists known as β-blockers prevent epinephrine and norepinephrine binding to their receptors. Why are β-blockers effective at treating high blood pressure?

310  C ha pter 10   Signaling 35.  A toxin secreted by the bacterium Vibrio cholerae catalyzes the covalent attachment of an ADP-ribose group to the α subunit of a G protein. This results in the inhibition of the intrinsic GTPase activity of the G protein. How does this affect the activity of adenylate cyclase? How are intracellular levels of cAMP affected?

essential for T cell activation. Describe the cell-signaling events that result in the activation of NFAT.

36.  Some G protein–linked receptors are associated with G proteins that inhibit rather than stimulate the activity of adenylate cyclase. A toxin secreted by the bacterium Bordetella pertussis (the causative agent of whooping cough) catalyzes the covalent attachment of an ADP-ribose group to the α subunit of the inhibitory G protein, preventing it from carrying out its normal function. How does this affect the activity of adenylate cyclase? How are intracellular levels of cAMP affected?

45.  Pathways that lead to the activation of protein kinase B (Akt) are considered to be anti-apoptotic (apoptosis is programmed cell death). In other words, protein kinase B stimulates a cell to grow and proliferate. Like all biological events, signaling pathways that are turned on must also be turned off. A phosphatase called PTEN plays a role in removing phosphate groups from proteins, but it is highly specific for removing a phosphate group from inositol trisphosphate. If PTEN is overexpressed in mammalian cells, do these cells grow or do they undergo apoptosis?

37.  Addition of the nonhydrolyzable analog GTPγS to cultured cells is a common practice in signal transduction experiments. What effect does GTPγS have on cellular cAMP levels for  a.  a stimulatory G protein (see Problem 35) and  b.  an inhibitory G protein (see Problem 36)?

O N O

O −O

P O−

S

P O−

O O

P

N O

CH2

O− H GTPγS

O

H

H

OH

OH

NH N

NH2

H

38.  Protein kinase G, a cGMP-dependent protein kinase, is the main mediator of cGMP signaling in the malaria parasite and is likely to be involved in many aspects of malarial infection. Elucidating this signaling pathway is an active area of research because of the potential to design drugs to target the pathway.  a.  Propose a name for the enzyme that, when activated by the G protein, catalyzes the synthesis of cGMP.  b.  Draw the structure of cGMP. 39.  How does the addition of a single phosphate group to a protein change that protein’s activity? 40. a.  Draw the reaction that shows the protein kinase A–catalyzed phosphorylation of a threonine residue on a target protein.  b. Draw the reaction that shows the phosphatase-catalyzed hydrolysis of the phosphorylated threonine.  c.  Some bacterial signaling systems involve kinases that transfer a phosphoryl group to a His side chain. Draw the structure of the phospho-His side chain. 41.  Phorbol esters, which are compounds isolated from plants, are structurally similar to diacylglycerol. How does the addition of phorbol esters affect the cellular signaling pathways of cells in culture? 42.  As described in the text, ligand binding to certain receptor tyrosine kinases results in the activation of a sphingomyelinase enzyme. Draw the reaction that shows the sphingomyelinase-catalyzed hydrolysis of sphingomyelin to ceramide. 43.  In unstimulated T cells, a transcription factor called NFAT (nuclear factor of activated T cells) resides in the cytosol in a phosphorylated form. When the cell is stimulated, the cytosolic Ca2+ concentration increases and activates a phosphatase called calcineurin. The activated calcineurin catalyzes the hydrolysis of the phosphate group from NFAT, exposing a nuclear localization signal that allows the NFAT to enter the nucleus and stimulate the expression of genes

44.  The immunosuppressive drug cyclosporine A is an inhibitor of calcineurin (see Problem 43). Why is cyclosporine A an effective immunosuppressant?

46.  Would you expect to find mutations in the gene for PTEN (see Problem 45) in human cancers? Explain why or why not. 47.  Nitric oxide (NO) is a naturally occurring signaling molecule (see Table 10.1) that is produced from the decomposition of arginine to NO and citrulline in endothelial cells. The enzyme that catalyzes this reaction, NO synthase, is stimulated by cytosolic Ca2+, which increases when acetylcholine binds to endothelial cells.  a. What is the source of the acetylcholine ligand?  b.  Propose a mechanism that describes how acetylcholine binding leads to the activation of NO synthase.  c.  NO formed in endothelial cells quickly diffuses into neighboring smooth muscle cells and binds to a cytosolic protein that catalyzes the formation of the second messenger cyclic GMP. Cyclic GMP then activates protein kinase G (see Problem 38), resulting in smooth muscle cell relaxation. How might protein kinase G bring about smooth muscle relaxation? 48.  As discussed in the text, any signal transduction event that is turned on must subsequently be turned off. Refer to your answer to Problem 47 and describe the events that would lead to the cessation of each step of the signaling pathway you described. 49.  NO synthase knockout mice (animals missing the NO synthase enzyme) have elevated blood pressure, an increased heart rate, and enlarged left ventricle chambers. Explain the reasons for these symptoms. 50.   Clotrimazole is a calmodulin antagonist (see Solution 47b). How does the addition of clotrimazole affect endothelial cells in culture? 51.  Nitroglycerin placed under the tongue has been used since the late nineteenth century to treat angina pectoris (chest pains resulting from reduced blood flow to the heart). Only recently have scientists elucidated its mechanism of action. Propose a hypothesis that explains why nitroglycerin placed under the tongue relieves the pain of angina.

O− O

N +

O

O

N +

O

O−

O −O

+

N

O

Nitroglycerin

52.  Viagra, a drug used to treat erectile dysfunction, is a cGMP phosphodiesterase inhibitor. Propose a mechanism that explains why the drug is effective in treating this condition.

Problems  311 53.  Propose an explanation why arrestin can be activated by the molecule shown here.

O HO

O −O

P

P

O

O

O

O−

O− O

OH

O O

P OH

O

O−

P

O

P

O−

OH

OH

O O

P

O−

OH

54.  Arrestin has been shown to bind to clathrin (see Fig. 9.23). What is the significance of this observation? 55.  Thrombin generated at the site of a blood vessel injury (see Section 6.5) binds to the thrombin receptor, a GPCR, on the surface of platelets, which activates the platelets, leading to their aggregation and the formation of a plug to help seal the broken blood vessel. Draw a diagram of the thrombin signaling pathway, given the following information: phospholipase C is activated; Ca2+ ions activate myosin-light chain kinase; the activated myosin light chain is responsible for platelet shape change; and protein kinase C stimulates the release of granules (substances essential in blood coagulation) when activated. 56.  A series of experiments showed that thrombin signaling (see Problem 55) is terminated when the GPCR is inactivated by phosphorylation catalyzed by a specific kinase called G protein coupled receptor kinase (GRK).  a.  Why does the receptor become radioactively labeled when [γ32P]-ATP is added to cells in culture?  b. Would this happen if Ser and Thr residues in the receptor were mutated?  c.  What is the expected result if GRK were overexpressed in cultured cells?  d.  Why does GRK activity decrease by 90% when a lysine residue in the kinase is mutated to an alanine?  e. Predict what would happen if high concentrations of thrombin were added to cultured cells. 57.  Bacillus anthracis, the causative agent of anthrax, produces a three-part toxin. One part facilitates the entry of the two other toxins into the cytoplasm of a mammalian cell. The toxin known as edema factor (EF) is an adenylate cyclase.  a.  Explain how EF could disrupt normal cell signaling.  b.  EF must first be activated by Ca2+-calmodulin binding to it. Explain how this requirement could also disrupt cell signaling.

10.3  Receptor Tyrosine Kinases 61.  Stimulation of the insulin receptor by ligand binding and autophosphorylation eventually leads to the activation of both protein kinase B (Akt) and protein kinase C. Protein kinase B phosphorylates glycogen synthase kinase 3 (GSK3) and inactivates it. (Active GSK3 inactivates glycogen synthase by phosphorylating it.) Glycogen synthase catalyzes synthesis of glycogen from glucose. In the presence of insulin, GSK3 is inactivated, so glycogen synthase is not phosphorylated and is active. Protein kinase C stimulates the translocation of glucose transporters to the plasma membrane by a mechanism not currently understood. One strategy for treating diabetes is to develop drugs that act as inhibitors of the phosphatases that remove phosphate groups from the phosphorylated tyrosines on the insulin receptor. Why might this be an effective treatment for diabetes? 62.  When insulin binds to its receptor, a conformational change occurs that results in autophosphorylation of the receptor on specific Tyr residues. In the next step of the signaling pathway, an adaptor protein called IRS-1 (insulin receptor substrate-1) docks with the phosphorylated receptor (the involvement of adaptor proteins in cell signaling is shown in Fig. 10.15). This step is essential for the downstream activation of protein kinases B and C (see Problem 61). If IRS-1 is overexpressed in muscle cells in culture, what effects, if any, would you expect to see on glucose transporter translocation and ­glycogen synthesis? 63.  The activity of Ras is regulated in part by two proteins, a guanine nucleotide exchange factor (GEF) and a GTPase activating protein (GAP). The GEF protein binds to Ras · GDP and promotes dissociation of bound GDP. The GAP protein binds to Ras · GTP and stimulates the intrinsic GTPase activity of Ras.  a.  How is downstream activity of a signaling pathway affected by the presence of GEF?  b.  By the presence of GAP? 64.  Mutant Ras proteins have been found in various types of cancers. What is the effect on a cell if the mutant Ras is able to bind GTP but is unable to hydrolyze it? 65.  As shown in Figure 10.15, Ras can activate a kinase cascade. The most common cascade is the MAP kinase pathway, which is activated when growth factors bind to cell surface receptors and activate Ras. This leads to the activation of transcription factors and other gene regulatory proteins and results in growth, proliferation, and differentiation. Use this information to explain why phorbol esters (see Problem 41) promote tumor development.

Ras Raf

58.  Bacillus anthracis (see Problem 57) also makes a toxin called lethal factor (LF). LF is a protease that specifically cleaves and inactivates a protein kinase that is part of a pathway for stimulating cell proliferation. Explain why the entry of LF into white blood cells promotes the spread of B. anthracis in the body. 59.  Ligand binding to some growth-factor receptors triggers kinase cascades and also leads to activation of enzymes that convert O2 to hydrogen peroxide (H2O2), which acts as a second messenger. Describe the likely effect of H2O2 on the activity of cellular phosphatases. 60.  Hydrogen peroxide acts as a second messenger, as described in Problem 59, and affects PTEN (see Problem 45) as well as other cellular phosphatases. Does H2O2 activate or inhibit PTEN?

Protein kinase C

MEK MAP kinase

Gene regulatory proteins

Transcription factors

Cell cycle proteins

66.  How might a signaling molecule activate the MAP kinase cascade (see Problem 65) via a G protein–coupled receptor rather than a receptor tyrosine kinase?

312  C ha pter 10   Signaling 67.  Mutations in proteins involved in the receptor tyrosine kinase pathway (see Fig. 10.15 and Problem 65) are found in many different types of cancers. How might such mutations alter signaling activity in a cancerous cell? Give several examples. 68.  Hsp90, a molecular chaperone that assists with protein folding (see Fig. 4.24) is overexpressed in cancer cells and is often found associated with proteins involved in signaling pathways. Why would cancer cells require additional assistance from chaperones? 69.  Ras can activate a cascade of reactions (see Fig. 10.15) involving membrane phosphatidylinositol (PI) (see Solutions 8.27 and 8.28). Ras activates PI 3-kinase (PI3K), which phosphorylates PI-4,5-­ bisphosphate (PIP2) to PI-3,4,5-trisphosphate (PIP3). PIP3 then activates Akt (see Problem 45). Why would PI3K inhibitors be effective at treating cancer? 70.  PTEN (see Problem 45) can catalyze the hydrolysis of PIP 3 to PIP2 (see Problem 69). Does PTEN promote cell survival or cell death? 71.  PKR is a protein kinase that recognizes double-stranded RNA molecules such as those that form during the intracellular growth of certain viruses. The structure of PKR includes the standard kinase domains as well as an RNA-binding module. In the presence of viral RNA, PKR undergoes autophosphorylation and is then able to phosphorylate cellular target proteins that initiate antiviral responses. Short ( 0) under standard conditions. In a living cell, reactants and products are almost never present at standard-state concentrations and the temperature may not be 25°C, yet reactions do occur with some change in free energy. Thus, it is important to distinguish the standard free energy change of a reaction from its actual free energy change, ΔG. ΔG is a function of the actual concentrations of the reactants and the temperature (37°C or 310 K in humans). ΔG is related to the standard free energy change for the reaction: ​[C][​​ D] ​​​  ​ΔG = ΔG°′ + RT ln ​ ______  [​ A][​​ B]​



(12.3)

Here, the bracketed quantities represent the actual, nonequilibrium concentrations of the reactants. The concentration term in Equation 12.3 is sometimes called the mass action ratio. When the reaction is at equilibrium, ΔG = 0 and [​​ C]​​  eq​​ [​​ D]​​  eq​​ ​ΔG°′​​​= −RT ln ​ _________ ​​  [​​ A]​​  eq​​ [​​ B]​​  eq​​

(12.4)

which is equivalent to Equation 12.2. Note that Equation 12.3 shows that the criterion for spontaneity for a reaction is ΔG, a property of the actual concentrations of the reactants, not the constant ΔG°′. Thus, a reaction with a positive standard free energy change (a reaction that cannot occur when the reactants are present at standard concentrations) may proceed in vivo, depending on the concentrations of reactants in the cell (see Sample Calculation 12.2). Keep in mind that thermodynamic spontaneity does not imply a rapid reaction. Even a substance with a strong tendency to undergo reaction (ΔG ≪ 0) will usually not react until acted upon by an enzyme that catalyzes the reaction.

SEE SAMPLE CALCULATION VIDEOS

SA MP L E CA LCULAT I O N 12. 2 Problem  The standard free energy change for the reaction catalyzed by phosphoglucomutase is –7.1 kJ · mol−1. Calculate the equilibrium constant for the reaction. Calculate ΔG at 37°C when the concentration of glucose-1-phosphate is 1 mM and the concentration of glucose-6-phosphate is 25 mM. Is the reaction spontaneous under these conditions? −2

HOCH2 H HO

O H OH H

H OH

H OPO2− 3

Glucose-1-phosphate

HO

O H OH H

​ ​K  ​eq​= ​e​​ −ΔG°′/RT​

​)/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​) ​(298 K)​

−1

​ = ​e​​ −(−7100 J · ​mol​ ​ = ​e ​2.87​= 17.6​ At 37°C, T = 310 K.

O3POCH2 H

Solution  The equilibrium constant Keq can be derived by rearranging Equation 12.2.

H OH

H OH

Glucose-6-phosphate

[​ ​glucose‐6‐phosphate​]​ ​ ΔG = Δ G°′ + RT  ln ​ __________________      ​​ ​[​glucose‐1‐phosphate​]​

​ = − 7100 J · ​mol​−1​ +  ​ (8.3145 J · ​K−1 ​ ​ · ​mol​−1​)(310 K)​ ln ​(0.025 / 0.001)​ ​ = − 7100 J · ​mol​−1​ + 8300 J · ​mol​−1​

​ = 1200 J · ​mol​−1​ = 1.2 kJ · ​mol​−1​ The reaction is not spontaneous because ΔG is greater than zero.

354  C ha pter 12   Metabolism and Bioenergetics

Unfavorable reactions are coupled to favorable reactions A biochemical reaction may at first seem to be thermodynamically forbidden because its free energy change is greater than zero. Yet the reaction can proceed in vivo when it is coupled to a second reaction whose value of ΔG is very large and negative so that the net change in free energy for the combined reactions is less than zero. ATP is often involved in such coupled pro­ cesses because its reactions occur with a relatively large negative change in free energy. Adenosine triphosphate (ATP) contains two phosphoanhydride bonds (Fig. 12.11). Cleavage of either of these bonds—that is, transfer of one or two of its phosphoryl groups to another molecule—is a reaction with a large negative standard free energy change (under physiological conditions, ΔG is even more negative). As a reference point, biochemists use the reaction in which the third phosphoryl group is transferred to water—in other words, hydrolysis of one phosphoanhydride bond: ​ATP + ​H2 ​ ​O → ADP + ​Pi ​​

This is a spontaneous reaction with a ΔG°′ value of –30 kJ · mol−1. Hydrolysis of ATP’s other phosphoanhydride bond is even more favorable (ΔG°′ = –45.6 kJ · mol−1). The following example illustrates the role of ATP in a coupled reaction. Consider the phosphorylation of glucose by inorganic phosphate (​HP​O​  42− ​  ​​ or Pi), a thermodynamically unfavorable reaction (ΔG°′ = +13.8 kJ · mol−1):

H HO

CH2OH O H

H

OH

H

H

OH

+

CH2OPO2− 3 O H H

H

Pi

OH

HO

Glucose

OH

H

H

OH

OH

Glucose-6-phosphate

NH2

O−

O− −

O

P γ O

O

N

N

phosphoanhydride bonds

O−

P O P O β α O O

CH2 H

H

HO

N

N O H

H

OH

Adenosine AMP ADP ATP   FIGURE 12.11   Adenosine triphosphate.  The three phosphate groups are sometimes described by the Greek letters α, β, and γ. The linkage between the first (α) and second (β) phosphoryl groups, and between the second (β) and third (γ), is a phosphoanhydride bond.

Question  How does hydrolysis of a phosphoanhydride bond affect the net charge of a nucleotide?

+ H2O

12.3  Free Energy Changes in Metabolic Reactions  355

When this reaction is combined with the ATP hydrolysis reaction, the values of ΔG°′ for each reaction are added: ΔG°′ glucose + Pi → glucose‐6‐phosphate + H2O ATP + H2O → ADP + Pi

glucose + ATP → glucose‐6‐phosphate + ADP

+13.8 kJ · mol−1 −30.5 kJ · mol−1 −16.7 kJ · mol−1

The net chemical reaction, the phosphorylation of glucose, is thermodynamically favorable (ΔG < 0). In vivo, this reaction is catalyzed by hexokinase (introduced in Section 6.3), and a phosphoryl group is transferred from ATP directly to glucose. The ATP is not actually hydrolyzed and there is no free phosphoryl group floating around the enzyme. However, writing out the two coupled reactions, as shown above, makes it easier to see what is going on thermodynamically. Some biochemical processes appear to occur with the concomitant hydrolysis of ATP to ADP + Pi, for example, the operation of myosin and kinesin (Section 5.4) or the Na,KATPase ion pump (Section 9.3). However, a closer look reveals that in all these processes, ATP actually transfers a phosphoryl group to a protein. Later, the phosphoryl group is transferred to water, so the net reaction takes the form of ATP hydrolysis. The same ATP “hydrolysis” effect applies to some reactions in which the AMP moiety of ATP (rather than a phosphoryl group) is transferred to a substance, leaving inorganic pyrophosphate (PPi). Cleavage of the phosphoanhydride bond of PPi also has a large negative value of ΔG°′. Because the ATP hydrolysis reaction appears to drive many thermodynamically unfavorable reactions, it is tempting to think of ATP as an agent that transfers packets of energy around the cell. This is one reason why ATP is commonly called the energy currency of the cell. The general role of ATP in linking exergonic ATP-producing processes to endergonic ATP-consuming processes can be diagrammed as

Nutrient

ADP + Pi

catabolism

Waste product

Product

anabolism

ATP

Precursor

In this scheme, it appears that the “energy” of the catabolized nutrient is transferred to ATP and then the “energy” of ATP is transferred to another product in a biosynthetic reaction. However, energy is not a tangible item and there is nothing magic about ATP. The two phosphoanhydride bonds of ATP are sometimes called “high-energy” bonds, but they are no different from other covalent bonds. What matters is that a reaction in which phosphoryl groups are transferred to another molecule—breaking these bonds—is a process with a large negative free energy change. Using the simple example of ATP hydrolysis, we can state that a large amount of free energy is released when ATP is hydrolyzed because the products of the reaction have less free energy than the reactants. It is worth examining two reasons why this is so. 1. The ATP hydrolysis products are more stable than the reactants. At physiological pH, ATP has three to four negative charges (its pK is close to 7) and the anionic groups repel each other. In the products ADP and Pi, separation of the charges relieves some of this unfavorable electrostatic repulsion. 2. A compound with a phosphoanhydride bond experiences less resonance stabilization than its hydrolysis products. Resonance stabilization reflects the degree of electron delocal­ ization in a molecule and can be roughly assessed by the number of different ways of depicting the molecule’s structure. There are fewer equivalent ways of arranging the bonds of the terminal phosphoryl group of ATP than there are in free Pi:

356  C ha pter 12   Metabolism and Bioenergetics Terminal phosphoryl group of ATP

Inorganic phosphate (Pi )

O

O

−

O

P

−

ADP

O

O−

O O

P

OH

O− 2−

P

ADP

O

O

3−

P

O ∙ H+

O   

O

To summarize, ATP functions as an energy currency because its hydrol­ysis reaction is highly exergonic (ΔG ≪ 0). The favorable ATP reaction (ATP → ADP) can therefore pull another, unfavorable reaction with it, provided that the sum of the free energy changes for both reactions is less than zero. In effect, the cell “spends” ATP to make another process happen.

Energy can take different forms ATP is not the only substance that functions as energy currency in the cell. Other compounds that participate in reactions with large negative changes in free energy can serve the same purpose. For example, a number of phosphorylated compounds other than ATP can give up their phosphoryl group to another molecule. Table 12.4 lists the standard free energy changes for some of these reactions in which the phosphoryl group is transferred to water. Although hydrolysis of the bond linking the phosphate group to the rest of the molecule could be a wasteful process (the product would be free phosphate, Pi), the values listed in the table are a guide to how such compounds would behave in a coupled reaction, such as the hexo­kinase reaction described above. For example, phosphocreatine has a standard free energy of hydrolysis of –43.1 kJ · mol−1: +

TA BLE 1 2 .4

 Standard Free Energy Change for Phosphate Hydrolysis

Compound

ΔG°′ (kJ · mol−1)

Phosphoenolpyruvate

–61.9

1,3-Bisphosphoglycerate

–49.4

ATP → AMP + PPi

Phosphocreatine

ATP → ADP + Pi

Glucose-1-phosphate

–45.6 –43.1 –30.5

PPi → 2 Pi

–20.9

Glucose-6-phosphate

–13.8

Glycerol-3-phosphate

 –9.2

–19.2

H2N

O

C

NH P

N

CH3 O−

+

O−

H2O

Pi

H2N C

NH2

N

CH3

CH2

CH2

COO−

COO−

Phosphocreatine

Creatine

Creatine has lower energy than phosphocreatine since it has two, rather than one, resonance forms; this resonance stabilization contributes to the large negative free energy change when phosphocreatine transfers its phosphoryl group to another compound. In muscles, phosphocreatine transfers a phosphoryl group to ADP to boost the production of ATP during periods of high demand. Like ATP, other nucleoside triphosphates have large negative standard free energies of hydrolysis. GTP rather than ATP serves as the energy currency for reactions that occur during cellular signaling (Section 10.2) and protein synthesis (Section 22.3). In the cell, nucleoside triphosphates are freely interconverted by reactions such as the one catalyzed by nucleoside diphosphate kinase, which transfers a phosphoryl group from ATP to a nucleoside diphosphate (NDP):

​ATP + NDP ⇌ ADP + NTP​

12.3  Free Energy Changes in Metabolic Reactions  357

Because the reactants and products are energetically equivalent, ΔG°′ values for these reactions are near zero, and the reactions can proceed in either direction. Another class of compounds that can release a large amount of energy upon hydrol­ ysis are thioesters, such as acetyl-CoA. Coenzyme A is a nucleotide derivative with a side chain ending in a sulfhydryl (SH) group (see Fig. 3.2a). An acyl or acetyl group (the “A” for which coenzyme A was named) is linked to the sulfhydryl group by a thioester bond. Hydrol­ysis of this bond has a ΔG°′ value of –31.5 kJ · mol−1, comparable to that of ATP hydrolysis: thioester bond

O CH3

C

O

H2O S

Acetyl-CoA

CoA

CH3

C

O− + HS

CoA

Hydrolysis of a thioester is more exergonic than the hydrolysis of an ordinary (oxygen) ester because thioesters have less resonance stability than oxygen esters, owing to the larger size of an S atom relative to an O atom. An acetyl group linked to coenzyme A can be readily transferred to another molecule because formation of the new linkage is powered by the favorable free energy change of breaking the thioester bond. We have already seen that in oxidation–reduction reactions, cofactors such as NAD+ and ubiquinone can collect electrons. The reduced cofactors are a form of energy currency because their subsequent reoxidation by another compound occurs with a negative change in free energy. Ultimately, the transfer of electrons from one reduced cofactor to another and finally to oxygen, the final electron acceptor in many organisms, releases enough energy to drive the synthesis of ATP. Keep in mind that free energy changes occur not just as the result of chemical changes such as phosphoryl-group transfer or electron transfer. As decreed by the first law of thermodynamics (Section 1.3), energy can take many forms. We will see that ATP production in cells depends on the energy of an electrochemical gradient, that is, an imbalance in the concentration of a substance (in this case, protons) on the two sides of a membrane. The energy released in dissipating this gradient (allowing the system to move toward equilibrium) is converted to the mechanical energy of an enzyme that synthesizes ATP. In photosynthetic cells, the chemical reactions required to generate carbohydrates are ultimately driven by the energy changes of reactions in which light-excited molecules return to a lower-energy state.

Regulation occurs at the steps with the largest free energy changes In a series of reactions that make up a metabolic pathway, some reactions have ΔG values near zero. These near-equilibrium reactions are not subject to a strong driving force to proceed in either direction. Rather, flux can go forward or backward, according to slight fluctuations in the concentrations of reactants and products. When the concentrations of metabolites change, the enzymes that catalyze these near-equilibrium reactions tend to act quickly to restore the near-equilibrium state. Reactions with large negative changes in free energy have a longer way to go to reach equilibrium; these are the reactions that experience the greatest “urge” to proceed forward. However, the enzymes that catalyze these reactions do not allow the reaction to reach equilibrium because they work too slowly. Often the enzymes are already saturated with substrate, so the reactions cannot go any faster (when [S] ≫ KM, ʋ ≈ Vmax; Section 7.2). The rates of these far-from-equilibrium reactions limit flux through the entire pathway because the reactions function like dams:

Free energy

358  C ha pter 12   Metabolism and Bioenergetics

∆G

pathway’s rate-limiting reaction near-equilibrium reactions

Metabolic pathway

Cells can regulate flux through a pathway by adjusting the rate of a reaction with a large free energy change. This can be done by increasing the amount of enzyme that catalyzes that step or by altering the intrinsic activity of the enzyme through allosteric mechanisms (see Fig. 7.17). As soon as more metabolite has gone past the dam, the near-equilibrium reactions go with the flow, allowing the pathway intermediates to move toward the final product. Most metabolic pathways do not have a single flow-control point, as the dam analogy might suggest. Instead, flux is typically controlled at several points to ensure that the pathway can work efficiently as part of the cell’s entire metabolic network.

Before Going On • Explain why free energy changes must be negative for reactions in vivo. • Relate the standard free energy change to a reaction’s equilibrium constant. • Describe the difference between ΔG and ΔG°′. • Explain why it is misleading to refer to ATP as a high-energy molecule. • Explain why cleavage of one of ATP’s phosphoanhydride bonds releases large amounts of free energy. • List the types of energy “currencies” used by cells. • Describe the reasons why cells control metabolic reactions that have large negative free energy changes.

Summary 12.1  Food and Fuel •  Polymeric food molecules such as starch, proteins, and triacylglycerols are broken down to their monomeric components (glucose, amino acids, and fatty acids), which are absorbed. These materials are stored as polymers in a tissue-specific manner. •  Metabolic fuels are mobilized from glycogen, fat, and proteins as needed.

12.2  Metabolic Pathways •  Series of reactions known as metabolic pathways break down and synthesize biological molecules. Several pathways make use of the same small molecule intermediates.

•  During the oxidation of amino acids, monosaccharides, and fatty acids, electrons are transferred to carriers such as NAD+ and ubiquinone. Reoxidation of the reduced cofactors drives the synthesis of ATP by oxidative phosphorylation. •  Metabolic pathways form a complex network, but not all cells or organisms carry out all possible metabolic processes. Humans rely on other organisms to supply vitamins and other essential materials.

12.3  Free Energy Changes in Metabolic Reactions •  The standard free energy change for a reaction is related to the equilibrium constant, but the actual free energy change also depends on the actual cellular concentrations of reactants and products.

Problems  359 •  A thermodynamically unfavorable reaction may proceed when it is coupled to a favorable process involving ATP, whose phosphoanhy­d ride bonds release a large amount of energy when cleaved.

•  Other forms of cellular energy currency include phosphorylated compounds, thioesters, and reduced cofactors. •  Cells regulate metabolic pathways at the steps that are farthest from equilibrium.

Key Terms catabolism anabolism metabolism chemoautotroph photoautotroph heterotroph lipoprotein adipose tissue metabolic fuel mobilization phosphorolysis diabetes mellitus lysosome proteasome

degron metabolic pathway intermediate metabolite glycolysis citric acid cycle oxidation reduction redox reaction cofactor coenzyme oxidative phosphorylation flux essential compound

transcriptomics transcriptome microarray (DNA chip) proteomics proteome metabolomics metabolome vitamin equilibrium constant (Keq) standard free energy change (ΔG°′) standard conditions mass action ratio resonance stabilization thioester

Bioinformatics Brief Bioinformatics Exercises

Bioinformatics Projects

12.1  Metabolism and the BRENDA and KEGG Databases

Metabolic Enzymes, Microarrays, and Proteomics

12.2  Cofactor Chemistry

Metabolomics Databases and Tools

Problems 12.1  Food and Fuel 1.  Classify the following organisms as chemoautotrophs, photoautotrophs, or heterotrophs: a. Hydrogenobacter, which converts molecular hydrogen and oxygen to water, b. Arabidopsis thaliana, a green plant, c. the nitrosifying bacteria, which oxidize NH3 to nitrite, d. Saccharomyces cerevisiae, yeast, e. Caenorhabditis ele­ gans, a nematode worm, f. the Thiothrix bacteria, which oxidize hydrogen sulfide, g. Cyanobacteria (erroneously termed “bluegreen algae” in the past). 2.  The purple nonsulfur bacteria obtain their cellular energy from a photosynthetic process that does not produce oxygen. These bacteria also require an organic carbon source. Using the terms in this chapter, coin a new term that describes the trophic strategy of this organism. 3.  Digestion of carbohydrates begins in the mouth, where salivary amylases act on dietary starch. When the food is swallowed and enters the stomach, carbohydrate digestion ceases (it resumes in the small intestine). Why does carbohydrate digestion not occur in the stomach? 4.  Pancreatic amylase, which is similar to salivary amylase, is secreted by the pancreas into the small intestine. The active site of pancreatic amylase accommodates five glucosyl residues and hydrolyzes the glycosidic bond between the second and third residues. The enzyme cannot accommodate branched chains. a. What type

of glycosidic bond is hydrolyzed by amylase? b. What are the main products of amylase digestion? c. What are the products of amylopectin digestion? 5.  Starch digestion is completed by the enzymes isomaltase (or α-dextrinase), which catalyzes the hydrolysis of α(1→6) glycosidic bonds, and maltase, which hydrolyzes α(1→4) bonds. Why are these enzymes needed in addition to α-amylase? 6.  White flour is highly processed to remove the germ and the bran from the grain, leaving the starch-rich endosperm. Whole-wheat flour is less processed and retains all of the elements of the whole grain. The starch in white bread is digested more quickly and efficiently than the starch in whole-wheat bread. Explain the reason for this observation. 7.  The KM and kcat values for isomaltase and maltase (see Problem 5) are shown in the table. Calculate the catalytic efficiency (see Section 7.2) of each enzyme. Considering only these parameters, which enzyme would make the greater contribution to starch digestion in the small intestine? KM (mM−1) kcat (s−1)

Maltase 0.4 63

Isomaltase 61 3.4

8.  You carry out an experiment in which you determine that isomaltase makes a greater contribution to starch digestion than maltase.

360  C ha pter 12   Metabolism and Bioenergetics Assuming that your results are correct, how can you explain these results along with the data in Solution 7?

acid promotes emulsification of fats in the stomach; that is, the products are more easily incorporated into micelles. Explain why.

9.  Monosaccharides, the products of polysaccharide and disaccharide digestion, enter the cells lining the intestine via a specialized transport system. What is the source of free energy for this transport process?

20.  Most of the fatty acids produced in the reaction described in Problem 19 form micelles and are absorbed as such, but a small percentage of fatty acids are free and are transported into the intestinal epithelial cells without the need for a transport protein. Explain why a transport protein is not required.

10.  Unlike the monosaccharides described in Problem 9, sugar alcohols such as sorbitol (see Solution 11-21c) are absorbed via passive diffusion. Why? Which process occurs more rapidly, passive diffusion or passive transport? 11.  Use what you know about the properties of alcohol (ethanol) to describe how it is absorbed in both the stomach and the small intestine. What effect does the presence of food have on the absorption of ethanol? 12.  Nucleic acids that are present in food are hydrolyzed by digestive enzymes. What type of mechanism most likely mediates the entry of the reaction products into intestinal cells? 13.  Hydrolysis of proteins begins in the stomach, catalyzed by the hydrochloric acid secreted into the stomach by parietal cells. a. How does the low pH of the stomach affect protein structure in such a way that the proteins are prepared for hydrolytic digestion? b. Draw the reaction that shows the hydrolysis of a peptide bond catalyzed by hydrochloric acid.

21.  The cells lining the small intestine absorb cholesterol but not cholesteryl esters. Draw the reaction catalyzed by cholesteryl esterase that produces cholesterol from cholesteryl stearate. 22.  Some cholesterol is converted back to cholesteryl esters in the epithelial cells lining the small intestine (the reverse of the reaction described in Problem 21). Both cholesterol and cholesteryl esters are packaged into particles called chylomicrons, which consist of lipid and protein. Use what you know about the physical properties of cholesterol and cholesteryl esters to describe their locations in the chylomicron particle. 23. a. Consider the physical properties of a polar glycogen molecule and an aggregation of hydrophobic triacylglycerols. On a per-weight basis, why is fat a more efficient form of energy storage than glycogen? b. Explain why there is an upper limit to the size of a glycogen molecule but there is no upper limit to the amount of triacylglycerols that an adipocyte can store.

14.  In addition to hydrochloric acid, the gastric protease pepsin catalyzes the hydrolysis of peptide bonds in dietary proteins. Like the serine proteases (see Section 6.4), pepsin is synthesized as a zymogen and is inactive at its site of synthesis, where the pH is 7. Pepsin becomes activated when secreted into the stomach, where it encounters a pH of ~2. Pepsinogen contains a “basic peptide” that blocks its active site at pH 7. The basic peptide dissociates from the active site at pH 2 and is cleaved, resulting in the formation of the active form of the enzyme. What amino acid residues are found in the active site of pepsin? Why does the basic peptide bind tightly to the active site at pH 7 and why does it dissociate at the lower pH?

24.  Glycogen can be expanded quickly, by adding glucose residues to its many branches, and degraded quickly, by simultaneously removing glucose from the ends of these branches. Are the enzymes that catalyze these processes specific for the reducing or nonreducing ends of the glycogen polymer? Explain.

15.  The hydrolysis of peptide bonds in the stomach is catalyzed by both hydrochloric acid (see Problem 13) and the gastric protease pepsin (see Problem 14). Peptide bond hydrolysis continues in the small intestine, catalyzed by the pancreatic enzymes trypsin and chymotrypsin. At what pH does pepsin function optimally; that is, at what pH is the Vmax for pepsin greatest? Is the pH optimum for pepsin different from that for trypsin and chymotrypsin? Explain.

26.  Hydrolytic enzymes encased within the membrane-bound lysosomes all work optimally at pH ∼5. This feature serves as a cellular “insurance policy” in the event of lysosomal enzyme leakage into the cytosol. Explain.

16.  Scientists have recently discovered why the botulinum toxin survives the acidic environment of the stomach. The toxin forms a complex with a second nontoxic protein that acts as a shield to protect the botulinum toxin from being digested by stomach enzymes. Upon entry into the small intestine, the two proteins dissociate and the botulinum toxin is released. What is the likely interaction between the botulinum toxin and the nontoxic protein, and why does the complex form readily in the stomach but not in the small intestine? 17.  Free amino acid transport from the intestinal lumen into intestinal cells requires Na+ ions. Draw a diagram that illustrates amino acid transport into these cells.

25.  The phosphorolysis reaction that removes glucose residues from glycogen in the liver yields as its product glucose-1-phosphate. Glucose-1-phosphate is isomerized to glucose-6-phosphate; then the phosphate group is removed in a hydrolysis reaction. Why is it necessary to remove the phosphate group before the glucose exits the cell to enter the circulation?

27.  Proteins destined for degradation by the proteasome are “tagged” by covalently linking the C-terminus of ubiquitin to a Lys side chain on the doomed protein. a. Draw the structure of this linkage. What type of linkage is this? b. Linkages between ubiquitin and other amino acids, such as Ser residues, have been discovered. Draw the structure of this linkage. What type of linkage is this? 28. a. One protease active site in the proteasome has an AspHis-Ser catalytic triad. What type of enzyme is this? b. Proteasomal proteases can be inhibited by compounds such as ixazomib (shown below). How does this compound inhibit protease activity? c. Ixazomib can be used as an anticancer drug. Why would inhibiting proteasomal proteases be detrimental to the cancer cell?

Cl

18.  In oral rehydration therapy (ORT), patients suffering from diarrhea are given a solution consisting of a mixture of glucose and electrolytes. Some formulations also contain amino acids. Why are electrolytes added to the mixture? 19.  Triacylglycerol digestion begins in the stomach. Gastric lipase catalyzes hydrolysis of the fatty acid from the third glycerol carbon. a. Draw the reactants and products of this reaction.  b.  Conversion of the triacylglycerol to a diacylglycerol and a fatty

CH3 O O

N H

HO Ixazomib

O

Problems  361

12.2  Metabolic Pathways 29.  The common intermediates listed in the table—acetyl-CoA, glyceraldehyde-3-phosphate (GAP), and pyruvate—appear as reactants or products in several pathways. Place a check mark in the box that indicates the appropriate pathway—glycolysis, citric acid cycle, fatty acid metabolism, triacylglycerol (TAG) synthesis, photosynthesis, and transamination—for each reactant.

Acetyl-CoA GAP Pyruvate

Glycolysis Citric acid cycle Fatty acid metabolism TAG synthesis Photosynthesis Transamination

a.  A reaction from the alcoholic fermentation pathway, a catabolic process:

C

CH3

C

H

C

C

C

SCoA

CH2

CH2

CH2

C

SCoA

O c.  A reaction associated with the citric acid cycle, which plays an important role in catabolism:

HO

COO−

COO−

C

C

H

O

CH2

CH2

COO–

COO–

d.  A reaction associated with the anabolic pentose phosphate pathway:

H OH

CH2OPO2– 3 O OH H OH H H H

OH

OH

CH2OPO2– 3

H

OPO2– 3

C

OH

CH2OPO2– 3

HPO2– 4

O

C

O–

C

O

CH3

H

C

O–

C

OH

CH3

34. a. For each of the reactions shown in Problem 33, identify the cofactor as NAD +, NADP +, NADH, or NADPH. b. The reaction shown in Problem 33a is one of the 10 steps in glycolysis. The reaction shown in Problem 33b occurs at the end of glycolysis if oxygen is unavailable. Why might these two reactions be linked to one another?

36.  Prokaryotic species A and B live close together because they are metabolically interdependent. a. If species A converts CH4 to CO2, what process would species B need to undertake: the conversion of 2− 2− S2− to ​SO​ 2− 4​  ​​or the conversion of ​​SO​ 4​  ​​ to S ? b. If species A converts CH4 to CO2, what process would species B need to undertake: the conversion of Fe3+ to Fe2+ or the conversion of Fe2+ to Fe3+?

H O R

C

C

I dentify the reaction component that undergoes b. oxidation and c. reduction.

H CH2

H

− CH4 + ​SO​ 2− 4​  ​​ → _______ + HS + H2O

CH3

b.  A reaction from the anabolic fatty acid synthesis pathway:

R

H

C

35.  A potential way to reduce the concentration of methane, a greenhouse gas, is to take advantage of sulfate-reducing bacteria. a. Complete the chemical equation for methane consumption by these organisms:

OH H

O

O

O

31.  For each of the (unbalanced) reactions shown below, tell whether the reactant is being oxidized or reduced.

H

a.   A reaction associated with the catabolic glycolytic pathway:

b.  A reaction that follows the catabolic glycolytic pathway if oxygen is unavailable:

30.  The pyruvate → acetyl-CoA reaction can proceed only in the direction indicated. Given this limitation, consult Figure 12.9 and tell which of the following transformations are possible: a. acetyl-CoA → glucose, b. acetyl-CoA → fatty acids,  c.  acetyl-CoA → alanine.

O

33.  For each of the (unbalanced) reactions shown below, tell whether the reactant is being oxidized or reduced.

H

CH2OPO2– 3 O H OH H

O

OH H

OH

32.  For each of the reactions shown in Problem 31, identify the cofactor as NAD+, NADP+, NADH, or NADPH.

37.  Vitamin B 12 is synthesized by certain gastrointestinal bacteria and is also found in foods of animal origin such as meat, milk, eggs, and fish. When vitamin B12–containing foods are consumed, the vitamin is released from the food and binds to a salivary vitamin B12–binding protein called haptocorrin. The haptocorrin–vitamin B12 complex passes from the stomach to the small intestine, where the vitamin is released from the haptocorrin and then binds to intrinsic factor (IF). The IF–vitamin B12 complex then enters the cells lining the intestine by receptor-mediated endocytosis. Using this information, make a list of individuals most at risk for vitamin B12 deficiency. 38.  A bodybuilding website sells the supplement reduced CoQ10, which is ubiquinone. The supplement is not FDA-approved, and most of the claims made by the company are not backed by solid scientific studies, although some claims are based on the known chemical properties of ubiquinone. Why might the company that sells the supplement claim that CoQ10 produces energy in a way that would enhance performance during exercise? 39.  A vitamin K–dependent carboxylase enzyme catalyzes the γ-carboxylation of specific glutamate residues in blood coagulation proteins. a. Draw the structure of a γ-carboxyglutamate residue. b. Why does this post-translational modification assist coagulation proteins in binding the Ca2+ ions required for blood clotting?

362  C ha pter 12   Metabolism and Bioenergetics 40.  Hartnup disease is a hereditary disorder caused by a defective transporter for nonpolar amino acids. a. The symptoms of the disease (photosensitivity and neurological abnormalities) can be prevented through dietary adjustments. What sort of diet would be effective? b. Patients with Hartnup disease often exhibit pellagra-like symptoms. Explain. 41.  Refer to Table 12.2 and identify the vitamin required to accomplish each of the following reactions:   a. 

O O O H H H NN C CC C CC N

O O O H H H NN C CC C CC N

OH OH OH COO–––– b.  COO COO

COO–––– COO COO O+ +ADP ADP+ +PPPi i CC O O + ADP + C i

COO O+ +ATP ATP+ +HCO HCO–––3–3 CC O O + ATP + HCO C 3 3

CH33 CH CH 33 c.

COO–––– COO COO

++H H3NN + + H333N

O+ + CC O O + C CH33 CH CH 33

CH22 CH CH 22 COO–––– COO COO

COO–– CH COO CH CH COO–– CH22 CH CH 2 2

CH22 CH CH 22 COO–––– COO COO

i

COO–––– COO COO–––– COO COO COO O CC O O C ++H H N CH + + + CH 3N CH + + H33N 3 CH22 CH CH 22 CH33 CH CH 33 CH CH222 CH 2

COO–––– COO COO

COO–––– d.  COO COO

O O O H3C C C CoA+ +CO CO22 H CoA + CO CC SSS CoA 3C H 33 22

COO O + + HS HS CoA CoA CC O O + HS CoA C CH33 CH CH 33

42.  Would you expect vitamin A to be more easily absorbed from raw carrots or from cooked carrots? Explain. 43.  The microbial enzyme lactate racemase contains a nickel-based prosthetic group. a. To which common coenzyme is this prosthetic group related? b. Which two amino acid side chains hold the group in place?

N Ni2+ S

O

O− −O

P

+

O

N H

N

O

O H

H

H

OH

OH

45. a. Calculate ΔG°′ for a reaction at 25°C when Keq = 0.25. b. If the reaction were carried out at 37°C, how would ΔG°′ change? c. Is the reaction spontaneous? 46. a. Calculate ΔG°′ for a reaction at 25°C when Keq = 4. b. If the reaction were carried out at 37°C, how would ΔG°′ change? c. Is the reaction spontaneous? d. How does this reaction compare to the reaction described in Problem 45?

47.  Consider two reactions: A ⇌ B and C ⇌ D. K eq for the A ⇌ B reaction is 10, and Keq for the C ⇌ D reaction is 0.1. You place 1 mM A in tube 1 and 1 mM C in tube 2 and allow the reactions to reach equilibrium. Without doing any calculations, determine whether the concentration of B in tube 1 will be greater than or less than the concentration of D in tube 2. 48.  Calculate the ∆G°′ values for the reactions described in Problem 47. Assume a temperature of 37°C.

49.  For the reaction E ⇌ F, Keq = 1. a. Without doing any calculations, what can you conclude about the ∆G°′ value for the reaction? b. You place 1 mM F in a tube and allow the reaction to reach equilibrium. Determine the final concentrations of E and F. 50.  Refer to the hypothetical reaction described in Problem 49. Determine the direction in which the reaction will proceed if you place 5 mM E and 2 mM F in a test tube. What are the final concentrations of E and F?

51.  For the reaction R ⇌ P, ∆G°′ = 4.1 kJ ⋅ mol−1. Calculate Keq for the reaction. 52. a. Calculate Keq for the reverse of the reaction described in Problem 51. b. Compare your answer with your answer to Problem 51. What is the relationship between the sign of ∆G°′ and the ratio of products to reactants at equilibrium?

53.  Calculate ∆G for the A ⇌ B reaction described in Problem 47 when the concentrations of A and B are 0.9 mM and 0.1 mM, respectively. In which direction will the reaction proceed? Assume a temperature of 37°C. 54.  Calculate ∆G for the C ⇌ D reaction described in Problem 47 when the concentrations of C and D are 0.9 mM and 0.1 mM, respectively. In which direction will the reaction proceed? 55.  Calculate ∆G for the E ⇌ F reaction described in Problem 49 under the conditions described in Problem 50. Assume a temperature of 37°C. Is the reaction spontaneous? Is this consistent with your answer to Problem 50?

56.  Calculate ∆G for the R ⇌ P reaction described in Problem 51 when the concentrations of R and P are 0.9 mM and 0.1 mM, respectively. Assume a temperature of 37°C. In which direction will the reaction proceed?

NH

S

12.3  Free Energy Changes in Metabolic Reactions

H

44.  Why is niacin technically not a vitamin?

57.  The ∆G°′ value for a hypothetical reaction is 10 kJ ⋅ mol−1 at 25°C. Compare the Keq for this reaction with the Keq for a reaction whose ∆G°′ value is twice as large. 58.  The ∆G°′ value for a hypothetical reaction is –10 kJ ⋅ mol−1 at 25°C. a. Compare the Keq for this reaction with the Keq for a reaction whose ∆G°′ value is twice as large. b. Compare your answer in part a to the answer to Problem 57. 59.  Calculate ΔG for the phosphoglucomutase reaction shown in Sample Calculation 12.2 at 37°C when the initial concentration of glucose-1-phosphate (G1P) is 5 mM and the initial concentration of glucose-6-phosphate (G6P) is 20 mM. Is the reaction spontaneous under these conditions?

Problems  363 60.  Calculate the ratio of the concentration of glucose-6-phosphate (G6P) to the concentration of glucose-1-phosphate (G1P) (see Problem 59) that gives a free energy change of –2.0 kJ · mol−1. Assume a temperature of 37°C. 61.  Use the standard free energies provided in Table 12.4 to calculate ∆G°′ for the isomerization of glucose-1-phosphate to glucose-6-phosphate. Is this value consistent with the value of ∆G°′ shown in Sample Calculation 12.2? Is this reaction spontaneous under standard conditions? 62.  Is the reaction described in Problem 61 spontaneous at 37°C when the concentration of glucose-6-phosphate is 5 mM and the concentration of glucose-1-phosphate is 0.1 mM? 63.  Use the standard free energies provided in Table 12.4 to calculate ∆G°′ for the synthesis of ATP from ADP when phosphoenolpyruvate (PEP) donates a phosphate group to ADP. a. Is this reaction favorable under standard conditions at 25°C? b. Is this reaction favorable when [PEP] = 10 mM, [ATP] = 5 mM, [ADP] = 5 µM, [pyruvate] = 20 mM, and the temperature is 37°C? 64.  Use the standard free energies provided in Table 12.4 to calculate ∆G°′ for the synthesis of glycerol-3-phosphate (G3P) when ATP serves as the phosphoryl donor. a. Is this reaction favorable under standard conditions at 25°C? b. Is this reaction favorable when [glycerol] = 3 mM, [ATP] = 5 mM, [ADP] = 5 µM, [G3P] = 1 mM, and the temperature is 37°C?

65.  Construct a graph plotting free energy versus the reaction coordinate for the following reactions: a. glucose + P i → glucose6-phosphate, b. ATP + H2O → ADP + Pi, and c. the coupled reaction. 66. a. Which of the compounds listed in Table 12.4 could be involved in a reaction coupled to the synthesis of ATP from ADP + Pi? b. Which of the compounds listed in Table 12.4 could be involved in a reaction coupled to the hydrolysis of ATP to ADP + Pi?

67.  ∆G°′ for the hydrolysis of ATP under standard conditions at pH 7 and in the presence of magnesium ions is −30.5 kJ ⋅ mol−1. a. How would this value change if ATP hydrolysis were carried out at a pH of less than 7? Explain. b. How would this value change if magnesium ions were not present?

68.  Creatine undergoes reversible phosphorylation in muscles: ATP + creatine ⇌ phosphocreatine + ADP. Some studies (but not all) show that creatine supplementation increases performance in high-intensity exercises lasting less than 30 seconds. Would you expect creatine supplements to affect endurance exercise? 69.  Calculate ∆G for the hydrolysis of ATP under cellular conditions, where [ATP] = 3 mM, [ADP] = 1 mM, and [Pi] = 5 mM. Assume a temperature of 37°C.

70.  The standard free energy change for the reaction catalyzed by triose phosphate isomerase is 7.9 kJ ⋅ mol−1.

H

H

H

C

O

C

OH

CH2OPO2– 3 Glyceraldehyde-3-phosphate

H

C

OH

C

O

CH2OPO2– 3 Dihydroxyacetone phosphate

a.  Calculate the equilibrium constant for the reaction at 25°C. b. Calculate ∆G at 37°C when the concentration of glyceraldehyde-3-phosphate (GAP) is 0.1 mM and the concentration of dihydroxyacetone phosphate (DHAP) is 0.5 mM. c. Is the reaction

spontaneous under these conditions? Would the reverse reaction be spontaneous?

71.  An apple contains about 72 Calories. Express this quantity in terms of ATP equivalents (that is, the number of ATP → ADP + Pi reactions). (Note: A nutritional “Calorie” is equivalent to 1 kilocalorie and 1 cal = 4.184 J.) 72.  A large hot chocolate with whipped cream purchased at a national coffee chain contains 760 Calories. Express this quantity in terms of ATP equivalents (see Problem 71). 73.  A moderately active adult female weighing 125 pounds must consume 2200 Calories of food daily a. If this energy is used to synthesize ATP, calculate the number of moles of ATP that would be synthesized each day under standard conditions (assuming 33% efficiency). b. Calculate the number of grams of ATP that would be synthesized each day. The molar mass of ATP is 505 g ⋅ mol−1. What is the mass of ATP in pounds (2.2 kg = 1 lb)? c. There is approximately 40 g of ATP in the adult 125-lb female. Consider this, and your answer to part b, and suggest an explanation that is consistent with these findings. 74. a. How many apples (see Solution 71) would be required to provide the amount of ATP calculated in Problem 73? b. How many large hot chocolate drinks (see Solution 72) would be required?

75.  The complete oxidation of glucose releases a considerable amount of energy. ∆G°′ for the reaction shown is –2850 kJ ⋅ mol−1. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

How many moles of ATP could be produced under standard conditions from the oxidation of one mole of glucose, assuming about 33% efficiency?

76.  The oxidation of palmitate, a 16-carbon saturated fatty acid, releases 9781 kJ ⋅ mol−1. C16H32O2 + 23 O2 → 16 CO2 + 16 H2O

a.  How many moles of ATP could be produced under standard conditions from the oxidation of one mole of palmitate, assuming 33% efficiency? b. Calculate the number of ATP molecules produced per carbon for glucose (see Solution 75) and palmitate. Explain the reason for the difference. 77.  Citrate is isomerized to isocitrate in the citric acid cycle (Chapter 14). The reaction is catalyzed by the enzyme aconitase. ∆G°′ for the reaction is 5 kJ ⋅ mol−1. The properties of the reaction are studied in vitro, where 1 M citrate and 1 M isocitrate are added to an aqueous solution of the enzyme at 25°C. a. What is the Keq for the reaction at 25°C? b. What are the equilibrium concentrations of the reactant and product? c. What is the preferred direction of the reaction under standard conditions? d. The aconitase reaction is the second step of an eight-step pathway and occurs in the direction shown in the figure. How can you reconcile these facts with your answer to part c?  

CH2 HO

C

COO–

COO–

CH2 Citrate

COO–

CH2

aconitase

HC HO

CH

COO–

COO– COO–

Isocitrate

78.  The equilibrium constant for the conversion of glucose6-phosphate (G6P) to fructose-6-phosphate (F6P) is 0.41. The reaction is reversible and is catalyzed by the enzyme phosphoglucose isomerase.

364  C ha pter 12   Metabolism and Bioenergetics

H HO

CH2OPO2– 3 CH2OPO2– 3 O phosphoglucose H2COH O H H isomerase OH H H HO OH H OH H

OH

Glucose-6-phosphate

OH

H

Fructose-6-phosphate

a.  What is ∆G°′ for this reaction? Would this reaction proceed in the direction written under standard conditions? b. What is ∆G for this reaction at 37°C when the concentration of glucose-6-phosphate is 2.0 mM and the concentration of fructose-6-phosphate is 0.5 mM? Would the reaction proceed in the direction written under these cellular conditions? 79.   The phosphorylation of glucose to glucose-6-phosphate, the first step of glycolysis (Chapter 13), can be described by the equation glucose + Pi ⇌ glucose-6-phosphate + H2O. a. Calculate the equilibrium constant for this reaction. b. What would the equilibrium concentration of glucose-6-phosphate (G6P) be under cellular conditions (both glucose and phosphate concentrations are 5 mM) if glucose were phosphorylated according to this reaction? c. Does this reaction provide a feasible route for producing glucose-6-phosphate for glycolysis? d. One way to increase the amount of product is to increase the concentrations of the reactants. This would decrease the mass action ratio (see Equation 12.3) and would theoretically make the reaction as written more favorable. What concentration of glucose would be required to achieve a glucose-6-phosphate concentration of 250 µM? Is this strategy physiologically feasible, given that the solubility of glucose in aqueous medium is less than 1 M? 80.  Another way to promote the formation of glucose-6-phosphate from glucose (see Problem 79) is to couple the phosphorylation of glucose to the hydrolysis of ATP as shown in Section 12.3. a. Calculate K eq for the reaction in which glucose is converted to ­glucose-6-phosphate with concomitant ATP hydrolysis. b. When the ATP-dependent phosphorylation of glucose is carried out, what concentration of glucose is needed to achieve a 250 µM intracellular concentration of glucose-6-phosphate when the concentrations of ATP and ADP are 5.0 mM and 1.25 mM, respectively? c. Which route is more feasible to accomplish the phosphorylation of glucose to glucose-6-phosphate: the direct phosphorylation by Pi (as described in Problem 79) or the coupling of this phosphorylation to ATP hydrolysis? Explain. 81.  Fructose-6-phosphate is phosphorylated to fructose-1,6-bisphosphate in the third step of the glycolytic pathway (Chapter 13). The phosphorylation of fructose-6-phosphate is described by the following equation:

fructose-6-phosphate + Pi ⇌ fructose-1,6-bisphosphate ∆G°′ = 47.7 kJ ⋅ mol−1

a.  What is the ratio of fructose-1,6-bisphosphate (F16BP) to ­fructose-6-phosphate (F6P) at equilibrium if the concentration of phosphate in the cell is 5 mM? b. Suppose that the phosphorylation of fructose-6-phosphate is coupled to the hydrolysis of ATP. Write the new equation that describes this process and calculate ∆G°′ for the reaction. c. What is the ratio of fructose-1,6-bisphosphate to fructose-6-phosphate at equilibrium for the reaction you wrote in part b if the equilibrium concentration of ATP = 3 mM and [ADP] = 1 mM? d. Compare your answers to parts a and c. What is the likely path for the cellular synthesis of fructose-1,6-bisphosphate?

82.  One can envisage two mechanisms for coupling ATP hydrolysis to the phosphorylation of fructose-6-phosphate (F6P) to ­fructose-1,6-bisphosphate (F16BP) (see Problem 81), yielding the same overall reaction: I: ATP is hydrolyzed as F6P is transformed to F16BP: F6P + Pi ⇌ F16BP + H2O

ATP + H2O ⇌ ADP + Pi

I I: ATP transfers its γ phosphate directly to F6P in one step, producing F16BP: F6P + ATP ⇌ F16BP + ADP

 hoose one of the above mechanisms as the more biochemically feaC sible and provide a rationale for your choice. 83.  Glyceraldehyde-3-phosphate (GAP) is converted to 3-phosphoglycerate (3PG) in one of the steps in the glycolytic pathway.

O

H H

C

O

C

OH

H

O–

C

OH

CH2OPO2– 3

CH2OPO2– 3 Glyceraldehyde-3-phosphate (GAP)

3-Phosphoglycerate (3PG)

O

H

C

C

OPO2– 3

C

OH

CH2OPO2– 3 1,3-Bisphosphoglycerate (1,3-BPG) Consider these two scenarios: I. GAP is oxidized to 1,3-BPG (∆G°′ = 6.7 kJ · mol−1), which is subsequently hydrolyzed to yield 3PG (∆G°′ = −49.3 kJ · mol−1). I I. GAP is oxidized to 1,3-BPG, which subsequently transfers its phosphate to ADP, yielding ATP (∆G°′ = −18.8 kJ · mol−1).  rite the overall equations for the two scenarios. Which is more W likely to occur in the cell, and why? 84.  The conversion of glutamate to glutamine is unfavorable. In order for this transformation to occur in the cell, it must be coupled to the hydrolysis of ATP. Consider two possible mechanisms: I : An ammonia group is transferred to glutamate and ATP is hydrolyzed:

glutamate + NH3 ⇌ glutamine ATP + H2O ⇌ ADP + Pi

I I: A phosphoryl group is transferred from ATP to glutamate, f­ ollowed by the transfer of an ammonia group to glutamate, with release of free phosphate:

glutamate + ATP ⇌ γ-glutamylphosphate + ADP

γ-glutamylphosphate + H2O + NH3 ⇌ glutamine + Pi

 rite the overall equation for the reaction for each mechanism. Is W one mechanism more likely than the other? Or are both mechanisms equally feasible for the conversion of glutamate to glutamine? Explain.

Chapter 12 Credits  365 85.  ∆G°′ for the formation of UDP–glucose from glucose-1phosphate and UTP is about zero. Yet the production of UDP–glucose is highly favorable. What is the driving force for this reaction? glucose-1-phosphate + UTP ⇌ UDP–glucose + PPi

86.  Palmitate is activated in the cell by forming a thioester bond to coenzyme A. ∆G°′ for the synthesis of palmitoyl-CoA from palmitate and coenzyme A is 31.5 kJ ⋅ mol−1.

O H3C

(CH2)14

C

O– + HSCoA O

H3C

(CH2)14

C

S

CoA + H2O

a.  What is the ratio of products to reactants at equilibrium? Is the reaction favorable? Explain.

b.  Suppose the synthesis of palmitoyl-CoA were coupled with ATP hydrolysis. Write the new equation for the activation of palmitate when coupled with ATP hydrolysis to ADP. Calculate ∆G°′ for the reaction. What is the ratio of products to reactants at equilibrium for the reaction? Is the reaction favorable? Compare your answer to the answer you obtained in part a. c.  Suppose the reaction described in part a were coupled with ATP hydrolysis to AMP. Write the new equation for the activation of palmitate when coupled with ATP hydrolysis to AMP. Calculate ∆G°′ for the reaction. What is the ratio of products to reactants at equilibrium for the reaction? Is the reaction favorable? Compare your answer to the answer you obtained in part b. d.  Pyrophosphate, PPi, is hydrolyzed to 2Pi. The activation of palmitate, as described in part c, is coupled to the hydrolysis of pyrophosphate. Write the equation for this coupled reaction and calculate ∆G°′. What is the ratio of products to reactants at equilibrium for the reaction? Is the reaction favorable? Compare your answer to the answers you obtained in parts b and c.

Selected Readings Coffey, R. and Ganz, T., Iron homeostasis: An anthropocentric perspective, J. Biol. Chem. 292, 12727–12734, doi: 10.1074/jbc. R117.781823 (2017). [Describes the storage and transport of iron and how these processes are regulated in humans.] Coll-Martínez, B. and Crosas, B., How the 26S proteasome degrades ubiquitinated proteins in the cell, Biomolecules 9, 395, doi: 10.3390/ biom9090395 (2019). [Provides structural and functional information about the proteasome, based on 30 years of research studies.] Kim, M.-S. et al., A draft map of the human proteome, Nature 509, 575–581 (2014). [Describes a dataset of 17,000 proteins representing 30 different human tissues.]

Ludwig, D.S., Willett, W.C., Volek, J.S., and Neuhouser, M.L., Dietary fat: From foe to friend? Science 362, 764–770, doi: 10.1126/science. aau2096 (2018). [Summarizes some of the controversies around high-carbohydrate and high-fat diets and the links to disease.] Martin, W.F. and Thauer, R.K., Energy in ancient metabolism, Cell 168, 953–955, doi: 10.1016/j.cell.2017.02.032 (2017). [Discusses ATP and other, more ancient energy currencies.] Wishart, D., Metabolomics for investigating physiological and pathophysiological processes, Physiol. Rev. 99, 1819–1875, doi: 10.1152/ physrev.00035.2018 (2019). [Reviews many practical aspects of metabolomics, including linking specific metabolites to diseases.]

Chapter 12 Credits Box 12.B Image of ferritin based on 1FHA. Lawson, D.M., Artymiuk, P.J., Yewdall, S.J., Smith, J.M., Livingstone, J.C., Treffry, A., Luzzago, A., Levi, S., Arosio, P., Cesareni, G., Thomas, C.D., Shaw, W.V., Harrison, P.M., Solving the structure of human H ferritin by genetically engineering intermolecular crystal contacts, Nature 349, 541–544 (1991).

Image of transferrin based on 1H76. Hall, D.R., Hadden, J.M., Leonard, G.A., Bailey, S., Neu, M., Winn, M., Lindley, P.F., The crystal and molecular structures of diferric porcine and rabbit serum transferrins at resolutions of 2.15 and 2.60A, respectively, Acta Crystallogr. D 58, 70 (2002).

photosomething/Adobe Stock

CHAPTER 13

Kimchi and other fermented foods take advantage of microbial metabolism. Salting draws water and nutrients out of raw vegetables and then sugar is added to initiate anaerobic glycolysis carried out by bacteria. The resulting lactate, along with other metabolites, gives the food its characteristic flavor and prevents the growth of organisms that would otherwise completely degrade and spoil the food.

Glucose Metabolism DO YOU REMEMBER? • Enzymes accelerate chemical reactions using acid–base catalysis, covalent catalysis, and metal ion catalysis (Section 6.2). • Glucose polymers include the fuel-storage polysaccharides starch and glycogen and the structural polysaccharide cellulose (Section 11.2). • Coenzymes such as NAD+ and ubiquinone collect electrons from compounds that become oxidized (Section 12.2). • A reaction with a large negative change in free energy can be coupled to another, unfavorable reaction (Section 12.3). • A reaction that breaks a phosphoanhydride bond in ATP occurs with a large change in free energy (Section 12.3). • Nonequilibrium reactions often serve as metabolic control points (Section 12.3).

Glucose occupies a central position in the metabolism of most cells. It is a major source of metabolic energy (in some cells, it is the only source), and it provides the precursors for the synthesis of other biomolecules. Recall that glucose is stored in polymeric form as starch in plants and as glycogen in animals (Section 11.2). The breakdown of these polymers provides glucose monomers that can be catabolized to release energy. In this chapter, we will examine the major metabolic pathways involving glucose, including the interconversion of the monosaccharide glucose with glycogen, the degradation of glucose to the three-carbon intermediate pyruvate, the synthesis of glucose from smaller compounds, and the conversion of glucose to the five-carbon monosaccharide ribose. For all the pathways, we will present the intermediates and some of the relevant enzymes. We will also examine the thermodynamics of reactions that release or consume large amounts of free energy and discuss how some of these reactions are regulated.

13.1 Glycolysis LEARNING OBJECTIVES Describe the substrates, products, and chemical reaction for each step of glycolysis. • Identify the energy-consuming and energy-generating steps of glycolysis. • List the flux-control points for the pathway. • Describe the metabolic uses of pyruvate. 366

13.1 Glycolysis  367

Pathways dealing with carbohydrates, highlighted in Figure 13.1, are part of the larger metabolic scheme introduced in Figure 12.10. Glycolysis, the conversion of glucose to pyruvate, is a good place to begin a study of metabolic pathways. As a result of many years of research, we know a great deal about the pathway’s intermediates and the enzymes that mediate their chemical transformations. We have also learned that glycolysis, along with other metabolic pathways, exhibits the following properties: 1. Each step of the pathway is catalyzed by a distinct enzyme.

BIOPOLYMERS Polysaccharides

1

MONOMERS Monosaccharides +

2. The energy consumed or released in certain reactions is transferred by molecules such as ATP and NADH. 3. The rate of the pathway can be controlled by altering the activity of individual enzymes. If metabolic processes did not occur via multiple enzyme-catalyzed steps, cells would have little control over the amount and type of reaction products and no way to manage energy. For example, the combustion of glucose and O2 to CO2 and H2O—if allowed to occur in one grand explosion—would release about 2850 kJ · mol −1 of energy all at once. In the cell, glucose combustion requires many steps so that the cell can recover its energy in smaller, more useful quantities. Glycolysis, representing the first ten steps of this process, appears to be an ancient metabolic pathway. The fact that it does not require molecular ­oxygen suggests that it evolved before photosynthesis increased the level of atmospheric O2. Overall, glycolysis is a series of enzyme-catalyzed steps in which a six-­carbon glu­cose molecule is broken down into two three-carbon pyruvate ­molecules. This catabolic pathway is accompanied by the phosphorylation of two molecules of ADP (to produce 2 ATP) and the reduction of two molecules of NAD+. The net equation for the pathway (ignoring water and protons) is glucose + 2 NAD+ + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 2 ATP

It is convenient to divide the 10 reactions of glycolysis into two phases. In the first (Reactions 1–5), the hexose is phosphorylated and cleaved in half. In the ­second (Reactions 6–10), the three-carbon molecules are converted to pyruvate (Fig. 13.2). As you examine each of the reactions of glycolysis described in the following pages, note how the reaction substrates are converted to products by the action of an enzyme (and note how the enzyme’s name often reveals its purpose). Pay attention also to the free energy change of each reaction.

NH4

NAD+

NAD+, Q

HO

H

OH

Glucose

OH

+ ATP

H hexokinase

HO

H

OH

NADH, QH2

citric acid cycle

photosynthesis

NADH, QH2 ADP

O2

oxidative phosphorylation ATP

H2O NAD+, Q

  FIGURE 13.1   Glucose metabolism in context.  (1) The polysaccharide glycogen is degraded to glucose, which is then catabolized by the glycolytic pathway (2) to the three-­ carbon intermediate pyruvate. Gluconeogenesis (3) is the pathway for the synthesis of glucose from smaller precursors. Glucose can then be reincorporated into glycogen (4). The conversion of glucose to ribose, a component of nucleotides, is not shown in this diagram.

CH2OPO2– 3 O H H + ADP OH H OH

Glucose-6-phosphate

NAD+

CO2

1. Hexokinase  In the first step of glycolysis, the enzyme hexokinase transfers a phosphoryl group from ATP to the C6 OH group of glucose to form glucose-6-phosphate: H

3

2- and 3-Carbon INTERMEDIATES

The early reactions of glycolysis can be considered as preparation for the later, energyproducing reactions. In fact, two of the early reactions require the investment of energy in the form of ATP.

H

2

NADH

Energy is invested at the start of glycolysis

CH2OH O H OH H

4

368  C ha pter 13   Glucose Metabolism   FIGURE 13.2   The reactions of

Glucose

glycolysis.  The substrates, products, and enzymes corresponding to the 10 steps of the pathway are shown. Shading indicates the substrates (blue) and products (green) of the pathway as a whole.

ATP

H HO

1 hexokinase

ADP

–2O

Glucose-6-phosphate

Question  Next to each reaction, write the term that describes the type of chemical change that occurs.

Fructose-6-phosphate

HO

–2O

ADP

Fructose-1,6-bisphosphate

–2O

Dihydroxyacetone phosphate triose phosphate 5 isomerase

+

2 NADH

glyceraldehyde-36 phosphate dehydrogenase

2 NAD+ + 2 H+

2 1,3-Bisphosphoglycerate

O H

H

H

OH

C C

OH

O

CH2OH

H

HO

HO

H

OH

2–

O

CH2OPO3

H

HO

HO

H

H

H

H OH

3POCH2

4 aldolase

+

OH

3POCH2

3 phosphofructokinase

OH

O

H

ATP

2 Pi

H

H

3POCH2

H

phosphoglucose 2 isomerase

Glyceraldehyde3-phosphate

CH2OH O H OH H

H

OH

2–

CH2OPO3 C

OH 2–

CH2OPO3

O H

O

CH2OH

2–

C C

OPO3 OH

2–

2 ADP 2 ATP

CH2OPO3

7

phosphoglycerate kinase

2 3-Phosphoglycerate

O H

C C

O– OH 2–

phosphoglycerate 8 mutase 2 2-Phosphoglycerate

CH2OPO3 O H

C C

O– 2–

OPO3

CH2OH

2 H2O

SEE GUIDED TOUR Glycolysis and the Citric Acid Cycle

9 enolase

O

C

O– 2–

2 Phosphoenolpyruvate

C

2 ADP

CH2

2 ATP

10 pyruvate kinase

2 Pyruvate

O

C C

OPO3

O– O

CH3

13.1 Glycolysis  369

A kinase is an enzyme that transfers a phosphoryl group to or from ATP (or another nucleoside triphosphate) to or from another substance. Recall from Section 6.3 that the hexokinase active site closes around its substrates so that a phosphoryl group is efficiently transferred from ATP to glucose. The standard free energy change for this reaction, which cleaves one of ATP’s phosphoanhydride bonds, is −16.7 kJ · mol−1 (ΔG, the actual free energy change for the reaction inside a cell, has a similar value). The magnitude of this free energy change means that the reaction proceeds in only one direction; the reverse reaction is extremely unlikely since its standard free energy change would be +16.7 kJ · mol−1. Consequently, hexokinase is said to catalyze a metabolically irreversible reaction that helps direct glucose toward its fate. Many metabolic pathways have a similar irreversible step near the start that commits a metabolite to proceed through the pathway.

2. Phosphoglucose Isomerase 

The second reaction of glycolysis is an isomerization reaction in which glucose-6-phosphate is converted to fructose-6-phosphate: –2

6

O3POCH2 O

5

H

H OH

4

HO

1

H

3

OH

2

H

–2

6

phosphoglucose isomerase

5

H

CH2OH 2

HO

H 4

OH

3

H

HO

OH

1

O

O3POCH2

H

Fructose-6-phosphate

Glucose-6-phosphate

Because fructose is a six-carbon ketose (Section 11.1), it forms a five-membered ring. This isomerization reaction is necessary in order to place a carbonyl group next to the C3—C4 bond that will be cleaved in step 4. The standard free energy change for the phosphoglucose isomerase reaction is +2.2  kJ · mol −1 , but the reactant concentrations in vivo yield a ΔG value of about −1.4 kJ · mol−1. A value of ΔG near zero indicates that the reaction operates close to equilibrium (at equilibrium, ΔG = 0). Such near-equilibrium reactions are considered to be freely reversible, since a slight excess of products can easily drive the reaction in reverse by mass action effects. In a metabolically irreversible reaction, such as the hexokinase reaction, the concentration of product could never increase enough to compensate for the reaction’s large value of ΔG.

3. Phosphofructokinase 

The third reaction of glycolysis consumes a second ATP  molecule in the phosphorylation of fructose-6-phosphate to yield fructose-1,6bisphosphate. –2

6

O3POCH2 5

H

H 4

HO

1

O

–2

CH2OH

HO 3

H

2

OH

Fructose-6-phosphate

6

O3POCH2

+ ATP

phosphofructokinase

5

H

H 4

HO

1

CH2OPO2– 3

O HO 3

2

OH

H

Fructose-1,6-bisphosphate

Phosphofructokinase operates in much the same way as hexokinase, and the reaction it catalyzes is irreversible, with a ΔG°′ value of −17.2 kJ · mol−1. In cells, the activity of phosphofructokinase is regulated. We have already seen how the activity of a bacterial phosphofructokinase responds to allosteric effectors (Section 7.3). ADP binds to the enzyme and causes a conformational change that promotes fructose-6phosphate binding, which in turn promotes catalysis. This mechanism is useful because the

+

ADP

370  C ha pter 13   Glucose Metabolism a.

b.

Glucose

Glucose

ATP Fructose-6-phosphate

Pi

Fructose-2,6bisphosphate

Fructose-6-phosphate

ADP Fructose-1,6-bisphosphate Fructose-1,6-bisphosphate Phosphoenolpyruvate Pyruvate Pyruvate

  FIGURE 13.3   Regulation of phosphofructokinase.  a. Regulation in bacteria. ADP, produced when ATP is consumed elsewhere in the cell, stimulates the activity of phosphofructokinase (green arrow). Phosphoenolpyruvate, a late intermediate of glycolysis, inhibits phosphofructokinase (red symbol), thereby decreasing the rate of the entire pathway.  b. Regulation in mammals.

concentration of ADP in the cell is a good indicator of the need for ATP, which is a product of glycolysis. Phosphoenolpyruvate, the product of step 9 of glycolysis, binds to bacterial phosphofructokinase and causes it to assume a conformation that destabilizes fructose-6phosphate binding, thereby diminishing catalytic activity. Thus, when the glycolytic pathway is producing plenty of phosphoenolpyruvate and ATP, the phosphoenolpyruvate can act as a feedback inhibitor to slow the pathway by decreasing the rate of the reaction catalyzed by phosphofructokinase (Fig. 13.3a). Citrate, an intermediate of the citric acid cycle (which completes glucose catabolism), is also a feedback inhibitor of phosphofructokinase. The most potent activator of phosphofructokinase in mammals is the compound fructose2,6-bisphosphate, which is synthesized from fructose-6-phosphate by an enzyme known as phosphofructokinase-2. (The glycolytic enzyme is therefore sometimes called phosphofructokinase-1.)

O

–2

O3POH2C

CH2OH

H

HO

HO

H

H

OH

Fructose-6-phosphate

ATP

ADP

phosphofructokinase-2

O

–2

O3POH2C

CH2OH

H

HO

HO

H

H

O PO2– 3

Fructose-2,6-bisphosphate

The activity of phosphofructokinase-2 is hormonally stimulated when the concentration of glucose in the blood is high. The resulting increase in fructose-2,6-bisphosphate concentration activates phosphofructokinase to increase the flux (rate of flow) of glucose through the glycolytic pathway (Fig. 13.3b). The phosphofructokinase reaction is the primary control point for glycolysis. It is the slowest reaction of glycolysis, so the rate of this reaction largely determines the flux of glu­cose through the entire pathway. In general, a rate-determining reaction—such as the phosphofructokinase reaction—operates far from equilibrium; that is, it has a large negative free energy change and is irreversible under metabolic conditions. The rate of the reaction can be altered by allosteric effectors but not by fluctuations in the concentrations of its substrates or products. Thus, it acts as a one-way valve. In contrast, a near-equilibrium reaction— such as the phosphoglucose isomerase reaction—cannot serve as a rate-determining step for a pathway because it can respond to small changes in reactant concentrations by operating in reverse.

13.1 Glycolysis  371

4. Aldolase  Reaction 4 converts the hexose fructose-1,6-bisphosphate to two three-­ carbon molecules, each of which bears a phosphate group. CH2OPO2– 3

CH2OPO2– 3

3

1

C

O

HO

C

H

H

C

OH

2

H

3 4

C

C

2

HO aldolase

OH

5

O

CH2 1

Dihydroxyacetone phosphate

+

O

C

H

1

H

C

2

OH

CH2OPO2– 3

CH2OPO2– 3

3

6

Glyceraldehyde3-phosphate

Fructose1,6-bisphosphate

This reaction is the reverse of an aldol (aldehyde–alcohol) condensation, so the enzyme that catalyzes the reaction is called aldolase. It is worth examining its mechanism. The active site of mammalian aldolase contains two catalytically important residues: a lysine residue that forms a Schiff base (imine) with the substrate and an ionized aspartate residue that acts as a base catalyst (Fig. 13.4). (Bacterial aldolase uses lysine and an ionized tyrosine residue.) Early studies of aldolase implicated a cysteine residue in catalysis because iodoacetate, a reagent that reacts with the cysteine side chain, also inactivates the enzyme:

C HC

O

HI

CH2

SH

+

NH

ICH2COO– Iodoacetate

C

O

HC

CH2

SCH2COO–

NH

Cys

Researchers used iodoacetate to help identify the intermediates of glycolysis: In the presence of iodoacetate, fructose-1,6-bisphosphate accumulates because the next step is blocked. Acetylation of the cysteine residue, which turned out not to be part of the active site, probably interferes with a conformational change that is necessary for aldolase activity. The ΔG°′ value for the aldolase reaction is +22.8 kJ · mol−1, indicating that the reaction is unfavorable under standard conditions. However, the reaction proceeds in vivo (ΔG is actually less than zero) because the products of the reaction are quickly whisked away by subsequent reactions. In essence, the rapid consumption of glyceraldehyde-3-phosphate and dihydroxyacetone phosphate “pulls” the aldolase reaction forward.

5. Triose Phosphate Isomerase  The products of the aldolase reaction are both phosphorylated three-carbon compounds, but only one of them—glyceraldehyde-3phosphate—proceeds through the remainder of the pathway. Dihydroxyacetone phosphate is converted to glyceraldehyde-3-phosphate by triose phosphate isomerase: H H

C

OH

C

O

CH2OPO2– 3 Dihydroxyacetone phosphate

O triose phosphate isomerase

H

C C

H OH

CH2OPO2– 3 Glyceraldehyde3-phosphate

372  C ha pter 13   Glucose Metabolism

(CH2)4

NH2 NH2

–

O

C

Asp

(CH2)4 Lys

–

O

O

C

CH2 2–

CH2OPO3

CH2

C

5. The Schiff base is hydrolyzed, releasing the second product and regenerating the enzyme.

1. Fructose-1,6-bisphosphate, shown in its linear form, binds to the enzyme.

H2O

2–

O

HO

C

H

H

C

O

H

C

H2N

H – OH O

CH2OPO2– 3 Fructose-1,6bisphosphate

C

(CH2)4

2–

CH2OPO3 HO

O

NH

C

H

–

O

2. The active-site Lys residue reacts

+

NH

HO

C

H

H

C

O

H

C

OH

–

2– CH2OPO3

C

O

4. The Asp residue gives up

its proton to the remainder of the substrate, re-forming a Schiff base.

(CH2)4 H

(CH2)4

CH2

with the substrate’s carbonyl group to form a Schiff base.

2– CH2OPO3

C

+

C H

CH2

H2O

O

CH2OH Dihydroxyacetone phosphate

CH2OPO3 C

O

O

C

O

CH2

3. The Asp side chain, acting as a base, accepts a proton from the substrate, and the bond between C3 and C4 breaks, releasing the first product, an aldehyde. The reaction is facilitated by the protonated Schiff base, which is a better electron-withdrawing group than the carbonyl group that was originally at C2.

2–

CH2OPO3 C HO

C H

NH

(CH2)4

H O

C

O

CH2 H SEE ANIMATED PROCESS DIAGRAM The enzymatic mechanism of aldolase

H

C C

O OH 2–

CH2OPO3 Glyceraldehyde-3-phosphate   FIGURE 13.4   The aldolase reaction.

Question  Does this reaction follow an ordered or ping pong mechanism (see Section 7.2)?

Triose phosphate isomerase was introduced in Section 7.2 as an example of a catalytically perfect enzyme, one whose rate is limited only by the rate at which its substrates can diffuse to its active site. The catalytic mechanism of triose phosphate isomerase may involve low-­barrier hydrogen bonds (which also help stabilize the transition state in serine proteases;­

13.1 Glycolysis  373

a.

b.

  FIGURE 13.5   Conformational changes in yeast triose phosphate isomerase.  a. The 11-residue

loop is highlighted in green.  b. The loop closes after substrate binds. In this model, the transition state analog 2-phosphoglycolate (orange) occupies the active site. Triose phosphate isomerase is actually a homodimer; only one subunit is pictured here.

see Section 6.3). In addition, the catalytic power of triose phosphate isomerase depends on the movement of a protein loop comprising residues 166–176. After a substrate binds, the loop closes over the active site and stabilizes the reaction’s transition state (Fig. 13.5). The standard free energy change for the triose phosphate isomerase reaction is slightly positive, even under physiological conditions (ΔG°′ = +7.9 · kJ · mol−1 and ΔG = +4.4 kJ · mol−1), but the reaction proceeds because glyceraldehyde-3-phosphate is quickly consumed in the next reaction, so more dihydroxyacetone phosphate is constantly being converted to glyceraldehyde-3-phosphate. Despite the catalytic “perfection” of triose phosphate isomerase, mistakes do occur. The interconversion of glyceraldehyde-3-phosphate and dihydroxyacetone phosphate proceeds via an enediol intermediate, which can break down to form methylglyoxal.

H

C C

OH OH

CH2OPO2– 3 Enediol intermediate

HPO2– 4

H

C C

O O

CH3 Methylglyoxal

Methylglyoxal, which is also a product of other enzymatic and nonenzymatic processes, can react with lipids, nucleotides, and the amino groups of proteins. This molecular damage seems to be an unavoidable consequence of carrying out glycolysis. Not surprisingly, most organisms have a set of highly conserved enzymes to detoxify methylglyoxal or eliminate its reaction products. In fact, many metabolic enzymes produce small amounts of toxic byproducts that must be dealt with by other “repair” enzymes.

ATP is generated near the end of glycolysis So far, the reactions of glycolysis have consumed 2 ATP, but this investment pays off later in glycolysis when 4 ATP are produced, for a net gain of 2 ATP. Reactions 6–10 all involve three-carbon intermediates, but keep in mind that each glucose molecule that enters the pathway yields two of these three-carbon units.

374  C ha pter 13   Glucose Metabolism

Some species convert glucose to glyceraldehyde-3-phosphate by different pathways than the one presented above. However, the last five reactions of glycolysis, which convert glyceraldehyde-3-phosphate to pyruvate, are the same in all organisms. This suggests that glycolysis may have evolved from the “bottom up”; that is, it first evolved as a pathway for extracting free energy from abiotically produced small molecules, before cells developed the ability to synthesize larger molecules such as hexoses.

6. Glyceraldehyde-3-Phosphate Dehydrogenase 

In the sixth reaction of glycolysis, glyceraldehyde-3-phosphate is both oxidized and phosphorylated:

O H

1C 2C

H

O + NAD+ + Pi

OH

2– 3CH2OPO3

H

glyceraldehyde-3-phosphate dehydrogenase

Glyceraldehyde3-phosphate

1C 2C

OPO2– 3 + NADH + H+

OH

2– 3CH2OPO3

1,3-Bisphosphoglycerate

Unlike the kinases that catalyze Reactions 1 and 3, glyceraldehyde-3-phosphate dehydrogen­ ase does not use ATP as a phosphoryl-group donor; it adds inorganic phosphate to the substrate. This reaction is also an oxidation–reduction reaction in which the aldehyde group of ­glyceraldehyde-3-phosphate is oxidized and the cofactor NAD+ is reduced to NADH. In effect, glyceraldehyde-3-phosphate dehydrogenase catalyzes the removal of an H atom (actually, a hydride ion); hence the name “dehydrogenase.” Note that the reaction product NADH must eventually be reoxidized to NAD+ or else glycolysis will come to a halt. In fact, the reoxidation of NADH, which is a form of “energy currency,” generates ATP in oxidative phosphorylation (Chapter 15). An active-site cysteine residue participates in the glyceraldehyde-3-phosphate dehydrogenase reaction (Fig. 13.6). The enzyme is inhibited by arsenate (As​​O​  43− ​  ​​), which competes with Pi (​​PO​  3− 4​  ​​) for binding in the enzyme active site.

7. Phosphoglycerate Kinase 

is an acyl phosphate.

The product of Reaction 6, 1,3-bisphosphoglycerate,

O

O

R C

O P

O–

O–

An acyl phosphate

The subsequent removal of its phosphoryl group releases a large amount of free energy in part because the reaction products are more stable (the same principle contributes to the large negative value of ΔG for reactions involving cleavage of ATP’s phosphoanhydride bonds; see Section 12.3). The free energy released in this reaction is used to drive the formation of ATP, as 1,3-bisphosphoglycerate donates its phosphoryl group to ADP:

O H

1C 2C

OPO2– 3 OH

O + ADP

2– 3CH2OPO3

1,3-Bisphosphoglycerate

phosphoglycerate kinase

H

1C 2C

O– OH

+ ATP

2– 3CH2OPO3

3-Phosphoglycerate

Note that the enzyme that catalyzes this reaction is called a kinase since it transfers a phosphoryl group between ATP and another molecule.

13.1 Glycolysis  375

NAD+ NAD+ S

S

H O

H 1. Glyceraldehyde-3-phosphate binds to the enzyme, whose active site already contains NAD+.

5. The inorganic phosphate attacks the thioester, forming 1,3-bisphosphoglycerate and regenerating the enzyme.

S

C

H

S

NAD+ H S

C

O –

C

O

4. NADH exits and is replaced by another molecule of NAD+. Pi enters the active site.

P

OH

O–

R

2. The SH group of the active-site Cys residue forms a covalent bond with glyceraldehyde-3-phosphate.

H

H+

O

O

R

+

R

NAD+

NAD+ H

C

OPO2– 3

NAD+ NADH NADH

3. NAD+ oxidizes the substrate, forming a thioester intermediate.

O

O–

S

C R

R   FIGURE 13.6   The glyceraldehyde-3-phosphate dehydrogenase reaction.

Question  Identify the reactant that undergoes oxidation and the reactant that undergoes reduction.

SEE ANIMATED PROCESS DIAGRAM

The standard free energy change for the phosphoglycerate kinase reaction is −18.8 kJ · mol−1. This strongly exergonic reaction helps pull the glyceraldehyde-3-phosphate dehydrogenase reaction forward, since its standard free energy change is greater than zero (ΔG°′ = +6.7 kJ · mol−1). This pair of reactions provides a good example of the coupling of a thermodynamically favorable and unfavorable reaction so that both proceed with a net decrease in free energy: −18.8 kJ · mol−1 + 6.7 kJ · mol−1 = −12.1 kJ · mol−1. Under physiological conditions, ΔG for the paired reactions is close to zero.

The enzymatic mechanism of GAPDH

8. Phosphoglycerate Mutase 

verted to 2-phosphoglycerate:

O

C1

O–

H

C2 OH

H

C3 OPO2– 3 H

In the next reaction, 3-phosphoglycerate is con-

3-Phosphoglycerate

O

phosphoglycerate mutase

C1

O–

H

C2 OPO2– 3

H

C3 OH H

2-Phosphoglycerate

Although the reaction appears to involve the simple intramolecular transfer of a phosphoryl group, the reaction mechanism is a bit more complicated and requires an enzyme active site that contains a phosphorylated histidine residue. The phospho-His transfers its phosphoryl

376  C ha pter 13   Glucose Metabolism

group to 3-phosphoglycerate to generate 2,3-bisphosphoglycerate, which then gives a phosphoryl group back to the enzyme, leaving 2-phosphoglycerate and phospho-His:

O

C

O–

H

C

OH

H

C

OPO2– 3

O

O –

His

O P O–

O–

C

O

H

C

OPO2– 3

H

C

OPO2– 3

His

H

H

3-Phosphoglycerate

C

O–

O

H

C

OPO2– 3

H

C

OH

–

O P

His

O–

H

2-Phosphoglycerate

As can be guessed from its mechanism, the phosphoglycerate mutase reaction is freely reversible in vivo.

9. Enolase 

Enolase catalyzes a dehydration reaction, in which water is eliminated:

O

C1

O–

O

H

C 2 OPO2– 3

H

C 3 OH

C

O–

C

enolase

H

H

+ H2O OPO2– 3

C H

2-Phosphoglycerate

Phosphoenolpyruvate

The enzyme active site includes an Mg2+ ion that apparently coordinates with the OH group at C3 and makes it a better leaving group. Fluoride ion and Pi can form a complex with the Mg2+ and thereby inhibit the enzyme. In early studies demonstrating the inhibition of glycolysis by F−, 2-phosphoglycerate, the substrate of enolase, accumulated. The concentration of 3-phosphoglycerate also increased in the presence of F− since phosphoglycerate mutase readily converted the excess 2-phosphoglycerate back to 3-phosphoglycerate.

10. Pyruvate Kinase 

The tenth reaction of glycolysis is catalyzed by pyruvate kinase, which converts phosphoenolpyruvate to pyruvate and transfers a phosphoryl group to ADP to produce ATP:

O

C C

O–

O

OPO2– 3

+ ADP

pyruvate kinase

CH2

C C

O– O + ATP

CH3 Pyruvate

Phosphoenolpyruvate

The reaction actually occurs in two parts. First, ADP attacks the phosphoryl group of phosphoenolpyruvate to form ATP and enolpyruvate:

O

C C

O–

O P

CH2

O

O

O O

–

O–

Phosphoenolpyruvate

+

ATP

–

O P

O

O– ADP

AMP

C C

H

+

O– OH

CH2 Enolpyruvate

13.1 Glycolysis  377

Removal of phosphoenolpyruvate’s phosphoryl group is not a particularly exergonic reaction. When written as a hydrolytic reaction (transfer of the phosphoryl group to water), the ΔG°′ value is −16 kJ · mol−1. This is not enough free energy to drive the synthesis of ATP from ADP + Pi (this reaction requires +30.5 kJ · mol−1). However, the second half of the pyruvate kinase reaction is highly exergonic. This is the tautomerization (isomerization through the shift of an H atom) of enolpyruvate to pyruvate.

O

C C

O–

O

C C

OH

CH2

O– O

CH3 Pyruvate

Enolpyruvate

ΔG°′ for this step is −46 kJ · mol−1, so ΔG°′ for the net reaction (hydrolysis of phosphoenol­ pyruvate followed by tautomerization of enolpyruvate to pyruvate) is −61.9 kJ · mol−1, more than enough free energy to drive the synthesis of ATP. Three of the ten reactions of glycolysis (the reactions catalyzed by hexokinase, phospho­ fructokinase, and pyruvate kinase) have large negative values of ΔG. In theory, any of these far-from-equilibrium reactions could serve as a flux-control point for the pathway. The other seven reactions function near equilibrium (ΔG ≈ 0) and can therefore accommodate flux in either direction. The free energy changes for the ten reactions of glycolysis are shown graphically in Figure 13.7. We have already discussed the mechanisms for regulating phosphofructokinase activity, the major control point for glycolysis. Hexokinase also catalyzes an irreversible reaction and is subject to inhibition by its product, glucose-6-phosphate. However, hexokinase cannot be the sole control point for glycolysis because glucose can also enter the pathway as glucose-6-phosphate, bypassing the hexokinase reaction. In addition, glycolysis is not the only fate for glucose-6-phosphate, as we will see. Finally, it would not be efficient for the pyruvate kinase reaction to be the primary regulatory step for glycolysis because it occurs at the very end of the 10-step pathway. Even so, pyruvate kinase activity can be adjusted. In some organisms, fructose-1,6-bisphosphate activates pyruvate kinase at an allosteric site. This is an example of feed-forward activation: Once a monosaccharide has entered glycolysis, fructose-1,6-bisphosphate helps ensure rapid flux through the pathway. The glycolytic pathway can be considered to be a sort of pipe, with an entrance for glu­ cose (or glucose-6-phosphate) and an exit for pyruvate. The control points for the pathway are

Glucose

1 2

Free energy

3 6+7 4

5

8

9 10 Pyruvate

  FIGURE 13.7   Graphical representation of the free energy changes of

glycolysis.  Three steps have large negative values of ΔG; the remaining steps are near equilibrium (ΔG ≈ 0). The height of each step corresponds to its ΔG value in heart muscle, and the numbers correspond to glycolytic enzymes. Keep in mind that ΔG values vary slightly among tissues.

378  C ha pter 13   Glucose Metabolism

like one-way valves that prevent backflow. Between those control points, intermediates can move in either direction. The pipe never runs dry because the pyruvate gets used up and more glucose is always available. In addition, intermediates can enter or leave the pathway at any point. Even the simplest cells contain many copies of each glycolytic enzyme, acting on a pool of millions of substrate molecules, so their collective behavior is what we refer to when we discuss flux through the pathway. To sum up the last five reactions of glycolysis: Glyceraldehyde-3-phosphate is converted to pyruvate with the synthesis of 2 ATP (in Reactions 7 and 10). Since each molecule of glu­ cose yields two molecules of glyceraldehyde-3-phosphate, Reactions 6–10 must be doubled, so the yield is 4 ATP. Two molecules of ATP are invested in Reactions 1 and 3, bringing the net yield to 2 ATP produced per glucose molecule. Two NADH are also generated for each glucose molecule. Monosaccharides other than glucose are metabolized in a similar fashion to yield ATP (Box 13.A).

Box 13.A Catabolism of Other Sugars A typical human diet contains many carbohydrates other than glucose and its polymers. For example, lactose, a disaccharide of glucose and galactose, is present in milk and food derived from it (Section 11.2). Lactose is cleaved in the intestine by the enzyme lactase, and the two monosaccharides are absorbed, transported to the liver, and metabolized. Galactose undergoes phosphorylation and isomerization and enters the glycolytic pathway as glucose6-phosphate, so its energy yield is the same as that of glucose. Sucrose, the other major dietary disaccharide, is composed of glucose and fructose (Section 11.2); it is present in a variety of foods of plant origin. Like lactose, sucrose is hydrolyzed in the small intestine, and its components, glucose and fructose, are absorbed. The monosaccharide fructose is also present in many foods, particularly fruits and honey. It tastes sweeter than sucrose, is more soluble, and is inexpensive to produce in the form of high-fructose corn syrup—all of which make fructose attractive to the manufacturers of soft drinks and other processed foods. This is the primary reason why the consumption of fructose in the United States rose dramatically from 1970 until about 2000 (it has been declining ever since). Some researchers have proposed that the overconsumption of fructose is contributing to the obesity epidemic. One possible explanation relates to the catabolism of fructose, which differs somewhat from the catabolism of glucose. Fructose is metabo­ lized primarily by the liver, but the form of hexokinase present in the liver (called glucokinase) has very low affinity for fructose. Fructose therefore enters glycolysis by a different route.

First, fructose is phosphorylated to yield fructose-1phosphate. The enzyme fructose-1-phosphate aldolase then splits the six-carbon molecule into two three-carbon molecules: glyceraldehyde and dihydroxyacetone phosphate (see diagram). Dihydroxyacetone phosphate is converted to glyceraldehyde3-phosphate by triose phosphate isomerase and can proceed through the second phase of glycolysis. The glyceraldehyde can be phosphorylated to glyceraldehyde-3-phosphate, but it can also be converted to glycerol-3-phosphate, a precursor of the backbone of triacylglycerols. This may contribute to an increase in fat dep­ osition. A second potential hazard of the fructose catabolic pathway is that fructose catabolism bypasses the phosphofructokinasecatalyzed step of glycolysis, a major regulatory point, and thus may disrupt the entire body’s fuel metabolism. However, fructose consumption has been decreasing for over 15 years, even while obesity rates rise. Furthermore, except in experimental settings, individuals typically consume about five times more glucose than fructose, so the actual fat-triggering culprit might simply be total sugar intake, not fructose in particular.

Question  When its concentration is extremely high, fructose is converted to fructose-1-phosphate much faster than it can be cleaved by the aldolase. How would this affect the cell’s ATP supply?

Dihydroxyacetone phosphate

CH2OPO2– 3 CH2OPO2– 3 HOCH2

O

CH2OH

H

HO

HO

H

H

Fructose

OH

ATP

ADP HOCH2

fructokinase

CH2OPO2– 3

O

H

HO

HO

H

H

OH

C

O

HO

C

H

H

C

OH

H

C

OH

CH2OH Fructose-1-phosphate

C HO

fructose-1phosphate aldolase

H H

O

CH2 +

C C

O OH

CH2OH Glyceraldehyde

13.1 Glycolysis  379 Glucose NAD+ NADH CO2

CO2 Pyruvate

Acetyl-CoA NADH

Oxaloacetate NADH

NAD+

NAD+

Lactate

Ethanol + CO2

  FIGURE 13.8   Fates of pyruvate.  Pyruvate can be converted to other 2-, 3-, and 4-carbon molecules.

Question  Beside each arrow, write the names of the enzymes that catalyze the process.

Some cells convert pyruvate to lactate or ethanol What happens to the pyruvate generated by the catabolism of glucose? It can be further broken down or used to synthesize other compounds. The fate of pyruvate depends on the cell type and the need for metabolic free energy and molecular building blocks. For example, pyruvate may be converted to a two-carbon acetyl group linked to the carrier coenzyme A. The acetyl group may be further broken down by the citric acid cycle or used to synthesize fatty acids. In muscle, pyruvate is reduced to lactate. Yeast degrade pyruvate to ethanol and CO2. Pyruvate can also be carboxylated to produce the four-carbon oxaloacetate. These options are summarized in Figure 13.8. During exercise, pyruvate may be temporarily converted to lactate. In a highly active muscle cell, glycolysis rapidly provides ATP to power muscle contraction, but the pathway also consumes NAD+ at the glyceraldehyde-3-phosphate dehydrogenase step. The two NADH molecules generated for each glucose molecule catabolized can be reoxidized in the presence of oxygen. However, this process is too slow to replenish the NAD+ needed for the rapid production of ATP by glycolysis. To regenerate NAD+, the enzyme lactate dehydrogenase reduces pyruvate to lactate:

O

C C

O– O

CH3 Pyruvate

H

H

O C

+ N R

O NH2

+ H+

lactate dehydrogenase

NADH

HO

C C

H

O– H

C +

CH3 Lactate

This reaction, sometimes called the eleventh step of glycolysis, allows the muscle to function anaerobically for a minute or two. The net reaction for anaerobic glucose catabolism is glucose + 2 ADP + 2 Pi → 2 lactate + 2 ATP

O

Lactate represents a sort of metabolic dead end: Its only options are to be eventually converted back to pyruvate (the lactate dehydrogenase reaction is reversible) or to be exported from the cell. The liver takes up lactate from the blood, oxidizes it back to pyruvate, and then converts the pyruvate to glucose. The fuel produced in this manner may eventually make its way back to the muscle to help power continued contraction. When the muscle is functioning aerobically, NADH produced by the glyceraldehyde-3-phosphate dehydrogenase reaction is reoxidized by oxygen and the lactate dehydrogenase reaction is not needed. Cancer cells sustain high rates of lactate production, even under aerobic conditions, as part of the metabolic alterations that allow rapid cell growth (Section 19.4).

+

N R

NAD+

NH2

380  C ha pter 13   Glucose Metabolism

Organisms such as yeast growing under anaerobic conditions can regenerate NAD+ by producing alcohol. In the mid-1800s, Louis Pasteur called this process fermentation, meaning life without air, although yeast also ferment sugars in the presence of O2. Alcoholic fermentation is a two-step process. First, pyruvate decarboxylase (an enzyme not present in animals) catalyzes the removal of pyruvate’s carboxylate group to produce acetaldehyde. Next, alcohol dehydrogenase reduces acetaldehyde to ethanol.

O

C C

O– O

CH3

CO2 pyruvate decarboxylase

Pyruvate

O

C

OH

+ H NADH NAD

CH3 Acetaldehyde

alcohol dehydrogenase

H

C

H

CH3 Ethanol

Ethanol is considered to be a waste product of sugar metabolism; its accumulation is toxic to other organisms (Box 13.B), including the yeast that produce it. This is one reason why the alcohol content of yeast-fermented beverages such as wine is limited to about 13%. “Hard” liquor must be distilled to increase its ethanol content. Yeast and other fungi, as well as bacteria, do more than convert glucose to ethanol or lactate. Microbial fermentation transforms a variety of animal and plant products to foods such as kimchi, cheese, salami, and soy sauce. Even Cacao seeds undergo fermentation during

Box 13.B Alcohol Metabolism Unlike yeast, mammals do not produce ethanol, although it is naturally present in many foods and is produced in small amounts by intestinal microorganisms. The liver is equipped to metabolize ethanol, a small, weakly polar substance that is readily absorbed from the gastrointestinal tract and transported by the bloodstream. First, alcohol dehydrogenase converts ethanol to acetaldehyde. This is the reverse of the reaction yeast use to produce ethanol. A second reaction converts acetaldehyde to acetate:

OH OH

+ + NAD NADH, NAD NADH, H+H+

H HC CH H CHCH 3 3

alcohol alcohol dehydrogenase dehydrogenase

Ethanol Ethanol – – + + OH NAD NADH, OH NAD NADH, H+H+ OO HH O O O–O– CC CC

CHCH 3 3 Acetaldehyde Acetaldehyde

acetaldehyde acetaldehyde dehydrogenase dehydrogenase

CHCH 3 3 Acetate Acetate

Note that both of these reactions require NAD+, a cofactor used in many other oxidative cellular processes, including glycolysis. The liver uses the same two-enzyme pathway to metabolize the excess ethanol obtained from alcoholic beverages. Ethanol itself is mildly toxic, and the physiological effects of alcohol also reflect the toxicity of acetaldehyde and acetate in tissues such as the liver and brain. Ethanol induces vasodilation, apparent as flushing (warming and reddening of the skin due to increased blood flow). At the same time, the heart rate and respiration rate become slightly lower. The kidneys increase the excretion of water as ethanol interferes with the ability of the hypothalamus (a region of the brain) to properly sense osmotic pressure.

Ethanol also stimulates signaling from certain neurotransmitter receptors that function as ligand-gated ion channels (Section 9.2) to inhibit neuronal signaling, producing a sedative effect. Sensory, motor, and cognitive functions are impaired, leading to delayed reaction time, loss of balance, and blurred vision. Some of the symptoms of ethanol intoxication can be experienced even at low doses, when the blood alcohol concentration is less than 0.05%. At high doses, usually at blood concentrations above 0.25%, ethanol can cause loss of consciousness, coma, and death. The mostly pleasant responses to moderate ethanol consumption are followed by a period of recovery, when the concentrations of ethanol metabolites are relatively high. The unpleasant symptoms of a hangover in part reflect the chemistry of producing acetaldehyde and acetate. Their production in the liver consumes NAD+, thereby lowering the cell’s NAD+:NADH ratio. Without sufficient NAD+, the liver’s ability to produce ATP by glycolysis is diminished (since NAD+ is required for the glyceraldehyde3-phosphate dehydrogenase reaction). Acetaldehyde itself can react with liver proteins, inactivating them. Acetate (acetic acid) production lowers blood pH. Long-term, excessive alcohol consumption exacerbates the toxic effects of ethanol and its metabolites. For example, a shortage of liver NAD+ slows fatty acid breakdown (like glycolysis, a process that requires NAD+) and promotes fatty acid synthesis, leading to fat accumulation in the liver. Over time, cell death causes permanent loss of function in the central nervous system. The death of liver cells and their replacement by fibrous scar tissue causes liver cirrhosis. Question  Normally, the liver converts lactate, produced mainly by muscles, back to pyruvate so that the pyruvate can be converted to glucose by gluconeogenesis (Section 13.2). How do the activities of alcohol dehydrogenase and acetaldehyde dehydrogenase contribute to ­hypoglycemia?

13.1 Glycolysis  381

chocolate production. In many cases, fermentation involves a community of organisms, acting together or in succession, to break down food molecules. In traditional cheesemaking, lactate (lactic acid) produced by bacterial fermentation of milk lowers the pH and causes the denaturation and solidification of milk proteins to produce soft, spreadable cheeses. Hard cheeses develop more slowly, as the proteins are catabolized by bacterial enzymes over months to produce more flavorful compounds. The interior of an aging cheese is usually anaerobic, but in veined cheeses like blue cheese, aerobic fungi grow—and produce characteristic by-products—in air channels within the cheese.

Pyruvate is the precursor of other molecules Although glycolysis is an oxidative pathway, its end product pyruvate is still a relatively reduced molecule. The further catabolism of pyruvate begins with its oxidative decarboxylation to form a two-carbon acetyl group linked to coenzyme A.

HSCoA NAD+ NADH CO2

O CH3

COO–

C

O CH3

pyruvate dehydrogenase

Pyruvate

C

SCoA

Acetyl-CoA

The resulting acetyl-CoA is a substrate for the citric acid cycle (Chapter 14), which converts the acetyl carbon atoms to CO 2. The complete oxidation of the six glucose carbons to CO2 (ΔG°′ = −2850 kJ · mol−1) releases much more free energy than the conversion of glucose to lactate (ΔG°′ = −196 kJ · mol−1). Much of this energy is recovered in the synthesis of ATP by the reactions of the citric acid cycle and oxidative phosphorylation (Chapter 15), pathways that require the presence of molecular oxygen. Pyruvate is not always destined for catabolism. Its carbon atoms provide the raw material for synthesizing a variety of molecules, including, in the liver, more glucose (discussed in the following section). Fatty acids, the precursors of triacylglycerols and many membrane lipids, can be synthesized from the two-carbon units of acetyl-CoA derived from pyruvate. This is how fat is produced from excess carbohydrate. Pyruvate is also the precursor of oxaloacetate, a four-carbon molecule that is an intermediate in the synthesis of several amino acids. It is also one of the intermediates of the citric acid cycle. Oxaloacetate is synthesized by the action of pyruvate carboxylase:

COO–

COO– C

O

+ CO2 + ATP

CH3

C pyruvate carboxylase

Pyruvate

O

CH2

+ ADP + Pi

COO–

Oxaloacetate

The pyruvate carboxylase reaction is interesting because of its unusual chemistry. The enzyme has a biotin prosthetic group that acts as a carrier of CO2. Biotin is considered a vitamin, but a deficiency is rare because it is present in many foods and is synthesized by intestinal bacteria. The biotin group is covalently linked to an enzyme lysine residue to form the functional biocytin cofactor:

O HN H

C

C

H2C

Lys residue

NH H C H

C S

O CH2

CH2

Biotin

CH2

CH2

C

O NH

(CH2)4

C CH NH

382  C ha pter 13   Glucose Metabolism 1. CO2 (as bicarbonate, HCO3–) reacts with ATP such that some of the free energy released in the removal of ATP’s phosphoryl group is conserved in the formation of the “activated” compound carboxyphosphate. –

O

ATP +

O

O C

HO

O

O

P

C

–

ADP

O

OH

OH

Carboxyphosphate

2. Like ATP, carboxyphosphate releases a large amount of free energy when its phosphoryl group is liberated. This free energy drives the carboxylation of biotin.

Pi Biotinyl-Lys

O

–

O N

C

NH

O

O–

O C C

3. The enzyme abstracts a proton from pyruvate, forming a carbanion.

O–

O

4. The carbanion attacks the carboxyl group attached to biotin, generating oxaloacetate.

C

O

C

O

Biotinyl-Lys

–

CH3

CH2

Pyruvate

R

S

O–

O C C

O

CH2 SEE ANIMATED PROCESS DIAGRAM The reaction mechanism of pyruvate carboxylase

C

–

O

O

Oxaloacetate   FIGURE 13.9   The pyruvate carboxylase reaction.

The lysine side chain and its attached biotin group form a 14-Å-long flexible arm that swings between two active sites in the enzyme. In one active site, a CO2 molecule is first “activated” by its reaction with ATP, and is then transferred to the biotin. The second active site transfers the carboxyl group to pyruvate to produce oxaloacetate (Fig. 13.9).

Before Going On • Write the net equation for glycolysis. • Draw the structures of the substrates and products of the 10 glycolytic reactions and name the enzyme that catalyzes each step. • List the glycolytic reactions that consume ATP or generate ATP. • Calculate the net yield of ATP and NADH per glucose molecule. • Identify the reactions that serve as flux-control points for glycolysis. • List the possible metabolic fates of pyruvate. • Describe the metabolic function of lactate dehydrogenase.

13.2 Gluconeogenesis  383

13.2 Gluconeogenesis LEARNING OBJECTIVES Describe the substrates, products, and reactions of gluconeogenesis. • List the enzymes that are unique to gluconeogenesis or are shared with glycolysis. • Explain how the rates of gluconeogenesis and ­glycolysis are related.

We have already alluded to the ability of the liver to synthesize glucose from noncarbohydrate precursors via the pathway of gluconeogenesis. This pathway, which also occurs to a limited extent in the kidneys, operates when the liver’s supply of glycogen is exhausted. Certain tissues, such as the central nervous system and red blood cells, which burn glucose as their primary metabolic fuel, rely on the liver to supply them with newly synthesized glucose. Gluconeogenesis is considered to be the reversal of glycolysis, that is, the conversion of two molecules of pyruvate to one molecule of glucose. Although some of the steps of gluconeo­ genesis are catalyzed by glycolytic enzymes operating in reverse, the gluconeogenic pathway contains several unique enzymes that bypass the three irreversible steps of glycolysis—the steps catalyzed by pyruvate kinase, phosphofructokinase, and hexokinase (Fig. 13.10). This principle applies to all pairs of opposing metabolic pathways: The pathways may share some near-equilibrium reactions but cannot use the same enzymes to catalyze thermodynamically favorable irreversible reactions. The three irreversible reactions of glycolysis are clearly visible in the “waterfall” diagram (see Fig. 13.7).

Four gluconeogenic enzymes plus some glycolytic enzymes convert pyruvate to glucose Pyruvate cannot be converted directly back to phosphoenolpyruvate because pyruvate kinase catalyzes an irreversible reaction (Reaction 10 of glycolysis). To get around this thermodynamic barrier, pyruvate is carboxylated by pyruvate carboxylase to yield the four-carbon compound oxaloacetate (the same reaction shown in Fig. 13.9). Next, phosphoenolpyruvate carboxykinase catalyzes the decarboxylation of oxaloacetate to form phosphoenolpyruvate:

O

C C

O–

HCO–3 + ATP

O

pyruvate carboxylase

CH3 Pyruvate

ADP + Pi

O

C C

O– O

CH2 –

O

C

O

Oxaloacetate

GTP

CO2 + GDP

phosphoenolpyruvate carboxykinase

O

C C

O– O

PO2– 3

CH2 Phosphoenolpyruvate

Note that the carboxylate group added in the first reaction is released in the second. The two reactions are energetically costly: Pyruvate carboxylase consumes ATP, and phosphoenol­ pyruvate carboxykinase consumes GTP (which is energetically equivalent to ATP). Cleavage of two phosphoanhydride bonds is required to supply enough free energy to “undo” the highly exergonic pyruvate kinase reaction. Amino acids (except for leucine and lysine) are the main sources of gluconeogenic precursors because they can all be converted to oxaloacetate and then to phosphoenolpyruvate. Thus, during starvation, proteins can be broken down and used to produce glucose to fuel the central nervous system. In mammals, fatty acids cannot serve as gluconeogenic precursors because they cannot be converted to oxaloacetate. (However, the three-carbon glycerol backbone of triacylglycerols is a gluconeogenic precursor.)

384  C ha pter 13   Glucose Metabolism

Glucose

  FIGURE 13.10   The reactions of gluconeogenesis.  The pathway uses the seven glycolytic enzymes that catalyze reversible reactions. The three irreversible reactions of glycolysis are bypassed in gluconeogenesis by the four enzymes that are highlighted in blue.

Pi

ATP hexokinase

glucose-6-phosphatase

ADP

H2O

Glucose-6-phosphate phosphoglucose isomerase

Fructose-6-phosphate Pi ATP

Question  Compare the ATP yield of glycolysis with the ATP consumption of gluconeogenesis.

phosphofructokinase

fructose bisphosphatase

ADP

H2O

Fructose-1,6-bisphosphate

glycolysis

gluconeogenesis

aldolase

Dihydroxyacetone phosphate

triose phosphate isomerase

Glyceraldehyde-3phosphate

NAD+ + Pi glyceraldehyde-3-phosphate dehydrogenase

NADH + H+

1,3-Bisphosphoglycerate ADP phosphoglycerate kinase

ATP 3-Phosphoglycerate phosphoglycerate mutase

2-Phosphoglycerate enolase

Phosphoenolpyruvate ADP pyruvate kinase

ATP

GDP + CO2 phosphoenolpyruvate carboxykinase

GTP Oxaloacetate ADP + Pi Pyruvate

pyruvate carboxylase

ATP + CO2

Two molecules of phosphoenolpyruvate are converted to one molecule of fructose-1,6-­ bisphosphate in a series of six reactions that are all catalyzed by glycolytic enzymes (steps 4–9 in reverse order). These reactions are reversible because they are near equilibrium (ΔG ≈ 0), and the direction of flux is determined by the concentrations of substrates and products. Note that the phosphoglycerate kinase reaction consumes ATP when it operates in the direction of gluconeogenesis. NADH is also required to reverse the glyceraldehyde-3-phosphate dehydrogenase reaction.

13.2 Gluconeogenesis  385

The final three reactions of gluconeogenesis require two enzymes unique to this pathway. The first step undoes the phosphofructokinase reaction, the irreversible reaction that is the major control point of glycolysis. In gluconeogenesis, the enzyme fructose bisphosphatase hydrolyzes the C1 phosphate of fructose-1,6-bisphosphate to yield fructose-6-phosphate. This reaction is thermodynamically favorable, with a ΔG value of −8.6 kJ · mol−1. Next, the glycolytic enzyme phosphoglucose isomerase catalyzes the reverse of step 2 of glycolysis to produce glucose-6-phosphate. Finally, the gluconeogenic enzyme glucose-6-phosphatase catalyzes a hydrolytic reaction that yields glucose and Pi. Note that the hydrolytic reactions cat­ alyzed by fructose bisphosphatase and glucose-6-phosphatase undo the work of two kinases in glycol­ysis (phosphofructokinase and hexokinase).

Gluconeogenesis is regulated at the fructose bisphosphatase step Gluconeogenesis is energetically expensive. Producing 1 glucose from 2 pyruvate consumes 6 ATP, 2 each at the steps catalyzed by pyruvate carboxylase, phosphoenolpyruvate carboxy­ kinase, and phosphoglycerate kinase. If glycolysis occurred simultaneously with gluconeogenesis, there would be a net consumption of ATP: glycolysis gluconeogenesis net

glucose + 2 ADP + 2 Pi → 2 pyruvate + 2 ATP 2 pyruvate + 6 ATP → glucose + 6 ADP + 6 Pi 4 ATP → 4 ADP + 4 Pi

To avoid this waste of metabolic free energy, gluconeogenic cells (mainly liver cells) ­c arefully regulate the opposing pathways of glycolysis and gluconeogenesis according to the cell’s energy needs. The major regulatory point is centered on the interconversion of fructose-6-­phosphate and fructose-1,6-bisphosphate. We have already seen that fructose2,6-­bisphosphate is a potent allosteric activator of phosphofructokinase, which catalyzes step 3 of glycolysis. Not surprisingly, fructose-2,6-bisphosphate is a potent inhibitor of fructose bisphosphatase, which catalyzes the opposing gluconeogenic reaction.

Fructose-6phosphate

glycolysis

PFK

F2,6P

FBPase gluconeogenesis

Fructose-1,6bisphosphate This mode of allosteric regulation is efficient because a single compound can control flux through two opposing pathways in a reciprocal fashion. Thus, when the concentration of ­fructose-2,6-bisphosphate is high, glycolysis is stimulated and gluconeogenesis is inhibited, and vice versa. Many cells that do not carry out gluconeogenesis do contain the gluconeogenic enzyme fructose bisphosphatase. What is the reason for this? When both fructose bisphosphatase (FBPase) and phosphofructokinase (PFK) are active, the net result is the hydrolysis of ATP: PFK fructose-6-phosphate + ATP → fructose-1,6-bisphosphate + ADP FBPase fructose-1,6-bisphosphate + H2O → fructose-6-phosphate + Pi net

ATP + H2O → ADP + Pi

386  C ha pter 13   Glucose Metabolism

This combination of metabolic reactions is called a futile cycle since it seems to have no useful result. However, Eric Newsholme realized that such futile cycles could actually provide a means for fine-tuning the output of a metabolic pathway. For example, flux through the phosphofructokinase step of glycolysis is diminished by the activity of fructose bisphosphatase. An allosteric compound such as fructose-2,6-bisphosphate modulates the activity of both enzymes so that as the activity of one enzyme increases, the activity of the other one decreases. This dual regulatory effect results in a greater possible range of net flux than if the regulator merely activated or inhibited a single enzyme. Similarly, a car’s speed is easier to control if it has both an accelerator and a brake.

Before Going On • List the reactions of gluconeogenesis that are catalyzed by glycolytic enzymes. • List the enzymes that are unique to gluconeogenesis and explain why they are needed. • Describe the fructose-6-phosphate futile cycle and explain its purpose.

13.3 Glycogen Synthesis and Degradation LEARNING OBJECTIVES Compare the processes of glycogen synthesis and degradation. • Identify the substrates and products for each process. • Compare the energy needs of each pathway. • List the metabolic fates of glucose-6-phosphate.

In animals, dietary glucose and the glucose produced by gluconeogenesis are stored in the liver and other tissues as glycogen. Later, glucose units can be removed from the glycogen polymer by phosphorolysis (see Section 12.1). Because glycogen degradation is thermodynamically spontaneous, glycogen synthesis requires the input of free energy. The two opposing pathways use different sets of enzymes so that each process can be thermodynamically favorable under cellular conditions.

Glycogen synthesis consumes the energy of UTP The monosaccharide unit that is incorporated into glycogen is glucose-1-phosphate, which is produced from glucose-6-phosphate (the penultimate product of gluconeogenesis) by the action of the enzyme phosphoglucomutase: –2

O3POCH2 H

HO

HOCH2 O

H OH

H

H

OH

H OH

Glucose-6-phosphate

H phosphoglucomutase

HO

O H OH

H

H

OH

H OPO2– 3

Glucose-1-phosphate

13.3  Glycogen Synthesis and Degradation  387

In mammalian cells, glucose-1-phosphate is then “activated” by reacting with UTP to form UDP–glucose (like GTP, UTP is energetically equivalent to ATP).

O SEE ANIMATED PROCESS DIAGRAM

HN O

O

O

O P

O P

O P

–

–

–

O

H HO

H

O

H

H

HO

H

N

The reaction catalyzed by UDP–glucose pyrophosphorylase

O

OH2C

–

O

CH2OH O H OH H

O

O

H

UTP

H

OH

O–

O P O–

OH

Glucose-1-phosphate UDP–glucose pyrophosphorylase

inorganic pyrophosphatase

PPi

2 Pi O

H HO

CH2OH O H OH H H

OH

HN H

O

O

O P

O P

O–

O O

OH2C

O–

H

UDP–glucose

N

H

HO

H

H

OH

This process is a reversible phosphoanhydride exchange reaction (ΔG ≈ 0). Note that the two phosphoanhydride bonds of UTP are conserved, one in the product PPi and one in UDP–­glucose. However, the PPi is rapidly hydrolyzed by inorganic pyrophosphatase to 2 Pi in a highly exergonic reaction (ΔG°′ = −19.2 kJ · mol−1). Thus, cleavage of a phospho­ anhydride bond makes the formation of UDP–glucose an exergonic, irreversible process— that is, PP i hydrolysis “drives” a reaction that would otherwise be near equilibrium. The hydrolysis of PPi by inorganic pyrophosphatase is a common strategy in biosynthetic reactions; we will see this reaction again in the synthesis of other polymers, namely DNA, RNA, and polypeptides. Finally, glycogen synthase transfers the glucose unit to the C4 OH group at the end of one of glycogen’s branches to extend the linear polymer of α(1 → 4)-linked residues.

UDP +

UDP–Glucose glycogen synthase

Glycogen

UDP

A separate enzyme, called a transglycosylase or branching enzyme, cleaves off a sevenresidue segment and reattaches it to a glucose C6 OH group to create an α(1 → 6) branch point.

388  C ha pter 13   Glucose Metabolism

The steps of glycogen synthesis can be summarized as follows:

UDP–glucose pyrophosphorylase glucose-1-phosphate + UTP ⇌ UDP–glucose + PPi pyrophosphatase PPi + H2O → 2 Pi glycogen synthase UDP–glucose + glycogen (n residues) → glycogen (n +1 residues) + UDP

net

glucose-1-phosphate + glycogen + UTP + H2O → glycogen + UDP + 2 Pi

The energetic cost of adding one glucose-1-phosphate molecule to glycogen is the cleavage of one phosphoanhydride bond of UTP. Adding a glucose molecule would cost two phosphoanhydride bonds, since ATP would be spent to convert glucose to glucose-1-phosphate. Nucleotides are also required for the synthesis of other saccharides. For example, lactose is synthesized from glucose and UDP–galactose. In plants, starch is synthesized using ADP– glucose, and cellulose is synthesized using CDP–glucose as starting materials. Glycogen synthase cannot build a new glycogen molecule from scratch; it can only extend a pre-existing chain of glucose residues. The first residues of a new chain are actually assembled by a pair of small proteins called glycogenin. Each glycogenin attaches one or two glucose residues (donated by ADP–glucose) to a tyrosine side chain on the other glycogenin protein. Then each glycogenin extends its own glucose chain by adding another dozen or so glucose residues. At this point, glycogen synthase and the branching enzyme take over, expanding the glycogen molecule by tens of thousands of glucose residues to form a spherical glycogen granule (Fig. 13.11).

Glycogen phosphorylase catalyzes glycogenolysis Glycogen breakdown follows a different set of steps than glycogen synthesis. In glycogenolysis, glycogen is phosphorolyzed, not hydrolyzed, to yield glucose-1-phosphate. However, a debranching enzyme removes α(1 → 6)-linked residues by hydrolysis. In the liver, phosphoglucomutase converts glucose-1-phosphate to glucose-6-phosphate, which is transported into the endoplasmic reticulum and hydrolyzed by glucose-6-phosphatase to release free glucose.

Pi glycogen

H2O

glycogen phosphorylase

glucose-1-phosphate

phosphoglucomutase

glucose-6-phosphate

Pi

glucose-6phosphatase

glucose

This glucose leaves the cell and enters the bloodstream. Only gluconeogenic tissues such as the liver can make glucose available to the body at large. Other tissues that store glycogen, such as muscle, lack glucose-6-phosphatase and so break down glycogen only for their own needs. In these tissues, the glucose-1-phosphate liberated by phosphorolysis of glycogen is converted to glucose-6-phosphate, which then enters glycolysis at the phosphoglucose isomerase reaction (step 2). The hexokinase reaction (step 1) is skipped, thereby sparing the consumption of ATP. Consequently, glycolysis using glycogen-derived glucose has a higher net yield of ATP than glycolysis using glucose supplied by the bloodstream. Because the mobilization of glucose must be tailored to meet the energy demands of a particular tissue or the entire body, the activity of glycogen phosphorylase is carefully

intermolecular glycosylation

intramolecular glycosylation

glycogen synthase, branching enzyme

Glycogenin

  FIGURE 13.11   Growth of a glycogen particle.  Glycogenin initiates glucose polymerization by intermolecular glycosylation followed by intramolecular glycosylation. Glycogen synthase then extends the chain, and branching enzyme creates branches.

Question  Draw the 1-O-linked glucose–tyrosine complex.

13.4  The Pentose Phosphate Pathway  389

regulated by a variety of mechanisms linked to hormonal signaling. Likewise, the activity of glycogen synthase is subject to hormonal control, so that the two opposing pathways of degradation and synthesis do not occur at the same time. In Chapter 19 we will examine some of the mechanisms for regulating different aspects of fuel metabolism, including glycogen synthesis and degradation. Some disorders of glycogen metabolism are discussed in Section 13.5.

Before Going On • Describe the role of UTP in glycogen synthesis. • Explain why protein appears in the core of every glycogen granule. • Explain the advantage of breaking down glycogen by phosphorolysis rather than hydrolysis. Explain why only some tissues contain glucose-6-phosphatase.

13.4 The Pentose Phosphate Pathway LEARNING OBJECTIVES Describe the substrates, products, and reactions of the pentose phosphate pathway. • Identify the oxidation–reduction reactions of the pentose phosphate pathway. • Explain how the pathway responds to the cell’s need for ribose groups.

We have already seen that glucose catabolism can lead to pyruvate, which can be further oxidized to generate more ATP or used to synthesize amino acids and fatty acids. Glu­cose is also a precursor of the ribose groups used for nucleotide synthesis. The pentose phosphate pathway, which converts glucose-6-phosphate to ribose-5-phosphate, is an oxidative pathway that occurs in all cells. But unlike glycolysis, the pentose phosphate pathway generates NADPH rather than NADH. The two cofactors are not interchangeable and are easily distinguished by degradative enzymes (which generally use NAD+ as an oxidizing agent) and biosynthetic enzymes (which generally use NADPH as a reducing agent). The pentose phosphate pathway is by no means a minor feature of glucose metabolism. As much as 30% of glucose in the liver may be catabolized by the pentose phosphate pathway. This pathway can be divided into two phases: a series of oxidative reactions followed by a series of reversible interconversion reactions.

The oxidative reactions of the pentose phosphate pathway produce NADPH The starting point of the pentose phosphate pathway is glucose-6-phosphate, which can be derived from free glucose, from the glucose-1-phosphate produced by glycogen phosphorolysis, or from gluconeogenesis. In the first step of the pathway, glucose-6-phosphate dehydrogenase catalyzes the metabolically irreversible transfer of a hydride ion from ­glucose-6-phosphate to NADP+, forming a lactone and NADPH:

H HO

H+ CH2OPO2– 3 + O NADP+ NADPH H H OH H glucose-6-phosphate OH dehydrogenase H

OH

Glucose-6-phosphate

H HO

CH2OPO2– 3 O H OH H H

OH

6-Phosphogluconoδ-lactone

O

390  C ha pter 13   Glucose Metabolism

The lactone intermediate is hydrolyzed to 6-phosphogluconate by the action of 6phosphogluconolactonase, although this reaction can also occur in the absence of the enzyme:

H HO

CH2OPO2– 3 O H OH H H

OH

O H2O O

H+

6-phosphogluconolactonase

6-Phosphogluconoδ-lactone

C

O–

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OPO2– 3 6-Phosphogluconate

In the third step of the pentose phosphate pathway, 6-phosphogluconate is oxidatively decarboxylated in a reaction that converts the six-carbon sugar to a five-carbon sugar and reduces a second NADP+ to NADPH:

O

C

O–

H

C

OH

HO

C

H

H

C

OH

H

C

OH

NADP+

CO2 + NADPH

6-phosphogluconate dehydrogenase

CH2OH C

O

H

C

OH

H

C

OH

CH2OPO2– 3 Ribulose-5-phosphate

CH2OPO2– 3 6-Phosphogluconate

The two molecules of NADPH produced for each glucose molecule that enters the pathway are used primarily for biosynthetic reactions, such as fatty acid synthesis and the synthesis of deoxynucleotides.

Isomerization and interconversion reactions generate a variety of monosaccharides The ribulose-5-phosphate product of the oxidative phase of the pentose phosphate pathway can isomerize to ribose-5-phosphate:

H

CH2OH

O

C

O

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

CH2OPO2– 3 Ribulose-5-phosphate

ribulose-5-phosphate isomerase

C

CH2OPO2– 3 Ribose-5-phosphate

Ribose-5-phosphate is the precursor of the ribose unit of nucleotides. In many cells, this marks the end of the pentose phosphate pathway, which has the net equation glucose-6-phosphate + 2 NADP+ + H2O → ribose-5-phosphate + 2 NADPH + CO2 + 2 H+

Not surprisingly, the activity of the pentose phosphate pathway is high in rapidly dividing cells that must synthesize large amounts of DNA. In fact, the pentose phosphate pathway not only produces

13.4  The Pentose Phosphate Pathway  391

ribose, it also provides a reducing agent (NADPH) required for the reduction of ribose to deoxy­ ribose. Ribonucleotide reductase carries out the reduction of nucleoside diphosphates (NDPs):

O

O

O P

O P

–

O–

O

O

H2C

O–

H NDP

base

H

H

OH

OH

O

O

O P

O P

–

O–

H

O

O

H2C

O–

H dNDP

base

H

H

OH

H

H

The enzyme, which is oxidized in the process, is restored to its original state by a series of reactions in which NADPH is reduced (Section 18.5). In some cells, however, the need for NADPH for other biosynthetic reactions is greater than the need for ribose-5-phosphate. In this case, the excess carbons of the pentose are recycled into intermediates of the glycolytic pathway so that they can be degraded to pyruvate or used in gluconeogenesis, depending on the cell type and its metabolic needs. Ribose-5-phosphate and xylulose-5-phosphate, formed from ribulose-5-phosphate by epimerization, participate in a set of reversible reactions that transform three five-carbon sugars into two six-carbon sugars (fructose-6-phosphate) and one three-carbon sugar (glyceraldehyde3-phosphate, Fig. 13.12). These molecular rearrangements are catalyzed by the enzymes transaldolase, which transfers a three-carbon segment, and transketolase, which transfers a

O

CH2OH C

O

HO

C

H

H

C

OH

H

O

C

H

C

H

CH2OH

OH

CH2OPO2– 3 Glyceraldehyde3-phosphate

CH2OPO2– 3 Xylulose-5-phosphate

C

a

CH2OH C

O

HO

C

H

C

O

HO

C

H

H

C

OH

H

C

OH

b

CH2OPO2– 3

Fructose-6-phosphate

CH2OH O

C

H

C

O

HO

C

H

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

CH2OPO2– 3 Ribose-5-phosphate

CH2OPO2– 3 Sedoheptulose7-phosphate

CH2OPO2– 3 Erythrose4-phosphate

c

CH2OPO2– 3 Fructose-6-phosphate

CH2OH C

O

HO

C

H

H

C

OH

CH2OPO2– 3 Xylulose-5-phosphate   FIGURE 13.12   Transaldolase and transketolase reactions of the pentose phosphate pathway.  Three five-carbon products of the oxidative phase of the pentose phosphate pathway are converted to other monosaccharides by reversible reactions. In Reactions a and c, transketolase transfers a two-carbon unit (boxed) and in Reaction b, transaldolase transfers a three-carbon unit (boxed).

O H

C C

H OH

CH2OPO2– 3 Glyceraldehyde3-phosphate Question  Make a similar color-coded diagram to show how glycolytic intermediates could be transformed to ribose-5-phosphate.

392  C ha pter 13   Glucose Metabolism

two-carbon segment (the reaction catalyzed by transketolase was introduced in Section 7.2). Because all these interconversions are reversible, glycolytic intermediates can also be siphoned from glycolysis or gluconeogenesis to synthesize ribose-5-phosphate. Thus, the cell can use some or all of the steps of the pentose phosphate pathway in order to generate NADPH, to produce ribose, and to interconvert other monosaccharides.

A summary of glucose metabolism Although our coverage of glucose metabolism is far from exhaustive, this chapter describes quite a few enzymes and reactions, which are compiled in Figure 13.13. As you examine this diagram, keep in mind the following points, which also apply to the metabolic pathways we will encounter in subsequent chapters:

Glycogen UTP Glucose ATP Glucose-1phosphate

NADPH Glucose-6phosphate

NADPH Ribulose-5phosphate

6-Phosphogluconate

Ribose-5phosphate

CO2 Fructose-6phosphate ATP Fructose-1,6bisphosphate

Dihydroxyacetone phosphate

Glyceraldehyde-3phosphate

NADH

NADH

1,3-Bisphosphoglycerate ATP

ATP

3-Phosphoglycerate

2-Phosphoglycerate

Phosphoenolpyruvate

Oxaloacetate

ATP Fatty acids

Pyruvate

Amino acids

CO2 ATP

Acetyl-CoA NADH citric acid cycle

GTP

NADH Lactate

  FIGURE 13.13   Summary of glucose metabolism.  This diagram includes the pathways of glycogen synthesis and degradation, glycolysis, gluconeogenesis, and the pentose phosphate pathway. Dotted lines are used where the individual reactions are not shown. Filled gold symbols indicate ATP production; shadowed gold symbols indicate ATP consumption. Filled and shadowed red symbols represent the production and consumption of the reduced cofactors NADH and NADPH.

13.5  Clinical Connection: Disorders of Carbohydrate Metabolism  393

1. A metabolic pathway is a series of enzyme-catalyzed reactions, so the pathway’s substrate is converted to its product in discrete steps. 2. A monomeric compound such as glucose is interconverted with its polymeric form (glycogen), with other monosaccharides (fructose-6-phosphate and ribose-5-phosphate, for example), and with smaller metabolites such as the three-carbon pyruvate. 3. Although anabolic and catabolic pathways may share some steps, their irreversible steps are catalyzed by enzymes unique to each pathway. 4. Certain reactions consume or produce energy in the form of ATP. In most cases, these are phosphoryl-group transfer reactions. 5. Some steps are oxidation–reduction reactions that require or generate a reduced cofactor such as NADH or NADPH.

Before Going On • List the products of the pentose phosphate pathway and describe how the cell uses them. • Explain why cells maintain a high NAD+/NADH ratio and a low NADP+/NADPH ratio. • Explain how the cell catabolizes excess ribose groups.

13.5 Clinical Connection: Disorders of Carbohydrate Metabolism LEARNING OBJECTIVES Relate enzyme deficiencies to defects in carbohydrate metabolism. • Explain why red blood cells are so susceptible to defects in glucose metabolic pathways. • Describe the symptoms of glycogen storage diseases affecting liver and muscle.

Although there is not enough space to describe all the known enzyme deficiencies that affect carbohydrate metabolic pathways in humans, a few disorders are worth highlighting. As with many metabolic diseases, the discovery and study of carbohydrate metabolic disorders has helped shed light on how the normal metabolic pathways function. Keep in mind that an enzyme deficiency may result from a genetic variation that limits production of the protein, directly impacts its catalytic power, or affects its regulation. Deficiencies of glycolytic enzymes usually have severe consequences, particularly in tissues that rely heavily on glycolysis for ATP production. For example, red blood cells, which lack mitochondria to produce ATP by oxidative phosphorylation, use the ATP generated by glycolysis to power the Na,K-ATPase that maintains the cells’ ion concentration gradients (Section 9.3). Low glycolytic activity reduces the supply of ATP, and the resulting ion imbalance leads to osmotic swelling and bursting of the red blood cells. In addition to anemia (the loss of red blood cells), other abnormalities may develop when glycolytic enzymes are defective. In a pyruvate kinase deficiency, several of the intermediates upstream of phosphoenolpyruvate (the pyruvate kinase substrate) accumulate because they are in equilibrium. Red blood cells normally convert one of these—1,3-bisphosphoglycerate— to 2,3-bisphosphoglycerate (BPG), which binds to hemoglobin to decrease its oxygen affinity (Section 7.1). The elevated concentrations of glycolytic intermediates resulting from a pyruvate kinase deficiency actually boost BPG production, allowing red blood cells to deliver O2 more efficiently, which helps offset the anemia caused by the enzyme deficiency. A shortage

394  C ha pter 13   Glucose Metabolism

of hexokinase, however, slows the entire glycolytic pathway, which limits the production of BPG in red blood cells and thereby reduces the amount of O2 delivered to tissues. Defects in pathways that metabolize sugars other than glucose have variable effects. Individuals with fructose intolerance lack fructose-1-phosphate aldolase (see Box 13.A). The resulting accumulation of fructose-1-phosphate ties up the liver’s phosphate supply, which hinders ATP production from ADP. As in the disorders already described, one of the first casualties is the Na,K-ATPase, whose inadequate activity leads to cell death. An inability to convert galactose to glucose can be deadly, especially in infancy, when the major carbohydrate source is lactose, a disaccharide of glucose and galactose. High concentrations of galactose that cannot be metabolized contribute to side reactions such as the addition of sugar derivatives to proteins. Damage to proteins in nerve cells causes growth retardation and abnormal brain development. Fortunately, an early diagnosis and a galactosefree diet can avoid such damage. A deficiency of glucose-6-phosphate dehydrogenase, the first enzyme of the pentose phosphate pathway, is the most common human enzyme deficiency. This defect decreases the cellular production of the reducing agent NADPH, which participates in certain oxidation– reduction processes and helps protect cells from oxidative damage. Red blood cells, where the O2 concentration is high, are most at risk. Glucose-6-phosphate dehydrogenase deficiency affects about 500 million people, mostly in Africa, tropical South America, and southeast Asia—areas with historically high rates of malaria. The enzyme defect causes anemia, but the release of heme from damaged red blood cells triggers anti-inflammatory responses that increase an individual’s chances of surviving malaria. The same effect is observed in individuals carrying the hemoglobin S variant (Section 5.2).

Glycogen storage diseases affect liver and muscle

TA BLE

Type GSD0 GSD1 GSD2 GSD3 GSD4 GSD5 GSD6 GSD7 GSD9 GSD10 GSD11 GSD12 GSD13 GSD14 GSD15

The glycogen storage diseases (GSDs) are a set of inherited disorders of glucose and glycogen metabolism, not all of which result in glycogen accumulation, as the name might suggest. The symptoms of the glycogen storage diseases vary, depending on whether the affected tissue is liver or muscle or both. In general, the disorders that affect the liver cause hypoglycemia (too little glucose in the blood) and an enlarged liver. Glycogen storage diseases that affect primarily muscle are characterized by muscle weakness and cramps. The incidence of glycogen storage diseases is estimated to be as high as 1 in 20,000 births, although some disorders are not apparent until adulthood. Table 13.1 lists the enzyme defi1 3 .1   Glycogen Storage Diseases ciency associated with each type of glycogen storage disease. A defect of glucose-6-phosphatase (GSD1) affects both gluconeogenEnzyme deficiency esis and glycogenolysis, since the phosphatase catalyzes the final step of Glycogen synthase gluconeogenesis and makes free glucose available from glycogenolysis. Glucose-6-phosphatase The enlarged liver and hypoglycemia can lead to a host of other symptoms, including irritability, lethargy, and, in severe cases, death. A related defect α-1,4-Glucosidase is the deficiency of the transport protein that imports glucose-6-phosphate Amylo-1,6-glucosidase into the endoplasmic reticulum, where the phosphatase is located. (debranching enzyme) Glycogen storage disease 3 results from a deficiency of the glycogen Amylo-(1,4 → 1,6)-transglycosylase debranching enzyme. This condition accounts for about one-quarter of all (branching enzyme) cases of glycogen storage disease and usually affects both liver and musMuscle glycogen phosphorylase cle. The symptoms include muscle weakness and liver enlargement due Liver glycogen phosphorylase to the accumulation of glycogen that cannot be efficiently broken down. Phosphofructokinase The symptoms of GSD3 often improve with age and disappear by early adulthood. Phosphorylase kinase The most common type of glycogen storage disease is GSD9. In this Phosphoglycerate mutase disorder, the kinase that activates glycogen phosphorylase is defective. Lactate dehydrogenase Symptoms range from severe to mild and may fade with time. The comAldolase plexity of this disease reflects the fact that the phosphorylase kinase conEnolase sists of four subunits, with isoforms that are differentially expressed in the liver and other tissues. Phosphoglucomutase In the past, glycogen storage diseases were diagnosed on the basis of Glycogenin symptoms, blood tests, and painful biopsies of liver or muscle to assess its

Key Terms  395

glycogen content. Current diag­nostic methods are centered on analyzing the relevant genes for mutations, a noninvasive approach. Treatment of glycogen storage diseases typically includes a regimen of frequent, small, carbohydrate-rich meals to alleviate hypoglycemia. However, because dietary therapy does not completely eliminate the symptoms of some glycogen storage diseases, and because the metabolic abnormalities, such as chronic hypoglycemia and liver damage, can severely impair physical growth as well as cognitive development, liver transplant has proved to be an effective treatment. The glycogen storage diseases are single-gene defects, which makes them attractive targets for gene therapy (see Section 3.3).

Before Going On • Make a list of enzyme deficiencies and their physiological effects. • Compare the impact of various disorders on red blood cells, liver cells, and muscle cells.

Summary 13.1  Glycolysis •  The pathway of glucose catabolism, or glycolysis, is a series of enzyme-catalyzed steps in which energy is conserved as ATP or NADH. •  The 10 reactions of glycolysis convert the six-carbon glucose to two molecules of pyruvate and produce two molecules of NADH and two molecules of ATP. The first and third reactions (catalyzed by hexokinase and phosphofructokinase) require the investment of two ATP. The irreversible reaction catalyzed by phosphofructokinase is the rate-determining step and the major control point for glycolysis. Later steps (catalyzed by phosphoglycerate kinase and pyruvate kinase) generate four ATP per glucose. •  Pyruvate may be reduced to lactate or ethanol, further oxidized by the citric acid cycle, or converted to other compounds.

13.2  Gluconeogenesis •  The pathway of gluconeogenesis converts two molecules of pyruvate to one molecule of glucose at a cost of six molecules of ATP. The pathway uses seven glycolytic enzymes, and the activities of pyruvate carboxylase, phosphoenolpyruvate carboxykinase, fructose bisphosphatase, and glucose-6-phosphatase bypass the three irreversible steps of glycolysis.

•  A futile cycle involving phosphofructokinase and fructose bisphosphatase helps regulate the flux through glycolysis and gluco­neogenesis.

13.3  Glycogen Synthesis and Degradation •  Glucose residues are incorporated into glycogen after first being activated by attachment to UDP. •  Phosphorolysis of glycogen produces phosphorylated glucose that can enter glycolysis. In the liver, this glucose is dephosphorylated and exported.

13.4  The Pentose Phosphate Pathway •  The pentose phosphate catabolic pathway for glucose yields NADPH and ribose groups. The five-carbon sugar intermediates can be converted to glycolytic intermediates.

13.5  Clinical Connection: Disorders of Carbohydrate Metabolism •  Disorders are associated with deficiencies of glycolytic enzymes and enzymes that metabolize fructose and galactose. •  Glycogen storage diseases cause hypoglycemia, muscle weakness, and liver damage.

Key Terms glycolysis kinase metabolically irreversible reaction near-equilibrium reaction rate-determining reaction

tautomerization feed-forward activation fermentation gluconeogenesis futile cycle

glycogenolysis pentose phosphate pathway glycogen storage disease (GSD)

396  C ha pter 13   Glucose Metabolism

Bioinformatics Brief Bioinformatics Exercises 13.1  Hexokinase Structure and Ligand Binding 13.2  Glycolysis and the KEGG Database

Problems 13.1  Glycolysis 1.  Which of the ten reactions of glycolysis are  a.  phosphorylations, b. isomerizations, c. oxidation–reductions, d. dehydrations, and  e.  carbon–carbon bond cleavages? 2.  Which reactions of glycolysis can be reversed? Which reactions are irreversible? What is the significance of the metabolically irreversible reactions? 3.  Except during starvation, the brain oxidizes glucose as its sole metabolic fuel and consumes up to 40% of the body’s circulating glucose. Brain hexokinase has a KM for glucose that is 100 times lower than the concentration of circulating glucose (5 mM). What is the advantage of this low KM? 4.  There are four isozymes of hexokinase termed hexokinases I, II, and III (which have KM values of ~0.02 mM) and hexokinase IV (which has a KM value of ~5 mM). Normal hepatocytes (liver cells) express low amounts of hexokinases I and II, but upon transformation to cancer cells, the expression of hexokinase II (and to some extent hexokinase I) increases while hexokinase IV expression is silenced. How does this strategy promote the survival of the cancer cell? 5.  Residue Asn 204 in the glucose binding site of hexokinase IV (see Problem 4) was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. What do these experiments reveal about the intermolecular interactions between the enzyme and the glucose substrate? 6.  The Vmax and KM values for an unusual hexokinase found in Trypansoma cruzi (the causative agent of Chagas disease) are shown in the presence and absence of a bisphonate inhibitor (structure shown). KM (mM) Vmax (μmol · min−1 · mL−1)

Without inhibitor 90 0.30

O− OH P O

P

With inhibitor 125 0.12

O O−

O−

O−

A bisphonate compound a.  What type of inhibitor is bisphonate?  b.  The parasite hexokinase, unlike the mammalian enzyme, is not inhibited by glucose-6-phosphate but is inhibited by pyrophosphate (PPi). Is this observation consistent with your answer to part a? c. Might bisphonate be a good candidate for a drug to treat the disease?

7.  1,5-Anhydroglucitol, a glucose derivative missing the hydroxyl group on the anomeric carbon, is present in small amounts in nearly all foods. Draw the structure of the product that results when hexo­ kinase uses 1,5-anhydroglucitol as a substrate. 8.  Mannose, the C2 epimer of glucose, enters mammalian cells via glucose transporters and becomes a substrate for hexokinase. Explain why treating cells with mannose can impair glycolysis, particularly in cells with low phosphomannose isomerase activity. 9.  What is the ratio of fructose-6-phosphate (F6P) to glucose-6-phosphate (G6P) under a. standard conditions at 25°C and b. cellular conditions at 37°C? In which direction does the reaction proceed under cellular conditions? 10.  The radioactively labeled compound [18F]fluorodeoxyglucose (FDG) is used to measure glucose uptake in cells. FDG, a derivative of glucose in which the C2 hydroxyl is replaced with fluorine, is phosphorylated to FDG-6-phosphate upon entering the cell but cannot proceed any further through glycolysis. The phosphorylated FDG remains in the cell and its presence can be detected and quantitated, providing information on the rate of glucose uptake. (Cancer cells take up glucose particularly rapidly.) a. Draw the structure of FDG. b. Why does phosphorylation trap the FDG in the cell? c. Why is FDG unable to proceed though glycolysis past the hexokinase reaction? (Hint: When glucose-6-phosphate is converted to fructose-6-phosphate, the ring opens, the isomerization reaction takes place, and then the ring closes again.) d. Is FDG taken up only by cancer cells? 11.  ADP stimulates the activity of phosphofructokinase (PFK), yet it is a product of the reaction and not a reactant. Explain this apparent contradictory regulatory strategy. 12.  The “T” and “R” nomenclature used to describe the low- and high-affinity forms of hemoglobin (Section 5.1) can also be used to describe the conformational changes that occur in allosteric enzymes like phosphofructokinase. Allosteric inhibitors stabilize the T form, which has a low affinity for its substrate, while activators stabilize the high-affinity R form. Do the following allosteric effectors stabilize the T form of PFK or the R form? a. ADP (bacteria), b. phosphoenolpyruvate (PEP; bacteria), c. fructose-2,6-­bisphosphate (mammals). 13.  Phosphofructokinase (PFK) isolated from the bacterium Bacillus stearothermophilus is a tetramer that binds fructose-6-phosphate with hyperbolic kinetics and a KM of 23 μM. What happens to the KM in the presence of phosphoenolpyruvate (PEP; see Fig. 7.15)? Use the T → R terminology described in Problem 12 to explain what happens. 14.  Refer to Figure 7.16. Why does the conformational change that results when Arg 162 changes places with Glu 161 result in a form of phosphofructokinase that has a low affinity for its substrate? 15.  Researchers isolated a yeast mutant that was deficient in phosphofructokinase. The mutant yeast was able to grow on glycerol as an energy source, but not glucose. Explain why.

Problems  397 16.  Researchers isolated a yeast phosphofructokinase mutant in which a serine at the fructose-2,6-bisphosphate (F26BP) binding site was replaced with an aspartate residue. The amino acid substitution completely abolished the binding of F26BP to PFK. There was a dramatic decline in glucose consumption and ethanol production in the mutant compared to control yeast. a. Propose a hypothesis that explains why the mutant PFK cannot bind F26BP. b. What does the decline of glucose consumption and ethanol production in the yeast reveal about the role of F26BP in glycolysis? 17.  Refer to the mechanism of aldolase shown in Figure 13.4. a. Is the pK of the Lys side chain higher, lower, or the same as it would be in the free amino acid? What is the role of the Lys side chain in catalysis? b. Does the pK of the Asp side chain change after formation of the Schiff base? What is the role of the Asp side chain in catalysis? c. The Asp residue is mutated to an Ala residue in a site-directed mutagenesis experiment. What is the effect on enzyme activity?

nonequilibrium conditions? Considering your answer to this question, how do you account for the fact that the conversion of DHAP to GAP occurs readily in cells? 22.  Biochemists use transition state analogs to determine the structure of a short-lived intermediate in an enzyme-catalyzed reaction. Because an enzyme binds tightly to the transition state, a compound that resembles the transition state should be a potent competitive inhibitor. Phosphoglycohydroxamate binds 150 times more tightly than dihydroxyacetone phosphate to triose phosphate isomerase. Based on this information, propose a structure for the intermediate of the triose phosphate isomerase reaction.

OH N

18.  Does the aldolase enzyme mechanism (see Fig. 13.4) use acid catalysis, base catalysis, covalent catalysis, or some combination of these strategies (see Section 6.2)? Explain. 19.  The aldolase reaction has a standard free energy change of 22.8 kJ · mol−1.  a. Is the reaction favorable under standard conditions? b. What is the actual free energy change, ΔG, at 37°C, if the concentration of fructose-1,6-bisphosphate (F16BP) is 14 μM, the concentration of glyceraldehyde-3-phosphate (GAP) is 3.0 μM, and the concentration of dihydroxyacetone phosphate (DHAP) is 16 μM? Is the forward or the reverse direction of this reaction favored under these conditions? Explain. 20.  Some bacteria degrade glucose via the Entner–Douderoff pathway, outlined below, rather than by the glycolytic pathway presented in the chapter. (The structure of 2-keto-3-deoxygluconate (KDG) is shown in Solution 11.23.) a. In which step does a bacterial aldolase catalyze a reaction similar to the aldolase reaction in glycolysis in humans? b. The compound shown below inhibits the bacterial aldolase. Propose a hypothesis to explain how the inhibitor interferes with the enzyme’s activity. c. The inhibitor has a KI of 250 μM. The KM for the substrate is 60 μM. Is the compound an effective inhibitor of the bacterial aldolase?

Glucose NADP+

ADP 2-Keto-3-deoxy-6-phosphogluconate (KDPG) Glyceraldehyde-3-phosphate 1,3-BPG 4 steps Pyruvate

H3C

C

C CH2

23.  Cancer cells have elevated levels of glyceraldehyde-3-phosphate dehydrogenase (GAPDH), which may account for the high rate of glycolysis in these cells. The compound methylglyoxal has been shown to inhibit GAPDH in cancer cells but not in normal cells. This observation may lead to the development of drugs for treating cancer. a. Propose a hypothesis to explain why GAPDH levels in cancer cells are elevated. b. Why might methylglyoxal inhibit GAPDH in cancer cells but not in normal cells?

24.  Glyceraldehyde-3-phosphate can react with arsenate, As​​O​  4 3− ​  ​​, a phosphate analog, in the GAPDH reaction to form 1-arseno-3-­ phosphoglycerate, which is unstable and spontaneously hydrolyzes to form 3-phosphoglycerate (3PG), as shown. What is the effect of arsenate on cells undergoing glycolysis?

O

C C

OAsO2− 3

H2O

AsO3− 4

OH

O− O

21.  What is the ratio of glyceraldehyde-3-phosphate (GAP) to dihydroxyacetone phosphate (DHAP) in cells at 37°C under

O H

C C

O− OH

CH2OPO2− 3 3PG

1-arseno-3-phosphoglycerate

H2O

O

Phosphoglycohydroxamate

CH2OPO2− 3

2-Keto-3-deoxygluconate (KDG) ATP

O

O−

CH2OPO2− 3

H

NADPH Gluconate

Pyruvate

C

25.  In several species of bacteria, GAPDH activity is controlled by the NADH/NAD+ ratio. Does the activity of GAPDH increase or decrease when the NADH/NAD+ ratio increases? Explain. Assume that only the forward direction of the reaction is relevant. 26.  The thermophilic archaebacterium Thermoproteus tenax expresses two GAPDH enzymes: one that carries out the ­reaction shown in the chapter (although NADP + is the reactant rather than NAD+) and a second GAPDH enzyme that catalyzes the reaction shown here. T. tenax Glyceraldehyde-3-phosphate + NAD+

GAPDH

3-phosphoglycerate + NADH + H+ a.  How do the properties of the two enzymes differ? b. In an experiment, the activity of the second T. tenax GAPDH enzyme was measured in the presence of various metabolites. The results are shown in the figure. Estimate the KM values for the enzyme in the presence and absence of effectors and classify the effectors as activators or inhibitors of the enzyme. c. Propose a hypothesis that explains why these effectors act as either activators or inhibitors.

Velocity (units . mg protein−1)

398  C ha pter 13   Glucose Metabolism a.  Calculate KM and Vmax in the presence and in the absence of the inhibitor. b. What type of inhibitor is oxalate? c. Calculate KI for the inhibitor.

25 20

x intercept (mM−1)

y intercept (units−1 · mg)

Without inhibitor

−0.44

0.0034

With inhibitor

−0.050

0.0034

15 10 5 0

0

5

NADP+

10 [NAD+] (mM) Control

AMP

15

20 Glucose-1-P

27.  Phosphoglycerate kinase in red blood cells is bound to the plasma membrane. This allows the kinase reaction to be coupled to the Na,K-ATPase pump. How does the proximity of the enzyme to the membrane facilitate the action of the pump? 28.  Vanadate, ​​VO​ 43− ​  ​​, inhibits GAPDH, not by acting as a phosphate analog (see Problem 24), but by interacting with essential —SH groups on the enzyme. What happens to cellular levels of phosphate, ATP, and 2,3-bisphosphoglycerate (see pathway below) when red blood cells are incubated with vanadate?

Glyceraldehyde-3-phosphate GAPDH

1,3-Bisphosphoglycerate PGK

2,3-Bisphosphoglycerate

3-Phosphoglycerate 29.  The mechanism of plant phosphoglycerate mutase is different from the mechanism of mammalian phosphoglycerate mutase presented in the chapter. 3-Phosphoglycerate (3PG) binds to the plant enzyme, transfers its phosphate to the enzyme, and then the enzyme transfers the phosphate group back to the substrate to form ­2-phosphoglycerate (2PG). What is the fate of the [32P] label when [ 32P]-labeled 3PG is added to a. cultured hepatocytes or  b.  plant cells? 30.  Several studies have shown that aluminum inhibits phospho­ fructokinase in liver cells. a. Compare the production of pyruvate by perfused livers in control and aluminum-treated rats using fructose as an energy source (see Box 13.A). b. What would the experimental results be if glucose were used instead of fructose? 31.  Investigators who wish to deplete ATP in cultured cells can add iodoacetate to the culture medium. Why does the addition of iodo­ acetate successfully deplete intracellular ATP? 32.  Enzymes occasionally display weak “side” activities. Draw the structure of the product of the reaction that results when pyruvate kinase, operating in reverse, uses lactate as a substrate. What are the other substrate and product of this reaction? 33.  The pyruvate kinase in red blood cells has a ΔG°ʹ value of −31.4 kJ · mol−1. a. What is the equilibrium constant, Keq, for the reaction at 25°C? b. Calculate the ΔG value for the reaction under the following conditions: [phosphoenolpyruvate, PEP] = 23 μM, [pyruvate] = 51 μM, [ATP] = 1.0 mM, and [ADP] = 0.1 mM. Is the reaction favorable under these conditions? 34.  In an experiment, the activity of a pyruvate kinase isoenzyme­ common in cancer cells was measured in the presence and in the absence of 1.6 mM oxalate. The kinetic data were graphed as a Lineweaver–Burk plot whose x and y intercepts are given in the table. 

35.  About 15% of ingested ethanol is metabolized by a cytochrome P450 (see Section 7.4) and chronic alcohol consumption induces the expression of this enzyme. Explain how this changes the effectiveness of therapeutic drugs. 36.  Drinking methanol can cause blindness and death, depending on the dosage. The causative agent is formaldehyde derived from methanol. a. Draw the balanced chemical reaction for the conversion of methanol to formaldehyde. b. Why would administering whiskey (ethanol) to a person poisoned with methanol be a good antidote? 37.  Assuming a standard free energy change of 30.5 kJ · mol−1 for the synthesis of ATP from ADP and Pi , how many molecules of ATP could theoretically be produced by the catabolism of glucose to lactate (ΔG°ʹ = −196 kJ · mol−1), assuming 33% efficiency? 38. a. How many molecules of ATP could theoretically be produced (see Problem 37) by the catabolism of glucose to CO2 (ΔG°ʹ = −2850 kJ · mol −1), assuming 33% efficiency? b. Compare your answer to Solution 37. Does this explain the Pasteur effect (the observation, first made by Louis Pasteur, that glucose consumption in yeast dramatically decreases in the presence of oxygen)? 39.  Studies have shown that the halophilic organism Halococcus saccharolyticus degrades glucose via the Entner–Doudoroff pathway (see Problem 20). a. What is the ATP yield per mole of glucose for this pathway? b. Describe (in general) what kinds of reactions would need to follow the Entner–Doudoroff pathway in this organism. 40.  Trypanosomes living in the bloodstream obtain all their energy from glycolysis. They take up glucose from the host’s blood and excrete pyruvate as a waste product. In this part of their life cycle, trypanosomes do not carry out any oxidative phosphorylation, but they do use another oxygen-dependent pathway, which is absent in mammals, to oxidize NADH. a. Why is the oxygen-dependent pathway necessary? b. Would this pathway be necessary if the trypanosome excreted lactate rather than pyruvate? c. Why would the oxygen-dependent pathway be a good target for antiparasitic drugs?

13.2  Gluconeogenesis 41.  Which of the 11 reactions of gluconeogenesis are a. phosphorylations, b. isomerizations, c. oxidation–reductions,  d.  hydrations,  e.  carbon–carbon bond formations,  f.  carboxylations,  g.  decarboxylations, and  h.  hydrolysis reactions? 42.  Which of the following substrates can be converted to glucose via the gluconeogenesis pathway? a. pyruvate, b. lactate, c. alanine,  d.  acetyl-CoA,  e.  oxaloacetate,  f.  leucine (which is degraded to acetyl-CoA),  g.  glycerol. 43.  Biotin deficiencies are rare because biotin is present in many foods and is also synthesized by intestinal bacteria. Biotin deficiencies have been observed, however, in individuals consuming large quantities of raw eggs. Explain why. (Hint: See Problem 4.93.) 44.  a.  Use the equations provided to determine ΔG°ʹ for the favorable synthesis of oxaloacetate (OAA): pyruvate + CO2 → H+ + OAA   ΔG°ʹ = 31.8 kJ · mol−1

ADP + Pi + H+ → ATP + H2O  ΔG°ʹ = 30.5 kJ · mol−1

Problems  399 b.  How is the synthesis of oxaloacetate from pyruvate accomplished in cells? c. Calculate ΔG for this reaction under the following conditions: 37°C, pH 7, [Pyruvate] = [CO2] = 4.0 mM, [OAA] = 2.0 mM, [ATP] = 3.5 mM, [Pi] = 5.0 mM, and [ADP] = 1.8 mM. Is the reaction favorable under these conditions? 45.  Flux through the opposing pathways of glycolysis and gluconeogenesis is controlled in several ways. a. Explain how the activation of pyruvate carboxylase by acetyl-CoA affects glucose metabolism. b. Pyruvate can undergo a reversible amino-group transfer reaction to yield alanine (see Section 12.2). Alanine is an allosteric effector of pyruvate kinase. Would you expect alanine to stimulate or inhibit pyruvate kinase? Explain. 46.  A physician diagnoses an infant patient with a pyruvate carboxylase deficiency in part by measuring the patient’s blood levels of lactate and pyruvate. a. How would the patient’s [lactate]/[pyruvate] ratio compare to the ratio in a normal infant? b. The diagnosis can be confirmed by giving the patient an alanine injection. What happens to the injected alanine in the patient and how does this differ in a normal patient? 47.  The activity of the fructose-1,6-bisphosphatase enzyme is measured at increasing concentrations of fructose-2,6-bisphosphate (F26BP), in the absence and in the presence of 25 µM AMP. The data are shown in the figure. How does F26BP affect enzyme activity in the absence of AMP? In its presence?

Relative activity (%)

100 80 60 40 20 0

0

1 2 3 4 [Fructose-2,6-bisphosphate] (μM) Without AMP

5

With AMP

48.  A liver biopsy of a four-year-old boy indicated that fructose1,6-bisphosphatase enzyme activity was 20% of normal. The patient’s blood glucose levels were normal at the beginning of a fast but then decreased suddenly. Pyruvate and alanine concentrations were also elevated, as was the glyceraldehyde-3-phosphate/dihydroxyacetone phosphate ([GAP]/[DHAP]) ratio. Explain the reason for these symptoms. 49.  Insulin is one of the major hormones that regulates gluconeogenesis. Insulin acts in part by decreasing the transcription of genes coding for certain gluconeogenic enzymes. For which genes would you expect insulin to suppress transcription? 50.  Type 2 diabetes is characterized by insulin resistance, in which insulin is unable to perform its many functions. What symptoms would you expect in a Type 2 diabetic patient if insulin is unable to perform the function described in Problem 49? 51.  The concentration of fructose-2,6-bisphosphate (F26BP) is regulated in the cell by a homodimeric enzyme with two catalytic activities: a kinase that phosphorylates fructose-6-phosphate to form F26BP and a phosphatase that catalyzes the hydrolysis of the C2 phosphate group. a. Which enzyme activity, the kinase or the phosphatase, would you expect to be active under fasting conditions? Explain. b. Which hormone is likely to be responsible for inducing this activity? c. Consult Section 10.2 and propose a mechanism for this induction.

52.  Brazilin, a compound found in aqueous extracts of sappan wood, has been used to treat diabetics in Korea. Brazilin increases the activity of the kinase enzyme that produces F26BP (see Problem 51), and the compound also stimulates the activity of pyruvate kinase. a. What is the effect of adding brazilin to hepatocytes (liver cells) in culture? b. Why would brazilin be an effective treatment for diabetes? 53.  Metformin is a drug that decreases the expression of phospho­ enolpyruvate carboxykinase. Explain why metformin would be ­helpful in treating diabetes. 54.  The “carbon skeletons” of most amino acids can be converted to glucose, a process that may require many enzymatic steps. Which amino acids can enter the gluconeogenic pathway directly after undergoing deamination (a reaction in which the carbon with the amino group becomes a ketone)? 55.  Draw a diagram that illustrates how alanine (see Problem 54) released from the muscle is converted back to glucose in the liver. What is the physiological cost if this cycle runs for a prolonged period of time? 56.  Draw a diagram that illustrates how lactate released from the muscle is converted back to glucose in the liver. What is the cost (in ATP) of running this cycle? 57.  The Leishmania parasite carries out a modified form of gluconeo­ genesis in a specialized organelle called the glycosome. Gluconeo­ genesis is essential to parasite survival, so understanding the pathway could assist in the development of drugs to treat parasitic infections. Draw a diagram of the reactions that occur in the glycosome, given the following information: PEPCK and fructose bisphosphatase are present; glucose-6-phosphatase and pyruvate carboxylase are not present. Pyruvate phosphate dikinase (PPDK), which converts pyruvate to phosphoenolpyruvate (PEP), is located in the glycosome. A glycerol-3-kinase (GK) enzyme is also located in the glycosome. Malate dehydrogenase (MDH), which catalyzes the reversible interconversion of malate and oxaloacetate (OAA), is found in the cytosol as well as the glycosome. 58.  The purpose of gluconeogenesis in the Leishmania parasite (see Problem 57) is to synthesize fructose-6-phosphate, a precursor for mannogen, a homopolysaccharide consisting of β(1 → 2) mannose residues. Using the diagram you constructed in Problem 57, explain how each of these compounds can be used to synthesize mannogen: a. glycerol, b. aspartate, and c. alanine. (Note: The glycosome includes transporters for glycerol, pyruvate, and malate.)

13.3  Glycogen Synthesis and Degradation 59.  ΔG°ʹ for the glycogen synthase reaction is −13.4 kJ · mol−1. ­Calculate Keq for this reaction at 25°C. 60.  a.  Use the information from Problem 59 as well as information in the chapter regarding the ΔG°ʹ values for the steps involved in glycogen synthesis to calculate a ΔG°ʹ value for the overall reaction shown below. b. Calculate Keq for this reaction at 25°C.

glucose-1-phosphate (G1P) + glycogen (n) + UTP + H2O →

glycogen (n + 1) + UDP + 2 Pi

61.  Beer is produced from raw materials such as wheat and barley. Explain why the grains are allowed to sprout, a process in which their starch is broken down to glucose, before fermentation begins. 62.  Some bread manufacturers add amylase to bread dough prior to the fermentation process. What role does this enzyme (see Section 12.1) play in the bread-making process? 63.  The equation for the degradation of glycogen is shown below.  a.  What is the ratio of [Pi]/[G1P] under standard conditions? b. What is the value of ΔG under cellular conditions when the [Pi]/[G1P] ratio

400  C ha pter 13   Glucose Metabolism is 50/1? c. What advantage does degradation by phosphorolysis have over a simple hydrolysis, which would produce g­ lucose instead of glucose-1-phosphate? phosphorylase glycogen (n) + Pi            glycogen (n − 1) + G1P ΔG°ʹ = +3.1 kJ · mol−1 64.  The mechanism of the phosphoglucomutase enzyme is similar to that of the plant mutase described in Problem 29 and is shown below. On occasion, the glucose-1,6-bisphosphate dissociates from the enzyme. Why does the dissociation of glucose-1,6-bisphosphate inhibit the enzyme?

O

Ser CH2

O

P

CH2

Glycolysis

OH

O−

H HO

CH2OPO2− 3 O H H OH H OH H

H HO

CH2OPO2− 3 O H H OH H OPO2− 3 H

OH

Glucose-6phosphate

Fat body Glycogen

Ser O−

desiccation and freezing. Its concentration in the hemolymph must be closely regulated. Trehalose is synthesized in the insect fat body, which plays a role in metabolism analogous to the vertebrate liver. Studies of the insect Manduca sexta show that during starvation, hemolymph glucose concentration decreases, which results in an increase in fat body glycogen phosphorylase activity and a decrease in the concentration of fructose-2,6-bisphosphate. What effect do these changes have on hemolymph trehalose concentration in the fasted insect?

OH

Glucose-1,6bisphosphate

Hemolymph Glucose

Glucose Trehalose

Trehalose

68.  The glycolytic pathway in the thermophilic archaebacterium Thermoproteus tenax differs from the pathway presented in this chapter. The phosphofructokinase (PFK) reaction in T. tenax is reversible and depends on pyrophosphate rather than ATP. In addition, T. tenax has two glyceraldehyde-3-phosphate dehydrogenase (GADPH) isozymes. The “phosphorylating GAPDH” is similar to the enzyme described in this chapter. The second isozyme is the irreversible “nonphosphorylating GAPDH,” which catalyzes the reaction described in Problem 26. T. tenax relies on glycogen stores as a source of energy. What is the ATP yield for one mole of glucose oxidized by the pathway that uses the nonphosphorylating GAPDH enzyme?

13.4  The Pentose Phosphate Pathway O

Ser CH2

O

O−

P O−

H HO

CH2OH O H OH H H

H OPO2− 3

OH

Glucose-1phosphate 65.  Glucagon, a hormone secreted by the pancreas in the fasted state, binds to G–protein coupled receptors in the liver. As a result of glucagon action, glycogen phosphorylase is phosphorylated. Consult Section 10.2 and describe the steps leading up to the phosphorylation of glycogen phosphorylase. Is the phosphorylated form of the enzyme active or inactive? 66.  In addition to phosphorylating glycogen phosphorylase, glucagon signaling (see Problem 65) results in the phosphorylation of glycogen synthase. Is the phosphorylated form of the enzyme active or inactive? 67.  Trehalose, a disaccharide consisting of two glucose residues (see Problem 11.40), is one of the major sugars in the insect hemolymph (the fluid that circulates through the insect’s body). Trehalose serves as a storage form of glucose and also helps protect the insect from

69.  Most metabolic pathways include an enzyme-catalyzed reaction that commits a metabolite to continue through the pathway. a. Identify the first committed step of the pentose phosphate pathway. Explain your reasoning. b. Hexokinase catalyzes an irreversible reaction at the start of glycolysis. Does this step commit glucose to continue through glycolysis? 70.  A given metabolite may follow more than one metabolic pathway. List all the possible fates of glucose-6-phosphate in a. a liver cell and b. a muscle cell. 71.  Experiments were carried out in cultured cells to determine the relationship between glucose-6-phosphate dehydrogenase (G6PD) activity and rates of cell growth. Cells were cultured in a medium supplemented with serum, which contains growth factors that stimulate G6PD activity. Predict how the cellular NADPH/NADP+ ratio would change under the following circumstances: a. Serum is withdrawn from the medium. b. DHEA, an inhibitor of glucose-6-phosphate dehydrogenase, is added. c. The oxidant H2O2 is added. d. Serum is withdrawn and H2O2 is added. 72.  Write a mechanism for the nonenzymatic hydrolysis of 6phosphogluconolactone to 6-phosphogluconate. 73.  Enzymes in the soil fungus Aspergillus nidulans use NADPH as a coenzyme when converting nitrate to ammonium ions. When the fungus was cultured in a growth medium containing nitrate, the activities of several enzymes involved in glucose metabolism increased. What enzymes are good candidates for regulation under these conditions? Explain. 74.  Enzymes of the pentose phosphate pathway are attractive drug targets to treat cancer. Explain why decreasing the activity of the pentose phosphate pathway would be detrimental to a cancer cell.

Problems  401 75.  Draw the structure of the product of the reaction that results when erythrose-4-phosphate is used as a substrate by glyceraldehyde3-phosphate dehydrogenase. 76.  Name and draw the structure of the product of the transaldolasecatalyzed condensation of dihydroxyacetone phosphate (DHAP) and erythrose-4-phosphate (E4P). 77.  Which pathways would “collaborate” to maximize a cell’s production of NADPH from glucose-6-phosphate, assuming that the cell needed NADPH but not ribose? 78.  Which pathways would “collaborate” to maximize a cell’s production of ribose-5-phosphate from glucose-6-phosphate, assuming that the cell needed ribose-5-phosphate but not NADPH? 79.  Several studies have shown that the metabolite glucose-1,6-­ bisphosphate (G16BP) regulates several pathways of carbohydrate metabolism by inhibiting or activating key enzymes. The effect of G16BP on several enzymes is summarized in the table below. What pathways are active when G16BP is present? What pathways are inactive? What is the overall effect? Explain. Enzyme Hexokinase Phosphofructokinase (PFK) Pyruvate kinase (PK) Phosphoglucomutase 6-Phosphogluconate dehydrogenase

Effect of G16BP Inhibits Activates Activates Activates Inhibits

80.  Xylulose-5-phosphate acts as an intracellular signaling molecule that activates kinases and phosphatases in liver cells. As a result of this signaling, there is an increase in the activity of the enzyme that produces fructose-2,6-bisphosphate, and the expression of genes for lipid synthesis is increased. What is the net effect of these responses?

13.5  Clinical Connection: Disorders of Carbohydrate Metabolism 81.  Red blood cells synthesize and degrade 2,3-bisphosphoglycerate (BPG) as a detour from the glycolytic pathway, as shown in Problem 28. BPG decreases the oxygen affinity of hemoglobin by binding in the central cavity of the deoxygenated form of hemoglobin (see Fig. 5.11). This encourages delivery of oxygen to tissues. A defect in one of the glycolytic enzymes may affect levels of BPG. The plot shows oxygen-binding curves for normal erythrocytes and for hexokinasedeficient and pyruvate kinase–deficient erythrocytes. Identify which curve corresponds to which enzyme deficiency.

90 Oxygen saturation (%)

83.  Galactose is metabolized by three enzymes, as shown below. Ultimately, galactose is converted to glucose-6-phosphate, which is connected to pathways of carbohydrate metabolism, as discussed in Solution 70. An inability to digest galactose could arise from a deficiency in any of the three enzymes shown. a. A deficiency in galactokinase results in the accumulation of galactose, which can serve as a substrate for a reductase enzyme that converts galactose to galactitol, a toxic derivative that is excreted in the urine. Using what you learned about monosaccharide derivatives in Section 11.1, draw the structure of galactitol. b. Galactose can also serve as a substrate for dehydrogenase enzymes that oxidize the aldehyde group to form galactonate, which is less toxic than galactitol. Draw its structure. c. A deficiency of the uridyltransferase leads to the accumulation of galactose-1-phosphate, which has been shown to inhibit glucose-6-phosphatase, phosphoglucomutase, and glycogen phosphorylase. What symptoms arise when these three enzymes are inhibited as a result of a uridyltransferase deficiency?

Galactose Galactokinase

ATP ADP

Galactose-1-Pi Uridyltransferase

UDP–Glucose Glucose-1-Pi

Glucose-6-Pi

UDP–Galactose Epimerase UDP–Glucose 84.  Individuals with fructose intolerance lack fructose-1-phosphate aldolase, a liver enzyme essential for catabolizing fructose. In the absence of fructose-1-phosphate aldolase, fructose-1-phosphate accumulates in the liver and inhibits glycogen phosphorylase and ­fructose-1,6-bisphosphatase. a. Explain why individuals with fructose ­intolerance exhibit hypoglycemia (low blood sugar). b. Administering glycerol and dihydroxyacetone phosphate does not alleviate the hypoglycemia, but administering galactose does relieve the hypoglycemia. Explain. 85.  Glycogen storage disease 7 (GSD7) results from a deficiency of muscle phosphofructokinase. Patients with this genetic disease may have muscle PFK levels that are 1–5% of normal. Why do these patients suffer from myoglobinuria (myoglobin in the urine) and muscle cramping during exercise?

100 80 70 60

Normal erythrocytes

50 40 30 20 10 0

Problem 81)? b. One of the symptoms of a pyruvate kinase deficiency is hemolytic anemia, in which red blood cells swell and eventually lyse. Explain why the enzyme deficiency brings about this symptom.

0

10

20

30 40 pO2 (torr)

50

60

82.  a.  What happens to the [ADP]/[ATP] and [NAD +]/[NADH] ratios in red blood cells with a pyruvate kinase deficiency (see

86.  Patients with McArdle’s disease have normal liver glycogen content and structure, but their muscles are easily fatigued. Identify the type of glycogen storage disease as listed in Table 13.1. 87.  A patient with McArdle’s disease (see Solution 86) performs ischemic (anaerobic) exercise for as long as he is able to do so. The patient’s blood is withdrawn every few minutes during the exercise period and tested for lactate. The patient’s samples are compared with control samples from a patient who does not suffer from a glycogen storage disease. The results are shown in the figure. Why does the lactate concentration increase in the normal patient? Why is there no corresponding increase in the patient’s lactate concentration?

Blood lactate concentration (mg · 100 mL−1)

402  C ha pter 13   Glucose Metabolism 91.  Would a small feeding of cornstarch administered at bedtime help relieve the symptoms of a child with glycogen storage disease 0 (GSD0, see Table 13.1)? Explain why or why not.

60

40

20

0

92.  Reduced glutathione (GSH), a tripeptide containing a Cys residue (see Solutions 4.17–4.18), is found in red blood cells, where it reduces organic hydroperoxides (ROOH) formed in cellular structures exposed to high concentrations of reactive oxygen. In converting the peroxides to less harmful alcohols (ROH), GSH becomes oxidized to GSSG.

Control

Patient

0

5

10

15 20 Time (min)

25

30

88.  During even mild exertion, individuals with McArdle’s disease (see Solution 86) experience painful muscle cramps. Yet the muscles in these individuals contain normal amounts of glycogen. What does this observation reveal about the pathways for glycogen degradation and glycogen synthesis? 89.  Patients with von Gierke’s disease (glycogen storage disease 1 or GSD1) have a deficiency of glucose-6-phosphatase. One of the most prominent symptoms of the disease is a protruding abdomen due to an enlarged liver. a. Explain why the liver is enlarged in patients with von Gierke’s disease. b. Some patients with von Gierke’s disease also have enlarged kidneys. Explain why. 90.  a.  Does a patient with McArdle’s disease (see Problems 86–88) suffer from hypoglycemia, hyperglycemia, or neither? b. Does a patient with von Gierke’s disease (see Problem 89) suffer from hypoglycemia, hyperglycemia, or neither?

2 GSH + R—O—OH

Glutathione peroxidase

Organic hydroperoxide

GSSG + R—OH + H2O

Reduced glutathione also plays a role in maintaining normal red blood cell structure and keeping the iron ion of hemoglobin in the +2 oxidation state. Oxidized glutathione (GSSG) is regenerated as shown in the following reaction: GSSG + NADPH + H+

Glutathione reductase

2 GSH + NADP+

 Seemingly unrelated activities, such as taking the antimalarial drug primaquine or consuming fava beans, stimulate peroxide formation. Why do individuals with a glucose-6-phosphate dehydrogenase deficiency suffer from hemolytic anemia if they take antimalarial drugs or consume fava beans?

Selected Readings Hannou, S.A., Haslam, D.E., McKeown, N.M., and Herman, M.A., Fructose metabolism and metabolic disease, J. Clin. Invest. 128, 545–555, doi: 10.1172/JCI96702 (2018). [Reviews the links between fructose and dysregulation of carbohydrate and lipid metabolism.] Kanungo, S., Wells, K., Tribett, T., and, El-Gharbawy, A., Glycogen metabolism and glycogen storage disorders, Ann. Transl. Med. 6, 474, doi: 10.21037/atm.2018.10.59 (2018). [Reviews the defects, symptoms, and treatments for genetic diseases of glucose and glycogen metabolism.] Lenzen, S., A fresh view of glycolysis and glucokinase regulation: History and current status, J. Biol. Chem. 289, 12189–12194 (2014). [Includes some history of glycolysis research, with an emphasis on regulation of Steps 1 and 3 of glycolysis.] Prats, C., Graham, T.E., and Shearer, J., The dynamic life of the ­glycogen granule, J. Biol. Chem. 293, 7089–7098, doi: 10.1074/jbc.

R117.802843 (2018). [Describes the synthesis, degradation, and structure of glycogen granules.] Rajas, F., Gautier-Stein, A., and Mithieux, G., Glucose-6-phosphate, a central hub for liver carbohydrate metabolism, Metabolites 9, 282, doi: 10.3390/metabo9120282 (2019). [Summarizes the metabolic fates of glucose-6-phosphate in health and disease.] Stincone, A., Prigione, A., Cramer, T., Wamelink, M.M.C., Campbell, K., Cheung, E., Olin-Sandoval, V., Grüning, N.-M., Krüger, A., Tauqeer Alam, M., Keller, M.A., Breitenbach, M., Brindle, M., Rabinowitz, J.D., and Ralser, M., The return of metabolism: Biochemistry and physiology of the pentose phosphate pathway, Biol. Rev. 90, 927–963, doi: 10.1111/brv.12140 (2015). [A thorough review of the pathway, its evolutionary history, regulation, and role in disease.]

Chapter 13 Credits Figure 13.5a Image based on 1YPI. Lolis, E., Alber, T., Davenport, R.C., Rose, D., Hartman, F.C., Petsko, G.A., Structure of yeast triosephosphate isomerase at 1.9-A resolution, Biochemistry 29, 6609–6618 (1990). Figure 13.5b Image based on 2YPI, Lolis, E., Petsko, G.A., Crystallographic analysis of the complex between triosephosphate isomerase

and 2-phosphoglycolate at 2.5-A resolution, implications for catalysis, Biochemistry 29, 6619–6625 (1990). Figure 13.7 Data from Newsholme, E.A. and Start, C., Regulation in Metabolism, p. 97, Wiley (1973).

The Citric Acid Cycle DO YOU REMEMBER? • Enzymes accelerate chemical reactions using acid–base catalysis, covalent catalysis, and metal ion catalysis (Section 6.2). • Coenzymes such as NAD+ and ubiquinone collect electrons from compounds that become oxidized (Section 12.2). • Metabolic pathways in cells are connected and are regulated (Section 12.2). • Many vitamins, substances that humans cannot synthesize, are components of coenzymes (Section 12.2).

Smith Collection/Gado/Getty Images

CHAPTER 14

Parasites, such as the roundworm shown here, produce and release a variety of metabolites, including the citric acid cycle intermediate succinate. Specialized epithelial cells in the host animal’s intestine, lungs, nose, and other locations detect the extracellular succinate and respond by increasing their antiparasite defenses.

• Pyruvate can be converted to lactate, acetyl-CoA, or oxaloacetate (Section 13.1).

The citric acid cycle logically follows glycolysis in an overview of cellular energy metabo­lism, but the cycle does much more than just continue the breakdown of glucose. Occupying a central place in the metabolism of most cells, the citric acid cycle processes the remnants of all types of metabolic fuels, including fatty acids and amino acids, so that their energy can be used to synthesize ATP. The citric acid cycle also operates anabolically, supplying the precursors for biosynthetic pathways. We will use pyruvate, the end product of glycolysis, as the starting point for our study of the citric acid cycle. We will then examine the eight steps of the citric acid cycle and discuss how this sequence of reactions might have evolved. Finally, we will consider the citric acid cycle as a multifunctional pathway with links to other metabolic processes.

14.1 The Pyruvate Dehydrogenase Reaction LEARNING OBJECTIVES Summarize the reactions carried out by the pyruvate dehydrogenase complex. • List the substrates, products, and cofactors of the pyruvate dehydrogenase reaction. • Explain the advantages of a multienzyme complex.

The end product of glycolysis is the three-carbon compound pyruvate. In aerobic organisms, these carbons are ultimately oxidized to 3 CO2 (although the oxygen atoms come not from molecular oxygen but from water and phosphate). The first molecule of CO2 is released when pyruvate is decarboxylated to an acetyl unit. The second and third CO 2 molecules are products of the citric acid cycle.

403

404  C ha pter 14   The Citric Acid Cycle

The pyruvate dehydrogenase complex contains multiple copies of three different enzymes

a.

The decarboxylation of pyruvate is catalyzed by the pyruvate dehydrogenase complex. In eukaryotes, this enzyme complex, and the enzymes of the citric acid cycle itself, are located inside the mitochondrion (an organelle surrounded by a double membrane and whose interior is called the mitochondrial matrix). ­Accordingly, pyruvate produced by glycolysis in the cytosol must first be imported into the mitochondria by a specific transport protein. For convenience, the three kinds of enzymes that make up the pyruvate dehydrogenase complex are called E1, E2, and E3. Together they catalyze the oxidative decarboxylation of pyruvate and the transfer of the acetyl unit to coenzyme A:

b.

  FIGURE 14.1   Models of the E2 core of the pyruvate dehydrogenase complex.  a. In the Azotobacter vinelandii complex, 24 E2 polypeptides are arranged in a cube.  b. The 60 subunits of the E2 core from B. stearothermophilus form a dodecahedron, a shape with 12 pentagonal faces.

​pyruvate + CoA + ​NAD​+​→ acetyl-CoA + ​CO​ 2​+ NADH​

The structure of coenzyme A, a nucleotide derivative containing the vitamin pantothenate, is shown in Figure 3.2a. In some bacteria, the 4600-kD pyruvate dehydrogenase complex consists of a cubic core of 24 E2 subunits (Fig. 14.1a), which are surrounded by an outer shell of 24 E1 and 12 E3 subunits. In mammals and some other bacteria, the enzyme complex is even larger, with 42–48 E1, 60 E2, and 6–12 E3 plus additional proteins that hold the complex together and regulate its enzymatic activity. The 60-subunit E2 core of the pyruvate dehydrogenase complex from Bacillus stearothermophilus is shown in Figure 14.1b.

Pyruvate dehydrogenase converts pyruvate to acetyl-CoA The operation of the pyruvate dehydrogenase complex requires several coenzymes, whose functional roles in the five-step reaction are described below.

−

O O

P

O−

O O

P

O−

O CH2

CH3 CH2

N H3C

NH2

CH3

CH2

+

+

N C

N

1. In the first step, which is catalyzed by E1 (also called pyruvate dehydrogen­ ase), pyruvate is decarboxylated. This reaction requires the cofactor ­thiamine pyrophosphate (TPP; Fig. 14.2). TPP attacks the carbonyl carbon of pyruvate, and the departure of CO2 leaves a hydroxyethyl group attached to TPP. This carbanion is stabilized by the positively charged thiazolium ring group of TPP:

S

H

Thiamine pyrophosphate   FIGURE 14.2   Thiamine pyrophosphate

(TPP).  This cofactor is the phosphorylated form of thiamine, also known as vitamin B1 (see Section 12.2). The central thiazolium ring (blue) is the active portion. An acidic proton (red) dissociates, and the resulting carbanion is stabilized by the nearby positively charged nitrogen. TPP is a cofactor for several different decarboxylases.

N

O

+

C

N

O− H

−

O

O

+

+ +

C

CH3

CH3

C S −

C

C C

O OH

S CH3

N

CO2 H

O

C

S

C− CH3

Hydroxyethyl–TPP

CH3 Pyruvate

2. The hydroxyethyl group is then transferred to E2 of the pyruvate dehydrogen­ ase complex. The hydroxyethyl acceptor is a lipoamide prosthetic group

14.1  The Pyruvate Dehydrogenase Reaction  405

(Fig. 14.3). The transfer reaction regenerates the TPP cofactor of E1 and oxidizes the hydroxyethyl group to an acetyl group:

CH3

H

O

C

N

S

H+

C− CH3

H

O

C C

N

S

H+

CH3

O

S

S

C− C

C

O CoA S

S

C

CH3

 HS

HS

HS  ecall that acetyl-CoA is a thioester, a form of energy currency (see Section 12.3). Some R of the energy released in the oxidation of the hydroxyethyl group to an acetyl group is conserved in the formation of acetyl-CoA. 4. The final two steps of the reaction restore the pyruvate dehydrogenase complex to its original state. E3 reoxidizes the lipoamide group of E2 by transferring electrons to a Cys–Cys disulfide group in the enzyme.

HS

FAD SH

 HS

SH

S  S

5. Finally, NAD+ reoxidizes the reduced cysteine sulfhydryl groups. This electron-transfer reaction is facilitated by an FAD prosthetic group (the structure of FAD, a nucleotide derivative, is shown in Fig. 3.2c).

FAD

CH2 C

O

NH (CH2)4 NH

CH

C

Lys

Lipoamide

Acetyl-CoA

CH3

S

CH2

O

CoA

S

Lipoic acid

CH2

CH3

Coenzyme A

FAD

CH2

S

3. Next, E2 transfers the acetyl group to coenzyme A, producing acetyl-CoA and leaving a reduced lipoamide group.

O

CH2

S

HS

Lipoamide

HS

CH2

S

HS

S

CH2

+

+

+

N

CH3

CH3

S

NAD

NADH  H

FAD

SH

S

SH

S

During the five-step reaction (summarized in Fig. 14.4), the long lipoamide group of E2 acts as a swinging arm that visits the active sites of E1, E2, and E3 within the multienzyme complex. The arm picks up an acetyl group from an E1 subunit and transfers it to coenzyme A in an E2 active site. The arm then swings to an E3 active site, where it is reoxidized. Some other multienzyme complexes also include swinging arms, often attached to hinged protein domains to maximize their mobility. A multienzyme complex such as the pyruvate dehydrogenase complex can carry out a multistep reaction sequence efficiently because the product of one reaction can quickly become

  FIGURE 14.3  Lipoamide.  This prosthetic group consists of lipoic acid (a vitamin) linked via an amide bond to the ε-amino group of a protein lysine residue. The active portion of the 14-Å-long lipoamide is the disulfide bond (red), which can be reversibly reduced.

406  C ha pter 14   The Citric Acid Cycle

NAD+ FAD SH SH S

OH CO2

CH3

C− TPP

1

O CH3

C

S

4

S

HS

Lipoamide 2

HS

O C

Pyruvate

NADH + H+ FAD

S

HydroxyethylTPP

5

O−

TPP

O CH3

C

3

O

S HS

Acetyl-dihydrolipoamide

CH3

C

S

CoA

Acetyl-CoA

CoA

  FIGURE 14.4   Reactions of the pyruvate dehydrogenase complex.  In these five reactions, an ­acetyl group from pyruvate is transferred to CoA, CO2 is released, and NAD+ is reduced to NADH.

Question  Without looking at the text, write the net equation for the reactions shown here. How many vitamin-derived cofactors are involved?

the substrate for the next reaction without diffusing away or reacting with another substance. There is also evidence that the individual enzymes of glycolysis and the citric acid cycle associate loosely with each other so that the close proximity of their active sites can increase flux through their respective pathways. Flux through the pyruvate dehydrogenase complex is regulated by product inhibition: Both NADH and acetyl-CoA act as inhibitors. The activity of the complex is also regulated by hormone-controlled phosphorylation and dephosphorylation, which suits its function as the gatekeeper for the entry of a metabolic fuel into the citric acid cycle.

Before Going On • Describe the functional importance of the coenzymes that participate in the reactions carried out by the pyruvate dehydrogenase complex. • Discuss the advantages of a multienzyme complex.

14.2 The Eight Reactions of the Citric Acid Cycle LEARNING OBJECTIVE

SEE GUIDED TOUR Glycolysis and the Citric Acid Cycle

Describe the substrate, product, and type of chemical reaction for each step of the citric acid cycle. The starting material for the citric acid cycle is an acetyl-CoA molecule that may be derived from a carbohydrate via pyruvate, as just described, or from another metabolic fuel.

14.2  The Eight Reactions of the Citric Acid Cycle  407   FIGURE 14.5   The citric acid cycle in context.  The citric acid cycle is a central metabolic pathway whose starting material is two-carbon acetyl units derived from amino acids, monosaccharides, and fatty acids. These are oxidized to the waste product CO2, with the reduction of the cofactors NAD+ and ubiquinone (Q).

The carbon skeletons of amino acids are broken down to either pyruvate or ­acetyl-CoA, and fatty acids are broken down to acetyl-CoA. In some tissues, the bulk of acetyl-CoA entering the citric acid cycle comes from fatty acids rather than carbohydrates or amino acids. Whatever their source, the citric acid cycle converts all these two-carbon acetyl groups into CO2 and therefore represents the final stage in fuel oxidation (Fig. 14.5). As the carbons become fully oxidized to CO2, their energy is conserved and subsequently used to produce ATP. The eight reactions of the citric acid cycle take place in the cytosol of prokar­ yotes and in the mitochondria of eukaryotes. Unlike a linear pathway such as glycolysis (see Fig. 13.2) or gluconeogenesis (see Fig. 13.10), the citric acid cycle always returns to its starting position, essentially behaving as a multistep catalyst. The cycle as a whole is highly exergonic, and free energy is conserved at several steps in the form of a nucleotide triphosphate (GTP) and reduced cofactors. For each acetyl group that enters the citric acid cycle, two molecules of fully oxidized CO2 are produced, representing a loss of four pairs of electrons. These electrons are transferred to 3 NAD+ and 1 ubiquinone (Q) to produce 3 NADH and 1 QH2. The net equation for the citric acid cycle is therefore ​acetyl-CoA + GDP + ​P​ i​+ 3 ​NAD​+​+ Q → 2 ​CO​ 2​+ CoA + GTP + 3 NADH + ​QH​ 2​

BIOPOLYMERS Proteins Nucleic acids Polysaccharides Triacylglycerols

MONOMERS Amino acids Nucleotides Monosaccharides Fatty acids +

NH4

NADH

NADH, QH2 2- and 3-Carbon INTERMEDIATES

NAD+, Q

In this section we examine the sequence of eight enzyme-catalyzed reactions of the citric acid cycle, focusing on a few interesting reactions. The entire pathway, summarized in Figure 14.6, is also known as the Krebs cycle (after Hans Krebs, who worked out the sequence of reactions in the 1930s) and the tricarboxylic acid cycle (because citrate has three carboxyl groups).

O

CH2

H2O HSCoA

 CH3

C

SCoA

COO Oxaloacetate

Acetyl-CoA

citrate synthase

C

ATP

H2O

CH2 HO

ADP oxidative phosphorylation

COO O

photosynthesis

NADH, QH2 O2

In the first reaction of the citric acid cycle, the acetyl group of acetyl-CoA condenses with the four-carbon compound oxaloacetate to produce the six-carbon compound citrate:

C

citric acid cycle

CO2

1. Citrate synthase adds an acetyl group to oxaloacetate

COO

NAD+

NAD+

COO

CH2 COO

Citrate

Citrate synthase, a dimer, undergoes a large conformational change when oxaloacetate binds. (Fig. 14.7). This shift creates a binding site for acetyl-CoA and explains why oxaloacetate must bind to the enzyme before acetyl-CoA can bind. Citrate synthase is one of the few enzymes that can synthesize a carbon–carbon bond without using a metal ion cofactor. Its mechanism is shown in Figure 14.8. The first reaction intermediate may be stabilized by the formation of low-barrier hydrogen bonds, which are stronger than ordinary hydrogen bonds (see Section 6.3). The coenzyme A released during the final step can be reused by the pyruvate dehydrogenase complex or used later in the citric acid cycle to synthesize the intermediate succinyl-CoA. The reaction catalyzed by citrate synthase is highly exergonic (ΔG°′ = –31.5 kJ · mol−1, equivalent to the free energy of hydrolyzing the thioester bond of acetyl-CoA). We will see later why the efficient operation of the citric acid cycle requires that this step have a large negative free energy change.

NAD+, Q

408  C ha pter 14   The Citric Acid Cycle

O

  FIGURE 14.6   Reactions of the citric acid cycle.

CH3

COO

Question  Identify the number of carbon atoms in each intermediate.

C

C

COO

SCoA

CH2

Acetyl-CoA

O

HSCoA

CH2

HO

COO

CH2

1 citrate synthase

COO

C

COO Citrate

Oxaloacetate

COO

NADH NAD



OH

CH2

HO

Malate

Isocitrate 3 isocitrate dehydrogenase

COO

H2O

CH

2 aconitase

8 malate dehydrogenase

COO CH

CH2

COO

7 fumarase

H

COO

NAD

CH2

COO

CH2

CH

C

HC

O

COO

COO

COO



Fumarate

QH2

C

COO

COO 6 succinate CH2 dehydrogenase

CH2 C

CH2 Q

α-Ketoglutarate

CH2



COO 5 succinyl-CoA Succinate synthetase

4 α-ketoglutarate dehydrogenase

NAD  HSCoA

O

SCoA Succinyl-CoA

NADH  CO2

GDP  Pi

GTP  HSCoA   FIGURE 14.7   Conformational changes in citrate synthase.  a. The enzyme in the absence of substrates. The two subunits of the dimeric enzyme are colored blue and green.  b. The enzyme with bound oxaloacetate (red, mostly buried) and an acetyl-CoA analog (orange).

a.

b.

NADH  CO2

14.2  The Eight Reactions of the Citric Acid Cycle  409 His 274

N

N N

N H

H OH 

Asp 375

O



C

OOC

C

CH2

COO

CH2

O

HO

COO

C

O

Citrate 1. Oxaloacetate and then acetyl-CoA bind to the enzyme.

H2O

N

OOC

Oxaloacetate

N

H

C

H O Acetyl-CoA

CH2 COO

H2C H



C

SCoA

O

O

C

OH 

C

OOC

H

N

OOC

C



H2C

C

SCoA

HO

COO

COO



Citrate

CH2

O

C

  FIGURE 14.8   The citrate synthase reaction.

C

COO

CH2

H2O

COO

CH

CH2

O

O

The second enzyme of the citric acid cycle catalyzes the reversible isomerization of citrate to isocitrate:

COO

C

3. The enolate attacks oxaloacetate to produce a citryl-CoA intermediate.

2. Aconitase isomerizes citrate to isocitrate

C

HO

SCoA

H

O

COO

H2O

C

N

CH2

CH2

CH2

COO



COO

O

CH2

Citryl-CoA

2. Asp 375, a base, removes a proton from the acetyl group to produce an enolate, which is stabilized by a hydrogen bond to His 274. This is the slowest, or rate-limiting, step of the reaction.

HO

N

N

O 

4. Hydrolysis releases CoA and citrate.

HSCoA  H

COO



Aconitate

The enzyme is named after the reaction intermediate.

H

C

COO

HO

C

H

COO Isocitrate

SEE ANIMATED PROCESS DIAGRAM Mechanism of citrate synthase

410  C ha pter 14   The Citric Acid Cycle

Citrate is a symmetrical molecule, yet only one of its two carboxymethyl arms (—CH2—COO−) undergoes dehydration and rehydration during the aconitase reaction. This stereochemical specificity long puzzled biochemists, including Hans Krebs, who first described the citric acid cycle. Eventually, Alexander Ogston pointed out that although citrate is symmetrical, its two carboxymethyl groups are no longer identical when it is bound to an asymmetrical enzyme (Fig. 14.9). In fact, a three-point attachment is not even necessary for an enzyme to distinguish two groups in a molecule such as citrate, which are related by mirror symmetry. You can prove this yourself with a simple organic chemistry model kit. By now you should appreciate that biological systems, including enzyme, are inherently chiral (also see Section 4.1).

  FIGURE 14.9  Stereochemistry of aconitase.  The threepoint attachment of citrate to the enzyme allows only one carboxymethyl group (shown in green) to react.

COO CH2 H HO O

The third reaction of the citric acid cycle is the oxidative decarboxylation of isocitrate to α-ketoglutarate. The substrate is first oxidized in a reaction accompanied by the reduction of NAD+ to NADH. Then the carboxylate group β to the ketone function (that is, two carbon atoms away from the ketone) is eliminated as CO2. An Mn2+ ion in the active site helps stabilize the negative charges of the reaction intermediate.

NAD

COO

H  NADH

CH2

O H

C

C

C

H O

C

3. Isocitrate dehydrogenase releases the first CO2



O

Isocitrate

O

COO O

C

C

C

O O

C

CH2

CO2 H



O

O

C

CH2

H

C C

COO H

O



O

O

C

H

C

O

C

O

-Ketoglutarate

The CO2 molecules generated by isocitrate dehydrogenase—along with the CO2 generated in the following reaction and the CO2 produced by the decarboxylation of pyruvate—diffuse out of the cell and are carried in the bloodstream to the lungs, where they are breathed out. Note that these CO2 molecules are produced through oxidation–reduction reactions. The carbons are oxidized, while NAD+ is reduced. O2 is not directly involved in this process.

4. α-Ketoglutarate dehydrogenase releases the second CO2 α-Ketoglutarate dehydrogenase, like isocitrate dehydrogenase, catalyzes an oxidative decarboxylation reaction. It also transfers the remaining four-carbon fragment to CoA:

COO CH2

COO HSCoA CO2

CH2 C

O

COO

α-Ketoglutarate

CH2 CH2

NAD NADH

C

O

S

CoA

Succinyl-CoA

The free energy of oxidizing α-ketoglutarate is conserved in the formation of the thioester succinyl-CoA. α-Ketoglutarate dehydrogenase is a multienzyme complex that resembles the pyruvate dehydrogenase complex in both structure and enzymatic mechanism. In fact, the same E3 enzyme is a member of both complexes.

14.2  The Eight Reactions of the Citric Acid Cycle  411

The isocitrate dehydrogenase and α-ketoglutarate dehydrogenase reactions both release CO2. These two carbons are not the ones that entered the citric acid cycle as acetyl-CoA; those acetyl carbons become part of oxaloacetate and are lost in subsequent rounds of the cycle (Fig. 14.10). However, the net result of each round of the citric acid cycle is the loss of two carbons as CO2 for each acetyl-CoA that enters the cycle.

Acetyl-CoA 1 Oxaloacetate

Citrate 2

5. Succinyl-CoA synthetase catalyzes substrate-level phosphorylation

Isocitrate

3 The thioester succinyl-CoA releases a large amount of free energy −1 when it is hydrolyzed (ΔG°′ = –32.6 kJ · mol ). This is enough free energy to drive the synthesis of a nucleoside triphosphate from a nucleoside diphosphate and Pi (ΔG°′ = 30.5 kJ · mol−1). The change in free energy for the net reaction is near zero, so the reaction is reversible. α-Ketoglutarate In fact, the enzyme is named for the reverse reaction. ­Succinyl-CoA 4 synthetase in the mammalian citric acid cycle generates GTP, whereas the plant and bacterial enzymes generate ATP (recall that GTP is Succinyl-CoA energetically equivalent to ATP). An exergonic reaction coupled to the transfer of a p ­ hosphoryl group to a nucleoside diphosphate is   FIGURE 14.10   Fates of carbon atoms in the citric termed s­ ubstrate-level phosphorylation to distinguish it from acid cycle.  The two carbon atoms that are lost as CO2 in oxidative phosphorylation (Section 15.4) and photophosphorylation the reactions catalyzed by isocitrate dehydrogenase (step 3) and α-ketoglutarate dehydrogenase (step 4) are not the same ­(Section 16.2), which are more indirect ways of synthesizing ATP. carbons that entered the cycle as acetyl-CoA (red). How does succinyl-CoA synthetase couple thioester cleavage to the synthesis of a nucleoside triphosphate? The reaction is a series of phosphoryl-group transfers that involve an active-site histidine residue (Fig. 14.11). After the succinyl group is phosphorylated, the phosphoryl group is transferred to the side chain of His 246 close by. A protein loop containing the phospho-His reaction intermediate must then move a large distance—about 35 Å—to deliver the phosphoryl group to the waiting nucleoside diphosphate (Fig. 14.12).

COO CH2

O  HO

CH2 C

P



O



1. A phosphate group displaces CoA in succinyl-CoA. The product, succinyl phosphate, is an acyl phosphate, which releases a large amount of free energy when hydrolyzed.

HSCoA

O

O

SCoA

  FIGURE 14.11   The succinylCoA synthetase reaction.

COO

Question  What is the fate of the free CoA molecule?

CH2 CH2 C

SEE ANIMATED PROCESS DIAGRAM

O

OPO2 3

Succinyl-CoA

The reaction catalyzed by succinylCoA synthetase

Succinyl phosphate

N

2. Succinyl phosphate donates its phosphoryl group to a His residue on the enzyme, producing a phospho-His intermediate and releasing succinate.

N H

GDP

PO2 3

GTP

COO CH2

N N H

His residue

CH2 GDP  H

3. The phosphoryl group is then transferred to GDP to form GTP.

N PO2 3 N

Phospho-His

 H

COO

Succinate

412  C ha pter 14   The Citric Acid Cycle

6. Succinate dehydrogenase generates ubiquinol The final three reactions of the citric acid cycle convert succinate back to the cycle’s starting substrate, oxaloacetate. Succinate dehydrogenase catalyzes the reversible dehydrogenation of succinate to fumarate. This ­oxidation–reduction reaction requires an FAD prosthetic group, which is reduced to FADH2 during the reaction:

Enz-FAD Enz-FADH2

COO H

C

H

H

C

H

  FIGURE 14.12   Substrate binding in succinyl-

H succinate dehydrogenase



COO

CoA synthetase.  The substrate succinyl-CoA is represented by coenzyme A (red), the side chain of His 246 is green, and the nucleoside diphosphate awaiting phosphorylation (ADP) is orange.

OOC

Succinate

C C

COO H

Fumarate

To regenerate the enzyme, the FADH2 group must be reoxidized. Since ­succinate dehydrogenase is embedded in the inner mitochondrial membrane (it is the only one of the eight citric acid cycle enzymes that is not soluble in the mitochondrial matrix), it can be reoxidized by the lipid-­soluble electron carrier ubiquinone (see Section 12.2) rather than by the soluble cofactor NAD+. ­Ubiquinone (abbreviated Q) acquires two electrons to become ubiquinol (QH2).

Q

QH2 Enzyme-FAD

Enzyme-FADH2

7. Fumarase catalyzes a hydration reaction In the seventh reaction, fumarase (also known as fumarate hydratase) catalyzes the reversible hydration of a double bond to convert fumarate to malate:

COO−

COO−

CH HC

H2O

H

C

OH

fumarase

H

C

H

COO−

COO−

Fumarate

Malate

8. Malate dehydrogenase regenerates oxaloacetate The citric acid cycle concludes with the regeneration of oxaloacetate from malate in an NAD+-dependent oxidation reaction:

COO H

C

OH

CH2 COO

Malate

NAD



NADH  H



malate dehydrogenase

COO C

O

CH2 COO

Oxaloacetate

The standard free energy change for this reaction is +29.7 kJ · mol−1, indicating that the reaction has a low probability of occurring as written. However, the product oxaloacetate is a substrate for the next reaction (Reaction 1 of the citric acid cycle). The highly exergonic—and therefore highly favorable—citrate synthase reaction helps pull the malate dehydrogenase

14.3  Thermodynamics of the Citric Acid Cycle  413

reaction forward. This is the reason for the apparent waste of free energy released by cleaving the thioester bond of acetyl-CoA in the first reaction of the citric acid cycle.

Before Going On • List the sources of the acetyl groups that enter the citric acid cycle. • Write the net equation for the citric acid cycle. • Draw the structures of the substrates and products of the eight reactions and name the enzyme that catalyzes each step. • Identify the steps that generate ATP, CO2, and reduced cofactors.

14.3 Thermodynamics of the Citric Acid Cycle LEARNING OBJECTIVES Explain how the citric acid cycle recovers energy for the cell. • Calculate the ATP yield for one round of the cycle. • Identify the irreversible steps that regulate flux through the cycle. • Describe how the cycle reactions can operate in reverse.

Because the eighth reaction of the citric acid cycle returns the system to its original state, the entire pathway acts in a catalytic fashion to dispose of carbon atoms derived from amino acids, carbohydrates, and fatty acids. Albert Szent-Györgyi discovered the catalytic nature of the pathway by observing that small additions of organic compounds such as succinate, fumarate, and malate stimulated O2 uptake in a tissue preparation. Because the O2 consumption was much greater than would be required for the direct oxidation of the added substances, he inferred that the compounds acted catalytically.

The citric acid cycle is an energy-generating catalytic cycle We now know that oxygen is consumed during oxidative phosphorylation, the process that reoxidizes the reduced cofactors (NADH and QH2) that are produced by the citric acid cycle. Although the citric acid cycle generates one molecule of GTP (or ATP), considerably more ATP is generated when the reduced cofactors are reoxidized by O2. Each NADH yields approximately 2.5 ATP and each QH 2 yields approximately 1.5 ATP (we will see in Section 15.4 why these values are not whole numbers). Every acetyl unit that enters the citric acid cycle can therefore generate a total of 10 ATP equivalents. The energy yield of a molecule of glucose, which generates two acetyl units, can be calculated (Fig. 14.13). A muscle operating anaerobically produces only 2 ATP per glucose, but under aerobic conditions when the citric acid cycle is fully

Glucose 2 NADH

glycolysis

5 ATP 2 ATP

2 Pyruvate 2 NADH

pyruvate processing

5 ATP

2 CO2

2 Acetyl-CoA

citric acid cycle

6 NADH

15 ATP

2 QH2

3 ATP 2 GTP

4 CO2   FIGURE 14.13   ATP yield from glucose.  Two rounds of the citric acid cycle are required to fully oxidize one molecule of glucose.

Question  Trace the fate of the six glucose carbon atoms.

414  C ha pter 14   The Citric Acid Cycle

functional, each glucose molecule generates about 32 ATP equivalents. This general phenomenon is called the Pasteur effect, after Louis Pasteur, who first observed that the rate of glucose consumption by yeast cells decreased dramatically when the cells were shifted from anaerobic to aerobic growth conditions.

The citric acid cycle is regulated at three steps Flux through the citric acid cycle is regulated primarily at the cycle’s three metabolically irreversible steps: those catalyzed by citrate synthase (Reaction 1), isocitrate dehydrogenase ­(Reaction 3), and α-ketoglutarate dehydrogenase (Reaction 4). The major regulators are shown in Figure 14.14. Neither acetyl-CoA nor oxaloacetate is present at concentrations high enough to saturate citrate synthase, so flux through the first step of the citric acid cycle depends largely on the substrate concentrations. The product of the reaction, citrate, inhibits citrate synthase (citrate also inhibits phosphofructokinase, thereby decreasing the supply of acetyl-CoA produced by glycolysis). Succinyl-CoA, the product of Reaction 4, inhibits the enzyme that produces it. It also acts as a feedback inhibitor by competing with acetyl-CoA in Reaction 1. The activity of isocitrate dehydrogenase is inhibited by its reaction product, NADH. NADH also inhibits α-ketoglutarate dehydrogenase and citrate synthase. Both dehydrogen­ ases are activated by Ca2+ ions, which generally signify the need to generate cellular energy. ADP, also representing the need for more ATP, activates isocitrate dehydrogenase. Changes in enzyme reaction rates also regulate the flow of acetyl carbons through the cycle by altering the concentrations of cycle intermediates. Because the entire cycle acts as a catalyst, more intermediates means that more acetyl groups can be processed, just as a city can move more commuters by adding buses during rush hour. Not surprisingly, citric acid cycle defects have serious consequences (Box 14.A).

Acetyl-CoA Oxaloacetate

Citrate

Malate

Isocitrate

ADP

NADH Ca2 Fumarate

-Ketoglutarate Ca2 Succinate

Succinyl-CoA

  FIGURE 14.14   Regulation of the citric acid cycle.  Inhibition is rep-

resented by red symbols, activation by green symbols.

14.3  Thermodynamics of the Citric Acid Cycle  415

Box 14.A Mutations in Citric Acid Cycle Enzymes Possibly because the citric acid cycle is a central metabolic pathway, severe defects in any of its components are expected to be incompatible with life. However, researchers have documented mutations in the genes for several of the cycle’s enzymes, including α-ketoglutarate dehydrogenase, succinyl-CoA synthetase, and succinate dehydrogenase. These defects, which are all rare, typically affect the central nervous system, causing symptoms such as movement disorders and neurodegeneration. A rare form of fumarase deficiency results in brain malformation and developmental disabilities. Some citric acid cycle enzyme mutations are linked to cancer. One possible explanation is that a defective enzyme contributes to carcinogenesis (the development of cancer, or uncontrolled cell growth) by causing the accumulation of particular metabolites, which are responsible for altering the cell’s activities. For example, normal cells respond to a drop in oxygen availability (hypoxia) by activating transcription factors known as hypoxia-­ inducible factors (HIFs; Section 19.2). These proteins interact with DNA to turn on the expression of genes for glycolytic enzymes and a growth factor that promotes the development of new blood vessels. When the fumarase gene is defective, fumarate

accumulates and inhibits a protein that destabilizes HIFs. As a result, the fumarase deficiency promotes glycolysis (an anaerobic pathway) and the growth of blood vessels. These two adaptations favor tumors, whose growth depends on high glycolytic flux and a constant supply of nutrients delivered by the bloodstream. Defects in isocitrate dehydrogenase also promote cancer in an indirect fashion. Many cancerous cells exhibit a mutation in one of the two genes for the enzyme, suggesting that the unaltered copy is necessary for maintaining the normal activity of the citric acid cycle, while the mutated copy plays a role in carcinogenesis. The mutated isocitrate dehydrogenase no longer carries out the usual reaction (converting isocitrate to α ­ -ketoglutarate) but instead converts α-ketoglutarate to 2-hydroxyglutarate in an NADPH-­d ependent manner. The mechanism whereby 2-­hydroxyglutarate contributes to carcinogenesis is not clear, but its involvement is bolstered by the observation that individuals who harbor other mutations that lead to 2-hydroxyglutarate accumulation have an increased risk of developing brain tumors. Question  How would a fumarase deficiency affect the levels of pyruvate, fumarate, and malate?

The citric acid cycle probably evolved as a synthetic pathway A circular pathway such as the citric acid cycle must have evolved from a linear set of pre-existing biochemical reactions. Clues to its origins can be found by examining the metabolism of organisms that resemble earlier life-forms. Such organisms emerged before atmospheric oxygen was available and may have used sulfur as their ultimate oxidizing agent, reducing it to H 2S. Their modern-day counterparts are anaerobic autotrophs that harvest free energy by pathways that are independent of the pathways of carbon metabolism. These organisms therefore do not use the citric acid cycle to generate reduced cofactors that are subsequently oxidized by molecular oxygen. However, all organisms must synthesize small molecules that can be used to build proteins, nucleic acids, carbohydrates, and so on. Even organisms that do not use the citric acid cycle contain genes for some citric acid cycle enzymes. For example, the cells may condense acetyl-CoA with oxaloacetate, leading to α-ketoglutarate, which is a precursor of several amino acids. They may also convert oxaloacetate to malate, proceeding to fumarate and then to succinate. Together, these two pathways resemble the citric acid cycle, with the right arm following the usual oxidative sequence of the cycle and the left arm following a reversed, reductive sequence (Fig. 14.15). The reductive sequence of reactions might have evolved as a way to regenerate the cofactors reduced during other catabolic reactions (for example, the NADH produced by the glyceraldehyde-3-phosphate dehydrogenase reaction of glycolysis; see Section 13.1).

Acetyl-CoA Oxaloacetate

Malate

Citrate

oxidation

Isocitrate

reduction Fumarate

-Ketoglutarate Succinate

  FIGURE 14.15   Pathways that might have given rise to the citric acid cycle.  The pathway starting from oxaloacetate and proceeding to the right is an oxidative biosynthetic pathway, whereas the pathway that proceeds to the left is a reductive pathway.

416  C ha pter 14   The Citric Acid Cycle

Pyruvate CO2 Acetyl-CoA

Citrate

Oxaloacetate

Malate

Isocitrate CO2 -Ketoglutarate

Fumarate Succinate

It is easy to theorize that the evolution of an enzyme to interconvert α-ketoglutarate and succinate could have connected the two “half pathways” (Fig. 14.15) to create a cyclic pathway similar to the modern citric acid cycle. Interestingly, E. coli, which uses the citric acid cycle under aerobic growth conditions, uses an interrupted citric acid cycle like the one diagrammed in Figure 14.15 when it is growing anaerobically. Since the final four reactions of the modern citric acid cycle are metabolically reversible, the primitive citric acid cycle might easily have accommo­ dated one-way flux in the clockwise direction, forming an oxidative cycle. If the complete cycle proceeded in the counterclockwise direction, the result would have been a reductive biosynthetic pathway (Fig. 14.16). This pathway, which would incorporate, or “fix,” atmospheric CO2 into biological molecules, may have preceded the modern CO2-fixing pathway found in green plants and some photosynthetic bacteria (described in Section 16.3).

Before Going On CO2

  FIGURE 14.16   A proposed reductive biosynthetic pathway based on the citric acid cycle.  This pathway might have operated to incorporate CO2 into biological molecules.

• Identify the products of the citric acid cycle that represent forms of energy currency for the cell. • Compare the ATP yield for glucose degradation via glycolysis and via glycolysis plus the citric acid cycle. • Describe how substrates and products of the citric acid cycle regulate flux through the pathway. • Explain how primitive oxidative and reductive biosynthetic pathways might have combined to generate a circular metabolic pathway.

14.4 Anabolic and Catabolic Functions of the Citric Acid Cycle LEARNING OBJECTIVES Explain how the citric acid cycle connects with other metabolic processes. • Identify the cycle intermediates that are precursors for the synthesis of other compounds. • Describe how citric acid cycle intermediates are replenished.

The citric acid cycle does not operate like a simple pipeline, where one substance enters at one end and another emerges from the opposite end. In mammals, six of the eight citric acid cycle intermediates (all except isocitrate and succinate) are the precursors or products of other pathways. For this reason, it is impossible to designate the citric acid cycle as a purely catabolic or anabolic pathway.

Citric acid cycle intermediates are precursors of other molecules Intermediates of the citric acid cycle can be siphoned off to form other compounds (Fig. 14.17). For example, succinyl-CoA is used for the synthesis of heme. The five-carbon

14.4  Anabolic and Catabolic Functions of the Citric Acid Cycle  417

Fatty acids, cholesterol

Glucose Oxaloacetate

Citrate

Pyruvate Malate

Isocitrate

Fumarate

-Ketoglutarate

Succinate

Amino acids, nucleotides

Succinyl-CoA Heme

  FIGURE 14.17   Citric acid cycle intermediates as biosynthetic precursors.

α-ketoglutarate (sometimes called 2-oxoglutarate) can undergo reductive amination by glutamate dehydrogenase to produce the amino acid glutamate:

COO

NADH  H

CH2 CH2 C



O

COO

α-Ketoglutarate

NH 4

COO

NAD

glutamate dehydrogenase

CH2 CH2 H

C



H2O

NH 3

COO

Glutamate

Glutamate is a precursor of the amino acids glutamine, arginine, and proline. Glutamine in turn is a precursor for the synthesis of purine and pyrimidine nucleotides. We have already seen that oxaloacetate is a precursor of monosaccharides (Section 13.2). Consequently, any of the citric acid cycle intermediates, which can be converted by the cycle to oxaloacetate, can ultimately serve as gluconeogenic precursors. Citrate produced by the condensation of acetyl-CoA with oxaloacetate can be transported out of the mitochondria to the cytosol by a specific transport protein. ATP-citrate lyase then catalyzes the reaction ​ATP + citrate + CoA → ADP + ​P​ i​+ oxaloacetate + acetyl-CoA​ The resulting acetyl-CoA is used for fatty acid and cholesterol synthesis, which take place in the cytosol. Because the rate of lipid synthesis is high in rapidly growing cancer cells, inhibitors of ATP-citrate lyase are being developed as anti-cancer drugs. Note that the ATP-citrate lyase reaction undoes the work of the exergonic citrate synthase reaction. This seems wasteful, but cytosolic ATP-citrate lyase is essential because acetyl-CoA, which is produced in the mitochondria, cannot cross the mitochondrial membrane to reach the cytosol, whereas citrate can. The oxaloacetate product of the ATP-citrate lyase reaction can be converted to malate by

418  C ha pter 14   The Citric Acid Cycle INNER MITOCHONDRIAL MEMBRANE MATRIX

CYTOSOL

Citrate

Citrate ATPcitrate lyase

citrate synthase

Acetyl-CoA

Acetyl-CoA

lipid biosynthesis

Oxaloacetate

malate dehydrogenase

Oxaloacetate

Malate

pyruvate carboxylase CO2

malic enzyme

Pyruvate

CO2

Pyruvate

  FIGURE 14.18   The citrate transport system.  This set of transport proteins and enzymes allows carbon atoms from mitochondrial acetyl-CoA to be transferred to the cytosol for the synthesis of fatty acids and cholesterol.

Question  Can the same CO2 molecule released in the malic enzyme reaction be used in the pyruvate carboxylase reaction?

a cytosolic malate dehydrogenase operating in reverse. Malate is then decarboxylated by the action of malic enzyme to produce pyruvate:

COO CH

NADP NADPH

OH

CH2 COO

Malate

malic enzyme

COO C

O



CO2

CH3 Pyruvate

Pyruvate can reenter the mitochondria, with the help of the pyruvate transporter, and be ­converted back to oxaloacetate to complete the cycle shown in Figure 14.18. In plants, iso­ citrate is diverted from the citric acid cycle in a biosynthetic pathway known as the g ­ lyoxylate ­pathway (Box 14.B).

Anaplerotic reactions replenish citric acid cycle intermediates Intermediates that are diverted from the citric acid cycle for other purposes can be replenished through anaplerotic reactions (from the Greek ana, “up,” and plerotikos, “to fill”; Fig. 14.19). One of the most important of these reactions is catalyzed by pyruvate carboxylase (this is also the first step of gluconeogenesis; Section 13.2): ​pyruvate + ​CO​ 2​+ ATP + ​H​ 2​O → oxaloacetate + ADP + ​P​ i​

14.4  Anabolic and Catabolic Functions of the Citric Acid Cycle  419

Box 14.B The Glyoxylate Pathway Plants and some bacterial cells contain certain enzymes that act together with some citric acid cycle enzymes to convert acetylCoA to oxaloacetate, a gluconeogenic precursor. Animals lack the enzymes to do this and therefore cannot undertake the net synthesis of carbohydrates from two-carbon precursors. In plants, the glyoxylate pathway includes reactions that take place in the mitochondria and the glyoxysome, an organelle that, like the peroxisome, contains enzymes that carry out some essential metabolic processes.

O CH3 C

SCoA

Acetyl-CoA

Citrate Oxaloacetate

COO CH2

Malate COO Fumarate

CH2 CH2

CH

COO

CH

OH

In the glyoxysome, acetyl-CoA condenses with oxaloacetate to form citrate, which is then isomerized to isocitrate, as in the citric acid cycle. However, the next step is not the isocitrate dehydrogenase reaction but a reaction catalyzed by the glyoxysome enzyme isocitrate lyase, which converts isocitrate to succinate and the two-carbon compound glyoxylate. Succinate continues as usual through the mitochondrial citric acid cycle to regenerate oxaloacetate. In the glyoxysome, the glyoxylate condenses with a second molecule of acetyl-CoA in a reaction catalyzed by the glyoxysome enzyme malate synthase to form the four-carbon compound malate. Malate can then be converted to oxaloacetate for gluconeo­ genesis. The two reactions that are unique to the glyoxylate pathway are shown in green in the figure; reactions that are identical to those of the citric acid cycle are shown in blue. In essence, the glyoxylate pathway bypasses the two CO2-generating steps of the citric acid cycle (catalyzed by iso­ citrate dehydrogenase and α-ketoglutarate dehydrogenase) and incorporates a second acetyl unit (at the malate synthase step). The net result of the glyoxylate pathway is the production of a four-carbon compound that can be used to synthesize glucose. This pathway is highly active in germinating seeds, where stored oils (triacylglycerols) are broken down to acetyl-CoA. The glyoxylate pathway thus provides a route for synthesizing glucose from fatty acids. Because animals lack isocitrate lyase and malate synthase, they cannot convert excess fat to carbohydrate. Question  Write the net equation for the glyoxylate cycle as shown here.

COO Isocitrate

isocitrate lyase

O

COO

Succinate

C

H

COO

COO

Glyoxylate

O CH3

C

SCoA

Acetyl-CoA

CH malate synthase

OH

CH2 COO

Oxaloacetate 

Malate

Acetyl-CoA activates pyruvate carboxylase, so when the activity of the citric acid cycle is low and acetyl-CoA accumulates, more oxaloacetate is produced. The concentration of oxalo­ acetate is normally low since the malate dehydrogenase reaction is thermodynamically unfavorable and the citrate synthase reaction is highly favorable. The replenished oxaloacetate is converted to citrate, isocitrate, α-ketoglutarate, and so on, so the concentrations of all the citric acid cycle intermediates increase and the cycle can proceed more quickly. Since the citric acid cycle acts as a catalyst, increasing the concentrations of its components increases flux through the pathway. The degradation of fatty acids with an odd number of carbon atoms yields the citric acid cycle intermediate succinyl-CoA. Other anaplerotic reactions include the pathways for the

gluconeogenesis

420  C ha pter 14   The Citric Acid Cycle

pyruvate, amino acids Oxaloacetate

Citrate

Malate

Isocitrate

Fumarate

-Ketoglutarate amino acids

amino acids Succinate

Succinyl-CoA amino acids, odd-chain fatty acids

  FIGURE 14.19   Anaplerotic reactions of the citric acid cycle.

degradation of some amino acids, which produce α-ketoglutarate, succinyl-CoA, fumarate, and oxaloacetate. Some of these reactions are transaminations, such as

COO 

H3N

C

H

CH2

COO





C

O

CH3

COO

Aspartate

Pyruvate

COO O

C

CH2

COO 

C

 H3N

H

CH3

COO

Oxaloacetate

Alanine

Because transamination reactions have ΔG values near zero, the direction of flux into or out of the pool of citric acid cycle intermediates depends on the relative concentrations of the reactants. In vigorously exercising muscle, the concentrations of citric acid cycle intermediates increase about three- to four-fold within a few minutes. This may help boost the energy-­ generating activity of the citric acid cycle, but it cannot be the sole mechanism, since flux through the citric acid cycle actually increases as much as 100-fold due to the increased activity of the three enzymes at the control points: citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase. The increase in citric acid cycle intermediates may actually be a mechanism for accommodating the large increase in pyruvate that results from rapid glycolysis at the start of exercise. Not all of this pyruvate is converted to lactate (Section 13.1); some is shunted into the pool of citric acid cycle intermediates via the pyruvate carboxylase reaction. Some pyruvate also undergoes a reversible reaction catalyzed by alanine aminotransferase:

COO

COO COO C

O

CH3 Pyruvate

COO

CH2 CH2

 H

C

NH 3

COO

Glutamate



H3N

C

H  CH2

CH3 Alanine

CH2 C

O

COO

α-Ketoglutarate

Bioinformatics  421

The resulting α-ketoglutarate then augments the pool of citric acid cycle intermediates, thereby increasing the ability of the cycle to oxidize the extra pyruvate. Note that any compound that enters the citric acid cycle as an intermediate is not itself oxidized; it merely boosts the catalytic activity of the cycle, whose net reaction is still the oxidation of the two carbons of acetyl-CoA.

Before Going On • List the citric acid cycle intermediates that are used for the synthesis of amino acids, glucose, and fatty acids. • Explain what the ATP-citrate lyase reaction accomplishes. • Explain why the concentration of oxaloacetate remains low. • Explain why synthesizing more oxaloacetate increases flux through the cycle. • Write equations for all the anaplerotic reactions.

Summary phosphorylation. In addition, succinyl-CoA synthetase yields one molecule of GTP or ATP.

14.1  The Pyruvate Dehydrogenase Reaction •  In order for pyruvate, the product of glycolysis, to enter the citric acid cycle, it must undergo oxidative decarboxylation catalyzed by the multienzyme pyruvate dehydrogenase complex, which yields acetyl-CoA, CO2, and NADH.

14.2  The Eight Reactions of the Citric Acid Cycle •  The eight reactions of the citric acid cycle function as a multistep catalyst to convert the two carbons of acetyl-CoA to 2 CO2.

14.3  Thermodynamics of the Citric Acid Cycle •  The electrons released in the oxidative reactions of the citric acid cycle are transferred to 3 NAD+ and to ubiquinone. The reoxidation of the reduced cofactors generates ATP by oxidative

•  The regulated reactions of the citric acid cycle are its irreversible steps, catalyzed by citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate dehydrogenase. •  The citric acid cycle most likely evolved from biosynthetic pathways leading to α-ketoglutarate or succinate.

14.4  Anabolic and Catabolic Functions of the Citric Acid Cycle •  Six of the eight citric acid cycle intermediates serve as precursors of other compounds, including amino acids, monosaccharides, and lipids. Anaplerotic reactions convert other compounds into citric acid cycle intermediates, thereby allowing increased flux of acetyl carbons through the pathway.

Key Terms mitochondrial matrix multienzyme complex citric acid cycle

substrate-level phosphorylation Pasteur effect carcinogenesis

Bioinformatics Brief Bioinformatics Exercises 14.1  Viewing and Analyzing the Pyruvate Dehydrogenase Complex 14.2  The Citric Acid Cycle and the KEGG Database

glyoxylate pathway anaplerotic reaction glyoxysome

422  C ha pter 14   The Citric Acid Cycle

Problems 14.1  The Pyruvate Dehydrogenase Reaction 1.  What are four possible transformations of pyruvate in mammalian cells? 2.  Determine which one of the five steps of the pyruvate dehydrogen­ ase complex reaction is metabolically irreversible and explain why. 3.  The product of the pyruvate dehydrogenase complex, acetyl-CoA, is released in step 3 of the overall reaction. What is the purpose of steps 4 and 5? 4.  Beriberi is a disease that results from a dietary lack of thiamine, the vitamin that serves as the precursor for thiamine pyrophosphate (TPP). There are two metabolites that accumulate in individuals with beriberi, especially after ingestion of glucose. Which metabolites accumulate and why? 5.  Most cases of pyruvate dehydrogenase deficiency disease that have been studied to date involve a mutation in the E1 subunit of the enzyme. The disease is extremely difficult to treat successfully, but physicians who identify patients with a pyruvate dehydrogen­ ase deficiency administer thiamine as a first course of treatment. Explain why. 6.  Arsenite is toxic in part because it binds to sulfhydryl compounds such as dihydrolipoamide, as shown in the figure. How would the presence of arsenite affect the citric acid cycle?

OH −O

As OH

Arsenite

−O

 HS HS

S As S

 2 H2O R

R

Dihydrolipoamide 7.  Using the pyruvate dehydrogenase complex reaction as a model, reconstruct the TPP-dependent yeast pyruvate decarboxylase reaction in alcoholic fermentation (see Box 13.B). 8.  The compound α-ketoisovalerate was shown to inhibit pyruvate dehydrogenase activity. Further investigation showed that when cultured cells were exposed to α-ketoisovalerate that was radioactively labeled at C2 (marked with * in the figure), the label became incorporated into the lipoamide prosthetic group of E2 of the pyruvate dehydrogenase complex. Draw the structure of the modified enzyme. What type of inhibitor is α-ketoisovalerate (see Section 7.3)?

O CH3

CH CH3

C *

9.  How is the activity of the pyruvate dehydrogenase complex affected by  a.  a high [NADH]/[NAD+] ratio or  b.  a high [acetylCoA]/[HSCoA] ratio? 10.  In addition to the allosteric regulation described in Problem 9, the activity of the pyruvate dehydrogenase complex is controlled by phosphorylation. Pyruvate dehydrogenase kinase (PDH kinase) catalyzes the phosphorylation of a specific Ser residue on the E1 subunit of the enzyme, rendering it inactive. Pyruvate dehydrogenase phosphatase (PDH phosphatase) reverses the inhibition by catalyzing the removal of this phosphate group. The PDH kinase is highly regulated and its activity is influenced by various cellular metabolites. Indicate whether the following would activate or inhibit PDH kinase:  a. NAD+,  b. NADH, c.  coenzyme A,  d. acetyl-CoA, e. ADP. 11.  The PDH kinase and PDH phosphatase enzymes (see Problem 10) are controlled by cytosolic Ca2+ levels. In the muscle, Ca2+ levels rise when the muscle contracts. Which of these two enzymes is inhibited by Ca2+ and which is activated by Ca2+? 12.  How would the activities of PDH kinase and PDH phosphatase enzymes (see Problem 10) be affected by the presence of a calmodulin antagonist (see Fig. 10.10)? 13.  Explain why PDH kinase–deficient mice (see Problem 10) have blood glucose levels that are lower than normal. 14.  Explain why the activity of the pyruvate dehydrogenase complex is suppressed in individuals consuming a low carbohydrate–high fat diet. 15.  Acetyl-CoA produced by the pyruvate dehydrogenase complex can enter the citric acid cycle or can be used to synthesize fatty acids (see Section 17.3). Hepatocytes (liver cells) in culture were incubated with fatty acids and the activity of PDH kinase (see Problem 10) was measured. Would you expect fatty acids to stimulate or inhibit the kinase? 16.  A second strategy to treat a pyruvate dehydrogenase deficiency disease (see Problem 5) involves administering dichloroacetate, a compound that inhibits PDH kinase (see Problem 10). How might this strategy be effective? 17.  Hypoxia inducible factor (HIF) is a transcription factor that stimulates the expression of various genes involved in metabolism during hypoxic (low oxygen) conditions (see Box 14.A). In an experiment, the amounts of PDH kinase (see Problem 10) and hexokinase (see Section 13.1) were measured by immunoblotting under both hypoxic and normoxic conditions (results are shown below). How does cellular oxygen concentration affect the expression of these two enzymes and which pathways are activated as a result? Hypoxia

Normoxia

PDH kinase

Hexokinase

COO−

α-Ketoisovalerate

18.  The experiment described in Problem 17 was repeated using cells in which the expression of HIF was “knocked out.” What do the results below reveal about the role of HIF in metabolism?

Problems  423 Hypoxia

OH

Normoxia

CoA

PDH kinase

14.2  The Eight Reactions of the Citric Acid Cycle 19.  Identify the “2- and 3-carbon intermediates” in Figure 14.5. What biomolecules are the source of these intermediates and what is their importance to the citric acid cycle? 20.  The combined actions of the pyruvate dehydrogenase reaction and the citric acid cycle result in the complete oxidation of the six carbons in glucose to CO2. At which steps is CO2 generated? 21.  Which steps of glycolysis and the citric acid cycle generate ATP (or its equivalent) via substrate-level phosphorylation? How much ATP is produced per glucose in this way? 22.  The combined actions of glycolysis, the pyruvate dehydrogenase reaction, and the citric acid cycle result in the loss of 24 protons and electrons from glucose. At which steps are electrons transferred to generate reduced cofactors? 23.  Does the citrate synthase enzyme mechanism (Fig. 14.8) use an acid catalysis strategy, a base catalysis strategy, a covalent catalysis strategy, or some combination of these strategies (see Section 6.2)? Explain. 24.  Site-directed mutagenesis techniques were used to synthesize a mutant citrate synthase enzyme in which the active site histidine was converted to an alanine. How would the activity of the mutant enzyme compare to the activity of the wild-type enzyme? Explain. 25.  Investigators interested in studying the effect of acetyl-CoA analogs on citrate synthase activity synthesized the compound S ­ -acetonyl-CoA from 1-bromoacetone and coenzyme A.  a.  Write the reaction for the formation of S-acetonyl-CoA.  b.  The ­Lineweaver–Burk plot for the citrate synthase reaction with and without S-acetonyl-CoA is shown. What type of inhibitor is S-acetonyl-CoA? Explain.

O C

CH2

S

1/v (nmol−1 . min)

O

with I

0.35 0.25

no I

0.15 0.05 −0.5

0

27.  Why is the isomerization of citrate to isocitrate a necessary step in the citric acid cycle? (Hint: Consider the chemical transformation that occurs in the following step.) 28.  The ΔG°′ value for the aconitase reaction is 5.8 kJ ∙ mol−1. What is the ratio of [isocitrate]/[citrate] under standard conditions? Given your answer, how can you account for the fact that this reaction occurs readily in cells? 29.  Kinetic studies with aconitase revealed that trans-aconitate is a competitive inhibitor of the enzyme if cis-aconitate is used as the substrate. However, if citrate is used as the substrate, trans-aconitate is a noncompetitive inhibitor. Propose a hypothesis that explains this observation. 30.  In a yeast mutant, the gene for aconitase is nonfunctional. What are the consequences for the cell, particularly with regard to energy production? 31.  During bacterial infections, macrophages (a type of white blood cell) produce large amounts of an immunoregulatory molecule called itaconate, which is generated by the decarboxylation of aconitate. Draw the structure of itaconate.

COO− CH2 C CH

COO− H+

COO−

Aconitate

32.  In addition to regulating immune responses, itaconate (see Solution 31) can condense with coenzyme A to yield itaconyl-CoA, which has antibacterial activity. Draw the structure of itaconyl-CoA.

34.  The crystal structure of isocitrate dehydrogenase shows a cluster of highly conserved amino acids in the substrate binding pocket— three arginines, a tyrosine, and a lysine. Why are these residues conserved and what is a possible role for their side chains in substrate binding?

0.45

−1.0

C

33.  The ΔG°′ value for the isocitrate dehydrogenase reaction is –21 kJ ∙ mol−1. What is Keq for this reaction at 25°C?

CoA

S-Acetonyl-CoA

0

CH2

Carboxymethyl-CoA (transition state analog)

Hexokinase

H3C

S

0.5

1.0

1/[acetyl-CoA] (μM−1)

26.  The compound carboxymethyl-CoA (shown below) is a competitive inhibitor of citrate synthase and is a proposed transition state analog. Propose a structure for the reaction intermediate derived from acetyl-CoA in the rate-limiting step of the reaction, just prior to its reaction with oxaloacetate.

35.  Cells contain several isozymes (see Section 6.1) of isocitrate dehydrogenase (IDH). IDH1 and IDH2 (located in the cytosol and mitochondrial matrix, respectively) catalyze the same reaction as IDH3 (the mitochondrial matrix enzyme discussed in this chapter) with the exception that NADP+ is used as a cofactor. IDH mutants are commonly found in cancer cells. In one common IDH1 mutant, histidine replaces an arginine in the active site, and in one common IDH2 mutant, lysine replaces an arginine. Would you expect the substrate-binding properties of these mutant enzymes to differ from the wild-type (see Solution 34)? Explain. 36.  The isocitrate dehydrogenase enzymes described in Problem 35 are “gain-of-function” mutants in which the normal catalytic properties of the enzyme are lost and a new biochemical activity is gained as a result of the mutation. The IDH1 mutant uses NADPH as a cofactor to catalyze the conversion of α-ketoglutarate to

424  C ha pter 14   The Citric Acid Cycle 2-hydroxyglutarate.  a.  Draw the structure of the product.  b. What is the source of the NADPH and what are the implications for the cell if this reaction occurs at a high rate?

labeled?  b.  If citrate becomes radioactively labeled, what can you conclude about the connection between glycosomal and mitochondrial pathways in the parasite?

37.  Using the pyruvate dehydrogenase complex reaction as a model, draw the intermediates of the α-ketoglutarate dehydrogenase reaction. Describe what happens in each of the five reaction steps.

14.3  Thermodynamics of the Citric Acid Cycle

38.  Using the mechanism you drew for Problem 37, explain how succinyl phosphonate (below) inhibits α-ketoglutarate dehydrogenase.

48.  Studies indicate that many animal tissues use circulating lactate, rather than pyruvate produced by intracellular glycolysis, as a substrate for the citric acid cycle. Compare the ATP yield for the complete oxidation of lactate and the complete oxidation of pyruvate by the citric acid cycle.

O O

P

O

C

O

49.  Flux through the citric acid cycle is regulated by the simple mechanisms of  a.  substrate availability,  b.  product inhibition, and  c.  feedback inhibition. Give examples of each.

CH2 CH2

50.  Predict the effect of the following metabolites on the activity of citrate synthase:  a. NADH, b. citrate, c.  succinyl-CoA, and  d. ATP.

COO

Succinyl phosphonate 39.  Succinyl-CoA synthetase is also called succinate thiokinase. Why is the enzyme considered to be a kinase? 40.  The His residue in the active site of succinyl-CoA synthetase (see Fig. 14.11) is in close proximity to a Glu residue (not shown in the figure) that participates in catalysis. Mutating this Glu residue to Asp had little effect on enzyme activity but mutating it to Gln abolishes enzyme activity. From this observation, what can you infer about the mechanism of the enzyme? 41.  Malonate is a competitive inhibitor of succinate dehydrogenase. What citric acid cycle intermediates accumulate if malonate is present in a preparation of isolated mitochondria? 42.  Malate dehydrogenase and lactate dehydrogenase can both use α-ketoglutarate as a substrate. Draw the product of those reactions. 43.  The ΔG°′ value for the fumarase reaction is −3.4 kJ · mol−1, but the ΔG value is close to zero. What is the ratio of fumarate to malate under cellular conditions at 37°C? Is this reaction likely to be a control point for the citric acid cycle? 44.  A mutant bacterial fumarase was constructed by replacing the Glu (E) at position 315 with Gln (Q). The kinetic parameters of the mutant and wild-type enzymes are shown in the table. Explain the significance of the changes.

Vmax (μmol · min−1 · mg−1) KM (mM) kcat/KM (M−1 · s−1)

47.  Which of the eight reactions of the citric acid cycle are  a. phosphorylations,  b. isomerizations, c.  oxidation–­reductions,  d. hydrations,  e.  carbon–carbon bond formations, and  f. decarboxylations?

Wild-type enzyme

E315Q mutant enzyme

345 0.21 5.6 × 106

32 0.25 4.3 × 105

45. a.  Oxaloacetate labeled at C4 with 14C is added to a suspension of respiring mitochondria. What is the fate of the labeled carbon?  b.  Acetyl-CoA labeled at C1 with 14C is added to a suspension of respiring mitochondria. What is the fate of the labeled carbon? 46.  The complex metabolic pathways in the parasite Trypanosoma brucei (the causative agent of sleeping sickness) were elucidated in part by adding radiolabeled metabolites to cultured parasites. In the parasite, glucose is converted to phosphoenolpyruvate (PEP) in the cytosol. PEP then enters an organelle called the glycosome and is converted to oxaloacetate (OAA); OAA is then converted to malate, and malate to fumarate. Fumarate reductase catalyzes the conversion of fumarate to succinate; the succinate is then secreted from the glycosome.  a.  If C1 of glucose is labeled, what carbons in succinate are

51.  Citrate competes with oxaloacetate for binding to citrate synthase. Isocitrate dehydrogenase is activated by Ca2+ ions, which are released when a muscle contracts. How do these two regulatory strategies assist the cell in making the transition from the rested state (low citric acid cycle activity) to the exercise state (high citric acid cycle activity)? 52.  Reactions 8 and 1 of the citric acid cycle can be considered to be coupled because the exergonic hydrolysis of the thioester bond of acetyl-CoA in Reaction 1 drives the regeneration of oxaloacetate in Reaction 8.  a.  Write the equation for the overall coupled reaction and calculate its ΔG°′.  b.  What is the equilibrium constant for the coupled reaction? Compare this equilibrium constant with the equilibrium constant of Reaction 8 alone. 53.  Administering high concentrations of oxygen (hyperoxia) is effective in treating lung injuries but at the same time can also be quite damaging.  a.  Lung aconitase activity is dramatically decreased during hyperoxia. How would this affect the concentration of citric acid cycle intermediates?  b.  The decreased aconitase activity and decreased mitochondrial respiration in hyperoxia are accompanied by elevated rates of glycolysis and the pentose phosphate pathway. Explain why. 54.  The scientists who carried out the hyperoxia experiments described in Problem 53 noted that they could mimic this effect by administering either fluoroacetate or fluorocitrate to cells in culture. Explain. (Hint: Fluoroacetate can react with coenzyme A to form fluoroacetyl-CoA.) 55.  In bacteria, isocitrate dehydrogenase is regulated by phosphorylation of a specific Ser residue in the enzyme active site. X-ray structures of the phosphorylated and the nonphosphorylated enzyme show no significant conformational differences.  a.  How does phosphorylation regulate isocitrate dehydrogenase activity?  b. To confirm their hypothesis, the investigators constructed a mutant enzyme in which the Ser residue was replaced with Asp. The mutant was unable to bind isocitrate. Are these results consistent with the hypothesis you proposed in part a? 56.  The expression of several enzymes changes when yeast grown on glucose are abruptly shifted to a 2-carbon food source such as acetate.  a.  Why does the level of expression of isocitrate dehydrogenase increase when the yeast are shifted from glucose to acetate?  b. The metabolism of a yeast mutant with a nonfunctional isocitrate dehydrogenase enzyme was compared to that of a wild-type yeast. The yeast were grown on glucose and then abruptly shifted to acetate as

Problems  425 the sole carbon source. The [NAD+]/[NADH] ratio was measured over a period of 48 hours. The results are shown below. Why does the ratio increase slightly at 36 hours for the wild-type yeast? Why is there a more dramatic increase in the ratio for the mutant?

succinate. These reactions can be combined with other standard citric acid cycle reactions to create a pathway from citrate to oxaloacetate. How does this alternative pathway compare to the standard citric acid cycle in its ability to make free energy available to the cell?

150

[NAD] / [NADH]

120

90

COO

CH2

CH2

CH2

CH2

C mutant

60

C

O

α-Ketoglutarate

0

12

24

Succinate

CO2

wild-type 36

Time (h)

57.  Malate dehydrogenase is more active in cells oxidizing glucose aerobically than in cells oxidizing glucose anaerobically. Explain why. 58.  Succinyl-CoA inhibits both citrate synthase and α-ketoglutarate dehydrogenase. How is succinyl-CoA able to inhibit both enzymes?

O

O

COO

30

0

COO

COO CH2 CH2 C

NADH NAD

O

H

Succinate semialdehyde

59.  Why is it advantageous for citrate, the product of Reaction 1 of the citric acid cycle, to inhibit phosphofructokinase, which catalyzes the third reaction of glycolysis?

14.4  Anabolic and Catabolic Functions of the Citric Acid Cycle

60.  Acetyl-CoA acts as an allosteric activator of pyruvate carboxyl­ ase. S-acetonyl-CoA (see Problem 25) does not activate pyruvate carboxylase, and it cannot compete with acetyl-CoA for binding to the enzyme. What does this tell you about the binding requirements for an allosteric activator of pyruvate carboxylase?

67. a.  Why is the reaction catalyzed by pyruvate carboxylase the most important anaplerotic reaction of the citric acid cycle?  b. Why is the activation of pyruvate carboxylase by acetyl-CoA a good regulatory strategy?

61.  Succinyl-CoA synthetase is a dimer of an α and a β subunit. A single gene encodes the α subunit protein. Two genes code for two different β subunit proteins. One β subunit, which is specific for ADP, is expressed in “catabolic tissues” such as brain and muscle, whereas the other β subunit, which is specific for GDP, is expressed in “anabolic tissues” such as liver and kidney. Propose a hypothesis to explain this observation. 62.  Individuals with a mutation in the gene for the α subunit of succinyl-CoA synthetase (see Problem 61) experience severe lactic acidosis and usually die within a few days of birth, but individuals with a mutation in the gene for the ADP-specific β subunit of the enzyme experience only moderately elevated concentrations of lactate and usually survive to their early twenties. Why is the prognosis for patients with the β subunit mutation better than for patients with the α subunit mutation? 63.  Why would a deficiency of succinate dehydrogenase lead to a shortage of coenzyme A? 64.  Individuals who are deficient in fumarase develop lactic acidosis. Explain why. 65.  A patient with an α-ketoglutarate dehydrogenase deficiency exhibits a small increase in blood pyruvate level and a large increase in blood lactate level, resulting in a [lactate]/[pyruvate] ratio that is many times greater than normal. Explain the reason for these symptoms. 66.  Certain microorganisms with an incomplete citric acid cycle decarboxylate α-ketoglutarate to produce succinate semialdehyde (shown below). A dehydrogenase then converts succinate semialdehyde to

68.  Many amino acids are broken down to intermediates of the citric acid cycle.  a.  Why can’t these amino acid “remnants” be completely oxidized to CO2 by the citric acid cycle?  b.  Explain why amino acids that are broken down to pyruvate can be completely oxidized by the citric acid cycle. 69.  Describe how the aspartate + pyruvate transamination reaction could function as an anaplerotic reaction for the citric acid cycle. 70.  A bacterial mutant with low levels of isocitrate dehydrogenase is able to grow normally when the culture medium is supplemented with glutamate. Explain why. 71.  Is net synthesis of glucose in mammals possible from the following compounds?  a.  The fatty acid palmitate (16:0), which is degraded to eight acetyl-CoA.  b.  The fatty acid pentadecanoate (15:0), which is degraded to six acetyl-CoA and one propionyl-CoA.  c. Glyceraldehyde-3-phosphate.  d.  Leucine, which is degraded to acetyl-CoA and acetoacetate (a compound that is metabolically equivalent to two acetyl-CoA groups).  e.  Tryptophan, which is degraded to alanine and acetoacetate.  f.  Phenylalanine, which is degraded to acetoacetate and fumarate. 72.  Pancreatic islet cells cultured in the presence of 1–20 mM glucose showed increased activities of pyruvate carboxylase and the E1 subunit of the pyruvate dehydrogenase complex proportional to the increase in glucose concentration. Explain why. 73.  A physician is attempting to diagnose a neonate with a pyruvate carboxylase deficiency. An injection of alanine normally leads to a gluconeogenic response, but in the patient no such response occurs. Explain.

426  C ha pter 14   The Citric Acid Cycle 74.  The physician treats the patient described in Problem 73 by administering glutamine. Explain why glutamine supplements are effective in treating the disease. 75.  Physicians often attempt to treat a pyruvate carboxylase deficiency by administering biotin. Explain why this strategy might be effective. 76.  Patients with a pyruvate dehydrogenase deficiency and patients with a pyruvate carboxylase deficiency (see Problems 73 to 75) both have high blood levels of pyruvate and lactate. Explain why.

86.  The plant metabolite hydroxycitrate is advertised as an agent that prevents fat buildup.  a.  How does this compound differ from citrate?  b.  Hydroxycitrate inhibits the activity of ATP-citrate lyase. What kind of inhibition is likely to occur?  c.  Why might inhibition of ATP-citrate lyase block the conversion of carbohydrates to fats?  d.  The synthesis of what other compounds would be inhibited by hydroxycitrate?

CH2

77.  Oxygen does not appear as a reactant in any of the citric acid cycle reactions, yet it is essential for the proper functioning of the cycle. Explain why. 78.  The activity of isocitrate dehydrogenase in E. coli is regulated by the covalent attachment of a phosphate group, which inactivates the enzyme. When acetate is the food source for a culture of E. coli, iso­ citrate dehydrogenase is phosphorylated.  a.  Draw a diagram s­ howing how acetate is metabolized in E. coli.  b.  When glucose is added to the culture, the phosphate group is removed from ­isocitrate dehydro­ genase. How does flux through the metabolic pathways change in E. coli when glucose is the food source instead of acetate? 79.  Yeast are unusual in their ability to use ethanol as a gluconeogenic substrate. Ethanol is converted to glucose with the assistance of the glyoxylate pathway. Describe how the ethanol → glucose conversion takes place. 80.  Animals lack a glyoxylate pathway and cannot convert fats to carbohydrates. If an animal is fed a fatty acid with all of its carbons replaced by the isotope 14C, some of the labeled carbons later appear in glucose. How is this possible? 81.  Succinate dehydrogenase is not considered to be part of the glyoxylate pathway (see Box 14.B), yet it is vital to the proper functioning of the pathway. Why? 82.  Although mammals do not possess the glyoxylate pathway (see Box 14.B), glyoxylate is generated as a metabolite in glycine and hydroxyproline metabolism. In the liver, glyoxylate is oxidized by lactate dehydrogenase. Draw the structure of the product. 83.  The activity of the purine nucleotide cycle (shown below) in muscles increases during periods of high activity. Explain how the cycle contributes to the ability of the muscle cell to generate energy during intense exercise. IMP is inosine monophosphate.

H2O

adenosine deaminase

AMP Fumarate adenylosuccinate lyase

HO

CH

Aspartate  GTP adenylosuccinate synthetase

GDP  Pi

84.  Metabolites in rat muscle were measured before and after exercising. After exercise, the rat muscle showed an increase in oxaloacetate concentration, a decrease in phosphoenolpyruvate concentration, and no change in pyruvate concentration. Explain. 85.  Various websites claim that taking supplemental B vitamins can provide an energy boost. Use your knowledge of the citric acid cycle to evaluate this claim.

COO COO

87.  Yeast cells that are grown on nonfermentable substrates and then abruptly switched to glucose exhibit substrate-induced inactivation of several enzymes. Which enzymes would glucose cause to be inactivated and why? 88.  Phagocytes protect the host against damage caused by invading microorganisms by engulfing a foreign microbe and forming a ­membrane-bound structure around it called a phagosome. The phagosome then fuses with a lysosome, a cellular organelle that contains proteolytic enzymes that destroy the pathogen. However, some microbes, such as Mycobacterium tuberculosis, survive dormant inside the phagosome for a prolonged period of time. In this scenario, levels of bacterial isocitrate lyase, malate synthase, citrate synthase, and malate dehydrogenase increase inside the phagosome to levels as much as 20 times above normal.  a.  What pathway(s) does M. ­tuberculosis employ while in the phagosome and why are these pathways essential to its survival?  b.  What might be good drug targets for treating a patient infected with M. tuberculosis? 89.  Bacteria and plants (but not animals) possess the enzyme phosphoenolpyruvate carboxylase (PPC), which catalyzes the reaction shown.  a.  What is the importance of this reaction to the organism?  b.  PPC is allosterically activated by both acetyl-CoA and fructose-1,6-bisphosphate. Explain these regulatory strategies.

COO

CH2 C

O

PO2– 3

Phosphoenolpyruvate

IMP Adenylosuccinate

C

Hydroxycitrate

COO

NH 4

HO

COO



HCO–3

PPC

CH2 C

O



Pi

COO

Oxaloacetate

90.  The pharmaceutical, cosmetics, and food industries synthesize succinic acid by “green” or environmentally responsible methods that involve bacteria instead of petrochemicals. Industrial succinate production by bacteria occurs under anaerobic conditions, in which malate dehydrogenase activity increases.  a.  Draw a reaction scheme outlining how phosphoenolpyruvate is converted to succinate. Include the names of all reactants, products, and enzymes.  b. Why is it essential that the production of succinate take place under anaerobic conditions? 91.  Experiments with cancer cells grown in culture show that glutamine is consumed at a high rate and used for biosynthetic reactions, aside from protein synthesis. One possible pathway involves the conversion of glutamine to glutamate and then to α-ketoglutarate. The α-ketoglutarate can then be used to produce pyruvate for gluconeogenesis.  a.  Describe the types of reactions that convert glutamine to

Chapter 14 Credits  427 α-ketoglutarate.  b.  Give the sequence of enzymes that can convert α-ketoglutarate to pyruvate. 92.  Many cancer cells carry out glycolysis at a high rate but convert most of the resulting pyruvate to lactate rather than to ­acetyl-CoA. Acetyl-CoA, however, is required for the synthesis of

fatty acids, which are needed in large amounts by rapidly growing cancer cells. In these cells, the isocitrate dehydrogenase reaction apparently operates in reverse. Explain why this reaction could facilitate the conversion of amino acids such as glutamate into fatty acids.

Selected Readings Cavalcanti, J.H.F., Esteves-Ferreira, A.A., Quinhones, C.G.S., PereiraLima, I.A., Nunes-Nesi, A., Fernie, A.R., and Araújo, W.L., ­Evolution and functional implications of the tricarboxylic acid cycle as revealed by phylogenetic analysis, Genome Biol. Evol. 6, 2830–2848, doi: 10.1093/gbe/evu221 (2014). [Includes descriptions of each of the cycle’s enzymes, from an evolutionary perspective.] Gray, L.R., Tompkins, S.C., and Taylor, E.B., Regulation of pyruvate metabolism and human disease. Cell Mol. Life Sci. 71, 2577–2604 (2014). [Includes useful diagrams of pathways involving pyruvate and discusses diseases related to them.] Martínez-Reyes, I. and Chandel, N.S., Mitochondrial TCA cycle metabolites control physiology and disease, Nature Communications

11, 102, doi: 10.1038/s41467-019-13668-3 (2020). [Summarizes the additional roles of citric acid cycle intermediates in signaling and regulating gene expression.] Muchowska, K.B., Varma, S.J., and Moran, J., Synthesis and breakdown of universal metabolic precursors promoted by iron, Nature 569, 104–107, doi: 10.1038/s41586-019-1151-1 (2019). [Describes a scenario for the prebiotic evolution of the citric acid cycle and glyoxylate cycle.] Patel, M.S., Nemeria, N.S., Furey, W., and Jordan, F., The pyruvate dehydrogenase complexes: Structure-based function and regulation, J. Biol. Chem. 289, 16615–16623 (2014). [Compares the structure and chemical reactions of the human and E. coli enzyme complexes.]

Chapter 14 Credits Figure 14.1a Image based on 1EAA. Mattevi, A., Hol, W.G.J., Atomic structure of the cubic core of the pyruvate dehydrogenase multienzyme complex, Biochemistry 32, 3887–3901 (1993). Figure 14.1b Image based on 1B5S. Izard, T., Aevarsson, A., Allen, M.D., Westphal, A.H., Perham, R.N., de Kok, A., Hol, W.G., Principles of quasi-equivalence and Euclidean geometry govern the assembly of cubic and dodecahedral cores of pyruvate dehydrogenase complexes, Proc. Natl. Acad. Sci. USA 96, 1240–1245 (1999). Figure 14.7a Image based on 5CSC. Liao, D.-I., Karpusas, M., Remington, S.J., Crystal structure of an open conformation of

citrate synthase from chicken heart at 2.8-A resolution, Biochemistry 30, 6031–6036 (1991). Figure 14.7b Image based on 5CTS. Karpusas, M., Branchaud, B., Remington, S.J., Proposed mechanism for the condensation reaction of citrate synthase: 1.9-A structure of the ternary complex with oxaloacetate and carboxymethyl coenzyme A, Biochemistry 29, 2213–2219 (1990). Figure 14.12 Image based on 1CQI. Joyce, M.A., Fraser, M.E., James, M.N., Bridger, W.A., Wolodko, W.T., ADP-binding site of Escherichia coli succinyl-CoA synthetase revealed by X-ray crystallography, Biochemistry 39, 17–25 (2000).

CHAPTER 15

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Oxidative Phosphorylation

To survive winter’s cold, honeybees form a cluster whose center is about 20°C higher than the outside. A high rate of oxidative metabolism in the innermost bees supports the shivering movements of their flight muscles, which generates heat to keep the colony warm.

DO YOU REMEMBER? • Living organisms obey the laws of thermodynamics (Section 1.3). • Transporters obey the laws of thermodynamics, providing a way for solutes to move down their concentration gradients or using ATP to move substances against their gradients (Section 9.1). • Coenzymes such as NAD+ and ubiquinone collect electrons from compounds that become oxidized (Section 12.2). • A reaction that hydrolyzes a phosphoanhydride bond in ATP occurs with a large negative change in free energy (Section 12.3).

The early stages of oxidation of metabolic fuels such as glucose, fatty acids, and amino acids, as well as the oxidation of acetyl carbons to CO2 via the citric acid cycle, yield the reduced cofactors NADH and ubiquinol (QH2). These compounds are forms of energy currency (see Section 12.3), not because they are chemically special but because their reoxidation— ultimately by molecular oxygen in aerobic organisms—is an exergonic process. That free energy is harvested to synthesize ATP, a phenomenon called oxidative phosphorylation. To understand oxidative phosphorylation, we need to first examine why and how electrons flow from reduced cofactors to O2. Then we can explore how the chemical energy of the redox reactions is conserved in the formation of a transmembrane gradient of protons, another type of energy that drives the rotation of ATP synthase so that it can build ATP from ADP and Pi.

The Thermodynamics of Oxidation–Reduction Reactions 15.1

LEaRNING OBJECtIVES Summarize the thermodynamics of oxidation–reduction reactions. • Use standard reduction potential and concentration to calculate a substance’s tendency to become reduced. • Predict the direction of electron transfer in a mixture of two substances. • Convert the change in reduction potential to the change in free energy for a reaction.

428

15.1  The Thermodynamics of Oxidation–­Reduction Reactions  429

In the scheme introduced in Figure 12.10, oxidative phosphorylation represents the final phase of the catabolism of metabolic fuels and the major source of the cell’s ATP (Fig. 15.1). Oxidative phosphorylation differs from the conventional biochemical reactions we have focused on in the last two chapters. In particular, ATP synthesis is not directly coupled to a single discrete chemical reaction, such as a kinase-­catalyzed reaction. Rather, oxidative phosphorylation is a more indirect process of energy transformation. The flow of electrons from reduced compounds such as NADH and QH 2 to an oxidized compound such as O2 is a thermodynamically favorable process. The free energy changes for the movements of electrons through a series of electron carriers can be quantified by considering the reduction potentials of the chemical species involved in each transfer. Oxidation–­reduction reactions (or redox reactions, introduced in Section 12.2) are similar to other chemical reactions in which a portion of a molecule—­ electrons in this case—­is transferred. In any oxidation–­reduction reaction, one reactant (called the oxidizing agent or oxidant) is reduced as it gains electrons. The other reactant (called the reducing agent or reductant) is oxidized as it gives up electrons:

BIOPOLYMERS Proteins Nucleic acids Polysaccharides Triacylglycerols

MONOMERS Amino acids Nucleotides Monosaccharides Fatty acids +

NH4

NAD+

NAD+ NADH

NADH, QH2

​A ​ oxidized​ + ​B​  reduced​ ⇌ ​A ​ reduced​ + ​B​  oxidized​

2- and 3-Carbon INTERMEDIATES

For example, in the succinate dehydrogenase reaction (step 6 of the citric acid cycle; see Section 14.2), the two electrons of the reduced FADH2 prosthetic group of the enzyme are transferred to ubiquinone (Q) so that FADH2 is oxidized and ubiquinone is reduced:

NAD+, Q

+ ​  Q​  ⇌ ​  FAD​ + ​  ​QH​ 2​  ​FADH​ 2 ​​  ​(​reduced​)​ ​(​oxidized​)​ ​(​reduced​)​ ​(​oxidized​)​

H

C

OH

reduction

C

O

Here, the carbon atom “loses” electrons to the more electronegative oxygen atom.

Reduction potential indicates a substance’s tendency to accept electrons

NADH, QH2

(the reverse equation would describe an oxidation half-­reaction).

ADP

O2

oxidative phosphorylation ATP

H2O NAD+, Q

  FIGURE 15.1   Oxidative phosphorylation in context.  The reduced cofactors NADH and QH2, which are generated in the oxidative catabolism of amino acids, monosaccharides, and fatty acids, are reoxidized by a process that requires molecular oxygen. The energy of this process is conserved in a manner that powers the synthesis of ATP from ADP + Pi.

The tendency of a substance to accept electrons (to become reduced) or to donate electrons (become oxidized) can be quantified. Although an oxidation–­reduction reaction necessarily requires both an oxidant and a reductant, it is helpful to consider just one substance at a time, that is, a half-­reaction. For example, the half-­reaction for ubiquinone (by convention, written as a reduction reaction) is ​Q + 2 ​H+ ​ ​+ 2 ​e​ −​⇌ ​QH​ 2​

photosynthesis

CO2

In this reaction, the two electrons are transferred as H atoms (an H atom consists of a proton and an electron, or H+ and e–). In oxidation–­reduction reactions involving the cofactor NAD+, the electron pair takes the form of a hydride ion (H–, a proton with two electrons). In biological systems, electrons usually travel in pairs, although, as we will see, they may also be transferred one at a time. Note that the change in oxidation state of a reactant may be obvious, such as when Fe3+ is reduced to Fe2+, or it may require closer inspection of the molecule’s structure, such as when succinate is oxidized to fumarate (Section 14.2). Often, oxidation can be recognized as the replacement of C—H bonds with C—O bonds, for example, oxidation

citric acid cycle

430  C ha pt er 15   Oxidative Phosphorylation

The affinity of a substance such as ubiquinone for electrons is its standard reduction potential (Ɛ °′), which has units of volts (note that the degree and prime symbols indicate a value under standard biochemical conditions where the pressure is 1 atm, the temperature is 25°C, the pH is 7.0, and all species are present at concentrations of 1 M). The greater the value of Ɛ °′, the greater the tendency of the oxidized form of the substance to accept electrons and become reduced. The standard reduction potentials of some biological substances are given in Table 15.1. Like a ΔG value, the actual reduction potential depends on the actual concentrations of the oxidized and reduced species. The actual reduction potential (Ɛ) is related to the standard reduction potential (Ɛ °′) by the Nernst equation: [​ ​A​  ]​ ​ ​ Ɛ​ ​= Ɛ °′ − ___ ​ RT  ​ ln ________  ​​  ​  reduced   nF ​[​A​ oxidized​]​

(15.1)

R (the gas constant) has a value of 8.3145 J · K−1 · mol−1, T is the temperature in Kelvin, n is the number of electrons transferred (one or two in most of the reactions we will encounter), and F is the Faraday constant (96,485 J · V−1 · mol−1; it is equivalent to the electrical charge of one mole of electrons). At 25°C (298 K), the Nernst equation reduces to ​[​A​ reduced​]​  ​​     ln ​ ________   ​ Ɛ = Ɛ °′ − _______ ​ 0.026 V n ​ [​ ​A​ oxidized]​ ​

(15.2)

In fact, for many substances in biological systems, the concentrations of the oxidized and reduced species are similar, so the logarithmic term is small (recall that ln 1 = 0) and Ɛ is close to Ɛ °′ (Sample Calculation 15.1).

TA B L E 1 5. 1   Standard Reduction Potentials of Some Biological Substances

Half-­reaction

Ɛ °′ (V)

1_​ ​  ​O​  ​+ 2 ​H+ ​ ​+ 2 ​e​ −​⇌ ​H​ 2​O​ 2 2

0.815

2− ​SO​ 2− 4​  ​+ 2 ​H​ ​+ 2 ​e​  ​⇌ ​SO​ 3​  ​+ ​H​ 2​O​

​ O​ − N 3​ ​+

+

−

+

−

2 ​H​ ​+ 2 ​e​  ​⇌

​ O​ − N 2​ ​+

0.48

​H​ 2​O​

Cytochrome a3 (Fe ) + e ​⇌​cytochrome a3 (Fe ) 3+

−

2+

Cytochrome a (Fe ) + e ​⇌​cytochrome a (Fe ) 3+

−

3+

−

2+

Cytochrome c (Fe ) + e ​⇌​cytochrome c (Fe ) 2+

Cytochrome c1 (Fe ) + e ​⇌​cytochrome c1 (Fe ) 3+

−

3+

−

2+

Cytochrome b (Fe ) + e ​⇌​cytochrome b (Fe ) (mitochondrial) Ubiquinone + 2 H+ + 2 e− ​⇌​ubiquinol

2+

Fumarate− + 2 H+ + 2 e− ​⇌​succinate− −

+

−

Pyruvate + 2 H + 2 e ​⇌​lactate −

+

−

−

S + 2 H + 2 e ​⇌​H2S +

0.235 0.22 0.077 0.045

− 0.166

−

− 0.185

Acetaldehyde + 2 H + 2 e ​⇌​ethanol +

0.29

~ 0.

Oxaloacetate + 2 H + 2 e ​⇌​malate −

0.385

0.031

FAD + 2 H + 2 e ​⇌​FADH2 (in flavoproteins) +

0.42

−

− 0.197

−

− 0.23

NAD+ + H+ + 2 e− ​⇌​NADH

− 0.29

Lipoic acid + 2 H+ + 2 e− ​⇌​dihydrolipoic acid

NADP + H + 2 e ​⇌​NADPH +

+

− 0.315

−

Acetoacetate + 2 H + 2 e ​⇌​3-hydroxybutyrate −

+

−

Acetate + 3 H + 2 e ​⇌​acetaldehyde + H2O −

+

−

−

− 0.320 − 0.346 − 0.581

15.1  The Thermodynamics of Oxidation–­Reduction Reactions  431

S AMP L E C A LCULATI O N 15. 1

​]​ [​ ​A​   ​  ​ln ______ ​  reduced   ​Ɛ = Ɛ °′ − _ ​ 0.026 V   n    ​[​A​  ​]​

Problem  Calculate the reduction potential of fumarate (Ɛ  °′ = 0.031 V) at 25°C when [fumarate] = 40 μM and [succinate] = 200 μM.

oxidized

(2 ×  ​10​ ​)     ​ ln ​ _ ​  = 0.031 V − _ ​ 0.026 V   −4

2

(4 × ​10​ ​) = 0.031 V − 0.021 V = 0.010 V​

Solution  Use Equation 15.2. Fumarate is the oxidized compound and succinate is the reduced compound.

The free energy change can be calculated from the change in reduction potential Knowing the reduction potentials of different substances is useful for predicting the movement of electrons between the two substances. When the substances are together in solution or connected by wire in an electrical circuit, electrons flow spontaneously from the substance with the lower reduction potential to the substance with the higher reduction potential. For example, in a system containing Q/QH2 and NAD+/ NADH, we can predict whether electrons will flow from QH2 to NAD+ or from NADH to Q. Using the standard reduction potentials given in Table 15.1, we note that Ɛ °′ for NAD+ (–0.315 V) is lower than Ɛ °′ for ubiquinone (0.045 V). Therefore, NADH will tend to transfer its electrons to ubiquinone; that is, NADH will be oxidized and Q will be reduced. A complete oxidation–­reduction reaction is just a combination of two half-­reactions. For the NADH–­u biquinone reaction, the net reaction is the ubiquinone reduction half-­r eaction (the half-­ −0.4 reaction as listed in Table 15.1) combined with the NADH oxidation half-­reaction (the reverse of the half-­reaction listed in Table 15.1). NADH Note that because the NAD+ half-­reaction has been reversed to indicate oxidation, we have also reversed the sign of its Ɛ °′ value: NADH ⇌ NAD + H + 2 e +

+

Q + 2 H + 2 e ⇌ QH2 +

Ɛ °′ = +0.315 V

NAD+

Complex I

net: NADH + Q + H ⇌ NAD + QH2 ΔƐ °′ = +0.360 V

ΔG°ʹ = −69.5 kJ · mol−1

Ɛ °′ = 0.045 V

+

When the two half-­reactions are added, their reduction potentials are also added, yielding a ΔƐ °′ value. Keep in mind that the reduction potential is a property of the half-­reaction and is independent of the direction in which the reaction occurs. Reversing the sign of Ɛ °′, as shown above, is just a shortcut to simplify the task of calculating ΔƐ °′. Another method for calculating ΔƐ °′ uses the following equation: ​Δ Ɛ °′ = ​Ɛ​ °′     (​ ​e​ −​ acceptor​)  ​​− ​Ɛ​ °′     (​ ​e​ −​ donor​)  ​​​

(15.3)

Not surprisingly, the larger the difference in Ɛ values (the greater the ΔƐ value), the greater the tendency of electrons to flow from one substance to the other, and the greater the change in free energy of the system. ΔG is related to ΔƐ as follows:

SEE SAMPLE CALCULATION VIDEOS

−0.2

−

−

+

−5

​ΔG °′ = −nFΔ Ɛ °′  or ΔG = −nFΔƐ​

(15.4)

Accordingly, an oxidation–­reduction reaction with a large positive ΔƐ value has a large negative value of ΔG (see Sample Calculation 15.2). Depending on the relevant reduction potentials, an oxidation–­ reduction reaction can release considerable amounts of free energy. This is what happens in the mitochondria, where the reduced cofactors generated by the oxidation of metabolic fuels are reoxidized. The free energy released in this process powers ATP synthesis by oxidative phosphorylation. Figure 15.2 shows the major mitochondrial electron

0

Q Complex III

Ɛ °ʹ (V)

0.2

ΔG°ʹ = −36.7 kJ · mol−1

cytochrome c

0.4 ΔG°ʹ = −112 kJ · mol−1

Complex IV 0.6

0.8

O2

H2O

  FIGURE 15.2   Overview of mitochondrial electron transport.  The reduction potentials of the key electron carriers are indicated. The oxidation–­reduction reactions mediated by Complexes I, III, and IV release free energy.

Question  Determine the total free energy change for the oxidation of NADH by O2.

432  C ha pt er 15   Oxidative Phosphorylation

S AMP L E C A LCULATI O N 1 5 . 2 Problem  Calculate the standard free energy change for the oxidation of malate by NAD+. Is the reaction spontaneous under standard conditions? Solution Method 1 Write the relevant half-­reactions, reversing the malate half-­ reaction (so that it becomes an oxidation reaction) and reversing the sign of its Ɛ  °′: malate → oxaloacetate + 2 H+ + e− NAD+ + H+ + 2 e− → NADH

Ɛ °′ = +0.166 V Ɛ °′ = −0.315 V

net: malate + NAD+  →  oxaloacetate + NADH + H+ ΔƐ    °′ = −0.149 V

Method 2 Identify the electron acceptor (NAD+) and electron donor (malate). Substitute their standard reduction potentials into Equation 15.3:

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​ΔƐ °′ = ​Ɛ  ​ °′  ​− ​Ɛ ​ °′  ​    ​(​e​ −​ acceptor​)​     (​ ​e​ −​ donor​)​  = −0.315 V − ​(​−0.166 V​)​ = −0.149 V​ Both Methods The ΔƐ °′ for the net reaction is –0.149 V. Use Equation 15.4 to calculate ΔG °′: ​ΔG °′ = −nFΔ Ɛ °′ ​ ​ · ​mol​−1​) (−0.149 V)​ = −(2) (96,485 J · ​V−1 = + 28,750 J · ​mol​−1​= + 28.8 kJ · ​mol​−1​ The reaction has a positive value of ΔG °′ and so is not spontaneous. (In vivo, this endergonic reaction occurs as step 8 of the citric acid cycle and is coupled to step 1, which is exergonic.)

transport components arranged by their reduction potentials. Each stage of electron transfer, from NADH to O2, the final electron acceptor, occurs with a negative change in free energy.

Before Going On • Explain why an oxidation–­reduction reaction must include both an oxidant and a reductant. • W  hen two reactants are mixed together, explain how you can predict which one will become reduced and which one will become oxidized. • E  xplain how adding the Ɛ °′ values for two half-­reactions yields a value of ΔƐ °′ and ΔG °′ for an oxidation–­reduction reaction. • S  elect the two half-­reactions from Table 15.1 that would be most likely to form a freely reversible (near-­equilibrium) redox reaction.

15.2 Mitochondrial Electron Transport LEA RNING OBJECTIVES Map the path of electrons through the redox groups of the electron transport pathway. • Explain why the mitochondrion includes a variety of transport systems. • Identify the sources of electrons for Complexes I, III, and IV. • Describe the mechanisms for transporting protons across the mitochondrial membrane. • Explain why respiratory complexes may not actually form a chain.

In aerobic organisms, the NADH and ubiquinol produced by glycolysis, the citric acid cycle, fatty acid oxidation, and other metabolic pathways are ultimately reoxidized by molecular oxygen, a process called cellular respiration. The standard reduction

15.2 Mitochondrial Electron Transport

433

potential of +0.815 V for the reduction of O2 to H2O indicates that O2 is a more effective oxidizing agent than any other biological compound (see Table 15.1). The oxidation of NADH by O2, that is, the transfer of electrons from NADH directly to O 2, would release a large amount of free energy, but this reaction does not occur in a single step. Instead, electrons are shuttled from NADH to O2 in a multistep process that offers several opportunities to conserve the free energy of oxidation. In eukaryotes, all the steps of oxidative phosphorylation are carried out by a series of electron carriers that include small molecules as well as the prosthetic groups of large integral membrane proteins in mitochondria (in prokaryotes, similar electron carriers are located in the plasma membrane). The following sections describe how electrons flow through this respiratory electron transport chain from reduced cofactors to oxygen.

Mitochondrial membranes define two compartments

a.

Gopal Murti/Medical Images.com

Don W. Fawcett/ Science Source

Carmen Mannella, Wadsworth Center, Albany, New York

In accordance with its origin as a bacterial symbiont, the mitochondrion (plural, mitochondria) has two membranes. The outer membrane, analogous to the outer membrane of some bacteria, is relatively porous due to the presence of porin-like proteins that permit the transmembrane diffusion of substances with masses up to about 10 kD (see Section 9.2 for an example of porin structure and function). The inner membrane has a convoluted architecture that encloses a space called the mitochondrial matrix. Because the Outer membrane inner mitochondrial membrane prevents the transmembrane movements of ions Inner membrane and small molecules (except via specific transport proteins), the composition of the matrix differs from that of the space between the inner and outer membranes. In fact, Intermembrane space the ionic composition of the intermembrane space is considered to be equivalent Matrix to that of the cytosol (Fig. 15.3). Mitochondria are customarily shown as bean-shaped organelles with the inner mitochondrial membrane forming a system of baffles called cristae (Fig. 15.4a). However, electron tomography, a technique for visualizing cellular structures in three dimensions by analyzing micrographs of sequential cell slices, reveals that mitochondria are highly variable structures. For example, the cristae may be irregular and bulbous rather than planar and may make several tubular connections with the FIGURE 15.3 Model of mitochonrest of the inner mitochondrial membrane (Fig. 15.4b). Moreover, a cell may contain drial structure. The relatively imperhundreds to thousands of discrete bacteria-shaped mitochondria, or a single tubumeable inner mitochondrial membrane lar organelle may take the form of an extended network with many branches and encloses the protein-rich matrix. The interinterconnections (Fig. 15.4c). Individual mitochondria can move around the cell and membrane space has an ionic composition undergo fusion (joining) and fission (separating). similar to that of the cytosol.

b.

FIGURE 15.4 Images of mitochondria. a. Conventional electron micrograph showing cristae as a system of planar baffles. b. Threedimensional reconstruction of a mitochondrion by electron tomography,

c. showing irregular tubular cristae. c. Fluorescence micrograph of cultured human cells. The cytosol contains a network of mitochondria (red). The nucleus is blue.

434  C ha pt er 15   Oxidative Phosphorylation

SEE GUIDED TOUR Oxidative Phosphorylation

Reflecting its ancient origin as a free-­living organism, the mitochondrion has its own genome and protein-­synthesizing machinery consisting of mitochondrially encoded rRNA and tRNA. The mitochondrial genome encodes 13 proteins, all of which are components of the respiratory chain complexes. This is only a small subset of the approximately 1500 proteins required for mitochondrial function; the other respiratory chain proteins, matrix enzymes, transporters, and so on are encoded by the cell’s nuclear genome, synthesized in the cytosol, and imported into the mitochondria via translocases in the outer membrane (the TOM complex) and inner membrane (the TIM complex). Lipids synthesized by the endoplasmic reticulum appear to be transferred to mitochondria at points where “tether” proteins temporarily hold the two organelles together. Much of the cell’s NADH and QH2 is generated by the citric acid cycle inside mitochondria. Fatty acid oxidation also takes place largely inside mitochondria and yields NADH and QH2. These reduced cofactors transfer their electrons to the protein complexes of the respiratory electron transport chain, which are tightly associated with the inner mitochondrial membrane. However, NADH produced by glycolysis and other oxidative processes in the cytosol cannot directly reach the respiratory chain. There is no transport protein that can ferry NADH across the inner mitochondrial membrane. Instead, “reducing equivalents” are imported into the matrix by the chemical reactions of systems such as the malate–­aspartate shuttle system (Fig. 15.5). Mitochondria also need a mechanism to export ATP and to import ADP and Pi, since most of the cell’s ATP is generated in the matrix by oxidative phosphorylation and is consumed in the cytosol. A transport protein called the ADP/ATP carrier, or adenine nucleotide translocase, exports ATP and imports ADP, binding one or the other and changing its conformation to release the bound nucleotide on the other side of the membrane (Fig. 15.6a). Inorganic phosphate, a substrate for oxidative phosphorylation, is imported from the cytosol in symport with H+ (Fig. 15.6b).

Complex I transfers electrons from NADH to ubiquinone The path electrons travel through the respiratory chain begins with Complex I, also called NADH:ubiquinone oxidoreductase or NADH dehydrogenase. This enzyme catalyzes the transfer of a pair of electrons from NADH to ubiquinone: ​ NADH + ​H​+​+ Q ⇌ ​NAD​+​+ ​QH​ 2​

+ H3N

COO−

COO−

C H

C O

CH2

CH2

COO−

COO−

Aspartate

Oxaloacetate

NADH + H+

NAD+

COO− HO C H

matrix malate dehydrogenase

CH2 COO−

Malate

MATRIX

  FIGURE 15.5   The malate–­

aspartate shuttle system.  Cytosolic oxaloacetate is reduced to malate for transport into mitochondria. Malate is then reoxidized in the matrix. The net result is the transfer of “reducing equivalents” from the cytosol to the matrix. Mitochondrial oxaloacetate can be exported back to the cytosol after being converted to aspartate by an aminotransferase.

CYTOSOL

+ H3N

COO−

COO−

C H

C O

CH2

CH2

COO−

COO−

Aspartate

Oxaloacetate

cytosolic malate dehydrogenase

COO− HO C H CH2

NADH + H+

NAD+

COO−

Malate

15.2  Mitochondrial Electron Transport  435   FIGURE 15.6   Mitochondrial transport systems. a. The ADP/ATP carrier binds either ATP or ADP and changes its conformation to release the nucleotide on the opposite side of the inner mitochondrial membrane. This translocase can therefore export ATP and import ADP.  b. A Pi –H+ symport protein permits the simultaneous movement of inorganic phosphate and a proton into the mitochondrial matrix.

ATP MATRIX

INTERMEMBRANE SPACE

H+

Pi

ADP

a.

b.

Question  Does the activity of either of these transporters contribute to the mitochondrial membrane potential?

Complex I is the largest of the electron transport proteins in the mitochondrial respiratory chain, with 45 different subunits and a total mass of about 970 kD in mammals. The crystal structure of a smaller (536-kD) bacterial Complex I reveals an L-­shaped protein with numerous transmembrane helices and a peripheral arm (Fig. 15.7). Fourteen subunits make up the core of Complex I and are conserved in all organisms. The additional 31 subunits in mammalian Complex I form a sort of shell around the core, stabilizing its structure and potentially helping to regulate its function. Electron transport takes place in the peripheral arm of Complex I, which includes several prosthetic groups that undergo reduction as they receive electrons and become oxidized as they give up their electrons to the next group. All these groups, or redox centers, appear to have reduction potentials approximately between the reduction potentials of NAD+ (Ɛ °′ = –0.315 V) and ubiquinone (Ɛ °′ = +0.045 V). This allows them to form a chain where the electrons travel a path of increasing reduction potential. The redox centers do not need to be in intimate contact with each other, as they would be if the transferred group were a larger chemical entity. An electron can move between redox centers up to 14 Å apart by “tunneling” through the covalent bonds of the protein.

a.   FIGURE 15.7   Structure of bacterial Complex I.  a. The 16 sub-

units are shown in different colors, with the redox centers in red (FMN) and orange (Fe–S clusters). The horizontal portion is embedded in the membrane (represented by black lines) and the “arm” p ­ rojects into the cytosol in bacteria (or mitochondrial matrix in e­ ukaryotes). 

b. b. Arrangement of the redox centers in Complex I. Atoms are color coded: C gray, N blue, O red, P orange, Fe gold, and S yellow. Electrons flow through the groups from upper left to lower right. Question  Identify the FMN group, the 4Fe–­4S clusters, and the 2Fe–­2S clusters in part b.

436  C ha pt er 15   Oxidative Phosphorylation

CH2OPO2− 3

CH2OPO2− 3 HO

C

H

HO

C

H

HO

C

H

HO

C

H

HO

C

H

HO

C

H

2 H+, 2 e−

CH2 H3C

N

H3C

N

N

O N

O

H

CH2

H

H3C

N

N

H3C

N

Flavin mononucleotide (FMN)

FMNH2

H

O N

O

H

  FIGURE 15.8   Flavin mononucleotide (FMN).  This prosthetic group resembles flavin ade-

nine dinucleotide (FAD; see Fig. 3.2c) but lacks the AMP group of FAD. The transfer of two electrons and two protons to FMN yields FMNH2.

Cys

S

Cys Fe

Cys

Fe S

Cys

2Fe–2S Cys Cys

S

Fe

Fe Fe

S S

Cys S

Fe

4Fe–4S

Cys

  FIGURE 15.9   Iron–­sulfur clusters.  Although some Fe–S clusters contain up to eight Fe atoms, the most common are the 2Fe–­2S and 4Fe–­4S clusters. In all cases, the iron–­sulfur clusters are coordinated by the S atoms of cysteine side chains. These prosthetic groups undergo one-­electron redox reactions.

The two electrons donated by NADH are first picked up by flavin mononucleotide (FMN; Fig. 15.8) near the far end of the Complex I arm. This noncovalently bound prosthetic group, which is similar to FAD, then transfers the electrons, one at a time, to a second type of redox center, an iron–­sulfur (Fe–S) cluster. Depending on the species, Complex I bears 8 to 10 of these prosthetic groups, which contain equal numbers of iron and sulfide ions (Fig. 15.9). Unlike the electron carriers we have introduced so far, Fe–S clusters are one-­electron carriers. They have an oxidation state of either +3 (oxidized) or +2 (reduced), regardless of the number of Fe atoms in the cluster (each cluster is a conjugated structure that functions as a single unit). Electrons travel between several Fe–S clusters before reaching ubiquinone. Like FMN, ubiquinone is a two-­electron carrier (see Section 12.2), but it accepts one electron at a time from an Fe–S donor. Iron–­sulfur clusters may be among the most ancient of electron carriers, dating from a time when the earth’s abundant iron and sulfur were major players in prebiotic chemical reactions. The ubiquinone binding site is located in the Complex I arm not far from the membrane surface. As electrons are transferred from NADH to ubiquinone, Complex I transfers four protons from the matrix to the intermembrane space. Comparisons with other transport proteins and detailed analysis of cryo-­EM and crystal structures indicate the presence of four proton-­ translocating “channels” in the membrane-­embedded arm of Complex I. When redox groups in the periph­eral arm are transiently reduced and reoxidized, the protein undergoes conformational changes that are transmitted from the peripheral arm to the membrane arm in part by a horizontally oriented helix that lies within the membrane portion of the complex. These conformational changes do not open passageways, as occurs in Na+ and K+ channel proteins (Section 9.2). Instead, each proton passes from one side of the membrane to the other via a proton wire, a series of hydrogen-­bonded protein groups plus water molecules that form a chain through which a proton can be rapidly relayed. (Recall from Fig. 2.13 that protons readily jump between water molecules.) Note that in this relay mechanism, the protons taken up from the matrix are not the same ones that are released into the intermembrane space. The reactions of Complex I are summarized in Figure 15.10.

Other oxidation reactions contribute to the ubiquinol pool The reduced quinone product of the Complex I reaction joins a pool of quinones that are soluble in the inner mitochondrial membrane by virtue of their long hydrophobic isoprenoid tails (see Section 12.2). The pool of reduced quinones is augmented by the activity of other oxidation–­reduction reactions. One of these is catalyzed by succinate dehydrogenase, which carries out step 6 of the citric acid cycle (see Section 14.2). ​succinate + Q ⇌ fumarate + ​QH​ 2​

15.2  Mitochondrial Electron Transport  437

Succinate dehydrogenase is the only one of the citric acid cycle NADH + H+ enzymes that is not soluble in the mitochondrial matrix; it is embedded in the inner membrane. Like the other respiratory complexes, it contains NAD+ several redox centers, including an FAD group. Succinate dehydrogenase is MATRIX also called Complex II of the mitochondrial respiratory chain. However, it is more like a tributary because it does not undertake proton translocation 2e− Q and therefore does not directly contribute the free energy of its oxidation–­ reduction reaction toward ATP synthesis. Nevertheless, it does feed reducComplex QH2 ing equivalents as ubiquinol into the electron transport chain (Fig. 15.11a). I A major source of ubiquinol is fatty acid oxidation, another energy-­ INTERMEMBRANE SPACE generating catabolic pathway that takes place in the mitochondrial matrix. A membrane-­bound fatty acyl-­CoA dehydrogenase catalyzes the oxida4 H+ tion of a C—C bond in a fatty acid attached to coenzyme A. The electrons removed in this dehydrogenation reaction are transferred to ubiquinone (Fig. 15.11b). As we will see in Section 17.2, the complete oxidation of a   FIGURE 15.10   Complex I function.  As two fatty acid also produces NADH that is reoxidized by the mitochondrial elecelectrons from the water-­soluble NADH are transferred tron transport chain, starting with Complex I. to the lipid-­soluble ubiquinone, four protons are transElectrons from cytosolic NADH can also enter the mitochondrial ubilocated from the matrix into the intermembrane space. quinol pool through the actions of a cytosolic and a mitochondrial glycerol-3-phosphate dehydrogenase (Fig. 15.11c). First, the cytosolic enzyme uses NADH to reduce dihydroxyacetone phosphate to glycerol-3-phosphate. Then the mitochondrial enzyme, embedded in the inner membrane, reoxidizes the glycerol-3-phosphate, transferring two electrons to ubiquinone. This system, which shuttles electrons from NADH to ubiquinol, bypasses Complex I.

Complex III transfers electrons from ubiquinol to cytochrome c Ubiquinol is reoxidized by Complex III, an integral membrane protein with 11 subunits in each of its two monomeric units. Complex III, also called ubiquinol:cytochrome c oxidoreductase or cytochrome bc1, transfers electrons to the peripheral membrane protein cytochrome c. Cytochromes R

R

H C H

Succinate

Fumarate

H C

H C H

C H

C O

C O

SCoA

SCoA

MATRIX

2e−

INTERMEMBRANE SPACE

succinate dehydrogenase (Complex II)

a.

2e−

Q

QH2

acyl-CoA dehydrogenase

mitochondrial dehydrogenase

Q QH2

Q QH2

2e−

b. H2C OH

H2C OH

C O

HO C H

CH2OPO32−   FIGURE 15.11   Reactions that contribute to the ubiquinol pool.  a. The succinate dehydrogenase (Complex II) reaction transfers electrons to the pool of reduced ubiquinone in the inner mitochondrial membrane.  b. The acyl-­CoA dehydrogenase reaction of fatty acid oxidation generates ubiquinol. R represents the hydrocarbon tail of the fatty acid.  c. In the glycerol-3-phosphate shuttle system, electrons from cytosolic NADH are transferred to the membrane ubiquinone pool.

CH2OPO32−

Dihydroxyacetone phosphate

Glycerol-3phosphate

cytosolic dehydrogenase

NADH + H+

c.

NAD+

438  C ha pt er 15   Oxidative Phosphorylation

CH2 H3C

CH

CH3 CH

N

N

Fe3+ H3C

N

N

CH3

CH2

CH2

CH2

CH2

COO−

COO−

Heme b

CH2

are proteins with heme prosthetic groups. The name cytochrome l­ iterally means “cell color”; cytochromes are largely responsible for the purplish-­brown color of mitochondria. Cytochromes are commonly named with a letter (a, b, or c) indicating the exact structure of the porphyrin ring of their heme group (Fig. 15.12). The structure of the heme group and the surrounding protein microenvironment influence the protein’s absorption spectrum. They also determine the reduction potentials of cytochromes, which range from about –0.080 V to about +0.385 V. Unlike the heme prosthetic groups of hemoglobin and myoglobin, the heme groups of cytochromes undergo reversible one-­electron reduction, with the central Fe atom cycling between the Fe3+ (oxidized) and Fe2+ (reduced) states. Consequently, the net reaction for Complex III, in which two electrons are transferred, includes two cytochrome c proteins: ​ ​QH​ 2​+ 2 cytochrome c (​Fe​3+​) ⇌ Q + 2 cytochrome c (​Fe​2+​) + 2 ​H+ ​​

Complex III itself contains two cytochromes (cytochrome b and cytochrome c ) that are integral membrane proteins. These two proteins, along with an iron–­ 1   FIGURE 15.12   The heme group of a b sulfur protein (also called the Rieske protein), form the functional core of Complex cytochrome.  The planar porphyrin ring III (these same three subunits are the only ones that have homologs in the corresurrounds a central Fe atom, shown here in its oxidized (Fe3+) state. The heme substituent sponding bacterial respiratory complex). Altogether, each monomer of Complex III groups that are colored blue differ in the a and c is anchored in the membrane by 14 transmembrane α helices (Fig. 15.13). cytochromes (the heme group of hemoglobin and The flow of electrons through Complex III is complicated, in part because myoglobin has the b structure; see Section 5.1). the two electrons donated by ubiquinol must split up in order to travel through a series of one-­electron carriers that includes the 2Fe–­2S cluster of the iron–­sulfur protein, cytochrome c1, and cytochrome b (which actually contains two heme groups with slightly different reduction potentials). Except for the 2Fe–­2S cluster, all the redox centers are arranged in such a way that electrons can tunnel from one to another. The iron–­sulfur protein must change its conformation by rotating and moving about 22 Å in order to pick up and deliver an electron. Further complicating the picture is the fact that each monomeric unit of Complex III has two active sites where quinone cofactors undergo reduction and oxidation. The circuitous route of electrons from ubiquinol to cytochrome c is described by the two-­ round Q cycle, diagrammed in Figure 15.14. The net result of the Q cycle is that two electrons from QH2 reduce two molecules of cytochrome c. In addition, four protons are translocated to the intermembrane space, two from QH2 in the first round of the Q cycle and two from QH2 in the second round. This proton movement contributes to the transmembrane proton gradient. The reactions of Complex III are summarized in Figure 15.15.

MATRIX

INTERMEMBRANE SPACE

a.   FIGURE 15.13   Structure of mammalian Complex III.  a. Backbone model. Eight transmembrane helices in each monomer of the dimeric complex are contributed by cytochrome b (light blue with heme groups dark blue). The iron–­sulfur protein (green with Fe–S clusters orange) and cytochrome c1 (pink with heme groups purple) project into the intermembrane space.  b.  Arrangement of

b. prosthetic groups. The two heme groups of each cytochrome b (blue) and the heme group of cytochrome c1 (purple), along with the iron–­ sulfur clusters (Fe atoms orange), provide a pathway for electrons between ubiquinol (in the membrane) and cytochrome c (in the intermembrane space).

15.2  Mitochondrial Electron Transport  439 1. In the first round, QH2 donates one electron to the iron–sulfur protein (ISP). The electron then travels to cytochrome c1 and then to cytochrome c.

Round 1 MATRIX

. Q− Q 3 2 e− e− 1

Q QH2

2. QH2 donates its other electron to cytochrome b. The two protons from QH2 are released into the intermembrane space.

cyt b

ISP

cyt c1

INTERMEMBRANE SPACE

2 H+

cyt c(Fe3+)

3. The oxidized ubiquinone diffuses to another quinone-binding site, where it accepts the electron from cytochrome b, becoming a half-reduced semiquinone (.Q −).

cyt c(Fe2+)

Round 2 2 H+

4. In the second round, a second QH2 surrenders its two electrons to Complex III and its two protons to the intermembrane space. One electron goes to reduce cytochrome c.

QH2 . Q− cyt b 5 e− e− 4

Q QH2

2 H+

ISP

cyt c1

cyt c(Fe3+)

5. The other electron goes to cytochrome b and then to the waiting semiquinone produced in the first part of the cycle. This step regenerates QH2, using protons from the matrix.

cyt c(Fe2+)

  FIGURE 15.14   The Q cycle.

Question  Write an equation for round 1 and for round 2.

Many prokaryotes lack a homolog of mitochondrial Complex III and instead use other machinery to oxidize quinol. Although the conventional and alternative complexes are not structurally similar, both include heme groups and iron–­sulfur clusters. Some alternative Complex III enzymes are able to transfer electrons directly to Complex IV without a mobile electron carrier such as cytochrome c.

Complex IV oxidizes cytochrome c and reduces O2 Just as ubiquinone ferries electrons from Complex I and other enzymes to Complex III in mitochondria, cytochrome c ferries electrons between Complexes III and IV, carrying just one at a time. Unlike ubiquinone and the other proteins of the respiratory chain, cytochrome c is soluble in the intermembrane space (Fig. 15.16). Because this small peripheral membrane protein is central to the metabolism of many organisms, analysis of its sequence has played a large role in elucidating evolutionary relationships. Complex IV, also called cytochrome c oxidase, is the last enzyme to deal with the electrons derived from the oxidation of metabolic fuels. Four electrons delivered by cytochrome c are consumed in the reduction of molecular oxygen to water: ​4 cytochrome c (​Fe​2+​) + ​O​ 2​+ 4​ H​+​  → 4 cytochrome c (​Fe​3+​) + 2 ​H​ 2​O​

2 H+ MATRIX

Complex III

QH2 + Q

2e−

2 QH2 INTERMEMBRANE SPACE

4 H+ 2 cyt c(Fe3+)

2 cyt c(Fe2+)

  FIGURE 15.15   Complex III function.  For every two electrons that pass from ubiquinol to cytochrome c, four protons are translocated to the intermembrane space.

Question  How does the proton-­translocating mechanism of Complex III differ from the one in Complex I?

440  C ha pt er 15   Oxidative Phosphorylation

  FIGURE 15.16  Cytochrome c.  The protein is shown as a gray transparent surface over its ribbon backbone. The heme group (pink) lies in a deep pocket.

The redox centers of mammalian Complex IV include heme groups and copper ions situated among the 13 subunits in each half of the dimeric complex (Fig. 15.17). Within each monomer, an electron travels from cytochrome c to the CuA redox center, which has two copper ions, and then to a heme a group. From there it travels to a binuclear center consisting of the iron atom of heme a3 and a copper ion (CuB). The four-­electron reduction of O2 occurs at the Fe–­Cu binuclear center, with help from a nearby tyrosine side chain. Note that the chemical reduction of O2 to H2O consumes four protons from the mitochondrial matrix. One possible mechanism, based on spectroscopic and other evidence, is shown in Figure 15.18. Cytochrome c oxidase also relays four additional protons from the matrix to the intermembrane space, most likely during steps 3, 4, 5, and 6 of the mechanism shown in Figure 15.18. In other words, two protons are translocated for every pair of electrons traveling through the electron transport chain. Complex IV appears to harbor two proton wires. One delivers H + ions from the matrix to the oxygen-­reducing active site. The other one spans the 50-Å distance between the matrix and intermembrane faces of the protein. Protons are relayed through the proton wires when the protein changes its conformation in response to changes in its oxidation state. The production of water and the proton relays both deplete the matrix proton concentration and thereby contribute to the formation of a proton gradient across the inner mitochondrial membrane (Fig. 15.19).

Cu+

Fe2+

YH

O2 Cu+

Fe3+

YH

O2− MATRIX

e−

INTERMEMBRANE SPACE

Fe4+

Cu2+

O2−

OH−

Y−

H+ a.

Fe4+

Cu2+

O2−

OH2

Y−

H+, e− Fe3+

Cu2+

Y−

OH− OH2 H+, e−

b.   FIGURE 15.17  Structure of cytochrome c oxidase.  a. The 13 subunits in each mono-

meric half of the mammalian complex comprise 28 transmembrane α helices.  b. The heme groups (C atoms gray, N blue, O red, and Fe gold) and copper ions (brown) from one half of the complex are shown in space-­filling form.

Question  Locate the copper ion and heme iron of the binuclear center where O2 is reduced.

Fe3+

H2O Cu+

YH

OH− H+, e− Fe2+

H2O Cu+

YH

  FIGURE 15.18   A proposed model for the cytochrome c oxidase reaction.  The Fe–­Cu binuclear center and tyrosine side chain (Y) are indicated.

15.2 Mitochondrial Electron Transport

1 2 MATRIX

O2 + 2 H+ H2O

Complex IV

441

FIGURE 15.19 Complex IV function. For every two electrons donated by cytochrome c, two protons are translocated to the intermembrane space. Two protons from the matrix are also consumed in the reaction _12 O 2 → H 2O (the full reduction of O2 requires four electrons).

2e −

INTERMEMBRANE SPACE

2 H+

2 cyt c(Fe3+) 2 cyt c(Fe2+)

Respiratory complexes associate with each other Complexes I–IV are located mostly in the planar portions of cristae, where they have some lateral mobility. However, the three complexes that contribute to the transmembrane proton gradient— Complexes I, III, and IV—are able to associate with each other in supercomplexes whose composition seems to vary with the type of cell and the physiological conditions. Supercomplexes containing a Complex III dimer along with Complex I or Complex IV have been described in detail. An assembly of all three complexes is known as a respirasome, since it is capable of catalyzing the complete process of electron transfer from NADH to O2 (Fig. 15.20). Supercomplexes have been observed in bacteria, yeast, and mammals, which argues that bringing respiratory enzymes close together is functionally significant. Although it is tempting to speculate that proximity increases the efficiency of electron transfer, the complexes are still somewhat separated, with membrane phospholipids between them. Moreover, the complexes appear to be dynamic. For example, the large Complex I is the least mobile and is found mostly in supercomplexes, whereas Complex IV exists largely as monomers but can form isolated dimers (as shown in Fig. 15.17) as well as supercomplexes. Supercomplex formation may be a way to stabilize the structures of the large proteins or to maximize the density of the electron-transporting machinery in the membrane. Even if the respiratory complexes do not actually form a chain, they provide a pathway for the reduced cofactors NADH and ubiquinol to surrender their electrons to O2. The oxidized cofactors rejoin the pool of cofactors in the cell, ready to accept additional electrons by participating in redox reactions such as those of the citric acid cycle. In this way, NAD+ and ubiquinone function as shuttles, accepting electrons (becoming reduced) then giving them up (becoming oxidized) over and over again. Because the electron carriers are regenerated with each reaction cycle, electron flow continues, provided there is plenty of metabolic fuel (the source of electrons) and O2 (the final electron acceptor). However, the electron transport system can also produce damaging by-products (Box 15.A).

FIGURE 15.20 A respirasome. This supercomplex of Complex I (green), a Complex II dimer (cyan), and monomeric Complex IV (blue) is based on the cyro-EM study of a respirasome from porcine heart.

Question Add lines to show the location of the mitochondrial membrane.

442

C ha ptER 15

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Box 15.a Reactive Oxygen Species Although electron transport and oxidative phosphorylation generate considerable amounts of ATP, processes that depend on O 2 have some disadvantages. Aerobic organisms— in addition to the challenges of acquiring O 2 from the air, water, or soil they live in— must also deal with the possibility of oxidative damage. The very machinery that makes aerobic metabolism possible also produces toxic molecules known as reactive oxygen species, including superoxide (·O 2−), hydrogen peroxide (H2O2), and the hydroxyl radical (·OH). Electrons occasionally leak from the electron transport chain, primarily in Complexes I and III, and react with O2 to generate reactive oxygen species, including free radicals. For example,

membranes and selectively oxidize Cys sulfhydryl groups, which often exist as thiolate groups (—S−), in cellular proteins. This modification may alter protein activity. Restoring the modified Cys residues and other intentionally or accidentally oxidized cellular components often involves the tripeptide glutathione (GSH, γ- Glu- Cys- Gly), in which the side chain of glutamate is linked to the amino group of cysteine:

O +

H3N

O 2 + e− → ·O − 2

A free radical is an atom or molecule with a single unpaired electron and is highly reactive as it seeks another electron to form a pair. Although a free radical is short-lived (the half-life of ·O 2− is 10–6 seconds), it can damage nearby proteins, lipids, and nucleic acids by stealing their electrons. Presumably, the most damage is inflicted on mitochondrial components. The accumulation of cellular damage from free radicals may be responsible for the degeneration of tissues that occurs with aging. The primary cellular defense against ·O 2− is the enzyme superoxide dismutase, which rapidly converts ·O − 2 to the less-toxic H2O2: + 2 ·O − 2 + 2 H → H 2O 2 + O 2

Other cellular components, such as ascorbate (see Box 5.B) and α-tocopherol (see Box 8.B) may protect cells from oxidative damage by scavenging free radicals. Findings such as these have led to the popular belief that dietary supplements with antioxidant activity can prevent aging. Following a similar line of reasoning, caloric restriction may lower free radical production by decreasing the availability of fuel molecules that undergo oxidative metabolism. Caloric restriction extends the lifespans of some experimental animals, but a similar effect has not been confirmed in humans. Furthermore, a high rate of fuel oxidation in active animal muscles is not always accompanied by high production of reactive oxygen species, which leaves the link between oxidative metabolism and free radicals in question. To make matters more complicated, many cells deliberately synthesize reactive oxygen and nitrogen species as signaling molecules. For example, an H2O2 molecule can diffuse through

C

CH

O–

CH2

Glutathione (GSH)

CH2 C

O

O

CH

C

HN

CH2

O HN

CH

C

O–

H

SH Enzymes called glutathione peroxidases use glutathione to convert hydrogen peroxide to water and to convert other organic peroxides (R—O—O—R) and sulfenic acids (R—S—O—H) to less toxic products. In the process, two glutathione tripeptides become linked through a disulfide bond between their Cys residues, to form GSSG (see Problem 4.18). This oxidized glutathione must then be reduced by glutathione reductase in a reaction that requires NADPH as an electron donor:

H2O2

2 GSH peroxidase

2 H2O

GSSG

NADP+ reductase

NADPH + H+

The high cellular concentration of glutathione—typically in the millimolar range—seems excessive for the job of scavenging free radicals and repairing other oxidized molecules, and it is possible that glutathione plays a more general role in regulating the overall redox balance in cells.

Before Going On • Describe the compartments of a mitochondrion. • List the transport proteins that occur in the inner mitochondrial membrane. • Draw a simple diagram showing the electron-transport complexes and the mobile carriers that link them. • List the different types of redox groups in the respiratory electron transport chain and identify them as one- or two-electron carriers. • Explain why O2 is the final electron acceptor in the chain. • Describe the operation of a proton wire. • Write an equation to describe the overall redox reaction carried out by each mitochondrial complex. • Compare the arrangement of an electron transport chain and a supercomplex.

15.3 Chemiosmosis  443

15.3 Chemiosmosis LEARNING OBJECTIVES Explain how the protonmotive force links electron transport and ATP synthesis. • Describe the formation of the proton gradient. • Relate the pH difference of the proton gradient to the free energy change.

The electrons collected from metabolic fuels during their oxidation are consumed in the reduction of O2 to H2O. However, their free energy has been conserved. How much free energy is potentially available? Using the ΔG values calculated from the standard reduction potentials of the substrates and products of Complexes I, III, and IV (presented graphically in Fig. 15.2), we can see that each of the three respiratory complexes theoretically releases enough free energy to drive the endergonic phosphorylation of ADP to form ATP (ΔG °′ = +30.5 kJ · mol−1).

Complex I: NADH → QH2

Complex III: QH2 → cytochrome c Complex IV: cytochrome c → O2

ΔG °′ =   −69.5 kJ · mol−1 ΔG °′ =   −36.7 kJ · mol−1 ΔG °′ = −112.0 kJ · mol−1

NADH → O2 ΔG °′ = −218.2 kJ · mol−1

Recall that energy cannot be created or destroyed, but it can be transformed. Understanding oxidative phosphorylation requires recognizing energy in several different forms along the way from metabolic fuels to ATP.

Chemiosmosis links electron transport and oxidative phosphorylation Until the 1960s, the connection between respiratory electron transport (measured as O 2 consumption) and ATP synthesis was a mystery. Many biochemists believed that fuel oxidation generated a “high-­energy” intermediate capable of phosphorylating ADP, but they were unable to identify any such compound. The actual connection between oxidation–­ reduction chemistry and ATP synthesis was discovered by Peter Mitchell, a somewhat unconventional scientist who was inspired by his work on mitochondrial phosphate transport and recognized the importance of compartmentation in biological systems. Mitchell’s chemiosmotic theory proposed that the proton-­translocating activity of the electron transport complexes in the inner mitochondrial membrane generates a proton gradient across the membrane. The protons cannot diffuse back into the matrix because the membrane is impermeable to ions. The imbalance of MATRIX + + + + + + + protons represents a source of energy, also called a protonmotive force, that can + + + + drive the activity of an ATP synthase. I III IV We now know that for each pair of electrons that flow through Complexes I, III, + ++ + + + ++ + + + + ++++ + ++ + + + + + + + ++ + + + +++ + + + ++ + and IV, 10 protons are translocated from the matrix to the intermembrane space (which + + + + + + + + + + + + + + is ionically equivalent to the cytosol). In bacteria, electron transport complexes in the + + + + + + + plasma membrane translocate protons from the cytosol to the cell exterior. Mitchell’s INTERMEMBRANE SPACE theory of chemiosmosis actually explains more than just aerobic respiration. It also applies to systems where the energy from sunlight is used to generate a transmembrane proton gradient (this aspect of photosynthesis is described in Section 16.2).   FIGURE 15.21   Generation of a proton

The proton gradient is an electrochemical gradient When the mitochondrial complexes translocate protons across the inner mitochondrial membrane, the concentration of H+ outside increases and the concentration of H+ inside decreases (Fig. 15.21). This imbalance of protons, a nonequilibrium state,

gradient.  During the oxidation–­reduction reactions catalyzed by mitochondrial Complexes I, III, and IV, protons (represented by positive charges) are translocated out of the matrix into the intermembrane space. This creates an imbalance in both proton concentration and electrical charge.

444  C ha pt er 15   Oxidative Phosphorylation

has an associated energy (the force that would restore the system to equilibrium). The energy of the proton gradient has two components, reflecting the difference in the concentration of the chemical species and the difference in electrical charge of the positively charged protons (for this reason, the mitochondrial proton gradient is referred to as an electrochemical gradient rather than a simple concentration gradient). The free energy change for generating the chemical imbalance of protons is ​[​H+ ​ ​]​ out​ ​ΔG = RT ln ​ _  ​  ​ ​[​H+ ​ ​]​ in​

(15.5)

The pH (–log [H+]) of the intermembrane space (out) is typically about 0.75 units less than the pH of the matrix (in). The free energy change for generating the electrical imbalance of protons is ​ΔG = ZFΔψ​

(15.6)

where Z is the ion’s charge (+1 in this case) and Δ​ψ​is the membrane potential caused by the imbalance in positive charges (see Section 9.1). For mitochondria, Δψ is positive, usually 150 to 200 mV. This value indicates that the intermembrane space or cytosol is more positive than the matrix (recall from Section 9.1 that for a whole cell, the cytosol is more negative than the extracellular space and Δψ is negative). Combining the chemical and electrical effects gives an overall free energy change for transporting protons from the matrix (in) to the intermembrane space (out): ​[​H​+​]​ out​ ​ΔG = RT ln ​ _  ​  + ZFΔψ​ ​[​H+ ​ ​]​ in​

(15.7)

Typically, the free energy change for translocating one proton out of the matrix is about +20 kJ · mol−1 (see Sample Calculation 15.3 for a detailed application of Equation 15.7). This is a thermodynamically costly event. Passage of the proton back into the matrix, following its electrochemical gradient, would have a free energy change of about –20 kJ · mol−1. This event is thermodynamically favorable, but it does not provide enough free energy to drive the synthesis of ATP. However, the 10 protons translocated for each pair of electrons transferred from NADH to O2 have an associated protonmotive force of over 200 kJ · mol−1, enough to drive the phosphorylation of several molecules of ADP.

SA MP L E C A LCULATI ON 1 5 . 3 Problem  Calculate the free energy change for translocating a proton out of the mitochondrial matrix, where pHmatrix = 7.8, pHcytosol = 7.15, Δψ = 170 mV, and T = 25°C. Solution  Since pH = −log [H+] (Equation 2.4), the logarithmic term of Equation 15.7 can be rewritten. Equation 15.7 then becomes

Substituting known values gives ​ΔG = 2.303(8.3145 J · ​K​−1​ · ​mol​−1​)(298 K) (7.8 − 7.15) + ​(1) ​(96,485 J · ​V−1 ​ ​ · ​mol​−1​)​ (0.170 V)​

= 3700 J · ​mol​−1​+ 16,400 J · ​mol​−1​



= + 20.1 kJ · ​mol​−1​

 ​ΔG = 2.303 RT (​pH​ in​ − ​pH​ out​) + ZFΔψ​

SEE SAMPLE CALCULATION VIDEOS

Before Going On • Describe the importance of mitochondrial structure for generating the protonmotive force. • Identify the source of the protons for the transmembrane gradient. • Explain why the proton gradient has a chemical and an electrical component.

15.4  ATP Synthase  445

15.4 ATP Synthase LEARNING OBJECTIVES Describe the structure and operation of ATP synthase. • Recognize the structural components of ATP synthase. • Identify the energy transformations that occur in ATP synthase. • Describe the binding change mechanism. • Explain why P:O ratios are nonintegral. • Explain why oxidative phosphorylation is coupled to electron transport.

The protein that taps the electrochemical proton gradient to phosphorylate ADP is known as the F-­ATP synthase (or Complex V). One part of the protein, called Fo, functions as a transmembrane channel that permits H+ to flow back into the matrix, following its gradient. The F1 component catalyzes the reaction ADP + Pi → ATP + H2O (Fig. 15.22). This section describes the structures of the two components of ATP synthase and shows how their activities are linked so that exergonic H+ transport can be coupled to endergonic ATP synthesis.

Proton translocation rotates the c ring of ATP synthase Not surprisingly, the overall structure of ATP synthase is conserved among different species, although the number of individual protein subunits varies. The F1 component contains three α and three β subunits surrounding a central shaft. The membrane-­embedded portion of ATP synthase includes a variety of subunits. Some of these make up a stalk that extends from the cytoplasmic side of Fo to the top of the F1 component in the matrix (Fig. 15.23). A number of membrane-­embedded c subunits—­ranging from 8 in mammalian mitochondria to as many as 17 in some bacteria—­form a ring that rotates within the lipid bilayer. In all species, proton transport through ATP synthase requires rotation of the c ring past the stationary a subunit. The carboxylate side chain of a highly conserved aspartate or glutamate residue on each c subunit serves as a proton binding site (Fig. 15.24). When properly positioned at the a subunit, a c subunit takes up a proton from the intermembrane space. Proton binding neutralizes the carboxylate group, freeing it from the electrostatic attraction of a positively charged arginine residue on the a subunit. As a result, the protonated c subunit moves away, and this slight rotation of the c ring brings another c subunit into position so that it can release its bound proton into the matrix. In this way, the c ring continues to rotate, translocating protons from the intermembrane space into the matrix (in a proATP ADP + Pi karyote, from outside the cell into the cytosol). The favorable thermodynamics of proton translocation force the c ring to keep moving in one direction. Experiments show that depending on the F1 relative concentrations of protons on the two sides of the membrane, the c ring can actually spin in either direction. Related proteins, known as P-­and H+ V-­ATPases, in fact function as active transporters that use the free energy of MATRIX the ATP hydrolysis reaction to drive ion movement across the membrane. Attached to the c ring and rotating along with it are three additional protein subunits. Two of these are small, but the one known as γ consists of Fo two long α helices arranged as a bent coiled coil that protrudes into the center of the globular F1 structure. The three α and three β subunits of F1 have INTERMEMBRANE similar tertiary structures and are arranged like the sections of an orange SPACE around the γ subunit (Fig. 15.25). Although all six subunits can bind adenine nucleotides, only the β subunits have catalytic activity (nucleotide binding to the α subunits may play a regulatory role).   FIGURE 15.22   ATP synthase function.  As The γ subunit interacts asymmetrically with the three pairs of αβ units. protons flow through the Fo component from the In fact, each αβ unit has a slightly different conformation, and the three units intermembrane space to the matrix, the F1 component cannot simultaneously adopt the same conformation. The three αβ pairs catalyzes the synthesis of ATP from ADP + Pi.

446

C ha ptER 15

Oxidative Phosphorylation

α

α

β

F1

MATRIX

c c

c

γ

c

c

c

c c

Fo

a

INTERMEMBRANE SPACE

a.

b.

Structure of ATP synthase. a. Diagram of the mammalian enzyme. The α and β subunits are connected via a central shaft to the membrane-embedded ring of 8 c subunits. A peripheral stalk links the a subunit to the catalytic domain. b. Ribbon model of the yeast enzyme, with subunits colored as in a. The yeast c ring contains 10 subunits. FIGURE 15.23

MATRIX

H+

c c

c  c c

c H+

c c H+

a

INTERMEMBRANE SPACE

H+

FIGURE 15.24 Mechanism of proton transport by ATP synthase. When a c subunit (pink) binds a proton from one side of the membrane, it moves away from the a subunit (blue). Because the c subunits form a ring, rotation brings another c subunit toward the a subunit, where it releases its bound proton to the opposite side of the membrane.

FIGURE 15.25 Structure of the F1 component of ATP synthase. The alternating α (blue) and β (green) subunits form a hexamer around the end of the γ shaft (purple). This view is looking from the matrix down onto the top of the ATP synthase.

change their conformations as the γ subunit rotates (it is like a shaft driven by the c ring “rotor”). The αβ hexamer itself does not rotate, since it is held in place by the peripheral stalk that is anchored to the a subunit (see Fig. 15.23a). As each proton moves across the membrane, the c ring and γ subunit rotate. Although the c ring may spin smoothly, the γ subunit lurches in steps of 120°, interacting successively with each of the three αβ pairs in one full rotation of 360°. Electrostatic interactions between the γ and β subunits apparently act as a catch that holds the γ subunit in place while translocation of protons builds up strain. The γ subunit is probably too stiff to absorb much rotational strain.

15.4  ATP Synthase  447

Instead, the strain seems to be transmitted to the α and β subunits, which wobble somewhat before the γ subunit suddenly snaps into position at the next β subunit. ATP synthase has 8 c subunits in mammals, 10 in yeast, and 15 in many bacteria. Thus, 8, 10, and 15 proton translocations are required for one complete rotation, and the c ring rotates in increments of 24° to 45° per proton translocated, depending on the number of c subunits.

The binding change mechanism explains how ATP is made At the start of the chapter, we pointed out that ATP synthase catalyzes a highly endergonic reaction (ΔG °′ = +30.5 kJ · mol−1) in order to produce the bulk of a cell’s ATP supply. This enzyme operates in an unusual fashion, using mechanical energy (rotation) to form a chemical bond (the attachment of a phosphoryl group to ADP). In other words, the enzyme converts mechanical energy to the chemical energy of ATP. The interaction between the γ subunit and the αβ hexamer explains this energy transduction. According to the binding change mechanism described by Paul Boyer, rotation-­driven ­conformational changes alter the affinity of each catalytic β subunit for an adenine nucleotide. At any moment, each catalytic site has a different conformation (and binding affinity), referred to as the open, loose, or tight state. ATP synthesis occurs as follows (Fig. 15.26): 1. The substrates ADP and Pi bind to a β subunit in the loose state. 2. The substrates are converted to ATP as rotation of the γ subunit causes the β subunit to shift to the tight conformation. 3. The product ATP is released after the next rotation, when the β subunit shifts to the open conformation.

i

ADP.Pi L O

AT P

ATP T

.Pi P

O

AD

L

ATP

ATP O

L

i

P .P

T

AD

P AT

  FIGURE 15.26   The binding change mechanism.  The diagram shows the catalytic (β)

subunits of the F1 component of ATP synthase from the same perspective as in Fig. 15.25. Each of the three β subunits adopts a different conformation: open (O), loose (L), or tight (T). The substrates ADP and Pi bind to a loose site, ATP is synthesized when the site becomes tight, and ATP is released when the subunit becomes open. The conformational shifts are triggered by the 120° rotation of the γ subunit, arbitrarily represented by the purple shape.

T

ATP

The P:O ratio describes the stoichiometry of oxidative phosphorylation Since the γ shaft of ATP synthase is attached to the c-­subunit rotor, 3 ATP molecules are synthesized for every complete c-­ring rotation. However, the number of protons translocated per ATP depends on the number of c subunits. For mammalian ATP synthase, which has 8 c subunits, the stoichiometry is 8 H+ per 3 ATP, or 2.7 H+ per ATP. Such

L

T

P .P

P AT

O

AD

Because the three β subunits of ATP synthase act cooperatively, they all change their conformations simultaneously as the γ subunit turns. A full rotation of 360° is required to restore the enzyme to its initial state, but each rotation of 120° results in the release of ATP from one of the three active sites. Theoretical calculations show that the rotation-­based mechanism of ATP synthase is more thermodynamically efficient than a back-­and-­forth mechanism like those used by other enzymes and transport proteins. This is because a spinning protein can move in increments, each step requiring a small free energy payment (the binding of one proton), whereas a conventional enzyme with an alternating-­conformation mechanism would require the simultaneous binding of ~3 protons, which has a higher free-­energy cost and would therefore be slower. Experiments with the isolated F1 component of ATP synthase show that in the absence of Fo, F1 functions as an ATPase, hydrolyzing ATP to ADP + Pi (a thermodynamically favorable reaction). In the intact ATP synthase, dissipation of the proton gradient is tightly coupled to ATP synthesis with near 100% efficiency. Consequently, in the absence of a proton gradient, no ATP is synthesized because there is no free energy to drive the rotation of the γ subunit. Agents that dissipate the proton gradient can therefore “uncouple” ATP synthesis from electron transport, the source of the proton gradient (Box 15.B).

ADP + Pi

448  C ha pt er 15   Oxidative Phosphorylation

Box 15.B Uncoupling Agents Prevent ATP Synthesis When the metabolic need for ATP is low, the oxidation of reduced cofactors proceeds until the transmembrane proton gradient builds up enough to halt further electron transport. When the protons reenter the matrix via the Fo component of ATP synthase, electron transport resumes. However, if the protons leak back into the matrix by a route other than ATP synthase, then electron transport will continue without any ATP being synthesized. ATP synthesis is said to be “uncoupled” from electron transport, and the agent that allows the proton gradient to dissipate in this way is called an uncoupler. Some small molecules that act as uncouplers are poisons, but physiological uncoupling does occur. Dissipating a proton gradient prevents ATP synthesis, but it allows oxidative metabolism to continue at a high rate. The by-­product of this metabolic activity is heat. Uncoupling for thermogenesis (heat production) occurs in specialized adipose tissue known as brown fat (its dark color is

due to the relatively high concentration of cytochrome-­containing mitochondria; ordinary adipose tissue is lighter). The inner membrane of the mitochondria in brown fat contains a transmembrane proton channel called a UCP (uncoupling protein). Protons translocated to the intermembrane space during respiration can reenter the mitochondrial matrix via the uncoupling protein, bypassing ATP synthase. The free energy of respiration is therefore given off as heat rather than used to synthesize ATP. Brown fat is abundant in hibernating mammals and newborn humans, and the activity of the UCP is under the control of hormones that also mobilize the stored fatty acids to be oxidized in the brown fat mitochondria. Question  Why would increasing the activity of UCP promote weight loss?

nonintegral values would be difficult to reconcile with most biochemical reactions, but they are consistent with the chemiosmotic theory: Chemical energy (from the respiratory oxidation–­ reduction reactions) is transduced to a protonmotive force, then to the mechanical movement of a rotary engine (the c ring and its attached γ shaft), and finally back to chemical energy in the form of ATP. Note that ATP synthase does not build an ATP molecule from scratch; it simply forms a bond between the second and third phosphate groups. When free energy–­requiring cellular reactions consume ATP, this same bond is most often the one that is broken. The relationship between respiration (the activity of the electron transport complexes) and ATP synthesis is traditionally expressed as a P:O ratio, that is, the number of phosphorylations of ADP relative to the number of oxygen atoms reduced. For example, the oxidation of NADH by O2 (carried out by the sequential activities of Complexes I, III, and IV) translocates 10 protons into the intermembrane space. The movement of these 10 protons back into the matrix via the Fo component would theoretically drive the synthesis of about 3.7 ATP since 1 ATP can be made for every 2.7 protons translocated, at least in mammalian mitochondria: 1 ATP   ​  × 10​ H​+​= 3.7​  ​_ 2.7 ​H+ ​​ Thus, the P:O ratio would be about 3.7 (3.7 ATP per ​​ _12  ​ ​O​  2​​​reduced). For an electron pair originating as QH2, only 6 protons would be translocated (by the activities of Complexes III and IV), and the P:O ratio would be approximately 2.2:  _ ​ 1 ATP+  ​ × 6 ​H+ ​ ​ = 2.2​ 2.7 H ​ ​​

In vivo, the P:O ratios are actually a bit lower than the theoretical values, because some of the protons translocated during electron transport do leak across the membrane or are consumed in other processes, such as the transport of Pi into the mitochondrial matrix (see Fig. 15.6). Consequently, experimentally determined P:O ratios are closer to 2.5 when NADH is the source of electrons and 1.5 for ubiquinol. These values are the basis for our tally of the ATP yield for the complete oxidation of glucose by glycolysis and the citric acid cycle (see Fig. 14.13).

The rate of oxidative phosphorylation reflects the need for ATP The processes of electron transport and oxidative phosphorylation, like other metabolic pathways, maintain a steady state. In animal cells, the proton gradient does not build up beyond the typical 0.75-pH-­unit difference across the membrane, because as soon as protons are pumped into the intermembrane space, they return to the matrix via ATP synthase. Similarly, the cell’s

15.4 ATP Synthase

449

concentration of ATP remains more or less constant because ATP is regenerated as quickly as it is converted to ADP. However, the rates of ATP consumption and synthesis may fluctuate dramatically, depending on the cell’s activity level (Box 15.C).

Box 15.C Powering Human Muscles Cells cannot stockpile ATP; its concentration remains remarkably stable (between 2 and 5 mM in most cells) under widely varying levels of demand. Human muscle cells illustrate the multiple ways that animal cells can generate ATP for different needs. For example, a single burst of activity by the actin–myosin system (Section 5.4) can be powered by the ATP already available in the cell. This ATP supply is boosted by phosphocreatine (introduced in Section 12.3). In resting muscles, when the demand for ATP is low, creatine kinase catalyzes the transfer of a phosphoryl group from ATP to creatine to produce phosphocreatine: ATP + creatine ⇌ phosphocreatine + ADP

This reaction runs in reverse when ADP concentrations rise, as they do when muscle contraction converts ATP to ADP + Pi. Phosphocreatine therefore acts as a sort of phosphoryl-group reservoir to maintain the supply of ATP. Without phosphocreatine, a muscle would exhaust its stock of ATP before it could be replenished by other, slower processes. Muscle activity lasting up to a few seconds requires phosphocreatine, but the amount of phosphocreatine itself is limited, so continued muscle contraction must rely on ATP produced by glycolysis, using glucose obtained from the muscle’s store of glycogen. The end product of this pathway is lactate, the conjugate base of a weak acid, and muscle pain sets in as the acid accumulates and the pH begins to drop. Up to this point, the muscle functions mostly anaerobically (without the participation of O2). Anaerobic systems

ATP

To continue its activity, the muscle must boost aerobic (O2-dependent) metabolism and further oxidize fuels via the citric acid cycle. Sources of acetyl- CoA for the citric acid cycle include additional glucose, delivered by the bloodstream, and fatty acids. Recall that the citric acid cycle generates reduced cofactors that must be reoxidized by molecular oxygen. Aerobic metabolism of glucose and fatty acids is slower than anaerobic glycolysis, but it generates considerably more ATP. Some forms of physical activity and the systems that power them are diagrammed below. A casual athlete can detect the shift from mostly anaerobic to mostly aerobic metabolism after about a minute and a half. In world-class athletes, the breakpoint occurs at about 150 seconds, which corresponds roughly to the finish line in a 1000-meter race. The muscles of sprinters have a high capacity for anaerobic ATP generation, whereas the muscles of distance runners are better adapted to produce ATP aerobically. Such differences in energy metabolism are visibly manifest in the flight muscles of birds. Migratory birds such as geese, which power their long flights primarily with fatty acids, have large numbers of mitochondria to carry out oxidative phosphorylation. The reddish-brown mitochondria give the flight muscles a dark color. Birds that rarely fly, such as chickens, have fewer mitochondria and lighter-colored muscles. When these birds do fly, it is usually only a short burst of activity that is powered by anaerobic mechanisms. Question Why do some athletes believe that creatine supplements boost their performance? Aerobic systems

high jump power lift shot put tennis serve Phosphocreatine

sprints football line play

Glycolysis

200 – 400 m race 100 m swim

race beyond 500 m

Oxidative phosphorylation

4s

10 s

1.5 min Duration of activity

3 min

450

C ha ptER 15

Oxidative Phosphorylation

FIGURE 15.27 IF1 bound to the F1 portion of ATP synthase. A 41-residue helical segment of IF1 (gold) inserts between an α (blue) and β (green) subunit. The view is from the bottom of F1 (as shown in Fig. 15.25), with the γ subunit in purple.

FIGURE 15.28 ATP synthase dimer. The orientation of the two enzymes (orange and yellow) induces a sharp bend in the membrane (black lines).

The close coupling between generation of the proton gradient and ATP synthesis allows oxidative phosphorylation to be regulated by the availability of reduced cofactors (NADH and QH 2) produced from metabolic fuels. This supply- and- demand system differs from other metabolic pathways, where control is exerted at one or a few irreversible, highly exergonic steps. The rate-limiting step of oxidative phosphorylation, the reaction catalyzed by Complex IV, is not a major control point, although there is evidence that ATP, the product of oxidative phosphorylation, allosterically inhibits Complex IV. Complexes I– IV and ATP synthase undergo phosphorylation catalyzed by protein kinases (Section 10.3), but the functional significance of these modifications is not yet understood. In eukaryotes, ATP synthase itself is regulated by a small protein called inhibitory factor 1 (IF1). Different forms of IF1—all less than 100 residues—are intrinsically disordered (see Section 4.3) at relatively high matrix pH values, when the electron transport chain is pumping protons into the intermembrane space and the proton gradient is steep. When the pH drops, a sign that the protonmotive force is waning, IF1 dimerizes and forms extended α helices that insert in between the α and β subunits of F1 and contact the γ shaft (Fig. 15.27). IF1 binding prevents ATP synthase from carrying out the binding change mechanism. IF1 binding to ATP synthase may also stabilize ATP synthase supercomplexes containing two or four copies of the enzyme. The V-shaped synthase dimer appears to cause the inner mitochondrial membrane to bend by about 70–90° (Fig. 15.28). Rows of ATP synthase dimers may be responsible for the sharply curved edges and tubular shapes of cristae (see Fig. 15.4). Alternatively, these arrays might be just an efficient way to pack ATP synthase—which accounts for about 20% of mitochondrial membrane protein—into a tight space.

Before Going On • Draw a simple diagram of ATP synthase and indicate which parts are stationary and which rotate. • Explain how ATP synthase dissipates the proton gradient. • Recount how the three conformational states of the β subunits of ATP synthase are involved in ATP synthesis. • Explain how ATP synthase could operate in reverse to hydrolyze ATP. • Explain why the number of protons translocated per ATP synthesized varies among species. • Explain why the availability of reduced substrates is the primary mechanism for regulating oxidative phosphorylation.

Problems  451

Summary 15.1  The Thermodynamics of Oxidation–­ Reduction Reactions •  The electron affinity of a substance participating in an oxidation–­reduction reaction, which involves the transfer of electrons, is indi­cated by its reduction potential, Ɛ  °′. •  The difference in reduction potential between substances undergoing oxidation and reduction is related to the free energy change for the reaction.

15.2  Mitochondrial Electron Transport •  Oxidation of reduced cofactors generated by metabolic reactions takes place in the mitochondrion. Shuttle systems and transport proteins allow the transmembrane movement of reducing equivalents, ATP, ADP, and Pi. •  The electron transport chain consists of a series of integral membrane protein complexes that contain multiple redox groups, including iron–­sulfur clusters, flavins, cytochromes, and copper ions, and that are linked by mobile electron carriers. Starting from NADH, electrons travel a path of increasing reduction potential through Complex I, ubiquinone, Complex III, cytochrome c, and then to Complex IV, where O2 is reduced to H2O.

15.3  Chemiosmosis •  The chemiosmotic theory describes how proton translocation during mitochondrial electron transport generates an electrochemical gradient whose free energy drives ATP synthesis.

15.4  ATP Synthase •  The energy of the proton gradient is tapped as protons sponta­ neously flow through ATP synthase. Proton transport allows rotation of a ring of integral membrane c subunits. The linked γ subunit thereby rotates, triggering conformational changes in the F1 portion of ATP synthase. •  According to the binding change mechanism, the three functional units of the F1 portion cycle through three conformational states to sequentially bind ADP and P i, convert the substrates to ATP, and release ATP. •  The P:O ratio quantifies the link between electron transport and oxidative phosphorylation in terms of the ATP synthesized and the O2 reduced. Because these processes are coupled, the rate of oxidative phosphorylation is controlled primarily by the availability of reduced cofactors from fuel metabolism.

•  As electrons are transferred, protons are translocated to the intermembrane space via proton wires in Complexes I and IV and by the action of the Q cycle associated with Complex III.

Key Terms oxidative phosphorylation oxidizing agent (oxidant) reducing agent (reductant) half-­reaction standard reduction potential (Ɛ °′) reduction potential (Ɛ) Nernst equation Faraday constant (F ) cellular respiration electron transport chain

mitochondrion mitochondrial matrix intermembrane space cristae electron tomography redox center proton wire cytochrome Q cycle supercomplex

respirasome reactive oxygen species free radical chemiosmotic theory protonmotive force Z binding change mechanism uncoupler thermogenesis P:O ratio

Bioinformatics Brief Bioinformatics Exercises 15.1  Viewing and Analyzing Complexes I–­IV

15.2  D iversity of the Electron-­Transport Chain and the KEGG Database

Problems 15.1  The Thermodynamics of Oxidation– Reduction Reactions

2.  Identify the oxidized and reduced forms from the following 4– pairs: a. Fe(CN)3– 6 /Fe(CN) 6 , b. H 2O 2/O 2, c. FMN/FMNH 2 (see Fig 15.8), d. α-ketoglutarate/isocitrate.

1.  Identify the oxidized and reduced forms from the following pairs: a. malate/oxaloacetate, b. pyruvate/lactate, c. NADH/ NAD+, d. fumarate/succinate.

3.  Identify the oxidized and reduced forms from the following pairs: a. O 2 /H 2 O, b. NO 2– /NO 3– , c. dihydrolipoic acid/lipoic acid, d. NADP+/NADPH.

452  C ha pt er 15   Oxidative Phosphorylation 4.  Identify the oxidized and reduced forms from the following pairs: a. ethanol/acetate, b. glucose-6-phosphate/6-phosphogluconate, c. 1,3-bisphosphoglycerate/glyceraldehyde-3-phosphate,  d.  ubiquinone/ubiquinol. 5.  Calculate the reduction potential of NADP + at 25°C when [NADP+] = 20 µM and [NADPH] = 200 µM. 6.  Calculate the reduction potential of FAD at 25°C when the concentration of FAD is 1 µM and the concentration of FADH2 is 14 µM. 7.  Calculate the reduction potential of fumarate at 37°C when [fumarate] = 80 µM and [succinate] = 100 µM. 8.  Calculate the reduction potential of ubiquinone (Q) at 37°C when the concentration of Q is 20 µM and the concentration of ubiquinol (QH2) is 5 µM. 9. a. Calculate the standard reduction potential of substance A when Ɛ = 0.47 V at 25°C, [Areduced] = 5 µM, and [Aoxidized] = 200 µM. Assume that n = 2. b. Consult Table 15.1. What is a possible identity of substance A? 10. a. Calculate the standard reduction potential of substance B when Ɛ = –0.62 V at 25°C, [Breduced] = 50 µM, and [Boxidized] = 2 µM. Assume that n = 2. b. Consult Table 15.1. What is a possible identity of substance B? 11. a. Calculate the standard reduction potential of substance C when Ɛ = 0.29 V at 25°C, [Creduced] = 5 µM, and [Coxidized] = 0.1 µM. Assume that n = 1. b. Consult Table 15.1. What is a possible identity of substance C? 12. a. Calculate the standard reduction potential of substance D when Ɛ = –0.334 V at 25°C, [Dreduced] = 0.1 mM, and [Doxidized] = 0.3 mM. Assume that n = 2. b. Consult Table 15.1. What is a possible identity of substance D? 13.  Calculate the standard free energy change for the reduction of pyruvate by NADH. Consult Table 15.1 for the relevant halfreactions. Is this reaction spontaneous under standard conditions? 14.  Calculate the standard free energy change for the oxidation of FADH2 by oxygen. Consult Table 15.1 for the relevant half-reactions. Is this reaction spontaneous under standard conditions? 15.  Calculate the standard free energy change for the reduction of aceto­ acetate by NADH. Consult Table 15.1 for the relevant ­half-reactions. Is this reaction spontaneous under standard conditions? 16.  Calculate the standard free energy change for the reduction of cytochrome c by cytochrome c1. Consult Table 15.1 for the relevant half-reactions. Is this reaction spontaneous under standard conditions? 17.  Calculate the standard free energy change for the reduction of cytochrome a3 by cytochrome a. Consult Table 15.1 for the relevant half-reactions. Is this reaction spontaneous under standard conditions? 18.  Calculate the standard free energy change for the reduction of oxygen by cytochrome a3. Consult Table 15.1 for the relevant halfreactions. Is this reaction spontaneous under standard conditions? 19.  Calculate the standard free energy change for the oxidation of malate by ubiquinone. Is the reaction spontaneous under standard conditions?

NAD+ reoxidizes E3. Calculate ΔG °′ for the electron transfer from dihydrolipoic acid to NAD+. 22.  Each electron from cytochrome c is donated to a CuA redox center in Complex IV. The Ɛ °′ value for the CuA redox center is 0.245 V. Calculate ΔG °′ for this electron transfer. 23.  Acetaldehyde may be oxidized to acetate. Would NAD+ be an effective oxidizing agent? Explain, supporting your answer with appropriate calculations. 24.  Acetoacetate may be reduced to 3-hydroxybutyrate. What serves as a better reducing agent, a. NADH or b. FADH2? Explain, supporting your answer with appropriate calculations. 25.  During the Q cycle, ubiquinol is oxidized and cytochrome c1 is reduced. This process can be summarized by the equation QH2 + 2 cyt c1 (Fe3+) → Q + 2 cyt c1 (Fe2+) + 2 H+

Calculate the free energy change for this process under standard conditions. 26.  Why is succinate oxidized by FAD instead of by NAD+? Explain, supporting your answer with the appropriate calculations. 27. a. What is the ΔƐ value for the oxidation of ubiquinol (QH2) by cytochrome c at 25°C when the ratio of QH2/Q is 10 and the ratio of cyt c (Fe3+)/cyt c (Fe2+) is 5? b. Calculate ΔG for the reaction described in part a. Is the reaction spontaneous under these conditions? 28. a. What is the ΔƐ value for the oxidation of cytochrome c by the CuA redox center in Complex IV (see Problem 22) when the ratio of cyt c (Fe2+)/cyt c (Fe3+) is 20 and the ratio of CuA (Cu2+)/CuA (Cu+) is 3? b. Calculate ΔG for the reaction described in part  a.  Is the reaction spontaneous under these conditions? 29.  An iron–sulfur protein in Complex III donates an electron to cytochrome c1. The reduction half-reactions and Ɛ °′ values are shown below. Write the balanced equation for the reaction and calculate the standard free energy change. How can you account for the fact that this reaction occurs spontaneously in the cell?   FeS (ox) + e− → FeS (red)    Ɛ °′ = 0.280 V

 cyt c1 (Fe3+) + e− → cyt c1 (Fe2+)   Ɛ °′ = 0.215 V

30.  If the ΔG value for the reaction described in Problem 29 is –6.0 kJ ∙ mol–1, what is the value of ΔƐ ?

15.2  Mitochondrial Electron Transport 31.  Calculate the overall efficiency of oxidative phosphorylation, assuming standard conditions, by comparing the a. free energy potentially available from the oxidation of NADH by O2 and b. the free energy required to synthesize 2.5 ATP from 2.5 ADP. 32. a. Write the equation for the overall reaction that occurs in Complex I. b. Calculate ΔG°′ for the reaction. c. Using the percent efficiency calculated in Problem 31, calculate the number of ATP generated by Complex I. 33. a. Write the equation for the overall reaction that occurs in Complex II. b. Calculate ΔG°′ for the reaction. c. Using the percent efficiency calculated in Problem 31, calculate the number of ATP generated by Complex II.

20.  In yeast, alcohol dehydrogenase reduces acetaldehyde to ethanol (Section 13.1). Calculate the free energy change for this reaction under standard conditions.

34. a. Write the equation for the overall reaction that occurs in Complex III. b. Calculate ΔG°′ for the reaction. c. Using the percent efficiency calculated in Problem 31, calculate the number of ATP generated by Complex III.

21.  In one of the final steps of the pyruvate dehydrogenase reaction (see Section 14.1), E3 reoxidizes the lipoamide group of E2, then

35. a. Write the equation for the overall reaction that occurs in Complex IV. b. Calculate ΔG °′ for the reaction. c. Using the percent

Problems  453 efficiency calculated in Problem 31, calculate the number of ATP generated by Complex IV.

46.  What side chains would you expect to find as part of a proton wire in a proton-translocating membrane protein?

36.  Use your answers to Problems 32–35 to determine the amount of ATP generated when a. NADH and b. FADH2 are oxidized in the electron transport chain. c. Do these values agree with those presented in Figure 14.13?

47.  Ubiquinone is not anchored in the mitochondrial membrane but is free to diffuse laterally throughout the membrane among the electron transport chain components. What aspects of its structure account for this behavior?

37.  The sequence of events in electron transport was elucidated in part by the use of inhibitors that block electron transfer at specific points along the chain. For example, adding rotenone (a plant toxin) or amytal (a barbiturate) blocks electron transport in Complex I; antimycin A (an antibiotic) blocks electron transport in Complex III; and cyanide (CN–) blocks electron transport in Complex IV by binding to the Fe2+ in the Fe–Cu binuclear center of cytochrome a3. What happens to oxygen consumption when these inhibitors are added to a suspension of respiring mitochondria?

48.  Explain why the ubiquinone-binding site of Complex I (Fig. 15.7) is located at the end of the peripheral arm closest to the membrane.

38.  What is the redox state of the electron carriers in the electron transport chain when each of the inhibitors described in Problem 37 is added separately to a mitochondrial suspension? 39. a. The investigators who first studied the effect of rotenone on electron transport (see Problem 37) noted that the inhibitor blocked the oxidation of malate but not succinate. How did these results assist in the identification of rotenone as a Complex I inhibitor? b. If the same experiments described in part a are carried out in the presence of antimycin A, what are the expected results? 40.  When the antifungal agent myxothiazol is added to a suspension of respiring mitochondria, the QH2/Q ratio increases. Where in the electron transport chain does myxothiazol inhibit electron transfer? 41.  What is the effect of added succinate on rotenone-blocked, antimycin A–blocked, or cyanide-blocked mitochondria (see Problem 37)? In other words, can succinate help “bypass” any of these blocks? Explain. 42. a. The compound tetramethyl-p-phenylenediamine (TMPD) donates a pair of electrons directly to Complex IV. Can TMPD act as a bypass for the rotenone-blocked, antimycin-blocked, or cyanide-blocked mitochondria described in Problem 37? b. Ascorbate (vitamin C) can donate a pair of electrons to cytochrome  c.  Can ascorbate act as a bypass for the rotenone-blocked, antimycin-blocked, or cyanide-blocked mitochondria described in Problem 37? 43.  If cyanide poisoning (see Problem 37) is diagnosed immediately, it can be treated by administering nitrites that can oxidize the Fe2+ in hemoglobin to Fe3+. Why is this treatment effective? 44.  The effect of the drug fluoxetine (Prozac®; Box 9.B) on isolated rat brain mitochondria was examined by measuring the rate of electron transport (units not given) in the presence of various combinations of substrates and inhibitors (see Problems 37–40). a. How do pyruvate, malate, and succinate serve as substrates for electron transport? b. What is the effect of fluoxetine on electron transport? Rate of electron transport [Fluoxetine] (mM)

pyruvate + malate

succinate + rotenone

ascorbate + TMPD

0

160

140

180

0.15

 80

130

120

45.  Complex I, succinate dehydrogenase, acyl-CoA dehydrogen­ ase, and glycerol-3-phosphate dehydrogenase (see Fig. 15.11) are all flavoproteins; that is, they contain an FMN or FAD prosthetic group. Explain the function of the flavin group in these enzymes. Why are the flavoproteins ideally suited to transfer electrons to ubiquinone?

49.  Cytochrome c is easily dissociated from isolated mitochondrial membrane preparations, but the isolation of cytochrome c1 requires the use of strong detergents. Explain why. 50.  Release of cytochrome c from the mitochondrion to the cytosol is one of the signals that induces apoptosis, a form of programmed cell death. What structural features of cytochrome c allow it to play this role? 51.  In coastal marine environments, high concentrations of nutrients from terrestrial runoff may lead to algal blooms. When the nutrients are depleted, the algae die and sink and are degraded by other microorganisms. The algal die-off may be followed by a sharp drop in oxygen in the depths, which can kill fish and bottom-dwelling invertebrates. Why do these “dead zones” form? 52.  Chromium is most toxic and highly soluble in its oxidized Cr(VI) state but is less toxic and less soluble in its more reduced Cr(III) state. Efforts to detoxify Cr-contaminated groundwater involve injecting chemical reducing agents underground. Another approach is bioremediation, which involves injecting molasses or cooking oil into the contaminated groundwater. Explain how these substances would promote the reduction of Cr(VI) to Cr(III). 53.  At one time, it was believed that myoglobin functioned simply as an oxygen-storage protein. New evidence suggests that myoglobin plays a much more active role in the muscle cell. The phrase myoglobin-facilitated oxygen diffusion describes myoglobin’s role in transporting oxygen from the muscle cell sarcolemma to the mitochondrial membrane surface. Mice in which the myoglobin gene was knocked out had higher tissue capillary density, elevated red blood cell counts, and increased coronary blood flow. Explain the reasons for these compensatory mechanisms in the knockout mice. 54.  The myoglobin and cytochrome c oxidase content were determined in several animals—hare, sheep, ox, and horse. The levels of both proteins were roughly correlated, i.e., the higher the myoglobin content, the greater the cytochrome c oxidase activity. Explain the relationship between these two proteins. 55.  Myoglobin has an affinity for membrane phospholipids.  a.  Which amino acid side chains are expected to interact favorably with phospholipid head groups? b. An experiment showed that the p50 value of myoglobin increased when myoglobin was incubated with a suspension of mitochondria. What is the physiological benefit of the p50 increase that occurs when myoglobin binds to phospholipids? 56.  Myoglobin is not confined to muscle cells. Tumor cells, which generally exist in hypoxic (low oxygen) conditions because of limited blood flow, express myoglobin. How does this adaptation increase the chances of tumor cell survival? 57.  Cancer cells, even when sufficient oxygen is available, produce large amounts of lactate. It has been observed that the concentration of fructose-2,6-bisphosphate is much higher in cancer cells than in normal cells. Why would this result in anaerobic metabolism being favored, even when oxygen is available? 58.  Hawkmoths are nectar-feeding insects with a high rate of aerobic metabolism. Some of the sugar they ingest is not catabolized by

454  C ha pt er 15   Oxidative Phosphorylation glycolysis to make ATP but instead is directed into the pentose phosphate pathway. Explain why this metabolic diversion helps the insect avoid damage from reactive oxygen species generated during electron transport. 59.  A group of elderly patients who did not exercise regularly were asked to participate in a 12-week exercise program. At the end of the study, total mitochondrial DNA and Complex II activity in the patients’ muscle cells had increased by 50%. The activity of the electron transport chain as a whole doubled over the 12-week period. Why did exercise intervention bring about these changes? 60.  Amyotrophic lateral sclerosis (ALS) is a neurodegenerative disease that causes muscle paralysis and eventually death. Researchers measured the activity of the electron transport chain complexes in various regions of the nervous system in patients with ALS. In a certain region of the spinal cord, Complex I showed decreased activity but not decreased concentration. How does this contribute to progression of the disease?

15.3  Chemiosmosis 61. a. How many protons are transferred from the matrix to the intermembrane space during the transport of two electrons through each of the complexes? b. What is the free energy change associated with the passage of a pair of electrons from NADH to O2 if the passage of one proton from the matrix to the intermembrane space costs 20.1 kJ ∙ mol–1, as shown in Sample Calculation 15.3? c. Compare your answer in part b to your answer to Problem 31a and comment on the significance of the comparison. 62. a. What is the free energy change associated with the passage of a pair of electrons from FADH2 to O2 if the passage of one proton from the matrix to the intermembrane space costs 20.1 kJ ∙ mol–1, as shown in Sample Calculation 15.3? b. Compare your answer in part a to your answer to Problem 14 and comment on the significance of the comparison. 63.  What is the free energy change for generating the electrical imbalance of protons in neuroblastoma cells, where Δψ is 81 mV? 64.  What is the free energy change for generating the electrical imbalance of protons in respiring mitochondria in culture, where Δψ is 150 mV? 65.  Calculate the free energy change for translocating a proton out of the mitochondrial matrix, where pHmatrix = 7.6, pHcytosol = 7.2, Δψ = 200 mV, and T = 37°C. 66.  Calculate the free energy change for transporting a proton out of the mitochondrial matrix when pHmatrix = 7.55, pHcytosol = 7.35, Δψ = 170 mV, and T = 37°C. 67.  What size of pH gradient (the difference between pHmatrix and pHcytosol) would correspond to a free energy change of 30.5 kJ ∙ mol–1? Assume that Δψ = 170 mV and T = 25°C. 68.  What size of pH gradient (the difference between pHmatrix and pHcytosol) would correspond to a free energy change of 19.2 kJ ∙ mol–1? Assume that Δψ = 170 mV and T = 37°C.

important? b. Could ATP still be synthesized if the membrane were permeable to other ions? 71.  Nigericin is an antibiotic that integrates into membranes and functions as a K+/H+ antiporter. Another antibiotic, valinomycin, is similar, but it allows the passage of K+ ions. When both antibiotics are added simultaneously to suspensions of respiring mitochondria, the electrochemical gradient completely collapses. a. Draw a diagram of a mitochondrion in which nigericin and valinomycin have integrated into the inner mitochondrial membrane, in a manner that is consistent with the experimental results. b. Explain why the electrochemical gradient dissipates. What happens to ATP synthesis? 72.  How does transport of inorganic phosphate from the intermembrane space to the mitochondrial matrix affect the pH difference across the inner mitochondrial membrane? 73.  Pioglitazone, a drug used to treat diabetes, causes some membrane-embedded portions of mitochondrial Complex I to separate from the rest of the protein that includes the matrix “arm.” Predict the effect of pioglitazone on electron transport and ATP production. 74.  Metformin, another diabetes drug, suppresses the activity of mitochondrial glycerol-3-phosphate dehydrogenase. Predict the effect of metformin on electron transport and ATP production.

15.4  ATP Synthase 75.  In experimental systems, the Fo component of ATP synthase can be reconstituted into a membrane. Fo can then act as a proton channel that is blocked when the F1 component is added to the system. What molecule must be added to the system in order to restore the proton-translocating activity of Fo? Explain. 76.  Oligomycin is an antibiotic that blocks proton transfer through the Fo proton channel of ATP synthase. What is the effect on a. ATP synthesis, b. electron transport, and c. oxygen consumption when oligomycin is added to a suspension of respiring mitochondria? 77.  Dicyclohexylcarbodiimide (DCCD) is a reagent that reacts with Asp or Glu residues. Explain why the reaction of DCCD with just one c subunit completely blocks both the ATP-synthesizing and ATP-­ hydrolyzing activity of ATP synthase. 78.  In the 1950s, experiments with isolated mitochondria showed that organic compounds are oxidized and O2 is consumed only when ADP is included in the preparation. When the ADP supply runs out, oxygen consumption halts. Explain these results. 79. a. Calculate the ratio of protons translocated to ATP synthesized for yeast ATP synthase, which has 10 c subunits, and for spinach chloroplast ATP synthase, which has 14 c subunits. b. Would you expect the yeast or the chloroplast to have a higher P:O ratio? 80.  The ATP synthase from bovine heart mitochondria has 8 c subunits. What is the P:O ratio for NADH? 81.  Experiments indicate that the c ring of ATP synthase spins at a rate of 6000 rpm. How many ATP molecules are generated each second?

69.  Several key experimental observations were important in the development of the chemiosmotic theory. Explain how each of these observations is consistent with the chemiosmotic theory as described by Peter Mitchell. a. The pH of the intermembrane space is lower than the pH of the mitochondrial matrix. b. Oxidative phosphorylation does not occur in mitochondrial preparations to which detergents have been added.

82.  The side chain of a glutamate residue normally has a pK value of about 5.0 in solution. The pH of the cytosol is about 7.0. a. Does this information explain how a Glu side chain on a c subunit of ATP synthase could acquire a cytosolic proton and then release it into the matrix (where the pH is about 8.0)? b. Within a few angstroms of the Glu side chain are two additional Glu side chains. Explain how these two residues could promote protonation of the first Glu residue.

70.  Mitchell’s original chemiosmotic hypothesis relies on the impermeability of the inner mitochondrial membrane to ions other than H +, such as Na + and Cl –. a. Why was this thought to be

83.  How much ATP can be obtained by the cell from the complete oxidation of one mole of glucose? Compare this value with the amount of ATP obtained when glucose is anaerobically converted to

Problems  455 lactate or ethanol. Do organisms that can completely oxidize glucose have an advantage over organisms that cannot?

conclude about the thermodynamic force that drives the two transport systems?

84.  During anaerobic fermentation in yeast, the majority of the available glucose is oxidized via the glycolytic pathway and the rest enters the pentose phosphate pathway to generate NADPH and ribose. This occurs during aerobic respiration as well, except that the percentage of glucose entering the pentose phosphate pathway is much greater in aerobic respiration than during anaerobic fermentation. Explain why.

96.  The ADP/ATP carrier, which exchanges cytoplasmic ADP and mitochondrial ATP, can also function as a passive proton transporter. a. Would the carrier protein augment or diminish the protonmotive force? b. Researchers found that nucleotide transport inhibits proton transport by the carrier protein. Could this competitive effect help link the rate of oxidative phosphorylation to the cell’s need for ATP?

85.  When cells cannot carry out oxidative phosphorylation, they synthesize ATP through substrate-level phosphorylation. a. Which enzymes of glycolysis and the citric acid cycle catalyze substrate-level phosphorylation? b. Compare the amount of ATP generated per mole of glucose via substrate-level phosphorylation and oxidative phosphorylation. 86.  The glycerol-3-phosphate shuttle can transport cytosolic NADH equivalents into the mitochondrial matrix (see Fig. 15.11c). In this shuttle, the protons and electrons are donated to FAD, which is reduced to FADH2. These protons and electrons are subsequently donated to coenzyme Q in the electron transport chain. a. How much ATP is generated per mole of glucose when the glycerol-3-­ phosphate shuttle is used? b. What is the P:O ratio for cytosolic NADH? 87.  A culture of yeast grown under anaerobic conditions is exposed to oxygen, resulting in a dramatic decrease in glucose consumption by the cells. This phenomenon is referred to as the Pasteur effect. a. Explain the Pasteur effect. b. The [NADH]/[NAD +] and [ATP]/[ADP] ratios also change when an anaerobic culture is exposed to oxygen. Explain how these ratios change and what effect this has on glycolysis and the citric acid cycle in the yeast.

97.  Hexokinase II, one of the four isozymes of hexokinase (see Problem 13.4), is upregulated in cancer cells. Recent evidence indicates that during the transformation process, the protein Akt (see Section 10.2) facilitates hexokinase binding to the outer mitochondrial membrane, where it then becomes closely associated with the ADP/ATP carrier. Explain why this process benefits the cancer cell. 98.  The compounds atractyloside and bongkrekic acid both bind tightly to and inhibit the ADP/ATP carrier. How do these compounds affect ATP synthesis? Electron transport? 99.  Lipid-soluble compounds such as dinitrophenol (DNP) uncouple electron transport and oxidative phosphorylation (see Box 15.B). The structure of dinitrophenol is shown below. The pK of the phenolic hydrogen is near neutral. a. How does DNP function as an uncoupler? (Hint: Draw the resonance structures for the phenolate anion.) b. Explain how the ability of substances like DNP to act as uncouplers is consistent with the chemiosmotic theory as described by Peter Mitchell.

OH NO2

88.  Experiments in the late 1970s attributed the Pasteur effect (see Problem 87) to the stimulation of hexokinase and phosphofructo­ kinase under anaerobic conditions. Upon exposure to oxygen, the stimulation of these enzymes ceases. Why are these enzymes more active in the absence of oxygen? 89.  A hormone signal leads to the activation of protein kinase A (see Fig. 10.7), which phosphorylates a Ser residue in IF1. Phosphorylated IF1 cannot bind to ATP synthase. Does the hormone signal instruct the cell to increase or decrease the rate of metabolic fuel oxidation? Explain. 90.  Some evidence suggests that about 2% of the time, ATP synthase translocates a K+ ion rather than a proton into the mitochondrial matrix. Do the cytosolic concentrations of K+ (see Fig. 2.12) and H+ (see Sample Calculation 15.3) support or contradict this claim? 91.  In addition to its effects on electron transport, fluoxetine (see Problem 44) can also inhibit ATP synthase. Why might long-term use of fluoxetine be a concern? 92.  Mutations that impair ATP synthase function are rare. Laboratory studies indicate that adding α-ketoglutarate boosts ATP production in ATP synthase–deficient cells, but only when aspartate is also added to the cells. Explain. 93.  The compound tetramethyl-p-phenylenediamine (TMPD) donates a pair of electrons directly to Complex IV (see Problem 42a). What is the P:O ratio for this compound? 94.  Ascorbate (vitamin C) can donate a pair of electrons to cytochrome c. What is the P:O ratio for ascorbate? 95.  Consider the ADP/ATP carrier and the Pi–H+ symport protein that import ADP and P i, the substrates for oxidative phosphorylation, into the mitochondrion (see Fig. 15.6). a. How does the activity of the ADP/ATP carrier affect the electrochemical gradient across the mitochondrial membrane? b. How does the activity of the Pi–H+ symport protein affect the gradient? c. What can you

NO2 Dinitrophenol 100.  The compound carbonylcyanide-p-trifluoromethoxy phenyl­ hydrazone (FCCP) is an uncoupler similar to DNP (see Problem 99). Describe how FCCP acts as an uncoupler.

N

C

N

C

H C

N

N

OCF3

FCCP

101.  Dinitrophenol (see Problem 99) was introduced as a “diet pill” in the 1920s. Its use was discontinued because the side effects were fatal in some cases. What was the rationale for believing that DNP would be an effective diet aid? 102.  A patient seeks treatment because her metabolic rate is twice normal and her temperature is elevated. A biopsy reveals that her muscle mitochondria are structurally unusual and not subject to normal respiratory controls. Electron transport takes place regardless of the concentration of ADP. a. What is the P:O ratio (compared to normal) for NADH that enters the electron transport chain in the mitochondria of this patient? b. Why are the patient’s metabolic rate and temperature elevated? c. Will this patient be able to carry out strenuous exercise? 103.  UCP1 is an uncoupling protein in brown fat (Box 15.B). Experiments were carried out using UCP1-knockout mice (animals missing the gene for UCP1). a. Oxygen consumption increased over twofold when a β3 adrenergic agonist that stimulates UCP1 was injected into normal mice. This was not observed when the agonist was injected

456  C ha pt er 15   Oxidative Phosphorylation into the knockout mice. Explain these results. b. In one experiment, normal mice and UCP1-knockout mice were placed in a cold (5°C) room overnight. The normal mice were able to maintain their body temperature at 37°C even after 24 hours in the cold. However, the body temperatures of the cold-exposed knockout mice decreased by 10°C or more. Explain. 104.  The Eastern skunk cabbage can maintain its temperature 15–35°C higher than ambient temperature during the months of February and March, when ambient temperatures range from −15 to +15°C. Thermogenesis in the skunk cabbage is critical to the survival of the plant since the spadix (a flower component) is not frost-resistant. An uncoupling protein is responsible for the observed thermogenesis. a. The spadix relies on the skunk cabbage’s massive root system, which stores appreciable quantities of starch. Why is a large quantity of starch required for the skunk cabbage to carry out sustained thermogenesis for weeks rather than hours? b. Oxygen consumption by the skunk cabbage increases as the temperature decreases, nearly doubling with every 10°C drop in ambient temperature. Oxygen consumption was observed to decrease during the day, when temperatures were close to 30°C, and increase at night. What is the biochemical explanation for these observations? 105.  The gene that codes for an uncoupling protein (see Box 15.B) in potatoes was isolated. The results of a Northern blot analysis (which detects mRNA) are shown below. What is your interpretation of these results? How does the mRNA level affect thermogenesis in the potato? 20°C Days:

0

1

2

4°C 3

0

1

2

3

106.  Glutamate can be used as an artificial substrate for mitochondrial respiration, as shown in the diagram. When ceramide is added to a mitochondrial suspension respiring in the presence of glutamate, respiration decreases, leading scientists to hypothesize that ceramide might regulate mitochondrial function in vivo.

Glutamate

glutamate transporter INTERMEMBRANE SPACE MATRIX

glutamate dehydrogenase α-Ketoglutarate Glutamate NAD+ NADH

a.  How does glutamate act as a substrate for mitochondrial respiration? b. Ceramide-induced inhibition of respiration could be due to several different factors. List several possibilities. c. Mitochondria treated with ceramide were exposed to an uncoupler, but the respiration rate did not increase. What site(s) of inhibition can be ruled out? d. In another experiment, mitochondria were subjected to a freeze–thaw cycle that rendered the inner mitochondrial membrane permeable to NADH. NADH could then be added to a mitochondrial suspension as a substrate for electron transport. When NADH was added, ceramide decreased the respiration rate to the same extent as when glutamate was the substrate. What site(s) of inhibition can be ruled out? 107.  In some organisms, starvation leads to an increase in α-ketoglutarate derived from the breakdown of muscle proteins. α-Ketoglutarate binds to and inhibits the β subunit of ATP synthase. How would this mechanism affect O2 consumption during starvation and why might this prolong the organism’s lifespan? 108.  Rapidly growing bacterial cells tend to rely on glycolysis and fermentation rather than oxidative phosphorylation to generate ATP, even when O2 is abundant. One group of researchers noted that glycolysis/fermentation requires fewer proteins than oxidative phosphorylation. Could this observation explain why rapidly growing cells prefer glycolysis over more efficient oxidative phosphorylation?

Selected Readings Ishigami, I., Lewis-­Ballester, A., Echelmeier, A., Brehm, G., Zatsepin, N.A., Grant, T.D., Coe, J.D., Lisova, S., Nelson, G., Zhang, S., Dobson, Z.F., Boutet, S., Sierra, R.G., Batyuk, A., Fromme, P., Fromme, R., Spence, J.C. H., Ros, A., Yeh, S-R., and Rousseau, D.L., Snapshot of an oxygen intermediate in the catalytic reaction of cytochrome c oxidase, Proc. Natl. Acad. Sci. USA 116, 3572–3577, doi: 10.1073/ pnas.1814526116 (2019). [Provides structural data in support of a mechanism in which O2 reduction is coupled to proton transport in Complex IV.] Milenkovic, D., Blaza, J.N., Larsson, N.-G., and Hirst, J., The enigma of the respiratory chain supercomplex, Cell Metabolism 25, 765–776, doi: 10.1016/j.cmet.2017.03.009 (2017). [Describes the supercomplexes and discusses possible reasons for their existence.] Parey, K., Brandt, U., Xie, H., Mills, D.J., Siegmund, K., Vonck, J., Kühlbrandt, W., and Zickermann, V., Cryo-­EM structure of respiratory complex I at work, eLife 7, doi: 10.7554/eLife.39213 (2018). [Presents detailed views of Complex I.]

Shinzawa-­Itoh, K., Sugimura, T., Misaki, T., Tadehara, Y., Yamamoto, S., Hanada, M., Yano, N., Nakagawa, T., Uene, S., Yamada, T., Ao­yama, H., Yamashita, E., Tsukihara, T., Yoshikawa, S., and Muramoto, K., Monomeric structure of an active form of bovine cytochrome c oxidase, Proc. Natl. Acad. Sci. USA 116, 19945–19951, doi: 10.1073/ pnas.1907183116 (2019). [Describes the structure and functional significance of the cytochrome c oxidase (Complex IV) monomer.] Srivastava, A.P., Luo, M., Zhou, W., Symersky, J., Bai, D., Chambers, M.G., Faraldo-­G ómez, J.D., Liao, M., and Mueller, D.M., High-­ resolution cryo-­EM analysis of the yeast ATP synthase in a lipid membrane, Science 360, doi: 10.1126/science.aas9699 (2018). [Pre­ sents the structure of ATP synthase and explains possible mechanisms for proton translocation.] Youle, R.J., Mitochondria—­S triking a balance between host and endosymbiont, Science 365, 655, doi: 10.1126/science.aaw9855 (2019). [Outlines the challenges for eukaryotic cells to mange mitochondrial gene expression and avoid mitochondrial dysfunction.]

Chapter 15 Credits  457

Chapter 15 Credits Table 15.1 Data mostly from Loach, P.A., in Fasman, G.D. (ed.), Handbook of Biochemistry and Molecular Biology (3rd ed.), Physical and Chemical Data, Vol. I, pp. 123–130, CRC Press (1976). Figure 15.7 Image based on 4HEA. Baradaran, R., Berrisford, J.M., Minhas, G.S., Sazanov, L.A., Crystal structure of the entire respiratory complex I, Nature 494, 443–448 (2013). Figure 15.13 Image based on 1BE3. Iwata, S., Lee, J.W., Okada, K., Lee, J.K., Iwata, M., Rasmussen, B., Link, T.A., Ramaswamy, S., Jap, B.K., Complete structure of the 11-subunit bovine mitochondrial cytochrome bc1 complex, Science 281, 64–71 (1998). Figure 15.16 Image based on 5CYT. Takano, T., Refinement of myoglobin and cytochrome c, in Methods and Applications in Crystallographic Computing, S.R. Hall and T. Shida, ed., 262 (1984). Figure 15.17 Image based on 2OCC. Yoshikawa, S., Shinzawa-Itoh, K., Nakashima, R., Yaono, R., Yamashita, E., Inoue, N., Yao, M., Fei, M.J., Libeu, C.P., Mizushima, T., Yamaguchi, H., Tomizaki, T., Tsukihara, T., Redox-coupled crystal structural changes in bovine heart cytochrome c oxidase, Science 280, 1723–1729 (1998).

Figure 15.20 Image based on 5GPN. Gu, J., Wu, M., Guo, R., Yan, K., Lei, J., Gao, N., Yang, M., The architecture of the mammalian respirasome, Nature 537, 639–643 (2016). Figure 15.23b Image based on 6CP6. Srivastava, A.P., Luo, M., Zhou, W., Symersky, J., Bai, D., Chambers, M.G., Faraldo-Gomez, J.D., Liao, M., Mueller, D.M., High-resolution cryo-EM analysis of the yeast ATP synthase in a lipid membrane, Science 360, 619 (2018). Figures 15.25 and 15.27 Images based on 4TSF. Bason, J.V., Montgomery, M.G., Leslie, A.G., Walker, J.E., Pathway of binding of the intrinsically disordered mitochondrial inhibitor protein to F1-ATPase, Proc. Natl. Acad. Sci. USA 111, 11305 (2014). Box 15.C Figure adapted from McArdle, W.D., Katch, F.I., and Katch, V.L., Exercise Physiology (2nd ed.), p. 348, Lea & Febiger (1986). Figure 15.28 Image based on 6J5K. Gu, J., Zhang, L., Zong, S., Guo, R., Liu, T., Yi, J., Wang, P., Zhuo, W., Yang, M., Cryo-EM structure of the mammalian ATP synthase tetramer bound with inhibitory protein IF1, Science 364, 1068–1075 (2019).

CHAPTER 16

Photosynthesis DO YOU REMEMBER? • Glucose polymers include the fuel-storage polysaccharides starch and glycogen and the structural polysaccharide cellulose (Section 11.2). • Coenzymes such as NAD+ and ubiquinone collect electrons from compounds that become oxidized (Section 12.2). Cells on the surface of the petals of the California poppy synthesize extra cellulose so that their outer cell walls form thick, triangular ridges. These ridges act like prisms to focus light onto pigments at the base of the cell, producing intense color with a silk-like reflectiveness that attracts pollinators.

• Electrons are transferred from a substance with a lower reduction potential to a substance with a higher reduction potential (Section 15.1). • The formation of a transmembrane proton gradient during electron transport provides the free energy to synthesize ATP (Section 15.3).

Every year, plants and bacteria convert an estimated 6 × 1016 grams of carbon to organic compounds by photosynthesis. About half of this activity occurs in forests and savannas, and the rest occurs in the ocean and under ice—wherever water, carbon dioxide, and light are available. The organic materials produced by photosynthetic organisms sustain them as well as the organisms that feed on them. We will begin by examining the absorption of light energy and then look at the electron transport complexes that convert solar energy to biologically useful forms of energy such as ATP and the reduced cofactor NADPH. Finally, we will see how plants use these energy currencies to synthesize carbohydrates.

16.1 Chloroplasts and Solar Energy LEARNING OBJECTIVES Describe the structure and purpose of pigment molecules. • Relate a pigment’s color to the energy of the light it absorbs. • List the ways that absorbed energy can be dissipated. • Explain how absorbed light energy is transferred to the reaction center.

458

The ability to use sunlight as an energy source evolved about 3.5 billion years ago. Before that, cellular metabolism probably centered around the inorganic reductive reactions associated with hydrothermal vents. The first photosynthetic organisms produced various pigments (light-absorbing molecules) to capture solar energy and thereby drive the reduction of metabolites. The descendants of some of these organisms are known today as purple bacteria and green sulfur bacteria. By about 2.5 billion years ago, the cyanobacteria had evolved. These organisms absorb enough solar energy to undertake the energetically costly oxidation

16.1 Chloroplasts and Solar Energy FIGURE 16.1 Photosynthesis in context. Photosynthetic organisms incorporate atmospheric CO2 into three-carbon compounds that are the precursors of biological molecules such as carbohydrates and amino acids. Photosynthesis requires light energy to drive the production of the ATP and NADPH consumed in the biosynthetic reactions.

of water to molecular oxygen. In fact, the dramatic increase in the level of atmospheric oxygen (from an estimated 1% to the current level of about 20%) around 2.1 to 2.4 billion years ago is attributed to the rise of cyanobacteria. Modern plants are the result of the symbiosis of early eukaryotic cells with cyanobacteria. Although the apparatus and reactions of photosynthesis are not found in all organisms, they can be placed in the context of the metabolic scheme outlined in Chapter 12 (Fig. 16.1). As you examine the harvest of solar energy and its use in incorporating CO2 into three-carbon compounds, you will see that significant portions of these processes resemble metabolic pathways that you have already encountered. Photosynthesis in green plants takes place in chloroplasts, discrete organelles that are descended from cyanobacteria. Like mitochondria, chloroplasts contain their own DNA, in this case coding for 100 to 200 chloroplast proteins. DNA in the cell’s nucleus contains close to a thousand more genes whose products are essential for photosynthesis. The chloroplast is enclosed by a porous outer membrane and an ionimpermeable inner membrane (Fig. 16.2). The inner compartment, called the stroma, is analogous to the mitochondrial matrix and is rich in enzymes, including those required for carbohydrate synthesis. Within the stroma is a membranous structure called the thylakoid. Unlike the planar or tubular mitochondrial cristae (see Fig. 15.4), the thylakoid membrane folds into stacks of flattened vesicles and encloses a compartment called the thylakoid lumen. The energy-transducing reactions of photosynthesis take place in the thylakoid membrane. The analogous reactions in photosynthetic bacteria typically take place in folded regions of the plasma membrane.

BIOPOLYMERS Proteins Nucleic acids Polysaccharides Triacylglycerols

MONOMERS Amino acids Nucleotides Monosaccharides Fatty acids +

NH4

NADH

NAD+, Q

hc E = ___ (16.1) λ where h is Planck’s constant (6.626 × 10–34 J · s), c is the speed of light (2.998 × 108 m · s–1), and λ is the wavelength (about 400 to 700 nm for visible light; see Sample Calculation 16.1). This energy is absorbed by the photosynthetic apparatus of the chloroplast and transduced to chemical energy.

NADH, QH2

Dr. Jeremy Burgess/Photo Researchers, Inc.

ADP oxidative phosphorylation ATP

H2O

Stroma Thylakoid membrane

b.

FIGURE 16.2 The chloroplast. a. Electron micrograph of a chloroplast from tobacco. b. Drawing of a chloroplast. The stacked thylakoid membranes are known as grana (singular, granum).

Question Compare these images to the images of a mitochondrion in Figure 15.4.

photosynthesis

CO2

Thylakoid lumen

a.

citric acid cycle

O2

Light can be considered as both a wave and a particle, the photon. The energy (E) of a photon depends on its wavelength, as expressed by Planck’s law:

Outer membrane

NADH, QH2 2- and 3-Carbon INTERMEDIATES

Pigments absorb light of different wavelengths

Inner membrane

NAD+

NAD+

NAD+, Q

459

460  C ha pter 16   Photosynthesis

S A MP L E C A LCU LATIO N 16. 1

SEE SAMPLE CALCULATION VIDEOS

Problem  Calculate the energy of a photon with a wavelength of 550 nm. Solution hc ​  ​ E = ​ ___ λ (6.626 × ​10​ −34​ J · s) (2.998 × ​10​ 8​m · ​s​ −1​) _________________________________      ​ E =      ​  550 × ​10​ −9​m E = 3.6 × ​10​ −19​J​



Chloroplasts contain a variety of light-absorbing groups called pigments or photoreceptors (Fig. 16.3). Chlorophyll is the principal photoreceptor. It appears green because it absorbs both blue and red light. Chlorophyll resembles the heme groups of hemoglobin and cytochromes (see Fig. 15.12), but it has a central Mg2+ rather than an Fe2+ ion; it includes a fused cyclopentane ring, and it has a long lipid side chain. The second most common

CH2

a.

CH H3C

CH3

N

CH2

N

CH3

Mg N

H3C

N

CH3

CH2 H3C

H3C

H3C

H3C

O

CH2

C

C

O

O

CH3

O

O

CH3 Chlorophyll a

b.

CH3

H3C CH3

H3C

CH3 CH3

β-Carotene

CH3 c.

–

CH3

H3C CH3

COO–

OOC

CH3

CH2 CH2

CH3 CH CH3

CH2 CH2

CH3 CH3 CH3

CH2

H O

N H

C H

N H

C H

N H

C H

N H

O

Phycocyanin   FIGURE 16.3   Some common chloroplast photoreceptors.   a. Chlorophyll a. In chlorophyll b, a

methyl group (blue) is replaced by an aldehyde group.  b. The carotenoid ­β-carotene, a precursor of vitamin A (see Box 8.B).  c.  Phycocyanin, a linear tetrapyrrole. It resembles an unfolded chlorophyll molecule.

16.1  Chloroplasts and Solar Energy  461

Chlorophyll a

Solar radiation

Chlorophyll b

Absorbance

Carotenoids

Phycocyanin

400

500 Wavelength (nm)

600

700

  FIGURE 16.4   Visible light absorption by some photosynthetic pigments.  The

wavelengths of absorbed light correspond to the peak of the solar energy that reaches the earth. Question  Use this diagram to explain the color of each type of pigment molecule.

pigments are the red carotenoids, which absorb blue light. Pigments such as phycocyanin, which absorb longer-wavelength red light, are common in aquatic systems because water absorbs blue light. Together, these and other types of pigments absorb all the wavelengths of visible light (Fig. 16.4). Each photosynthetic pigment is a highly conjugated molecule. When it absorbs a photon of the appropriate wavelength, one of its delocalized electrons is promoted to a higher-energy orbital and the molecule is said to be excited. The excited molecule can return to its lowenergy, or ground, state by several mechanisms (Fig. 16.5): 1. The absorbed energy can be lost as heat. 2. The energy can be given off as light, or fluorescence. For thermodynamic reasons, the emitted photon has a lower energy (longer wavelength) than the absorbed photon. 3. The energy can be transferred to another molecule. This process is called exciton transfer (an exciton is the packet of transferred energy) or resonance energy transfer, since the molecular orbitals of the donor and recipient groups must be oscillating in a coordinated manner in order to transfer energy. 4.  An electron from the excited molecule can be transferred to another molecule. In this process, called photooxidation, the excited molecule becomes oxidized and the recipient molecule becomes reduced. A different electronChlorophyll transfer reaction is required to restore the photooxidized molecule to its original reduced state.

excitation

All of these energy-transferring processes occur in chloroplasts to some extent, but exciton transfer and photooxidation are the most important for photosynthesis.

Chlorophyll*

Xox

X

Light-harvesting complexes transfer energy to the reaction center The primary reactions of photosynthesis occur at specific chlorophyll molecules called reaction centers. However, chloroplasts contain many more chlorophyll molecules

heat

Chlorophyll

light (fluorescence) Chlorophyll

exciton transfer

photooxidation Xred

X* Chlorophyll

Chlorophyll+

  FIGURE 16.5   Dissipation of energy in a photoexcited molecule. 

A pigment molecule such as chlorophyll is excited by absorbing a photon. The excited molecule (chlorophyll*) can return to its ground state by one of several mechanisms.

462  C ha pter 16   Photosynthesis

a.

b.

  FIGURE 16.6   A light-harvesting complex from Rhodopseudomonas acidophila.  The nine pairs of subunits (light and dark gray) are mostly buried in the membrane and form a scaffold for two rings

  FIGURE 16.7   Function of light-­

harvesting complexes.  A typical photosynthetic system consists of a reaction center (dark green) surrounded by light-harvesting complexes (light green), whose multiple pigments absorb light of different wavelengths. Exciton transfer funnels this captured solar energy to the chlorophyll at the reaction center.

c.

of chlorophyll molecules (yellow and green) and carotenoids (red).  a. Side view. The extracellular side is at the top.  b. Top view.  c. Top view showing only the chlorophyll molecules.

and other pigments than reaction centers. Many of these extra, or antenna, pigments are located in membrane proteins called light-­harvesting complexes. Over 30 different kinds of light-harvesting complexes have been characterized, and they are remarkable for their regular geometry. For example, one light-­harvesting complex in purple photosynthetic bacteria consists of 18 polypeptide chains holding two concentric rings of chlorophyll molecules, plus carotenoids (Fig. 16.6). This artful arrangement of light-absorbing groups is essential for the function of the light-harvesting complex. The pigments are all within a few angstroms of each other and overlap, so that excitation energy can be delocalized over the entire ring. The protein microenvironment of each photoreceptor influences the wavelength (and therefore the energy) of the photon it can absorb ( just as the cytochrome protein structure influences the reduction potential of its heme group; see Section 15.2). Consequently, the various light-harvesting complexes with their multiple ­pigments can absorb light of many different wavelengths. Within a light-harvesting complex, the precisely aligned pigment molecules can quickly transfer their energy to other pigments. Exciton transfer eventually brings the energy to the chlorophyll at the reaction center (Fig. 16.7). Without light-harvesting complexes to collect and concentrate light, the reaction center chlorophyll could collect only a small fraction of the incoming solar radiation. Even so, a leaf captures only about 1% of the available solar energy. During periods of high light intensity, some accessory pigments may function to dissipate excess solar energy as heat so that it does not damage the photosynthetic apparatus by inappropriate photooxidation. Various pigment molecules may also act as photosensors to regulate the plant’s growth rate and shape and to coordinate the plant’s activities—such as germination, flowering, and dormancy—according to daily or seasonal light levels.

Before Going On • Correlate a photon’s energy to its wavelength. • Explain why it is advantageous for photosynthetic pigments to absorb different colors of light. • Use nonscientific terms to explain why leaves are green. • Describe the four mechanisms by which a photoexcited molecule can return to its ground state. • Explain the function of a light-harvesting complex.

16.2  The Light Reactions  463

16.2 The Light Reactions LEARNING OBJECTIVES Trace the energy transfor­mations that take place during the light reactions of photosynthesis. • Recount the changes in reduction potential and free energy that occur during photooxidation. • Describe the substrates, products, and driving force for the water-splitting reaction. • List the order of electron carriers from H2O to NADP+. • Describe the events of photophosphorylation. • Compare linear and cyclic electron flow.

In plants and cyanobacteria, the energy captured by the antenna pigments of the light-­ harvesting complexes is funneled to two photosynthetic reaction centers. Excitation of the reaction centers drives a series of oxidation–reduction reactions whose net results are the oxidation of water, the reduction of NADP+, and the generation of a transmembrane proton gradient that powers ATP synthesis. These events are known as the light reactions of photosynthesis. (Most photosynthetic bacteria undertake similar reactions but have a single reaction center and do not produce oxygen.) The two photosynthetic reaction centers that mediate light energy transduction are part of protein complexes called Photosystem I and Photosystem II. These, along with other integral and peripheral proteins of the thylakoid membrane, operate in a series, much like the mitochondrial electron transport chain.

Photosystem II is a light-activated oxidation–reduction enzyme In plants and cyanobacteria, the light reactions begin with Photosystem II (the number indicates that it was the second to be discovered). This integral membrane protein complex is dimeric, with more bulk on the lumenal side of the thylakoid membrane than on the stromal side. The cyanobacterial Photosystem II monomer contains at least 19 subunits (14 of them integral membrane proteins). Its numerous prosthetic groups include light-absorbing pigments and redox-active cofactors (Fig. 16.8). Plant Photosystem II contains additional proteins but has the same core subunits and the same overall structure as the cyanobacterial system. In the thylakoid membrane, Photosystem II is surrounded by light-harvesting complexes to form a supercomplex (Fig. 16.9).

  FIGURE 16.8   Structure of cyanobacterial Photosystem II.  The proteins are shown as gray ribbons, and the various prosthetic groups are shown as stick models and color-coded: chlorophyll, green; pheophytin, orange; β-carotene, red; heme, purple; and quinone, blue. The stroma is at the top and the thylakoid lumen at the bottom.

Question  Add lines to show the location of the lipid bilayer.

464  C ha pter 16   Photosynthesis

The fully assembled supercomplex houses about 400 ­pigment molecules, mostly chlorophylls and carotenoids. Some ­light-harvesting proteins are mobile within the membrane and may dissociate from Photosystem II as light conditions change. Several dozen chlorophyll molecules in the core of Photosystem II function as internal antennas, funneling energy to the two reaction centers, each of which includes a pair of chlorophyll molecules known as P680 (680 nm is the wavelength of one of their absorption peaks). The reaction center chlorophylls overlap so that they are electronically coupled and function as a single unit. When P680 is excited, as indicated by the notation P680*, it quickly gives up an electron, dropping to a lower-energy state, P680+. In other words, light has oxidized P680. The photooxidized chlorophyll molecule must be reduced in order to return to its original state.

  FIGURE 16.9   Plant Photosystem II supercomplex.  The core dimeric photosystem, viewed from the stromal side, is gray, and the various types of light-harvesting complexes are shown in color.

P680*

e−

P680+

P680

e−

The two P680 groups are located near the lumenal side of Photosystem II. The electron given up by each photooxidized P680 travels through several redox groups (Fig. 16.10). Although the prosthetic groups in Photosystem II are arranged more or less symmetrically, they do not all directly participate in electron transfer. First, the electron travels to one of two pheophytin groups, which are essentially chlorophyll molecules without the central Mg2+ ion. Next, the electron is transferred to a tightly bound plastoquinone molecule and then to a loosely bound plastoquinone on the stromal side of Photosystem II. An iron atom may assist the final electron transfer. Plastoquinone (PQ) is similar to mammalian mitochondrial ubiquinone (see Section 12.2).

O H3C

H

H3C

[CH2

CH3 CH

C

CH2]n

H

O

Plastoquinone

  FIGURE 16.10  Arrangement of prosthetic groups in Photo­ system II.  The green chlorophyll groups constitute the photooxidizable P680. The two “accessory” chlorophyll groups (yellow) do not undergo oxidation or reduction. Pheophytin is orange, plastoquinone is blue, and an iron atom is red. The lipid tails of the prosthetic groups are not shown.

It functions in the same way as a two-electron carrier. The fully reduced plastoquinol (PQH2) joins a pool of plastoquinones that are soluble in the thylakoid membrane. Two electrons (two photooxidations of P680) are required to fully reduce plastoquinone to PQH2. This reaction also consumes two protons, which are taken from the stroma.

The oxygen-evolving complex of Photosystem II oxidizes water O2, a waste product of photosynthesis, is generated from H2O by a lumenal portion of Photosystem II called the oxygen-evolving center. This reaction can be written as ​ 2 ​H​ 2​O → ​O​ 2​+ 4 ​H​ +​+ 4 ​e​ −​

The electrons derived from H2O are used to restore photooxidized P680 to its reduced state.

16.2  The Light Reactions  465

The catalyst for the water-splitting reaction is a cofactor with the composition Mn4CaO5 that is held in place by Asp, Glu, and His side chains (Fig. 16.11). This unusual inorganic cofactor occurs in all Photosystem II complexes, which suggests a unique chemistry that has remained unaltered for about 2.5 billion years. No synthetic catalyst can match the manganese cluster in its ability to extract electrons from water to form O2. The water-splitting reaction is rapid, with about 50 O2 produced per second per Photosystem II, and generates most of the earth’s atmospheric O2. During water oxidation, the manganese cluster undergoes multiple changes in its oxidation state, somewhat reminiscent of the changes in the Fe–Cu binuclear center of cytochrome c oxidase (mitochondrial Complex IV; see Fig. 15.18), which carries out the reverse reaction. The four water-derived protons are released into the thylakoid lumen, contributing to a drop in pH relative to the stroma. A tyrosine radical (Y·) in Photosystem II transfers each of the four water-derived electrons to P680+ (a tyrosine radical also plays a role in electron transfer in cytochrome c oxidase; see Section 15.2).

O

  FIGURE 16.11   Structure of the Mn4CaO5 cluster.  Atoms are ­color-coded: Mn purple, Ca green, and O red. One or more of the four H2O molecules associated with the cluster (oxygen atoms in blue) may be substrates for the water-splitting reaction.

Tyrosine radical

The oxidation of water is a thermodynamically demanding reaction because O 2 has an extremely high reduction potential (+0.815 V) and electrons spontaneously flow from a group with a lower reduction potential to a group with a higher reduction potential (see Section 15.1). In fact, photooxidized P680 is the most powerful biological oxidant, with a reduction potential of about +1.15 V. Upon photoexcitation, the reduction potential of P680 (now P680*) is dramatically diminished, to about –0.8 V. This low reduction potential allows P680* to surrender an electron to a series of groups with increasingly positive reduction potentials (Fig. 16.12). Recall −1.4 −1.2 −1.0 P680*

−0.8 −0.6

Pheophytin

−0.4

Ɛ°′(V)

−0.2

PQA

0 PQB

+0.2 +0.4 +0.6 +0.8 +1.0 +1.2

H2O

.

Y

P680

  FIGURE 16.12   Reduction potential and electron flow in Photosystem II. 

Electrons flow spontaneously from a group with a lower reduction potential to a group with a higher reduction potential. The transfer of electrons from H2O to plastoquinone is made possible by the excitation of P680 (wavy arrow), which dramatically lowers its reduction potential. Question  Will pheophytin be in the oxidized or reduced state when it is dark?

466  C ha pter 16   Photosynthesis

4 H+ STROMA

Photosystem II 4e− LUMEN

2 PQH2

that the lower the reduction potential, the higher the energy. The overall result is that the input of solar energy allows an electron to travel a thermodynamically favorable path from water to plastoquinone. Four photo­ oxidation events in Photosystem II are required to oxidize two H2O molecules and produce one O2 molecule. Figure 16.13 summarizes the functions of Photosystem II.

2 PQ

Cytochrome b6f links Photosystems I and II

After they leave Photosystem II as plastoquinol, electrons reach a second membrane-bound protein complex known as cytochrome b6 f. This complex resembles mitochondrial Complex III (also called cytochrome bc1)—from the entry of electrons in the form of a reduced quinone, through the circular 2 H2O 4 H+ O2 flow of electrons among its redox groups, to the final transfer of electrons to a mobile electron carrier.   FIGURE 16.13   Photosystem II function.  For every oxygen molecule produced, two plastoquinone The cytochrome b6 f complex contains eight subunits in each of its molecules are reduced. monomeric halves (Fig. 16.14). Three subunits bear electron-transporting prosthetic groups. One of these subunits is cytochrome b6, which is homologous to mitochondrial cytochrome b. The second is cytochrome f, whose heme group is actually of the c type. Although it shares no sequence homology with mitochondrial cytochrome c1, it functions similarly. The photosynthetic complex also contains a Rieske iron– sulfur protein with a 2Fe–2S group that behaves like its mitochondrial counterpart. However, the cytochrome b6 f complex also contains subunits with prosthetic groups that are absent in the mitochondrial complex: a chlorophyll molecule and a β-carotene. These light-absorbing molecules do not appear to participate in electron transfer and may instead help regulate the activity of cytochrome b6 f by registering the amount of available light. Electron flow in the cytochrome b6 f complex follows a cyclic pattern that is probably identical to the Q cycle in mitochondrial Complex III (see Fig. 15.14). However, in chloroplasts, the final electron acceptor is not cytochrome c but plastocyanin, a small protein with an active-site copper ion (Fig. 16.15). Plastocyanin functions as a one-electron carrier by cycling between the Cu+ and Cu2+ oxidation states. Like cytochrome c, plastocyanin is a peripheral Oxygenevolving complex

  FIGURE 16.14   Structure of spinach cytochrome b6  f.  Each sub-

unit of the dimeric complex is a different color. The prosthetic groups are not shown. Question  Compare this structure to the functionally similar mitochondrial cytochrome bc1 (Complex III) in Figure 15.13.

  FIGURE 16.15  Plastocyanin.  The redox-active copper ion (green) is coordinated by a Cys, a Met, and two His residues (yellow).

16.2  The Light Reactions  467

membrane protein; it picks up electrons at the lumenal surface of cytochrome b6 f and delivers them to another integral membrane protein complex, in this case Photosystem I. The net result of the cytochrome b6 f Q cycle is that for every two electrons emanating from Photosystem II, four protons are released into the thylakoid lumen. Since the oxidation of 2 H2O is a four-electron reaction, the production of one molecule of O2 causes the cytochrome b6 f complex to produce eight lumenal H+ (Fig. 16.16). The resulting pH gradient between the stroma and the lumen is a source of energy that drives ATP synthesis, as described below.

4 H+ STROMA

Cytochrome b6f

2 PQH2 + 2 PQ

4e−

4 PQH2 LUMEN

8 H+

A second photooxidation occurs at Photosystem I

4 Plastocyanin (Cu2+)

4 Plastocyanin (Cu+)

  FIGURE 16.16  Cytochrome b6  f function. 

The stoichiometry shown for the cytochrome b6 f Q Photosystem I, like Photosystem II, is a large protein complex containing cycle reflects the four electrons released by the multiple pigment molecules. The Photosystem I in the cyanobacterium oxygen-evolving complex of Photosystem II. Synechococcus is a symmetric trimer with 12 proteins in each monomer (Fig. 16.17). Ninety-six chlorophyll molecules and 22 carotenoids operate as a built-in light-harvesting complex. In plants, Photosystem I is a monomer that shares a core structure with the cyanobacterial complex. Four Photosystem I–specific light-harvesting complexes associate with one face of the plant photosystem to form a supercomplex (Fig. 16.18). A light-harvesting complex normally associated with Photosystem II can dock to the opposite face of Photosystem I under some conditions. Photosystem I requires slightly longer-wavelength light than Photosystem II, which can lead to unequal excitation, yet the two complexes must work in concert. When ­Photosystem II is overexcited, the level of oxidized plastoquinone decreases. This activates a kinase to phosphorylate the Photosystem II light-harvesting complexes, some of which then move through the membrane to function as antenna complexes for Photosystem I. In the core of Photosystem I, a pair of chlorophyll molecules constitute the photoactive group known as P700 (it has a slightly longer-wavelength absorbance maximum than P680). Like P680, P700 undergoes exciton transfer from an antenna pigment. P700* gives up an

  FIGURE 16.17   Structure of cyanobacterial Photosystem I monomer.  The protein is shown as a gray ribbon, and the various prosthetic groups are color-coded: chlorophyll, green; β-carotene, red; phylloquinone, blue; and Fe–S clusters, orange. In this side view, the stroma is at the top.

  FIGURE 16.18   Plant Photosystem I supercomplex.  The photosystem, viewed from the stroma, is gray and the four copies of the light-harvesting complex are in color.

468  C ha pter 16   Photosynthesis

electron to achieve a low-energy oxidized state, P700+. The group is then reduced by accepting an electron donated by plastocyanin.

P700*

e−

P700

Plastocyanin (Cu2+)   FIGURE 16.19  Prosthetic groups in Photosystem I.  The groups include P700 (the green chlorophyll molecules), “accessory” chlorophylls (yellow), quinones (marked by blue spheres), and 4Fe–4S clusters (orange). The lipid tails of the prosthetic groups are not shown.

P700+

Plastocyanin (Cu+)

P700 is not a particularly good reducing agent (its reduction potential is relatively high, about +0.45 V). However, excited P700 (P700*) has an extremely low Ɛ °′ value (about –1.3 V), so electrons can spontaneously flow from P700* to the other redox groups of Photosystem I. These groups include four additional chlorophyll molecules, quinones, and iron– sulfur clusters of the 4Fe–4S type (Fig. 16.19). As in Photosystem II, these prosthetic groups are arranged with approximate symmetry. However, in Photosystem I, all the redox groups appear to undergo oxidation and reduction. Each electron given up by photooxidized P700 eventually reaches ferredoxin, a small peripheral protein on the stromal side of the thylakoid membrane. Ferredoxin undergoes a one-electron reduction at a 2Fe–2S cluster (Fig. 16.20). Reduced ferredoxin participates in two different electron transport pathways in the chloroplast, which are known as linear and cyclic electron flow. In linear electron flow, ferredoxin serves as a substrate for ferredoxin–NADP+ reductase. This stromal enzyme uses two electrons (from two separate ferredoxin molecules) to reduce NADP+ to NADPH (Fig. 16.21). The net result of linear electron flow is therefore the transfer of electrons from water, through Photosystem II, cytochrome b 6 f, Photosystem I, and then on to NADP+. For every two water molecules converted to O2 by Photosystem II, two

2 NADP+ + 2 H+

2 NADPH

4 Ferredoxinox 4 Ferredoxinred STROMA

Photosystem I

4e−

LUMEN

4 Plastocyanin (Cu+)

  FIGURE 16.20  Ferredoxin.  The 2Fe–2S cluster is shown in orange.

4 Plastocyanin (Cu2+)

  FIGURE 16.21   Linear electron flow through Photosystem I.  Electrons donated by plastocyanin are transferred to ferredoxin and used to reduce NADP+. The stoichiometry reflects the four electrons released by the oxidation of 2 H2O in Photosystem II.

16.2  The Light Reactions  469 −1.4

  FIGURE 16.22   The Z-scheme of photosynthesis.  The major components are positioned according to their reduction potentials (the individual redox groups within Photosystem II, cytochrome b6 f, and Photosystem I are not shown).

P700*

−1.2 −1.0 P680*

−0.8

Ferredoxin

−0.6 −0.4

Ɛ°′(V)

−0.2

NADP+

Plastoquinone

0 Cytochrome b6f

+0.2 +0.4

Plastocyanin P700

+0.6 +0.8 +1.0

H2O P680

+1.2

NADPH are produced. Photosystem I does not contribute to the transmembrane proton gradient except by consuming stromal protons in the reduction of NADP+ to NADPH. When plotted according to reduction potential, the electron-carrying groups of the pathway from water to NADP+ form a diagram called the Z-scheme of photosynthesis (Fig. 16.22). The zigzag pattern is due to the two photooxidation events, which markedly decrease the reduction potentials of P680 and P700. The input of light energy ensures that electrons follow a thermodynamically favorable pathway to groups with increasing reduction potential. Note that the four-electron process of producing one O2 and two NADPH is accompanied by the absorption of eight photons (four each at Photosystem II and Photosystem I). In cyclic electron flow, electrons from Photosystem I do not reduce NADP+ but instead return to the cytochrome b6 f complex. Formation of a supercomplex—containing Photosystem I, its light-harvesting complexes, and cytochrome b6 f—may facilitate this detour. From cytochrome b6 f, the electrons are transferred to plastocyanin and flow back to Photosystem I to reduce photooxidized P700+. Meanwhile, plastoquinol molecules circulate between the two quinone-binding sites of cytochrome b6 f so that protons are translocated from the stroma to the lumen, in the Q cycle (Fig. 16.23). Cyclic electron flow requires the input of light energy at Photosystem I but not Photosystem II. During cyclic flow, no free energy is recovered in the form of the reduced cofactor NADPH, but free energy is conserved in the formation of a transmembrane proton gradient by the activity of the cytochrome b6 f complex. Consequently, cyclic electron flow augments ATP generation by chemiosmosis (in some bacteria with just a single reaction center, electrons flow through a similar pathway that does not produce O 2 or NADPH). By varying the proportion of electrons that follow the linear and cyclic pathways through Photosystem I, a photosynthetic cell can vary the proportions of ATP and NADPH produced by the light reactions.

Chemiosmosis provides the free energy for ATP synthesis Chloroplasts and mitochondria use the same mechanism to synthesize ATP: They couple the dissipation of a transmembrane proton gradient to the phosphorylation of ADP. In photosynthetic organisms, this process is called photophosphorylation. Chloroplast ATP synthase is

Question  Compare the redox changes depicted here with those of the mitochondrial electron transport chain in Figure 15.2.

470  C ha pter 16   Photosynthesis

ADP + Pi

ATP

CF1 H+

Ferredoxin

STROMA

2H+ STROMA

Cytochrome b6f

CFo

Photosystem I PQ PQH2

e−

LUMEN

LUMEN

2H+ Plastocyanin   FIGURE 16.23   Cyclic electron flow.  Electrons circulate between Photosystem I and the cytochrome b6 f complex. No NADPH or O2 is produced, but the activity of cytochrome b6 f  builds up a proton gradient that drives ATP synthesis.

  FIGURE 16.24  Photophosphoryl­ ation.  As protons traverse the CFo component of chloroplast ATP synthase (following their concentration gradient from the lumen to the stroma), the CF1 component carries out ATP synthesis.

highly homologous to mitochondrial and bacterial ATP synthases. The CF1CFo complex (“C” indicates chloroplast) consists of a proton-translocating integral membrane component (CFo) mechanically linked to a soluble CF1 component where ATP synthesis occurs by a binding change mechanism (as described in Fig. 15.26). The movement of protons from the thylakoid lumen to the stroma provides the free energy to drive ATP synthesis (Fig. 16.24). As in mitochondria, the proton gradient has both chemical and electrical components. In chloroplasts, the pH gradient (about 3.5 pH units) is much larger than in mitochondria (about 0.75 units). However, in chloroplasts, the electrical component is less than in mitochondria because of the permeability of the thylakoid membrane to ions such as Mg2+ and Cl–. Diffusion of these ions tends to minimize the difference in charge due to protons. Assuming linear electron flow, 8 photons are absorbed (4 by Photosystem II and 4 by Photosystem I) to generate 4 lumenal protons from the oxygen-evolving complex and 8 protons from the cytochrome b6 f complex. Theoretically, these 12 protons can drive the synthesis of about 3 ATP, which is consistent with experimental results showing approximately 3 ATP generated for each molecule of O2. The c ring of chloroplast ATP synthase contains 15 subunits, almost twice the number in mammalian mitochondrial ATP synthase (8). ATP synthesis in chloroplasts therefore requires more protons, but this inefficiency may also allow ATP to be produced more quickly than in mitochondria.

Before Going On • Summarize the functions of Photosystem II, the oxygen-evolving complex, plastoquinone, the cytochrome b6 f complex, plastocyanin, Photosystem I, and ferredoxin. • Explain how photon absorption drives electron transfer from water to plastoquinone. • Describe the Z-scheme of photosynthesis and explain its zigzag shape. • Add an arrow to Fig. 16.22 to represent cyclic electron flow. • Discuss the yields of O2, NADPH, and ATP in cyclic and linear electron flow. • Compare and contrast the chloroplast light reactions and mitochondrial electron transport. • Compare photophosphorylation and oxidative phosphorylation. • Draw a diagram to explain the interdependence of photosynthesis and cellular respiration.

16.3  Carbon Fixation  471

16.3 Carbon Fixation LEARNING OBJECTIVES Describe the steps of carbon fixation by the Calvin cycle. • Distinguish the two activities of rubisco. • Identify the functions of the other Calvin cycle enzymes. • Explain how the “dark” reactions are linked to the light reactions. • List the metabolic fates of newly synthesized glyceraldehyde-3-phosphate.

The production of ATP and NADPH by the photoactive complexes of the thylakoid membrane (or bacterial plasma membrane) is only part of the story of photosynthesis. The rest of this chapter focuses on the use of the products of the light reactions in the so-called dark reactions. These reactions, which occur in the chloroplast stroma, incorporate atmos­ pheric carbon dioxide in biologically useful organic molecules, a process called carbon fixation. Photosynthetic organisms have evolved six different pathways for fixing CO2, but the focus here is on the most widely used pathway.

Rubisco catalyzes CO2 fixation Carbon dioxide is fixed by the action of ribulose bisphosphate carboxylase/oxygenase, or rubisco. This enzyme adds CO2 to a five-carbon sugar and then cleaves the product to two three-carbon units (Fig. 16.25). This reaction itself does not require ATP or NADPH, but the

CH2OPO2– 3

1

O

C

2

H 3C

OH

H 4C

OH

CH2OPO2– 3

1. The enzyme abstracts a proton from C3 of ribulose-1,5-bisphosphate. An active-site Mg 2 ion may help stabilize the developing negative charge.

H+

CH2OPO2– 3 –

O H

C C

O

C

OH

H

CH2OPO2– 3

5

Ribulose-1,5-bisphosphate

2. The enediolate intermediate nucleophilically attacks CO2 .

CO2

Enediolate

CH2OPO2– 3 HO H

CH2OPO32– HO

C

H

COO– + COO– H

C

OH

CH2OPO32– 2 3-Phosphoglycerate

CH2OPO2– 3 4. The 6-carbon product splits to yield two molecules of 3-phosphoglycerate. This step provides much of the free energy for the reaction, since it yields two resonance-stabilized carboxylate products.

HO

C

COO–

HO

C

O–

H

C

OH

CH2OPO2– 3

  FIGURE 16.25   The rubisco carboxylation reaction.

Question  Before this reaction was understood, scientists believed that carbon fixation involved the reaction of CO2 with a two-carbon molecule. Explain.

H2O H+

C

COO–

C

O

C

OH

CH2OPO2– 3

3. H2O attacks C3.

472  C ha pter 16   Photosynthesis

  FIGURE 16.26   Spinach rubisco.  The complex has a mass of approximately 550 kD. The eight catalytic sites are located in the large subunits (dark colors). Only four of eight small subunits (light colors) are visible in this image.

reactions that transform the rubisco reaction product, 3-phosphoglycerate, to the three-carbon sugar glyceraldehyde-3-phosphate require both ATP and NADPH, as we will see. Three-carbon compounds are the biosynthetic precursors of monosaccharides, amino acids, and—indirectly—nucleotides. They also give rise to the two-­carbon ­a cetyl units used to build fatty acids. The metabolic importance of these small ­molecular building blocks is one reason why the scheme shown in Figure 16.1 presents photosynthesis as a process in which CO2 is converted to two- and three-carbon intermediates. Rubisco is a notable enzyme, in part because its activity directly or indirectly sustains most of the earth’s biomass. Plant chloroplasts are packed with the enzyme, which accounts for about half of the chloroplast’s protein content. One reason for its high concentration is that it is not a particularly efficient enzyme. Its catalytic output is only about three CO2 fixed per second. Bacterial rubisco is usually a small dimeric enzyme, whereas the plant enzyme is a large multimer of eight large and eight small subunits (Fig. 16.26). In some archaebacteria, rubisco has 10 identical subunits. Enzymes with multiple catalytic sites typically exhibit cooperative behavior and are regulated allosterically, but this does not seem to be true for plant rubisco, whose eight active sites operate independently. Multimerization may simply be an efficient way to pack more active sites into the limited space of the chloroplast. Despite its metabolic importance, rubisco is not a highly specific enzyme. It also acts as an oxygenase (as reflected in its name) by reacting with O2, which chemically resembles CO2. The products of the oxygenase reaction are a three-carbon and a two-carbon compound:

CH2OPO2– 3

CH2OPO2– 3 O

C

H

C

OH + O2

H

C

OH

–

O

H2O

CH2OPO2– 3 Ribulose bisphosphate

C

O

2-Phosphoglycolate

+

rubisco

COO– H

C

OH

CH2OPO2– 3 3-Phosphoglycerate

The 2-phosphoglycolate product of the rubisco oxygenation reaction is subsequently metabolized by a pathway that consumes ATP and NADPH and produces CO2. This process, called photorespiration, uses the products of the light reactions and therefore wastes some of the free energy of captured photons. Oxygenase activity is a feature of all known rubisco enzymes and must play an essential role that has been conserved throughout plant evolution. Photorespiration apparently provides a mechanism for plants to dissipate excess free energy under conditions where the CO2 supply is insufficient for carbon fixation. Photorespiration may consume significant amounts of ATP and NADPH at high temperatures, which favor oxygenase activity. Some plants have evolved a mechanism, called the C4 pathway, to minimize photorespiration (Box 16.A).

The Calvin cycle rearranges sugar molecules If rubisco is responsible for fixing CO2, what is the origin of its other substrate, ribulose bisphosphate? The answer—elucidated over many years by Melvin Calvin, James Bassham, and

16.3  Carbon Fixation  473

Box 16.A The C4 Pathway On hot, bright days, the light reactions produce O2, the substrate for photorespiration, and CO2 supplies are low as plants close their stomata (pores in the leaf surface) to avoid evaporative water loss. This combination of events can bring photosynthesis to a halt. Some plants avoid this possibility by stockpiling CO2 in four-carbon molecules so that photosynthesis can proceed even while stomata are closed. The mechanism for storing carbon begins with the conden­ sation of bicarbonate (​​HCO​ 3−​  ​​) with phosphoenolpyruvate to yield oxaloacetate, which is then reduced to malate (see diagram). These four-carbon acids give the C4 pathway its name. The subsequent oxidative decarboxylation of malate regenerates CO2 and NADPH to be used in the Calvin cycle. The three-carbon ­remnant, pyruvate, is recycled back to phosphoenolpyruvate. Because the C4 pathway and the rubisco reaction compete for CO2, they take place in different types of cells or at different times of day. For example, in some plants, carbon accumulates in mesophyll cells, which are near the leaf surface and lack rubisco. The C4 compounds then enter bundle sheath cells in the leaf interior, which contain abundant rubisco. In other

plants, the C 4 pathway occurs at night, when the stomata are open and water loss is minimal, and carbon is fixed by rubisco during the day. The C4 pathway is energetically expensive, so it requires lots of sunlight. Consequently, C4 plants grow more slowly than conventional, or C3, plants when light is limited, but they have the advantage in hot, dry climates. About 5% of the earth’s plants, including the economically important maize (corn), sugarcane, and sorghum, use the C4 pathway. The recognition of climate change has led to predictions that C4 “weeds” may overtake economically important C3 plants as temperatures increase. In fact, the increase in atmospheric CO2 that is driving global warming appears to promote the growth of C3 plants, which can obtain CO2 more easily without losing too much water via their stomata. However, if water is limited, C4 plants may still have a competitive edge, as they are adapted not just for hot environments but for arid ones. Question  Operation of the C4 pathway consumes ATP. Using the pathway shown here, indicate where this occurs.

COO–

COO–

OPO2– 3

C

C

CH3

CH2

Pyruvate

Phosphoenolpyruvate

CO2

carbonic anhydrase

HCO–3 Pi

phosphoenolpyruvate carboxylase

COO– C

O

O

CH2 COO– Oxaloacetate

malic enzyme

NADPH + CO2 NADP+

COO–

malate dehydrogenase

CHOH +

NADPH NADP

CH2 COO–

Malate

Andrew Benson—is a metabolic pathway known as the Calvin cycle. Early experiments to study the fate of 14C-labeled CO2 in algae showed that, within a few minutes, the cells had synthesized a complex mixture of sugars, all containing the radioactive label. Rearrangements among these sugar molecules generate the five-carbon substrate for rubisco. The Calvin cycle actually begins with a sugar monophosphate, ribulose-5-phosphate, which is phosphorylated in an ATP-dependent reaction (Fig. 16.27). The resulting bisphosphate is a substrate for rubisco, as we have already seen. Each 3-phosphoglycerate product of the rubisco reaction is then phosphorylated, again at the expense of ATP. This phosphorylation reaction (step 3 of Fig. 16.27) is identical to the phosphoglycerate kinase reaction of glycolysis (see ­Section 13.1). Next, bisphosphoglycerate is reduced by the chloroplast enzyme glyceraldehyde-3-­ phosphate dehydrogenase, which resembles the glycolytic enzyme but uses NADPH rather than NADH. The NADPH is produced during the light reactions of photosynthesis.

Calvin cycle

474  C ha pter 16   Photosynthesis

CH2OPO2– 3 CH2OPO2– 3

CH2OH C

O

H

C

OH

H

C

OH

ATP

C

O

H

C

OH

H

C

OH

ADP

phosphoribulokinase

1

CH2OPO2– 3

HO

H

COO–

CO2

+

rubisco

COO–

2

CH2OPO2– 3

Ribulose-5-phosphate

C

C

H

Ribulose-1,5-bisphosphate

OH

CH2OPO2– 3 2 3-Phosphoglycerate

regeneration

3 biosynthesis

4

H

O

glyceraldehyde-3-phosphate dehydrogenase

C H

C

ATP

phosphoglycerate kinase

OH

CH2OPO2– 3

Pi + NADP+ NADPH

2 Glyceraldehyde-3-phosphate

ADP

OPO2– 3

O C H

C

OH

CH2OPO2– 3 2 1,3-Bisphosphoglycerate

  FIGURE 16.27   Initial reactions of the Calvin cycle.  Note that ATP and NADPH, products of the light-dependent reactions, are consumed in the process of converting CO2 to glyceraldehyde3-phosphate.

Some glyceraldehyde-3-phosphate is siphoned from the Calvin cycle for metabolic fates such as glucose or amino acid synthesis. Recall from Section 13.2 that the pathway from glyceraldehyde-3-phosphate to glucose consists of reactions that require no further input of ATP. Glyceraldehyde-3-phosphate can also be converted to pyruvate and then to oxaloacetate, both of which can undergo transamination to generate amino acids. Additional reactions lead to other metabolites. The glyceraldehyde-3-phosphate that is not used for biosynthetic pathways enters a series of isomerization and group-transfer reactions that regenerate ribulose-5-phosphate to complete the Calvin cycle (Fig. 16.28). These interconversion reactions are similar to those of the pentose phosphate pathway (Section 13.4). Consequently, if the Calvin cycle starts with three five-carbon ribulose molecules, so that three CO2 molecules are fixed (Fig. 16.27), the products are six three-carbon glyceraldehyde-3-phosphate molecules, five of which are recycled to form three ribulose molecules (Fig. 16.28), leaving the sixth (representing the three fixed CO2) as its net product. The net equation for the Calvin cycle, including the ATP and NADPH cofactors, is

​ 3 ​CO​ 2​+ 9 ATP + 6 NADPH → glyceraldehyde-3-phosphate + 9 ADP + 8 ​P​ i​+ 6 ​NADP​ +​

Fixing a single CO2 therefore requires 3 ATP and 2 NADPH—approximately the same quantity of ATP and NADPH produced by the absorption of eight photons. The relationship between the number of photons absorbed and the amount of carbon fixed or oxygen released is known as the quantum yield of photosynthesis. Keep in mind that the exact number of carbons

16.3  Carbon Fixation  475

H

O C H

C

H

O C

OH

H

CH2OPO2– 3 Glyceraldehyde3-phosphate

H

O C

C

OH

H

C

CH2OPO2– 3

H

OH

CH2OPO2– 3

Glyceraldehyde3-phosphate

Glyceraldehyde3-phosphate

CH2OPO2– 3

2

Dihydroxyacetone phosphate

O

HO

C

H

H

C

OH

H

C

OH

CH2OPO2– 3

C

3

CH2OH C

H

O

C

CH2OH

C

C

OH

H

Glyceraldehyde3-phosphate

O

CH2OPO2– 3

3

H

C

OH

C

O

HO

C

H

CH2OH

H

C

OH

C

O

C

O

HO

C

H

H

C

OH

HO

C

H

H

C

OH

H

C

OH

H

C

OH

Sedoheptulose7-phosphate

Fructose-6-phosphate

CH2OH

C

O

C

O

H

C

OH

H

C

OH

H

C

OH

H

C

OH

CH2OPO2– 3 Ribulose-5-phosphate

C

OH

H

C

OH

H

C

OH

CH2OPO2– 3 Ribose-5-phosphate

5

Xylulose-5-phosphate

CH2OH

4

H

CH2OPO2– 3

CH2OPO2– 3

Xylulose-5-phosphate

H C

CH2OH

CH2OPO2– 3

Glyceraldehyde3-phosphate

O

OH

Erythrose4-phosphate

CH2OPO2– 3

2

C

CH2OH

OH

1

H

CH2OPO2– 3

C

CH2OPO2– 3

Dihydroxyacetone phosphate

CH2OH O

H

O

C

1 C

H

O

CH2OPO2– 3 Ribulose-5-phosphate

  FIGURE 16.28   Regeneration reactions of the Calvin cycle.  Reaction 1 is catalyzed by triose phosphate isomerase, 2 by aldolase and an isomerase, 3 by transketolase (see Section 7.2), 4 by an ­epimerase, and 5 by an isomerase.

Question  Add boxes around the 2-carbon fragments transferred by transketolase.

fixed per photon absorbed depends on factors such as the number of protons translocated per ATP synthesized by the chloroplast ATP synthase and the ratio of cyclic to linear electron flow in Photosystem I.

The availability of light regulates carbon fixation Plants must coordinate the light reactions with carbon fixation. During the day, both processes occur. At night, when the photosystems are inactive, the plant turns off the “dark” reactions to conserve ATP and NADPH while it turns on pathways to regenerate these cofactors by metabolic pathways such as glycolysis and the pentose phosphate pathway. It would

CH2OH

4

C

O

H

C

OH

H

C

OH

CH2OPO2– 3 Ribulose-5-phosphate

476  C ha pter 16   Photosynthesis

be wasteful for these catabolic processes to proceed simultaneously with the Calvin cycle. Thus, the “dark” reactions do not actually occur in the dark! All the mechanisms for regulating the Calvin cycle are directly or indirectly linked to the availability of light energy. Some of the regulatory mechanisms are highlighted here. For example, a catalytically essential Mg2+ ion in the rubisco active site is coordinated in part by a carboxylated lysine side chain that is produced by the reaction of CO 2 with the ɛ-amino group: —​​(​ ​CH​ 2​)​​ 4​​—​NH​ 2​+ ​CO​ 2​⇌ — ​(​CH​ 2​)​ 4​ —NH— ​COO​ −​+ ​H​ +​

​ Lys​

By forming the Mg2+-binding site, this “activating” CO2 molecule promotes the ability of rubisco to fix additional substrate CO2 molecules. The carboxylation reaction is favored at high pH, a signal that the light reactions are working (depleting the stroma of protons) and that ATP and NADPH are available for the Calvin cycle. Magnesium ions also directly activate rubisco and several of the Calvin cycle enzymes. During the light reactions, the rise in stromal pH triggers the flux of Mg2+ ions from the lumen to the stroma (this ion movement helps balance the charge of the protons that are translocated in the opposite direction). Some of the Calvin cycle enzymes are also activated when the ratio of reduced ferredoxin to oxidized ferredoxin is high, another signal that the photosystems are active.

Calvin cycle products are used to synthesize sucrose and starch Many of the three-carbon sugars produced by the Calvin cycle are converted to sucrose or starch. The polysaccharide starch is synthesized in the chloroplast stroma as a temporary storage depot for glucose. It is also synthesized as a long-term storage molecule elsewhere in the plant, including leaves, seeds, and roots. In the first stage of starch synthesis, two molecules of glyceraldehyde-3-phosphate are converted to glucose-6-phosphate by reactions analogous to those of mammalian gluconeogenesis (see Fig. 13.10). Phosphoglucomutase then carries out an isomerization reaction to produce glucose-1-phosphate. Next, this sugar is “activated” by its reaction with ATP to form ADP–glucose:

P



Glucose-1-phosphate

P

P

P

Adenosine

ATP

P

P

ADP–glucose

Adenosine



P

P PPi

(Recall from Section 13.3 that glycogen synthesis uses the chemically related nucleotide sugar UDP–glucose.) Starch synthase then transfers the glucose residue to the end of a starch polymer, forming a new glycosidic linkage.

ADP + ADP–glucose starch synthase

Starch

ADP

The overall reaction is driven by the exergonic hydrolysis of the PPi released in the formation of ADP–glucose. Thus, one phosphoanhydride bond is consumed in lengthening a starch molecule by one glucose residue. Sucrose, a disaccharide of glucose and fructose, is synthesized in the cytosol. ­Glyceraldehyde-3-phosphate or its isomer dihydroxyacetone phosphate is transported out of the c­ hloroplast by an antiport protein that exchanges phosphate for a phosphorylated three-carbon sugar. Two of these sugars combine to form fructose-6-phosphate, and two others combine to form glucose-1-phosphate, which is subsequently activated by UTP. Next, fructose-6-­phosphate reacts with UDP–glucose to produce sucrose-6-phosphate. Finally, a phosphatase converts the phosphorylated sugar to sucrose:

16.3  Carbon Fixation  477

HOCH2 H HO

O H OH

H

H

OH

O

2–

O3POH2C

H UDP

+

H

H

CH2OH

HO

OH

OH

H

Fructose-6-phosphate

UDP–Glucose

UDP

H HO

CH2OH O H OH H H

H HOCH2

O

H

O

HO

CH2OPO2– 3

H

OH

OH

H

Sucrose-6-phosphate

H2O Pi

H HO

CH2OH O H OH H H

H HOCH2 O

O

H OH

OH

H

HO

CH2OH

H

Sucrose

Sucrose can then be exported to other plant tissues. This disaccharide probably became the preferred transport form of carbon in plants because its glycosidic linkage is insensitive to amylases (starch-digesting enzymes) and other common hydrolases. Also, its two anomeric carbons are tied up in the glycosidic bond and therefore cannot react nonenzymatically with other substances. Cellulose, the other major polysaccharide of plants, is also synthesized from UDP–­ glucose (cellulose is described in Section 11.2). Plant cell walls consist of almost-crystalline cables, each containing approximately 36 cellulose polymers, which are embedded in an amorphous matrix of other polysaccharides (see Box 11.B; synthetic materials such as fiberglass are built on the same principle). Unlike starch in plants or glycogen in mammals, cellulose is synthe­sized by enzyme complexes in the plant plasma membrane and is extruded into the extracellular space.

Before Going On • List the reactants and products of the two reactions catalyzed by rubisco. • Compare the physiological implications of carbon fixation and photorespiration. • Explain how the carbon from a molecule of fixed CO2 becomes incorporated into other compounds such as monosaccharides. • Identify the source of the ribulose-1,5-bisphosphate used for carbon fixation by rubisco. • Compare the Calvin cycle to the pentose phosphate pathway. • Describe some of the mechanisms for regulating the activity of the “dark” reactions. • Summarize the role of nucleotides in the synthesis of starch and sucrose.

478  C ha pter 16   Photosynthesis

Summary 16.1  Chloroplasts and Solar Energy •  Plant chloroplasts contain pigments that absorb photons and release the energy, primarily by transferring it to another molecule (exciton transfer) or giving up an electron (photooxidation). Light-harvesting complexes act to capture and funnel light energy to the photosynthetic reaction centers.

16.2  The Light Reactions •  In the so-called light reactions of photosynthesis, electrons from the photooxidized P680 reaction center of Photosystem II pass through several prosthetic groups and then to plastoquinone. The P680 electrons are replaced when the oxygen-evolving complex of Photo-­ system II converts water to O2, a four-electron oxidation reaction. •  Electrons flow next to a cytochrome b6 f complex that carries out a proton-translocating Q cycle, and then to the protein plastocyanin. •  A second photooxidation at P700 of Photosystem I allows electrons to flow to the protein ferredoxin and finally to NADP+ to produce NADPH.

•  The free energy of light-driven electron flow, particularly cyclic flow, is also conserved in the formation of a transmembrane proton gradient that drives ATP synthesis in the process called photophosphorylation.

16.3  Carbon Fixation •  The enzyme rubisco “fixes” CO2 by catalyzing the carboxylation of a five-carbon sugar. Rubisco also acts as an oxygenase in the process of photorespiration. •  The reactions of the Calvin cycle use the products of the light reactions (ATP and NADPH) to convert the product of the rubisco reaction to glyceraldehyde-3-phosphate and to regenerate the five-carbon carboxylate receptor. These “dark” reactions are regulated according to the availability of light energy. •  Chloroplasts convert the glyceraldehyde-3-phosphate product of photosynthesis into glucose residues for incorporation into starch, sucrose, and cellulose.

Key Terms photosynthesis chloroplast stroma thylakoid photon Planck’s law photoreceptor fluorescence

exciton transfer photooxidation reaction center antenna pigment light-harvesting complex light reactions linear electron flow Z-scheme

cyclic electron flow photophosphorylation dark reactions carbon fixation photorespiration C4 pathway Calvin cycle quantum yield

Bioinformatics Brief Bioinformatics Exercises 16.1  Viewing and Analyzing Photosystems I and II 16.2  Photosynthesis and the KEGG Database

Problems 16.1  Chloroplasts and Solar Energy 1.  Indicate with a C or an M whether the following occur in chloroplasts, mitochondria, or both:  a.  proton translocation,  b. photo­ phosphorylation,  c. photooxidation, d. quinones, e. oxygen reduction,  f.  water oxidation,  g.  electron transport,  h. oxidative phosphorylation,  i.  carbon fixation,  j.  NADH oxidation,  k. Mn

cofactor,  l.  heme groups,  m.  binding change mechanism,  n. iron– sulfur clusters,  o. NADP+ reduction. 2.  Compare and contrast the structures of chloroplasts and mitochondria. 3.  The thylakoid membrane contains some unusual lipids. One of these is galactosyl diacylglycerol; a β-galactose residue is attached

Problems  479 to the first glycerol carbon. Draw the structure of β-galactosyl diacylglycerol. 4.  Thylakoid membranes contain lipids with a high degree of unsaturation. What does this tell you about the character of the thylakoid membrane? 5.  Calculate the energy per photon and per mole of photons with a wavelength of 680 nm. 6.  Calculate the energy per photon and per mole of photons with wavelengths of  a.  400 nm and  b.  700 nm. What is the relationship between wavelength and energy? 7.  Phycocyanins absorb light at 620 nm.  a.  What color is this light? What color are phycocyanins?  b.  Calculate the energy per photon of light at this wavelength. 8.  Chlorophyll a absorbs light at 430 nm and 662 nm.  a.  What colors are associated with each of these wavelengths?  b.  What color is chlorophyll a?  c.  Calculate the energy per photon of light at these wavelengths. Which wavelength of light has greater energy? 9.  At what wavelength would a mole of photons have an energy of 250 kJ? 10.  At what wavelength would a mole of photons have an energy of 400 kJ? 11.  Assuming 100% efficiency, calculate the number of moles of ATP that could be generated per mole of photons with the energy calculated in Problem 5. 12.  Assuming 100% efficiency, calculate the number of moles of ATP that could be generated per mole of photons with the energies calculated in Problem 6. 13.  Assuming 100% efficiency, calculate the number of moles of ATP that could be generated per mole of photons with the energy calculated in Problem 7. 14.  Assuming 100% efficiency, calculate the number of moles of ATP that could be generated per mole of photons with the energies calculated for the  a.  430 nm light and  b.  662 nm light in Problem 8.  c.  What is the relationship between wavelength and amount of ATP synthesized? 15.  Red tides result from algal blooms that cause seawater to become visibly red. In the photosynthetic process, red algae take advantage of wavelengths not absorbed by other organisms. Describe the photosynthetic pigments of the red algae. 16.  Some photosynthetic bacteria live in murky ponds where visible light does not penetrate easily. What wavelengths might the photosynthetic pigments in these organisms absorb? 17.  Compare the structures of chlorophyll a (Fig. 16.3) and the reduced form of heme b (Fig. 15.12). 18.  You are investigating the functional similarities of chloroplast cytochrome f and mitochondrial cytochrome c1.  a.  Which would ­provide more useful information: the amino acid sequences of the proteins or models of their three-dimensional shapes? Explain.  b. Would it be better to examine high-resolution models of the two apoproteins (the polypeptides without their heme groups) or low-resolution models of the holoproteins (polypeptides plus heme groups)? 19.  Under conditions of very high light intensity, excess absorbed solar energy is dissipated by the action of “photoprotective” proteins in the thylakoid membrane. Explain why it is advantageous for these proteins to be activated by a buildup of a proton gradient across the membrane. 20.  Of the four mechanisms for dissipating light energy shown in Figure 16.5, which would be best for “protecting” the photosystems from excess light energy as described in Problem 19?

16.2  The Light Reactions 21.  The three electron-transporting complexes of the thylakoid membrane can be called plastocyanin–ferredoxin oxidoreductase, ­plastoquinone–plastocyanin oxidoreductase, and water–­plastoquinone oxidoreductase. What are the common names of these enzymes and in what order do they act? 22.  Photosystem II is located mostly in the tightly stacked regions of the thylakoid membrane, whereas Photosystem I is located mostly in the unstacked regions (see Fig. 16.2). Why might it be important for the two photosystems to be separated? 23.  The herbicide 3-(3,4-dichlorophenyl)-1,1-dimethylurea (DCMU) blocks electron flow from Photosystem II to Photosystem I. What is the effect on oxygen production and photophosphorylation when DCMU is added to plants? 24. a.  When the antifungal agent myxothiazol is added to a suspension of chloroplasts, the QH2/Q ratio increases. Where does myxothiazol inhibit electron transfer?  b.  Compare the effect of myxothiazol on chloroplasts (part a) and mitochondria (Problem 15.40). 25. a.  Plastoquinone is not firmly anchored to any thylakoid membrane component but is free to diffuse laterally throughout the membrane among the photosynthetic components. What aspects of its structure account for this behavior?  b.  What component of the mitochondrial electron transport chain exhibits similar behavior? 26.  Photosystem II includes a protein called D1 that contains the PQB binding site. D1 in the single-celled alga Chlamydomonas reinhardtii is predicted to have five hydrophobic membrane-spanning helical segments. A loop between the fourth and fifth segments is located in the stroma along the membrane surface and contains several highly conserved amino acid residues. D1 proteins with mutations at the Ala 251 position were evaluated for photosynthetic activity and herbicide susceptibility. The results are shown in the table. What are the essential properties of the amino acid at position 251 in D1? Mutant

Ala → Cys Ala → Ser Ala → Ile

Ala → Arg

Characteristics Similar to wild-type Photosynthesis impaired Photoautotrophic growth and photosynthesis impaired; resistant to herbicides Not photosynthetically competent

27.  Use Equation 15.4 to calculate the free energy change for transforming one mole of P680 to P680*. 28.  Determine the wavelength of the photons whose absorption would supply the free energy to transform one mole of P680 to P680* (see Problem 27). 29.  Calculate the free energy of translocating a proton out of the stroma when the lumenal pH is 3.5 units lower than the stromal pH, the temperature is 25°C, and Δψ is −50 mV. 30. a.  Compare the free energy of proton translocation calculated in Problem 29 to the free energy of translocating a proton out of a mitochondrion where the pH difference is 0.75 units and Δψ is 200 mV. Compare both types of translocation. Are the processes exergonic or endergonic?  b.  Which contributes a larger component of the free energy for each process, the pH difference or the membrane potential? 31.  Calculate the standard free energy change for the oxidation of one molecule of water by NADP+. 32.  Calculate the energy available in two photons of wavelength 600 nm. Compare this value with the standard free energy change you calculated in Problem 31. Do two photons supply enough energy to drive the oxidation of one molecule of water by NADP+?

480  C ha pter 16   Photosynthesis 33.  Photophosphorylation in chloroplasts is similar to oxidative phosphorylation in mitochondria. What is the final electron acceptor in photosynthesis? What is the final electron acceptor in mitochondrial electron transport? 34.  If radioactively labeled water (H 218O) is provided to a plant, where does the label appear? 35.  In the 1940s, Robert Hill conducted an experiment in which he illuminated chloroplasts in the absence of CO2 but in the presence of the electron acceptor ​​Fe(CN)​ 63− ​  ​​. Oxygen was evolved. What was the other product of the reaction? What is the significance of this experiment? 36.  Cyclic electron flow in some organisms can occur by more than one mechanism. For example, electrons may travel from ferredoxin to plastoquinone via a dehydrogenase complex that resembles mitochondrial Complex I in overall structure and function.  a. Sketch a diagram, similar to Figure 16.12, that shows the relative reduction potentials of the components in this pathway.  b.  Explain the advantage of using the dehydrogenase complex rather than a single oxidoreductase enzyme to transfer electrons from ferredoxin to plastoquinone. 37.  Predict the effect of an uncoupler such as dinitrophenol (see Box 15.B) on production of  a.  ATP and  b.  NADPH by a chloroplast. 38.  Antimycin A (an antibiotic) blocks electron transport in Complex III of the electron transport chain in mitochondria. How would the addition of antimycin A to chloroplasts affect chloroplast ATP synthesis and NADPH production? 39.  Does the quantum yield of photosynthesis increase or decrease for systems where  a.  the CFo component of ATP synthase contains more c subunits and  b.  the proportion of cyclic electron flow through Photosystem I increases? 40.  Oligomycin inhibits the proton channel (Fo) of the ATP synthase enzyme in mitochondria but does not inhibit CFo. When oligomycin is added to plant cells undergoing photosynthesis, the cytosolic ATP/ADP ratio decreases whereas the chloroplastic ATP/ADP ratio is unchanged or even increases. Explain these results. 41.  The compound paraquat accepts electrons from ferredoxin and then reduces oxygen to superoxide (· ​​O− 2​  ​  ​). What are the consequences for a plant treated with paraquat? 42.  A chemical reagent is used to modify the Lys and Arg residues in an isolated preparation of CF1. What are the results, compared to controls, when ADP is added to the preparation?

16.3  Carbon Fixation 43.  Defend or refute this statement: The “dark” reactions are so named because these reactions occur only at night. 44.  Examine the net equation for the light and “dark” reactions of photosynthesis—that is, the incorporation of one molecule of CO2 into carbohydrate, which has the chemical formula (CH2O)n. CO2 + 2 H2O → (CH2O) + O2 + H2O

How would this equation differ for a bacterial photosynthetic system in which H2S rather than H2O serves as a source of electrons?

45.  Melvin Calvin and his colleagues noted that when 14CO2 was added to algal cells, a single compound was radiolabeled within 5 seconds of exposure. What is the compound and where does the radioactive label appear? 46.  As described in the chapter, rubisco is not a particularly specific enzyme. Scientists have wondered why millions of years of evolution failed to produce a more specific enzyme. What advantage would be

conferred upon a plant that evolved a rubisco enzyme that was able to react with CO2 but not oxygen? 47.  A tiny acorn grows into a massive oak tree. Using what you know about photosynthesis, what accounts for the increase in mass? 48.  The ΔG°′ value for the rubisco reaction is −35.1 kJ · mol−1 and the ΔG value is −41.0 kJ · mol−1. What is the ratio of products to reactants under normal cellular conditions? Assume a temperature of 25°C. 49.  Efforts to engineer a more efficient rubisco, one that could fix CO2 more quickly than three per second, could improve farming by allowing plants to grow larger and/or faster. Explain why the engineered rubisco might also decrease the need for nitrogen-containing fertilizers. 50.  Some plants synthesize the sugar 2-carboxyarabinitol-1-­ phosphate (CA1P, shown below). This compound binds very tightly to rubisco, with a Kd of 32 nM, which inhibits enzyme activity.  a. What is the probable mechanism of action of the inhibitor?  b. Why do plants synthesize CA1P at night and break it down during the day?  c.  The enzyme that breaks down CA1P is a phosphatase that catalyzes hydrolysis of the phosphate at C1 of CA1P. How would NADPH, ­r ibulose-5-phosphate, and P i affect the activity of the phosphatase?

CH2OPO2– 3 HO

C

COO–

H

C

OH

H

C

OH

CH2OH 2-Carboxyarabinitol1-phosphate 51.  An “activating” CO2 reacts with a lysine side chain on rubisco to carboxylate it. This modification is essential for rubisco activity.  a.  Explain why the carboxylation reaction is favored at high pH.  b.  How does this activation step help link carbon fixation to the light reactions?  c.  How does the activation step “cooperate” with CA1P (see Problem 50) to regulate the activity of rubisco? 52.  Chloroplast phosphofructokinase (PFK) is inhibited by ATP and NADPH. How does this observation support the hypothesis that the light reactions are linked to the regulation of glycolysis? 53.  Crabgrass, a C4 plant, remains green during a long spell of hot dry weather when C3 grasses turn brown. Explain this observation. 54.  Chloroplasts contain thioredoxin, a small protein with two ­cysteine residues that can form an intramolecular disulfide bond. The sulfhydryl/disulfide interconversion in thioredoxin is catalyzed by an enzyme known as ferredoxin–thioredoxin reductase. This enzyme, along with some of the Calvin cycle enzymes, also includes two Cys residues that undergo sulfhydryl/disulfide transitions. Show how disulfide interchange reactions involving thioredoxin could coordinate the activity of Photosystem I with the activity of the ­Calvin cycle. 55.  The inner chloroplast membrane is impermeable to large polar and ionic compounds such as NADH and ATP. However, the membrane has an antiport protein that facilitates the passage of dihydroxy­ acetone phosphate or 3-phosphoglycerate in exchange for Pi. This system permits the entry of Pi for photophosphorylation and the exit of the products of carbon fixation. Draw a diagram that shows how the same antiport could “transport” ATP and reduced cofactors from the chloroplast to the cytosol.

Chapter 16 Credits  481 56.  The sedoheptulose bisphosphatase (SBPase) enzyme in the Calvin cycle catalyzes the removal of a phosphate group from C1 of sedoheptulose-1,7-bisphosphate (SBP) to produce sedoheptulose7-phosphate (S7P). The ΔG°′ value for this reaction is −14.2 kJ · mol−1 and the ΔG value is −29.7 kJ · mol−1. What is the ratio of products to reactants under normal cellular conditions? Is this enzyme likely to be regulated in the Calvin cycle? Assume a temperature of 25°C. 57.  Phosphoenolpyruvate carboxylase (PEPC) catalyzes the carboxylation of phosphoenolpyruvate (PEP) to oxaloacetate (OAA). The enzyme is commonly found in plants but is absent in animals. The reaction is shown.  a.  Why is PEPC referred to as an anaplerotic enzyme?  b.  Acetyl-CoA is an allosteric regulator of PEPC. Does ­acetyl-CoA activate or inhibit PEPC? Explain.

COO– C

– OPO2– 3 + HCO3

CH2

PEP

PEPC

COO– C

O + HPO2– 4

CH2

61.  The Monsanto Company constructed a transgenic “Roundup Ready” soybean cultivar in which the bacterial gene for the enzyme EPSPS is inserted into the plant genome. EPSPS catalyzes an important step in the synthesis of aromatic amino acids, as shown in the diagram. The herbicide Roundup® contains glyphosate, a compound that competitively inhibits plant EPSPS but not the bacterial form of the enzyme. Explain the strategy for using glyphosate-­containing herbicide on a soybean crop to kill weeds.  b.  Is Roundup® toxic to humans? Explain why or why not.

Shikimate ATP Shikimate-3-P PEP

EPSPS

5-Enolshikimate-3-P –

COO

OAA

58.  In germinating oil seeds, triacylglycerols are rapidly converted to sucrose and protein. What is the role of PEPC (see Problem 57) in this process? 59.  The enzyme that “activates” glucose for starch synthesis is called ADP-glucose pyrophosphorylase (AGPase). Use the following observations to explain the role of inorganic phosphate (Pi) in regulating the activity of AGPase:  a.  Plants grown in a phosphate-deficient medium showed a ten-fold increase in starch accumulation.  b. Leaf discs incubated with mannose (a C2 epimer of glucose) showed a ­fifteen-fold increase in starch accumulation. 60.  AGPase (see Problem 59) is allosterically regulated by 3-­phosphoglycerate (3PG). Is 3PG an allosteric activator or an inhibitor? Explain.

Chorismate Phenylalanine

Tyrosine

62.  Novartis constructed a corn cultivar that contains a gene from the bacterium Bacillus thuringiensis (Bt) that codes for an endotoxin protein. When insects of the order Lepidoptera (which includes the European corn borer but also unfortunately includes the Monarch butterfly) eat the corn, the ingested endotoxin enters the high pH environment of the insect’s midgut. Under these conditions, the endotoxin forms a pore in the membrane of the cells lining the midgut, causing ions to flow into the cells and ultimately resulting in the death of the organism. Is the Bt endotoxin toxic to humans? Explain why or why not.

Selected Readings Bathellier, C., Tcherkez, G., Lorimer, G.H., and Farquhar, G.D., Rubisco is not really so bad, Plant Cell Environ. 41, 705–716, doi: 10.1111/ pce.13149 (2018). [Argues that rubisco’s speed, rate enhancement, and substrate specificity are in line with other enzymes.] Cardona, T., Shao, S., and Nixon, P.J., Enhancing photosynthesis in plants: The light reactions, Essays Biochem. 62, 85–94, doi: 10.1042/ EBC20170015 (2018). [Discusses ways to increase the efficiency of photosynthesis through genetic engineering and also includes an overview of the light-dependent reaction pathway.]

Johnson, M.P., Photosynthesis, Essays Biochem. 60, 255–273, doi: 10.1042/EBC20160016 (2016). [A very readable summary, with helpful diagrams.] Martin, W.F., Bryant, D.A., and Beatty, J.T., A physiological perspective on the origin and evolution of photosynthesis, FEMS Microbiol. Rev. 42, 205–231, doi: 10.1093/femsre/fux056 (2018). [Outlines the features of early photosynthetic electron transport systems.]

Chapter 16 Credits Figure 16.6 Image based on 1KZU. Prince, S.M., Papiz, M.Z., Freer, A.A., McDermott, G., Hawthornthwaite-Lawless, A.M., Cogdell, R.J., Isaacs, N.W., Apoprotein structure in the LH2 complex from Rhodopseudomonas acidophila strain 10050: Modular assembly and protein pigment interactions, J. Mol. Biol. 268, 412–423 (1997).

Figures 16.8 and 16.10 Images based on 1S5L. Ferreira, K.N., Iverson, T.M., Maghlaoui, K., Barber, J., Iwata, S., Architecture of the photosynthetic oxygen-evolving center, Science 303, 1831–1838 (2004). Figure 16.9 Image based on 5XNL. Su, X., Ma, J., Wei, X., Cao, P., Zhu, D., Chang, W., Liu, Z., Zhang, X., Li, M., Structure and assembly

482  C ha pter 16   Photosynthesis mechanism of plant C2S2M2-type PSII-LHCII supercomplex, Science 357, 815–820 (2017). Figure 16.11 Image based on 3WU2. Umena, Y., Kawakami, K., Shen, J.-R., Kamiya, N., Crystal structure of oxygen-evolving photosystem II at a resolution of 1.9 A, Nature 473, 55–60 (2011). Figure 16.14 Image based on 6RQF. Malone, L.A., Qian, P., Mayneord, G.E., Hitchcock, A., Farmer, D.A., Thompson, R.F., Swainsbury, D.J.K., Ranson, N.A., Hunter, C.N., Johnson, M.P., Cryo-EM structure of the spinach cytochrome b6f complex at 3.6 Angstrom resolution, Nature 575, 535–539 (2019). Figure 16.15 Image based on 1PLC. Guss, J.M., Bartunik, H.D., Freeman, H.C., Accuracy and precision in protein structure analysis: Restrained least-squares refinement of the structure of poplar plastocyanin at 1.33 A resolution, Acta Crystallogr. B 48, 790–811 (1992). Figures 16.17 and 16.19 Images based on 1JB0. Jordan, P., Fromme, P., Witt, H.T., Klukas, O., Saenger, W., Krauss, N., Three-dimensional

structure of cyanobacterial Photosystem I at 2.5 A resolution, Nature 411, 909–917 (2001). Figure 16.18 Image based on 4XK8. Qin, X., Suga, M., Kuang, T., Shen, J.R., Structural basis for energy transfer pathways in the plant PSILHCI supercomplex, Science 348, 989–995 (2015). Figure 16.20 Image based on 1CZP. Morales, R., Charon, M.H., Hudry-Clergeon, G., Petillot, Y., Norager, S., Medina, M., Frey, M., Refined X-ray structures of the oxidized, at 1.3 A, and reduced, at 1.17 A, [2Fe-2S] ferredoxin from the cyanobacterium Anabaena PCC7119 show redox-linked conformational changes, Biochemistry 38, 15764–15773 (1999). Figure 16.26 Image based on 1RCX. Taylor, T.C., Andersson, I., The structure of the complex between rubisco and its natural substrate ribulose 1,5-bisphosphate, J. Mol. Biol. 265, 432–444 (1997).

CHAPTER 17

Lipid Metabolism DO YOU REMEMBER? • Lipids are predominantly hydrophobic molecules that can be esterified but cannot form polymers (Section 8.1). • M  etabolic fuels can be mobilized by breaking down glycogen, triacylglycerols, and proteins (Section 12.1). • A  few metabolites appear in several metabolic pathways (Section 12.2). • C  ells also use the energy of other phosphorylated compounds, thioesters, reduced cofactors, and electrochemical gradients (Section 12.3).

About half the mass of an almond seed is lipid, 80% of which contains monounsatuated fatty acids such as oleate. Monounsaturated fatty acids are associated with a healthy balance among circulating lipoproteins, and inside cells, these fatty acids interact with regulatory molecules that promote mitochondiral biogenesis and oxidative metabolism.

Like carbohydrates, lipids are metabolic fuels and therefore can be examined in terms of their synthesis, storage, mobilization, and catabolism—pathways that intersect with the processes we have already studied. In this chapter, we will investigate the breakdown and synthesis of fatty acids and related molecules. Unlike other molecules, lipids are insoluble in water, so we will begin by looking at how they travel between organs.

17.1 Lipid Transport LEARNING OBJECTIVES Summarize the roles of lipoproteins in lipid metabolism. • Explain why lipoproteins are needed to transport lipids. • Relate a lipoprotein’s density to its lipid content. • Describe the functions of LDL and HDL in cholesterol transport.

Approximately half of all deaths in the United States are linked to the vascular disease athero­ sclerosis (a term derived from the Greek athero, “paste,” and sclerosis, “hardness”). Atherosclerosis is a slow progressive disease in which white blood cells known as macrophages take up circulating lipids—particularly oxidized lipids—and lodge themselves in the walls of large blood vessels. The engorged macrophages participate in an inflammatory response and produce signals that recruit additional white blood cells, thereby perpetuating the inflammation. The damaged vessel wall forms a plaque with a core of cholesterol, cholesteryl esters, and remnants of dead macrophages, surrounded by proliferating smooth muscle cells that 483

C hA pTER 17

Lipid Metabolism

James Cavallini/BSIP/Phototake

484

FIGURE 17.1 An atheroscle­ rotic plaque in an artery. Note the thickening of the vessel wall.

FIGURE 17.2 Model of a lipo­ protein. In this computer simulation of an early HDL, two copies of the APOA1 protein (purple and blue) encircle a 100-Å disk containing 20 cholesterol molecules and 200 phospholipids (with C green, H white, O red, and P orange).

Question Explain why the proteins are amphipathic.

may undergo calcification, as occurs in bone formation. This accounts for the “hardening” of the arteries. Although a very large plaque can occlude the lumen of the artery (Fig. 17.1), blood flow is usually not completely blocked unless the plaque ruptures, triggering formation of a blood clot that can prevent circulation to the heart (a heart attack) or brain (a stroke). The lipids that lead to atherosclerosis are packaged in lipoproteins known as LDL (for low-density lipoproteins). Lipoproteins (particles consisting of lipids and specialized proteins) are the primary form of circulating lipid in animals (Fig. 17.2). The proteins wrap around a core of lipids in a flexible arrangement that accommodates changes in the size of the particle. The proteins help target the particle to cell surfaces and modulate the activities of enzymes that act on the component lipids. The various types of lipoproteins differ in size, lipid composition, protein composition, and density (a function of the relative proportions of lipid and protein). Recall from Section 12.1 that dietary lipids travel from the intestine to other tissues as chylomicrons. These lipoproteins are relatively large (1000 to 5000 Å in diameter) with a protein content of only 1% to 2%. Their primary function is to transport dietary triacylglycerols to adipose tissue and cholesterol to the liver. The liver repackages the cholesterol and other lipids—including triacylglycerols, phospholipids, and cholesteryl esters—into other lipoproteins known as VLDL (very-low-density lipoproteins). VLDL have a triacylglycerol content of about 50% and a diameter of about 500 Å. As they circulate in the bloodstream, VLDL give up triacylglycerols to the tissues, becoming smaller, denser, and richer in cholesterol and cholesteryl esters. After passing through an intermediate state (IDL, or intermediate-density lipoproteins), they become LDL, about 200 Å in diameter and about 45% cholesteryl ester (Table 17.1). High concentrations of circulating LDL, measured as serum cholesterol (popularly called “bad cholesterol”), are a major factor in atherosclerosis. Some high-fat diets (especially those rich in saturated fats) may contribute to atherosclerosis by boosting LDL levels, but genetic factors, smoking, and infection also increase the risk of atherosclerosis. The disease is less likely to occur in individuals who consume low-cholesterol diets and who have high levels of HDL (high-density lipoproteins, sometimes called “good” cholesterol). HDL particles are even smaller and denser than LDL (see Table 17.1), and their primary function is to transport the body’s excess cholesterol back to the liver. HDL therefore counter the atherogenic tendencies of LDL. The roles of the various lipoproteins are summarized in Figure 17.3. Briefly, the large chylomicrons, which are mostly lipid, transport dietary lipids to the liver and other tissues. The liver produces triacylglycerol-rich VLDL and releases them into the circulation. As the VLDL give up their triacylglycerols, they become cholesterol-rich LDL, which are taken up by cells. High-density lipoproteins (HDL), the smallest and densest of the lipoproteins, transport cholesterol from the tissues back to the liver. All cells in the body can synthesize cholesterol (Section 17.4), which is an essential membrane component, but LDL are also a major source of this lipid. When LDL proteins dock with the LDL receptor on the cell surface, the lipoprotein–receptor complex undergoes endocytosis (Section 9.4). Inside the cell, the lipoprotein is degraded and cholesterol enters the cytosol. TAB L E 17. 1

Characteristics of Lipoproteins

Lipoprotein

Diameter (Å)

Chylomicrons

Density (g · cm–3)

% Protein

% Triacylglycerol

% Cholesterol and cholesteryl ester

1000–5000

100 proteins (gray), three of the five snRNAs (green), and a the number of phosphodiester bonds, the cutting-and-pasting portion of an intron (orange). 5′ splice site

5′

3′ splice site

branch point

AG GUAAGU

A

CAG G

3′

20–50 bases 5′ exon

intron

3′ exon

  FIGURE 21.23   Consensus sequence at eukaryotic mRNA splice sites.  Nucleotides

shown in bold are invariant.

648  C ha pter 21   Transcription and RNA

2′ OH 5′

pG

3′

A

1. In the first transesterification reaction, the 2′ OH group of the branch point adenosine residue (which lies within the intron) attacks the phosphate (p) at the 5′ end of the intron. This frees the 3′ end of the first exon and generates a lariatshaped intermediate. Gp 3′ OH

A

p

2. In the second transesterification reaction, the free 3′ OH group of the first exon attacks the 5′ phosphate of the second exon. This reaction forms a phosphodiester bond to unite the two exons. The excised intron diffuses away and is degraded.

  FIGURE 21.24   mRNA splicing.

process, which is catalyzed by the snRNAs, needs no external source of free energy. However, proteins are essential for the overall reaction, which includes ATP hydrolysis–driven conformational changes to position the segments of RNA being processed. Some types of introns, particularly in prokaryotic and protozoan rRNA genes, undergo self-splicing; that is, they catalyze their own transesterification reactions without the aid of proteins. These rRNA molecules were the first RNA enzymes (ribozymes) to be described, in 1982. One hypothesis for the evolutionary origin of splicing suggests that introns, and the splicing machinery itself, are the result of RNA molecules that spliced themselves into mRNA molecules, which were converted to DNA by the action of a reverse transcriptase (see Box 20.A) and then incorporated into the genome through recombination. Introns typically comprise over 90% of a gene’s total length, which means that a lot of RNA must be transcribed and then discarded. Moreover, a cell must spend energy to synthe­ size the RNA and proteins that make up the spliceosome and that destroy intronic RNA and incorrectly spliced transcripts. Finally, the complexity of the splicing process creates many opportunities for things to go wrong: A majority of mutations linked to inherited diseases involve defective splicing. So just what is the advantage of arranging a gene as a set of exons separated by introns? One answer to this question is that splicing allows cells to increase variation in gene expression through alternative splicing. At least 95% of human protein-coding genes exhibit splice variants. Variation may result from selecting alternative sequences to serve as 5′ or 3′ splice sites, from skipping an exon, or from retaining an intron. Thus, certain exons present in the gene may or may not be included in the mature RNA transcript (Fig. 21.25). The signals that govern exon selection and splice sites probably involve RNA-binding proteins that recognize sequences or secondary structures within introns as well as exons. As a result of alterna­tive splicing, a given gene can generate more than one protein product, and gene expression can be finely tailored to suit the needs of different types of cells. The evolutionary advantage of this regulatory flexibility clearly outweighs the cost of making the machinery that cuts and pastes RNA sequences. Alternative splicing also explains why humans are vastly more complex than organisms such as roundworms, which contain a comparable number of genes (Table 21.2).

21.3  RNA Processing  649 gene mRNA transcripts Striated muscle Smooth muscle Nonmuscle/ fibroblast Brain   FIGURE 21.25   Alternative splicing.  The rat gene for the muscle protein α-tropomyosin (top) encodes 12 exons. The mature mRNA transcripts in different tissues consist of different combinations of exons (some exons are found in all transcripts), reflecting alternative splicing pathways.

TA B L E 21.2   Protein-Coding Genes and Protein Count

Organism

Protein-coding genes

Estimated proteins

Saccharomyces cerevisiae ( yeast)

  6000

  6000

Arabidopsis thaliana ( plant)

25,300

48,300

Drosophila melanogaster (fruit fly)

13,000

30,600

Homo sapiens

21,000

118,400

Source: Data from NCBI https://www.ncbi.nlm.nih.gov/genome/.

mRNA turnover and RNA interference limit gene expression Although mRNA accounts for only about 5% of cellular RNA (rRNA accounts for about 80% and tRNA for about 15%), it continuously undergoes synthesis and degradation. The lifespan of a given mRNA molecule is another regulated aspect of gene expression: mRNA molecules decay at different rates. In mammalian cells, mRNA lifespans range from less than an hour to about 24 hours. In eukaryotes, mRNA molecules must also be transported out of the nucleus (Box 21.C). The rate of mRNA turnover depends in part on how rapidly its poly(A) tail is shortened by the activity of deadenylating exonucleases. Poly(A) tail shortening is followed by decapping, which allows exonucleases access to the 5′ end of the transcript and eventually leads to

Box 21.C The Nuclear Pore Complex Messenger RNA is synthesized and processed in the nucleus of eukaryotes, but ribosomes, where the mRNA is translated into protein, are in the cytosol. mRNA exits the nucleus via the nuclear pore complex, an assembly of proteins that spans the double membrane surrounding the nucleus. The yeast complex is built from about 550 copies of 30 different proteins collectively known as nucleoporins. The nuclear pore complexes in vertebrates have a similar structure but are larger. Some structural details suggest that the nuclear pore components share an evolutionary origin with clathrin and other membrane coat proteins (Section 9.4).

The nuclear pore complex consists largely of three symmetric protein rings: one facing the cytoplasm, one facing the nucleo­ plasm, and an inner ring positioned where the inner and outer nuclear membranes fuse together. The inner ring is reinforced by diagonal columns connected by cables, which apparently give the overall structure some flexibility. Not shown in the model here are fibrous extensions that protrude into the cytoplasm and nucleus to form open cagelike structures. These structures may help organize traffic into and out of the nucleus.

CYTOPLASM

side view

top view

NUCLEUS

200 Å

The center of the nuclear pore complex is occupied by intrinsically disordered protein segments contributed by an assortment of nucleoporins. The highly concentrated polypeptides, which are rich in Phe-Gly repeats, form a sort of gel that prevents the contents of the nucleus from spilling out into the cytoplasm. Molecules with a mass less than about 40 kD can freely diffuse through the the gel-filled pore. Larger molecules must be escorted by carrier proteins, called transport factors. The carrier proteins may find their way by tracking along more rigid elements of the pore structure; they may also interact with the Phe-Gly repeats to “melt” into the gel. Without a transport factor to accompany them, large molecules are excluded from the pore, which is the only conduit between the nucleus and cytosol. A single nucleus may harbor a few thousand nuclear pore complexes.

Kim, S. J. et al 2018/Springer Nature

650  C ha pter 21   Transcription and RNA

The 40-nm inner diameter of the pore is large enough for several macromolecular “cargoes” to move simultaneously in the same or opposite directions. Diffusion is inherently random, so directionality is imposed on the transport process by additional proteins that use the free energy of ATP or GTP. By incorporating a nucleotide-hydrolysis step into the transport cycle, the movement becomes one-way. For example, proteins moving into the nucleus rely on the activity of a small G protein called Ran (other types of G proteins are described in Section 10.2). With the appropriate transport factors, proteins can also move from the nucleus to the cytosol. Messenger RNA moving out of the nucleus uses protein cofactors that hydrolyze ATP. Presumably, the transport factors that facilitate mRNA export are able to recognize fully processed mRNA, possibly when poly(A) polymerase finishes its work, so that only mature messages are sent to the cytosol to be translated into protein.

destruction of the entire message (Fig. 21.26). In vivo, the RNA cap and tail are close together because a protein involved in translation binds to both ends of the mRNA, effectively circularizing it. RNA-binding regulatory proteins are almost certainly involved in monitoring RNA integrity. For example, transcripts with a premature stop codon are preferentially degraded, thereby avoiding the waste of synthesizing a nonfunctional truncated polypeptide. AAAAAAAAAAAAA 3′ Poly(A) tail

5′ Cap

AAAAAA Deadenylase

Decapping enzyme

5′ exonuclease

3′ exonuclease

  FIGURE 21.26   mRNA decay.  A mature mRNA bears a 5′ cap and a 3′ poly(A) tail. After a deadenylase has shortened the tail, a decapping enzyme removes the ­methylguanosine cap at the 5′ end. The mRNA can then be degraded by exonucleases from both ends.

21.3  RNA Processing  651

Sequence-specific degradation of certain RNAs, a phenomenon called RNA interference (RNAi), provides another mechanism for regulating gene expression after transcription has occurred. RNA interference was discovered by researchers who were attempting to boost gene expression in various types of cells by introducing extra copies of genetic information in the form of RNA. They observed that instead of increasing gene expression, the RNA— particularly if it was double-stranded—actually blocked production of the gene’s product. This interference or gene-silencing effect results from the ability of the introduced RNA to target a complementary cellular mRNA for destruction. Endogenously produced RNAs, known as small interfering RNAs (siRNAs) and micro RNAs (miRNAs), appear to mediate RNA interference in virtually all types of eukaryotic cells, including human cells. For siRNA, the RNA interference pathway begins with the production of double-stranded RNA, which may result when a single polynucleotide strand folds back on itself in a hairpin. A ribonuclease called Dicer cleaves the double-stranded RNA to generate segments of 20 to 25 nucleotides with a two-nucleotide overhang at each 3′ end (Fig. 21.27). These siRNAs bind to a multiprotein complex called the RNA-induced silencing complex (RISC), where one strand of the RNA (the “passenger” strand) is separated from the other by a helicase and/or degraded by a nuclease. The remaining strand serves as a guide for the RISC to identify and bind to a complementary mRNA molecule. The “Slicer” activity of the RISC, a protein known as Argonaute, then cleaves the mRNA, rendering it unfit for translation. Like RNA splicing, RNA interference at first appears wasteful, but it provides cells with a mechanism for eliminating mRNAs—which could otherwise be translated into protein many times over—in a highly specific manner. It is believed that RNA interference originally evolved as an antiviral defense, since many viral life cycles include the formation of ­double-stranded RNAs. In the miRNA pathway, RNA hairpins containing imperfectly paired nucleotides are processed by Dicer and other enzymes to double-stranded miRNAs that bind to the RISC. The passenger RNA strand is ejected and the remaining strand helps the RISC lo­c ate

1. The endonuclease Dicer cleaves double-stranded RNA into 20- to 25-nucleotide segments.

Dicer

siRNA RISC

2. siRNA binds to the RNA-induced silencing complex (RISC).

RISC cleaved mRNA

5. The Argonaute component of RISC cleaves the mRNA so that it cannot be translated. mRNA

4. The guide strand of the siRNA directs the RISC to bind a complementary mRNA. mRNA

  FIGURE 21.27   RNA interference.  The steps involving siRNA are shown. The miRNA pathway for inactivating mRNA is similar.

3. RISC unwinds and degrades the passenger RNA strand.

SEE ANIMATED PROCESS DIAGRAM Mechanism of RNA interference

652  C ha pter 21   Transcription and RNA

complementary target mRNAs. Whereas an siRNA specifically seeks and destroys an mRNA that is perfectly complementary, an miRNA can bind to a large number of target mRNAs—possibly hundreds—because it forms base pairs with a stretch of only six or seven nucleotides. The captured mRNAs are unavailable for translation and are susceptible to the standard mechanisms for RNA degradation diagrammed in Figure 21.26. In addition to serving as a powerful laboratory technique for silencing genes in order to explore their functions, the RNA interference system is being exploited for practical purposes. RNA-based drugs have been developed to block the expression of genes responsible for some inherited diseases. siRNAs could also be used to turn off viral genes that are necessary for viral replication. Other diseases in which gene silencing would be desirable, such as cancer, are amenable to RNAi therapy, provided that the siRNA can be delivered selectively to cancerous cells. In general, introducing exogenous RNAs into cells is challenging, since nucleic acids don’t easily cross cell membranes, and the presence of extracellular RNA may trigger the body’s innate RNA-degrading antiviral defenses. Apples and potatoes have been engineered to use RNAi to prevent the synthesis of the oxi­dative enzymes that cause browning. It is hoped that these crops will be more acceptable to ­c onsumers who avoid traditional genetically modified foods that contain foreign genes (see Box 3.B).

rRNA and tRNA processing includes the addition, deletion, and modification of nucleotides rRNA transcripts, which are generated mainly by RNA polymerase I in eukaryotes, must be processed to produce mature rRNA molecules. rRNA processing and all but the final stages of ribosome assembly take place in the nucleolus, a discrete liquid phase in the nucleus. The initial eukaryotic rRNA transcript is cleaved and trimmed by endo- and exonucleases to yield three rRNA molecules (Fig. 21.28). The rRNAs are known as 18S, 5.8S, and 28S rRNAs for their sedimentation coefficients (large molecules have larger sedimentation coefficients, a measure of how quickly they settle in an ultra-high-speed centrifuge). rRNA transcripts may be covalently modified (in both prokaryotes and eukaryotes) by the conversion of some uridine residues to pseudouridine and by the methylation of certain bases and ribose 2′ OH groups.

O

H

N

O

N Ribose

Uridine

H O

O Ribose

N N H

Pseudouridine (ψ)

45S RNA 5′

3′

18S

5.8S

28S

  FIGURE 21.28   Eukaryotic rRNA processing.  The initial transcript of about 13.7 kb has a sedimentation coefficient of 45S. Three smaller rRNA molecules (18S, 5.8S, and 28S) are derived from it by the action of nucleases.

This last type of modification is guided by a multitude of small nucleolar RNA molecules (called snoRNAs) that recognize and pair with specific 15-base segments in the rRNA sequences, thereby directing an associated protein methylase to each site. Without the snoRNAs to mediate sequence-specific ribose methylation, the cell would require many different methylases in order to recognize all the different nucleotide sequences to be modified. A rapidly growing mammalian cell may synthesize as many as 7500 rRNA transcripts each minute, each of which associates with about 150 different snoRNAs. The processed rRNAs eventually combine with some 80 different ribosomal proteins to

21.3  RNA Processing  653

H3C

NH

NH2 N

+

N N

N

O

NH2 +

N N Ribose

3-Methylcytidine (m3C)

H (CH3)2N

CH3

C

CH3

N

N

Ribose

N 6-Isopentenyladenosine (i 6A)

1-Methyladenosine (m1A)

H3C

CH

N

N

Ribose

CH2

O H

N

N N

N

O Ribose

N 2,N 2-Dimethylguanosine (m22G)

O N N

H H H H

Ribose Dihydrouridine (D)

  FIGURE 21.29   Some modified nucleotides in tRNA molecules.  The parent nucleo-

tide is in black; the modification is shown in red.

generate fully functional ribosomes, a task that requires careful coordination between RNA synthesis and ribosomal protein synthesis. tRNA molecules, produced by the action of RNA polymerase III in eukaryotes, undergo nucleolytic processing and covalent modification. The initial tRNA transcripts are trimmed by ribonuclease P (see below). Some tRNA transcripts undergo splicing to remove introns. In some bacteria, newly made tRNAs end with a 3′ CCA sequence, which serves as the attachment point for an amino acid that will be used for protein synthesis. In most organisms, however, the three nucleotides are added to the 3′ end of the immature molecule by the action of a nucleotidyl transferase. Up to 25% of the nucleotides in tRNA molecules are covalently modified. The alterations range from simple additions of methyl groups to complex restructuring of the base. Some of the 200 or so known nucleotide modifications are shown in Figure 21.29. These are yet more examples of how cells alter genetic information as it is transcribed from relatively inert DNA to highly variable and much more dynamic RNA molecules.

RNAs have extensive secondary structure Some of the modified nucleotides common in tRNAs also occur in other types of RNA, including mRNA. For example, N 6-methyladenosine occurs at thousands of highly conserved sites in mammalian mRNAs, which suggests that it has some functional significance. This modified nucleotide seems to function like the epigenetic marks in DNA, as it has reader, writer, and eraser enzymes associated with it. RNA methylation and other modifications could affect protein binding, RNA secondary structure, or liquid–liquid phase separation in the nucleus or cytosol. Unlike DNA, whose conformational flexibility is considerably constrained by its d ­ oublestranded nature, single-stranded RNA molecules can adopt highly convoluted shapes through base pairing between different segments. In addition to the standard (Watson–Crick) types of base pairs, RNA accommodates nonstandard base pairs as well as hydrogen-bonding interactions among three bases (Fig. 21.30). Base stacking stabilizes the RNA tertiary structure, achieving the same sort of balance between rigidity and flexibility exhibited by protein enzymes. A folded RNA molecule can then bind substrates, orient them, and stabilize the transition state of a chemical reaction.

654  C ha pter 21   Transcription and RNA

O A R

N H

N

N O

N

R R

A

N

H H

G

H

N

N N R

R

O

N

N N

H N

R

R

N

N N G

H

NH2

O

N H

N N

O C

H

R

N

H N

O

H N

N H N

O

N

N

H

G

A

N

O

N N

N N

N N

N

N

N

H H

U

H N

H

H

NH2

N

N R

R

N

N

A

U

  FIGURE 21.30   Some nonstandard base pairs.  R represents the ribose–phosphate backbone.

  FIGURE 21.31   RNase P.  In this model of the Ther-

motoga maritima enzyme, the 347-nucleotide RNA is gold and the product tRNA is red. The small protein component (117 amino acids) is green.

All cells contain two essential ribozymes: the tRNA-processing RNase P and the ribosomal RNA that catalyzes peptide bond formation during protein synthesis. There are at least six other naturally occurring types of catalytic RNAs (such as those involved in splicing) and many more synthetic ribozymes. Self-splicing introns are not true catalysts, since they cannot participate in more than one reaction cycle, but the RNA component of RNase P, with its numerous base-paired stems and compact protein-like structure, is a true catalyst (Fig. 21.31). At one time it was believed that the enzyme’s RNA molecule merely helped align the tRNA substrate for the protein to cleave, but the bacterial RNase P RNA is able to cleave its substrate in the absence of the RNase P protein. The existence of RNA enzymes such as RNase P lends support to the theory of an early RNA world when RNA functioned as a reposi­ tory of biological information (like modern DNA) as well as a catalyst (like modern proteins). Experiments with synthetic RNAs in vitro have demonstrated that RNA can catalyze a wide variety of chemical reactions, including the biologically relevant synthesis of glycosidic bonds (the type of bond that links the base and ribose in a nucleoside) and RNA-template–directed RNA synthesis. Apparently, most ribozymes that originated in an early RNA world were later supplanted by protein catalysts, leaving only a few examples of RNA’s catalytic abilities.

Before Going On • Describe the structure and function of the mRNA 5′ cap and 3′ tail. • Draw diagrams to recount the events of RNA splicing. • Explain why splicing does not require free energy input. • List the advantages of arranging genes as sets of exons and introns. • Summarize the steps of RNA interference. • Compare RNA polymerase, poly(A) polymerase, and the tRNA CCA-adding enzyme with respect to substrates, products, and requirement for a template. • Explain why the products of transcription exhibit much more variability than the genes that encode them.

Bioinformatics  655

Summary generate an RNA chain that forms a short double helix with the template DNA.

21.1  Initiating Transcription •  Transcription is the process of converting a segment of DNA into RNA. An RNA transcript may represent a protein-coding gene or it may partic­ipate in protein synthesis or other activities, including RNA processing.

•  The polymerase acts processively along the DNA template but reverses to allow the excision of a mispaired nucleotide. •  The elongation phase of transcription in eukaryotes is triggered by phosphorylation of the C-terminal domain of RNA polymerase II.

•  Gene expression may be regulated by altering histones through acetylation, phosphorylation, and methylation, by methylating DNA, and by rearranging nucleosomes. •  Transcription begins at a DNA sequence known as a promoter. A gene to be transcribed must be recognized by a regulatory factor such as the σ factor in prokaryotes. •  In eukaryotes, a set of general transcription factors interact with DNA at the promoter to form a complex that recruits RNA polymerase and may further alter chromatin structure.

•  Transcription termination in prokaryotes involves destabilization of the DNA–RNA hybrid helix. In eukaryotes, transcription termination is linked to RNA cleavage.

21.3  RNA Processing •  mRNA transcripts undergo processing that includes the addition of a 5′ cap structure and a 3′ poly(A) tail. mRNA splicing, carried out by RNA–protein complexes called spliceosomes, joins exons and eliminates introns.

•  Regulatory DNA sequences may affect transcription through binding proteins that interact with RNA polymerase via the Mediator complex.

•  RNA interference is a pathway for inactivating mRNAs according to their ability to pair with a complementary siRNA or miRNA.

•  The bacterial lac operon illustrates the regulation of transcription by a repressor protein.

•  rRNA and tRNA transcripts are processed by nucleases and enzymes that modify particular bases.

21.2  RNA Polymerase

•  The chemical and structural variability of RNA molecules makes it possible for some to function as enzymes.

•  Eukaryotic RNA polymerase II transcribes protein-coding genes. It requires no primer and polymerizes ribonucleotides to

Key Terms transcription gene messenger RNA (mRNA) ribosomal RNA (rRNA) transfer RNA (tRNA) operon RNA processing RNA polymerase noncoding RNA (ncRNA) histone code CpG island imprinting epigenetics

promoter consensus sequence TATA box general transcription factor enhancer activator Mediator silencer repressor nucleolus cap poly(A) tail intron

exon splicing spliceosome small nuclear RNA (snRNA) ribozyme nuclear pore complex RNA interference (RNAi) small interfering RNA (siRNA) micro RNA (miRNA) small nucleolar RNA (snoRNA) RNA world

Bioinformatics Brief Bioinformatics Exercises

21.2  RNA Polymerase, Transcription, and the KEGG Database

21.1  Viewing and Analyzing RNA Polymerase

21.3  Viewing and Analyzing the lac Repressor

656  C ha pter 21   Transcription and RNA

Problems 21.1  Initiating Transcription 1.  Why does the genome contain so many more genes for rRNA than mRNA?

residues in DNA. If a demethylase carries out a hydrolytic reaction to restore cytosine residues, what is the other reaction product?

2.  Why is it effective for a bacterial cell to organize genes for related functions as an operon? How do eukaryotes achieve the same benefits?

CH3

5.  Draw the structures of the amino acid side chains that correspond to the following histone modifications:  a.  acetylation of lysine, b.  phosphorylation of serine,  c.  phosphorylation of histidine.  d.  How do these modifications change the character of their respective side chains? How is the binding affinity between the histone and DNA affected as a result? 6.  A specific type of histone methyltransferase (HMT) catalyzes the methylation of a single lysine or a single arginine in a histone protein (usually H3 or H4). Draw the structures of methylated lysine and methylated arginine residues. 7.  A protein involved in reading the “histone code” binds to a trimethylated lysine in histone H3.  a.  Draw the structure of a trimethylated lysine residue.  b.  Describe the protein’s putative binding site, given the observation that the protein binds to histone H3 only when the lysine is trimethylated. 8.  The reversal of histone arginine methylation converts the methylated arginine to citrulline in a reaction that consumes H2O. Draw the resulting amino acid residue. What is the other product of the reaction? 9.  Histones may be modified by the attachment of a single ubiquitin molecule to a lysine side chain. As described in Section 12.1, proteins destined for the proteasome are also tagged with ubiquitin. How does ubiquitin tagging compare with the modification of a histone with ubiquitin? Are ubiquitinated histones marked for proteolytic destruction by the proteasome? 10.  Enzymes that catalyze histone acetylation (histone acetyltransferases, or HAT) are closely associated with transcription factors, which are proteins that promote transcription. Why is this a good biochemical strategy? 11.  DNA methylation requires the methyl group donor S-adenosylmethionine, which is produced by the condensation of methionine with ATP (see Box 18.B). The sulfonium ion’s methyl group is used in methyl-group transfer reactions. The demethylated S-adenosylmethionine is then hydrolyzed to produce adenosine and a nonstandard amino acid. Draw the structure of this amino acid.  a. How does the cell convert this compound back to methionine to regenerate ­S -adenosylmethionine?  b.  The proper regulation of gene expression requires methylation as well as demethylation of cytosine

S

H CH2

CH2

3.  Proteins can interact with DNA through relatively weak forces, such as hydrogen bonds and van der Waals interactions, as well as through stronger electrostatic interactions such as ion pairs. Which types of interactions predominate for sequence-specific DNA-­ binding proteins and for sequence-independent binding proteins? 4.  Certain proteins that stimulate expression of a gene bind to DNA in a sequence-specific manner and induce conformational changes in the DNA. Describe the purpose of these two modes of interaction with the DNA.

+

H

CH2

C

COO–

+

NH3

O

H HO

Adenine

H

H

OH

S-Adenosylmethionine 12.  Explain why epigenetic marks such as methylation of cytosine residues are almost completely erased during the early stages of embryogenesis. 13.  5-Methylcytosine residues can be converted to 5-formylcytosine and 5-carboxylcytosine.  a.  Draw the structures of these modified bases.  b.  Demethylation of 5-methylcytosine residues in DNA is indirect; instead of removing the methyl group, the base is oxidized and then the entire formylcytosine is removed and replaced by the base excision repair pathway (see Fig. 20.17). List the enzymes that are involved in this process. 14.  Mice that are homozygous for a mutation that renders the DNA methyltransferase enzyme nonfunctional usually die in utero. Why does this mutation have such serious consequences? 15.  Despite their structural similarity, butyrate and ­β-hydroxybutyrate have different physiological roles. Butyrate is a short-chain fatty acid produced by colonic bacteria dining on dietary fermentable fiber, whereas β-hydroxybutyrate is a ketone body.  a.  Draw the structures of these two compounds.  b.  An embryonic kidney cell line was used as a model to investigate the hypothesis that these compounds act as histone deacetyl­ ase (HDAC) inhibitors. Cells were treated with 5 mM butyrate and increasing concentrations of β ­ -hydroxybutyrate. Cell lysates were analyzed by immunoblotting in which histone H3 acetylated at Lys 9 and Lys 14 (H3K9/14Ac) was detected by the ­binding of specific anti-acetyl-Lys antibodies. The results are shown in the figure. Is the hypothesis correct? Hydroxybutyrate, mM 5 mM Control Butyrate

10

20

30

40

H3K9/14Ac

Total H3

c.  In a second experiment, butyrate and β-hydroxybutyrate were investigated for their ability to influence the transcription of the genes PGC1α and CPT1b. (PGC1α is a transcription factor involved in mitochondrial biogenesis, and CPT1b is a carnitine acyltransferase; see Fig. 17.5.) The results are shown in the table. How do butyrate

Problems  657 and β-hydroxybutyrate influence transcription of these two genes and what are the physiological implications? How is an HDAC inhibitor able to influence gene transcription?

Control Butyrate β-Hydroxybutyrate

Relative gene expression PGC1α CPT1b 1.5 1.2 7.9 1.9 1.0 1.3

22.  A rational drug design project was undertaken to identify small molecules that interfere with the binding of σ factor to RNA polymerase. Why would these small molecules serve as effective drugs to treat diseases caused by bacterial pathogens?

16.  The structurally similar compounds butyrate and ­β -hydroxybutyrate (see Problem 15) were investigated for their ­a bility to influence transcriptional changes resulting in the secretion of ­c ytokines CCL2, IL-6, IL-8, and IL-1β. These ­p ro-­i nflammatory p ­ roteins are implicated in diseases such as atherosclerosis and ­a utoimmunity.  a.  Use the results below to evaluate the ability of butyrate and ­β -hydroxybutyrate to elicit a pro-­inflammatory response.  b.  How might these results be related to the ability of butyrate or β-hydroxybutyrate to act as an HDAC inhibitor (see ­Problem 15)?  c.  How might you respond to the claim that the currently p ­ opular “keto” diet (see Problems 17.69−17.72) is anti-inflammatory?

Relative gene expression

3.0 Control Butyrate (5 mM) β-Hydroxybutyrate (20 mM)

2.5 2.0 1.5 1.0 0.5 CCL2

IL-6

IL-8

IL-1β

17. a.  The sense (coding) strand for the E. coli promoter for the rrnA1 gene is shown below. The transcription initiation site is shown by +1. Identify the −35 and −10 regions for this gene.  b. Which region is AT-rich and why is this region composed of A:T and not G:C base pairs?

+1

AATGCTTGACTCTGTAGCGGGAAGGCGTATTATCCAACA

18.  Predict the effect of a mutation in one of the bases in either the −35 or the −10 region of the promoter. 19.  Sp1 is a sequence-specific human DNA-binding protein that binds to a region on the DNA called the GC box, a promoter element with the sequence GGGCGG. Binding of Sp1 to the GC box enhances RNA polymerase II activity 50- to 100-fold. How would you use affinity chromatography (see Section 4.6) to purify Sp1? 20.  Identify possible eukaryotic promoter elements in the sequence of the mouse β globin gene shown below. The first nucleotide to be transcribed is indicated by +1.

+1

21.  One challenge for regulating gene expression is that transcription factors must locate a specific sequence of DNA.  a. About how many times would a unique 10-bp sequence appear in the yeast genome? (Assume that the genome has a random sequence.)  b. About how many times could you expect to find a certain 16-bp sequence in the human genome?

GAGCATATAAGGTGAGGTAGGATCAGTTGCTCCTCACATTT

23.  A histone lysine methyltransferase is upregulated in various types of cancers. How does this upregulation affect Lys 27 of histone 3? How does this affect the transcription of genes associated with this histone? 24.  Three human TBP-associated factors (TAFs) contain protein domains that are homologous to those found in histones H2B, H3, and H4. Why is this finding not surprising? 25.  The T7 bacteriophage RNA polymerase recognizes specific promoter sequences and melts open the DNA to form a transcription bubble without the need for transcription factors.  a.  Dissociation constants (Kd) were measured for the interaction between the polymerase and DNA segments containing the promoter sequences. In some cases, the DNA contained a bulge, caused by a mismatch of one, four, or eight bases, to mimic the intermediates in the formation of a transcription bubble. To which DNA segment does the polymerase bind most tightly? Explain in terms of the DNA structure. DNA promoter segment Fully base-paired One-base bulge Four-base bulge Eight-base bulge

Kd (nM) 315 0.52 0.0025 0.0013

b.  Use the data in the table to calculate the ∆G°′ for the binding of T7 RNA polymerase to fully base-paired DNA and to DNA with an eight-base bulge. (Note: The Kd is the inverse of Keq.) Assume a temperature of 25°C.  c.  What do these results reveal about the thermodynamics of melting open a DNA helix for transcription? What is the approximate free energy cost of forming a transcription bubble equivalent to eight base pairs? 26.  In bacteria, the core RNA polymerase binds to DNA with a dissociation constant of 5 × 10−12 M. The polymerase in complex with its σ factor has a dissociation constant of 10−7 M. Explain the reason for the difference in binding affinities. 27.  One of the genes expressed by the lac operon is lacY, which encodes a lactose permease transporter that allows lactose to enter the cell. Why does the expression of this gene assist in the expression of the operon? 28.  The genes of the lac operon are not expressed when the lac repressor binds to the operator. However, removal of the lac repressor is not sufficient to allow gene expression—a protein called catabolite activator protein (CAP) is also required to assist RNA polymerase and facilitate transcription. CAP can interact with the lac promoter only when it binds to its ligand cAMP. In E. coli, the intracellular concentration of cAMP falls when glucose is present. Describe the activity of the lac operon in each of the following scenarios:  a.  Both lactose and glucose are present.  b.  Glucose is present but lactose is absent.  c.  Both glucose and lactose are absent.  d.  Lactose is present and ­glucose is absent.

658  C ha pter 21   Transcription and RNA

31.  The compound phenyl-β-d-galactose (phenyl-Gal) is not an inducer of the lac operon because it is unable to bind to the repressor. However, it can serve as a substrate for β-galactosidase, which cleaves phenyl-Gal to phenol and galactose. How can the addition of ­p henyl-Gal to growth medium distinguish between wild-type ­bacterial cells and cells that have a mutation in the lacI gene? 32.  In bacterial cells, the genes that code for the enzymes of the tryptophan biosynthetic pathway are organized in an operon, as shown below. Another gene encodes a repressor protein that binds tryptophan. How does the repressor protein control the expression of the genes in the trp operon?

P

trpL

trpE

trpD

trpC

trpB

trpA

33.  RNA polymerase makes an error once in 106 bases while DNA polymerase makes an error once in 109 bases.  a.  Explain why the error rate for DNA polymerase is much lower than that of RNA polymerase.  b.  Why is the cell not harmed by the lower fidelity of RNA polymerase? 34.  The promoters for genes transcribed by eukaryotic RNA polymerase I exhibit little sequence variation, yet the promoters for genes transcribed by eukaryotic RNA polymerase II are highly variable. Explain. 35.  Explain why the adenosine derivative cordycepin inhibits RNA synthesis.

NH2 N

N

N

N HOCH2 H

O

H

H

H

OH

H

Cordycepin

36.  How does your answer to Problem 35 provide evidence to support the hypothesis that transcription occurs in the 5′→3′ direction, not the 3′→5′ direction? 37.  The activity of RNA polymerase II is inhibited by the mushroom toxin α-amanitin (Kd = 10−8 M). By contrast, RNA polymerase III is only moderately inhibited by the toxin (Kd = 10−6 M), and RNA ­polymerase I is not affected at all. What would be the effect of adding 10 nM α-amanitin to cells in culture?

Polymerase activity Absorbance

1500

1.0

II

1000

0.5

500

15

21.2  RNA Polymerase

I

2000

III 20

25

30 35 Fraction number

40

45

39.  Actinomycin D, an antibiotic isolated from Streptomyces bacteria, intercalates between DNA bases and prevents the progression of RNA polymerases. Eukaryotic RNA polymerases (RNAP) differ in their sensitivity to actinomycin D, as shown below. Which enzyme is most sensitive? Which is least sensitive? Compare these results to the sensitivities of the eukaryotic RNA polymerases to α-amanitin (see Problem 37). IC50 (μg ⋅ mL−1) RNAP I RNAP II RNAP III

0.05 0.5 5.0

40.  A newly developed cancer drug selectively inhibits RNA polymerase I without affecting RNA polymerases II and III. Why might this drug be successful at treating cancer? 41.  Triptolide is a cancer drug that inhibits the ATPase activity of TFIIH. Describe the consequences of treating cancer cells with ­triptolide. Does the drug affect transcription by all RNA polymerases equally? 42.  Radioactively labeled γ-[32P]GTP is added to a bacterial culture undergoing transcription. Is the resulting RNA labeled? If so, where? 43.  In addition to its role in transcription, TFIIH participates in nucleotide excision repair (see Fig. 20.20). Using what you know about the mechanism of action of TFIIH, describe its role in the repair process. 44.  The antibiotic rifampicin binds to the β subunit of bacterial RNA polymerase. In the presence of rifampicin, cultured bacterial cells are capable of synthesizing only short RNA oligomers.  a.  At what point in the transcription process does rifampicin exert its inhibitory effect? b.  Why is rifampicin used to treat bacterial infections? 45.  Draw the structure of the heptad repeat in RNA polymerase II, in which Ser 2 and Ser 5 (and possibly Ser 7) are phosphorylated.

Absorbance at 280 nm

30.  A bacterial strain expresses a mutant lac repressor protein that retains its ability to bind to the operator but cannot bind lactose. What is the effect on gene expression in these mutants? What happens when lactose is added to the growth medium?

38.  The three different eukaryotic RNA polymerases were discovered in the 1970s by researchers who loaded cell extracts onto a DEAE ion-exchange column (see Section 4.6) and then eluted the proteins with a salt gradient. Collected fractions were assayed for RNA polymerase activity in the presence and in the absence of Mg2+ ions and in the presence of the mushroom toxin α-amanitin. In addition to the difference in α-amanitin sensitivity described in Problem 37, the investigators noted that RNA polymerase I was fully active in the presence of 5 mM Mg2+ ions whereas polymerases II and III were only 50% active. How did these results support the conclusion that the three peaks constituted three different forms of RNA polymerase?

Polymerase activity (units)

29.  Researchers have isolated bacterial cells with mutations in various segments of the lac operon. What is the effect on gene expression if a mutation in the operator occurs so that the repressor cannot bind? What happens when lactose is added to the growth medium of these mutants?

Problems  659 46. a.  The C-terminal domain of RNA polymerase II projects away from the globular portion of the protein. Why?  b.  In an experiment, RNA polymerase II is truncated to produce a protein with a missing C-terminal domain (CTD). How would this affect cells? 47.  The DNA sequence of a hypothetical E. coli terminator is shown below. N is an abbreviation for any of the four nucleotides.  a. Write the sequence of the mRNA transcript that is made using the top strand as the coding strand.  b.  Draw the hairpin structure that would form in this RNA transcript. 5′ · · · NNAAGCGCCGNNNNCCGGCGCT T T T T TNNN · · · 3′ 3′ · · · NNT TCGCGGCNNNNGGCCGCGAAAAAANNN · · · 5′ 48.  Inosine triphosphate (ITP; Section 18.5) is added to a culture of bacteria, which use it in place of GTP. Inosine (I) base-pairs with cytidine (C), forming two hydrogen bonds.  a.  Write the sequence of the mRNA transcript that is made using the top strand of the gene shown in Problem 47 as the coding strand.  b.  Draw the hairpin structure that would form in this RNA transcript. Compare the stabilities of this RNA hairpin and the RNA hairpin you drew in Solution 47b. How is termination of transcription affected by the ITP substitution? 49.  Formation of an RNA hairpin cannot be the sole factor in the termination of transcription in prokaryotes. Why? 50.  The addition of β,γ-imido nucleoside triphosphates to cells in culture has been shown to inhibit Rho-dependent termination. Explain why.

–

O

O

O

P

HN P

O

–

O O –

O

Base

P

O

CH2

O

–

O

β,γ-Imido nucleoside triphosphate

H

H

H

OH

OH

H

51.  In bacteria, the organization of functionally related genes in an operon allows the simultaneous regulation of expression of those genes. If the operon consists of genes encoding the enzymes for a biosynthetic pathway, then the pathway activity as a whole can be feedback-inhibited when the concentration of the pathway’s final product accumulates. a.  In one mode of feedback regulation, a repressor protein binds to a site in the operon (called the operator) to decrease the rate of transcription only when the repressor has bound a molecule representing the operon’s ultimate metabolic product. Draw a diagram showing how such a regulatory system would work.  b.  Feedback regulation of gene expression can also occur after RNA synthesis has begun. In this case, the presence of the operon’s ultimate product causes transcription to terminate prematurely or leads to an mRNA that cannot be translated. Draw a diagram illustrating this control mechanism. Assume that the feedback mechanism includes a protein to which the product binds.  c.  How would the feedback inhibition system in part b differ if no protein were involved? 52.  In some bacteria, several genes required for the biosynthesis of the redox cofactor flavin adenine dinucleotide (FAD; Fig. 3.2c) are arranged in an operon. Comparisons of the sequences of this operon in different species reveal a conserved sequence in the untranslated region at the 5′ end of the operon’s mRNA. The tertiary structure of an RNA molecule typically includes regions of base pairing and unpaired loops (stem−loop structures). By examining an RNA sequence and noting which positions are most conserved, it is possible to predict the stem−loop structure of the RNA. A portion of a

conserved mRNA sequence called RFN, which regulates the expression of the FAD-synthesizing operon, is shown here. ···GAUUCAGUUUAAGCUGAAGC··· a.  Draw the stem−loop structure for this RNA segment.  b.  In order to function as an FAD sensor, the RFN element (which consists of about 165 nucleotides) must alter its conformation when FAD binds. How could researchers assess RNA conformational changes?  c. FAD can be considered as a derivative of flavin mononucleotide (FMN; a coenzyme that resembles FAD but lacks its AMP moiety), which in turn is derived from riboflavin (Fig. 3.2c). The ability of FAD, FMN, and riboflavin to bind to the RFN element was measured as a dissociation constant, Kd; results are shown in the table. Which compound is the most effective regulator of FAD biosynthesis in the cell? What portion of the FAD molecule is likely to be important for interacting with the mRNA? Compound FAD FMN Riboflavin

Kd (nM)  300   5 3000

53.  A number of human neurological diseases result from the presence of trinucleotide repeats in certain protein-coding genes. The severity of each disease is correlated with the number of repeats, which may increase due to the slippage of DNA polymerase during replication.  a.  The most common repeated triplet is CAG, which is almost always located within an open reading frame. What amino acid is encoded by this triplet (see Table 3.3) and how would the repeats affect the protein?  b.  To test the effect of CAG repeats on transcription, researchers used a yeast expression system with genes engineered to contain CAG repeats. In addition to the expected transcripts corresponding to the known lengths of the genes, RNA molecules up to three times longer were obtained. Based on your knowledge of RNA synthesis and processing, what factors could account for longer-than-expected transcripts of a given gene?  c.  Unexpectedly long transcripts could result from slippage of RNA polymerase II during transcription of the CAG repeats. In this scenario, the polymerase temporarily ceases polymerization, slides backward along the DNA template, and then resumes transcription, in effect re-transcribing the same sequence. Slippage may be triggered by the formation of secondary structure in the DNA template strand. Draw a diagram showing how a DNA strand containing CAG repeats could form a secondary structure that might prevent the advance of RNA polymerase. 54.  In E. coli, replication is several times faster than transcription. Occasionally, the replication fork catches up to an RNA pol­y­merase that is moving in the same direction as replication fork movement. When this occurs, transcription stops, and the RNA polymerase is displaced from the template DNA. DNA polymerase can use the existing RNA transcript as a primer to continue replication.  a. Draw a diagram of this process, showing how such collisions would produce a discontinuous leading strand.  b.  Explain why most E. coli genes are oriented such that replication and transcription proceed in the same direction.

21.3  RNA Processing 55.  In E. coli, mRNA degradation is carried out by an endonuclease, but the mRNA must first be modified by a 5′ pyrophosphohydrolase. What reaction does this enzyme catalyze?

660  C ha pter 21   Transcription and RNA 56.  The bacterial enzyme polynucleotide phosphorylase (PNPase) is a 3′→5′ exoribonuclease that degrades mRNA. The enzyme c­ atalyzes a phosphorolysis reaction, as does glycogen phosphorylase (see ­Section 13.3), rather than hydrolysis.  a.  Write an equation for the mRNA phosphorolysis reaction.  b.  In vitro, PNPase also catalyzes the reverse of the phosphorolysis reaction. What does this reaction accomplish and how does it differ from the reaction carried out by RNA polymerase?  c.  PNPase includes a binding site for long polyribonucleotides, which may promote the enzyme’s processivity. Why would this be an advantage for the primary activity of PNPase in vivo? 57.  Construct a table showing the template, substrates, and product for the following polymerases:  a.  DNA polymerase,  b.  human telomerase,  c.  RNA polymerase,  d.  poly(A) polymerase, and  e. tRNA CCA-adding enzyme. 58.  Tell whether the following elements are found on DNA or RNA: a. cap, b.  CpG islands,  c.  −35 region,  d.  −10 region,  e. poly(A) tail,  f.  TATA box,  g. enhancer, h.  3′ CCA sequence,  i.  5′ and 3′ splice sites,  j. promoter, k.  branch point. 59.  Why are only mRNAs capped and polyadenylated? Why do these post-transcriptional modifications not take place on rRNA or tRNA? 60.  In eukaryotes, enzymes involved in capping the 5′ end and extending the tail on the 3′ end of the nascent RNA bind to the ­C-terminal domain (CTD) of RNA polymerase II. Why is this a good cellular strategy? 61.  Explain why capping the 5′ end of an mRNA molecule makes it resistant to 5′→3′ exonucleases. Why is it necessary for capping to occur before the mRNA has been completely synthesized? 62.  Some short regulatory bacterial RNAs are capped at the 5′ end by the structure shown below.  a.  What metabolite is the source of the capping group?  b.  How might the cap structure link the activity of the attached RNA to the cell’s overall metabolic state?  c. Would the cap help protect the RNA from degradation by 5′ endonucleases?

O NH2 O H

H OH

H2C H OH

O

O H O

P

N

N

O–

P

N

N O

CH2

O

–

O

H

H

H

O

OH

66.  The only mRNA transcripts that lack poly(A) tails are those encoding histones. Why do these mRNA transcripts not require poly(A) tails? 67.  ATP can be labeled with 32P at any one of its three phosphate groups, designated α, β, and γ (see Fig. 12.11). A eukaryotic cell carrying out transcription and RNA processing is incubated with labeled ATP. Where will the radioactive isotope appear in RNA if the ATP is labeled with 32P at the  a.  α position,  b.  β position, and  c.  γ position? 68.  Genetic engineers must modify eukaryotic genes so that they can be expressed in bacterial host cells. Explain why the DNA from a eukaryotic gene cannot be placed directly into the bacteria but is first transcribed to mRNA and then reverse-transcribed back to cDNA. 69.  Introns in eukaryotic protein-coding genes may be quite large, but almost none are smaller than about 65 bp. What are some reasons for this minimum intron size? 70.  Introns are removed co-transcriptionally rather than post-transcriptionally. Why is this a good cellular strategy? 71.  A portion of the β globin gene is shown below. The sequence on the left includes the 5′ splice site and the sequence on the right includes the 3′ splice site for the intron between exons 1 and 2. Identify the 5′ splice site and the 3′ splice site. · · · GGCAGGTTGGTA · · · · ACCCTTAGGCTGCT · · · 72.  The β globin gene contains three exons, so two introns must be removed from the primary mRNA transcript (see Problem 71). What types of mutations would result in splicing errors? How would the mutations affect the protein translated from the improperly spliced mRNA? 73.  The gene for the ovalbumin Y protein contains seven exons. The sequence on the left includes the 5′ splice site and the sequence on the right includes the 3′ splice site for the first exon−intron junction. a.  Identify the 5′ splice site and the 3′ splice site. · · · GAAGAAGGTAAGTT · · · · CTTGCAGGTTCTT · · ·

NH2

O

+

N

65.  A poly(A)-binding protein (PABP) has an affinity for RNA molecules with poly(A) tails. What is the effect of adding PABP to a cellfree system containing mRNA and RNases?

H

RNA 63.  The complex of enzymes that cleaves newly synthesized RNA and attaches a poly(A) tail includes phosphatases. What do the phosphatases do and why is this important for transcription? 64.  The poly(A) polymerase that modifies the 3′ end of mRNA molecules differs from other polymerases. The active sites of DNA and RNA polymerases are large enough to accommodate a double-stranded polynucleotide, but the active site of poly(A) polymerase is much narrower.  a.  Explain why.  b.  Explain how the substrate specificity of poly(A) polymerase differs from that of a conventional RNA polymerase.

b.  Why does intron removal from ovalbumin Y mRNA not require the energy input of ATP hydrolysis? 74.  The choice of exons in alternative mRNA splicing may reflect the rate of transcription. Propose an explanation that links the pace of RNA polymerase to the use of “strong” (common) or “weak” (rare) splice sites by the spliceosomes. 75.  The ribozyme known as RNase P processes certain immature tRNA and rRNA molecules. Comparisons of RNase P RNAs from different species reveal conserved features that appear to be involved in endonuclease activity. For example, an unpaired uridine at position 69 is universally conserved. This residue does not pair with another nucleotide but forms a bulge in the RNA secondary structure. To test whether the identity or the geometry of U69 is critical for RNase P activity, several mutants were constructed, and their endonuclease activities measured. The results for each mutant are given as a rate constant (k) for catalysis and a dissociation constant (Kd) for substrate binding. RNase P RNA Wild-type U69 U69 → G69 U69 → C69 U69 deletion U69 + U70

k (min−1) 0.26 0.0034 0.0056 0.0056 0.0054

Kd (nM) 1.7 73 3 7 181

Chapter 21 Credits  661 a.  What is the effect of mutating U69 to a G or a C residue? Do these data reveal whether the U69 bulge is more important for substrate binding or catalysis?  b.  What is the effect of increasing the size of the bulge by adding a second U residue (the U69 + U70 mutant) or removing the bulge by deleting the U69 residue? c. Like proteins, RNAs have primary, secondary, and tertiary structures. Use what you have learned about the primary, secondary, and tertiary structures of proteins (see Chapter 4) to describe the effect of base substitution on the structure and activity of the ribozyme RNase P. 76.  Explain why the vast majority of nucleic acids with catalytic activity are RNA rather than DNA. 77.  What are the consequences to the cell if mRNA leaves the nucleus before it is “export ready”? 78.  Compare and contrast transcription in prokaryotic and eukaryotic cells. 79.  RNA interference was investigated as a method to silence the gene for vascular endothelial growth factor (VEGF), a protein required for angiogenesis (development of blood vessels) in most cancers. The addition of siRNA specific for the VEGF gene almost completely eliminates the secretion of VEGF from prostate cancer

cells in culture. A portion of the gene sequence (bases 189−207) is shown here. Design an siRNA targeted to this region of the gene. 5′ · · · GGAGTACCCTGATGAGATC · · · 3′ 80.  What are some general concerns regarding the use of siRNA as a drug to inhibit protein synthesis to treat disease (see Problem 79)? 81.  Some tRNA molecules include sulfur-containing nucleotides. Draw the structures of 4-thiouridine and 2-thiocytidine. 82.  tRNA modifications are important for temperature adaptations in thermophiles (organisms that live in extreme heat) and psychrophiles (organisms that live in extreme cold). Provide a hypothesis that explains this observation. 83.  Biochemistry textbooks published a few decades ago often used the phrase “one gene, one protein.”  a.  Why is this phrase no longer accurate?  b.  Is your answer to part a consistent with the data shown in Table 21.2 in which the number of proteins expressed in humans is far greater than the number of protein-coding genes? 84.  Based on the information in this chapter, give at least three reasons why a silent mutation in a gene (that is, a mutation that does not alter the amino acid sequence of the encoded protein) could decrease the amount of protein expressed.

Selected Readings Corbett, A.H., Post-transcriptional regulation of gene expression and human disease, Curr. Opin. Cell Biol. 52, 96–104, doi: 10.1016/ j.ceb.2018.02.011 (2018). [Reviews the steps of post-transcriptional mRNA processing and disorders related to defects in specific proteins.] Cramer, P., Organization and regulation of gene transcription, Nature 573, 48–54, doi: 10.1038/s41586-019-1517-4 (2019). [A succinct review of transcription that includes information about RNA polymerases I and III and emphasizes the importance of separate liquid phases for initiation and elongation.] Gayon, J., From Mendel to epigenetics: History of genetics, C. R. Biol. 339, 225–230, doi: 10.1016/j.crvi.2016.05.009 (2016). [Provides some historical context around the difficulty of defining gene.] Haberle, V. and Stark, A., Eukaryotic core promoters and the functional basis of transcription initiation, Nat. Rev. Mol. Cell Biol. 19, 621–637, doi: 10.1038/s41580-018-0028-8 (2018). [Discusses

promoters, RNA polymerase pausing, and other events of transcription initiation.] Murakami, K.S., Structural biology of bacterial RNA polymerase, Biomolecules 5, 848–864, doi: 10.3390/biom5020848 (2015). [Summarizes key features of the structure and function of the bacterial enzyme.] Stewart, M., Polyadenylation and nuclear export of mRNAs, J. Biol. Chem. 294, 2977–2987, doi: 10.1074/jbc.REV118.005594 (2019). [Reviews the events of transcription termination and movement of processed mRNA through the nuclear pore.] Wilkinson, A.C., Nakauchi, H., and Göttgens, B., Mammalian transcription factor networks: Recent advances in i­ nterrogating biological complexity, Cell Systems 5, 319–331, doi: 10.1016/ j.cels.2017.07.004 (2017). [Describes some practical and theoretical approaches to understanding the role of transcription factors in ­regulating gene expression.]

Chapter 21 Credits Figure 21.1 Based on a diagram by Ali Shilatifard, St. Louis University School of Medicine. Figure 21.2 Image based on 3UVW. Filippakopoulos, P., Picaud, S., Mangos, M., Keates, T., Lambert, J.P., Barsyte-Lovejoy, D., Felletar, I., Volkmer, R., Muller, S., Pawson, T., Gingras, A.C., Arrowsmith, C.H., Knapp, S., Histone recognition and large-scale structural analysis of the human bromodomain family, Cell 149, 214–231 (2012).

Figure 21.3 Image from Wagner, F.R., Dienemann, C., Wang, H., Stützer, A., Tegunov, D., Urlaub, H., & Cramer, P., Structure of SWI/ SNF chromatin remodeller RSC bound to a nucleosome. Nature, 579(7799), 448–451. © 2020 Springer Nature. Figure 21.7 Image based on 1YTB. Kim, Y., Geiger, J.H., Hahn, S., Sigler, P.B., Crystal structure of a yeast TBP/TATA-box complex, Nature 365, 512–520 (1993).

662  C ha pter 21   Transcription and RNA Box 21.A Image of λ repressor based on 1LMB. Beamer, L.J., Pabo, C.O., Refined 1.8 A crystal structure of the lambda repressoroperator complex, J. Mol. Biol. 227, 177–196 (1992). Box 21.A Image of Zif268 based on 1AAY. Elrod-Erickson, M., Rould, M.A., Nekludova, L., Pabo, C.O., Zif268 protein-DNA complex refined at 1.6 A: a model system for understanding zinc finger-DNA interactions, Structure 4, 1171–1180 (1996). Box 21.A Image of GCN4 based on 1DGC. Konig, P., Richmond, T.J., The X-ray structure of the GCN4-bZIP bound to ATF/CREB site DNA shows the complex depends on DNA flexibility, J. Mol. Biol. 233, 139–154 (1993). Box 21.C Image from Kim, S.J., Fernandez-Martinez, J., Nudelman, I., Shi, Y., Zhang, W., Raveh, B., Herricks, T., Slaughter, B.D., Hogan, J.A., Upla, P., Chemmama, I.E., Pellarin, R., Echeverria, I., Shivaraju, M., Chaudhury, A.S., Wang, J., Williams, R., Unruh, J.R., Greenberg, C.H., Rout, M.P., Integrative structure and functional anatomy of a nuclear pore complex. Nature, 555(7697), 475–482. © 2018 Springer Nature. Figure 21.10 Image based on 5OQM. Schilbach, S., Hantsche, M., Tegunov, D., Dienemann, C., Wigge, C., Urlaub, H., Cramer, P., Structures of transcription pre-initiation complex with TFIIH and Mediator, Nature 551, 204–209 (2017). Figure 21.11 Image based on 1LBG. Lewis, M., Chang, G., Horton, N.C., Kercher, M.A., Pace, H.C., Schumacher, M.A., Brennan, R.G.,

Lu, P., Crystal structure of the lactose operon repressor and its complexes with DNA and inducer, Science 271, 1247–1254 (1996). Figure 21.13 Image based on 5FLM. Bernecky, C., Herzog, F., Baumeister, W., Plitzko, J.M., Cramer, P., Structure of transcribing mammalian RNA polymerase II, Nature 529, 551 (2016) Figure 21.14a Image based on 5IYD. He, Y., Yan, C., Fang, J., Inouye, C., Tjian, R., Ivanov, I., Nogales, E., Near-atomic resolution visualization of human transcription promoter opening, Nature 533, 359–365 (2016). Figure 21.14b Based on a drawing by Roger Kornberg. Figure 21.20 Image based on 1CVJ. Deo, R.C., Bonanno, J.B., Sonenberg, N., Burley, S.K., X-Ray crystal structure of the poly(A)-binding protein in complex with polyadenylate RNA, Cell 98, 835–845 (1999). Figure 21.22 Image based on 3JB9. Yan, C., Hang, J., Wan, R., Huang, M., Wong, C., Shi, Y.. Structure of a yeast spliceosome at 3.6-Angstrom resolution, Science 349, 1182–1191 (2015). Figure 21.25 Based on Breitbart, R.E., Andreadis, A., and NadalGinard, B., Annu. Rev. Biochem. 56, 481 (1987). Table 21.2 Data from NCBI, https://www.ncbi.nlm.nih.gov/genome/. Figure 21.31 Image based on 3Q1Q. Reiter, N.J., Osterman, A., TorresLarios, A., Swinger, K.K., Pan, T., Mondragon, A., Structure of a bacterial ribonuclease P holoenzyme in complex with tRNA, Nature 468, 784–789 (2010).

Protein Synthesis DO YOU REMEMBER? • DNA and RNA are polymers of nucleotides, each of which consists of a purine or pyrimidine base, deoxyribose or ribose, and phosphate (Section 3.2). • The biological information encoded by a sequence of DNA is transcribed to RNA and then translated into the amino acid sequence of a protein (Section 3.3). • Amino acids are linked by peptide bonds to form a polypeptide (Section 4.1). • Protein folding and protein stabilization depend on noncovalent forces (Section 4.3).

Science Photo Library-Steve Gschmeissner/Brand X Pictures/ Getty Images

CHAPTER 22

Each of these rapidly growing E. coli cells contains about 50,000 ribosomes, which account for about one-­third of the cell’s dry mass. These numbers reflect the fact that protein manufacturing is typically a cell’s main job; they also mean that cells spend a significant part of their energy budget not only on synthesizing proteins but also on building the infrastructure to make those proteins.

• rRNA and tRNA transcripts are modified to produce functional molecules (Section 21.3).

In the decade that followed the 1953 elucidation of DNA structure, nearly all the components required for expressing genetic information—­that is, making a protein—­were identified, including mRNA, tRNA, and ribosomes. We can examine how these molecules participate in protein synthesis by considering one step at a time, beginning with the attachment of a specific amino acid to the appropriate tRNA molecule. We can then look at how the tRNAs align with an mRNA sequences in a ribosome so that peptide bonds can link the amino acids in the order specified by the mRNA. We will also look at some of the steps required to convert a newly made polypeptide to a fully functional protein.

22.1 tRNA and the Genetic Code LEARNING OBJECTIVES Describe the role of tRNA in reading the genetic code. • Explain why the genetic code is redundant, unambiguous, and nonrandom. • Identify the structural features of tRNAs. • Describe the substrates, products, and catalytic activities of aminoacyl–­tRNA synthetases. • Explain how one tRNA anticodon can pair with more than one mRNA codon.

In protein synthesis, the final step of the central dogma of molecular biology, a sequence of nucleotides (the first language) is translated to a sequence of amino acids (the second language). Soon after Francis Crick helped work out the structure of DNA, he hypothesized that translation required “adaptor” molecules (subsequently identified as tRNA) that carried amino acids and recognized genetic information in the form of nucleotides. The 663

664  C ha pter 22   Protein Synthesis

Thr

Ile

Pro His

Ser

Ser Leu

correspondence between DNA sequences and protein sequences was indisputable, but it required some biochemical detective work to discover the nature of the genetic code. Ultimately, the genetic code was shown to be based on three-­nucleotide codons that are read in a sequential and nonoverlapping manner.

Arg

Arg Glu

Glu

The genetic code is redundant

Ser

A triplet code is a mathematical necessity, since the number of possible combinations of three nucleotides of four different kinds (43, or 64) is more than enough to specify the 20 amino acids found in polypeptides (a doublet code, with 42, or 16, possibilities, would be inadequate). Genetic experiments with mutant bacteriophages demonstrated that triplet codons are read sequentially. For example, a mutation resulting from the deletion of a nucleotide within a gene can be corrected by a second mutation that inserts another nucleotide into the gene. The second mutation can restore gene function because it maintains the proper reading frame for translation. Since a given nucleotide sequence in an mRNA molecule can potentially have three different reading frames (Fig. 22.1), the selection of the proper one depends on the precise identification of a translation start site. In rare cases, a single RNA can be translated in different ways. The genetic code, shown in Table 22.1, is said to be redundant because several mRNA codons may correspond to the same amino acid. In fact, most amino acids are specified by two or more codons (arginine, leucine, and serine each have six codons). Only methionine and tryptophan have only one codon each (they are also among the amino acids that occur least frequently in polypeptides; see Fig. 4.3). The methionine codon also functions as a translation initiation point. Three codons, known as stop or nonsense codons, signal translation termination. In Table 22.1, codons are shaded according to the overall hydrophobic, polar, or ionic character of the corresponding amino acid (using the scheme introduced in Fig. 4.2). Codons for chemically similar amino acids exhibit the same residue at the second position; for example, U at the second position invariably specifies a hydrophobic amino acid. This apparently nonrandom pattern of codon–­amino acid correspondence suggests that the genetic code might have evolved from a simpler system involving only two nucleotides and a handful of amino acids.

  FIGURE 22.1   Reading frames.  Even with a nonoverlapping genetic code based on nucleotide triplets, a given nucleotide sequence has three possible reading frames.

TAB L E 22. 1   The Standard Genetic Code

First position (5ʹ end)

U

U

UUU Phe

UCU Ser

UUC Phe UUA Leu C

A

G

Second position G

Third position (3ʹ end)

UAU Tyr

UGU Cys

U

UCC Ser

UAC Tyr

UGC Cys

C

UCA Ser

UAA Stop

UGA Stop

A

UUG Leu

UCG Ser

UAG Stop

UGG Trp

G

CUU Leu

CCU Pro

CAU His

CGU Arg

U

CUC Leu

CCC Pro

CAC His

CGC Arg

C

CUA Leu

CCA Pro

CAA Gln

CGA Arg

A

CUG Leu

CCG Pro

CAG Gln

CGG Arg

G

AUU Ile

ACU Thr

AAU Asn

AGU Ser

U

AUC Ile

ACC Thr

AAC Asn

AGC Ser

C

AUA Ile

ACA Thr

AAA Lys

AGA Arg

A

AUG Met

ACG Thr

AAG Lys

AGG Arg

G

GUU Val

GCU Ala

GAU Asp

GGU Gly

U

GUC Val

GCC Ala

GAC Asp

GGC Gly

C

GUA Val

GCA Ala

GAA Glu

GGA Gly

A

GUG Val

GCG Ala

GAG Glu

GGG Gly

G

C

A

22.1  tRNA and the Genetic Code  665

The genetic code is essentially universal (there are only a few minor variations in mitochondria and some unicellular eukaryotes). The common genetic code makes genetic engineering possible: a bacterium decodes a human gene in the same way a human cell does. The universal code also allows scientists to deduce evolutionary relationships based on DNA sequence differences (Section 1.4). This would not be possible if each organism had its own way of interpreting genetic information.

tRNAs have a common structure Each tRNA interacts specifically with one codon via its anticodon sequence. A bacterial cell typically contains 30 to 40 different tRNAs, and humans produce over 300 different tRNAs in a tissue-­specific manner. These numbers illustrate redundancy in biological systems, since only 20 different amino acids are routinely incorporated into polypeptides. tRNAs that bear the same amino acid but have different anticodons are called isoacceptor tRNAs. The hundreds of tRNAs in mammals that have the same anticodons but differ in other segments are known as isodecoder tRNAs. Nevertheless, the structures of all tRNA molecules are similar—­even those that carry different amino acids. Each tRNA molecule contains about 76 nucleotides (the range is 54 to 100), of which an average of 13 are post-­transcriptionally modified (the structures of some of these modified nucleotides are shown in Fig. 21.29). Many of the tRNA bases pair intramolecularly, generating the short stems and loops of what is commonly called a cloverleaf secondary structure (Fig. 22.2a). A segment at the 5′ end of the tRNA pairs with bases near the 3′ end to form the acceptor stem (an amino acid attaches to the 3′ end). Several other base-­paired stems end in small loops. The D loop often contains the modified base dihydrouridine

SEE GUIDED TOUR Protein Synthesis

3′

5′ P

D G A

G C G G A U U U

A C C A C G C U U A A

OH

Acceptor stem

G A C A C

C U A

G U GUG C D loop T ψ C C G G A G C U A G G G G A C G G Variable loop C G A U G C A ψ C Anticodon loop A U Y G A A D

C U C G

a.

A

TψC loop

Anticodon

  FIGURE 22.2   Structure of yeast tRNAPhe. a. Secondary

s­ tructure. The 76 nucleotides of this tRNA molecule, which can carry a phenylalanine residue at its 3′ end, form four base-­paired stems arranged in a cloverleaf pattern. Invariant bases are shown in boldface. ψ is pseudouridine and Y is a guanosine derivative. Some C and G residues in this structure are methylated.  b. Tertiary structure,

b. with the various structures colored as in part a. The long arm of the L consists primarily of the anticodon loop and D loop, and the short arm is primarily made up of the TψC loop and acceptor stem. Question  Why is it important that the bases in the anticodon loop point outward?

666  C ha pter 22   Protein Synthesis

(abbreviated D), and the TψC loop usually contains the indicated sequence (ψ is the symbol for the nucleotide pseudouridine; see Section 21.3). The variable loop, as its name implies, ranges from 3 to 21 nucleotides in different tRNAs. The anticodon loop includes the three nucleotides that pair with an mRNA codon. The various elements of tRNA secondary structure fold into a compact L shape that is stabilized by extensive stacking interactions and nonstandard base pairs (Fig. 22.2b). Virtually all the bases are buried in the interior of the tRNA molecule, except for the anticodon triplet and the CCA sequence at the 3′ end. The narrow elongated structure of tRNA molecules allows them to align side-­by-­side so that they can interact with adjacent mRNA codons during translation. However, the tRNA anticodon is located a considerable distance (about 75 Å) from the 3′ aminoacyl group, whose identity is specified by that anticodon.

tRNA aminoacylation consumes ATP Aminoacylation, the attachment of an amino acid to a tRNA, is catalyzed by an aminoacyl–­ tRNA synthetase (AARS). To ensure accurate translation, the synthetase must attach the appropriate amino acid to the tRNA bearing the corresponding anticodon. Many AARSs interact with the tRNA anticodon as well as the aminoacylation site at the other end of the tRNA molecule. An AARS catalyzes the formation of an ester bond between an amino acid and an OH group of the ribose at the 3′ end of a tRNA to yield an aminoacyl–­tRNA:

O O

P O

O

CH2

−

H

O

H

H

3′

tRNA

H

2′

O

H

Adenine

OH

C

O

C

R

amino acid

NH+ 3 Aminoacyl–tRNA

The tRNA molecule is then said to be “charged” with an amino acid. The aminoacylation reaction has two steps and requires the free energy of ATP (Fig. 22.3). The overall reaction is amino acid + tRNA + ATP → aminoacyl−tRNA + AMP + PPi Similar enzyme-­catalyzed reactions use ATP to “activate” fatty acids for oxidation (Section 17.2) and ribose groups for nucleotide synthesis (Section 18.5). Most cells contain 20 different AARS enzymes, corresponding to the 20 standard amino acids (isoacceptor tRNAs are recognized by the same AARS). Although all AARSs catalyze the same reaction, they do not exhibit a conserved size or quaternary structure. Nevertheless, the enzymes fall into two groups based on several shared structural and functional features (Table 22.2). For example, the class I enzymes attach an amino acid to the 2′ OH group of the tRNA ribose, whereas the class II enzymes attach an amino acid to the 3′ OH group (this distinction is ultimately of no consequence, as the 2′-aminoacyl group shifts to the 3′ position before it takes part in protein synthesis). Some bacteria appear to lack the full complement of 20 AARSs. The enzymes most commonly missing are GlnRS and AsnRS (which aminoacylate tRNAGln and tRNAAsn). In these organisms, Gln–­tRNAGln and Asn–­tRNAAsn are synthesized indirectly. First, GluRS and AspRS with relatively low tRNA specificity charge tRNAGln and tRNAAsn with their corresponding acids (glutamate and aspartate). Next, an amidotransferase converts Glu–­tRNAGln and Asp–­tRNAAsn to Gln–­tRNAGln and Asn–­tRNAAsn using glutamine as an amino-­group donor. In some microorganisms, this is the only pathway for producing asparagine.

22.1  tRNA and the Genetic Code  667

H O O–

C

R C

+

  FIGURE 22.3   The aminoacyl–­tRNA synthetase reaction.

ATP

NH+ 3 Amino acid

1. The amino acid reacts with ATP to form an aminoacyl –adenylate (aminoacyl –AMP). The subsequent hydrolysis of the PPi product makes this step irreversible in vivo.

PPi H O R C

C

O O

NH+ 3

P

O

Question  How many “high-­energy” phosphoanhydride bonds break in this process? How many “high-­energy” acyl-­ phosphate bonds form?

TA B L E 22. 2

Ribose–Adenine

 Classes of Aminoacyl–­ tRNA Synthetases

Amino acids

O– Aminoacyl–adenylate (aminoacyl–AMP)

tRNA AMP

Class I

2. The amino acid, which has been “activated” by its adenylylation, reacts with tRNA to form an aminoacyl–tRNA and AMP. Class II

H O R C

C

O

tRNA

NH+ 3 Aminoacyl–tRNA

Arg

Leu

Cys

Met

Gln

Trp

Glu

Tyr

Ile

Val

Ala

Lys

Asn

Pro

Asp

Phe

Gly

Ser

His

Thr

An AARS probably binds a tRNA relatively quickly and nonspecifically at first, via electrostatic interactions between positively charged side chains and tRNA phosphate groups. Specific enzyme–­tRNA contacts form more slowly and involve discrete nucleotides (often chemically modified) or other structural features. Not all AARSs interact with their tRNA anticodons. For example, SerRS must be able to attach serine to tRNAs bearing six different anticodons; for SerRS, the main recognition feature lies in the variable arm of the tRNASer molecules. The structure of a complex of E. coli GlnRS and its cognate (matching) tRNA (tRNAGln) shows the extensive interaction between the protein and the concave face of the tRNA molecule (the inside of the L; Fig. 22.4). Most AARSs can activate an amino acid in the absence of a tRNA molecule, but GlnRS, GluRS, and ArgRS require a cognate tRNA molecule for aminoacyl–­AMP formation. This suggests that the anticodon-­recognition site and the aminoacylation active site somehow communicate with each other, which might help guarantee the attachment of the correct amino acid to the tRNA.

Editing increases the accuracy of aminoacylation The accuracy of the AARS reaction depends on several factors. First, the amino acid binding sites of individual AARS enzymes may be tailored precisely to the geometry and electrostatic properties of a particular amino acid, making it less likely that one of the other 19 amino acids would be activated or transferred to a tRNA molecule. For example, TyrRS (the enzyme responsible for synthesizing Tyr–­tRNATyr) can distinguish between tyrosine and phenylalanine, which have similar shapes, because only tyrosine can form hydrogen bonds with the protein. PheRS uses the opposite strategy: An active-­site Ala residue favors interaction with phenylalanine over tyrosine. The active site of ThrRS includes a zinc atom, which can interact with threonine but not with the similarly sized valine. Kinetic studies indicate that even when different amino acids exhibit similar KM values for a given AARS, the correct amino acid has a higher k cat value; in

  FIGURE 22.4   Structure of GlnRS with tRNAGln.  In this complex, the synthetase is green and the cognate tRNA is red. Both the 3′ (acceptor) end of the tRNA (top right) and the anticodon loop (lower left) are buried in the protein. ATP at the active site is shown in yellow.

668  C ha pter 22   Protein Synthesis

other words, the wrong amino acid might enter the enzyme’s active site but reacts too slowly to generate a significant amount of misacylated tRNA molecules. In some cases, the accuracy of tRNA aminoacylation may be enhanced through proof-­ reading by the AARS. For example, both valine and isoleucine—­which differ only by a single methylene group—­can easily fit into the active site of IleRS. To prevent the synthesis of Val–­tRNAIle, the enzyme relies on two active sites that operate by a “double-­sieve” mechanism. The first active site activates isoleucine and presumably other amino acids that are chemically similar to and smaller than isoleucine (such as valine, alanine, and glycine) but excludes larger amino acids (such as phenylalanine and tyrosine). The second active site, which hydrolyzes aminoacylated tRNAIle, admits only aminoacyl groups that are smaller than isoleucine. Thus, the activating and editing active sites together ensure that IleRS produces only Ile–­tRNAIle (Val–­tRNAIle is made very rarely, about once every 50,000 reactions). The two active sites are on separate domains of the synthetase, so a newly aminoacylated tRNA must visit the proofreading hydrolytic active site before dissociating from the enzyme. About half of the AARS enzymes use some sort of editing mechanism to ensure the high fidelity of aminoacylation. The editing activity may be part of the AARS protein, as with IleRS, or it may be a separate enzyme. For example, d aminoacyl–­tRNA deacylases target tRNAs that have been charged with d amino acids, so that they can be destroyed before they reach a ribosome. Bacterial AARSs are solitary proteins, but in eukaryotes, these enzymes tend to form complexes. In humans, the “multisynthetase” complex includes nine synthetases (ProRS and GluRS are actually on a single polypeptide chain) and three accessory proteins. The advantages of this arrangement might include an enhanced ability to match aminoacylation—­and therefore protein synthesis—­to the cell’s amino acid supply or to the cell’s overall metabolic state. In vertebrates, some AARSs exhibit a variety of functions apart from charging tRNAs; these “moonlighting” enzymes might provide additional ways to coordinate the cell’s activities.

tRNA anticodons pair with mRNA codons During translation, tRNA molecules align with mRNA codons, base pairing in an antiparallel fashion, for example

tRNA anticodon

3′

– A – A – G–

5′

mRNA codon

5′

–U–U– C–

3′

At first glance, this sort of specific pairing would require the presence of 61 different tRNA molecules, one to recognize each of the “sense” codons listed in Table 22.1. In fact, many isoacceptor tRNAs can bind to more than one of the codons that specify their amino acid. For example, yeast tRNAAla has the anticodon sequence 3′–CGI–5′ (I represents the purine nucleotide inosine, a deaminated form of adenosine) and can pair with the alanine codons GCU, GCC, and GCA.

TA BLE 2 2 .3

tRNA anticodon

– C –G– I –

– C –G– I –

– C –G– I –

mRNA codon

– G – C –U–

–G–C–C–

–G–C–A–

 Allowed Wobble Pairs at the Third Codon–­ Anticodon Position

5′ Anticodon base

3′ Codon base

C

G

A

U

U

A, G

G

U, C

I

U, C, A

As Crick originally proposed in the wobble hypothesis, the third codon position and the 5′ anticodon position experience some flexibility, or wobble, in the geometry of their hydrogen bonding. The base pairs permitted by wobbling are given in Table 22.3. The wobble hypothesis explains why many bacterial cells can bind all 61 codons with a set of fewer than 40 tRNAs. The wobble position is a particularly popular location for modified bases in mammalian tRNAs. A few variant tRNAs allow nonstandard amino acids to be occasionally incorporated into polypeptides at positions corresponding to stop codons (Box 22.A).

22.2  Ribosome Structure  669

Box 22.A The Genetic Code Expanded In addition to the 20 standard amino acids listed in Figure 4.2, some amino acid variants can be incorporated into proteins during translation (keep in mind that a mature protein may contain a number of modified amino acids, but these changes almost always take place after the protein has been synthesized). Addition of a nonstandard amino acid during protein synthesis requires a dedicated tRNA and a stop codon that can be reinterpreted. The expanded genetic code includes two naturally occurring amino acids, selenocysteine and pyrrolysine, plus a number of amino acids produced in the laboratory. Selenocysteine occurs in a few proteins in both prokaryotes and eukaryotes, which explains why selenium is an essential trace element. Humans may produce as many as two dozen selenoproteins.

NH CH C

CH2

Se

H

O

Selenocysteine (Sec) residue Selenocysteine (Sec), which resembles cysteine, is generated from serine that has been attached to tRNASec by the action of SerRS. A separate enzyme then converts Ser–­tRNASec to Sec–­tRNASec. This charged tRNA has an ACU anticodon (reading in the 3′ → 5′ direction), which recognizes a UGA codon. Normally, UGA functions as a stop codon, but a hairpin secondary structure in the selenoprotein’s mRNA provides the contextual signal for selenocysteine to be accepted by the ribosome at that point.

A few prokaryotic species incorporate pyrrolysine (Pyl) into certain proteins. Synthesis of these proteins requires a 21st type of AARS that directly charges the tRNAPyl with pyrrolysine.

NH CH C

O CH2

CH2

CH2

Pyrrolysine (Pyl) residue The Pyl–­tRNAPyl recognizes the stop codon UAG, which is reinterpreted as a Pyl codon with the help of a protein that recognizes secondary structure in the mRNA bound to the ribosome. In the laboratory, proteins containing unnatural amino acids can be synthesized in bacterial, yeast, and mammalian cells. These experimental systems rely on a pair of genetically engineered components: a tRNA that can “read” a stop codon and an AARS that can attach the unnatural amino acid to the tRNA. When the cell translates an mRNA containing the stop codon, the novel amino acid is incorporated at that codon. Dozens of amino acid derivatives with fluoride, reactive acetyl and amino groups, fluo­ rescent tags, and other modifications have been introduced into specific proteins using this technology. Because the novel amino acids are genetically encoded, they appear only at the expected positions in the translated protein—a more reliable outcome than chemically modifying a protein in a test tube. Question  What is the disadvantage for a cell to have a variant tRNA that can insert an amino acid at a stop codon?

• Explain why a universal genetic code is useful for genetic engineers. • Draw a simple diagram of a tRNA molecule and label its parts. • Write an equation for each step of the aminoacyl–­tRNA synthetase reaction and identify the energy-­requiring step. • Explain why accurate aminoacylation is essential for accurate translation. • Explain why cells do not require 61 different codons. • Identify some codons whose meanings would change following a single-­nucleotide substitution. Identify some codons whose meanings would not change.

Recognize the major features of the ribosome. • Explain the importance of ribosomal RNA. • Identify the three tRNA binding sites in the ribosome.

CH3

C N

• Summarize the features of the genetic code.

LEARNING OBJECTIVES

NH

O

Before Going On

22.2 Ribosome Structure

CH2

670  C ha pter 22   Protein Synthesis

TA B L E 22. 4   Ribosome Components

RNA

Polypeptides

E. coli ribosome (70S)   Small subunit (30S)

16S

21

  Large subunit (50S)

23S, 5S

31

  Small subunit (40S)

18S

33

  Large subunit (60S)

28S, 5.8S, 5S

47

Mammalian ribosome (80S)

In order to synthesize a protein, genetic information (in the form of mRNA) and amino acids (attached to tRNA) must get together so that the amino acids can be covalently linked in the specified order. This is the job of the ribosome, and it is a huge job: A large mammalian cell may contain a few billion protein molecules.

a.

b.

  FIGURE 22.5   Structure of the 30S ribosomal subunit from Thermus thermophilus.  a. The 30S subunit with the rRNA in gray and the proteins in purple.  b. Structure of the 16S rRNA alone.

The ribosome is mostly RNA The ribosome is a large complex containing both RNA and protein. At one time, ribosomal RNA (rRNA) was believed to serve as a structural scaffolding for ribosomal proteins, which presumably carried out protein synthesis, but it is now clear that rRNA itself is central to ribosomal function. A typical bacterial cell contains tens of thousands of ribosomes, a yeast cell around 200,000, and a large mammalian cell up to 10 million. These numbers account for the observation that at least 80% of a cell’s RNA is located in ribosomes (tRNA comprises about 15% of cellular RNA; mRNA accounts for only a few percent of the total). A ribosome consists of a large and a small subunit containing rRNA molecules, all of which are described in terms of their sedimentation coefficients, S. Thus, the 70S bacterial ribosome has a large (50S) and a small (30S) subunit (the sedimentation coefficient indicates how quickly a particle settles during ultracentrifugation; it is related to the particle’s mass). The 80S eukaryotic ribosome is made up of a 60S large subunit and a 40S small subunit. The compositions of prokaryotic and eukaryotic ribosomes are listed in Table 22.4. Regardless of its source, about two-­thirds of the mass of a ribosome is due to the rRNA; the remainder is due to dozens of different proteins (80 unique proteins in eukaryotes). The core of the ribosome is probably the most highly conserved structure across all forms of life. The structures of intact ribosomes from both prokaryotes and eukaryotes have been elucidated by X-­ray crystallography—a monumental undertaking, given the ribosome’s large size (about 2500 kD in bacteria and about 4300 kD in eukaryotes). The small ribosomal subunit from the heat-­tolerant bacterium Thermus thermophilus is shown in Figure 22.5. The overall shape of the subunit is defined by the 16S rRNA (1542 nucleotides in E. coli), which has numerous base-­paired stems and loops that fold into several domains. This multidomain structure appears to confer some conformational flexibility on the 30S subunit—a requirement for protein synthesis. Twenty-­one small polypeptides dot the surface of the structure. Compared to the 30S subunit, the prokaryotic 50S subunit is solid and immobile. Its 23S rRNA (2904 nucleotides in E. coli) and 5S rRNA (120 nucleotides) fold into a single mass (Fig. 22.6). As in the small subunit, the ribosomal proteins associate with the surface of the rRNA, but the surfaces of the large and small subunits that make contact in the intact 70S ribosome are largely devoid of protein. This highly conserved rRNA-­rich subunit interface is the site where mRNA and tRNA bind during protein synthesis. Eukaryotic ribosomes are about 40–50% larger than those from bacteria and contain many additional proteins and more extensive rRNA. The RNA sequences that have no counterparts in bacterial ribosomes are known as expansion segments; these

22.2  Ribosome Structure  671

structures, along with the unique eukaryotic protein components, surround a core structure that is shared with the simpler bacterial ribosome (Fig. 22.7). Eukaryotic rRNA includes a higher percentage of modified nculeotides than prokaryotic rRNA. In eukaryotes, the large and small ribosomal subunits are assembled mostly in the nucleolus, a dense membraneless organelle inside the nucleus, where clustered repeats of several hundred rRNA genes are localized. In mammals, over 250 proteins participate in ribosome assembly by cleaving newly made rRNA (see Fig. 21.28), modifying the rRNA with the help of small nucleolar RNAs (snoRNAs), and arranging the ribosomal proteins (which must be imported into the nucleus via the nuclear pore complex; Box 21.C). Theoretical studies suggest that a structure based on a few relatively large rRNA molecules plus numerous smaller proteins of similar size maximizes the efficiency of ribosome assembly. The cell can manufacture many small proteins in less time than it would take to make one large protein. The rRNAs are synthesized more quickly than the proteins, so there is no constraint on their size. The large and small ribosomal subunits exit the nucleus and finish maturing in the cytosol. As described below, the two subunits must be separate before translation can begin. Excess ribosomes are destroyed by autophagy (Section 9.4), particularly during periods of low nutrient intake when protein synthesis is curtailed and the ribosomal proteins are needed as a source of amino acids for fuel.

  FIGURE 22.6   Structure of the 50S ribosomal subunit from Haloarcula marismortui.  The 50S subunit is shown with rRNA in gray and proteins in green. Most of the ribosomal proteins are not visible in this view. The protein-­free central area forms the interface with the 30S subunit.

Three tRNAs and one mRNA bind to the ribosome

Ben-Shem, A., et al. 2011/AAAS

Up to three tRNA molecules may bind to the ribosome at a given time (Fig. 22.8). The binding sites are known as the A site (for aminoacyl), which accommodates an incoming aminoacyl–­ tRNA; the P site (for peptidyl), which binds the tRNA with the growing polypeptide chain; and the E site (for exit), which temporarily holds a deacylated tRNA after peptide bond formation. The anticodon ends of the tRNAs extend into the 30S subunit to pair with mRNA codons, while their aminoacyl ends extend into the 50S subunit, which catalyzes peptide bond formation. In bacteria, the two ribosomal subunits and the various tRNAs are held in place mainly by RNA–­RNA contacts, with a number of stabilizing Mg2+ ions. In eukaryotic ribosomes, numerous proteins form intersubunit bridges. In both cases, the mRNA, which threads through the 30S subunit, makes a sharp bend between the codons in the A site and P site,

The exit tunnel

  FIGURE 22.7   Eukaryotic ribosomal subunits.  The solvent-­exposed surfaces of the 40S subunit (left) and 60S subunit (right) are shown with the conserved ribosomal core structures in gray. Proteins that are unique to eukaryotes are shown in transparent yellow, and rRNA expansion segments are red. Newly synthesized proteins emerge from the ribosome through the exit tunnel.

Question  Which areas of the ribosome appear to be most conserved between bacteria and eukaryotes? Why?

Schmeing, T et al. 2009/Springer Nature.

672  C ha pter 22   Protein Synthesis

  FIGURE 22.8   Model of the complete bacterial ribosome.  The large subunit is shown in shades of gold (rRNA) and brown (proteins), and the small subunit in shades of blue (rRNA) and purple (proteins). The three tRNAs are colored magenta (A site), green (P site), and yellow (E site). An mRNA molecule is shown in dark gray. Note that the anticodon ends of the tRNAs contact the mRNA in the small subunit, while their aminoacyl ends are buried in the large subunit, where peptide bond formation occurs.

a.

  FIGURE 22.9   Cryo-electron tomography of a human cell.  The image in part a, from a cancerous HeLa cell, has been color-­coded in part b. Gold = chromatin, light purple = endoplasmic reticulum, red = actin filaments, green = microtubules, blue = large ribosomal subunits, yellow = small ribosomal subunits, and dark purple = nuclear pore complexes. The arrowhead marks the nuclear envelope.

b.

Wolfgang Baumeister, Max Planck Institute of Biochemistry

where an Mg2+ ion interacts with mRNA backbone phosphate groups (see Fig. 22.8). The kink allows two tRNAs to fit side-­by-­side while interacting with consecutive mRNA codons. It may also help the ribosome maintain the reading frame by preventing it from slipping along the mRNA. In a prokaryotic cell, DNA, mRNA, and ribosomes are in the same cellular compartment. But in eukaryotic cells, mRNA is synthesized and processed inside the nucleus and then exported across the double nuclear membrane via the nuclear pore. Ribosomes in the cytoplasm then translate the mRNA into protein. Cryo-electron tomography, a method for reconstructing three-­dimensional cellular structures by analyzing frozen slices by electron microscopy, has been used to visualize ribosomes and other structures (Fig. 22.9). The ribosomes appear to cluster near the nuclear envelope in a network of actin filaments and microtubules (see Section 5.3).

22.3 Translation  673

Ribosomes are not all identical and interchangeable; some show a preference for producing certain types of proteins, which are often related to a particular cellular function. Ribosome specialization might depend on the location of the ribosome or on the presence or absence of particular ribosomal proteins, which also exhibit variable chemical modifications. In mammals, defects in ribosomal proteins cause tissue-­specific abnormalities, which would not be the case if all ribosomes in all cells were equally affected. Ribosomes that translate a preferred set of mRNAs, perhaps by recognizing specific mRNA sequences or structural features, could produce proteins at different rates or in different parts of the cell and would provide a way to coordinate the production of subunits in proteins with quaternary structure.

Before Going On • Describe the overall structure of a ribosome and its three tRNA binding sites. • Summarize the structural importance of ribosomal RNA and ribosomal proteins. • Compare bacterial and eukaryotic ribosomes.

22.3 Translation LEARNING OBJECTIVES Summarize the events of translation initiation, elongation, and termination. • Identify steps carried out by proteins and steps carried out by RNA. • Explain why transpeptida­tion does not require free energy input. • Explain how the ribosome maximizes the accuracy of translation. • Describe the role of GTP in translation. • List some factors that affect the translation rate. Like DNA replication and RNA transcription, protein synthesis can be divided into separate phases for initiation, elongation, and termination. These stages require an assortment of accessory proteins that bind to tRNA and to the ribosome in order to enhance the speed and accuracy of translation.

Initiation requires an initiator tRNA In both prokaryotes and eukaryotes, protein synthesis begins at an mRNA codon that specifies methionine (AUG). In bacterial mRNAs, this initiation codon lies about 10 bases downstream of a conserved sequence at the 5′ end of the mRNA, called a Shine–­Dalgarno sequence (Fig. 22.10). This sequence base pairs with a complementary sequence at the 3′ end of the 16S rRNA, thereby positioning the initiation codon in the ribosome. Eukaryotic mRNAs lack a Shine–­Dalgarno sequence that can pair with the 18S rRNA. Instead, translation usually begins at the first AUG codon of an mRNA molecule. In all types of organisms, a start codon may also have a CUG or GUG sequence. The initiation codon is recognized by an initiator tRNA that has been charged with methio­ nine. This tRNA does not recognize other Met codons that occur elsewhere in the coding sequence of the mRNA. In bacteria, chloroplasts, and mitochondria, the methionine attached to the initiator tRNA is modified by the transfer of a formyl group from tetrahydrofolate

674  C ha pter 22   Protein Synthesis mRNA A A A A C C

A G G A G

C U A U U U A A U G G C A A C A

U C C U C 3′ A U

C A C U A G

rRNA   FIGURE 22.10   Alignment of a Shine–­Dalgarno sequence with 16S rRNA.  A segment of the

rRNA (purple) pairs with the Shine–Dalgarno sequence (blue) in the mRNA, just upstream of the start codon (green). Note that the Shine–Dalgarno sequence shown is a consensus sequence that varies slightly from gene to gene.

(see Section 18.2). The resulting aminoacyl group is designated f Met, and the initiator tRNA is known as tRN​​A​  Met f​  ​​:

CH3 S CH2 O HC

CH2 O NH

CH

C

O

tRNA fMet

N-Formylmethionine–tRNA fMet (f Met–tRNA fMet )

SEE ANIMATED PROCESS DIAGRAM Translation initiation in E. coli

Because the amino group of f Met is derivatized, it cannot form a peptide bond. Consequently, f Met can be incorporated only at the N-­terminus of a polypeptide. Later, the formyl group or the entire f Met residue may be removed. In eukaryotic and archaebacterial cells, the initiator tRNA, designated tRN​​A​  Met i​  ​​, is charged with methionine but is not formylated. Initiation in E. coli requires three initiation factors (IFs) called IF-1, IF-2, and IF-3. IF-3 binds to the small ribosomal subunit to promote the dissociation of the large and small subunits. f  Met–­tRN​​A​  fMet ​  ​​binds to the 30S subunit with the assistance of IF-2, a GTP-­binding protein. IF-1 sterically blocks the A site of the small subunit, thereby forcing the initiator tRNA into the P site. An mRNA molecule may bind to the 30S subunit either before or after the initiator tRNA has bound, indicating that a codon–­anticodon interaction is not essential for initiating protein synthesis. After the 30S–­mRNA–­f  Met–­tRN​​A​  Met f​  ​​complex has assembled, the 50S subunit associates with it to form the 70S ribosome. This change causes IF-2 to hydrolyze its bound GTP to GDP + Pi and dissociate from the ribosome. The ribosome is now poised—­with f Met–­tRN​​A​ fMet ​  ​​ at the P site—­to bind a second aminoacyl–­tRNA in order to form the first peptide bond (Fig. 22.11). Similar events occur during translation initiation in eukaryotes, when a 40S and a 60S subunit associate following the binding of Met–­tRN​​A​ iMet ​  ​​ to an initiation codon. However, eukaryotes require at least 12 distinct initiation factors. Among these are proteins that recognize the 5′ cap and poly(A) tail of the mRNA (see Section 21.3) and interact so that the mRNA actually forms a circle. Initiation may also require the RNA helicase activity of an initiation factor called eIF-­4A to remove secondary structure in the mRNA that would impede translation. The 40S subunit scans the mRNA in an ATP-­dependent manner until it encounters the first AUG codon, which is typically 50 to 70 nucleotides downstream of the 5′ cap (Fig. 22.12). The initiation factor eIF2 (the e signifies eukaryotic) hydrolyzes its bound GTP and dissociates, and the 60S subunit then joins the 40S subunit to form the intact 80S ribosome. IF-2 and eIF2 operate much like the heterotrimeric G proteins that participate in intracellular signal transduction pathways (see Section 10.2). In each case, GTP hydrolysis induces conformational changes that trigger additional steps of the reaction sequence.

22.3 Translation

FIGURE 22.11 Summary of translation initiation in E. coli. For simplicity, the ribosomal E site is not shown.

30S GTP

Question How must this diagram be modified in order to illustrate initiation in a eukaryote?

IF-2 mRNA

1. mRNA and fMet–tRNAfMet in complex with IF-2–GTP bind to the small (30S) ribosomal subunit.

tRNA Met f GTP IF-2 5′

675

fMet

mRNA

AUG

50S

2. Association of the large (50S) subunit with the 30S subunit triggers IF-2 to hydrolyze its bound GTP.

GDP IF-2

A site

P site

AUG

The tRNA bearing the initial fMet is positioned in the P site of the ribosome.

The appropriate tRNAs are delivered to the ribosome during elongation All tRNAs have the same size and shape so that they can fit into small slots in the ribosome. In each reaction cycle of the elongation phase of protein synthesis, an aminoacyl– tRNA enters the A site of the ribosome (the initiator tRNA is the only one that enters the P site without first binding to the A site). After peptide bond formation, the tRNA moves to the P site, and then to the E site. As Figure 22.8 shows, there is not much room to spare. In addition, all tRNAs must be able to bind interchangeably with protein cofactors. Aminoacyl–tRNAs are delivered to the ribosome in a complex with a GTP-binding elongation factor (EF) known as EF-Tu in E. coli. EF-Tu is one of the most abundant E. coli proteins (about 100,000 copies per cell, enough to bind all the aminoacyl– tRNA molecules). An aminoacyl–tRNA can bind on its own to a ribosome in vitro, but EF-Tu increases the rate in vivo. Because EF-Tu interacts with all 20 types of aminoacyl–tRNAs (representing more than 20 different tRNA molecules), it must recognize common elements of tRNA structure, primarily the acceptor stem and

initiation factors cap-binding protein eIF-2

poly(A) tail 40S

tRNA Met i AUG

stop codon

mRNA FIGURE 22.12 Circularization of eukaryotic mRNA at translation initiation. A number of initiation factors form a complex that links the 5′ cap and 3′ poly(A) tail of the mRNA. The small (40S) ribosomal subunit binds to the mRNA and locates the AUG start codon.

676  C ha pter 22   Protein Synthesis

a.

b.

Ogle, J. M. et al. 2001/AAAS.

  FIGURE 22.13   Structure of an EF-­Tu–­tRNA complex.  The protein (blue) interacts with the acceptor end and TψC loop of an aminoacyl–­tRNA (red).

one side of the TψC loop (Fig. 22.13). A highly conserved protein pocket accommodates the aminoacyl group. Despite the differing chemical properties of their amino acids, all aminoacyl–­tRNAs bind to EF-­Tu with approximately the same affinity (uncharged tRNAs bind only weakly to EF-­Tu). Apparently, the protein interacts with aminoacyl–­tRNAs in a combinatorial fashion, offsetting less-­than-­optimal binding of an aminoacyl group with tighter binding of the acceptor stem and vice versa. This ­allows EF-­Tu to deliver and surrender all 20 aminoacyl–­tRNAs to a ribosome with the same ­efficiency. EF-­Tu does not know in advance which amino acid will be needed by the ribosome. In a cell, the incoming aminoacyl–­tRNA is selected on the basis of its ability to recognize a complementary mRNA codon in the A site. Due to competition among all the aminoacyl–­tRNA molecules in the cell, this is the rate-­limiting step of protein synthesis. Before the 50S subunit catalyzes peptide bond formation, the ribosome must verify that the correct aminoacyl–­tRNA is in place. This is accomplished by interactions between the rRNA, mRNA, and tRNA. For example, when tRNA binds to the A site of the 30S subunit, two highly conserved residues (A1492 and A1493) of the 16S rRNA “flip out” of an rRNA loop in order to form hydrogen bonds with various parts of the mRNA codon as it pairs with the tRNA anticodon. These interactions physically link the two rRNA bases with the first two base pairs of the codon and anticodon so that they can sense a correct match between the mRNA and tRNA (Fig. 22.14). Incorrect base pairing at the first or second codon position would prevent this three-­way mRNA–­tRNA–­rRNA interaction. As expected from the wobble hypothesis (Section 22.1), the A1492/A1493 sensor can accommodate nonstandard base pairing at the third codon position. As the rRNA nucleotides shift to confirm a correct codon–­anticodon match, the conformation of the ribosome changes in such a way that the G protein EF-­Tu is induced to hydrolyze its bound GTP. As a result of this reaction, EF-­Tu dissociates from the ribosome, leaving behind the tRNA with its aminoacyl group to be incorporated into the growing polypeptide chain. However, if the tRNA anticodon is not properly paired with the A-­site codon, the 30S conformational change and GTP hydrolysis by EF-­Tu do not occur. Instead, the aminoacyl–­ tRNA, along with EF-­Tu–­GTP, dissociates from the ribosome. Because a peptide bond cannot form until after EF-­Tu hydrolyzes GTP, EF-­Tu ensures that polymerization does not occur unless the correct aminoacyl–­tRNA is positioned in the A site. The energetic cost of proofreading at the decoding stage of translation is the free energy of GTP hydrolysis (catalyzed by EF-­Tu). The function of EF-­Tu is summarized in Figure 22.15. In eukaryotes, elongation factor eEF1α performs the same service as the prokaryotic EF-­Tu. The functional correspondence between some prokaryotic and eukaryotic translation cofactors is given in Table 22.5.

  FIGURE 22.14   The ribosomal sensor for proper codon–­anticodon pairing.  These images

show the A site of the 30S subunit in the  a. absence and  b. presence of mRNA and tRNA analogs. The rRNA is gray, with the “sensor” bases in red. The mRNA analog, representing the A-­site codon, is purple, and the tRNA analog (labeled ASL) is gold. A ribosomal protein (S12) and two Mg2+ ions (magenta) are also visible. Note how rRNA bases A1492 and A1493 flip out to sense the codon–­anticodon interaction.

22.3 Translation Correct aminoacyl–tRNA

GTP EF-Tu

GDP GTP

EF-Tu

EF-Tu

A site

1. EF-Tu–GTP delivers an aminoacyl–tRNA to the A site of the ribosome.

GTP

GTP

EF-Tu

EF-Tu

Incorrect aminoacyl–tRNA

2. If the tRNA anticodon matches the mRNA codon, EF-Tu hydrolyzes its GTP and dissociates from the ribosome, leaving the aminoacyl–tRNA in the A site.

3. If the tRNA anticodon and mRNA codon are mismatched, the aminoacyl–tRNA dissociates before EF-Tu hydrolyzes GTP.

GTP EF-Tu

Ribosome with a mismatched tRNA

FIGURE 22.15

Function of EF-Tu in translation elongation in E. coli.

TA B L E 22.5

Prokaryotic and Eukaryotic Translation Factors

Prokaryotic protein

Eukaryotic protein

Function

IF-2

eIF2

Delivers initiator tRNA to P site of ribosome

EF-Tu

eEF1α

Delivers aminoacyl–tRNA to A site of ribosome during elongation

EF-G

eEF2

Binds to A site to promote translocation following peptide bond formation

RF-1, RF-2

eRF1

Binds to A site at a stop codon and induces peptide transfer to water

The ribosome itself performs a bit of proofreading. The departure of EF-Tu– GDP leaves behind the aminoacyl–tRNA, whose acceptor end can now slip all the way into the A site of the 50S ribosomal subunit. The 30S subunit closes in around the tRNA, but at this point, the only interactions that hold the aminoacyl–tRNA in place are codon–anticodon contacts. If there is still a slight mismatch, such as a G:U base pair at the first or second position, which is not detectable by the A1492/A1493 sensor, the strain of not being able to form a perfect Watson– Crick pair will be felt by the ribosome and possibly the tRNA itself, and the tRNA will slip out of the A site. In this way, the ribosome verifies correct codon–anticodon pairing twice for each aminoacyl–tRNA: when EF-Tu first delivers it to the ribosome and after EF-Tu departs. Ribosomal proofreading helps limit the error rate of translation to about 10–4 (one mistake for every 104 codons).

The peptidyl transferase active site catalyzes peptide bond formation When the ribosomal A site contains an aminoacyl– tRNA and the P site contains a peptidyl–tRNA (or, prior to formation of the first peptide bond, an initiator tRNA), the peptidyl

Ribosome ready for transpeptidation

677

678

C hA pTER 22

Protein Synthesis P site

A site

P site

A site

NH

P site

A site

E site

CH

O

C

NH

NH

R CH

R CH

O

O

C NH2

NH Aminoacyl–tRNA arrives

R

C NH

R

CH

R

CH

R

CH

O

C

O

C

O

C

O

O

tRNA

tRNA

Peptidyl–tRNA

Aminoacyl–tRNA

OH tRNA Uncharged tRNA

O tRNA Peptidyl–tRNA

FIGURE 22.16 The peptidyl transferase reaction. Note that the nucleophilic attack of the aminoacyl group on the peptidyl group produces a free tRNA in the P site and a peptidyl–tRNA in the A site. transpeptidation

translocation

FIGURE 22.17 Movement of tRNAs through the ribosome. The aminoacyl–tRNA enters the A site and, after transpeptidation, moves to the P site. The tRNA that previously carried the peptidyl group moves from the P site to the E site, displacing the tRNA from the previous cycle.

Question Sketch a diagram to show the next three steps of the process.

Question Compare this reaction to the condensation reaction of two amino acids shown in Section 4.1.

transferase activity of the large subunit catalyzes a transpeptidation reaction in which the free amino group of the aminoacyl–tRNA in the A site attacks the ester bond that links the peptidyl group to the tRNA in the P site (Fig. 22.16). This reaction lengthens the peptidyl group by one amino acid at its C-terminal end. Thus, a polypeptide grows in the N → C direction. No external source of free energy is required for transpeptidation because the free energy of the broken ester bond of the peptidyl–tRNA is comparable to the free energy of the newly formed peptide bond. (Recall, though, that ATP was consumed in charging the tRNA with an amino acid.) The peptidyl transferase active site lies in a highly conserved region of the bacterial 50S subunit, and the newly formed peptide bond is about 18 Å away from the nearest protein. Thus, the ribosome is a ribozyme (an RNA catalyst). How does rRNA catalyze peptide bond formation? Two highly conserved rRNA nucleotides, G2447 and A2451 in E. coli, do not function as acid–base catalysts, as was initially proposed. Rather, these residues help position the substrates for reaction, an example of induced fit (Section 6.3). Binding of a tRNA at the A site triggers a conformational change that exposes the ester bond of the peptidyl–tRNA in the P site. At other times, the ester bond must be protected so that it does not react with water, a reaction that would prematurely terminate protein synthesis. Proximity and orientation effects in the ribosome increase the rate of peptide bond formation about 107-fold above the uncatalyzed rate. Some antibiotics exert their effects by binding to the peptidyl transferase active site to directly block protein synthesis (Box 22.B). During transpeptidation, the peptidyl group is transferred to the tRNA in the A site, and the P-site tRNA becomes deacylated. The new peptidyl–tRNA then moves into the P site, and the deacylated tRNA moves into the E site. The mRNA, which is still base paired with the peptidyl–tRNA anticodon, advances through the ribosome by one codon (Fig. 22.17). Experiments designed to assess the force exerted by the ribosome demonstrate that formation of a peptide bond causes the ribosome to loosen its grip on the mRNA. Cryo-EM studies of ribosomes frozen at different stages are revealing how events at the peptidyl transferase site in the large subunit are communicated to the mRNA decoding site in the small subunit.

22.3 Translation  679

Box 22.B Antibiotic Inhibitors of Protein Synthesis Antibiotics interfere with a variety of cellular processes, including cell-­wall synthesis, DNA replication, and RNA transcription. Some of the most effective antibiotics, including many in clinical use, target protein synthesis. Because bacterial and eukaryotic ribosomes and translation factors differ, these antibiotics can kill bacteria without harming their mammalian hosts. Puromycin, for example, resembles the 3′ end of Tyr–­tRNA and competes with aminoacyl–­tRNAs for binding to the ribosomal A site. Transpeptidation generates a puromycin–­peptidyl group that cannot be further elongated because the puromycin “amino acid” group is linked by an amide bond rather than an ester bond to its “tRNA” group. As a result, peptide synthesis comes to a halt.

NH2 N

N

N

N tRNA H3C

N

CH3

C

O

OH

H

O CH2

Tyrosyl-tRNA

H

OH O

CH

H

NH+ 3 H

HN

H

CH

O

H

O

C

N

N

H

H

N

N

HOCH2

O3POCH2

CH2

NH+ 3 Puromycin

OCH3

OH

The antibiotic chloramphenicol interacts with the active-­site nucleotides, including the catalytically essential A2451, to ­prevent transpeptidation.

OH CH2OH O O2N

C

C

H

H

NH

C

CHCl2

Chloramphenicol Other antibiotics with more complicated structures interfere with protein synthesis through different mechanisms. For example, erythromycin physically blocks the tunnel that conveys the nascent polypeptide away from the active site. Six to eight peptide bonds form before the constriction of the exit tunnel blocks further chain elongation. Streptomycin kills cells by binding tightly to the backbone of the 16S rRNA and stabilizing an error-­prone conformation of the ribosome. In the presence of the antibiotic, the ribosome’s affinity for aminoacyl–­tRNAs increases, which increases the likelihood of codon–­anticodon mispairing and therefore increases the error rate of translation. Presumably, the resulting burden of inaccurately synthesized proteins kills the cell. The drugs described here, like all antibiotics, lose their effectiveness when their target organisms become resistant to them. For example, mutations in ribosomal components can prevent anti­ biotic binding. Alternatively, an antibiotic-­susceptible organism may acquire a gene, often present on an extrachromosomal ­plasmid, whose product inactivates the antibiotic. Acquisition of an acetyltransferse gene leads to the addition of an acetyl group to chloramphenicol, which prevents its binding to the ribosome. Acquisition of a gene for an ABC transporter (Section 9.3) can hasten a drug’s export from the cell, rendering it useless. Question  Explain why some of the side effects of antibiotic use in ­humans can be traced to impairment of mitochondrial ­function.

The movement of tRNA and mRNA, which allows the next codon to be ­translated, is known as translocation. This dynamic process requires the G protein called elongation factor G (EF-G) in E. coli. EF-G bears a striking resemblance to the EF-­Tu–­tRNA complex (Fig. 22.18), and the ribosomal binding sites for the two proteins overlap. Structural studies show that the EF-G–­ GTP complex physically displaces the peptidyl–­tRNA in the A site, causing it to translocate to the P site. This movement causes the deacylated tRNA in the P site to shift to the E site. EF-G binding to the ribosome stimulates its GTPase activity. After EF-G hydrolyzes its bound GTP, it dissociates from the ribosome, leaving a vacant A site available for the arrival of another aminoacyl–­tRNA and another round of transpeptidation. The GTPase activity of G proteins such as EF-­Tu and EF-G allows the ribosome to cycle efficiently through all the steps of translation elongation. Because GTP hydrolysis is irreversible, the elongation reactions—­aminoacyl–­tRNA binding, transpeptidation, and translocation—­proceed   FIGURE 22.18   Structure of EF-G from T. thermophilus. 

Question  Compare the size and shape of this complex to the size and shape of the complex containing EF-Tu and an aminoacyl–tRNA (see Fig. 22.13).

680

C hA pTER 22

Protein Synthesis

FIGURE 22.19 The E. coli ribosomal elongation cycle.

GTP EF-Tu

GDP

A site

EF-Tu

delivery of aminoacyl–tRNA

SEE ANIMATED PROCESS DIAGRAM

P site

mRNA

GDP

Elongation cycle in E. coli ribosomes

transpeptidation EF-G–GDP

GTP

translocation

GTP

EF-G–GTP

unidirectionally. The E. coli ribosomal elongation cycle is shown in Figure 22.19. Experimental evidence shows that bacterial RNA polymerase can sometimes bind to the ribosome, thereby feeding mRNA from the polymerase active site directly into the ribosome’s decoding center. Not only does this help synchronize transcription and translation, but the one-way movement of the ribosome helps prevent backtracking by the RNA polymerase. Eukaryotic cells contain elongation factors that function similarly to EF-Tu and EF-G (Table 22.5). These G proteins are continually recycled during protein synthesis. In some cases, accessory proteins help replace bound GDP with GTP to prepare the G protein for another reaction cycle.

Release factors mediate translation termination

FIGURE 22.20 Structure of RF-1. The RF-1 protein is shown as a purple ribbon with its anticodon-binding Pro–Val–Thr (PVT) sequence in red and its peptidyl transferase–binding Gly–Gly– Gln (GGQ) sequence in orange. This image shows the structure of RF-1 as it exists when bound to a ribosome.

As the peptidyl group is lengthened by the transpeptidation reaction, it exits the ribosome through a tunnel in the center of the large subunit. The tunnel, about 100 Å long and 15 Å in diameter in the bacterial ribosome, shelters a polypeptide chain of up to 30 residues. The tunnel is defined by ribosomal proteins as well as the 23S rRNA. A variety of groups—including rRNA bases, backbone phosphate groups, and protein side chains—form a mostly hydrophilic surface for the tunnel. There are no large hydrophobic patches that could potentially impede the exit of a newly synthesized peptide chain. Translation ceases when the ribosome encounters a stop codon (see Table 22.1). With a stop codon in the A site, the ribosome cannot bind an aminoacyl–tRNA but instead binds a protein known as a release factor (RF). In bacteria such as E. coli, RF-1 recognizes stop codons UAA and UAG, and RF-2 recognizes UAA and UGA. In eukaryotes, one protein—called eRF1— recognizes all three stop codons. The release factor must specifically recognize the mRNA stop codon; it does this via an “anticodon” sequence of three amino acids, such as Pro–Val–Thr in RF-1 and Ser–Pro–Phe in RF-2, that interact with the first and second bases of the stop codon. At the same time, a loop of the release factor with the conserved sequence Gly– Gly– Gln projects into the peptidyl transferase site of the 50S subunit (Fig. 22.20). The amide group of the Gln residue promotes transfer of the

22.3 Translation 1. A release factor (RF-1 or RF-2) recognizes a stop codon in the A site.

2. RF-1/RF-2 causes the ribosome to transfer the peptidyl group to water to release the polypeptide chain.

3. GTP hydrolysis by RF-3 allows the release factors to dissociate. Additional steps are required to prepare the ribosome for another round of translation.

RF Polypeptide mRNA Stop codon

FIGURE 22.21

Translation termination in E. coli.

GTP

RF-3–GTP

peptidyl group from the P-site tRNA to water, apparently by stabilizing the transition state of this hydrolysis reaction. The product of the reaction is an untethered polypeptide that can exit the ribosome. At one time, release factors were believed to act by mimicking tRNA molecules, as EF-G does. However, it now appears that release factors undergo conformational changes upon binding to the ribosome, so they don’t operate straightforwardly as tRNA surrogates. In E. coli, an additional RF (RF-3), which binds GTP, promotes the binding of RF-1 or RF-2 to the ribosome. In eukaryotes, eRF3 performs this role. Hydrolysis of the GTP bound to RF-3 (or eRF3) allows the release factors to dissociate (Fig. 22.21). This leaves the ribosome with a bound mRNA, an empty A site, and deacylated tRNAs in both the P and E sites. In bacteria, preparing the ribosome for another round of translation is the responsibility of a ribosome recycling factor (RRF) that works in concert with EF-G. RRF apparently slips into the ribosomal A site. EF-G binding then translocates RRF to the P site, thereby displacing the deacylated tRNAs. Following GTP hydrolysis, EF-G and RRF dissociate from the ribosome, leaving it ready for a new round of translation initiation. In eukaryotes, a protein with ATPase activity binds to the ribosome containing eRF3 and helps dissociate the large and small subunits. Initiation factors appear to bind during this process, so translation termination and reinitiation are tightly linked.

Translation is efficient and dynamic A ribosome can extend a polypeptide chain by approximately 20 amino acids every second in bacteria and by an average of about 6 amino acids every second in eukaryotes. At these rates, most protein chains can be synthesized in under a minute. As we have seen, various G proteins trigger conformational changes that keep the ribosome operating efficiently through many elongation cycles. Cells also maximize the rate of protein synthesis by forming polysomes. These structures include a single mRNA molecule simultaneously being translated by multiple ribosomes (Fig. 22.22). As soon as the first ribosome has cleared the initiation codon, a second ribosome can assemble and begin translating the mRNA. The circularization of eukaryotic mRNAs (see Fig. 22.12) may promote repeated rounds of translation. Because the stop codon at the 3′ end of the coding sequence may be relatively close to the start codon at the 5′ end, the ribosomal components released at termination can be easily recycled for reinitiation. The preceding sections have already mentioned some variations in the process of protein synthesis, such as ribosome heterogeneity and alternative start codons in mRNA. In fact, a large number of factors can impact the outcome of translation. Some of the dynamics of ribosome activity have been revealed through various structural and kinetic methods and by ribosome profiling, a technique based on sequencing and identifying mRNA segments that are being translated and are therefore protected from RNase digestion by the presence of an active ribosome. In addition to chemical modifications to mRNA nucleotides, which is sometimes referred to as “RNA epigenetics,” RNA secondary structure can affect the progress of a ribosome. Riboswitches illustrate how mRNA structure responds to the presence of a ligand to determine

GDP

RF-3–GDP

SEE ANIMATED PROCESS DIAGRAM Translation termination in E. coli

681

C hA pTER 22

Protein Synthesis

a. Oscar L. Miller, Jr. and Steven L. McKnight, University of Virginia

b.

a.

A polysome. a. In this electron micrograph, a single mRNA strand encoding silkworm fibroin is studded with ribosomes. Arrows indicate growing fibroin polypeptide chains. b. In this cryo-electron microscopy–based reconstruction of a FIGURE 22.22

Brandt, F. et al. 2009/Elsevier

682

b.

polysome, ribosomes are blue and gold, and the emerging polypeptides are red and green. Question Identify the polysomes in Fig. 22.9.

whether translation can occur. A riboswitch is a segment of mRNA, usually located near the 5′ end, that can change its conformation, for example, to expose a ribosome binding site (Fig. 22.23). Riboswitches occur in all types of organisms and may also affect transcription termination (in prokaryotes) or mRNA splicing (in eukaryotes). The ligands for the most common riboswitches are coenzymes, whose presence prevents translation of a protein involved in the synthesis of that coenzyme (the example in Fig. 22.23 shows a ligand promoting translation). Once it begins translating an mRNA, the progress of the ribosome is not necessarily uniform. The ribosome may slow or pause for a number of reasons, such as when it encounters a rarely used codon. Recall that each amino acid is specified by one to six different codons, which appear in genes at different frequencies, depending on the type of organism. Ribosome pausing may simply slow the rate of translation or it may lead to degradation of the mRNA. This outcome explains why many “silent” mutations that do not change the meaning of a codon can result in significant changes in the amount of a protein in an organism. Sequencing studies show that the proteins with the highest expression levels tend to use the most common codons, which favors fast translation, whereas less-abundant proteins tend to include more rare codons. The sequence of the polypeptide product can also affect the rate of translation. For example, the transfer of a peptidyl group to proline is relatively slow, so the rate of transpeptidation

3′

GUA

Adenine

3′ A U A C U U C G G 5′

UA G U U GA A A G C U C UC A G A A

A A G G

A U A C U U C G G 5′

U A U G A A G U C

FIGURE 22.23 A bacterial adenine riboswitch. In the absence of adenine, the Shine–Dalgarno sequence (blue) and start codon (green) are inaccessible; adenine binding stabilizes an alternate structure that allows translation to proceed.

22.4  Post-­Translational Events 

drops when the ribosome reaches a stretch of proline codons. The nascent polypeptide may also interact with groups in the ribosome’s exit channel, slowing its emergence into the cytosol. In rare cases, a ribosome may slip, particularly at a run of repeating nucleotides, resuming in a different reading frame and producing a polypeptide with a different sequence past that point. The examples above are ample evidence that cells use a variety of ways to regulate gene expression, in addition to transcriptional control. As a result, some genes are rarely expressed while some gene products are made in large amounts. For example, in eukaryotes, an average of 1500 different proteins account for 90% of the total protein mass; in prokaryotes, the average is about 300. Many of these proteins are involved in energy-­harvesting pathways and protein production. Surprisingly, about a quarter of the most abundant proteins have no assigned function. The activities of many low-­abundance proteins are likewise mysterious. Curiously, ribosomes often translate mRNA sequences outside of the canonical coding sequence defined by the start and stop codons. As many as half of mammalian mRNAs include an additional open reading frame—a sequence that may start with an alternative start codon and that potentially codes for a polypeptide—­on the 5ʹ side of the main start codon. Translation starting from the upstream start codon can yield so-­called microproteins that may have discrete functions. An alternative translation start site may also tie up ribosomes in a way that prevents them from translating the rest of the mRNA molecule.

Before Going On • Draw a diagram of a gene with and without introns and label the promoter, transcription termination site, start codon, and stop codon. • Make a list of all the substances that must interact with a ribosome in order to produce a polypeptide. • Compare the way a ribosome recognizes a start codon and a stop codon. • Describe the functions of bacterial IF-2, EF-­Tu, EF-G, RF-1, and RRF. • Explain how the ribosome ensures that the correct amino acid is positioned in the ribosomal A site. • Compare translation initiation, elongation, and termination in bacteria and eukaryotes. • Summarize the different types of RNA–­RNA interactions that occur during translation. • Without looking at the text, draw a diagram of a polysome, add labels for mRNA, polypeptides, and ribosomes, and indicate which ribosome was the first to arrive. • Describe the factors that affect the rate of translation or the amount or type of protein produced.

22.4 Post-­Translational Events LEARNING OBJECTIVES List the events that occur during post-­translational processing. • Describe what chaperones do and how they work. • Recount the steps of producing a membrane or secreted protein. • Recognize different types of post-­translational modifications.

The polypeptide released from a ribosome is not yet fully functional. For example, it must fold to its native conformation; it may need to be transported to another location inside or outside the cell; and it may undergo post-­translational modification, or processing. As we will see, many of these events are actually co-­translational, not strictly post-­translational, and the ribosome plays a part.

683

684  C ha pter 22   Protein Synthesis

Nenad Ban, Eidgenossische Technische Hochschule Honggerberg, Zurich

Chaperones promote protein folding Studies of protein folding in vitro have revealed numerous insights into the pathways by which proteins (usually relatively small ones that have been chemically denatured) assume a compact globular shape with a hydrophobic core and a hydrophilic surface (see Section 4.3). Protein folding in vivo is not fully understood. For one thing, a protein can begin to fold as soon as its N-­terminus emerges from the ribosome, even before it has been fully synthesized. The rate of translation can affect the packing options for newly formed polypeptide segments by making certain segments available at the proper intervals. In addition, a polypeptide must fold in an environment crowded with other proteins with which it might interact unfavorably. Finally, for proteins with quaternary structure, individual polypeptide chains must assemble with the proper stoichiometry and orientation. All of these processes may be facilitated in a cell by proteins known as molecular chaperones. To prevent improper associations within or between polypeptide chains, chap  FIGURE 22.24   Trigger factor bound erones bind to exposed hydrophobic patches on the protein surface (recall that hydroto a ribosome.  Trigger factor, with its phobic groups tend to aggregate, which could lead to nonnative protein structure or domains shown in different colors, binds protein aggregation and precipitation). Many chaperones are ATPases that use the free to the 50S ribosomal subunit where a polyenergy of ATP hydrolysis to drive conformational changes that allow them to bind and peptide (magenta helix) emerges. PT reprelease a polypeptide substrate while it assumes its native shape. Some chaperones were resents the peptidyl transferase active site. originally identified as heat-­shock proteins (Hsp) because their synthesis is induced by high temperatures—­conditions under which proteins tend to denature (unfold) and aggregate. The first chaperone a bacterial protein meets, called trigger factor, is poised just outside the ribosome’s polypeptide exit tunnel, bound to a ribosomal protein (Fig. 22.24). When trigger factor binds to the ribosome, it opens up to expose a hydrophobic patch facing the exit tunnel. Hydrophobic segments of the emerging polypeptide bind to this patch and are thereby prevented from sticking to each other or to other cellular components. Trigger factor may ­dissociate from the ribosome but remain associated with the nascent (newly formed) polypeptide until another chaperone takes over. Eukaryotes lack trigger factor, although they have other small heat-­shock proteins that function in the same manner to protect newly made proteins. These chaperones are extremely abundant in cells, so there is at least one per ribosome. Trigger factor may hand off a new polypeptide to another chaperone, such as DnaK in E. coli (it was named when it was believed to participate in DNA synthesis). DnaK and other heat-­shock proteins in prokaryotes and eukaryotes interact with newly synthesized polypeptides and with existing cellular proteins and therefore can prevent as well as reverse improper folding. These chaperones, in a complex with ATP, bind to a short extended polypeptide segment with exposed hydrophobic groups (they do not recognize folded proteins, whose hydrophobic groups are sequestered in the interior). ATP hydrolysis causes the chaperone to release the polypeptide. As the polypeptide folds, the heat-­shock protein may repeatedly bind and release it. Ultimately, protein folding may be completed by multisubunit chaperones, called chaperonins, which form cagelike structures that physically sequester a folding polypeptide. The best-­known chaperonin complex is the GroEL/GroES complex of E. coli. Fourteen GroEL subunits form two rings of seven subunits, with each ring enclosing a 45-Å-­diameter chamber that is large enough to accommodate a folding polypeptide. Seven GroES subunits form a domelike cap for one GroEL chamber (Fig. 22.25). The GroEL ring nearest the cap is called the cis ring, and the other is the trans ring. Each GroEL subunit has an ATPase active site. All seven subunits of the ring act in concert, hydrolyzing their bound ATP and undergoing conformational changes. The two GroEL rings of the chaperonin complex act in a reciprocating fashion to promote the folding of two polypeptide chains in a safe environment (Fig. 22.26). Note that seven ATP   FIGURE 22.25   The GroEL/GroES are consumed for each 10-second protein-­folding opportunity. If the released substrate chaperonin complex.  The two seven-­ has not yet achieved its native conformation, it may rebind to the chaperonin complex. subunit GroEL rings, viewed from the Only about 10% of bacterial proteins seem to require the GroEL/GroES chaperonin side, are colored red and yellow. A seven-­ complex, and most of these range in size from 10 to 55 kD (a protein larger than about subunit GroES complex (blue) caps the so-­called cis GroEL ring. 70 kD probably could not fit inside the protein-­folding chamber). Immunocytological

22.4  Post-­Translational Events  1. A GroEL ring binds 7 ATP and an unfolded polypeptide, which associates with hydrophobic patches on the GroEL subunits.

685

2. A GroES cap binds, triggering a conformational change that retracts the hydrophobic patches, thereby releasing the polypeptide into the GroEL chamber, where it can fold.

ATP

ATP

3. Within about 10 seconds, the cis GroEL ring hydrolyzes its 7 ATP.

ADP

7 Pi

7 ATP GroES

6. The trans GroEL ring can now bind a GroES cap, and steps 3 –5 repeat.

4. A second polypeptide 7ATP substrate and 7 ATP bind to the trans GroEL ring.

ADP

7 ADP ATP

5. The cis ring releases its GroES cap, the 7 ADP, and the folded substrate polypeptide.

  FIGURE 22.26   The chaperonin reaction cycle.

studies indicate that some proteins never stray far from a ­chaperonin complex, perhaps because they tend to unfold and must periodically restore their native structures. In eukaryotes, a chaperonin complex known as CCT or TRiC functions analogously to bacterial GroEL, but it has eight subunits in each of its two rings, and its fingerlike projections take the place of the GroES cap. One of its primary functions is to fold the abundant cytoskeletal proteins actin and tubulin (Section 5.3).

The signal recognition particle targets some proteins for membrane translocation For a cytosolic protein, the journey from a ribosome to the protein’s final cellular destination is straightforward (in fact, the journey may be short, since some mRNAs are directed to specific cytosolic locations before translation commences). In contrast, an integral membrane protein or a protein that is to be secreted from the cell follows a different route since it must pass partly or completely through a membrane. These proteins account for about one-­third of the proteins made by a cell. In both prokaryotic and eukaryotic cells, most membrane and secretory proteins are synthesized by ribosomes that dock with the plasma membrane (in prokaryotes) or endoplasmic reticulum (in eukaryotes). The protein that emerges from the ribosome is inserted into or through the membrane co-­translationally. The membrane translocation system is fundamentally similar in all cells and requires a ribonucleoprotein known as the signal recognition particle (SRP). As in other ribonucleoproteins, including the ribosome and the spliceosome (see Section 21.3), the RNA component of the SRP is highly conserved and is essential for SRP function. The E. coli SRP consists of a single multidomain protein and a 4.5S RNA. The mammalian SRP contains a larger RNA and six different proteins, but its core is virtually identical to the bacterial SRP. The RNA component of the SRP includes several nonstandard base pairs, including G:A,

ATP

Batey, R. T. et al. 2000/AAAS.

686  C ha pter 22   Protein Synthesis

signal sequence binding groove

MALWMRLLPLLALLALWGPDPAAAFVN.... The hydrophobic segment and a flanking arginine residue are shaded in green and pink. The signal peptide binds to the SRP in a pocket formed mainly by a methionine-­ peptide binding domain.  This rich protein domain. The flexible side chains of the hydrophobic Met residues allow model shows the molecular surface of a the pocket to accommodate helical signal peptides of variable sizes and shapes. In portion of the E. coli signal recognition particle. The protein is magenta addition to the Met residues, the SRP binding pocket contains a segment of RNA, with hydrophobic residues in yellow. whose negatively charged backbone interacts electrostatically with the positively Adjacent RNA phosphate groups are red, charged N-­terminus and basic residue of the signal peptide (Fig. 22.27). Thus, the sigand the rest of the RNA is dark blue. A nal peptide makes hydrophobic as well as electrostatic contacts with the SRP protein signal peptide binds in the SRP groove. and RNA. Electron microscopic studies indicate that the SRP binds to the ribosome at the polypeptide exit tunnel. The SRP recognizes the signal peptide as it emerges, most likely competing with trigger factor or its eukaryotic counterpart. SRP binding to the signal peptide halts transSEE ANIMATED PROCESS lation elongation, possibly via conformational changes in the ribosome resulting from SRP DIAGRAM RNA–­rRNA interactions. The stalled ribosome–­SRP complex then docks at a receptor on the membrane in a process that requires GTP hydrolysis by the SRP. When translation resumes, The secretory pathway the growing polypeptide is translocated through the membrane via a structure known as a translocon. The translocon proteins, called SecY in prokaryotes and Sec61 in eukaryotes, form a transmembrane channel with a constricted pore that limits the diffusion of other substances across the membrane (Fig. 22.28). Insertion of the signal peptide itself nudges aside two Sec helices in order to open up the pore wide enough to allow a polypeptide segment to pass through. In addition to allowing a secreted protein to pass entirely across a membrane in this ­manner, the translocon has a lateral opening that allows one or more hydrophobic protein segments to move sideways into the lipid bilayer, which explains how transmembrane proteins ­(Section 8.3) become incorporated into the membrane. The driving force for co-­translational protein translocation is provided by the ribosome: the process of polypeptide synthesis simply pushes the chain through the channel. In situations where a completed polypeptide is inserted across or into a membrane post-­ translationally (after it has left the ribosome), the transport system—­which may involve the Sec translocon or other machinery—­relies on some sort of motor protein that uses the free energy of the ATP hydrolysis reaction. When the signal peptide emerges on the far side of the membrane, it may be cleaved off by an integral membrane protein known as a signal peptidase. This enzyme recognizes extended polypeptide segments such as those flanking the hydrophobic segment of a signal peptide, but it does not recognize α-­helical structures, which are common in mature membrane proteins. The steps of translocating a eukaryotic secretory protein are summarized in Figure 22.29. After translocation in eukaryotic cells, chaperones and other proteins in the endoplasmic reticulum may help the polypeptide fold into its native conformation, form disulfide bonds, and assemble with other protein subunits. Extracellular proteins are then transported from the endoplasmic reticulum through the Golgi apparatus and to the   FIGURE 22.28   Ribosome bound to plasma membrane via vesicles. The proteins may undergo processing (described next) Sec61.  In this cryo-EM-­based image, en route to their final destination. a yeast ribosome (partially cut away) is Not all membrane and secreted proteins follow the pathway outlined above. For bound to Sec 61 (pink), which is positioned example, proteins whose C-­terminal tail is anchored in a membrane (about 5% of at the end of the polypeptide exit tunnel. eukaryotic membrane proteins) are synthesized by ribosomes in the cytosol, and chapPortions of the nascent polypeptide (NC, erones direct the proteins to the appropriate membrane, where a translocon inserts green) are visible in the exit tunnel.   FIGURE 22.27  The SRP signal

Becker, T. et al. 2009/AAAS.

G:G, and A:C, and interacts with several protein backbone carbonyl groups (most RNA–­ protein interactions involve protein side chains rather than the backbone). How does the SRP recognize membrane and secretory proteins? Such proteins typically have an N-­terminal signal peptide consisting of an α-­helical stretch of 6 to 15 hydrophobic amino acids preceded by at least one positively charged residue. For example, human proinsulin (the polypeptide precursor of the hormone insulin) has the following signal sequence:

22.4  Post-­Translational Events 

the protein into the lipid bilayer. Some proteins destined for the extracellular environment are packaged in exosomes (see Box 9.C), whose membranes eventually disintegrate and release the protein. Approximately 900 of the 1000 mitochondrial proteins move from the cytosol across the outer membrane via a β-­barrel transporter known as TOM (translocase of the outer membrane) and from there can then pass through TIM (translocase of the inner membrane).

Many proteins undergo covalent modification

687

Signal peptide Ribosome mRNA SRP

ER LUMEN

1. The SRP binds to the signal peptide when it emerges from the ribosome. Polypeptide elongation halts.

2. The SRP delivers the ribosome to the endoplasmic reticulum membrane. Translation resumes, and the Translocon polypeptide crosses the membrane ER membrane via a translocon.

CYTOSOL

Proteolysis is part of the maturation pathway of many proteins. For example, after it has entered 3. Signal peptidase removes the signal peptide from the growing polypeptide. the endoplasmic reticulum lumen and had its signal peptide removed and its cysteine side Signal peptidase chains cross-­linked as disulfides, the insulin precursor undergoes proteolytic processing. The prohormone is cleaved at two sites to generate 4. The rest of the polypeptide is the mature hormone (Fig. 22.30). Over 200 other translocated into the ER lumen. types of post-­t ranslational modification have From here it is transported by vesicles Polypeptide been documented. to the extracellular space. Many extracellular eukaryotic proteins are glycosylated at asparagine, serine, or threonine side chains to generate glycoproteins (see Section 11.3). The short sugar chains (oligosaccharides) attached to glycoproteins may protect the proteins from degradation or mediate molecular recognition events. Methyl,   FIGURE 22.29   Membrane translocation of a eukaryotic secretory protein. acetyl, and propionyl groups may be added to various side chains. N-­terminal groups are frequently modified by acetylation (up to 80% of human proteins), and C-­terminal groups by amidation. Fatty acyl chains and other lipid groups are added to proteins to anchor them to membranes (Section 8.3). The addition and removal of phosphoryl groups is a powerful mechanism for allosterically regulating cellular signaling components (Section 10.2) and metabolic pathways (Section 19.2). C chain Modification of a protein by covalently attaching another protein occurs during protein degradation, when ubiquitin is P P G E L A L A Q L G G L G G G L E VA G A Q P N Q covalently linked to a target protein (Section 12.1). A related Q A K protein, called SUMO (for small ubiquitin-­like modifier), is also E R covalently attached to a target protein’s lysine side chains, but A chain R G S S rather than mark the protein for degradation, as ubiquitin does, I R C V E Q C C T S I C S LY Q L E N YC N SUMO is involved in various other processes, including protein A S S transport into the nucleus. K S S All the post-­translational modifications mentioned above P T N F V N Q H L C G S H LV E A LY LV C G E R G F F Y are catalyzed by specific enzymes, which act more or less B chain ­reliably depending on the nature of the modification and the cellular context. One consequence of post-­translational process-   FIGURE 22.30   Conversion of proinsulin to insulin.  The ing therefore is that proteins may exhibit a great deal of varia- prohormone, with three disulfide bonds, is proteolyzed at two tion beyond the sequence of amino acids that is specified by the bonds (indicated by arrows) to eliminate the C chain. The mature genetic code. insulin hormone consists of the disulfide-­linked A and B chains.

688  C ha pter 22   Protein Synthesis

Before Going On • Compare protein folding as catalyzed by heat-­shock proteins and by the chaperonin complex. • Explain how the SRP recognizes a membrane or secretory protein. • Describe the functions of the translocon and the signal peptidase. • List the types of reactions that a polypeptide may undergo following its synthesis.

Summary and delivers them to the A site of the ribosome. Correct pairing between the mRNA codon and the tRNA anticodon allows the EF-­Tu to hydrolyze its bound GTP and dissociate from the ribosome.

22.1  tRNA and the Genetic Code •  The sequence of nucleotides in DNA is related to the sequence of amino acids in a protein by a triplet-­based genetic code that must be translated by tRNA adaptors. •  tRNA molecules have similar L-­shaped structures with a three-­ base anticodon at one end and an attachment site for a specific amino acid at the other end. •  Attachment of an amino acid to a tRNA is catalyzed by an aminoacyl–­t RNA synthetase in a reaction that requires ATP. Various proofreading mechanisms ensure that the correct amino acid becomes linked to the tRNA.

22.2  Ribosome Structure •  The ribosome, the site of protein synthesis, consists of two subunits containing both rRNA and protein. The ribosome includes a binding site for mRNA and three binding sites (called the A, P, and E sites) for tRNA.

22.3  Translation •  Translation of mRNA requires an initiator tRNA bearing methio­nine (formyl-­methionine in bacteria). Proteins known as initiation factors facilitate the separation of the ribosomal subunits and their reassembly with the initiator tRNA and an mRNA to be translated. •  During the elongation phase of protein synthesis, an elongation factor (EF-­Tu in E. coli) interacts with aminoacyl–­tRNAs

•  Transpeptidation, or formation of a peptide bond, is catalyzed by rRNA in the large ribosomal subunit. The growing polypeptide chain becomes attached to the tRNA in the A site, which then moves to the P site. This movement is assisted by a GTP-­binding protein elongation factor (EF-G in E. coli). •  Translation terminates when a release factor recognizes an mRNA stop codon in the A site of the ribosome. Additional factors prepare the ribosome for another round of translation. •  The efficiency of translation is affected by factors such as polysome formation, the presence of riboswitches, and the use of rare codons and alternative start codons.

22.4  Post-­Translational Events •  Chaperone proteins bind newly synthesized polypeptides to facilitate their folding. Large chaperonin complexes form a barrel-­shaped structure that encloses a folding protein. •  Proteins to be secreted must pass through a membrane. An RNA–­protein complex called a signal recognition particle directs polypeptides bearing an N-­terminal signal sequence to a membrane for translocation. •  Additional modifications to newly synthesized proteins include proteolytic processing and the attachment of carbohydrate, lipid, or other groups.

Key Terms translation codon reading frame anticodon isoacceptor tRNA isodecoder tRNA wobble hypothesis ribosome nucleolus A site

P site E site initiation factor (IF) G protein elongation factor (EF) transpeptidation translocation release factor (RF) ribosome recycling factor (RRF) polysome

ribosome profiling riboswitch post-translational processing molecular chaperone signal recognition particle (SRP) signal peptide translocon glycosylation

Problems  689

Bioinformatics Brief Bioinformatics Exercises 22.1  Viewing and Analyzing Transfer RNA 22.2  Ribosomes, Protein Processing, and the KEGG Database

Problems 22.1  tRNA and the Genetic Code 1.  How many combinations of four nucleotides are possible with a hypothetical quadruplet code? 2.  Synthetic biologists at the Scripps Institute in La Jolla, CA expanded the genetic repertoire by adding two new bases into living bacterial cells. The two bases are named d5SICS and dNaM, and they pair with one another. How many different amino acids could theoretically be encoded by a synthetic nucleic acid containing six different nucleotides? 3.  Cells have a mechanism for tagging and destroying proteins containing a C-terminal poly(Lys) sequence. How are these proteins synthesized and why is destroying them helpful for the cell? 4.  Cystic fibrosis is caused by a mutation in the 250,000-bp CFTR gene. The mature CFTR mRNA is only 6129 nucleotides. a. Why is the mature mRNA so much shorter than the gene from which it was transcribed? b. What is the minimum number of nucleotides required to encode the 1480-residue CFTR protein? 5.  The CFTR gene (see Problem 4) contains the sequence ···ATCATCTTTGGTGTT···, which codes for residues 506–510 of the protein. a. Identify the residues in this segment of the protein. b. In the most common mutated form of the gene, this same segment of DNA has the sequence ···ATCATTGGTGTT···. What type of mutation has occurred and how does it affect the sequence of the encoded protein? 6.  One portion of the normal CFTR gene has the sequence ···AATATAGATACAG···. In some individuals with cystic fibrosis, this portion of the gene has the sequence ···AATAGATACAG···. How has the DNA sequence changed and how does this affect the encoded protein? 7.  A portion of DNA from a phage genome is shown. a. How many reading frames are possible? Specify the amino acid sequences for all the possible reading frames. b. Which frames represent open reading frames? 5′  C G A T G A G C C T T T C A G C A C C G C T T A G T G A G G T T G 3′  G C T A C T C G G A A A G T C G T G G C G A A T C A C T C C A A C 8.  Explain why a redundant genetic code helps protect an organism from the effects of mutations. 9.  Marshall Nirenberg and his colleagues deciphered the genetic code in the early 1960s. Their experimental strategy involved constructing RNA templates by using various ribonucleotides and polynucleotide phosphorylase, an enzyme that links available nucleotides together in random order. What protein sequence was obtained when the following templates were added to a cell-free translation system? a. poly(U), b. poly(C), c. poly(A).

10.  The experimental strategy described in Problem 9 was used to synthesize poly(UA). Because the polymerization of nucleotides by polynucleotide phosphorylase is random, all possible codons containing U and A could occur in the RNA template. a. What amino acids would be incorporated into a polypeptide synthesized by a cellfree translation system using this template? b. What amino acids would be incorporated into a protein when poly(UC) is used as a template? c. Repeat the exercise for poly (UG). d. How do these results show that the genetic code is redundant? 11.  The elucidation of the genetic code was completed using H. Gobind Khorana’s method of synthesizing polynucleotides with precise rather than random sequences, as described in Problem 10.  a.  What polypeptides are synthesized from a nonrandom template with the sequence poly(UAUC)? b. Explain why a single RNA template can yield more than one polypeptide. 12.  What polypeptide is synthesized from a nonrandom poly(AUAG) template? Compare your results with your solution to Problem 11. 13.  Organisms differ in their codon usage. For example, in yeast, only 25 of the 61 amino acid codons are used with high frequency. Cells contain high concentrations of the tRNAs that pair best with those codons, that is, form standard Watson–Crick base pairs. Explain how a point mutation in a gene, which does not change the identity of the encoded amino acid, could decrease the rate of synthesis of the protein corresponding to that gene. 14. a. Would you expect that the highly expressed genes in a yeast cell would have sequences corresponding to the cell’s set of 25 preferred codons (see Problem 13)? Would you expect this to be the case for genes that are expressed only occasionally? b. The genomes of many bacterial species appear to contain genes acquired from other species, including mammals. Even when a gene’s function cannot be identified, the gene’s nonbacterial origin can be recognized. Explain. 15.  Like the yeast described in Problem 13, E. coli also has codon preferences and does not use all 64 possible codons with the same frequency. Investigators interested in over-expressing human interleukin-2 (IL-2) in E. coli designed a synthetic DNA sequence that produces the correct amino acid sequence of the human protein but contains the codons preferred by E. coli rather than human cells. Shown below is a partial E. coli codon usage chart, with the fraction of codon use per amino acid. The human IL-2 gene sequence corresponding to the first 13 amino acids is also shown. a. Design a synthetic DNA that uses codons preferred by E. coli in order to maximize the protein yield in the expression system. b. What is the sequence of the protein produced by your synthetic DNA and is it the same as coded by the original gene sequence? c. If you wanted to maximize protein yield with the original DNA sequence, what alternative strategy could you use?

690  C ha pter 22   Protein Synthesis ···GCACCTACTTCAAGTTCTACAAAGAAAACACAGCTACAA···

24.  Draw the “wobble” base pair that forms between inosine (I) and adenosine.

UCU 0.11 CUA 0.05 CUG 0.46

N

UCA 0.15 CCU 0.17

CAA 0.30

CCG 0.55 ACU 0.16

CAG 0.70 AAA 0.73

AGU 0.14

ACC 0.47

AAG 0.27

AGC 0.33

O

N

ACA 0.13 GCC 0.31 GCA 0.21 16.  The genetic code is nearly universal, but there are some exceptions in the genetic code used in mitochondria, where UGA codes for Trp and CUN (where N is any nucleotide) codes for Thr. Shown below is a partial DNA sequence for subunit 1 of cytochrome c oxidase (Cox1), a protein synthesized in yeast mitochondria. a. What is the sequence of the protein produced by this DNA sequence? b. The human mitochondrial genome produces 13 proteins, as well as the required cognate tRNAs. The remaining ­proteins required by the mitochondrion must be imported. Why is it advantageous for the mitochondrion to synthesize Cox1? c. Why does the mitochondrion need to express tRNAs in addition to the 13 proteins it expresses? …TGATCACATCATATGTATATTGTAGGATTAGATGCAGATCTTAGA… 17.  IleRS uses a double-sieve mechanism to accurately produce Ile– tRNAIle and prevent the synthesis of Val–tRNAIle. Which other pairs of amino acids differ in structure by a single carbon and might have AARSs that use a similar double-sieve proofreading mechanism? 18.  An examination of AlaRS (the enzyme that attaches alanine to tRNAAla) suggests that the enzyme’s aminoacylation active site cannot discriminate between Ala, Gly, and Ser. a. List all the products of the AlaRS reaction. b. A double-sieve mechanism like the one in IleRS cannot entirely solve the problem of misacylated tRNAAla. Explain. c. Many organisms express a protein called AlaXp, which is a soluble analog of the AlaRS editing domain that is able to hydrolyze Ser–tRNAAla but not Gly–tRNAAla. What is the purpose of AlaXp? 19.  Why doesn’t GlyRS need a proofreading domain? 20.  A tRNA molecule cannot be aminoacylated unless it bears a 3′ CCA sequence. Many tRNA precursors are synthesized without this sequence, so a CCA-adding enzyme must append the three nucleotides to the 3′ end of the immature tRNA molecule. a. The CCA-adding enzyme does not require a polynucleotide template. What does this imply about the mechanism for adding the CCA sequence? b. What can you conclude about the substrate specificity of the CCA-adding enzyme? c. Most CCA-adding enzymes consist of a single polymerase domain, but in one species of bacteria the enzyme has two polymerase domains. Explain how this CCA-adding enzyme operates. 21.  Name the amino acid that is attached to the 3′ end of the E. coli tRNA molecule with the anticodon sequence a. GUG, b. GUU, and c. CGU. 22.  Three E. coli tRNA molecules with the anticodon sequences CGG, GGG, and UGG are charged with the same amino acid. What is the amino acid? 23.  The 5′ nucleotide of a tRNA anticodon is often a nonstandard nucleotide such as a methylated guanosine. Why doesn’t this interfere with the ribosome’s ability to read the genetic code?

N H N Inosine Arg

  25.  E. coli produces a ​​ t RNA​  ACG  ​​​ (ACG is its 5′→3′ anticodon sequence). a. What codon does it recognize? b. What happens during protein synthesis if cellular enzymes convert the adenosine to inosine (see Problem 24)?

26.  What anticodon would allow a tRNA molecule to recognize both lysine codons in E. coli? 27.  A new tRNA discovered in E. coli contains a uridine modified to form uridine-5′-oxyacetic acid (cmo5U). The modified uridine can base pair with G, A, and U. What mRNA codons are recognized by t​​RNA​  Leu ​  cmo 5UAG​​? 28.  Some RNA transcripts are substrates for an adenosine deamin­ ase. This “editing enzyme” converts adenosine residues to inosine residues, which can base pair with guanosine residues. Explain how the action of the deaminase could potentially increase the number of gene products obtained from a given gene. 29.  Predict the effect on a protein’s structure and function for all possible nucleotide substitutions at the first position of a Lys codon in the gene encoding the protein. 30.  Protein engineers who study the effects of nonstandard amino acids on protein structure and function can use a cell-based system for synthesizing proteins containing nonstandard amino acids. In theory, a polypeptide containing the amino acid norleucine could be produced by cells growing in media containing high concentrations of norleucine and lacking norleucine’s standard counterpart, leucine. Experimental results have shown that peptides containing norleucine in place of leucine are not produced unless the cells contain a mutant LeuRS. Explain these results.

COO– H

C

CH2

CH2

CH2

CH3

NH+ 3 Norleucine

22.2  Ribosome Structure 31.  The sequences for all the ribosomal proteins in E. coli have been elucidated and have been found to contain large amounts of lysine and arginine residues. Why is this finding not surprising? What kinds of interactions are likely to form between the ribosomal proteins and the ribosomal RNA? 32.  Newly synthesized proteins emerge from the ribosome via an exit tunnel (see Figs. 22.7 and 22.24), which is very narrow (about the width of an α helix) and is lined with hydrophilic amino acids. What does this observation tell you about the extent of protein folding of the nascent protein? Why is the tunnel lined with hydrophilic and not hydrophobic amino acids? 33.  In eukaryotes, the primary rRNA transcript is a 45S rRNA that includes the sequences of the 18S, 5.8S, and 28S rRNAs separated by short spacers (see Fig. 21.28). What is the advantage of this operon-like arrangement of rRNA genes?

Problems  691 34.  The sequence of ribosomal RNA is highly conserved, even though there are many rRNA genes in the genome. How does this observation argue for a functional (not just structural) role for rRNA?

46.  An E. coli phage replicase gene has the mRNA initiation sequence shown below. Indicate the correct start codon and the Shine–Dalgarno sequence upstream of it.

35.  Ribosomal inactivating proteins (RIPs) are RNA N-glycosidases found in plants. They catalyze the hydrolysis of specific adenine residues in RNA. RIPs are highly toxic but might be useful as anti­tumor drugs because ribosome synthesis is upregulated in transformed cells. Give a general explanation that describes how RIPs inactivate ribosomes.

5′–UAACUAGGAUGAAAUGCAUGUCUAAG · · ·

36.  What is the effect of adding EDTA, a chelating agent specific for divalent cations, to a bacterial cell extract carrying out protein synthesis? 37.  In an experiment, >95% of the proteins are extracted from the 50S ribosomal subunit. Only the 23S rRNA and some protein fragments remain. Peptidyl transferase activity is unaffected. What can you conclude from these results? Propose a role for the protein fragments left behind. Why might it have been difficult to remove them? 38.  Ribosomal proteins can be separated by two-dimensional electrophoresis (a technique that separates proteins based on both charge and size). One of the proteins in the large ribosomal subunit is sometimes acetylated at its N-terminus. Explain why two-dimensional electrophoresis of ribosomal proteins yields two spots corresponding to this protein. 39.  Like their protein counterparts, RNA molecules fold into a variety of structural motifs. Ribosomal proteins contain a so-called RNA-recognition motif. Rho factor and the poly(A) binding protein contain this same motif. Why is this observation not surprising? 40.  Propose at least one hypothesis to explain why eukaryotic ribosomes are more complex than prokaryotic ribosomes.

22.3  Translation 41.  Indicate the substrates, product, template, primer, enzyme, and cellular location for the following eukaryotic processes: a. replication, b. transcription, and c. translation. 42.  In this chapter, we have frequently compared eukaryotic translation to bacterial translation rather than to prokaryotic translation in general. This distinction is intentional, because the protein-synthesizing machinery in Archaea is more similar to the eukaryotic system than to the system in Bacteria. Is this consistent with the evolutionary scheme outlined in Figure 1.15? 43.  The direction of protein synthesis was determined by carrying out an experiment in a cell-free system in which the mRNA consisted of a polymer of A residues with C at the 3′ end, as shown.  a.  What polypeptide was synthesized? b. What would the result be if the mRNA were read in the 3′ → 5′ direction? c. How does this directionality allow prokaryotes to begin translation before transcription is complete? 5′–A A A A · · · A A A C–3′ 44. a. Mycobacteriophage genes occasionally begin with GUG or, even more rarely, UUG (rather than AUG). Which amino acids correspond to these codons? b. Yeast genes occasionally begin with CUG, ACG, or AUC. Which amino acids correspond to these codons? 45.  The translation initiation sequence for the bacterial ribosomal protein L10 is shown below. Draw a diagram that shows how the Shine–Dalgarno sequence aligns with the appropriate sequence on the 16S rRNA. Identify the initiation codon. 5′–CUACCAGGAGCAAAGCUAAUGGCUUUA–3′

47.  Why does eukaryotic translation initiation require proteins that recognize the 5′ cap and the poly(A) tail on the RNA? 48.  S1, a protein in the small ribosomal subunit, has a high affinity for single-stranded RNA and has been shown to be important in initiation. What role might S1 play during initiation? 49.  What happens when colicin E3, which cleaves on the 5′ side of A1493 in the 16S rRNA, is added to a bacterial culture? 50.  Explain why modification or mutagenesis of prokaryotic 16S rRNA at position 1492 or 1493 increases the error rate of translation. 51.  A mutation in yeast eIF2 that stimulates premature GTPase activity results in initiation at non-AUG start codons. Explain how this observation clarifies the role of eIF2 in translation initiation. 52.  In eukaryotes, cellular stressors such as misfolded proteins, oxidative stress, and nutrient shortages may lead to phosphorylation of a serine residue in the eIF2 protein, which blocks its ability to exchange GDP for GTP. How would this affect protein synthesis? How does this strategy ensure the survival of a cell under stress? 53.  The pairing of an aminoacyl–tRNA with EF-Tu offers an opportunity for proofreading during translation. EF-Tu binds all 20 aminoacyl–tRNAs with approximately equal affinity so that it can deliver them and surrender them to the ribosome with the same efficiency. Based on the experimentally determined binding constants for EF-Tu and correctly charged and mischarged aminoacyl–tRNAs (shown here), explain how the tRNA–EF-Tu recognition system prevents the incorporation of the wrong amino acid in a protein. Aminoacyl–tRNA Ala-tRNAAla Gln-tRNAAla Gln-tRNAGln Ala-tRNAGln

Dissociation constant (nM) 6.2 0.05 4.4 260

54.  The affinity of a ribosome for a tRNA in the P site is about 50 times higher than for a tRNA in the A site. Explain why this promotes translational accuracy. 55.  The bacterial elongation factors EF-Tu and EF-G are essential for translation in vivo, but bacterial ribosomes can translate mRNA into protein in vitro in the absence of EF-Tu and EF-G. Why are these factors not required in vitro? How does their absence affect the accuracy of translation? 56.  Predict the effect on protein synthesis if EF-Tu were able to recognize and form a complex with fMet−tR​​NA​  Met f​  ​​. 57.  Identify the peptide encoded by the DNA sequence shown below (the top strand is the coding strand): C G A T A A T G T C C G A C C A A G C G A T C T C G T A G C A G C T A T T A C A G G C T G G T T C G C T A G A G C A T C G T 58.  The sequence of a portion of the coding strand of a gene is shown, along with the sequence of a mutant form of the gene. a. Give the polypeptide sequence that corresponds to the wild-type gene. b. How does the mutant gene differ from the wild-type gene and how does the mutation affect the encoded polypeptide? wild-type  ACACCATGGTGCATCTGACT mutant    ACACCATGGTTGCATCTGAC

692  C ha pter 22   Protein Synthesis

ACAGACACCATGGTGCACCTGACTCCTG 60.  Complete the same exercise described in Problem 59 when the DNA sequence is mutated as shown below. How does the mutation affect translation of the gene if the DNA sequence is a. the coding strand or b. the noncoding strand? ACAGACACCATTGTGCACCTGACTCCTG 61.  The ribosome incorporates the wrong amino acid into a polypeptide chain at a rate of about 10–4. For an average-sized protein containing 400 amino acids, what percentage of protein molecules have the wrong sequence? 62.  Why are rare codons often located at positions corresponding to the boundary between two protein structural domains? 63.  Calculate the approximate energetic cost (in kJ) for the ribosomal synthesis of one mole of a 20-residue polypeptide. Why is the actual energetic cost in vivo probably higher than this value? 64.  All cells contain an enzyme that hydrolyzes peptidyl–tRNA molecules that are not bound to a ribosome. Cells that are deficient in peptidyl–tRNA hydrolase grow very slowly. What is the function of this enzyme, which hydrolyzes the peptidyl group from the tRNA? What does this tell you about the ability of ribosomes to carry out protein synthesis? 65.  In an experimental system, the rate of the peptidyl transferase reaction varies with the identity of the amino acid residue attached to the tRNA in the P site, as shown in the table. a. Explain why transpeptidation involving Pro is much slower than for other amino acids. b. What can you conclude about the electrostatic environment of the peptidyl transferase active site? c. For the nonpolar amino acids, what factor seems to facilitate transpeptidation? Peptidyl group Ala Asp Lys Phe Pro

Rate of peptidyl transfer (s–1) 57 8 100 16 0.14

66.  The peptidyl transferase center of the ribosome actually catalyzes two reactions involving the ester bond linking a polypeptide to the tRNA: aminolysis and hydrolysis. Explain. 67.  The rate of the peptidyl transferase reaction increases as the pH increases from 6 to 8. Use your knowledge of the peptidyl transferase reaction mechanism to explain this observation. 68.  The events of transpeptidation resemble peptide bond hydrolysis in reverse (see Fig. 6.10). a. Draw the “tetrahedral intermediate” of the transpeptidation reaction. b. Researchers hypothesized that residue A2451 protonated at position N1 stabilizes the reaction intermediate. Draw this protonated adenine and explain how it might stabilize the tetrahedral intermediate. Would this catalytic mechanism be enhanced as the pH increases? (Note: This hypothesis turned out to be incorrect, as described in the chapter.) 69.  Identify which terms are associated with transcription and which with translation: a. Promoter, b. TATA box, c. Shine–Dalgarno sequence, d. σ factor, e. –35 region, f. AUG codon, g. downstream promoter element (DPE).

70.  Using what you have learned in the first three sections of this chapter, compare and contrast bacterial and eukaryotic translation. 71.  In 1961, Howard Dintzis carried out a series of experiments demonstrating that translation occurs in the N → C terminal direction. He added [3H]Leu to immature reticulocytes (cells that actively synthesize hemoglobin), then determined the extent of labeling in each isolated β-globin chain. Results of one of his experiments are shown in the graph. The data points represent protein fragments, with their distance from the N-terminus (expressed as a residue number) indicated on the x axis. a. How do the results confirm that translation occurs in the N → C direction? b. Why was it important to carry out the experiment in 4 minutes, which is less time than it takes the ribosome to synthesize the entire polypeptide chain? c. What would the curve look like if the reticulocytes were incubated with [3H]Leu for longer than it takes to synthesize the entire polypeptide chain? 100 Relative amount of [3H] Leu

59.  A DNA sequence from the beginning of a gene is shown. a. If this sequence is the coding strand, choose the appropriate reading frame and provide the sequence of the encoded peptide. b. What is the correct reading frame and the peptide sequence if the sequence is the noncoding strand?

80 60

Time of incubation: 4 minutes

40 20 0

0

20

40

60 80 100 Amino acid number

120

140

72.  A group of investigators working around the same time as Dintzis (see Problem 71) carried out translation in reticulocytes and obtained results that supported his conclusion that translation proceeds in the N → C direction. The investigators added [14C]Val to reticulocytes in order to label globin chains that were being synthesized. They chose valine because it is the N-terminal residue of both α and β globins (see Fig. 5.5). After a short labeling period, the cells were lysed, and the ribosomes were isolated and then incubated with non-labeled valine in a cell-free protein translation system. The nascent hemoglobin chains were then isolated and analyzed to see what parts of the globin chains were radioactive. What were the results? What results would have been obtained had protein synthesis occurred in the C → N terminal direction? 73.  A “nonsense suppressor” mutation results from a mutation in a tRNA anticodon sequence so that the tRNA can pair with a stop codon (also called a nonsense codon). a. What would be the effect of a nonsense suppressor mutation on protein synthesis in the cell? b. Would all of the cell’s proteins be affected by the mutation? c. Could the aminoacyl–tRNA synthetase that aminoacylates the nonmutated tRNA play a role in minimizing the effects of a nonsense suppressor mutation? 74.  Oxazolidinones are synthetic antibiotics that interfere with bacterial protein translation. An E. coli lacZ expression system (lacZ codes for β-galactosidase; see Section 21.1) was used to study the mechanism of inhibition. Various mutations were introduced into the lacZ gene and the effect on translation was determined by measuring β-galactosidase activity. a. A stop codon was inserted near the N-terminal region of the enzyme and, as expected, the level of β-galactosidase activity was low. In the presence of an oxazolidinone, β-galactosidase activity increased eight-fold. How does the oxazolidinone affect translation of the mutant β-galactosidase gene? b. A lacZ gene containing either a one-nucleotide insertion or a one-nucleotide deletion was constructed. In the presence of the

Problems  693 oxazolidinone, β-galactosidase activity levels increased 15–25 times relative to controls. What does this suggest about the action of the oxazolidinone? c. Next, the researchers measured β-galactosidase activity using a lacZ gene in which the codon for an active-site glutamate residue was mutated to alanine. β-Galactosidase activity of this mutant was similar to that of the control. What does this reveal about the action of the oxazolidinone? d. How does the oxazolidinone inhibit bacterial growth? 75.  Some bacterial genes that code for enzymes involved in the pathway for synthesizing an amino acid generate mRNAs whose 5′ untranslated segment includes a binding site for a tRNA corresponding to that amino acid. When the tRNA that interacts with the binding site is charged with the corresponding amino acid, translation is inhibited. Describe how this regulatory system works and why it makes sense for the bacterial cell. 76.  How might a small molecule that inhibited the formation of a riboswitch serve as a drug to treat a human disease caused by a bacterial pathogen?

22.4  Post-Translational Events 77.  In prokaryotes, translation can begin even before an mRNA transcript has been completely synthesized. Why is co-transcriptional translation not possible in eukaryotes? 78.  The CFTR gene (see Problem 4) codes for the cystic fibrosis transmembrane conductance regulator. This protein functions as a chloride ion transporter in the cell membrane. What can you conclude about the identity of the first 20 or so amino acids in the newly synthesized CFTR polypeptide? 79.  In 1957, Christian Anfinsen carried out a denaturation experiment in vitro with ribonuclease, a pancreatic enzyme consisting of a single 124-residue chain cross-linked by four disulfide bonds (see Problem 4.59). Urea (a denaturing agent) and 2-mercaptoethanol (a reducing agent) were added to a solution of purified ribonuclease, resulting in protein unfolding with a concomitant loss of biological activity. When urea and 2-mercaptoethanol were removed, the ribonuclease spontaneously folded back to its native conformation and regained full enzymatic activity. Why could proper protein folding occur in this experiment in the absence of molecular chaperones? 80.  In another set of experiments, the lysine side chains on the surface of ribonuclease (see Problem 79) were covalently attached to an eight-residue chain of polyalanine. The presence of these polyalanine chains did not affect the ability of the ribonuclease to fold properly. What do these experiments reveal about the driving force for protein folding? 81.  Chaperones are located not just in the cytosol but in the mitochondria as well. What is the role of mitochondrial chaperones? 82.  The chaperone Hsp90 interacts with the tyrosine kinase domain of a growth factor receptor that has lost its ligand-binding domain but retains its tyrosine kinase domain (see Box 10.B). The antibiotic geldanamycin inhibits Hsp90 function. What is the effect of adding geldanamycin to cells expressing the abnormal growth factor receptor? 83.  Multidomain proteins tend to fold better inside cagelike chaperonin structures (such as GroEL/GroES in E. coli) than with cytosolic chaperones. Explain why. 84.  In immature red blood cells, globin synthesis is carefully regulated. The genes for α and β globin (see Section 5.1) are located on separate chromosomes, and there are two α globin genes for every β globin gene. If too many β chains are produced, they form a functionally useless tetrameric hemoglobin. Excess α chains tend to precipitate and damage red blood cells. a. Explain why it is advantageous for the cell to synthesize a slight excess of α chains. b. Red

blood cells express a protein that appears to stabilize the α chains and prevents their precipitation. Why is the stabilizing protein necessary? 85.  The disease β thalassemia results from a defect in a β globin gene. Heterozygotes (who have one normal and one abnormal allele) may develop mild anemia, but homozygotes (who have two defective β globin alleles) exhibit severe anemia. a. Explain why an extra copy of an α globin gene in an individual lacking one β globin gene would result in more severe anemia (see Problem 84). b. Explain why a mutation in an α globin gene would reduce the severity of anemia in β thalassemia. 86.  Immature red blood cells (see Problem 84) produce a kinase that phosphorylates the ribosomal initiation factor eIF2. The phosphorylated eIF2 is unable to exchange bound GDP for GTP. a. How does this affect the rate of protein synthesis in the cell? b. In the presence of heme, the kinase is inactive. How does this mechanism regulate hemoglobin synthesis? 87.  The N-terminal sequence of the secreted protein bovine proalbumin is shown below. Identify the essential features of the signal peptide in this protein. signal peptidase cleavage site

MKWVTFISLLLLFSSAYSRGV 88.  The mammalian signal recognition particle (SRP) consists of one molecule of RNA and six proteins. In addition to interacting with the signal peptide, what might be the role of the RNA in the SRP? 89.  One of the six proteins in the mammalian signal recognition particle (SRP) contains a cleft lined with hydrophobic amino acids. Propose a role for this protein in the SRP. 90.  Why does the mammalian signal recognition particle (SRP) bind to the nascent polypeptide as it emerges from the ribosome? Why doesn’t the SRP wait until translation is complete before escorting the new polypeptide to the ER membrane? 91.  A eukaryotic “cell-free” translation system contains all the components required for protein synthesis—ribosomes; tRNAs; aminoacyl– tRNA synthetases; initiation, elongation, and termination factors; amino acids; GTP; and Mg2+ ions. An exogenous mRNA added to this mixture can direct protein synthesis in vitro. When an mRNA encoding a secretory protein is added to this cell-free system, along with the SRP, the entire protein is synthesized. When microsomes (sealed vesicles derived from ER membranes) are subsequently added, the protein is not translocated into the microsomal lumen and the signal sequence is not removed. What does this observation reveal about the role of SRP in the synthesis of secretory proteins? 92.  The elongation factor EF-Tu performs proofreading by monitoring codon–anticodon base pairing during translation. When a match is confirmed, a conformational change occurs, and EF-Tu hydrolyzes its bound GTP (see Fig. 22.15). Could the SRP use a similar mechanism to perform a proofreading function? 93.  Polyglutamine diseases are neurodegenerative disorders caused by mutations in the DNA that produce CAG repeats (See Problem 21.53). Proteins translated from the mutated genes contain long stretches of glutamine residues that interfere with protein folding. Polyglutamine proteins tend to aggregate and form cytosolic and nuclear inclusion bodies. The result is a loss of neuron function, although the mechanism is unknown. Recent studies show that post-translational modification of the polyglutamine proteins may play a role in the progression of the disease. Interestingly, some post-translational modifications are neurotoxic while others are protective. a. Protein kinase B (see Section 10.2) phosphorylates

694  C ha pter 22   Protein Synthesis a polyglutamine protein on an essential serine residue, resulting in decreased toxicity. Draw the structure of a phosphorylated serine residue. b. Proteins are marked for degradation by the attachment of the protein ubiquitin to a lysine side chain on the condemned protein (see Section 12.1). An isopeptide bond forms between the lysine side chain and the carboxyl terminal group of the ubiquitin. Ubiquitination enhances the degradation of the inclusion body protein. Draw the structure of the linkage between ubiquitin and a polyglutamine protein. c. Polyglutamine proteins interact with a histone acetyltransferase (see Section 21.1), sequestering the enzyme in inclusion bodies and hastening its degradation. What are the cellular consequences of this interaction? 94.  The protein c-Myc is a leucine zipper protein (see Box 21.A) that regulates gene expression in cell proliferation and differentiation. Its activity was known to be regulated by phosphorylation of a specific

threonine residue, but later studies showed that this same threonine could be modified by an N-acetylglucosamine residue and that phosphorylation and glycosylation were competitive processes. The specific threonine is mutated in some human lymphomas. Draw the structure of the O-glycosylated threonine residue. 95.  In the N-myristoylation process, the fatty acid myristate (14:0) is attached to an N-terminal glycine residue of a protein during translation. Draw the structure of a myristoylated N-terminal glycine residue. 96.  In the palmitoylation process, the fatty acid palmitate (16:0) is attached to the side chain of an internal cysteine residue of a protein. Draw the structure of a palmitoylated cysteine residue. What proteins involved in cell signaling pathways include this modification, and what is the role of the palmitate?

Selected Readings Choi, J., Grosely, R., Prabhakar, A., Lapointe, C.P, Wang, J., and Puglisi, J.D., How mRNA and nascent chain sequences regulate translation elongation, Annu. Rev. Biochem. 87, 421–449, doi: 10.1146/annurev-­biochem-060815-014818 (2018). [Describes the approaches for studying translation dynamics and reviews the steps of elongation.] Emmott, E., Jovanovic, M., and Slavov, N., Ribosome stoichiometry: From form to function, Trends Biochem. Sci. 44, 95–109, doi: 10.1016/j.tibs.2018.10.009 (2019). [Discusses approaches for exploring ribosome specialization.] Hayer-­Hartl, M., Bracher, A., and Hartl, F.U., The GroEL-­GroES chaperonin machine: A nano-­cage for protein folding, Trends Biochem. Sci. 41, 62–76, doi: 10.1016/j.tibs.2015.07.009 (2016). [Summarizes the key features of chaperonin action.] Rubio Gomez, M.A. and Ibba, M., Aminoacyl-­tRNA synthetases, RNA 26, 910–936, doi: 10.1261/rna.071720.119 (2020). [A thorough

review of the enzymes that link amino acids to tRNAs, including editing steps.] Saier, M.H, Jr., Understanding the genetic code, J. Bacteriol. 201, e00091-19, doi: 10.1128/JB.00091-19 (2019). [A highly readable discussion of the evolution and conservation of the genetic code, with a focus on the importance of the second position.] Stein, K.C. and Frydman, J., The stop-­and-­go traffic regulating protein biogenesis: How translation kinetics controls proteostasis, J. Biol. Chem. 294, 2076–2084, doi: 10.1074/jbc.REV118.002814 (2019). [Addresses the determinants of efficient translation and co-­translational protein folding.] Yusupova, G. and Yusupov, M., Crystal structure of eukaryotic ribosome and its complexes with inhibitors. Phil. Trans. R. Soc. B 372, 20160184, doi: 10.1098/rstb.2016.0184 (2017). [A brief and readable summary of ribosome structure and the effects of some inhibitors on ribosome function.]

Chapter 22 Credits Figure 22.2b Image based on 4TRA. Westhof, E., Dumas, P., Moras, D., Restrained refinement of two crystalline forms of yeast aspartic acid and phenylalanine transfer RNA crystals Acta Crystallogr. A 44, 112–123 (1988). Figure 22.4 Image based on 1QRT. Arnez, J.G., Steitz, T.A., Crystal structures of three misacylating mutants of Escherichia coli glutaminyl-tRNA synthetase complexed with tRNA(Gln) and ATP, Biochemistry 35, 14725–14733 (1996). Figure 22.5 Image based on 1J5E. Wimberly, B.T., Brodersen, D.E., Clemons Jr., W.M., Morgan-Warren, R.J., Carter, A.P., Vonrhein, C.,

Hartsch, T., Ramakrishnan, V., Structure of the 30S ribosomal subunit, Nature 407, 327–339 (2000). Figure 22.6 Image based on 1JJ2. Klein, D.J., Schmeing, T.M., Moore, P.B., Steitz, T.A., The kink-turn, a new RNA secondary structure motif, EMBO J. 20, 4214–4221 (2001). Figure 22.7 Image from Ben-Shem, A., Garreau de Loubresse, N., Melnikov, S., Jenner, L., Yusupova, G., & Yusupov, M., The Structure of the Eukaryotic Ribosome at 3.0 A Resolution. Science 334(6062), 1524–1529. © 2011 American Association for the Advancement of Science.

Chapter 22 Credits  695 Figure 22.8 Image from Schmeing, T., Ramakrishnan, V. What recent ribosome structures have revealed about the mechanism of translation. Nature 461, 1234–1242. © 2009 Springer Nature. Figure 22.13 Image based on 1TTT. Nissen, P., Kjeldgaard, M., Thirup, S., Polekhina, G., Reshetnikova, L., Clark, B.F., Nyborg, J., Crystal structure of the ternary complex of Phe-tRNAPhe, EF-Tu, and a GTP analog, Science 270, 1464–1472 (1995). Figure 22.14 Image from Ogle, J.M., Recognition of Cognate Transfer RNA by the 30S Ribosomal Subunit. Science 292(5518), 897–902. © 2001 American Association for the Advancement of Science. Figure 22.18 Image based on 2BV3. Hansson, S., Singh, R., Gudkov, A.T., Liljas, A., Logan, D.T., Crystal structure of a mutant elongation factor G trapped with a GTP analogue, FEBS Lett. 579, 4492 (2005). Figure 22.20 Image based on 3D5A Laurberg, M., Asahara, H., Korostelev, A., Zhu, J., Trakhanov, S., Noller, H.F., Crystal structure of a translation termination complex formed with release factor RF2, Proc. Natl. Acad. Sci. USA 105, 19684–19689 (2008).

Figure 22.22b Image from Brandt, F., Etchells, S.A., Ortiz, J.O., Elcock, A.H., Hartl, F.U., & Baumeister, W., The Native 3D Organization of Bacterial Polysomes. Cell 136(2), 261–271. © 2009 Elsevier. Figure 22.25 Image based on 1AON. Xu, Z., Horwich, A.L., Sigler, P.B., The crystal structure of the asymmetric GroEL-GroES-(ADP)7 chaperonin complex, Nature 388, 741–750 (1997). Figure 22.27 Image from Batey, R.T., Crystal Structure of the Ribonucleoprotein Core of the Signal Recognition Particle. Science 287(5456), 1232–1239. © 2000 American Association for the Advancement of Science. Figure 22.28 Image from Becker, T., Bhushan, S., Jarasch, A., Armache, J.-P., Funes, S., Jossinet, F., Gumbart, J., Mielke, T., Berninghausen, O., Schulten, K., Westhof, E., Gilmore, R., Mandon, E. C., & Beckmann, R., Structure of Monomeric Yeast and Mammalian Sec61 Complexes Interacting with the Translating Ribosome. Science 326(5958), 1369–1373. © 2009 American Association for the Advancement of Science.

Glossary Numbers and Greek letters are alphabetized as if they were spelled out. A A-DNA.  A conformation of DNA in which the double helix is wider than the standard B-DNA helix and in which base pairs are inclined to the helix axis. A site.  The ribosomal binding site that accommodates an aminoacyl–tRNA. Abasic site.  The deoxyribose residue remaining after the removal of a base from a DNA strand. ABC transporter.  A member of a family of structurally similar transmembrane proteins that use the free energy of ATP to drive conformational changes that move substances across the membrane. Acid.  A substance that can donate a proton. Acid–base catalysis.  A catalytic mechanism in which partial proton transfer from an acid or partial proton abstraction by a base lowers the free energy of a reaction’s transition state. Acid catalysis.  A catalytic mechanism in which partial proton transfer from an acid lowers the free energy of a reaction’s transition state. Acid dissociation constant (Ka).  The dissociation constant for an acid in water. Acidic solution.  A solution whose pH is less than 7.0 ([H+] > 10–7 M). Acidosis.  A pathological condition in which the pH of the blood drops below its normal value of 7.4. Actin filament.  A 70-Å-diameter cytoskeletal element composed of ­polymerized actin subunits. Also called a microfilament. Action potential.  The momentary reversal of membrane potential that occurs during transmission of a nerve impulse. Activation energy (free energy of activation, ΔG‡).  The free ­energy of the transition state minus the free energies of the reactants in a chemical reaction. Activator.  A protein that binds at or near a gene so as to promote its transcription. Active site.  The region of an enzyme in which catalysis takes place. Active transport.  The transmembrane movement of a substance from low to high concentrations by a protein that couples this endergonic transport to an exergonic process such as ATP hydrolysis. See also secondary active transport. Acyl group.  A portion of a molecule with the formula —COR, where R is an alkyl group. Adipose tissue.  Tissue consisting of cells that are specialized for the storage of ­triacylglycerols. See also brown adipose tissue. Aerobic.  Occurring in or requiring oxygen. Affinity.  A measure of the strength of binding between two molecules, often expressed as a dissociation constant. Affinity chromatography.  A procedure in which a molecule is isolated by its ability to bind specifically to a second immobilized molecule.

Agonist.  A substance that binds to a receptor so as to evoke a cellular response. Aldose.  A sugar whose carbonyl group is part of an aldehyde. Alkalosis.  A pathological condition in which the pH of the blood rises above its normal value of 7.4. Allele.  An alternate form of a gene; a diploid organism may contain two alleles for each gene. Allosteric protein.  A protein in which the binding of ligand at one site affects the binding of other ligands at other sites. Some enzymes are allosteric proteins. See also cooperative binding. Allosteric regulation.  Binding of an activator or inhibitor to one subunit of a multisubunit enzyme, which increases or decreases the catalytic activity of all the subunits. See also posi­tive effector and negative effector. α-amino acid.  See amino acid. α anomer.  A sugar in which the OH substituent of the anomeric ­carbon is on the oppo­site side of the ring from the CH2OH group of the chiral center that designates the d or l configuration. α carbon.  See Cα. α helix.  A regular secondary structure of polypeptides, with 3.6 residues per righthanded turn and hydrogen bonds between each backbone CO group and the backbone N—H group that is four residues further. Amino acid (α-amino acid).  A compound consisting of a carbon atom to which are attached a primary amino group, a carboxylate group, a side chain (R group), and an H atom. Amphiphilic (amphipathic).  Having both polar and nonpolar ­regions and therefore being both hydrophilic and hydrophobic. Amyloid deposit.  An accumulation of certain types of insoluble protein aggregates in tissues (e.g., in the brain in Alzheimer’s disease). Anabolism.  The reactions by which ­biomolecules are synthesized from simpler components. Anaerobic.  Occurring independently of oxygen. Anaplerotic reaction.  A reaction that replenishes the intermediates of a metabolic pathway. Anemia.  A condition caused by the insufficient production of or loss of red blood cells. Anneal.  To allow base pairing between complementary single polynucleotide strands so that double-stranded segments form. Anomers.  Sugars that differ only in the configuration around the carbonyl carbon that becomes chiral when the sugar cyclizes. Antagonist.  A substance that binds to a receptor but does not elicit a cellular response. Antenna pigment.  A molecule that transfers its absorbed energy to other pigment molecules and eventually to a photosynthetic reaction center. Antibody.  See immunoglobulin.

Anticodon.  The sequence of three nucleotides in a tRNA that recognizes an mRNA codon through complementary base pairing. Antigen.  A substance, usually foreign to the body, that binds to a receptor on a B or T lymphocyte to trigger an immune response. Antiparallel.  Running in opposite directions. Antiparallel β sheet.  See β sheet. Antiport.  Transport that involves the simultaneous transmembrane movement of two molecules in opposite directions. Antisense strand.  See noncoding strand. Apoptosis.  Programmed cell death that ­re­sults from extracellular or i­ ntracellular ­signals and involves the activation of enzymes that ­selectively degrade cellular structures. Aquaporin.  A membrane protein that facili­ tates the transmembrane movement of water molecules. Archaea.  One of the two major groups of prokaryotes. Atherosclerosis.  A disease characterized by the formation of ­cholesterol-containing fibrous plaques in the walls of blood vessels. ATPase.  An enzyme that catalyzes the hydrolysis of ATP to ADP + Pi. Autoactivation.  A process by which the product of an activation reaction also acts as a catalyst for the same reaction, so that it appears that the compound catalyzes its own activation. Autolysosome.  The organelle resulting from the fusion of an autophagosome and a lysosome during autophagy, inside of which unneeded cellular components are hydrolyzed. Autophagosome.  The double-membraned compartment that fully encloses cellular materials during autophagy. Autophagy.  The process of engulfing and degrading unneeded materials within a eukaryotic cell. Autophosphorylation.  The phosphorylation of a kinase by another molecule of the same kinase. Avidity.  A measure of the strength of an intermolecular interaction, such as antigen binding to an immunoglobin, that reflects the presence of multiple binding sites. Axon.  The extended portion of a neuron that conducts an action potential from the cell body to a synapse with a target cell. B B-DNA.  The standard conformation of double-helical DNA. B lymphocyte.  A type of white blood that responds to the presence of an antigen by releasing immunoglobulins, which are soluble versions of the cell’s antigen receptor. Backbone.  The atoms that form the repeating linkages between successive residues of a polymeric molecule, exclusive of the side chains.

G-1

G-2  GLOSSA RY Bacteria.  One of the two major groups of prokaryotes. Bacteriophage.  A virus specific for bacteria. Also known as a phage. Base.  (1) A substance that can accept a proton. (2) A purine or pyrimidine component of a nucleoside, nucleotide, or nucleic acid. Base catalysis.  A catalytic mechanism in which partial proton abstraction by a base lowers the free energy of a reaction’s transition state. Base excision repair.  A DNA repair pathway in which a damaged base is removed by a gly­cosylase so that the resulting abasic site can be repaired. Base pair.  The specific hydrogen-bonded association between nucleic acid bases. The standard base pairs are A:T and G:C. See also bp. Basic solution.  A solution whose pH is greater than 7.0 ([H+] < 10–7 M). β anomer.  A sugar in which the OH substituent of the anomeric carbon is on the same side of the ring as the CH2OH of the chiral center that designates the d or l configuration. β barrel.  A protein structure consisting of a β sheet rolled into a cylinder. β oxidation.  A series of enzyme-catalyzed reactions in which fatty acids are progressively degraded by the removal of two-carbon units as acetyl-CoA. β sheet.  A regular secondary structure in which extended polypeptide chains form interstrand hydrogen bonds. In parallel β sheets, the polypeptide chains all run in the same direction; in antiparallel β sheets, neighboring chains run in opposite directions. Bilayer.  An ordered, two-layered arrangement of amphiphilic molecules in which polar segments are oriented toward the two solvent-exposed surfaces and the nonpolar segments associate in the center. Bile acid.  A cholesterol derivative that acts as a detergent to solubilize lipids for digestion and absorption. Bimolecular reaction.  A reaction involving two molecules, which may be identical or different. Binding change mechanism.  The mechanism whereby the subunits of ATP synthase adopt three successive conformations to convert ADP + Pi to ATP as driven by the dissipation of the transmembrane proton gradient. Biofilm.  A complex of bacterial cells and a protective extracellular ­matrix containing polysaccharides. Bioinformatics.  The use of computers in collecting, storing, accessing, and analyzing biological data, such as molecular sequences and structures. Biomineralization.  The association of organic molecules such as proteins with inorganic crystals to form materials such as bones and shells. Bisubstrate reaction.  An enzyme-catalyzed reaction involving two ­substrates. Blunt ends.  The fully base-paired ends of a DNA fragment that are generated by a restriction endonuclease that cuts both strands at the same point. Bohr effect.  The decrease in O2 binding affinity of hemoglobin in response to a decrease in pH.

bp.  Base pair, the unit of length used for DNA molecules. Brown adipose tissue.  A type of adipose tissue in which fatty acid oxidation is uncoupled from ATP production so that the free energy of the fatty acids is released as heat. Buffer.  A solution of a weak acid and its conjugate base, which resists changes in pH upon the addition of acid or base. C C-terminus.  The end of a polypeptide that has a free carboxylate group. Cachexia.  The severe loss of body mass that occurs in cancer and some other chronic diseases. Cα.  The alpha carbon, the carbon of an amino acid whose substituents are an amino group, a carboxylate group, an H atom, and a variable R group. Calvin cycle.  The sequence of photosynthetic reactions in which ribulose-5-phosphate is carboxylated, converted to three-carbon carbohydrate precursors, and regenerated. cAMP.  Cyclic AMP, an intracellular second messenger. Cap.  A 7-methylguanosine residue that is post-transcriptionally added to the 5′ end of a eukaryotic mRNA. Capsid.  The protein shell surrounding the nucleic acid of a virus. Carbanion.  A compound that bears a nega­ tive charge on a carbon atom. Carbohydrate.  A compound with the formula (CH2O)n, where n ≥ 3. Also called a saccharide. Carbon fixation.  The incorporation of CO2 into biologically useful organic molecules. Carcinogen.  An agent that causes a mutation in DNA that leads to cancer. Carcinogenesis.  The process of developing cancer. Catabolism.  The degradative metabolic reactions in which nutrients and cell constituents are broken down for energy and raw ­materials. Catalyst.  A substance that promotes a chemical reaction without ­undergoing permanent change. A catalyst increases the rate at which a reaction approaches equilibrium but does not affect the free energy change of the reaction. Catalytic constant (kcat ).  The ratio of the maximal velocity (Vmax) of an enzymecatalyzed reaction to the enzyme concentration. Also called a turnover number. Catalytic perfection.  A state achieved by an enzyme that operates at the diffusion-controlled limit. Catalytic triad.  The hydrogen-bonded Ser, His, and Asp residues that participate in catalysis in serine proteases. cDNA.  See complementary DNA. Cellular respiration.  The metabolic phenomenon whereby organic molecules are oxi­ dized, with the electrons eventually transferred to molecular oxygen. Central dogma of molecular biology.  The idea that genetic information in the form of DNA is rewritten, or transcribed, to RNA, which then is translated to direct protein synthesis. Information passes from DNA to RNA to protein. Centromere.  The region of a replicated eu­kar­yotic chromosome where the two identical DNA molecules are attached and

where the spindle fibers attach during cell division. C4 pathway.  A photosynthetic process used in some plants to concentrate CO2 by incorporating it into oxaloacetate (a C4 compound). Channeling.  The transfer of an intermediate product from one enzyme active site to another in such a way that the intermediate remains in contact with the protein. Chaperone.  See molecular chaperone. Chemical labeling.  A technique for identifying functional groups in a macromolecule by treating the molecule with a reagent that reacts with those groups. Chemiosmotic theory.  The postulate that the free energy of electron transport is conserved in the formation of a transmembrane proton gradient that can be subsequently used to drive ATP synthesis. Chemoautotroph.  An organism that obtains its building materials and free energy from inorganic compounds. Chirality.  The asymmetry or “handedness” of a molecule such that it cannot be superimposed on its mirror image. Chloroplast.  The plant organelle in which photosynthesis takes place. Chromatin.  The complex of DNA and protein that comprises the ­eukaryotic chromosomes. Chromatography.  A technique for separating the components of a mixture of molecules based on their partition between a mobile solvent phase and a porous matrix (stationary phase), often performed in a column. Chromosome.  The complex of protein and a single DNA molecule that comprises some or all of an organism’s genome. Citric acid cycle.  A set of eight enzymatic reactions, arranged in a cycle, in which energy in the form of ATP, NADH, and QH2 is recovered from the oxidation of the acetyl group of acetyl-CoA to CO2. Clinical trial.  A three-phase series of tests of a drug’s safety and effectiveness in human subjects. Clone.  An organism or collection of identical cells derived from a single parental cell. Coagulation.  The process of forming a blood clot. Coding strand.  The DNA strand that has the same sequence (except for the replacement of U with T) as the transcribed RNA; it is the nontemplate strand. Also called the sense strand. Codon.  The sequence of three nucleotides in DNA or RNA that specifies a single amino acid. Coenzyme.  A small organic molecule that is required for the catalytic activity of an enzyme. A coenzyme may be tightly associated with the enzyme as a prosthetic group. Cofactor.  A small organic molecule (coenzyme) or metal ion that is required for the catalytic activity of an enzyme. Coiled coil.  An arrangement of polypeptide chains in which two α helices wind around each other. Competitive inhibition.  A form of enzyme inhibition in which a substance competes with the substrate for binding to the enzyme active site and thereby appears to increase KM.

G LO SSARY  G-3 Complement.  (1) A molecule that pairs in a reciprocal fashion with another. (2) A set of circulating proteins that sequentially activate each other and lead to the formation of a pore in a microbial cell membrane. Complementary DNA (cDNA).  A segment of DNA synthesized from an RNA template. Condensation reaction.  The formation of a covalent bond between two molecules, during which the elements of water are lost. Conformation.  The three-dimensional shape of a molecule that it attained through rotation of its bonds. Conjugate base.  The compound that forms when an acid donates a proton. Consensus sequence.  A DNA or RNA sequence showing the nucleotides most commonly found at each position. Conservative substitution.  A change of an amino acid residue in a protein to one with similar properties (e.g., Leu to Ile or Asp to Glu). Constant domain.  A structural domain of an immunoglobulin heavy or light chain, whose sequence is the same in all immunoglobulins of the same class. Constitutive.  Being expressed at a continuous, steady rate rather than induced. Convergent evolution.  The independent development of similar characteristics in unrelated species. Cooperative binding.  A situation in which the binding of a ligand at one site on a macromolecule affects the affinity of other sites for the same ligand. See also allosteric protein. Cori cycle.  A metabolic pathway in which lactate produced by glycolysis in the muscles is transported via the bloodstream to the liver, where it is used for gluconeogenesis. The re­sulting glucose returns to the muscles. Covalent catalysis.  A catalytic mechanism in which the transient formation of a covalent bond between the catalyst and a reactant lowers the free energy of a reaction’s transition state. CpG island.  A cluster of CG sequences that often marks the beginning of a gene in a mammalian genome. CRISPR.  Clustered regularly interspersed short palindromic repeats, short DNA segments involved in bacterial defense against bacteriophages. CRISPR-Cas9 system.  A gene-editing tool that uses the bacterial endonuclease Cas9, which is directed to cut a specific DNA segment that is complementary to a CRISPR-like guide RNA. The CRISPR-Cas9 system can be designed to inactivate a target gene or to replace it with an altered version of the gene. Cristae.  The infoldings of the inner mitochondrial membrane. Cross-talk.  The interactions of different signal transduction pathways through activation of the same signaling components. Cryo-electron microscopy (cryo-EM).  An application of electron microscopy in which high-resolution images of molecular structures are collected at very low temperatures. Cyclic electron flow.  The light-driven circulation of electrons between Photosystem I and cytochrome b6 f, which leads to the production of ATP but not NADPH.

Cytochrome.  A protein that carries electrons via a prosthetic Fe-­containing heme group. Cytokinesis.  The splitting of the cell into two following mitosis. Cytoskeleton.  The network of intracellular fibers that gives a cell its shape and structural rigidity. D d sugar.  A monosaccharide isomer in which the asymmetric carbon farthest from the carbonyl group has the same spatial arrangement as the chiral carbon of d-glyceraldehyde. Dark reactions.  The photosynthetic reactions in which NADPH and ATP produced by the light reactions are used to incorporate CO2 into carbohydrates. Deamination.  The hydrolytic or oxidative removal of an amino group. Degron.  An amino acid sequence that functions as a signal for the polypeptide to be degraded by a proteasome or by autophagy. ΔG.  See free energy. ΔG‡.  See activation energy. ΔG°′.  See standard free energy change. ΔGreaction.  The difference in free energy between the reactants and products of a chemical reaction; ΔGreaction = ΔGproducts – ΔGreactants. Δψ.  See membrane potential. Denaturation.  The loss of ordered structure in a polymer, such as the disruption of native conformation in an unfolded polypeptide or the unstacking of bases and separation of strands in a nucleic acid. Denitrification.  The conversion of nitrate (NO−3) to nitrogen (N2 ). Deoxyhemoglobin.  Hemoglobin that does not contain bound oxygen or is not in the oxygen-binding conformation. Deoxynucleotide.  A nucleotide in which the pentose is 2ʹ-­deoxyribose. Deoxyribonucleic acid.  See DNA. Desensitization.  A cell’s adaptation to longterm stimulation through a reduced response to the stimulus. Diabetes mellitus.  A disease caused by a deficiency of insulin or the inability to respond to insulin, and characterized by elevated levels of glucose in the blood. Diazotroph.  A bacterium that carries out nitrogen fixation, the conversion of N2 to NH3. Dielectric constant.  A measure of the ability of a substance to interfere with electrostatic interactions; a solvent with a high dielectric constant is able to dissolve salts by shielding the attractive electrostatic forces that would otherwise bring the ions together. Diffraction pattern.  The record of the radiation scattered from an object, for example, in X-ray crystallography. Diffusion-controlled limit.  The theoretical maximum rate of an ­enzymatic reaction in solution, about 108 to 109 M–1 · s–1. Dimer.  An assembly consisting of two monomeric units. Diploid.  Having two equivalent sets of chromosomes. Dipole–dipole interaction.  A type of van der Waals interaction ­between two strongly polar groups.

Disaccharide.  A carbohydrate consisting of two monosaccharides. Discontinuous synthesis.  A mechanism whereby the lagging strand of DNA is synthe­ sized as a series of fragments that are later joined. Dissociation constant (Kd).  The ratio of the products of the concentrations of the dissociated species to those of their parent compounds at equilibrium. Disulfide bond.  A covalent —S—S— linkage, often between two Cys residues in a protein. Divergent evolution.  The accumulation of changes in species that share an ancestor. DNA (Deoxyribonucleic acid).  A polymer of deoxynucleotides whose sequence of bases encodes genetic information in all living cells. DNA barcoding.  A technique for using species-specific gene sequences to identify different organisms represented in a sample of DNA. DNA chip.  See microarray. DNA fingerprinting.  A technique for distinguishing individuals on the basis of DNA polymorphisms, such as the number of short tandem repeats. DNA ligase.  An enzyme that catalyzes the formation of a phosphodiester bond to join two DNA segments. DNA polymerase.  See polymerase. Domain.  A stretch of polypeptide residues that fold into a globular unit with a hydrophobic core. Dysbiosis.  A change in the number of species or their relative abundance that affects the functionality of the microbiota. E Ɛ.  See reduction potential. Ɛ °′.  See standard reduction potential. E site.  The ribosomal binding site that accommodates a deacylated tRNA before it dissociates from the ribosome. Edman degradation.  A procedure for the stepwise removal and identification of the N-terminal residues of a polypeptide. EF.  See elongation factor. Effector function.  The activity of an immunoglobulin, following antigen binding to the Fab region(s), that depends on the Fc region of the protein. Ehlers–Danlos syndrome.  A genetic disease characterized by elastic skin and joint hyperextensibility, caused by mutations in genes for collagen or collagen-processing proteins. EI complex.  The noncovalent complex that forms between an enzyme and a reversible inhibitor. Eicosanoids.  Compounds derived from the C20 fatty acid arachidonic acid, which act in or near the cells that produce them and mediate pain, fever, and other physiological responses. Electron tomography.  A technique for reconstructing three-dimensional structures by analyzing electron micrographs of consecutive tissue slices. Electron transport chain.  The series of small molecules and protein prosthetic groups that transfer electrons from reduced cofactors such as NADH to O2 during cellular respiration.

G-4  GLOSSA RY Electronegativity.  A measure of an atom’s affinity for electrons. Electrophile.  A compound containing an electron-poor center. An electrophile (electronlover) reacts readily with a nucleophile (nucleus-lover). Electrophoresis.  A procedure in which macromolecules are separated on the basis of charge or size by their differential migration through a gel-like matrix under the influence of an applied electric field. In polyacrylamide gel electrophoresis (PAGE), the matrix is cross-linked polyacrylamide. In SDS-PAGE, the detergent sodium dodecyl sulfate is used to denature proteins. Electrostatic catalysis.  A catalytic mechanism in which sequestering the reacting groups away from the aqueous solvent lowers the free energy of a reaction’s transition state. Elongation factor (EF).  A protein that interacts with tRNA and/or the ribosome during polypeptide synthesis. Enantiomers.  Stereoisomers that are nonsuperimposable mirror ­images of one another. Endergonic reaction.  A reaction that has an overall positive free ­energy change (a nonspontaneous process). Endocytosis.  The inward folding and budding of the plasma membrane to form a new intracellular vesicle. See also receptormediated endocytosis and pinocytosis. Endonuclease.  An enzyme that catalyzes the hydrolysis of the phosphodiester bonds between two nucleotide residues within a polynucleotide strand. Endopeptidase.  An enzyme that catalyzes the hydrolysis of a peptide bond within a polypeptide chain. Endosome.  The membrane-enclosed vesicle that results from the infolding of the plasma membrane during endocytosis. Endothermic reaction.  A reaction that absorbs heat from the surroundings so that its change in enthalpy (H ) is greater than zero. Enhancer.  A eukaryotic DNA sequence located some distance from the transcription start site, where an activator of transcription may bind. Enthalpy (H).  A thermodynamic quantity that is taken to be equivalent to the heat content of a biochemical system. Entropy (S).  A measure of the degree of randomness or disorder of a system. Enzyme.  A biological catalyst. Most enzymes are proteins; a few are RNA. Epigenetics.  The inheritance of patterns of gene expression mediated by chromosomal modifications that do not alter the DNA sequence. Epimers.  Sugars that differ only by the configuration at one C atom (excluding the anomeric carbon). Equilibrium constant (Keq).  The ratio, at equilibrium, of the product of the concentrations of reaction products to that of the reactants. ES complex.  The noncovalent complex that forms between an enzyme and its substrate in the first step of an enzyme-catalyzed reaction. Essential compound.  An amino acid, fatty acid, or other compound that an animal ­cannot synthesize and must therefore obtain in its diet. Euchromatin.  The transcriptionally active, relatively uncondensed chromatin in a eukar­ yotic cell.

Eukarya.  See eukaryote. Eukaryote.  An organism consisting of a cell (or cells) whose genetic material is contained in a membrane-bounded nucleus. Evolution.  Change over time, often driven by the process of natural selection. Exciton transfer.  A mode of decay of an energetically excited molecule, in which electronic energy is transferred to a nearby unexcited molecule. Exergonic reaction.  A reaction that has an overall negative free ­energy change (a sponta­ neous process). Exocytosis.  The fusion of an intracellular vesicle with the plasma membrane in order to release the contents of the vesicle outside the cell. Exome.  The set of exons, or expressed protein-coding gene sequences, in an organism’s genome. Exon.  A portion of a gene that appears in both the primary and mature mRNA transcripts. Exonuclease.  An enzyme that catalyzes the hydrolytic excision of a nucleotide residue from the end of a polynucleotide strand. Exopeptidase.  An enzyme that catalyzes the hydrolytic excision of an amino acid residue from one end of a polypeptide chain. Exosome.  A vesicle released from a cell, possibly as a form of intercellular communication. Also known as a microvesicle. Exothermic reaction.  A reaction that releases heat to the surrounding so that its change in enthalpy (H) is less than zero. Extracellular matrix.  The extracellular proteins and polysaccharides that fill the space between cells and form connective tissue in animals. Extrinsic protein.  See peripheral membrane protein. F F.  See Faraday constant. F-actin.  The polymerized form of the protein actin. See also G-actin. Fab fragment.  One of the two antigenbinding portions of an immunoglobulin. Factory model of replication.  A model for DNA replication in which DNA polymerase and associated proteins remain stationary while the DNA template is spooled through them. Faraday constant (F ).  The charge of one mole of electrons, equal to 96,485 coulombs · mol–1 or 96,485 J · V –1 · mol–1. Fatty acid.  A carboxylic acid with a longchain hydrocarbon side group. Fc fragment.  The portion of an immunoglobulin that is responsible for the protein’s effector functions but does not bind antigen. Feed-forward activation.  The activation of a later step in a reaction sequence by the ­product of an earlier step. Feedback inhibitor.  A substance that inhibits the activity of an enzyme that catalyzes an early step of the substance’s synthesis. Fermentation.  An anaerobic catabolic process. Fibrous protein.  A protein characterized by a stiff, elongated conformation, that tends to form fibers. First-order reaction.  A reaction whose rate is proportional to the concentration of a single reactant.

Fischer projection.  A graphical convention for specifying molecular configuration in which horizontal lines represent bonds that extend above the plane of the paper and vertical bonds extend below the plane of the paper. 5ʹ end.  The terminus of a polynucleotide whose C5ʹ is not esterified to another nucleotide residue. Flip-flop.  See transverse diffusion. Flippase.  A translocase that catalyzes the movement of membrane lipids from the non-cytoplasmic leaflet to the cytoplasmic leaflet. Floppase.  A translocase that catalyzes the movement of membrane lipids from the cytoplasmic leaflet to the non-cytoplasmic leaflet. Fluid mosaic model.  A model of biological membranes in which integral membrane proteins float and diffuse laterally in a fluid lipid layer. Fluorescence.  A mode of decay of an excited molecule, in which electronic energy is emitted in the form of a photon. Flux.  The rate of flow of metabolites through a metabolic pathway. Fractional saturation (Y ).  The fraction of a protein’s ligand-binding sites that are occupied by ligand. Free energy (G).  A thermodynamic quantity whose change indicates the spontaneity of a process. For spontaneous processes, ΔG < 0, whereas for a process at equilibrium, ΔG = 0. Free energy of activation.  See activation energy (ΔG‡). Free radical.  A molecule with an unpaired electron. Futile cycle.  Two opposing metabolic ­reactions that function together to provide a control point for regulating metabolic flux. G G.  See free energy. G-actin.  The monomeric form of the protein actin. See also F-actin. G protein.  A guanine nucleotide–binding and –hydrolyzing protein, involved in a process such as signal transduction or protein synthesis, that is inactive when it binds GDP and active when it binds GTP. G protein–coupled receptor.  A transmembrane protein that binds an extracellular ligand and transmits the signal to the cell interior by interacting with an intracellular G protein. Gametes.  The haploid cells produced by meiosis. Gas constant (R).  A thermodynamic constant equivalent to 8.3145 J · K–1 · mol–1. Gated channel.  A transmembrane channel that opens and closes in response to a signal such as changing voltage, ligand binding, or mechanical stress. Gel filtration chromatography.  See size­ exclusion chromatography. Gene.  A unique sequence of nucleotides that encodes a polypeptide or RNA; it may include nontranscribed and nontranslated sequences that have regulatory functions. Gene expression.  The transformation by transcription and translation of the information contained in a gene to a functional RNA or protein product. Gene therapy.  The manipulation of genetic information in the cells of an individual in order to produce a therapeutic effect.

G LO SSARY  G-5 General transcription factor.  One of a set of eukaryotic proteins that are typically required for the synthesis of mRNAs. See also transcription factor. Genetic code.  The correspondence between the sequence of nucleotides in a nucleic acid and the sequence of amino acids in a polypeptide; a series of three nucleotides (a codon) specifies an amino acid. Genome.  The complete set of genetic instructions in an organism. Genome-wide association study (GWAS).  An attempt to correlate genetic variations with a trait such as a particular disease. Genomics.  The study of the size, organization, and gene content of organisms’ genomes. Globin.  The polypeptide component of myoglobin and hemoglobin. Globular protein.  A water-soluble protein characterized by a compact, highly folded structure. Glucogenic amino acid.  An amino acid whose degradation yields a gluconeogenic precursor. See also ketogenic amino acid. Gluconeogenesis.  The synthesis of glucose from noncarbohydrate precursors. Glucose–alanine cycle.  A metabolic pathway in which pyruvate ­produced by glycolysis in the muscles is converted to alanine and transported to the liver, where it is converted back to pyruvate for gluconeogenesis. The re­sulting glucose returns to the muscles. Glycan.  See polysaccharide. Glycerophospholipid.  An amphipathic lipid in which two hydrocarbon chains and a polar phosphate derivative are attached to a glycerol backbone. Glycogen storage disease (GSD).  An inherited defect in an enzyme or transporter that affects the formation, structure, or ­degradation of glycogen. Glycogenolysis.  The enzymatic degradation of glycogen to glucose-1-phosphate. Glycolipid.  A lipid to which carbohydrate is covalently attached. Glycolysis.  The 10-reaction pathway by which glucose is broken down to 2 pyruvate with the concomitant production of 2 ATP and the reduction of 2 NAD+ to 2 NADH. Glycomics.  The systematic study of the structures and functions of carbohydrates, including large glycans and the small oligosaccharides of glycoproteins. Glycoprotein.  A protein to which carbohydrate is covalently attached. Glycosaminoglycan.  An unbranched polysaccharide consisting of alternating residues of an amino sugar and a sugar acid. Glycosidase.  An enzyme that catalyzes the hydrolysis of glycosidic bonds. Glycoside.  A molecule containing a saccharide linked to another molecule by a glycosidic bond to the anomeric carbon. Glycosidic bond.  The covalent linkage between two monosaccharide units in a polysaccharide, or the linkage between the anomeric carbon of a saccharide and an alcohol or amine. Glycosylation.  The attachment of carbohydrate chains to a protein through N- or Oglycosidic linkages. Glycosyltransferase.  An enzyme that catalyzes the addition of a monosaccharide residue to a polysaccharide.

Glyoxylate pathway.  A variation of the citric acid cycle in plants that allows acetyl-CoA to be converted quantitatively to gluconeogenic ­precursors. Glyoxysome.  A membrane-bounded plant organelle in which the reactions of the glyoxylate pathway take place. Gout.  An inflammatory disease, usually caused by impaired uric acid excretion and characterized by painful deposition of uric acid in the joints. GPCR.  See G protein–coupled receptor. Gram-negative bacteria.  Bacterial cells that have an outer membrane and are unable to retain the crystal violet stain. Gram-positive bacteria.  Bacterial cells that have a thick peptidoglycan cell wall and retain the crystal violet stain. GSD.  See glycogen storage disease. Gustation.  The sense of taste, involving the perception of acidity and saltiness by ion channels, or the perception of sweet, bitter, and umami substances by receptors in specialized cells. GWAS.  See genome-wide association study. H H.  See enthalpy. Half-reaction.  The single oxidation or reduction process, involving the reduced and oxidized forms of a substance, that must be combined with another half-reaction to form a complete oxidation–­reduction reaction. Haploid.  Having one set of chromosomes. Haworth projection.  A drawing of a sugar ring in which ring bonds that project in front of the plane of the paper are represented by heavy lines and ring bonds that project behind the plane of the paper are represented by light lines. Helicase.  An enzyme that unwinds DNA. Heme.  A protein prosthetic group that binds O2 (in myoglobin and hemoglobin) or undergoes redox reactions (in cytochromes). Henderson–Hasselbalch equation.  The mathematical expression of the relationship between the pH of a solution of a weak acid and its pK: pH = pK + log ([A–]/[HA]). Hetero-.  Different. In a heteropolymer, the subunits are not all identical. Heterochromatin.  Highly condensed, nonexpressed eukaryotic DNA. Heterotroph.  An organism that obtains its building materials and free energy from organic compounds produced by other organisms. Hexose.  A six-carbon sugar. High-performance liquid chromatography (HPLC).  An automated chromatographic procedure for fractionating molecules using high pressure and computer-controlled solvent delivery. Highly repetitive DNA.  See tandemly repeated DNA. Histone code.  The correlation between patterns of covalent modification of histone proteins and the transcriptional activity of the associated DNA. Histones.  Highly conserved basic proteins that form a core to which DNA is bound in a nucleosome. Homeostasis.  The maintenance of constant internal conditions. Homo-.  The same. In a homopolymer, all the subunits are identical.

Homologous chromosomes.  Chromosomes with similar sequences in diploid cells; the result of combination of two haploid sets of chromosomes at fertilization. Homologous genes.  Genes that are related by evolution from a ­common ancestor. Homologous proteins.  Proteins that are related by evolution from a common ancestor. Horizontal gene transfer.  The transfer of genetic material between ­species. Hormone.  A substance that is secreted by one tissue and induces a physiological response in other tissues. Hormone response element.  A DNA sequence to which an intracellular hormone– receptor complex binds so as to regulate gene expression. HPLC.  See high-performance liquid chromatography. Hydration.  The molecular state of being surrounded by and interacting with solvent water molecules; that is, solvated by water. Hydrogen bond.  A partly electrostatic, partly covalent interaction ­between a donor group such as O—H or N—H and an electronegative acceptor atom such as O or N. Hydrolysis.  The cleavage of a covalent bond accomplished by adding the elements of water; the reverse of a condensation. Hydronium ion.  A proton associated with a water molecule, H3O+. Hydrophilic.  Having high enough polarity to readily interact with ­water molecules. Hydrophilic substances tend to dissolve in water. Hydrophobic.  Having insufficient polarity to readily interact with water molecules. Hydrophobic substances tend to be insoluble in water. Hydrophobic effect.  The tendency of water to minimize its contacts with nonpolar substances, thereby inducing the substances to aggregate. Hypercholesterolemia.  Elevated levels of cholesterol in the blood. Hyperglycemia.  Elevated levels of glucose in the blood. Hypervariable loop.  One of six surface loops with highly variable amino acid sequences that make up the unique antigenbinding site of an immunoglobulin. I IC50.  The concentration of an inhibitory substance necessary to achieve a 50% reduction in activity. IF.  See initiation factor. Illumina sequencing.  A procedure for determining the sequence of nucleotides in DNA by detecting the presence of a fluorescent marker attached to one of the four nucleotides added to a growing DNA strand. Imine.  A molecule with the formula C NH. Imino group.  A portion of a molecule with the formula C N . Immunization.  The intentional exposure of the immune system to an antigen representing a pathogen in order to establish memory cells that can respond and prevent illness if the pathogen is later encountered. Also known as vaccination. Immunoglobulin.  A soluble version of the antigen receptor of a B lymphocyte, consisting of two heavy and two light chains, which can bind

G-6  GLOSSA RY two antigen molecules as well as trigger additional responses. Also known as an antibody. Imprinting.  A heritable variation in the level of expression of a gene according to its parental origin. In vitro.  In the laboratory (literally, in glass). In vivo.  In a living organism. Indel.  An insertion or deletion of nucleo­ tides in DNA, often as the result of an unrepaired DNA polymerase error during replication. Induced fit.  An interaction between a protein and its ligand that induces a conformational change in the protein that enhances the protein’s interaction with the ligand. Inhibition constant (KI).  The dissociation constant for the complex ­between an enzyme and a reversible inhibitor. Initiation factor (IF).  A protein that interacts with mRNA and/or the ribosome and that is required to initiate translation. Insulin resistance.  The inability of cells to respond to insulin. Integral membrane protein.  A membrane protein that is embedded in the lipid bilayer. Also called an intrinsic protein. Intermediate.  See metabolite. Intermediate filament.  A 100-Å-diameter cytoskeletal element consisting of coiled-coil polypeptide chains. Intermembrane space.  The compartment between the inner and outer mitochondrial membranes, which is equivalent to the cytosol in ionic composition. Interspersed repetitive DNA.  Sequences of DNA that are present at hundreds of thousands of copies in the human genome. Formerly known as moderately repetitive DNA. Intrinsic protein.  See integral membrane protein. Intrinsically disordered protein.  A protein whose tertiary structure includes highly flexible extended segments that can adopt different conformations. Intron.  A portion of a gene that is transcribed but excised by splicing prior to translation. Invariant residue.  A residue in a protein that is the same in all evolutionarily related proteins. Ion exchange chromatography.  A fractionation procedure in which charged molecules are selectively retained by a matrix bearing oppositely charged groups. Ion pair.  An electrostatic interaction between two ionic groups of opposite charge. Ionic interaction.  An electrostatic interaction between two groups that is stronger than a hydrogen bond but weaker than a covalent bond. Ionization constant of water (Kw).  A quantity that relates the concentrations of H+ and OH– in pure water: K w = [H+][OH–] = 10–14. Irregular secondary structure.  A segment of a polymer in which each residue has a different backbone conformation; the opposite of regular secondary structure. Irreversible inhibitor.  A molecule that binds to and permanently inactivates an enzyme.

Isoacceptor tRNA.  A tRNA that carries the same amino acid as ­another tRNA but has a different codon. Isodecoder tRNA.  A set of tRNAs that differ in sequence and structure but share the same anticodon and carry the same aminoacyl group. Isoelectric point (pI).  The pH at which a molecule has no net charge. Isopeptide bond.  An amide linkage between two amino acids involving an amino or carboxyl group in a side chain rather than at the α position. Isoprenoid.  A lipid constructed from fivecarbon units with an isoprene skeleton. Also called a terpenoid. Isozymes.  Different proteins that catalyze the same reaction. K K.  See dissociation constant. k.  See rate constant. Ka.  See acid dissociation constant. kb.  Kilobase pairs; 1000 base pairs. kcat.  See catalytic constant. kcat/KM.  The apparent second-order rate constant for an enzyme-­catalyzed reaction; it indicates the enzyme’s overall catalytic efficiency. Kd.  See dissociation constant. Keq.  See equilibrium constant. Ketogenesis.  The synthesis of ketone bodies from acetyl-CoA. Ketogenic amino acid.  An amino acid whose degradation yields compounds that can be converted to fatty acids or ketone bodies but not to glucose. See also glucogenic amino acid. Ketone bodies.  Compounds (acetoacetate and 3-hydroxybutyrate) that are produced from ­acetyl-CoA by the liver and used as metabolic fuels in other tissues when glucose is unavailable. Ketose.  A sugar whose carbonyl group is part of a ketone. KI.  See inhibition constant. Kinase.  An enzyme that transfers a phosphoryl group between ATP and another molecule. Kinetics.  The study of chemical reaction rates. Kinetochore.  A complex of proteins that is anchored to a chromosome and slides along a fraying microtubule, thereby pulling the chromosome toward one pole of a dividing cell. KM.  See Michaelis constant. Kw.  See ionization constant of water. Kwashiorkor.  A form of severe malnutrition resulting from inadequate protein intake; marked by abdominal swelling and reddish hair. See also marasmus. L l sugar.  A monosaccharide isomer in which the asymmetric carbon ­farthest from the carbonyl group has the same spatial arrangement as the chiral carbon of l-glyceraldehyde. Lagging strand.  The DNA strand that is synthesized as a series of discontinuous fragments that are later joined. Lateral diffusion.  The movement of a membrane component within one leaflet of a ­bilayer.

Le Châtelier’s principle.  The observation that a change in concentration, temperature, volume, or pressure in a system at equilibrium causes the equilibrium to shift in order to counteract the change. Leading strand.  The DNA strand that is synthesized continuously during DNA replica­tion. Ligand.  (1) A small molecule that binds to a larger molecule. (2) A molecule or ion bound to a metal ion. Ligase.  See DNA ligase. Light-harvesting complex.  A pigmentcontaining protein that collects light energy in order to transfer it to a photosynthetic reaction center. Light reactions.  The photosynthetic reactions in which light energy is absorbed and used to generate NADPH and ATP. Linear electron flow.  The light-driven linear path of electrons from water through Photosystems II and I, which leads to the production of O2, NADPH, and ATP. Also known as noncyclic electron flow. Lineweaver–Burk plot.  A rearrangement of the Michaelis–Menten equation that permits the determination of KM and Vmax from a linear plot. Lipid.  Any member of a broad class of macromolecules that are largely or wholly hydrophobic and therefore tend to be insoluble in water but soluble in organic solvents. Lipid bilayer.  See bilayer. Lipid-linked protein.  A protein that is anchored to a biological membrane via a covalently attached lipid. Lipolysis.  The degradation of a triacylglycerol so as to release fatty ­acids. Lipoprotein.  A globular particle, containing lipids and proteins, that transports lipids between tissues via the bloodstream. Lock-and-key model.  An early model of enzyme action, in which the substrate fit the enzyme like a key in a lock. London dispersion forces.  The weak van der Waals interactions between nonpolar groups as a result of fluctuations in their electron distributions that create a temporary separation of charge (polarity). Low-barrier hydrogen bond.  A short, strong hydrogen bond in which the proton is equally shared by the donor and acceptor atoms. Lysosome.  A membrane-bounded organelle in a eukaryotic cell that contains a battery of hydrolytic enzymes and that functions to digest ingested material and to recycle cell components. M Maillard reaction.  The reaction of a monosaccharide carbonyl group with an unprotonated amino group to yield a reactive product that can generate additional reaction products, including substances that give flavor and color to cooked foods. Major groove.  The wider of the two grooves on a DNA double helix. Marasmus.  Body wasting due to inadequate intake of all types of foods. See also kwashiorkor. Mass action ratio.  The ratio of the product of the concentrations of reaction products to that of the reactants.

G LO SSARY  G-7 Mass spectrometry.  A technique for identifying molecules by measuring the mass-tocharge ratios of gas-phase ions, such as peptide fragments. Matrix.  See mitochondrial matrix. Mediator.  A protein complex that interacts with transcription factors and RNA polymerase to regulate gene expression in eukaryotes. Meiosis.  A variation of mitosis to generate gametes with a haploid set of chromosomes. Melting temperature (Tm).  The midpoint temperature of the melting curve for the thermal denaturation of a macromolecule. For a lipid, the temperature of transition from an ordered crystalline state to a more fluid state. Membrane potential (Δψ).  The difference in electrical charge across a membrane. Membraneless organelle.  An intracellular aggregation of molecules, including proteins or RNA, with a consistency that differs from the surrounding solution and that results from liquid–liquid phase separation. Memory cell.  A B or T lymphocyte that originates during an immune response to an antigen and that can mount a more effective response during a subsequent encounter with the same antigen. Messenger RNA (mRNA).  A ribonucleic acid whose sequence is complementary to that of a protein-coding gene in DNA. Metabolic acidosis.  A low blood pH caused by the overproduction or retention of hydrogen ions. Metabolic alkalosis.  A high blood pH caused by the excessive loss of hydrogen ions. Metabolic fuel.  A molecule that can be oxi­ dized to provide free energy for an organism. Metabolic pathway.  A series of enzyme-catalyzed reactions by which one ­substance is transformed into another. Metabolic syndrome.  A set of symptoms related to obesity, including inulin resistance, atherosclerosis, and hypertension. Metabolically irreversible reaction.  A reaction whose value of ΔG is large and negative so that the reaction cannot proceed in reverse. Metabolism.  The total of all degradative and biosynthetic cellular ­reactions. Metabolite.  A reactant, intermediate, or product of a metabolic reaction. Metabolome.  The complete set of metab­ olites produced by a cell or tissue. Metabolomics.  The study of all the metab­ olites produced by a cell or tissue. Metagenomics.  The analysis of all the DNA present in a sample in order to identify the genomes of the different species present. Metal ion catalysis.  A catalytic mechanism that requires the presence of a metal ion to lower the free energy of a reaction’s transition state. Metamorphic protein.  A protein that has more than one stable tertiary structure and can easily switch between them. Metastasis.  The spread of cancerous cells from the site of a primary tumor to a secondary site. Micelle.  A globular aggregate of amphiphilic molecules in aqueous ­solution that are oriented such that polar segments form the surface of the aggregate and the nonpolar segments form a core that is out of contact with the solvent.

Michaelis constant (KM).  For an enzyme that follows the Michaelis–Menten model, KM = (k–1 + k2)/k1; KM is equal to the substrate concentration at which the reaction velocity is half-maximal. Michaelis–Menten equation.  A mathematical expression that describes the activity of an enzyme in terms of the substrate concentration ([S]), the enzyme’s maximal velocity (Vmax), and its Michaelis constant (KM): v0 = Vmax[S]/(KM + [S]). Micro RNA (miRNA).  A 20- to 25nucleotide double-stranded RNA that binds to and inactivates a number of complementary mRNA molecules in RNA interference. Microarray.  A collection of DNA sequences that hybridize with RNA molecules and that can therefore be used to identify active genes. Also called a DNA chip. Microbiome.  The collective genetic material belonging to the microbiota. Microbiota.  The collection of microorganisms that live in or on the human body. Microenvironment.  A group’s immediate neighbors, whose chemical and physical properties may affect the group. Microfilament.  See actin filament. Microtubule.  A 240-Å-diameter cytoskeletal element consisting of a hollow tube of polymerized tubulin subunits. Microvesicle.  See exosome. Minor groove.  The narrower of the two grooves on a DNA double ­helix. (–) end.  The end of a polymeric filament where growth is slower. See also (+) end. miRNA.  See micro RNA. Mismatch repair.  A DNA repair pathway that removes and replaces mispaired nucleotides on a newly synthesized DNA strand. Mitochondrial matrix.  The gel-like solution of enzymes, substrates, ­cofactors, and ions in the interior of the mitochondrion. Mitochondrion (pl. mitochondria).  The double-membrane-­enveloped eukaryotic organelle in which aerobic metabolic reactions occur, including those of the citric acid cycle, fatty acid oxidation, and oxidative phosphorylation. Mitosis.  The process of allocating equivalent sets of chromosomes to daughter cells during eukaryotic cell division. Mixed inhibition.  A form of enzyme inhibition in which an inhibitor binds to the enzyme such that it causes the apparent Vmax to decrease and the apparent KM to increase or decrease. Mobilization.  The process in which polysaccharides, triacylglycerols, and proteins are degraded to make metabolic fuels available. Moderately repetitive DNA.  See interspersed repetitive DNA. Molecular chaperone.  A protein that binds to unfolded or misfolded proteins in order to promote their normal folding. Monoclonal antibody.  One of the identical immunoglobulins produced from a single B lymphocyte or from clones (copies) of that cell. Monogenetic disease.  A disease linked to a defect in a single gene. Monomer.  A structural unit from which a polymer is built up. Monomorphic protein.  A protein with a single stable tertiary structure.

Monosaccharide.  A carbohydrate consisting of a single sugar molecule. Motor protein.  An intracellular protein that couples the free energy of ATP hydrolysis to molecular movement relative to another protein that often acts as a track for the linear movement of the motor protein. mRNA.  See messenger RNA. Multienzyme complex.  A group of noncovalently associated enzymes that catalyze two or more sequential steps in a metabolic pathway. Multifunctional enzyme.  A protein that carries out more than one chemical reaction. Mutagen.  An agent that induces a mutation in an organism. Mutation.  A heritable alteration in an organism’s genetic material. Myelin sheath.  The multilayer coating of sphingomyelin-rich membranes that insulates a mammalian neuron. N N-linked oligosaccharide.  An oligosaccharide linked to the amide group of a protein Asn residue. N-terminus.  The end of a polypeptide that has a free amino group. Nanopore sequencing.  A method that identifies each residue in a polynucleotide strand by monitoring how the current changes as the strand passes through a protein pore in a polarized membrane. Native structure.  The fully folded conformation of a macromolecule. Natural selection.  The evolutionary process by which the continued existence of a replicating entity depends on its ability to survive and reproduce under the existing conditions. ncRNA.  See noncoding RNA. Near-equilibrium reaction.  A reaction whose ΔG value is close to zero, so that it can operate in either direction depending on the substrate and product concentrations. Negative effector.  A substance that diminishes an enzyme’s activity through allosteric inhibition. Nernst equation.  An expression of the relationship between the actual (Ɛ) and standard reduction potential (Ɛ °′) of a substance A: Ɛ = Ɛ °′ – (RT/nF ) ln([Areduced ]/[Aoxidized ]). Neurotransmitter.  A substance released by a nerve cell to alter the activity of a target cell. Neutral solution.  A solution whose pH is equal to 7.0 ([H+] = 10–7 M). Nick.  A single-strand break in a doublestranded nucleic acid. Nick translation.  The progressive movement of a single-strand break (nick) in DNA through the actions of an exonuclease that removes residues followed by a polymerase that replaces them. Nitrification.  The conversion of ammonia (NH3) to nitrate (NO−3). Nitrogen cycle.  A set of reactions, including nitrogen fixation, nitrification, and denitrification, for the interconversion of different forms of nitrogen. Nitrogen fixation.  The process by which atmospheric N2 is ­converted to a biologically useful form such as NH3.

G-8  GLOSSA RY NMR spectroscopy.  See nuclear magnetic resonance spectroscopy. Noncoding RNA (ncRNA).  An RNA mol­ ecule that is not translated into protein. Noncoding strand.  The DNA strand that has a sequence complementary (except for the replacement of U with T) to the transcribed RNA; it is the template strand. Also called the antisense strand. Noncompetitive inhibition.  A form of enzyme inhibition in which an inhibitor binds to an enzyme such that the apparent Vmax decreases but KM is not affected. Noncyclic electron flow.  See linear electron flow. Nonessential amino acid.  An amino acid that an organism can synthesize from common intermediates. Nonhomologous end-joining.  A ligation process that repairs a double-stranded break in DNA. Nonreducing sugar.  A saccharide with an anomeric carbon that has formed a glycosidic bond and cannot therefore act as a reducing agent. Nonspontaneous process.  A thermodynamic process that has a net increase in free energy (ΔG > 0) and can occur only with the input of free energy from outside the system. See also endergonic reaction. Nuclear magnetic resonance (NMR) spectroscopy.  A spectroscopic method in which the signals emitted by atomic nuclei in a magnetic field can be used to determine the three-dimensional structure of a molecule. Nuclear pore complex.  A large assembly of nucleoporin proteins that form a three-ring structure spanning the inner and outer nuclear membranes, with a gel-like pore that permits transport factors to escort proteins and RNA between the nucleus and cytosol. Nuclease.  An enzyme that degrades nucleic acids. Nucleic acid.  A polymer of nucleotide residues. The major nucleic acids are deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Also known as a polynucleotide. Nucleolus.  A dense membraneless organelle inside nucleus, where ribosomal RNA genes are transcribed and ribosomal proteins are assembled with rRNA to make the large and small subunits of the ribosome. Nucleophile.  A compound containing an electron-rich group. A nucleophile (nucleus-lover) reacts with an electrophile (electron-lover). Nucleoside.  A compound consisting of a nitrogenous base linked to a five-carbon sugar (ribose or deoxyribose). Nucleosome.  The disk-shaped complex of a histone octamer and DNA that represents the fundamental unit of DNA organization in eukaryotes. Nucleotide.  A compound consisting of a nucleoside esterified to one or more phosphate groups. Nucleotides are the monomeric units of nucleic acids. Nucleotide excision repair.  A DNA repair pathway in which a ­damaged single-stranded segment of DNA is removed and replaced with normal DNA. O O-linked oligosaccharide.  An oligosaccharide linked to the hydroxyl group of a protein Ser or Thr side chain.

Odorant.  The ligand for an olfactory (smell) receptor. Okazaki fragments.  The short segments of DNA formed in the discontinuous laggingstrand synthesis of DNA. Olfaction.  The sense of smell, mediated by olfactory receptors in sensory neurons in the nose. Oligonucleotide.  A polynucleotide consisting of a few nucleotide ­residues. Oligopeptide.  A polypeptide consisting of a few amino acid residues. Oligosaccharide.  A polymeric carbohydrate containing a few monosaccharide residues. In glycoproteins, the groups are known as N-linked and O-linked oligosaccharides. Omega-3 fatty acid.  A fatty acid with a double bond starting at the third carbon from the methyl (omega) end of the molecule. Oncogene.  A mutant gene that interferes with the normal regulation of cell growth and contributes to cancer. Open reading frame (ORF).  A portion of the genome that potentially codes for a protein. Operon.  A prokaryotic genetic unit that consists of several genes with related functions that are transcribed as a single mRNA molecule. Opioid.  A derivative or analog of morphine, which blocks pain perception but may also depress respiration. Ordered mechanism.  A multisubstrate reaction with a compulsory ­order of substrate binding to the enzyme. ORF.  See open reading frame. Organelle.  A membrane-enclosed compartment, with a specialized function, inside a eukaryotic cell. Orientation effects.  See proximity and orientation effects. Osmosis.  The movement of solvent from a region of low solute concentration to a region of high solute concentration. Osteogenesis imperfecta.  A disease caused by mutations in collagen genes and characterized by bone fragility and deformation. Oxidant.  See oxidizing agent. Oxidation.  A reaction in which a substance loses electrons. Oxidative phosphorylation.  The process by which the free energy obtained from the oxidation of metabolic fuels is used to generate ATP from ADP + Pi. Oxidizing agent.  A substance that can accept electrons, thereby becoming reduced. Also called an oxidant. Oxyanion hole.  A cavity in the active site of a serine protease that accommodates the reac­ tants during the transition state and thereby lowers its energy. Oxyhemoglobin.  Hemoglobin that contains bound oxygen or is in the oxygen-binding conformation. P P site.  The ribosomal binding site that accommodates a peptidyl–tRNA. Palindrome.  A segment of DNA that has the same sequence on each strand when read in the 5ʹ → 3ʹ direction. Parallel β sheet.  See β sheet. Partial oxygen pressure (pO2).  The concentration of gaseous O2 in units of torr.

Passive transport.  The thermodynamically spontaneous protein-mediated transmembrane movement of a substance from high to low concentration. Pasteur effect.  The greatly increased sugar consumption of yeast grown under anaerobic conditions compared to that of yeast grown ­under aerobic conditions. Pathogen.  An infectious agent, such as a virus, bacterium, or microscopic eukaryote, that causes illness. PCR.  See polymerase chain reaction. Pentose.  A five-carbon sugar. Pentose phosphate pathway.  A pathway for glucose degradation that yields ribose-5phosphate and NADPH. Peptide.  A short polypeptide. Peptide bond.  An amide linkage between the α-amino group of one amino acid and the α-carboxylate group of another. Peptide bonds link the amino acid residues in a polypeptide. Peptidoglycan.  The cross-linked polysaccharides and polypeptides that form bacterial cell walls. Peripheral membrane protein.  A protein that is weakly associated with the surface of a biological membrane. Also called an extrinsic protein. Peroxisome.  A eukaryotic organelle with specialized oxidative functions, including fatty acid degradation. p50.  The ligand concentration (or pressure for a gaseous ligand) at which a binding protein such as hemoglobin is half-saturated with ligand. pH.  A quantity used to express the acidity of a solution, equivalent to –log[H+]. Phage.  See bacteriophage. Phagophore.  The membrane-enclosed compartment that forms a cuplike shape around materials to be degraded by autophagy. Phagosome.  See autophagosome. Pharmacokinetics.  The behavior of a drug in the body, including its metabolism and excretion. φ (phi).  The torsion angle describing rotation around the N—Cα bond in a peptide group. Phosphatase.  An enzyme that hydrolyzes phosphoryl ester groups. Phosphodiester bond.  The linkage in which a phosphate group is esterified to two alcohol groups (e.g., two ribose units that join the adjacent nucleotide residues in a polynucleotide). Phosphoinositide signaling system.  A signal transduction pathway in which hormone binding to a cell-surface receptor induces phospholipase C to catalyze the hydrolysis of phosphatidylinositol bisphosphate to yield the second messengers inositol trisphosphate and diacylglycerol. Phospholipase.  An enzyme that hydrolyzes one or more bonds in a glycerophospholipid. Phosphorolysis.  The cleavage of a chemical bond by the substitution of a phosphate group rather than water. Photoautotroph.  An organism that obtains its building materials from inorganic compounds and its free energy from sunlight. Photon.  A packet of light energy. Photooxidation.  A mode of decay of an excited molecule, in which oxidation occurs through the transfer of an electron to an acceptor ­molecule.

G LO SSARY  G-9 Photophosphorylation.  The synthesis of ATP from ADP + Pi coupled to the dissipation of a proton gradient that has been generated through light-driven electron transport. Photoreceptor.  A light-absorbing molecule, or pigment. Photorespiration.  The consumption of O2 and evolution of CO2 by plants (a dissipation of the products of photosynthesis), a consequence of the competition between O2 and CO2 for ribulose bisphosphate ­carboxylase. Photosynthesis.  The light-driven incorporation of CO2 into organic compounds. pI.  See isoelectric point. Pi.  Inorganic phosphate or a phosphoryl group: HPO–3 or PO32–. Pigment.  See photoreceptor. Ping pong mechanism.  An enzymatic reaction in which one or more products are released before all the substrates have bound to the enzyme. Pinocytosis.  Endocytosis of small amounts of extracellular fluid and solutes. pK.  A quantity used to express the tendency for an acid to donate a proton (dissociate); equal to –log K, where K is the dissociation constant. Planck’s law.  An expression for the energy (E ) of a photon: E = hc/λ, where c is the speed of light, λ is its wavelength, and h is Planck’s constant (6.626 × 10–34 J · s). Plasmid.  A small circular DNA molecule that autonomously replicates and may be used as a vector for recombinant DNA. (+) end.  The end of a polymeric filament where growth is faster. See also (–) end. P:O ratio.  The ratio of the number of molecules of ATP synthesized from ADP + Pi to the number of atoms of oxygen reduced. Point mutation.  The substitution of one base for another in DNA, arising from mispairing during DNA replication or from chemical alterations of existing bases. Polarity.  Having an uneven distribution of charge. Poly(A) tail.  The sequence of adenylate residues that is post-transcriptionally added to the 3′ end of eukaryotic mRNAs. Polygenic disease.  A disease linked to variations in more than one gene. Polymer.  A molecule consisting of numerous smaller units that are linked together in an organized manner. Polymerase.  An enzyme that catalyzes the addition of nucleotide residues to a polynucleotide. DNA is synthesized by DNA polymerase, and RNA is synthesized by RNA polymerase. Polymerase chain reaction (PCR).  A procedure for amplifying a segment of DNA by repeated rounds of replication centered between primers that hybridize with the two ends of the DNA segment of interest. See also quantitative PCR (qPCR) and reverse transcriptase quantitative PCR (RT-qPCR). Polynucleotide.  See nucleic acid. Polypeptide.  A polymer consisting of amino acid residues linked in linear fashion by peptide bonds. Polyprotein.  A polypeptide that undergoes proteolysis after its synthesis to yield several separate protein molecules. Polyprotic acid.  A substance that has more than one acidic proton and therefore has multiple ionization states.

Polysaccharide.  A polymeric carbohydrate containing multiple monosaccharide residues. Also called a glycan. Polysome.  An mRNA transcript bearing multiple ribosomes in the process of translating the mRNA. Porin.  A β barrel protein in the outer membrane of bacteria, mitochondria, or chloroplasts that forms a weakly solute-selective pore. Positive effector.  A substance that boosts an enzyme’s activity through allosteric activation. Post-translational processing.  The removal or derivatization of amino acid residues following their incorporation into a polypeptide. pO2.  See partial oxygen pressure. Primary structure.  The sequence of residues in a polymer. Primase.  The enzyme that synthesizes a segment of RNA to be extended by DNA polymerase during DNA replication. Primer.  An oligonucleotide that base pairs with a template polynucleotide strand and is extended through template-directed polymerization. Prion.  An infectious protein that causes its cellular counterparts to misfold and aggregate, thereby leading to the development of a disease such as transmissible spongiform encephalopathy. Probe.  A labeled single-stranded DNA or RNA segment that can hybridize with a DNA or RNA of interest in a screening procedure. Processing.  See RNA processing and posttranslational processing. Processivity.  A property of a motor protein or other enzyme that ­undergoes many reaction cycles before dissociating from its track or substrate. Product inhibition.  A form of enzyme inhibition in which the reaction product acts as a competitive inhibitor. Prokaryote.  A unicellular organism that lacks a membrane-bounded nucleus. All bacteria and archaea are prokaryotes. Promoter.  The DNA sequence at which RNA polymerase binds to initiate transcription. Proofreading.  An additional catalytic activity of an enzyme, which acts to correct errors made by the primary enzymatic activity. Prosthetic group.  An organic group (such as a coenzyme) that is permanently associated with a protein. Protease.  An enzyme that catalyzes the hydrolysis of peptide bonds. Protease inhibitor.  An agent, often a protein, that reacts incompletely with a protease so as to inhibit further proteolytic activity. Proteasome.  A multiprotein complex with a hollow cylindrical core in which cellular proteins are degraded to peptides in an ATP-­ dependent process. Protein.  A macromolecule that consists of one or more polypeptide chains. Proteoglycan.  An extracellular aggregate of protein and glycosaminoglycans. Proteome.  The complete set of proteins synthesized by a cell. Proteomics.  The study of all the proteins synthesized by a cell. Protofilament.  One of the linear polymers of tubulin subunits that forms a microtubule.

Proton jumping.  The rapid movement of a proton among hydrogen-bonded water molecules. Proton wire.  A series of hydrogen-bonded water molecules and ­protein groups that can relay protons from one site to another. Protonmotive force.  The free energy of the electrochemical proton gradient that forms during electron transport. Proximity and orientation effects.  A catalytic mechanism in which reacting groups are brought close together in an enzyme active site to accelerate the reaction. Pseudogene.  A nonfunctional genomic copy of a true gene, which has lost its function through evolution. ψ (psi).  The torsion angle describing rotation around the Cα—C bond in a peptide group. Purine.  A derivative of the compound purine, such as the nucleotide base adenine or guanine. Purinosome.  A complex of enzymes that catalyze the reactions of purine nucleotide synthesis. Pyrimidine.  A derivative of the compound pyrimidine, such as the nucleotide base cytosine, uracil, or thymine. Q Q cycle.  The cyclic flow of electrons involving a semiquinone intermediate in Complex III of mitochondrial electron transport and in photosynthetic electron transport. qPCR.  See quantitative PCR. Quantitative PCR (qPCR).  A variation of the polymerase chain reaction in which the amount of amplified DNA is monitored as it is generated. Also known as real-time PCR. Quantum yield.  The ratio of carbon atoms fixed or oxygen molecules produced to the number of photons absorbed by the photosynthetic machinery. Quaternary structure.  The spatial arrangement of a macromolecule’s individual subunits. Quorum sensing.  The ability of cells to monitor population density by detecting the concentrations of extracellular substances. R R.  See gas constant. R group.  A symbol for a variable portion of a molecule, such as the side chain of an amino acid. R state.  One of two conformations of an allosteric protein; the other is the T state. Raft.  An area of a lipid bilayer with a distinct lipid composition and near-crystalline consistency. Random mechanism.  A multisubstrate reaction without a compulsory ­order of substrate binding to the enzyme. Rate constant (k).  The proportionality constant between the velocity of a chemical reaction and the concentration(s) of the reactant(s). Rate-determining reaction.  The slowest step in a multistep sequence, such as a metabolic pathway, whose rate determines the rate of the entire sequence. Rate equation.  A mathematical expression for the time-dependent progress of a reaction as a function of reactant concentration. Rational drug design.  The synthesis of more effective drugs based on detailed knowledge of the target molecule’s structure and function.

G-10  GLOSSARY Reactant.  One of the starting materials for a chemical reaction. Reaction center.  A chlorophyll-containing protein where photooxidation takes place. Reaction coordinate.  A line representing the progress of a reaction, part of a graphical presentation of free energy changes during a reaction. Reaction specificity.  The ability of an enzyme to discriminate between possible substrates and to catalyze a single type of chemical reaction. Reactive oxygen species.  Free radicals and their reaction products, such as ​​·O​  − 2​  ​​, H2O2, and ​ ·​OH, derived enzymatically and nonenzymatically from O2. Reading frame.  The grouping of nucleotides in sets of three whose sequence corresponds to a polypeptide sequence. Real-time PCR.  See quantitative PCR. Receptor.  A binding protein that is specific for its ligand and elicits a discrete biochemical effect when its ligand is bound. Receptor-mediated endocytosis.  Endocytosis of an extracellular component as a result of its specific binding to a cell surface receptor. Receptor tyrosine kinase.  A cell-surface receptor whose intracellular domains become active as Tyr-specific kinases as a result of extracellular ligand binding. Recombinant DNA.  A DNA molecule containing DNA segments from different sources. Recombination.  The exchange of polynucleotide strands between ­separate DNA segments; recombination is one mechanism for ­repairing damaged DNA by allowing a homologous segment to serve as a template for replacement of the damaged bases. Redox center.  A group that can undergo an oxidation–reduction ­reaction. Redox reaction.  A chemical reaction in which one substance is ­reduced and another substance is oxidized. Reducing agent.  A substance that can donate electrons, thereby becoming oxidized. Also called a reductant. Reducing sugar.  A saccharide with an anomeric carbon that has not formed a glycosidic bond and can therefore act as a reducing agent. Reductant.  See reducing agent. Reduction.  A reaction in which a substance gains electrons. Reduction potential (Ɛ).  A measure of the tendency of a substance to gain electrons. Regular secondary structure.  A segment of a polymer in which the backbone adopts a regularly repeating conformation; the opposite of irregular secondary structure. Release factor (RF).  A protein that recognizes a stop codon and causes a ribosome to terminate polypeptide synthesis. Renaturation.  The refolding of a denatured macromolecule so as to regain its native conformation. Replication.  The process of making an identical copy of a DNA molecule. During DNA replication, the parental polynucleotide strands separate so that each can direct the synthesis of a complementary daughter strand, resulting in two complete DNA double helices. Replication fork.  The point in a replicating DNA molecule where the two parental strands

separate in order to serve as templates for the synthesis of new strands. Replisome.  The complex of proteins that unwind the DNA helix at the replication fork and synthesize complementary DNA strands using both parental strands as templates. Repressor.  A protein that binds at or near a gene so as to prevent its transcription. Residue.  A term for a monomeric unit after it has been incorporated into a polymer. Resonance stabilization.  The effect of delocalization of electrons in a molecule that cannot be depicted by a single structural diagram. Respirasome.  A respiratory supercomplex containing Complexes I, III, and IV of the mitochondrial electron transport chain, which carries out the entire process of electron transfer from NADH to O2. Respiration.  See cellular respiration. Respiratory acidosis.  A low blood pH caused by insufficient elimination of CO2 (carbonic acid) by the lungs. Respiratory alkalosis.  A high blood pH caused by the excessive loss of CO2 (carbonic acid) from the lungs. Restriction digest.  The generation of a set of DNA fragments by the action of a restriction endonuclease. Restriction endonuclease.  A bacterial enzyme that cleaves a specific DNA sequence. Reverse transcriptase.  A DNA polymerase that uses RNA as its ­template. Reverse transcriptase quantitative PCR (RT-qPCR).  A variation of the polymerase chain reaction in which cellular messenger RNA is first reverse-transcribed to DNA before being amplified and quantified by PCR; this method provides information about expressed DNA sequences. RF.  See release factor. Ribonucleic acid.  See RNA. Ribosomal RNA (rRNA).  The RNA molecules that comprise most of the mass of the ribosome and catalyze peptide bond formation. Ribosome.  The RNA-and-protein particle that synthesizes polypeptides under the direction of mRNA. Ribosome profiling.  A technique based on sequencing and identifying mRNA segments undergoing translation, which are protected from ribonuclease digestion by the presence of an active ribosome. Ribosome recycling factor (RRF).  A protein that binds to a ribosome after protein synthesis to prepare it for another round of translation. Riboswitch.  A segment of mRNA that changes conformation in response to binding a ligand in order to expose a sequence that can affect transcription termination (in prokaryotes), splicing (in eukaryotes), or some other aspect of gene expression, such as translation initiation. Ribozyme.  An RNA molecule that has catalytic activity. RNA (Ribonucleic acid).  A polymer of ribonucleotides, such as messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). RNA interference (RNAi).  A phenomenon in which short RNA segments direct the degradation of complementary mRNA, thereby inhibiting gene expression.

RNA polymerase.  See polymerase. RNA processing.  The addition, removal, or modification of nucleotides in an RNA molecule that is necessary to produce a fully functional RNA. RNA world.  A hypothetical time before the evolution of DNA or ­protein, when RNA stored genetic information and functioned as a catalyst. RRF.  See ribosome recycling factor. rRNA.  See ribosomal RNA. RT-qPCR.  See reverse transcriptase quantitative PCR. S S.  See entropy. Saccharide.  See carbohydrate. Salvage pathway.  A pathway that reincorporates an intermediate of nucleotide degra­dation into a new nucleotide, thereby mini­mizing the need for the nucleotide biosynthetic pathways. Saturated fatty acid.  A fatty acid that does not contain any double bonds in its hydrocarbon chain. Saturation.  The state in which all of a macromolecule’s ligand-binding sites are occupied by ligands. Schiff base.  An imine that forms between an amine and an aldehyde or ketone. Scissile bond.  The bond that is to be cleaved during a proteolytic ­reaction. Scramblase.  A translocase that catalyzes the equilibration of membrane lipids between the two bilayer leaflets. SDS-PAGE.  A form of polyacrylamide gel electrophoresis in which denatured polypeptides are separated by size in the presence of the detergent sodium dodecyl sulfate. See electrophoresis. Second messenger.  An intracellular ion or molecule that acts as a signal for an extracellular event such as ligand binding to a cell-­ surface ­receptor. Second-order reaction.  A reaction whose rate is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. Secondary active transport.  Transmembrane transport of one substance that is driven by the free energy of an existing gradient of a ­second substance. Secondary structure.  The local spatial arrangement of a polymer’s backbone atoms without regard to the conformations of its substituent side chains. Selection.  A technique for distinguishing cells that contain a particular feature, such as resistance to an antibiotic. Semiconservative replication.  The mechanism of DNA duplication in which each new molecule contains one strand from the ­parent molecule and one newly synthesized strand. Sense strand.  See coding strand. Serine protease.  A peptide-hydrolyzing enzyme that has a reactive Ser residue in its active site. Set-point.  A body weight that is maintained through regulation of fuel metabolism and that resists change when an individual attempts to alter fuel consumption or expenditure. Signal peptide.  A short sequence in a membrane or secretory protein that binds

G LO SSARY  G-11 to the signal recognition particle in order to direct the translocation of the protein across a membrane. Signal recognition particle (SRP).  A complex of protein and RNA that recognizes membrane and secretory proteins and mediates their binding to a membrane for translocation. Signal transduction.  The process by which an extracellular signal transmits information to the cell interior by binding to a cell-surface receptor such that binding triggers a series of intracellular events. Silencer.  A DNA sequence some distance from the transcription start site, where a repressor of transcription may bind. Single-nucleotide polymorphism (SNP).  A nucleotide sequence variation in the genomes of two individuals from the same species. siRNA.  See small interfering RNA. Sister chromatids.  The two identical molecules that are the products of eukaryotic DNA replication and that remain together until cell division. Size-exclusion chromatography.  A procedure in which macromolecules are separated on the basis of their size and shape. Also called gel filtration chromatography. Small interfering RNA (siRNA).  A 20to 25-nucleotide double-­stranded RNA that targets for destruction a fully complementary mRNA molecule in RNA interference. Small nuclear RNA (snRNA).  Highly conserved RNAs that participate in eukaryotic mRNA splicing. Small nucleolar RNA (snoRNA).  RNA molecules that direct the sequence-specific methylation of eukaryotic rRNA transcripts. snoRNA.  See small nucleolar RNA. SNP.  See single-nucleotide polymorphism. snRNA.  See small nuclear RNA. Solute.  The substance that is dissolved in water or another solvent to make a solution. Solvation.  The state of being surrounded by solvent molecules. Specificity pocket.  A cavity on the surface of a serine protease, whose chemical characteristics determine the identity of the substrate residue on the N-terminal side of the bond to be cleaved. Sphingolipid.  An amphipathic lipid containing an acyl group, a palmitate derivative, and a polar head group attached to a serine backbone. In sphingomyelins, the head group is a phosphate ­derivative. Sphingomyelin.  See sphingolipid. Spliceosome.  A complex of protein and snRNA that carries out the splicing of immature mRNA molecules. Splicing.  The process by which introns are removed and exons are joined to produce a mature RNA transcript. Spontaneous process.  A thermodynamic process that has a net decrease in free energy (ΔG < 0) and occurs without the input of free energy from outside the system. See also exergonic reaction. SRP.  See signal recognition particle. Stacking interactions.  The stabilizing van der Waals interactions between successive (stacked) bases in a polynucleotide.

Standard conditions.  A set of conditions including a temperature of 25°C, a pressure of 1 atm, and reactant concentrations of 1 M. ­Biochemical standard conditions include a pH of 7.0 and a water concentration of 55.5 M. Standard free energy change (ΔG°′).  The force that drives reactants to reach their equilibrium values when the system is in its biochemical standard state. Standard reduction potential (Ɛ °′).  A measure of the tendency of a substance to gain electrons (to be reduced) under standard conditions. Steady state.  A set of conditions under which the formation and degradation of individual components are balanced such that the system does not change over time. Stereocilia.  Microfilament-stiffened cell processes on the surface of cells in the inner ear, which are deflected in response to sound waves. Sticky ends.  Single-stranded extensions of DNA that are complementary, often because they have been generated by the action of the same restriction endonuclease. Stroma.  The gel-like solution of enzymes and small molecules in the interior of the chloroplast; the site of carbohydrate synthesis. Substrate.  A reactant in an enzymatic reaction. Substrate-level phosphorylation.  The transfer of a phosphoryl group to ADP that is di­rectly coupled to another chemical reaction. Subunit.  One of several polypeptide chains that make up a protein. Sugar–phosphate backbone.  The chain of (deoxy)ribose groups linked by phosphodiester bonds in a polynucleotide chain. Suicide substrate.  A molecule that chemi­ cally inactivates an enzyme only after undergoing part of the normal catalytic reaction. Supercoiling.  A topological state of DNA in which the helix is underwound or overwound so that the molecule tends to writhe or coil up on itself. Supercomplex.  An assembly of mitochondrial electron-transport proteins that may contain variable numbers of Complexes I, III, and IV. Symport.  Transport that involves the simultaneous transmembrane movement of two molecules in the same direction. Synaptic vesicle.  A vesicle loaded with neurotransmitters to be released from the end of an axon. T T lymphocyte.  A type of white blood cell that responds to the presence of an antigen in part by releasing signals that help regulate immunoglobulin production in B lymphocytes. T state.  One of two conformations of an allosteric protein; the other is the R state. Tandemly repeated DNA.  Clusters of short DNA sequences that are repeated side-by-side and are present at millions of copies in the human genome. Formerly known as highly repetitive DNA. Tastant.  The ligand for a receptor that detects sweet, bitter, or umami substances. TATA box.  A eukaryotic promoter element with an AT-rich sequence located upstream from the transcription start site. Tautomer.  One of a set of isomers that differ only in the positions of their hydrogen atoms.

Telomerase.  An enzyme that uses an RNA template to polymerize deoxynucleotides and thereby extend the 3′-ending strand of a eukar­ yotic chromosome. Telomere.  The end of a linear eukaryotic chromosome, which consists of tandem repeats of a short G-rich sequence on the 3′-ending strand and its complementary sequence on the 5′-ending strand. Terpenoid.  See isoprenoid. Tertiary structure.  The entire threedimensional structure of a single-chain polymer, including the conformations of its side chains. Tetramer.  An assembly consisting of four monomeric units. Tetrose.  A four-carbon sugar. Thalassemia.  A hereditary disease caused by insufficient synthesis of hemoglobin, which results in anemia. Thermogenesis.  The process of generating heat by muscular contraction or by metabolic reactions. Thick filament.  A muscle cell structural element that is composed of several hundred myosin molecules. Thin filament.  A muscle cell structural element that consists primarily of an actin filament. Thioester.  A compound containing an ester linkage to a sulfur rather than an oxygen atom. Thioester bond.  An ester linkage to a sulfur rather than an oxygen atom. 3ʹ end.  The terminus of a polynucleotide whose C3ʹ is not esterified to another nucleotide residue. Thylakoid.  The membranous structure in the interior of a chloroplast that is the site of the light reactions of photosynthesis. Tm.  See melting temperature. Topoisomerase.  An enzyme that alters DNA supercoiling by breaking and resealing one or both strands. Torsion angle.  The angle described by successive bonds in a polymeric chain. The torsion angles φ and ψ describe the conformation of a polypeptide backbone. Trace element.  An element that is present in small quantities in a living organism. Transamination.  The transfer of an amino group from an amino acid to an α-keto acid to yield a new α-keto acid and a new amino acid. Transcription.  The process by which RNA is synthesized using a DNA template, thereby transferring genetic information from the DNA to the RNA. Transcription factor.  A protein that promotes the transcription of a gene by binding to DNA sequences at or near the gene or by interacting with other proteins that do so. See also general transcription factor. Transcriptome.  The set of all the RNA ­molecules produced by a cell. Transcriptomics.  The study of the genes that are transcribed in a certain cell type or at a certain time. Transfer RNA (tRNA).  The small L-shaped RNAs that deliver specific amino acids to ribosomes according to the sequence of a bound mRNA. Transgenic organism.  An organism that stably expresses a foreign gene.

G-12  GLOSSARY Transition mutation.  A point mutation in which one purine (or pyrimidine) is replaced by another purine (or pyrimidine). Transition state.  The point of highest free energy, or the structure that corresponds to that point, in the reaction coordinate diagram of a chemical reaction. Transition state analog.  A stable substance that geometrically and electronically resembles the transition state of a reaction and that therefore may inhibit an enzyme that catalyzes the reaction. Translation.  The process of transforming the information contained in the nucleotide sequence of an RNA to the corresponding amino acid sequence of a polypeptide as specified by the genetic code. Translocase.  An enzyme that catalyzes the movement of a substance from one side of a membrane to the other. Translocation.  The movement of tRNA and mRNA, relative to the ribosome, that occurs following formation of a peptide bond and that allows the next mRNA codon to be translated. Translocon.  The complex of membrane proteins that mediates the transmembrane movement of a polypeptide. Transmissible spongiform encephalopathy (TSE).  A fatal neurodegenerative disease caused by infection with a prion. Transpeptidation.  The ribosomal process in which the peptidyl group attached to a tRNA is transferred to the aminoacyl group of another tRNA, forming a new peptide bond and lengthening the polypeptide by one residue at its C-terminus. Transposable element.  A segment of DNA, sometimes including genes, that can move (be copied) from one position to another in a genome. Transverse diffusion.  The movement of a membrane component from one leaflet of a bilayer to the other. Also called flip-flop. Transversion mutation.  A point mutation in which a purine is replaced by a pyrimidine or vice versa. Treadmilling.  The addition of monomeric units to one end of a polymer and their removal from the opposite end such that the length of the polymer remains unchanged. Triacylglycerol.  A lipid in which three fatty acids are esterified to a glycerol backbone. Also called a triglyceride.

Triglyceride.  See triacylglycerol. Trimer.  An assembly consisting of three monomeric units. Triose.  A three-carbon sugar. Triple helix.  The right-handed helical structure formed by three left-handed helical polypeptide chains in collagen. Trisaccharide.  A carbohydrate consisting of three monosaccharides. tRNA.  See transfer RNA. TSE.  See transmissible spongiform encephalopathy. Tumor suppressor gene.  A gene whose loss or mutation may lead to cancer. Turnover number.  See catalytic constant. U Uncompetitive inhibition.  A form of enzyme inhibition in which an inhibitor binds to an enzyme–substrate complex such that the apparent Vmax and KM are both decreased to the same extent. Uncoupler.  A substance that allows the proton gradient across a membrane to dissipate without ATP synthesis so that electron transport proceeds without oxidative phosphorylation. Unimolecular reaction.  A reaction involving one molecule. Uniport.  Transport that involves transmembrane movement of a single molecule. Unsaturated fatty acid.  A fatty acid that contains at least one double bond in its hydrocarbon chain. Urea cycle.  A cyclic metabolic pathway in which amino groups are converted to urea for disposal. Usher syndrome.  A genetic disease characterized by profound deafness and retinitis pigmentosa that leads to blindness, caused in some cases by a defective myosin protein. V v.  Velocity (rate) of a reaction. Vaccination.  See immunization. van der Waals interaction.  A weak noncovalent association between molecules that arises from the attractive forces between polar groups (dipole–dipole interactions) or between nonpolar groups whose fluctuating electron distribution gives rise to temporary dipoles (London dispersion forces). van der Waals radius.  The distance from an atom’s nucleus to its effective electronic surface.

Variable domain.  A structural domain of an immunoglobulin heavy or light chain, whose highly variable surface loops define the antigen-binding site. Variable residue.  A position in a polypeptide that is occupied by different residues in evolutionarily related proteins; its substitution has little or no effect on protein function. Vector.  A DNA molecule, such as a plasmid, that can accommodate a segment of foreign DNA. Vesicle.  A fluid-filled sac enclosed by a lipidbilayer membrane. Virion.  A single virus particle. Virus.  A nonliving infectious particle consisting of genetic information (DNA or RNA) inside a protein capsule that may be surrounded by a lipid membrane. A virus requires a host cell in order to replicate. Vitamin.  A metabolically required substance that cannot be synthesized by an animal and must therefore be obtained from the diet. Vmax.  Maximal velocity of an enzymatic reaction. v0.  Initial velocity of an enzymatic reaction. W Warburg effect.  The increased rate of glycolysis observed in cancerous tissues. Wobble hypothesis.  An explanation for the nonstandard base pairing between tRNA and mRNA at the third codon position, which allows a tRNA to recognize more than one codon. X X-Ray crystallography.  A method for determining three-dimensional molecular structures from the diffraction pattern produced by exposing a crystal of a molecule to a beam of X-rays. Y Y.  See fractional saturation. Z Z.  The net charge of an ion. Z-scheme.  A Z-shaped diagram indicating the electron carriers and their reduction potentials in the photosynthetic electron transport system of plants and cyanobacteria. Zinc finger.  A protein structural motif consisting of 20–60 residues, including Cys and His residues to which one or two Zn2+ ions are tetrahedrally coordinated. Zymogen.  The inactive precursor (proenzyme) of a proteolytic ­enzyme.

Odd-Numbered Solutions Chapter 1

21.

1. a.  carboxylic acid;  b. amine; c. ester; d. alcohol. 3. a. aldehyde;  b. imine; c. thioester; d.  diphosphoric acid. 5.

Hydroxyl group OH N

C

CH2

CH2

SH

O

CH2

Hydroxyl HO group H3C

Amino group

C

O

C

H

C

CH3 O

CH2

NH2 N

N O

P O−

O O

P

N

N CH2 O

O

−O

H

Phosphoanhydride linkage

−O

H

H

O

OH

P

H

OH

H

OH

Hydroxyl group

27.  As described in the text, palmitate and cholesterol are highly nonpolar and are therefore insoluble in water. Both are highly aliphatic. Alanine is water soluble because its amino group and c­ arboxylate groups are ionized, which render the molecule “saltlike.” Glucose is also water soluble because its carbonyl group and many hydroxyl groups are able to form hydrogen bonds with water.

Sulfhydryl group

NH

H

25. a. glycosidic; b. peptide; c. phosphoester.

Amido group

CH2

OH

23.  Uracil has a carbonyl functional group, whereas cytosine has an amino functional group.

  [From Li, S.-Y. et al., Eur. J. Med. Chem. 71, 36–45 (2014).] HN

H

H

CH2 OH

CH3

Ether linkage 7. 

C

HO

Imino group

O

O

Carbonyl group

O−

O

H

Hydroxyl group

Phosphoryl group

9.  Amino acids, monosaccharides, nucleotides, and lipids are the four types of biological small molecules. Amino acids, monosaccharides, and nucleotides can form polymers of proteins, polysaccharides, and nucleic acids, respectively. 11. a.  11-Cyclohexylundecanoate is a lipid.  b.  Selenocysteine is an amino acid.  c.  Galactose is a monosaccharide.  d.  O6-Methylguanosine monophosphate is a nucleotide. 13. a.  C and H plus some O.  b.  C, H, and O.  c.  C, H, O, and N plus small amounts of S. 15.  You should measure the nitrogen content, since this would indicate the presence of protein (neither lipids nor carbohydrates contain appreciable amounts of nitrogen). 17.  A diet high in protein results in a high urea concentration, since urea is the body’s method of ridding itself of extra nitrogen. Nitrogen is found in proteins but is not found in significant amounts in lipids or carbohydrates. A low-protein diet provides the patient with just enough protein for tissue repair and growth. In the absence of excess protein consumption, urea production decreases, and this puts less strain on the patient’s weakened kidneys. 19.  Serine has a hydroxyl group and lysine does not.

29.  DNA forms a more regular structure because DNA consists of only four different nucleotides, whereas proteins are made up of as many as 20 different amino acids. In addition, the 20 amino acids have much more individual variation in their structures than do the four nucleotides. Both of these factors result in a more regular structure for DNA. The cellular role of DNA relies on the sequence of the nucleotides that make up the DNA, not on the overall shape of the DNA molecule itself. Proteins, on the other hand, fold into unique shapes, as illustrated by endothelin in Figure 1.4. The ability of proteins to fold into a wide variety of shapes means that proteins can also serve a wide variety of biochemical roles in the cell. According to Table 1.2, the major roles of proteins in the cell are to carry out metabolic reactions and to support cellular structures. 31.  Pancreatic amylase is unable to digest the glycosidic bonds that link the glucose residues in cellulose. Figure 1.6 shows the structural differences between starch and cellulose. Pancreatic amylase binds to starch prior to catalyzing the hydrolysis of the glycosidic bond; thus the enzyme and the starch must have shapes that are complementary. The enzyme would be unable to bind to the cellulose, whose ­structure is very different from that of starch. 33.  A positive entropy change indicates that the system has become more disordered; a negative entropy change indicates that the system has become more ordered:  a. negative; b. positive; c. positive; d. positive; e. negative.  35.  The polymeric molecule is more ordered and thus has less entropy. A mixture of constituent monomers has a large number of different arrangements (like the balls scattered on a pool table) and thus has greater entropy. 37.  The dissolution of ammonium nitrate in water is a highly endothermic process, as indicated by the positive value of ∆H. This means that when ammonium nitrate dissolves in water, the system absorbs heat from the surroundings and the surroundings become cold. The plastic bag containing the ammonium nitrate becomes cold and can be used as a cold pack to treat an injury.

S-1

S-2  ODD- N UM B ER ED S O LUT I O NS 39.  The dissolution of urea in water is an endothermic process and has a positive ∆H value. In order for the process to be spontaneous, the process must also have a positive ∆S value in order for the free energy change of the process to be negative. Solutions have a higher degree of entropy than the solvent and solute alone. 41.  Calculate ∆H and ∆S, as described in Sample Calculation 1.1: ∆H = HB − HA ∆H = 60 kJ · mol−1 − 54 kJ · mol−1 ∆H = 6 kJ · mol−1 ∆S = SB − SA

∆S = 43 J · K−1 · mol−1 − 22 J · K−1 · mol−1 ∆S = 21 J · K−1 · mol−1 43.  ∆G = ∆H − T∆S   0 > 15,000 J · mol−1 − (T) (51 J · K−1 · mol−1)   −15,000 > −(T) (51 K−1)   15,000 < (T) (51 K−1)   294 K < T The reaction is favorable at temperatures of 21°C and higher. 45.  Since the process is spontaneous, ∆G must be negative. As the value of ∆H is positive, the value of ∆S must be sufficiently positive in order to result in a negative value of ∆G. This is consistent with the observation that ions in solution have a greater entropy than ionic solids.

55. a. oxidized; b. oxidized; c. oxidized; d. reduced. 57.  Retinoic acid is most oxidized, followed by retinal and then retinol, which is most reduced. 59. a.  Palmitate’s carbon atoms, which have the formula —CH2—, are more reduced than CO2, so their reoxidation to CO2 releases free energy.  b.  Because the —CH2— groups of palmitate are more reduced than those of glucose (—HCOH—), their conversion to the fully oxidized CO2 would be even more thermodynamically favorable (have a larger negative value of ∆G) than the conversion of glucose carbons to CO2. Therefore, palmitate carbons provide more energy than glucose carbons. 61.  The folded protein is more ordered than the random coil. However, the solution in which the protein folding takes place is less ordered. When the entire system is considered, a loss of entropy results, consistent with the laws of thermodynamics. 63.  Morphological differences, which are useful for classifying large organisms, are not useful for bacteria, which often look alike. Furthermore, microscopic organisms do not leave an easily interpreted imprint in the fossil record, as vertebrates do. Thus, molecular information is often the only means for tracing the evolutionary history of bacteria. 65.

A

B C

  ∆G = ∆H − T∆S   0 > 26.4 kJ · mol−1 − (273 + 25 K) (∆S)   −26.4 kJ · mol−1 > − (298 K) (∆S)   89 J · K−1 · mol−1 > ∆S

67.  The antibiotics kill some of the bacteria normally growing in the intestine, creating an opportunity for the pathogenic species to grow.

∆S must be greater than 89 J · K−1 · mol−1 to be consistent with the prediction.

Chapter 2

47.  ∆G = ∆H − T∆S   0 > −14.3 kJ · mol−1 − (273 + 25 K) (∆S)   14.3 kJ · mol−1 > − (298 K) (∆S)   −48 J · K−1 · mol−1 > ∆S

1.  The water molecule is not perfectly tetrahedral because the electrons in the nonbonding orbitals repel the electrons in the bonding orbitals more than the bonding electrons repel each other. The angle between the bonding orbitals is therefore slightly less than 109°.

∆S could be any positive value or it could be a negative number with a magnitude less than −48 J · K−1 · mol−1.

3.  Water has the higher boiling point because, although each molecule has the same geometry and can form hydrogen bonds with its neighbors, the hydrogen bonds formed between water molecules are stronger than those formed between H2S molecules. The electronegativity difference between H and O is greater than that between H and S and results in greater differences in the partial charges on the atoms in the water molecule.

49. a.  Entropy decreases when the antibody–protein complex binds because the value of ∆S is negative. b.  ∆G = ∆H − T∆S   ∆G = −87,900 J · mol−1 − (298 K) (−118 J · K−1 · mol−1)   ∆G = −52.7 kJ · mol−1  he negative value of ∆G indicates that the complex forms T spontaneously.

5.  The arrows point toward hydrogen acceptors and away from hydrogen donors.

c.  The second antibody binds to cytochrome c more readily than the first because the change in free energy of binding is a more negative value. [From Raman, C.S. et al., Biochemistry 34, 5831–5838 (1995).] 51. a.  The conversion of glucose to glucose-6-phosphate is not favorable because the ∆G value for the reaction is positive, indicating an endergonic process.

O +H

3N

C

CH2

b.  If the two reactions are coupled, the overall reaction is the sum of the two individual reactions. The ∆G value would be the sum of the ∆G values for the two individual reactions.    

CH

CH2 O N H

COO−

CH

53.  C (most oxidized), A, B (most reduced).

CH3

O

Aspartame

ATP + glucose → ADP + glucose-6-phosphate ∆G = −16.7 kJ · mol−1

Coupling the conversion of glucose to glucose-6-phosphate with the hydrolysis of ATP converts an unfavorable reaction to a favorable reaction. The ∆G value of the coupled reaction is negative, which indicates that the reaction as written is favorable.

C

H

O

S

O N H

O H2N

N

N

O

NH2

O

Sulfanilamide

H

N H

Uric acid

OD D -NUMB ER ED SO LUTI O NS  S-3 7.  Identical hydrogen bonding patterns in the two molecules are shown as open arrows in Solution 6. 9.  [From Puschner, P. et al., J. Vet. Diagn. Invest. 19, 616–624 (2007).] O

HN

H

NH

O

N

O

H

H

H

N

N

N

N H

25.  The waxed car is a hydrophobic surface. To minimize its interaction with the hydrophobic molecules (wax), each water drop minimizes its surface area by becoming a sphere (the geometrical shape with the lowest possible ratio of surface to volume). Water does not bead on glass, because the glass presents a hydrophilic surface with which the water molecules can interact. This allows the water to spread out.

H

N N

more consistent with glycine’s physical properties as a white crystalline solid with a high melting point. While structure A could be water soluble because of its ability to form hydrogen bonds, the high solubility of glycine in water is more consistent with an ionic compound whose positively and negatively charged groups are hydrated in aqueous solution by water molecules.

27.  Polar and nonpolar regions of the detergents are indicated.

H

CH3

Melamine cyanurate 11.  The greater an atom’s electronegativity, the more polar its bond with H and the greater its ability to act as a hydrogen bond acceptor. Thus, N, O, and F, which have relatively high electronegativities, can act as hydrogen bond acceptors, whereas C and S, whose electronegativities are only slightly greater than that of hydrogen, cannot. 13.  There are many possibilities. One example is H

O

R

C

δ− δ+

δ−

H

δ+

O

H3C

(CH2)15

N+ CH3 CH3

Nonpolar

Hexadecyltrimethylammonium

CH3

OH CH3

Polar O– O

CH3

R HO

C

R

H

Nonpolar

Polar

OH

Cholate

15. a.  van der Waals forces (dipole–dipole interactions);  b. ­hydrogen bonding;  c.  van der Waals forces (London dispersion forces);  d.  ionic interactions. 17.  Solubility in water decreases as the number of carbons in the alcohol increases. The hydroxyl group of the alcohol is able to form hydrogen bonds with water, but water cannot interact favorably with the hydrocarbon chain. Increasing the length of the chain increases the number of potentially unfavorable interactions of the alcohol with water and solubility decreases as a result. 19.  Aquatic organisms that live in the pond are able to survive the winter. Since the water at the bottom of the pond remains in the liquid form instead of freezing, the organisms are able to move around. The ice on top of the pond also serves as an insulating layer from the cold winter air.

29.  Polar and nonpolar regions of the detergents are indicated.

Polar

Polar O H3C

(CH2)9

Nonpolar

P

CH3

H

CH3

HO

Dimethyldecylphosphine oxide

CH2OH O H OH H H

O H

(CH2)7 CH3

Nonpolar

OH

n-Octylglucoside

31.  Compounds A and D are amphiphilic, compound B is nonpolar, 21.  The positively charged ammonium ion is surrounded by a shell and compounds C and E are polar. of water molecules that are oriented so that the partially negatively 33. a.  In the nonpolar solvent, AOT’s polar head group faces the charged oxygen atoms interact with the positive charge on the ammointerior of the micelle and its nonpolar tails face the solvent: nium ion. Similarly, the negatively charged sulfate ion is hydrated with water molecules oriented so that the partially positively charged CH2CH3 O hydrogen atoms interact with the negative charge on the sulfate H3C (CH2)3 CH CH2 O C CH2 O Polar anion. (Not shown in the diagram is the fact that the ammonium ions Nonpolar head outnumber the sulfate ions by a 2:1 ratio. Also note that the exact tails H3C (CH2)3 CH CH2 O C CH S O number of water molecules shown is unimportant.) O O CH2CH3

H H

H +

O NH4 O H H

O

H H

H

H

H

O

H

O

H

H

O

AOT O

H

2–

SO4

H

H

H

H

H

O

b.  The protein, which contains numerous polar groups, interacts with the polar AOT groups in the micelle interior. O

H

O

23.  Structure A depicts a polar compound, while structure B depicts an ionic compound similar to a salt like sodium chloride. This is

Water

Isooctane

Protein

35. a.  The nonpolar core of the lipid bilayer helps prevent the passage of water since the polar water molecules cannot easily penetrate

S-4  ODD- N UM B ER ED S O LUT I O NS the hydrophobic core of the bilayer.  b.  Most human cells are surrounded by a fluid containing about 150 mM Na+ and slightly less Cl− (see Fig. 2.12). A solution containing 150 mM NaCl mimics the extracellular fluid and therefore helps maintain the isolated cells in near-normal conditions. If the cells were placed in pure water, water would tend to enter the cells by osmosis; this might cause the cells to burst.

​ Kw​ = 1.0 × ​​10​​  −14​​ = [​​H​​  +​​][​​OH​​  −​​]

37.  Compounds a and b are polar and d is ionic; as these substances are highly hydrated, none of them would be able to cross a lipid bilayer. Compound c is nonpolar and would be able to cross a bilayer.



pH = −log [​​H​​  +​​]



pH = −log (3.4 × ​​10​​  −13​​)



pH = 12.5



−14 [​​H​​  +​​] = _________ ​​  1.0 × ​1−0​​   ​​​    [​OH​​  ​]



−14 [​​H​​  +​​] = _________ ​​  1.0 × ​10​​   ​​​  (0.029 M)



[​​H​​  +​​] = 3.4 × ​​10​​  −13​​ M

39.  Substances present at high concentration move to an area of low concentration spontaneously, or “down” a concentration gradient in a process that increases their entropy. The export of Na+ ions from the cell requires that the sodium ions be transported from an area of low concentration to an area of high concentration. The same is true for potassium transport. Thus, these processes are not spontaneous and an input of cellular energy is required to accomplish the transport.

55. a.  The final concentration of HBr is (0.010 L)(0.50 M) ÷ 0.510 L = 0.0098 M. Since HBr is a strong acid and dissociates completely, the added [H+] is equal to [HBr]. (The existing hydrogen ion concentration in the water itself, 1.0 × 10−7 M, can be ignored because it is much smaller than the hydrogen ion concentration contributed by the HBr.)

41. a.  In a high-solute medium, the cytoplasm loses water and therefore its volume decreases.  b.  In a low-solute medium, the cytoplasm gains water and therefore its volume increases.



pH = −log [H+]



pH = −log (0.0098)



pH = 2.0

43.  Since the molecular mass of H2O is 18.0 g · mol−1, a given volume (for example, 1 L or 1000 g) has a molar concentration of 1000 g · L−1 ÷ 18.0 g · mol−1 = 55.5 M. By definition, a liter of water at pH 7.0 has a hydrogen ion concentration of 1.0 × 10−7 M. Therefore, the ratio of [H2O] to [H+] is 55.5 M ÷ (1.0 × 10−7 M) = 5.55 × 108. 45.  The HCl is a strong acid and dissociates completely. This means that the concentration of hydrogen ions contributed by the HCl is 1.0 × 10−9 M. However, the concentration of the hydrogen ions contributed by the dissociation of water is 100-fold greater than this: 1.0 × 10−7 M. The concentration of the hydrogen ions contributed by the HCl is negligible in comparison. Therefore, the pH of the solution is equal to 7.0. 47.

−

b.  The final concentration of NaOH is (0.025 L)(0.25 M) ÷ 0.525 L = 0.012 M. Since NaOH dissociates completely, the added [OH−] is equal to the [NaOH]. (The existing hydroxide ion concentration in the water itself, 1.0 × 10−7 M, can be ignored because it is much smaller than the hydroxide ion concentration contributed by the NaOH.) Kw = 1.0 × 10−14 = [H+][OH−]



−14

[H+] = _________ ​​  1.0 × ​1−0​   ​​​    ​[OH​  ​] −14 ​ ​[​H​ +]​ ​= _________ ​  1.0 × ​10​   ​​​    0.012 M

[H+] = 8.3 × 10−13 M

49.  The stomach contents have a low pH due to the contribution of gastric juice (pH 1.5–3.0). When the partially digested material enters the small intestine, the addition of pancreatic juice (pH 7.8–8.0) neutralizes the acid and increases the pH.



pH = −log [H+]



pH = −log (8.3 × 10−13)



pH = 12.1

57.

CH2



pH = −log [H+]



pH = −log (0.038)



pH = 1.4

b.  The final concentration of KOH is (0.015 L)(1.0 M) ÷ 0.515 L = 0.029 M. Since KOH dissociates completely, the added [OH−] is equal to the [KOH]. (The existing hydroxide ion concentration in the water itself, 1.0 × 10−7 M, can be ignored because it is much smaller than the hydroxide ion concentration contributed by the KOH.)

COOH

CH2

H N+

COOH

H

Citric acid

2− 2− 2− 3− 3− 2− 51.  a. ​​C​ 2​​​​​O2− 4​  ​  ​​  b. ​​SO​ 3​  ​​  c. ​​HPO​ 4​  ​​  d. ​​CO​ 3​  ​​  e. ​​AsO​ 4​  ​​  f. ​​PO​ 4​  ​​  g. ​​O​ 2​  ​​

53. a.  The final concentration of HNO3 is (0.020 L)(1.0 M) ÷ 0.520 L = 0.038 M. Since HNO3 is a strong acid and dissociates completely, the added [H+] is equal to [HNO3]. (The existing hydrogen ion concentration in the water itself, 1.0 × 10−7 M, can be ignored because it is much smaller than the hydrogen ion concentration contributed by the nitric acid.)

C

HO

COOH

Piperidine

O +

H2N H

CH

C

OH

HO

O

O

C

C

CH2

O

CH2 CH2

NH

CH2

O

H2N+ H O O

N H

O

Barbituric acid

Lysine NH+

OH

Oxalic acid

CH2

CH2

S

O–

O

4-Morphine ethanesulfonic acid (MES)

OD D -NUMB ER ED SO LUTI O NS  S-5 59.  At pH 7.4, imidazole is protonated (there is more than one way to draw the structure). +

[​A​ −​] ​​ _____   ​    = 1.7 ​4 or [A​ −​] = 1.74 [HA]​ [HA] [A−] = 1.74[HA] = 1.74(0.05 M − [A−])



HN NH

61.   First, calculate the final concentrations of boric acid (HA) and borate (A−). Note that sodium is a spectator ion. The final volume of the solution is 500 mL + 10 mL + 20 mL = 530 mL = 0.53 L. (​ 0.010 L)​(0.050 M)​ ​ ​[HA]​= ________________    ​      ​ = 9.4 × ​10​ −4​M​ (​ 0.53 L)​ (​ 0.020 L)(​ 0.020 M)​ [​ ​A​ −​]​= ________________    ​     ​  ​ = 7.5 × ​10​ −4​M​ (​ 0.53 L)​

Next, substitute these values into the Henderson–Hasselbalch equation using the pK for boric acid given in Table 2.4 and as shown in Sample Calculation 2.2: [​A​ −​]  ​​  ​ pH = pK + log ​ _____  [HA] 7.5 × ​10​ −4 ​​​  ​= 9.24 + log ​ _________ 9.4 × ​10​ −4​ = 9.24 − 0.10 = 9.14 63.  First, calculate the final concentrations of ​H2​PO​4 −​ ​ (HA) and​ − + HPO​2− 4  ​  ​ (A ). Note that K is a spectator ion. (​ 0.025 L)(​ 2.0 M)​ [​ HA]​= ______________    ​   ​  = 0.25 M​ ​ (​ 0.200 L)​ (​ 0.050 L)(​ 2.0 M)​ [​ ​A​ −]​ ​= ______________    ​   ​  = 0.50 M​ ​ (​ 0.200 L)​

Next, substitute these values into the Henderson–Hasselbalch equation using the pK value for dihydrogen phosphate given in Table 2.4 and as shown in Sample Calculation 2.2: ​[A​ −​]  ​​  pH = pK + log ​​ _____  [HA] 0.50 ​​  ​ pH = 6.82 + log ​ ____ 0.25





pH = 6.82 + 0.30



pH = 7.12

65.  Since the acid is only 1% dissociated, [A−]/[HA] = 0.01. Use the Henderson–Hasselbalch equation and the pK value for formic acid given in Table 2.4 to calculate the pH as shown in Sample ­Calculation 2.2:

​[A​ −​] pH = pK + log ​​ _____   ​​  [HA]



= 3.75 + log (0.01)



= 3.75 + (−2)



= 1.75

[A−] = 0.087 − 1.74[A−] 2.74[A−] = 0.087 M [A−] = 0.032 M or 32 mM 69. 

CH2 CH2

CH2

COOH

CH2

pH 2

COOH

CH2

COO–

COO–

CH2

COO–

pH 5

pH 7

One of the carboxylic acid groups has a pK of 4.21; the second has a pK of 5.64. Compare the pK values for each group with the pH (as demonstrated in Sample Calculation 2.4). At pH 2, both groups are protonated since the pH is below both pK values. At pH 5, the pH is above pK 1 and below pK 2, so the group with the lower pK is ionized and the group with the higher pK is protonated. At pH 7, the pH is greater than both pK values, so both groups are ionized. 71.  The pK of the fluorinated compound would be lower (it is 9.0); that is, the compound becomes less basic and more acidic. This occurs because the F atom, which is highly electronegative, pulls on the nitrogen’s electrons, loosening its hold on the proton. 73. a.  10 mM glycinamide buffer, because its pK is closer to the desired pH.  b.  20 mM Tris buffer, because the higher the concentration of the buffering species, the more acid or base it can neutralize.  c.  Neither. Both a weak acid and a conjugate base are required buffer constituents. Neither the weak acid alone (boric acid) nor the conjugate base alone (sodium borate) can serve as an effective buffer. 75. a.  The three ionizable protons of phosphoric acid have pK values of 2.15, 6.82, and 12.38 (Table 2.4). The pK values are the midpoints of the titration curve: pK3 Midpoint three [HPO42– ] = [PO43– ]

14 12

PO43–

pK2 Midpoint two [H2PO4– ] = [HPO42– ]

10 8 pH 6 4

pK1 Midpoint one [H3PO4] = [H2PO4– ]

2

[From Dussourd, D.E. et al., PLoS ONE 14, e0218994 (2019).]

0

67.  Use the Henderson–Hasselbalch equation to determine the ratio of acetate (A −) to acetic acid (HA) at pH 5.0, using the pK value for acetic acid given in Table 2.4 and as shown in Sample ­Calculation 2.3:

H3PO4

[​A​ −​] ​ pH = pK + log ​ _____   ​​  [HA] [​A​ −​] ​ log ​ _____   ​    = pH − pK = 5.0 − 4.76 = 0.24​ [HA]

COOH

HPO42–

H2PO–4 0.5

1.0 1.5 2.0 H+ ions dissociated

2.5

3.0

b.  ​​H​ 3​​​​​PO​ 4​​​ ⇌ ​ ​​​H​​  +​​ + ​​H2​  ​​​​​PO​ − 4​  ​​ ​ ​​​H​​  +​​ + ​​HPO​ 2− ​​H​ 2​​​​​PO​ − 4​  ​​ ⇌ 4​  ​​ ​ ​​​H​​  +​​ + ​​PO​ 3− ​​HPO​ 2− 4​  ​​ ⇌ 4​  ​​ c.  The dissociation of the second proton has a pK of 6.82, which is closest to the pH of blood. Therefore, the weak acid present in blood is H2​PO​ 4−​  ​​ and the weak base is ​​HPO​ 42− ​  ​​.  d.  The dissociation of the

S-6  ODD- N UM B ER ED S O LUT I O NS third proton has a pK of 12.38. Therefore, a buffer solution at pH 11 would consist of the weak acid ​HP​O4​ 2− ​  ​and its conjugate base ​PO​4 3− ​  ​ (supplied as the sodium salts Na2HPO4 and Na3PO4).

A− is (1.0 L)(0.10 mol · L−1) = 0.10 moles. After the HCl is added, the amount of A− will be 0.10 moles − x and the amount of HA will be x. Consequently,

77.  Use the pK value in Table 2.4 and the Henderson–Hasselbalch equation to calculate the ratio of the concentrations of imidazole (A−) to the imidazolium ion (HA):

​[A​ −​] ​​ _____      ​​ = 2.82 = ____________ ​​  0.10 mole − x ​  x ​ [HA]

[​ ​A​ −]​ ​ ​ pH = pK + log ​ _____   ​​  [​ HA]​ ​[​A​ −]​ ​  ​ = pH − pK​ ​ log ​ _____  [​ HA]​ [​ ​A​ −]​ ​ ​​ _____   ​ = ​10​ ​(​pH −pK​)​ [​ HA]​

​[​A​ −]​ ​ ​  2.5 ​ ​  ​ = ​10​ ​(​7.4−7.0​)​= ___ ​​ _____  [​ HA]​ 1

2.82  x = 0.10 mol − x 3.82  x = 0.10 mol

The amount of 6.0 M HCl to add is 0.0262 mol ÷ 6.0 mol · L−1 = 0.0044 L = 4.4 mL. To make the buffer, dissolve 26 g of HEPES salt (see part c) in less than 1.0 L. Add 4.4 mL of 6.0 M HCl, then add water to bring the final volume to 1.0 L. 85. a.

[​ ​A​​  −​]​ ​pH = pK + log ​ _____   ​​  ​[HA]​

N

(CH2)2

(H2C)2

N

N

CH2OH

Conjugate base (A−)

Since [​​A​​  −​​] + [HA] = 0.10 M, [​​A​​  −​​] = 0.10 M − [HA], (0.10 M − [HA]) and ______________ ​​      ​​  = 0.79 [HA]

0.79 [HA] = 0.10 M − [HA]



1.79 [HA] = 0.10 M [HA] = ______ ​​  0.10 M  = 0.056 M = 56 mM  ​​  1.79



SO–3

[​​A​​  −​​] + [HA] = 0.10 M = 100 mM, so [​​A​​ −​​] = 44 mM

d.  When HCl is added, an equivalent amount of Tris base (A−) is converted to Tris acid (HA). Let x = moles of H+ added = (0.0015 L) (3.0 mol · L−1) = 0.0045 moles = 4.5 mmol. The final amount of A− is 44 mmol − 4.5 mmol = 39.5 mmol. The final amount of HA is 56 mmol + 4.5 mmol = 60.5 mmol. Use the Henderson–Hasselbalch equation to calculate the new pH:

Weak acid (HA) HO

NH2 + H+

−

The volume of glacial acetic acid needed is 0.057 mol ÷ 17.4 mol · L−1 = 0.0033 L = 3.3 mL. The addition of 3.3 mL to a 500-mL solution dilutes the solution by less than 1%, which does not introduce significant error. HN

Weak acid (HA)

C

[HA]

[HA] = 0.057 moles

(H2C)2

CH2OH

CH2OH HOH2C

[​A​​  ​] _____ ​​     ​​ = ​​10​​  (pH−pK)​​ = ​​10​​  (8.2−8.3)​​ = 10−0.1 = 0.79

[​ A​ −​] ____ ​​ _____   ​​ = ​​  1.74  ​​   1 [HA]    ​​  [HA] = _________ ​​  0.10 moles 1.74

HO

NH+ 3

c.  Rearranging the Henderson–Hasselbalch equation gives

81.  The ratio of acetate (A−) to acetic acid (HA) at pH 5.0 is 1.74 (see Solution 67). The moles of acetate (A−) already present is determined: (0.50 L)(0.20 mol · L−1) = 0.10 moles acetate. Next, calculate the moles of acetic acid needed, based on the calculated ratio as shown in Sample Calculation 2.5:

+

C

b.  The pK of Tris is 8.30; therefore, its effective buffering range is 7.30–9.30.

​[​A​​  −]​ ​  ​ = pH − pK​ ​log ​ _____  [​ HA]​ [​ ​A​​  −]​ ​  ​ = ​10​​  ​(​​pH−pK​)​​ ​​ _____  [​ HA]​ ​[​A​​  −]​ ​ ​  0.79 ​​ _____   ​ = ​10​​  ​(​​10.5−10.6​)​​​ = ____  ​  ​ [​ HA]​ 1

83. a.

CH2OH HOH2C

79.  Use the pK value in Table 2.4 and the Henderson–­Hasselbalch equation to calculate the ratio of methylammonia (A − ) to the ­methylammonium ion (HA) as shown in Sample Calculation 2.5:



x = 0.10 mol ÷ 3.82 = 0.0262 mol

(CH2)2

SO–3 + H+

Conjugate base (A) b.  The pK for HEPES is 7.55 (see Table 2.4); therefore, its effective buffering range is 6.55–8.55. 260.3 g  × ______  = 26 g​ c.  ​1.0 L × _________ ​  0.10 mole  ​  ​   ​  L mol Weigh 26 g of the HEPES salt and add to a beaker. Dissolve in slightly less than 1.0 liter of water (leave “room” for the HCl solution that will be added in the next step). d.  At the final pH, −

​[A​  ​] _____ ​​     ​​ = 10(pH − pK) = 10(8.0 − 7.55) = 100.45 = 2.82 [HA]

For each mole of HCl added, x, one mole of HEPES salt (A−) will be converted to a mole of HEPES acid (HA). The starting amount of



[​A​​  −​] pH = pK + log ​​ _____   ​​  [HA]



39.5 mmol ÷ 1001.5 mL ​​ pH = 8.3 + log ​​ ____________________    60.5 mmol ÷ 1001.5 mL



pH = 8.3 + (−0.2)



pH = 8.1

The buffer is effective. The pH decreases about 0.1 unit (from 8.2 to 8.1) with the addition of the strong acid. In comparison, the addition of the same amount of acid to water, which is not buffered, results in a pH change from approximately 7.0 to 2.3 (see Solution 54a). e.  When NaOH is added, an equivalent amount of Tris acid (HA) is converted to Tris base (A−). Let x = moles of OH− added = (0.0015 L) (3.0 mol · L−1) = 0.0045 moles = 4.5 mmol. The final amount of A− is 44 mmol + 4.5 mmol = 48.5 mmol. The final amount of HA is 56 mmol − 4.5 mmol = 51.5 mmol.

OD D -NUMB ER ED SO LUTI O NS  S-7 Use the Henderson–Hasselbalch equation to calculate the new pH:

[​A​​  −​]  ​​  pH = pK + log ​​ _____  [HA] 48.5 mmol ÷ 1001.5 mL ​​    pH = 8.3 + log ​​ ____________________ 51.5 mmol ÷ 1001.5 mL



pH = 8.3 + (−0.026)



pH = 8.27

The buffer is effective. The pH increases only 0.07 units (from 8.2 to 8.27) with the addition of the strong base. In comparison, the addition of the same amount of base to water, which is not buffered, results in a pH change from approximately 7.0 to 11.6 (see Solution 54b). 87. a.  The amount of imidazole required is 1.0 L × 0.10 mol · L−1 × 68.1 g · mol−1 = 6.81 g. Weigh 6.81 g of imidazole and add to a beaker. Dissolve in slightly less than 1.0 liter of water (leave “room” for the HCl solution that will be added in the next step).  b.  At the final pH, the ratio of imidazole (A−) to the imidazolium ion (HA) is 2.5:1 (see Solution 77). For each mole of HCl added, x, one mole of imidazole (A−) will be converted to a mole of imidazolium (HA). The starting amount of A− is (1.0 L)(0.10 mol · L−1) = 0.10 moles. After the HCl is added, the amount of A− will be 0.10 moles − x and the amount of HA will be x. Consequently, [​ ​​ ​A​​  −]​​ ​​ ____ − x  ​​ _____  ​   ​ = ​  2.5 ​  = ___________ ​  0.10 mol x ​ 1 [​ ​​HA​]​​ ​2.5 x = 0.10 mol − x​

​3.5 x = 0.10 mol​ ​ x = ________ ​ 0.10 mol    ​  = 0.029 mol​ 3.5 The amount of 6.0 M HCl to add is 0.029 mol ÷ 6.0 mol · L−1 = 0.0048 L = 4.8 mL. To make the buffer, dissolve 6.81 g of imidazole (see part a) in less than 1.0 L. Add 4.8 mL of 6.0 M HCl, then add water to bring the final volume to 1.0 L. 89. a. 

​​H2​  ​​​​​CO​ 3​​​ ⇌ ​ ​​​H​​  ​​ + +

​​ CO​ − H 3​  ​​

​ ​​​H​​  +​​ + ​​CO​ 2− ​​HCO​ − 3​  ​​ ⇌ 3​  ​​  .  The pK of the first dissociation is closer to the pH; therefore the weak b acid present in blood is H2CO3 and the conjugate base is ​​HCO​ − 3​  ​​. c.

−

​[​​HC​O3​  ​  ​​]​​   ​ H = pK + log ​ _______ p  ​​ 

​[​​HC​O− 3​  ​  ​​]​​    ​  = pH − pK​ ​log ​ _______ ​[​​​H2​  ​​C​O3​  ​​​]​​

​[​​​H2​  ​​C​O3​  ​​​]​​

[​ ​​HC​O−3​  ​  ​​]​​    ​  = 7.4 − 6.1 = 1.3​ ​log ​ _______ ​[​​​H2​  ​​C​O3​  ​​​]​​

95.  Ammonia and ammonium ions are in equilibrium: N ​ ​H4+​  ​ ​ ⇌ H+ + NH3. Carbonic acid and bicarbonate ions are in equilibrium: H2CO3 ⇌ H+ + ​HC​O​3 −​ ​. Phosphate ions are in equilibrium: H2​PO​ 4−​  ​​ ⇌ H+ + ​HP​O42− ​  ​  ​. In metabolic acidosis, the concentration of protons increases, so the equilibrium shifts to form ​H2​  ​P​O4​ −​ ​, carbonic acid, and ammonium ions. In order to bring the pH back to normal, the kidney excretes H 2​PO​ 4−​  ​​ and ammonium ions and reabsorbs bicarbonate ions. The result is a decrease in the concentration of protons and an increase in blood pH. 97.  The concentrations of both Na+ and Cl− are greater outside the cell than inside (see Fig. 2.12). Therefore, the movement of these ions into the cell is thermodynamically favorable. Na+ movement into the cell drives the exit of H+ via an exchange ­protein in the plasma membrane (the favorable movement of Na+ into the cell “pays for” the unfavorable movement of H+ out of the cell). Similarly, the movement of Cl− into the cell drives the m ­ ovement of ​​HCO​ − 3​  ​​out of the cell through another exchange protein. 99.  The cell-surface carbonic anhydrase catalyzes the conversion of H+ + ​​HCO​ 3−​  ​​ to CO2, which can then diffuse into the cell (the ionic H+ and ​​HCO​ 3−​  ​​cannot cross the hydrophobic lipid bilayer on their own). Inside the cell, carbonic anhydrase converts the CO2 back to H+ + ​​HCO3−​  ​  ​​.

Chapter 3 1.  The heat treatment destroys the polysaccharide capsule of the wild-type Pneumococcus, but the DNA survives the heat treatment. The DNA then “invades” the mutant Pneumococcus and supplies the genes encoding the enzymes needed for capsule synthesis that the mutant lacks. The mutant is now able to synthesize a capsule and has the capacity to cause disease, which results in the death of the mice and the appearance of encapsulated Pneumococcus in the mouse tissue. 3.  Some of the labeled “parent” DNA appears in the progeny, but none of the labeled protein appears in the progeny. This indicates that the bacteriophage DNA is involved in the production of progeny bacteriophages, but bacteriophage protein is not required. 5. a.  purine;  b. purine;  c.  pyrimidine;  d. pyrimidine. 7.  Thymine (5-methyluracil) contains a methyl group attached to C5 of the pyrimidine ring of uracil. 9.  The base, 5-chlorouracil, is a substitute for thymine (5methyluracil). 11.  [From Jordheim, L.P. et al., Natl. Rev. Drug Discov. 12, 447–464 (2013).] NH2

[​ ​​HC​O−3​  ​  ​​]​​   ​  20 ​​   ​  = ​10​​  1.3​ = ___ ​​ _______ 1 ​[​​​H2​  ​​C​O3​  ​​​]​​ −3 __________      ​ = ___ ​  20 ​ ​ ​ ​  24 × ​10​​  ​M 1 ​[​​​H2​  ​C​O3​  ​​​]​

93.  During hyperventilation, too much CO2 (which is equivalent to H+ in the form of carbonic acid) is given off, resulting in respiratory alkalosis. By repeatedly inhaling the expired air, the individual can recover some of this CO2 and restore acid–base balance.

Cl N

N CH2OH O

​ [​ ​H2​  ​C​O3​  ]​ ​= 1.2 × ​10​​ −3​= 1.2 mM​

91.  Mechanical hyperventilation removes CO2 from the patient’s lungs. Carbonic acid in the blood produces more water and CO2 to make up for the loss of CO2. This in turn causes additional hydrogen ions and bicarbonate ions to form more carbonic acid. The loss of hydrogen ions results in an increased pH, bringing the patient’s pH back to normal.

N

N

H

H

H

H

OH OH

8-Chloroadenosine 13.

H3C O

O N N H

S-8  ODD- N UM B ER ED S O LUT I O NS 15.  [From Ryu, H. et al., Nucleic Acids Res. 46, 9160–9169 (2018).] O OH

HN HO

O CH2 O H

H

it contains uracil and not thymine. The number of A + G residues = 14,755 and the number of C + U residues = 15,056. Since A + G ≠ C + T, Chargaff’s rules do not apply and the genome consists of singlestranded RNA. [From Sah, R. et al., Microbiol. Resour. Announc. 9:e00169-20 (2020).] 27. 

N

N H

H

H

N

OH OH

5-Hydroxyuridine 17.  If the dinucleotide were DNA, it would lack OH groups at each ribose C2ʹ position.

H

O

N

H

NH N

N

N

Hypoxanthine

H

O

N H

N

Cytosine

P

O

CH2 O H

H

H

−O

P

H

N

H

N H

N N

H

OH OH

19.

O O −O

N

P

N N

N

A

H O

H

H

5′CH2

O

N NH2

3′

H

CH2

5′

OH O H

G

O

H

N

O H 2′

N

NH NH2

H

OH O P

N

N

Hypoxanthine

N

O

O

H

NH N

N

Hypoxanthine

29.  The statement is false because the greater stability of GC-rich DNA is due to the stronger stacking interactions involving G:C base pairs and does not depend on the number of hydrogen bonds in the base pairs.

NH2

CH2 O

H

Uracil

O

O

H

N H

OH

O

Phosphodiester bond

N

O

N

H

N

O

N

O

NH

Adenine

NH2

O−

O

21.  The genome must also contain 19% A (since [A] = [T] according to Chargaff’s rules) and 62% C + G (or 31% C and 31% G, since [C] = [G]). Each cell is a diploid, containing 60,000 kb or 6 × 107 bases. Therefore, [A] = [T] = (0.19)(6 × 107 bases) = 1.14 × 107 bases [C] = [G] = (0.31)(6 × 107 bases) = 1.86 × 107 bases 23. a.  Using Chargaff’s rules (see Solution 21), the number of C residues must also be 24,182. Subtracting (2 × 24,182) from 97,004 yields 48,640 (A + T) residues. Dividing this number by 2 yields 24,320 residues each of A and T.  b.  GenBank reports only the sequence of a single strand of DNA, since the sequence of the complementary strand can easily be deduced, as A pairs with T and G pairs with C. 25.  The SARS-CoV-2 genome is 29.9% A (8903 ÷ 29811 = 0.299), 18.4% C (5482 ÷ 29811 = 0.184), 19.6% G (5852 ÷ 29811 = 0.196), and 32.1% U (9574 ÷ 29811 = 0.321). The genome is RNA, not DNA, since

31.  The sugar–phosphate backbone is on the outside of the molecule. The polar sugar groups can form hydrogen bonds with the surrounding water molecules. The negatively charged phosphate groups interact favorably with positively charged ions. The nonpolar nitrogenous bases are found on the inside of the molecule and interact favorably via stacking interactions. In this way, contact with the aqueous solution is minimized, as described by the hydrophobic effect. 33. a.  The Tm is approximately 72°C. b.  1.4 Relative absorbance at 260 nm

−O

H

N

N H O−

N

O

1.3 D. discoideum

S. albus

1.2 Tm

1.1

1.0

30

50

70

Tm

90

Temperature (°C)

35.  The DNA from the organisms that thrive in hot environments would contain more G and C than DNA from species living in a more temperate environment. The higher GC content increases the stability of DNA at high temperatures. 37. a.  You should increase the temperature to melt out imperfect matches between the probe and the DNA.  b.  You should decrease the temperature to increase the chances that the two strands will align, despite the mismatch.

OD D -NUMB ER ED SO LUTI O NS  S-9 39. a.  An inherited characteristic could be determined by more than one gene.  b.  Some sequences of DNA encode RNA molecules that are not translated into protein (for example, rRNA and tRNA). c.  Some genes are not transcribed during a cell’s lifetime. This can occur if the gene is expressed only under certain environmental conditions or in certain specialized cells in a multicellular organism.

again will pair with T in the new strand, and the existing T will pair with A. As a result, two cells will have DNA with the normal G:C base pair, one cell will have DNA with a G:T base pair, and one cell will have DNA with an A:T pair.

41. a.  The top strand is the coding strand and the bottom strand is the noncoding strand.  b.  Only coding strands are published because the mRNA sequence is identical to the sequence of the coding strand, with the exception that U replaces T in the mRNA. 43. a.  A poly-Phe polypeptide was produced.  b.  Poly A produces poly-Lys; poly C yields poly-Pro; and poly G yields poly-Gly. 45.  The number of possible sequences of four different nucleotides taken n at a time is 4n; here n is the number of nucleotides in the sequence.  a.  41 = 4,  b. 42 = 16,  c. 43 = 64,  d. 44 = 256. At least three nucleotides are necessary to code for 20 amino acids.

G T DNA with G:T mismatch G C Original DNA G C Normal DNA

47. a.  First reading frame: AGG TCT TCA GGG AAT GCC TGG CGA GAG GGG AGC AGC -Arg - Ser - Ser - Gly - Asn - Ala - Trp - Arg - Glu - Gly - Ser - SerSecond reading frame: A GGT CTT CAG GGA ATG CCT GGC GAG AGG GGA GCA GC - Gly - Leu - Gln - Gly - Met - Pro - Gly - Glu - Arg - Gly - Ala - AlaThird reading frame: AG GTC TTC AGG GAA TGC CTG GCG AGA GGG GAG CAG C - Val - Phe - Arg - Glu - Cys - Leu - Ala - Arg - Gly - Glu - Glnb.  The second reading frame, which produces a protein in which every third amino acid is Gly, is the correct reading frame. 49. a.  The first reading frame has the sequence -Phe-Gln-Stop-LeuSer-Ser-Leu-Ser-Arg-Glu-. The second reading frame has the sequence -Ser-Asn-Asp-Stop-Val-Leu-Ser-Leu-Glu-. The third reading frame has the sequence -Phe-Met-Thr-Glu-Phe-Ser-Leu-Stop-Arg-.  b.  The reverse (5ʹ → 3ʹ) complement of the strand is CTCTCTAGAGAGAGAACTCAGTCATTGGAA. Using this strand, the first reading frame produces -Leu-Ser-Arg-Glu-Arg-Thr-Gln-Ser-Leu-Glu-, the second reading frame produces -Ser-Leu-Glu-Arg-Glu-Leu-Ser-His-Trp-, and the third reading frame produces -Leu-Stop-Arg-Glu-Asn-SerVal-Ile-Gly-.  c.  Either the first or the second reading frame on the reverse complement strand is potentially correct, as the other translation products contain stop codons. 51. a.  Asparagine has two codons, AAU and AAC (see Table 3.3). An A → G mutation at the second position could generate a codon for serine (AGU or AGC).  b.  The CGA codon codes for the amino acid arginine; the mutation (C → T in the DNA) converts the codon to a stop codon. When the mRNA for the gene is translated, protein synthesis terminates prematurely and the truncated protein is nonfunctional.

53. a.  The normal protein sequence is ···Glu-Asn-Ile-Ile-Phe-GlyVal-Ser-Tyr···. The mutant protein sequence is the same except the Phe at position 508 is deleted. Note that although the deletion of Phe affects codons 507 and 508, the redundancy of the genetic code means that the Ile at position 507 is not affected, and the amino acids downstream of the mutation are also unaffected.  b.  The deletion of Phe (the one letter code for Phe is F) explains why this form of the disease is referred to as ΔF508 (Δ is delta, which means “deletion”). 55.  One daughter cell will contain DNA with a G:T mismatch and the other daughter cell will have DNA with the normal G:C base pair. When the abnormal DNA is replicated, the existing tautomerized G

G T DNA with G:T mismatch A T Altered DNA G C Normal DNA G C Normal DNA

57.  C. ruddii, with such a small genome and only 182 genes, must be some sort of parasite rather than a free-living bacterium. (In fact, C. ruddii is an insect symbiont.) 59.  The 35 million differences out of 3.0 billion total nucleotides represent approximately 1%, or a bit less than the original claim. (This number reflects single-base differences and does not account for insertions and deletions of multiple bases.) 61.  The coding sequence begins with the start codon ATG (AUG in mRNA), which corresponds to Met (Table 3.3). A CCA CCC CAA CTG AAA ATG TCT CTC TCT GAT GC Met - Ser - Leu - Ser - Asp63. a.  The first reading frame is the longest ORF. First reading frame: TAT GGG ATG GCT GAG TAC AGC ACG TTG AAT GAG -Ty r - G l y - M e t - A l a- G l u - Ty r - S e r - T h r -L e u-A s n-G l u GCG ATG GCC GCT GGT GAT G -Ala - Met - Ala - Ala - Gly - AspSecond reading frame: T ATG GGA TGG CTG AGT ACA GCA CGT TGA ATG AGG -Met-G ly - Trp -Leu- S er - T h r - Ala -Ar g-Stop-M e t -A r gCGA TGG CCG CTG GTG ATG -Arg -Trp - P ro - Leu-Va l-Met Third reading frame: TA TGG GAT GGC TGA GTA CAG CAC GTT GAA TGA GGC -Trp - Asp -Gly-Stop-Val - Gln- His -Val - Glu-Stop-GlyGAT GGC CGC TGG TGA TG -Asp -Gly -Arg -Trp - Stop b.  Assuming the reading frame has been correctly identified, the most likely start site is the first Met residue in the first ORF. 65.  If an SNP occurs every 300 nucleotides or so and if there are about 3 million kb (3 × 109 bp) in the genome (see Table 3.4), then there are (3 × 109 bp ÷ 300 bp/SNP) = 1 × 107 (10 million) total SNPs in the human genome. [Source: ghr.nlm.nih.gov/handbook/ genomicresearch/snp.] 67. a.  The strongest associations are located between positions 67,370,000 and 67,470,000.  b.  Gene B contains SNPs associated with the disease whereas Genes A and C do not. [From Duerr, R.H. et al., Science 314, 1461–1463 (2006).]

S-10  ODD- N UM B ER E D S O LUT I O NS

Chapter 4

OH

1.

α-amino group

COO–

+H

3N

C

H

Chiral carbon

CH2 CH2

NH

ε-amino group

CH2

O

N

H2C

CH2

CH

C

N

Ser

L-Lysine

COO– +H N 3

C

H

H

C

OH

O +H

H

C

NH+ 3

HO

C

H

H

C

NH+ 3

CH2

HO

C

H

H

C

OH

CH2

CH3

C HN

5.  Leucine, isoleucine and valine have “branched chain” R groups.

CH2

CH2

C

β-Ala Mg2+

COO– H

C

CH2

Glutathione (GSH)

O

O

CH

C

CH

HN

C

O–

H

19.  +H

3N

CH CH2

O C

H N

O CH

C

H N

CH2

COO– CH2

O

SH O–

CH3

α-Ala

O–

CH2

O

C

OCH3

CH2

C

CH

H

CH2

C

O +H N 3

COO

NH+ 3

CH

Aspartame (Asp–Phe–OMe)

C

O–

O

O–

17.

O

C

Gly

O

3N

7.

C

C

–

CH3

9. a. 

CH

H N

CH2

CH3

COO–

NH+ 3

3N

COO–

CH3

+H

15.

O

CH2

b Tyr

NH+ 3

3.  Threonine has two chiral carbons; therefore, four stereoisomers are possible.

c

CH

OH

COO–

NH+ 3

b.  In the mouth cavity, the glutamate and the Mg2+ ion separate and the amino group remains protonated (since the pH of saliva, ~7, is less than the pK value of the amino group, ~9.0). Therefore, magnesium glutamate and monosodium glutamate yield the same form of glutamate.

O CH

C

H N

O CH

CH2

CH2

CH2

COO–

CH2 OH

C

H N

COO– CH CH2

4

CH2 CH2

CH2

CH2

NH+ 3

NH+ 3

21. a.  At pH 6.0, groups with pK values less than 6.0 are mostly deprotonated, and groups with pK values greater than 6.0 are mostly protonated. The dipeptide has a net charge of –1.

11.  Histones contain an abundance of the positively charged amino acids lysine and arginine. The positive charges of these amino acid side chains form ion pairs with the negatively charged phosphate groups on the backbone of the DNA molecule in order to minimize charge–charge repulsion of the negatively charged phosphate groups.

Group

13. a.  The three amino acids are Ser, Tyr, and Gly.  b.  Cyclization of the polypeptide backbone occurs between the carbonyl carbon of Ser and the amide nitrogen of Gly.  c.  Oxidation results in a double bond in the Tyr side chain between Cα and Cβ (the second carbon of the side chain).

Charge

N-terminus

+1

Glu

–1

Tyr

 0

C-terminus

–1

Net charge

–1

b.  At pH 7.0, groups with pK values less than 7.0 are mostly deprotonated and groups with pK values greater than 7.0 are mostly protonated. The tripeptide has a net charge of –3:

OD D -NUMB ER ED SO LUTI O NS  S-11 Group

37.

Charge

N-terminus

+1

3 Asp

–3

C-terminus

–1

Net charge

–3

H

H O Cα

+1

His

 0

Lys

+1

Glu

–1

C-terminus

–1

Net charge

 0

N

Cα

H O H

39. a.  The peptide group’s C—N bond would be shorter than the N—Cα single bond because the peptide group’s C—N bond has a partial double bond character.  b.  The peptide group’s CO bond would be longer than the CO bond commonly found in aldehydes and ketones because the CO bond in the peptide group has partial single bond character.

Charge

N-terminus

C H

c.  At pH 8.0, groups with pK values less than 8.0 are mostly deprotonated and groups with pK values greater than 8.0 are mostly protonated. The tripeptide has a net charge of 0.

Group

O

41.  Both the DNA helix and the α helix turn in the right-handed direction. Both helices have tightly packed interiors; in DNA the interior of the helix is occupied by nitrogenous bases and in the α helix the atoms of the polypeptide backbone contact one another. In the α helix, the side chains extend outward from the helix; no such structure exists in the DNA helix.

23.  All of the amino acid side chains in oxytocin are neutral. Therefore, the only charges associated with the peptide are the N-terminus (+1) and the C-terminus (–1), which make the net charge zero at pH 7.

43.  The amino group of Pro is linked to its side chain (see Fig. 4.2), which limits the conformational flexibility of a peptide bond involving the amino group. The geometry of this peptide bond is incompatible with the bond angles required for a polypeptide to form an α helix.

25. a.  The net charge is 0, since the Pro side chains are neutral, the Tyr side chains are protonated and neutral, and there is no free N-terminus or C-terminus in the cyclic molecule.  b.  If the molecule were linear, it would contain a free amino group (+1 charge) and a free carboxylate group (–1 charge), but its net charge would still be 0.

45.  The nonpolar amino acids are shown in red: FWGALAKGALKLIPSLFS. The hydrophobic side of the helix interacts favorably with the nonpolar membrane, disrupting its structure and killing the cell. [The investigators found that the peptide was able to kill the cell by forming a pore in the membrane, which disrupts the ionic balance. From Corzo, G. et al., Biochem. J. 359, 35–45 (2001).]

27.  A poly-aspartate molecule is more soluble at neutral pH, because the side chains are unprotonated and negatively charged. At a low pH, the side chains are protonated, which decreases their polarity and makes the molecule less soluble in water.

47.  The highlighted region, consisting of mainly nonpolar residues, is about the right length to span the membrane (3 nm thickness × 1 amino acid/0.15 nm = 20 amino acids to span the membrane).

29.

C S Y

E

G

51. a. The resonance structures for the Arg side chain are shown below.

C N

S S C

R

49.  Triose phosphate isomerase is an example of an α/β protein.

S S

C

H

HHFSEPEITLIIFGVMAGVIGTILLISYGIRRLIKKSPSD

P A

31.  In a free amino acid, the charged amino and carboxylate groups, which are separated only by the α carbon, electronically influence each other. When the amino acid forms a peptide bond, one of these groups is neutralized, thereby altering the electronic properties of the remaining group. 33.  There are six possible sequences: HPR, HRP, PHR, PRH, RHP, and RPH. 35. a. tertiary b. quaternary c. primary d. secondary.

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

CH2

NH

NH

NH+

C

NH+ 2

NH2

C

NH2

NH+ 2

C

NH2

NH2

b.  Although Arg has a polar side chain, the resonance-stabilized guanidinium group has a delocalized positive charge (whereas the Lys ε-amino group bears a full positive charge). Thus the Arg side is less polar than the Lys side chain and can be buried in the hydrophobic core of a protein more readily. [From Pace, C.N. et al., J. Biol. Chem. 284, 13285–13289 (2009).]

S-12  ODD- N UM B ER E D S O LUT I O NS 53.  It is possible that the ligand has a positive charge and forms an ion pair with the negatively charged Glu on the receptor. When the Glu is mutated to Ala, the negative charge on the receptor is lost and the ion pair between the receptor and the ligand can no longer form. 55.  There are many possible answers for this question. An example is shown for each.

a.

H N

O CH CH2 CH2

A Lys-Glu ion pair CH2 CH2 NH+ 3

C

b.

H N

CH

C

C

c.

H N

CH2

H O

CH2 HN

CH

Hydrogen bonding between Tyr and Ser O H

COO– CH2 O

O

CH2 O HN

CH

C

O CH

C

CH

CH3

CH2 CH3 CH3 H3C

CH

O

HN

CH

C

van der Waals forces between Ile and Val

57.  A polypeptide synthesized in a living cell has a sequence that has been optimized by natural selection so that it folds properly (with hydrophobic residues on the inside and polar residues on the outside). The random sequence of the synthetic peptide cannot direct a coherent folding process, so hydrophobic side chains on different molecules aggregate, causing the polypeptide to precipitate from solution. 59.  Anfinsen’s ribonuclease experiment demonstrated that a protein’s primary structure dictates its three-dimensional structure. Although some proteins, like ribonuclease, can renature spontaneously in vitro, most proteins require the assistance of molecular chaperones to fold properly in vivo. 61.  When the temperature increases, the vibrational and rotational energy of the atoms making up the protein molecules also increases, which increases the chance that the proteins will denature. Increasing the synthesis of chaperones under these conditions allows the cell to renature, or refold, proteins denatured by heat. 63.  Proline does not fit well into the structure of the α helix, both because of its geometry (see Solution 43) and the absence of a peptide —NH group to contribute to hydrogen bonding. This amino acid substitution produces a protein with decreased stability, which would affect the ability of red blood cells to deform in order to squeeze through capillaries. The cells would become damaged and would be removed from circulation, causing anemia. [From Johnson, C.P. et al., Blood 109, 3538–3543 (2007).] 65.  The chaperones help maintain the β/γ-crystallins in the proper conformation, which keeps the lens transparent. Without chaperones, misfolded crystallins would tend to precipitate or form fibers that would block light transmission. 67.  Serine, glutamine, and asparagine have side chains that are relatively small, not charged, and not hydrophobic. Glycine has no side chain at all. These residues can fit into tight spaces because they are small and do not electrostatically repel each other. They are polar enough to be located on solvent-exposed polypeptide segments without irreversibly aggregating as a result of the hydrophobic effect. 69.  If the proteins were homodimers, they would be more likely to have two identical sites to interact with their palindromic recognition

sites in the DNA. Heterodimeric proteins would likely lack the necessary symmetry. (In fact, these enzymes are homodimeric.) 71.  The positively charged Arg residues and the negatively charged Asp residues are likely to be found on the surface of the monomer. These residues are likely to form ion pairs that stabilize the dimeric form. When these residues were mutated to residues with neutral side chains, the ion pairs could not form between the dimers and the equilibrium shifted in favor of the monomers. [From Huang, Y. et al., J. Biol. Chem. 283, 32800–32888 (2008).] 73. a.  The experimental results indicate that an ion pair formed by His 142 and Asp 568 is essential for tetramer formation.  b. The experimental results indicate the importance of hydrophobic interactions in the stabilization of tetramer structure. [From Hsieh, J.Y. et al., J. Biol. Chem. 284, 18096–18105 (2009).] 75.  The extra copy of chromosome 21 increases the amount of the precursor protein and therefore contributes to a higher concentration of the amyloid-β fragment. As a result, amyloid fibers begin forming in the brains of these individuals at a younger age. 77. a.  The extensive α-helical secondary structure in myoglobin makes it unlikely to easily adopt the all-β conformation necessary for amyloid formation.  b.  This result suggests that any polypeptide— even one whose native conformation is all α-helical—can assume the β conformation if conditions permit. 79.  The mutation disrupts the secondary structure of the protein, which in turn alters the tertiary structure of each monomer, disrupting the hydrophobic contacts and hydrogen bonds that hold the tetramer together. [From Yang, M. et al., Protein Sci. 12, 1222–1231 (2003).] 81. a.  In order for alanine to have no net charge, its α-carboxylate group (pK ≈ 3.5) must be unprotonated (negatively charged) and its α-amino group (pK ≈ 9.0) must be protonated (positively charged): pI = ½ (3.5 + 9.0) = 6.3. b. In order for glutamate to have no net charge, its α-carboxylate group must be unprotonated (negatively charged), its side chain must be protonated (neutral), and its α-amino group must be protonated (positively charged). Because protonation of the α-carboxylate group or deprotonation of the side chain would change the amino acid’s net charge, the pK values of these groups (~3.5 and ~4.1) should be used to calculate the pI: pI = ½ (3.5 + 4.1) = 3.8. c. In order for lysine to have no net charge, its α-carboxylate group must be unprotonated (negatively charged), its α-amino group must be unprotonated (neutral), and its side chain must be protonated (positively charged). Because protonation of the α-amino group or deprotonation of the side chain would change the amino acid’s net charge, the pK values of these groups (~9.0 and ~10.5) should be used to calculate the pI: pI = ½ (9.0 + 10.5) = 9.8. 83. a.  The protein must contain groups that undergo protonation/ deprotonation at pH values near 4.3. The only amino acids with side chain pK values in this range are Asp and Glu (Table 4.1), so the protein is likely to contain an abundance of these residues.  b.  The protein must contain groups that undergo protonation/deprotonation at pH values near 11.0. The only amino acids with side chain pK values in this range are Lys and Arg (Table 4.1), so the protein is likely to contain an abundance of these residues. 85.  At pH 7.0, the peptide is likely to have a net positive charge since Arg (R) and Lys (K) outnumber Asp (D) and Glu (E). Therefore, the peptide is likely to bind to CM groups but not to DEAE groups. 87.  The SDS-PAGE gel is shown below. Lane 1 shows a ladder of proteins with known molecular weights. Lane 2 shows TGF-β in the presence of 2-mercaptoethanol. Reduction of the disulfide bond separates the protein into its two identical subunits of 12.5 kD. Lane 3 shows TGF-β in the absence of 2-mercaptoethanol; here the intact

OD D -NUMB ER ED SO LUTI O NS  S-13 protein has a molecular mass of 25 kD. [From Assoian, R.K. et al., J. Biol. Chem. 258, 7155–7160 (1983)].

100

1.  Globin lacks an oxygen-binding group and therefore cannot bind O2. Heme alone is easily oxidized and therefore cannot bind O2. The bound heme gives a protein such as myoglobin the ability to bind O2. In turn, the protein helps prevent oxidation of the heme Fe atom.

63

3.  The mitochondrial proteins accept O2 from hemoglobin and thus must be able to bind O2. Therefore, these proteins, like myoglobin and hemoglobin, must have a heme group as part of their structures.

40

5.  Equation 5.4 can be used to calculate the fractional saturation (Y) at 20 torr and 80 torr, letting K = p50 = 2.8 torr:

e2

e1

Lan

Lan

Mass (kD)

Chapter 5

e3

Lan

p​O​ 2​ ​      a.  ​Y = ​ _________ ​p​ 50​+ p​O​ 2​

20 torr   ​   Y = ​ ______________    2.8 torr + 20 torr   Y = 0.88​ p​O​ 2​ ​ b.  ​Y = _________ ​       ​p​ 50​+ p​O​ 2​

25

16

10

89. a.  The phosphoproteins bear negatively charged phosphate groups that bind electrostatically to positively charged iron atoms.  b.  Nucleic acids also have numerous phosphate groups that can also bind to the matrix during affinity chromatography. 91.  Cation exchange chromatography could be used, with a mobile phase buffer adjusted to a pH less than the pI value of lysozyme but greater than the pI values of the other three egg white proteins. This renders lysozyme positively charged while the other proteins are negatively charged. The other proteins would be repelled by the negatively charged beads and would elute first. A buffer change to a higher pH would likely be required to elute the lysozyme. [Based on Abeyrathne, E. et al., Poult. Sci. 12, 3292– 3299 (2013).] 93.  Biotin could be coupled to a chromatography matrix and act as a ligand to purify avidin using affinity chromatography. As none of the major egg white proteins bind to biotin, they would flow through the affinity column. A change in salt concentration or pH would be required to elute the avidin, as it binds very tightly to biotin. 95.  The sequence of the fragment is IGTGLVGALTKVYSYRFVWWAISTAAM. Highlighted amino acids indicate points of cleavage by trypsin; underlined amino acids indicate points of cleavage by chymotrypsin. [Based on Khorana, H.G. et al., Proc. Natl. Acad. Sci. USA 76, 5046–5050 (1979).] 97.  Leu and Ile are isomers and have the same mass; therefore, mass spectrometry cannot distinguish them. 99.

Phe

Val H3C

+

H3N

Ala CH3

O N H

Asp

H N O

O– O O–

O N CH3 H

O

a.  Dashed lines indicate broken bonds. The smallest charged fragment is the N-terminal residue (Phe), which has a mass of approximately 149 D (9 C + 1 N + 1 O + 11 H).  b.  The next smallest fragment is the Phe–Val dipeptide. The difference in mass between the smallest and next smallest fragments is the mass of the Val residue, or approximately 99 D.

80 torr   ​      Y = ​ ______________ 2.8 torr + 80 torr   Y = 0.97​

7.  Equation 5.4 can be rearranged to solve for pO2 at Y = 0.25 (25% saturated) and Y = 0.90 (90% saturated), letting K = p50 = 2.8 torr: p​O​ 2​ a.   ​Y = _________ ​  ​      ​p​ 50​+ p​O​ 2​

​p​  ​× Y   p​O​ 2​= _______ ​  50  ​  1−Y 2.8 torr × 0.25   p​O​ 2​= ____________   ​  ​    (​ ​1 − 0.25​)​

  p​O​ 2​= 0.93 torr​

p​O​ 2​ ​      b.    ​Y = _________ ​  ​p​ 50​+ p​O​ 2​ ​p​  ​× Y   p​O​ 2​= _______ ​  50  ​  1−Y 2.8 torr × 0.90   p​O​ 2​= ____________   ​  ​    (​ ​1 − 0.90​)​   p​O​ 2​= 25 torr​

9.  Equation 5.4 can be used to calculate the fractional saturation (Y) at 1.5 torr and 3.5 torr, letting K = p50 = 2.8 torr: p​O​ 2​ ​​ ​Y = _________ ​       ​p​ 50​+ p​O​ 2​ 1.5 torr   ​​ ​=​0.35  t 1.5 torr, ​Y = _______________ A ​    2.8 torr + 1.5 torr

3.5 torr   ​ = 0.56​ At 3.5 torr, ​Y = _______________ ​    2.8 torr + 3.5 torr

 t 1.5 torr, 35% of Mb has oxygen bound, but at 3.5 torr, 56% of myoA globin has oxygen bound. Thus a small change in oxygen partial pressure can lead to a large change in the percentage of oxygenated myoglobin in the muscle cell. 11.  Ferric myoglobin must be reduced to ferrous myoglobin because only the ferrous form is able to bind O2 (the primary function of myoglobin). 13. a. The p50 values from the oxygen binding curve are 0.6, 1.0, and 1.4 torr for the tuna, bonito, and mackerel, respectively.  b.  The tuna has the lowest p50 value and the highest O2 binding affinity; the mackerel has the highest p50 value and the lowest O2 affinity. Interestingly, the p50 values are the same when adjusted for the body temperatures of the fish habitats, which average 25°C, 20°C and 13°C for the tuna, bonito, and mackerel, respectively. [From Marcinek, D.J. et al., Am. J. Physiol. Regul. Integr. Comp. Physiol. 280, R1123–R1133 (2001).]

S-14  ODD- N UM B ER E D S O LUT I O NS 15.  Both of these amino acid residues are highly conserved. In Figure 5.5, the ValE11 residue is shaded in purple, indicating that it is invariant in all vertebrate myoglobin and hemoglobin chains, while PheCD1 is identical in myoglobin and hemoglobin α and β chains. This indicates the importance of a hydrophobic heme pocket to the oxygen binding function of myoglobin (and hemoglobin). 17.  The invariant residues are shaded in blue; structurally similar residues are shaded in yellow. The single variant residue is not shaded. [From Vigna, R., Gurd, L.J., and Gurd, F.R.N., J. Biol. Chem. 249, 4144–4148 (1974).]

His F8

His E7

promotes the formation of the ion pair between the two side chains. The increased pK value means that the imidazole ring’s affinity for protons increases. [From Berenbrink, M., Respir. Physiol. Neurobiol. 154, 165–184 (2006).] 29. a. ​​ HCO​  −3​  ​​is formed when CO2 reacts with water to form carbonic acid (H2CO3), a reaction in the blood catalyzed by carbonic anhydrase. The H2CO3 dissociates to form protons and ​​HCO​  3−​  ​​ (see Fig. 2.17).  b.  Both curves are sigmoidal. According to the investigators, the p50 for crocodile hemoglobin is 6.8 torr in the absence of ​​HCO​  3−​  ​​ and 44 torr in its presence. The higher p50 value in the presence of​​ HCO​  3−​  ​​indicates that the crocodile hemoglobin has a lower affinity for oxygen.

Human SEDLKKHGATVLTALGGILKKKGH HEAEIKPLAQSHA Harbor SEDLRKHGKTVLTALGGILKKKGH HDAELKPLAQSHA seal



w/o HCO3 Y



w/ HCO3

19.  Use Equation 5.4 to calculate the fractional saturation (Y) for hyperbolic binding, letting K = p50 = 26 torr: p​O​ 2​ ​​ ​Y = _________ ​       ​p​ 50​+ p​O​ 2​

30 torr   ​ = 0.54​ ​At 30 torr, Y = ______________ ​    26 torr + 30 torr 100 torr ​At 100 torr, Y = _______________ ​     ​ = 0.79​   26 torr + 100 torr

 herefore, if hemoglobin exhibited hyperbolic oxygen-binding behavT ior, it would be only 79% saturated in the lungs (where pO2 ≈ 100 torr) and would exhibit a loss of saturation of only 25% (79% − 54%) in the tissues (where pO2 ≈ 30 torr). Hemoglobin’s sigmoidal binding behavior allows it to bind more O2 in the lungs so that it can deliver relatively more O2 to the tissues (for an overall change in saturation of about 40%; see Fig. 5.7).

(​ ​p​O​ 2​)n​ ​ 21.  ​Y = _____________ ​        ​ ​(​p​ 50​)​n​+ (​ ​p​O​ 2​)​n​

​(​25 torr​)3​ ​   Y = ​ ___________________        ​ (​ ​40 torr​)3​ ​+ (​ ​25 torr​)3​ ​   Y = 0.20

​(​p​O​ 2​)n​ ​       ​   Y = ​ _____________ ​(​p​ 50​)n​ ​+ (​ ​p​O​ 2​)n​ ​

(​ ​120 torr​)3​ ​   Y = ____________________ ​        ​ (​ ​40 torr​)3​ ​+ (​ ​120 torr​)3​ ​

  Y = 0.96​

23.  A patient with a light complexion would have flushed, reddened skin due to the color of the Hb · CO complex. The patient would also suffer from dizziness, headaches, and shortness of breath due to lack of oxygen. 25.  In the T form, the His residue in the β chain could potentially interact with the Thr residue by hydrogen bonding and dipole–dipole interactions and to the Pro residue via London dispersion forces. In the R form, the His residue could potentially interact with both Thr residues on the α chain via hydrogen bonds and dipole–dipole interactions. 27. a.  His 146, which is protonated, and Asp 94 form an ion pair in the deoxy form of hemoglobin. The protons are produced by cellular respiration. As described by the Bohr effect, an increase in hydrogen ion concentration favors the deoxy form of hemoglobin so that O2 can be delivered to the tissues.  b.  The presence of the negatively charged Asp 94 increases the pK value of the imidazole ring of His and

pO2

c.  The positively charged Lys side chain forms an ion pair with the negatively charged ​​HCO​  3−​  ​​. The phenolate oxygens on the two tyrosine side chains act as hydrogen bond acceptors for the bicarbonate hydrogen. [From Komiyama, N.H. et al., Nature 373, 244–246 (1995).] 31.  Negatively charged Glu side chains on the surface of the oxygenated lamprey hemoglobin monomer resist association due to the charge–charge repulsion. However, when the pH decreases, excess protons bind to the Glu side chains, neutralizing them. The monomers associate to form the deoxygenated tetramer. In this manner, O2 is delivered to lamprey tissue when the pH decreases in metabolically active tissue. [From Qiu, Y. et al., J. Biol. Chem. 275, 13517–13528 (2000).] 33.  At the high altitude where the bar-headed goose resides, less O2 is available to bind to hemoglobin in the lungs. The bar-headed goose hemoglobin has a lower p50 value and a higher oxygen affinity than the plains-dwelling grelag goose hemoglobin, so the bar-headed goose hemoglobin can more easily bind oxygen in order to deliver it to the tissues. [From Jessen, T.-H. et al., Proc. Natl. Acad. Sci. USA 88, 6519–6522 (1991).] 35. a.  A person hyperventilates in order to obtain more O2. Excessive removal of CO2 from the lungs during hyperventilation causes the H2CO3 ⇌ H2O + CO2 equilibrium to shift to the right, resulting in a decrease in blood H 2CO3 concentration. This causes the H+ + ​​HCO​  3−​  ​​ ⇌ H2CO3 equilibrium to shift to the right, which results in a decrease in the concentration of hydrogen ions, which results in a basic blood pH (see Fig. 2.17).  b.  The decrease in alveolar pCO2 concentration can be explained by the hyperventilation, as described in part a. The concentration of 2,3-BPG increases in order to convert more of the hemoglobin molecules to the low affinity T form so that oxygen can be effectively delivered to the cells. 37.  A decrease in pH diminishes hemoglobin’s affinity for O2 (the Bohr effect), thereby favoring deoxyhemoglobin. Since only deoxyhemoglobin S polymerizes, sickling of cells is most likely to occur when the parasite-induced drop in pH promotes the formation of deoxyhemoglobin. 39.  The loss of the His residue at position 146 decreases the ability of Hb Cowtown to bind protons. Therefore, the oxygen binding affinity of Hb Cowtown is less sensitive to pH changes than normal

OD D -NUMB ER ED SO LUTI O NS  S-15 hemoglobin. [From Perutz, M.F. et al., Proc. Natl. Acad. Sci. USA 81, 4781–4784 (1984).] 41. a.  Hb A has a sigmoidal curve, which means that the binding and release of O2 from normal hemoglobin is cooperative. The hyperbolic binding curve of Hb Great Lakes indicates that there is little cooperativity in O2 binding and release. b.  Hb Great Lakes has a higher affinity for O2. More than 60% of the mutant hemoglobin has bound O2. Hb A is about 30% oxygenated.  c.  Both hemoglobins are essentially 100% oxygenated and therefore have equal affinities.  d.  Normal hemoglobin is more efficient at O2 delivery. It delivers about 70% of its bound O2 (since it is 100% oxygenated at 75 torr and 30% oxygenated at 20 torr). Hb Great Lakes is less efficient since less than 40% of its bound oxygen is delivered to the tissues. [From Rahbar, S. et al., Blood 58, 813–817 (1981).] 43. a.  Yes, because the negatively charged Glu side chain would be replaced by the positively charged Lys side chain.  b.  Residue 43 of the β chain is located in the loop that connects helices C and D. The amino acid side chain at position 43 may face the solvent rather than participate in protein-stabilizing interactions with other protein groups, so the substitution of Lys for Glu may have only a minor effect on the protein’s structure. 45.  The substitution of the bulky, largely nonpolar Trp side chain for the Cys side chain at the β93 position in Hb Santa Giusta Sardegna decreases the p 50 value and decreases cooperative oxygen binding, as evidenced by the decreased n value (see Problem 20). A likely explanation is that the large Trp residue prevents the His β146 side chain from forming ion pairs with Asp 94 (see Solution 27) and prevents the C-terminal from forming an ion pair with a Lys residue on a neighboring chain (see Solution 28). If these interactions cannot form to stabilize the T state, then the R form is stabilized, which explains the decreased p 50 value and the increased O 2 binding affinity. The mutation also affects the ability of oxygen to bind and dissociate cooperatively, which may also contribute to the stabilization of the R form. [From Fais, A. et al., Hemoglobin 36, 151–156 (2012).] 47. a.  The oxygen affinity of Hb Providence is greater. The substitution of the neutral Asn for the positively charged Lys results in decreased binding of BPG in the central cavity of hemoglobin, since Lys forms an ion pair with the negatively charged BPG. BPG binds to the T form but not the R form of hemoglobin. Therefore, decreased BPG binding means the R form is favored and the oxygen affinity of the mutant is increased. b. 

O

O HN

CH

+

C

NH4

C

O

NH2

HN

CH

C

CH2

CH2 H2O

C

O

O

proton-binding affinity. Since His 146 is located in the central cavity, BPG would bind less readily, which favors the oxygenated or R form of hemoglobin. 51.   Both types of molecules are proteins and consist of polymers of amino acids. Both contain elements of secondary structure. However, globular proteins are water-soluble and nearly spherical in shape. Examples include proteins such as hemoglobin and myoglobin as well as enzymes. Their cellular role involves participating in the chemical reactions of the cell in some way. In contrast, fibrous proteins tend to be water-insoluble and have an elongated shape. Their cellular role is structural—as elements of the cytoskeleton of the cell and the connective tissue matrix between cells. 53.  Actin filaments and microtubules consist entirely of subunits that are assembled in a head-to-tail fashion, so the polarity of the subunits (actin monomers and tubulin dimers) is preserved in the fully assembled fiber. In intermediate filament assembly, only the initial step (dimerization of parallel helices) maintains polarity. In subsequent steps, subunits align in an antiparallel fashion, so in a fully assembled intermediate filament each end contains heads and tails. 55.  Because phalloidin binds to F-actin but not to G-actin, the addition of phalloidin fixes actin in the filamentous form. This impairs cell motility because cell movement requires both actin polymerization at the leading edge of the cell and depolymerization at the trailing edge. In the presence of phalloidin, depolymerization does not occur and cell movement is not possible. 57.  During rapid microtubule growth, β-tubulin subunits containing GTP accumulate at the (+) end because GTP hydrolysis occurs following subunit incorporation into the microtubule. In a slowly growing microtubule, the (+) end will contain relatively more GTP that has already been hydrolyzed to GDP. A protein that preferentially binds to (+) ends that contain GTP rather than GDP could thereby distinguish fast- and slow-growing microtubules. 59.  Polymers composed of β-tubulin molecules allowed to polymerize in the presence of a nonhydrolyzable analog of GTP are more stable. When the β-tubulin subunits are exposed to GTP in solution, the GTP binds to the β-tubulin and then is hydrolyzed to GDP, which remains bound to the β-tubulin. Additional αβ heterodimers are then added. The microtubule ends with GDP bound to the β-tubulin are less stable than those bound to GTP because protofilaments with GDP bound are curved rather than straight and tend to fray. If a nonhydrolyzable analog is bound, it will resemble GTP and the protofilament will be straight rather than curved. It is less likely to fray and the resulting protofilament is more stable as a result. 61.  Microtubules form the mitotic spindle during cell division. Because cancer cells divide rapidly and hence undergo mitosis at a rate more rapid than in most other body cells, drugs that target tubulin and interfere with the formation of the mitotic spindle in some way will slow the growth of cancerous tumors.

c.  The oxygen affinity of Hb Providence Asp is even greater than that of Hb Providence Asn. The presence of the negatively charged Asp repels the negatively charged phosphate groups of BPG, resulting in an even greater decrease in affinity for BPG. Since BPG binds only to deoxyhemoglobin, the inability of BPG to bind to Hb Providence Asp results in a stabilization of the R form and an increase in oxygen affinity. [From Bonaventura, J. et al., J. Biol. Chem. 251, 7563–7571 (1976).]

63.  Colchicine, which promotes microtubule depolymerization, inhibits the mobility of the neutrophils because cell mobility results from polymerization and depolymerization of microtubules.

49.  The substitution at the C-terminus could affect the position of His F8 in such a way that O2 either binds more readily or dissociates less readily. The substitution is quite near His 146, which binds protons (see Problem 27), and may decrease this side chain’s

67. a.  The first and fourth side chains are buried in the coiled coil, but the remaining side chains are exposed to the solvent and therefore tend to be polar or charged.  b.  Although the residues at positions 1 and 4 in both sequences are hydrophobic, Trp and Tyr are much larger

65.  As shown in Figure 5.25, microtubules link replicated chromosomes to two points at opposite sides of the cell. Vinblastine’s ability to stabilize the microtubules at the (+) end while destabilizing the (–) end disrupts this linkage. Mitosis slows down or completely halts as a result. [From Panda, D. et al., J. Biol. Chem. 271, 29807–29812 (1996).]

S-16  ODD- N UM B ER E D S O LUT I O NS than Ile and Val and would therefore not fit as well in the area of contact between the two polypeptides in a coiled coil (see Fig. 5.29). 69.  The reducing agent breaks the disulfide bonds (—S—S—) between keratin molecules. Setting the hair brings the reduced Cys residues (with their —SH groups) closer to new partners on other keratin chains. When the hair is then exposed to an oxidizing agent, new disulfide bonds form between the Cys residues and the hair retains the shape of the rollers. 71. a.  Actin’s primary structure is its amino acid sequence. Its secondary structure includes its α helices, β sheets, and other conformations of the polypeptide backbone. Its tertiary structure is the arrangement of its backbone and all its side chains in a globular structure. Monomeric actin by definition has no quaternary structure. However, when actin monomers associate to form a filament, the arrangement of subunits becomes the filament’s quaternary structure. Thus, actin is an example of a protein that has quaternary structure under certain conditions.  b.  Collagen’s primary structure is its amino acid sequence. Its secondary structure is the left-handed helical conformation characteristic of the Gly–Pro–Hyp repeating sequence. Its tertiary structure is essentially the same as its secondary structure, since most of the protein consists of one type of secondary structure. Collagen’s quaternary structure is the arrangement of its three chains in a triple helix. It is also possible to view the triple helix as a form of tertiary structure, with the quaternary structure referring to the association of collagen molecules. 73.  

O NH

CH

2

C 1

CH2

3

CH2

CH

OH

CH2

6

NH+ 3

75. 

H NH

CH

87.  Since individuals with severe cases of osteogenesis imperfecta do not survive to a reproductive age, their particular genetic defect is not passed on. Hence, most cases arise from new mutations.

C O O

H N

83. a. (Pro–Pro–Gly)10 has a melting temperature of 41°C, while (Pro–Hyp–Gly) 10 has a melting temperature of 60°C. (Pro–Hyp– Gly)10 and (Pro–Pro–Gly)10 both have an imino acid content of 67%, but (Pro–Hyp–Gly)10 contains hydroxyproline, whereas (Pro–Pro– Gly)10 does not. Hydroxyproline therefore has a stabilizing effect relative to proline.  b. (Pro–Pro–Gly)10 and (Gly–Pro–Thr(Gal))10 have the same melting point, indicating that they have equal stabilities. This is interesting because (Pro–Pro–Gly)10 has an imino acid content of 67%, whereas (Gly–Pro–Thr(Gal))10 has an imino acid content of only 33%. The glycosylated threonine must have a stabilizing effect similar to that of proline. It is possible that the galactose, which contains many hydroxyl groups, provides additional sites for hydrogen bonding and would thus contribute to the stability of the triple helix.  c.  The inclusion of (Gly–Pro–Thr) 10 is important because the results show that this molecule does not form a triple helix. This molecule is included as a control to show that the increased stability of the (Gly–Pro–Thr(Gal))10 is due to the galactose, not to the threonine residue itself. [From Bann, J.G. et al., FEBS Lett. 473, 237–240 (2000).] 85.  Because collagen has such an unusual amino acid composition (nearly two-thirds of the protein’s residues are Gly and Pro or Pro derivatives), it contains relatively few of the other amino acids and is therefore not as good a source of amino acids as proteins containing a greater variety of amino acids. In particular, gelatin lacks tryptophan and contains only small amounts of methionine.

4 5

81. a.  Collagen B is from the rat and collagen A is from the sea urchin.  b.  The stability of each of these collagens is correlated with their hydroxyproline content. The higher the percentage of hydroxyproline, the more regular the structure and the more difficult it is to melt, resulting in more stable collagen. The rat has a more stable collagen and the sea urchin, which lives in cold water, has a less stable collagen. It is important to note that the melting temperature of each collagen molecule is higher than the temperature at which each organism lives. Thus, each organism has stable collagen at the temperature of its environment. [From Mayne, J. and Robinson, J.J., J. Cell. Biochem. 84, 567–574 (2001).]

CH

C

H

77. a.  The patients all suffer from scurvy, a disease resulting from the lack of vitamin C, or ascorbate, in the diet.  b.  Ascorbic acid is necessary for the formation of hydroxyproline residues in newly synthesized collagen chains. Underhydroxylated collagen is less stable, so tissues containing the defective collagen are less sound, leading to bruising, joint swelling, fatigue, and gum disease.  c.  Patients with a gastrointestinal disease may actually be consuming foods with vitamin C, but the disease impairs absorption. Patients suffering from poor dentition and alcoholism may have overall difficulties with food intake. Patients following various fad diets might consume diets that are so unusual or restrictive that their intake of vitamin C is insufficient to support healthy collagen synthesis. [From Olmedo, J.M. et al., Int. J. Dermatol. 45, 909–913 (2006).] 79.  The bacterial enzymes degrade collagen, the major protein in connective tissue. Treatment of the tissue with these enzymes degrades the collagen in the extracellular matrix without harming the cells themselves and thus facilitates the preparation of cells for culturing. [Source: Worthington Biochemical Corporation.]

89.  Myosin is both fibrous and globular. Its two heads are globular, with several layers of secondary structure. Its tail, however, consists of a single fibrous coiled coil. 91. a.  Diffusion is a random process. It tends to be slow (especially for large substances and over long distances). Because it is random, it operates in three dimensions (not linearly) and has no directionality.  b.  An intracellular transport system must have some sort of track (for linear movement of cargo) and an engine that moves cargo along the track by converting chemical energy to mechanical energy. The engine must operate irreversibly to promote rapid movement in one direction. Finally, some sort of addressing system is needed to direct cargo from its source to a particular destination. 93.  When muscles contract, myosin heads bind and release actin in a process that requires ATP for the physical movement of myosin along the actin filament. At the time of death, cellular processes that generate ATP cease. Myosin heads remain bound to actin, but in the absence of ATP, the conformational change that causes myosin to release the actin does not occur, and stiffened muscles are the consequence. 95.  Normal bone development involves the formation of bone tissue in response to stresses placed on the bone. When muscle activity is impaired, as in muscular dystrophy, the forces that shape bone development are also abnormal, leading to abnormal bone growth.

OD D -NUMB ER ED SO LUTI O NS  S-17 97. a.  There are four immunoglobulin-fold domains in each Fab fragment. b.  There are six hypervariable loops: three from the heavy chain variable domain and three from the light chain variable domain. 99. a.  There are eight immunoglobulin-fold domains: two from each of the two heavy chains and two from each of the two light chains.  b.  There are 12 hypervariable loops, twice as many as in an Fab fragment. 101.

Antigen-binding sites

k = 0.693/t1/2. [From Radzicka, A. and Wolfenden, R., Science 267, 90–93 (1995).] 365 days _ ​7.3 years × _ ​          ​× ​ 24 hours     ​× _ ​ 3600 s    ​= 2.3 × ​10​ 8​ s year  day hour 0.693    k​ = ​ _    ​= 3.0 × ​10​ −9​ s​ −1​ ​= ____________ ​  0.693  ​t​ 1/2​ 2.3 × ​10​ 8​ s 7. a. Since k = 0.693/t1/2, 0.693   ​= 2.2 × ​10​ 9​ s ​    ​t​ 1/2​= _______________ 3.2 × ​10​ −10​ ​s​ −1​  day year 2.2 ​ × ​10​ 9​ s × _ ​  hour              ​× _ ​  ​× _ ​  ​= 70 years​ 3600 s 24 hour 365 days b.  The rate enhancement is calculated as in Solution 3:

Region recognized by macrophage

−1

300​ s​  ​   ​= 9.4 × ​10​ 11​ _______________ ​    −10

​ s​ −1​

3.2 × ​10​ 

[From Radzicka, A. and Wolfenden, R., Science 267, 90–93 (1995).] 9.  For adenosine deaminase: −1

370 ​s​  ​   ​= 2.1 × ​10​ 12​ ​ _____________   

103. 

1.8 × ​10​ −10​ ​s​ −1​

Antibody

For triose phosphate isomerase: Antigen

−1

4300 ​s​  ​  ​= 1.0 × ​10​ 9​ _____________ ​    −6 −1

4.3 × ​10​  ​ ​s​  ​

105.  The immunoglobulins are proteins. If they are ingested, they will be denatured in the acidic pH of the stomach and hydrolyzed by the digestive enzymes in the stomach and small intestine. Delivering the proteins directly into the blood guarantees that the molecules can circulate to all areas of the body. 107. a.  The antibodies that bind to the second strain are recognized by the macrophages’ Fc receptors, so the cells readily take up the bound virus and are efficiently infected. (This is known as antibody-dependent enhancement.)  b.  The same phenomenon occurred when antibodies were formed in response to the vaccine (the first exposure to the viral antigens). When the children later encountered the actual virus, the existing antibodies promoted virus uptake by macrophages, leading to more massive infection. [Arredondo-García, J.L. et al., Clin. Microbiol. Infect. 24, 755–763 (2018).]

Chapter 6 1.  A globular protein can bind substrates in a sheltered active site and can support an arrangement of functional groups that facilitates the reaction and stabilizes the transition state. Most fibrous proteins are rigid and extended and therefore cannot surround the substrate to sequester it or promote its chemical transformation. 3.  The rate enhancement is calculated as the ratio of the catalyzed rate to the uncatalyzed rate as shown below. [From Sreedhara, A. et al., J. Am. Chem. Soc. 122, 8814–8824 (2000).] −1

3.57 ​h​  ​  ​  = 9.9 × ​10​ 7​ ___________ ​    3.6 × ​10​ −8​ ​h​ −1​

5.  The rate constant for the uncatalyzed reaction is calculated by converting the half-life to seconds and then using the relationship

The rate of the uncatalyzed reaction is slower for the adenosine deaminase reaction than for the triose phosphate isomerase reaction. However, adenosine deaminase is able to catalyze its reaction so that it occurs more quickly than the reaction catalyzed by triose phosphate isomerase. Therefore, the rate enhancement for the adenosine deaminase reaction is greater. 11. a.  Bonds hydrolyzed in the presence of a peptidase enzyme are indicated by the arrows below.  b.  A peptidase belongs to the hydrolase class of enzymes. H N

N

+

H3N

O N H

O

O

H N

O O–

O

13. a.  Pyruvate decarboxylase is a lyase. During the elimination of the carboxylate group (—COO−) of pyruvate, a double bond is formed in CO2 (O=C=O).  b.  Alanine aminotransferase is a transferase. The amino group is transferred from alanine to α-ketoglutarate.  c. Alcohol dehydrogenase is an oxidoreductase. Acetaldehyde is reduced to ethanol or ethanol is oxidized to acetaldehyde.  d.  Hexokinase is a transferase. The phosphate group is transferred from ATP to glucose to form glucose-6-phosphate.  e.  Chymotrypsin is a hydrolase. Chymotrypsin catalyzes the hydrolysis of peptide bonds. 15.  The aromatic l-amino acid decarboxylase (AADC) catalyzes a reaction in which a carboxylate group is released as CO2. O +H N 3

CH CH2

C

O– AADC H+

CO2

+H N 3

CH2 CH2

S-18  ODD- N UM B ER E D S O LUT I O NS

CH

39.  His 57 abstracts a proton from Ser 195, thus rendering the serine oxygen a better nucleophile (see Fig. 6.10). When Ser 195 is modified by formation of a covalent bond with DIP, the proton is no longer available and Ser 195 is unable to function as a nucleophile.

CH

41. 

17.  a. 

COO– H

C

H

H

C

H

COO–

succinate dehydrogenase

COO–

COO–

Succinate b. 

COO– H

C

OH

CH2 COO–

Malate

Acetylcholinesterase

Fumarate

C

+

F

CH2OH

COO–

malate dehydrogenase

O

CH3

O CH2

COO–

Oxaloacetate

23. a.  2 H2O2 ⇌ O2 + 2 H2O  b.  2 glutathione + H2O2 ⇌ 2 gluta­ thione disulfide + H2O

29.  ΔG‡1, the energy of activation required for the substrate to be converted to the intermediate, is greater than ΔG‡2, the energy of activation required for the intermediate to be converted to product. Therefore, the first step is the slower of the two steps in this reaction. 31.  Yes. An enzyme decreases the activation energy barrier for both the forward and the reverse directions of a reaction. 33. a.  Gly, Ala, and Val have side chains that lack the functional groups required for acid–base or covalent catalysis.  b. Mutating one of these residues may alter the conformation at the active site enough to disrupt the arrangement of other groups that are involved in catalysis. 35. a.  In order for any molecule to act as an enzyme, it must be able to recognize and bind a substrate specifically, it must have the appropriate functional groups to carry out a chemical reaction, and it must be able to position those groups for reaction.  b.  Functional groups on the nitrogen bases can participate in chemical reactions in much the same way as amino acid side chains on proteins. For example, the amino groups on adenine, guanine, and cytosine bases could act as nucleophiles and could also act as proton donors.  c.  DNA, as a double-stranded molecule, has limited conformational freedom. RNA, which is single-stranded, is able to assume a greater range of conformations. This flexibility allows it to bind to substrates and carry out chemical transformations. 37.  The aldolase reaction uses both base catalysis and covalent catalysis. The fructose-1,6-bisphosphate substrate forms a covalent bond to the enzyme’s active site Lys residue in the form of a Schiff base. The enzyme’s active site Tyr residue acts as a base and accepts a proton from the substrate in order to facilitate the breakage of the scissle bond, releasing the first product.

CH

H3C

O

P

O

CH3

43.  His residues are often involved in proton transfer. A carboxymethylated His would be unable to donate or accept protons. [From Shapiro, R. et al., Proc. Natl. Acad. Sci. USA 84, 8783–8787 (1987).] O HN

25.  Chymotrypsin catalyzes the hydrolysis of peptide bonds on the carboxyl side of Tyr, Trp, Phe, and Leu side chains. 27.  Every 10-fold increase in rate corresponds to a decrease of about 5.7 kJ · mol−1 in ΔG‡.  a.  For the nuclease, with a rate enhancement on the order of 1014, ΔG‡ is lowered about 14 × 5.7 kJ · mol−1, or about 80 kJ · mol−1.  b.  For the isomerase, with a rate enhancement on the order of 109, ΔG‡ is lowered about 9 × 5.7 kJ · mol−1, or about 50 kJ · mol−1.

Modified enzyme

O

H3C

c.  Both succinate dehydrogenase and malate dehydrogenase are oxidoreductases.

21. a.  Reaction 4,  b.  Reaction 1,  c.  Reaction 3,  d.  Reaction 2.

Sarin

HF

CH2

19. a. Chymotrypsin, b.  triose phosphate isomerase,  c. adenosine deaminase,  d.  staphylococcal nuclease.

P

CH3 CH O CH3

CH CH2

O

C

HN

HBr

CH

C

CH2

+ BrCH2COO–

HN

N N

N

CH2COO–

45.  Cys 278 is highly exposed and unusually reactive compared to other cysteines in creatine kinase. Cys 278, because of its high reactivity, is probably one of the catalytic residues in the enzyme. The other cysteine residues are not as reactive because they may not be directly involved in catalysis and/or because they are shielded in some way that prevents them from reacting with NEM. 47.

1st tetrahedral 2nd tetrahedral intermediate intermediate E+S G ES covalent intermediate

E+P

Reaction coordinate 49.  At very low pH values, His would be protonated and unable to form a hydrogen bond with Ser. Asp would also be protonated and unable to form a hydrogen bond with His. At very high pH values, Ser would be unprotonated and unable to form a hydrogen bond with His. 51.  [From Kong, L. et al., J. Biol. Chem. 290, 7160–7168 (2015).] Asp CH2 C O–

His

Cys

CH2

O HN

CH2 NH+

RC

H N

S C RN

O–

OD D -NUMB ER ED SO LUTI O NS  S-19 53.

57. 

His 124 HN

CH

Asp 70

O C

HN

CH2

CH

(CH2)2

C

N

O

H

O

H

H

RNA Substrate

Asp 32 C

H

O

O– RC

OH

H N

O

O O

H

P

C

O

O

O (CH2)2

C

RN

O–

C

Glu 224

[From Li, L. et al., Biochemistry 39, 2399–2405 (2000).] 59. a.  The deamidation reaction for asparagine is shown. The deamidation reaction for glutamine is similar.

H RN

Asp 32

O CH

NH

O + H2O

C

CH

NH

CH2

CH2

C

C

O

H OH RN N C RC

Aspartate

CH2

CH2 O

NH2

Asp 215

H

C

A

Asp 32

CH2

O

CH2

O–

OH C

OH O

R CNH2

RN

C O–

O

CH2

HA

CH2

1

CH2

2

O

HO

NH+ 3 H

C

O

O–

CH2 C

OH

O

C

+ NH+ 4

C

b.

O

C

RC

His His Glu

NH2

O–

H2N

O

Asparagine

OH

OH

Zn2+

Asp 215

C

RC

O–

–

O

O

C

O

N

C

His His Glu

O

H

RN

OH

H

Zn2+

C

b.  His 124 most likely has a pK of less than 5.0 because the His shown in the relay is unprotonated. It must be unprotonated in order to accept a proton from the water molecule. If His 124 had a higher pK value, it would be more likely to be partially protonated at physiological pH and less able to accept a proton.  c. Alanine has an aliphatic side chain and is unable to accept a proton in the relay, as shown in part a. [From Oda, Y. et al., J. Biol. Chem. 268, 88–92 (1993).] 55. a. 

C

OH

Glu 224

O

CH2

HN

O

O

C

O–

NH+ 3 A–

H

3

CH2

4 NH+ 4

NH3

CH2 C

O

O

H

O

Asp 215

b.  Renin uses an acid–base catalytic strategy. In the first step, Asp 215 acts as an acid and donates a proton while Asp 32 acts as a base and accepts a proton. In the second step, these roles are reversed.  c.  The pK value of Asp 32 is lower than that of Asp 215 since the Asp 32 side chain is unprotonated and the Asp 215 side chain is protonated.

c.  Ser and Thr residues could stabilize the transition state. They could also serve as bases (if unprotonated) and accept a proton from water to form a hydroxide ion that would act as the attacking nucleophile. Ser and Thr (in the unprotonated form) could also act as attacking nucleophiles themselves.  d.  The mechanism for the deamidation of an amino terminal Gln residue is shown below. Amino terminal Asn residues are not deamidated because the product would be a four-membered ring, which is unstable. [From Wright, H.T., Crit. Rev. Biochem. Mol. Biol. 26, 1–52 (1991).]

S-20  ODD- N UM B ER E D S O LUT I O NS O NH2

CH

C

NH

CH2 O

O

CH2 C

O

NH2

NH

CH

C

CH2 CH2

C

NH

+ NH3 + HA H

A

61.  The ability of an enzyme to accelerate a reaction depends on the free energy difference between the enzyme-bound substrate and the enzyme-bound transition state. As long as this free energy difference is less than the free energy difference between the unbound substrate and the uncatalyzed transition state, the enzyme-mediated reaction proceeds more quickly. 63.  The Asp side chain is negatively charged, which would repel the oxyanion rather than stabilizing it. The decrease in transition state stabilization lowers the activity of the mutated enzyme. [From Wan, Y. et al., J. Biol. Chem. 294, 11391–11401 (2019).] 65.  In a serine protease, there is no need to exclude water from the active site, since it is a reactant for the hydrolysis reaction catalyzed by the enzyme. By contrast, hexokinase must prevent water from entering the active site, which could potentially hydrolyze ATP. 67.  The zinc ion participates in catalysis by polarizing the water molecule so that its proton is more easily abstracted by Glu 224. The positively charged zinc ion stabilizes the negatively charged oxygen in the transition state. 69.  The transition state structure is likely tetrahedral at position 6 on the purine ring, since adenosine is planar whereas 1,6-dihydropurine is tetrahedral at this position. Enzymes bind the transition state much more tightly than the substrate. 71.  Because enzymes bind more tightly to their transition states than to their substrates, a drug designed to treat a disease by inhibiting a particular enzyme would be a more effective inhibitor if its structure resembled that of the transition state. Transition state analogs inhibit their target enzymes effectively at low concentrations, allowing a low dose to be used, which would be less likely to cause side effects. 73.  A mutation can increase or decrease an enzyme’s catalytic activity, depending on how it affects the structure and activity of groups in the active site.

amide bond is an arginine side chain, which would not fit into chymotrypsin’s specificity pocket. 81. a.  Persistent activation of trypsinogen to trypsin also results in the activation of chymotrypsinogen to chymotrypsin (see Solution 77) and causes proteolytic destruction of the pancreatic tissue.  b. Since trypsin is at the “top of the cascade,” it makes sense to inactivate it by using a trypsin inhibitor. [From Hirota, M. et al., Postgrad. Med. J. 82, 775–778 (2006).] 83.  Serine protease inhibitors would interfere with the digestion of proteins in the small intestine. Undigested proteins could not be absorbed by the cells of the small intestine, and thus their nutritional value would be lost. Gastric upset is also a possibility. 85.  The function of factor IXa is to activate factor X in order to promote thrombin activation and fibrin formation. Factor VIIa can also activate factor X, so the physiological effect is similar. 87.  Factor IXa leads to the activation of thrombin, so the absence of factor IX delays clot formation, causing bleeding. Although factor XIa also leads to thrombin production, factor XI plays no role until it is activated by thrombin itself. By this point, coagulation is already well under way, so a deficiency of factor XI may not significantly delay coagulation. 89.  O CH2

CH2

NH2  NH+ 3

Factor XIIIa

CH2

CH2

CH2

CH2

+

NH4

CH2

C

NH

CH2

CH2

CH2

CH2

91.  In DIC, the high rate of coagulation actually leads to depletion of platelets and the various coagulation factors. This prevents normal coagulation from occurring, so the patient bleeds. 93.  Heparin must enter the bloodstream (via intravenous administration) in order to act with antithrombin as an anticoagulant. If consumed orally, it will not enter the circulation but will be degraded to its monosaccharide components.

Chapter 7

77.  Chymotrypsin activation is a cascade mechanism, since chymotrypsinogen is activated by trypsin, which is in turn activated by enteropeptidase.

d[S] 3.  ​v = − ​ ___ ​   dt

1.  The hyperbolic shape of the velocity versus substrate concentration curve suggests that the enzyme and substrate physically combine so that the enzyme becomes saturated at high concentrations of substrate. The lock-and-key model describes the interaction between an enzyme and its substrate in terms of a highly specific physical association between the enzyme (lock) and the substrate (key).

0.025 M v = − ​______________________________           ​ 440 y × 365 d ⋅ ​y​ −1​× 24 h ⋅ ​d​ −1​ 3600 s ⋅ ​h​ −1​ v = −1.8 × 10−12 M ⋅ ​s​ −1​

Trypsinogen Chymotrypsinogen

Trypsin

d[S] 5.  ​v = − ​_      ​ dt

Chymotrypsin

79.  No, the compound shown in Problem 12 would not be hydrolyzed by chymotrypsin. The side chain on the carboxyl side of the

CH2

O

75. a.  Carboxypeptidase A and chymotrypsin have similar substrate specificities; therefore, the specificity pocket of carboxypeptidase A is likely to be similarly hydrophobic.  b.  Both Lys and Arg side chains will be able to form ion pairs and hydrogen bonds with the negatively charged Asp in the enzyme’s specificity pocket. The Leu and Ile residues can interact with the nonpolar portions of the Lys and Arg side chains via London dispersion forces. [From Akparav, V. et al., Acta Cryst. F71, 1335–1340 (2015).]

Enteropeptidase

C

0.025 M v = − ​___________________________________            ​ 6.6 × ​10​ 6​y × 365 d ⋅ ​y​ −1​× 24 h ⋅ ​d​ −1​× 3600 s ⋅ ​h​ −1​

v = − 1.2 × ​10​ −16​M ⋅ ​s​ −1​

OD D -NUMB ER ED SO LUTI O NS  S-21 25.

d[S]   7.  ​v = − ​ ___ ​ dt v​ = − ​_____________   0.5 M15      ​= 2.0 × ​10​ −16​M ∙ ​s​ −1​ 2.5 × ​10​  ​ s

v0

d[P] __________________________ 25 × 10  M      ​    ​ = ​    9.       ​ v = ​ _ –6

dt

zero order

50 d × 24 h ⋅ d−1 × 3600 s ⋅ h−1

first order

v = 5.8 × ​10​ −12​M ⋅ ​s​ −1​ [From Bryant, R.A.R. and Hansen, D.E., J. Am. Chem. Soc. 118, 5498–5499 (1996).] 11.  ​(5.8 × ​10​ −12​M ⋅ ​s​ −1​) (4.7 × ​10​ 11​) = 2.7 M ⋅ ​s​ −1​

(65 μmol ⋅ ​min​ −1​) (1.0 μM) ​          v​​  0​= ____________________ (0.135 μM) + (1.0 μM)

d[S] 13.  ​v = − ​ _    ​ dt     ​  v = − ​_  0.065 M 60 s −3 v = − 1.1 × ​10​  ​M ⋅ ​s​ −1​ 15. a. Reaction A→B+C A+B→C 2A→B 2A→B+C

v​​  0​= 57 μmol ⋅ ​min​ −1​

Molecularity Unimolecular Bimolecular Bimolecular Bimolecular Units of k s−1 M−1 · s−1 M−1 · s−1 M−1 · s−1

29.  Rearrange Equation 7.21 to solve for KM, then substitute the values for v0, substrate concentration, and Vmax:

Rate equation v = k [A] v = k [A][B] v = k [A]2 v = k [A]2 Reaction velocity proportional to… [A] [A] and [B]   [A] squared   [A] squared  

​V​  ​ [S] ​  max        ​v​ 0​= _ ​KM ​  ​ + [S] [S ] (​Vmax ​  ​−  ​v​ 0​) ​      K​ ​   ​= ____________ M ​v​   ​

Order First Second Second Second

17.  ​v = k[sucrose]

v = (5.0 × ​10​ −11​​s​ −1​) (0.050 M)

K​ ​   ​= 5.6 mM​ M

[From Doğru, Y.Z. and Erat, M., Food Res. Int. 49, 411–415 (2012).] 31.  Rearrange Equation 7.21 to express v0 and Vmax as a ratio: ​V​  ​ [S] ​  max      ​v​ 0​= _ ​KM ​  ​+ [S] ​v​ 0​ [S] _         ​​  ​= ________ ​   ​ ​Vmax ​  ​

​KM ​  ​+ [S]

0.10 mM   ​0.9 = ______________ ​    ​KM ​  ​+ 0.10 mM K​ ​   ​= 0.011 mM​ M

v = 2.5 × ​10​ −12​M ⋅ ​s​ −1​

[From Zhu, X. et al., Biochemistry 34, 2560–2565 (1995).]

19.  v = k[trehalose]

33.  Rearrange Equation 7.21 to solve for substrate concentration, then substitute the values for v0, KM and Vmax:

v = (3.3 × ​10​ −15​​s​ −1​) (0.050 M) v​ = 1.7 × ​10​ −16​M ⋅ ​s​ −1​ 21.  Using Equation 7.2 and the k value provided in Problem 18:

​V​  ​[S] ​  max      ​ v​ 0​= _ ​KM ​  ​+ [S] ​v​  ​​K​  ​ ​ [S] = _ ​  0 M    ​Vm ​  ax​ − ​v​ 0​

​v = k[sucrose] [sucrose] = _ ​ v ​ k

(0.3 μmol ⋅ ​min​ −1​ ⋅ ​mL​ −1​) (0.6 mM) ​ = _____________________________     [S] ​        ​ (1.1 − 0.3 μmol ⋅ ​min​ −1​ ⋅ ​mL​ −1​ )

× ​10​ −3​M ⋅ ​s ​ −1​​    [sucrose]​= _____________ ​ 5.0   1.0 × ​10​ 4​​s​ −1​

[S] = 0.2 mM​

[sucrose] = 5.0 × ​10​ −7​M = 0.50 μM​

[From Botman, D. et al., J. Histochem. Cytochem. 62, 813–826 (2014).]

23. a.  v = k [Enzyme][Pi]

35.  The Vmax is approximately 30 μM · s –1 and the KM is approximately 5 μM.

b.  Using Equation 7.3:

37. a.  When v0 = 0.75Vmax,

​v = k [Enzyme][​P​ i​]

v = (3.9 × ​10​  ​​M​  ​ ⋅ ​s​  ​) (15 × ​10​  −1

(10 mM) (0.36 − 0.23 U ⋅ ​min​ −1​ ⋅ ​mL​ −1​) K​ ​   ​= _______________________________ ​            M 0.23 U ⋅ ​min​ −1​ ⋅ ​mL​ −1​ 0

b.  The simultaneous collision of three molecules (E, A, and B) is an unlikely event. It is much more likely that the enzyme binds first one and then the other substrate. For example, the first bimolecular reaction might be E + A → EA, and the second would be EA + B → EAB.

6

[S]

​ ​  ​ [S] V 27.  ​v​ 0​= _ ​  max      ​KM ​  ​+ [S]

−1

−12

−3

​M) (50 × ​10​  ​M)

v = 2.9 × ​10​ −6​M ⋅ ​s​ −1​

[From Meadow, N.D. et al., J. Biol. Chem. 280, 41872–41880 (2005).]

​V​  ​ [S] ​  max      ​ ​0.75​Vm ​  ax​= _ [S] + ​KM ​  ​ Vmax cancels out on both sides.

S-22  ODD- N UM B ER E D S O LUT I O NS [S] ​0.75 = ______ ​     ​  [S] + ​KM ​  ​ 0.75​([S] + ​KM ​  ​)​= [S] 0.75​KM ​  ​= 0.25[S] 3​KM ​  ​= [S]​ Thus, the substrate concentration is three times as high as the KM. b.  Using the same logic as in part a: [S]    ​  ​0.9 = ​ ______ [S] + ​KM ​  ​ ​  ​)​ = [S] 0.9​([S] + ​KM 0.9​KM ​  ​= 0.1[S] 9​KM ​  ​= [S]​ Thus, the substrate concentration is nine times as high as the KM. 39.  The apparent KM would be greater than the true KM because the experimental substrate concentration would be less than expected if some of the substrate precipitated out of solution during the reaction. 41. a.  N-Acetyltyrosine ethyl ester, with its lower KM value, has a higher affinity for chymotrypsin. The aromatic tyrosine residue more easily fits into the nonpolar “pocket” (see Fig. 6.17) on the enzyme than does the smaller aliphatic valine residue.  b.  The value of Vmax is not related to the value of KM, so no conclusion can be drawn. ​V​  ​ 43. a.         ​​kcat ​  ​= _ ​  max   [​E]​ T​

Wild-type:

​kcat ​  ​ ___________ 21 ​s​ −1​   ​= 1.1 × ​10​ 4​​M​ −1​ ⋅ ​s​ −1​  ​_   ​= ​    ​KM ​  ​ 1.9 × ​10​ −3​M

​kc​  at​ ___________ 120 ​s​ −1​  ​= 6.0 × ​10​ 4​​M​ −1​ ⋅ ​s​ −1​ Mutant: ​ _   ​= ​    ​KM ​  ​ 2.0 × ​10​ −3​M Residue 31 is near the Asp residue of the Asp–His–Ser catalytic triad of the subtilisin enzyme. In the catalytic mechanism, histidine abstracts a proton from serine and becomes positively charged (see Fig. 6.10). The role of the Asp is to stabilize the positively charged imidazole ring, so Leu in some way may enable the Asp to fulfill this function better than isoleucine. Another possibility is that the substitution of leucine for isoleucine alters the three-dimensional protein structure such that the catalytic triad residues are closer to one another and thus proton transfer is facilitated.  c.  Subtilisin removes protein stains by hydrolyzing the peptide bonds of the protein and releasing amino acids or short peptides as products, which could be easily washed away from the clothing. [From Takagi, H. et al., J. Biol. Chem. 263, 19592–19596 (1988).] 49.  The V max can be calculated by taking the reciprocal of the y intercept: V​max   ​= _ ​  1     y int 1    ​    V​ ​  max​= __________________ 4.41 × ​10​ −4​μ​M​ −1​ ⋅ h ​ V​ ​   ​= 2.27 × ​10​ 3​μM ⋅ ​h​ −1​ max The KM can be determined by first calculating the x intercept and then taking its reciprocal: b ​x int = − ​ _ m  ​

4.41 × ​10​ −4​μ​M​ −1​ ⋅ h     ​ x int = − ​_________________      0.26 μ​M​ −1​ ⋅ h ⋅ μM x int = − 1.70 × ​10​ −3​μ​M​ −1

4.0   × ​10​ −7​M ⋅ ​s​ −1 k​​   ​= ______________ ​      ​ cat 1.0 × ​10​ −7​M k​​   ​= 4.0 ​s​ −1​ cat

K​ ​   ​= − ​_   1     M x int 1        K​ ​   ​= − ​_______________ M − 1.70 × ​10​ −3​μ​M​ −1​ K​ ​   ​= 590 μM​ M

The kcat is the turnover number, which is the number of catalytic cycles per unit time. Each molecule of the enzyme therefore undergoes four catalytic cycles per second.

45. a.  Use the kcat for dopamine from Solution 44a and the KM from Solution 29: ​kcat ​  ​ ___________ 0.30 ​s​ −1​  ​= 5.4 × ​10​ 1​​s​ –1​ ⋅ ​M​ −1​  ​_   ​= ​    ​KM ​  ​ 5.6 × ​10​ −3​M

b.  Use the kcat for catechol from Solution 44b and the KM given in the problem: ​kc​  at​ ___________ 0.33 ​s​  ​  ​= 4.2 × ​10​ 2​​s​ –1​ ⋅ ​M​ −1​   ​= ​    ​ _ −1

​KM ​  ​

7.9 × ​10​ −4​M

The enzyme has a greater catalytic efficiency when catechol is the substrate. 47. a.  The enzyme catalyzes the hydrolysis of the peptide bond on the carboxyl side of the Phe residue. One of the products, ­p-nitrophenolate, is bright yellow, and the rate of its appearance was monitored spectrophotometrically (see Section 6.1).  b. The KM values are nearly identical, which means that each enzyme has the same affinity for its substrate. The kcat value for the Leu 31 enzyme is nearly six times greater than the kcat value for the wild-type enzyme, which means that the mutant enzyme has a greater catalytic efficiency and a higher turnover rate of substrate converted to product per second. The kcat/KM ratio reflects the specific reactivity of the substrate AAPF with the enzyme, and this ratio is larger in the mutant enzyme than in the wild type, as shown below:

51. a.  The collagenase binds a single substrate and is likely to obey Michaelis–Menten kinetics.  b.  Hexokinase binds two substrates and requires a conformational change in order to bind the ­second substrate, so it is unlikely to obey Michaelis–Menten kinetics.  c.  Threonine dehydratase binds a single substrate—either Thr or Ser—and is likely to obey Michaelis–Menten kinetics.  d. ATCase is allosterically regulated (see Problem 89) and is unlikely to obey Michaelis–Menten kinetics. 53.  Calculate the reciprocals of [S] and v0 and construct a plot of 1/v0 versus 1/[S] (shown below). The intercept on the 1/[S] axis is –0.10 mM–1, which is equal to –1/KM. Therefore, KM = 10 mM. The intercept on the 1/v0 axis is 0.05 mM–1 · s, which is equal to 1/Vmax. Therefore, Vmax = 20 mM · s–1. 0.6 0.5 1/v0 (mM1. s)

​k​  ​ 4.0 ​s​ −1​   ​= 2.9 × ​10​ 4​​s​ –1​ ⋅ ​M​ −1​ _ b.  ​ ​  cat  ​= ​ ___________    ​KM ​  ​ 1.4 × ​10​ −4​M

y  0.50x  0.05

0.4 0.3 0.2 0.1

0.2

0

0.2

0.4 1/[S] (mM1)

0.6

0.8

1.0

OD D -NUMB ER ED SO LUTI O NS  S-23

57. a. competitive, b. uncompetitive, c.  pure noncompetitive,  d. noncompetitive, e. noncompetitive. 59.

Type of inhibition

I binds to E, ES or both?

I binds to active site?

Competitive

E

Yes

Uncompetitive

ES

No

Noncompetitive

Both

No

NADP+ as a cofactor. The differences in Vmax are not as great. [From Hansen, T. et al., FEMS Microbiol. Lett. 216, 249–253 (2002).] 71. a. The KM increases by a factor α (see Equation 7.28). The value of KM with the inhibitor is four times greater than the value of KM without the inhibitor (40 μM ÷ 10 μM). Rearrange Equation 7.29 to solve for KI as shown in Solution 65: 30 μM ​KI​ ​= _ ​   ​  = 10 μM ​ 4−1 b.  The KM increases fourfold in the presence of the inhibitor while the Vmax remains the same; these are the characteristics of a competitive inhibitor. [From Gross, R.W. and Sobel, B.E., J. Biol. Chem. 258, 5221–5226 (1983).] 73. a.  Lineweaver–Burk plots are shown below. The KM is calculated from the x intercept, and the Vmax from the y intercept (see Solution 49). 0.40 1/v0 (nmol1 . min)

55. a.  If an irreversible inhibitor is present, the enzyme’s activity would be exactly 100 times lower when the sample is diluted 100-fold. Dilution would not change the degree of inhibition.  b.  If a reversible inhibitor is present, dilution would lower the concentrations of both the enzyme and the inhibitor enough that some inhibitor would dissociate from the enzyme. The enzyme’s activity would therefore not be exactly 100 times less than the diluted sample; it would be slightly greater because the proportion of uninhibited enzyme would be greater at the lower concentration.

61.  Without the inhibitor, the Vmax is estimated to be 250 nmol · min–1 and the KM is estimated to be 1.5 mM. In the presence of the inhibitor, the Vmax is estimated to be 110 nmol · min–1 and the KM is estimated to be 3.0 mM. Because the Vmax decreases and the KM increases, this is noncompetitive inhibition. 63. a.  Since the structures of the inhibitor and the substrate are similar, the inhibitor is competitive. Competitive inhibitors compete with the substrate for binding to the active site, so the structures of the inhibitor and the substrate must be similar.  b.  Yes, the inhibition can be overcome. If large amounts of substrate are added, the substrate will be able to effectively compete with the inhibitor such that very little inhibitor will be bound to the active site. The substrate “wins” the competition when it is in excess.  c.  Like all competitive inhibitors, the inhibitor binds reversibly. [From Andersen, H.S. et al., J. Biol. Chem. 275, 7101–7108 (2000).] 65.  The KM increases by a factor α (see Equation 7.28). The value of KM with the inhibitor is two times greater than the value of KM without the inhibitor (3 mM ÷ 1.5 mM; see Solution 61). Next, rearrange Equation 7.29 to solve for KI: [I] ​KI​ ​= _____ ​    ​​  α − 1 5.0 μM ​KI​ ​= _ ​     ​= 5.0 μM ​ 2−1 67.  Calculate α using Equation 7.29, then use α to compare the KM values with and without the inhibitor: Note that the “apparent” KM in app the presence of the inhibitor is denoted as ​K    ​M​  ​. [I] α​ = 1 + __ ​   ​​  ​KI​ ​

y  3.434x  0.038 0.30 0.20 y  1.899x  0.035

0.10

0.02

0

0.02 0.04 0.06 1/[S] (M1)

0.08

0.10

Without p6* With p6*

x intercept (μM–1) KM (μM) y intercept (nmol–1 · min) Vmax (nmol · min–1)

Without p6* –0.0184 54 0.035 28.6

With p6* –0.011 90 0.038 26.3

b.  The inhibitor is a competitive inhibitor. The Vmax is essentially the same in the presence and absence of the inhibitor (within experimental error), but the KM increased nearly twofold, indicating that p6* is competing with the substrate for binding to the active site of the enzyme. c.  The KM increases by a factor α (see Equation 7.28). The value of KM with the inhibitor is 1.67 times greater than the value of KM without the inhibitor (90 μM ÷ 54 μM = 1.67). Rearrange Equation 7.29 to solve for KI as shown in Solution 65: 10 μM ​    ​KI​ ​= _  ​  = 15 μM​ 1.67 − 1 [From Paulus, C. et al., J. Biol. Chem. 274, 21539–21543 (1999).] 75. a.  The Lineweaver–Burk plot is shown. The KM is calculated from the x intercept, and the Vmax from the y intercept (see Solution 49).

4 μM ​α = 1 + ​ _   ​= 3 2 μM

1.50

app ​K ​ M​  ​ _

1/v0 (nM1. s)

α​ = ​   ​ = 3 ​KM ​  ​ app

​K ​ M​  ​     3​ = ​ _ 10 μM app

K​ ​   ​  ​= 30 μM​ M

y  9.137x  0.035

1.00

0.50 y  0.912x  0.035

+

69. a.  NADPH is structurally similar to NADP and is likely to be a competitive inhibitor.  b. The Vmax would be the same in the presence and in the absence of NADPH because inhibition can be overcome at high substrate concentrations. The KM would increase because a higher concentration of substrate would be needed to achieve half-maximal activity in the presence of an inhibitor.  c. The KM is 400 times greater for NAD+, indicating that the enzyme prefers

0.05

0

0.05

0.10

1/[FDP] (M1) Without vanadate With vanadate

0.15

S-24  ODD- N UM B ER E D S O LUT I O NS

x intercept (μM−1) KM (μM) y intercept (nM−1 · s) Vmax (nM · s−1)

Without vanadate −0.038 26 0.035 28.6

With vanadate −0.0038 260 0.035 28.6

b.  The inhibitor is a competitive inhibitor. The Vmax is the same in the presence and absence of the inhibitor, but the K M increases tenfold in the presence of the inhibitor, indicating that vanadate is competing with the substrate for binding to the active site of the enzyme.

x intercept (mM–1) KM (mM) y intercept (OD−1 · min) Vmax (OD · min−1)

200 y = 72.8x + 49.7 1/v (∆A−1 . min)

79.  The structure of coformycin resembles the proposed transition state for adenosine deaminase and this supports the proposed structure. However, 1,6-dihydroinosine has a K I of 1.5 × 10 –13 M (see Section 7.3) whereas coformycin’s KI is about 0.25 µM; thus 1,6-dihydroinosine more closely resembles the transition state than does coformycin.

1/v0 (pmol1 . min)

y  1066x  200

1200 1000

0.2

800 600

y  795x  148

400 200 0

0.2 0.4 0.6 0.8 1/[Pregnenolone] (M1)

1.0

No inhibitor With inhibitor

x intercept (µM–1) KM (µM) y intercept (pmol–1 · min) Vmax (pmol · min–1)

Without inhibitor –0.186 5.4 148 6.8 × 10–3

–1.0

–0.5

150

100 y = 18.4x + 12.5

50

0.0

With inhibitor –0.188 5.3 200 5.0 × 10–3

b.  Troglitzaone inhibits the hydroxylase competitively (see Solution 74) and the dehydrogenase noncompetitively. The drug binds to the active site of the hydroxylase, but at a site other than the active site on the dehydrogenase. [From Arlt, W. et al., J. Biol. Chem. 276, ­16767–16771 (2001).] 83.  The KM is calculated from the x intercept; the Vmax from the y intercept (see Solution 49). In the presence of the inhibitor, the Vmax and the KM values decrease to a similar extent, as shown in the table. Dodecyl gallate is an uncompetitive inhibitor and binds to the enzyme–substrate complex. [From Kubo, I. et al., Food Chem. 81, 241–247 (2003).]

0.5 1/[F6P]

Without inhibitor

x intercept (mM−1) KM (mM) y intercept (ΔA−1 · min) Vmax (ΔA · min−1)

1400

With inhibitor –2.70 0.37 4.27 0.23

85. a.  The Lineweaver–Burk plot is shown. The KM is calculated from the x intercept, and the Vmax from the y intercept (see Solution 49).

77.  The compound is a transition state analog (it mimics the planar transition state of the reaction) and therefore acts as a competitive inhibitor.

81. a.  The Lineweaver–Burk plot is shown below. The KM is calculated from the x intercept, and the Vmax from the y intercept (see Solution 49). Troglitazone is a noncompetitive inhibitor and does not bind to the active site of the dehydrogenase. The KM values are the same in the presence and absence of the inhibitor and the Vmax is about 25% lower in the presence of the inhibitor.

Without inhibitor –0.99 1.01 1.51 0.66

1.0

1.5

2.0

(mM−1) With inhibitor

Without suramin −0.68 1.5 12.5 0.080

With suramin −0.68 1.5 49.7 0.020

b.  Suramin is a pure noncompetitive inhibitor. The Vmax decreases in the presence of the inhibitor but the KM remains the same. The inhibitor binds to the enzyme at a site other than the active site and interferes with the activity of the enzyme in some way.  c. Because the KM value does not change in the presence of the inhibitor, the value of α is equal to 1, and the Vmax decreases by a factor αʹ (see Table 7.2). The value of Vmax with the inhibitor is fourfold lower than the value of Vmax without the inhibitor (0.080 ΔA · min−1 ÷ 0.020 ΔA · min−1 = 4). Rearrange Equation 7.29 to solve for KI as shown in Solution 65: ​ 5 nM   ​= 1.7 nM ​ ​KI​ ​= _ 4−1 d.  The suramin inhibitor has a KI of 1.7 nM for the parasitic enzyme and 40 nM for the mammalian enzyme. This 20-fold difference indicates that the inhibitor has a greater affinity for the parasitic enzyme and has the potential to be an effective drug because it likely will selectively inhibit the parasitic enzyme and not the mammalian enzyme. [From Sharma, B., Parasite. Vectors 4, 227 (2011).] 87.  It is difficult to envisage how an inhibitor that interferes with the catalytic function (represented by kcat or Vmax) of amino acid side chains at the active site would not also interfere with the binding (represented by KM) of a substrate to a site at or near those same amino acid side chains. 89. a.  ATCase is an allosteric enzyme because its activity versus [S] curve has a sigmoidal shape.  b.  CTP is a negative effector, or inhibitor, because when CTP is added, the KM increases and thus the

OD D -NUMB ER ED SO LUTI O NS  S-25 affinity of the enzyme for the substrate decreases. CTP is the eventual product of the pyrimidine biosynthesis pathway; thus, when the concentration of CTP is sufficient for the needs of the cell, CTP inhibits an early enzyme in the synthetic pathway, ATCase, by feedback inhibition.  c.  ATP is a positive effector, or activator, because when ATP is added, the KM decreases and thus the affinity of the enzyme for its substrate increases. ATP is a reactant in the reaction sequence, so it serves as an activator. ATP is also a purine nucleotide, whereas CTP is a pyrimidine nucleotide. Stimulation of ATCase by ATP encourages CTP synthesis when ATP synthesis is high, thus balancing the cellular pool of purine and pyrimidine nucleotides. 91.  The formation of a disulfide bond under oxidizing conditions, or its cleavage under reducing conditions, could act as an allosteric signal by altering the conformation of the enzyme in a way that affects the groups at the active site.

9.  [From Daenen, L.G.M., JAMA Oncol. 1, 350–358 (2015).]

16:4n –3 11. 

97.  Because cytochrome P450 enzymes can modify warfarin to hasten its excretion, it is helpful to know which P450 enzyme variants are present. The clinician can then use this information to predict how quickly the drug will be modified and can select the appropriate dose to achieve the desired anticoagulant effect.

O O H3C

(H2C)14

C

CH2 O

O

C

(CH2)14

CH3

(CH2)14

CH3

(CH2)10

CH3

(CH2)10

CH3

O

CH CH2

C

O

Tripalmitin 13. 

O O

93.  A drug candidate’s small size and limited hydrogen-bonding capacity indicate that the compound would be able to diffuse across biological membranes in order to enter cells to exert its effects. 95.  Digestive enzymes in the stomach and small intestine may destroy the drug before it has a chance to be absorbed by the body. Therefore, some drugs must bypass the digestive system and be delivered directly to the bloodstream.

CH)4CH2CH2COO–

CH3(CH2CH

H3C

(CH2)12

C

CH2 O

C

O

C

O

C

O

H

CH2

15.  [From Innis, S.M., Adv. Nutr. 2, 275–283 (2011).] O H3C

(CH2)14

O

O

C

CH2

O

C

C

(CH2)7

CH

CH

(CH2)6

(CH2CH

(CH2)7

CH3

H

CH2 C

O

CH)2

(CH2)4

CH3

O

Chapter 8

17. a.

1. a.

(CH2)12

H3C

COO–

O

Myristate (14:0) b.

H3C

CH

CH

(CH2)5

H3C

COO–

(CH2)7

(CH2)14

O

CH

(CH

CH2

CH2)3

α-Linolenate (18:3n-3) d.

H3C

CH

CH

(CH2)7

COO–

(CH2)13

Nervonate (24:1n -9)

19.

3.  [From Sayanova, O. et al., Plant Physiol. 144, 455–467 (2007).] H3C

(CH2)4

CH

CH

CH2

CH

CH

(CH2)4

CH

CH

(CH2)3

(CH2)13

CH

cis, b. H3C

(CH2)4

(CH2)2 CH CH 5,9 cis-∆ -Tetracosadienoate

(CH

CH

CH

CH2)2 (CH2)2

(CH2)3 CH

CH

CH

all cis-∆5,9,15,18-Tetracosatetraenoate H

7.  H3C

(CH2)7

C

C

(CH2)7

COOH

H

Elaidic acid (trans-∆9-octadecenoic acid)

CH2 HO

C

(CH2)14

CH3

OH

OH

O + 3 Na+ –O

H

CH2 COO–

Sciadonate (all-cis-∆5,11,14-eicosatrienoate) 5. a. H3C

C

b.  The monoacylglycerol and fatty acid products are both amphi­ pathic molecules, with a polar head group and one nonpolar tail. These molecules form micelles as a consequence of the hydrophobic effect (see Section 2.2).

COO–

(CH2)6

O + 2 –O

H

C

CH2

Palmitoleate (16:1n-7) c. H3C

OH

CH2

C

C

(CH2)10

CH3

Soap

OH

Glycerol 21.  [From Chao-Mei, Y. et al., Can. J. Chem. 74, 730–735 (1996).]

(CH2)3

COO– CH2

CH

(CH2)3

O

COO–

HO

C

O

C

(CH2)7

H

CH3

C

C

(CH2)7

CH3

H

CH2

OH

23.  All except phosphatidylcholine have hydrogen-bonding head groups. 25.  The polar head group consists of the phosphate derivative (shown in red in Section 8.1) and the glycerol backbone (shown in black). The two nonpolar tails are shown in blue.

S-26  ODD- N UM B ER E D S O LUT I O NS 27.  The polar head group is indicated in blue; the two nonpolar tails in red. R1 R2 O

C

C

O

O

H2C

O

CH

39.  Vitamin A, and the compound from which it is derived, βcarotene, are lipid-soluble molecules. The vegetables in a typical salad do not contain large amounts of lipid. Adding lipid-rich avocado to a salad provides a means to solubilize the β-carotene and thus increase its absorption. [From Unlu, N.Z. et al., J. Nutr. 135, 431–436 (2005).]

CH2 O

O

P

O–

O HO H

H

29. 

H

H HO OH H

OH H

OH

O– O

P

CH3 O

(CH2)2

O H2C O

O

O

C

C

N+ CH3 CH3

CH2

CH

(CH2)14 (CH2)14 CH3

31. a.  Glucose HO

CH3

O

CH

CH

NH

CH

C

49. a.  A hydrocarbon chain is attached to the glycerol backbone at position 1 by a vinyl ether linkage. In a glycerophospholipid, an acyl group is attached by an ester linkage.  b.  The presence of this plasmalogen would not have a great effect since it has the same head group and same overall shape as phosphatidylcholine.

O

(CH2)12 (CH2)26 CH3

CH2 O

C

(CH2)6(CH2CH

CH)2(CH2)4CH3

O

b. 

HO HO

CH2

CH

CH

CH

NH

CH

C

O

(CH2)12 (CH2)26 CH3

NH

CH2 O

C O

43. a.  The traditional definition of a vitamin (a substance that an organism requires but cannot synthesize) implies that vitamins must be obtained in the diet. Because vitamin D3 can be produced from cholesterol, it is not strictly a vitamin. However, its production does require ultraviolet light, which an individual must obtain through exposure to the sun.  b.  Because vitamin D synthesis requires exposure to sunshine, which is less abundant at higher latitudes, the prevalence of the autoimmune disease increases with increasing latitude.

47.  b is polar, d is nonpolar, and a, c, and e are amphipathic.

CH2

CH

41.  Oxidation of retinal (an aldehyde) yields retinoic acid (a carboxylic acid).

45.  Bacteria provide about half of an individual’s daily vitamin K requirement. Prolonged use of antibiotics may kill these beneficial vitamin K–producing bacteria as well as the disease-causing bacteria, resulting in a vitamin K deficiency.

DPPC

O

37.  The spicy ingredient in the food is a powder made from peppers that contains the hydrophobic compound capsaicin (see Section 8.1). Yogurt containing whole milk also contains hydrophobic ingredients that can cleanse the palate of the irritating capsaicin. Water is polar, so it does not dissolve the capsaicin and cannot cleanse the palate.

(CH2)2

CH C

O

33.  Both DNA and phospholipids have exposed phosphate groups that are recognized by the antibodies. 35.  Archaeal lipids consist of a glycerol backbone with fatty acyl chains attached via an ether linkage rather than an ester linkage. The ether linkage is not hydrolyzed as easily as an ester linkage, which accounts for the greater stability of the archaeal lipids at high temperatures.

51.  Lipids that form bilayers are amphipathic, whereas triacylglycerols are nonpolar. Amphipathic molecules orient themselves so that their polar head groups face the aqueous medium on the inside and outside of the cell. Also, triacylglycerols, which lack a large head group, are cone-shaped rather than cylindrical and thus would not fit well in a bilayer structure. 53.  Two factors that influence the melting point of a fatty acid are the number of carbons and the number of double bonds. Double bonds are a more important factor than the number of carbons, since a significant change in structure (a “kink”) occurs when a double bond is introduced. An increase in the number of carbons increases the melting point, but the change is not nearly as dramatic. For example, the melting point of palmitate (16:0) is 63.1°C, whereas the melting point of stearate (18:0) is only slightly higher at 69.1°C. However, the melting point of oleate (18:1) is 13.4°C, a dramatic decrease with the introduction of a double bond. 55.  The melting point of elaidic acid is higher than the melting point of oleic acid, since the trans double bond in elaidic acid gives it an elongated shape, whereas the cis double bond in oleic acid gives it a bent shape. 57.  The melting point for 9-octadecynoic acid is 48°C, which is less than the melting point of stearic acid (69.1°C) and greater than the melting point of oleic acid (13.4°C). The triple bond does not produce

OD D -NUMB ER ED SO LUTI O NS  S-27 a kink in the fatty acid the way a cis double bond does, but the presence of the triple bond affects the shape of the molecule so that it can’t pack together with its neighbors as effectively as stearic acid. Thus the triple-bonded fatty acid has a melting point between that of stearic acid and oleic acid, but closer to that of stearic acid. 59.  In general, animal triacylglycerols contain longer and/or more saturated acyl chains than plant triacylglycerols, since these chains have higher melting points and are more likely to be in the crystalline phase at room temperature. The plant triacylglycerols contain shorter and/or less saturated acyl chains in order to remain fluid at room temperature. 61.  The lipids from the meat of the reindeer slaughtered in February contained fewer unsaturated acyl chains than their healthy counterparts. Unsaturated fatty acids have a lower melting point due to the presence of double bonds that prevent them from packing together tightly. These lipids therefore help membranes remain fluid even at low temperatures as the reindeer walks through the snow. Saturated fatty acids, which pack together more efficiently and have higher melting points, decrease membrane fluidity at low temperatures. The decreased percentage of lipids with unsaturated fatty acyl chains results in decreased membrane fluidity and may compromise the ability of the animal to survive a cold winter. [From Soppela, P. and Nieminen, M., Comp. Biochem. Physiol. 128, 63–72 (2001).] 63.  The cyclopropane ring in lactobacillic acid produces a bend in the aliphatic chain and thus its melting point is closer to the melting point of oleate, which also has a bend due to the double bond. The presence of bends decreases the opportunity for London dispersion forces to act among neighboring molecules. Less heat is required to disrupt the intermolecular forces, resulting in a melting point that is lower than that of a saturated fatty acid with a similar number of carbons. Therefore stearate has the highest melting point (69.6°C), followed by lactobacillic acid (28°C) and oleate (13.4°C). 65.  Cholesterol’s planar ring system interferes with the movement of acyl chains and thus tends to decrease membrane fluidity. At the same time, cholesterol prevents close packing of the acyl chains, which tends to prevent their crystallization. The net result is that cholesterol helps the membrane resist melting at high temperatures and resist crystallization at low temperatures. Therefore, in a membrane containing cholesterol, the shift from the crystalline form to the fluid form is more gradual than it would be if cholesterol were absent. 67.  The cold temperatures might induce the activation of desaturase enzymes in the plant that convert the 18:0, 18:1, and 18:2 fatty acids to 18:3 fatty acids. Of these four fatty acids, 18:3 is the most unsaturated and has the lowest melting point. Membranes composed of phospholipids containing unsaturated fatty acids maintain their fluidity in cold temperatures. [From Shi, Y. et al., Protoplasma 232, 173–181 (2008).] 69. a.  Detergents are required to solubilize a transmembrane protein because the protein domains that interact with the nonpolar acyl chains are highly hydrophobic and would not form favorable interactions with water.  b.  A schematic diagram of the detergent SDS interacting with a transmembrane protein is shown below. The polar head group of SDS is represented by a circle and the nonpolar tail as a wavy line. The nonpolar tails of SDS interact with the nonpolar regions of the protein, effectively masking these regions from the polar solvent. The polar head groups of SDS interact favorably with water. The presence of the detergent effectively solubilizes the transmembrane protein so that it can be purified.

SDS

71.  a.

O HN

O

CH

C

b.

O HN

CH

CH2

CH2

O

O

C (CH2)7 CH

O

C

C (CH2)6 CH3

CH (CH2)5 CH3

73.  The membrane-spanning segment is a stretch of 19 residues (highlighted) that are all uncharged and mostly hydrophobic. [Source: UniProt] YPVTNFQKHMIGICVTLTVIIVCSVFIYKIFKID 75.  A steroid is a hydrophobic lipid that can easily cross a membrane to enter the cell. It does not require a cell-surface receptor, as does a polar molecule such as a peptide. 77. a.  The extracellular domain of the protein is hydrolyzed by trypsin, but the remainder of the protein is protected by the membrane. The SDS-PAGE gel shows a higher-molecular-weight protein band from the control cells than from the trypsin-treated cells.  b.  The cytosolic protein is protected from trypsin digestion because trypsin cannot cross the membrane. The control cells and the treated cells show identical results in SDS-PAGE.  c.  This protein is also protected from trypsin digestion and the results are the same as described for part b. 79. a.  Alcohols, ether, and chloroform are nonpolar molecules and can easily pass through the nonpolar portion of the lipid bilayer, the aliphatic acyl chains of the phospholipids. Salts, sugars, and amino acids are highly polar and would not be able to traverse the nonpolar portion of the membrane.  b.  Cells contain integral membrane proteins that serve as transporters. Proteins that transport water, known as aquaporins, transport the highly polar water molecule across the membrane. [From Kleinzeller, A., News Physiol. Sci. 12, 49–54 (1997).] 81.  A is a lipid-linked protein, B is an integral membrane protein, C and D are peripheral membrane proteins, and E is a GPI-anchored protein.

S-28  ODD- N UM B ER E D S O LUT I O NS 83.  After fusion, the green and red markers were segregated because they were bound to cell-surface proteins derived from the two different kinds of cells. Over time, the cell-surface proteins that could diffuse in the lipid bilayer became distributed randomly over the surface of the hybrid cell, so the green and red markers were intermingled. At 15°C, the lipid bilayer was in a gel-like rather than a fluid state, which prevented membrane protein diffusion. Edidin’s experiment supports the fluid mosaic model by demonstrating the ability of proteins to diffuse through a fluid membrane. [From Edidin, M., Nat. Rev. Mol. 4, 414–418 (2003).] 85. a.  Glycosphingolipids pack together loosely because their very large head groups do not allow tight association.  b.  The lipid raft is less fluid, because of the presence of both cholesterol and the saturated fatty acyl chains, which pack together more tightly than unsaturated acyl chains. [From Pike, L., J. Lipid Res. 44, 655–667 (2003).] 87. a.  PS and PE both contain amino groups.  b.  PC and SM both contain choline groups.  c.  PE, PC, and SM are all neutral, but PS carries an overall negative charge. Since PS is exclusively found on the cytosolic leaflet, this side of the membrane is more negatively charged than the other side.

Chapter 9 1. a.  Use Equation 9.2 and substitute the values for [Na+]in and [Na+]out: [Na+] ​ Δψ = 0.058 V log _________ ​  + in   ​  [Na ]out 10 × ​10​ −3​M  ​    Δψ = 0.058 V log ______________ ​    100 × ​10​ −3​M Δψ = −0.058 V = −58 mV​ [Na+]in b. ​ Δψ = 0.058 V log ​ _________    ​  [Na+]out −3    ​ Δψ = 0.058 V log _____________ ​ 40 × ​10​ −3​M  25 × ​10​  ​M Δψ = 0.012 V = 12 mV​ 3.  Use Equation 9.2 and solve for [Na +] in as shown in Sample ­Calculation 9.1: Δψ ​log ​​[Na+]in ​​  ​​ = _ ​     ​  ​​  ​​​ + log ​​[Na+]out 0.058 V ​​  ​​ = _ ​ −0.055 V    ​+ log ​(0.440)​ ​log ​​[Na+]in 0.058 V ​​  ​​ = −0.948 − 0.357 = −1.30​ ​log ​​[Na+]in ​​  ​​ ​​​[Na+]in

= 0.050 M = 50 mM​

[From Xu, N., J. Membrane Biol. 246, 75–90 (2013).] 5.  Use Equation 9.2 and substitute the value of the membrane potential: [Na+]in    ​  ​ Δψ = 0.058 V log ​ _________ [Na+]out [Na+]  ​  −0.070 V = 0.058 V log _________ ​  + in  [Na ]out [Na+]in −1.20 = log ​ _________    ​  [Na+]out

+

[Na ] ​  + in   ​  ​10​ −1.20​= _________ [Na ]out



+

[Na ]in _ ​ 0.062    ​​      ​ = _________ ​  1

+

[Na ]out

7.  Use Equation 9.4 and substitute the value of the ratio calculated in Solution 5: [Na+]in ​ ΔG = RT ln _________ ​     ​  + ZFΔψ [Na+]out

​ 0.062     ​  ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (310 K) ln _ 1 + (1) (96,485 × ​10​ −3​kJ  · ​V​​ −1​ · ​mol​​  −1​) (−0.070 V)



ΔG = −7.2 kJ · ​mol​​  −1​ − 6.8 kJ · ​mol​​  −1​



ΔG = −14 kJ · ​mol​​  −1​

 t the resting potential, the movement of Na+ ions into the cell is a A favorable process. 9.  Use Equation 9.4 and let Z = 1 for Na+ and 2 for Ca2+ and T = 293 K: [Na+] ​ ΔG = RT ln _________ ​  + in   ​  + ZFΔψ [Na ]out



10 × ​10​ −3​M  ​    ​    ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (293 K) ln ___________ 450 × ​10​ −3​M + (1) (96,485 × ​10​ −3​kJ · ​V​ −1​ · ​mol​​  −1​) (−0.070 V)



ΔG = −9.27 kJ · ​mol​​  −1​ − 6.75 kJ · ​mol​​  −1​



ΔG = −16.0 kJ · ​mol​​  −1​



[Ca2+]in ΔG = RT ln ​ __________    ​  + ZFΔψ [Ca2+]out





 −6​M  ​     ​ 0.1 × ​10​ ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (293 K) ln ______________ −3 4 × ​10​  ​M + (2) (96,485 × ​10​ −3​kJ · ​V​ −1​ · ​mol​​  −1​) (−0.070 V)



ΔG = −25.8 kJ · ​mol​​  −1​ − 13.5 kJ · ​mol​​  −1​



ΔG = −39.3 kJ · ​mol​​  −1​



 ince the concentrations of both ions are greater outside the cell than S inside and the cell potential is negative, the passive movement of the ions will be from outside the cell to inside. In order to maintain the ion concentrations given in the problem, energy-consuming active transport processes are required. 11.  Use Equation 9.4 and let Z = 2 and T = 310 K: [Ca2+]in a.  ​ΔG = RT ln ​ __________    ​  + ZFΔψ [Ca2+]out

 −6​M  ​     ​ 0.1 × ​10​ ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (310 K) ln ______________ −3 1 × ​10​  ​M + (2) (96,485 × ​10​ −3​kJ  · ​V​​ −1​ · ​mol​​  −1​) (−0.05 V)



ΔG = −23.7 kJ · ​mol​​  −1​ − 9.6 kJ · ​mol​​  −1​



ΔG = −33.3 kJ · ​mol​​  −1​

 he negative value of ∆G indicates a thermodynamically favorable T process, indicating that movement of Ca2+ ions is more favorable from the ER to the cytosol. [Ca2+]in b.  ​ΔG = RT ln ​ __________    ​  + ZFΔψ [Ca2+]out

 −6​M  ​     ​ 0.1 × ​10​ ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (310 K) ln ______________ 1 × ​10​ −3​M + (2) (96,485 × ​10​ −3​kJ ·​V​​ −1​ · ​mol​​  −1​) (+0.05 V)



ΔG = −23.7 kJ · ​mol​​  −1​ + 9.6 kJ · ​mol​​  −1​



ΔG = −14.1 kJ · ​mol​​  −1​

 he negative value of ∆G indicates a thermodynamically favorable T process, but not as favorable as in part a.

OD D -NUMB ER ED SO LUTI O NS  S-29 13.  Use Equation 9.4 and let Z = 1 and T = 293 K:



[K+] ​ ΔG = RT ln ________ ​  + in   ​  + ZFΔψ [K ]out

 −3​M  ​  ΔG = (8.3145 × ​10​ −3​kJ  · ​K​​  −1​ · ​mol​​  −1​) (310 K) ln ___________    ​ 0.5 × ​10​ 4 × ​10​ −3​M



ΔG = −5.4 kJ · ​mol​​  −1​

−3    ​ ​ 50 × ​10​ −3​M  ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (293 K) ln _____________ 15 × ​10​  ​M + (1) (96,485 × ​10​ −3​kJ ·​V​​ −1​ · ​mol​​  −1​) (−0.05 V)



−1

−1



ΔG = 2.9 kJ · ​mol​​  ​ − 4.8 kJ · ​mol​​  ​



ΔG = −1.9 kJ · ​mol​​  −1​

 + ions move spontaneously from the outside of the cell to the K cytosol. 15.  Use Equation 9.4 and let Z = –1 and T = 310 K: −

[​Cl​​  ​]in ​ ΔG = RT ln _______ ​     ​  + ZFΔψ [​Cl​​  −​]out



5 × ​10​ −3​M  ΔG = (8.3145 × ​10​ −3​kJ · ​K​ −1​ · ​mol​​  −1​) (310 K) ln ______________  ​ ​    120 × ​10​ −3​M + (−1) (96,485 × ​10​ −3​kJ  · ​V​​ −1​ · ​mol​​  −1​) (−0.05 V)



ΔG = −8.2 kJ · ​mol​​  −1​ + 4.8 kJ · ​mol​​  −1​



ΔG = −3.4 kJ · ​mol​​  −1​

This process is spontaneous. 17. a.  Since all the terms on the right side of Equation 9.1 are constant, except for T, the following proportion for the two temperatures (310 K and 313 K) applies: Δψ _ ​ ​ −70 mV        ​= ​ _ ​ 310 K 313 K

Δψ = −70.7 mV​

 he difference in membrane potential at the higher temperature T would not significantly affect the neuron’s activity.  b.  It is more likely that an increased temperature would increase the fluidity of cell membranes (see Section 8.2). This in turn might alter the activity of membrane proteins, including ion channels and pumps, which would have a more dramatic effect on membrane potential than temperature alone.

23.  The less polar a substance, the faster it can diffuse through the lipid bilayer. From slowest to fastest: C, A, B. 25. a.  Glucose has a slightly larger permeability coefficient than mannitol and therefore moves across the synthetic bilayer more easily.  b.  Both solutes have higher permeability coefficients for the red blood cell membrane, indicating that transport is occurring via a protein transporter rather than diffusion through the membrane. The transporter binds glucose specifically and transports it rapidly across the membrane whereas it is less specific for mannitol and transports it less effectively. 27. a.  Phosphate ions are negatively charged and lysine side chains most likely carry a full positive charge at physiological pH. It is possible that an ion pair forms between the phosphate and the lysine side chains and that the lysine side chains serve to funnel the phosphate ions through the porin.  b.  If the hypothesis described in part a is correct, the replacement of lysines with the negatively charged glutamates would abolish phosphate transport by the porin, due to charge– charge repulsion. Possibly the mutated porin might even transport positively charged ions instead of phosphate. [From Sukhan, A. and Hancock, R.E.W., J. Biol. Chem. 271, 21239–21242 (1996).] 29. a.  Acetylcholine binding triggers the opening of the channel, an example of a ligand-gated transport protein.  b. Na+ ions flow into the muscle cell, where their concentration is low.  c.  The influx of positive charges causes the membrane potential to increase. 31.  The transfer to pure water increases the influx of water by osmosis, and the cell begins to swell. Swelling, which puts pressure on the cell membrane, causes mechanosensitive channels to open. As soon as the cell’s contents flow out, the pressure is relieved and the cell can return to its normal size. Without these channels, the cell would swell and burst. 33. a.

H N

 −3​M  ​     ​ 0.5 × ​10​ ΔG = (8.3145 × ​10​ −3​kJ  · ​K​​  −1​ · ​mol​​  −1​) (293 K) ln ___________ −3 5 × ​10​  ​M ΔG = −5.6 kJ · ​mol​​  −1​ [glucose]in b. ​ΔG = RT ln ____________ ​       ​ [glucose]out 5 × ​10​ −3​M   ​ ​    ΔG = (8.3145 × ​10​ −3​kJ  · ​K​​  −1​ · ​mol​​  −1​) (293 K) ln ___________ 0.5 × ​10​ −3​M ΔG = 5.6 kJ · ​mol​​  −1​ 21.  Use Equation 9.3: [Glucose]in a. ​ΔG = RT ln ____________ ​       ​ [Glucose]out  −3​M ​     ​ 0.5 × ​10​−3 ΔG = (8.3145 × ​10​ −3​kJ  · ​K​​  −1​ · ​mol​​  −1​) (310 K) ln ___________ 15 × ​10​  ​M ΔG = −8.8 kJ · ​mol​​  −1​ [Glucose]in b. ​ ΔG = RT ln ​ ____________      ​ [Glucose]out

CH

C

CH2 δ– O

19.  Use Equation 9.3 and substitute the values for extracellular and cytosolic glucose: [glucose]in a. ​ΔG = RT ln ​ ____________      ​ [glucose]out

O

H δ+ δ– N

H δ+

δ– O

–

Cl

CH2

H δ+

b.  The hydroxyl, phenol, and amido groups act as proton donors to coordinate the negatively charged chloride ion. The hydrogens are bonded to the more electronegative oxygen and nitrogen atoms and are therefore partially positively charged (denoted as δ+ in the diagram) and can attract the chloride ion. Cations could not interact with the hydrogens and so would be excluded from the ion channel. 35. a.  A transport protein, like an enzyme, carries out a chemical process (in this case, the transmembrane movement of glucose) but is not permanently altered in the process. Because the transport protein binds glucose, its rate does not increase in direct proportion to increasing glucose concentration, and it becomes saturated at high glucose concentrations.  b.  The transport protein has a maximum rate at which it can operate (corresponding to Vmax, the upper limit of the curve; see the diagram). It also binds glucose with a characteristic affinity (corresponding to KM, the glucose concentration at half-maximal velocity). The estimated Vmax for this transporter is about 0.8 × 106 mM ∙ cm−2 ∙ s−1 and the KM is about 0.5 mM (see diagram).

Glucose flux (mM . cm−2 . s−1 × 106 )

S-30  ODD- N UM B ER E D S O LUT I O NS Vmax

x intercept (mM−1)

1.0

KM

0

Without inhibitor

–0.119

8.4

1.6

With phlorizin

–0.047

21.2

1.5

b.  Phlorizin is a competitive inhibitor. The Vmax is nearly unchanged while the KM increases nearly threefold in the presence of the inhibitor. Phlorizin competes with glucose for binding to the transporter. [From Betz, A.L. et al., Biochim. Biophys. Acta 406, 505–515 (1975).]

0.5

0

Vmax KM (mM) (µmol · min−1 · g−1)

2

6 4 [Glucose] (mM)

8

10

47.

H N

O C

CH CH2

37.  Intracellular exposure of the glucose transporter to trypsin indicates that there is at least one cytosolic domain of the transport protein that is essential for glucose transport. Hydrolysis of one or more peptide bonds in this domain(s) abolishes glucose transport. However, extracellular exposure of the ghost transporter to trypsin has no effect, so there is no trypsin-sensitive extracellular domain that is essential for transport. This experiment also shows that the glucose transporter is asymmetrically arranged in the erythrocyte membrane. 39.  As the glutamate (charge –1) enters the cell, four positive charges also enter (3 Na+, 1 H+) for a total of three positive charges. Since 1 K+ exits the cell at the same time, two positive charges are added to the cell for each glutamate transported inside. 41. a.  CO2 produced by respiring tissues enters the red blood cell and combines with water to form carbonic acid, which then dissociates to form H+ and ​​HCO​ 3−​  ​​ ions. The ​​HCO​ 3−​  ​​ions are transported out of the cell by Band 3 in exchange for Cl− ions, which enter the cell. The ​​HCO​ − 3​  ​​ions travel through the circulation to the lungs, where they recombine with H+ ions to form carbonic acid, which subsequently dissociates to form water and CO2. The CO2 is then exhaled in the lungs.  b.  When red blood cells pass through tissues, ​​HCO​ 3−​  ​​ ions are transported out of the cell by Band 3. However, in the lungs, ​​HCO​ 3−​  ​​ ions are transported into red blood cells by Band 3 (in exchange for Cl− ions, which exit the cell). The intracellular ​​HCO​ 3−​  ​​ ions combine with H+ ions that dissociate from hemoglobin (promoting conversion of hemoglobin from the deoxy T form to the oxy R form), forming carbonic acid. The carbonic acid dissociates into water and CO 2, which diffuses out of the cell and is exhaled. 43. a.  Carbonic anhydrase catalyzes the conversion of CO2 and H2O to carbonic acid, which dissociates to form H+ and ​​HCO​ 3−​  ​​. The ​​HCO​ 3−​  ​​ enters the cell via the bicarbonate transporter to buffer the intracellular pH and to provide the alkaline environment preferred by the cancer cell.  b.  Drug targets include the GLUT1 transport protein, which transports glucose into the cell (however, this transporter is also active in normal cells), depriving the cancer cell of needed glucose. Inhibiting the MCT4 protein that exports lactic acid results in a decrease in intracellular pH that is unfavorable to the cancer cell. Inhibiting carbonic anhydrase (again, this enzyme functions in normal cells) prevents the formation of ​​ HCO​ −3​  ​​. Blocking the bicarbonate transporter prevents ​​HCO​ −3​  ​​ from entering the cell to produce the favorable intracellular alkaline environment. [From McDonald, P.D. et al., Oncotarget 3, 84–97 (2012).] 45. a.  The values are shown in the table below. The KM is calculated by taking the negative reciprocal of the x intercept (which is equal to –b/m; see Solution 7.49). The Vmax is the reciprocal of the y intercept.

C

O

O –O

P

O

O–

Aspartyl phosphate 49.  The proline transporter can bind l-hydroxyproline but not d-proline, indicating that the transporter is stereoselective. The transporter does not co-transport proline with H+ ions. Because the rate of transport is not affected by Na+ or K+ it is unlikely that a secondary active transport mechanism is used. The proline transporter does not have a binding site for ouabain as the Na,K-ATPase does (see ­Problem 48). The proline transporter likely uses an active transport mechanism that requires energy in the form of ATP, since intracellular ATP depletion decreases the rate of transport. [From L ­ ’Hostis, C.L. et al., Biochem. J. 291, 297–301 (1993).] 51. a.  Glucose uptake increases as sodium concentration increases in pericytes. In endothelial cells, glucose uptake is constant regardless of sodium ion concentration.  b.  The shape of the curve for the pericytes indicates that a protein transporter is involved. Glucose uptake initially increases as sodium ion concentration increases, then reaches a plateau at high sodium ion concentration, indicating that the transporter is saturated and is operating at its maximal capacity.  c.  It is likely that the pericytes use secondary active transport to import glucose. Sodium ions and glucose molecules enter the cell in symport. The sodium ions are then ejected from the cell by the Na,K-ATPase transporter. 53.  The ABC transporters bind ATP, then undergo a conformational change as the ATP is hydrolyzed and P i is released, leaving ADP. Vanadate, a phosphate analog, may serve as a competitive inhibitor by binding to the phosphate portion of the ATP binding site. With ATP unable to bind, the necessary conformational change cannot occur and the transporter is inhibited. 55.  Both transporters are examples of secondary active transport. The H+/Na + exchanger uses the free energy of the Na+ gradient (established by the Na,K-ATPase) to remove H+ from the cell as Na+ enters. Similarly, a preexisting Cl− gradient (see Fig. 2.12) allows the − cell to export ​​HCO​ − 3​  ​​ as Cl enters. 57.  Di- and tripeptides enter the cell in symport with H+ ions. The H+ ions leave in exchange for Na+ via the antiport protein. The Na+ ions are ejected via the Na,K-ATPase (see diagram). This is an example of secondary active transport in which the expenditure of ATP by the Na,K-ATPase pump is the driving force for peptide entry into the cell.

OD D -NUMB ER ED SO LUTI O NS  S-31

Peptide

H+

H+

cone-shaped, thereby increasing bilayer curvature, which is a necessary step in the formation of a new vesicle by budding.

Na+

71.  Autophagy degrades cellular macromolecules to their monomeric units. In this way, cells can convert existing proteins and stored polysaccharides to small molecules that can be used as metabolic fuels, rather than building blocks, in order to keep the cell alive. H+

73.  Receptor-mediated endocytosis reaches a maximum when all receptor sites are occupied, whereas the rate of pinocytosis is proportional to the amount of substance internalized. Receptor-mediated endocytosis is a far more efficient way to deliver substances to the interior of the cell.

Na+

3 Na+

2 K+

Concentration of substance internalized

Peptide

H+

ATP

3 Na+

2 K+

59.  Acetylcholinesterase inhibitors prevent the enzyme from breaking down acetylcholine. This increases the concentration of acetylcholine in the synaptic cleft and increases the chances that acetyl‑ choline will bind to a dwindling number of receptors in the postsynaptic cell. [From Thanvi, B.R. and Lo, T.C.N., Postgrad. Med. J., 80, 690–700 (2004).] 61.

Receptor-mediated endocytosis

Pinocytosis

Time

CH(CH3)2 O

Acetylcholinesterase +

F

CH2OH

P

O

DIPF

O CH(CH3)2 HF

(H3C)2HC

O

P

1.  Signaling molecules that are lipids (cortisol and thromboxane) or very small (nitric oxide) can diffuse through lipid bilayers and do not need a receptor on the cell surface. 3.  Because 90% of the receptors are occupied, [R · L] = 22.5 mM and [R] = 2.5 mM. Use Equation 10.1 to calculate Kd: [R] [L] ​   ​  ​ ​K ​ d​= _ [R · L]

CH2 O

Chapter 10

Modified enzyme O

O CH(CH3)2

63.  Serotonin recycling depends on a transporter that uses the free energy of the Na+ gradient to move serotonin back into the cell. The Na+ gradient is established through the action of the Na,K-ATPase. 65.  No, preventing serotonin reuptake would prolong its signaling potential (thereby boosting serotonin’s mood-enhancing effects), but blocking its receptor would be expected to have the opposite effect (it would not act as an antidepressant).



(2.5 × ​10​ −3​ M) (125 × ​10​ −6​ M ) ​K ​ d​= _____________________   ​ ​        22.5 × ​10​ −3​ M



​K ​ d​= 1.4 × ​10​ −5​M = 14 μM​

5.  Use Equation 10.1 and solve for [R · L]: [R] [L] ​   ​  ​ ​K ​ d​= _ [R · L]

[R] [L] [R · L] = _ ​      ​  K d



(5 × ​10​ −3​ M) (18 × ​10​ −3​ M)   ​    [R · L] = ___________________ ​     3 × ​10​ −3​ M



[R · L] = 30 × ​10​ −3​M = 30 mM​

67.  The tetanus toxin cleaves the SNAREs, which are required for the fusion of synaptic vesicles with the neuronal plasma membrane. This prevents the release of acetylcholine, interrupting communication between nerves and muscles and causing paralysis.

7.  Let [R · L] = x and [R] = 0.010 – x.

69.  By adding a phosphate group, the kinase increases the size and negative charge of the lipid head group, which then occupies a larger volume and more strongly repels neighboring negatively charged lipid head groups. The phosphatidylinositol would become more



[R] [L] ​   ​  ​ ​K ​ d​= _ [R · L]



[R] [L] [R · L] = x = _ ​        ​ ​K ​ d​  (0.010 − x) (2.5 × ​10​ −3​ M)   ​        x = ___________________ ​  1.5 × ​10​ −3​ M

S-32  ODD- N UM B ER E D S O LUT I O NS (0.000025 − 0.0025x) x = _______________    ​       ​ 0.0015



0.0015x = 0.000025 − 0.0025x





0.0040x = 0.000025





[L] 136,000 ____________ _ ​       ​  ​= ​   



x = 0.00625 = 6.25 mM = [R · L]​

 he percentage of receptors occupied by ligand is 6.25 mM ÷ 10 mM T or 62.5%. 9.  The Kd estimated from the curve is about 0.1 mM.

160,000

0.35 μM + [L] 0.85(0.35 μM​+ [L]) = [L] 0.2975 μM + 0.85[L] = [L]



0.2975 μM = 0.15[L]



[L] = 2.0 μM​

17.  The Kd is obtained from the x-intercept in the double-reciprocal plot. 10

1/[R . L] (μM−1)

[R. L]/[R]T

1.0

0.6

Kd

0.2 0.1

−1/Kd

0.2

0.3 0.4 [L] (mM)

0.5

0.6





−4

−3

−2

−1

([R]T − [R · L] ) [L] ​        ​ Kd = _____________ [R · L] Kd[R · L] = [R]T [L] − [R · L] [L] (Kd[R · L]) + ([R · L] [L]) = [R]T [L]



[R · L] (Kd + [L]) = [R]T [L]



0.10([L] + 1.0 × ​10​  ​  ​M) = [L] 0.10[L] + 1.0 × ​10​ −​ 11​M = [L]



−10 

[L] + 1.0 × ​10​ 



2

3

4

Kd = 0.5 μM​



19.  ​ slope = −  _ ​  1   ​ ​Kd​  ​ ​  1   ​ − 0.067 n​M​ −1​= −  _ ​Kd​  ​

​Kd​  ​= 15 nM​

[ From Hubbard, M.J. and Klee, C.B. J. Biol. Chem. 262, 15062–15070 (1987).] ​ M

− 10



1

(μM−1)

− 2 μ​M​ −1​= − _ ​  1   ​ Kd





[L] _  ​ ​  100      ​= ______________ ​    1000

1/[L]



[L] [R · L] _ _    ​= ​       ​ ​  [R]T Kd + [L]

[L] [R · L] _ _    ​= ​       ​ 13.  ​ ​  [R]T [L] + Kd

0

​ x-int = − _ ​  1    ​ K d

[R] = [R]T − [R · L]



4

0.7

[R]T = [R] + [R · L]



6

2

[R] [L] 11.  ​ ​Kd​  ​= _ ​   ​  [R · L]

8

1.0 × ​10​ −11​M = 0.9[L] [L] = 1.11 × ​10​ −11​M​

15. a. The binding site with a Kd of 0.35 µM is the high-affinity binding site and the site with a Kd of 7.9 µM is the low-affinity binding site. Kd is the ligand concentration at which the receptor is half saturated with ligand; therefore the lower the Kd, the lower the concentration of ligand required to achieve half-saturation.  b.  The high-affinity binding site with a Kd of 0.35 µM is most effective in the 0.1–0.5 µM range, because at the upper limit of this range, the high-affinity binding sites will be more than 50% occupied, whereas the low-affinity sites will be less than 50% occupied.  c.  Both of these agonists can compete at high concentrations but the methylthio-ADP has a lower Kd and will be a more effective inhibitor at low concentrations.  d.  Use the equation derived in Solution 11. [From Jefferson, J.R. et al., Blood 71, 110–116 (1988).] [R · L] _ [L] ​ ​ _    ​= ​       ​ [R]T Kd + [L]

21.  Cell-surface receptors are difficult to purify because they are usually integral membrane proteins and require the addition of detergents to dissociate them from the membrane (see Solution 8.69). The receptor proteins constitute a very small proportion of all of the proteins in the cell; this makes it difficult for the experimenter to isolate the receptor protein from other cellular proteins. 23.  The different types of G protein–linked receptors are found in different types of cells. The cellular response elicited when a ligand binds to a receptor depends on how that particular cell integrates and processes the signal. Different cells have different intracellular components, which results in different responses to what appears to be the same signal. 25.  If receptors are removed from the cell surface, the ligand cannot bind and an intracellular response cannot occur. 27.  a. 

H N

O CH

C

CH2 S C O

(CH2)14

CH3

OD D -NUMB ER ED SO LUTI O NS  S-33 b.  The GPCR is a lipid-linked protein. The palmitoyl group interacts with the fatty acyl chains of membrane phospholipids, anchoring the receptor to the membrane.  c.  If the Cys residue is mutated to a Gly, the palmitoyl chain cannot be attached. This might interfere with the localization of the receptor to the membrane and result in the loss of its function. 29.  The synthesis of the GPCR may be impaired in such a way that it is not targeted to the cell membrane; the receptor may be present on the cell membrane but a mutation alters the binding site so that the ligand fails to bind; the mutation may render the GPCR unable to interact with a G protein, or the GPCR may be more sensitive to phosphorylation by the GPCR kinase, allowing arrestin to bind more easily. 31.  Stimulation of GTPase activity by RGS accelerates the hydrolysis of GTP to GDP, converting the receptor-associated G protein into its inactive form more rapidly. This shortens the duration of signaling. 33.  Both hormones lack tyrosine’s carboxylate group and have hydroxyl groups attached to the ring and to the β carbon. In epinephrine, the amino nitrogen bears a methyl group. 35.  The inhibition of the intrinsic GTPase activity results in a continuously active G protein. This increases the activity of adenylate cyclase, which results in an increase in the concentration of intracellular cAMP. In intestinal cells, the increase cAMP concentration leads to the loss of water and electrolytes from the cells and results in diarrhea that can be fatal. 37. a. GTP γS can bind to a G protein, but since it cannot be hydrolyzed, the G protein is in a persistently active state. If GTP γS binds to a stimulatory G protein, adenylate cyclase is continually active, which increases the cellular cAMP concentration.  b.  If GTP γS binds to an inhibitory G protein, adenylate cyclase is continually inhibited and the cellular cAMP concentration decreases. 39.  The phosphate group is large and bulky and may cause a conformational change that alters the activity of the protein. The negatively charged phosphate group can interact with other amino acid side chains in the protein, forming either hydrogen bonds or ion pairs that change protein structure. The phosphate group may also serve as a recognition site that allows other proteins to bind to the phosphorylated protein. 41.  Because of their structural similarity to diacylglycerol, phorbol esters stimulate protein kinase C, as diacylglycerol does. Increased protein kinase C activity leads to an increase in the phosphorylation of the kinase’s cellular targets. Because protein kinase C phosphorylates proteins involved in cell division and growth, the addition of phorbol esters can have profound effects on the rates of cell division and growth when added to cells in culture. 43.  The T cell is stimulated when an extracellular ligand binds to a G protein–linked receptor and activates phospholipase C. The activated phospholipase C catalyzes the hydrolysis of phosphatidylinositol bisphosphate, yielding diacylglycerol and inositol trisphosphate. The inositol trisphosphate binds to channel proteins in the endoplasmic reticulum and allows calcium ions to flow into the cytosol. Calcium ions then bind to calmodulin, causing a conformational change that allows it to bind and activate calcineurin. The activated calcineurin then activates NFAT as described in the problem. 45.  Overexpression of PTEN in mammalian cells promotes apoptosis. PTEN removes a phosphate group from inositol trisphosphate, and when this occurs, inositol trisphosphate is no longer able to activate protein kinase B. In the absence of protein kinase B, cells are not stimulated to grow and proliferate and instead undergo apoptosis. 47. a. Upon stimulation by an action potential, acetylcholinecontaining synaptic vesicles in neurons fuse with the plasma

membrane and release their contents into the synaptic cleft (see Section 9.4). Acetylcholine then diffuses across the synaptic cleft to the endothelial cell.  b.  Acetylcholine binding to G protein–linked cell-surface receptors in the endothelial cell leads to activation of phospholipase C, which hydrolyzes phosphatidylinositol bisphosphate to diacylglycerol and inositol trisphosphate. The inositol trisphosphate binds to calcium channels in the endoplasmic reticulum, which opens the channels and floods the cell with Ca2+. Calcium ions bind to calmodulin, changing its conformation and allowing it to bind to NO synthase to activate the enzyme.  c. Guanylate cyclase catalyzes the formation of cGMP from GTP (see Solution 38). It is possible that cGMP activates protein kinase G in a manner analogous to cAMP activating protein kinase A; that is, cGMP binding could displace regulatory subunits from protein kinase G to release active catalytic subunits. The active protein kinase G would next phosphorylate proteins involved in the muscle contraction process, perhaps myosin or actin, resulting in smooth muscle relaxation. 49.  If NO synthase is missing, the signaling pathway described in Solution 47 cannot be completed. NO cannot be synthesized, and subsequent steps, including the production of the second m ­ essenger cGMP and activation of protein kinase G, do not occur. Protein kinase G acts on muscle in such a way that the muscle relaxes. If this does not occur, muscles lining the blood vessels are constricted, resulting in high blood pressure. This makes it more difficult for the heart to pump blood through the circulatory system, leading to an increased heart rate and increased size of the ventricular chambers. 51.  Nitroglycerin decomposes to form NO, which passes through cell membranes in tissues of the tongue to enter the bloodstream. NO activates guanylate cyclase in smooth muscle cells, as described in Solution 47, producing cyclic GMP, which subsequently activates protein kinase G. The kinase phosphorylates proteins involved in muscle contraction, which leads to the relaxation of the smooth muscle cell. This increases blood flow to the heart and relieves the pain associated with angina. 53.  This molecule, inositol hexakisphosphate (inositol with six attached phosphoryl groups; also known as phytate), mimics a phosphorylated G protein–coupled receptor, binding to the Arg and Lys side chains of arrestin and activating it. 55.  Thrombin

Gα GPCR

PLC

DAG

PIP2

ER

PKC Ca2+

MLCK

MLC

Pi

Pi

Granules

platelet aggregation

exocytosis

57. a. Adenylate cyclase generates the second messenger cAMP in response to activation of G proteins by G protein–coupled receptors. The EF toxin generates large amounts of cAMP in the absence of any specific hormone signal.  b.  When Ca2+-calmodulin is bound to EF,

S-34  ODD- N UM B ER E D S O LUT I O NS it is not available to activate any other Ca2+-sensitive proteins that might be involved in normal cell signaling. 59.  Since the growth factor stimulates kinase activity, the H2O2 second messenger is likely to produce similar responses, so it must inactivate the phosphatases. 61.  Phosphatases that remove phosphate groups from the insulin receptor would turn the insulin signaling pathway off and protein kinases B and C would not be activated. Without active protein kinase B, glycogen synthase is inactive and glycogen cannot be synthesized from glucose. Without protein kinase C, glucose t­ ransporters are not translocated to the membrane and glucose is not brought into the cell but remains in the blood. Drugs that act as inhibitors of these ­phosphatases would potentiate the action of the insulin receptor, allowing the receptor to remain active with a lower concentration of ligand, and thus are potentially effective treatments for diabetes.

the release of arachidonic acid from the membrane by C1P increases the concentration of substrate available for prostaglandin synthesis. One of the enzymes that catalyzes the first step in the production of prostaglandins is COX-2, which is stimulated by S1P. Both C1P and S1P can potentially increase production of prostaglandins, which accounts for their inflammatory properties, as shown in the diagram. 77.  S1P might use a variety of mechanisms to activate Ras, either through receptor tyrosine kinases or activation of protein kinase C. Ras then activates the MAP kinase pathway (see Problem 65), which leads to the phosphorylation of transcription factors that promote the expression of proteins involved in the cell cycle, ultimately leading to cell survival. 79. a. 

63. a. In the presence of GEF, the activity of the signaling pathway increases, since GEF promotes dissociation of bound GDP, and Ras · GDP is inactive. Once GDP has dissociated, GTP can bind and activate Ras. b. The opposite is true in the presence of GAP. Ras · GTP is active, but when the GTP is hydrolyzed to GDP, Ras is converted from the active to the inactive form. 65.  As noted in Problem 41, phorbol esters are diacylglycerol analogs that can activate protein kinase C. According to the information in this problem, protein kinase C activates the MAP kinase cascade, which leads to the phosphorylation of proteins that influence gene expression. When these genes are expressed, progression through the cell cycle is altered and cells are stimulated to grow and proliferate, a characteristic of tumor cells. 67.  Mutations in the receptor itself could result in a receptor that is continually phosphorylated. Mutations in Ras that do not allow GTP hydrolysis result in a persistently active Ras (see Solution 64). Mutations that inactivate Ras GAP or stimulate GEF (see Problem 63) activate Ras. Mutations in MEK or MAP kinase (see Problem 65) could alter the expression of gene regulatory proteins, transcription factors, and cell cycle proteins that allow the cancer cell to proliferate. Mutations in any of these proteins have the potential to activate the signaling pathway in the absence of growth factors. 69.  A PI3K inhibitor prevents the formation of PIP3 from PIP2. In the absence of PIP3, Akt (protein kinase B), which promotes cell survival, is not activated and the cells undergo apoptosis (programmed cell death). Thus, inhibition of PI3K results in the death of the cancer cell and is an effective treatment. 71.  In order to become activated, two inactive PKR proteins must come close enough to phosphorylate each other (autophosphorylation). A long RNA molecule can bind two PKR proteins simultaneously, holding them in close proximity so that they can activate each other. Short RNA molecules prevent PKR activation because when a short RNA molecule occupies the PKR RNA-binding site, the PKR cannot bind to another RNA where it might encounter a second PKR and be phosphorylated. [From Nallagatla, S.R. et al., Curr. Opin. Struct. Biol. 21, 119–127 (2011).] 73.  Substances cannot enter the nucleus unless they possess a nuclear localization signal, a sequence that interacts with the nuclear pore and allows entry into the nucleus. The nuclear localization signal on the progesterone receptor must be exposed, even when ligand is not bound. But the nuclear localization signal on the glucocorticoid receptor must be masked. When ligand binds, a conformational change occurs that unmasks the nuclear localization signal, and the complex can pass through the nuclear pore and enter the nucleus. 75.  Arachidonic acid is the substrate for the production of prostaglandins, many of which have inflammatory properties. Stimulating

O

Enzyme

C

OH COCH3

+ CH2OH

O

Acetylsalicylic acid O

Acetylated enzyme

C CH2O

C

+

CH3

OH

OH

O

Salicylic acid

b.  Without knowing the mechanism of the enzyme, it is not possible to say for certain why acetylating the serine inhibits cyclooxygenase activity. However, it is possible that the acetylation alters the structure of the active site such that the arachidonic acid substrate is unable to bind. It’s also possible that serine participates in catalysis, possibly as a nucleophile, as in chymotrypsin. An acetylated serine would be unable to function as a nucleophile, which would explain why the modified enzyme is catalytically inactive.  c.  Because a covalent bond forms between the acetyl group and the Ser side chain, aspirin is an irreversible inhibitor (see Section 7.3). 81.  Phospholipase A 2 catalyzes the release of arachidonate from membrane phospholipids. Blocking this reaction would prevent the COX-catalyzed conversion of arachidonate to proinflammatory prostaglandins. 83.

High osmolarity (salt or glucose)

Ras cAMP

Raf MEK

PKA

MAP kinase

PFK2

Pi

Activation of glycolysis

Accumulation of glycerol

OD D -NUMB ER ED SO LUTI O NS  S-35

Chapter 11 3.  Coenzyme A, NAD, and FAD all contain ribose residues.

19.  All the sugar molecules will be converted to product because the α and β anomers are in equilibrium. Depletion of molecules in the α form will cause more of the β anomers to convert to α anomers, which will then be converted to product.

5.  Sugars b and d are epimers.

21. a. 

1. a. aldotetrose, b. ketopentose, c.  aldohexose,  d.  ketopentose.

7. a. 

CHO



b. 

COO–

CHO

H

C

OH

HO

C

H

HO

C

H

HO

C

H

HO

C

H

HO

C

H

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

H

OH

C

O–

OH

CH2O

CH2OH

P

CH2OH C

O

HO

C

H

HO

C

H

H

C

OH



b. 

O

H

C

OH

H

C

OH

HO

C

H

H

c. 



23.            

11.  [From Oku, T. et al., Nutr. Res. 34, 961–967 (2014).]

13. HO H

CH2OH O H OH H H

H

C

OH

HO

C

H

H

C

25. a. 

CHO OH

OH

HO

C

H

CH2OH

HO

C

H

H

C

OH

CH2OH O H OH H

HO H

H

OH

α-D-Galactose

OH

H

H

H

OH

H

H

H

OH

H

H

OH OH

α-D-Ribose

β-D-Ribose

b.  β-d-Ribose is found in RNA (see Section 3.1).

H HO

O H OH HO H

O–

P

H

H

H

C

OH

H

C

OH

C

O

C

H

H

C

OH

H

C

OH

O– H OH



COO–

b.  HO H

O H OH

H

H

OH

H OH

2–

CH2OPO3 C

O

HO

C

H

H

C

OH

H

C

OH

H

C

OH 2–

CH2OPO3

29. a.  The epimerization reaction converts a d sugar to an l sugar. CHO

O CH2O

C

H

27.           

OH

OH OH

17.         

HO

COO–

β-D-Galactose

HOCH2 O

H

OH

Galacturonate

15. a.  A five-membered ring results. HOCH2 O

C

CH2OH

C

OH

H

KDG

H

H

CH2OH

COO–

CH2OH O

OH

CH2OH

c.  Tagatose is less effectively absorbed in the small intestine because the transport proteins in the epithelial cells lining the small intestine do not bind and transport tagatose as efficiently as they do sugars that are more naturally and commonly present in the diet.

C

CH2OH O H OH H

Sorbitol

CH2OH

CH2OH

HO

Gluconate

O

CH2OH C

H

CH2OH

O–

9. a. 

b. 



H

C

OH

HO

C

H

H

C

OH

HO

C

H

COO–

b. 

H H HO

O COO– OH H H –O

OH H

O S O

O

S-36  ODD- N UM B ER E D S O LUT I O NS 31. a. 

CH2OH O H OH H

H HO

H

H

H

CH3I

H3CO

OH

H

OH

α-D-Glucose

b.

H3CO

H

43.       H HO

OCH3

CHO H

+,

H

C

OCH3

H3CO

C

H

H

C

OCH3

H

C

OH

H2O

CH3OH

OCH3

45. a.  H HO

H

HO

H

O

(CH2)15

CH3

H

OH

OH H2C

H HO

H

H O H

CH2OH O H OH H H

OH

OH H

OH

41.  Trehalase digestion produces glucose, which exists in solution as a mixture of the α and β anomers.

H HO

H

OH

H OH

H HO

CH2OH O H OH H H

H

OH

OH

OH H

H

CH2OH

HO

OH

H

b.  Isomaltulose is a reducing sugar. The anomeric carbon on fructose is not involved in a glycosidic bond, so the ring is free to open. Under the conditions of the Benedict’s test, fructose isomerizes to form an aldehyde that reduces Cu2+.  c.  Isomaltose contains the same residues as sucrose (glucose and fructose) but in sucrose, the two monosaccharides are linked by a (1→2) bond, whereas in isomaltulose the glycosidic bond is α(1→6). In sucrose, the fructose is in the β form, but in isomaltulose, the fructose is free to interconvert between anomers. [From Lina, B.A.R. et al., Food Chem.Toxicol. 40, 1375–1381 (2002).] 47.  The disaccharide is maltose.

H HO

CH2OH O H OH H H

 ellobiose is a reducing sugar. The anomeric carbon of the glucose C on the right side of the structure shown above is free to reverse the cyclization reaction to re-form the aldehyde functional group, which can be reduced.

CH2OH O H OH H

H

O

HO

O

H

37.  Lactose is a reducing sugar because it has an anomeric carbon (C1 of the glucose residue) that is free to reverse the cyclization reaction to re-form the aldehyde group, which can be reduced. Sucrose is not a reducing sugar because the anomeric carbons of both glucose and fructose are involved in forming the glycosidic bond. CH2OH O H OH H

H OH

OH

35.  The Maillard reaction (See Box 11.A) occurs between a carbohydrate carbonyl group and an unprotonated amino group. The alkaline dip helps ensure that reactive amino groups are in their basic form.

39.    

H H

O

33.  [From Hashim, R. et al., Liq. Cryst. (2003).]

H

CH2

H

2,3,4,6-Tetra-Omethyl-D-glucose

CH2OH O H OH H

O

OH

CH2OH O H OH H

CH2OCH3

1,2,3,4,6-Penta-Omethyl-D-glucose

CH2OH O H OH H H

1,2,3,4,6-Penta-Omethyl-D-glucose

CH2OCH3 O H H OCH3 H OCH3

H

CH2OCH3 O H H OCH3 H OCH3

H

H O

CH2OH O H OH H H

OH

OH

51. a.  H HO

CH2OH O H OH H H

OH

H O

H

HO

CH2OH O H H OH

H O

H

HO

CH2OH O H H H

H

OH

49.  Four. Both starch and glycogen contain glucose residues linked by α(1→4) and α(1→6) glycosidic bonds. Sucrose consists of an α-glucose linked by a (1→2) glycosidic bond to β-fructose. Lactose consists of a β-galactose linked by a β(1→4) bond to glucose.

H OH

H

H OH

OH

b.  Humans are not able to digest the α(1→3) glycosidic bonds found in nigerotriose. 53.  Starch, glycogen, cellulose, and chitin are homopolymers. Peptido­ glycan and chondroitin sulfate are heteropolymers.

OD D -NUMB ER ED SO LUTI O NS  S-37 55. a.        

CH2OH O H

H

67.   

O

HO

H

HO

OH

HOCH2

CH2

59.  The pectin polysaccharide consists of galacturonate residues— some of which are methylated—linked via α(1→4) glycosidic bonds. Pectin also contains side chains composed of neutral sugars (not shown in the problem), such as rhamnose (see Box 11.B). H

H O

O

O

H OH

H

H

OH

O

H

H

P

O

H

OH

HO

CH2OH O H OH H H

O

R

NH

CH

CH C

H

75.          H HO

CH2OH O H OH HO H

O

H

H

H

O H OH

H

H

OH

O H

+

NH3

77. H HO

H O

OH

H

HN O P O–

NH C

H

O

CH2OH O H OH H H

O

CH2OH O H OH

n

CH2

n

H

HO

CH2

OH

R = H (Ser) R = CH3 (Thr)

O

CH3

H

CH2

H

NH C

OH

O

H

H

H

O –O

H

HO

73.  [From Schirm, M. et al., Anal. Chem. 77, 7774–7782 (2005).]

O

63.  Celery is mainly cellulose and water, neither of which provides nutritive calories. Humans do not have β-glucosidase enzymes and cannot hydrolyze the β-glycosidic bonds linking the glucose resides in cellulose. Since cellulose is not digested, the body does not spend any energy to further process it. Foods like celery contribute fiber to the diet, but these foods neither provide nor cost the body much in the way of energy. 65.        

H

71.  The N-linked saccharide is N-acetylglucosamine, and the bond has the β configuration. The O-linked oligosaccharide is N-acetylgal­ actosamine, and the bond has the α configuration.

57.  Three enzymes are required to digest the xyloglucan to its constituent monosaccharides. An α-galactosidase enzyme is needed to hydrolyze the α(1→2) bond between the galactose and xylose, a second enzyme is required to hydrolyze the α(1→6) bond between xylose and glucose, and a cellulase-like enzyme is required to hydrolyze the β(1→4) bonds between glucose residues.

H

H

O

O

69.  Heparin consists of sugar residues with negatively charged sulfate groups and therefore has a polyanionic character that is similar to that of DNA, so proteins that bind to DNA may also bind to heparin.

CH2

H OH

H H

b.  Humans do not have the enzymes to digest β(2→1) glycosidic bonds (although the bacteria that inhabit the small intestine possess the necessary enzymes and do have this capability). Food manufacturers often classify nondigestible carbohydrates as “fiber”; thus inulin extracted from chicory root is often added to processed foods to boost their perceived fiber content.

H

H

H

HO

H

61.       

HOCH2

O

O

H

OH H CH2OH O O H

 

O

CH3

O O

P

O O

CH2

O– H

N

O

H

H

OH

OH

H

79.  The residues of the disaccharide are glucuronate linked by a β(1→3) glycosidic bond to N-acetylgalactosamine-4-sulfate. Disaccharides are linked to each other by β(1→4) bonds. 81. a. The monosaccharide is N-acetylglucosamine.  b.  The amide bond forms between Ala and the carboxylate group on the C3 substituent in the disaccharide. The arrow indicates the linkage site.

S-38  ODD- N UM B ER E D S O LUT I O NS

H

CH2OH O H OH H H

NH

H O H C O

CH2OH O H H H

CH3

O

NH

b.   C

H CH3

O

n

CHCOO–

CH3

O– + +H3N

C

O

O

C

HCl

+ H2O

NH

O

15.  The pH optimum for pepsin is ~2, which is the pH of the stomach. The pH optimum for trypsin and chymotrypsin is ~7–8, as the small intestine is slightly basic (see Table 2.3). Each enzyme functions optimally in the conditions of its environment. 17.  Amino acids enter the cells lining the small intestine via secondary active transport. This system is similar to the process for glucose absorption shown in Figure 9.18. 

83. a.  Sialic acid is negatively charged. The presence of sialic acid on their surfaces weakens the attachment of the tumor cells to each other and may promote the detachment process.  b.  A drug could act as an inhibitor of one of the enzymes in the biochemical pathway for synthesizing sialic acid. Alternatively, foreign sialic acid precursors could be administered that would be taken up by the tumor cells and used for sialic acid synthesis. The use of a foreign precursor would result in the synthesis of a sialic acid derivative that is more immunogenic. [From Fuster, M.M. and Esko, J.D., Nat. Rev. Cancer 5, 526–542 (2005).]

Na+

Amino acid

Na-amino acid transporter

INTESTINAL SPACE

INTESTINAL CELL

Na+

Amino acid

K+ ATP

Chapter 12

ADP + Pi

1. a. chemoautotroph, b. photoautotroph, c. chemoautotroph,  d.  heterotroph, e. heterotroph, f. chemoautotroph, g. photoautotroph. 3.  The pH of the stomach is ~2 (see Table 2.3). At this pH, the salivary amylase is denatured and can no longer catalyze the hydrolysis of glycosidic bonds in dietary carbohydrates. 5.  Maltase is required to hydrolyze the α(1→4) glycosidic bonds in maltotriose and maltose (see Solution 4). Isomaltase is needed to hydrolyze the α(1→6) glycosidic bonds in the limit dextrins because α-amylase only catalyzes the hydrolysis of α(1→4) glycosidic bonds and cannot accommodate branch points (see Problem 4). These enzymes are required to hydrolyze the starch completely to its component monosaccharides, since only monosaccharides can be absorbed.

Amino acid transporter

Na,K-ATPase

19. a.  

O R2

9.  Monosaccharides enter the cells lining the intestine via secondary active transport (see Fig. 9.18). Na+ ions and glucose enter the cell in symport: the Na+ ions along a concentration gradient and the glucose against a concentration gradient. The Na+ ions are subsequently pumped out of the cell by the Na,K-ATPase transporter, which uses the free energy of ATP hydrolysis to eject the Na+ ions against their concentration gradient. 11.  Ethanol is water soluble but also is small and has sufficient nonpolar character to pass through the membranes of the cells lining the stomach and the small intestine without requiring a transport protein. Ethanol absorption occurs quickly on an empty stomach but more slowly if ethanol is consumed with food. 13. a. The low pH denatures the protein, unfolding it so peptide bonds are more accessible to proteolytic digestion by stomach enzymes.

K+

O H2C

C

O

O

gastric lipase

O

CH

H2C

R1

C

O

R3

C

O

H2O

Triacylglycerol

R3

O–

C

Fatty acid O O

7.  The catalytic efficiencies for maltase and isomaltase are 1.6 × 105 M−1· s−1 and 5.6 × 101 M−1· s−1, respectively. Considering only these data, maltase has the greater catalytic efficiency and makes a greater contribution to starch digestion than isomaltase. ​kcat ​  ​ ___________ 63 ​ s​−1​   ​ = 1.6 × ​10​5​ ​ M​−1​ · ​s−1 ​​ ___ ​ ​  ​ = ​     ​KM ​  ​ 0.4 × ​10​−3​  M −1 ​kc​  at​ ___________ = 5.6 × ​10​1​ ​ M​−1​ · ​s−1 ​ ​ ​​ ___  ​ = ​  3.4 ​ s​ ​  ​  ​KM ​  ​ 61 × ​10​−3​  M

Na+

Amino acid

R2

C

H2C O

O

C

R1

CH

H2C

OH

Diacylglycerol b.  Both diacylglycerol and fatty acids are amphipathic molecules— they have both hydrophilic and hydrophobic domains. These molecules can form micelles that emulsify the dietary triacylglycerols, which are nonpolar and are unable to form micelles. 21. 

O H3C

Cholesteryl ester

C (H2C)17 O

Cholesteryl esterase

O H3C

(H2C)17

C

O–

Fatty acid

HO

Cholesterol

OD D -NUMB ER ED SO LUTI O NS  S-39 23. a. The polar glycogen molecule is fully hydrated, so its weight reflects a large number of closely associated water molecules. Fat is stored in anhydrous form. Therefore, a given weight of fat stores more free energy than the same weight of glycogen. b. Because it must be hydrated, a glycogen molecule occupies a large effective volume of the cytoplasm, which it shares with other glycogen molecules, enzymes, organelles, etc. Because hydrophobic fat molecules are sequestered from the bulk of the cytoplasm, they do not have the same potential for interfering with other cellular constituents, so their collective volume is virtually unlimited. 25.  The phosphorylated glucose molecule is not recognized by the glucose transporter. The negatively charged phosphate group makes it more difficult for the glucose to exit the cell by passive diffusion. Removal of the phosphate group allows the glucose to leave the cell more easily. 27. a. An amide linkage forms (see Table 1.1). It is often referred to as an “isopeptide” linkage because of its resemblance to a peptide bond.

b.  The additional carboxylate group on the glutamate residue confers a −2 charge on the side chain, generating a high-affinity binding site for the Ca2+ ions essential for blood clotting. 41. a. vitamin C, b. biotin, c. pyridoxine, d. pantothenic acid. 43. a. The prosthetic group is related to NAD+, as it contains a nico­ tinamide–ribose–phosphate group in which the amide group is a thioamide. b. A histidine and a lysine side chain hold the prosthetic group in place.

NH N

O

O−

−O

P

O

O

Ubiquitin

NH

C

NH

(CH2)4

H

CH C

O

Ubiquitin

C

OH

OH

H

a.  ​ΔG°ʹ = −RT ln ​Keq ​  ​ ΔG°ʹ = −(8.3145 J · ​K−1 ​ ​ · ​mol​−1​ ) (298 K ) ln 0.25

NH O

H

nicotinamide

45.  Use Equation 12.2:

b.  An ester linkage forms (see Table 1.1). O

H

Lys N H

+

N

ribose

O O

Ni2+ S

S

phosphate

His

CH2

ΔG°ʹ = 3400 J · ​mol​−1​= 3.4 kJ · ​mol​−1​

CH C

b.  ​ΔG°ʹ = −RT ln ​Ke​  q​

O

ΔG°ʹ = −(8.3145 J · ​K−1 ​ ​ · ​mol​−1​ ) (310 K ) ln 0.25 ΔG°ʹ = 3600 J · ​mol​−1​= 3.6 kJ · ​mol​−1​

29.  Acetyl-CoA Glycolysis Citric acid cycle

✔

Fatty acid metabolism

✔

GAP

Pyruvate

✔

✔

TAG synthesis

✔

Photosynthesis

✔

Transamination

✔

31. a. reduced, b. reduced, c. oxidized, d. oxidized. 33. a. oxidized, b. reduced.

35. a. CH4 + ​​SO​42− ​  ​​ → ​​HCO​3−​  ​​ + HS− + H2O, b. CH4 is oxidized to​​ − 2− HCO​3​  ​​, c. ​​SO​4​  ​​is reduced to HS−. 37.  Individuals with gastrointestinal disorders might have a gastrointestinal tract that is not colonized by the appropriate vitamin B12− synthesizing bacteria. A deficiency in haptocorrin or intrinsic factor would be manifested as a vitamin B12 deficiency, since these proteins are essential for absorption of the vitamin. Vegetarians and vegans who consume no animal products would also be at risk for a deficiency of vitamin B12. 39. H N

O CH

C

CH2 –OOC

CH

c.  The ΔG°′ values for both reactions are positive, so the reaction is not spontaneous at either temperature. 47.  Because Keq is the ratio of the product concentration to the reactant concentration at equilibrium, the reaction with the larger K eq will have a higher concentration of product. Therefore, the concentration of B in tube 1 will be greater than the concentration of D in tube 2. 49. a. Since Keq = 1, ln Keq = 0 and ∆G°′ is also equal to zero (Equation 12.2). b. Since Keq = 1, the concentrations of reactants and products must be equal at equilibrium. If the reaction started with 1 mM F, the equilibrium concentrations will be 0.5 mM E and 0.5 mM F. 51.  The equilibrium constant Keq can be derived by rearranging Equation 12.2 as shown in Sample Calculation 12.2: ​ ​Keq ​  ​= ​e​ −ΔG°ʹ/RT​

​/(8.3145 J · ​K−1 ​ ​ · mo​l−1 ​ ​) (310 K)

−1

​ ​Keq ​  ​= ​e​​  −4100 J · mo​l​

​​

​ ​Keq ​  ​= ​e​ −1.59​= 0.20​

53.  Use Equation 12.3 and the ∆G°′ value calculated in Solution 48: [B] ΔG = ​ΔG°ʹ​ + RT ln ​ ___  ​  [A] −3 ​  0.1 × ​10​ ​ M ​  ​ ΔG = −5900 J · ​mol​−1​+ (8.3145 J · ​K​−1​ · ​mol​−1​) (310 K) ln​ __________ (0.9 × ​10​−3​ M )

ΔG = −11,600 J · ​mol​−1​= −11.6 kJ · ​mol​−1​ COO–

γ-Carboxyglutamate

​​The reaction will proceed as written, with A converted to B until the ratio of [B]/[A] = 10/1.

S-40  ODD- N UM B ER E D S O LUT I O NS 55.  Use Equation 12.3 and the ∆G°′ value determined in Solution 49a:

65. 

a

[​ F]​ ​ G = ΔG​°′ ​ + RT ln ​ ___  ​​ Δ ​[E]​

G

−3 ​ΔG = 0 + (8.3145 J · ​K−1 ​ ​ · mo​l−1 ​ ​) (310 K) ln​ __________ ​  2 × ​10​−3​ M  ​  ( 5 × ​10​ ​ M )

c

​ ​ ΔG ​ = −2400 J · mo​l−1 ​ ​= −2.4 kJ · mo​l−1 ​​Yes, the reaction is spontaneous and will proceed in the forward direction, until the ratio of [F]/[E] = 1/1. This is consistent with Solution 50, which predicted that the reaction would proceed until [E] = [F] = 3.5 mM. 57.  The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): ​ ​Keq ​  ​= ​e​ −ΔG°ʹ/RT​

​/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​)(298 K)

−1



​Keq ​  ​= ​e​ −10,000 J · ​mol​



−4.04

​Keq ​  ​= ​e​ 



​Ke​  q​= ​e​ −ΔG°ʹ/RT​



​Keq ​  ​= ​e​ −20,000 J · ​mol​



​Keq ​  ​= ​e​ −8.07​= 0.00031

​

​= 0.018

​/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​)(298 K)

−1

​

​ mall changes in ∆G°′ result in large changes in Keq. Doubling the ∆G°′ S value (a positive, unfavorable value) leads to a ~60-fold decrease in Keq. 59.  Use Equation 12.3 and the value of ∆G°′ given in Sample Calculation 12.2: [G6P] ΔG = ​ΔG°ʹ​ + RT ln ​ ______   ​ [G1P]

−3 ΔG = −7100 J · ​mol​−1​+ (8.3145 J · ​K−1 ​ ​ · ​mol​−1​) (310 K) ln​ __________ ​  20 × ​10​ ​ M ​  ​ ( 5 × ​10​−3​ M )

ΔG = −7100 J · ​mol​−1​+ 3600 J · ​mol​−1​ −1

b Reaction coordinate

67. a. The phosphate groups on the ATP molecule would be less negative at a lower pH. Therefore, there would be less charge–charge repulsion and therefore less energy released upon hydrolysis. The ∆G° value would be less negative at a lower pH. b. Magnesium ions are positively charged and form ion pairs with the negatively charged phosphate groups. Thus, magnesium ions decrease the charge– charge repulsion associated with the phosphate groups. In the absence of magnesium ions, the charge–charge repulsion is greater; thus, more free energy is released upon the removal of one of the phosphate groups. This results in a ∆G°′ value that is more negative. 69.  The complete reaction is ATP + H2O → ADP + Pi. Use Equation 12.3 and the value of ∆G°′ from Table 12.4. The concentration of water is assumed to be equal to 1. ​[ADP][P​ i​ ] ​ΔG = ΔG°ʹ + RT ln ​ _________      ​ [ATP] ΔG = −30,500 J · ​mol​−1​+

(1 × ​10​−3​ M) (5 × ​10​−3​ M) (8.3145 J · ​K−1 ​ ​ · ​mol​−1​ ) (310 K) ln​ ____________________  ​ ​        ​ ( ) 3 × ​10​−3​ M

ΔG = −30,500 J · ​mol​−1​+ (−16,500 J · ​mol​−1​) = −47 kJ · ​mol​−1​

−1

ΔG = −3500 J · ​mol​ ​= −3.5 kJ · ​mol​ ​ ​​The ∆G is negative, indicating that the reaction is spontaneous under these conditions. ​Glucose + ​P​ i​ → Glucose-6-phosphate + ​H​ 2​O Glucose-1-phosphate + ​H​ 2​O → Glucose + ​P​ i​

61. 

71.  First, convert Calories to joules: 72,000 cal × 4.184 J· cal−1 = 300,000 J or 300 kJ. Since the ATP → ADP + Pi reaction releases 30.5 kJ · mol−1, the apple contains 300 kJ ÷ 30.5 kJ · mol−1 or the equivalent of about 10 moles of ATP.

ΔG°ʹ =     13.8 kJ · ​mol​−1​ 73. a.  ADP + Pi → ATP + H2O ∆G°′ = 30.5 kJ · mol−1 ΔG°ʹ = −20.9 kJ · mol​−1​ 2200 Cal  × _________        ​  ​  1 mol ATP  ​  × ________ ​  4.184 kJ  ​  × 0.33 = 100 mol ATP · ​d−1 ​ ​ ​​ ________ d 30.5 kJ 1 Cal Glucose-1-phosphate → Glucose-6-phosphate ΔG°ʹ =      −7.1 kJ · mol​−1​ ​​The value of ∆G°′ obtained above is the same as the value presented in Sample Calculation 12.2. Free energy is independent of path, so the values of ∆G°′ are the same whether the reaction is visualized as occurring in one step or two steps. ∆G°′ is negative, which indicates that the reaction is spontaneous under standard conditions.

63. P ​ EP + ​H​ 2​O ⇌ ​pyruvate + P​ i​ ADP + ​P​ i​⇌ ATP + ​H​ 2​O



PEP + ADP ⇌ pyruvate + ATP

ΔG°ʹ = −61.9 kJ · ​mol​−1​ ΔG°ʹ = 30.5 kJ · ​mol​−1​

505 g ______ b.  ​100 mol ATP × _____ ​   ​ × ​  1 lb  ​  = 23 lb​ mol 2200 g c.  ATP does not accumulate but instead is constantly recycled. As ATP is used, its hydrolysis products, ADP and Pi, serve as reactants for ATP synthesis in the process of oxidative phosphorylation. 75.  The synthesis of ATP from ADP requires 30.5 kJ · mol −1 of energy: ​ADP + ​P​ i​→ ATP + ​H​ 2​O  ΔG°ʹ = 30.5 kJ · ​mol​−1​

ΔG°ʹ = −31.4 kJ · ​mol​−1

−1

kJ · ​mol​  ​​ × 0.33 = 30.8 ATP​ _____________       ​  2850

[pyruvate][ATP] ΔG ​ = ΔG°′ + RT ln ​ ______________        ​​ [PEP][ADP] ​ΔG = −31,400 J · mo​l−1 ​ ​+

(​ 20 × ​10​ ​ M)​ (​ 5 × ​10​ ​ M)​  (8.3145 J · ​K−1 ​ ​ · mo​l−1 ​ ​) (310 K) ln​ ​ ______________________         ​ ​​ ( ​(10 × ​10​−3​ M)​ ​(5 × ​10​−6​ M)​) −3

​ ​ ​ΔG = −11,800 J · mo​l​−1​= −11.8 kJ · mo​l−1 ​​Yes, the reaction is spontaneous under these conditions.

−3

30.5 kJ · ​mol​−1​

77. a. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): ​ ​Ke​  q​= ​e​ −ΔG°′/RT​

​/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​) (298 K)

−1

​Ke​  q​= ​e​ −5000 J · ​mol​ −2.02

​Keq ​  ​= ​e​ 

​= 0.133​

​

OD D -NUMB ER ED SO LUTI O NS  S-41 b.  Since



[​ isocitrate]​ ​ ​K​  ​= __________ ​     ​  = 0.133 eq



[citrate]

[F16BP] 4.4 × ​10​−9​= _______________ ​       ​ [F6P](5 × ​10​−3​ M)

[F16BP] _________ 0​−11​      ​ _______  ​  = ​  2.2 × ​1 ​ ​ 1 [F6P]

[isocitrate ] = 0.133 [citrate]​

The total concentration of isocitrate and citrate is 2 M, so

b.

F6P + ​P​ i​⇌ F16BP + ​H​ 2​O

ATP + ​H​ 2​O ⇌ ADP + ​P​ i​

  [isocitrate] = 2 M − [citrate]



Combining the two equations gives



​ 0.133 [citrate ] = 2 M − [citrate]

c.  ​    ​Keq ​  ​= ​e​ −ΔG°′/RT​



1.133 [citrate ] = 2 M



−6.94

[isocitrate ] = 2 M − 1.77 M = 0.23 M​

c.  The preferred direction under standard conditions is toward the formation of citrate. d. The reaction occurs in the direction of iso­ citrate synthesis because standard conditions do not exist in the cell. Also, the reaction is the second step of an eight-step pathway, so iso­ citrate is removed as soon as it is produced in order to serve as the reactant for the next step of the pathway. 79. a. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2). The ΔG°′ value for the reaction given is obtained by reversing the ΔG°′ value for the hydrolysis of glucose-6-phosphate (G6P) in Table 12.4: ​ ​Keq ​  ​= ​e​ −ΔG°′/RT​

​/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​) (298 K)

−1



​Keq ​  ​= ​e​ −13,800 J · ​mol​



​Keq ​  ​= ​e​ −5.57​= 0.0038​

​

b.  Use the equilibrium constant expression and the Keq calculated in part a to solve for the equilibrium concentration of G6P: [G6P] ​        ​ ​   K = ____________ eq

[glucose] [​P​ i​]



[G6P] 0.0038 = ​ ____________________        ​



[G6P] = 9.5 × ​10​−8​ M​

(5 × ​10​−3​ M) (5 × ​10​−3​ M)

c.  Under the given conditions, the reaction would produce only 9.5 × 10−8 M glucose-6-phosphate from 5 mM glucose and thus is not a feasible route for generating this compound for the glycolytic pathway. d. 

250 × ​10​ ​ M 0.0038 = __________________ ​      ​    [glucose](5 × ​10​−3​ M)

​​Driving the reaction to the right using this method is not feasible because it is impossible to achieve a concentration of 13 M glucose inside the cell. 81.  The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2):

​Keq ​  ​= ​e​​  −ΔG°′/RT​

​/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​)(298 K)

−1

​Keq ​  ​= ​e​ −47,700 J · ​mol​

​/(8.3145 J · ​K−1 ​ ​ · ​mol​−1​)(298 K)

​

−4

​K   ​  ​= ​e​  ​= 9.7 × ​10​ ​ eq [F16BP][ADP]  ​Keq ​  ​= _____________ ​        ​ [F6P][ATP] [F16BP](1 × ​10​−3​ M)  ​ 9.7 × ​10​−4​= _________________ ​       [F6P](3 × ​10​−3​ M) [F16BP] ________ ​10​−3​   ​ _______    ​  = ​  2.9 ×  ​ ​  1 [F6P]

d.  The conversion of fructose-6-phosphate to fructose-1,6-bisphosphate is unfavorable. The ratio of products to reactants at equilibrium is 2.2 × 10−11 under standard conditions. However, if the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate is coupled with the hydrolysis of ATP, the reaction becomes more favorable and the ratio of fructose-1,6-bisphosphate to fructose-6-phosphate increases to 2.9 × 10−3, a change of eight orders of magnitude. 83.  ​ I. GAP ⇌ 1,3BPG ΔG°ʹ​= 6.7 kJ · ​mol​−1​​​ 1,3BPG + ​H​ 2​O ⇌ 3PG + ​P​ i​ ΔG°ʹ​= −49.3 kJ · ​mol​−1​

GAP + ​H​ 2​O ⇌ 3PG + ​P​ i​ ΔG°ʹ​= −42.6 kJ · ​mol​−1​



GAP + ADP ⇌ 3PG + ATP ΔG°ʹ​= −12.1 kJ · ​mol​−1​

II.  GAP ⇌ 1,3BPG ΔG°ʹ​= 6.7 kJ · ​mol​−1​ 1,3BPG + ADP ⇌ 3PG + ATP ΔG°ʹ​= −18.8 kJ · ​mol​−1​

​​The second scenario is more likely. The first coupled reaction is more exergonic, but the second coupled reaction “captures” some of this free energy in the form of ATP, which the cell can use. 85.  The hydrolysis of pyrophosphate releases 19.2 kJ · mol−1 (see Table 12.4). These two reactions are coupled, and the overall ∆G value for the coupled reactions is negative; thus UDP–glucose formation occurs spontaneously.

1. a. Reactions 1, 3, 7, and 10;  b.  Reactions 2, 5, and 8;  c.  Reaction 6; d.  Reaction 9;  e.  Reaction 4.

[glucose ]  = 13 M​

a.  ​

ΔG°ʹ = 17.2  kJ · ​mol​−1​

Chapter 13

−6



[G6P] ​Keq ​  ​= ____________ ​        ​ [glucose ] [​P​ i​]

ΔG°ʹ = −30.5  kJ · ​mol​−1​

−1

 ​Keq ​  ​= ​e​ −17,200 J · ​mol​

[citrate ] = 1.77 M



F6P + ATP ⇌ F16BP + ADP

ΔG°ʹ = 47.7  kJ · ​mol​−1​



​Keq ​  ​= ​e​ −19.2​= 4.4 × ​10​−9​



[F16BP] ​Keq ​  ​= ________ ​     ​  ​[F6P][P​ i​ ]

​

3.  The low KM means that the enzyme will be saturated with glucose and will therefore operate at maximum velocity. Even if the concentration of glucose were to fluctuate slightly, the brain’s ability to catabolize glucose would not be affected. 5.  The amide functional group of the Asn side chain can form hydrogen bonds with the hydroxyl groups of the glucose substrate and can potentially function as either a hydrogen bond donor or a hydrogen bond acceptor. The methyl group of Ala cannot participate in hydrogen bond formation, which explains the decrease in glucose affinity as indicated by the higher KM for the mutant enzyme. The side chain of Asp could potentially serve as a hydrogen bond acceptor, but the higher KM value for this mutant indicates that the substrate binds even less well than when the hydrogen bonding is abolished

S-42  ODD- N UM B ER E D S O LUT I O NS at this site. This indicates that the —NH2 group of the Asn side chain functions as a hydrogen bond donor when interacting with the —OH groups of the glucose substrate and that this interaction is vitally important for substrate binding. [From Pilkis, S.J. et al., J. Biol. Chem. 269, 21925–21928 (1994).] 7.

–2O

3POCH2

O

H HO

H OH

H

H

OH

19. a.  No, the ∆G°′ value is positive and therefore the reaction is not favorable under standard conditions.

H

b.  Use Equation 12.3.

H

[GAP][DHAP] ​ΔG = ΔG​°′ ​+ RT ln __________    ​   ​​  [F16BP]

9. a. Rearrange Equation 12.2 to solve for Keq, as shown in Sample Calculation 12.2, and then use the equilibrium constant expression to determine the [F6P]/[G6P] ratio:

−1

Keq = ​e​ −​ 2200 J ⋅ ​mol​ 

​/(8.3145 J ⋅ ​K​ −1​⋅ mol​ −1​)(298 K)

​

−0.88

​

[F6P] ​ 0.41 ​ = ​ _  Keq = _ ​ 1 [G6P] b.  Rearrange Equation 12.3 to solve for the [F6P]/[G6P] ratio: [F6P]   ​ΔG = ΔG​°′ ​+ RT ln _ ​  [G6P]

−1

(−1400 J ⋅ ​mol​  ​−  2200 J ⋅ ​mol​  ​)/(8.3145 J ⋅ ​K​​  e​​ ​

−1

The reaction will proceed in the direction written. 21.  Rearrange Equation 12.3 to solve for the [GAP]/[DHAP] ratio: [GAP] ​ΔG = ΔG°ʹ + RT ln ​ _  [DHAP] [GAP] (ΔG−Δ​G​ ∘′​  ​)/RT e​​ ​ ​= _ ​    [DHAP] −1

−1

−1

(4400 J ⋅ ​mol​  ​− 7900 J ⋅ ​mol​  ​)/(8.3145 J ⋅ ​K​  e​​ ​

[DHAP]

[F6P] ​    e​​  (ΔG−Δ​​G​​  ​ ′ ​)/RT​= _____ [G6P]​ [F6P] ​ = _____ ​    ​ [G6P]​

​ ⋅ mo​l​​  −1​)(310 K)

[F6P] _____   ​ 0.25 ​  ​  ​ = ​e​ −1.4​= _ [G6P]​

(3.0 × ​10​ −6​M) (16 × ​10​ −6​M) (8.3145 J ⋅ ​K​ −1​⋅ mo​l​ −1​) (310 K ) ln​ _______________________ ​        ​​  ​ ( ) 14 × ​10​ −6​M

[GAP] ​=_ ​    [DHAP]

​ ⋅ ​mol​ −1​) (310K)

[GAP] _  ​   ​ 0.26 ​  ​= ​e​ −1.36​= _

∘

−1

​ΔG = 22,800 J ⋅ mo​l​ −1​+

​ΔG = − 9600 J ⋅ mo​l​ −1​= − 9.6 k J ⋅ mo​l​ −1​

∘

​Keq = ​e​​  −Δ​​G​​  ​ ′ ​/RT​

Keq = ​e​ 

is cleaved and the first substrate is released.  c.  Because Ala cannot function as an acid catalyst, the enzyme mechanism could not continue past the formation of the Schiff base between the substrate and the enzyme. The aldehyde product is not released and the enzyme is “frozen” with a covalent bond between the substrate and the Lys side chain on the enzyme.

1

 he reaction proceeds in the forward direction until [F6P]/[G6P] T reaches 0.41. 11.  One might expect the product of a reaction to inhibit the enzyme that catalyzes the reaction, while the reactant would act as an activator. Although it is true that ADP is a direct product of the PFK reaction, PFK is sensitive to the ATP needs of the cell as a whole. Rising ADP concentrations are an indication that ATP is needed; the subsequent stimulation of PFK increases glycolytic flux and generates ATP as a final pathway product. 13.  In the presence of the inhibitor, the curve is sigmoidal (indicating cooperative binding) and the KM increases dramatically (nearly 10-fold, to 200 µM), indicating that a greater quantity of substrate is required to achieve ½ Vmax. PEP stabilizes the T form of PFK (see Solution 12b). 15.  Glycerol can serve as an energy source because it can be c­ onverted to glyceraldehyde-3-phosphate, which can then enter the glycolytic pathway “below” the phosphofructokinase step. The mutants cannot grow on glucose because glucose enters the g­ lycolytic pathway by first being converted to glucose-6-phosphate, then ­fructose-6-phosphate. The next step, conversion to ­f ructose-1,6-bisphosphate, requires phosphofructokinase. Thus, glycerol is a suitable substrate for this mutant, but glucose is not. 17. a.  The pK of the Lys side chain is lower than it would be in the free amino acid. The side chain is deprotonated in order to serve as a nucleophile; the —NH2 nucleophilically attacks the electrophilic carbonyl group of the F16BP substrate to begin the reaction.  b.  The Asp side chain is unprotonated at the beginning of the reaction, but after formation of the Schiff base, the side chain pK increases and the Asp acts as a base to accept a proton from the substrate. The scissile bond

1

 he ratio of [GAP] to [DHAP] is 0.26/1, which seems to indicate that T the formation of DHAP, not the formation of GAP, is favored. However, GAP, the product of the triose phosphate isomerase reaction, is the substrate for the glyceraldehyde-3-phosphate dehydrogenase reaction that occurs next in the pathway. The continuous removal of the product GAP by the action of the dehydrogenase shifts the equilibrium toward formation of GAP from DHAP. 23. a.  The cancer cells may express the GAPDH protein at higher levels (i.e., transcription of the GAPDH gene and translation of its mRNA may occur at a higher rate). Another possibility is that GAPDH is degraded more slowly.  b.  The structure of the active site in GAPDH from cancer cells may differ from the structure in normal cells such that the binding of methylglyoxal is permitted, which then precludes the binding of the substrate. Another possibility is that the altered GAPDH might have a binding site for methylglyoxal elsewhere on the protein, which causes a conformational change in the protein that alters the substrate binding site so that the substrate can no longer bind. [From Ray, M. et al., Mol. Cell. Biochem. 177, 21–26 (1997).] 25.  As the NADH/NAD + ratio increases, the activity of GAPDH decreases and less 1,3-bisphosphoglycerate is produced from glyceraldehyde-3-phosphate. NAD+ is a reactant and NADH is a product of the reaction, so as NAD+ becomes less available and NADH accumulates, the ratio of [product]/[reactant] increases and the activity of the enzyme decreases. 27.  Phosphoglycerate kinase catalyzes the conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate with concomitant production of ATP from ADP. The kinase can generate the ATP required by the ion pump, and the ADP produced when the pump is phosphorylated can serve as a substrate in the kinase reaction. 29. a.  In hepatocytes, the phospho-His on the phosphoglycerate mutase transfers its phosphate to the C2 position of 3PG to form 2,3-­bisphosphoglycerate. The [32P]-labeled phosphate on the C3 p ­ osition is transferred back to the enzyme to form the 2PG product, so initially the enzyme would be labeled. In the next round of catalysis, the labeled

OD D -NUMB ER ED SO LUTI O NS  S-43 phosphate on the enzyme is transferred to the C2 position of the next molecule of 3PG substrate, so 2PG becomes labeled. ­Eventually, this phosphate is transferred to ADP to form ATP, so ATP is labeled.  b. In the plant, the labeled phosphate is transferred to C2 to form 2PG, so 2PG is labeled and then eventually ATP. The plant enzyme is not labeled. 31.  Iodoacetate inactivates aldolase, which leads to the accumulation of fructose-1,6-bisphosphate, a metabolite that “costs” 2 ATP to produce. If glycolysis cannot be completed, the early steps consume cellular ATP, but the “energy-payoff” stage does not occur and no new ATP is generated. Thus, the addition of iodoacetate depletes the cells of ATP. 33. a.  The equilibrium constant Keq can be derived by rearranging Equation 12.2 as shown in Sample Calculation 12.2: ​Keq ​  ​= ​e​​  −ΔG°ʹ/RT −1

K​ ​   ​= ​e​​  −(−31,400 J ⋅ mo​l​  eq

​)/(8.3145 J ⋅ ​K ​ −1​ ⋅ mo​l​ −1​) (298 K )

​​

​Keq ​  ​= ​e​ 12.7​= 3.2 × ​10​ 5​ b.  Use Equation 12.3. [pyruvate][ATP] ​ΔG = ΔG​°′ ​+ RT ln ______________ ​         ​​ [PEP][ADP] ​ΔG = − 31,400 J ⋅ mo​l​ −1​+

(51 × ​10​ −6​M)(1.0 × ​10​ −3​M) (8.3145 J ⋅ ​K​ −1​⋅ mo​l​ −1​) (310 K) ln​ ​ ________________________      ​ ​​ ( (23 × ​10​ −6​M)(0.1 × ​10​ −3​M) )

​ΔG = − 23,400 J ⋅ mo​l​ −1​= − 23.4 k J ⋅ mo​l​ −1​ The reaction will proceed in the direction written. 35.  Cytochrome P450 enzymes are responsible for detoxifying substances that enter the body, including drugs. Increased expression of cytochrome P450 due to long-term alcohol consumption allows more rapid modification and excretion of the drug molecules, thereby making them less effective. −1

37.  The glucose–lactate pathway releases 196 kJ · mol of free energy, enough theoretically to drive the synthesis of (196 ÷ 30.5) × 0.33, or about 2 ATP. 39. a.  One mole of ATP is invested when KDG is converted to KDPG. One mole of ATP is produced when 1,3-bisphosphoglycerate is converted to 3PG. One mole of ATP is produced when phosphoenolpyruvate is converted to pyruvate. Therefore, the net yield of this pathway (per mole of glucose) is one mole of ATP.  b.  In order to keep the pathway going, subsequent reactions would need to reoxidize the NADPH that is produced when ­glucose is converted to gluconate and the NADH that is produced by GAPDH. [From Johnsen, U. et al., Arch. Microbiol. 175, 52–61 (2001).] 41. a.  The conversion of 3-phosphoglycerate (3PG) to 1,3-bisphosphoglycerate (1,3BPG) is a phosphorylation.  b.  The conversion of 2-phosphoglycerate (2PG) to 3PG, the conversion of glyceraldehyde-3-phosphate (GAP) to dihydroxyacetone phosphate (DHAP), and the conversion of fructose-6-phosphate (F6P) to glucose-6-phosphate (G6P) are all isomerizations.  c.  The conversion of 1,3BPG to GAP is an oxidation–reduction reaction.  d.  The conversion of phosphoenolpyruvate (PEP) to 2PG is a hydration.  e.  A carbon– carbon bond is formed when GAP and DHAP react to form fructose-1,6-bisphophate (FBP).  f.  A carboxylation reaction occurs when pyruvate is converted to oxaloacetate (OAA).  g.  A decarboxylation reaction occurs when OAA is converted to PEP.  h.  Both the conversion of FBP to F6P and the conversion of G6P to glucose are hydrolysis reactions. 43.  Raw eggs contain avidin, which binds tightly to biotin (see Problem 4.93). In the digestive tract, formation of an avidin–biotin complex makes biotin unavailable for carboxylase enzymes such as

pyruvate carboxylase. Cooking destroys avidin, so the consumption of cooked eggs does not result in a biotin deficiency. 45. a.  Acetyl-CoA produced from pyruvate is a substrate for the citric acid cycle, an energy-producing pathway. When the cell’s need for energy is low, acetyl-CoA accumulates and activates pyruvate carboxylase, which catalyzes the first step of gluconeogenesis. As a result, the cell can synthesize glucose when the need to catabolize fuel is low.  b.  The deamination of alanine produces pyruvate, a substrate for gluconeogenesis. By inhibiting pyruvate kinase, alanine suppresses glycolysis so that flux through the shared steps of glycolysis and gluconeogenesis will favor gluconeogenesis. 47.  The activity of the enzyme decreases with increasing F26BP in the absence of AMP. In the presence of AMP, the decrease in activity is even greater, indicating that AMP also inhibits the enzyme, and in a way that is synergistic with F26BP. [From Van Schaftingen, E. and Hers, H.-G., Proc. Natl. Acad. Sci. USA 78, 2861–2863 (1981).] 49.  Insulin, the hormone of the fed state, might be expected to suppress the transcription of the gluconeogenic enzymes pyruvate carboxylase, PEPCK, fructose-1,6-bisphosphatase, and glucose6-phosphatase. [In fact, insulin has been shown to suppress the transcription of PEPCK and glucose-6-phosphatase.] 51. a.  The phosphatase enzyme is active under fasting conditions. The phosphatase removes the phosphate group from F26BP, forming fructose-6-phosphate. Thus, F26BP is not present to stimulate glycolysis (or to inhibit gluconeogenesis); therefore, gluconeogenesis is active.  b.  The hormone of the fasted state is glucagon. c. When glucagon binds to its receptors, cellular cAMP levels rise, as described in Section 10.2. This activates protein kinase A, which phosphorylates the bifunctional enzyme, resulting in the activation of the phosphatase activity and the inhibition of the kinase activity. 53.  Phosphoenolpyruvate carboxykinase catalyzes an essential step of gluconeogenesis. Lower expression of this enzyme decreases the gluconeogenic output of the liver in a manner similar to that described in Solution 52, which helps decrease the level of circulating glucose in patients with diabetes. 55.  These transformations comprise the glucose–alanine cycle, which is shown in Figure 19.4. Prolonged operation of this cycle results in muscle breakdown, since muscle protein is the source of the amino acids required in the transamination of pyruvate to alanine. 57. Glucose-6-phosphate Fructose-6-phosphate

FBPase Fructose-1,6-BP Glycerol-3-Pi Glycerol

GK

GAP + DHAP 1,3-BPG

PPDK pyruvate

PEP

PEPCK OAA

MDH malate

S-44  ODD- N UM B ER E D S O LUT I O NS 59. a.  The equilibrium constant Keq can be derived by rearranging Equation 12.2 as shown in Sample Calculation 12.2: ​ ​Keq ​  ​= ​e​​  −ΔG°ʹ/RT​

−1

​ ​Keq ​  ​= ​e​​  −(−13,400 J ⋅ mo​l​  5.4

​)/(8.3145 J ⋅ ​K​ −1​ ⋅ mo​l​ −1​)(298 K )

​

2

​ ​Keq ​  ​= ​e​  ​= 2.2 × ​10​  ​ 61.  The glycosidic bonds in the starch in the grains must be hydrolyzed to glucose because the yeast use glucose as a substrate for fermentation. 63. a.  See Sample Calculation 12.2. Keq can be calculated by rearranging Equation 12.2, then the equilibrium constant expression can be used to solve for the [Pi]/[G1P] ratio. The values for glycogen (n – 1 residues) and glycogen (n residues) are not substantially different from one another and can be set equal to 1. ​Keq ​  ​= ​e​​  −Δ​G° ′ ​/RT K​ ​   ​= ​e​​  −(3100 J ⋅ mo​l​​  eq

−1

​)/(8.3145 J ⋅ ​K​​  −1​ ⋅ mo​l​​  −1​)(298 K)

K​ ​   ​= ​e​​  −1.25 eq [glycogen (n ​− 1)][G1P] K​ ​   ​= _ ​ 0.29 ​ = ​ ______________________       eq 1 [glycogen (n​)][Pi] 1[G1P] _ ​ = ​ _   0.29 ​ 1 1[Pi] [P ] 3.5  ​= _ ​  i   ​  _ ​ 1 [G1P] b.  Use Equation 12.3; substitute in values of 1 for glycogen (n – 1 residues) and glycogen (n residues), and 50/1 for the [Pi]/[G1P] ratio given in the problem: [glycogen (n ​− 1)][G1P] ​ΔG = ΔG​°′ ​+ RT ln ______________________    ​    ​ [glycogen (n​)][Pi]

(1)(1) ​    ​ = 3100 J ⋅ ​mol​ −1​+ (8.3145 J ⋅ ​K​ −1​⋅ ​mol​ −1​) (310 K) ln _ (1)(50) −1 −1 −1 = 3100 J ⋅ ​mol​  ​+ (−10,100 J ⋅ ​mol​  ​) = − 7.0 kJ ⋅ ​mol​  ​

c.  Production of glucose-1-phosphate requires only an isomerization reaction catalyzed by phosphoglucomutase to convert it to glucose-6-phosphate, which can enter glycolysis. This skips the hexo­ kinase step and saves a molecule of ATP. Hydrolysis, which produces glucose, would require expenditure of an ATP to phosphorylate glucose to glucose-6-phosphate. 65.  Glucagon binds to its G protein–coupled receptor and activates the G protein, whose α subunit swaps GDP for GTP and dissociates from the β and γ subunits. The G protein activates adenylate cyclase, which catalyzes the conversion of ATP to cyclic AMP (cAMP). Four molecules of cAMP bind to the regulatory subunits of protein kinase A, releasing catalytically active subunits. The activation of protein kinase A leads to the phosphorylation of glycogen phosphorylase, which activates the enzyme. This promotes glycogen degradation in the liver and the release of glucose to the blood during the fasted state. 67.  An increase in the activity of glycogen phosphorylase in the fat body increases degradation of glycogen to glucose. Because fructose-2,6-bisphosphate concentrations are low, glycolysis will not be stimulated (F26BP is a potent activator of the glycolytic enzyme PFK). Instead, glucose can be used to synthesize trehalose, which leaves the fat body and enters the hemolymph. In this way, the fat body produces sugars for use by other tissues in the fasting insect. [From Meyer-Fernandes, J.R. et al., Insect Biochem. Mol. Biol. 31, 165–170 (2001).]

69. a.  The first committed step of the pentose phosphate pathway is the first reaction, which is catalyzed by glucose-6-phosphate dehydrogenase and is irreversible. Once glucose-6-phosphate has passed this point, it has no other fate than conversion to a pentose phosphate. b.  The hexokinase reaction does not commit glucose to the glycolytic pathway, since the product of the reaction, glucose-6-phosphate, can enter the pentose phosphate pathway and other pathways, such as glycogen synthesis. 71. a.  Serum withdrawal decreases G6PD activity and therefore the NADPH/NADP + ratio decreases. NADPH is a product of the G6PD reaction and NADP + is a reactant, so decreased enzyme activity increases the oxidized form of the coenzyme and decreases the reduced form.  b.  In the presence of the G6PD inhibitor DHEA, the NADPH/NADP + ratio decreases for the reasons outlined in part a.  c.  Adding H2O2 should not affect the ratio, since G6PD can increase its activity to produce more NADPH to react with H 2 O 2 .  d.  In the absence of the serum containing growth factors, G6PD cannot handle the increased load of H 2O2, and the ratio decreases. [From Tian, W.-N. et al., J. Biol. Chem. 273, 10609– 10617 (1998).] 73.  If NADPH is used as a coenzyme for nitrate reduction, it is oxidized to NADP+ and will need to be regenerated. It is likely that the NADPH-producing enzymes of the pentose phosphate pathway, glucose-6-phosphate dehydrogenase and 6-phosphogluconate dehydrogenase, are stimulated by nitrate. [From Hankinson, O. and Cove, D.J., J. Biol. Chem. 249, 2344–2353 (1974).] 75.

O

C

OPO2– 3

H

C

OH

H

C

OH 2–

CH2OPO3

1,4-Bisphosphoerythronate 77.  Glucose-6-phosphate is first oxidized to ribulose-5-phosphate in the oxidative phase of the pentose phosphate pathway, generating 2 NADPH and 1 CO2. Some of the ribulose-5-phosphate is then converted to ribose-5-phosphate and some to xylulose-5-phosphate. Next, these 5-carbon sugars are converted to fructose-6-phosphate (F6P) and glyceraldehyde-3-phosphate (GAP) via the transaldolase and transketolase reactions shown in Figure 13.12. Then F6P and GAP enter gluconeogenesis to produce glucose-6-phosphate, which can enter the oxidative phase of the pentose phosphate pathway. This cycle occurs a total of six times, yielding a total of 12 NADPH and 6 CO2. Thus, the carbons in glucose-6-phosphate are completely oxidized to maximize the cellular yield of NADPH. 79.  G16BP inhibits hexokinase but activates PFK and pyruvate kinase. This means that glycolysis will be active, but only if the substrate is glucose-6-phosphate, since glucose cannot be phosphorylated in the absence of hexokinase activity. The pentose phosphate pathway is inactive, since 6-phosphogluconate dehydrogenase is inhibited. Phosphoglucomutase is activated, which converts glucose-1-phosphate (the product of glycogenolysis) to glucose-6-phosphate. Thus, in the presence of G16BP, glycogenolysis is active and produces substrate for glycolysis but not the pentose phosphate pathway. This is a more efficient process than using glucose taken up from the blood, which would need to be phosphorylated at the expense of ATP. [From Beitner, R., Trends Biol. Sci. 4, 228–230 (1979).] 81.  Hexokinase-deficient erythrocytes (purple curve) have low levels of all glycolytic intermediates, since hexokinase catalyzes the first

OD D -NUMB ER ED SO LUTI O NS  S-45 step of glycolysis. Therefore, the concentration of BPG in the erythrocyte is decreased as well, favoring the oxygenated form of hemoglobin and decreasing its p50 value. Pyruvate kinase–deficient erythrocytes (blue curve) have high levels of BPG since pyruvate kinase catalyzes the last step of glycolysis. Blockade at the last step increases the concentrations of all of the intermediates “ahead” of the block. Thus, the oxygen affinity of hemoglobin decreases with increased BPG concentration, and the p50 value increases. CH2OH   b. 

83. a. 

91.  Yes, this is an effective treatment strategy. Patients with GSD0 lack glycogen synthase and are unable to synthesize glycogen in the fed state. For that reason, the liver lacks glycogen that would normally be degraded and released as glucose to the bloodstream during the fasted state. A feeding of cornstarch helps relieve the symptoms of overnight hypoglycemia in this patient. Salivary and pancreatic amylases hydrolyze the glycosidic bonds in the starch (see Fig. 12.2), releasing glucose and alleviating hypoglycemia.

COO–

H

C

OH

H

C

OH

Chapter 14

HO

C

H

HO

C

H

HO

C

H

HO

C

H

H

C

OH

H

C

OH

1.  In mammalian cells, pyruvate can be converted to lactate by lactate dehydrogenase. Pyruvate can also be transformed into oxaloacetate; this reaction is catalyzed by pyruvate carboxylase. Pyruvate can be converted to acetyl-CoA by the pyruvate dehydrogenase complex. Pyruvate can be converted to alanine by transamination.

CH2OH

Galactitol

CH2OH

Galactonate

c.  Inhibition of these enzymes leads to the disruption of both the pentose phosphate and the glycogen degradation pathways. Pentose phosphate pathway dysfunction leads to low levels of ribose-5phosphate and NADPH, impairing nucleic acid synthesis. A lack of NADPH may also limit the capacity of the individual to deal with reactive oxygen species. Inhibiting glycogen phosphorylase decreases the ability of the liver to degrade glycogen to glucose for release into the bloodstream in the fasted state. An individual suffering from a uridyltransferase deficiency is likely to show symptoms of hypoglycemia (low blood sugar) along with liver damage that occurs due to decreased NADPH levels.

3.  The purpose of steps 4 and 5 is to regenerate the enzyme. In step 3, the product acetyl-CoA is released with concomitant reduction of the lipoamide prosthetic group of E2. In step 4, the E3 reoxidizes the lipoamide group by accepting the protons and electrons from the reduced lipoamide. In step 5, the enzyme is reoxidized by NAD+. The product NADH then diffuses away. 5.  The E1 subunit of the pyruvate dehydrogenase complex requires TPP, the phosphorylated form of thiamine, as a cofactor. Administering large doses of thiamine might be a successful treatment option if the E1 mutation weakens the interactions between the protein and thiamine. 7. 

O

85.  The low PFK activity means that glycolysis cannot generate the amount of ATP required by myosin for muscle contraction (see Section 5.4) and the patient experiences cramping as a result. The effort of exercise may also damage muscle cells, causing muscle pain and releasing myoglobin into the blood, which subsequently appears in the urine. 87.  Normally, muscle glycogen is degraded to glucose-6-phosphate, which enters glycolysis to be oxidized to yield ATP for the active muscle. In anaerobic conditions, pyruvate, the end product of glycolysis, is converted to lactate, which is released from the muscle into the blood and enters the liver to be converted back to glucose via g­ luconeogenesis (this is the Cori cycle discussed in ­Solution 56). The patient’s muscle cells are unable to degrade glycogen to ­glucose-6-phosphate; thus, there is no glucose-6-phosphate to enter glycolysis and lactate formation does not occur. [From Stanbury, J.B. et al., The Metabolic Basis of Inherited Disease, pp. 151–153, ­McGraw-Hill, New York (1978).] 89. a.  Glucose-6-phosphatase catalyzes the last reaction in gluconeogenesis (and glycogenolysis) in the liver. Glucose-6-phosphate is converted to glucose, and the glucose transporters export glucose to the circulation, where it is available to other body tissues that do not carry out gluconeogenesis and that do not store glycogen. In the absence of this enzyme, glucose-6-phosphate cannot be converted to glucose and instead accumulates in the liver and is converted to glucose-1-phosphate, which is used for glycogen synthesis. Glycogen synthesis is therefore elevated in the livers of patients with this disease. The accumulation of glycogen enlarges the liver and causes the abdomen to protrude.  b.  Although the kidney stores a limited amount of glycogen, it does participate in both gluconeogenesis and glycogenolysis, so the lack of glucose-6-phosphatase causes the kidney to enlarge for the same reasons as described in part a.

R O–

C O

C

S

CH3

Pyruvate

TPP

O H

C

R

H+

H+



CH3

N+

–C

1

4

CH3

Acetaldehyde O H

H R

CH3

N+

C

S

CH3

S

CH3

R

2

R

R N C

N+

C

CO2

HO H3C

O R

C

R

3

H+

O

HO

C

CH3

O

CH3

HO C

C S

R

H3C

N+

CH3

C S

R

Resonance-stabilized carbanion 9.  Both  a.  a high [NADH]/[NAD +] ratio and  b.  a high [acetylCoA]/[HSCoA] ratio result from the activity of the pyruvate dehydrogenase complex. The rising concentrations of NADH and acetyl-CoA (both are products of the reaction) decrease pyruvate dehydrogenase activity by competing with NAD+ and HSCoA for binding sites on the enzyme.

S-46  ODD- N UM B ER E D S O LUT I O NS 11.  Ca2+ inhibits PDH kinase and activates PDH phosphatase. This activates the pyruvate dehydrogenase complex which can then funnel substrates from glycolysis into the citric acid cycle to provide ATP needed by myosin (see Section 5.4) for muscle contraction. 13.  If PDH kinase is deficient, PDH is not phosphorylated and inactivated, resulting in higher-than-normal levels of PDH. This diverts pyruvate to the citric acid cycle and away from gluconeogenesis (see Section 13.2). Decreased gluconeogenesis activity results in a decreased output of glucose from the liver to the blood, which accounts for the decreased blood glucose levels. 15.  Fatty acids would stimulate the PDH kinase, which catalyzes the phosphorylation and inactivation of the PDH complex. Acetyl-CoA is not required for fatty acid synthesis if the cellular concentration of fatty acids is high. Instead, pyruvate can be used as a reactant in other metabolic pathways, as described in Solution 1. 17.  The expression of PDH kinase, to a large extent, and hexokinase, to a lesser extent, are upregulated under hypoxic (low oxygen) conditions. PDH kinase phosphorylates and inactivates the pyruvate dehydrogenase complex (see Problem 10) and hexokinase catalyzes the first reaction in glycolysis (see Section 13.1). The result is that glycolysis is upregulated, and glucose is converted to lactate rather than acetyl-CoA. This allows the cell to obtain energy in the form of ATP mainly from glycolysis. Converting pyruvate to acetyl-CoA requires an active citric acid cycle, which produces reduced coenzymes for oxidative phosphorylation, which cannot occur in the absence of oxygen (see Fig. 14.5). 19.  The 3-carbon intermediate is pyruvate, the product of glycolysis. Pyruvate is converted to the 2-carbon intermediate acetyl-CoA via the pyruvate dehydrogenase reaction. Both carbons in acetyl-CoA are oxidized to CO2 in the citric acid cycle. Digestion of biopolymers found in food (see Section 12.1) generates pyruvate and acetyl-CoA to fuel the citric acid cycle. Poly-, oligo-, and monosaccharides are metabolized to glucose, which enters glycolysis to form pyruvate (see Section 13.1). Dietary triacyclglycerols are digested to glycerol (which can be converted to glyceraldehyde-3-phosphate to enter glycolysis) and fatty acids, which are catabolized to acetyl-CoA (see Chapter 17). Amino acids from protein digestion are degraded to products that can be converted either to pyruvate or acetyl-CoA, depending on the amino acid (see Chapter 18). 21.  Production of ATP via substrate-level phosphorylation occurs in glycolysis at Steps 7 and 9 when 1,3-bisphophoglycerate (13BPG) and phosphoenolpyruvate (PEP), respectively, donate a phosphate group to ADP to form ATP. One glucose produces two 13BPG and two PEP, resulting in the production of four ATP, but the net yield for the pathway is two ATP due to the energy investment required to phosphorylate glucose and fructose-6-phosphate. One GTP is produced via substrate-level phosphorylation in the citric acid cycle when succinyl-CoA is converted to succinate. Two turns of the citric acid cycle are required for every glucose molecule. Taken together, one glucose molecule yields four ATP equivalents via substrate-level phosphorylation. 23.  Citrate synthase uses a base catalyst strategy. In the rate-limiting step of the reaction, an unprotonated Asp residue acts as a base by accepting a proton from the acetyl group of the acetyl-CoA to form an enolate anion. 25. a. 

of the inhibitor. The KM increases in the presence of the inhibitor, indicating that a higher substrate concentration is required to reach half-maximal velocity when the inhibitor is present. The inhibitor is able to compete with acetyl-CoA for binding to the citrate synthase active site because its structure resembles that of acetyl-CoA. 27.  When citrate is converted to isocitrate, a tertiary alcohol is converted to a secondary alcohol. This is a necessary prerequisite for the next step, in which the secondary alcohol is oxidized to a carbonyl group. Tertiary alcohols cannot be oxidized. 29.  Cis-aconitate is an intermediate in the reaction when citrate is converted to isocitrate by aconitase. Trans-aconitate structurally resembles cis-aconitate and would be expected to compete with cis-aconitate for binding to the enzyme. However, because trans-­ aconitate is a noncompetitive inhibitor when citrate is used as the substrate, the citrate binding site must be distinct from the aconitate binding site. Citrate and trans-aconitate do not compete for binding and can bind to the enzyme simultaneously, but when both substrate and inhibitor are bound, the substrate cannot be converted to product. [From Villafranca, J.J., J. Biol. Chem. 249, 6149–6155 (1974).] 31.

CH2 CH2

COO−

Itaconate

33.  See Sample Calculation 12.2. Keq can be calculated by rearranging Equation 12.2: ​ ​K​ eq​= ​e​ −ΔG​°′ ​/  RT​

CH2

C

−1



​K​ eq​= ​e​​  −(−21,000 J · mol​ 



​K​ eq​= ​e​ 8.5​



​K​ eq​= 4.8 × ​10​ 3​



HS

HBr H3C

COO

COO CO2

CH2 C

CH2

O

 tep 2. The succinyl group is then transferred to the lipoamide prosS thetic group of E2 of the α-ketoglutarate dehydrogenase complex. COO−

C−

HO CoA

S-Acetonyl-CoA b.  S-acetonyl-CoA is a competitive inhibitor. The Vmax of the citrate synthase reaction is the same in the absence and in the presence

TPP

α-Ketoglutarate

CH2 S

C

HO

COO

O CH2

CH2

CH2

CoA C

​

37.  Step 1. In the first step, α-ketoglutarate is decarboxylated, a process that requires TPP. The carbon of the carbonyl group becomes a carbanion, which forms a bond with TPP.

TPP

CH2 CH3

​)/(8.3145 J · ​K​ −1​· ​mol​ −1​) (298 K)

35.  The substrate binding properties of the mutant enzymes are not expected to differ dramatically from the wild-type enzyme. The substitution of arginine for either a histidine or a lysine side chain preserves the positive charges in the active site and presumably allows formation of the ion pairs that are important for stabilizing the transition state (see Solution 34).

O Br

COO−

C

COO− CH2

CH2 C

S

O

S

TPP S

R

Lipoamide

HS

R

OD D -NUMB ER ED SO LUTI O NS  S-47  tep 3. The succinyl group is transferred to coenzyme A, and the S lipoamide group is reduced. COO− CH2

HS

Reduced lipoamide HS

CH2 C

O

CH2 R

CH2 C

S HS

COO−

Coenzyme A

O

CoA

Succinyl-CoA

R

 teps 4 and 5. The last two steps are the same as for the pyruvate S dehydrogenase complex. E3 reoxidizes the lipoamide when its disulfide group accepts two protons and two electrons. The NAD+ reoxidizes the enzyme, and the NADH and H+ products diffuse away. 39.  When operating in reverse, succinyl-CoA synthetase catalyzes a kinase-type reaction, the transfer of a phosphoryl group from a nucleoside triphosphate (GTP or ATP). 41.  Succinate accumulates because it cannot be converted to fumarate. Succinyl-CoA also accumulates because the succinyl-CoA synthetase reaction is reversible. However, the succinyl-CoA ties up some of the cell’s CoA supply, so the α-ketoglutarate dehydrogenase reaction, which requires CoA, slows. As a result, α-ketoglutarate accumulates.

available citrate will inhibit citrate synthase, while the low concentrations of Ca2+ ions ensure that the activity of isocitrate dehydrogenase is also low. Upon beginning exercise, the Ca2+ concentration increases in muscle cells, which increases the activity of isocitrate dehydrogenase. This depletes the cellular concentration of citrate, depriving citrate synthase of its inhibitor, increasing the activity of citrate synthase. Thus, when the cell transitions from resting to exercise mode, the activities of the enzymes involved in the citric acid cycle increase in order to meet the increased demand for ATP in the working muscle. 53. a.  Aconitase catalyzes the reversible isomerization of citrate to isocitrate. Because this reaction is followed by and preceded by irreversible reactions, the inhibition of aconitase leads to an accumulation of citrate. The concentrations of other citric acid cycle intermediates are decreased.  b.  If the citric acid cycle and mitochondrial respiration are not functioning, the cell turns to glycolysis to produce the ATP for its energy needs. Consequently, flux through glycolysis increases. The increase in the rate of the pentose phosphate pathway is required to meet the increased demand for reducing equivalents in the form of NADPH during hyperoxia. [From Allen, C.B. et al., Am. J. Physiol. 274 (3 Pt. 1), L320–L329 (1998).]

55. a.  Usually, phosphorylation of an enzyme causes a conformational change in the protein that subsequently alters its activity. For the bacterial isocitrate dehydrogenase, however, phosphorylation of an active site Ser residue introduces negative charges that repel the negatively charged isocitrate and prevent it from binding.  b. Con43.  Rearrange Equation 12.3 to solve for the [malate]/[fumarate] struction of the mutant supports this hypothesis. The introduction of ratio: the negatively charged Asp residue in place of the Ser residue simi[malate] __________ ​ΔG = ΔG°ʹ + RT ln ​     ​  larly introduces a negative charge to the active site and prevents iso­ [fumarate] citrate binding in the same manner. [From Dean, A.M. et al., J. Biol. Chem. 264, 20482–20486 (1989).] [malate] (ΔG−ΔG°′)/RT __________ ​ e​​  ​= ​     ​  [fumarate] 57.  When cells obtain energy in the absence of oxygen, glucose is oxidized to pyruvate, which is subsequently reduced to lactate with [malate] (0 J ⋅ ​mol​​  −1​ + 3400 J ⋅ ​mol​​  −1​)/8.3145 J  ⋅ ​K​​  −1​  ⋅ ​mol​​  −1​)(310 K) ​e​​  ​ = __________ ​     ​  concomitant regeneration of NAD+. The citric acid cycle is not active [fumarate] and the cells respond by down-regulating the activity of the citric [malate] ​ __________   ​  3.7 ​​   ​  = ​e​​  1.32​ = ___ acid cycle enzyme malate dehydrogenase. In the presence of oxygen, 1 [fumarate] pyruvate is converted to acetyl-CoA, which enters the citric acid cycle The ratio of malate to fumarate is 3.7 to 1, indicating that the reaction so that ATP can be produced by oxidative phosphorylation. Under proceeds in the direction of formation of malate. This is not a control aerobic conditions, malate dehydrogenase is absolutely essential for point for the citric acid cycle because the ΔG is close to zero, indicatthe regeneration of oxaloacetate as part of the citric acid cycle; thereing that it is a near-equilibrium reaction. fore, activity levels are much higher. 45. a.  The isotopic label on C4 of oxaloacetate is released as 14CO2 in 59.  The phosphofructokinase reaction is the major rate-control the α-ketoglutarate dehydrogenase reaction.  b.  The isotopic label on point for the pathway of glycolysis. Inhibiting phosphofructokinase C1 of acetyl-CoA is scrambled at the succinyl-CoA synthetase step. slows the glycolytic pathway, so the production of pyruvate and then Because succinate is symmetrical, C1 and C4 are chemically equiva­acetyl-CoA can be decreased when the citric acid cycle is operating at lent, so in a population of molecules, both C1 and C4 would appear maximum capacity and the citrate concentration is high. to be labeled (half the label would appear at C1 and half at C4). Con61.  Succinyl-CoA synthetase catalyzes the only substrate-level phossequently, one round of the citric acid cycle would yield oxaloacephorylation reaction in the citric acid cycle. The enzyme with the tate with half the labeled carbon at C1 and half at C4. Both of these ADP-specific β subunit could produce ATP in the brain and muscle to labeled carbons would be lost as 14CO2 in a second round of the citric meet the energy needs of these tissues. The enzyme with the GDP-speacid cycle. cific β subunit could produce GTP needed by phospho­enolpyruvate 47.  Reaction 5 is a. a phosphorylation; Reaction 2 is b. an isomerization; Reactions 3, 4, and 6 are c. oxidation–reductions; Reaction 7 is d. a hydration; Reaction 1 involves e. formation of a carbon-­carbon bond; and Reactions 3 and 4 are f. decarboxylations.

49. a.  Substrate availability: Acetyl-CoA and oxaloacetate levels regulate citrate synthase activity.  b.  Product inhibition: Citrate inhibits citrate synthase, NADH inhibits isocitrate dehydrogenase and α-ketoglutarate dehydrogenase, and succinyl-CoA inhibits α-ketoglutarate dehydrogenase.  c.  Feedback inhibition: NADH and succinyl-CoA inhibit citrate synthase. 51.  During the resting state, citric acid cycle activity is low as a result of the low activities of the two enzymes described. Any

carboxykinase for gluconeogenesis in the liver and kidney.

63.  A deficiency of succinate dehydrogenase would result in the accumulation of succinate. Because the succinyl-CoA synthetase reaction is reversible, succinate would be converted to succinyl-CoA at a greater rate than normal. The conversion of succinate to ­succinyl-CoA requires coenzyme A as a reactant, so if the reaction occurs at a greater rate, cellular levels of coenzyme A will decrease. 65.  Because of the enzyme deficiency, the citric acid cycle cannot be completed and glucose must be oxidized anaerobically. Lactate is a product of anaerobic oxidation of glucose and thus lactate levels are elevated in the patients. The pyruvate → lactate transformation is reversible, so if lactate levels rise due to increased glycolytic

S-48  ODD- N UM B ER E D S O LUT I O NS activity, pyruvate levels rise as well. However, lactate levels rise more (because there are other options for pyruvate as described in Solution 1), so the [lactate]/[pyruvate] ratio is increased in the patient. [From Bonnefont, J.-P. et al., J. Pediatrics 121, 255–258 (1992)]. 67. a.  Pyruvate carboxylase converts pyruvate to oxaloacetate, one of the reactants for the first reaction of the citric acid cycle. If the first reaction of the cycle cannot take place, the remaining reactions cannot proceed.  b.  Both reactants for the first reaction of the citric acid cycle can be produced from pyruvate. Pyruvate dehydrogenase converts pyruvate to acetyl-CoA while pyruvate carboxylase converts pyruvate to oxaloacetate. Stimulation of pyruvate carboxylase by acetyl-CoA occurs when excess acetyl-CoA is present and more oxaloacetate is needed. This regulatory strategy ensures that there are sufficient amounts of both of these reactants to begin the citric acid cycle. Note that because oxaloacetate acts catalytically as part of the citric acid cycle, its concentration does not need to match the acetyl-CoA concentration: a small amount of oxaloacetate, which is regenerated with each turn of the cycle, can process a much larger amount of acetyl-CoA. 69.  The amino acid aspartate and the glycolytic product pyruvate undergo a transamination in which aspartate’s amino group is transferred, leaving oxaloacetate, which is a citric acid cycle intermediate. 71.  Any metabolite that can be converted to oxaloacetate can enter gluconeogenesis and serve as a precursor for glucose. Biological molecules that are degraded to acetyl-CoA cannot be used as glucose precursors because acetyl-CoA enters the citric acid cycle and its two carbons are oxidized to carbon dioxide. Thus,  b. pentadecanoate,  c. ­glyceraldehyde-3-phosphate,  e. tryptophan, and  f.  phenylalanine can serve as gluconeogenic substrates because at least one of their breakdown products can be converted to oxaloacetate.  a. ­Palmitate and  d.  leucine are not glucogenic because their breakdown products are acetyl-CoA or one of its derivatives. ­Acetyl-CoA cannot be converted to pyruvate in mammals. 73.  Alanine can be converted to pyruvate via a reversible transamination reaction. In gluconeogenesis, pyruvate is converted to oxaloacetate via the pyruvate carboxylase reaction, then oxaloacetate is converted to phosphoenolpyruvate, and so on to produce glucose. If pyruvate carboxylase is deficient, alanine is converted to pyruvate, but the gluconeogenic pathway can go no further. 75.  Pyruvate carboxylase requires biotin as a cofactor (see Table 12.2). If the pyruvate carboxylase defect decreases the enzyme’s affinity for biotin, administering large doses of the vitamin might help treat the disease. Biotin treatment is ineffective for more severe forms of the disease in which the mutation does not occur in the biotin binding region or in which the enzyme is expressed at extremely low levels or not at all. 77.  As the citric acid cycle oxidizes acetyl carbons to CO2, NAD+ and ubiquinone collect the electrons and become reduced. Oxygen is required as the final electron acceptor in the electron transport chain that reoxidizes NADH and QH2. 79.  Ethanol is converted to acetaldehyde and then to acetate, and acetate is converted to acetyl-CoA. Acetyl-CoA then enters the glyoxylate pathway. The first step is the synthesis of citrate from acetyl-CoA and oxaloacetate. Citrate is isomerized to isocitrate. Isocitrate lyase then splits isocitrate into succinate and glyoxylate. Succinate leaves the glyoxysome and enters the mitochondrion, where it participates in the citric acid cycle. The glyoxylate fuses with a second molecule of acetyl-CoA to yield malate. Malate leaves the glyoxysome and enters the cytosol, where it is converted to oxaloacetate via the malate dehydrogenase reaction. Oxaloacetate can then enter gluconeogenesis to form glucose. 81.  Isocitrate lyase, a glyoxylate pathway enzyme, catalyzes the conversion of isocitrate to succinate and glyoxylate. The glyoxylate

product of this reaction goes on to form malate, which is eventually converted to glucose via gluconeogenesis. The succinate product is not part of this pathway and is “disposed of” by mitochondrial succinate dehydrogenase, which regenerates the oxaloacetate needed to keep the glyoxylate pathway operating. 83.  The concentration of AMP in muscle cells rises during periods of high muscle activity, indicating the need for ATP production via glycolysis and the citric acid cycle. AMP is converted to IMP by adeno­ sine deaminase, as shown in the figure. Breakdown of muscle protein produces aspartate, which joins with IMP to produce adenylosuccinate. This substrate is lysed to form AMP and fumarate. Fumarate is a citric acid cycle intermediate, and increasing its concentration leads to greater citric acid cycle activity. Thus, the purine nucleotide cycle acts as an anaplerotic mechanism for the citric acid cycle (at the expense of muscle protein). 85.  B vitamins are critical to the proper functioning of the citric acid cycle (see Table 12.2). Several vitamins, precursors to coenzymes used by citric acid cycle enzymes, are needed only in small amounts and by themselves do not provide energy. Biotin (B7) is a cofactor for pyruvate carboxylase, which catalyzes the most important anaplerotic reaction of the citric acid cycle. Niacin (B3) is required for synthesis of NAD+, a cofactor for pyruvate dehydrogenase and several other citric acid cycle dehydrogenases. Pantothenic acid (B5) is a component of coenzyme A. Riboflavin (B2) is a component of FAD, a cofactor required by succinate dehydrogenase. An individual with a vitamin B deficiency would not have sufficient citric acid cycle function to obtain energy from the catabolism of metabolic fuels. However, a varied diet provides ample B vitamins, so supplementation is not necessary, and B vitamins are water-soluble (see Table 12.2) so vitamins ingested in excess of what is needed are excreted. 87.  Enzymes of the glyoxylate pathway, particularly malate synthase and isocitrate lyase (which are unique to this pathway), would be inactivated. The glyoxylate pathway produces glucose from noncarbohydrate sources and is not required when glucose is available. Enzymes required for gluconeogenesis that are not involved in glycolysis would also be inactivated, mainly phosphoenolpyruvate carboxykinase and fructose-1,6-bisphosphatase. 89. a.  This reaction is an anaplerotic reaction in bacteria and plants, analogous to the pyruvate carboxylase reaction in animals. PPC produces oxaloacetate for the citric acid cycle to ensure its continued operation as a pathway for oxidizing fuel molecules and for producing intermediates for biosynthetic reactions.  b.  Acetyl-CoA and oxaloacetate are both required for the citrate synthase reaction that begins the citric acid cycle. If the concentration of acetyl-CoA rises, the concentration of oxaloacetate will need to increase as well, so acetyl-CoA stimulates the enzyme that produces its co-substrate. The activation by fructose-1,6-bisphosphate appears to be a feed-forward mechanism to ensure that sufficient oxaloacetate is present to condense with the acetyl-CoA produced by glycolysis and the pyruvate dehydrogenase reaction. Note that because oxaloacetate acts catalytically as part of the citric acid cycle, its concentration does not need to match the acetyl-CoA concentration: a small amount of oxaloacetate, which is regenerated with each turn of the cycle, can process a much larger amount of acetyl-CoA. 91.  Glutamine is converted to glutamate and ammonium by deamidation and then glutamate can react with any α-keto acid, resulting in the production of α-ketoglutarate and another amino acid. If α-ketoglutarate reacts with alanine, the products are glutamate and pyruvate. Pyruvate can be used as a substrate for gluconeogenesis.  b.  α-Ketoglutarate dehydrogenase, succinyl-CoA synthetase, succinate dehydrogenase, fumarase, and malic enzyme act in sequence to convert α-ketoglutarate to pyruvate.

OD D -NUMB ER ED SO LUTI O NS  S-49

The ​ΔG °′​value is negative and the reduction of pyruvate by NADH

Chapter 15 1.  a.  Oxaloacetate is oxidized and malate is reduced,  b.  pyruvate is oxidized and lactate is reduced,  c. NAD + is oxidized and NADH is reduced,  d.  fumarate is oxidized and succinate is reduced (see Table 15.1). 3.  a. O2 is oxidized and H2O is reduced,  b.  NO3– is oxidized and NO2– is reduced,  c.  lipoic acid is oxidized and dihydrolipoic acid is reduced,  d. NADP+ is oxidized and NADPH is reduced (see Table 15.1). 5.  NADP+ is the oxidized compound, NADPH is the reduced compound, and Ɛ °′ = –0.320 V (Table 15.1). Use Equation 15.2 as shown in Sample Calculation 15.1:

(see Section 13.1) is spontaneous under standard conditions.

15.  The electron acceptor is acetoacetate and the electron donor is NADH. Substitute their standard reduction potentials into Equation 15.3 as shown in Sample Calculation 15.2. Method 2 is illustrated here. ​ΔƐ °′ = Ɛ ​°′​ ​(​e​​  −​  acceptor)​​​ − Ɛ °′​ ​(​e​​  −​  donor)​​​ ΔƐ °′ = − 0.346 V − (​ − 0.315 V)​= − 0.031 V​

Use Equation 15.4 to calculate ​Δ​G °′ for this reaction: ​ΔG °′ = −nFΔℰ°′ ΔG °′ = −(2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(​ − 0.031 V)​

[NADPH] V  ​Ɛ = Ɛ °′ − _ ​ 0.026  ​  ​ln _________ ​  n    [​NADP​ +​]

ΔG °′ = 6000 J ⋅​  mol​​  −1​= 6.0 kJ ⋅​  mol​​  −1​    ​

(200 × ​10​ −6​M)       ​ ​ ln ​ ___________ Ɛ = − 0.320 V − _ ​ 0.026  V  2 (20 × ​10​ −6​M)

The ​Δ​G °′ value is positive and the reaction is not spontaneous under

Ɛ = − 0.320 V − 0.030 V = − 0.350 V​

17.  The electron acceptor is cytochrome a3 and the electron donor is cytochrome a. Substitute their standard reduction potentials into Equation 15.3 as shown in Sample Calculation 15.2. Method 2 is illustrated here.

standard conditions.

7.  Fumarate is the oxidized compound, succinate is the reduced compound, and Ɛ °′ = 0.031 V (Table 15.1). Use Equation 15.1: [succinate] ​   ​  ​Ɛ = Ɛ °′ − _ ​ RT   ​ln ________ nF [fumarate]

​ΔƐ °′ = Ɛ °′​ ​(​e​​  −​  acceptor)​​​ − Ɛ °′​ (​ ​e​​  −​  donor)​​​

(8.3145 J ⋅ ​K​ −1​ ⋅ ​mol​ −1​) (310 K) ___________ (100 × ​10​ −6​ M)     Ɛ = 0.031 V − ________________________ ​      ​  ​ ln    ​     −1 −1 (2)(96,485 J ⋅ ​V​  ​ ⋅ ​mol​  ​) (80 × ​10​ −6​M)

Δℰ °′ = 0.385 V − 0.29 V = 0.095 V​

Use Equation 15.4 to calculate ​Δ​G °′ for this reaction:

Ɛ = 0.031 V − (0.0134 V)(0.223) = 0.028 V​

​ΔG °′ = −nFΔℰ °′ 

9.  a.  Rearrange Equation 15.2 to solve for Ɛ °′, then enter the values given in the problem: ​[​A​ reduced]​ ​   ​ ln _ ​   ​  ​Ɛ = Ɛ °′ − _ ​ 0.026 V n    [​ ​A​ oxidized]​ ​ ​[​A​ reduced]​ ​ Ɛ °′ = Ɛ + _ ​ 0.026 V   ​ ln _______ ​   ​  n    [​ ​A​ oxidized]​ ​

ΔG °′ = −(1) (96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.095 V) ΔG °′ = − 9200 J ⋅ ​mol​​  −1​= −9.2 kJ ⋅ ​mol​​  −1​    ​

The ​Δ​G °′ value is negative and the reaction is spontaneous under standard conditions. 19.  Reverse the malate half-reaction and the sign of its Ɛ °′ to indicate oxidation, then combine the half-reactions and their Ɛ °′ values as shown in Sample Calculation 15.2. Method 1 is illustrated here. (Note that ubiquinone is Q and ubiquinol is QH2.)

(5 × ​10​ −6​ M) Ɛ °′ = 0.47 V + _ ​ 0.026 V          ​ln __________ ​     ​ 2 (200 × ​10​ −6​ M) Ɛ °′ = 0.47 V + (− 0.05 V) = 0.42 V​

​Q + ​2  H​​  +​+ 2 ​e​​  −​ → ​QH​  2​​

b.  A possible identity for substance A is nitrate. 11. a. Rearrange Equation 15.2 to solve for Ɛ °′ as shown in Solution 9, then enter the values given in the problem: ​[​C​  reduced]​​ ​ 0.026 V ​Ɛ °′ = Ɛ +  ​ _ ​ ln  ​ _   ​  n    ​[​C​  oxidized​​]​

−

+

−

malate → oxaloacetate + ​2  H​​  ​+ 2 ​e​​  ​

Ɛ °′ = 0.166 V

Q + malate− → ​QH​  2​​ + oxaloacetate− ΔƐ °′ = 0.211 V​ Use Equation 15.4 to calculate ​Δ​G °′ for this reaction:

(5 × ​10​​  −6​ M) 0.026 V Ɛ °′ = 0.29 V +  ​ _  ln ___________     ​  ​     ​ 1 (0.1 × ​10​​  −6​ M) Ɛ = 0.29 + 0.10 V = 0.39 V​

​ΔG °′ = − nFΔƐ °′ ΔG °′ = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.211 V) ΔG °′ = − 40,700 J ⋅ ​mol​​  −1​= − 40.7 kJ ⋅ ​mol​​  −1​​

c.  A possible identity for substance C is cytochrome a3. 13.  Pyruvate is reduced and NADH is oxidized in the reaction. Reverse the NADH half-reaction and the sign of its Ɛ °′ to indicate oxidation, then combine the half-reactions and their Ɛ °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here. ​pyruvate− + 2 ​H​​  +​+ 2 ​e​​  −​ → lactate−        Ɛ °′ = − 0.185 V NADH → ​NAD​​  +​ + ​H​​  +​+ 2 ​e​​  −​               Ɛ °′ =   0.315 V



Ɛ °′ =  0.130 V​ NADH + pyruvate− + ​H​​  +​ → ​NAD​​  +​+ lactate−    Δ 

The reduction of oxidation of malate by ubiquinone is spontaneous under standard conditions.

21.  The relevant reactions and their Ɛ °′ values are obtained from Table 15.1. Reverse the NAD+ half-reaction and the sign of its Ɛ °′ to indicate oxidation, then combine the half-reactions and their Ɛ °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here. ​Dihydrolipoic acid → lipoic acid + ​2 H​​ +​+ 2 ​e​​  −​  

Use Equation 15.4 to calculate ​Δ​G °′ for this reaction:

+

+

Ɛ °′ =

0.29 V

−

​NAD​​  ​ ​+ H​​  ​+ 2 ​e​​  ​ → NADH         Ɛ °′ = − 0.315 V

​ ΔG °′ = − nFΔ Ɛ °′ ΔG °′ = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.130 V) −1

Ɛ °′ = 0.045 V −

−1

ΔG °′ = − 25,100 J ⋅ ​mol​​  ​= − 25.1 kJ ⋅ ​mol​​  ​​

​Dihydrolipoic acid + NAD​​  +​ → lipoic acid + ​NADH + H​​ +​ ΔƐ °′= − 0.025 V​

S-50  ODD- N UM B ER E D S O LUT I O NS Use Equation 15.4 to calculate ​Δ​G °′ for this reaction: ​ΔG °′ = − nFΔ Ɛ °′ −1

−1

ΔG °′ = − (2)(96,485 J ⋅ ​V​​  ​  ⋅ ​mol​​  ​)(−  0.025 V) −1

−1

ΔG °′ = 4800 J ⋅ ​mol​​  ​= 4.8 kJ ⋅ ​mol​​  ​​

23.  Consult Table 15.1 for the relevant half-reactions involving acetaldehyde and NAD+. Reverse the acetaldehyde half-reaction and the sign of its Ɛ °′ value to indicate oxidation, then combine the halfreactions and their Ɛ °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here. ​acetaldehyde + ​H​  2​​O → acetate− + 3 ​H​​  +​+ 2 ​e​​  −​

Ɛ °′ = 0.581 V

​NAD​​  +​ + ​H​​  +​+ 2 ​e​​  −​ → NADH

Ɛ °′ = − 0.315 V

+

−

+

acetaldehyde + ​H​  2​​O + ​NAD​​  ​ → NADH + acetate + ​2  H​​  ​ ΔƐ °′ =  0.266 V​ Use Equation 15.4 to calculate ​Δ​G °′ for this reaction: ​ΔG °′ = − nFΔƐ °′ ΔG °′ = − (2) (96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)  (0.266 V) ΔG °′ = − 51,300 J ⋅ ​mol​​  −1​= − 51.3 kJ ⋅ ​mol​​  −1​​

The oxidation of acetaldehyde by NAD+ is spontaneous, as shown by

​​QH​  2​​ → Q + 2 ​H​​  +​+ 2 ​e​​  −​   

Ɛ = − 0.015 V

2 cyt c  ​(​Fe​​ 3+)​ ​+ 2 ​e​​  −​ → 2 cyt c  ​(​Fe​​ 2+)​ ​ 

Ɛ=

​QH​  2​​+ 2 cyt c  ​(​Fe​​ 3+)​ ​ → Q + 2 ​H​​  +​+ 2 cyt c  ​(​Fe​​ 2+)​ ​    ΔƐ =    0.241 V​ b.  Use Equation 15.4 to calculate ​Δ​G for this reaction: ​ΔG = − nFΔƐ  ΔG = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.241 V) ΔG = − 46,500 J ⋅ ​mol​​  −1​= − 46.5 kJ ⋅ ​mol​​  −1​​

The ΔG value is negative and the reaction is spontaneous under these conditions.

29.  Reverse the half-reaction for the iron–sulfur protein and the sign of its Ɛ °′ value to indicate oxidation, then combine the half-reactions and their Ɛ °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here. ​FeS (​ red)​ → FeS (​ ox)​+ ​e​​  −​  

Ɛ °′ = − 0.280 V

cyt ​c​ 1​​ (​ ​Fe​​ 3+)​ ​+ ​e​​  −​ → cyt ​c​ 1​​ (​ ​Fe​​ 2+)​ ​ 

Ɛ °′ = 0.215 V

FeS (​ red)​+ cyt ​c​ 1​​ ​(​Fe​​ 3+​)​ → FeS (​ ox)​+ cyt ​c​ 1​​ ​(​Fe​​ 2+​)​ ΔƐ °′ = − 0.065 V​ Use Equation 15.4 to calculate ΔG °′ for this reaction:

the negative ​Δ​G °′ value, so NAD+ is an effective oxidizing agent.

​ΔG °′ = − nFΔƐ °′

25.  Consult Table 15.1 for the relevant half-reactions involving ubiquinol and cytochrome c1. Reverse the ubiquinol half-reaction and the sign of its Ɛ °′ value to indicate oxidation, multiply the coefficients in the cytochrome c1 equation by 2 so that the number of electrons transferred is equal, and then combine the half-reactions and their Ɛ °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here.

ΔG °′ = −(1) (96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​) (− 0.065 V)

​​QH​  2​​ → Q + 2 ​H​​  +​+ 2 ​e​​  −​ 3+)

Ɛ °′ = − 0.045  V 2+)

2 cyt ​c​ 1​​ ​(​Fe​​  ​ ​+ 2 ​e​​  ​ → 2 cyt  ​c​ 1​​ ​(​Fe​​  ​ ​        Ɛ °′ = 0.22 V −

​QH​  2​​+ 2 cyt  ​c​ 1​​ ​(​Fe​​ 3+​)​ → Q + 2 cyt ​c​ 1​​ ​(​Fe​​ 2+​)​  + 2 ​H​​  +​  ΔƐ °′ =   0.175 V​ Use Equation 15.4 to calculate ​Δ​G °′ for this reaction: ​ΔG °′ = − nFΔƐ °′ ΔG °′ = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.175 V) −1

−1

ΔG °′ = − 33,800 J ⋅ ​mol​​  ​= − 33.8 kJ ⋅ ​mol​​  ​​

The ​Δ​G °′ value is negative and the reaction is spontaneous under

standard conditions.

27. a. Use Equation 15.2 to determine the Ɛ  values for these two half-reactions: +

−

Q + ​2  H​​  ​+ 2 ​e​​  ​→ ​QH​  2​​ Ɛ °′ = 0.045 V Q ​ H​  ​​ ​ ​ [ ] 2 0.026 V Ɛ = Ɛ °′ − ​ _       ​ ln _ ​   ​ n    [Q] ​ 0.026 V _ Ɛ = 0.045 V − ​      10  ​ ln  2 ​ Ɛ = 0.015 V  ​​          ​  ​  ​  ​ ​ ​  ( 2+​  )​   ​​​           3+ ​ )  ​ − ( 2 cyt c  ​ ​Fe​​  ​ ​+ 2 ​e​​  ​ → 2 cyt c  ​ ​Fe​​  ​ ​ Ɛ °′ = 0.235 V 2+ [​ cyt c  (​ ​Fe​​  ​)​]​ V  ___________  ​  Ɛ = Ɛ °′ − _ ​  0.026        ​ ln ​  n ​[cyt c  (​ ​Fe​​ 3+​)​]​ V  1 ​    Ɛ  = 0.235 V − _ ​ 0.026 ​  ln ​ _ 5 2 Ɛ = 0.256 V

​ ​ ​

0.256 V

ΔG °′ = 6300 J ⋅ ​mol​​  −1​= 6.3 kJ ⋅ ​mol​​  −1​​

The positive ΔG °′ value indicates that the electron transfer is unfa-

vorable under standard conditions. However, cellular conditions are not necessarily standard conditions, and the ΔG value for this reaction is likely to be negative. Also, since this reaction occurs as part of the electron transport chain, the electrons gained by cytochrome c1 will be passed along to Complex IV, in effect coupling the two reactions, which would also tend to make the process more favorable than the ΔG °′ value indicates.

31. a. Consult Table 15.1 for the relevant half-reactions involving O2 and NADH. Reverse the NADH half-reaction and the sign of its Ɛ  °′ value to indicate oxidation, then combine the half reactions and their Ɛ  °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here. ​​1 ⁄ 2​ ​O​  2​​ + 2 ​H​​  +​+ 2 ​e​​  −​ → ​H​  2​​O +

+

Ɛ °′ = 0.815 V −

NADH → ​NAD​​  ​ + ​H​​  ​+ 2 ​e​​  ​

Ɛ °′ = 0.315 V

NADH + 1​  ⁄ 2​ ​O​  2​​ + ​H​​  +​ → ​NAD​​  +​ + ​H​  2​​OΔƐ °′ = 1.130 V​

Use Equation 15.4 to calculate ΔG °′ for this reaction: ​ΔG °′ = − nFΔƐ °′ ΔG °′ = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(1.130 V) ΔG °′ = − 218,000 J ⋅ ​mol​​  −1​= − 218 kJ ⋅ ​mol​​  −1​​ b.  The synthesis of 2.5 ATP requires a free energy investment of 2.5 × 30.5 kJ​ ⋅ ​mol−1, or 76.3 kJ​ ⋅ ​mol−1. The efficiency of oxidative phosphorylation is therefore 76.3 ÷ 218 = 0.35, or 35%. 33. a. Electrons are transferred from succinate to ubiquinone (Q) in Complex II, yielding fumarate and ubiquinol (QH2): succinate + Q ⇌ fumarate + QH2. b.  The relevant reactions and their Ɛ °′ values are obtained from Table 15.1. Reverse the succinate half-reaction and the sign of its Ɛ °′ value to indicate oxidation, then combine the half reactions and their Ɛ °′ values, as shown in Sample Calculation 15.2. Method 1 is illustrated here.

OD D -NUMB ER ED SO LUTI O NS  S-51 ​succinate− → fumarate− + ​2  H​​  +​+ 2 ​e​​  −​

Ɛ °′ = − 0.031 V

​Q  +  2  H​​  +​+ 2 ​e​​  −​ → ​QH​  2​​ 

Ɛ °′ =  0.045 V

succinate− + Q → fumarate− + ​QH​  2​​  ΔƐ °′ =   0.014 V​

Use Equation 15.4 to calculate ΔG °′ for this reaction: ​ΔG °′ = − nFΔƐ °′ ΔG °′ = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)  (0.014 V) ΔG °′ = − 2,700 J ⋅ ​mol​​  −1​= − 2.7 kJ ⋅ ​mol​​  −1​​ c.  There is not enough free energy to drive ATP synthesis under standard conditions (ΔG °′ = 30.5 kJ​  ⋅ ​mol–1). 35. a. Electrons are transferred from reduced cytochrome c to molecular oxygen in Complex IV, yielding oxidized cytochrome c and water. Four cytochrome c molecules are required to deliver four electrons to reduce O2 to water: ​4 cytochrome c  ​(​Fe​​ 2+)​ ​ + ​O​  2​​ + 4​ H​​  +​ ⇌ 4 cytochrome c  (​ ​Fe​​ 3+)​ ​ + 2​  H​  2​​O​ b.  Consult Table 15.1 for the relevant half-reactions involving cytochrome c and oxygen. Reverse the cytochrome c half-reaction and the sign of its Ɛ °′ value to indicate oxidation, multiply the coefficients by 2 to obtain an equal number of electrons transferred, and then combine the half-reactions and their Ɛ °′ values. (Note that the equation shown in part a is divided by 2 in order to calculate the free energy associated with a pair of electrons as shown in Solutions 32–34.) ​2 cyt c  ​(​Fe​​ 2+)​ ​ → 2 cyt c  ​(​Fe​​ 3+)​ ​+ 2 ​e​​  −​ 1​  ⁄ 2​ ​O​ 

+ − 2​​ + 2 ​H​​  ​+ 2 ​e​​  ​ → ​H​  2​​O

Ɛ °′ = − 0.235 V Ɛ °′ = 0.815 V

2 cyt c  ​(​Fe​​ 2+​)​+ 1​  ⁄ 2​ ​O​  2​​ + 2​  H​​  +​ → 2 cyt c  ​(​Fe​​ 3+​)​ + ​H​  2​​O   ΔƐ °′ = 0.580 V​

Use Equation 15.4 to calculate ΔG °′ for this reaction: ​ΔG °′ = − nFΔƐ °′ ΔG °′ = − (2)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.580 V) ΔG °′ = − 112,000 J ⋅ ​mol​​  −1​= − 112 kJ ⋅ ​mol​​  −1​​ c.  The phosphorylation of ADP to ATP requires 30.5 kJ · mol−1 of free energy. Assuming that this reaction is 35% efficient (see Solution 31), (112 ÷ 30.5) × 0.35 = 1.3 mol ATP can be synthesized. 37. a. Since all of these inhibitors interfere with electron transfer somewhere in the electron transport chain, oxygen consumption decreases when any of the inhibitors are added to a suspension of respiring mitochondria. Adding any of these inhibitors prevents electrons from being transferred to oxygen, the final electron acceptor. 39.   a.  Malate is oxidized to oxaloacetate, with concomitant production of NADH (see Fig. 14.6), which enters the electron transport chain at Complex I. The oxidation of malate is blocked if NADH cannot donate its protons to the electron carriers in Complex I. Succinate is oxidized to fumarate (see Fig. 14.6), with concomitant production of FADH2, which enters the electron transport chain at Complex II. Complex I is bypassed, so the transfer of electrons from succinate is unaffected in the presence of a Complex I inhibitor. b. Antimycin A blocks electron transfer in Complex III (see Problem 37). Oxidation of both malate and succinate would be inhibited, as both of these compounds rely on Complex III. 41.  Adding succinate to rotenone-blocked mitochondria effectively bypasses the block as succinate donates its electrons to ubiquinone and electron transport resumes. Adding succinate is not an effective

bypass for antimycin A–blocked or cyanide-blocked mitochondria because succinate donates its electrons upstream of the block. 43.  Cyanide binds to the Fe2+ in the Fe–Cu center of cytochrome a3 (see Problem 37). When the iron in hemoglobin is oxidized from Fe2+ to Fe3+, cytochrome a3 can donate an electron to reduce the hemoglobin to Fe2+. This oxidizes the iron in cytochrome a3 to Fe3+. Cyanide does not bind to Fe3+, so it is released and Complex IV can again function normally. The cyanide binds to the Fe2+ in hemoglobin, where it does not interfere with respiration (although it does interfere with oxygen delivery). HbO2 (Fe2+) HbO2 (Fe3+) cytochrome a3 (Fe2+

Cu)

cytochrome a3 (Fe3+

Cu)

CN –

HbO2 (Fe2+)

45.  All of these enzymes catalyze reactions in which electrons are transferred from reduced substances, such as NADH, to ubiquinone. The flavin group, whose reduction potential is lower than that of ubiquinone and higher than that of NADH (Table 15.1), is ideally suited to shuttle electrons between the reduced NADH and the oxidized ubiquinone. 47.  Like the lipids that comprise the membrane, coenzyme Q is amphiphilic, with a hydrophilic head and a hydrophobic tail. The hydrophobic tail of coenzyme Q forms favorable van der Waals interactions with the hydrophobic acyl chains of the membrane phospholipids. Coenzyme Q literally dissolves in the membrane, which facilitates rapid diffusion. 49.  Cytochrome c is a water-soluble, peripheral membrane protein and is easily dissociated from the membrane by adding salt solutions that interfere with the ionic interactions that tether it to the inner mitochondrial membrane. Cytochrome c1 is an integral membrane protein and is largely water-insoluble due to the nonpolar amino acids that interact with the hydrophobic acyl chains of the membrane lipids. Detergents are required to dissociate cytochrome c1 from the membrane because amphiphilic detergents can disrupt the membrane and coat membrane proteins, acting as substitute lipids in the solubilization process. 51.  The dead algae are a source of food for aerobic microorganisms lower in the water column. As the growth of these organisms increases, the rates of respiration and O2 consumption increase to the point where the concentration of O2 in the water becomes too low to sustain larger aerobic organisms. 53.  In the absence of myoglobin, the mice developed several compensatory mechanisms to ensure adequate oxygen delivery to tissues. The symptoms described all involve increasing the amount of available hemoglobin. In this manner, hemoglobin takes over some of the functions usually performed by myoglobin. 55. a. Positively charged Lys and Arg side chains would be expected to form favorable ion pairs with the negatively charged head groups of phospholipids. b. An increase in the p 50 value indicates a decreased affinity for oxygen. When myoglobin binds to phospholipids in the mitochondrial membrane, its affinity for oxygen decreases, which facilitates the delivery of oxygen to the electron transport chain. [From Postnikova, G.B. and Shekhovtsova, E.A., Biochemistry (Moscow) 83, 168–183 (2018).] 57.  Fructose-2,6-bisphosphate is a potent stimulator of PFK (see Section 13.2). This results in greater flux through the glycolytic pathway and generation of large amounts of pyruvate. The citric acid cycle cannot accommodate the increased pyruvate. Consequently, pyruvate

S-52  ODD- N UM B ER E D S O LUT I O NS is converted to lactate instead of being converted to acetyl-CoA and entering the citric acid cycle. 59.  As a result of the exercise program, the number of mitochondria in the muscle cells of the participants increased, as indicated by an increase in DNA content. The increase in Complex II activity is of similar magnitude, which might be the result of the increased number of mitochondria. However, the total electron transport chain activity is twofold greater after the exercise intervention, indicating that mitochondrial function increased as well, to meet the cellular demands for ATP required by the exercise regimen. These results suggest that even though evidence indicates that oxidative damage to mitochondria increases with age, exercise can help maintain or increase mitochondrial function. [From Menshikova, E.V. et al., J. Gerontol. A-Biol. 61, 534–540 (2006).] 61. a. During electron transport, four protons are transferred from the matrix to the intermembrane space in Complex I, none in Complex II, four protons in Complex III, and two protons in Complex IV. b. The passage of two electrons from NADH to oxygen utilizes Complexes I, III, and IV and results in the translocation of 10 protons. If the translocation of each proton costs 20.1 kJ ∙ mol–1, the cost of translocating 10 protons is 201 kJ ∙ mol–1.  c.  This is about the same as the free energy released when NADH is oxidized by oxygen (−218 kJ ∙ mol–1; see Solution 31a). The free energy released in the reaction is captured in the formation of a proton gradient, which is used to drive ATP synthesis. 63.  The free energy change for generating the electrical imbalance is calculated using Equation 15.6: ​ΔG = ZFΔψ ΔG = (1)(96,485 J ⋅ ​V​​  −1​  ⋅ ​mol​​  −1​)  (0.081 V) ΔG = 7800 J ⋅ ​mol​​  −1​= 7.8 kJ ⋅ ​mol​​  −1​​ 65.  Use the rearrangement of Equation 15.7 as shown in Sample Calculation 15.3: ​ΔG = 2.303​RT ​(​pH​  in​​ − ​pH​  out​​)​ + ZFΔψ ΔG = 2.303 (8.3145 J ⋅ ​K​​  −1​  ⋅ ​mol​​  −1​)(310 K)(7.6 − 7.2) ​​  +  (1)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)  (0.200 V)​ ΔG = 2400 J ⋅ ​mol​​  −1​+ 19,300 J ⋅ ​mol​​  −1​= 21.7 kJ ⋅ ​mol​​  −1​​

71. a. 

K+

Nigericin

H+

K+ Valinomycin

MATRIX INTERMEMBRANE SPACE CYTOSOL

b.  Potassium ions enter the matrix with the assistance of valinomycin. These ions are then exported by nigericin in exchange for protons. Importing protons into the mitochondrial matrix dissipates the proton gradient. Since the proton gradient serves as the energy reservoir that drives ATP synthesis, no ATP can be synthesized. 73.  The partial disassembly does not interfere with the redox reactions in the “arm” portion of Complex I but weakens the link between the redox reactions and proton translocation (which occurs in the membrane-embedded portion). As a result, the protonmotive force is decreased, leading to less ATP produced by ATP synthase. [From García-Ruiz, I. et al., BMC Biology 11, 88 (2013).] 75.  Fo acts as a proton channel as the c ring rotates, feeding protons through the a subunit (see Fig. 15.24). The addition of F1 blocks proton movement because the γ shaft rotates along with the c ring. In this system, the γ subunit and the c ring can rotate only when the binding change mechanism is in operation, that is, when the β subunits are binding and releasing nucleotides. ATP or ADP + Pi must be added to the system in order for the γ subunit to move. 77.  Inactivation of one c subunit by DCCD blocks all proton translocation by Fo since the movement of protons across the membrane requires continuous rotation of the c ring. Without this rotation, the γ subunit of F1 cannot move, and therefore the β subunits cannot undergo the conformational changes necessary to synthesize or hydrolyze ATP by the binding change mechanism.

ΔG − ZFΔψ ​pH​  in​​ − ​pH​  out​​ = ______________ ​        2.303RT

79. a. For yeast, the ratio is 10 H+ per 3 ATP, or 3.33. For the chloroplast, the ratio is 14 H+ per 3 ATP, or 4.67. b. Since it has 10 c subunits, the yeast enzyme can theoretically produce 3 ATP for every 10 protons translocated. In the chloroplast, 3 ATP are synthesized for every 14 protons. Thus, the bacterium is more efficient in its use of the proton gradient established during electron transport and has a higher ratio of ATP produced per oxygen consumed.

mitochondrial membrane is required for oxidative phosphorylation to take place. Without an intact membrane, an electrochemical gradient, which is the energy reservoir that drives ATP synthesis, cannot be established, and ATP synthesis does not occur.

85. a. Substrate-level phosphorylation is catalyzed by phosphoglycerate kinase and pyruvate kinase in glycolysis and by succinyl-CoA synthetase in the citric acid cycle. b. Most ATP is generated by oxidative phosphorylation. Of the 32 ATP generated per mole of glucose,

67.  Use the rearranged form of Equation 15.7 shown in Sample Calculation 15.3 and solve for (pHin – pHout), and then enter the values given in the problem: ​ΔG = 2.303​RT ​(​pH​  in​​ − ​pH​  out​​)​ + ZFΔψ

81.  One complete revolution of the c ring generates 3 ATP. (30,500 J ⋅ ​mol​​  −1​) − (1)(96,485 J ⋅ ​V​​ −1​  ⋅ ​mol​​  −1​)(0.170 V) ​ p H​  in​​ − ​pH​  out​​ = _____________________________________________       ​        ​ 2.303(8.3145 J ⋅ ​K​​  −1​  ⋅ ​mol​​  −1​)(298  K) 6000 revolutions ______________ ​​     ​  ×_ ​  3 ATP  ​  ×_ ​  1 min ​ = 300 ATP ⋅ ​s​​  −1​​ −1 60 s min revolution ​ 14,100 J ⋅ ​ m ol​​  ​pH​  in​​ − ​pH​  out​​ = _________________ ​        ​= 2.5​ −1 5700 J ⋅ ​mol​​  ​ 83.  A total of 32 ATP is obtained from the exergonic oxidation of glucose under aerobic conditions (see Fig. 14.13), whereas only 2 69. a. The pH of the intermembrane space is lower than the pH of the ATP are produced when glucose is oxidized in the absence of oxygen mitochondrial matrix because protons are pumped out of the matrix, to lactate or ethanol (see Section 13.1). Organisms that can oxidize across the inner membrane, and into the intermembrane space. The glucose in the presence of oxygen have an advantage over anaerobic increase in concentration of protons in the intermembrane space organisms because these organisms can extract more energy per gludecreases the pH; the deficit of protons in the matrix results in an cose. This may have been important in evolution. increase in pH. b. Detergents disrupt membranes. An intact inner

OD D -NUMB ER ED SO LUTI O NS  S-53 4 ATP are generated via substrate level phosphorylation (2 ATP in glycolysis and 2 ATP equivalents—usually in the form of GTP— in the citric acid cycle) and 28 ATP via oxidative phosphorylation (see Fig. 14.13). 87. a. Aerobic oxidation of glucose yields 32 ATP per glucose whereas alcoholic fermentation of glucose by the yeast yields only 2 ATP per glucose (see Solution 83). Assuming that the energy needs of the yeast cell remain constant under both aerobic and anaerobic conditions, the catabolism of glucose by the yeast will be 16-fold greater in the absence of oxygen than in the presence of oxygen in order to produce the same amount of ATP. Thus, the rate of conOHexposed to oxygen sumption of glucose decreases when the cells are H+ 2 the same because fewer glucose molecules must be oxidized to NO yield amount of ATP. b. Both ratios will initially increase, as the citric acid cycle (which does not operate under anaerobic conditions) produces H+ more NADH equivalents for electron transport. [ATP]/[ADP] NOThe 2 ratio will also increase since aerobic oxidation of glucose produces more ATP per mole of glucose than anaerobic oxidation (as described in part a). ATP and NADH will “reset” the equilibrium by inhibiting the regulatory enzymes of glycolysis and the citric acid cycle, slowing down these processes. Eventually, the [NADH]/[NAD+] and [ATP]/ [ADP] ratios return to their “original” values. 89.  The hormone instructs the cell to increase its rate of fuel oxidation. At the same time, the hormone signal activates protein kinase A to phosphorylate and inactivate IF1. ATP synthase will be more active and able to convert the energy of fuel oxidation to the energy of ATP. 91.  By decreasing both the rate of electron transport (see Solution 44) and ATP synthesis, fluoxetine decreases the rate of ATP production in the brain. The brain relies on a constant source of ATP for proper function, so decreased ATP production could lead to an impairment of brain function. 93.  The donation of a pair of electrons to Complex IV results in the synthesis of about 1.3 ATP per atom of oxygen (½ O2; see Solution 35c). Therefore, the P:O ratio for this compound is 1.3. 95. a. The import of ADP (net charge –3) and the export of ATP (net charge –4) represent a loss of negative charge inside the mitochondria. This decreases the difference in electrical charge across the membrane since the outside is positive due to the translocation of protons during electron transport. Consequently, the gradient is diminished by the activity of the ADP/ATP carrier. b. The activity of the Pi–H+ symport protein diminishes the proton gradient by allowing protons from the intermembrane space to reenter the matrix.  c.  Both transport systems are driven by the free energy of the electrochemical proton gradient. 97.  Locating hexokinase II on the outer mitochondrial membrane is strategic because ATP produced by oxidative phosphorylation is exported from the mitochondrial matrix to the cytosol via the ADP/ ATP carrier. Hexokinase II is poised to capture one of its reactants— ATP—so the phosphorylation of glucose to glucose-6-phosphate (G6P) is especially rapid. This allows the tumor cell to funnel G6P into cellular pathways needed to sustain the cell, such as glycolysis and the pentose phosphate pathway. Akt (protein kinase B) also promotes cell growth and survival in other ways. [From Mathupala, S.P. et al., Oncogene 25, 4777–4786 (2006).] 99. a. Lipid-soluble dinitrophenol crosses the inner mitochondrial membrane and loses a proton in the high pH environment of the mitochondrial matrix, forming the dinitrophenolate ion. The ion can cross the inner mitochondrial membrane to return to the matrix, despite its negative charge, because it is resonance-stabilized, which

delocalizes the negative charge (see figure). In the low-pH environment of the intermembrane space, the dinitrophenolate anion becomes re-protonated and the cycle begins again. OH

O–

H+

NO2

O– N+

O–

O

N+

O

O–

H+ NO2 O–

O– N+

O

N+

O O

O–

N+

O–

O–

N+

O

O

N+

O O

O–

N+

N+

O–

–

O

N+

–

O– O O–

O–

b.  Uncouplers that can ferry protons across the inner mitochondrial membrane, as described in part a, dissipate the proton gradient established by electron transport. In the presence of the uncoupler, electron transport still occurs, but the free energy released by the process is dissipated as heat instead of being harnessed to synthesize ATP. This lends support to the chemiosmotic theory, which describes the establishment of a proton gradient during electron transport as being crucial to the production of ATP. 101.  Dinitrophenol (DNP) uncouples electron transport from oxidative phosphorylation by dissipating the proton gradient. Electron transport still occurs, but the energy released by electron transport is dissipated as heat instead of being harnessed to synthesize ATP. One might think that DNP would be an effective diet aid because the sources of the electrons for the electron transport chain are dietary carbohydrates and fatty acids. If the energy of these compounds is dissipated as heat instead of used to synthesize ATP (which would then be used for, among other processes, the synthesis of fatty acids in adipocytes), weight gain from the ingestion of food could theoretically be prevented. 103. a. When UCP1 is stimulated in normal mice, oxidative phosphorylation is uncoupled from electron transport. This means that the ATP yield per substrate molecule oxidized decreases since ATP must then be produced via substrate level phosphorylation. The cell’s energy needs are not met, which increases the rate of glycolysis and the citric acid cycle, and eventually electron transport, in a vain attempt to synthesize more ATP. Since oxygen is the final electron acceptor in electron transport, oxygen consumption increases in order to keep up with the increased rate of electron transport. In knockout mice, since there is no UCP1, oxidative phosphorylation is not uncoupled. Therefore, ATP can be synthesized via oxidative phosphorylation. Since the energy needs of the cell are being met, the rate of electron transport and oxygen consumption do not increase. b. In normal mice, the cold temperature stimulates the uncoupling protein. As a result, the energy of electron transport is dissipated as heat rather than used to synthesize ATP. This helps the mice maintain normal body temperature. However, the UCP1-knockout mice lack the uncoupling protein and are unable to uncouple oxidation from phosphorylation. Thus, they could not generate “extra” heat and their body temperatures decreased as a result. [From Enerbäck, S. et al., Nature 387, 90–94 (1997).] 105.  The synthesis of the uncoupling protein increases with decreasing temperature, presumably by an increase in transcription of the mRNA that codes for the uncoupling protein (although a decrease

S-54  ODD- N UM B ER E D S O LUT I O NS in the rate of mRNA degradation would produce the same results). The increased amount of mRNA likely results in an increase in the concentration of the uncoupling protein so that the mitochondrial proton gradient would be dissipated. Thus, at low temperatures, the potato could generate heat rather than ATP. [From Laloi, M. et al., Nature 389, 135–136 (1997).] 107.  A starving animal has less metabolic fuel to be oxidized, which would lower O2 consumption, but the α-ketoglutarate effect on ATP synthase represents an additional mechanism to lower O2 consumption, since fuel oxidation (electron transport) is coupled to the action of ATP synthase. Inhibition of ATP production could prolong life­ span by favoring maintenance over growth, which delays aging. Lower oxygen consumption would also decrease the production of oxygen free radicals that damage the organism over time. [From Chin, R.M. et al., Nature 510, 397–401 (2014).]

Chapter 16 1. a.  proton translocation C, M,  b. photophosphorylation C,  c.  photooxidation C,  d.  quinones C, M,  e.  oxygen reduction M,  f.  water oxidation C,  g.  electron transport C, M,  h. oxidative phosphorylation M,  i.  carbon fixation C,  j.  NADH oxidation M,  k.  Mn cofactor C,  l.  heme groups C, M,  m.  binding change mechanism C, M,  n.  iron–sulfur clusters C, M,  o. NADP + reduction C. 3. HO H

CH2OH O H OH H H

OH

O H

O

CH2 HC

O

C

R1 O

CH2

O

C

R2

5.  Use Planck’s law (Equation 16.1) to determine the energy of a single photon (see Sample Calculation 16.1), then multiply by Avogadro’s number (N) to calculate the energy of a mole of photons: ​ E = ___ ​ hc ​  λ (6.626 × ​10​​  −34​ J  ·  s)  (2.998 × ​10​​  8​ m  · ​s​​  −1​) ________________________________     E =      ​   ​ 680 × ​10​​  −9​ m E = 2.9 × ​10​​ −19​ J

E = (2.9 × ​10​​ −19​  J  · ​photon​​  −1​)(6.022 × ​10​​  23​ photons  · ​mol​​  −1​) 5

−1

−1

E = 1.8 × ​10​​  ​ J  · ​mol​​  ​= 180 kJ · ​mol​​  ​​

7. a.  The color of the light is orange, so phycocyanins are blue.  b.  Use Planck’s law (Equation 16.1) to determine the energy of a single photon (see Sample Calculation 16.1): hc ​  ​ E = ​ ___ λ (6.626 × ​10​​  −34​ J  ·  s)  (2.998 × ​10​​  8​ m  · ​s​​  −1​)           ​ E = _______________________________ ​  620 × ​10​​  −9​ m

E = 3.2 × ​10​​ −19​ J​

9.  Multiply Planck’s law (Equation 16.1) by Avogadro’s number (N ), then solve for λ: ​ E = ___ ​ hc ​ × N λ   λ = ____ ​ hcN  ​ E



(6.626 × ​10​​  −34​ J  ·  s)  (2.998 × ​10​​  8​  m  · ​s​​  −1​)  (6.022 × ​10​​  23​ ​mol​​  −1​)  ​            λ = ________________________________________________ ​  250 × ​10​​  3​ J  · ​mol​​  −1​



λ = 4.8 × ​10​​ −7​m = 480 nm​

11.  If the synthesis of each ATP requires 30.5 kJ · mol−1, then 5.9 mol ATP (180 ÷ 30.5) could be synthesized. 13.  First, convert the energy per photon to the energy contained in a mole of photons: 6.022 × ​10​ 23​photons 3.2 × ​10​ −19 ​ ​J  ​​ __________      × __________________ ​   ​  = 190,000 J · ​mol​​  −1​ = 190 kJ · ​mol​​  −1​ ​​​ photon mol I f the synthesis of each ATP requires 30.5 kJ · mol−1, then 6.2 mol ATP (190 ÷ 30.5) could be synthesized. 15.  Because the algae appear red, red light is transmitted rather than absorbed. Therefore, the photosynthetic pigments in the red algae do not absorb red light but absorb light of other wavelengths. 17.  A

D

N

N

N

N

B

C

 he central metal ion in chlorophyll a is Mg2+, whereas in heme b T the central metal ion is Fe2+. In chlorophyll a, there is a cyclopentanone ring fused to ring C. Ring B in chlorophyll a has an ethyl side chain; the chain in heme b is unsaturated. The propionyl side chain in ring D of chlorophyll a is esterified to a long, branchedchain alcohol. 19.  The buildup of the proton gradient indicates a high level of activity of the photosystems. A steep gradient could therefore trigger photoprotective activity to prevent further photooxidation when the proton-translocating machinery is operating at maximal capacity. 21.  The order of action is water–plastoquinone oxidoreductase (Photosystem II), then plastoquinone–plastocyanin oxidoreductase (cytochrome b6 f ), then plastocyanin–ferredoxin oxidoreductase ­(Photosystem I). 23.  If electrons cannot be transferred to Photosystem I, then Photosystem II remains reduced and cannot be reoxidized. The photosynthetic production of oxygen ceases. No proton gradient is generated, so ATP synthesis does not occur in the presence of DCMU. 25. a.  Like the lipids that compose the membrane, plastoquinone is amphiphilic, with a hydrophilic head and a hydrophobic tail. “Like dissolves like,” so plastoquinone literally dissolves in the thylakoid membrane, which facilitates rapid diffusion.  b. Coenzyme Q in the mitochondrial electron transport chain is highly lipophilic and exhibits similar behavior in the inner mitochondrial membrane. 27.  The difference in reduction potential between P680* and P680 is −0.8 V − 1.15 V = −1.95 V (see Fig. 16.22). ​ΔG°′ = −nFΔƐ °′ ΔG°′ = −(1) (96,485 J · ​V​​  −1​  · ​mol​​  −1​)  (−1.95 V) ΔG°′ = 188,000 J · ​mol​​  −1​= 188 kJ · ​mol​​  −1​​

OD D -NUMB ER ED SO LUTI O NS  S-55 29.  Use Equation 15.7, as applied in Sample Calculation 15.3. The matrix and the stroma are both in. ​ΔG = 2.303 RT (​pH​ in​​ − ​pH​ out​​) + ZFΔψ ΔG = 2.303(8.3145 J · ​K​​  −1​  · ​mol​​  −1​) (298 K) (3.5) + (1) (96,485 J · ​V​​ −1​  · ​mol​​  −1​) (− 0.05 V) −1

−1

−1

ΔG = 20,000 J · ​mol​​  ​− 4800 J · ​mol​​  ​= 15.2 kJ · ​mol​​  ​​

31.  Consult Table 15.1 for the reduction potentials of the relevant half-reactions, reversing the sign for the water oxidation half-­ reaction, as shown in Sample Calculation 15.2: ​ ​H2​  ​O → __ ​  2 1  ​  ​O2​  ​+ 2 ​H​ +​+ 2 ​e​ −​



Ɛ °′ = − 0.815 V

​NADP​ +​+ ​H​ +​+ 2 ​e​ −​→ NADPH

Ɛ °′ = − 0.320 V

​H2​  ​O + ​NADP​ +​→ __ ​  2 1  ​  ​O2​  ​+ NADPH + ​H​ +​ ΔƐ °′ = − 1.135 V​

Use Equation 15.4 to calculate ΔG°′: ​ΔG°′ = − nFΔƐ °′ −1

−1

ΔG°′ = − (2) (96,485 J · ​V​​  ​  · ​mol​​  ​) (− 1.135 V) ΔG°′ = 219,000 J · ​mol​​  −1​​  ivide by Avogadro’s number to obtain the free energy change per D molecule: −1

219,000 J · ​mol​  ​ _________________________    ​​      ​ = 3.6 × ​10​ −19​J · ​molecule​ −1​ 23

6.022 × ​10​  ​molecules · ​mol​ −1​

33.  The final electron acceptor in photosynthesis is NADP+. The final electron acceptor in mitochondrial electron transport is oxygen. 35.  The other reaction product is Fe​​(CN)​ 64− ​  ​​. The experiment was significant because it demonstrated that CO2 is not the electron acceptor in the light reactions. The final electron acceptor was later found to be NADP+ (see Solution 33). [From Walker, D.A., Photosynth. Res. 73, 51–54 (2002).] 37. a.  An uncoupler dissipates the transmembrane proton gradient by providing a route for translocation other than ATP synthase. Therefore, chloroplast ATP production would decrease.  b. The uncoupler would not affect NADP + reduction since light-driven ­electron-transfer reactions would continue regardless of the state of the proton gradient. 39. a. More c subunits mean that more protons are required to rotate the ATP synthase through one ATP-synthesizing step. Therefore, more photons must be absorbed to drive the translocation of more protons, so the quantum yield decreases.  b.  Cyclic electron flow contributes to the proton gradient and therefore leads to ATP synthesis. However, carbon fixation by the Calvin cycle requires NADPH also, so the additional photons that drive cyclic flow do not lead to more carbon fixed. Consequently, the quantum yield decreases. 41.  Paraquat is an herbicide. The electrons accepted from ferredoxin by paraquat are not transferred to NADP+ to generate NADPH. NADP+ is the final electron acceptor in photosynthesis (see Solution 33). The lack of NADPH decreases carbon fixation, and the production of the reactive superoxide damages the plant. Eventually the plant dies. 43.  This statement is false. The “dark” reactions do not require darkness in order to proceed. Sometimes the “dark” reactions are called “light-independent” reactions in order to specify that these reactions do not directly require light energy. This term is also misleading

because the “dark” reactions of the Calvin cycle do require the products of the light reactions—ATP and NADPH—in order to proceed. Thus, for a majority of plants, the “dark” reactions actually occur during the day when the light reactions are operational and can produce the needed ATP and NADPH. 45.  3-Phosphoglycerate is the first stable radioactive intermediate that forms when algal cells are exposed to 14CO2. The radioactive label is found on the carboxyl group of the compound. 47.  The increase in mass comes from carbon dioxide obtained from the atmosphere. CO2 is the carbon source for cellulose, a major structural component of the tree. Water also contributes to the increase in mass. Soil nutrients contribute a very small percentage of the mass of the full-grown oak tree. 49.  Normally, plants must synthesize large quantities of rubisco, a protein whose constituent amino acids all contain nitrogen. If rubisco had greater catalytic activity, the plant might produce less of the enzyme, thereby decreasing its need for nitrogen. 51. a.  The unprotonated Lys side chain serves as a nucleophile when reacting with CO2. At high pH, a higher percentage of ε-amino groups are in the unprotonated form.  b.  During the light reactions, protons from the stroma are released into thylakoid lumen, resulting in a proton gradient. The proton movement results in a lower concentration of protons in the stroma and a higher pH, which favors the unprotonated form of the Lys side chain.  c.  During the day, the light reactions supply NADPH, which stimulates the activity of the CA1P phosphatase (see Solution 50). This allows the phosphatase to hydrolyze the phosphate group from CA1P, removing the inhibitor from the rubisco active site and allowing the carboxylase enzyme access to modify the essential Lys residue. 53.  Grasses turn brown because they undergo photorespiration in hot dry conditions. Rubisco reacts with oxygen to form 2-phosphoglycolate, which subsequently consumes large amounts of ATP and NADPH. CO2 concentrations are low because the plants close their stomata in order to avoid loss of water when the weather is hot and dry (see Box 16.A). Without CO2, photosynthesis does not occur and the grass turns brown. However, C4 plants such as crabgrass can generate CO2 from oxaloacetate, which can enter the Calvin cycle. Carbon fixation occurs and the crabgrass thrives even in hot dry weather. 55. Glyceraldehyde 3-phosphate

Pi + NADP+

NADPH

1,3-Bisphosphoglycerate ADP ATP 3-Phosphoglycerate

Dihydroxyacetone phosphate STROMA

CYTOSOL

Dihydroxyacetone phosphate

3-Phosphoglycerate ATP

Glyceraldehyde 3-phosphate

NAD+ + Pi

ADP 1,3-Bisphosphoglycerate NADH

S-56  ODD- N UM B ER E D S O LUT I O NS 57. a.  PEPC catalyzes the formation of oxaloacetate, one of the two reactants for the first reaction of the citric acid cycle. Anaplerotic reactions are important because they replenish citric acid cycle intermediates (see Fig. 14.19). If oxaloacetate is unavailable, the citric acid cycle cannot continue.  b.  Acetyl-CoA is an allosteric activator of PEPC. When the concentration of acetyl-CoA rises, additional oxaloacetate will be required to react with it in the first reaction of the citric acid cycle. Activation of PEPC by acetyl-CoA will lead to increased production of the required oxaloacetate. 59.  Phosphate inhibits AGPase. Pyrophosphate, a direct product of the reaction, is subsequently hydrolyzed to 2 P i, so it is likely that Pi acts as a feedback inhibitor of AGPase. This is supported by the data.  a.  Plants grown in medium lacking the phosphate inhibitor were able to increase their rate of starch synthesis.  b.  The same result is observed in plants incubated with mannose, because cellular phosphate levels are depleted by the conversion of mannose to mannose-6-phosphate. [From Kleczkowski, L.A., Annu. Rev. Plant Physiol. Mol. Biol. 45, 339–368 (1994).] 61. a.  Glyphosate herbicides are effective at killing weeds because glyphosate inhibits the plant EPSPS enzyme required for the synthesis of aromatic amino acids. The transgenic crops are protected from the herbicide because these crops contain the bacterial enzyme that is not subject to inhibition by glyphosate. Using this strategy allows for weed eradication while preserving the desired crop.  b. Glyphosate is not toxic to humans because humans do not contain the enzyme EPSPS. This pathway was lost during evolution and the result is that phenylalanine is an essential amino acid that must be obtained in the diet (see Table 12.1).

formation of atherosclerotic plaques) and transporting it to the liver for excretion. 11. 

Intestine Dietary lipids Chylomicrons TAGs

Cholesterol TAGs

VLDL

Cholesterol

Liver

IDL

Peripheral tissues HDL

LDL

13.  During fasting or exercise, epinephrine binds to a G protein– coupled receptor on the surface of the adipocyte and activates a G protein. The α subunit of the G protein swaps GDP for GTP, and the βγ dimer dissociates from the α subunit. The α subunit with GTP bound activates adenylate cyclase, which converts intracellular ATP to cAMP. The cAMP then activates protein kinase A, which phosphorylates and activates hormone-sensitive lipase. The lipase hydrolyzes fatty acids from the stored triacylglycerols; the fatty acids then leave the adipocyte, bind to albumin, and are transported to other ­tissues through the circulation. 15. a.  ​fatty acid + CoA + ATP → ​fatty acyl-CoA + AMP + PP​ i​ 

ΔG°′ =

 0 kJ ⋅ ​mol​ −1

PP​ ​  i​+ ​H​ 2​O → 2​P​ i​  ΔG°′ = − 19.2 kJ ⋅ ​mol​ −1​

Chapter 17

fatty acid + CoA + ATP + ​H​ 2​O → fatty acyl-CoA + AMP + ​2 P​ i​ 

1.  The lipoproteins increase in density as the percentage of protein content increases and the percentage of lipid content decreases. Thus, chylomicrons have the lowest density and HDL particles have the highest density.



3. a. Cholesterol and  d.  phospholipids are amphipathic;  b. cholesteryl esters and  c.  triacylglycerols are nonpolar. 5. a. The total serum cholesterol value does not indicate the distribution of the cholesterol between the LDL and HDL particles. The HDL/LDL ratio provides this information.  b.  A high HDL/LDL cholesterol ratio is desirable. HDL particles remove cholesterol from cells and are therefore anti-atherogenic. LDL particles deposit lipids in arterial walls, forming the base of an atherosclerotic plaque that can lead to a heart attack or stroke if the plaque ruptures. A high concentration of HDL and a low concentration of LDL results in a high HDL/LDL cholesterol ratio and indicates that the patient is at low risk for atherosclerosis. 7.  Mutations could occur in the protein component of LDL that recognizes the receptor, in the receptor itself to preclude LDL binding, in proteins that position the LDL receptor on the plasma membrane, or in the protein machinery that allows the LDL particle to enter the cell. 9.  Whereas lipoproteins such as VLDL are secreted from the liver into the bloodstream to deliver lipids to peripheral body tissues, HDL particles do the opposite by removing excess cholesterol from peripheral tissues (including the macrophages that initiate the

ΔG°′ = − 19.2 kJ ⋅ ​mol​ −1​

b.  The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): ​​Keq ​  ​​ = ​e​​  −ΔG°′/RT −(−19,200 J ⋅ ​mol​​  K​ ​ eq​​ = ​e​​ 

−1

​)/(8.3145 J  ⋅ ​K​​  −1​  ⋅ ​mol​​  −1​)(298 K)

7.75 ​ = 2.3 × ​10​​  3​​ K​ ​ eq​​ = ​e​​ 

c.  The reaction is spontaneous under standard conditions because the value of ΔG°′ is negative and Keq is a large number (greater than 1), indicating that the products are favored. 17.  The fatty acyl-CoA, once delivered to the mitochondrial matrix, enters β oxidation. The continual removal of this product keeps the entire transport process running in the direction of acyl-CoA delivery to the mitochondrial matrix. 19.  If carnitine is deficient, fatty acid transport from the cytosol to the mitochondrial matrix (the site of β oxidation) is impaired. Fatty acid oxidation generates a great deal of ATP to power the muscle, so in the absence of fatty acid oxidation the muscle relies on stored glycogen or uptake of circulating glucose to obtain the necessary ATP. Muscle cramping is exacerbated by fasting because the concentration of circulating glucose is decreased and glycogen stores are depleted. Exercise also increases muscle cramping because the demand for ATP by the muscle is greater.

OD D -NUMB ER ED SO LUTI O NS  S-57 21. a. Medium-chain acyl-CoA (4–12 carbons) accumulates in individuals with an MCAD deficiency because the conversion of fatty acyl-CoA to enoyl-CoA is blocked. Acylcarnitine esters would also accumulate.  b.  In the absence of medium-chain acyl-CoA dehydrogenase, a large portion of the fatty acids in the individual’s diet would be unable to be metabolized. The body turns to carbohydrates as an energy source (this is similar to the carnitine acyltransferase deficiency described in Solution 18), resulting in increased catabolism that results in hypoglycemia. Carbohydrate oxidation yields fewer ATP per carbon than fatty acid oxidation (see Solution 12.76), so the lethargy indicates that carbohydrate oxidation alone is insufficient to meet the energy needs of the individual.

27. a.  Palmitate goes through seven cycles of β oxidation. The first six cycles each produce 1 QH 2, 1 NADH, and 1 acetyl-CoA. The seventh cycle produces 1 QH 2, 1 NADH, and 2 acetyl-CoA. Each QH2 generates 1.5 ATP in the electron transport chain, each NADH generates 2.5 ATP in the electron transport chain, and each acetylCoA generates a total of 10 ATP (1 QH2 × 1.5 = 1.5 ATP; 3 NADH × 2.5 = 7.5 ATP; 1 GTP = 1 ATP for a total of 10 ATP per acetyl-CoA). The total is 108 ATP. Two ATP must be subtracted to account for the ATP spent in activating palmitate to palmitoyl-CoA. This yields a total of 106 ATP.  b.  The same logic is used for stearate, except that stearate goes through eight cycles of β oxidation. The total is 120 ATP.

23.  The similarities are summarized in the diagram. The conversion of fatty acyl-CoA to enoyl-CoA is similar to the conversion of succinate to fumarate because both reactions involve oxidation of the substrate and concomitant reduction of FAD to FADH2 (see Section 14.2). The conversion of enoyl-CoA to hydroxyacyl-CoA is similar to the conversion of fumarate to malate because both reactions involve the addition of water across a trans double bond. The conversion of hydroxyacyl-CoA to ketoacyl-CoA is similar to the conversion of malate to oxaloacetate because both reactions involve the oxidation of an alcohol to a ketone with concomitant reduction of NAD+ to NADH.

29.  Palmitoleate is a 16-carbon fatty acid. If it were completely saturated, 106 ATP would be generated via β oxidation (see Problem 27a). The double bond at the 9,10 position must be converted from the cis to the trans isomer. β Oxidation continues after this isomerization step, but with the loss of 1 QH2. Thus, 1.5 ATP should be subtracted from the total to yield 104.5 ATP. (Subtract 2 ATP for activation from the total shown in the figure.)

O H3C

β Oxidation Fatty acyl-CoA

Citric acid cycle Succinate

Enoyl-CoA

Fumarate

Hydroxyacyl-CoA

Malate

Ketoacyl-CoA

Oxaloacetate

(CH2)5

C H

C (CH2)7 H

3(1 QH2 , 1 NADH, 1 acetyl-CoA) = 3(1.5 + 2.5 + 10) = 42 ATP

3 cycles of β oxidation

O H3C

(CH2)5

C

C

H

H

CH2

α COO–

Benzoate

CH2

CH2

H3C

C

S

CoA

Acetyl-CoA

β CH2

α COO–

O COO–

Phenylacetate

+

H3C

(CH2)6

O

C

S

C

C

C

SCoA

H

1 cycle of β oxidation (acyl-CoA dehydrogenase reaction is bypassed) (CH2)6

3 cycles of β oxidation

Phenylbutyrate

CH2

H3C

H3C O

+

SCoA

1 NADH, 1 acetyl-CoA = (2.5 + 10) = 12.5 ATP O

Phenylpropionate

COO–

C

enoyl-CoA isomerase

25. a.  Benzoate was produced when the dogs were fed phenylpro­ pionate (see diagram).  b.  Phenylacetate was produced when the dogs were fed phenylbutyrate. β CH2

SCoA

Palmitoleoyl-CoA

H

CH2

C

CoA

Acetyl-CoA

C

SCoA

3(1 QH2 , 1 NADH, 1 acetyl-CoA) = 3(1.5 + 2.5 + 10) = 42 ATP

Acetyl-CoA (10 ATP)

31.  Linoleate is an 18-carbon fatty acid and would yield 120 ATP via β oxidation if it were completely saturated (see Problem 27b). The double bond at the 9,10 position must be converted from the cis to the trans isomer, resulting in the loss of 1 QH2 (1.5 ATP). In addition, an NADPH-dependent reduction of the 12,13 double bond costs 2.5 ATP. This reduces the ATP total by 4, so oxidation of linoleate yields 116 ATP. (Subtract 2 ATP for activation from the total shown in the figure.)

S-58  ODD- N UM B ER E D S O LUT I O NS O H3C

(CH2)4

CH

CH2

CH

CH

CH

(CH2)7

C

SCoA

Linoleoyl-CoA 3 (1 QH2, 1 NADH, 1 acetyl-CoA) = 3(1.5 + 2.5 + 10) = 42 ATP

3 cycles of β oxidation

O H3C

(CH2)4

CH

CH

CH2

C

C

H

H

CH2

C

H

O

C

C

CoA

enoyl-CoA isomerase

H3C

(CH2)4

CH

CH

CH2

CH2

C

CoA

H

1 cycle of β oxidation (acyl-CoA dehydrogenase reaction is bypassed)

1 NADH, 1 acetyl-CoA = (2.5 + 10) = 12.5 ATP O

H3C

(CH2)4

CH

CH2

CH

acyl-CoA dehydrogenase

H3C

(CH2)4

CH

CH2

C

S

CoA

1 QH2 = 1.5 ATP

CH

C

H

O

C

C

S

CoA

H NADPH NADP+

2,4-dienoyl-CoA reductase

Costs 2.5 ATP O

H3C

(CH2)4

CH2

CH

CH

CH2

C

S

CoA

enoyl-CoA isomerase H3C

(CH2)4

CH2

CH2

1 cycle of β oxidation (acyl CoA dehydrogenase reaction is bypassed) 3 cycles of β oxidation

C

H

O

C

C

S

CoA

H 1 NADH, 1 acetyl-CoA = (2.5 + 10) = 12.5 ATP 3 (1 QH2, 1 NADH, 1 acetyl-CoA) = 3(1.5 + 2.5 + 10) = 42 ATP

Acetyl-CoA (10 ATP)

33. a. A C 17 fatty acid goes through seven cycles of β oxidation. The first six cycles each produce 1 QH 2 , 1 NADH, and 1 acetyl-CoA. The seventh cycle produces 1 QH 2, 1 NADH, 1 acetyl-CoA, and 1 propionyl-CoA. Each QH 2 generates 1.5 ATP in the electron transport chain, each NADH generates 2.5 ATP, and each acetyl-CoA generates a total of 10 ATP (1 QH 2 × 1.5 = 1.5 ATP; 3 NADH × 2.5 = 7.5 ATP; 1 GTP = 1 ATP for a total of 10 ATP per acetyl-CoA). The total is 98 ATP. Propionyl-CoA is metabolized to succinyl-CoA (at a cost of 1 ATP; see Fig. 17.7) and enters the citric acid cycle. Conversion of succinyl-CoA to succinate yields 1 GTP (which offsets the cost of the propionyl-CoA → succinyl-CoA conversion), and conversion of succinate to fumarate yields 1 QH2

(equivalent to 1.5 ATP). Fumarate is converted to malate and then malate is converted to pyruvate, yielding 1 NADPH (equivalent to 2.5 ATP). The pyruvate dehydrogenase reaction converts pyruvate to acetyl-CoA (which is subsequently oxidized by the citric acid cycle to yield 10 ATP) and 1 NADH, which yields 2.5 ATP. Therefore, oxidation of propionyl-CoA yields an additional 16.5 ATP. The total is 98 ATP + 16.5 ATP = 114.5 ATP. Two ATP must be subtracted from this total to account for the ATP spent in activating the C17 fatty acid to a fatty acyl-CoA. This yields a final total of 112.5 ATP.  b.  A C17 fatty acid yields more ATP than palmitate (106 ATP, see Solution 27a) and less than oleate (118.5 ATP, see Solution 30).  c.  Fatty acids are normally not glucogenic, as the breakdown product is acetyl-CoA, which is not a substrate for gluconeogenesis. However, the degradation of an odd-chain fatty acid produces propionyl-CoA in the last step of β oxidation. Propionyl-CoA can be converted to pyruvate as explained in part a, which can subsequently enter gluconeogenesis. 35.  The patient could be treated with injections of vitamin B 12 directly into the bloodstream. Alternatively, the patient could be treated with high doses of oral vitamin B12. In the presence of high concentrations of the vitamin, sufficient amounts may be absorbed even in the absence of intrinsic factor. 37.  A fatty acid cannot be oxidized until it has been activated by its attachment to coenzyme A in an ATP-requiring step. The first phase of glycolysis also requires the investment of free energy in the form of ATP (see Fig. 13.2). Consequently, neither β oxidation nor glycolysis can produce any ATP unless some ATP is already available to initiate these catabolic pathways. 39. a.  Fatty acid degradation occurs in the mitochondrial matrix, synthesis in the cytosol.  b.  The acyl carrier in degradation is coenzyme A; for synthesis it is the acyl-carrier protein.  c. During degradation, ubiquinone and NAD+ accept electrons to become ubiquinol and NADH; during synthesis NADPH donates electrons and becomes oxidized to NADP +.  d.  Degradation requires one ATP (and the hydrolysis of two phosphoanhydride bonds) to activate the fatty acid; synthesis consumes one ATP per two carbons incorporated into the growing fatty acyl chain.  e.  Degradation produces two-carbon units (acetyl-CoA); synthesis requires a C3 intermediate (malonyl-CoA).  f.  The hydroxyacyl intermediate in the degradation pathway has the l configuration; in the synthetic pathway the configuration is d.  g.  Both synthesis and degradation take place at the thioester end of the fatty acyl chain. 41.  [From Jitrapakdee, S. et al., Biochem. J. 413, 369–387 (2008).]

Phase I Adenosine

O

O

O O

P

P

O–

ATP

O–

–O

O +

P

O

O–

OH

O HO

P O–

O

O ADP C

Pi

O C

biotinyl-enzyme

O–

Carboxyphosphate O

O

C O

+

NH

HN

O S

(CH2)4

C

NH

Biotinyl-enzyme

(CH2)4

E

HO

P

O

Pi

O C

O–

biotinyl-enzyme

O–

OD D -NUMB ER ED SO LUTI O NS  S-59

Carboxyphosphate O C O

concentration of citrate would be required to increase the activity of the enzyme.

O +

NH

HN

C

(CH2)4

S

C

NH

N

O S

Phase II Acetyl-CoA O

(CH2)4

NH

Acetyl-CoA enolate O–

CoAS

CH2

O

H

(CH2)4

O

+

C

CH2

CH2 C

NH

Carboxybiotinyl-enzyme

O

O– +

+ O–

BH

O

O

CoAS

C

+

O

E

Malonyl-CoA

C

B

N

C

Carboxybiotinyl-enzyme

C

C

E

(CH2)4

Biotinyl-enzyme

O

O–

NH

O

O–

CoAS

47.  Cytosolic ACC1 generates malonyl-CoA as a substrate for lipid biosynthesis (which occurs in the cytosol), whereas mitochondrial ACC2 generates malonyl-CoA to serve as an inhibitor for carnitine acyltransferase (see Fig. 17.15) that transports fatty acyl-CoAs into the mitochondrial matrix for β oxidation. In this manner, the two isoenzymes “cooperate” to stimulate the synthesis of fatty acids (ACC1) while inhibiting breakdown (ACC2).

O

N

O NH

HN

NH

Biotinyl-enzyme

43.  Epinephrine signaling via an adrenergic receptor activates a G protein, which activates adenylate cyclase to produce cyclic AMP to activate cellular kinases, including protein kinase A (PKA). PKA phosphorylates its substrates, including acetyl-CoA carboxylase, thereby inactivating the enzyme. As a result of lower acetyl-CoA carboxylase activity, the rate of fatty acid synthesis drops. Epinephrine signaling also leads to phosphorylation and activation of glycogen phosphorylase (see Section 13.3), which mobilizes glucose from glycogen. These responses are consistent: When the cell needs to mobilize metabolic fuel (for example, by glycogenolysis), storage of fuel (for example, by fatty acid synthesis) is inhibited. 45. a.  Citrate is an activator. A high cytosolic citrate concentration is correlated with a high rate of fatty acid synthesis. The citrate transport system (see Fig. 17.11), which helps move acetyl groups from the mitochondrial matrix to the cytosol, transports citrate to the cytosol. There, ATP-citrate lyase converts citrate to acetyl-CoA and oxaloacetate. Acetyl-CoA is then available for fatty acid synthesis in the cytosol. Since all of the acetyl units that will be used to synthesize fatty acids must be transported out of the mitochondrial matrix into the cytosol using this system, citrate levels are high when fatty acid synthesis is occurring at a high rate.  b.  The phosphorylated form of acetyl-CoA carboxylase is less active than the dephosphorylated form (see Solution 43); therefore, a greater

49.  The fungus does not store energy in the form of triacylglycerols, but acetyl-CoA carboxylase is essential because it is required for the synthesis of fatty acids, which appear in structural components of the organism, such as cell membranes (see Section 17.4). 51.  The synthesis of palmitate from acetyl-CoA costs 42 ATP. Seven rounds of the synthase reaction are required. ATP is required to convert each of 7 acetyl-CoA to malonyl-CoA for a total of 7 ATP. Two NADPH are required for seven rounds of synthesis, which is equivalent to 2 × 7 × 2.5 = 35 ATP. 53.  Mammalian fatty acid synthase differs structurally from bacterial fatty acid synthase; thus, triclosan inhibits the bacterial enzyme but not the mammalian enzyme. Mammalian fatty acid synthase is a multifunctional enzyme made up of two identical polypeptides (see Fig. 17.12). In bacteria, the enzymes of the fatty acid synthetic pathway are separate proteins. In fact, triclosan inhibits the bacterial enoyl-ACP reductase. The enzymes of the mammalian multifunctional enzyme could be arranged in such a way as to preclude the binding of triclosan to the active site of the mammalian enoyl-ACP reductase. 55.  In gluconeogenesis, an input of free energy is required to reverse the exergonic pyruvate kinase reaction of glycolysis. Pyruvate is carboxylated to produce oxaloacetate and then oxaloacetate is decarboxylated to produce phosphoenolpyruvate. Each of these reactions requires the hydrolysis of one phosphoanhydride bond (in ATP and GTP, respectively). In fatty acid synthesis, ATP is consumed in the acetyl-CoA carboxylase reaction, which produces malonyl-CoA. The decarboxylation reaction is accompanied by hydrolysis of a thioester bond, which releases a similar amount of free energy as phospho­ anhydride bond hydrolysis. 57.

OH COO– H3C

(CH2)46

CH β

CH α

(CH2)25

CH3

59.  Yes, by blocking fatty acid synthesis and by promoting fatty acid oxidation, cerulenin decreases the levels of fatty acids available in the cell, thereby preventing fungal growth. 61.  Fatty acids that cannot be synthesized from palmitate using cellular elongases and desaturases are essential fatty acids and must be obtained from the diet. Mammals do not have a desaturase enzyme that can introduce double bonds beyond C9.  a.  Oleate and  d.  palmitoleate, each with a double bond at the 9,10 position, are not essential fatty acids (see figure).  b.  Linoleate has a second double bond at the 12,13 position and therefore is an essential fatty acid.  c. α-Linolenate has double bonds at the 9,10 position, the 12,13 position, and the 15,16 position and is also essential.

S-60  ODD- N UM B ER E D S O LUT I O NS Fig. 14.19). Restoring citric acid cycle function would consume acetyl-CoA, which would decrease its rate of conversion to ketone bodies.  e.  Weight loss could be due to glycogen depletion from the liver and muscle, loss of water (needed to dissolve excess ketone bodies excreted in the urine), loss of body fat that results from increased lipolysis when glycogen stores are depleted, as well as the weight loss due to the caloric restriction that is customary in any “diet” regimen. (In addition, ketone bodies have an appetite suppressant effect.)

O C

O–

(CH2)7 CH

O C

–

O

CH

(CH2)7

CH2

CH

O –

C

O

(CH2)7

O

CH

CH

CH

CH2

CH2

C

–

O

(CH2)7

CH

CH

CH

CH

CH

CH

CH

CH

71.  The fresh vegetables in Individual A’s diet provide essential vitamins, minerals, and fiber, while the avocado and olive oil provide “healthy” unsaturated fats (see Box 17.A). Individual B’s diet lacks these beneficial nutrients and contains a large amount of “unhealthy” saturated fat as well.

CH3

73.  In the absence of pyruvate carboxylase, pyruvate cannot be converted to oxaloacetate. Without sufficient oxaloacetate to react with acetyl-CoA in the first reaction of the citric acid cycle, acetyl-CoA accumulates. The excess acetyl-CoA is converted to ketone bodies.

a.       b.         c.         d.

75. a.  Because a thioesterase can reverse the inhibition, the modification must involve the acylation of a cysteine side chain, as shown here.

(CH2)7 CH3

Oleate

(CH2)4 CH3

Linoleate

63.

CH2 CH3

α-Linolenate OH

H3C

(CH2)5

(CH2)5

CH

Palmitoleate

O (CH2)8

C

O–

65. a.  Acetoacetate is a ketone body (see Fig. 17.16). It is converted to acetyl-CoA, which can be oxidized by the citric acid cycle to supply free energy to the cell.  b.  Intermediates of the citric acid cycle are also substrates for other metabolic pathways, but unless they are replenished, the catalytic activity of the cycle is diminished. Ketone bodies are metabolic fuels, but they cannot be converted to citric acid cycle intermediates. A three-carbon glucose-derived compound such as pyruvate can be converted to oxaloacetate to increase the pool of citric acid cycle intermediates and keep the cycle operating at a high rate. 67.  The synthesis of the ketone body acetoacetate does not require the input of free energy (the thioester bonds of 2 acetyl-CoA are hydrolyzed; see Fig. 17.16). Conversion of acetoacetate to ­3-hydroxybutyrate consumes NADH (which could otherwise generate 2.5 ATP by oxidative phosphorylation). However, the conversion of ­3-hydroxybutyrate back to 2 acetyl-CoA regenerates the NADH (see Fig. 17.17). This pathway also requires a CoA group donated by ­succinyl-CoA. The conversion of succinyl-CoA to succinate by the citric acid cycle enzyme succinyl-CoA synthetase generates GTP from GDP + Pi, so the conversion of ketone bodies to acetyl-CoA has a free energy cost equivalent to one phosphoanhydride bond. 69. a.  Oxidation of fatty acids in the diet yields acetyl-CoA, which normally enters the citric acid cycle by reacting with oxaloacetate. Oxaloacetate is produced by the action of pyruvate carboxylase on pyruvate produced by glycolysis. Without glucose metabolism to generate pyruvate, the citric acid cycle cannot proceed. Excess acetyl-CoA is converted to ketone bodies, which can be used as an energy source in place of glucose by some tissues (with the exception of the liver, which lacks an enzyme required to metabolize ketone bodies, and red blood cells, which lack mitochondria).  b.  Ketogenic (described in part a) and glucogenic pathways are active in the individual adhering to the keto diet. Gluconeogenesis uses amino acids from dietary protein as substrates to produce glucose not available in the diet. (Acetyl-CoA is not a substrate for gluconeogenesis.)  c.  The keto diet mimics the fasting state, so as a result, glycogen and fatty acid synthetic pathways are inactive.  d.  A high protein intake provides amino acids, which are substrates for gluconeogenesis. Several amino acids can be degraded in anaplerotic reactions that provide citric acid cycle intermediates (see

CH2 S C

O

(CH2)n CH3

b.  The acylation of several Cys residues alters the three-dimensional structure of the protein in some way that prevents the active site from either binding substrate or converting it to product. c. The long-chain fatty acids fuel β oxidation. Inhibition of PFK by the fatty acids inhibits glycolysis, allowing β oxidation to generate acetyl-CoA for the citric acid cycle. Under these conditions, the citrate transport system (see Fig. 17.11) leads to an increase in the concentration of cytosolic citrate, which also inhibits PFK (see Section 13.1). 77.  All cells can obtain glycerol-3-phosphate from glucose, because glucose enters glycolysis to form dihydroxyacetone phosphate, which is subsequently transformed to glycerol-3-phosphate via the glycerol3-phosphate dehydrogenase reaction. Cells capable of carrying out gluconeogenesis, such as liver cells, can transform pyruvate to dihydroxyacetone phosphate. (Interestingly, adipocytes do not carry out gluconeogenesis but do express PEPCK and are thus able to convert pyruvate to glycerol-3-phosphate.) 79.  A hydrolysis reaction removes a fatty acyl group from a triacylglycerol, leaving a diacylglycerol. The intent is to reduce the total fatty acid content of the oil, thereby reducing its caloric value, without drastically altering the fluidity of the product. 81. a.  The reaction is catalyzed by a kinase.  b.  The hydrolysis of PPi drives the reaction to completion.  c.  Three phosphoanhydride bonds must be hydrolyzed in order to provide sufficient free energy to synthesize phosphatidylcholine from choline and diacylglycerol. Choline is activated by conversion to phosphocholine, with a phosphate group donated by ATP. The phosphocholine is subsequently converted to CDP–choline, a reaction in which a second phosphoanhydride bond is hydrolyzed. Pyrophosphate is the leaving group in this reaction, and the third phosphoanhydride bond is hydrolyzed when the pyrophosphate is hydrolyzed to orthophosphate. 83.  The first enzyme in the pathway shown in Figure 17.19 is choline kinase, which catalyzes the ATP-dependent phosphorylation of

OD D -NUMB ER ED SO LUTI O NS  S-61 choline to produce phosphocholine, which is needed for phosphatidyl­ choline synthesis. Increasing the activity of the enzyme that catalyzes the first step of the pathway allows the cancer cell to synthesize the membrane lipids required for cellular growth and proliferation. 85. a.  Acyl-CoA transferase, step 3;  b.  3-ketosphinganine synthase, step 1;  c.  dihydroceramide dehydrogenase, step 4;  d. 3-ketosphinganine reductase, step 2. 87.  The experimental results indicate that HMG-CoA reductase is regulated by phosphorylation. In normal cells, delivery of LDL particles results in an increase in cellular cholesterol, which stimulates a kinase that phosphorylates HMG-CoA reductase on the essential serine residue. The phosphorylated form of the enzyme is less active. In the mutant cells, the alanine cannot be phosphorylated, so the enzyme activity is unaffected by the presence of increasing cellular cholesterol. 89. a.  Fumonisin inhibits ceramide synthase. The concentration of the final product, ceramide, decreases, while other lipid synthetic pathways are not affected. The first enzyme in the pathway, serine palmitoyl transferase, is not the target of fumonisin B1 because there is no significant decrease in its product, 3-ketosphinganine. The second enzyme in the pathway, 3-ketosphinganine reductase, is also not a target because, if it were, the substrate of this reaction would accumulate in the presence of fumonisin. Accumulation of sphinganine indicates that ceramide synthase is inhibited because sphinganine cannot be converted to dihydroceramide.  b.  Fumonisin likely acts as a competitive inhibitor. It is structurally similar to sphingosine and its derivatives and thus can bind to the active site and prevent substrate from binding. Alternatively, fumonisin may act as a substrate and be converted to a product that cannot be subsequently converted to ceramide. [From Wang, E. et al., J. Biol. Chem. 266, 14486–14490 (1991).] 91. a.  Because cholesterol is water-insoluble, it is commonly found associated with other lipids in cell membranes. Only an integral membrane protein would be able to recognize cholesterol, which has a small —OH head group and is mostly buried within the lipid bilayer.  b. Proteolysis releases a soluble fragment of the SREBP that can travel from the cholesterol-sensing site to other areas of the cell, such as the nucleus.  c.  The DNA-binding portion of the protein might bind to a DNA sequence near the start of certain genes to mark them for transcription. In this way, the absence of cholesterol could stimulate the expression of proteins required to synthesize or take up cholesterol.

b.  Plants need a source of nitrogen in the form of ammonia or nitrate, so the organisms that carry out the processes marked with an asterisk in part a can potentially support plant growth. 7.  ​NH​4 +​ ​ is the protonated form of the NH3 product of the reaction and would be expected to act as an inhibitor. 9.  Leghemoglobin, like myoglobin, is an O2-binding protein. Its presence decreases the concentration of free O2 that would otherwise inactivate the bacterial nitrogenase. 11. a. 

Glutamine synthetase Glutamine synthetase (more active) (less active) CH2

OH

1. a. +5, b. +3, c. 0, d. ‒3, e. ‒3, f. +2  g. +1. 3.  The ATP-­induced conformational change must decrease ℰ ​ ​°′ ​  from −0.29 V to about −0.40 V. The decrease in reduction potential allows the protein to donate electrons to N2, since electrons flow spontaneously from a substance with a lower reduction potential to a substance with a higher reduction potential. Without the conformational change, nitrogenase would not be able to reduce N2. 5.  a. 

N2

*

*

+

NH4

* NO− 2

NO− 3

*

+ PPi

O −O

P

O

CH2

O H

O

Adenine

H

H

OH

OH

H

b.  α-­Ketoglutarate would inhibit adenylylation, preserving the enzyme in its unmodified, active form. Transamination reactions involving α-­ketoglutarate produce glutamate, a reactant for the glutamine synthetase reaction. − − −− −c.  − − − − COO − − COO − COO− − COO d.  13. a.  b.  − COO COO COOCOO COO− COO COO COO COO COO COO COO

C H

O C OC C C O O CO OC C O OC CO C COO OC C

O O C

O C

H CH CH CH CH22CH2 2CH H 2 CHH 2 2 CH22CHCH 2 2 CH CH2 CH2CH OH CH2OH OH 2 CH2 CH2CH2

CH2

NH NH NH

NH

C

Chapter 18

CH2

+ ATP

+ NH C+ 2 NH C 2NH+ 2C

NH2 NH2NH2

O

CH2

OH

NH+ 2

NH2

15. a. leucine, b. methionine, c. tyrosine,  d. valine. 17.  In a ping pong mechanism, one substrate binds and one product is released before the other substrate binds and the second product is released. In the transamination reaction, the amino acid binds first and then the first α-­keto acid is released. Next, the second α-­keto acid binds and finally the second amino acid is released. 19. a. Cancer cells upregulate glutamine membrane transport proteins, which results in the effective uptake of glutamine from the circulation. Another strategy used by cancer cells is upregulation of glutamine synthetase, which converts glutamate to glutamine. b. Glutamine analogs (compounds that are structurally similar to glutamine) could be used to either block the transport proteins or to bind to the enzyme and act as competitive inhibitors to interfere with substrate binding. 21. a. Both reactions are reversible. The ALT reaction is shown in Section 18.2; the AST reaction is shown below.

S-62  ODD- N UM B ER E D S O LUT I O NS

+H N 3

CH

C

CH2 COO−

Aspartate

side chain is phosphorylated whereas in the asparagine synthetase reaction, the amino acid is linked to an AMP moiety.

COO−

O O−

C

O

+ CH2

COO−

AST

CH2

H

COO−

CH2

C

+NH 3

α-Ketoglutarate

O

CH2

O Ribose PO− 3

O

Adenine

O

COO− C

C

+H N 3

C

O−

CH2

+

CH2

COO−

Oxaloacetate

CH

COO−

Glutamate

b.  Cytosolic malate dehydrogenase acts on the oxaloacetate product of the AST reaction and NADH to produce malate and NAD+ (Fig. 15.5). The NAD+ is regenerated for glycolysis and the malate is transported into the mitochondrial matrix, where the reaction is reversed by mitochondrial malate dehydrogenase, providing NADH for electron transport. The resulting oxaloacetate, along with glutamate, reacts to form aspartate and α-­ketoglutarate, as catalyzed by mitochondrial matrix AST. Aspartate leaves the mitochondrial matrix via a transport protein to complete the cycle. c. Alanine produced in the muscle is released to the bloodstream and taken up by the liver, where it can react with α-­ketoglutarate to form pyruvate and glutamate, as catalyzed by ALT (the reverse of the reaction shown in the chapter). Pyruvate can then enter gluconeogenesis. [From Botros, M. and Sikaris, K.A., Clin. Biochem. Rev. 34, 117–130 (2013).]

27. a. 3-Phosphoglycerate is derived from glycolysis, illustrating the importance of glycolytic intermediates in amino acid biosynthesis. b. The first reaction of the serine biosynthetic pathway is a redox reaction catalyzed by a dehydrogenase enzyme. The second reaction is catalyzed by a transaminase. The third reaction is catalyzed by a phosphatase. 29.  Cells that undergo rapid cell division (and thus need to replicate their DNA) require a rapid rate of nucleotide synthesis. Examples include skin cells, cells lining the gastrointestinal tract, and bone marrow cells. Tumor cells also have a high rate of cell division. 31.

Pyruvate

transamination

3-Phosphoglycerate

Ala

Ser

Gly

Cys

α-Ketoglutarate

transamination

Gln synthetase

Glu

Gln

Oxaloacetate transamination Asp

Asn synthetase

23. a.

Asn

​Glu + ATP + ​NH​+ 4  ​ ​→ Gln + ​Pi​ ​+ ADP

Glu

Arg

Pro

Gln + α-­ketoglutarate + NADPH → 2 Glu + ​NADP​ +​ Glu + α-­keto acid → α-­ketoglutarate + amino acid ​ H​+ N 4  ​ ​+ ATP + NADPH + α-­keto acid →  ADP + ​Pi​ ​+ ​NADP​ +​+ amino acid + b.  ​ α-­ketoglutarate + ​NH​+ 4  ​ ​+ NAD​(P)​H → Glu + ​H2​  ​O + NAD​(P)​  ​

Glu + α-­keto acid → α-­ketoglutarate + amino acid + α-­keto acid + NAD(P)H + ​NH​+ 4  ​ ​→ NAD(P) + amino acid + ​H2​  ​O​

25. a. In the glutamine synthetase reaction, ATP donates a phosphoryl group to glutamate, and the phosphoryl group is then ­displaced by an ammonium ion, producing glutamine and phosphate. The ammonium ion is the nitrogen source. The asparagine synthetase reaction also requires ATP as an energy source, but the nitrogen donor is glutamine, not an ammonium ion. Aspartate is converted to asparagine, and the glutamine becomes glutamate after donating an amino group. ATP is hydrolyzed to AMP and pyrophosphate (PPi) instead of ADP and phosphate as in the glutamine synthetase reaction. b. The structure of the intermediate for the asparagine synthetase reaction is shown below. Both reaction intermediates are mixed anhydrides, but in the glutamine synthetase reaction the amino acid

33. a. Valine, b. leucine, and c. isoleucine are produced. The α-­keto acid product is α-­ketoglutarate. 35.  Cysteine is the source of taurine. The cysteine sulfhydryl group is oxidized to a sulfonic acid, and the amino acid is decarboxylated. 37. 

O −O

C

CH +NH 3

CH2

CH2

C

NH

CH2

CH3

O

39.  If an essential amino acid is absent from the diet, then the rate of protein synthesis drops significantly, since most proteins contain an assortment of amino acids, including the deficient one. The other amino acids that would normally be used for protein synthesis are therefore broken down and their nitrogen excreted. The decrease in protein synthesis, coupled with the normal turnover of body proteins, leads to the excretion of nitrogen in excess of intake.

OD D -NUMB ER ED SO LUTI O NS  S-63 41.

+

H3N

(CH2)5

+

+

NH3

H3N

Cadaverine

(CH2)4

+

+

53. a. 

NH3

Putrescine

NH3 CH3

CH2

Isoleucine 43. a. The values of KM and Vmax are shown in the table.

–1

–1

Vmax (μmol · mg · min ) KM (mM)

CH

CH

COO−

CH3

1 transaminase

Thr only

Thr + Ile

Thr + Val

210

180

220

8

75

6

b.  Isoleucine is an allosteric inhibitor of threonine deaminase and binds to the T form of the enzyme. Velocity decreases by about 15%, but the nearly 10-fold increase in KM is more dramatic. The decrease in velocity and increase in K M indicate that isoleucine, the end product of the pathway, acts as a negative allosteric regulator of the enzyme that catalyzes an early, committed step of its own synthesis. The velocity versus substrate concentration curve obtained for threo­ nine deaminase in the presence of isoleucine has greater sigmoidal character, which means that binding of threonine is even more cooperative in the presence of the inhibitor. c. Valine stimulates threo­ nine deaminase by binding to the R form. The maximal velocity is somewhat increased, but the KM is decreased, indicating that the threo­nine substrate has a higher affinity for the enzyme in the presence of valine. The cooperative binding of threonine to threonine deaminase is abolished in the presence of valine, however, as indicated by the hyperbolic shape of the curve. [From Eisenstein, E., J. Biol. Chem. 266, 5801–5807 (1991).]

O CH3

49. a. Arginine residues are converted to citrulline residues by a process of deamination (water is a reactant and ammonia is a product). Note that free citrulline produced by the urea cycle (Section 18.4) or in the generation of nitric oxide is not incorporated into polypeptides by ribosomes since there is no codon for this nonstandard amino acid. b. The nonstandard amino acid citrulline is not normally ­incorporated into polypeptides, so its presence appears foreign to the immune system, increasing the risk of triggering an autoimmune response. 51.  Threonine catabolism yields glycine and acetyl-­CoA, as shown in the chapter. The acetyl-­CoA is a substrate for the citric acid cycle, which ultimately provides ATP for the rapidly dividing cell. Glycine is a source of one-­carbon groups, which become incorporated into methylene-­tetrahydrofolate (THF) via the glycine cleavage system. THF delivers one-­carbon groups for the synthesis of purine nucleotides and for the methylation of dUMP to produce dTMP; nucleotides are needed in large amounts in rapidly dividing cells (see Section 18.5). [From Wang, J. et al., Science 325, 435–439 (2009).]

CH CH3

2

COO−

C

branched-chain α-keto-acid dehydrogenase O

CH3

CH2

CH

C

SCoA

CH3

3 acyl-CoA dehydrogenase O CH3

CH

C

C

SCoA

CH3

4 enoyl-CoA hydratase O

H CH3

C

CH

C

SCoA

OH CH3

5 hydroxyacyl-CoA dehydrogenase

45. a. Conversion of tyrosine to dopamine requires a hydroxylase and a decarboxylase, conversion of dopamine to norepinephrine requires a hydroxylase, and conversion of norepinephrine to epinephrine requires a methyltransferase. b. Conversion of tryptophan to serotonin requires a hydroxylase and a decarboxylase, and conversion of serotonin to melatonin requires an N-­acetyltransferase and a methyltransferase. 47.  Pyruvate can be transaminated to alanine, carboxylated to oxaloacetate, or oxidized to acetyl-­CoA to enter the citric acid cycle. ­α-­Ketoglutarate, succinyl-­CoA, fumarate, and oxaloacetate are all citric acid cycle intermediates; they can also all enter g­ luconeogenesis. Acetyl-­CoA can enter the citric acid cycle, be converted to aceto­ acetate (a ketone body), or be used for fatty acid synthesis.

CH2

O

O CH3

C

CH

C

SCoA

CH3

6 thiolase O CH3

C

O SCoA + CH3

Acetyl-CoA

CH2

C

SCoA

Propionyl-CoA

b.  Reaction 2 is analogous to the pyruvate dehydrogenase reaction. The dehydrogenase catalyzes the release of carbon dioxide and the formation of a thioester bond with coenzyme A.  c.  Reaction 3 is analogous to the acyl-­CoA dehydrogenase enzyme of fatty acid biosynthesis. Hydrogens are removed from the α and β carbons of the substrate. d. The fate of propionyl-­CoA produced by degradation of isoleucine is identical to that of propionyl-­CoA produced in the oxidation of odd-­chain fatty acids (see Fig. 17.7). Propionyl-­CoA is converted to (S)-methylmalonyl-­CoA by propionyl-­CoA carboxylase. A racemase converts the (S)-methylmalonyl-­CoA to the (R) form. A mutase converts the (R)-methylmalonyl- ­CoA to succinyl-­CoA, which enters the citric acid cycle. 55. a. Acetyl-­CoA can enter the citric acid cycle if sufficient oxalo­ acetate is available; if not, excess acetyl-­CoA is converted to ketone bod­ ies. Leucine differs from isoleucine in that leucine catabolism yields a ketone body and a ketone body precursor upon degradation. Isoleucine produces propionyl-­CoA along with acetyl-­CoA; the former can be converted to succinyl-­CoA (see Solution 53d) and then to glucose. Thus,

S-64  ODD- N UM B ER E D S O LUT I O NS isoleucine is glucogenic as well as ketogenic whereas leucine is exclusively ketogenic. b. Persons deficient in HMG-­CoA lyase are unable to degrade leucine and must restrict this amino acid in their diets. A low-­fat diet is also recommended because this same enzyme is involved in the production of ketone bodies (see Reaction 3 in Fig. 17.16). A diet high in fat would generate a high concentration of acetyl-­CoA, which could not be converted to ketone bodies in the absence of this enzyme. 57. a. Insulin inhibits the enzyme, whereas glucagon stimulates the enzyme. b. In the presence of phenylalanine, the activity of the enzyme increases dramatically, more so in the presence of glucagon. Phenylalanine acts as an allosteric activator of phenylalanine hydroxylase and plays a role in converting the enzyme from the inactive dimeric form (T) to the active tetrameric (R) form. c. The incorporation of phosphate into the active form of phenylalanine hydroxylase indicates that the enzyme is regulated by phosphorylation as well as allosteric control. Glucagon signaling is likely to lead to phosphorylation of the enzyme (see Section 10.2). d. Phenyl­ alanine hydroxylase is most active when the glucagon concentration is high, corresponding to the fasting state. Under these circumstances, phenylalanine is degraded to produce acetoacetate (a ketone body) and fumarate (which can be converted to glucose); both ­compounds provide necessary resources in the fasted state. 59.  Persons with NKH lack a functioning glycine cleavage system. This is the major route for the disposal of glycine, and in its absence, glycine accumulates in body fluids. The presence of excessive glycine, a neurotransmitter, in the cerebrospinal fluid explains the effects on the nervous system. 61.  The reaction of serine and homocysteine to produce cysteine and α-­ketobutyrate, the reaction catalyzed by asparaginase, the conversion of serine to pyruvate, the conversion of cysteine to pyruvate, the glycine cleavage system, the glutamate dehydrogenase reaction, and the catabolism of the pyrimidine breakdown products β-­ureidopropionate and β-­ureidoisobutyrate (see Section 18.5) all generate free ammonia. 63.  The post-­excitatory movements of both Ca2+ and Na+ ions from the post-­synaptic cell occur against their concentration gradients and are therefore ATP-­dependent processes. The Na+ ions are ejected from the cell via the Na,K-­ATPase (see Fig. 9.15), a transport protein that requires phosphorylation by ATP to drive the required conformational changes. Import of Ca2+ ions into the mitochondrial matrix also requires an energy source, most likely provided by the membrane potential generated during electron transport. Use of the membrane potential to drive Ca2+ transport decreases the overall yield of ATP obtained by oxidative phosphorylation. 65.  The glutamate dehydrogenase reaction converts α-­ketoglutarate to glutamate. In the presence of excess ammonia, α-­ketoglutarate in the brain could be depleted, diminishing flux through the citric acid cycle. 67.  Arginine allosterically stimulates N-­acetylglutamate synthase. Arginine is the substrate for the urea cycle enzyme arginase, which catalyzes the hydrolysis of arginine, producing urea and regenerating ornithine so that the urea cycle can continue. Allosteric stimulation of carbamoyl phosphate synthetase by N-­acetylglutamate provides carbamoyl phosphate that enters the urea cycle, and the presence of arginine under these conditions ensures that sufficient ornithine will be available to react with the carbamoyl phosphate. 69.  One nitrogen atom is derived from ammonia that is incorporated into carbamoyl phosphate for entrance into the urea cycle. The other nitrogen atom comes from aspartate, which serves as a substrate in the argininosuccinase reaction. Ultimately, both nitrogen atoms that appear in urea originated from excess dietary protein.

71. a. Deficiencies of enzymes that are part of the urea cycle or are associated with it decrease the rate at which nitrogen can be eliminated as urea. Since the source of nitrogen for urea synthesis includes free ammonia, low urea cycle activity leads to high levels of ammonia in the body. b. A low-­protein diet might reduce the amount of nitrogen to be excreted. 73.  Carbamoyl phosphate synthetase catalyzes the reaction in which ammonia is incorporated into carbamoyl phosphate, the substrate for the first reaction of the urea cycle. If this reaction cannot take place, ammonia cannot be converted into urea for excretion. 75.  OTC catalyzes the condensation of carbamoyl phosphate with ornithine to form citrulline in the first reaction of the urea cycle. If the enzyme activity is deficient, citrulline cannot be produced. Both citrulline and arginine (see Solution 72) act as anaplerotic substrates to restore urea cycle function in the absence of OTC activity. A low-­ protein diet places less demand on the urea cycle. Restricting glutamine, the major nitrogen carrier, reduces the synthesis of amino acids for which glutamine is a nitrogen donor, which may also decrease demand on the urea cycle. 77. a. A deficiency of argininosuccinase results in the accumulation of argininosuccinate, the enzyme’s substrate, as well as citrulline, the previous intermediate in the urea cycle. b. The argininosuccinase deficiency causes a deficiency of arginine, the product of the reaction. c. Adding arginine would increase flux through the urea cycle. 79.  An individual consuming a high-­protein diet uses amino acids as metabolic fuels. As the amino acid skeletons are converted to glucogenic or ketogenic compounds, the amino groups are disposed of as urea, leading to increased flux through the urea cycle. During starvation, proteins (primarily from muscle) are degraded to provide precursors for gluconeogenesis. Nitrogen from these protein-­derived amino acids must be eliminated, which demands a high level of urea cycle activity. 81. a. ​g                                     lutaminase ⟶​   glutamate + ​NH​+   ​ glutamine  ​ ⎯⎯    4  ​ ​ glutamate ​ dehydrogenase​

            ⟶ glutamate + NAD​(P)​​  +​  ​⎯      ​   α-­ketoglutarate + ​NH​+ 4  ​ ​+ NAD(P)H ⎯→ glutamine + NAD​(P)​ +​ ​    ​ α-­ketoglutarate + 2 ​NH​+ 4  ​ ​+ NAD(P)H​ b.  Glutamine is degraded to glutamate with the release of ammonia, which can bind protons to form ​​NH​ 4+​  ​​. This helps counteract the acidosis that occurs when the concentration of acidic ketone bodies in the blood increases during starvation. 83. a. ADP and GDP both serve as allosteric inhibitors of ribose phosphate pyrophosphokinase. b. PRPP, the substrate of the amidophosphoribosyl transferase, stimulates the enzyme by feed-­forward activation. AMP, ADP, ATP, GMP, GDP, and GTP are all products and inhibit the enzyme by feedback inhibition. 85.  Inhibiting HGPRT blocks production of IMP, which is a precursor of AMP and GMP. In order to be an effective drug target, HGPRT must be essential for parasite growth; that is, the parasite cannot synthesize its own purine nucleotides de novo but instead must rely solely on salvage reactions using the host cell’s hypoxanthine. 87.  Lymphocytes that did not fuse with a myeloma cell are unable to use the de novo synthetic pathway because it is blocked by amino­ pterin. These cells are able to use the HGPRT salvage pathway, but the cells will not survive beyond 7–10 days. Myeloma cells cannot survive in HAT medium because the aminopterin blocks the de novo pathway and these cells lack HGPRT and cannot use the s­ alvage pathway. Only hybridomas that result from the fusion of a lymphocyte (which

OD D -NUMB ER ED SO LUTI O NS  S-65 can carry out the salvage pathway) and a myeloma cell (which can divide in culture indefinitely) will survive in HAT medium. 89.  This is an example of divergent evolution, as the three types of enzymes likely diverged from a common ancestor and have retained their basic structure and mechanism (see Section 6.4). 91.  In an autoimmune disease, the body’s own white blood cells become activated to mount an immune response that leads to pain, inflammation, and tissue damage. The activity of the white blood cells, which proliferate rapidly, can be diminished by methotrexate, as occurs in rapidly dividing cancer cells, using the same mechanism described in the chapter.

Chapter 19 1.  Glycolysis, the pyruvate dehydrogenase reaction, glycogen synthesis, fatty acid synthesis, and triacylglycerol synthesis are all active in the liver. b. Glycolysis and triacylglycerol synthesis are active in adipose tissue. 3.  Phosphorylation of glucose decreases the concentration of free glucose in the hepatocytes and promotes the continuous transport of glucose into the cells. Phosphorylation also prevents glucose from leaving the cells, because glucose-6-phosphate is not recognized by GLUT2 transporters. In addition, the negatively charged phosphate group decreases the likelihood that glucose will leave the hepatocytes via simple diffusion through the cell membrane. 5. a.  Glycogen synthesis is active in the muscle in the fed state. b. Glycogenolysis, glycolysis, anaerobic fermentation, β oxidation, ketone body oxidation, and protein degradation are all active in the muscle, depending on the intensity and the duration of the exercise. 7.  The Na,K-ATPase pump requires ATP to expel Na + ions while importing K+ ions, both against their concentration gradients (see Fig. 9.15). Inhibition by oubain reveals that the brain devotes half of its ATP production solely to power this pump. Because the brain does not store much glycogen, it must obtain glucose from the circulation. Glucose is oxidized aerobically to maximize ATP production. 9. a.  The reaction is probably a near-equilibrium reaction because the reactants and products have the same total number of phosphoanhydride bonds. b. In highly active muscle, ATP is rapidly converted to ADP. Adenylate kinase catalyzes the conversion of two ADP to ATP and AMP as a way to generate additional ATP to power the actin–myosin contractile mechanism (see Section 5.4). 11.  Glucose enters glycolysis, which produces the ATP required for malonyl-CoA biosynthesis (see Section 17.3). The end product of glycolysis is pyruvate, which is converted to the starting material for fatty acid synthesis—acetyl-CoA—via the pyruvate dehydrogenase reaction (see Section 14.1). Glucose also serves as a source of NADPH required for fatty acid biosynthesis via the pentose phosphate pathway (see Section 13.4). 13.  Glycolysis produces two moles of ATP per mole of glucose. Synthesis of one mole of glucose via gluconeogenesis costs six moles of ATP. Therefore, the cost of running one round of the Cori cycle is four ATP. The extra ATP is generated from the oxidation of fatty acids in the liver. 15. a.  Lactate is an anionic molecule that cannot enter mitochondria by diffusing through a membrane; therefore, it requires a transport protein. b. Lactate dehydrogenase, which uses NAD+, and pyruvate carboxylase, which uses ATP and CO2.

c. 

CYTOSOL

Lactate

MITOCHONDRION Lactate NAD+ NADH

lactate dehydrogenase

Pyruvate ATP + CO2 Oxaloacetate

ADP + Pi

pyruvate carboxylase

Oxaloacetate

d.  The mitochondrial LDH uses NAD+, which is produced by the action of Complex I of the electron transport chain. The ATP and CO2 required in the next reaction are also products of mitochondrial pathways. By using mitochondrial LDH, liver cells can avoid using cytosolic LDH, which would consume NAD+ that might be needed for other processes, and still generate oxaloacetate for gluconeogenesis. [From Passarella, S. and Schurr, A., Front. Oncol. 8, 120 (2018).] 17. a.  Since pyruvate carboxylase catalyzes the carboxylation of pyruvate to oxaloacetate, a deficiency of the enzyme would result in increased pyruvate levels and decreased oxaloacetate levels. Some of the excess pyruvate would also be converted to alanine, so alanine levels would be elevated. b. Some of the excess pyruvate is converted to lactate, which explains the lactic acidosis. Decreased oxaloacetate levels result in a decrease of the activity of the citrate synthase reaction, the first step of the citric acid cycle. This causes the accumulation of acetyl-CoA, which forms ketone bodies that accumulate in the blood, resulting in ketosis. c. Acetyl-CoA stimulates pyruvate carboxylase activity (see Section 14.4). Adding acetyl-CoA would allow the investigators to determine if there was a slight amount of pyruvate carboxylase activity that could be detected by adding this activator. 19.  Increased levels of citrulline indicate that the cytosolic argininosuccinate synthetase reaction in the urea cycle (see Section 18.4) is not occurring normally. This reaction requires aspartate as a reactant in addition to citrulline. The accumulation of citrulline may occur because there is a shortage of aspartate. A deficiency of pyruvate carboxylase results in a decreased concentration of oxaloacetate that could otherwise be transaminated to aspartate. Hyperammonemia is the result of urea cycle impairment, since ammonia is not being converted to urea for excretion by the kidneys. [From Coude, F.X. et al., Pediatrics 68, 914 (1981).] 21.  Fatty acid oxidation is a major source of metabolic energy during metamorphosis. 23.  Step 1 is an isomerization catalyzed by phosphoglucose iso­ merase; step 2 is an amination catalyzed by glutamine:­f ructose6-phosphate aminotransferase; step 3 is an acetylation catalyzed by glucosamine phosphate N-acetyltransferase; step 4 is an isomerization catalyzed by phosphoglucomutase; step 5 is uridylylation ­catalyzed by UDP-N-acetylhexosamine pyrophosphorylase. 25.  The KM for hexokinase is about 0.1 mM, which is lower than the fasting blood concentration of glucose. This means that hexokinase is saturated with glucose at all physiological glucose concentrations. The KM for glucokinase is about 5 mM, which is in the range of fasting blood glucose concentration and lower than the blood glucose concentration immediately after a meal. This means that glucokinase is about half-saturated with substrate during fasting and more than half-saturated with substrate after a meal. 27.  Insulin binding to its receptor stimulates the tyrosine kinase activity of the receptor. Proteins whose tyrosine residues are phosphorylated by the receptor tyrosine kinase can then interact with

S-66  ODD- N UM B ER E D S O LUT I O NS additional components of the signaling pathway. These interactions cannot occur if a tyrosine phosphatase catalyzes the removal of the phosphoryl groups attached to the Tyr residues. 29.  In the fed state, insulin increases the expression of GLUT4 transporters, allowing glucose to enter the cell. This is consistent with the observation that glucose is the major fuel used by muscle cells in the fed state. In the fasted state, insulin is not released and GLUT4 transporters return to intracellular vesicles (see Fig. 19.9). Glucose can no longer be taken up and muscle cells switch to using fatty acids as their major fuel during the fasted state. 31.  Phosphorolytic cleavage yields glucose-1-phosphate, which is negatively charged due to its phosphate group and cannot exit the cell via the glucose transporter. In addition, glucose-1-phosphate can be isomerized to glucose-6-phosphate (and can enter glycolysis) without the expenditure of ATP. Hydrolytic cleavage yields neutral glucose, which can leave the cell via the glucose transporter. Converting free glucose to glucose-6-phosphate so that it can enter glycolysis requires the expenditure of ATP in the hexokinase reaction. 33.  As a glycogen precursor, glucose-6-phosphate (G6P) is an allosteric activator of glycogen synthase. High concentrations of G6P indicate that substrates for glycogen synthesis are abundant. G6P is isomerized to glucose-1-phosphate (G1P), which serves as the substrate for glycogen synthase. G6P allosterically inhibits glycogen phosphorylase. In the presence of abundant G6P, degradation of glycogen (which yields G1P and then G6P as a product) is unnecessary. 35.  Imidazole propionate is derived from histidine by deamination. 37.  Phosphorylation of glycogen synthase by GSK3 inactivates the enzyme so that glycogen synthesis does not occur, but when insulin activates protein kinase B, the kinase phosphorylates GSK3. Phosphorylated GSK3 is inactive and unable to phosphorylate glycogen synthase. Dephosphorylated glycogen synthase is active and glycogen synthesis can occur. In this manner, insulin promotes glycogen synthesis in the fed state. 39. a.  AMPK increases the expression of GLUT4, which facilitates entry of glucose into the muscle cell. Glucose subsequently enters glycolysis and produces ATP. This is consistent with AMPK’s cellular role in activating ATP-producing catabolic pathways. b. AMPK decreases expression of glucose-6-phosphatase. This enzyme catalyzes the last step of gluconeogenesis, a pathway that consumes ATP. This is consistent with AMPK’s cellular role in inhibiting ATP-­ consuming pathways. 41.  Phosphorylase kinase is regulated by phosphorylation, which causes a conformational change that activates the enzyme. The enzyme is phosphorylated by protein kinase A (Fig. 19.13), so the activity of the enzyme therefore depends somewhat on the G ­protein– coupled receptor pathway activated by either glucagon or epinephrine. However, phosphorylase kinase is not fully active until calmodulin (a calcium-binding protein that is part of its structure) binds calcium and undergoes its own conformational change. The intracellular concentration of calcium increases when the phosphoinositide signaling system is activated (see Section 10.2), so the activation of phosphorylase kinase depends on this signaling pathway as well. 43.  Ingesting large amounts of glucose stimulates the β cells of the pancreas to release insulin, which causes liver and muscle cells to use the glucose to synthesize glycogen and causes adipose tissue to synthesize fatty acids. Insulin also inhibits the breakdown of metabolic fuels. The body is in a state of resting and digestion and is not prepared for running. 45. a. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O. b. C16H32O2 + 23 O2 → 16 CO2 + 16 H2O. c. For the combustion of glucose, one CO2 is produced for each oxygen, resulting in a respiratory exchange ratio of 1.0. For the combustion of palmitate, the 16 CO2 produced and 23 O2 consumed give a ratio of about 0.7 (16 CO2 ÷ 23 O2 = 0.7).

47. a.  Normally, glucagon binds to cell-surface receptors on the liver, stimulating adenylate cyclase to produce cAMP to activate protein kinase A, which subsequently activates glycogen phos­ phorylase via phosphorylation. Glycogen phosphorylase catalyzes the degradation of glycogen to glucose, which is released into the bloodstream. Blood glucose concentrations should rise shortly after an intravenous injection of glucagon. Glycogen degradation in the patient’s liver thus appears to be normal. b. Glycogen metabolism in the liver appears to be normal, since glycogen content is normal and the patient’s response to the glucagon test is normal as described in part a. Elevated muscle glycogen suggests a defect in muscle glycogen metabolism, but the normal structure of the muscle glycogen indicates that muscle glycogen synthesis is not impaired and the difficulty most likely involves glycogen degradation. A deficiency in the muscle glycogen phosphorylase enzyme (GSD5) is the most probable explanation. 49.  Glucose-6-phosphate (G6P) enters the lumen of the endoplasmic reticulum (ER) via a specific transport protein. The glucose-6phosphatase enzyme, an integral membrane protein with its active site located in the ER lumen, catalyzes the reaction in which G6P is hydrolyzed to yield glucose and inorganic phosphate (Pi), each of which leaves the ER lumen for the cytosol via specific transport proteins. Glucose-6phosphatase G6P

Glucose

G6P

Glucose

Pi

CYTOSOL

ER MEMBRANE

ER LUMEN

+

Pi

51.  Lactate, alanine, and glycerol (and metabolites that can be converted to these compounds) are the major gluconeogenic substrates. Note that acetyl-CoA is not a gluconeogenic substrate. 53.  Several days into a fast, muscle and liver glycogen are depleted. In the absence of dietary glucose, the main source of endogenous glucose is gluconeogenesis. Citric acid cycle intermediates are converted to oxaloacetate, which enters the gluconeogenic pathway. Catalytic amounts of citric acid cycle intermediates are required for proper functioning of the cycle, so when these intermediates are diverted to gluconeogenesis, the citric acid cycle cannot function properly. 55.  During starvation, muscle proteins are broken down to produce gluconeogenic precursors. The amino groups of the amino acids are transferred to pyruvate via transamination reactions. The resulting alanine travels to the liver, which can dispose of the nitrogen via the urea cycle and produce glucose from the alanine skeleton (pyruvate) and other amino acid skeletons. This glucose circulates not just to the muscles but to all tissues that need it, so the metabolic pathway is not truly a cycle involving just the liver and muscles. 57.  In the first few days of a fast, body protein is degraded to amino acids that serve as substrates for gluconeogenesis in order to meet the needs of tissues such as the brain, which have a high demand for glucose. The conversion of amino acids to gluconeogenic substrates generates ammonium ions, which are converted to urea for excretion. The result is a doubling of urea cycle activity. If the fast becomes prolonged, however, continued rapid degradation of protein would be detrimental to the survival of the organism. The body switches to using fatty acids, which are converted to ketone bodies and can be

OD D -NUMB ER ED SO LUTI O NS  S-67 used by body tissues as an alternative energy source. The use of stored fat as an energy source preserves body protein as described in Solution 56. Amino acids are not used as gluconeogenic substrates, decreasing the demand on the urea cycle, whose activity declines as a result. 59. a.  Leptin stimulates glucose uptake by skeletal muscle.  b. ­ Glycogenolysis is inhibited, probably by direct inhibition of glycogen phosphorylase. c. Leptin increases the activity of cAMP phosphodiesterase, resulting in a decrease in cellular cAMP concentration. In this way, leptin acts as a glucagon antagonist in the same manner that insulin does; glucagon’s signal transduction pathway leads to an increase in cAMP concentration. 61. a.  Norepinephrine binds to G protein coupled–receptors on the plasma membrane of adipocytes. The G protein is activated when GDP is swapped for GTP and the α subunit dissociates from the β and γ subunits. The activated G protein activates adenylate cyclase, which catalyzes the formation of cyclic AMP from ATP. Cyclic AMP then activates protein kinase A, which activates a lipase that catalyzes the hydrolysis of fatty acids from triacylglycerols, releasing the fatty acids and making them available for β oxidation. b. Fatty acids are shuttled into the mitochondrial matrix via the carnitine transporter and undergo β oxidation. The products of β oxidation are NADH and QH2 (which enter electron transport) and acetyl-CoA, which enters the citric acid cycle to produce additional NADH and QH2 for electron transport. In this manner, electron transport is supplied with ample reduced coenzymes and occurs at a high rate. However, in the presence of the uncoupling protein (UCP), the protons pumped out of the mitochondrial matrix into the intermembrane space during electron transport are returned to the matrix via the UCP and not via ATP synthase. The synthase is not active, ATP is not synthesized, and the free energy of electron transport is released as heat. 63.  Both observations indicate a decrease in the ability of the mitochondrion to oxidize fatty acids. Carnitine acyltransferase is required to transport fatty acids from the cytosol to the mitochondrial matrix (see Fig. 17.5), the site of β oxidation. Products of β oxidation—the reduced coenzymes NADH and QH2—are reoxidized by the electron transport chain, with concomitant production of ATP. If fatty acid oxidation cannot occur, fatty acids will be used to synthesize triacylglycerols for storage in adipose tissue instead. 65.  Endogenous fatty acid synthesis is required when dietary fatty acid intake is insufficient. Stimulation of acetyl-CoA carboxylase ensures that the body will have enough fatty acids in the absence of dietary lipids (although essential fatty acids will still be lacking under these circumstances). During starvation (and untreated diabetes, which is similar to the starved state), body tissues do not have the resources to synthesize fatty acids, so acetyl-CoA carboxylase is inhibited. In the starved state, fatty acids are mobilized to provide fuel to body tissues. 67. a.  Insulin stimulates the activity of ACC2 in the muscle cells of normal mice, which promotes fatty acid synthesis and inhibits fatty acid oxidation (due to increased malonyl-CoA levels). The muscle cells in the knockout mice lack ACC2 and are not subject to insulin-mediated control. Fatty acid synthesis does not occur, ­malonyl-CoA levels do not rise, and fatty acid oxidation proceeds normally, even in the presence of insulin. b. Knockout mice are leaner because their heart and muscle tissue cannot synthesize fatty acids, so triacylglycerols are mobilized to provide fatty acids for these tissues. Knockout mice have a higher rate of fatty acid oxidation and a lower rate of synthesis, as described in Solution 66, which also accounts for their lower weight gain despite the increased caloric intake. c. Molecular modeling techniques could be used to design a drug that inhibits the enzyme activity of ACC2 but not ACC1. The drug would need to be designed so that it would be delivered to the mitochondrial matrix, where ACC2 is located. [From Abu-Elheiga, L. et al., Science 291, 2613–2616 (2001).]

69.  The drugs activate the intracellular tyrosine kinase domains of the insulin receptor, bypassing the need for insulin to bind to the receptor. 71.  AMPK phosphorylates and activates phosphofructokinase-2, the enzyme that catalyzes the synthesis of fructose-2,­6-bisphosphate. This metabolite is a potent activator of the glycolytic enzyme phosphofructokinase and an inhibitor of fructose-1,6-bisphosphatase, which catalyzes the opposing reaction in gluconeogenesis. Stimulation of AMPK increases the concentration of fructose2,6-bisphosphate and would therefore stimulate glycolysis and inhibit gluconeogenesis. The increase in glucose utilization and decrease in glucose production lowers the level of glucose in the blood in the diabetic patient. [From Hardie, D.G. et al., J. Physiol. 574, 7–15 (2006).] 73. a.  H HO

CH2OH O H OH H H

H

H OH

2-Deoxy-D-glucose

ATP

ADP H

hexokinase

HO

CH2OPO2– 3 O H H OH H OH H

H

2-Deoxy-D-glucose-6-phosphate

b.  Cancer cells utilize anaerobic metabolism even in the presence of oxygen, as described by Warburg. The cells obtain ATP mainly from glycolysis, which explains why a glycolytic inhibitor interferes with ATP production whereas an inhibitor of electron transport does not. 75. a.  Glutamine would be a good substitute for glucose, since cancer cells take up large amounts of this amino acid. b. Cancer cells take up large amounts of glutamine because it can be deaminated to form glutamate, a precursor for several nonessential amino acids. Glutamine is a nitrogen donor in the synthesis of nucleotides, which are required for DNA replication. Cancer cells undergo rapid proliferation, a process that requires a high rate of DNA synthesis. 77.  The transfer of a phosphate group from phosphoenolpyruvate to the active site histidine residue in the mutase produces pyruvate, even in the absence of pyruvate kinase activity. The subsequent spontaneous hydrolysis of the phospho-His allows the mutase to remain active. This alternative conversion of phosphoenolpyruvate to pyruvate occurs without concomitant ATP production. This strategy allows the cancer cell to use glycolysis as a source of biosynthetic intermediates and not as an ATP-generating pathway. [From Vander Heiden, M.G. et al., Science 329, 1492–1499 (2010).] 79.  Cancer cells produce lactate from pyruvate, even in the presence of oxygen. Lactate is released from the tumor cells and is taken up by the liver, which converts it to glucose via gluconeogenesis. Glucose leaves the liver and can be taken up from the bloodstream by the tumor cells. 81.  Glutamine and aspartate serve as the nitrogen donors for nucleotide biosynthesis and can be produced from citric acid cycle intermediates. Glutamate synthase catalyzes the conversion of ­α-ketoglutarate to glutamate; glutamine is synthesized from glutamate via the glutamine synthetase reaction. Aspartate can be produced from oxaloacetate via a transamination reaction. 83. a.  Fumarase catalyzes the transformation of fumarate to malate in the citric acid cycle. A fumarase deficiency results in the accumulation of the fumarate substrate and a decrease in the concentration of the malate product. Pyruvate levels increase because the citric acid cycle cannot be completed in the absence of fumarate. Pyruvate is transformed to lactate when the citric acid cycle is not functioning.  b.  Succinate accumulates in a patient with a succinate dehydrogenase deficiency because succinate cannot be converted to fumarate in the absence of this enzyme. Succinate also accumulates in the patient with a fumarase deficiency because the succinate dehydrogen­ase reaction is reversible.

S-68  ODD- N UM B ER E D S O LUT I O NS 85. a.  There are two possibilities—a CGT codon can be changed to a CAT codon, or a CGC codon can be changed to a CAC codon. In both cases, a G → A change occurs in the DNA. b. The structure of 2-hydroxyglutarate is shown. c. Normal isocitrate dehydrogenase catalyzes an oxidative decarboxylation reaction to yield α-ketoglutarate (2-oxoglutarate), while the mutant enzyme catalyzes a decarboxylation reaction to yield 2-hydroxyglutarate. d. Production of this compound in the glioma cells depletes the cell’s reduced cofactors, since a reduced coenzyme is required as a proton and an electron donor to form the hydroxyl group at the C2 position. [From Reitman, Z. and Yan, H., J. Natl. Cancer Inst. 102, 932–941 (2010).] C

5. a. Yes. By moving along a single DNA strand, the helicase acts as a wedge to push apart the double-­stranded DNA ahead of it. b. The free energy of dTTP hydrolysis is similar to the free energy of ATP hydrolysis. Each hydrolysis reaction drives the helicase along two to three bases of DNA. c. The T7 helicase is probably a processive enzyme. Its hexameric ring structure is reminiscent of the clamp structure that promotes the processivity of DNA polymerase (see Fig. 20.8). [From Kim, D.-E. et al., J. Mol. Biol. 321, 807–819 (2002).] 7.  The eukaryotic helicase binds to the leading-­strand template. The leading strand is synthesized in the 5ʹ → 3ʹ direction, the same direction the replication fork opens up, so its template has the opposite polarity, 3ʹ → 5ʹ. The bacterial helicase binds to the lagging-­strand template.

COO− H

3.  The origin is more likely to be richer in A:T base pairs, since these experience fewer stacking interactions (see Section 3.2) and are more easily separated, which allows easier access for the replication proteins.

OH

CH2 CH2 COO−

2-Hydroxyglutarate

Chapter 20 1.  Parental 15 N-­labeled DNA strands are shown in black, and newly synthesized 14N DNA strands are shown in gray. The original 15 N-­labeled parental DNA strands persist throughout succeeding generations, but their proportion of the total DNA decreases as new DNA is synthesized.

15N

DNA (heavy)

9. a. SSB, which coats single-­stranded DNA exposed at the replication fork, has a relatively low affinity for DNA because it is displaced as the polymerase proceeds. The polymerase, along with accessory proteins such as the clamp, has a relatively high affinity for the DNA. b. The cell most likely contains large amounts of SSB that coats the single-­stranded DNA to prevent it from forming secondary structures. Although the cell requires multiple copies of SSB for each replication fork, in theory, only two DNA polymerase enzymes are required (one for the leading strand and one for the lagging strand). 11.  The E. coli SSB tetramer is more likely to dissociate into monomers at the nonpermissive temperature, making it less likely to bind to DNA. This leaves DNA single strands either more susceptible to degradation by nucleases or more likely to reanneal. In either case, bacterial growth slows and then stops because the cells have difficulty replicating their DNA.

13.  The drug would inhibit DNA synthesis because the polymerization reaction is accompanied by the release and hydrolysis of inorganic pyrophosphate (PPi; see Fig. 20.5). Failure to hydrolyze the PPi removes the thermodynamic driving force for the overall process.

+

First generation

Hybrid DNA

15. a. The positively charged Mg2+ or Mn2+ ion could decrease the affinity of the DNA’s 3ʹ O atom for H, thereby increasing the nucleophilicity of the 3ʹ O atom (making it behave more like an O−­ ion). The positively charged metal ion could also help neutralize the negative charge on the incoming nucleotide. b. The metal ion could stabilize the negative charge that develops on the pentacovalent transition state. (DNA) C3 O

O

Second generation

+

3 4

Third generation

+

+

+

14N

DNA (light)

O

P O

O

P O

O

O O

P

O

C5 (NTP)

O

17.  DNA polymerase α is least processive because it synthesizes only a short DNA segment before polymerase δ or ε takes over. DNA polymerase ε is most processive because it synthesizes the leading strand continuously. DNA polymerase δ has intermediate processivity because it synthesizes Okazaki fragments (~200 bp long).

1 4

19.  In 10 hours, 6 × 105 bp of DNA can be replicated by one pair of DNA polymerases moving in one direction from an origin:

Fourth generation

7 8

+

1 8

​10​ 3​ bp 60 min ​10 h × ​ _     ​ × ​ _    ​ = 6 × ​10​ 5​bp​ min h  eplication proceeds in both directions, so each origin can support R the replication of 1.2 × 106 bp of DNA. The haploid human genome is

OD D -NUMB ER ED SO LUTI O NS  S-69 3.279 × 103 Mb (see Table 3.6), so the diploid set of 46 chromosomes is 6.56 × 109 bp, which would require about 5500 origins.

31. a. 

H H

9

6.56 × ​10​  ​ bp ________________ ​      ​  = 5.47 × ​10​ 3​ 1.2 × ​10​ 6​ bp

O

23.  The DNA molecule (chromosome) is much longer than an ­Okazaki fragment and must be condensed and packaged in some way to fit inside the nucleus (in a eukaryote) or cell (in a prokaryote). If the cell waited until the entire DNA molecule had been replicated, the newly synthesized lagging strand, in the form of many Okazaki fragments, might already be packaged and inaccessible to the endonuclease, polymerase, and ligase enzymes necessary to produce a continuous lagging strand. 25.  The DNase creates a nick (a break in one strand). Then the exonuclease activity of DNA polymerase I removes nucleotides on the 5ʹ side of the nick. At the same time, the polymerase active site adds radioactive deoxynucleotides to the 3ʹ side of the nick. The removal and replacement of nucleotides translates (moves) the nick in the 5ʹ → 3ʹ direction. DNA ligase then seals the nick between the original DNA and the newly synthesized radioactive segment.

O E

Lys

NH2 + N

R

O

E

Lys

P –O

1

O P

O

NAD+

O

R A

O–

NMN+

O

+

NH2

P

O

OH

OH

P

O –O

+ E

O

R 3

P

O

O

N

N

H H

H

b.  A

T

T

A G

G

G

G

G

G

N

N

H

N

N

N H

T

G

G

H

H

N N

G T

A T

T G G G A T

T

33.  Yes, the resulting telomeres will have a sequence complementary to the mutated sequence of the telomerase-­associated RNA template. This experiment was important because it established the mechanism of the enzyme and verified the role of the RNA template in extending chromosome length. 35.  The telomere length is 4 to 15 kb, and the cell can divide 35 to 50 times. For the shortest telomeres and the highest rate of cell division, the loss would be 4000 bp ÷ 50 divisions or 80 bp/division, and for the longest telomeres and the lowest rate of cell division, the loss would be 15,000 bp ÷ 35 divisions or 429 bp/division.

NH2 H

O–

Lys

O N

N

O

NH2

A

AMP

O O

N

H

39. a. 

O P

P O

N

H

N

N

O

O–

29. a.  DNA polymerase, b. reverse transcriptase or telomerase, c. primase or RNA polymerase.

H

N

N

H N

O–

O

O –O

O

O N

N

H

N

N

H

37.  Base changes in a, b, e, and f are transversions. Base changes in c and d are transitions.

R A

O–

2

H

H

21.  If the proofreading mechanism allows the polymerase to correct only the most recently incorporated base, the error rate will be higher because the polymerase keeps moving. Being able to correct a mismatched base even after the polymerase has moved on increases accuracy.

27.

N

N

H

N

N

N

N

N

Adenine H

8-Oxoguanine

b.  A G → T transversion results if the damage is not repaired. c. The base excision repair pathway is used to repair the damage. An endonuclease nicks the backbone, DNA polymerase fills in the gap, and ligase seals the nick. d. The cancer cells typically have A:T base pairs at points where G:C base pairs should be. 41. a.  N N

H

  b. 

O

H

NH N

Hypoxanthine

O N N

N N

Hypoxanthine

H

N N O

N

Cytosine

S-70  ODD- N UM B ER E D S O LUT I O NS c.  An A:T base pair is converted to a C:G base pair. 43. a. The structures of the intercalating agents resemble A:T and G:C base pairs, which explains why they can slip in between the stacked base pairs of DNA. This creates what appears to the replication machinery as an “extra” base pair. An extra base incorporated into the newly synthesized DNA may eventually lead to a frameshift mutation (in which the additional nucleotide causes the translation apparatus to read a different set of successive three-­nucleotide codons). b. A frameshift mutation caused by ethidium bromide has disastrous consequences when the mutation occurs in a protein-­ coding region. The frameshift mutation causes the translation apparatus to read a different set of codons and produce a completely different sequence of amino acids in the resulting protein. Nitrous acid causes the deamination of cytosine to uracil, adenine to hypo­ xanthine (Problem 41), and guanine to xanthine (Problem 42). These changes generate altered base pairs when the DNA is replicated. If the point mutation occurs in a protein-­coding region, the identity of the encoded amino acid may change, but there is no change in the reading frame. 45.  Bromouracil causes an A:T to G:C transition. O

O

Br

Br

NH

N H

5-Bromouracil (keto tautomer) 47. a. 

N

N N

O

5-Bromouracil (enol tautomer)

H

N

H

Guanine

H N

H

N

N N N

H

O

N

N

N H

2-Aminopurine

Cytosine b. 

H

N

O

N H

N

O

H

H N

H

N N

H H

N



N

O

N

N

N

H H N N N

H H

N

N

O H

N

N

N

H

[From Sowers, L.C. et al., Biochemistry 29, 7613–7620 (2000).] 49.  During the course of the reaction, a methyl group from the O 6 -methylguanine is transferred to a cysteine residue in the enzyme’s active site, inactivating the enzyme. Normally, enzymes are

regenerated after a cycle of catalysis. Apparently repairing this type of damage is so critical for cell functioning that it is worth the use of cellular resources to synthesize an enzyme that will be used only once. 51.  The thymine–­thymine dimer is most common, since this lesion forms upon exposure of the DNA to ultraviolet light. 53.  The deaminations all generate bases that are foreign to DNA; therefore, they can be quickly spotted and repaired before DNA has fully replicated and the damage passed on to the next generation. 55.  The mutant bacteria are unable to repair deaminated cytosine (uracil). In these cells, the rate of change of G:C base pairs to A:T base pairs is much greater than normal. 57. a. The human genome contains 3.279 × 109 bp (see Table 3.6), which means that every cell contains twice this much DNA, and each DNA molecule has two strands to be replicated. An error rate of 1/22,000 would generate (2)(2)(3.279 × 109)/(22,000) = 5.96 × 105 errors. b. Reducing the error rate 100-fold would reduce the number of errors to about 5960. 59.  DNA polymerase III replicates DNA until a thymine dimer is encountered. Polymerase III is accurate but cannot quickly bypass the damage. Polymerase V, which can more quickly proceed through the damaged site, does so, but at the cost of misincorporating G rather than A opposite T. Thus, replication can continue at a high rate. The tendency for DNA polymerase V to continue to introduce errors is minimized by its low processivity: Soon after passing the thymine dimer, it dissociates and the more accurate polymerase III can continue replicating the DNA with high fidelity. 61.  Without functional DNA repair enzymes, additional mutations may arise in genes that are involved in regulating cell growth. In the absence of proper growth controls, cells may begin to proliferate at an accelerated rate. 63. a. Normally, the Rb protein acts as a tumor suppressor by preventing the cell from synthesizing DNA, a pre-­requisite for cell division. A mutation in the Rb protein may yield a protein that is unable to bind to and inhibit its target transcription factor. The transcription factor is thus free to induce the expression of genes required for DNA synthesis. This may result in the loss of control of the cell cycle, converting a non-­cancerous cell to a cancerous cell. b. Mutations in these proteins do not necessarily lead to cancer. According to the multiple-­hit hypothesis, more than one genetic change is required to initiate tumor growth. In addition, individuals with a defective allele for an oncogene or tumor suppressor gene may also have another normal allele for the gene that helps protect them from carcinogenesis. [From Giacinti, C. and Giordano, A., Oncogene 25, 5220–5227 (2006).] 65.  The ring shape of PCNA allows it to slide along the DNA helix without making sequence-­specific contacts. A protein with a similar structure could likewise slide along the DNA. Distortions in the DNA helix caused by nicks, gaps, missing bases, or bulky chemical adducts could halt the progress of the sensor and allow it to recruit DNA repair proteins. 67.  According to the pathway described in the problem, overactivation of Ras would lead to a decrease in ubiquitinated p53. This would decrease the rate of p53 degradation, leaving more available to halt the cell cycle. This would actually counteract the growth-­promoting activity of the Ras pathway. 69.  p53 increases the production of cytochrome c oxidase, the terminal enzyme of the electron transport chain, which consumes oxygen and contributes to the proton gradient that powers ATP synthesis (Section 15.3). In the absence of p53, less cytochrome c oxidase is

OD D -NUMB ER ED SO LUTI O NS  S-71

73.  Novobiocin and ciprofloxacin are useful as antibiotics because they inhibit prokaryotic DNA gyrase but not eukaryotic topoisomerases. They can kill disease-­causing prokaryotes without harming host eukaryotic cells. Doxorubicin and etoposide inhibit eukaryotic topoisomerases and can be used as anticancer drugs. Although these drugs inhibit topoisomerases from both cancer cells and normal cells, cancer cells have a higher rate of DNA replication and are more susceptible to the effects of the inhibitors than are normal cells. 75.  DNA gyrase is a type II topoisomerase in E. coli. It can introduce negative supercoils into the DNA ahead of the replication fork. In the absence of DNA gyrase, strand separation would cause overwinding of the DNA ahead of the replication fork, generating positive supercoils that would hinder DNA unwinding. 77. a. The side chains of lysine and arginine residues have high pK values and are positively charged at physiological pH. The positively charged groups can form ion pairs with the negatively charged phosphate groups on the backbone of the DNA molecule. b. A solution of 0.5 M NaCl effectively dissociates the positively charged histone proteins from the negatively charged DNA by disrupting the ionic bonds that hold these two chromatin components together. The charged sodium and chloride ions can form substitute ionic interactions with the histone protein and DNA. 79.  Acetylated lysine residues are more likely to be found in the transcriptionally active euchromatin. Acetylation removes the positive charge on the lysine side chain (see Solution 10) and weakens the electrostatic interactions between the histone protein and the negatively charged DNA (see Solution 77), which allows greater access of the transcriptional machinery to the DNA. 81. a. Because the protamine–­DNA complexes occupy less volume than nucleosomes, DNA can be more easily packed into the small ­volume of the sperm nucleus. b. DNA in nucleosomes is less compact and therefore more accessible for transcription, so one would expect nucleosomal DNA to contain genes that would need to be expressed immediately following fertilization, that is, genes essential for early embryogenesis. 83.  Okazaki fragments are longer in bacterial cells because prokaryotic DNA is not associated with nucleosomes. About 146 bp of DNA winds around each nucleosome, which must be unwrapped prior to the synthesis of an Okazaki fragment. Thus Okazaki fragments isolated from eukaryotic cells are 100–200 bp long, which is about the length of the DNA associated with each nucleosome. 85.  MspI, AsuI, EcoRI, PstI, SauI, and NotI generate sticky ends. AluI and EcoRV generate blunt ends. 87.  AluI cleaves the sequence at two locations and EcoRI and NotI cleave the sequence at one location each. The other enzymes listed in Table 20.1 do not cleave this segment of the plasmid DNA. EcoRI

AluI

AluI

NotI

TCCGAATTCGAGCTCCGTCGACAAGCTTGCGGCCGCACT

Circular

71.  Topoisomerase I reactions are driven by the free energy change of DNA shifting from a supercoiled conformation to a relaxed conformation, so no external source of free energy is needed. The enzyme merely accelerates a reaction that is already favorable. Topoisomerase II reactions involve more extensive mechanical intervention because both strands of the DNA are cleaved and held apart while another segment of DNA passes through the break. This process requires the free energy of ATP hydrolysis, since it is not thermodynamically favorable on its own.

89. a. A HaeII digest of the linear 2743-bp pGEM-­3Z plasmid (see Fig. 20.36) produces four bands, as shown below (1–323 produces a 323-bp band, 324–693 produces a 370-bp band, 694–2564 produces a 1871-bp band, and 2565–2743 produces a 179-bp band). b. A digest of the circular plasmid produces the 370-bp band and the 1871-bp band but also produces a 502-bp band (2565–323) in place of the 323-bp band and the 179-bp band. [From Promega.] Linear

made, so the cell relies less on aerobic respiration as a source of ATP and relies more on glycolysis.

3000 bp 1500 bp 1000 bp

500 bp

100 bp

91.  Polymerization occurs in the 5ʹ→3ʹ direction and a 3ʹ OH group must be available, so the primer must be complementary to the sequence as shown. 5ʹ-AGTCGATCCCTGATCGTACGCTACGGTAACGT-3ʹ 3ʹ-TGCCAT TGCA-5ʹ 93.  The primers are shown in red below. 5ʹ-ATGATTCGCCTCGGGGCTCCCC · · · AGTCGCTGGTGCTGCTGACGCTGCTCGTCG-3ʹ ← 3ʹ-ACGACTGCGACGAGCAGC-5ʹ

5 ʹ-ATGAT TCGCC T CGGGGC T-3ʹ → 3 ʹ-TACTAAGCGGAGCCCCGAGGGGTCAGCGACC · · ·

ACGACGACTGCGACGAGCAGC-5ʹ 95.  Primers with high GC content have high T m values. If the annealing temperature is much lower than the melting temperature, improper base pairing may occur (see Fig. 3.9) and the desired gene fragment may not be amplified. 97.  Taq polymerase is commonly used in PCR. Because the enzyme lacks exonuclease activity, it cannot proofread its work. As a result, it may introduce errors while amplifying a segment of DNA. If an error occurs in an early cycle, it will affect a larger proportion of the DNA product than if an error occurs in a later cycle. 99.  To amplify the protein-­c oding DNA sequence, the primers should correspond to the first three and last three residues of the protein (each amino acid represents three nucleotides, so the primers would each be nine bases long). Use Table 3.3 to find the codons that correspond to the first three residues: Met AUG

Gly

Ser

GGU GGC GGA GGG

UCU UCC UCA UCG AGU AGC

S-72  ODD- N UM B ER E D S O LUT I O NS  sing just the topmost set of codons, a possible DNA primer would have U the sequence 5ʹ-ATGGGTTCT-3ʹ. This primer could base-­pair with the gene’s noncoding strand, and its extension from its 3ʹ end would yield a copy of the coding strand of the gene (see Fig. 20.37). The other primer must correspond to the last three amino acids of the protein: Val

Ser

Pro

GUU GUC GUA GUG

UCU UCC UCA UCG AGU AGC

CCU CCC CCA CCG

7. a. 

O NH

CH

C

CH2 CH2 CH2 CH2 H3C

 gain, considering just the topmost set of codons, a probable DNA A coding sequence would be 5ʹ-GTTTCTCCT-3ʹ. This sequence cannot be used as a primer. However, a suitable primer would be the complementary sequence 5ʹ-AGGAGAAAC-3ʹ, which can then be extended from its 3ʹ end to yield a copy of the noncoding strand of the gene. The number of possible primer pairs is quite large, because all but one of the amino acids has more than one codon. For the first primer, there are 1 × 4 × 6 = 24 possibilities; for the second, 4 × 6 × 4 = 96 possibilities. There are 24 × 96 = 2304 different pairs of primers that could be used to amplify the gene by PCR. 101. a.

modifications decrease the affinity of the histone for DNA, resulting in less condensed chromatin and facilitating the assembly of the transcription machinery on the DNA.

ATTGTTCCCACAGACCG

CGGCGAAGCATTGTTCC        ACCGTGTTTCCGACCG TTGTTCCCACAGACCGTG b.  Portions of the sequence where three fragments align are more accurate, and you can be confident that the sequence was determined correctly.

N+ CH3 CH3

b.  The additional methyl groups shield the positive charge and make the side chain more nonpolar. The binding site that recognizes the trimethylated lysine is likely to be lined with nonpolar amino acid side chains. An unmodified Lys side chain, with a strong positive charge on its amino group, would be unlikely to bind favorably to such a binding site. 9.  No, histones modified with ubiquitin are not marked for proteolytic destruction by the proteasome because the amino acid side chains of the histone proteins have only one ubiquitin attached. In order to enter the proteasome for degradation, a chain of at least four ubiquitin molecules is required (see Section 12.1). 11. a. The nonstandard amino acid is homocysteine, which can accept a methyl group donated by methyl-tetrahydrofolate to regenerate methionine (see Box 18.B). COO– +

H3N

1.  Messenger RNA is translated into protein, and a single mRNA can be used to translate many proteins. In this way, the mRNA is “amplified.” Ribosomal RNA performs a structural role and is not amplified, so many more rRNA genes are needed to express sufficient rRNA to meet the needs of the cell. 3.  Sequence-specific interactions require contact with the bases of DNA, which can participate in hydrogen bonding and van der Waals interactions with protein functional groups. Electrostatic interactions involve the ionic phosphate groups of the DNA backbone and are therefore sequence independent. b. 

CH2

CH2 CH2 CH2 C

O –O

P O–

c. 

13.  a. 

O H

O

NH2

C

N N

HO O

N –O P

R

O

CH3

d.  Acetylation of lysine removes the residue’s positive charge and produces a neutral side chain; phosphorylation of the serine and histidine side chains produces a side chain with two negative charges. The acetylated Lys is not able to form an ion pair with the phosphodiester backbone of the DNA as the unmodified Lys side chain does. Phosphorylation of the His and Ser side chains introduces negative charges that repel the negatively charged DNA. Both types of

N O

R

5-Carboxylcytosine

b.  A DNA glycosylase recognizes the oxidized base and an e­ ndonuclease then hydrolyzes the phosphodiester backbone. DNA ­polymerase then fills in the gap, and ligase seals the nick (see Fig. 20.17). O CH2

CH2

Butyrate O–

NH2

C N

5-Formylcytosine

H3C

N O

b.  The other product is methanol, CH3OH.

15. a. 

CH2

O

NH

CH2 SH

Chapter 21

CH2

H

CH2

103.  If the replacement gene has the same sequence as the original, it too will be recognized by the guide RNA, and Cas9 will cleave it.

5.  a.    

C

C

OH O−

H3C

CH

O CH2

C

O−

β-Hydroxybutyrate

b.  The results indicate that butyrate is an HDAC inhibitor but ­β-hydroxybutyrate is not. When the cells are incubated with 5 mM butyrate, the amount of acetylated Lys is greater than in the control, which is consistent with an inactive deacetylase. By contrast, even when the concentration of β-hydroxybutyrate is increased to 40 mM, the level of acetylated Lys residues remains the same as the control. c.  Incubating cells with butyrate increases the expression of both PGC1α (which is involved in mitochondrial biogenesis) and CPT1b (which shuttles fatty acids into the mitochondrial matrix for β ­oxidation). Increasing the expression of these two genes is likely to result in an increase in fatty acid catabolism. Because butyrate is an HDAC inhibitor, it may exert its effects by promoting the acetylation of histone Lys side chains, which neutralizes the positive charge of Lys and weakens the interaction

OD D -NUMB ER ED SO LUTI O NS  S-73 between the protein and the DNA. This makes the DNA more available for transcription. By contrast, ­β-hydroxybutyrate treatment results in a small decrease in PGC1α expression and no change in CPT1b e­ xpression. These results indicate that hydroxybutyrate is unlikely to bring about changes in fatty acid catabolism. 17. a. The promoter region is shaded. b. The –10 region is AT-rich. A:T base pairs have weaker stacking interactions (see Section 3.2) and are easier to melt apart than G:C base pairs, which have stronger stacking interactions. This facilitates DNA unwinding to expose the template for transcription.

+1

AATGCTTGACTCTGTAGCGGGAAGGCGTATTATCCAACA

19.  Affinity chromatography takes advantage of the ability of the desired protein to bind to a specific ligand (see Section 4.6). To purify Sp1, the GGGCGG oligonucleotide is covalently attached to tiny beads, which constitute the stationary phase of a chromatography column. A cellular extract containing the Sp1 protein is loaded onto the column, and a buffer (the mobile phase) is washed through the column to elute proteins that do not bind to the oligonucleotide ligand. Next, a high-salt buffer is applied to the column to disrupt the strong interactions between the Sp1 and the GC box, and the Sp1 protein is eluted. [From Kadonaga, J.T. et al., Trends Biochem. Sci. 11, 20–23 (1986) and Kadonaga, J.T. and Tjian, R., Proc. Natl. Acad. Sci. USA 83, 5889–5893 (1986).] 21. a. There are four choices for each base pair in the sequence, so the chance of finding a certain 10-bp sequence is one in 410 or one in 1.05 × 106. The yeast genome has 1.22 × 107 bp (see Table 3.6), so the sequence would be expected to appear about 12 times. 7

1.22 × ​10​  ​ bp ______________ ​      ​= 11.6​ 1.05 × ​10​ 6​ bp

b.  There are 416 or 4.29 × 109 possible 16-bp sequences. In the 3.28 × 109 bp human genome (see Table 3.6), a specific sequence could be expected to show up about once. 9

3.28 × ​10​  ​ bp ______________    ​   ​= 0.76​ 9

4.29 × ​10​  ​ bp

23.  Upregulating the methylase increases the extent of methylation of Lys 27 in histone 3. Since trimethylated Lys 27 is normally associated with transcriptionally inactive chromatin, these genes are hypersilenced in cancer cells. [From Stark, G.R. et al., Cell Res., 21, 375–380 (2011).] 25. a.  The polymerase binds most tightly to the DNA segment with the largest bulge. This DNA mimics the transcription bubble, in which the DNA strands are separated. b. Since Kd is a dissociation ­constant, the apparent equilibrium constant for the binding is 1/Kd. The equilibrium constant can be determined by using Equation 12.2. For fully based-paired DNA: ​ΔG​°′ ​= − RT ln ​(1/​Kd​  ​)​ ΔG​°′ ​= − (8.3145​J ⋅ K​ −1​⋅ ​mol​ −1​) (298 K) ln (1/315 × 1​ 0​ −9​) ΔG​°′ ​= − 37,000 J ⋅ ​mol​ −1​= − 37 k J ⋅ ​mol​ −1​ For the eight-base bulge, ​ΔG​°′ ​= − RT ln ​(1/​Kd​  ​)​ ΔG​°′ ​= − (8.3145​J ⋅ K​ −1​⋅ ​mol​ −1​) (298 K) ln (1/0.0013 × ​10​ −9​) ΔG​°′ ​= − 68,000 J ⋅ ​mol​ −1​= − 68 k J ⋅ ​mol​ −1​  olymerase binding to the eight-base bulge DNA (a mimic of melted P DNA) is more favorable than binding to fully base-paired DNA. c. Melting open a DNA helix is thermodynamically unfavorable. Some of the favorable free energy of binding the polymerase to the DNA is spent in forming the transcription bubble. When the ­transcription bubble is preformed (for example, in the DNA with an eight-base

bulge), this energy is not spent and is reflected in the apparent energy of polymerase binding. The difference in ∆G°′ ­values for polymerase binding to double-stranded DNA and to the eight-base bulge is −68 − (−37) = −31 kJ ⋅ mol−1. This value is an estimate of the free energy cost (+31 kJ ⋅ mol−1) of melting open eight base pairs of DNA. [From Bandwar, R.P., and Patel, S.S., J. Mol. Biol. 324, 63–72 (2002).] 27.  The lactose permease allows lactose to enter the cell, which increases the intracellular lactose concentration. Lactose can then serve as a substrate to form allolactose, which binds to the repressor protein to remove it from the operator. The presence of additional lactose assists in the full expression of the operon. 29.  If the repressor cannot bind to the operator, the genes of the lac operon are constitutively expressed; that is, the genes are expressed irrespective of whether lactose is present or absent in the growth medium. Adding lactose has no effect on gene expression. 31.  Wild-type cells cannot grow in the presence of phenyl-Gal. The wild-type cells produce a small amount of β-galactosidase in the absence of lac operon expression, but not enough to be able to hydrolyze phenyl-Gal to phenol and galactose. The lacI mutants, however, will thrive in this growth medium. The mutation in the lacI gene results in the expression of a nonfunctional repressor (or perhaps no repressor); in any case, the lac operon is constitutively expressed and β-galactosidase is produced in sufficient amounts to act on p ­ henyl-Gal to release galactose. The use of this growth medium permits selection of repressor mutants since the mutants survive while the wild-type cells do not. 33. a. If RNA polymerase incorporates the wrong base, the DNA– RNA hybrid is distorted, polymerase activity halts temporarily, and the nascent RNA backs out of the active site. TFIIS stimulates endonuclease activity, which excises the incorrect base, and polymerization then resumes. By contrast, DNA polymerase, in addition to several proofreading mechanisms, relies on repair processes such as the mismatch repair system, which combine to yield an error rate for DNA polymerase that is 1000-fold lower than that of RNA polymerase.  b.  The accurate transmission of genetic information from one generation to the next requires a high degree of fidelity in DNA replication. A higher rate of error in RNA transcription is permitted because the cell’s survival usually does not depend on accurately synthesized RNA. If translated, an RNA transcript containing an error may lead to a defective protein, which is likely to be destroyed by the cell before it can do much damage. The gene can be transcribed again and again to generate accurate transcripts. 35.  Cordycepin, which resembles adenosine, can be phosphorylated and used as a substrate by RNA polymerase. However, it blocks further RNA polymerization because it lacks a 3′ OH group, which is required to nucleophilically attack the 5′ phosphate of the incoming nucleotide. 37.  If α-amanitin were added to cells in culture, the synthesis of mRNA would be inhibited, but the synthesis of all other types of RNA would be relatively unaffected. RNA polymerase II is responsible for mRNA synthesis and is the most sensitive to inhibition by α-amanitin. Experiments with this toxin permitted investigators to determine the types of RNA synthesized by each polymerase. 39.  RNA polymerase I is the most sensitive to actinomycin D, RNA polymerase III is least sensitive, and RNA polymerase II has intermediate sensitivity. By contrast, α-amanitin mainly inhibits RNA polymerase II (see Solution 37). 41.  TFIIH is the helicase used by RNA polymerase II to unwind the DNA prior to transcription. TFIIH, like many helicases, requires the energy of ATP hydrolysis to melt the DNA; thus, cancer cells treated with triptolide will be unable to carry out RNA polymerase II–catalyzed transcription of mRNA. RNA polymerases I and III have intrinsic helicase activity and do not use TFIIH, so they would not be expected to be affected by triptolide. [From Laham-Karam, N. et al., Front. Chem. 8, 276 (2020).]

S-74  ODD- N UM B ER E D S O LUT I O NS 43.  The nucleotide excision repair pathway is responsible for repairing large-scale DNA damage. In the first step, an approximately 30-bp segment containing the damaged nucleotides is removed. This requires the ATP-fueled helicase activity provided by TFIIH. The repair process is completed by a DNA polymerase that fills in the gap and ligase that seals the nick. 45. O O O HN

CH

C

NH

CH

CH2

N

C

CH

b. 

Product absent mRNA translation DNA

Operon

C Product present

CH2 O –O

P

Product O

RNA-binding protein

O–

OH O NH

O NH

CH

C

CH

OH

CH3 –O

CH

C

O N

CH

C

CH

C

CH2

CH2

O

O

P

–O

O

O–

P

N

N

N

O

O–

N C C G G C G C U U UUUUNNN․․․3′

G C C G C G A A 5′․․․NN

terminate transcription or block translation DNA

47. a. 5′ . . . NNAAGCGCCGNNNNCCGGCGCUUUUUUNNN . . . 3′ b. 

mRNA

O NH

c.  If no protein were involved, the operon’s product would have to interact directly with the mRNA. 53. a.  CAG codes for glutamine, so the resulting protein would contain a series of extra Gln residues. These polar residues would most likely be located on the protein surface but could interfere with protein folding, stability, interactions with other proteins, and catalytic activity.  b.  The longer transcripts could be due to transcription initiating upstream of the normal site or failing to stop at the usual termination point. Longer mRNA molecules could also result from the addition of an abnormally long poly(A) tail or the failure to undergo splicing. c. 

49.  As transcription proceeds, the nascent RNA forms a variety of secondary structures as portions of the transcript form complementary base pairs. The formation of these secondary structures may cause transcription to pause but not necessarily terminate. 51. a. 

Product absent

G A C G A C G A C CAG

C

A G C A G C A G CAG

[From Fabre, E. et al., Nucleic Acids Res. 30, 3540–3547 (2002).]

Repressor RNA polymerase

transcription

DNA

Operator Operon

Product present

RNA polymerase no transcription Product

Repressor

55.  Bacterial mRNAs have a 5′ triphosphate group. The pyrophosphohydrolase removes two of the phosphoryl groups as pyrophosphate (PPi), leaving a 5′ monophosphate. This apparently makes the mRNA a better substrate for the endonuclease. 57.  Polymerase

Template

Substrates

Product

DNA polymerase

DNA

dATP, dCTP, dGTP, dTTP

DNA

Human telomerase

RNA

dATP, dGTP, dTTP

telomere DNA

RNA polymerase

DNA

ATP, CTP, GTP, UTP

RNA

Poly(A) polymerase

None

ATP

poly(A) tail of mRNA

tRNA CCAadding enzyme

None

ATP, CTP

3′ CCA on tRNA

59.  Messenger RNAs are transcribed only by RNA polymerase II. The phosphorylated tail of RNA polymerase II recruits the enzymes needed for capping and polyadenylation. Other types of RNAs are synthesized

OD D -NUMB ER ED SO LUTI O NS  S-75 by different RNA polymerases that do not have phosphorylated tails and cannot recruit enzymes involved in post-transcriptional modification. Thus, only mRNAs are capped and polyadenylated. 61.  The capped mRNA has a 5′–5′ triphosphate linkage, which is not recognized by exonucleases (which normally cleave 5′–3′ phosphodiester bonds). Capping must occur as soon as the 5′ end of the mRNA emerges from the RNA polymerase so that the message will not be degraded by exonucleases. 63.  The phosphatases remove the phosphate groups from the ­C-terminal domain of RNA polymerase, which transforms the RNA polymerase back to the unphosphorylated form that is capable of binding Mediator for the next transcription initiation event. 65.  The PABP binds to the poly(A) tails and protects the mRNA from degradation by the nucleases. Increasing the concentration of PABP extends the half-lives of the mRNAs bound to this protein. 67. a.  The phosphate groups of the phosphodiester backbone of RNA will be labeled wherever α-[32P]-ATP is used as a substrate by RNA polymerase.  b.  32P will appear only at the 5′ end of RNA molecules that have A as the first residue (this residue retains its α and β phosphates). In all other cases where β-[32P]-ATP is used as a substrate for RNA synthesis, the β and γ phosphates are released as PPi (see Fig. 20.5).  c. No 32P will appear in the RNA chain. During polymerization, the β and γ phosphates are released as PPi. The terminal (γ) phosphate of an A residue at the 5′ end of an RNA molecule is removed during the capping process.

structure in proteins refers to the overall three-dimensional shape of the macromolecule; similarly, for the ribozyme, the tertiary structure refers to the three-dimensional shape of the molecule. In this study, changing the identity of a base (or adding a base) alters the primary structure of the RNase P in a way that changes the secondary and ­tertiary structure of the ribozyme. These changes affect substrate binding and catalysis to different extents. [From Kaye, N.M. et al., J. Mol. Biol. 324, 429–442 (2002).] 77.  If the mRNA is missing a cap, the translation machinery will not recognize the RNA as mRNA. If introns are not spliced out prior to export, they will be translated, potentially disrupting protein function. An mRNA missing a poly(A) tail is more sensitive to degradation by endonucleases. 79.  The siRNA is shown in the bottom strand. 5′ . . . GGAGUACCCUGAUGAGAUC . . . 3′ 3′ . . . CCUCAUGGGACUACUCUAG . . . 5′ [From Takei, Y. et al., Cancer Res. 64, 3365–3370 (2004).] 81.

NH2

S HN O

N N

Ribose 4-Thiouridine

S

N

Ribose 2-Thiocytidine

69.  The splicing reactions are mediated by the spliceosome, a large RNA–protein complex. The intron must be large enough to include spliceosome binding site(s). In addition, the formation of a lariatshaped intermediate (see Fig. 21.24) requires a segment of RNA long enough to curl back on itself without strain.

83. a.  The mRNA from a gene may be alternatively spliced to yield ­several different types of proteins.  b.  This increases the diversity of the proteins produced by the cell without a correspondingly large ­number of genes.

71. 

Chapter 22

5′ splice site

. . . GGCAGGTTGGTA . . .  

3′ splice site . . . ACCCTTAGGCTGCT . . .

73. a.  5′ splice site 3′ splice site . . . GAAGAAGGTAAGTT . . .   . . . CTTGCAGGTTCTT . . .

1.  A hypothetical quadruplet code would have 44, or 256, possible combinations.

[From Heilig, R. et al., Nucleic Acids Res. 10, 4363–4382 (1982).]

3.  The poly(Lys) peptide results from translation of the poly(A) tail of a mRNA whose stop codon is missing as the result of a mutation or faulty transcription (AAA is a codon for lysine). With the addition of the poly(Lys) segment, the protein is likely to be nonfunctional, so it is best to destroy it and reuse its amino acids.

b.  In the two-step transesterification process (see Fig. 21.24), phosphodiester bonds are exchanged such that the number of these bonds is the same before and after the splicing process. Thus, additional energy in the form of ATP hydrolysis is not necessary.

5. a. This segment has the sequence Ile–Ile–Phe–Gly–Val. b. The mutation is the deletion of three nucleotides (CTT), affecting codons 507 and 508. The resulting protein segment has the sequence Ile–Ile– Gly–Val, which is missing Phe 508.

75. a.  Mutating U69 to G or C dramatically decreases activity, as shown by the nearly 50-fold decrease in the value of the rate constant k. Mutating the U residue to the similar-sized C residue has little effect on binding, but a mutation to the larger G residue has a much greater effect on binding, as evidenced by the larger Kd for the U69G mutant. In general, catalysis seems to be affected more than substrate binding, indicating the importance of the bulge to ­catalysis.  b.  Deleting the U69 residue, which deletes the bulge, dramatically decreases activity, as evidenced by the decrease in the rate constant k, but has little effect on substrate binding. Increasing the size of the bulge dramatically decreases both the activity and substrate binding affinity. These results confirm that the presence of the bulge is essential for catalysis and that the geometry of the bulge is important for substrate binding.  c.  The primary structure of a protein refers to its sequence of amino acids; in the RNase P RNA, this corresponds to the sequence of 417 nucleotides. Secondary structure in proteins refers to the conformation of the backbone groups, which often forms α helices and β sheets. In the ribozyme, secondary structure refers to the base-paired stem-and-loop structures. Tertiary

7. a. There are six reading frames: three for the top strand and three for the bottom strand. The amino acid sequences are indicated below; a * denotes a stop codon. b. Two of the six reading frames do not contain stop codons are therefore open reading frames. R * A F Q H R L V R L D E P F S T A * * G M S L S A P L S E V Q P H * A V L K G S S N L T K R C * K A H T S L S G A E R L I 9. a. poly(Phe), b. poly(Pro), c. poly(Lys). 11. a. A polypeptide consisting of a repeating Tyr–Leu–Ser–Ile tetra­ peptide will be produced. b. Depending on the reading frame, the polypeptide may begin with Tyr, Ile, or Ser. 13.  Because the tRNAs that match the common codons are most abundant in the yeast cell, protein synthesis is normally efficient. If

S-76  ODD- N UM B ER E D S O LUT I O NS a mutation alters a codon so that it is not one of the 25 commonly used codons, it is likely that the isoacceptor tRNA for that codon is relatively scarce. Consequently, waiting for the appropriate tRNA to deliver the amino acid to the ribosome would result in a lower rate of protein synthesis, even though the sequence of the protein is unchanged. 15. a. The redesigned sequence is shown with the altered codons underlined. ···GCCCCGACCAGCAGCAGCACCAAAAAAACCCAGCTGCAG···

ribosomal RNAs have specific residues that are essential to their function; damaging these residues causes loss of activity. 37.  The peptidyl transferase activity lies entirely within the 23S rRNA; that is, 23S rRNA is a ribozyme. The proteins might be necessary to assist the 23S rRNA in forming the necessary threedimensional structure required for catalytic activity, just as the proper conformation is required for protein enzymes. Extremely strong intermolecular interactions between the proteins and the rRNA confirm the importance of the proteins and explain why the extraction process failed to remove them.

c.  An alternative strategy would be to overexpress the tRNAs in E. coli that correspond to the human preferred codons. [From ­Williams, D.P. et al., Nucleic Acids Res. 16, 10453–10467 (1988) and the Codon Usage Database.]

39.  All three types of proteins have an RNA recognition motif because all of them bind to RNA. The ribosomal proteins bind to rRNA to form the ribosome. Rho factor is a bacterial transcriptional terminator that acts as a helicase to pry the nascent RNA away from its DNA template (see Section 21.2). The eukaryotic poly(A) binding protein binds to the poly(A) tail at the 3′ end of mRNA (see Fig. 21.20).

17.  Gly and Ala; Val and Leu; Ser and Thr; Asn and Gln; and Asp and Glu.

Process

b.  Using one-letter abbreviations, the sequence of both proteins is APTSSSTKKTQLQ.

19.  Gly is the smallest amino acid, so the aminoacylation site in GlyRS can be small enough to prevent the entry of any other amino acid.

41. Replication

Substrates dATP, dCTP,   dGTP, dTTP

Transcription

Translation

ATP, CTP,   GTP, UTP

20 different   amino acids,   each linked   to a tRNA

21. a. His, b. Asn, c. Thr [From tRNAdb, http://trnadb.bioinf.unileipzig.de/DataOutput/]

Product

23.  The 5′ nucleotide is at the wobble position, which can participate in non-Watson–Crick base pairings with the 3′ nucleotide of an mRNA codon. Because the first two codon positions are more impor­ tant for specifying an amino acid (see Table 22.1), wobble at the third position may not affect translation.

Template both parent   or guide   DNA strands   are used as  templates

one strand of   DNA is used   as a template

mRNA sequence   specifies the   order of amino  acids

25. a. The tRNA recognizes CGU. b. If adenosine is modified to inosine (the latter can base pair with U, C, and A), the modified tRNA can recognize CGC and CGA in addition to CGU, all of which code for arginine.

Primer

RNA primer

no primer  needed

Met attached to   an initiator  tRNA

Enzyme

DNA poly merase

RNA polymerase ribosomal  peptidyl  transferase  (rRNA)

27.  CUG, CUA, and CUU. [From Sørensen, M.A. et al., J. Mol. Biol. 354, 16–24 (2005).] 29.  The two Lys codons are AAA and AAG. Substitution with C would yield CAA and CAG, which code for Gln; substitution with G would yield GAA and GAG, which code for Glu; and substitution with U would yield UAA and UAG, which are stop codons. Replacing a Lys codon with a stop codon would terminate protein synthesis prematurely, most likely producing a nonfunctional protein. Replacing Lys with Glu or Gln could disrupt the protein’s structure and therefore its function if the Lys residue were involved in a structurally essential interaction such as an ion pair in the protein interior. If the Lys residue were on the surface of the protein, replacing it with Glu or Gln, both of which are hydrophilic, might not have much impact on the protein’s structure or function. 31.  Like other nucleic acid–binding proteins (histones are one example), ribosomal proteins contain positively charged lysine and arginine residues that interact favorably with the polyanionic RNA. The most important interactions between the protein and the nucleic acid are likely to be ion pairs. 33.  The assembly of functional ribosomes requires equal amounts of the rRNA molecules. Therefore, it is advantageous for the cell to synthesize the rRNAs all at once. 35.  Ribosomal inactivating proteins catalyze the removal of adenine bases from ribosomal RNA. This is analogous to removing a side chain from an amino acid residue in a protein. Like proteins,

two identical a single-stranded a polypeptide   double helices   RNA molecule  chain

Cellular nucleus  location

nucleus

cytosol

43. a. The polypeptide would contain all lysine residues with aspara­ gine at the C-terminus. b. If the mRNA were read from 3′ → 5′, the polypeptide would consist of an N-terminal glutamine followed by a series of lysine residues. c. Transcription and translation both take place in the 5′ → 3′ direction; this allows bacterial cells to begin translating nascent mRNA before transcription is complete. If translation took place in the 3′ → 5′ direction, the ribosome would have to wait for mRNA synthesis to be completed before translation could begin. 45.  The sequence on the 16S rRNA that aligns with the Shine–Dalgarno sequence is shown. The initiation codon is highlighted in gray. 5′· · · CUACCAGGAGCAAAGCUAAUGGCUUUA · · ·3′ 3′· · · UCCUC · · · 5′ 47.  The presence of a 5′ cap and a 3′ poly(A) tail indicates that the RNA is messenger RNA that was transcribed by RNA polymerase II (see Section 21.2). These modifications help the ribosome to distinguish mRNA, which is a template for translation, from other types of RNA, which are not.

OD D -NUMB ER ED SO LUTI O NS  S-77

51.  The eukaryotic initiation factor eIF2 delivers the initiator tRNA to the P site on the ribosome, ensures a proper codon–anticodon match, and then hydrolyzes its bound GTP to GDP before departing. If the GTPase activity is activated prematurely, the selection of nonAUG start codons is more likely. This observation supports the role of eIF2 in the selection of the correct AUG start codon. 53.  The correctly charged tRNAs (Ala–tRNAAla and Gln–tRNAGln) bind to EF-Tu with approximately the same affinity, so they are delivered to the ribosomal A site with the same efficiency. The mischarged Ala–tRNAGln binds to EF-Tu more loosely, indicating that it may dissociate from EF-Tu before it reaches the ribosome. The mischarged Gln–tRNA Ala binds to EF-Tu much more tightly, indicating that EF-Tu may not be able to dissociate from it at the ribosome. These results suggest that either a higher or a lower binding affinity could affect the ability of EF-Tu to carry out its function, which would decrease the rate at which mischarged aminoacyl–tRNAs bind to the ribosomal A site during translation. 55.  In a living cell, EF-Tu and EF-G enhance the rate of protein synthesis by rendering various steps of translation irreversible. They also promote the accuracy of protein synthesis through proofreading. In the absence of the elongation factors, translation would be too slow and too inaccurate to support life. These constraints do not apply to an in vitro translation system, which can proceed in the absence of EF-Tu and EF-G. However, the resulting protein is likely to contain more misincorporated amino acids than a protein synthesized in a cell. 57.  The mRNA has the sequence CGAUAAUGUCCGACCAAGCGAUCUCGUAGCA The start codon and stop codon are highlighted. The encoded protein has the sequence Met–Ser–Asp–Gln–Ala–Ile–Ser. 59. a. The sequence of mRNA transcribed from the given strand of DNA (assumed to be the coding strand) is shown with the AUG start codon shaded. The second reading frame contains a stop codon, and neither the second nor the third reading frames contains an AUG start codon. The partial sequence of the protein is Met–Val–His–Leu–Thr–Pro–. ACAGACACCAUGGUGCACCUGACUCCUG b.  The sequence of the mRNA transcribed from the given strand of DNA (assumed to be the noncoding strand) is shown (written in the 5′ → 3′ direction) with the AUG start codon shaded. This is the third reading frame. Neither the first nor the second reading frames contains an AUG start codon. The partial sequence of the protein is Met–Val–Ser–. CAGGAGUCAGGUGCACCAUGGUGUCUGU 1 error 61.  ​​ ______________        ​​ × 400 amino acids × 100 = 4% 10,000 amino acids 63.  The number of phosphoanhydride bonds (about 30 kJ × mol–1 each) that are hydrolyzed in order to synthesize a 20-residue polypeptide can be calculated as follows (the relevant ATP- or GTPhydrolyzing proteins are indicated in parentheses):

Aminoacylation (AARS)

2 × 20 ATP

Translation intiation (IF-2)

1 GTP

Positioning of each aminoacyl–tRNA (EF-Tu)

19 GTP

Translocation after each transpeptidation (EF-G)

19 GTP

Termination (RF-3)

1 GTP Total: 80 ATP equivalents

Thus, approximately 80 × 30 kJ · mol–1, or 2400 kJ · mol–1, is required. In a cell, proofreading during aminoacylation and during translation requires the hydrolysis of additional phosphoanhydride bonds, making the cost of accurately synthesizing the 20-residue polypeptide greater than 2400 kJ · mol–1. 65. a. The ribosome positions the peptidyl group for reaction with the incoming aminoacyl group, so a peptidyl group with a constrained geometry, like Pro, is unable to react optimally. b. Because Lys (with a positively charged side chain) reacts much faster than Asp (with a negatively charged side chain), the active site must be more accommodating of cationic groups than anionic groups. c. Transpeptidation of Ala is faster than for Phe, so for nonpolar amino acids, small size is more favorable. [From Wohlgemuth, I. et al., J. Biol. Chem. 283, 32229–32235 (2008).] 67.  Transpeptidation involves the nucleophilic attack of the amino group of the aminoacyl–tRNA on the carbonyl carbon of the peptidyl–tRNA (see Fig. 22.16). The higher the pH, the more nucleo­ philic the amino group (the less likely it is to be protonated). 69.  Terms listed in a, b, d, e, and g describe transcription; terms listed in c and f describe translation. 71. a. When [3H]Leu is added to the reticulocytes, the cells’ ribosomes are in various stages of synthesizing polypeptides, so the radioactive label is incorporated into the growing polypeptide chain as soon as the next Leu codon is encountered. This results in a greater amount of radioactivity near the carboxyl end of the chain, as shown in the graph, indicating that the carboxyl terminal is synthesized last and that translation occurs in the N → C direction. b. Because 4 minutes is less than the amount of time required to synthesize the entire β globin chain, only a few polypeptides are completely synthesized and released from the ribosome by the end of the experiment. A longer time period would allow more polypeptide chains to incorporate the label and to be completely synthesized. If the time period is sufficiently long, all the chains would be completely labeled, and it would be difficult to determine the direction of translation. c. This is illustrated in the graph, which shows data Dintzis collected at 60 minutes. [From Naughton, M.A. and Dintzis, H.M., Proc. Natl. Acad. Sci. USA 48, 1822–1830 (1962).]

Relative amount of [3H] Leu

49.  Colicin E3 is lethal to the cells because it prevents accurate and efficient translation. Cleavage of the 16S rRNA at A1493 destroys the part of the 30S ribosomal subunit that verifies codon–anticodon pairing (see Fig. 22.14). As a result, the ribosome is less able to incorporate the correct aminoacyl group into a growing polypeptide. In addition, EF-Tu hydrolysis of GTP is slow, because EF-Tu does not receive a signal from the ribosome that an mRNA–tRNA match has occurred, so the speed of translation decreases.

100 Time of incubation: 60 minutes 80 60 Time of incubation: 4 minutes 40 20 0

0

20

40

60 80 100 Amino acid number

120

140

S-78  ODD- N UM B ER E D S O LUT I O NS 73. a. The mutation would allow the aminoacylated tRNA rather than a release factor to enter the ribosome and pair with a stop codon. The result would be incorporation of an amino acid into a polypeptide rather than translation termination, so the ribosome would continue to read mRNA codons and produce elongated polypeptides. The inability of the mutated tRNA to recognize its amino acid–specifying codons would have a minor impact on protein synthesis, since the cell likely contains other isoacceptor tRNAs that can recognize the same codons. b. Not all proteins would be affected. Only proteins whose genes include the stop codon that is read by the mutated tRNA would be affected. Proteins whose genes include one of the other two stop codons would be synthesized normally. c. Aminoacyl–tRNA synthetases usually recognize both the anticodon and acceptor ends of their tRNA substrates. A mutation in the tRNA anticodon, such as a nonsense suppressor mutation, might interfere with tRNA recognition, so the mutated tRNA molecule might not undergo aminoacyl­ ation. This would minimize the ability of the mutated tRNA to insert the amino acid at a position corresponding to a stop codon. 75.  The 5′ segment of the mRNA functions as a riboswitch. When the amino acid is plentiful, the tRNA is charged. The aminoacyl– tRNA binds to the mRNA, preventing synthesis of an enzyme that is not needed. When the amino acid is scarce, the tRNA is uncharged and cannot bind to the site on the mRNA. Translation proceeds in order to produce the enzyme needed to synthesize the amino acid. 77.  In prokaryotes, both mRNA and protein synthesis take place in the cytosol, so a ribosome can assemble on the 5′ end of an mRNA, even while RNA polymerase is synthesizing the 3′ end of the transcript. In eukaryotes, RNA is produced in the nucleus, but ribosomes are located in the cytosol. Because transcription and translation occur in separate compartments, they cannot occur simultaneously. A eukaryotic mRNA must be transported from the nucleus to the cytosol before it can be translated. 79.  Anfinsen’s ribonuclease experiment demonstrated that a protein’s primary structure dictates its three-dimensional structure. Molecular chaperones assist in the protein-folding process but do not contribute any additional information regarding the tertiary structure of the protein. The purified ribonuclease was able to refold without the assistance of chaperones because other cellular components were absent. Chaperones are required in vivo because they prevent the interaction and aggregation of proteins and other cellular components. 81.  Mitochondria contain ribosomes that synthesize proteins encoded by mitochondrial DNA. Like cytosolic proteins, mitochondrial proteins require the assistance of chaperones for proper folding. Other proteins are synthesized in the cytosol and are transported partially unfolded through pores in the mitochondrial membranes; these proteins also require the assistance of chaperones to fold properly once they reach their destination. 83.  The different domains in a multidomain protein associate with one another via van der Waals forces, since the domain interfaces eventually end up in the interior of the protein. A cagelike chaperonin

structure allows these proteins to fold in a protected environment where the hydrophobic regions of the protein are not exposed to other intracellular proteins with which they could potentially aggregate. 85. a. When a deficiency of β chains is coupled with an excess of α chains, the α chains precipitate and destroy the red blood cells, worsening the anemia that results from the lack of β chains. b. The imbalance between the amounts of α and β chains is minimized when the synthesis of both globins is depressed due to mutations in both an α globin gene and a β globin gene. 87.  The basic residue is highlighted in gray; the hydrophobic core is underlined. MKWVTFISLLLLFSSAYSRGV 89.  The hydrophobic cleft of this particular protein might allow it to recognize the hydrophobic core of the signal sequence. Other proteins in the SRP might be involved in pausing translation and docking the ribosome with the endoplasmic reticulum. 91.  In the cell-free system, the SRP can bind to the exit tunnel of the ribosome, but translation is not arrested when no membrane is present. This indicates that the SRP must interact with both the nascent polypeptide and the ER membrane in order to pause translation. When microsomal membranes are subsequently added, the protein is not translocated, indicating that translocation must occur co-translationally, not post-translationally. Proteins that are not translocated retain their signal sequences because they do not have access to the signal peptidase, which is located in the microsomal lumen. 93. a. 

O    b.  NH

O

CH

C

O NH

CH

C

CH2

CH2

O

CH2

P

O–

Polyglutamine protein

CH2

O–

CH2 HN

C

Ubiquitin

O

c.  The acetyltransferase acetylates lysine residues in histones, neutralizing the positive charge of the lysine side chain and weakening its interaction with the DNA, so that transcriptional activity is increased. If the acetyltransferase is inactive, the DNA will be less transcriptionally active and certain genes will not be expressed. The loss of transcriptional activity could contribute to the progression of the polyglutamine disease. [From Pennuto, M. et al., Hum. Mol. Gen. 18, R40–R47 (2009).] 95. 

O H3C

(CH2)12

C O

NH

CH H

C

Index Page references followed by T indicate tables. Page references followed by F indicate figures.

3ʹ end, 62 5ʹ caps, 645–646 5ʹ end, 62 7-transmembrane (7TM) receptors, 292 (–)/(+) end, 137 A AARS see aminoacyl tRNA synthetase abasic sites, 597 ABC transporters, 271 ABO blood groups, 326–327 acceptors, hydrogen bonding, 28 acetaldehyde, 177, 380 acetaminophen (paracetamol), 220–221, 306 acetoacyl-CoA, 508 acetylcholine, 273 acetyl-CoA, 343 cholesterol synthesis, 508 citric acid cycle, 404–406, 407–409 fatty acid synthesis, 496–497 ketone body synthesis, 503–504 acetyl-CoA carboxylase, 496–497, 566 acetylsalicylate (aspirin), 306 acid–base catalysis, 174–175 acid–base chemistry, 37–49 bicarbonate buffer system, 46–49 buffers, 44–46 functional groups, 43 Henderson–Hasselbalch equation, 42 human body, 46–49 pK values, 40–46 acid catalysis, 174 acid dissociation constant (Ka), 40–46 acidic, 38 acidification, oceanic, 39 acidosis, 48–49 acids definition, 39 pK values, 41T aconitase, 409–410 ACP see acyl carrier protein actin, 136–139 myosin, 148–151 action potentials, 260, 261F activation allosteric regulation, 218–219 chymotrypsin, 185–186 fatty acids, 487–488 fatty acid synthesis, 501–502 G proteins, 293–294 kinases, 290, 294–295 protein kinase A, 294–295 receptor tyrosine kinases, 302 activation energy (ΔG‡), 172–173 activators, 634 active sites catalytic triads, 177–179 DNA polymerases, 587–588 induced fit, 182–183 low-barrier hydrogen bonds, 181 oxyanion holes, 180–181 peptidyl transferase, 677–680 proximity and orientation effects, 181–182 RNA polymerase, 639–641 serine proteases, 177–179 specificity pockets, 184–185 active transport, 262, 269–272 acyclovir (Zovirax), 220 acyl carrier protein (ACP), 496F acyl-CoA synthetase, 505–507 acyl groups, 235 acyl phosphates, 374 adenine, 58–60, 59F, 62 adenosine diphosphate (ADP), 218

adenosine monophosphate (AMP), 59F, 541–542 adenosine signaling, 288–289 adenosine triphosphate (ATP), 6F actin, 137–138 citric acid cycle, 413 coupled reactions, 354–356 gluconeogenesis, 383–385 glucose phosphorylation, 367–369 glycolysis, 367–369, 373–376 kinesin, 151–153 myosin, 149–151 oxidative phosphorylation, 445–450 photosynthesis, 469–470, 474–475 receptor tyrosine kinases, 302 succinyl-CoA synthetase, 411 yield from glucose, 413F see also ATP synthase adenylate cyclase, 294 adipocytes, 339–340, 564, 569–570 adiponectin, 565 adipose tissue, 339–340, 557, 561T A-DNA, 64 β2-adrenergic receptor, 292–293 advanced glycation end products, 320 aerobic glycolysis, 573–574 aerobic organisms, 16–17 affinity antibodies, 155, 157–158 ligand–receptor, 288 affinity chromatography, 113 agonists, 288 alanine, 5F, 88F, 89 alanine aminotransferase, 170 alcohol dehydrogenase, 380 alcohol metabolism, 380 alcohols, 4T aldehydes, 4T aldolase, 371–372 aldose reductase, 571 aldoses, 316 alkalosis, 48–49 alleles, 68–70 allosteric effectors, 369–370 allosteric enzymes, 208–209 allosteric proteins, 132, 218 allosteric regulation, 216–219 covalent modification, 562–563 gluconeogenesis, 385 urea cycle, 539 α-amino acids, 88 α anomers, 317 α helices, 96–97 coiled coils, 142–143 membrane proteins, 245–246 triple helices, 144–146 α-ketoglutarate, 410–411, 417, 520, 537 α-ketoglutarate dehydrogenase, 410–411, 414 α-tocopherol (vitamin E), 240 alternative splicing, 648–649 Alzheimer’s disease, 109–110 Amadori products, 319–320 amides, 4T amino acids, 5, 8, 10, 86–95 abbreviations, 88F acid–base catalysis, 175F aromatic, 526–529, 527–528, 534–535 biosynthesis, 523–531 branched chain, 526, 534 charged, 88F, 90 chirality, 89 covalent catalysis, 177 degradation, 532–540 essential, 349T, 524T functional groups, 87

glucogenic, 532, 533T gluconeogenesis, 383 hydrophobic, 88F, 89, 100–102 ketogenic, 532–534, 533T metabolic diseases, 535–536 neurotransmitter synthesis, 530–531 nonessential, 524T nucleotide degradation, 546 peptide bonds, 91–94 percent occurrence, 92F pK values, 92 polar, 88F, 90, 100–102 R groups, 87 sequence, 93–94 structure, 88F, 88–90 aminoacylation, transfer RNAs, 666–667 aminoacyl–tRNA synthetase (AARS), 666–668 γ-aminobutyric acid (GABA), 530 amino groups, 4T acid–base chemistry, 43 Maillard reaction, 319–320 aminotransferases, 521–523 ammonia, assimilation, 519–521 AMP see adenosine monophosphate AMP-dependent protein kinase (AMPK), 566 amphipathic, 35 see also amphiphilic molecules amphiphilic molecules hydrophobic effect, 35–37 lipids, 237–244 amphotericin B, 265 AMPK see AMP-dependent protein kinase amplification, DNA, 614–618 amyloid deposits, 109–110 amylopectin, 322 amylose, 322 anabolism, 337 anaerobic respiration, 17, 379–381, 449 analysis, protein structure, 111–117 anaplerotic reactions, 416–421 anemia, 129 annealing, 66 anomers, 317–318 antagonists, 289 antenna, 462 antibiotic resistance, 613 antibodies, 154–158 affinity, 155, 157–158 domains, 155–156 effector functions, 157 monoclonal, 157–158 anticodons, 665, 668 antidepressants, 275 antidiabetic drugs, 571, 572T antigens, 154 antiparallel, 63 antiparallel β sheets, 96–98 antiporters, 271 antithrombin, 188 apoptosis, 604 aquaporins, 266–268 aqueous chemistry, 27–56 acid–base, 37–49 amphiphilic molecules, 35–37 bicarbonate buffer system, 46–49 buffers, 44–46 carbon dioxide, 39 cellular, 32–33 equilibrium constants, 40–46 Henderson–Hasselbalch equation, 42 hydrogen bonding, 27–33 hydronium ions, 37 hydrophobic effect, 33–37

ionization constant, 38–39 pH, 38–46 pK values, 40–46 proton jumping, 37–38 solution characteristics, 31–33 arachidonate, 241, 305 archaea, 17–18 lipids, 238–239 microbiome, 19 arginase, 538–539 arginine, 88F, 90, 92F, 525, 531, 538 argininosuccinase, 538–539 argininosuccinate synthetase, 538–539 aromatic amino acids, 527–528 degradation, 534–535 synthesis, 526–529 arrestin, 293, 296 ascorbate (vitamin C), 144 asparagine, 5F, 88F, 90, 92F, 524, 532 aspartate, 88F, 90, 526 aspartate aminotransferase (AST) assays, 523 aspirin (acetylsalicylate), 306 assembly actin filaments, 137–138 microtubules, 139–140 assimilation of nitrogen, 519–521 AST see aspartate aminotransferase asymmetry, membranes, 243–244, 250–251 ataxia telangiectasia, 606 atherosclerosis, 483–485, 500 ATM, 606 Atorvastatin (Lipitor), 508–509 ATP see adenosine triphosphate ATPases, 269–271 ATP-binding cassette (ABC) proteins, 271 ATP-citrate lyase, 417 ATP synthase, 445–450 binding change mechanism, 447 chloroplasts, 469–470 proton translocation, 445–448 regulation, 450 structure, 445–448 uncoupling agents, 448 autoactivation, 185 autolysosome, 277–278 autophagosomes, 277–278 autophagy, 277–279 autophosphorylation, 302 avidity, 155 axons, 260, 261F B bacteria, 17–18 biofilms, 324 cell walls, 328–330 glyoxysomes, 419 microbiome, 19 bacteriophages, 78 bacteriorhodopsin, 246F baking, 104 base catalysis, 174 base excision repair, 598–599 base pairs (bp), 62–64 bases definition, 39 nucleotides, 58–60 base stacking, RNA, 653–654 basic, 38 B-DNA, 63F, 64 beeswax, 241 beriberi, 349 β2-adrenergic receptor, 292–293 β-amyloid, 109–110 β anomers, 317

I-1

I-2  IN DEX β barrels, 246, 263–264 β-carotene, 460F β cells, 560–561 β-galactosidase, 637 β oxidation, 488–495 β sheets, 96–98 transmembrane proteins, 246 bicarbonate buffer system, 46–49 C4 pathway, 473 fatty acid synthesis, 497 bilayers, 35 bile acids, 509 bimolecular reactions, 201 binding change mechanism, 447 bioenergetics, 352–358 biofilms, 324 biofuels, 323–324 bioinformatics, 1 biological fluids, pH, 39T biological markers, 326–327 biomineralization, 106–107 biomolecules, 3–10 amino acids, 5, 8, 10, 86–95 carbohydrates, 5, 9–10, 315–336 functional groups, 4T lipids, 6 nucleic acids, 8–9, 10, 57–85 nucleotides, 6, 8–9, 10 polymers, 6–10 polysaccharides, 9–10 proteins, 8, 10, 86–233 standard reduction potentials, 430T biopolymers, 6–10 actin, 136–139 digestion, 338–339 functions, 10T nucleic acids, 8–9, 10, 57–85 polysaccharides, 9–10 proteins, 8, 10, 86–233 tubulin, 139–142 biosynthesis amino acids, 523–531 anaplerotic reactions, 416–421 cholesterol, 508–510 citric acid cycle products, 416–418 fatty acids, 235, 495–504 purines, 542–543 pyrimidines, 542–543 triacylglycerols, 505–507 biotin, 381, 497 1,3-bisphosphoglycerate, 374 2,3-bisphosphoglycerate (BPG), 133 bisubstrate reactions, 207 bleeding disorders, 188–189 blood coagulation cascade, 187–188 pH, 39, 46–48 blunt ends, 612 B lymphocytes, 154, 156–157 body weight, 569–570 Bohr effect, 132 bonds disulfide, 93, 102 by functional group, 4T glycosidic, 9, 476 hydrogen, 27–33 isopeptide, 102 peptide, 8, 91–94, 95 phosphodiester, 8, 61–62 relative strength, 30F thioester, 102–103 bone defects, 147 bp see base pairs BPG see 2,3-bisphosphoglycerate branched chain amino acids degradation, 534 synthesis, 526 brown adipose tissue, 569–570 buffers, 44–46 C Ca2+-ATPase, 270–271 caffeine, 289, 297 calcium ions phosphoinositide signaling system, 297–298 synaptic transmission, 273, 274F

Calvin cycle, 472–475 cancer genetics, 604–607 metabolism, 573–575 nucleotide synthesis, 545 oncogenes, 303 telomeres, 596 Warburg effect, 573–574 capsaicin, 241 capsid, 78 carbamoyl phosphate, 537 carbamoyl phosphate synthetase, 537 carbanions, 175 carbohydrates, 5, 9–10, 315–336 anomers, 317–318 biofilms, 324 Calvin cycle, 472–475 chirality, 316–317 derivatization, 318–320 disaccharides, 321 fuel-storage molecules, 321–322 gluconeogenesis, 383–386 glycolysis, 366–382 glycomics, 320–321 glycoproteins, 325–329 metabolic disorders, 393–395 metabolism, 366–402 monosaccharides, 315–320 phosphorylation, 319 polysaccharides, 320–324 structural molecules, 322–324 carbon, oxidation states, 14T carbon atoms, citric acid cycle fates, 411F carbon dioxide (CO2) aqueous chemistry, 39 bicarbonate buffer system, 47–48 citric acid cycle, 403–404, 408F, 410–411, 413 fixation, 471–477 ocean acidification, 39 carbon fixation, 471–477 carbonic acid, 39, 46–48 carbonic anhydrase, 473 carbon monoxide poisoning, 130 carboxylic acids, 4T, 43 carcinogenesis, 415, 604–607 carcinogens, 598 carnitine shuttle system, 488 β-carotene, 460F CAR-T cells, 158 Cas9, 602 catabolism, 337 amino acids, 532–540 glucose, 366–378 other sugars, 378 see also degradation catalysis/catalysts acid–base, 174–175 activation energy, 173 covalent, 175–177 electrostatic, 183 enzymatic methods, 14, 171–179 metal ions, 177 Michaelis–Menten equation, 201–209 oxygen-evolving complex, PSII, 465 see also enzymes catalytic constant (kcat), 204–205 catalytic efficiency, 205 catalytic perfection, 205 catalytic triads, 177–179 catechols, 530 CATH system, 99, 100F cell division, 141 cell immortality, 596 β cells, 560–561 cells aqueous chemistry, 32–33 components by mass, 7F, 28 evolution, 14–20 ionic concentrations, 36F cellular respiration, 432–433 cellulose, 322–324, 477 cellulosic biofuels, 323–324 cell walls cellulose, 322–323 peptidoglycans, 328–330

central dogma, 67–74 centromeres, 67–68, 593 ceramides, 238 cerebrosides, 238 cerulenin, 502 C4 pathway, 472, 473 channeling, 529 chaperones, 103–104, 684–685 Chargaff’s rules, 62–63 charged amino acids, 88F, 90 chemical labeling, 177 chemiosmosis, 443–444, 469–470 chemiosmotic theory, 443–444 chemoautotrophs, 338 Chesapeake hemoglobin, 136 chirality amino acids, 89 carbohydrates, 316–317 chitin, 324 chloride channel proteins, 265 chlorophyll a, 460F chloroplasts, 18, 458–462 ATP synthase, 469–470 chemiosmosis, 469–470 see also photosynthesis cholate, 509 cholesterol, 6F, 239, 508–510 chorismate, 528 chromatin, 610–611 chromatin remodeling, 629–630 chromatography, proteins, 111–113 chromosomes, 57 meiosis, 68–70, 141 mitosis, 68, 141 chylomicrons, 484 chymotrypsin, 168–169, 177–179, 181–182, 183–186 chymotrypsinogen, 185–186 citrate, 370, 407–410 citrate synthase, 407–409, 414, 496 citrate transport system, 416–418, 418F, 496–497 citric acid cycle, 403–427 aconitase, 409–410 anaplerotic reactions, 416–421 ATP generation, 413 carbon atom fates, 411F carbon dioxide, 403–404, 408F, 410–411, 413 citrate synthase, 407–409, 414 evolution, 415–416 fumarase, 412 intermediates as precursors, 416–418 isocitrate dehydrogenase, 410, 414, 415 α-ketoglutarate dehydrogenase, 410–411, 414 malate dehydrogenase, 412–413 mutations, 415 pyruvate dehydrogenase, 403–406 reactions, 408F regulation, 414–415 succinyl-CoA synthetase, 411, 412F thermodynamics, 413–416 clamp-loading complexes, 588 classes of proteins, 99, 100F classification, enzymes, 170T clathrin, 277 CLC proteins, 265 climate change, 39 clinical trials, 221 clones, 613 clotting, 187–188 clustered regularly interspersed short palindromic repeats (CRISPRs), 602 CoA see coenzyme A coagulation cascade, 187–188 cobalamin (vitamin B12), 492–494 Cockayne syndrome, 600 coding strands, 70 codons, 70–71, 71T, 75, 663–669 amino acid variants, 669 coenzyme A (CoA), 60, 496F coenzymes, 174, 345 cofactors, 173–174, 345, 519 coiled coils, 142–143

colchicine, 141 collagen, 144–147 covalent modification, 145–147 defects, 147 structure, 144–146 compartmentation, 555–559 competitive inhibition, 210–213, 216T complement, 16, 64, 265 Complex I, 434–436 Complex III, 437–439 Complex IV, 439–441 concentration cells, 32–33 coagulation factors, 188 extracellular, 36F ionic, 36F condensation reactions, 91, 235 conformation, 8 conformational changes gated channels, 265–266 hemoglobin, 131–132 transport proteins, 265–266, 268–271 conjugate bases, 41 consensus sequences, 631–632, 647–648 conservative replication, 583 conservative substitution, 128 constant domains, 156 constitutive genes, 603 convergent evolution, 184 cooperative binding, 129–132, 208–209 Cori cycle, 557–558 cortisol, 304 coupled reactions, 13, 354–356 covalent catalysis, 175–177 covalent modification collagen, 145–147 DNA, 631 lipid attachment, 246–247 proteins, 104–105, 246–247, 325–328, 687 COVID-19 see SARS-CoV-2 COX see cyclooxygenase CpG islands, 631 creatine, 356, 449 CRISPR-Cas9 system, 602, 613 CRISPRs see clustered regularly interspersed short palindromic repeats cristae, 433 crops, transgenic, 73 cross-linking, collagen, 145–146 cross-talk, 298, 303 C-terminus, 91 CTP see cytidine triphosphate cyclic AMP (cAMP) G protein–coupled receptor activation, 294–296 hydrolysis, 296–297 cyclic AMP phosphodiesterase, 296–297 cyclic electron flow, 469 cyclization, monosaccharides, 318 cyclooxygenases (COX), 305–306 inhibition, 306 cysteine, 5F, 88F, 90, 92F, 93, 525, 533 cytidine triphosphate (CTP), 59F, 505 cytochrome b, 438F cytochrome b6   f, 466–467 cytochrome c, 437–440 cytochrome c oxidase, 440 cytochrome P450, 220 cytochromes, 437–440 cytokinesis, 150 cytosine, 58–59, 59F, 62, 597 cytoskeleton, 137 actin, 136–139 tubulin, 139–142 D damage, DNA, 596–604 dark reactions, 471–477 deafness, 151 deamination, 532, 597 degradation amino acids, 532–540 fatty acids, 487–495 glycogen, 388–389 nucleotides, 545–547 degrons, 341

I NDEX  I-3 ΔG see free energy change ΔG°ʹ see standard free energy change denaturation DNA, 64–65 proteins, 103–104 deoxyhemoglobin, 131–132 deoxyribonucleic acid see DNA deoxyribonucleotides, biosynthesis, 543–544 depolymerization, microtubules, 140–141 derivatization, carbohydrates, 318–320 desaturases, 500 desensitization, 296 DHA, 236 diabetes mellitus, 340, 560, 570–572 diacylglycerol, 298 diazotrophs, 519 dielectric constants, 31–32 solvents, 32T diet, 500, 558 differentiation, B lymphocytes, 157 diffraction patterns, 116 diffusion lateral, 244 transverse, 244 diffusion-controlled limit, 205 digestion, 338–339 dihydrobiopterin, 534–535 dihydrofolate, 545 dihydroxyacetone, 315 dihydroxyacetone phosphate, 371–373, 505 diisopropylphosphofluoridate (DIPF), 177 dimers coiled coils, 142–143 hormone response elements, 304 DIPF see diisopropylphosphofluoridate diphosphoric acid esters, 4T diphosphoryl groups, 4T diploid, 68 dipole–dipole interactions, 30 disaccharides, 315, 321 discontinuous replication, 586 diseases advanced glycation end products, 320 amino acid metabolism, 535–536 ataxia telangiectasia, 606 atherosclerosis, 483–485 cancer, 573–575 carbohydrate metabolism, 393–395 citric acid cycle enzyme mutations, 415 Cockayne syndrome, 600 collagen defects, 147 diabetes, 570–572 DNA repair, 600 fuel metabolism, 568–575 genetic, 71–72 hemoglobin variants, 134–135, 136F hypercholesterolemia, 485 kwashiorkor, 568 marasmus, 568 metabolic syndrome, 572–573 obesity, 569–570 protein misfolding, 109–111 xeroderma pigmentosum, 600 disorder, proteins, 105–107 dissociation constants, 40–46 buffers, 44–46 Henderson–Hasselbalch equation, 42 dissociation constants (Kd), ligand–receptor, 288 disulfide bonds, 93, 102 divergent evolution, 184 DNA barcoding, 79 DNA-binding proteins, 635–636 DNA chips, 348 DNA (deoxyribonucleic acid), 57–59 A form, 64 base excision repair, 598–599 B form, 63F, 64 central dogma, 67–74 covalent modification, 631

damage, 596–604 double-strand breaks, 601–604 epigenetic modifications, 631 gene editing, 602 genetic code, 70–71, 71T, 663–669 helicases, 584–585 hydrogen bonding, 62–66 imprinting, 631 major features, 63–64 manipulation, 602, 611–618 melting temperature, 65–66 mismatch repair, 598 mutation, 71–72, 596–598 noncoding, 77 nucleosomes, 610 nucleotide excision repair, 599–600 nucleotides, 58–59 packing, 607–611 plasmids, 613 primers, 585–586 proofreading, 589–590 recombinant, 72–74 renaturation, 65–66 repair, 598–604 replication, 67–70, 71–72, 582–596 RNA binding, 64 stacking interactions, 64–65 structure, 61–64 supercoiling, 607–609 tandem repeats, 77 telomeres, 593–596 topoisomerases, 584–585 transcription, 627–645 transposable elements, 77 viruses, 78 DNA fingerprinting, 614 DNA gyrase, 609 DNA ligase, 590–592 DNA polymerase, 583–584, 585–590 active sites, 587–588 processivity, 588 proofreading, 589–590 structure, 587–588 DNA sequencing, 615–618 domains, 100–103 antibodies, 155–156 DNA-binding, 635–636 G protein–coupled receptors, 292–293 helix-turn-helix motifs, 635–636 receptor tyrosine kinases, 300–302 donors, hydrogen bonding, 28 dopamine, 530 double helix, 62–66 double-strand breaks, 601–604 drugs clinical trials, 221 commonly prescribed, 222T development, 219–222 pharmacokinetics, 220–221 see also individual compounds d sugars, 316–317 dual specificity phosphatases, 297 dynamics, translation, 681–683 dysbiosis, 558 E Edman degradation, 114 EF see elongation factors effector functions, 157 Ehlers–Danlos syndrome, 147 EI complexes see enzyme–inhibitor eicosanoids, 305 elastase, 184 electrochemical gradients, 443–444 electronegativity, 29–31 electronic structure, water, 28 electron tomography, 433 electron transport chain, 432–444 Complex I, 434–436 Complex III, 437–439 Complex IV, 439–441 cytochrome c, 437–440 proton translocation, 436, 438–441 Q cycle, 438, 439F ubiquinol, 434–438 electrophiles, 177 electrophoresis, 113, 114F

electrostatic catalysis, 183 elements, 3, 30T elongation RNA transcription, 642–643 translation, 675–677 elongation factors (EF), 675–677, 679–680 elongation of fatty acids, 500–501 enantiomers, 316–317 endergonic reactions, 12 endocytosis, 276–277 endonucleases, 64 endopeptidases, 91 endosomes, 276–277 endothermic reactions, 11 enediolate, 472 enediol intermediates, 373 energy, 10–14 enthalpy/entropy, 11 metabolic, 354–358 spontaneous processes, 12 enhancers, 634 enolase, 376 enol-keto tautomerization, 174 enolpyruvate, 376–377 enoyl-CoA isomerase, 491 enteropeptidase, 185 enthalpy (H), 11–12 entropy (S), 11–12 hydrophobic effect, 33–34 membranes, 35–36 enzyme–inhibitor (EI) complexes, 211–212 enzymes, 14, 167–233 acid–base catalysis, 174–175 activation, 185–186 active sites, 177–183 allosteric regulation, 208–209, 216–219 ATP synthase, 445–450 blood coagulation, 187–188 catalytic constants, 204–205, 204T catalytic efficiency, 205 catalytic mechanisms, 171–179 catalytic triads, 177–179 chymotrypsin, 168–169, 177–179, 181–182, 183–186 classification, 170T competitive inhibition, 210–213, 216T concepts, 167–170 cooperative binding, 208–209 covalent catalysis, 175–177 glycosylation, 325–326 inhibition, 186, 188, 209–222 irreversible inhibition, 209–210 kinases, 290 kinetics, 198–233 Lineweaver–Burk plots, 206–207 low-barrier hydrogen bonds, 181 Michaelis–Menten equation, 201–209 multifunctional, 497–500 multistep reactions, 208 multisubstrate reactions, 207–208 naming, 170–171 nonhyperbolic reactions, 208–209 oxyanion holes, 180–181 product inhibition, 212 proximity and orientation effects, 181–182 rate enhancements, 169T saturation, 200 specificity pockets, 184 substrate specificity, 169, 184–185 transition state analogs, 212–214 transition state stabilization, 180–181 translocases, 250–251 uncompetitive inhibition, 213–216, 216T unique properties, 180–183 zymogens, 185 see also individual enzymes… enzyme–substrate (ES) complex, 200 epigenetic modifications, 631 epimers, 317 epinephrine, 293, 530, 564 ℰ see reduction potentials ℰ°ʹ see standard reduction potentials

equilibrium constants (Keq), 40–46, 352 ES see enzyme–substrate essential amino acids, 523, 524T essential fatty acids, 347–349, 501 esters, 4T, 235 ethanol, acetaldehyde conversion, 177 ethers, 4T euchromatin, 610 eukarya, 17–18 eukaryotes, 17–19, 18–19F definition, 17 evolution, 18–19 genomes, 75–77 evolution cells, 14–20 citric acid cycle, 415–416 conservative substitution, 128 convergent, 184 divergent, 184 eukaryotic cells, 18–19 homology, 128–129 invariant residues, 128–129 natural selection, 16–17 prebiotic, 15–17 RNA world, 654 selection, 613 variable positions, 128 exciton transfer, 461–462 exergonic reactions, 12 exocytosis, 273–276 exome, 75 exons, 75, 646 exonucleases, 64 exopeptidases, 91 exoskeletons, 324 exosomes, 278 exothermic reactions, 11 expression, 67 activators, 634 consensus sequences, 631–632 enhancers, 634 histone code, 629–630 Mediator complexes, 634–636, 642–643 mRNA turnover, 649–650 operons, 628, 636–637 promoters, 631–632 regulation, 631, 633–638, 649–652 ribosomes, 669–673 RNA interference, 651–652 transcription factors, 302, 304 translation, 663–695 see also transcription extracellular concentrations, ionic, 36F extracellular matrix, 144 extracellular signaling, 288T cross-talk, 298, 303 cyclic AMP, 294–296 G protein–coupled receptors, 289–290, 291–300 insulin, 300–302, 560–563 lipid hormones, 303–306 phosphoinositide, 297–298 receptor tyrosine kinases, 290, 300–303 second messengers, 289–290, 294–296, 298 extrinsic membrane proteins, 245, 246–247 F F see Faraday constant Fab fragments, 155–157 F-actin, 137–138 factor VIIa, 187–188 factor IXa, 188 factor X, 187–188 factor Xa, 187–188 factor XIa, 188 factory model of replication, 583–584 FAD see flavin adenine dinucleotide Faraday constant (F ), 259, 430

I-4  IN DEX fatty acids, 235–237, 236T activation, 501–502 adipose tissue, 339–340 degradation, 488–495 desaturases, 500 elongation, 500–501 essential, 347–349, 349T, 501 odd-chain, 492–494 oxidation, 486–495 synthesis, 495–504 fatty acid synthase, 497–500 Fc fragments, 155–157 feedback inhibitors, 217–219 feed-forward activation, 377 fermentation, 380 ferredoxin, 468 ferritin, 351 fertilization, 68–69 Fe–S clusters see iron–sulfur clusters fibrinogen, 187–188 fibrous proteins, 99 first-order reactions, 201 Fischer projections, 317 fixation carbon, 471–477 nitrogen, 519 flavin adenine dinucleotide (FAD), 60, 412 flavin mononucleotide (FMN), 436 flip-flop, 244 flippase, 250 floppase, 250–251 fluidity of membranes, 242–243 fluid mosaic model, 248–851 fluorescence, 461 fluorescent probes, oligonucleotide, 66 fluorine, solubility enhancements, 31 5-fluorodeoxyuridylate, 545 5-fluorouracil, 210 fluoxetine (Prozac), 31, 275 flux, 346, 385 FMN see flavin mononucleotide folding, proteins, 103–107, 109–111, 684–685 fractional saturation, 127 free energy change (ΔG), 11–14 activation energy, 172–173 coupled reactions, 354–356 electrochemical gradients, 443–444 hydrophobic effect, 33–34 metabolic reactions, 352–358 phosphate hydrolysis, 356 protein folding, 105–106 reduction potentials, 431–432 free radicals, 442 fructose, 316, 321 catabolism, 378 intolerance, 394 fructose bisphosphatase, 384F, 385–386 fructose-1,6-bisphosphate, 371–372 fructose-2,6-bisphosphate, 370 fructose-6-phosphate, 207–208, 319, 369–370, 384F, 385, 476–477 nucleotide synthesis, 389–391 fructose-1-phosphate aldolase, 378 fuel metabolism regulation, 555–581 AMP-dependent protein kinase, 566 cancer, 573–575 epinephrine, 564 glucagon, 564–565 hypoxia-inducible factor, 567 insulin, 560–564 leptin, 565 metabolic syndrome, 572–573 redox balance, 566–567 starvation, 568–569 fuel-storage molecules, 321–322, 339–340 fumarase, 412 fumarate, 412, 538–539 functional groups acid–base chemistry, 43 amino acids, 87 common, 4T hydrogen bonding, 30 functions biopolymers, 10T lipids, 239–241 nucleic acids, 67–79

organs, 556–557, 557F protein disorder, 106–107 proteins, 125–166 antibodies, 154–158 motor, 148–153 oxygen binding, 126–135 structural, 136–147 fungi, microbiome, 19 furosine, 320 fusion, membranes, 272–279 futile cycle, 386 G G3P see glyceraldehyde-3-phosphate ΔG‡ see activation energy ΔG see free energy change ΔG°ʹ see standard free energy change GABA see γ-aminobutyric acid G-actin, 137–138 galactose, 321, 394 β-galactosidase, 637 gametes, 68–70 γ-aminobutyric acid (GABA), 530 gangliosides, 238 gas constant (R), 259 gated channels, 260, 264–265 GDP see guanosine diphosphate gene editing, 602 general transcription factors, 633 genes, 67 cancer, 303 consensus sequences, 631–632 definition, 628 DNA sequencing, 615–618 exons, 75 horizontal transfer, 79 imprinting, 631 introns, 75 mutation, 71–72 open reading frames, 75 tumor suppressor, 605 gene therapy, 72–74, 74T genetically modified organisms (GMOs), 73 genetic code, 70–71, 71T, 75, 663–669 genetic diseases, 71–72 genetics cancer, 604–607 codons, 70–71, 70T horizontal transfer, 79 Mendelian inheritance, 68–70 mutation, 71–72 oncogenes, 303 genomes, 70 DNA sequencing, 615–618 gene editing, 602 human, 77 mitochondrial, 434 sizes, 75–77 viruses, 78 genome-wide association studies (GWAS), 72 genomics, 74–79 applications, 77–79 geraniol, 241 ghrelin, 565 Gleevec (imatinib), 303 globins, 128–135, 136F variants, 134–135, 136F globular proteins, 99–103 glucagon, 564–565 glucocorticoid receptor, 304 glucogenic amino acids, 532, 533T glucokinase, 560–561 gluconeogenesis, 383–386, 384F, 568–569 glucosamine, 319F glucose-1-phosphate, 386–388 glucose-6-phosphatase, 385, 394 glucose-6-phosphate, 367–369, 386–388, 476 glucose-6-phosphate dehydrogenase, 389, 394 glucose, 5F, 9, 316–319 ATP yield, 413F disaccharides, 321 fuel-storage molecules, 321–322 GLUT transporter, 268, 271–272 insulin, 560–565

metabolism, 366–402, 560–565 methanol reactivity, 318F phosphorylation, 367–369 structural molecules, 322–324 see also cellulose; starch glucose–alanine cycle, 559 glucuronate, 319F glutamate, 88F, 90, 91, 92F, 417, 525, 537–538 glutamate dehydrogenase, 417, 537, 539 glutamate receptors, 299 glutamate synthase, 519–521 glutamine, 88F, 90, 533, 574–575 glutamine synthetase, 519–521 glutathione (GSH), 442 gluten, 104 GLUT4, 562 GLUT transporter, 268, 271–272 glycans, 320 glyceraldehyde, 315 glyceraldehyde-3-phosphate dehydrogenase, 374 glyceraldehyde-3-phosphate (G3P), 319, 343, 371–373, 476 glycerol-3-phosphate, 505 glycerophospholipids, 237–244 biosynthesis, 507 membranes, 241–244 polar head groups, 237–239 glycine, 88F, 90, 525 glycine cleavage system, 534 glycogen, 321–322 metabolism, 562–563 phosphorolysis, 340–341 synthesis, 386–393 glycogenolysis, 388–389 glycogen phosphorylase, 388–389 glycogen storage diseases (GSDs), 394–395 glycogen synthase, 562–563 glycolipids, 250 glycolysis, 343–344, 366–382 aldolase, 371–372 ATP generation, 373–376 enolase, 376 glyceraldehyde-3-phosphate dehydrogenase, 374 hexokinase, 367–369 phosphofructokinase, 369–370 phosphoglucose isomerase, 369 phosphoglycerate kinase, 374–375 phosphoglycerate mutase, 375–376 reactions, 368F triose phosphate isomerase, 371–373 glycomics, 320–321 glycoproteins, 250, 325–329 glycosaminoglycans, 327–328 glycosidases, 325 glycosides, 318, 320–329 glycoproteins, 324–329 polysaccharides, 320–324 glycosidic bonds, 9, 318, 476 glycosylation, 687 glycosyltransferases, 325–326 glyoxylate pathway, 419 glyoxysomes, 419 glyphosate, 528 GMOs see genetically modified organisms GMP see guanosine monophosphate gout, 141 G protein–coupled receptors (GPCRs) arrestin, 293, 296 desensitization, 296 eicosanoid signaling, 305 GDP/GTP binding, 289, 294 inactivation, 296–297 ligand binding, 292–293 phosphoinositide, 297–298 phosphorylation, 296 sensory signaling, 298–300 signaling pathways, 291–300 structure, 292–293 G proteins, 293–294 Gram-negative bacteria, 329 Gram-positive bacteria, 329 G-rich sequences, 594 GroEL, 684–685

GSDs see glycogen storage diseases GSH see glutathione GTP see guanosine triphosphate guanine, 58–59, 59F, 62, 597 guanosine diphosphate (GDP), 59F, 289, 294, 411 guanosine monophosphate (GMP), 541–542 guanosine triphosphate (GTP) gluconeogenesis, 383–384 G protein–coupled receptors, 289, 294 purine synthesis, 541–542 Ras, 302–303 tubulin, 140 gustation, 298–300 GWAS see genome-wide association studies H H see enthalpy hair cells, 151 half-reactions, 429–430 haploid, 68 Haworth projections, 317 HDL see high-density lipoproteins hearing, 151 heart disease, 500 helicases, 584–585 α helices, 96–97 coiled coils, 142–143 membrane proteins, 245–246 triple helices, 144–146 helix-turn-helix (HTH) motifs, 635–636 heme, 126–128, 438 hemoglobin, 126, 129–135 2,3-bisphosphoglycerate, 133 carbon monoxide poisoning, 130 cooperative binding, 129–132 homology, 128–129 pH, 132–133 variants, 134–135, 136F Henderson–Hasselbalch equation, 42 heparin, 188 heterochromatin, 610 heterotrophs, 338 hexokinase, 182, 367–369, 394, 560–561 hexosamine pathway, 559 hexoses, 316 HIF-α see hypoxia-inducible factor high-density lipoproteins (HDL), 484–485 high-performance liquid chromatography (HPLC), 113 histidine, 88F, 90, 92F, 529 histone code, 629–630 histones, 610–611 HIV protease inhibitors, 214 HIV reverse transcriptase, 595 HMG-CoA reductase, 508–510 homeostasis, 2 bicarbonate buffer system, 46–49 hormones, 560–572 insulin, 560–565 homocysteine, 525–526 homodimers, 107 homologous chromosomes, 68 homologous genes, 75 homologous proteins, 128 homology, RNA polymerases, 639 homotetramers, 107 homotrimers, 107 horizontal gene transfer, 79 hormone, obesity, 569–570 hormone response elements, 304 hormones, 288, 303–306 adiponectin, 565 epinephrine, 564 fuel metabolism, 560–572, 565T glucagon, 564–565 G protein–coupled receptors, 289–290, 291–300 gut-secreted, 565 insulin, 560–565 leptin, 565 receptor tyrosine kinases, 290, 300–303 hormone-sensitive lipase, 564

I NDEX  I-5 HPLC see high-performance liquid chromatography HTH see helix-turn-helix motifs human body acid–base chemistry, 46–49 microorganisms, 19 organs, 556–559 see also diseases human genome, 77 hydrated, 32 hydrogen bonding, 27–33 collagen, 145 electronegativity, 29–31 length, 29 low-barrier, 181 nucleic acids, 62–66 polarity, 28–29 hydrogen ions, hemoglobin binding, 132–133 hydrolysis adenosine triphosphate, 137, 354–356 enzymatic, 167–169, 177–179, 181–188 lipids, 237–238 peptide bonds, 91 phosphate free energy, 356 phosphocreatine, 356 hydronium ion, 37 hydrophilic, 33 hydrophobic, 33 amino acids, 88F, 89, 100–102 hydrophobic cores, 100–102 hydrophobic effect, 33–37, 101–102 hydroxyapatite, 147 hydroxyproline, 144–145 hypercholesterolemia, 485 hyperglycemia, 571–572 hypervariable loops, 155–156 hypoxanthine, 541 hypoxia-inducible factor (HIF-α), 567 I ibuprofen, 306 IDL see intermediate-density lipoproteins IF1 see inhibitory factor 1 IFs see initiation factors Illumina sequencing, 616 imatinib (Gleevec), 303 iminium ions, 176 imino groups, 4T, 145, 176 acid–base chemistry, 43 Maillard reaction, 319–320 immunity, self-recognition, 326–327 immunization, 157 immunofluorescence microscopy, 157 immunoglobulin G (IgG), 154–156 immunoglobulins, 154–158 see also antibodies IMP see inosine monophosphate imprinting, 631 incretins, 565 indels, 589 induced fit, 182–183 inheritance diseases, 71–72 horizontal transfer, 79 Mendel’s laws, 68–71 mutation, 71–72 inhibition, 209–222 allosteric regulation, 216–219 coagulation cascade, 188 common prescription drugs, 222T competitive, 210–213, 216T cyclooxygenases, 306 drug development, 219–222 fatty acid synthesis, 501–502 feedback, 217–219 HIV protease, 214 HMG-CoA reductase, 508–509 irreversible, 209–210 mTOR, 563 noncompetitive, 216, 216T nucleotide biosynthesis, 545 products, 212 proteases, 186, 188 transition state analogs, 212–214 uncompetitive, 213–216, 216T

inhibition constant (KI), 211 inhibitory factor 1 (IF1), 450 initial velocity (v0), 203 initiation transcription, 627–638 translation, 673–675 initiation factors (IFs), 674 inorganic phosphate (Pi), 137 inosine monophosphate (IMP), 541–542 insulin, 560–565 signaling, 300–302 insulin resistance, 570–572 integral membrane proteins, 245–246 ATP synthase, 445–450 electron transport chain, 432–442 G protein–coupled receptors, 291–300 receptor tyrosine kinases, 290, 300–303 SNAREs, 272–276 intermediate-density lipoproteins (IDL), 484 intermediate filaments, 137, 142–144 intermembrane space, 433, 443–444 interspersed repetitive sequences, 77 intestinal microbiota, 558 intestinal peptide hormones, 565 intracellular concentrations, ionic, 36F intrinsically disordered proteins, 106 intrinsic membrane proteins, 245–246 introns, 75, 646, 648 invariant residues, 128–129 Invirase (Saquinavir), 214 in vitro, 12 in vivo, 12 iodoacetate, 371 ion channels, 264–266 ion exchange chromatography, 112–113 ionic interactions, 30 ionization constant of water (Kw), 38 ion pairs, 101–102 iron metabolism, 351 iron–molybdenum cofactors, 519 iron–sulfur (Fe–S) clusters, 436, 438, 466, 468, 519 irregular secondary structure, 98 irreversible inhibition, 209–210 islets of Langerhans, 561 isocitrate, 409–410 isocitrate dehydrogenase, 410, 414, 415 isodecoder tRNAs, 665 isoelectric point (pI), 112–113 isoleucine, 88F, 89, 534 isomerization, unsaturated fatty acids, 491 isoniazid, 502 isopentenyl pyrophosphate, 508 isopeptide bonds, 102 isoprenoids, 239 isozymes, 171 J jasmonate, 305 K Ka see acid dissociation constant Kansas hemoglobin, 136 kb see kilobase pairs K+ channels, 264–265, 266F keratin, 142–144 α-keto acid transamination, 521–523 ketogenesis, 503–504 ketogenic amino acids, 532–534, 533T α-ketoglutarate, 410–411, 417, 520, 537 α-ketoglutarate dehydrogenase, 410–411, 414 ketohexoses, 316 ketone bodies, 503–504, 568–569 ketones, 4T ketone tautomerization, 174 ketosamines, 319–320 ketoses, 316 KI see inhibition constant kidneys, 557 bicarbonate buffer system, 47–49 kilobase pairs (kb), 64

kinases, 182, 290, 369 activation, 290, 294–295 mTOR, 563–564 purine synthesis, 541 kinesin, 151–153 kinetics, 198–233 allosteric regulation, 216–219 catalytic constants, 204–205 competitive inhibition, 210–213, 216T inhibition, 209–222 initial velocity, 203 Lineweaver–Burk plots, 206–207 maximum reaction velocity, 203 Michaelis–Menten equation, 201–209 multistep reactions, 208 multisubstrate reactions, 207–208 noncompetitive inhibition, 216, 216T nonhyperbolic reactions, 208–209 product inhibition, 212 uncompetitive inhibition, 213–216, 216T velocity, 199, 203–205 kinetochore, 141 KM see Michaelis–Menten constant Ku, 601 Kw see ionization constant of water kwashiorkor, 568 L lac repressor, 637 lactate, 379, 557–558, 573–574 lactate dehydrogenase, 379 lactose, 321 lactose permease, 637 lacZ, 613 lagging strand, 586, 590–592 lamins, 142 lateral diffusion, 244 LDL see low-density lipoproteins leading strand, 586 length, hydrogen bonds, 29 leptin, 565, 569 leucine, 88F, 89, 534 life, thermodynamics, 12–14 ligand binding, G protein–coupled receptors, 292–293 ligands heme, 126, 132 receptor interactions, 287–290 light DNA damage, 597–598 exciton transfer, 461–462 photons, 459–460 pigment absorption spectra, 461F light-harvesting complexes, 461–462 light reactions, 463–470 limonene, 241 linear electron flow, 468–469 Lineweaver–Burk plots, 206–207 linoleate, 491 lipid bilayers, 35–36, 241–251 asymmetry, 243–244, 250–251 fluidity, 242–243 fluid mosaic model, 248–851 lateral diffusion, 244 membrane proteins, 244–248, 249–250 oligosaccharides, 326–328 translocases, 250–251 transverse diffusion, 244 see also membrane… lipid hormones, 303–306 lipid-linked proteins, 245, 246–247 lipids, 6, 234–257 aqueous chemistry, 35–37 archaea, 238–239 fatty acids, 235–237, 236T fatty acid synthase, 497–500 glycolipids, 238 hormones, 303–306 hydrolysis, 237–238 melting points, 242–243 metabolism, 483–517, 511F physiological functions, 239–241 polar head groups, 237–239 sphingolipids, 238

transport, 483–485 vitamins, 240 see also lipid bilayers Lipitor (Atorvastatin), 508–509 lipoamide, 405 lipolysis, 564 lipoprotein lipase, 486 lipoproteins, 339, 484–485 liver, 556–559, 561T lock-and-key model, 180 London dispersion forces, 30–31 loops, 98 lovastatin (Mevacor), 508–509 low-barrier hydrogen bonds, 181 low-density lipoproteins (LDL), 484–485 l sugars, 316–317 lungs respiratory acidosis, 48–49 respiratory alkalosis, 49 lysine, 88F, 90, 92F, 534 lysosomes, 341 M macromolecules, 2 Maillard reaction, 319–320 major groove, 63 malaria, 135 malate, 412, 418 malate–aspartate shuttle system, 434 malate dehydrogenase, 412–413, 473 malnutrition, 568–569 malonyl-CoA, 497 mammals, metabolic regulation, 555–581 manganese clusters, 465 marasmus, 568 mass action ratio, 353 mass spectrometry, proteins, 114–116 matrix, 433 maximum reaction velocity (Vmax), 203 mechanosensitive channels, 266 Mediator complexes, 634–636, 642–643 meiosis, 68–70 kinetochore, 141 melatonin, 531 melting points of membranes, 242–243 melting temperatures (Tm), DNA, 65–66 membrane fusion, 272–279 membraneless organelle, 107 membrane potential, 259–262 membrane proteins, 244–248 active transport, 262, 269–272 ATP synthase, 445–450 electron transport chain, 432–442 G protein–coupled receptors, 291–300 integral, 245–246 lipid-linked, 245, 246–247 orientation, 249–250 passive transport, 260–262, 263–269 peripheral, 245, 246–247 receptor tyrosine kinases, 290, 300–303 selectivity filters, 264–265 signal peptides, 686 transporters, 263–272 membranes asymmetry, 243–244, 250–251 fluid mosaic model, 248–851 glycosaminoglycans, 327–328 hydrophobic effect, 35–36 lipid bilayers, 241–251 lipid diffusion, 244 melting points, 242–243 mitochondrial, 433–441, 445–446 oligosaccharides, 326–328 rafts, 243 thylakoid, 459 translocases, 250–251 membrane solubility, fluorine, 31 membrane-spanning β barrels, 246 membrane-spanning α helices, 245–246 membrane translocation, 685–686

I-6  IN DEX membrane transport, 258–286 active, 262, 269–272 endocytosis, 276–277 exocytosis, 273–276 fusion, 272–279 passive, 260–262, 263–269 thermodynamics, 258–262 memory cells, 157 Mendelian inheritance, 68–71 messenger RNA (mRNA), 70–71, 628 caps and tails, 645–646 codons, 70–71, 71T processing, 645–649 riboswitches, 682 splicing, 646–649 translation, 671–683 turnover, 649–652 metabolic acidosis, 48 metabolic alkalosis, 48 metabolically irreversible reactions, 369 metabolic fuels, 340 metabolic syndrome, 572–573 metabolism, 2, 10–14, 337–581 alcohol, 380 amino acids, 523–540 ATP synthase, 445–446 bioenergetics, 352–358 cancer, 573–575 cellular locations, 536F chemiosmosis, 443–444, 469–470 citric acid cycle, 344, 403–427 compartmentation, 555–559 complexity, 346–347 Cori cycle, 557–558 coupled reactions, 13–14, 354–356 definition, 337 electron transport chain, 432–444 flux, 346 free energy changes, 352–358 gluconeogenesis, 383–386 glucose, 366–402, 560–565 glucose–alanine cycle, 559 glycogen synthesis, 386–393 glycolysis, 343–344, 366–382 hormonal control, 560–575 iron, 351 lipids, 483–517, 511F mammalian regulation, 555–581 nitrogen, 518–554 nucleotides, 540–547 overview, 347F oxidative phosphorylation, 346, 428–457 pathways, 343–351 pentose phosphate pathway, 389–392 photosynthesis, 458–482 principles, 337–342 Q cycle, 438, 439F redox reactions, 344–346 regulation, 357–358, 555–581 respiratory complexes, 441 vitamins, 347–351, 492–494 metabolites Cori cycle, 557–558 glucose–alanine cycle, 559 hexosamine pathway, 559 transport, 557–559 metabolome, 348 metabolomics, 348 metagenomics, 79 metal ion catalysis, 177 metamorphic proteins, 106 metastasis, 139 methanol, glucose reaction, 318F methionine, 88F, 89, 525–526 methotrexate, 545 methylation, 631 methylene-tetrahydrofolate, 545 methylglyoxal, 373 methylmalonyl-CoA mutase, 492–494 Mevacor (Lovastatin), 508–509 mGlu5 receptor, 292 micelles, 35 Michaelis–Menten constant (KM), 203–204 Michaelis–Menten equation, 201–209 microarrays, 348

microbiome, 19 microbiota, 19, 558 microenvironment, 92, 181–184, 462 micro RNAs (miRNAs), 651–652 microtubules, 137, 139–142 assembly, 139–140 depolymerization, 140–141 kinesin, 151–153 microvesicles, 278 Milledgeville hemoglobin, 136 minor groove, 63 miRNAs see micro RNAs misfolded proteins, 109–111 mismatch repair, 598 mitochondria, 18 ATP synthase, 445–450 carnitine shuttle system, 488 chemiosmosis, 443–444 Complex I, 434–436 Complex III, 437–439 Complex IV, 439–441 cytochrome c, 437–440 electron transport chain, 432–444 genome, 434 intermembrane space, 433, 443–444 membranes, 433–441, 445–446 β oxidation, 488–495 reactive oxygen species, 442 respiratory complexes, 441 structure, 433–434 transport systems, 434–435 see also citric acid cycle mitochondrial matrix, 404, 433 mitosis, 68 kinetochore, 141 mixed inhibition, 216, 216T mobilization, metabolic fuels, 340 molecular biology, central dogma, 67–74 molecular chaperones, 103–104, 684–685 molybdenum–iron cofactors, 519 monoclonal antibodies, 157–158 monogenetic diseases, 72 monomers, 7 monomorphic proteins, 105 monosaccharides, 5, 315–320 anomers, 317–318 chirality, 316–317 derivatization, 318–320 Maillard reaction, 319–320 monosodium glutamate, 91 motor proteins, 148–153, 584–585 mRNA see messenger RNA mTOR, 563–564 multienzyme complexes, pyruvate dehydrogenase, 404–406 multifunctional enzymes, 497–500 multistep reactions, kinetics, 208 multisubstrate reactions, kinetics, 207–208 muscle cells, 556–557 actin–myosin interactions, 149 Cori cycle, 557–558 creatine, 449 insulin effects, 561T muscles, neuronal signaling, 273, 274F mutagens, 598 mutation, 71–72 conservative substitution, 128 invariant residues, 128 variable positions, 128 mutations, 596–598 citric acid cycle enzymes, 415 myosin, 151 myelin sheath, 260, 261F myoglobin, 126–129 myosin, 148–151 ATP binding, 149–151 mechanism, 149–151 structure, 148–149 myristoylation, 246–247 N NAD+ see nicotinamide adenine dinucleotide NADH see nicotinamide adenine dinucleotide

NADH:ubiquinone oxidoreductase, 434–436 NADP+ see nicotine adenine dinucleotide phosphate NADPH see nicotine adenine dinucleotide phosphate Na,K-ATPase, 269–271, 270F, 272F naming, enzymes, 170–171 nanopore sequencing, 617–618 native structure, 103 natural selection, 16–17 ncRNA see noncoding RNA NDP see nucleoside diphosphates near-equilibrium reactions, 369 negative effectors, 218 negative elongation factor (NELF), 642 NELF see negative elongation factor Nernst equation, 430 nerve–muscle synapses, 273–274 neurodegenerative diseases, 109–111 neuromuscular junction, 273, 274F neurons action potentials, 260, 261F axons, 260, 261F membrane potentials, 258–262 myelin sheath, 260, 261F synaptic transmission, 273, 274F neurotransmitters, 273, 275, 530–531 neutral, 38 niacin, 60, 349–350 nicks, 590 nick translation, 590 nicotinamide adenine dinucleotide (NAD+/NADH), 60, 345 alcohol metabolism, 380 citric acid cycle, 404, 407, 408F, 410 Complex I, 434–436 glycolysis, 367, 374, 378 lactate synthesis, 379 oxidative phosphorylation, 429F pyruvate dehydrogenase, 404 nicotine adenine dinucleotide phosphate (NADP+/NADPH), 345 fatty acid synthesis, 498–500 pentose phosphate pathway, 389–391 nitric oxide, 531 nitrification, 519 nitrite reductase, 519 nitrogenase, 519 nitrogen cycle, 519, 520F nitrogen metabolism, 518–554 amino acid degradation, 532–540 amino acid synthesis, 523–531 assimilation, 519–521 fixation, 519 nucleotide metabolism, 540–547 urea cycle, 536–540, 538F N-linked oligosaccharides, 325–327 NMR see nuclear magnetic resonance spectroscopy noncoding DNA, 77 noncoding RNA (ncRNA), 75, 628 noncoding strands, 70 noncompetitive inhibition, 216 nonessential amino acids, 523–531, 524T nonhomologous end joining, 601 nonhyperbolic reactions, 208–209 nonpolar molecules, hydrophobic effect, 33–37 nonreducing sugars, 318 nonspontaneous reactions, 12 nonstandard base pairs, 654 norepinephrine, 293, 530, 570 N-terminus, 91 nuclear magnetic resonance (NMR) spectroscopy, proteins, 116 nuclear pore complex, 649–650 nucleases, 64 nucleic acids, 8, 8–9, 10, 57–85 central dogma, 67–74 complementarity, 64 denaturation, 64–65 genetic code, 70–71, 71T genomics, 74–79 horizontal gene transfer, 79 open reading frames, 75

stacking interactions, 64–65 structure, 8–9, 61–66 viruses, 78 see also DNA; RNA nucleolus, 643 nucleophiles, 177 nucleoside diphosphates (NDP), 391 nucleosides, 58–59, 59F nucleosomes, 610, 630 nucleotide excision repair, 599–600 nucleotides, 6, 8–9, 10, 57–61 degradation, 545–547 metabolism, 540–547 phosphodiester bonds, 61–62 salvage pathways, 545–547 synthesis, 389–392, 541–545 O obesity, 569–570 ocean acidification, 39 odd-chain fatty acids, 492–494 odorants, 298 Okazaki fragments, 586, 590–592 oleate, 491 olfaction, 298 oligonucleotides, 64, 66 oligopeptides, 93 oligosaccharides, 325–330 O-linked oligosaccharides, 325 omega-3 fatty acids, 236 OmpF porin, 263 oncogenes, 303 one-carbon chemistry, 526 open reading frames (ORF), 75 operons, 628, 636–637 opioids, 299 ordered mechanism, 208 ORF see open reading frames organelles, 19 membraneless, 107 peroxisomes, 494–495 see also chloroplasts; mitochondria organisms, organization, 2F organs, 556–559 AMPK effects, 566T Cori cycle, 557–558 glucose–alanine cycle, 559 insulin effects, 561T orientation, membrane proteins, 249–250 orientation effects, 182 ornithine transcarbamoylase, 538–539 osmosis, 266 osteogenesis imperfecta, 147 oxaloacetate, 344, 381–382, 383–384, 407–409, 413, 417–418, 420–421, 532 oxidant, 429 oxidation, 13–14, 344–345 fatty acids, 486–495 light-mediated, 461 β oxidation, 488–495 oxidation–reduction reactions, 344–346 free energy, 431–432 photosystem I, 468–469 photosystem II, 463–466 reduction potentials, 429–431 thermodynamics, 428–432 oxidation states, carbon, 14T oxidative phosphorylation, 346, 428–457 ATP synthase, 445–450 binding change mechanism, 447 chemiosmosis, 443–444 electron transport chain, 432–444 P:O ratio, 447–448 proton translocation, 445–448 redox thermodynamics, 428–432 oxidizing agent, 429 9-oxoguanine, 597 oxyanion holes, 180–181 oxygen-binding proteins, 126–135 2,3-bisphosphoglycerate, 133 carbon monoxide poisoning, 130 cooperative binding, 129–132 heme, 126–128 hemoglobin, 126, 129–135, 136F

I NDEX  I-7 homology, 128–129 myoglobin, 126–129 pH, 132–133 oxygen debt, 379, 449, 557–558 oxygen-evolving complex, 464–466 oxygen levels, fuel metabolism, 566–567 oxygen radicals, 442 oxyhemoglobin, 131–132 P p53, 606–607 P680, 464 P700, 467–468 P:O ratio, 447–448 Pi see inorganic phosphate packing of DNA, 607–611, 628–631 paclitaxel, 142 palindromic sequences, 612 palmitate, 6F palmitoylation, 247 pancreatic islet cells, 560–561 paracetamol see acetaminophen parallel β sheets, 96–98 partial pressure of oxygen, 127 passive transport, 260–262, 263–269 aquaporins, 266–268 ion channels, 264–266 porins, 263–264 pathogens, 154 PCR see polymerase chain reaction pellagra, 349–350 pentose phosphate pathway, 389–392 pentoses, 316 peptide bonds, 8, 91–94 resonance, 95 peptides, 93 peptide sequencing, 114–116 peptidoglycans, 328–330 peptidyl transferase, 677–680 peripheral membrane proteins, 245, 246–247 peroxisomes, 494–495 pH, 38–46 acidic, 38 basic, 38 biological fluids, 39T buffers, 44–46 hemoglobin, 132–133 Henderson–Hasselbalch equation, 42 neutral, 38 physiological, 39 phagocytes, 157 phagophores, 277–278 pharmaceuticals clinical trials, 221 common prescriptions, 222T rational design, 220 see also individual compounds… pharmacokinetics, 220–221 phenylalanine, 88F, 89, 527–528, 534–535 phenylketonuria (PKU), 535–536 phosphatases, 297 phosphatidylcholine, 505–507 phosphatidylethanolamine, 505–507 phosphatidylserine, 507 phosphoanhydride linkages, 4T phosphocholine, 238 phosphocreatine, 356 phosphodiester bonds, 8, 61–62 phosphoenolpyruvate, 217–218, 370, 376–378, 383–384, 473 phosphoenolpyruvate carboxykinase, 383–384 phosphoenolpyruvate carboxylase, 473 phosphoester linkages, 4T phosphoethanolamine, 238 phosphofructokinase, 216–218, 369–370, 414, 566 phosphoglucomutase, 386, 476 6-phosphogluconate, 390 6-phosphogluconate dehydrogenase, 390 6-phosphogluconolactonase, 390 6-phosphoglucono-δ-lactone, 389–390

phosphoglucose isomerase, 369, 385 2-phosphoglycerate, 375–376 3-phosphoglycerate, 374–375 2,3-bisphosphoglycerate, 471–472 phosphoglycerate kinase, 374–375 phosphoglycerate mutase, 375–376 phosphoinositide signaling system, 297–298 phospholipase C, 297, 297–298 phospholipases, 237–238 phospholipids, biosynthesis, 505–507 5-phosphoribosyl pyrophosphate (PRPP), 529, 541 phosphoric acid esters, 4T phosphorolysis, 340–341 phosphorylase kinase, 564 phosphorylation glucose, 367–369 glycogen synthase, 562 G protein–coupled receptors, 296 protein kinase A, 295 receptor tyrosine kinases, 302 RNA polymerase II, 642 substrate-level, 411 sugars, 319 see also ATP synthase; photophosphorylation phosphoryl groups, 4T acid–base chemistry, 43 photoautotrophs, 338 photons, 459–460 photooxidation, 461 photosystem I, 466–469 photosystem II, 464–466 photophosphorylation, 469–470 photoreceptors, 460 photorespiration, 472 photosynthesis, 458–482 Calvin cycle, 472–475 carbon fixation, 471–477 chemiosmosis, 469–470 C4 pathway, 472, 473 cytochrome b6  f, 466–467 dark reactions, 471–477 exciton transfer, 461–462 light-harvesting complexes, 461–462 light reactions, 463–470 oxygen-evolving complex, 464–466 photosystem I, 466–469 photosystem II, 463–466 pigments, 459–461 quantum yield, 474–475 ribulose bisphosphate, 471–475 rubisco, 471–472 Z-scheme, 469 photosystem I, 466–469 photosystem II, 463–466 phycocyanin, 460F phylloquinone (vitamin K), 240 physiological functions, lipids, 239–241 physiological pH, 39, 46–48 bicarbonate buffer system, 46–49 pI see isoelectric point pigments, 459–461 ping pong mechanism, 208 pinocytosis, 276–277 PKA see protein kinase A PKC see protein kinase C PKU see phenylketonuria pK values, 40–46 acids, 41T amino acids, 92 buffers, 44–46 plant cells cellulose, 322–324, 477 chloroplasts, 458–462 C4 pathway, 472, 473 photosynthesis, 458–482 plasmids, 613 plastocyanin, 466–467 plastoquinone (PQ), 464 PLP see pyridoxal-5ʹ-phosphate point mutations, 597 polar amino acids, 88F, 90, 100–101 polar head groups, 237–239 polarity, water, 28–29 poly(A) tails, 646, 649–650

polygenic diseases, 72 polymerase chain reaction (PCR), 614–615 polymerases, 64 polymers, 7 polynucleotides, 8 see also DNA; nucleic acids; RNA polypeptides, 8, 87 see also proteins polyproteins, 214 polyprotic acids, 41 polysaccharides, 9–10, 320–324 biofilms, 324 disaccharides, 321 fuel-storage, 321–322 structural molecules, 322–324 polysomes, 681 polyubiquitin, 341 porins, 263–264 positive effectors, 218 post-translational processing, 104–105, 683–687 lipid attachment, 246–247 potassium ions membrane potential, 259–262 Na,K-ATPase, 269–271, 270F, 272F PQ see plastoquinone prebiotic evolution, 15–17 prenylation, 247 preparation, buffers, 45–46 prescription drugs, common, 222T primary structure, 94–95 primase, 585–586 primer excision, 590–591 primers, 585–586, 590–591, 614–615 prions, 110–111 probes, oligonucleotide, 66 processing, 628 tRNAs, 665–668 see also post-translational processing processive enzymes, 588, 641–642 processivity, 152–153 product inhibition, 212 production, bicarbonate, 47–48 prokaryotes, 17–18, 18F definition, 17 genomes, 75–77 horizontal gene transfer, 79 proline, 88F, 89, 144–145, 525 promoters, 631–632, 631–634 proofreading, 589–590, 642 propionyl-CoA, 492–493 prostaglandins, 305 prosthetic groups, 126 proteases, 114, 115T autoactivation, 185 catalytic triads, 177–179 chymotrypsin, 168–169, 177–179, 181–182, 183–186 coagulation, 187–188 inhibition, 186, 188 proteasomes, 341–342 protein kinase A (PKA), 294–295, 564 protein kinase C (PKC), 298 proteins, 8, 10, 86–233 classes, 99, 100F composition, 93–95, 94T degradation, 341 denaturation, 103–104 disorder, 105–107 disulfide bonds, 93, 102 fibrous, 99 folding, 103–107, 109–111, 684–685 function, 125–166 antibodies, 154–158 motor, 148–153 oxygen binding, 126–135 structural, 136–147 globular, 99–103 glycosylation, 325–328 α helices, 96–97 homology, 128 hydrophobic cores, 100–102 ion pairs, 101–102 isopeptide bonds, 102 lipid-linked, 245, 246–247 loops, 98

membrane translocation, 685–686 microenvironment, 92, 462 misfolding, 109–111 peptide bonds, 91–94 post-translational processing, 104–105, 246–247, 325–328, 683–687 primary structure, 94–95 prions, 110–111 quaternary structure, 94, 107–109 secondary structure, 94, 95–98 sequence, 93–94 sequencing, 114–116 β sheets, 96–98 stability, 99–107 structure, 86–124 amino acids, 86–95 analysis, 111–117 disorder, 105–107 folding, 103–107, 109–111 primary, 94–95 quaternary, 94, 107–109 secondary, 94, 95–98 tertiary, 94, 99–107 synthesis, 663–995 tertiary structure, 94, 99–107 thioester bonds, 102–103 torsion angles, 95–96 transmembrane, 245–246 see also enzymes protein synthesis, 663–995 chaperones, 684–685 dynamics, 681–683 genetic code, 663–669 nonstandard amino acids, 669 peptidyl transferase, 677–680 post-translational processing, 683–687 release factors, 680–681 ribosomes, 669–673 signal recognition particle, 685–687 transfer RNAs, 665–669, 671, 675–677 proteoglycans, 327–328 proteome, 106, 348 proteomics, 348 protofilaments, 140–141 proton gradients ATP synthase, 445–446 chemiosmosis, 443–444 formation, 436, 438–441 proton jumping, 37–38 protonmotive force, 443–447 proton translocation ATP synthase, 445–448 Complex I, 436 Complex IV, 439–441 Q cycle, 438, 439F Providence hemoglobin, 136 proximity and orientation effects, 182 Prozac (fluoxetine), 31, 275 PrPC, 110–111 PRPP see 5-phosphoribosyl pyrophosphate pseudogenes, 75 P-type ATPases, 271 purines, 58–59 biosynthesis, 541–542 degradation, 546 purinosome, 541 pyralline, 320 pyridoxal-5ʹ-phosphate (PLP), 521–522 pyridoxine (vitamin B6), 521 pyrimidines, 58–59 biosynthesis, 542–543 degradation, 546 pyrophosphoryl groups, 4T pyrrolysine, 669 pyruvate, 343 anaerobic respiration, 379–381 cancer cells, 574 Cori cycle, 558 gluconeogenesis, 383–384 glycolysis, 376–378 as a precursor, 381–382 transport, 417–418, 418F pyruvate carboxylase, 381–382, 418–419

I-8  IN DEX pyruvate decarboxylase, 170 pyruvate dehydrogenase, 403–406 pyruvate kinase, 376–378, 393, 574 PYY3-36, 565 Q Q see ubiquinone Q cycle, 438, 439F QH2 see ubiquinol QH· see ubisemiquinone qPCR see quantitative PCR quantitative PCR (qPCR), 614 quantum yield, 474–475 quaternary structure, 94, 107–109 quorum sensing, 288 R R see gas constant Rad51, 601 rafts, 243 random mechanism, 207–208 rapamycin, 563 Ras, 302–303 RAS oncogene, 303 rate constant, 201 rate-determining reaction, 370 rate equation, 201 rational drug design, 220 reabsorption, bicarbonate, 47 reaction centers, 461–462 reaction coordinates, 172 reaction mechanisms, 175 reactions, free energy, 11–14 reactive oxygen species, 442 reading frames, 664 real-time quantitative PCR (RT-qPCR), 614 RecA, 601–603 receptor-mediated endocytosis, 276 receptors, 287–314 cross-talk, 298, 303 desensitization, 291 G protein–coupled, 289–290, 291–300 hormone response elements, 304 ligand interactions, 287–290 low-density lipoproteins, 484–485 second messengers, 289–290, 294–296 tyrosine kinases, 290, 300–303 receptor tyrosine kinases, 290, 300–303 recombinant DNA, 72–74 recombination, 601–604 antibodies, 156–157 red blood cells, ABO blood groups, 326–327 redox balance, fuel metabolism, 566–567 redox centers, 435 redox reactions, 344–346 free energy, 431–432 photosystem I, 468–469 photosystem II, 463–466 reduction potentials, 429–431 thermodynamics, 428–432 reducing agent, 429 reducing sugars, 318 reductant, 429 reduction, 13–14, 344–346 unsaturated fatty acids, 491–492 reduction potentials (ε), 429–431, 465F redundancy, 664–665 regular secondary structure, 96–98 regulation allosteric, 216–219 ATP synthase, 450 Calvin cycle, 475–476 citric acid cycle, 414–415 enzymes, 212, 216–219, 219 fuel metabolism, 560–575 gene expression, 631, 633–638, 649–652 gluconeogenesis, 385–386, 568–569 glycogen synthase, 562 metabolism, 357–358

phosphofructokinase, 369–370 protein kinase A, 294–295 pyruvate dehydrogenase complex, 406 transcription factors, 302, 304, 567 release factors (RF), 680–681 renaturation DNA, 65–66 proteins, 103 repair of DNA, 598–604 base excision, 598–599 cancer, 605–607 double-strand breaks, 601–604 mismatch, 598 nucleotide excision, 599–600 replication, 15, 67–70, 71–72, 582–596 DNA ligase, 590–592 factory model, 583–584 forks, 583–584 helicases, 584–585 indels, 589 lagging strands, 586, 590–592 meiosis, 68–70 mitosis, 68 Okazaki fragments, 586, 590–592 polymerase chain reaction, 614–615 prebiotic RNA, 17 primases, 585–586 proofreading, 589–590 RNases, 590–592 semi-conservative, 583 single-strand binding proteins, 584–585 telomeres, 593–596 topoisomerases, 584–585 replisome, 583 repressors, 635–636 residues, 61–62, 91 resonance, peptide bonds, 95 resonance stabilization, 355–356 respirasomes, 441 respiratory acidosis, 48–49 respiratory alkalosis, 49 respiratory complexes, 441 restriction digests, 612–613 restriction endonucleases/enzymes, 612 retinoate, 304 retinol (vitamin A), 240 reverse transcriptase, 594–595 RF see release factors R groups, 87 ribonucleic acid see RNA ribonucleotide reductase, 391, 543–544 ribose, 316, 319 ribose-5-phosphate, 390–391 ribosomal RNA (rRNA), 70, 628, 670–671 processing, 652–653 ribosome profiling, 681 ribosomes, 70, 669–80 components, 670T peptidyl transferase, 677–680 release factors, 680–681 structure, 669–673 translocation, 679–680 riboswitches, 682 ribozymes, 648 ribulose bisphosphate, 471–475 ribulose-5-phosphate, 390 ribulose-5-phosphate isomerase, 390–391 Rieske iron–sulfur proteins, 466 RNA-dependent RNA polymerase, 640 RNA interference (RNAi), 651–652 RNA polymerase, 638–644 active sites, 639–641 elongation complexes, 642–643 mediator complexes, 634–636 proofreading, 642 RNA (ribonucleic acid) central dogma, 67, 70–71 DNA binding, 64 genetic code, 70–71, 71T, 663–669 hydrogen bonding, 64 noncoding, 75

nucleotides, 58–59 processing, 628, 645–654, 665–668 reverse transcription, 595 secondary structure, 653–654 self-replication, 17 splicing, 646–649 structure, 64 transcription, 627–645 translation, 663–695 viruses, 78 RNase P, 654 RNases, 590–592 RNA world, 654 rRNA see ribosomal RNA RT-qPCR see real-time quantitative PCR rubisco, 471–472 S S see entropy salvage pathways, 545–547 Saquinavir (Invirase), 214 SARS-CoV-2 (COVID-19), 78 saturated fatty acids, 235 saturation enzymes, 200 myoglobin, 127 SCAP, 510 Schiff bases see imino groups scissile bonds, 177–179 scramblase, 250 scurvy, 144, 349 SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis), 113, 114F secondary active transport, 271–272 secondary structure proteins, 94, 95–98 RNA, 653–654 second messengers, 289–290 cyclic AMP, 294–296 diacylglycerol, 298 second-order reactions, 201 selection, 613 selectivity, ion channels, 264–265 selectivity filters, 264–265 selenocysteine, 669 selenoproteins, 669 self-recognition, 326–327 self-replication, RNA, 17 self-splicing, 648 semiconservative replication, 583 sensory receptors, G protein–coupled receptors, 298–300 sequence amino acids, 93–94 determination, proteins, 114–116 genetic code, 70–71, 71T, 75 invariant residues, 128–129 sequencing, 615–618 serine, 88F, 90, 525, 533 serine proteases activation, 185–186 catalytic triads, 177–179 chymotrypsin, 168–169, 177–179, 181–182, 183–186 coagulation, 187–188 inhibition, 188 specificity pockets, 184 serine/threonine kinases, 295, 563–564, 566 serotonin, 275, 530–531 serotonin reuptake inhibitors (SSRIs), 275 sertraline, 275 set-points, 569 β sheets, 96–98, 246 shelterin, 594 Shine–Dalgarno sequence, 673–674 sickle cell hemoglobin, 134–135, 136F signaling, 287–314 cross-talk, 298, 303 cyclic AMP, 294–296 desensitization, 291 diacylglycerol, 298 effects, 290–291

eicosanoids, 305 extracellular, 288T general features, 287–891 G protein–coupled receptors, 291–300 insulin, 300–302, 560–563 lipid hormones, 303–306 mTOR, 563–564 neurotransmitters, 530–531 nitric oxide, 531 phosphoinositide, 297–298 receptor tyrosine kinases, 290, 300–303 second messengers, 289–290, 294–296, 298 transcription factors, 302, 304 signal peptides, 686 signal recognition particle (SRP), 685–687 signal transduction cessation, 296–297 cross-talk, 298, 303 cyclic AMP, 294–296 effects, 290–291 eicosanoids, 305 G protein–coupled receptors, 289–290, 291–300 insulin, 300–302 lipid hormones, 303–306 phosphoinositide, 297–298 principles, 289–291 receptor tyrosine kinases, 290, 300–303 second messengers, 289–290, 294–296, 298 transcription factors, 302, 304 silencers, 635–636 Singapore hemoglobin, 136 single-nucleotide polymorphisms (SNPs), 72 single-strand binding proteins (SSB), 584–585 siRNAs see small interfering RNAs sister chromatids, 67–68 size double helix, 62–63 genomes, 75–77 ribosomes, 670T size-exclusion chromatography, 112 small interfering RNAs (siRNAs), 651 small nuclear RNAs (snRNAs), 647 small nucleolar RNA molecules (snoRNAs), 652–653 smell see olfaction SNAREs, 272–276 snoRNAs see small nucleolar RNA molecules SNPs see single-nucleotide polymorphisms snRNAs see small nuclear RNAs sodium dodecyl sulfate polyacrylamide gel electrophoresis see SDS-PAGE sodium ions membrane potential, 259–262 Na,K-ATPase, 269–271, 270F, 272F solubility, fluorine, 31 solute, 32 solutions, pH, 38–46 solvated, 32 solvents dielectric constants, 32T water as, 31–33 sorbitol, 571 specificity pockets, 184 sphingolipids, 238, 298, 507 sphingomyelins, 238, 298 spliceosome, 647–649 splicing, mRNA, 646–649 spontaneous processes, 12 sports drinks, 36 SREBP see sterol regulatory element binding protein SSB see single-strand binding proteins SSRIs see serotonin reuptake inhibitors stability, proteins, 99–107

I NDEX  I-9 stabilization, transition states, 180–181 stacking interactions nucleic acids, 64–65 RNA, 653–654 standard conditions, 353 standard free energy change (ΔG°ʹ), 352–353 standard prefixes, 7 standard reduction potentials (ℰ°ʹ), 430–431, 430T starch, 321–322 starvation, 568–569 statins, 508–509 steady state, 202 stereocilia, 151 sterol regulatory element binding protein (SREBP), 510 sticky ends, 612–613 stroma, 459 structural polysaccharides, 322–324 structural proteins, 136–147 actin, 136–139 collagen, 144–147 keratin, 142–144 treadmilling, 138–139 tubulin, 139–142 structure actin filaments, 138 amino acids, 88–90, 88F ATP synthase, 445–448 collagen, 144–146 Complex I, 435 DNA, 61–66 immunoglobulins, 154–158 keratin, 142–143 microtubules, 139–140 mitochondria, 433–434 myosin, 148–149 nucleic acids, 61–66 proteins, 86–124 amino acids, 86–95 analysis, 111–117 disorder, 105–107 folding, 103–107, 109–111 primary, 94–95 quaternary, 94, 107–109 secondary, 94, 95–98 tertiary, 94, 99–107 ribosomes, 669–673 RNA, 64 RNA polymerase, 639–641 substrate, 169 substrate-level phosphorylation, 411 substrates competitive inhibition, 210–213 suicide, 209–210 substrate specificity, 169, 184–185 subunits, 107 ATP synthase, 445–448 succinate, 411–412 succinate dehydrogenase, 412, 436–437 succinyl-CoA, 410–411 succinyl-CoA synthetase, 411, 412F sucrose, 321 Maillard reaction, 320 synthesis, 476–477 sugar–phosphate backbone, 62–63 d and l sugars, 316–317 suicide substrates, 209–210 SUMO, 687 supercoiling, 607–609 supercomplexes, 441 sweet receptors, 299, 300F symporters, 269, 271–272 synaptic vesicles, 273–274 Syracuse hemoglobin, 136 T T3 see triiodothyronine T4 see thyroxine Tm see melting temperatures tandemly repeated DNA, 77 tastants, 299 taste see gustation TATA-binding protein (TBP), 633 TATA boxes, 632

Tau peptides, 110 tautomerization, 377 tautomers, 174 TBP see TATA-binding protein telomerase, 594–596 telomeres, 593–596 temperature, 12 termination, transcription, 644 terpenoids, 239 Ter sequences, 590 tertiary structure, 94, 99–107, 653–654 tetrahydrobiopterin, 534–535 tetrahydrofolate (THF), 525 tetroses, 316 TFIIB/TFIIS, 642 thalassemias, 135, 136F thalidomide, 89 thermodynamics citric acid cycle, 413–416 life, 12–14 membrane transport, 258–262 metabolism, 352–358 redox reactions, 428–432 thermogenesis, 448 THF see tetrahydrofolate thiamine (vitamin B1), 349 thiamine pyrophosphate (TPP), 404 thiazolidinedione pharmaceuticals, 571, 572T thick filaments, 149 thin filaments, 149 thioester bonds, 102–103 thioesters, 4T, 357 thiols, 4T threonine, 88F, 90, 533 thrombin, 187–188 thylakoid membranes, 459 thymidine, 544–545 thymidylate synthase, 210 thymine, 58–59, 59F, 62, 597 thyroxine (T4), 304 TIM complex, 434 tissues, 556–559 titration, buffers, 44–46 T lymphocytes, 154 α-tocopherol (vitamin E), 240 topoisomerases, 584–585, 608–609 torsion angles, 95–96 TPP see thiamine pyrophosphate trace elements, 3 transaldolase, 391 transamination, 521–523 transcription, 67, 70, 627–645 activators, 634 antibodies, 156 consensus sequences, 631–632 elongation complexes, 642–643 enhancers, 634 initiation, 627–638 Mediator complexes, 634–636, 642–643 promoters, 631–632 proofreading, 642 regulation, 631, 633–638, 649–652 RNA polymerase, 638–644 termination, 644 transcription factors, 302, 304, 563, 567, 633–637 transcriptome, 348 transcriptomics, 348 transferrin, 351 transfer RNA (tRNA), 64, 70, 628, 665–669, 671, 675–677 processing, 652–653 transgenic organisms, 73 transition mutations, 597 transition state analogs, 212–214 transition states, 172–173 catalyst effects, 173 chymotrypsin mechanism, 179 stabilization, 180–181 transketolase, 208–209, 391 translation, 67, 70–71, 70T, 663–695 dynamics, 681–683 efficiency, 681–683 elongation, 675–677

genetic code, 663–669 initiation, 673–675 peptidyl transferase, 677–680 release factors, 680–681 ribosomes, 669–673 transfer RNAs, 665–669, 671, 675–677 translocation, 679 translocases, 250–251 translocation, 679 translocon, 686 transmembrane proteins, 245–246 active transport, 262, 269–272 G protein–coupled receptors, 291–300 Na,K-ATPase, 269–271, 272F passive transport, 260–262, 263–269 receptor tyrosine kinases, 290, 300–303 selectivity filters, 264–265 transmissible spongiform encephalopathies (TSEs), 110–111 transpeptidation, 678–680 transport carnitine, 488 glucose–alanine cycle, 559 hexosamine pathway, 559 iron, 351 lipids, 483–485 messenger RNA, 649–650 metabolites, 557–559 mitochondrial, 434–435 vesicles, 151–153 transposable elements, 77 transverse diffusion, 244 transversion, 597 treadmilling, 138–139 triacylglycerols, 235–237 adipose tissue, 339–340 biosynthesis, 505–507 triclosan, 502 triglycerides see triacylglycerols triiodothyronine (T3), 304 trimers, 144–146, 293–294 triose phosphate isomerase, 99F, 199, 371–373 trioses, 316 triple helices, 144–146 trisaccharides, 315 tRNA see transfer RNA trypsin, 184, 185 tryptophan, 88F, 89, 350, 527, 529, 530 TSEs see transmissible spongiform encephalopathies tubulin, 139–142 tumor suppressor genes, 605 tunneling, 435 turnover number, 204–205 tyrosine, 88F, 90, 92F, 465, 527–528, 534–535 tyrosine kinases, activation, 290 tyrosine radicals, 465 U ubiquinol pool, 436–437, 437F ubiquinol (QH2), 346 citric acid cycle, 412 electron transport chain, 434–438 oxidative phosphorylation, 429, 429F Q cycle, 438, 439F ubiquinone (Q), 239, 346 citric acid cycle, 412 Complex I, 434–436 oxidative phosphorylation, 429, 429F Q cycle, 438, 439F ubiquitin, 341, 687 ubisemiquinone (QH·), 346 UCP see uncoupling protein UDP–glucose, 476–477 ultraviolet light, DNA damage, 597–598 umami receptors, 299, 300F UMP see uridine monophosphate uncompetitive inhibitors, 213–219

uncoupling agents, 448 uncoupling protein (UCP), 570 uniporters, 271 units, 7, 64 unsaturated fatty acids, 235–236, 491–492 uracil, 58–59, 59F urate, 545–546 urea cycle, 536–540, 538F urease, 540 uridine monophosphate (UMP), 542–543 uridine triphosphate (UTP), 386–388 Usher syndrome, 151 UTP see uridine triphosphate V v see velocity v0 see initial velocity Vmax see maximum reaction velocity vaccination, 157 valine, 88F, 89, 534 van der Waals interactions, 30, 64–65 van der Waals radii, 29 variable domains, 156 variable loops, 665–666 variable positions, 128 variants, hemoglobin, 134–135, 136F vectors, 613 velocity (v), 199 very-low-density lipoproteins (VLDL), 484 vesicles, 35–36 membrane fusion, 272–279 synaptic, 273–274 transport, 151–153 viral enzyme inhibitors, 214 virions, 78 viruses, 78 reverse transcription, 595 RNA-dependent RNA polymerases, 641 vitamin A, 240 vitamin B1, 349 vitamin B6, 521 vitamin B12, 492–494 vitamin C (ascorbate), 144 vitamin D, 240 vitamin E, 240 vitamin K, 240 vitamins, 60 functions, 350T metabolism, 347–351 VLDL see very-low-density lipoproteins voltage-gated K+ channel, 260, 264–265, 266F W Warburg effect, 573–574 water aquaporins, 266–268 dielectric constant, 31–32 electronic structure, 28 hydrogen bonding, 27–33 hydronium ions, 37 ionization constant, 38–39 oxygen evolution, 464–466 polarity, 28–29, 31–33 proton jumping, 37–38 as a solvent, 31–33 see also aqueous chemistry waxes, 241 wobble hypothesis, 668–669 X xeroderma pigmentosum, 600 X-ray crystallography, proteins, 116–117 xylitol, 319F xylulose-5-phosphate, 391 Z zinc fingers, 101, 636 Zovirax (acyclovir), 220 Z-scheme, 469 zymogens, 185

Nucleic Acid Bases, Nucleosides, and Nucleotides Base Formula

Base Nucleoside (X = H) (X = ribose or deoxyribose)

Nucleotide (X = ribose phosphate or deoxyribose phosphate)

NH2



N

N

N

N

Adenine (A)

Adenosine

Adenosine monophosphate (AMP)

Guanine (G)

Guanosine

Guanosine monophosphate (GMP)

Cytosine (C)

Cytidine

Cytidine monophosphate (CMP)

Thymine (T)

Thymidine

Thymidine monophosphate (TMP)

Uracil (U)

Uridine

Uridine monophosphate (UMP)

X

O H



N

N N

H 2N

N X

NH2



N O

N X

H



O N

O

N

CH 3

X

H



O

O N N X

The Standard Genetic Code

U

C

A

G

Third Position (3′ end)

U

UUU Phe UUC Phe UUA Leu UUG Leu

UCU Ser UCC Ser UCA Ser UCG Ser

UAU Tyr UAC Tyr UAA Stop UAG Stop

UGU Cys UGC Cys UGA Stop UGG Trp

U C A G

C

CUU Leu CUC Leu CUA Leu CUG Leu

CCU Pro CCC Pro CCA Pro CCG Pro

CAU His CAC His CAA Gln CAG Gln

CGU Arg CGC Arg CGA Arg CGG Arg

U C A G

A

AUU Ile AUC Ile AUA Ile AUG Met

ACU Thr ACC Thr ACA Thr ACG Thr

AAU Asn AAC Asn AAA Lys AAG Lys

AGU Ser AGC Ser AGA Arg AGG Arg

U C A G

G

GUU Val GUC Val GUA Val GUG Val

GCU Ala GCC Ala GCA Ala GCG Ala

GAU Asp GAC Asp GAA Glu GAG Glu

GGU Gly GGC Gly GGA Gly GGG Gly

U C A G

First Position (5′ end)

Second Position

Common Functional Groups and Linkages in Biochemistry

Structure a

Compound name

Functional group

RNH 3 R2NH 2 R3NH

Amine b

RNH2 or R2NH or R3N or

Alcohol

ROH

OH (hydroxyl group)

Thiol

RSH

SH (sulfhydryl group)

Ether

ROR

N

O

O R

Aldehyde

C

R

C

H

C

Carboxylic (Carboxylate)

C

R

(carbonyl group), R

O

OH or

C O

OH (carboxyl group) or

R

C

O

C

O (carboxylate group)

C

R

C

(acyl group)

C

(carbonyl group), R

C O

R

(acyl group)

C

R

O OR

C

O

(ester linkage)

O

(thioester linkage)

C

N

(amido group)

C

N

S

S Thioester

O

O

O Ester

(ether linkage)

O

O acid b

(amino group)

N

O

O Ketone



or

C

OR

O

Amide

Imine b

R

C O

NH2

R

C O

NHR

R

C

NR2

R R

NH or NR or

O

R R

NH 2 NHR

O R

O

P

R

O

P

OH or

O O

O



P

O

O

P

O O

R

O

P O

O

H

(imino group)

(phosphoester linkage)

OH

O

OH or

P

P

OH or

O

P

O O



O

O

P OH

O (phosphoryl group)

O O

OH O

OH O

Diphosphoric acid ester b, d

P

OH

O R



N

O

OH O

Phosphoric acid ester b, c

C

or

P OH O

O O O P

P

O

OH OH or

OH

(phosphoanhydride linkage)

O P O

O O

P O

(diphosphoryl group, pyrophosphoryl group) a  R represents any carbon-containing group. In a molecule with more than one R group, the groups may be the same or different. b 

Under physiological conditions, these groups are ionized and hence bear a positive or negative charge.

c 

2− When R = H, the molecule is inorganic phosphate (abbreviated Pi ), usually ​​H2PO​  − 4​  ​​ or ​​HPO​  4​  ​​.

d 

When R = H, the molecule is pyrophosphate (abbreviated PPi ).

O

Useful Constants Avogadro’s number

6.02 × 1023 molecules · mol–1

Gas constant (R)

8.314 J · K –1 · mol–1

Faraday (F )

96,485 J · V –1 · mol–1

Kelvin (K)

°C + 273

Key Equations Henderson–Hasselbalch equation [A−]  ​​  ​pH = pK + log ​ _  [HA] Michaelis–Menten equation ​V​  ​​  [S] ​  max   ​​v​ 0​​ = _  ​​  ​KM ​  ​​ + [S] Lineweaver–Burk equation ( ​Vm ​  ax​​ ) [S]

​KM ​  ​​ _ _   ​​  ​​ ​​  1   ​ + _ ​​  1  ​ = ​​ ​​ ​ _ ​  1   ​​  v​​ 0​​

​Vm ​  ax​​

Nernst equation [​A​  reduced​​] [​A​  reduced​​] 0.026 V ​RT ​​     ​   ln  ​ _         ​   ln  ​ _  ​   ​  ​​Ɛ = Ɛ°' − ​​​_ ​  or Ɛ = Ɛ°' − ​​​ ​ ​​ _ ​​ n n F [​A​  oxidized​​] [​A​  oxidized​​] Thermodynamics equations ΔG = ΔH − TΔS ​ΔG°'​​  ​ = −RT ln ​K ​  eq​​           ​ [C ] [D] ​ ​​ ​ ​​ ΔG = ΔG°' + RT ln  ​ _ ​  [A ] [B] ​ ​ ​ΔG°'​​  = −n FΔƐ°' ​​​​

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