Essential biochemistry [Fourth edition.] 9781119012375, 1119012376, 9781119444374, 1119444373

Table of contents :
Cover
Essential_Biochemistry
©
About the Authors
Brief Contents
Contents
Preface
Acknowledgments
Part 1 Foundations
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
ΔG is less than zero for a spontaneous process
Life is thermodynamically possible
1.4 The Origin and Evolution of Life
The prebiotic world
Origins of modern cells
Box 1.B How Does Evolution Work?
2 Aqueous Chemistry
2.1 Water Molecules and Hydrogen Bonds
Hydrogen bonds are one type of electrostatic force
Box 2.A Why Do Some Drugs Contain Fluorine?
Water dissolves many compounds
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 diff usion
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
3 From Genes to Proteins
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
DNA must be decoded
A mutated gene can cause disease
3.4 Genomics
Gene number is roughly correlated with organismal complexity
Genes are identified by comparing sequences
Genomic data reveal biological functions
3.5 Tools and Techniques: Manipulating DNA
Cutting and pasting generates recombinant DNA
The polymerase chain reaction amplifies DNA
Box 3.A Genetically Modified Organisms
Box 3.B DNA Fingerprinting
DNA sequencing uses DNA polymerase to make a complementary strand
DNA can be altered
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 a helix exhibits a twisted backbone conformation
The ß sheet contains multiple polypeptide strands
Proteins also contain irregular secondary structure
4.3 Tertiary Structure and Protein Stability
Proteins have hydrophobic cores
Protein structures are stabilized mainly by the hydrophobic effect
Other interactions help stabilize proteins
Protein folding begins with the formation of secondary structures
Some proteins have more than one conformation
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
Protein structures are determined by X-ray crystallography, electron crystallography, and NMR spectroscopy
Box 4.C Mass Spectrometry Applications
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
Some drugs aff ect 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
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
Eff icient 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
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
Box 7.A Inhibitors of HIV Protease
Other types of inhibitors aff ect Vmax
Allosteric enzyme regulation includes inhibition and activation
Several factors may influence enzyme activity
7.4 Clinical Connection: Drug Development
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 a helix can cross the bilayer
A transmembrane ß sheet forms a barrel
Lipid-linked proteins are anchored in the membrane
8.4 The Fluid Mosaic Model
Membrane glycoproteins face the cell exterior
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 ß barrel proteins
Ion channels are highly selective
Box 9.A Pores Can Kill
Gated channels undergo conformational changes
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
Box 9.B Antidepressants Block Serotonin Transport
SNAREs link vesicle and plasma membranes
Endocytosis is the reverse of exocytosis
Box 9.C Exosomes
10 Signaling
10.1 General Features of Signaling Pathways
A ligand binds to a receptor with a characteristic affinity
Box 10.A Bacterial Quorum Sensing
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
Adenylate cyclase generates the second messenger cyclic AMP
Cyclic AMP activates protein kinase A
Signaling pathways are also switched off
The phosphoinositide signaling pathway generates two second messengers
Calmodulin mediates some Ca2+ signals
10.3 Receptor Tyrosine Kinases
The insulin receptor dimer binds one insulin
The receptor undergoes autophosphorylation
Box 10.B Cell Signaling and Cancer
10.4 Lipid Hormone Signaling
Eicosanoids are short-range signals
Box 10.C Aspirin and Other Inhibitors of Cyclooxygenase
11 Carbohydrates
11.1 Monosaccharides
Most carbohydrates are chiral compounds
Cyclization generates a and ß anomers
Monosaccharides can be derivatized in many different ways
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.A Cellulosic Biofuel
Bacterial polysaccharides form a biofilm
11.3 Glycoproteins
Oligosaccharides are N-linked or O-linked
Oligosaccharide groups are biological markers
Box 11.B The ABO Blood Group System
Proteoglycans contain long glycosaminoglycan chains
Bacterial cell walls are made of peptidoglycan
Part 3 Metabolism
12 Metabolism and Bioenergetics
12.1 Food and Fuel
Cells take up the products of digestion
Box 12.A Dietary Guidelines
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
Box 12.B The Transcriptome, the Proteome, and the Metabolome
Human metabolism depends on vitamins
12.3 Free Energy Changes in Metabolic Reactions
The free energy change depends on reactant concentrations
Unfavorable reactions are coupled to favorable reactions
Free energy can take different forms
Box 12.C Powering Human Muscles
Regulation occurs at the steps with the largest free energy changes
13 Glucose Metabolism
13.1 Glycolysis
Reactions 1–5 are the energy-investment phase of glycolysis
Reactions 6–10 are the energy-payoff phase of glycolysis
Box 13.A Catabolism of Other Sugars
Pyruvate is converted to other substances
Box 13.B Alcohol Metabolism
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 free 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 aff ect liver and muscle
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. a-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
The citric acid cycle probably evolved as a synthetic pathway
Box 14.A Mutations in Citric Acid Cycle Enzymes
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
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
Box 15.A Free Radicals and Aging
15.3 Chemiosmosis
Chemiosmosis links electron transport and oxidative phosphorylation
The proton gradient is an electrochemical gradient
15.4 ATP Synthase
ATP synthase rotates as it translocates protons
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 depends on the rate of fuel catabolism
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
Box 16.A The C4 Pathway
The Calvin cycle rearranges sugar molecules
The availability of light regulates carbon fixation
Calvin cycle products are used to synthesize sucrose and starch
17 Lipid Metabolism
17.1 Lipid Transport
17.2 Fatty Acid Oxidation
Fatty acids are activated before they are degraded
Each round of ß 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
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 diff icult to synthesize
Box 18.B Glyphosate, the Most Popular Herbicide
Amino acids are the precursors of some signaling molecules
Box 18.C Nitric Oxide
18.3 Amino Acid Catabolism
Amino acids are glucogenic, ketogenic, or both
Box 18.D 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 uric acid or amino acids
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 Microbiome Contributes to Metabolism
19.2 Hormonal Control of Fuel Metabolism
Insulin is released in response to glucose
Insulin promotes fuel use and storage
Glucagon and epinephrine trigger fuel mobilization
Additional hormones influence fuel metabolism
AMP-dependent protein kinase acts as a fuel sensor
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
The metabolic syndrome links obesity and diabetes
19.4 Clinical Connection: Cancer Metabolism
Aerobic glycolysis supports biosynthesis
Cancer cells consume large amounts of glutamine
Part 4 Genetic Information
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
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
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
Enhancers and silencers act at a distance from the promoter
Box 21.A DNA-Binding Proteins
Prokaryotic operons allow coordinated gene expression
21.2 RNA Polymerase
RNA polymerases have a common structure and mechanism
RNA polymerase is a processive enzyme
Transcription elongation requires a conformational change 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
rRNA and tRNA processing includes the addition, deletion, and modification of nucleotides
RNAs have extensive secondary structure
22 Protein Synthesis
22.1 tRNA and the Genetic Code
The genetic code is redundant
tRNAs have a common structure
tRNA aminoacylation consumes ATP
Some synthetases have proofreading activity
tRNA anticodons pair with mRNA codons
Box 22.A The Genetic Code Expanded
22.2 Ribosome Structure
The ribosome is mostly RNA
Three tRNAs 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 in vivo
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

Citation preview

AMINO ACID STRUCTURES AND ABBREVIATIONS Hydrophobic amino acids COO⫺ CH3

COO⫺ H

CH3

C

H

NH⫹ 3

Alanine (Ala, A)

COO⫺ H

C

CH2

NH⫹ 3

CH

C

NH⫹ 3

COO⫺ H

CH

CH3

CH

C

CH2

CH3

H

Leucine (Leu, L)

CH2

C

CH2

N H

Tryptophan (Trp, W)

COO⫺ CH2 H C CH2 H2N⫹ CH2

CH3

S

NH⫹ 3

Isoleucine (Ile, I)

CH2

C

NH⫹ 3

COO⫺

NH⫹ 3

CH3

H

Phenylalanine (Phe, F)

COO⫺ CH3 H

CH2

C

NH⫹ 3

Valine (Val, V)

CH3

COO⫺

Methionine (Met, M)

Proline (Pro, P)

Polar amino acids COO⫺ CH3

COO⫺ H

C

CH2

OH

H

H

C

CH2

OH

H

NH⫹ 3

Serine (Ser, S)

COO⫺

CH

C

NH⫹ 3

COO⫺

COO⫺

C

NH2

H

CH2

C

NH⫹ 3

OH

H

CH2

NH2

C

H

NH⫹ 3

Glutamine (Gln, Q)

COO⫺

N

CH2

C

SH

Cysteine (Cys, C)

COO⫺

O

CH2

C

NH⫹ 3

Tyrosine (Tyr, Y)

NH⫹ 3

Asparagine (Asn, N)

CH2

C

NH⫹ 3

Threonine (Thr, T)

O

COO⫺

H

N H

C

H

NH⫹ 3

Histidine (His, H)

Glycine (Gly, G)

Charged amino acids COO⫺ H

C

CH2

COO⫺

O C

O⫺

NH⫹ 3

Aspartate (Asp, D)

H

C

CH2

COO⫺

O CH2

C

NH⫹ 3

Glutamate (Glu, E)

O⫺

H

C

COO⫺

CH2

CH2

CH2

NH⫹ 3

CH2

NH⫹ H 3

C

NH2

CH2

CH2

CH2

NH

NH⫹ 3

Lysine (Lys, K)

Arginine (Arg, R)

C

NH⫹ 2

Essential Biochemistry Fourth Edition

CHARLOT T E W. PRATT Seattle Pacific University

KATHLEEN CORNELY Providence College

VICE PRESIDENT, PUBLISHER SPONSORING EDITOR ASSOCIATE DEVELOPMENT EDITOR SENIOR MARKETING MANAGER PRODUCT DESIGNER SENIOR DESIGNER SENIOR PHOTO EDITOR EDITORIAL ASSISTANT PRODUCT DESIGN ASSISTANT MARKETING ASSISTANT SENIOR PRODUCTION EDITOR

Petra Recter Joan Kalkut Laura Rama Kristine Ruff Sean Hickey Thomas Nery Billy Ray Mili Ali Alyce Pellegrino Maggie Joest Elizabeth Swain

Cover image: The active site of RNA polymerase, based on a structure determined by Yuan He, Chunli Yan, Jie Fang, Carla Inouye, Robert Tjian, Ivaylo Ivanov, and Eva Nogales (pdb 5IYD). This book was typeset by Aptara and printed and bound by Quad Graphics Versailles. The cover was printed by Quad Graphics Versailles. Copyright © 2018, 2014, 2011, 2004 John Wiley and Sons, Inc. 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 978-1-119-31933-7 The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct. Printed in the United States of America 10

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7

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4

3

2

1

About the Authors C H A R L O T T E PR 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 L E E N COR 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 biochemistry from Cornell University. She currently serves as a Professor in the Department of 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.

iii

Brief Contents PRE FAC E

xi Part 3

Part 1

Foundations

Metabolism

12

Metabolism and Bioenergetics 301

1

The Chemical Basis of Life 1

13

Glucose Metabolism 329

2

Aqueous Chemistry 24

14

The Citric Acid Cycle 362

15

Oxidative Phosphorylation 385

16

Photosynthesis 411

17

Lipid Metabolism 432

18

Nitrogen Metabolism 464

19

Regulation of Mammalian Fuel Metabolism 497

Part 2

Molecular Structure and Function

3

From Genes to Proteins 52

4

Protein Structure 85

5

Protein Function 119

6

How Enzymes Work 154

7

Enzyme Kinetics and Inhibition 183

8

Lipids and Membranes 215

20

DNA Replication and Repair 519

9

Membrane Transport 235

21

Transcription and RNA 551

10

Signaling 260

22

Protein Synthesis 580

11

Carbohydrates 283

iv

Part 4

Genetic Information

Contents PRE FAC E

Part 1

xi

Foundations

3.4

3.5

10 10

The Origin and Evolution of Life 14

The prebiotic world 14 Origins of modern cells 16 Box 1.B How Does Evolution Work? 17

2 Aqueous Chemistry 2.1

24

The Hydrophobic Effect

26

29

Acid–Base Chemistry

4.1

Tools and Techniques: Manipulating DNA

68

85

Molecular Structure and Function

3 From Genes to Proteins Nucleotides

4.2

52

Secondary Structure: The Conformation of the Peptide Group 94 The α helix exhibits a twisted backbone conformation 95 The β sheet contains multiple polypeptide strands 95 Proteins also contain irregular secondary structure 96

4.3

Tertiary Structure and Protein Stability 97 Proteins have hydrophobic cores 98 Protein structures are stabilized mainly by the hydrophobic effect 99 Other interactions help stabilize proteins 100 Protein folding begins with the formation of secondary structures 101 Some proteins have more than one conformation 102

4.4 4.5

52

Nucleic acids are polymers of nucleotides 53 Some nucleotides have other functions 54

Amino Acids, the Building Blocks of Proteins 86

The 20 amino acids have different chemical properties 87 Box 4.A Does Chirality Matter? 88 Box 4.B Monosodium Glutamate 90 Peptide bonds link amino acids in proteins 90 The amino acid sequence is the first level of protein structure 93

Tools and Techniques: Buffers 40 Clinical Connection: Acid–Base Balance in Humans 42

Part 2

3.1

64

Cutting and pasting generates recombinant DNA 69 The polymerase chain reaction amplifies DNA 71 Box 3.A Genetically Modified Organisms 72 Box 3.B DNA Fingerprinting 74 DNA sequencing uses DNA polymerase to make a complementary strand 74 DNA can be altered 76

33

[H+] and [OH–] are inversely related 33 The pH of a solution can be altered 34 Box 2.C Atmospheric CO2 and Ocean Acidification 35 A pK value describes an acid’s tendency to ionize 36 The pH of a solution of acid is related to the pK 37

2.4 2.5

63

Water Molecules and Hydrogen Bonds 24

Amphiphilic molecules experience both hydrophilic interactions and the hydrophobic effect 31 The hydrophobic core of a lipid bilayer is a barrier to diffusion 31 Box 2.B Sweat, Exercise, and Sports Drinks 32

2.3

Genomics

4 Protein Structure

Hydrogen bonds are one type of electrostatic force Box 2.A Why Do Some Drugs Contain Fluorine? 28 Water dissolves many compounds 28

2.2

59

61 DNA must be decoded 62

Gene number is roughly correlated with organismal complexity 65 Genes are identified by comparing sequences 66 Genomic data reveal biological functions 67

6

Enthalpy and entropy are components of free energy ∆G is less than zero for a spontaneous process 11 Life is thermodynamically possible 12

1.4

The Central Dogma

A mutated gene can cause disease

What Is Biochemistry? 1 Biological Molecules 3

Energy and Metabolism

3.3 1

Cells contain four major types of biomolecules 3 There are three major kinds of biological polymers Box 1.A Units Used in Biochemistry 7

1.3

Nucleic Acid Structure 56 DNA is a double helix 56 RNA is single-stranded 59 Nucleic acids can be denatured and renatured

1 The Chemical Basis of Life 1.1 1.2

3.2

4.6

Quaternary Structure 104 Clinical Connection: Protein Misfolding and Disease 105 Tools and Techniques: Analyzing Protein Structure 107 Chromatography takes advantage of a polypeptide’s unique properties 107

v

vi

CONTE NTS

109

Mass spectrometry reveals amino acid sequences Protein structures are determined by X-ray crystallography, electron crystallography, and NMR spectroscopy 110 Box 4.C Mass Spectrometry Applications 110

5 Protein Function 5.1

Chymotrypsin is activated by proteolysis 171 Protease inhibitors limit protease activity 172

6.5

Inhibition

119

Myoglobin and Hemoglobin: Oxygen-Binding Proteins 120

Clinical Connection: Hemoglobin Variants Structural Proteins 130 Actin filaments are most abundant 130

173

7 Enzyme Kinetics and 7.1 7.2

Oxygen binding to myoglobin depends on the oxygen concentration 120 Myoglobin and hemoglobin are related by evolution 121 Oxygen binds cooperatively to hemoglobin 123 A conformational shift explains hemoglobin’s cooperative behavior 124 Box 5.A Carbon Monoxide Poisoning 124 H+ ions and bisphosphoglycerate regulate oxygen binding to hemoglobin in vivo 126

5.2 5.3

Clinical Connection: Blood Coagulation

183

Introduction to Enzyme Kinetics 183 Derivation and Meaning of the Michaelis–Menten Equation 186 Rate equations describe chemical processes 186 The Michaelis–Menten equation is a rate equation for an enzyme-catalyzed reaction 187 KM is the substrate concentration at which velocity is half-maximal 189 The catalytic constant describes how quickly an enzyme can act 190 kcat /KM indicates catalytic efficiency 190 KM and Vmax are experimentally determined 191 Not all enzymes fit the simple Michaelis–Menten model 192

127 7.3

Enzyme Inhibition

194

Actin filaments continuously extend and retract 131 Tubulin forms hollow microtubules 132 Some drugs affect microtubules 134 Keratin is an intermediate filament 135 Collagen is a triple helix 136 Box 5.B Vitamin C Deficiency Causes Scurvy 137 Collagen molecules are covalently cross-linked 139 Box 5.C Bone and Collagen Defects 139

Some inhibitors act irreversibly 195 Competitive inhibition is the most common form of reversible enzyme inhibition 195 Transition state analogs inhibit enzymes 197 Box 7.A Inhibitors of HIV Protease 198 Other types of inhibitors affect Vmax 199 Allosteric enzyme regulation includes inhibition and activation 200 Several factors may influence enzyme activity 203

5.4

7.4

Motor Proteins

141

Myosin has two heads and a long tail 141 Myosin operates through a lever mechanism 142 Kinesin is a microtubule-associated motor protein 143 Box 5.D Myosin Mutations and Deafness 144 Kinesin is a processive motor 146

6 How Enzymes Work 6.1

What Is an Enzyme?

154

154

Enzymes are usually named after the reaction they catalyze 157

6.2

Chemical Catalytic Mechanisms

6.3

Unique Properties of Enzyme Catalysts 166 Enzymes stabilize the transition state 166 Efficient catalysis depends on proximity and orientation effects 168 The active-site microenvironment promotes catalysis 168

6.4

8 Lipids and Membranes 8.1

Chymotrypsin in Context 169 Not all serine proteases are related by evolution 170 Enzymes with similar mechanisms exhibit different substrate specificity 170

Lipids

204

215

215

Fatty acids contain long hydrocarbon chains 216 Box 8.A Omega-3 Fatty Acids 216 Some lipids contain polar head groups 217 Lipids perform a variety of physiological functions 219 Box 8.B The Lipid Vitamins A, D, E, and K 221

8.2

The Lipid Bilayer

222

The bilayer is a fluid structure 223 Natural bilayers are asymmetric 224

158

A catalyst provides a reaction pathway with a lower activation energy barrier 159 Enzymes use chemical catalytic mechanisms 160 Box 6.A Depicting Reaction Mechanisms 162 The catalytic triad of chymotrypsin promotes peptide bond hydrolysis 164

Clinical Connection: Drug Development

8.3

Membrane Proteins

225

Integral membrane proteins span the bilayer 225 An α helix can cross the bilayer 226 A transmembrane β sheet forms a barrel 226 Lipid-linked proteins are anchored in the membrane

8.4

The Fluid Mosaic Model 228 Membrane glycoproteins face the cell exterior

9 Membrane Transport 9.1

227

229

235

The Thermodynamics of Membrane Transport 235 Ion movements alter membrane potential 237 Membrane proteins mediate transmembrane ion movement 237

CONTENTS

9.2

Passive Transport

9.3

Active Transport

Starch and glycogen are fuel-storage molecules 288 Cellulose and chitin provide structural support 289 Box 11.A Cellulosic Biofuel 290 Bacterial polysaccharides form a biofilm 291

240

Porins are β barrel proteins 240 Ion channels are highly selective 241 Box 9.A Pores Can Kill 242 Gated channels undergo conformational changes Aquaporins are water-specific pores 243 Some transport proteins alternate between conformations 244

242

245

The Na,K-ATPase changes conformation as it pumps ions across the membrane 246 ABC transporters mediate drug resistance 247 Secondary active transport exploits existing gradients 247

9.4

Membrane Fusion

248

291

Oligosaccharides are N-linked or O-linked 292 Oligosaccharide groups are biological markers 293 Box 11.B The ABO Blood Group System 293 Proteoglycans contain long glycosaminoglycan chains 294 Bacterial cell walls are made of peptidoglycan 295

Part 3

Metabolism

12.1 Food and Fuel

260

301

12.2 Metabolic Pathways

A ligand binds to a receptor with a characteristic affinity 261 Box 10.A Bacterial Quorum Sensing 262 Most signaling occurs through two types of receptors The effects of signaling are limited 264

263

10.2 G Protein Signaling Pathways 265 G protein–coupled receptors include seven transmembrane helices 265 The receptor activates a G protein 265 Adenylate cyclase generates the second messenger cyclic AMP 266 Cyclic AMP activates protein kinase A 267 Signaling pathways are also switched off 267 The phosphoinositide signaling pathway generates two second messengers 269 Calmodulin mediates some Ca2+ signals 270

10.3 Receptor Tyrosine Kinases

274

Eicosanoids are short-range signals 275 Box 10.C Aspirin and Other Inhibitors of Cyclooxygenase 276

11 Carbohydrates

306

Some major metabolic pathways share a few common intermediates 307 Many metabolic pathways include oxidation– reduction reactions 308 Metabolic pathways are complex 310 Box 12.B The Transcriptome, the Proteome, and the Metabolome 311 Human metabolism depends on vitamins 312

12.3 Free Energy Changes in Metabolic Reactions

314

The free energy change depends on reactant concentrations 314 Unfavorable reactions are coupled to favorable reactions 316 Free energy can take different forms 318 Box 12.C Powering Human Muscles 320 Regulation occurs at the steps with the largest free energy changes 321

270

The insulin receptor dimer binds one insulin 271 The receptor undergoes autophosphorylation 271 Box 10.B Cell Signaling and Cancer 273

10.4 Lipid Hormone Signaling

301

Cells take up the products of digestion 302 Box 12.A Dietary Guidelines 303 Monomers are stored as polymers 304 Fuels are mobilized as needed 304

10.1 General Features of Signaling Pathways 260

283

11.1 Monosaccharides 283 Most carbohydrates are chiral compounds 284 Cyclization generates α and β anomers 285 Monosaccharides can be derivatized in many different ways 286

11.2 Polysaccharides

11.3 Glycoproteins

12 Metabolism and Bioenergetics

Box 9.B Antidepressants Block Serotonin Transport 250 SNAREs link vesicle and plasma membranes 251 Endocytosis is the reverse of exocytosis 252 Box 9.C Exosomes 253

10 Signaling

vii

287

Lactose and sucrose are the most common disaccharides 288

13 Glucose Metabolism 13.1 Glycolysis

329

330

Reactions 1–5 are the energy-investment phase of glycolysis 330 Reactions 6–10 are the energy-payoff phase of glycolysis 336 Box 13.A Catabolism of Other Sugars 340 Pyruvate is converted to other substances 341 Box 13.B Alcohol Metabolism 342

13.2 Gluconeogenesis

344

Four gluconeogenic enzymes plus some glycolytic enzymes convert pyruvate to glucose 345 Gluconeogenesis is regulated at the fructose bisphosphatase step 346

13.3 Glycogen Synthesis and Degradation

347

Glycogen synthesis consumes the free energy of UTP 348 Glycogen phosphorylase catalyzes glycogenolysis 349

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CONTE NTS

13.4 The Pentose Phosphate Pathway 350 The oxidative reactions of the pentose phosphate pathway produce NADPH 350 Isomerization and interconversion reactions generate a variety of monosaccharides 351 A summary of glucose metabolism 352

13.5 Clinical Connection: Disorders of Carbohydrate Metabolism 353 Glycogen storage diseases affect liver and muscle 354

14 The Citric Acid Cycle

362

14.1 The Pyruvate Dehydrogenase Reaction

362

The pyruvate dehydrogenase complex contains multiple copies of three different enzymes 363 Pyruvate dehydrogenase converts pyruvate to acetyl-CoA 363

14.2 The Eight Reactions of the Citric Acid Cycle 365 1. Citrate synthase adds an acetyl group to oxaloacetate 366 2. Aconitase isomerizes citrate to isocitrate 368 3. Isocitrate dehydrogenase releases the first CO2 369 4. α-Ketoglutarate dehydrogenase releases the second CO2 369 5. Succinyl-CoA synthetase catalyzes substrate-level phosphorylation 370 6. Succinate dehydrogenase generates ubiquinol 370 7. Fumarase catalyzes a hydration reaction 371 8. Malate dehydrogenase regenerates oxaloacetate 371

14.3 Thermodynamics of the Citric Acid Cycle 372 The citric acid cycle is an energy-generating catalytic cycle 372 The citric acid cycle is regulated at three steps 372 The citric acid cycle probably evolved as a synthetic pathway 373 Box 14.A Mutations in Citric Acid Cycle Enzymes 373

14.4 Anabolic and Catabolic Functions of the Citric Acid Cycle 375 Citric acid cycle intermediates are precursors of other molecules 375 Anaplerotic reactions replenish citric acid cycle intermediates 376 Box 14.B The Glyoxylate Pathway 377

15 Oxidative Phosphorylation

385

15.1 The Thermodynamics of Oxidation–Reduction Reactions 385 Reduction potential indicates a substance’s tendency to accept electrons 386 The free energy change can be calculated from the change in reduction potential 388

15.2 Mitochondrial Electron Transport 390 Mitochondrial membranes define two compartments 390 Complex I transfers electrons from NADH to ubiquinone 392 Other oxidation reactions contribute to the ubiquinol pool 394

Complex III transfers electrons from ubiquinol to cytochrome c 394 Complex IV oxidizes cytochrome c and reduces O2 Box 15.A Free Radicals and Aging 398

397

15.3 Chemiosmosis 399 Chemiosmosis links electron transport and oxidative phosphorylation 400 The proton gradient is an electrochemical gradient 400

15.4 ATP Synthase

401

ATP synthase rotates as it translocates protons 401 The binding change mechanism explains how ATP is made 403 The P:O ratio describes the stoichiometry of oxidative phosphorylation 403 Box 15.B Uncoupling Agents Prevent ATP Synthesis 404 The rate of oxidative phosphorylation depends on the rate of fuel catabolism 404

16 Photosynthesis

411

16.1 Chloroplasts and Solar Energy

411

Pigments absorb light of different wavelengths 412 Light-harvesting complexes transfer energy to the reaction center 414

16.2 The Light Reactions

415

Photosystem II is a light-activated oxidation– reduction enzyme 416 The oxygen-evolving complex of Photosystem II oxidizes water 417 Cytochrome b6f links Photosystems I and II 418 A second photooxidation occurs at Photosystem I Chemiosmosis provides the free energy for ATP synthesis 421

16.3 Carbon Fixation

419

422

Rubisco catalyzes CO2 fixation 422 Box 16.A The C4 Pathway 424 The Calvin cycle rearranges sugar molecules 424 The availability of light regulates carbon fixation 426 Calvin cycle products are used to synthesize sucrose and starch 426

17 Lipid Metabolism

432

17.1 Lipid Transport 432 17.2 Fatty Acid Oxidation 435 Fatty acids are activated before they are degraded 435 Each round of β oxidation has four reactions 436 Degradation of unsaturated fatty acids requires isomerization and reduction 439 Oxidation of odd-chain fatty acids yields propionyl-CoA 440 Some fatty acid oxidation occurs in peroxisomes 442

17.3 Fatty Acid Synthesis

443

Acetyl-CoA carboxylase catalyzes the first step of fatty acid synthesis 444 Fatty acid synthase catalyzes seven reactions 445 Other enzymes elongate and desaturate newly synthesized fatty acids 447

ix

CONTENTS

Box 17.A Fats, Diet, and Heart Disease 448 Fatty acid synthesis can be activated and inhibited 449 Box 17.B Inhibitors of Fatty Acid Synthesis 450 Acetyl-CoA can be converted to ketone bodies 450

17.4 Synthesis of Other Lipids 452 Triacylglycerols and phospholipids are built from acyl-CoA groups 452 Cholesterol synthesis begins with acetyl-CoA 454 A summary of lipid metabolism 457

Additional hormones influence fuel metabolism 505 AMP-dependent protein kinase acts as a fuel sensor 506

19.3 Disorders of Fuel Metabolism 507 The body generates glucose and ketone bodies during starvation 507 Box 19.B Marasmus and Kwashiorkor 507 Obesity has multiple causes 508 Diabetes is characterized by hyperglycemia 509 The metabolic syndrome links obesity and diabetes 511

19.4 Clinical Connection: Cancer Metabolism 511 Aerobic glycolysis supports biosynthesis 512

18 Nitrogen Metabolism

18.1 Nitrogen Fixation and Assimilation Nitrogenase converts N2 to NH3 465

464

Part 4

Ammonia is assimilated by glutamine synthetase and glutamate synthase 466 Transamination moves amino groups between compounds 467 Box 18.A Transaminases in the Clinic 469

Helicases convert double-stranded DNA to single-stranded DNA 521 DNA polymerase faces two problems 522 DNA polymerases share a common structure and mechanism 523 DNA polymerase proofreads newly synthesized DNA An RNase and a ligase are required to complete the lagging strand 525

20.2 Telomeres

477

531

20.3 DNA Damage and Repair

531 DNA damage is unavoidable 531

Glutamate supplies nitrogen to the urea cycle 481 The urea cycle consists of four reactions 482

485

Purine nucleotide synthesis yields IMP and then AMP and GMP 485 Pyrimidine nucleotide synthesis yields UTP and CTP 486 Ribonucleotide reductase converts ribonucleotides to deoxyribonucleotides 487 Thymidine nucleotides are produced by methylation 488 Nucleotide degradation produces uric acid or amino acids 489

Repair enzymes restore some types of damaged DNA Base excision repair corrects the most frequent DNA lesions 533 Nucleotide excision repair targets the second most common form of DNA damage 535 Double-strand breaks can be repaired by joining the ends 535 Recombination also restores broken DNA molecules

536

539

20.5 DNA Packaging 540

19 Regulation of Mammalian

DNA is negatively supercoiled 541 Topoisomerases alter DNA supercoiling 541 Eukaryotic DNA is packaged in nucleosomes 543

497

19.1 Integration of Fuel Metabolism 498 Organs are specialized for different functions Metabolites travel between organs 499 Box 19.A The Intestinal Microbiome Contributes to Metabolism 500

533

20.4 Clinical Connection: Cancer as a Genetic Disease 538 Tumor growth depends on multiple events 538 DNA repair pathways are closely linked to cancer

Fuel Metabolism

525

528

Telomerase extends chromosomes 529 Box 20.A HIV Reverse Transcriptase 530 Is telomerase activity linked to cell immortality?

476

18.4 Nitrogen Disposal: The Urea Cycle 480

18.5 Nucleotide Metabolism

519

20.1 The DNA Replication Machinery 519 Replication occurs in factories 520

Several amino acids are easily synthesized from common metabolites 470 Amino acids with sulfur, branched chains, or aromatic groups are more difficult to synthesize 471 Box 18.B Glyphosate, the Most Popular Herbicide 473 Amino acids are the precursors of some signaling molecules 475 Box 18.C Nitric Oxide 476 Amino acids are glucogenic, ketogenic, or both Box 18.D Diseases of Amino Acid Metabolism 480

Genetic Information

20 DNA Replication and Repair

18.2 Amino Acid Biosynthesis 469

18.3 Amino Acid Catabolism

512

Cancer cells consume large amounts of glutamine

464

21 Transcription and RNA

498

21.1 Initiating Transcription What is a gene? 552

19.2 Hormonal Control of Fuel Metabolism 501 Insulin is released in response to glucose 502 Insulin promotes fuel use and storage 503 Glucagon and epinephrine trigger fuel mobilization

504

551

552

DNA packaging affects transcription 553 DNA also undergoes covalent modification 555 Transcription begins at promoters 555 Transcription factors recognize eukaryotic promoters

557

x

CONTE NTS

Enhancers and silencers act at a distance from the promoter 558 Box 21.A DNA-Binding Proteins 558 Prokaryotic operons allow coordinated gene expression 560

21.2 RNA Polymerase

562

RNA polymerases have a common structure and mechanism 562 RNA polymerase is a processive enzyme 564 Transcription elongation requires a conformational change in RNA polymerase 564 Transcription is terminated in several ways 565

21.3 RNA Processing

567

Eukaryotic mRNAs receive a 5′ cap and a 3′ poly(A) tail 567 Splicing removes introns from eukaryotic RNA 567 mRNA turnover and RNA interference limit gene expression 569 rRNA and tRNA processing includes the addition, deletion, and modification of nucleotides 572 RNAs have extensive secondary structure 573

22 Protein Synthesis

22.2 Ribosome Structure 586 The ribosome is mostly RNA 586 Three tRNAs bind to the ribosome 587

22.3 Translation

589

Initiation requires an initiator tRNA 589 The appropriate tRNAs are delivered to the ribosome during elongation 590 The peptidyl transferase active site catalyzes peptide bond formation 593 Box 22.B Antibiotic Inhibitors of Protein Synthesis 594 Release factors mediate translation termination 595 Translation is efficient in vivo 596

22.4 Post-Translational Events

G-1

ODD-NUMBERED SOLUTIONS

22.1 tRNA and the Genetic Code

580 The genetic code is redundant 581 tRNAs have a common structure 581

INDE X

582

597

Chaperones promote protein folding 597 The signal recognition particle targets some proteins for membrane translocation 599 Many proteins undergo covalent modification 600 G LO SS A RY

580

tRNA aminoacylation consumes ATP

Some synthetases have proofreading activity 584 tRNA anticodons pair with mRNA codons 584 Box 22.A The Genetic Code Expanded 585

I-1

S-1

Preface Several years ago, we set out to write a short biochemistry textbook that combined succinct, clear chapters with extensive problem sets. We believed that students would benefit from a modern approach involving broad but not overwhelming coverage of biochemical facts, focusing on the chemistry behind biology, and providing students with practical knowledge and problem-solving opportunities. Our experience in the classroom continues to remind us that effective learning also requires students to become as fully engaged with the material as possible. To that end, we have embraced a strategy of posing questions and suggesting study activities throughout each chapter, so that students will not simply read and memorize but will explore and discover on their own—a truer reflection of how biochemists approach their work in the laboratory or clinic. As always, we view our textbook as a guidebook for students, providing a solid foundation in biochemistry, presenting complete, up-to-date information, and showing the practical aspects of biochemistry as it applies to human health, nutrition, and disease. We hope that students will develop a sense of familiarity and comfort as they encounter new material, explore it, and test their understanding through problem solving.

With the same goal of making it easy for students to navigate complex topics, some material within sections has been reorganized, and several new sections of text now focus on key content areas: 14.3 Thermodynamics of the Citric Acid Cycle, 17.1 Lipid Transport, 18.5 Nucleotide Metabolism, 20.5 DNA Packaging, 21.1 Initiating Transcription, and 22.1 tRNA and the Genetic Code. Above all, the focus of the fourth edition is ease of use, particularly for students and instructors taking advantage of new ways to assess student understanding. New Learning Objectives at the start of every section are based on verbs, giving students an indication of what they need to be able to do, not just know. Before You Go On study hints at end of each section reinforce the activities that support learning. The endof-chapter problem sets have been refreshed, with a total of 1,624 problems (averaging 74 per chapter, an increase of 18% over the previous edition). Problems are grouped by section and offered in pairs, with the answers to odd-numbered problems provided in an appendix.

New to This Edition

Traditional Pedagogical Strengths

Many details in the text and illustration program have been updated, with virtually no section left untouched. Some significant changes are worth mentioning: Chapter 3 includes an updated discussion of genomics and a completely new presentation of DNA sequencing technologies and the use of CRISPR-Cas to edit genes. Other new items include a discussion of archaeal lipids, details on the GLUT membrane transport protein, a box on exosomes, new illustrations of respiratory cilia and bacterial peptidoglycan, new molecular graphics of mitochondrial respiratory complexes, an updated presentation of the ribonucleotide reductase mechanism, and more information on the microbiome, cancer, and obesity. Descriptions of DNA replication and transcription have been extensively modified, with numerous new diagrams to present a more realistic picture of these processes. The histone code and readers, writers, and erasers are explained. New details on RNA splicing and protein translocation round out the revised text. Eight health-related topics that were previously confined to short boxes have been updated and expanded to Clinical Connection sections to give them the appropriate attention: 2.5 Acid–Base Balance in Humans, 4.5 Protein Misfolding and Disease, 5.2 Hemoglobin Variants, 6.5 Blood Coagulation, 7.4 Drug Development, 13.5 Disorders of Carbohydrate Metabolism, 19.4 Cancer Metabolism, and 20.4 Cancer as a Genetic Disease.

• “Do You Remember?” review questions start each chapter, to help students tie new topics to what they have already studied. • Figure Questions that accompany key tables and figures prompt students to inspect information more closely. • Key sentences summarizing main points are printed in italics to assist with quick visual identification. • Tools and Techniques Sections appear at the end of Chapters 2, 3, and 4, to showcase practical aspects of biochemistry and provide an overview of experimental techniques that students will encounter in their reading or laboratory experience. • 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 major section headings, highlight important concepts to guide students to the most important points within each section. • Key terms are in boldface. Their definitions are also included in the Glossary. • An annotated list of Selected Readings for each chapter includes recent short papers, mostly reviews, that students are likely to find useful as sources of additional information. xi

xii

PRE FACE

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, Fourth 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, From Genes to Proteins). This is followed by chapters on protein structure (Chapter 4) and protein function (Chapter 5), with coverage of myoglobin and hemoglobin, and cytoskeletal and motor proteins. 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. WileyPLUS is packed with interactive study tools and resources, including the complete online textbook. NEW Ten 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. NEW Assignable End-of-Chapter Questions, over 20 per chapter, can be assigned to students through WileyPLUS. NEW Twenty-four Sample Calculation Videos walk students through each step of the sample calculations. NEW 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. NEW Do You Remember Practice Quizzes help students prepare for new material by reinforcing relevant topics from previous chapters. NEW Concept Check Questions for each section allow students to test their knowledge. NEW Discussion Questions are thought-provoking questions that serve as a point of departure for student discussion and engagement with content. NEW Twenty-three Animated Process Diagrams bring multi-step figures to life. NEW ORION Biology and Chemistry Refresher offers ORIONS’s diagnostics and adaptive practice for foundational topics, to support Biochemistry students who come to the course with differing levels of background knowledge. UPDATED Bioinformatics Projects, written by Paul Craig at Rochester Institute of Technology, provide guidance

P REFAC E

for 12 extended explorations of online databases, with questions, many open-ended, for students to learn on their own.

Additional Instructor Resources in WileyPLUS • PowerPoint Art Slides. • Exercise Questions with immediate descriptive feedback updated for the fourth edition 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. • Practice and Pre-Lecture Questions by Steven Vik, Southern Methodist University, and Mary Peek, Georgia Institute of Technology. • PowerPoint Lecture Slides with Answer 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 We would like to thank everyone who helped develop Essential Biochemistry, Fourth Edition, including Biochemistry Editor Joan Kalkut, Product Designer Sean Hickey, Associate Development Editor Laura Rama, Senior Production Editor Elizabeth Swain, Senior Designer Tom Nery, and Senior Photo Editor Billy Ray. 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, Fourth Edition.

Fourth Edition Reviews Arkansas Anne Grippo, Arkansas State University California Rakesh Mogul, California State Polytechnic University, Pomona Brian Sato, University of California, Irvine Colorado Andrew Bonham, Metropolitan State University of Denver Hawaii Jon-Paul Bingham, University of Hawaii-Manoa, College of Tropical Agriculture and Human Resources Florida David Brown, Florida Gulf Coast University Georgia Chavonda Mills, Georgia College Rich Singiser, Clayton State University Illinois Jon Friesen, Illinois State University Stanley Lo, Northwestern University Kristi McQuade, Bradley University Indiana Mohammad Qasim, Indiana University Purdue, Fort Wayne Kansas Peter Gegenheimer, The University of Kansas Louisiana Jeffrey Temple, Southeastern Louisiana University Michigan Kathleen Foley, Michigan State University Deborah Heyl-Clegg, Eastern Michigan University Mississippi Arthur Chu, Delta State University

xiii

Missouri Nuran Ercal, Missouri University of Science and Technology Nebraska Jodi Kreiling, University of Nebraska, Omaha New Jersey Bryan Speigelberg, Rider University Yufeng Wei, Seton Hall University New York Sergio Abreu, Fordham University Susan Rotenberg, Queens College of CUNY Oregon Jeannine Chan, Pacific University Pennsylvania Mahrukh Azam, West Chester University of Pennsylvania Robin Ertl, Marywood University Amy Hark, Muhlengerg College Sandra Turchi-Dooley, Millersville University Laura Zapanta, University of Pittsburgh Rhode Island Erica Oduaran, Roger Williams University South Carolina Weiguo Cao, Clemson University Kerry Smith, Clemson University Tennessee Meagan Mann, Austin Peay State University Texas Johannes Bauer, Southern Methodist University Utah Craig Thulin, Utah Valley University

Previous Edition Reviews 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

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PRE FAC E

Gregg Jongeward, University of the Pacific Pavan Kadandale, University of California, Irvine Paul Larsen, University of California, Riverside Rakesh Mogul, California State Polytechnic University, Pomona 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 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 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 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 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 Calgary

Astrid & Hanns-Frieder Michler / Science Source Images

CHAPTER 1 The Chemical Basis of Life

While no one has yet succeeded in reproducing all of a cell’s chemical reactions in a test tube, it is possible to identify and quantify the thousands of molecules present in a cell, such as this amoeba. Understanding the structures and functions of those molecules is key to understanding how cells live, move, grow, and reproduce.

This first chapter offers a preview of the study of biochemistry, 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 discussion of the origin of self-replicating life-forms and their evolution 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 encountered in subsequent chapters.

1.1

What Is 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 organisms. 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 immunology; 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 isolated from cells. A thorough understanding of each molecule’s physical structure and chemical 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). But just as a clock completely disassembled no longer resembles a clock, information about a multitude 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.

LEARNING OBJECTIVE Recognize the main themes of biochemistry.

1

2

CH APTE R 1 The Chemical Basis of Life

Organism

Organ

Cell

Organelle

Liver Hepatocyte

Mitochondrion FIGURE 1.1 Levels of

organization in a living organism. Biochemistry focuses on the structures and functions of molecules. Interactions between molecules give rise to higher-order structures (for example, organelles), which may themselves be components of larger entities, leading ultimately to the entire organism. [Photodisc/Rubberball/

Molecules

Human

Citrate synthase

DNA

Citrate

Ubiquinone

Getty Images]

Chapters 3 through 22 of this book are divided into three groups that roughly correspond to three major themes of biochemistry: 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 dramatically alter its shape or metabolic activities, but it does so within certain limits. And in order to remain in a steady, non-equilibrium 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.

Biological Molecules

LEARNING OBJECTIVES

Biological Molecules

1.2

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, biological molecules are constructed from H, N, O, P, and S. Most of these molecules belong to one of a few structural 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. 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 common 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 C

NH⫹ 3 Asparagine

H NH2

C

CH2

SH

NH⫹ 3 Cysteine

1 H 11 Na 19 K

12 Mg 20 Ca

5 B 13 Al 23 V

3

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

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.

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. • Distinguish the main structural features of carbohydrates, amino acids, nucleotides, and lipids. • Identify the monomers and linkages in polysaccharides, polypeptides, and nucleic acids. • Summarize the biological functions of the major classes of biological molecules.

4

CH APTE R 1 The Chemical Basis of Life

TA BLE 1 .1

Common Functional Groups and Linkages in Biochemistry

COMPOUND NAME

STRUCTURE a

Amineb

RNH2 or R2NH or R3N or

FUNCTIONAL GROUP

RNH⫹ 3 R2NH⫹ 2 R3NH⫹

N

⫹

or

N

(amino group)

Alcohol

ROH

OH (hydroxyl group)

Thiol

RSH

SH (sulfhydryl group)

Ether

ROR

O

Aldehyde

Ketone

R

R

O

O

C H

C

O

O

C

R

C

O Carboxylic acidb (Carboxylate)

O R

(carbonyl group),

C O

R

(carbonyl group),

C

R

C O

OH or

C O

OH (carboxyl group) or

R

C

O⫺

C

O⫺ (carboxylate group)

R

C

(acyl group)

(acyl group)

O

O

O Ester

(ether linkage)

OR

C

O

(ester linkage)

C

N

(amido group)

C

N

O Amide

Imineb

R

C O

NH2

R

C O

NHR

R

C

NR2

R R

NH or NR or

O

R R

NH⫹ 2 NHR⫹

O Phosphoric acid esterb

R

O

P

O

P

OH or

O

O

P

O⫺

O

P O⫺

O

(imino group)

(phosphoester linkage)

P

O OH or

P

O O

P

O OH or

P

P

O O

OH

OH O O

O O

O⫺

O⫺

P OH

O⫺ (phosphoryl group, Pi )

O⫺

OH

OH O R

H

OH

O R

P

O

O⫺

Diphosphoric acid esterb

⫹

N

O

OH O R

C

or

P

O

OH O

O

(phosphoanhydride linkage)

P OH

O OH or

P O⫺

O O

P

O⫺

O⫺

(diphosphoryl group, pyrophosphoryl group, PPi ) a b

R represents any carbon-containing group. In a molecule with more than one R group, the groups may be the same or different. Under physiological conditions, these groups are ionized and hence bear a positive or negative charge.

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

Biological Molecules

5

COO⫺ H

C

CH3

NH⫹ 3 (a) In a structural formula, some bonds, such as the C—O and N—H bonds, are implied. Around the central carbon, the horizontal bonds extend slightly above the plane of the page, and the vertical bonds extend slightly behind it.

(b)

(c)

The atoms are color-coded by convention: C gray, N blue, O red, and H white. A balland-stick representation reveals the identities of the atoms and their positions in space.

In a space-filling model, each atom is presented as a sphere whose radius (the van der Waals radius) corresponds to the distance of closest approach by another atom.

accurately depicted in the ball-and-stick model (b), although the relative sizes and electrical charges of atoms are not shown. A spacefilling model (c) best represents the actual shape of the molecule but may obscure some of its atoms and linkages.

FIGURE 1.3 Representations of alanine. The structural formula

(a) indicates all the atoms and the major bonds. Because the central carbon atom has tetrahedral geometry, its four bonds do not lie flat in the plane of the paper. This tetrahedral arrangement is more

2. Carbohydrates Simple carbohydrates (also called monosaccharides or just sugars) 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): H

O C H C

OH

HO

C

H

H

C

OH

H C

OH

H HO

CH2OH

CH2OH O H OH H H

H OH

OH

Glucose

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 molecules allow them to be easily recognized as carbohydrates.

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

P O⫺

O O

P O⫺

O O

N

N

Triphosphate

P

O

CH2

O⫺ H

N

N O

H

H

OH

OH

Adenosine triphosphate (ATP)

Adenine

H

Ribose

6

CH APTE R 1 The Chemical Basis of Life

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 compounds 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 conditions. The anionic lipid is therefore called palmitate. O CH2

CH2 H3C

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 CH2

CH3

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 macromolecules 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 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 materials. 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

Polymer

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 without units. Here it is understood that the mass is expressed relative to one-twelfth the mass of an atom of the common carbon isotope 12C (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 expressing the minute concentrations of biomolecules in living cells. Concentrations are usually given as moles per liter (mol ⋅ L–1 or M), with the appropriate prefix such as m, μ, or n:

mega (M) kilo (k) milli (m) micro (μ)

106 103 10 –3 10–6

nano (n) pico (p) femto (f)

10 –9 10 –12 10 –15

For example, the concentration of the sugar glucose in human blood is about 5 mM, but many intracellular molecules are present 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. Q The diameter of a typical spherical bacterial cell is about 1 μm. What is the cell’s volume?

The linkage between monomeric units is characteristic of each type of polymer. The monomers are called residues after they have been incorporated into the polymer. Strictly speaking, lipids do not form polymers, although they do tend to aggregate to form larger structures such as cell membranes.

1. Proteins Polymers of amino acids are called polypeptides or proteins. Twenty different 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).

R1 O ⫹

H3N C H

R2

C N

C

H H

(a)

O C O⫺

Because the side chains of the 20 amino acids have different sizes, shapes, and chemical properties, 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 accommodate the functional groups of its amino acid residues (Fig. 1.4). The 20 different amino acids can be combined in almost any order and in almost any proportion 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. Polymerization involves the phosphate and sugar groups of the nucleotides, which become linked by phosphodiester bonds.

(b) FIGURE 1.4 Structure of human endothelin. The 21 amino acid residues of this polypeptide, shaded from blue to red, form a compact structure. In (a), each amino acid residue is represented by a sphere. The ball-and-stick model (b) shows all the atoms except hydrogen. [Structure (pdb 1EDN) determined by B. A. Wallace and R. W. Jones.]

8

CH APTE R 1 The Chemical Basis of Life

O⫺ ⫺

O

P

O

O

CH2 O H

Base H

H

H O

Phosphodiester bond ⫺O

H

P

O

O

CH2 O H

Base H

H

H OH

CGUACG (a)

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 nucleotide residues rather than in their three-dimensional shape (Fig. 1.5). Nevertheless, many nucleic acids do bend and fold into compact globular shapes, as proteins do.

3. Polysaccharides Polysaccharides usually contain only one or a few different types of monosaccharide residues, so even though a cell may synthesize dozens of different 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 residues (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):

(b) FIGURE 1.5 Structure of a nucleic acid. (a) Sequence of nucleotide residues, using one-letter abbreviations. (b) Ball-and-stick model of the polynucleotide, showing all atoms except hydrogen (this structure is a six-residue segment of RNA). [Structure (pdb ARF0108) determined by R. Biswas, S. N. Mitra, and M. Sundaralingam.]

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. 1.6). The starch and cellulose polymers differ in the arrangement 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 particles where protein synthesis takes place. Under certain conditions,

Biological Molecules

Glucose

Starch

Cellulose

FIGURE 1.6 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.

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. 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.

TA B L E 1.2

Functions of Biopolymers ENCODE INFORMATION

CARRY OUT METABOLIC REACTIONS

STORE ENERGY

SUPPORT CELLULAR STRUCTURES

Proteins

—

✔

✓

✔

Nucleic acids

✔

✓

—

✓

Polysaccharides

✓

—

✔

✔

BIOPOLYMER

✔ major function ✓ minor function

9

10

CH APTE R 1 The Chemical Basis of Life

L EARNING 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.

1.3

Energy and Metabolism

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 thermodynamics (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 created 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.

Enthalpy and entropy are components of free energy The energy relevant to biochemical systems is called the Gibbs free energy (after the scientist 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 dispersed 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. 1.7). Free energy, enthalpy, and entropy are related by the equation [1.1]

G = H  TS

where T represents temperature in Kelvin (equivalent to degrees Celsius plus 273). Temperature 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

(a)

(b)

FIGURE 1.7 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.

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

Energy and Metabolism

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 [1.2]

∆G = ∆H  T∆S

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).

S A MPLE CALCULATIO N 1.1 Problem Use the information below to calculate the change in enthalpy and the change in entropy for the reaction A → B.

A B

Enthalpy (kJ · mol1)

Entropy (J · K1 · mol1)

60 75

22 97

Solution ∆H = HB – HA = 75 kJ ⋅ mol–1 – 60 kJ ⋅ mol–1 = 15 kJ ⋅ mol–1 = 15,000 J ⋅ mol–1

∆S = SB – SA = 97 J ⋅ K–1 ⋅ mol–1 – 22 J ⋅ K–1 ⋅ 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 ∆G > 0 Nonspontaneous

B→A ∆G < 0 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 concentrations of the reacting molecules, the temperature, and the presence of a catalyst.) When a reaction, 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.

SEE SAMPLE CALCULATION VIDEOS

11

12

CH APTE R 1 The Chemical Basis of Life

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 temperature, 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 reaction can offset an unfavorable decrease in entropy (see Sample Calculation 1.2).

SAMP LE CA LCU LAT I O N 1 . 2 SEE SAMPLE CALCULATION VIDEOS

Problem Use the information given in Sample Calculation 1.1 to determine whether the reaction A → B is spontaneous at 25°C.

Solution Substitute the values for ∆H and ∆S, calculated in Sample Calculation 1.1, into Equation 1.2. To express the temperature in Kelvin, add 273 to the temperature in degrees Celsius: 273 + 25 = 298 K. ∆G = ∆H – T∆S = 15,000 J ⋅ mol–1 – 298 K (75 J ⋅ K–1 ⋅ mol–1) = 15,000 – 22,400 J ⋅ mol–1 = –7400 J ⋅ mol–1 = –7.4 kJ ⋅ mol–1 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.

Life is thermodynamically possible In order to exist, life must be thermodynamically spontaneous. Does this hold at the molecular level? When analyzed in a test tube (in vitro, literally “in glass”), many of a cell’s metabolic 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 B→C

∆G = +15 kJ ⋅ mol1 (nonspontaneous) ∆G = –20 kJ ⋅ mol1 (spontaneous)

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 A→C

∆G = (15 kJ ⋅ mol1) + (–20 kJ ⋅ mol1) ∆G = –5 kJ ⋅ mol1

This phenomenon is shown graphically in Figure 1.8. 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,

Energy and Metabolism

B

13

B

Free energy (G) A C

Reaction coordinate

TA B L E 1. 3

FIGURE 1.8 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.

Oxidation States of Carbon

COMPOUND a

FORMULA

Carbon dioxide most oxidized (least reduced )

O

C

O

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

H Acetic acid

depends only on the initial and final states of the system, without regard to the specific chemical 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 chemical reactions occur with a net negative change in free energy. These thermodynamically favorable (spontaneous) reactions are coupled to the unfavorable synthesis of monosaccharides from atmospheric CO2 (Fig. 1.9). In this process, the carbon is reduced. Reduction, 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 animal 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 electrons 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 coupled to energy-requiring processes such as the synthesis of building blocks and their polymerization to form macromolecules. Virtually all metabolic processes occur with the aid of catalysts called enzymes, most of which are proteins (a catalyst greatly increases the rate of a reaction without itself undergoing any net change). For example, specific enzymes catalyze the formation of peptide, phosphodiester, and glycosidic linkages during polymer synthesis. Other enzymes catalyze cleavage of these bonds to break the polymers into their monomeric units.

H

C

O C OH

H Carbon monoxide

C

O

Formic acid

H

C

O OH

Acetone

H H C H

O C

H Acetaldehyde

H

C

H C H H O

C H

H H Formaldehyde

C

O

H Acetylene

H

C

C

H

H H Ethanol

H C

C

OH

H H H

H

Light energy C

O

Reduction (unfavorable)

H

C

OH

Carbon of a monosaccharide

Oxidation (favorable)

C

H

H H H

Ethane O

C

Ethene

Free energy

O

C

O

Reduction and reoxidation of carbon compounds. The sun provides 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.

H

C

C

H

H H H Methane least oxidized (most reduced )

FIGURE 1.9

a

H

C

H

H

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

14

CH APTE R 1 The Chemical Basis of Life

A living organism—with its high level of organization of atoms, molecules, and larger structures—represents a state of low entropy relative to its surroundings. Yet the organism can maintain this thermodynamically unfavorable state as long as it continually obtains free energy from its food. Thus, living organisms do indeed obey the laws of thermodynamics. When the organism ceases to obtain a source of free energy from its surroundings or exhausts its stored food, the chemical reactions in its cells reach equilibrium (∆G = 0), which results in death.

BEFORE GOING ON • Make up values for ∆H and ∆S to generate ∆G values corresponding to a spontaneous and a nonspontaneous reaction. • Show how increasing temperature affects ∆G when ∆H and ∆S are constant. • 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.

L EARNING OBJECTIVES Summarize the evolutionary history of cells. • List the events that must have occurred during prebiotic evolution. • Name the three domains of life. • Distinguish prokaryotic and eukaryotic cells.

1.4

The Origin and Evolution of Life

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 organism’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 example, nucleic acids participate in the storage and transmission of genetic information in all organisms, and the oxidation of glucose 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.

The prebiotic world A combination of theory and experimental data leads to several plausible scenarios for the emergence of life from nonbiological (prebiotic) materials on the early earth. In one scenario, inorganic compounds such as H2, H2O, NH3, and CH4—which may have been present in the early atmosphere—could have given rise to simple biomolecules, such as amino acids, when sparked by lightning. Laboratory experiments with the same raw materials and electrical discharges to simulate lightning do in fact yield these molecules (Fig. 1.10). Other experiments suggest that hydrogen cyanide (HCN), formaldehyde (HCOH), and phosphate could have been converted to nucleotides with a similarly modest input of energy. Over time, simple molecular building blocks could have accumulated and formed larger structures, particularly in shallow waters where evaporation would have had a concentrating effect. Eventually, conditions would have been ripe for the assembly of functional, living cells. Charles Darwin proposed that life might have arisen in some “warm little pond”; however, the early earth was probably a much more violent place, with frequent meteorite impacts and volcanic activity.

The Origin and Evolution of Life

15

Electrodes

Mixture of gases

Condenser Water Heat source Stopcock for removing samples FIGURE 1.10 Laboratory synthesis of biological molecules. A mixture of gases—H2, H2O, NH3, and CH4—is subject to an electrical discharge. Newly formed compounds, such as amino acids, accumulate in the aqueous phase as water vapor condenses. Samples of the reaction products can be removed via the stopcock.

In an alternative scenario, supported by studies of the metabolism of some modern bacteria, the first cells could have developed at deep-sea hydrothermal vents, some of which are characterized by temperatures as high as 350°C and clouds of gaseous H2S and metal sulfides (giving them the name “black smokers”; Fig. 1.11). In the laboratory, incubating a few small molecules in the presence of iron sulfide and nickel sulfide at 100°C yields acetic acid, an organic compound with a newly formed CC bond: CH3SH Methyl thiol

+

FeS, NiS

CO + Carbon monoxide

H2O ⏤⏤⟶ CH3COOH Acetic acid

+

H2S

Under similar conditions, amino acids spontaneously form short polypeptides. Although the high temperatures that are necessary for their synthesis also tend to break them down, these compounds would have been stable in the cooler water next to the hydrothermal vent. Regardless of how they formed, the first biological building blocks would have had to polymerize. This process might have been stimulated when the organic molecules—often bearing anionic (negatively charged) groups—aligned themselves on a cationic (positively charged) mineral surface. − 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 matter how stable or chemically versatile, such molecules would never have given rise to anything larger or more complicated: The probability of assembling a fully functional cell from

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. [B. Murton/Southhampton Oceanography Centre/Science Photo Library/Photo Researchers.]

16

CH APTE R 1 The Chemical Basis of Life

A A A A

A Original

U U

U

U U

A A A A A

U U U U U

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).

2. The two polymer chains separate. A A A A A

U U U U U

+

Original Complement A

A

A

A A

A A A A A

+

U U U U U

A A A A A

+

Origins of modern cells 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. The first cells were probably able to “fix” CO2 —that is, convert it to reduced organic compounds—using the free energy released in the oxidation of readily available inorganic 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 blue-green algae) used the sun’s energy to fix CO2: CO2 + H2O → (CH2O) + O2

U U U U U A A A A A

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

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).

The concomitant oxidation of H2O to O2 dramatically increased the concentration of atmospheric O2, about 2.4 billion years ago, and made it possible for aerobic (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 organisms; these reactions proceed in the absence of oxygen. Now that the earth’s atmosphere contains about 18% oxygen, anaerobic organisms have not disappeared, but they have been restricted to microenvironments where O2 is scarce, such as the digestive systems of animals or underwater sediments. The earth’s present-day life-forms are of two types, which are distinguished by their cellular architecture:

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

Original Copy FIGURE 1.12 Possible mechanism for the selfreplication of a primitive RNA molecule. For simplicity, the RNA molecule is shown as a polymer of adenine nucleotides, A.

Q Draw a diagram showing how polyU would be replicated.

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

The Origin and Evolution of Life

Box 1.B How Does Evolution Work? Documenting evolutionary change is relatively straightforward, but the mechanisms whereby evolution occurs are prone to misunderstanding. 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 alterations (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 population 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 parent but most likely developed bit by bit, over many generations, by modification of a gill or 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 actually fly in pursuit of food or to evade predators. Although we tend to think of evolution as an imperceptibly slow process, occurring on a geological time scale, it is ongoing and accessible to observation in the laboratory. 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 progress 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 as it adapts to the new conditions.

Q Why can’t acquired (rather than genetic) characteristics serve as the raw material for evolution?

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. The evolutionary history of eukaryotes is complicated by the fact that eukaryotic cells exhibit characteristics of both bacteria and archaea. Eukaryotic cells also contain organelles that are almost certainly the descendants of free-living prokaryotic cells. Specifically, the chloroplasts of plant cells, which carry out photosynthesis, closely resemble the photosynthetic cyanobacteria. The mitochondria of plant and animal cells, which are the site of

FIGURE 1.14 A eukaryotic cell. The paramecium, a one-celled organism, contains a nucleus and other membranebounded compartments. [Dr. David Patterson/ FIGURE 1.13 Prokaryotic cells. These

single-celled Escherichia coli bacteria lack a nucleus and internal membrane systems. [E. Gray/Science Photo Library/Photo Researchers.]

Science Photo Library/Photo Researchers.]

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

17

18

CH APTE R 1 The Chemical Basis of Life

Prokaryotes

Eukaryotes Animals

Bacteria

Archaea

Eukarya Fungi Plants

Nucleus

DNA 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.

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.

[After Wheelis, M. L., Kandler, O., and Woese, C. R., Proc. Natl. Acad. Sci. USA, 89, 2930–2934 (1992).]

much of the eukaryotic cell’s aerobic metabolism, resemble certain bacteria. In fact, both chloroplasts and mitochondria contain their own genetic material and protein-synthesizing machinery. It is likely that an early eukaryotic cell developed gradually from a mixed population of prokaryotic cells. Over many generations of living in close proximity and sharing each other’s metabolic products, some of these cells became incorporated within a single larger cell. This arrangement would account for the mosaic-like character of modern eukaryotic cells (Fig. 1.16). At some point, cells in dense populations might have traded their individual existence for a colonial lifestyle. This would have allowed for a division of labor as cells became specialized and would have eventually produced multicellular organisms. The earth currently sustains about 9 million different species (although estimates vary widely). Perhaps some 500 million species have 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. And 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 insects), prokaryotic metabolic strategies are amazingly varied. Nevertheless, by documenting characteristics that are common to all species, we can derive far-reaching conclusions about what life is made of, what sustains it, and how it has developed over the eons. 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.

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 biological polymers are proteins, nucleic acids, and polysaccharides.

Problems

19

1.3 Energy and Metabolism

1.4

The Origin and Evolution of Life

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

• The earliest cells may have evolved in concentrated solutions of molecules or near hydrothermal vents.

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

• Eukaryotic cells contain membrane-bounded organelles. Prokaryotic cells, which are smaller and simpler, include the bacteria and the archaea.

Key Terms protein peptide bond conformation polynucleotide nucleic acid phosphodiester bond polysaccharide glycosidic bond free energy (G) enthalpy (H) entropy (S) exothermic reaction

bioinformatics homeostasis trace element amino acid carbohydrate monosaccharide nucleotide lipid monomer polymer residue polypeptide

evolution complement natural selection aerobic anaerobic prokaryote bacteria archaea eukaryote eukarya

endothermic reaction ∆G spontaneous process exergonic reaction nonspontaneous process endergonic reaction in vitro in vivo reduction oxidation enzyme replication

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 Structure of Vitamin C

Problems 1.2 Biological Molecules 1. Use Table 1.1 to assign the appropriate compound name to each molecule.

2. Use Table 1.1 to assign the appropriate compound name to each molecule. a. H3C

O

O a. H3C

(CH2)14 C

b. H3C

CH2

OH OH

C

d. H3C

CH2

b. H3C

O

O

P O

NH2 c. H3C

O c. H3C

CH3

CH2

SH

O O

CH2 OH

CH3

d. H3C

C

CH3

CH3

20

CH APTE R 1 The Chemical Basis of Life

3. Investigators synthesized a series of compounds that showed promise as drugs for the treatment of Alzheimer’s disease. The structure of one of the compounds is shown below. Identify the functional groups in this compound.

OH

8. A compound present in many foods has the formula C44H86O8NP. To which class of molecules does this compound belong? Explain your answer.

N O

CH3

4. The structures of several molecules are shown below. Identify the functional groups in each structure.

O

OH O

O

C

HO

OH

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

C

CH2OH

OH

H

9. 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? 10. The structures of three compounds are shown below. Based on your answer to Problem 9, 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. O

N

Vitamin C

Nicotinic acid (niacin)

O

H C

O H3CO

H C

CH3 (CH2

H3CO

CH3 CH

C

N

OH

CH2OPO32⫺

5. 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?

H HO

H

NH

H C

CH3 NH2

O

O

O H2N

H

CH2OH

CH2

C

O

HO

C

H

H

C

OH

H

C

OH

H H

OH CH2

NH2

14. Consider the monosaccharide fructose. a. How does its molecular formula differ from that of glucose? b. How does its structure differ from the structure of glucose?

N

H

c. HS

C

13. The “straight-chain” structure of glucose is shown in Section 1.2. What functional groups are present in the glucose molecule?

CH2 O

O⫺

C

11. 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 consume a low-protein diet?

N

O

CH NH⫹ 3

OH

CH2OH Fructose

COO⫺

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

d.

NH2

O HN O

O R

C

O

O

12. 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?

OH

b.

P

C

Urea

O

⫺O

NH2

B

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

CH2OH O H OH H

O⫺

O⫺ A

Coenzyme Q

C

CH2

N

H2N

CH CH2

N

CH2)10H

O

a.

⫹H N 3

NH2

N N H

Uracil

O

N H Cytosine

Problems

16. What are the structural components of the biological molecules called nucleotides? 17. Compare the solubilities in water of alanine, glucose, palmitate, and cholesterol, and explain your reasoning. 18. Cell membranes are largely hydrophobic structures. Which compound will pass through a membrane more easily, glucose or 2,4-dinitrophenol? Explain.

OH NO2

21

28. Campers carry hot packs with them, especially when camping during the winter months or at high altitudes. The design is similar to that described in Problem 27, except that calcium chloride is used in place of the ammonium nitrate. The equation for the dissolution of calcium chloride in water is shown below. How does the hot pack work? H2O

CaCl2 (s) ⏤⟶ Ca2+ (aq) + 2 Cl– (aq)

ΔH = –81 kJ ⋅ mol–1

29. Urea (NH2CONH2) dissolves readily in water; i.e., this is a spontaneous process. The 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? 30. For the reaction in which reactant A is converted to product B, tell whether this process is favorable at a. 4°C and b. 37°C.

NO2 2,4-Dinitrophenol 19. 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. 20. What are the two major biological roles of polysaccharides? 21. 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. 22. The complete digestion of starch in mammals yields 4 kilocalories per gram (see Problem 21). What is the energy yield for cellulose?

1.3 Energy and Metabolism 23. What is the sign of the entropy change for each of the following processes? a. Water freezes. b. Water evaporates. c. Dry ice sublimes. d. Sodium chloride dissolves in water. e. Several different types of lipid molecules assemble to form a membrane. 24. Does entropy increase or decrease in the following reactions in aqueous solution? a. COO⫺

C

COO⫺

O ⫹ CO2(g)

C

O

CH2

CH3

COO⫺ b. COO⫺

C

H

O ⫹ H⫹

C

CH3

O ⫹ CO2(g)

CH3

25. Which has the greater entropy, a polymeric molecule or a mixture of its constituent monomers? 26. How does the entropy change when glucose undergoes combustion? C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 27. 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) + NO3– (aq) ΔH = 26.4 kJ ⋅ mol–1

A B

H (kJ · mol–1)

S (J · K–1 · mol–1)

54 60

22 43

31. For a given reaction, the value of ∆H is 15 kJ ⋅ mol–1 and the value of ∆S is 51 J ⋅ K–1 ⋅ mol–1. Above what temperature will this reaction be spontaneous? 32. Which of the following processes are spontaneous? a. A reaction 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. 33. 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 ∆S for this reaction? 34. 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 ∆S is 190 J ⋅ K–1 ⋅ mol–1. What is the value of ∆H? Is heat absorbed from or released to the surroundings? 35. A monoclonal antibody binds to the protein cytochrome c. The ∆H value for binding at 25°C is −87.9 kJ ⋅ mol–1 and the ∆S is −118 J ⋅ K–1 ⋅ mol–1. a. Does entropy increase or decrease when the antibody binds to the protein? b. Calculate ∆G for the formation of the antibody−protein complex. Does the complex form spontaneously? c. The ∆G 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? 36. Phosphofructokinase catalyzes the transfer of a phosphate group (from ATP) to fructose-6-phosphate to produce fructose-1, 6-bisphosphate at 37°C. The ∆H value for this reaction is −9.5 kJ ⋅ mol–1 and the ∆G is −17.2 kJ ⋅ mol–1. a. Is heat absorbed from or released to the surroundings? b. What is the value of ∆S 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 ∆H or ∆S value? Comment on the significance of this observation. 37. Glucose can be converted to glucose-6-phosphate: glucose + phosphate → glucose-6-phosphate + H2O ∆G = 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 ∆G for the coupled reaction. Is the

22

CH APTE R 1 The Chemical Basis of Life

conversion of glucose to glucose-6-phosphate favorable under these conditions? Explain. ΔG = –30.5 kJ ⋅ mol–1

ATP + H2O → ADP + phosphate

38. Glyceraldehyde-3-phosphate (GAP) is converted to 1,3-bisphosphoglycerate (1,3BPG) as shown. GAP + Pi + NAD+ → 1,3BPG + NADH

ΔG = +6.7 kJ ⋅ mol–1

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

42. For each of the reactions in Problem 41, tell whether an oxidizing agent or a reducing agent is needed to accomplish the reaction. 43. 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 Fig. 1.9. b. On a per-carbon basis, which would make more free energy available for metabolic reactions: palmitate or glucose? 44. Which yields more free energy when completely oxidized, stearate or α-linolenate?

Write the equation for the overall conversion of GAP to 3PG. Is the coupled reaction favorable?

H3C

H

C

H

C

H

O

C

B

1.4

C

40. Identify the process described in the following statements as an oxidation or reduction process. a. Monosaccharides are synthesized from carbon dioxide by plants during photosynthesis. b. An animal eats the plant and breaks down the monosaccharide in order to obtain energy for cellular processes. 41. Given the following reactions, tell whether the reactant is being oxidized or reduced. Reactions may not be balanced.

O

O (CH2)14

O⫺

C

8 CH3

C

S

CoA

COO⫺

b. COO⫺

CH2 CH

CH2

(CH

CH2 OH

C

COO⫺

␣-Linolenate

(CH2)6

The Origin and Evolution of Life

45. Why is molecular information so important for classifying and tracing the evolutionary relatedness of bacterial species but less important for vertebrate species? 46. The first theories to explain the similarities between bacteria and mitochondria or chloroplasts suggested that an early eukaryotic cell actually 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? 47. Draw a simple evolutionary tree that shows the relationships between species A, B, and C based on the DNA sequences given here. Species A

TCGTCGAGTC

Species B

TGGACTAGCC

Species C

TGGACCAGCC

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

COO⫺

CH

CH2

CH

CH

COO⫺

COO⫺

OH H15

O

d. ⫹H N 3

CH

C

H7

O⫺

H10

CH2 S S

C O

H3

O ⫹ H2

CH2 ⫺O

COO⫺

O

COO⫺

c. COO⫺

CHCH2)3

O

H A

a. CH3

H3 C

H OH

COO⫺

Stearate

39. Place these molecules in order from the most oxidized to the most reduced.

O

(CH2)16

CH NH⫹ 3

2 ⫹H3N

CH CH2 SH

C

O⫺

H4 H14

Selected Readings

23

Selected Readings Koonin, E. V., The origin and early evolution of eukaryotes in the light of phylogenomics, Genome Biol. 11, 209 (2010). [Discusses several models for the origin of eukaryotic cells.]

Nee, S., More than meets the eye, Nature 429, 804–805 (2004). [A brief commentary about appreciating the metabolic diversity of microbial life.]

Koshland, D. E., Jr., The seven pillars of life, Science 295, 2215–2216 (2002). [Describes some of the essential attributes of all organisms, including a DNA program, ability to mutate, compartmentalization, need for energy, ability to regenerate, adaptability, and seclusion.]

Nisbet, E. G., and Sleep, N. H., The habitat and nature of early life, Nature 409, 1083–1091 (2001). [Explains some of the hypotheses regarding the early earth and the origin of life, including the possibility that life originated at hydrothermal vents.]

Mora, C., Tittensor, D. P., Adl, S., Simpson, A. G. B., and Worm, B., How many species are there on earth and in the ocean? PLoS Biol 9(8): e1001127. doi:10.1371/journal.pbio.1001127 (2011). [Shows how statistical analysis of databases can be used to estimate the total number of known and not-yet-described species.]

Tinoco, I., Jr., Sauer, K., Wang, J. C., Puglisi, J. C., Harbison, G., and Rovnyak, D., Physical Chemistry. Principles and Applications in Biological Sciences (5th ed.), Chapters 2–4, Prentice-Hall (2014). [This and other physical chemistry textbooks present the basic equations of thermodynamics.]

Lisa Collins/robertharding/Getty Images, Inc.

CHAPTER 2 Aqueous Chemistry

Plotosus japonicus, a species of catfish, locates food by detecting the subtle change in pH caused by the release of carbon dioxide from hidden prey organisms.

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

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. And 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.

L EARNING 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.

24

2.1

Water Molecules and Hydrogen Bonds

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% by weight water, most of it in the extracellular fluid (the fluid surrounding cells) and inside cells:

Intracellular water (40%)

Non-water (40%)

Extracellular water (15%)

Water in circulatory system (5%)

Water Molecules and Hydrogen Bonds

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+

25

H

O

δ+ H

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:

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.

δ− δ+

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” a hydrogen 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 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 six-membered ring shown here:

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 (25ºC). At the same time,

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.)

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

26

CH APTE R 2 Aqueous Chemistry

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 FIGURE 2.3 A water strider supported by the surface tension of water. [Hermann Eisenbeiss/Photo Research, Inc.]

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. But 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. 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). Hydrogen bonds, despite their partial covalent nature, are classified as a type of electrostatic interaction. At about 1.8 Å, they 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). (a) O

H

Covalent bond

1Å

(b) O

H

Hydrogen bond

1.8 Å

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

400

Bond strength (kJ . mol−1)

350 300 250 200

TA B L E 2. 1

150

ELEMENT

ELECTRONEGATIVITY

100

C

2.55

50

F

3.98

H

2.20

N

3.04

O

3.44

S

2.58

0

Covalent bond

Ionic Hydrogen interaction bond

van der Waals interaction

FIGURE 2.4 Relative strengths of bonds in biological

molecules.

Electronegativities of Some Elements

Water Molecules and Hydrogen Bonds

H

H O

H

H

O

R O

R

27

H N

H

Water–alcohol

R Water–amine

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

Other electrostatic interactions occur between particles that are polar but not actually charged, for example, two carbonyl groups: δ−

C

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



H



C



H

H





H

H



C



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. [Hulton Archive/Getty Images.]

28

CH APTE R 2 Aqueous Chemistry

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 only rarely 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?

H N

O

CH3 F3C

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. 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 intermolecular 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.

Prozac (Fluoxetine) In designing an effective drug, pharmaceutical scientists often intentionally introduce F in order to alter the drug’s chemical or

Q Identify the hydrogen-bonding groups in Prozac.

Water dissolves many compounds

Na+

Cl–

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

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,

The Hydrophobic Effect

TAB L E 2. 2

29

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

Q Compare the hydrogen-bonding ability of these solvents.

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 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. 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

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.

FIGURE 2.6 Portion of a Dictyostelium 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. [Courtesy Wolfgang Baumeister, Max Planck Institute for Biochemistry.]

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.

30

CH APTE R 2 Aqueous Chemistry Nonpolar molecule

Layer of constrained water molecules

Water

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.

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 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). But 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 tend to clump together, removing themselves 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 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

(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.

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

The Hydrophobic Effect

aqueous phase by the unfavorable entropy cost of individually hydrating them. The hydrophobic effect governs the structures and functions of many biological molecules. 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.

31

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 hydrogen-bonding capacity of the polar head groups is satisfied through interactions 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.

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

32

CH APTE R 2 Aqueous Chemistry

INTRACELLULAR

EXTRACELLULAR

(b)

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 barrier, 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). FIGURE 2.11

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. But 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 don’t 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. Q Compare the ion concentrations of sweat and extracellular fluid.

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.

Acid–Base Chemistry

2.3

LEARNING OBJECTIVES

Acid–Base Chemistry

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: H2O ⇌ 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

H

O

O

or

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.

[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: K=

33

[H + ][OH − ] [H2O]

[2.1]

The square brackets represent the molar concentrations of the indicated species.

+

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.

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.

34

CH APTE R 2 Aqueous Chemistry 1

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

[H+] (M)

[OH−] (M)

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

FIGURE 2.14 −14

equal to 10

Because the concentration of H2O (55.5 M) is so much greater than [H+] or [OH−], it is considered to be constant, and K is redefined as Kw , the ionization constant of water: K w = K [H2O] = [ H + ][OH − ]

[2.2]

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

Basic

Neutral

Acidic

[2.3]

pH

[H+] (M)

14

10−14

13

10−13

12

10−12

11

10−11

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:

10

10−10

pH = −log [H + ]

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

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.

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

[2.4]

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).

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

TA B L E 2. 3 FLUID

pH Values of Some Biological Fluids pH

Pancreatic juice

7.8–8.0

Blood

7.4

Saliva

6.4–7.0

Urine

5.0–8.0

Gastric juice

1.5–3.0

Acid–Base Chemistry

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): CO2 + H2O ⇌ H2CO3 ⇌ H + + HCO3− 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 (CO32−) to construct protective shells of calcium carbonate (CaCO3). However, carbonate ions can combine with H+ to form bicarbonate:

+ − CO2− 3 + H ⇌ HCO3

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: CaCO3 + H + ⇌ HCO3− + Ca2+ This could have disastrous consequences for these species-rich ecosystems. Q 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.

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 + H2O → H3O + + 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 + H3O + → Na + + 2 H2O 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. Calculating the pH of a solution of strong acid or base is straightforward (see Sample Calculation 2.1).

S A MPLE CALCULATIO N 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) = 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).

(a) The final concentration of HCl is

SEE SAMPLE CALCULATION VIDEOS

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CH APTE R 2 Aqueous Chemistry

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+]. K w = 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: CH3COOH + H2O ⇌ CH3COO − + H3O + The equilibrium constant for this reaction takes the form K=

[CH3COO − ][H3O + ] [CH3COOH][H2O]

[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: K a = K [H2O] =

[CH3COO − ][H + ] [CH3COOH]

[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 K a

[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]

The larger an acid’s Ka, the smaller its pK and the greater its “strength” as an acid. Consider an acid such as the ammonium ion, NH4+: NH4+ ⇌ NH3 + 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 NH4+, readily accepts a proton. The pK values of some compounds are listed in Table 2.4. A polyprotic acid, a compound with

Acid–Base Chemistry

TA B L E 2.4

pK Values of Some Acids

NAME

FORMULAa

pK

Trifluoroacetic acid

CF3COOH

0.18

Phosphoric acid

H3PO4

2.15b

Formic acid

HCOOH

3.75

Succinic acid

HOOCCH2CH2COOH

4.21b

Acetic acid

CH3COOH

4.76 −

5.64c

Succinate

HOOCCH2CH2COO

Thiophenol

C6H5SH

6.60

Phosphate

H2PO4−

6.82c

N-(2-acetamido)-2-aminoethanesulfonic acid (ACES)

H2NCOCH2NH2CH2CH2SO3−

+

6.90



NH

Imidazolium ion

NO2

p-Nitrophenol

HO

N-2-hydroxyethylpiperazineNʹ-2-ethanesulfonic acid (HEPES)

HOCH2CH2HN



+

Glycinamide

a

7.00

N H

7.24

NCH2CH2SO 3

7.55

H3NCH2CONH2

8.20

+

Tris(hydroxymethyl)aminomethane (Tris)

(HOCH2 ) 3CNH3

8.30

Boric acid

H3BO3

9.24

Ammonium ion

NH +4

9.25

Phenol

C6H5OH

9.90

Methylammonium ion

CH3NH3+

10.60

Phosphate

HPO2− 4

12.38d

The acidic hydrogen is highlighted in red; bpK1; cpK2; dpK3.

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 + In other words, the final pH depends on the equilibrium between HA and A−, Ka =

[A − ][H + ] HA

[2.9]

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so that [H + ] = K a

[ HA] [A − ]

[2.10]

We can express [H+] as a pH, and Ka as a pK, which yields −log[H + ] = −log K a − log

[ HA] [A − ]

[2.11]

or pH = pK + log

[A − ] [ HA]

[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).

SAMP LE CA LCU LAT I O N 2 . 2 SEE SAMPLE CALCULATION VIDEOS

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) = 0.0089 M 1.011 L (0.005 L) (0.4 M) = 0.0020 M [ A−] = 1.011 L

[ HA] =

Next, substitute these values into the Henderson–Hasselbalch equation using the pK for acetic acid given in Table 2.4: pH = pK + log

[ A−] [ HA]

pH = 4.76 + log

0.0020 0.0089

= 4.76 − 0.65 = 4.11

SAMP LE CA LCU LAT I O N 2 . 3 SEE SAMPLE CALCULATION VIDEOS

Problem Calculate the concentration of formate in a 10-mM solution of formic acid at pH 4.15.

Acid–Base Chemistry

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. pH = pK + log log

[ A−] [ HA]

[ A−] = 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

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).

S A MPLE CALCULATIO N 2.4 Problem Determine which molecular species of phosphoric acid predominates at pH values of (a) 1.5, (b) 4, (c) 9, and (d) 13.

Solution From the pK values in Table 2.4, we know that: Below pH 2.15, the fully protonated H3PO4 species predominates. At pH 2.15, [ H3PO4] = [H2PO4− ]. Between pH 2.15 and 6.82, the H2PO4− species predominates. At pH 6.82, [ H2PO4− ] = [HPO2− 4 ]. Between pH 6.82 and 12.38, the HPO2− 4 species predominates. 2− 3− At pH 12.38, [HPO4 ] = [PO4 ]. Above pH 12.38, the fully deprotonated PO3− 4 species predominates. Therefore, the predominant species at the indicated pH values are (a) H3PO4, (b) H2PO4− , 3− (c) HPO2− 4 , and (d) PO4 .

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CH APTE R 2 Aqueous Chemistry

The pH-dependent ionization of biological molecules is also key to understanding their structures and functions. Many 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 (NH3+ ). 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  COOH

4 < pH < 10  COO−

NH3+

NH3+

pH > 10  COO− 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. 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 Fig. 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.

L EARNING 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.

2.4

Tools and Techniques: Buffers

When a strong acid such as HCl is added to pure water, all the added acid contributes directly to a decrease in pH. But when HCl is added to a solution containing a weak acid 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 + 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 − + H2O 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,

Tools and Techniques: Buffers

8 End point 7

Midpoint [CH3COOH] = [CH3COO−]

6 pK

Effective buffering range

5 pH

4 [CH3COOH] > [CH3COO−]

3 2

[CH3COOH] < [CH3COO−]

Start point

1

0

0.5 H+ ions dissociated

1

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.

Q Sketch the titration curve for ammonia.

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. 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, 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.

S A MPLE CALCULATIO 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?

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Solution Rearrange the Henderson–Hasselbalch equation to isolate the [A−]/[HA] term: pH = pK + log log

[ A−] [ HA]

[ A−] = pH − pK [ HA] [ A−] = 10(pH−pK ) [ HA]

Substitute the known pK (from Table 2.4) and the desired pH: [ A−] = 10(9.45−9.24) = 100.21 = 1.62 [ HA]

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.

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 (producing more H2PO4− ) or added base (producing more HPO2− 4 ). pK = 6.82 H2PO 4

H  HPO42

Acid added

H2PO 4

H  HPO42

Base added

H2PO 4

H  HPO42

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.

L EARNING 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.

Clinical Connection: Acid–Base Balance in Humans 2.5

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

Clinical Connection: Acid–Base Balance in Humans

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

Hⴙ  HCO 3 H  HCO 3 H  HCO 3

43

• Describe the roles of the lungs and kidneys in maintaining blood pH homeostasis. • Summarize the causes and treatments of acidosis and alkalosis.

Insufficient acid

H  HCO 3 H  HCO 3 Hⴙ  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.

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 CO2 + H2O ⇌ H2CO3 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 ionizes to bicarbonate, HCO3− (see Box 2.C): H2CO3 ⇌ H + + HCO3− so that the overall reaction is CO2 + H2O ⇌ H + + HCO3− The pK for this process is 6.1 (the ionization of HCO3− to Ion exchanger CO32− occurs with a pK of 10.3 and is therefore not significant at physiological pH). Na 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 1 system is augmented by the fact that excess hydrogen ions can HCO H 3 not only be buffered but can also be eliminated from the body. This is possible because after the H+ combines with HCO3− to 2 re-form H2CO3, which rapidly equilibrates with CO2 + H2O, some of the CO2 can be given off as a gas in the lungs. If it CO2 becomes necessary 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). FILTRATE Changes in pulmonary function can adequately adjust blood pH on the order of minutes to hours; however, longerterm adjustments of hours to days are made by the kidneys, which use a variety of mechanisms to excrete or retain H+, bicarbonate, and other ions. In fact, the kidneys 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 oxidation of glucose and fatty acids, and the ingestion of acidic groups in the form of phosphoproteins and phospholipids. The HCO3− 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 the urine (Fig. 2.18). In addition to reabsorbing filtered HCO3−, the kidneys also generate additional HCO3− to offset losses due to the buffering of metabolic acids and the exhalation of CO2. Metabolic activity

Na

HCO 3

H 3 CO2

KIDNEY CELL

FIGURE 2.18 Bicarbonate reabsorption. H+ leaves the kidney cells in exchange for Na+ (step 1). The expelled H+ combines with HCO3− in the filtrate, forming CO2 (step 2). Because it is nonpolar, the CO2 can diffuse into the kidney cell, where it is converted back to H+ + HCO3− (step 3).

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CH APTE R 2 Aqueous Chemistry

H pump

Ion exchanger Cl

H

HCO3

H 1

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).

in the kidney cells produces CO2, which is converted to H+ + HCO3−. The cells actively secrete the H+, which is lost via the urine, accounting for the mildly acidic pH of normal urine. The bicarbonate remaining in the cell is returned to the bloodstream in exchange for Cl− (Fig. 2.19). Some HCO3− also accumulates as a result of the metabolism of the amino acid glutamine in the kidneys. COO H

C

CH2

O CH2



C NH2

NH3 Glutamine

The two amino groups are removed as ammonia (NH3), which is ultimately excreted in the urine. Because the ammonium ion (NH4+) 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 HCO3−. 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 (NaHCO3). 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.

Summary

45

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 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 NH4+ 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.

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.

2.2 The Hydrophobic Effect

2.4

• 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.

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

• Amphiphilic molecules, which contain both polar and nonpolar groups, may aggregate to form micelles or bilayers.

2.5 Clinical Connection: Acid–Base Balance in Humans

2.3 Acid–Base Chemistry

• 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.

• The dissociation of water produces hydroxide ions (OH−) and protons (H+) whose concentration can be expressed as a pH value. The pH of

Tools and Techniques: Buffers

46

CH APTE R 2 Aqueous Chemistry

Key Terms ionization constant of water (Kw) neutral solution acidic solution basic solution pH acid base acid dissociation constant (Ka) pK conjugate base polyprotic acid

hydration hydrophilic hydrophobic hydrophobic effect amphiphilic amphipathic micelle bilayer vesicle hydronium ion proton jumping

polarity hydrogen bond ionic interaction van der Waals radius electronegativity van der Waals interaction dipole–dipole interaction London dispersion forces dielectric constant solute solvation

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 CH4 molecule is 109º. Explain why the HO H bond angle in water is only about 104.5º. 2. Each CO bond in CO2 is polar, yet the whole molecule is nonpolar. Explain.

O 

H3N

CH

CH2 O

C

3. Which compound has a higher boiling point, H2O or H2S? Explain.

CH2

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?

COO

N

CH

Water, H2O Ammonia, NH3 Methane, CH4

Melting point (°C)

18.0 17.0 16.0

0 −77 −182

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.

O

CH3

H

S

O

O

NH2

N

N

Aspartame

H 2N

H

O H

O Molecular weight (g · mol–1)

C

O N

N

H

H

Uric acid

Sulfanilamide 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.

Problems

H

H H3C

N

N⫹

N H

N

N

O H

N

N H

H

R

N

H MTX

a. N

N

N H

H 2C

8. In 2007, pets consuming a brand of pet food containing melamine (Problem 1.10) 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 form hydrogen bonds with each other.

H

N

H

N

N N H

O H

N N

HN O

NH N

N

H

C

H 3C

H CH2

OH

D

H 3C

CH2

Cl and H O H

E

H3C

CH2

O

C

CH3

O b. H3C

C

c. H3C

CH2

d. CsCl

NH2 CH3

(CH2)4 Lys

⫺

OOC CH2 Asp

CH2 Phe

13. 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). But as the size of the alcohol increases, solubility decreases. Explain.

Alcohol

Solubility in water (mmol · 100 g–1)

Propanol Butanol Pentanol

Miscible 0.11 0.05

CH2

CH3 and H

15. 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?

N H

H

11. What are the most important intermolecular interactions in the following molecules?

a. H3C

NH⫹ 3

14. Is the solubility trend for the alcohols in Table 2.2 related to their dielectric constants?

N

O

b.

Cyanuric acid

O

B

Ser

NH2

CH3 Ala

O

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

C

Asn

HO CH2

c.

9. Examine Table 2.1. a. Rank the six atoms listed in order of increasing electronegativity. b. What is the relationship between an atom’s electronegativity and its ability to participate in hydrogen bonding?

H 3C

C

H

H

Melamine

A

O

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.)

H

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

N

N

H

H

R

47

16. 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? 17. Ammonium sulfate, (NH4)2SO4, is a water-soluble salt. Draw the structures of the hydrated ions that form when ammonium sulfate dissolves in water. 18. 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? 19. 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 high water solubility. Why does

48

CH APTE R 2 Aqueous Chemistry

structure B more accurately represent the structure of glycine than structure A? H 2N



COOH

CH2

H3N

Structure A

CH2

COO

Structure B

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

26. Which compounds in Problem 25 are capable of forming micelles? Which are capable of forming bilayers? 27. The compound bis-(2-ethylhexyl)sulfosuccinate (abbreviated AOT) is capable of forming “reverse” micelles in the hydrocarbon solvent 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?

22. 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?

2.2

CH2CH3

The Hydrophobic Effect

H 3C

(CH2)3

CH

CH2

O

C

CH2 O

H3C

(CH2)3

CH

CH2

O

C

CH

H3C (H2C)15

OH CH3

N CH3 CH3

CH3

O

CH3

O

Hexadecyltrimethylammonium

Cholate

O (CH2)9

CH3

P

CH2OH O OH

H

H

OH

OH

CH3

O

(CH2)7 CH3

H

25. Consider the structures of the molecules below. Are these molecules polar, nonpolar, or amphiphilic? CH3 A

H3C

N

(CH2)11

CH2COO

H3C

C

H3C

(CH2)11

H3C

CH3 Cl N CH3

CH3

CH3 CH2

D

HC HO E

H3C

CH2 CH OH

O

O

S

ONa

O Sodium dodecyl sulfate (SDS) 29. 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 typically contains about 150 mM NaCl? What would happen if the cells were placed in pure water? 30. 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 29?

a. CO2 b.

c.

CHO O C

O

(CH2)11

31. Which of the following substances might be able to cross a bilayer? Which substances could not? Explain your answer.

CH3 B

C

O

O

n-Octylglucoside

Dimethyldecylphosphine oxide

O

28. Many household soaps are amphiphilic substances, often the salts of long-chain fatty acids, that form water-soluble micelles. An example is sodium dodecyl sulfate (SDS), an anionic detergent. a. Identify the polar and nonpolar portions of the SDS molecule. b. Draw the structure of the micelle formed by SDS. c. Explain how the SDS micelles “wash away” water-insoluble substances such as cooking grease.

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

H

S

Bis-(2-ethylhexyl)sulfosuccinate (AOT)

OH

HO

H3C

O

CH2CH3

23. 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. CH3

O

(CH2)11 (CH2)11

CH3 CH3

H

OH

HO H H

H OH OH

O

CH2OH 

Glucose

COO

d. Ca

2+

OH NO2

NO2 2,4-Dinitrophenol (DNP)

Problems

32. 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? 33. 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 energyrequiring process? Explain. 34. 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? 35. 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. low salt concentration?

How does the pH of the partially digested mixture change as it passes from the stomach into the intestine (see Table 2.3)? 44. 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? 45. Give the conjugate base of the following acids: a. HC2O4−; b. HSO−3; c. H2PO−4; d. HCO−3; e. HAsO42−; f. HPO42−; g. HO−2. 46. Give the conjugate acid of the species listed in Problem 45. 47. 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. 48. What is 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? 49. Identify the acidic hydrogens in the following compounds.

CH2 HO

C

COOH

COOH

CH2

Periplasmic space

E. coli



N H2

COOH

Citric acid

Piperidine

O Cytoplasm

Cytoplasmic membrane

49



H3N

CH

Cell wall

C

OH

HO

O

O

C

C

CH2 CH2

O

CH2

36. Under growth conditions, E. coli regulates cytoplasmic K+ content (in addition to water; see Problem 35) 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 low-salt medium? What happens when E. coli is grown in a high-salt medium?

OH

Oxalic acid

NH

CH2

O



NH3

O

N H

Barbituric acid

Lysine

O

2.3 Acid–Base Chemistry

O



NH

CH2

CH2

+

37. Compare the concentrations of H2O and H in a sample of pure water at pH 7.0 at 25ºC. 38. Like all equilibrium constants, the value of Kw is temperature dependent. Kw is 1.47 × 10−14 at 30ºC. What is “neutral” pH at this temperature?

41. Draw a diagram similar to Fig. 2.13 showing how a hydroxide ion appears to jump through an aqueous solution. 42. Fill in the blanks of the following table:

Blood Saliva Urine Gastric juice

Acid, base, or neutral?

pH

[H+] (M)

[OH–] (M)

______ neutral ______ ______

7.42 ______ ______ ______

______ ______ ______ 7.9 × 10–3

______ ______ 6.3 × 10–8 ______

43. Several hours after a meal, partially digested food leaves the stomach and enters the small intestine, where pancreatic juice is added.

O

O 4-Morphine ethanesulfonic acid (MES) 50. Rank the following according to their strength as acids: Acid

39. What is the pH of a solution of 1.0 × 10−9 M HCl? 40. What is the pH of a solution of 1.0 × 10−9 M NaOH?

S

A B C D E

citrate succinic acid succinate formic acid citric acid

Ka

pK

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

4.76 ______ ______ ______ 3.13

51. A solution is made by mixing 50 mL of a stock solution of 2.0 M K2HPO4 and 25 mL of a stock solution of 2.0 M KH2PO4. The solution is diluted to a final volume of 200 mL. What is the final pH of the solution? 52. Calculate the ratio of imidazole to the imidazolium ion in a solution at pH 7.4. 53. 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. 54. Calculate the concentration of acetate in a 50-mM solution of acetic acid buffer at pH 5.0.

50

CH APTE R 2 Aqueous Chemistry

55. What is the volume (in mL) of glacial acetic acid (17.4 M) that would have to be added to 500 mL of a solution of 0.20 M sodium acetate in order to achieve a pH of 5.0? 56. What is the mass of NaOH that would have to be added to 500 mL of a solution of 0.20 M acetic acid in order to achieve a pH of 5.0?

of 0.10 M HEPES. (HEPES is supplied by the chemical company as a sodium salt with a molecular weight of 260.3 g · mol–1.) d. What is the volume (in mL) of a stock solution of 6.0 M HCl that must be added to the 0.1 M HEPES to achieve the desired pH of 8.0? Describe how you will make the buffer.

57. 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 6.35; the pK for the second ionizable proton is 10.33. 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 7.40.

66. One liter of a 0.10 M Tris buffer (see Table 2.4) is prepared and adjusted to a pH of 8.2.

58. The pK of CH3CH2NH3+ is 10.7. Would the pK of FCH2CH2NH3+ be higher or lower?

Pyruvic acid

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 48a 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 48b in which the same amount of base was added to the same volume of pure water.

60. 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 predominate at pH 2, 7, and 10.

67. What volume (in mL) of a 1.0 M solution of imidazolium chloride must be added to a 500 mL solution of 10 mM imidazole in order to obtain a pH of 6.5?

59. The structure of pyruvic acid is shown. 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

O

COOH

CH2OH HOH2C

C

NH2

CH2OH Tris(hydroxymethyl)aminomethane

61. 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.

68. 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.

62. 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.

2.5 Clinical Connection: Acid–Base Balance in Humans

2.4

Tools and Techniques: Buffers

63. 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? 64. 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)

65. 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 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 until the desired pH is achieved. Describe how you will make 1.0 L

69. 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. 70. Metabolic acidosis often occurs in patients with impaired circulation from cardiac arrest. Mechanical hyperventilation (a standard treatment for acidosis) cannot be used with these patients because they often have acute lung injury (ALI). A group of physicians at San Francisco General Hospital advocated using tris(hydroxymethyl)aminomethane (Tris) to treat the metabolic acidosis of these patients. a. How would mechanical hyperventilation help alleviate acidosis in patients who do not have ALI? b. Why would treatment with sodium bicarbonate be effective in treating metabolic acidosis? Why would this treatment also be unacceptable for patients with ALI? c. Explain how Tris (see Problem 66) works to treat metabolic acidosis. Why is this treatment acceptable for patients with ALI? 71. 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? 72. A patient who has taken an 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 HCO−3 to H2CO3 in the patient’s blood at this pH. How does this compare to the ratio

Selected Readings

of HCO−3 to H2CO3 in normal blood? Can the H2CO3/HCO−3 system work effectively as a buffer in this patient under these conditions? 73. 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, H2PO4−; ammonium, NH4+; or bicarbonate, HCO3−. Which ions are excreted and which ions are reabsorbed in metabolic acidosis? Explain, using relevant chemical equations. 74. 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 73)? 75. 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.

51

76. In uncontrolled diabetes, the body converts fats to the so-called ketone bodies acetoacetate and 3-hydroxybutyrate, which accumulate in the bloodstream. O CH3

C

OH CH2

COO

CH3

C

CH2

COO

H Acetoacetate

3-Hydroxybutyrate

Do the ketone bodies contribute to acidosis or alkalosis? How might the body compensate for this acid–base imbalance? 77. 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? 78. 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.] Ellis, R. J., Macromolecular crowding: obvious but underappreciated, Trends Biochem. Sci. 26, 597–603 (2001). [Discusses how the large number of macromolecules in cells could affect reaction equilibria and reaction rates.] Gerstein, M. and Levitt, M., Simulating water and the molecules of life, Sci. Am. 279(11), 101–105 (1998). [Describes the structure of water and how water interacts with other molecules.]

Kropman, M. F. and Bakker, H. J., Dynamics of water molecules, Science 291, 2118–2120 (2001). [Describes how water molecules in solvation shells move more slowly than bulk water molecules.] Yucha, C., Renal regulation of acid–base balance, Nephrol. Nursing J. 31, 201–206 (2004). [A short review of physiological buffer systems and the role of the kidney.]

Kletr/Shutterstock

CHAPTER 3 From Genes to Proteins

The Anopheles mosquito is not deadly on its own, but as the carrier of the malaria parasite, it is linked to the deaths of several hundred thousand people every year. Researchers are investigating whether the latest genetic engineering techniques can produce a mosquito population that is unable to transmit the parasite.

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).

All the structural components of cells 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. The Tools and Techniques section of this chapter includes some of the methods used to study and manipulate DNA in the laboratory.

L EARNING OBJECTIVES Recognize the structures of nucleotides. • Identify the base, sugar, and phosphate groups of nucleotides. • Recognize nucleotide derivatives.

52

3.1

Nucleotides

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

Nucleotides

53

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

O H

N

N

N

N

N1 2

N

N

H2 N

N

N

H

N

6 3

9

8

N

N

H

Adenine

FIGURE 3.1 Human chromosomes from amniocentesis. In this image, the chromosomes have been stained with fluorescent dyes.

7

5 4

H

Guanine

[Dr. P. Boyer/Photo Researchers, Inc.]

Purine

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

NH2

O H

N

CH3

N

4

N3 2

O

O

N

1

N

H

5 6

N

H

Cytosine

Thymine

Pyrimidine

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

O H N O

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

Ribose

4

H

H 3

OH

5

base

HOCH2 O

H 1 2

OH

H

base

HOCH2 O 2-Deoxyribose

4

H

H 3

OH

H 1 2

H

H

54

CH APTE R 3 From Genes to Proteins

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



H

O

O

H2N

N

O P O P O H 2C

O

H

H

H

O



O



H

Adenosine monophosphate (AMP)

N

O

O

O

O O



O P O P O P O H 2C O

H

H

OH OH

N

N

N



O

O P O H2C O

N

N



H

N

N O

NH2

O

NH2

O

H

O

H

H

H

OH OH

OH OH Guanosine diphosphate (GDP)

H

N

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 deoxyadenosine 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.

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.

TAB 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.

Nucleotides

55

O HN

CH2

C

CH2

C

Nicotinamide

SH



O

N O

CH2 CH2

CH2 O H

Ribose

COO H

H

H

NH

Pantothenic acid residue

HO

NH2

C

O

HO

C

H

H3C

C

CH3

CH2

NH2

N

N

O

O

O

O P

O

O

P

O

CH2

O

N

N O

H

O

H

O

Coenzyme A (CoA) (a)

O

H O P

P

O

O

N

N CH2 O

H

Niacin

N

N

H

OH

N

NH2

O

O

Adenosine

H

P

OH

Adenosine

H H

HO OH Nicotinamide adenine dinucleotide (NAD) (b)

O

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.

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.

NH2 N

N O

O CH2 HO

C C

H

HO

C

H

O

P O

H

HO Riboflavin

O

P

OCH2

O

N

N O

H

Adenosine

H H

H OH

OH

CH2 H3C

N

H 3C

N

N

O NH

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

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.

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

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.

56

CH APTE R 3 From Genes to Proteins

L EARNING 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.

3.2

Nucleic Acid Structure

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

P

O

NH2

O

N

5 CH2

SEE GUIDED TOUR

Adenine

O

N

Nucleic Acid Structure 4

H

H

H

3

O

N N

H

O

2

H

N

5

Phosphodiester bond

O

P O

O

CH2 4

H

O H

N H

3

O

NH N

Guanine

NH2

H

O

2

H3 C

H

NH

5

O

P O

O

CH2 4

H

O H

3

O

O

N H

Thymine

H

NH2

2

H

N

5

O

P O

O

CH2 4

O H

N

Cytosine

O

H

H

H 3

2

OH

H

3 end

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:

Nucleic Acid Structure

H N H

N Adenine

N

N

O

CH3

H N

Thymine

N

N O

H N Guanine

O

N

H N

N H

N

N H

O

Cytosine

N

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

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.)

57

58

CH APTE R 3 From Genes to Proteins 5 end 3 end

Major groove 34 Å (10 bp)

Minor groove 3 end 5 end 20 Å

(a)

(b)

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.

Q How many nucleotides are shown in this double helix?

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). 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

Nucleic Acid Structure

59

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 exonuclease 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 3.3, 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 conformations 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.

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. [Structure (pdb 4TRA) determined by E. Westhoff, P. Dumas, and D. Moras.]

Nucleic acids can be denatured and renatured 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 complementary 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 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. The stacking interactions between neighboring G:C base pairs are stronger than those of A:T base pairs (this is not related to the fact that G:C base pairs have one more hydrogen bond than A:T base pairs). Consequently, a DNA helix that is rich in G and C is harder to disrupt than DNA with a high proportion of A and T. These differences can be quantified in the melting temperature (Tm ) of the DNA. To determine the melting point of a sample of DNA, the temperature is slowly increased. At a sufficiently high temperature, the base pairs begin to unstack, hydrogen bonds break, and the two strands begin to separate. This process continues as the temperature rises, until the two strands come completely apart. The melting, or denaturation, of the DNA can be recorded as

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). [Structure (pdb 1FIX) determined by N. C. Horton and B. C. Finzel.]

60

CH APTE R 3 From Genes to Proteins

Melting begins

Relative absorbance at 260 nm

1.4

+

1.3

1.2

1.1

1.0

FIGURE 3.6 Axial view of DNA base pairs. A view down the central axis of the DNA helix shows the overlap of neighboring base pairs (only the first two nucleotide pairs are highlighted).

Melting ends

Tm

30

50

70 Temperature (°C)

90

FIGURE 3.7 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.

Q Locate the base and sugar in the nucleotides with the blue bases.

a melting curve (Fig. 3.7) by monitoring an increase in the absorbance of ultraviolet (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 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. 3.8). 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 3.5). 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.

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

[Data from Brown, T. A. (ed.), Molecular Biology LabFax, vol. I., Academic Press (1998), pp. 233–237.]

The Central Dogma

61

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

High heat (melting)

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

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

FIGURE 3.8 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.

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

LEARNING OBJECTIVES

The Central Dogma

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. 3.9). 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. Thus, genetic information—in the form of a sequence of nucleotide residues—is transmitted each time a cell divides. A similar phenomenon is responsible for the expression of that genetic information, a process in which the information is 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 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

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.

62

CH APTE R 3 From Genes to Proteins Old Old

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.

DNA must be decoded

New

New

Even in the simplest organisms, DNA is an enormous molecule, and many organisms contain several different DNA molecules (for example, the chromosomes of eukaryotes). 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 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 (the template strand is called the noncoding strand).

5′ Old New

New Old

FIGURE 3.9 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 doublehelical DNA molecules.

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

3′

DNA

A

G G T A G C C A T C

C

G C T

5′ T

DNA

A

G

T

Coding strand (nontemplate)

A

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 protein. The protein’s amino acid sequence therefore ultimately depends on the nucleotide sequence of the DNA.

5 DNA 3

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

5

Transcription

mRNA 5 tRNAs

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

Ser

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, and the remaining 61 represent, with some redundancy, the 20 standard amino acids found

The Central Dogma

TA B L E 3.3 FIRST POSITION (5ʹ END)

The Standard Genetic Codea SECOND POSITION

U

C

A

G

THIRD POSITION (3ʹ 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 Vae 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

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.

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

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, even by examining one gene at a time. Thousands of genes have been identified through studies of the genes’ protein products, and millions more have been catalogued through genome-sequencing projects (discussed below in Section 3.4). Although the functions of many genes are not yet understood, some genes have come to light through the study of inherited diseases. In a traditional approach, researchers have used the defective protein associated with a particular disease to track down the relevant genetic defect. For example, the variant hemoglobin protein that causes sickle cell disease results from the substitution of the amino acid glutamate (Glu) by valine (Val). In the gene for that protein chain, the normal GAG codon has been mutated (altered) to GTG. 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 · · · Many human diseases are polygenic, so sorting out all the genetic variations that contribute to the disease is difficult. But for almost 5000 monogenic diseases, such as sickle cell disease and cystic fibrosis, a defect in a specific gene explains the molecular basis of the disease.

63

64

CH APTE R 3 From Genes to Proteins

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.

BEFORE GOING ON • • • •

L EARNING OBJECTIVES Identify the types of information provided by genomic analysis. • Compare the genomes of different species. • Explain how genes are identified. • Describe the usefulness of identifying genetic variations between individuals.

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. List some reasons why knowing a gene’s sequence might be useful.

3.4

Genomics

The ability to sequence large tracts of DNA has made it possible to study entire genomes, from the small DNA molecules of parasitic bacteria to the enormous multichromosome genomes of plants and mammals. Sequence data are customarily deposited in a public database such as GenBank. The data can be accessed electronically in order to compare a given sequence to sequences from other genes. Some of the thousands of organisms whose genomes have been partially or fully sequenced are listed in Table 3.4. This list includes species that are widely used as model organisms for different types of biochemical studies (Fig. 3.10).

TAB L E 3. 4

Genome Size and Gene Number of Some Organisms

ORGANISM

GENOME SIZE (kb)

NUMBER OF GENES

Bacteria Mycoplasma genitalium Haemophilus influenzae Synechocystis PCC6803 Escherichia coli

580 1,830 3,947 4,639

525 1,740 3,618 4,289

Archaea Methanocaldococcus jannaschii Archaeoglobus fulgidus

1,740 2,178

1,830 2,486

Fungi Saccharomyces cerevisiae (yeast)

12,070

6,034

119,200 389,000 ∼2,400,000

∼26,000 ∼35,000 ∼32,000

97,000 139,500 3,038,000

19,099 13,061 ∼21,000

Plants Arabidopsis thaliana Oryza sativa (rice) Zea mays (corn) Animals Caenorhabditis elegans (nematode) Drosophila melanogaster (fruit fly) Homo sapiens [Data from NCBI Genome Project.]

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

Genomics

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

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

(b) Baker’s yeast, Saccharomyces cerevisiae, is one of the simplest eukaryotic organisms, with just over 6000 genes.

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

FIGURE 3.10 Some model organisms. [Dr. Kari Lounatmaa/Science Photo Library/Photo Researchers; Andrew Syred/Science Photo Library/Photo Researchers; Sinclair Stammers/Science Photo Library/Photo Researchers; Dr. Jeremy Burgess/Science Photo Library/Photo Researchers.]

Gene number is roughly correlated with organismal complexity Not surprisingly, organisms with the simplest lifestyles tend to have the least amount of DNA and the fewest genes. For example, M. genitalium and H. influenzae (see Table 3.4) are human parasites that depend on their host to provide nutrients; these organisms do not contain as many genes as free-living bacteria such as Synechocystis (a photosynthetic bacterium). Multicellular organisms generally have even more DNA and more genes, presumably to support the activities of their many specialized cell types. Interestingly, humans contain about as many genes as nematodes, suggesting that organismal complexity results not just from the raw number of genes but from how the genes are transcribed and translated into protein. Note that humans and many other organisms are diploid (having two sets of genetic information, one from each parent), so that each human cell contains roughly 6 billion base pairs of DNA. For simplicity, genetic information usually refers to the haploid state, equivalent to one set of genetic instructions.

65

66

CH APTE R 3 From Genes to Proteins Unique sequence

Moderately repetitive

Protein-coding

Highly repetitive

FIGURE 3.11 Coding and noncoding portions of the human genome. Approximately 1.4% of the genome codes for proteins. Moderately repetitive sequences make up 45% of the genome and highly repetitive sequences about 3%, so that roughly half of the human genome consists of unique DNA sequences of unknown function. Up to 80% of the genome may be transcribed, however.

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. 3.11). Much of the noncoding DNA consists of repeating sequences with no known function. The presence of repetitive DNA helps explain why certain very large genomes actually include only a modest number of genes. For example, the maize (corn) and rice genomes contain about the same number of genes, but the maize genome is at least 6 times larger than the rice genome. Over half of the maize genome appears to be composed of transposable elements, short segments of DNA that are copied many times and inserted randomly into the chromosomes. The human genome contains several types of repetitive DNA sequences, including the inactive remnants of transposable elements. About 45% of human DNA consists of moderately 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. Highly repetitive sequences account for another 3% of the human genome. These segments of 2 to 10 bases are present in millions of copies. They are repeated tandemly (side by side), sometimes thousands of times. The number of repeats of a given sequence often varies between individuals, even in the same family, so this information can be analyzed to produce a DNA “fingerprint” (see Section 3.5).

Genes are identified by comparing sequences For many genomes, 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 DNA sequence for an open reading frame (ORF), that is, a stretch of nucleotides that can potentially be transcribed and translated. For a protein-coding gene, 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 proteins. 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). Other socalled ab initio (“from the beginning”) gene-identifying methods scan the DNA for other features that characterize the beginnings and endings of genes. Another method for identifying genes in a genome relies on sequence comparisons with known genes. Such genome-to-genome comparisons are possible because of the universal nature of the genetic code and the relatedness of all organisms through evolution (Section 1.4). Genes with similar functions in different species tend to have similar sequences; such genes

Genomics

FIGURE 3.12 Portion of the human genome. The black line represents a 500,000-bp segment of human chromosome 6 that includes six protein-coding genes (purple boxes) distributed unevenly and separated by stretches of noncoding DNA. [Based on information from the NCBI Map Viewer, www.ncbi.nlm.nih. gov/mapview.]

Q Why are the genes pointing in two directions?

are said to be homologous. Even an inexact match can still indicate a protein’s functional category, such as enzyme or hormone receptor, although its exact role in the cell may not be obvious. Genes that appear to lack counterparts in other species are known as orphan genes. The RNA or protein products of many genes have not yet been isolated or otherwise identified. For example, about 20% of the genes in the well-studied organism E. coli have not yet been assigned functions. Genome maps, such as the one shown in Figure 3.12, indicate the arrangement of genes on each strand of DNA in a chromosome. Human protein-coding genes are typically much longer than bacterial genes (27,000 bp versus 1000 bp on average), since they contain sequences removed from the mRNA transcripts before translation. In addition, the spaces between genes are much larger in the human genome. In addition to the approximately 20,000 protein-coding genes in the human genome are a comparable number of genes that correspond to noncoding RNA (ncRNA) molecules. Many of these transcripts, which come in a variety of sizes, appear to be involved in regulating the expression of other genes. Gene-mapping projects have uncovered some interesting aspects of evolution, including horizontal gene transfer. This occurs when a gene is transferred between species rather than from parent to offspring of the same species (vertical gene transfer). Horizontal gene transfer may be mediated by viruses, which can pick up extra DNA as they insert and excise themselves from the host’s chromosomes. This activity can generate, for example, what appears to be a mammalian gene inside a bacterial genome. The ease with which many bacterial organisms trade their genes has given rise to the idea that groups of bacteria should be viewed as a continuum of genomic variations instead of separate species with discrete genomes.

Genomic data reveal biological functions Genomics, the study of genomes, has a number of practical applications. For one thing, the number of genes and their putative functions provide a rough snapshot of the metabolic capabilities of a given organism. For example, humans and fruit flies differ in the number of genes that code for developmental pathways and immune system functions (Fig. 3.13). An unusual number of genes belonging to one category might indicate some unusual biological property in an organism. This sort of knowledge can be useful for developing drugs to inhibit the growth of a pathogenic organism according to its unique metabolism. Studies in which protein-coding genes are inactivated, one at a time, suggest that relatively simple organisms, such as Saccharomyces (yeast), have about 1000 essential genes. Humans likewise seem to absolutely require only about 10%—roughly 2000—of their genes. This core set of genes encodes proteins that are produced in abundance and carry out the most basic cellular activities. But clearly, the other 90% of genes are not useless. An analysis of 17,000 genes in 44 different human tissues has revealed that about half are expressed in all locations. Genomic analysis also reveals variations among individuals, some of which can be linked to an individual’s chance of developing a particular disease. In addition to genetic changes that are clearly associated with a single-gene disorder, millions more sequence variations have been catalogued. On average, the DNA of any two humans differs at 3 million sites, or about once every thousand base pairs. These single-nucleotide polymorphisms (SNPs, instances where the DNA sequence differs among individuals) are compiled in databases. Some of the factors that can alter DNA are discussed in Section 20.3. A person begins life with an average of 60 new genetic changes that were not present in either parent.

67

68

CH APTE R 3 From Genes to Proteins Development Immune system

(a) Human

(b) Drosophila

FIGURE 3.13 Functional classification of genes. This diagram is based on 19,184 human genes (a) and 10,787 Drosophila genes (b), grouped according to the biochemical function of the gene product. Humans devote a larger proportion of genes to development (7.7% versus 3.9% in Drosophila) and to the immune system (4.3% versus 2.0% in Drosophila). [Data from the Protein Analysis through Evolutionary Relationships classification system, www.pantherdb.org/.]

Researchers have attempted to correlate SNPs with disorders, such as cardiovascular disease or cancer, that likely depend on the contributions of many genes. Genome-wide association studies (GWAS) have identified, for example, 39 sites that are associated with type 2 diabetes and 71 that are associated with Crohn’s disease, an autoimmune disorder. The risk tied to any particular genetic variant is low, but the entire set of variations can explain up to 50% of the heritability of the disease. Although the SNPs are only proxies for disease genes, these data should provide a starting point for researchers to explore the DNA near the SNPs to discover the genes that are directly involved in the disease. Several commercial enterprises offer individual genome-sequencing services, but until genetic information can be reliably translated into effective disease-prevention or treatment regimens, the practical value of “personal genomics” is somewhat limited.

BEFORE GOING ON • • • •

L EARNING OBJECTIVES

Describe the rough correlation between gene number and organismal lifestyle. List some ways in which the human genome differs from a bacterial genome. Describe the approaches used to identify genes. Explain the value and limitations of genome-wide association studies.

Tools and Techniques: Manipulating DNA 3.5

Describe how researchers manipulate DNA in vitro and in vivo. • Summarize the roles of restriction enzymes, DNA ligase, and vectors in generating recombinant DNA molecules.

Molecular biologists have devised clever procedures for manipulating DNA in the laboratory and in living organisms. Many of the techniques take advantage of naturally occurring enzymes that cut, copy, and link nucleic acids. These techniques also exploit the ability of nucleic acids to interact with complementary molecules. In this section we focus on currently used methods for generating recombinant DNA, amplifying specific DNA segments, determining the sequence of DNA, and altering the DNA.

Tools and Techniques: Manipulating DNA

Cutting and pasting generates recombinant DNA Long DNA molecules tend to break from mechanical stress during laboratory manipulations. However, such randomly fragmented DNA is not always useful, so researchers often employ enzymes that cut DNA in defined ways. Bacteria produce DNA-cleaving enzymes known as restriction endonucleases (or restriction enzymes) that catalyze the breakage of phosphodiester bonds at specific nucleotide sequences. These enzymes can thereby destroy foreign DNA that enters the cell, such as bacteriophage (viral) DNA. In this way, the bacterial cell “restricts” the growth of the phage. The bacterial cell protects its own DNA from endonucleolytic digestion by methylating it (adding a CH3 group) at the same sites recognized by its restriction endonucleases. Hundreds of these enzymes have been discovered; some are listed in Table 3.5 along with their recognition sequences and cleavage sites. Restriction enzymes typically recognize a 4- to 8-base sequence that is identical, when read in the same 5ʹ → 3ʹ direction, on both strands. DNA with this form of symmetry is said to be palindromic (words such as madam and noon are palindromes). One restriction enzyme isolated from E. coli is known as EcoRI (the first three letters are derived from the genus and species names). Its recognition sequence is

5 — G A A T T C — 3 3 — C T T A A G — 5 The arrows indicate the phosphodiester bonds that are cleaved. Note that the sequence reads the same on both strands. Because the EcoRI cleavage sites are symmetrical but staggered, the enzyme generates DNA fragments with single-stranded extensions known as sticky ends:

—G A A TTC — — CTT A A G— In contrast, the E. coli restriction enzyme known as EcoRV cleaves both strands of DNA at the center of its 6-bp recognition sequence so that the resulting DNA fragments have blunt ends:

5 — G A T A T C — 3 3 — C T A T A G — 5

TAB L E 3. 5

— GA T — CTA

ATC— T AG—

Recognition and Cleavage Sites of Some Restriction Endonucleases

ENZYME

RECOGNITION/CLEAVAGE SITE a

AluI

AG | CT

MspI

C | CGG

AsuI

G | GNCCb

EcoRI

G | AATTC

EcoRV

GAT | ATC

PstI

CTGCA | G

SauI

CC | TNAGG

NotI

GC | GGCCGC

a

The sequence of one of the two DNA strands is shown.

The vertical bar indicates the cleavage site. b

N represents any nucleotide.

[An exhaustive source of information on restriction enzymes is available through the Restriction Enzyme Database: rebase.neb.com/rebase/rebase.html.]

69

• Explain how researchers clone a gene using a plasmid. • Describe how DNA polymerase makes PCR and DNA sequencing possible. • Summarize the steps involved in sequencing DNA. • Explain how genes are edited using the CRISPR-Cas9 system. • List some practical applications of transgenic organisms, PCR, and CRISPR-Cas9 gene editing.

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CH APTE R 3 From Genes to Proteins

− 23130

cut

anneal

ligate

9416 6557 4361

2322 2027

+ FIGURE 3.14 Digestion of bacteriophage λ DNA by the restriction enzyme HindIII. The restriction enzyme cleaves the DNA to produce eight fragments of defined size, six of which are large enough to be separated by electrophoresis in an agarose gel. The DNA was applied to the top of the gel, and the negatively charged DNA fragments moved downward through the gel toward the positive electrode. The fragments were visualized by binding a fluorescent dye. The numbers indicate the number of base pairs in each fragment. [Reprinted from www.neb.com, http:// www.neb.com; © 2012 with permission from New England Biolabs.]

Q Explain why the bands at the top appear brighter than the bands at the bottom.

SEE ANIMATED PROCESS DIAGRAM Construction of a recombinant DNA molecule

Vector

Recombinant DNA Foreign DNA

FIGURE 3.15 Production of a recombinant DNA molecule. A circular vector and a sample of DNA are cut with the same restriction enzyme, generating complementary sticky ends, so that the fragment of foreign DNA can be ligated into the vector.

Restriction enzymes have many uses in the laboratory. For example, they are indispensable for reproducibly breaking large pieces of DNA into smaller pieces of manageable size. Restriction digests of well-characterized DNA molecules, such as the 48,502-bp E. coli bacteriophage λ, yield fragments of predictable size that can be separated by electrophoresis, a procedure in which molecules move through a gel-like matrix such as agarose or polyacrylamide under the influence of an electric field. Because all the DNA segments have a uniform charge density, they are separated on the basis of their size (the smallest molecules move fastest; Fig. 3.14). A segment of DNA cut by a restriction enzyme or obtained by a method such as chemical synthesis or PCR (see below) can be joined to another DNA molecule. When different samples of DNA are digested with the same sticky end–generating restriction endonuclease, all the fragments have identical sticky ends. If the fragments are mixed together, the sticky ends can find their complements and re-form base pairs. The discontinuities in the sugar– phosphate backbone can then be mended by a DNA ligase (an enzyme that forms new phosphodiester bonds between adjacent nucleotide residues). DNA ligases that act on blunt-ended DNA segments are also available. The cutting-and-pasting process allows a desired DNA segment to be incorporated into a carrier DNA molecule called a vector, leaving an unbroken, fully base-paired recombinant DNA molecule (Fig. 3.15). Some commonly used vectors are derived from plasmids, which are small, circular DNA molecules present in many bacterial cells. A single cell may contain multiple copies of a plasmid, which replicates independently of the bacterial chromosome and usually does not contain genes essential for the host’s normal activities. However, plasmids often do carry genes for specialized functions, such as resistance to certain antibiotics (these genes often encode proteins that inactivate the antibiotics). An antibiotic resistance gene allows the selection of cells that harbor the plasmid: Only cells that contain the plasmid can survive in the presence of the antibiotic. A recombinant plasmid—one that contains a foreign DNA sequence—can be introduced into a bacterial host, where it multiplies as the bacteria multiply. This is one way to produce large amounts of a desired DNA segment. (The segment can be harvested by recovering the plasmids and treating them with the same restriction enzyme used to insert the foreign DNA.) DNA that is copied in this manner is said to be cloned. Note that a clone is simply an identical copy of an original. The term is used to refer either to a gene that has been amplified, as described here, or to a cell or organism that is genetically identical to its parent. An example of a plasmid used as a cloning vector is shown in Figure 3.16. This plasmid contains a gene (called ampR) for resistance to the antibiotic ampicillin and a gene (called lacZ) encoding the enzyme β-galactosidase, which catalyzes the hydrolysis of certain galactose derivatives. The lacZ gene has been engineered to contain several restriction sites, any one of which can be used as an insertion point for a piece of foreign DNA with compatible sticky ends. Interrupting the lacZ gene with foreign DNA prevents the synthesis of the β-galactosidase protein. Colonies of bacterial cells harboring the intact plasmid can be detected when their β-galactosidase cleaves a galactose derivative that generates a blue dye. Colonies of bacterial cells in which a foreign DNA insert has interrupted the lacZ gene are unable to cleave the galactose derivative and therefore do not turn blue (Fig. 3.17). A single white colony can then be removed from the culture plate and grown in order to harvest its recombinant DNA. Other screening techniques rely on signals such as fluorescence.

Tools and Techniques: Manipulating DNA

ampR

lacZ

EcoRI SacI KpnI AvaI XmaI SmaI BamHI XbaI SalI AccI HincII PstI SphI HindIII

FIGURE 3.16 Map of a cloning vector. This 2743-bp circular DNA molecule, called pGEM-3Z, has a gene for resistance to ampicillin so that bacterial cells containing the plasmid can be selected by their ability to grow in the presence of the antibiotic. The plasmid also has a site comprising recognition sequences for 14 different restriction enzymes. Inserting a foreign DNA segment at this site interrupts the lacZ gene, which encodes the enzyme β-galactosidase.

A wide variety of vectors have been developed to accommodate different sizes of DNA inserts and to deliver that DNA to different types of hosts, including human cells. If a gene that has been isolated and cloned in a host cell is also expressed (transcribed and translated into protein), it may affect the metabolism of that cell. The functions of some gene products have been assessed in this way. Sometimes a specific combination of vector and host cell are chosen so that large quantities of the gene product can be isolated from the cultured cells or from the medium in which they grow. This is an economical method for producing certain proteins that are difficult to obtain directly from human tissues (Table 3.6). Recombinant DNA can also be introduced into a different species to create a transgenic organism (Box 3.A).

The polymerase chain reaction amplifies DNA Cloning DNA in cultured cells is a laborious process. A much more efficient technique for amplifying a particular DNA sequence was developed by Kary Mullis in 1985: the polymerase chain reaction (PCR). One of the advantages of PCR over traditional cloning techniques is that the starting material need not be pure (this makes the technique ideal for analyzing complex mixtures such as tissues or biological fluids). PCR takes advantage of the enzyme DNA polymerase, which catalyzes the polymerization of nucleotides in the order determined by their base-pairing with a single-stranded DNA template. But because DNA polymerase cannot begin a new nucleotide strand (it can only

TA B L E 3.6

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

Coagulation factors IX and X

Treat hemophilia b and other bleeding disorders

Tissue plasminogen activator

Promote clot lysis following myocardial infarction or stroke

Colony stimulating factor

Promote white blood cell production after bone marrow transplant

71

FIGURE 3.17 Culture dish with recombinant bacteria. Blue colonies arise from cells whose plasmids have an intact β-galactosidase gene. White colonies arise from cells whose plasmids contain an inserted DNA that interrupts the β-galactosidase gene. Colonies that lack the plasmid (and its ampR gene) would also be white, but including ampicillin in the culture medium prevents their growth. [Courtesy S. Kopczak and D. P. Snustad, University of Minnesota.]

72

CH APTE R 3 From Genes to Proteins

Box 3.A Genetically Modified Organisms Introducing a foreign gene into a single host cell via a vector alters the genetic makeup of that cell and all its descendants. But 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, the modified DNA must be injected 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 produced by introducing recombinant DNA into a few cells, which can often develop into an entire plant whose cells all contain the foreign DNA. Desirable traits, such as resistance to insect pests and resistance to weed-killing herbicides, have been introduced into a number of important crop species. Approximately 80 to 90% of the U.S. corn (maize), soybean, and cotton harvest is genetically modified. Concerns about the safety of foods containing foreign genes have limited their acceptance by consumers. Transgenic organisms also present some biological risks. For example, genes that code for insecticides (such as the bacterial toxin intended to kill the insect larvae that would otherwise feast on corn 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. Transgenic plants have also been engineered for better nutrition. Golden Rice, for example, contains foreign plant genes that

encode the enzymes necessary to synthesize β-carotene (an orange pigment that is the precursor of vitamin A) and a gene for the iron-storage protein ferritin. This genetically modified rice could help alleviate vitamin A deficiencies (which afflict some 400 million people) and iron deficiencies (an estimated 30% of the world’s population suffers from iron deficiency).

Golden Rice, whose high β-carotene content gives the normally white rice grains a yellow hue. [phloen/Alamy Limited]

extend a preexisting chain), a short single-stranded primer that base pairs with the template strand is added to the mixture. This means that the polymerase copies only what it is “told” to copy. The PCR reaction mixture contains a DNA sample, DNA polymerase, all four deoxynucleotide substrates for the polymerase, and two synthetic oligonucleotide primers that are complementary to the 3ʹ ends of the two strands of the target DNA sequence. Keep in mind that the reaction mixture actually contains millions of molecules of each of these substances. In the first step of PCR, the sample is heated to ∼95°C to separate the DNA strands. Next, the temperature is lowered to about 55°C, cool enough for the primers to hybridize with the DNA strands. The temperature is then increased to about 75°C, and the DNA polymerase synthesizes new DNA strands by extending the primers (Fig. 3.18). The three steps—strand separation, primer binding, and primer extension—are repeated as many as 40 times. Because the primers represent the two ends of the target DNA, this sequence is preferentially amplified so that it doubles in concentration with each reaction cycle. For example, 20 cycles of PCR can theoretically yield 220 = 1,048,576 copies of the target sequence in a matter of hours. PCR relies on bacterial DNA polymerases that can withstand the high temperatures required for strand separation (these temperatures inactivate most enzymes). Commercial PCR kits usually contain DNA polymerase from Thermus aquaticus (“Taq,” which lives in hot springs) or Pyrococcus furiosus (“Pfu,” which inhabits geothermally heated marine sediments), since their enzymes perform optimally at high temperatures. One of the limitations of PCR is that choosing appropriate primers requires some knowledge of the DNA sequence to be amplified; otherwise the primers will not anneal with complementary sequences in the DNA and the polymerase will not be able to make any new DNA strands. However, since no new DNA is synthesized unless the primers can bind to the DNA sequence, PCR can be used to verify the presence of that sequence. Practical applications of PCR include identifying human genetic defects and diagnosing infections. PCR is the most efficient way to detect the presence of bacteria and viruses that are difficult or dangerous to work with, such as Ebola virus. Blood banks use PCR to test for the human immunodeficiency

Tools and Techniques: Manipulating DNA 3′ 5′ Target sequence

separate strands

Primers

anneal primers

Cycle 1 extend primers

separate strands

anneal primers Cycle 2 extend primers

separate strands

anneal primers Cycle 3 extend primers

FIGURE 3.18 The polymerase chain reaction. Each cycle consists of separation of DNA strands, binding of primers to the 3ʹ ends of the target sequence, and extension of the primers by DNA polymerase. The target DNA doubles in concentration with each cycle.

Q Indicate the temperature at which each step takes place.

virus (HIV), hepatitis viruses, and West Nile virus. Scientists have used PCR to amplify DNA from ancient bones, and, in the forensics laboratory, to analyze DNA of more recent origin (Box 3.B). If merely detecting a specific DNA sequence does not provide enough information, realtime PCR (also known as quantitative PCR, qPCR) can be employed. In this technique, the polymerase chain reaction continually generates new DNA sequences that bind to fluorescent probes, so that the amount of the DNA sequence can be monitored over time (rather than at the end of many reaction cycles, as in standard PCR). This approach is useful for quantifying

SEE ANIMATED PROCESS DIAGRAM The polymerase chain reaction

73

CH APTE R 3 From Genes to Proteins

Box 3.B DNA Fingerprinting Individuals can be distinguished by examining polymorphisms in their DNA. Current DNA fingerprinting methods use PCR to examine segments of repetitive DNA sequences, most often short tandem repeats of four nucleotides. The exact number of repeats varies among individuals, and each set of repeats, or allele, is small enough (usually less than 500 bp) that alleles differing by just one four-residue repeat can be easily differentiated. Because the first step of fingerprinting is PCR, only a tiny amount of DNA is needed—about 1 μg, or the amount present on a coffee cup or a licked envelope. And since the target segment is short, the purity and integrity of the DNA sample is usually not an issue. The locus, or region of DNA containing the repeats, is PCR-amplified using fluorescent primers that are complementary to the unique (nonrepeating) sequences flanking the repeats. In the example below, the two DNA segments have seven and eight tandem repeats of the AATG sequence. The PCR primers hybridize with sequences flanking the repeats (shaded blue), which are the same in all individuals. The amplified products are then separated according to size by electrophoresis and detected by their fluorescence. The results are compared to reference standards containing a known number of AATG repeats, from 5 to 10 in this example. Sample A comes from an individual with two copies of the 7-repeat allele and sample B from an individual with one copy of the 7-repeat allele and one copy of the 8-repeat allele (recall that humans are diploid, with two copies of each “gene”). Each of the loci that have been selected for forensic use generally have 7 to 30 different alleles. In a single sample of DNA, multiple loci can be amplified by PCR simultaneously, provided that the sizes of the PCR products are sufficiently different that they will not overlap during electrophoresis. Alternatively, each PCR primer can bear a different fluorescent dye. The probability of two individuals having matching DNA fingerprints depends on the number of loci examined and the number of possible alleles at each locus. For example, assume that one locus has 20 alleles and that each allele has a frequency

in the population of 5% (1 in 20, or 1/20). Another locus has 10 alleles, and each has a frequency of 10% (1 in 10, or 1/10). The probability that two individuals would match at both sites is 1/20 × 1/10 = 1/200 (the probabilities of independent events are multiplied). If multiple loci are examined, the probabilities can reach the range of 1 in a million or more. For this reason, most courts now consider DNA sequences to be unambiguous identifiers of individuals.

Sample A

Fluorescence

74

Sample B

Reference

5

6

7

8

9

10

Size of PCR product

Q Would a child’s DNA fingerprint match the parents’ DNA fingerprints?

AATG AATG AATG AATG AATG AATG AATG

AATG AATG AATG AATG AATG AATG AATG AATG

infectious agents such as bacteria and viruses. Real-time PCR methods are also used to assess the level of gene expression in cells: Cellular mRNA is first reverse-transcribed to DNA, then a specific DNA sequence is amplified by PCR. A gene’s level of expression is sometimes reported relative to that of a gene that encodes a protein such as actin (Section 5.3), which is typically produced at constant levels in cells and can therefore serve as a sort of benchmark.

DNA sequencing uses DNA polymerase to make a complementary strand Determining the sequence of nucleotides in a DNA molecule involves using DNA polymerase to make copies. Various methods have been devised to detect the newly made DNA, such as labeling it with radioactive or fluorescent tags. The key, of course, is to identify the added

75

Tools and Techniques: Manipulating DNA Single-strand DNA template C G A T C G A A T C G T C G G A C A G

G C T A

Primer

G ATP

PPi

1. DNA polymerase incorporates a complementary nucleotide into the new DNA chain (extends the primer).

2. Sulfurylase converts the PPi reaction product to ATP.

light

3. Luciferase uses the ATP to power a reaction that produces light.

Bead FIGURE 3.19 Pyrosequencing. PCR-amplified copies of a single-stranded template DNA are attached to a plastic bead, and a primer, DNA polymerase, and other components are added. Nucleotide solutions are passed, one at a time, over the beads. Polymerization of a nucleotide leads to a flash of light.

nucleotides one at a time in order to reconstruct the sequence of the original (template) DNA. The newest, most efficient DNA sequencing methods require specialized instruments, but they operate extremely quickly, sequence thousands of DNA segments simultaneously, and can handle huge tasks, such as sequencing an organism’s entire genome within hours or days. TGGCTTTGTTGCAGTTG In pyrosequencing (also known as 454 sequencing), DNA T 2G C 3T segments to be sequenced are immobilized on tiny plastic beads, one segment per bead, and then a variation of PCR is used to make single-stranded copies that are attached to the same bead. DNA polymerase, a primer, and other reaction components are added to this template DNA. A solution of one of the four deoxynucleotides is then introduced to each bead, and if the nucleotide is a match—that is, if it can pair with the nucleotide in the immobilized template strand—it is incorporated into a new DNA chain. This polymerization reaction releases pyrophosphate (the diphosphate portion of the nucleotide, represented as PPi). An enzyme T C AG T C A G T C called sulfurylase converts the PPi product to ATP, which is used by another enzyme, firefly luciferase, in a reaction that produces a flash of light (Fig. 3.19). A detector records the light generated by each bead as solutions of each of the four nucleotides are successively washed over the beads. If an incoming nucleotide does not pair with the template DNA, no polymerization reaction occurs and no light is produced. If an incoming nucleotide is incorporated into the new DNA strand more than once, the flash of light is proportionately brighter. The record of light produced by each bead reveals the sequence of nucleotides corresponding to the template DNA strand attached to that bead (Fig. 3.20). The Illumina sequencing method uses a slightly different approach. First, segments of DNA are attached to a glass surface and amplified in place. After the addition of DNA polymerase and a primer, a solution containing all four deoxynucleotides, each with a different fluorescent group, is introduced. Unreacted nucleotides are washed away, and the nucleotides that remain—those that DNA polymerase incorporated into a new strand complementary to the template DNA—can be identified by their fluorescence. Before the next batch of nucleotides is delivered, the fluorescent groups on the growing DNA strands are detached so that new fluorescent signals can be detected in the next round of polymerization. The order of

G 2T

G

C A G 2T

G

A G T C A GT C A G TC A G FIGURE 3.20 Results of pyrosequencing. Nucleotides were added to the reaction system in the order shown at the bottom. A flash of light (peak) indicates that the nucleotide was polymerized. A double- or triple-height peak indicates multiple incorporations of a nucleotide. The absence of a peak means that no reaction occurred, because the complementary nucleotide did not occur in the template DNA. The deduced sequence of the newly made DNA strand is shown at the top.

Q What is the sequence of the template DNA in this example?

76

CH APTE R 3 From Genes to Proteins Single-strand DNA template C G A T C G A A T C G T C G G A C A G

G C T A

Primer A G T C

1. DNA polymerase incorporates a complementary nucleotide into the new DNA chain and unreacted nucleotides are washed away.

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

G C T A G

2. The incorporated nucleotide is detected by its fluorescence, then the fluorescent group is removed.

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

G C T A G

A G T C

3. A new solution of nucleotides is introduced.

Glass FIGURE 3.21 Illumina sequencing. DNA polymerase uses immobilized single strands of DNA as templates to add a fluorescent nucleotide to a new strand. A detector records the fluorescence to identify the nucleotide added in that round of polymerization.

appearance of fluorescent signals corresponds to the nucleotide sequence of the growing DNA chain (Fig. 3.21). Both of the sequencing methods described here generate “reads” of no more than a few hundred nucleotides, so deducing the sequence of a longer segment of DNA requires that the original sample be randomly broken into multiple short segments that are individually sequenced (this is known as a “shotgun” approach). The original sequence can then be reconstructed by computer analysis of the overlapping sequences. Original sequence T G C C G A T A A G C T G C G G A T T C C G T A A G T A C A G

Overlapping fragments T G C C G A T A A G C T G C G G A T T C C G T A A G T A C A G T G C C G A T A A G C T G C G G A T T C C G T A A G T A C A G T G C C G A T A A G C T G C G G A T T C C G T A A G T A C A G

The ability to rapidly sequence a huge amount of DNA, such as the genome of one individual, makes it possible to compare the DNA of children and parents to identify a genetic defect present only in the children. This approach has likely already saved lives in cases where no other diagnostic tests were available. Genomic data also provide the raw material for exploring the vast amounts of DNA that do not encode proteins and for comparing the genomes of different species in order to elucidate their evolutionary relationships.

DNA can be altered In addition to rearranging, copying, and sequencing DNA, genetic engineers can modify the DNA in order to study how genes work, to make new gene products, and even to change the genetic makeup of an entire organism. DNA is constantly and randomly changing as a result of natural processes, but alterations can be introduced much more quickly and in a targeted manner in the laboratory. For example, researchers can produce genes with specific mutations

Tools and Techniques: Manipulating DNA

so that after the genes are transcribed and translated in a cell, the encoded proteins exhibit altered structures and functions. One method for such site-directed mutagenesis of a gene is to perform a variation of PCR using oligonucleotide primers that are synthesized with the desired nucleotide changes. Several rounds of PCR yield a population of newly synthesized DNA molecules that contain the mutations specified by the primers. The altered DNA can then be introduced into a host cell. An even more efficient technique for modifying genes takes advantage of a bacterial system that evolved to destroy bacteriophages. A bacterial cell carries traces of its ancestors’ encounters with various bacteriophages in the form of multiple short segments of phage DNA that have been inserted into the bacterial chromosome among a set of clustered regularly interspersed short palindromic repeats, or CRISPRs. Remnants of bacteriophage DNA

CRISPR repeats

When this DNA is transcribed, the resulting RNA is cleaved into short segments, each corresponding to about 30 nucleotides of phage DNA. These RNAs then bind to a nuclease called Cas9 (for CRISPR-associated) and guide it to complementary DNA sequences, which presumably represent bacteriophages trying to infect the cell. The Cas9 nuclease cleaves both strands of this DNA, destroying the incoming bacteriophage. Without the CRISPR RNA guide, Cas9 cannot cleave DNA, but it will cleave whatever DNA is complementary to the guide RNA. This creates the ability to use the CRISPR-Cas9 system as a gene-editing tool. Using recombinant DNA technology, the Cas9 gene and an engineered guide RNA “gene” are introduced into a host cell. The cell makes the encoded Cas9 protein and the guide RNA, which act to cleave the DNA of the gene that is complementary to the guide RNA (Fig. 3.22).

Cas9 Cas Ca Cas9

Guid Guide RNA ȗ
ȗ
ȗ ȗ
ȗ
ȗ ȗ
ȗ
ȗ ȗ
ȗ
ȗ

DNA cleavage ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

DNA repair

DNA repair Replacement DNA

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

ȗ
ȗ
ȗ

Inactivated gene

ȗ
ȗ
ȗ ȗ
ȗ
ȗ

Altered gene

FIGURE 3.22 The CRISPR-Cas9 gene-editing system. A guide RNA containing a 22–23-bp segment that is complementary to the target DNA positions Cas9 to cleave the DNA at the points marked by arrowheads. This leads to permanent inactivation of the gene (left) or, if an altered DNA segment is provided, leads to replacement of the original gene sequence with the altered sequence (right).

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CH APTE R 3 From Genes to Proteins

Cells have various mechanisms for repairing broken DNA (described in more detail in Section 20.3). For example, the severed pieces are typically knit back together—but imperfectly, so the repair does not restore the original DNA sequence. When this occurs in the middle of a gene, the result is almost always an unreadable gene. The CRISPR-Cas9 system can therefore “knock out” a specific gene in an organism. To “knock in” a gene, the target gene is replaced by a modified version of the gene. In this case, researchers deploy the CRISPR-Cas9 system to cleave the target gene but also introduce an altered segment of that same gene. The cell then undertakes a more elaborate repair process to rebuild the cleaved DNA using the intact—but altered—sequence as a blueprint. The result is a functional gene containing a modified DNA sequence. Because the CRISPR-Cas9 gene-editing system can permanently disable a defective gene or replace a defective gene with a normal copy at the editing site, it has become the preferred approach for genetically modifying organisms. For example, CRISPR-Cas9 technology can generate animal models for human genetic diseases. It is also an option for gene therapy, which introduces a functional gene into an individual in order to compensate for or correct a malfunctioning gene. In traditional gene therapy, an extra gene is delivered to a patient’s cells, usually by a viral vector (viruses are already efficient at delivering DNA into mammalian cells). The first successful gene therapy trials treated children with severe combined immunodeficiency (SCID), a normally fatal condition caused by a single-gene defect. Bone marrow cells were removed from each patient and cultured in the presence of a viral vector containing a normal version of the defective gene. When the modified cells were infused into the patient, they differentiated into functional immune system cells. Some of the diseases treated by gene therapy are listed in Table 3.7. Despite over 30 years of effort, the list of successes is still small. For one thing, the viral vectors used in gene therapy 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. Another challenge is to deliver the therapeutic gene to a sufficient number of the appropriate type of host cells in order to “correct” the disease over the long term. The CRISPR-Cas9 method shares some of the limitations of traditional gene therapy, such as the need for a suitable vector that can deliver its payload to the intended tissues. In addition, Cas9 sometimes cleaves DNA sequences that are not perfectly complementary to its guide RNA. Such “off target” gene inactivation could have disastrous unintended consequences. However, CRISPR-Cas9 offers some clear advantages. This gene-editing system can actually inactivate a misbehaving gene as well as introduce a replacement gene, and it can permanently correct the host cell’s DNA. Possible applications of CRISPR-Cas9–based gene therapy include genetic diseases such as those in Table 3.7 as well as conditions such as Huntington’s disease, where the traditional approach of adding a normal gene would be ineffective because the defective gene is active and must be disabled. HIV infection could potentially be treated by inactivating the CCR5 gene for the cell-surface protein that the virus latches onto when it infects a cell. As with other developments in genetic engineering, researchers and clinicians must carefully weigh the ethical implications of this new gene-editing technology. A major challenge is anticipating the off-target effects of Cas9-mediated DNA cleavage. And because the genetic changes are permanent, the research community recognizes the need to proceed cautiously.

TAB L E 3. 7

Some Hereditary Diseases Treated by Gene Therapy

DISEASE

SYMPTOMS

Adrenoleukodystrophy

Neurodegeneration

Hemophilia

Bleeding

Leber’s congenital amaurosis

Blindness

Severe combined immunodeficiency (SCID)

Immunodeficiency

β-Thalassemia

Anemia

Wiskott-Aldrich syndrome

Immunodeficiency

Key Terms

79

Because a cell can transmit its modified DNA to its daughter cells when the cell divides, it is prudent to limit gene alterations to somatic (body) cells and avoid altering the DNA in reproductive cells that pass DNA to the next generation. BEFORE GOING ON • Make a list of the reagents and equipment needed to carry out each of the procedures outlined in this section. • Summarize the steps involved in each process. • Explain why DNA sequencing and DNA fingerprinting rely on PCR. • Using Fig. 3.18, draw the products of the fourth PCR cycle. • Sketch the pyrosequencing results (as in Fig. 3.20) for the DNA shown in Fig. 3.19. • Make a list of the commercial and therapeutic applications of the techniques described in this section.

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.

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.

• Genetic variations can be linked to human diseases even when specific disease genes have not been identified.

3.5

Tools and Techniques: Manipulating DNA

• DNA molecules can be reproducibly fragmented by the action of restriction enzymes, which cleave DNA at specific sequences. • DNA fragments can be joined to each other to generate recombinant DNA molecules that are then introduced into host cells. • A DNA segment can be amplified many times by the polymerase chain reaction, in which a DNA polymerase makes complementary copies of the target DNA. • Modern techniques for determining the sequence of nucleotides in a segment of DNA immobilize DNA segments, amplify them, use DNA polymerase to add nucleotides to a new complementary chain, and detect each added nucleotide by a flash of light or a fluorescent signal.

• The sequence of nucleotides in a segment of DNA can reveal mutations that cause disease.

• The gene-editing CRISPR-Cas9 system uses the bacterial Cas9 nuclease to cut target DNA at a sequence complementary to a guide RNA. Subsequent DNA repair yields an inactivated gene or a gene with an altered sequence.

3.4 Genomics

• In gene therapy, a normal gene is introduced or a gene is edited in order to cure a genetic disease.

• Genomes contain variable amounts of repetitive and other forms of noncoding DNA in addition to genes, which are identified by their sequence characteristics or similarity to other genes.

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

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CH APTE R 3 From Genes to Proteins

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 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 polygenic disease monogenic disease diploid haploid transposable element moderately repetitive DNA

highly repetitive DNA open reading frame (ORF) homologous genes orphan gene genome map noncoding RNA (ncRNA) horizontal gene transfer genomics single-nucleotide polymorphism (SNP) genome-wide association study (GWAS) restriction endonuclease bacteriophage palindrome sticky ends blunt ends restriction digest electrophoresis DNA ligase

vector recombinant DNA plasmid selection clone polymerase chain reaction (PCR) transgenic organism primer real-time PCR DNA fingerprinting allele locus pyrosequencing PPi Illumina sequencing site-directed mutagenesis CRISPR CRISPR-Cas9 system gene therapy

Bioinformatics Brief Bioinformatics Exercises

3.4 Analysis of Genomic DNA

3.1 Drawing and Visualizing Nucleotides

3.5 Restriction Enzyme Mapping

3.2 The DNA Double Helix 3.3 Melting Temperature and the GC Content of Duplex DNA

Bioinformatics Project Databases for the Storage and “Mining” of Genome Sequences

Problems 3.1

Nucleotides

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

Problems

5. Describe the chemical difference between uracil and thymine. 6. Many cellular signaling pathways involve the conversion of ATP to cyclic AMP, in which a single phosphate group is esterified to both C3ʹ and C5ʹ. Draw the structure of cyclic AMP.

22. The adenine derivative hypoxanthine can base pair with cytosine, adenine, and uracil. Show the structures of these base pairs.

N

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

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-2ʹ-deoxyuridine was recently reported. Draw the structure of this compound. For what base is this compound a substitute?

3.2 Nucleic Acid Structure 13. a. What kind of linkage joins the two nucleotides in the dinucleotides NAD and FAD (see Fig. 3.2)? b. How do the adenosine groups in FAD and CoA differ? 14. Draw a CA (ribo)dinucleotide and label the phosphodiester bond. How would the structure differ if it were DNA? 15. 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 2ʹ-OH of GMP and the 5ʹ-phosphate of AMP and the other between the 3ʹ-OH of AMP and the 5ʹ-phosphate of GMP. Draw this dinucleotide. 16. Another intracellular signaling molecule is cyclic ADP-ribose (cADPR), in which the two phosphate groups at the 5ʹ position of ADP are linked to a second ribose at its 5ʹ carbon. This second ribose in turn is covalently linked to N1 of the ADP. Draw this nucleotide. 17. 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. 18. A well-studied bacteriophage has 97,004 base pairs in its complete 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? 19. Do Chargaff’s rules hold true for RNA? Explain why or why not. 20. The complete genome of a virus contains 1578 T residues, 1180 G residues, 1609 A residues, and 1132 C residues. What can you conclude about the structure of the viral genome, given this information? 21. Identify the base pair highlighted in blue in Figure 3.6.

NH

O

7. In some organisms, DNA is modified by addition of methyl groups. Draw the structure of 5-methylcytosine. 8. In certain pathogenic bacteria, the methylation of certain adenines in DNA is required in order for the bacteria to cause disease. a. Draw the structure of N 6-methyladenine. b. Why might scientists be interested in studying the bacterial enzyme N 6-DNA methyltransferase, which catalyzes the transfer of methyl groups to adenine?

81

N

N

H Hypoxanthine

23. 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. 24. Hydrogen bonding does not make a significantly large contribution to the overall stability of the DNA molecule. Explain. 25. How can the hydrophobic effect (Section 2.2) explain why DNA adopts a helical structure? 26. a. Would you expect proteins to bind 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. Why do histones have a high affinity for DNA?

O HN

CH

C

O NH

CH

CH2

CH2

CH2

CH2

CH2

CH2

CH2

NH

NH 3

C

Lysine

C

NH 2

NH2

Arginine

27. a. What is the Tm of the DNA sample whose melting curve is shown in Figure 3.7? b. Draw melting curves that would be obtained from the DNA of Dictyostelium discoideum and Streptomyces albus (see Table 3.2). 28. a. Using Table 3.2 as a guide, estimate the melting temperature of the DNA from an organism whose genome contains equal amounts of all four nucleotides. b. To what temperature would you have to cool the DNA to allow it to reanneal? 29. What might you find in comparing the GC content of DNA from Thermus aquaticus or Pyrococcus furiosus and DNA from bacteria in a typical backyard pond? 30. Explain why the melting temperature of a sample of doublehelical DNA increases when the Na+ concentration increases. 31. 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? 32. In the laboratory technique known as fluorescence in situ hybridization (FISH), a fluorescent oligonucleotide probe is allowed 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.

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3.3

CH APTE R 3 From Genes to Proteins

The Central Dogma

33. 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. 34. The semiconservative nature of DNA replication (as shown in Fig. 3.9) was proposed by Watson and Crick in 1953, but it wasn’t 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? 35. 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. 36. A portion of a gene is shown below. 5ʹ-ATGATTCGCCTCGGGGCTCCCCAGTCGCTGGTGCT3ʹ -TACTAAGCGGAGCCCCGAGGGGTCAGCGACCACGAGCTGACGCTGCTCGTCG-3ʹ CGACTGCGACGAGCAGC-5ʹ The sequence of the mRNA transcribed from this gene has the following 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 sequences of amino acids 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 assist you in your identification of the correct reading frame? AGGTCTTCAGGGAATGCCTGGCGAGAGGGGAGCAGCTGGTATCGCTGGGCCCAAAGGC 42. 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. 43. 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. 44. Is it possible for the same segment of DNA to encode two different proteins? Explain. 45. A portion of the nucleotide sequence from the DNA coding strand of the chick ovalbumin gene is shown. What is the partial amino acid sequence of the encoded protein? CTCAGAGTTCACCATGGGCTCCATCGGTGCAGCAAGCATGGAA-(1104 bp)-TTCTTTGGCAGATGTGTTTCCCCTTAAAAAGAA

5ʹ-AUGAUUCGCCUCGGGGCUCCCCAGUCGCUGGUGCUGCUGACGCUGCUCGUCG-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. 37. 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? 38. Har Gobind Korana extended Nirenburg’s work 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 37? b. Do these results help to decipher the identities of the codons involved? 39. How many different codons are possible in nucleic acids containing four different nucleotides if a codon consisted 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? 40. 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? 41. An open reading frame (ORF) is a portion of the genome that potentially codes for a protein. A given mRNA nucleotide sequence

46. A type of gene therapy called RNA interference (RNAi) is being investigated to treat Huntington’s disease. This disease is caused by a mutation in the DNA that results in the synthesis of an altered protein that leads to nervous system defects. To treat this disease, scientists synthesize short sequences of RNA (siRNA, or small interfering RNA) that form base pairs with the mRNA that codes for the mutated protein. a. Design an siRNA that will interfere with the synthesis of the protein shown in Problem 45. b. Explain how the addition of the siRNA will prevent the synthesis of the protein. c. What are the difficulties that must be overcome in order for RNA interference to be an effective technique for treating the disease? 47. The disease cystic fibrosis is the result of a mutation in the gene that encodes the cystic fibrosis transmembrane regulator (CFTR), a channel 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. normal gene sequence

504 505 506 507 508 509 510 511 512 · · · GAA AAT ATC ATC TTT GGT GTT TCC TAT · · ·

mutated gene · · · GAA AAT ATC AT– – –T GGT GTT TCC TAT · · · sequence

a. What are the sequences of the normal and mutated proteins? b. In another patient with a less severe form of the disease, the CF gene has the sequence ...AAT AGA TAC AG... (the normal sequence of the CF gene in this region is ...AAT ATA GAT ACA G...). How has the DNA sequence changed and how does this affect the encoded protein? 48. 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

Problems

. . . CTGGCCGGCTCAATGGAGAGTCC . . .

3.4 Genomics 49. 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? 50. 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. 51. 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? 52. When genomes of various organisms were sequenced, biologists expected that the DNA content (the C-value) would always

56. 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. 57. 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 –1og10 P-value of 7 or greater is assumed to be associated with the disease. a. What are the locations on the chromosome that show the strongest correlation with the disease? b. Which genes contain SNPs associated with the disease and which do not? gene A

gene B

gene C

14 12 10 8 6 4 2 0

−log10 P-value

receptor, resulting in hypercholesterolemia that 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.

83

67,300,000

67,400,000

67,500,000

58. 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 –1og10 P-value of 5 was used as the cutoff.

22.5 20.0

−og10 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

Chr12

Chr22

Chrx

be positively correlated with organismal complexity. But 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?

3.5

Tools and Techniques: Manipulating DNA

53. 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?

61. The sequence of a segment of the pET28 plasmid is shown below. Which of the restriction enzymes shown in Table 3.5 could cleave this DNA segment to insert a foreign gene into the plasmid?

59. Which restriction enzymes in Table 3.5 generate sticky ends? Blunt ends? 60. Which is more likely to be called a “rare cutter”: a restriction enzyme with a four-base recognition sequence or a restriction enzyme with an eight-base recognition sequence?

CGCGGATCCGAATTCGAGCTCCGTCGACAAGCTTGCGTATGGGATGGCTGAGTACAGCACGTTGAATGAGGCGAT-

GCCGCACTCGAG

GGCCGCTGGTGATG 54. The bacteriophage DNA described in Problem 53 contains 59 kb and 105 ORFs. None of the ORFs codes for tRNAs. How does the bacteriophage replicate its DNA and synthesize the structural proteins necessary to replicate itself? 55. If each person’s genome contains a SNP every 300 nucleotides or so, how many SNPs are in that person’s genome?

62. A sample of the DNA segment shown below is treated with the restriction enzyme MspI. Next, the sample is incubated with an exonuclease that acts only on single-stranded polynucleotides. What mononucleotides will be present in the reaction mixture? TGCT TAGCCGGAACGA ACGAATCGGCCT T GCT

84

CH APTE R 3 From Genes to Proteins

63. The plasmid pGEM-3Z shown in Figure 3.16 has three restriction sites for the enzyme HaeII, which cleaves at locations 323, 693 and 2564. Draw the result of agarose gel electrophoresis (see Figure 3.14) if the sample contains a. linear or b. circular plasmid digested with HaeII. 64. Restriction enzymes are used to construct restriction maps of DNA. These are diagrams of specific DNA molecules that show the sites where the restriction enzymes cleave the DNA. To construct a restriction map, purified samples of the DNA are treated with restriction enzymes, either alone or in combination, and then the reaction products are separated by agarose gel electrophoresis. Use the results of the agarose gel electrophoresis shown here to construct a restriction map for the sample of DNA. Control EcoRI

PstI

EcoRI + PstI

20 kb

15 kb

11 kb 9 kb 8 kb 7 kb

3 kb 2 kb

65. Design a 10-bp primer that could be used to amplify the following sequence of DNA: 5ʹ-AGTCGATCCCTGATCGTACGCTACGGTAACGT-3ʹ 66. The primer used in sequencing a cloned DNA segment often includes the recognition sequence for a restriction endonuclease. Explain. 67. Refer to the DNA sequence shown in Problem 36. Design two 18-bp primers that could be used to amplify this gene segment. 68. Refer to the DNA sequence shown in Problem 45. a. Design two 18-bp PCR primers that could be used to amplify this gene. b. Suppose you wanted to add EcoRI restriction sites to each end of the gene. How would you modify the sequences of the PCR primers you designed in part a to amplify the gene? 69. The “reverse” primer, which pairs with the coding strand, for the DNA segment shown in Problem 36 has a GC content of 67% (see

Solution 67). Why might a primer with a high GC content be problematic in PCR? 70. Could you perform PCR with an ordinary DNA polymerase, that is, one that is destroyed by high temperatures? What modifications would you make in the PCR protocol? 71. Examine the sequence of the protein in Solution 45. Assume that the corresponding DNA sequence is not known. Using the amino acid sequence as a guide, design a pair of nine-base deoxynucleotide primers that could be used for PCR amplification of the protein-coding portion of the gene. (Hint: DNA polymerase can extend a primer only from its 3ʹ end.) How many different pairs of primers could you choose from? 72. A researcher trying to identify the gene for a known protein might begin by looking closely at the protein’s sequence in order to design a single-stranded oligonucleotide probe that will hybridize with the DNA of the gene. Why would the researcher focus on a segment of the protein containing a methionine (Met) or tryptophan (Trp) residue (see Table 3.3)? 73. The following DNA fragments were produced by using the “shotgun” method. Show how the fragments should be aligned to determine the sequence of the original DNA. ACCGTGTTTCCGACCG ATTGTTCCCACAGACCG CGGCGAAGCATTGTTCC TTGTTCCCACAGACCGTG 74. A group of investigators is interested in studying the gp41 protein from the human immunodeficiency virus (HIV). In order to do this, they use site-directed mutagenesis to synthesize a series of truncated proteins. A partial sequence of the gp41 protein is shown. Design an 18-bp “mismatched primer” that could be used to synthesize a truncated protein that would terminate after the Leu residue at position 700.

700 . . . WYIKLFIMIVGGLVGLRIVFAVLSIVNRVRGGYSP . . . 75. In a CRISPR-Cas9 gene knock-in experiment, the replacement DNA sequence must differ from the original DNA sequence. Explain. 76. Researchers have devised a way to “turn on” specific genes by linking a transcription-activating protein to a version of Cas9 that cannot cleave DNA (it is called dCas9, where d stands for dead). What else is needed to turn a target gene on?

Selected Readings Dickerson, R. E., DNA structure from A to Z, Methods Enzymol. 211, 67–111 (1992). [Describes the various crystallographic forms of DNA.] Doudna, J.A. and Charpentier, E., The new frontier of genome engineering with CRISPR-Cas9, Science 346, 1047 (2014) and DOI: 10.1126/science.1258096. [A review by two of the scientists who developed the CRISPR-Cas9 method; the full article is available at http://dx.doi.org/10.1126/science.1258096.] International Human Genome Sequencing Consortium, Initial sequencing and analysis of the human genome, Nature 409, 860–921 (2001) and Venter, J. C., et al., The sequence of the human genome, Science 291, 1304–1351 (2001). [These and other papers in the same issues of Nature and Science describe the data that constitute the draft sequence of the human genome and discuss how this information can be used in understanding biological function, evolution, and human health.]

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.] Naldini, L. Gene therapy returns to centre stage, Nature 526, 351– 360 (2015). [Describes the various approaches for gene therapy and provides details on methods that have been successful in patients.] Rigden, D. J., Fernández-Suárez, X. M., and Galperin, M. Y., The 2016 database issue of Nucleic Acids Research and an updated molecular biology database collection. Nucl. Acids Res. 44, D1– D6 (2016). [Introduces the annual database issue, which includes a summary of bioinformatics resources and some additions to the list of 1685 molecular biology databases. The entire issue is available at http://nar.oxfordjournals.org/.] Thieffry, D., Forty years under the central dogma, Trends Biochem. Sci. 23, 312–316 (1998). [Traces the origins, acceptance, and shortcomings of the idea that nucleic acids contain biological information.]

CHAPTER 4

The Asahi Shimbun/Getty Images

Protein Structure

Many species of tree frog protect their newly laid eggs with the help of proteins called ranaspumins, which are usually secreted by the female and whipped into a frothy coating by the male. Foaming is an unusual characteristic for proteins, since most proteins lose their function when this occurs.

DO YOU REMEMBER? • 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 3.2).

Proteins are the workhorses of the cell. 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 the protein chain folds into a unique three-dimensional shape. Finally, the Tools and Techniques section of this chapter examines some of the procedures for purifying and sequencing proteins and determining their structures.

85

86

CH APTE R 4 Protein Structure

L EARNING 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.

Amino Acids, the Building Blocks of Proteins 4.1

Proteins come in a huge variety of shapes and sizes (Fig. 4.1), but they are all built 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)

Maltoporin (E. coli) Permits sugars to cross the bacterial cell membrane (more in Section 9.2) 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. [Structure of insulin (pdb 1ZNI) determined by M. G. W. Turkenburg, J. L. Whittingham, G. G. Dodson, E. J. Dodson, B. Xiao, and G. A. Bentley; structure

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

Phosphoglycerate kinase (yeast) Catalyzes one of the central reactions in metabolism (more in Section 13.1) of maltoporin (pdb 1MPM) determined by R. Dutzler and T. Schirmer; structure of phosphoglycerate kinase (pdb 3PGK) determined by P. J. Shaw, N. P. Walker, and H. C. Watson; structure of DNA polymerase (pdb 1KFS) determined by C. A. Brautigan and T. A. Steitz; and structure of plastocyanin (pdb 1PND) determined by B. A. Fields, J. M. Guss, and H. C. Freeman.]

Amino Acids, the Building Blocks of Proteins

87

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: 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.

Hydrophobic amino acids COO⫺ CH3

COO⫺ H

C

CH3

H

NH⫹ 3

COO⫺ H

C

H

CH3

H

CH3

H

NH⫹ 3

CH3

Leucine (Leu, L)

H

C

CH2

CH2

CH2

S

N H

Tryptophan (Trp, W) COO⫺ CH2 H C CH2 H2N⫹ CH2

CH3

NH⫹ 3

Isoleucine (Ile, I)

C

NH⫹ 3

COO⫺

CH CH2

C

CH2

Phenylalanine (Phe, F)

COO⫺ CH3

CH3

C

COO⫺

NH⫹ 3

Valine (Val, V)

CH2 CH

NH⫹ 3

CH

C NH⫹ 3

Alanine (Ala, A)

COO⫺

Methionine (Met, M)

Proline (Pro, P)

Polar amino acids COO⫺ CH3

COO⫺ H

C

CH2 OH

H

NH⫹ 3

H

C

OH

H

NH⫹ 3

COO⫺

O

CH2 C

NH2

H

NH⫹ 3

C

CH2 CH2

CH2

OH

H

C

NH2

H

C

Glutamine (Gln, Q)

SH

COO⫺ N

CH2

NH⫹ 3

CH2

Cysteine (Cys, C)

COO⫺

O

C

NH⫹ 3

Tyrosine (Tyr, Y)

NH⫹ 3

Asparagine (Asn, N)

C

COO⫺

NH⫹ 3

Threonine (Thr, T)

Serine (Ser, S) COO⫺

CH

C

COO⫺

H

N H

C

H

NH⫹ 3

Histidine (His, H)

Glycine (Gly, G)

Charged amino acids COO⫺ H

C

COO⫺

O

CH2 C

O

NH⫹ 3

Aspartate (Asp, D)

⫺

H

C

COO⫺

O

CH2 CH2 C

NH⫹ 3

Glutamate (Glu, E)

O

⫺

H

C

COO⫺

CH2 CH2

CH2

CH2

NH⫹ 3

H

NH⫹ 3

C

NH2

CH2 CH2 CH2 NH C

NH⫹ 3

Lysine (Lys, K)

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.

Q Identify the functional groups in each amino acid. Refer to Table 1.1.

Arginine (Arg, R)

NH⫹ 2

88

CH APTE R 4 Protein Structure

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

O N

O

H

O

C 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. 3.4). The lessons learned from thalidomide have made drugs more costly to develop and test but have also made them safer.

O Thalidomide

Q Which of the 20 amino acids is not chiral?

Nineteen of the twenty 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⫺ H3N⫹

C␣ CH3

H

COO⫺ H

C␣

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 the 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. 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),

Amino Acids, the Building Blocks of Proteins

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

CH2

HC

N H

NH⫹ 3

COO⫺

H⫹

N

H

⫹

NH⫹

CH2 N H

NH⫹ 3

Base

Acid

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

SH

HC 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⫺ HC

COO⫺

CH2

S

S

CH2

NH⫹ 3

CH 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).

89

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CH APTE R 4 Protein Structure

Box 4.B Monosodium Glutamate processed foods as a flavor enhancer. For example, a low-salt food item can be made more appealing by adding MSG to it. According to some popular accounts, “Chinese restaurant syndrome” can be attributed to the consumption of excess MSG added to prepared foods or present in soy sauce (MSG is also naturally present in many other foods, including cheese and tomatoes). The symptoms of the syndrome reportedly include muscle tingling, headache, and drowsiness—all of which could potentially reflect the role of glutamate in the nervous system. However, a definitive link between MSG intake and neurological symptoms has not been demonstrated in scientific studies and therefore remains mostly anecdotal.

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. But 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 the form of monosodium glutamate (MSG) is sometimes added to

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

Peptide bonds link amino acids in proteins SEE GUIDED TOUR Protein Structure

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):

R1 ⫹

H3N

C

H

O ⫹

C

H N

⫺

O

H

R2

⫹

O

C

C O⫺

H H H2O

⫹

H3N

R1 O

R2

C

C

C N

H 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). 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):

R1 N-terminus

⫹

H3N

O

R2

O

R3

O

R4

O

CH C N CH C N CH C N CH C H

Residue 1

H Residue 2

O⫺

C-terminus

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

Amino Acids, the Building Blocks of Proteins

pK Values of Ionizable Groups in Amino Acids

TA B L E 4.1 GROUP a

C-terminus

pK

COOH

3.5

O CH2

Asp

C

OH

3.9

O Glu

CH2

His

CH2

CH2

C

OH

4.1

NH⫹ 6.0

N H Cys

CH2

N-terminus

NH⫹ 3

Tyr

CH2

Lys

CH2

SH

8.4 9.0

OH CH2

CH2

10.5

NH⫹ 3

CH2

10.5

NH2 CH2

Arg

CH2

CH2

NH

C

NH⫹ 2

12.5

a

The ionizable proton is indicated in red.

polypeptide therefore depend primarily on the identities of the side chains (R groups) that project out from the polypeptide backbone. 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.

91

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CH APTE R 4 Protein Structure

SAMP LE CA LCU LAT I O N 4 . 1 SEE SAMPLE CALCULATION VIDEOS

Problem Estimate the net charge of the polypeptide chain below at physiological pH (7.4) and at pH 5.0. 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 N-terminus Arg His Asp C-terminus net charge

Charge +1 +1 0 –1 –1 0

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

Group N-terminus Arg His Asp C-terminus net charge

Charge +1 +1 +1 –1 –1 +1

Most polypeptides contain between 100 and 1000 amino acid residues, although some contain thousands of amino acids (Table 4.2). Short polypeptides are often called oligopeptides (oligo is Greek for “few”) or just peptides. 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 20100 or 1.27 × 10130 possible amino acid sequences. This number is clearly unattainable in nature, since there are only about 1079 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. And, 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).

Amino Acids, the Building Blocks of Proteins

TA B L E 4.2

Composition of Some Proteins

PROTEIN

NUMBER OF AMINO ACID RESIDUES

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 folds up to form a more compact shape, usually 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 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.

93

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CH APTE R 4 Protein Structure

Primary structure The sequence of amino acid residues

Secondary structure The spatial arrangement of the polypeptide backbone

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

–Lys–Val–Asn–Val–Asp–

FIGURE 4.4 Levels of protein structure in hemoglobin. [Structure of human hemoglobin (pdb 2HHB) determined by G. Fermi and M. F. Perutz.]

L EARNING 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.

Secondary Structure: The Conformation of the Peptide Group 4.2

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

N

H

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): H N

O C␣

C

H N H

C␣

C O

N

O C␣

C

N H

C␣

C O

Here the H atom and R group attached to Cα are not shown. The polypeptide backbone can still rotate around the N— Cα and Cα— C bonds, although rotation is somewhat limited. For example, a sharp bend at Cα could bring the carbonyl oxygens of neighboring residues too close:

Here the atoms are color-coded: C gray, O red, N blue, and H white, and the van der Waals surfaces of the O atoms are shown.

Secondary Structure: The Conformation of the Peptide Group

95

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 folds so that it can satisfy as many of these hydrogen-bonding requirements as possible. 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). Most α helices are about 12 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).

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.

FIGURE 4.5 The α helix. In this conformation, the polypeptide backbone twists in a right-handed 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.]

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

(a)

(b)

FIGURE 4.6 An α helix from myoglobin. In (a) a ball-and-stick model and (b) a space-filling model of residues 100–118 of myoglobin, the backbone atoms are green and the side chains are gray. [Structure of sperm whale myoglobin (pdb 1MBD) determined by S. E. V. Phillips.]

96

CH APTE R 4 Protein Structure

N

N

C

C O

O

C

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

O

O

C

C

H N

H N

H N

C

C O

C

O

C N H

N H C

C C

O

O

N H C C O

O

O

H N

C

C N H

O

C N H

C C

C

C

H N

C C

C

C

N H

O

H N

C

C

N H

O

C

C O

C C

O

H N

C

C

N H

C

C

C

C

C

O

N H

H N

C C

O

C

C

C

N H

C C

O

O

C O

O

C C

N H

C

O

H N

C

N H

C

O

O

C

N H

N H

C

C N H

C

O

H N C

C

C C

O

H N C

O

C C

O

H N

C

N

C C

N H

C C

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.

O N H

C

N C

C

N H

FIGURE 4.7 β sheets.

N

C C

N Antiparallel

Parallel

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).

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).

(a)

(b)

FIGURE 4.8 Side view of two parallel strands of a β sheet. In (a) a ball-and-stick model and (b) a space-filling model of a β sheet from carboxypeptidase A, the backbone atoms are green. The amino acid side chains (gray) point alternately to each side of the β sheet. [Structure of carboxypeptidase (pdb 3CPA) determined by W. N. Lipscomb.]

Tertiary Structure and Protein Stability

97

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):

Usually, the loops that link β strands or α helices consist of residues with irregular secondary structure; that is, the polypeptide does not adopt a defined secondary structure in which successive residues have the same backbone conformation. Note that “irregular” does not mean “disordered”: The peptide backbone almost always adopts a single, unique conformation. Most proteins contain a combination of regular and irregular secondary structure. On average, 31% of residues are in α helices, 28% are in β sheets, and most of the remainder are in irregular loops of different sizes. However, many proteins deviate from the average in dramatic ways. 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

The three-dimensional shape of a protein, known as its tertiary structure, includes its 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. In a fully folded protein under physiological conditions, virtually every atom has a designated place. One of the most powerful techniques for probing the atomic structures of macromolecules, including proteins, is X-ray crystallography (Section 4.6). The structure of myoglobin, the first protein structure to be determined by X-ray crystallography, came to light in 1958 through the efforts of John Kendrew, who painstakingly determined the conformation of every backbone and side-chain group. Kendrew’s results—coming just a few years after Watson and Crick had published their elegant model of DNA—were a bit of a disappointment. The myoglobin structure lacked the simplicity and symmetry of a molecule such as DNA and was more irregular and complex than expected (myoglobin structure is examined in detail in Section 5.1). The structure of another well-studied protein, the enzyme triose phosphate isomerase, is shown in Figure 4.9a. This enzyme is a globular protein (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. The tertiary structure of triose phosphate isomerase can be simplified by showing just the peptide backbone (Fig. 4.9b). Alternatively, the structure can be represented by a ribbon that passes through Cα of each residue (Fig. 4.9c). This rendering makes it easier to identify elements of secondary structure. However, it is important to keep in mind that proteins, like all molecules, are solid objects with no empty space inside.

LEARNING OBJECTIVES Describe the forces that stabilize protein structure. • Analyze protein structures presented in different styles. • Explain how the hydrophobic effect stabilizes protein structures. • Describe the intramolecular interactions that can stabilize protein structures. • Summarize the process of protein folding. • Recognize the inherent flexibility in protein structures.

98

CH APTE R 4 Protein Structure

Commonly used systems for analyzing protein structures are based on the presence of elements of regular secondary structure. For example, four classes are recognized by the CATH system (the name refers to a hierarchy of organizational levels: Class, Architecture, Topology, and Homology). 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. Structural data—including the three-dimensional coordinates for each atom—for over a hundred thousand proteins and other macromolecules are available in online databases. The use of software for visualizing and manipulating such structures provides valuable insight into molecular structure and function. (a)

(b)

Proteins have hydrophobic cores Globular proteins typically contain at least two layers of 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 solvent; in the core, these groups are sequestered from the solvent. 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 core 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 regular secondary structure. This is because the formation of α helices and β sheets, which are internally hydrogen-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 to water molecules.

(c) FIGURE 4.9 Triose phosphate isomerase. (a) Space-filling 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. [Structure (pdb

(b)

(a)

1YPI) determined by T. Alber, E. Lolis, and G. A. Petsko.]

(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. [Structure (pdb 1HGU) determined by L. Chantalat, N. Jones, F. Korber, J. Navaza, and A. G. Pavlovsky.] (b) γβ-Crystallin, an all-β protein. [Structure (pdb 1GCS) determined by S. Najmudin, P. Lindley, C. Slingsby, O. Bateman, D. Myles, S. Kumaraswamy, and I. Glover.] (c) Flavodoxin, an α/β protein. [Structure (pdb 1CZN) determined by W. W. Smith, K. A. Pattridge, C. L. Luschinsky, and M. L. Ludwig.] (d) Tachystatin, a protein with little secondary structure. [Structure (pdb 1CIX) determined by N. Fujitani, S. Kawabata, T. Osaki, Y. Kumaki, M. Demura, K. Nitta, and K. Kawano.]

Tertiary Structure and Protein Stability

The requirement for a hydrophobic core and a hydrophilic surface also places constraints on amino acid sequence. The location of a particular side chain in a protein’s tertiary structure is related to its hydrophobicity: The greater a residue’s hydrophobicity, the more likely it is to be located in the protein interior. In the protein interior, side chains pack together, leaving essentially no empty space or space that could be occupied by a water molecule. Table 4.3 lists two scales for assessing the hydrophobicity of amino acid side chains. Such information is useful for predicting the locations of amino acid residues within proteins. For example, highly hydrophobic residues such as Phe and Met are almost always buried. Polar side chains, like hydrogen-bonding backbone groups, can participate in hydrogen bonding in the protein interior, which helps “neutralize” their polarity and allows them to be buried in a nonpolar environment. When a charged residue occurs in the protein interior, it is almost always located near another residue with the opposite charge, so the two groups can interact electrostatically to form an ion pair. By color-coding the amino acid residues of myoglobin according to their hydrophobicity, it is easy to see that hydrophobic side chains cluster in the interior while hydrophilic side chains predominate on the surface (Fig. 4.12).

Protein structures are stabilized mainly by the hydrophobic effect Surprisingly, the fully folded conformation of a protein is only marginally more stable than its unfolded form. The difference in thermodynamic stability amounts to about 0.4 kJ ⋅ mol–1 per TAB L E 4. 3 RESIDUE

a b

99

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. [Structure (pdb 1GPD) determined by D. Moras, K. W. Olsen, M. N. Sabesan, M. Buehner, G. C. Ford, and M. G. Rossmann.]

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

Hydrophobicity Scales SCALE Aa

SCALE B b

Phe

2.8

3.7

Met

1.9

3.4

Ile

4.5

3.1

Leu

3.8

2.8

Val

4.2

2.6

Cys

2.5

2.0

Trp

–0.9

1.9

Ala

1.8

1.6

Thr

–0.7

1.2

Gly

–0.4

1.0

Ser

–0.8

0.6

Pro

–1.6

–0.2

Tyr

–1.3

–0.7

His

–3.2

–3.0

Gln

–3.5

–4.1

Asn

–3.5

–4.8

Glu

–3.5

–8.2

Lys

–3.9

–8.8

Asp

–3.5

–9.2

Arg

–4.5

–12.3

Scale A is from Kyte, J., and Doolittle, R.F., J. Mol. Biol. 157, 105–132 (1982).

Scale B is from Engelman, D.M., Steitz, T.A., and Goldman, A., Annu. Rev. Biophys. Chem. 15, 321–353 (1986).

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

100

CH APTER 4 Protein Structure Folded

Unfolded

Unfavorable solvation FIGURE 4.13 The hydrophobic effect in protein folding. In a folded protein, hydrophobic regions (represented by blue 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.

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 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. Nevertheless, most proteins do fold into a unique and 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 fine-tune a folded conformation that is already largely stabilized by the hydrophobic effect.

Other interactions help stabilize proteins Once folded, many polypeptides appear to maintain their shapes through various other types of interactions, the most common being ion pairs, disulfide bonds, and interactions with zinc ions. An ion pair can form between oppositely charged side chains or the N- and C-terminal groups of a polypeptide (Fig. 4.14). Although the resulting electrostatic interaction is strong, 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 color-coded: 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.

Q Identify residues that could form two other types of ion pairs.

Tertiary Structure and Protein Stability

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 side chains become fixed in the ion pair. For a buried ion pair, there is the additional energetic cost of desolvating the charged groups in order for them to enter the hydrophobic core. 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. 4.15). Here, the bonds may help prevent protein unfolding under relatively harsh extracellular conditions. Domains containing cross-links called 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. 4.16). Protein domains 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).

Protein folding begins with the formation of secondary structures The crowded nature of the cell interior (see Fig. 2.6) demands that proteins and other macromolecules assume compact shapes. In the cell, a newly synthesized polypeptide begins to fold as soon as it emerges from the ribosome, so part of the polypeptide 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 use full-length polypeptides 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 protein to unfold. When the solutes are removed, the proteins renature. Protein renaturation experiments demonstrate that protein folding is not a random process. That is, the protein does not just happen upon its most stable tertiary structure (its native structure) by trial and error but approaches this conformation in a more coherent manner. A newly made polypeptide chain has high energy and high entropy and approaches a low-energy, low-entropy state as it folds. The progress of folding can be diagrammed as a funnel-shaped diagram where the wide mouth of the funnel represents high energy and high entropy and the

(a)

(b) 2+

FIGURE 4.16 Zinc fingers. (a) A zinc finger with one Zn (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. [Structure of transcription factor IIIA (pdb 1TF6) determined by R. T. Nolte, R. M. Conlin, S. C. Harrison, and R. S. Brown; structure of GAL4 (pdb 1D66) determined by R. Marmorstein and S. Harrison.]

101

FIGURE 4.15 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. [Structure (pdb 1E8L) determined by H. Schwalbe, S. B. Grimshaw, A. Spencer, M. Buck, J. Boyd, C. M. Dobson, C. Redfield, and L. J. Smith.]

102

CH APTER 4 Protein Structure Entropy

narrow end respresents low energy and low entropy (Fig. 4.17). The folding process is not necessarily smooth and linear, since a partially folded polypeptide has many possible conformations in the folding “landscape” and sometimes gets stuck temporarily in a “valley” before reaching its most stable conformation. 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 Energy as a “hydrophobic collape,” when nopolar groups are forced out of contact with water. Once the hydrophobic core of the protein begins to take shape, the hydrogen bonding responsible for regular secondary structure can occur, and other structures and side chains jostle themselves into their final positions in the tertiary structure. In the laboratory, certain small proteins can be repeatedly denatured and renaNative structure tured, but in the cell, protein folding is more complicated and may require the assistance of other proteins. Some of these are known as molecular chaperones and are described in more detail in Section 22.4. FIGURE 4.17 A protein folding funnel. An The pathway to full functionality may require additional steps beyond polyunfolded polypeptide has high energy and peptide folding. Some proteins contain several polypeptide chains, which must high entropy, but in its native conformation fold individually before assembling. In addition, many proteins undergo posthas low energy and low entropy. translational processing before reaching their mature forms. 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 SEE ANIMATED groups usually have a discrete biological function and may also help stabilize the folded conPROCESS DIAGRAM formation of the protein. Finally, some proteins become functional only after associating with metal ions or a specific organic molecule. Denaturation and renaturation of RNase A

Some proteins have more than one conformation 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 given amino acid sequence will form an α helix, β sheet, or irregular secondary structure. This 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). In some cases, a protein can adopt more than one conformation. Nicknamed “transformer proteins,” these molecules have two stable conformations that exist in dynamic equilibrium. A change in cellular conditions, such as pH or oxidation state, or the presence of a binding partner can tip the balance toward one conformation or the other (Fig. 4.19). Because structural biologists have traditionally examined proteins with fixed, stable shapes, the number of proteins known to have conformational flexibility is likely to increase. Even more intriguingly, many proteins appear to include polypeptide segments that are relatively extended and have no fixed conformation at all (Fig. 4.20). Calculations suggest that up to 40% of the proteins in eukaryotes contain such segments, which comprise at least (b)

(a)

(c)

O HOCH2 H

O H3C

(CH2)14

C

S

CH2

O

O H OH H

NH

C

CH2 O⫺

H

⫺

H

HN C

CH3

O

P

O

CH2

O

O 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 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.

Tertiary Structure and Protein Stability

103

30 residues and tend to be rich in hydrophilic side chains. A few proteins may even be completely disordered from end to end. These intrinsically unstructured proteins or protein segments are particularly well suited for regulatory roles, in which one protein must interact with several others. The lack of a fixed structure makes it easier for a protein to inspect multiple potential binding partners and then, as it latches onto one, to adopt a defined tertiary structure that maximizes the strength of that interaction. Of course, all proteins are inherently flexible, since they are built from polypeptide chains with some degree of rotational freedom and are stabilized by relatively weak nonco(a) (b) valent forces. Rather than existing as rigid blocks, proteins FIGURE 4.19 A protein with two stable conformations. One form of are surprisingly dynamic, undergoing minor movements as the protein lymphotactin consists of a three-stranded β sheet and an α individual bonds rotate and stretch. In addition to these small helix (a). This form interconverts rapidly with an alternate form (b) with random motions, other structural changes are necessary for an all-β structure. [Structures (a, pdb 1J9O and b, pdb 2JP1) determined by B. F. many proteins to carry out their biological functions. For ex- Volkman.] ample, in order 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. Although many dramatic protein conformational changes have been documented, most movements are more subtle adjustments that involve a few side chains or a loop of polypeptide, reposition an α helix or β strand, or allow one protein domain to tilt relative to another.

BEFORE GOING ON • Explain why a folded polypeptide assumes a shape with 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.

Catalytic B-chain

Auto-inhibitory

CaM binding

Regulatory

FIGURE 4.20 An intrinsically unstructured protein. The structure of calcineurin includes domains with well-defined tertiary structure (orange and yellow) as well as a 95-residue disordered segment (blue and gold). [Courtesy of David Weis, University of Kansas.]

104

CH APTER 4 Protein Structure

L EARNING 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.

4.4

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- 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, 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. 4.21). 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. 4.21c). A few multimeric proteins are known to adopt more than one possible quaternary structure and apparently shift among the alternatives by dissociating, adopting different tertiary structures, and reassembling. The advantages of multisubunit protein structure are many. For starters, the cell can construct extremely large proteins through the incremental addition of small building blocks that are encoded by a single gene (we will see examples of this in the next chapter). This is clearly an asset 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

FIGURE 4.21 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. [Structure (pdb 1AS8) determined by M. E. P Murphy, E. T. Adams, and S. Turley.] (b) E. coli

fumarase, a homotetrameric enzyme. [Structure (pdb 1FUQ) determined by T. Weaver and L. Banaszak.] (c) Human hemoglobin,

a heterotetramer with two α subunits (blue) and two β subunits (red). [Structure (pdb

(a)

(b)

(c)

(d)

2HHB) determined by G. Fermi and M. F. Perutz.] (d) Bacterial methane

hydroxylase, whose two halves (right and left in this image) each contain three kinds of subunits. [Structure (pdb 1MMO) determined by A. C. Rosenzweig, C. A. Frederick, S. J. Lippard, and P. Nordlund.]

Q Which of the proteins shown in Fig. 4.1 has quaternary structure?

Clinical Connection: Protein Misfolding and Disease

105

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.

Clinical Connection: Protein Misfolding and Disease 4.5

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 faulty protein folding, more specifically, to the cell’s failure to deal with the misfolded protein. Normally, the chaperones that help new proteins to fold can also help misfolded proteins to refold. If the protein cannot be salvaged in this way, it is usually degraded to its component amino acids (Fig. 4.22). The operation of this quality control system could explain why some mutated proteins, which fold incorrectly, never reach their intended cellular destinations. Other diseases result when misfolded proteins are not immediately refolded or degraded but instead aggregate, often as long insoluble fibers. Although the fibers can occur throughout the body, they appear to be deadliest when they occur in the brain. Among the human diseases characterized by such fibrous deposits are Alzheimer’s disease, Parkinson’s disease, and the transmissible spongiform encephalopathies, which lead to neurological abnormalities and loss of neurons (nerve cells). The aggregated proteins, a different type in each disease, are commonly called amyloid deposits (a name originally referring to their starchlike appearance). Alzheimer’s disease, the most common neurodegenerative disease, is accompanied by both intracellular “tangles” and extracellular “plaques” (Fig. 4.23).The fibrous material inside cells is made of a protein called tau, which is involved in the assembly of microtubules, a component of the cytoskeleton (Section 5.3). Tau deposits also appear in some other neurodegenerative diseases, and tau’s role in Alzheimer’s disease is not clear. The extracellular amyloid material in Alzheimer’s disease consists primarily of a protein called amyloid-β, a 40or 42-residue fragment of the much larger amyloid precursor protein, which is a membrane protein. Enzymes called β-secretase and γ-secretase cleave the precursor protein to generate amyloid-β. Normal brain tissue contains some extracellular amyloid-β, but neither its function

Newly synthesized polypeptide

LEARNING OBJECTIVES Explain how misfolded proteins cause disease. • List the possible fates of a misfolded protein. • Describe how an amyloid fiber forms.

misfolding aggregation

folding

Misfolded structure

degradation

Native structure

refolding

FIGURE 4.22 The fates of a misfolded protein.

FIGURE 4.23 Brain tissue from Alzheimer’s disease. Amyloid deposits (large red areas) are surrounded by intracellular tangles (smaller dark shapes). [Courtesy of Dennis Selkoe and Marcia Podlisny, Harvard University Medical School.]

106

CH APTER 4 Protein Structure

nor the function of its precursor protein is completely understood. However, excess amyloid-β is clearly linked to Alzheimer’s disease. The neurodegeneration of Alzheimer’s disease may begin many years before symptoms such as memory loss appear, and studies of the disease in animal models indicate that both of these signs can be detected before amyloid fibers accumulate in the brain. These and other experimental results support the hypothesis that the early stages of amyloid-β misfolding and aggregation are toxic to neurons and are the ultimate cause of Alzheimer’s disease. The accumulation of extracellular fibers may actually be a protective mechanism to deal with excessive amyloid-β production. In Parkinson’s disease, neurons in a portion of the brain accumulate fragments of a protein known as α-synuclein. Like amyloid-β, α-synuclein’s function is not fully known, but it appears to play a role in neurotransmission. α-Synuclein is a small soluble protein (140 amino acid residues) with an extended conformation, part of which appears to form α helices upon binding to other molecules. The intrinsic disorder of the protein may contribute to its propensity to form amyloid deposits, which are characterized by a high content of β secondary structure. Accumulation of this material is associated with the death of neurons, leading to the typical symptoms of Parkinson’s disease, FIGURE 4.24 Model for amyloid formation. Five which include tremor, muscular rigidity, and slow movements. Mutations in aggregated polypeptides are shown in different colors. The vertical arrow indicates the long axis of the the gene for α-synuclein that lead to increased expression of the protein or amyloid fiber. [From Meier, B., et al., Science, 319, promote its self-aggregation are responsible for some hereditary forms of 1523–1526 (2008). Reprinted with permission from AAAS.] Parkinson’s disease. Mad cow disease is the best known of the transmissible spongiform encephalopathies (TSEs), a group of disorders that also includes scrapie in sheep and Creutzfeldt–Jakob disease in humans. These fatal diseases, which give the brain a spongy appearance, were once thought to be caused by a virus. However, extensive investigation has revealed that the infectious agent is a protein called a prion. Interestingly, normal human brain tissue contains the same protein, named PrPC (C for cellular), which occurs on neural cell membranes and appears to play a role in normal brain function. The scrapie form of the prion protein, PrPSc, has the same 253–amino acid sequence as PrPC but includes more β secondary structure. Introduction of PrPSc into cells apparently triggers PrPC, which contains more α-helical structure, to assume the PrPSc conformation and thereby aggregate with it. How is a prion disease transmitted? One route seems to be by ingestion, as illustrated by the incidence of bovine TSE in cows consuming feed prepared from scrapie-infected sheep. (This feeding practice has been discontinued, but not before a number of people—who presumably consumed the infected beef—developed a variant form of Creutzfeldt–Jakob disease.) PrPSc must be absorbed, without being digested, and transported to the central nervous system. Here, it causes neurodegeneration, possibly through the loss of PrPC as it converts to PrPSc or through the toxic effects of the PrPSc as it accumulates. Despite the lack of sequence or structural similarities among amyloid-β, α-synuclein, and PrPSc, their misfolded forms are all rich in β structure, and this seems to be the key to the formation of amyloid deposits. Studies of a fungal polypeptide similar to PrPSc show how amyloid formation might occur. In its original state, the protein is mostly α-helical. A segment of polypeptide shifts to an all-β conformation, which allows single molecules to stack on top of each other in parallel to form a triangular fiber stabilized by interchain hydrogen bonding (Fig. 4.24). Presumably, other amyloid-forming proteins undergo similar conformational changes. Significant aggregation may not occur until the protein concentration reaches some critical threshold (this may explain why the amyloid diseases, even the TSEs, take years to develop). After a few β structures have assembled, they act to trigger further protein misfolding, and the amyloid fibers rapidly propagate. Once formed, the fibers are resistant to degradation by cellular enzymes. BEFORE GOING ON • List some diseases that are linked to faulty protein folding. • Explain why the symptoms of protein misfolding diseases do not appear immediately. • Describe the common structural features of the proteins in amyloid deposits.

Tools and Techniques: Analyzing Protein Structure

Tools and Techniques: Analyzing Protein Structure 4.6

Like nucleic acids (Section 3.5), proteins can be purified and analyzed in the laboratory. In this section, we will examine some commonly used approaches to isolating proteins and determining their sequence of amino acids and 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 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. 4.25). 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 derivatized (a)

(b)

Protein mixture

Small proteins

Large proteins

FIGURE 4.25 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 size-exclusion 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 the bottom of the column. In this way, a mixture of proteins can be separated according to size.

107

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. • Describe how the threedimensional arrangement of atoms in a protein can be deduced by measuring the diffraction of X-rays or electrons or by analyzing nuclear magnetic resonance.

108

CH APTER 4 Protein Structure

FIGURE 4.26 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).

High salt

Protein mixture

High salt

Bound proteins Uncharged proteins Cationic proteins

with positively charged diethylaminoethyl (DEAE) groups or negatively charged carboxymethyl (CM) groups:

DEAE

CH2

CH2 CH2 NH⫹ CH2

CH3 CM

CH2

COO⫺

CH3

Negatively charged proteins will bind tightly to the DEAE groups, while uncharged and 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. 4.26). 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 2̷ (pK1 + pK2) [4.1] 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.

SAMP LE CA LCU LAT I O N 4 . 2 SEE SAMPLE CALCULATION VIDEOS

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 2̷ (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

Tools and Techniques: Analyzing Protein Structure

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 molecules move through a gel-like matrix such as polyacrylamide under the influence of an electric field (Section 3.5). 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. 4.27).

109

FIGURE 4.27 SDS-PAGE. Following electrophoresis, the gel was stained with Coomassie blue dye so that each protein is visible as a blue band. Lanes 1 and 8 contain proteins of known molecular mass that serve as standards for estimating the masses of other proteins. [Courtesy

Mass spectrometry reveals amino acid sequences A standard approach to sequencing a protein has several steps:

of Bio-Rad Laboratories, ©2012.]

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.

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.

L EARNING 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.

9.4

Membrane Fusion

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

Membrane Fusion

249

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.

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.

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

250

CH APTER 9 Membrane Transport

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 a 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⫹

H2 O

⫹

N(CH3)3

acetylcholinesterase

Acetylcholine

O H3C

C

O⫺ ⫹ HO

CH2

Acetate

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).

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

H N

O

NH2

CF3

HO

Fluoxetine

N H

H N H 3C

H H

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®).

Cl Cl

Sertraline

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. Q What types of food could contribute to serotonin production in the body?

Membrane Fusion

251

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. 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 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 fourhelix complex. This action necessarily brings the two membranes close together (Fig. 9.21). Because the formation of the SNARE complex is thermodynamically favorable, membrane fusion is also spontaneous. 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’ 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.

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. [Structure (pdb 1SFC) determined by R. B. Sutton and A. T. Brunger.]

Removal of head group

Removal of lipid tail

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

Vesicle

SNAREs

Plasma membrane

SNARE complex

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.

Q Why does disassembly of the SNARE complex require ATP?

SEE ANIMATED PROCESS DIAGRAM Model for SNAREmediated vesicle fusion

252

CH APTER 9 Membrane Transport Vesicle

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

Endocytosis is the reverse of exocytosis 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 mechanism is simply the reverse of the process shown in Figure 9.22. Soon after the vesicle and plasma membranes fuse, the “fusion pore” closes and the vesicle re-forms. The empty vesicle can later 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 (Figure 9.22 in reverse) 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 solutes happen to be present. In receptormediated 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. The vesicle 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 and then maintaining that structure as the vesicle makes its way to another site in the cell. 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

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 a model derived from electron micrographs. Multiple subunits assemble to form a lattice that surrounds and defines the shape of an intracellular vesicle. [Structure (pdb 1XI4) determined by A. Fotin, Y. Cheng, P. Sliz, N. Grigorieff, S. C. Harrison, T. Kirchhausen, and T. Walz.]

Summary

253

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. Most animal cells release exosomes, which are typically 50 to 150 nm in diameter and contain an assortment of the cell’s proteins, lipids, and several types of RNA. 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 stimulation.

Neighboring cells take up the exosomes and their cargo, so this transport system may be a mechanism for sharing information about what’s going on inside the cells. The RNA components may be particularly informative, since mRNAs reflect what 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 or other materials that promote tissue healing after injury.

CYTOSOL

Exosomes

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). 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 changes in bilayer curvature are necessary for endocytosis and exocytosis.

Summary 9.1 The Thermodynamics of Membrane Transport

9.2

• The transmembrane movements of ions generate changes in membrane potential, Δψ, during neuronal signaling.

• Passive transport proteins such as porins allow the transmembrane movement of substances according to their concentration gradients. Aquaporins mediate the transport of water molecules.

• 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.

Passive Transport

• 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.

254

CH APTER 9 Membrane Transport

• Membrane proteins such as the passive glucose transporter undergo conformational changes that alternately expose ligand-binding sites on each side of the membrane.

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.

9.4

Membrane Fusion

• 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.

• Secondary active transport allows the favorable movement of one substance to drive the unfavorable transport of another substance.

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 exosome (microvesicle)

Bioinformatics Brief Bioinformatics Exercise 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. 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. 3. 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 2 and what is the significance of the change? 4. 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 2). Assume the temperature is 37°C. Is this process favorable? 5. Use Equation 9.4 to calculate the free energy change for the movement of Na+ into the depolarized nerve cell described in Problem 3. Assume the temperature is 37°C. How does this compare with the value you calculated in Problem 4 and what is the significance of the difference?

6. 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. 7. 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? 8. 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? 9. 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? 10. Calculate the free energy change for the movement of K+ into a cell when the K+ concentration outside is 15 mM and the cytosolic

Problems

K+ concentration is 50 mM. Assume that T = 20°C and Δψ = –50 mV (inside negative). Is this process spontaneous?

listed in the table. How do the permeability coefficients assist you in ranking the rate of the transmembrane diffusion of these compounds?

11. 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? 12. 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? 13. 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? 14. 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.

255

Permeability coefficient (cm · s–1) 9 × 10–6 1 × 10–5 1 × 10–7

Acetamide Butyramide Urea

21. The permeability coefficients (see Problem 20) 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)

Synthetic lipid bilayer Red blood cell membrane

Glucose

Mannitol

2.4 × 10–10 2.0 × 10–4

4.4 × 10–11 5.0 × 10–9

CHO

15. 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?

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

16. 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.

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

17. Glucose absorbed by the epithelial cells lining the intestine leaves these cells and travels to the liver via the portal vein. After a highcarbohydrate 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.

9.2

18. 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. 19. Rank the rate of transmembrane diffusion of the following compounds:

O H3C

C

O NH2

H3C

A. Acetamide

CH2

CH2

C

NH2

B. Butyramide O H2N

C NH2 C. Urea

20. 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 19 are

Passive Transport

23. 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. 24. 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? 25. 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? 26. 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?

CH APTER 9 Membrane Transport

27. 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 death. 28. 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 light of what you know about aquaporin, does this surprise you? 29. The selectivity filter in the bacterial chloride channel ClC is formed in part by the hydroxyl groups of Ser and Tyr side chains and main chain NH groups. Explain how these groups would allow Cl– but not cations to pass through. 30. 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? 31. The kinetics of transport through protein transporters can be described using the language of Michaelis and Menten. 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. Glucose flux (mM . cm . s−1 × 106)

1.0

0.5

35. Glutamate acts as a neurotransmitter in the brain, and it 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? 36. 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. What is the net charge imbalance generated by this system? b. Why do you suppose the investigators who elucidated this mechanism referred to the process as a “hydrogen pump”? ⫺OOC

COO⫺

decarboxylase

Oxalate

H H⫹

COO⫺

Formate

CO2

37. 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 HCO3– ions in exchange for Cl– ions. What role does Band 3 play in transporting CO2 to the lungs where it can be exhaled? 38. Band 3 is unusual in that it can transport HCO3– in exchange for Cl– ions, in either direction, via passive transport. Why is it important that Band 3 operate in both directions? 39. A group of investigators was interested in the mechanism of lactate transport in Plasmodium vivax, a causative agent of 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. 15

0

0

2

4

6

8

10

[Glucose] mM

32. The compound 6-O-benzyl-D-galactose competes with glucose for binding to the glucose transporter. Sketch a curve similar to the one shown in Problem 31 that illustrates the kinetics of glucose transport in the presence of this inhibitor. 33. 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. But 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? 34. 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. Another experiment showed that if a propyl group is added to the hydroxyl group 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 tell you about the mechanism of passive glucose transport?

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

256

pH 7.1

pH 5.5

10

5

0

0

1 Time (min)

2

40. 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 37), 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. 41. 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

Problems

radioactively labeled choline by the transfected cells was measured at increasing choline concentrations.

CH3 (CH2)2

N

⫹

CH3

Choline

7 6 5 4 3 2 100

50

150

200

[Choline] (μM)

b. Plasma concentrations of choline range from 10 to 80 μM, although the concentration may be higher in 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 (TEA) ions inhibit choline transport.

CH3 CH2 H3C

CH2

N

⫹

CH2

CH3

CH2 CH3 Tetraethylammonium (TEA) 42. 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 in the presence and absence of phlorizin. b. What kind of inhibitor is phlorizin? Explain.

7 1/v (g . min . μmol−1)

46. 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 cytoplasm? 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? 47. 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? 1.6 1.4

Endothelial cells

1.2 1.0

Pericytes

0.8 0.6 0.4 0.2

0

50

100

150

200

Concentration of Na+ ions (mM)

48. Use the information provided in the figure to estimate the KM and Vmax (see Problem 31) for glucose uptake by the pericyte transporter described in Problem 47.

8

0.30

6

With inhibitor y = 13.8x + 0.65

5 4 3 2

No inhibitor y = 5.4x + 0.64

1 −0.2 −0.1

45. The parasite Trypanosoma brucei, the causative agent of 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 44). Incubation of the cells with iodoacetate (which decreases intracellular ATP) decreases the rate of proline transport into the parasite. What do these experiments tell you about the nature of the proline transporter in trypanosomes?

0

0.1

0.2

0.3

1/[Glucose] (mM−1)

0.4

0.5

1/v (mg . h . nmol−1)

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

a. Use the language of Michaelis and Menten as described in Problem 31 to estimate KM and Vmax from the curve shown.

0

43. The Na,K-ATPase first binds sodium ions, then reacts with ATP to form an aspartyl phosphate intermediate. Draw the structure of the phosphorylated Asp residue. 44. 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

1 0

Active Transport

Glucose uptake (nmol . mg−1 . h−1)

HO

9.3

257

0.25 0.20 0.15 0.10 0.05 0 −0.5

0

0.5

1

1/[S] (mM−1)

1.5

2

258

CH APTER 9 Membrane Transport

49. Many ABC transporters are inhibited by vanadate, a phosphate analog. Why is vanadate an effective inhibitor of these transporters? 50. 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? 51. Kidney cells include two antiport proteins, an H+/Na+ exchanger and a Cl–/HCO3– exchanger (see Fig. 2.18). What is the source of free energy that drives the transmembrane movement of all these ions? 52. Many cells have a mechanism for exporting ammonium ions. Describe how this could occur through secondary active transport. 53. 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) a Na+–H+ antiporter (H+/Na+ exchanger), and (c) a Na,K-ATPase. Draw a diagram that illustrates how these three transporters work together to transport peptides into the cell. 54. The X-ray structure of a Ca2+-ATPase was determined by crystallizing the enzyme along with adenosine-5′-(βγ-methylene) triphosphate (AMPPCP), an ATP analog (structure shown below). How did the co-crystallization strategy assist the crystallographers in imaging this protein?

59. 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. 60. 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. 61. In addition to being an SSRI, fluoxetine may also bind to and block signaling by one type of serotonin receptor. Would you expect this activity to be consistent with fluoxetine’s ability to treat the symptoms of depression? 62. 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. 63. 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. 64. The drug known as Botox is a preparation of botulinum toxin, which is similar to the tetanus toxin (Problem 63). 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. 65. Phosphatidylinositol is a membrane glycerophospholipid whose head group includes a monosaccharide (inositol) group. A certain kinase adds another phosphate group to a phosphorylated phosphatidylinositol:

NH2 N

O ⫺

O

P O⫺

O CH2

P O⫺

O O

P

N

O

CH2

O⫺ H

N

H OH

O

P OH

O H2C

H OH

O

N

O

H

O⫺

O

CH

CH2

O

H

H

H HO OH H H

R1 R2

O

OH

kinase

H O⫺ P

O⫺

O

Adenosine-5⬘-(␤,␥-methylene) triphosphate (AMPPCP)

O⫺

9.4

O

Membrane Fusion

55. Myasthenia gravis is an autoimmune disorder characterized by muscle weakness and fatigue. Patients with the disease generate antibodies against the acetylcholine receptor in postsynaptic cells; 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? 56. Lambert–Eaton syndrome is another autoimmune muscle disorder (see Problem 55), 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. 57. 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. Draw the structure of the enzyme’s catalytic residue modified by DIPF. 58. Parathion and malathion are organophosphorus compounds similar to DIPF (see Problem 57). These compounds are sometimes used as insecticides. Why are these compounds deadly poisons?

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 Why might this activity be required during the production of a new vesicle, which forms by budding from an existing membrane? 66. 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. 67. In autophagy (literally, “self-eating”), a damaged or unneeded cellular organelle becomes enclosed in a compartment called an autophagosome. Autophagosome formation begins with an assembly of lipids and proteins that grows by acquiring additional lipids until it forms a small bilayer-enclosed compartment called a phagophore.

Selected Readings

Developing phagophore

0.2 LDL bound (μg)

This structure continues to expand, encircling the damaged organelle. A membrane fusion event closes off the autophagosome. Complete the diagram of autophagosome formation shown here. How many membranes now separate the damaged organelle from the rest of the cell?

0.1

0 Damaged organelle

69. Draw a graph (similar to the one shown in Problem 31) that compares the kinetics of substance delivery into the cell via pinocytosis and receptor-mediated endocytosis (see Fig. 9.23).

20

0

20

40

60

80

100

40

60

80

100

40

60

80

100

LDL (μg . mL−1)

0.5

0

Cholesterol synthesis (nmol . h−1)

70. Low-density lipoprotein (LDL, see Problem 8.70 for a description) enters the cell via receptor-mediated endocytosis by binding to a specific cell-surface receptor that localizes to an invaginated 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 are shown below. Redraw these graphs to show how the normal process would be altered in a. a patient with a defective receptor that could not bind LDL, and b. a patient with an LDL receptor that does not localize to a clathrin-coated pit.

0

0.1 LDL internalized (μg)

68. After an autophagosome has formed (Problem 67), a lysosome fuses with it to deliver hydrolytic enzymes that will eventually degrade the damaged organelle inside the autophagosome. Using the diagram you made for Problem 63 as a starting point, show that the lysosomal enzymes must degrade a lipid bilayer before they can begin to degrade the organelle.

259

LDL (μg . mL−1)

0.4

0.2

0

0

20

LDL (μg . mL−1)

Selected Readings Gouaux, E. and MacKinnon, R., Principles of selective ion transport in channels and pumps, Science 310, 1461–1465 (2005). [Compares several transport proteins of known structure and discusses the selectivity of Na+, K+, Ca2+, and Cl– transport.]

Kreida, S. and Tömroth-Horsefield, S., Structural insights into aquaporin selectivity and regulation, Curr. Opin. Struct. Biol. 33, 126–134 (2015). [Provides up-to-date information on aquaporin structures and functions.]

Higgins, C. F., Multiple molecular mechanisms for multidrug resistance transporters, Nature 446, 749–757 (2007). [Describes some general features of ABC transporters and others that function in drug resistance.]

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.]

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.]

Hysteria/Shutterstock

CHAPTER 10 Signaling

Tetrahydrocannabinol, the active ingredient in Cannabis sativa, mimics endogenous (natural) endocannabionoid signaling molecules such as anandamide, which act on receptors in the central nervous system. Substances that bind to the endocannabinoid receptors are useful because they stimulate appetite and reduce anxiety, but they also interfere with neural development and short-term memory.

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).

L EARNING 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. 260

All cells, including prokaryotes, must have mechanisms for sensing external conditions 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

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 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.

General Features of Signaling Pathways

TA B L E 10.1

261

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

A ligand binds to a receptor with a characteristic affinity Extracellular signals can take many forms, including amino acids and their derivatives, peptides, lipids, and other small molecules (Table 10.1). Some are formally called hormones, which are substances produced in one tissue that affect the functions of other tissues, but many signals go by other names. Keep in mind that other stimuli—such as light, mechanical stress, odorants, and tastants—also serve as signals for cells, although we will not discuss them here. Some bacterial signal molecules are described in Box 10.A. 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 Biochemists express the strength of receptor–ligand binding as a dissociation constant, Kd, which is the reciprocal of the association constant. For this reaction, Kd =

[ R ][ L ] [ R·L ]

[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 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. The dissociation constant Kd is the ligand concentration at which half of the receptor molecules have bound ligand.

1.0

[R . L] [R]T 0.5 Kd

[L]

Q Compare this graph to Figures 5.3 and 7.5.

262

CH APTER 1 0 Signaling

Box 10.A Bacterial Quorum Sensing Even free-living single-celled organisms need to communicate with others of their kind. In bacteria, one form of intercellular communication is known as quorum sensing, which allows the cells to monitor population density and adjust gene expression accordingly. As a result, groups of cells can undertake communal endeavors such as producing the polysaccharides and other materials that form a protective matrix called a biofilm (discussed in Section 11.2). Quorum sensing also prepares cells to take up or exchange DNA, an occasional necessity for asexually reproducing organisms. Pathogenic bacteria may use quorum sensing to wait until their numbers are sufficiently high before synthesizing toxins and other proteins needed to attack a host organism. The essence of quorum sensing is that cells respond to some signal molecule that increases in concentration as cell density increases. Only a few types of these molecules have been identified. One type consists of acyl homoserine lactones. The acyl chains of these molecules, which may include from 4 to 18 carbons, are derived either from fatty acids within the cell or from exogenous lipids.

O H3C

A number of different acyl chains can be attached to the homoserine lactone group, making this class of compounds highly diverse. These lipids are poorly soluble in water, so they may be released from cells in the form of lipid vesicles (Section 2.2) that also contain other types of molecules. Because they are hydrophobic, the molecules can diffuse across the membranes of recipient cells and combine with a receptor protein in the cytosol. The receptor– ligand complex then binds to DNA to turn on the expression of certain genes. A given bacterial species may produce dozens of different molecules that could be used for quorum sensing. Some of these molecules appear to do double duty as toxins for other species of bacteria, thus coordinating the growth of one species at the expense of others. In retaliation, some bacteria have evolved mechanisms to degrade the signals produced by other species or to synthesize molecules that competitively inhibit receptor binding by the other signals.

O

O N H

O

An acyl homoserine lactone

Q Microbiologists have proposed that drugs that interfere with quorum sensing would be useful as antibiotics, particularly since there would be little selective pressure for individual bacteria to evolve resistance to the drug. Explain.

in cardiac muscle slows the heart, and adenosine signaling in the brain leads to a decrease in neurotransmitter release, producing a sedative effect.

NH2 N

N

N

N

Adenosine

O

HOH2C H

H

H

H OH

OH

SAMP LE CA LCU LAT I O N 1 0 . 1 SEE SAMPLE CALCULATION VIDEOS

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

General Features of Signaling Pathways

Caffeine is an antagonist of the adenosine receptor because it binds to the receptor but does not trigger a response. 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.

O H3C

O

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 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)

Ligand

Ligand

(b) G protein– coupled receptor

Receptor tyrosine kinase

Enzyme

P

G protein (inactive)

Substrate G protein (active)

Second messenger Kinase 1 (inactive)

Target proteins

P

ATP 

Kinase 2 (inactive)

ADP 

Kinase 1 (active) P

ATP 

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.

Q Which pathway components are enzymes?

CH3 N

N N CH3 Caffeine

263

N

264

CH APTER 1 0 Signaling

SEE GUIDED TOUR Signal Transduction

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 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.

G Protein Signaling Pathways

10.2

265

LEARNING OBJECTIVES

G Protein Signaling Pathways

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 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 a GPCR is defined by a portion of the helical core of the protein as well as its extracellular loops. Aside from this general location, there are few similarities in the structures of the binding sites in different GPCRs, which is consistent with the observation that each receptor is specific for only one or a few of many possible ligands, which may be large or small, polar or nonpolar substances. 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.

Describe signaling via G protein–coupled receptors. • Recount the events of signaling via a G protein. • Summarize the roles of nucleotides in the signaling pathway. • Describe how a kinase is activated. • List the mechanisms that terminate the G protein signaling pathway. • Compare the adenylate cyclase pathway and the phosphoinositide 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.

CH3  H2N



CH2 CH

HO

H3N OH

CH2 CH

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 binding site for a specific G protein (Fig. 10.4a). The G protein is presumably already close to the receptor, since it is lipid-linked. The trimeric G proteins associated with GPCRs consist of three subunits, designated α, β, and γ

FIGURE 10.3 The β2-adrenergic 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. [Structure (pdb 2RH1) determined by V. Cherezov, D. M. Rosenbaum, M. A. Hanson, S. G. F. Rasmussen, F. S. Thian, T. S. Kobilka, H. J. Choi, P. Kuhn, W. I. Weis, B. K. Kobilka, and R. C. Stevens.]

Q Explain why the receptor for a polar hormone must be a transmembrane protein.

266

CH APTER 1 0 Signaling

FIGURE 10.4 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. [Structure (pdb 3SN6) determined by S. G. F. Rasmussen et al.] (b) The β

subunit (green) has a propellerlike 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. The α and β subunits are covalently attached to lipids, so they are anchored to the cytosolic leaflet of the plasma membrane near the receptor. [Structure (pdb 1GP2) determined by M. A. Wall and S. R. Sprang.].

(a)

(b)

(Fig. 10.4b; other types of G proteins do not have this three-part structure). In the resting state, GDP is bound to the α subunit, but association with the hormone–receptor complex causes the G protein to release 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: GTP + H2O → GDP + Pi Hydrolysis of the GTP allows the α and βγ units to reassociate as an inactive trimer (Fig. 10.5). 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).

Adenylate cyclase generates the second messenger cyclic AMP Cells contain a number of different types of 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 the activated G protein 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 in the cell.

SEE ANIMATED PROCESS DIAGRAM Overview of heterotrimeric G protein signaling

NH2 N

N O

O O

P O

O

P O

O O

P

H O

O

H H HO ATP

N

PPi

H

adenylate cyclase

H H OH

N

N

H N O

H2C

NH2

H2 C O

O P

O

H

H N

N O H O

H

H

OH

Cyclic AMP (cAMP)

G Protein Signaling Pathways

Cyclic AMP activates protein kinase A

αGDPβγ

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. 10.6). 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

4 cAMP

R R

inactive

+

+

C

active

inactive GDP Pi

O

PO2 3

Phospho-Ser

GTP

C

αGTP

active

active

Consequently, cAMP functions as an allosteric activator of the kinase, and the level of cAMP determines the level of activity of protein kinase A. 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

CH

267

O

PO23

CH3

Phospho-Thr

The substrates for the reaction bind in a cleft defined by the two lobes of the protein (Fig. 10.7a). Other kinases share this same core structure but often have additional domains that determine their subcellular location or provide additional regulatory functions. 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. 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 conformational changes that affect catalysis. For example, the negatively charged phospho-Thr interacts with a positively charged arginine residue in the active site. Efficient catalysis requires that this arginine residue and an adjacent aspartate residue be re-positioned for phosphoryl-group transfer from ATP to a protein substrate (Fig 10.7b). Among the many targets of protein kinase A are 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. C 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.

Signaling pathways are also switched off What happens after a ligand binds to a receptor, a G protein responds, a second messenger is produced, an effector enzyme such as protein kinase A 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. We have already seen that the intrinsic

+

βγ

active

FIGURE 10.5 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.

R

C

R

FIGURE 10.6 Inactive protein kinase A. In the inactive complex, the two regulatory subunits (R) block the active sites of the two catalytic subunits (C). [Structure (pdb 3TNP) determined by P. Zhang, E. V. Smith-Nguyen, M. M. Keshwani, M. S. Deal, A. P. Kornev, and S. S. Taylor.]

268

CH APTER 1 0 Signaling

(a)

(b)

ATP phospho-Thr Arg

Asp substrate

FIGURE 10.7 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 stick form. A peptide that mimics a target protein is blue. (b) Close-up view of the active site region. When 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. [Structure (pdb 1ATP) determined by J. Zheng, E. A. Trafny, D. R. Knighton, N.-H. Xuong, S. S. Taylor, L. F. Teneyck, and J. M. Sowadski.]

GTPase activity of G proteins limits their activity. And 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 P

O

H

N

O

H2O phosphodiesterase O

H

H

OH cAMP

N

N

H N O

H

NH2

O P O

H O

H N

N O

H2C H H HO

H H OH

AMP

Caffeine, 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.

G Protein Signaling Pathways

Ultimately, 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. The phosphorylated receptor is then recognized by a protein known as arrestin (Fig. 10.8), which includes lysine and arginine residues that bind the phosphoryl group. This binding halts signaling, by blocking interactions with a G protein (hence the name arrestin), and promotes the movement of the receptor from the cell surface to an intracellular compartment by the process of endocytosis. Interestingly, arrestins also serve as scaffold proteins for organizing components of other signaling pathways. Experimental evidence suggests that some GPCRs, including the β2-adrenergic receptor, interact with arrestin so that they can initiate an intracellular response without ever recruiting a G protein.

269

EXTRACELLULAR SPACE

CYTOSOL

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-adrenergic receptor, which is part of 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 O

P O

CH2

CH

O

O

C

O C

R1

O

H CH2

OH H OH HO H

O

OPO2 3

H 2O

H OPO2 3

H

Phosphatidylinositol bisphosphate

H2O phospholipase C

3PO

H

[Structure (pdb 4ZWJ) determined by Y. Kang et al.]

Q Compare this figure and Figure 10.4a.

OPO2 3

OH H OH HO H

FIGURE 10.8 A receptor– arrestin complex. Rhodopsin, a model G protein–coupled receptor, is shown as a purple ribbon, and the intracellular arrestin protein as a pink ribbon.

H OPO2 3

H

Inositol trisphosphate

R2

The highly polar inositol trisphosphate is a second messenger that triggers the opening of calcium channels in the endoplasmic reticulum membrane, allowing Ca2+ ions to flow into the cytosol. The flux of Ca2+ initiates numerous events in the cell, including the activation of a Ser/Thr kinase known as protein kinase B or Akt. Inositol trisphosphate also directly activates kinases and can undergo additional phosphorylations to generate a series of second messengers containing up to eight phosphoryl groups. 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 (which shares about 40% sequence identity), 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

OH CH2

CH

 O

O

C

O C

R1

CH2

O

R2 Diacylglycerol

270

CH APTER 1 0 Signaling

(a)

(b)

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. Certain compounds that mimic diacylglycerol can activate protein kinase C, leading eventually to the uncontrolled growth characteristic of cancer. 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 C inside the cell, and it apparently exits the cell via an ABC transport protein (Section 9.3), then binds to a G protein–coupled receptor to trigger additional cellular responses.

Calmodulin mediates some Ca2+ signals FIGURE 10.9 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. [Structure of calmodulin (pdb 3CLN) determined by Y. S. Babu, C.E. Bugg, and W. J. Cook. Structure of calmodulin bound to a 26-residue target (pdb 2BBM) determined by G. M. Clore, A. Bax, M. Ikura, and A. M. Gronenborn.]

L EARNING 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.

In some cases where Ca2+ ions elicit a change in 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. 10.9a). Free calmodulin has an extended shape, but in the presence of Ca2+ and a target protein, the helix partially unwinds and calmodulin bends in half to grasp the target protein to activate or inhibit it (Fig. 10.9b). 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. • Draw a diagram to illustrate the phosphoinositide signaling pathway. • For each pathway, identify points where signaling activity can be switched off. • 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.

10.3

Receptor Tyrosine Kinases

A number of hormones and other signaling molecules that regulate cell growth, division, and immune responses bind to cell-surface receptors that operate as tyrosine kinases. These receptors are typically built from two protein subunits, each with a single membrane-spanning segment. At one time, ligand binding was believed to bring two monomeric receptors together to form a catalytically active dimer, but more recent evidence indicates that these receptors are already dimerized in the inactive state. When a ligand molecule binds, the receptor is activated by conformational changes that lead to a scissor-like rearrangement of the receptor’s extracellular and intracellular domains.

Receptor Tyrosine Kinases 







tyrosine kinase domains

(a)

(b)

FIGURE 10.10 The insulin receptor. (a) Schematic diagram. The receptor consists of two α and two β subunits joined by disulfide bonds (horizontal lines). The α subunits bind one molecule of insulin, and the β subunits each include a membrane-spanning segment (coil) and a cytoplasmic tyrosine kinase domain. (b) The extracellular portion of the insulin receptor. The cell surface is at the bottom. One αβ pair of subunits is shown in space-filling form, and the other is shown as a backbone trace. Insulin binds to one of the two binding sites indicated by arrows. [Structure (pdb 2DTG) determined by M. C. Lawrence and V. A. Streltsov.]

Although the receptor’s two halves are identical and would be expected to have identical ligand-binding sites, only one ligand molecule appears to bind to the dimer. In some cases, binding to a site on one half of the dimer, and then to the other half, causes the dimer to become structurally and functionally asymmetric. However, in all cases, ligand binding to the extracellular portion of the receptor leads to tyrosine kinase activity in the intracellular portion. Due to extensive analysis, the insulin receptor serves as a useful model for the approximately 60 other receptor tyrosine kinases in humans.

The insulin receptor dimer binds one insulin Insulin, a 51-residue polypeptide hormone that regulates many aspects of fuel metabolism in mammals, binds to receptors that are present in most of the body’s tissues. The receptor is constructed from two long polypeptides that are cleaved after their synthesis, so the mature receptor has an α2β2 structure in which all four polypeptide segments are held together by disulfide bonds (Fig. 10.10a). The extracellular portion of the insulin receptor has an inverted V shape with multiple structural domains (Fig. 10.10b). Segments of the α subunits define the two insulin-binding sites, but the sites are too far apart (∼65 Å) to simultaneously bind a single hormone molecule. Instead, biochemical evidence indicates that binding to just one site pulls the two α subunits together in such a way that the second binding site cannot bind insulin. Interdomain interactions, along with the disulfide linkages, suggest that the receptor is rigid, and this feature is believed to be important in transducing the message (insulin binding to the extracellular α subunits) to the intracellular signaling apparatus (the tyrosine kinase domains of the β subunits). As the ligand-binding domains come together, the intracellular tyrosine kinase domains separate (hence the analogy to scissors) but in a way that switches on their catalytic activity.

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 receptors appear to phosphorylate themselves, this process is termed autophosphorylation. Each tyrosine kinase domain has the

271

272

CH APTER 1 0 Signaling

FIGURE 10.11 Activation of the insulin receptor tyrosine kinase. The backbone structure of the inactive tyrosine kinase domain of the insulin receptor is shown in light blue, with the activation loop in dark blue. The structure of the active tyrosine kinase domain is shown in light green, with the activation loop in dark green. Note that three Tyr side chains in the activation loop have been phosphorylated (a result of each tyrosine kinase domain phosphorylating the other) so that the activation loop has swung aside to better expose the active site. The phospho-Tyr side chains are shown in stick form with atoms color coded: C green, O red, and P orange. [Structure of the inactive kinase domain (pdb 1IRK) determined by S. R. Hubbard, L. Wei, L. Ellis, and W. A. Hendrickson. Structure of the active kinase domain (pdb 1IR3) determined by S. R. Hubbard.]

Q How would a phosphatase affect the activity of the insulin receptor?

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 a conformational change that 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 (Fig. 10.11). However, some receptor tyrosine kinases do not phosphorylate any other proteins but activate their intracellular targets 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 a bridge between Ras and the phospho-Tyr residues of the receptor (Fig. 10.12). These proteins also stimulate Ras to release GDP and bind GTP. Like other G proteins, Ras is active as long as it has GTP bound to it. The Ras · GTP complex allosterically activates a Ser/Thr kinase, which becomes active and phosphorylates another kinase, activating it, and so on. Cascades of several kinases can therefore amplify the initial growth factor signal. The ultimate targets of Ras-dependent signaling cascades are nuclear proteins, which, when phosphorylated, bind to specific sequences of DNA to induce (turn on) or repress (turn off) gene expression. The altered activity of these transcription factors means that the original hormonal signal not only alters the activities of cellular enzymes on a short time scale (seconds to minutes) 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. Like the other signaling pathways we have examined, the receptor tyrosine kinase 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).

Receptor Tyrosine Kinases

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.

273

The active Ras can now initiate a kinase cascade.

Ligand

GDP

P

P

GDP

Ras (inactive)

P

P

GTP

Adapter proteins

P

P

Ras (active)

ATP 

Kinase (inactive)

GTP

ADP 

Kinase (active)

P

FIGURE 10.12 The Ras pathway. Ras links receptor–hormone binding to an intracellular kinase cascade.

Q Describe the role of protein conformational changes in each step of the pathway.

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-binding 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 anticancer 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

N H3C

N N C

CH3

O

N 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. Q Using Figures 10.2 and 10.12 as a guide, identify other types of signaling proteins that are potential anticancer drug targets.

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.

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CH APTER 1 0 Signaling

L EARNING 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.

10.4

Lipid Hormone 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 receptors. For example, retinoic acid and the thyroid hormones thyroxine (T4) and triiodothyronine (T3) belong to this class of hormones (Fig. 10.13). Retinoic acid (retinoate), a compound that regulates cell growth and differentiation, particularly in the immune system, is synthesized from retinol, a derivative of β-carotene (Box 8.B). The thyroid hormones, which generally stimulate metabolism, are derived from a large precursor protein called thyroglobulin: Tyrosine side chains are enzymatically iodinated, then two of these residues undergo condensation, and the hormones are liberated from thyroglobulin by proteolysis. The 27-carbon cholesterol, introduced in Section 8.1, is the precursor of a large number of hormones that regulate metabolism, salt and water balance, and reproductive functions. Androgens (which are primarily male hormones) have 19 carbons, and estrogens (which are primarily female hormones) contain 18 carbons. Cortisol, a C21 glucocorticoid hormone, affects the metabolic activities of a wide variety of tissues.

O CH3

HO

OH OH

CH3

O Cortisol

Retinoic acid, thyroid hormones, and steroids are all hydrophobic molecules that are carried in the bloodstream either by specific carrier proteins or by albumin, a sort of all-purpose binding protein. The receptors to which the lipid hormones bind are located inside the appropriate target cell, either in the cytoplasm or the nucleus. Ligand binding often—but not always—causes the receptors to form dimers. Each receptor subunit is constructed CH3 CH3 O from several modules, which include a ligand-binding domain and a H3C CH3 DNA-binding domain. The ligand-binding domains are as varied as their O hormone ligands, but the DNA-binding domains exhibit a common structure that includes two zinc fingers, which are cross-links formed by the CH3 interaction of four cysteine side chains with a Zn2+ ion (Section 4.3). In Retinoate the absence of a ligand, the receptor cannot bind to DNA. Following ligand binding and dimerization, the receptor moves to the O I nucleus (if it is not already there) and binds to specific DNA sequences called hormone response elements. Although the hormone response I HO O elements vary for each receptor–ligand complex, they are all composed NH of two identical 6-bp sequences separated by a few base pairs. Simulta3 O I neous binding of the two hormone response element sequences explains why many of the lipid hormone receptors are dimers (Fig. 10.14). I The receptors function as transcription factors so that the genes near Thyroxine (T4) the hormone response elements may experience higher or lower levels of expression. For example, glucocorticoids such as cortisol stimulate O I the production of phosphatases, which dampen the stimulating effects of I HO  kinases. This property makes cortisol and its derivatives useful as drugs O to treat conditions such as chronic inflammation or asthma. However, NH 3 because so many tissues respond to glucocorticoids, the side effects of O these drugs can be significant and tend to limit their long-term use. I The changes in gene expression triggered by steroids and other lipid Triiodothyronine (T3) hormones require many hours to take effect. However, cellular responses FIGURE 10.13 Some lipid hormones. to some lipid hormones are evident within seconds or minutes, suggesting

Lipid Hormone Signaling

275

that the hormones also participate in signaling pathways with shorter time courses, such as those centered on G proteins and/or kinases. In these instances, the receptors must be located on the cell surface.

Eicosanoids are short-range signals Many of the hormones discussed in this chapter are synthesized and stockpiled to some extent before they are released, but some lipid hormones are synthesized as a response to other signaling events (sphingosine-1-phosphate is one example; Section 10.2). The lipid hormones called eicosanoids are produced when the enzyme phospholipase A2 is activated by phosphorylation and FIGURE 10.14 The glucocorticoid receptor–DNA complex. of the glucocorticoid by the presence of Ca2+. One substrate of the phospholipase is The two DNA-binding zinc finger domains receptor are blue and green. The Zn2+ ions are shown as gray the membrane lipid phosphatidylinositol. In this lipid, cleavage spheres. The hormone response element sequences of the DNA of the acyl chain attached to the second glycerol carbon often (bottom) are colored red. Two protein helices make sequencereleases arachidonate, a C20 fatty acid (the term eicosanoid specific contacts with nucleotides. [Structure (pdb 1GLU) determined comes from the Greek eikosi, meaning “twenty”). by B. F. Luisi, W. X. Xu, Z. Otwinowski, L. P. Freedman, K. R. Yamamoto, Arachidonate, a polyunsaturated fatty acid with four double and P. B. Sigler.] bonds, is further modified by the action of enzymes that catalyze Q What is the surface charge of the receptor’s DNA-binding cyclization and oxidation reactions (Fig. 10.15). A wide variety domain? of eicosanoids can be produced in a tissue-dependent fashion, and their functions are similarly varied. Eicosanoids regulate such things as blood pressure, blood coagulation, inflammation, pain, and fever. The eicosanoid thromboxane, for example, helps activate platelets (cell fragments that participate in blood coagulation) and induces vasoconstriction. Other eicosanoids have the opposite effects: They prevent platelet activation and promote vasodilation. The use of aspirin as a “blood thinner” stems from its ability to inhibit the enzyme that initiates the conversion of arachidonate to thromboxane (see Fig. 10.15). A number of other drugs interfere with the production of eicosanoids by blocking the same enzymatic step (Box 10.C). COO The receptors for eicosanoids are G protein–coupled receptors that trigger cAMP-dependent and phosphoinositide-dependent responses. However, eicosanoids are degraded relatively quickly. This instability, along with their hydrophobicity, means that their Arachidonate effects are relatively limited in time and space. Eicosanoids tend cyclooxygenase to elicit responses only in the cells that produce them and in nearby cells. In contrast, many other hormones travel throughout O the body, eliciting effects in any tissue that exhibits the appropriCOO ate receptors. For this reason, eicosanoids are sometimes called local mediators rather than hormones. O Similar signaling molecules operate in plants. For example, OH the lipid hormone jasmonate is synthesized as a result of localized Prostaglandin H2 herbivore damage and quickly spreads to other parts of the plant to trigger defensive responses.

O

CH3

Jasmonate

O

O

Volatile derivatives of jasmonate can travel through the air to spread the alarm to neighboring plants.

COO

OH

HO



other biologically active lipids

O OH Thromboxane

FIGURE 10.15 Arachidonate conversion to eicosanoid signal molecules. The first step is catalyzed by cyclooxygenase. Only two of the many dozens of eicosanoids are shown here.

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CH APTER 1 0 Signaling

Box 10.C Aspirin and Other 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 acetylsalicylate, 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 universal 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. 10.15). 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 H3C

CH2

CH

O HO

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 the drugs Celebrex and Vioxx. These compounds bind only to the active site of COX-2 (they are too large to fit into the COX-1 active site) and therefore can selectively block the production of pro-inflammatory eicosanoids without damaging gastric tissue. Unfortunately, the side effects of these drugs include an increased risk of heart attacks, through a mechanism that is not fully understood, and as a result Vioxx has been taken off the market and the use of Celebrex is limited. 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

NH

C

CH3

Acetaminophen Q Which of the drugs shown here is chiral (see Section 4.1), with two different configurations?

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.

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.

Problems

10.2

G Protein Signaling Pathways

• A ligand such as epinephrine binds to a G protein–coupled 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. • 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 phospholipase C generate inositol trisphosphate and diacylglycerol second messengers, which activate protein kinase B and protein kinase C, respectively.

277

• 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.

10.3

Receptor Tyrosine Kinases

• Many receptor tyrosine kinases are dimeric molecules with a single ligand-binding site. Extracellular ligand binding activates the cytoplasmic tyrosine kinase domains 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 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

Problems 10.1 General Features of Signaling Pathways 1. Which of the signal molecules listed in Table 10.1 would not require a cell-surface receptor? 2. The structure of the drug prednisone is shown. What kind of molecule is this and by what pathway is it likely to exert its effects?

OH O O CH3

O

CH3

OH

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.

278

CH APTER 1 0 Signaling

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.

16. In the study described in Problem 15, 160,000 high-affinity binding sites were identified on each platelet. What is the concentration of ADP required to achieve 85% binding? Use the equation you derived in Problem 11. 17. Like the Michaelis–Menten equation, the equation derived in Problem 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. 10

1.0

8

[R . L] 0.6 [R]T

1/[R . L] (μM1)

6 4 2

0.2 4 0.1

0.2

0.3

0.4

0.5

0.6

3

2

0.7

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 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-methythioADP 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?

1

(μM1)

2

3

4

18. A Scatchard plot is another method of representing ligand binding data using a straight line. 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 estimate a value for Kd for calmodulin binding to calcineurin. 7 6

y = 0.067x  6.890

5

[R . L]/[L]

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 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 how to proceed.)

0

1/[L]

[L] (mM) 10. 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 was repeated in the presence of caffeine?

1

4 3 2 1 20

40

60

80

100

[R . L] (nM) 19. Why might it be difficult to purify cell-surface receptors using the techniques described in Section 4.6? 20. 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. 21. 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. 22. In the liver, both 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? 23. 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?

Problems

24. 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

O N O

A active O

Binactive

P O

Bactive

O S

P O

O O

P

N O

O GTPS

Cactive

Dinactive

Dactive

10.2 G Protein Signaling Pathways 25. 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? 26. 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. 27. 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. 28. Mutations in a GPCR gene (see Problem 27) occasionally result in increased receptor activity. List the types of mutations in a GPCR that would result in gain of function of the receptor. 29. 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? 30. 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? 31. How do epinephrine and norepinephrine differ from tyrosine, their parent amino acid? 32. Why are the antagonists known as β-blockers effective at treating high blood pressure? 33. 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? 34. 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 ADPribose 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? 35. 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?

NH N

NH2

CH2 O H H

Cinactive

279

H H

OH

OH

36. 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. 37. How does the addition of a single phosphate group to a protein change that protein’s activity? 38. 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. 39. 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? 40. 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 sphingomyelinasecatalyzed hydrolysis of sphingomyelin to ceramide. 41. 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 essential for T cell activation. Describe the cell-signaling events that result in the activation of NFAT. 42. The immunosuppressive drug cyclosporine A is an inhibitor of calcineurin (see Problem 41). Why is cyclosporine A an effective immunosuppressant? 43. 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? 44. Would you expect to find mutations in the gene for PTEN (see Problem 43) in human cancers? Explain why or why not. 45. 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

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CH APTER 1 0 Signaling

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, resulting in smooth muscle cell relaxation. How might protein kinase G bring about smooth muscle relaxation? 46. As discussed in the text, any signal transduction event that is turned on must subsequently be turned off. Refer to your answer to Problem 45 and describe the events that would lead to the cessation of each step of the signaling pathway you described. 47. 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. 48. Clotrimazole is a calmodulin antagonist (see Solution 45b). How does the addition of clotrimazole affect endothelial cells in culture? 49. 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). But 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

O

O

O O



N

N 

O

Nitroglycerin 50. 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.

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? 56. 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. 10.12). This step is essential for the downstream activation of protein kinases B and C (see Problem 55). 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? 57. 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. How is downstream activity of a signaling pathway affected by the presence of GEF? By the presence of GAP? 58. 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? 59. As shown in Figure 10.12, 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 39) promote tumor development.

51. Bacillus anthracis, the cause 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. 52. Bacillus anthracis also makes a toxin called lethal factor (LF; see Problem 51). 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. 53. 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. 54. Hydrogen peroxide acts as a second messenger, as described in Problem 53, and affects PTEN (see Problem 43) as well as other cellular phosphatases. Does H2O2 activate or inhibit PTEN?

10.3

Receptor Tyrosine Kinases

55. Stimulation of the insulin receptor by ligand binding and autophosphorylation eventually leads to the activation of both protein

Ras

Protein kinase C

Raf MEK MAP kinase Gene regulatory proteins

Transcription factors

Cell cycle proteins

60. How might a signaling molecule activate the MAP kinase cascade (see Problem 59) via a G protein–coupled receptor rather than a receptor tyrosine kinase? 61. Ras can activate a cascade of reactions (see Fig. 10.12) involving membrane phosphatidylinositol (PI) lipids (see Solutions 8.19 and 8.20). 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 43). Why would PI3K inhibitors be effective at treating cancer?

Problems

62. PTEN (see Problem 43) can catalyze the hydrolysis of PIP3 to PIP2 (see Problem 61). Does PTEN promote cell survival or cell death? 63. 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: ∆G = ∆G °ʹ + RT ln

[C] [D] [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. TA B L E 12. 3 When the reaction is at equilibrium, ΔG = 0 and ∆G°ʹ = −RT ln

[C]eq [D]eq [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

Biochemical Standard State

Temperature

25°C (298 K)

Pressure

1 atm

Reactant concentration

1M

pH

7.0 ([H+] = 10–7 M)

Water concentration

55.5 M

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CH APTER 1 2 Metabolism and Bioenergetics

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.

SAMP LE CA LCU LAT I O N 1 2 . 2 SEE SAMPLE CALCULATION VIDEOS

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 glucose6-phosphate is 25 mM. Is the reaction spontaneous under these conditions? ⫺2O POCH 3 2

HOCH2 H HO

O H OH

H

H

OH

H

H OPO2⫺ 3

HO

Glucose-1-phosphate

O H OH

H

H

OH

H OH

Glucose-6-phosphate

Solution The equilibrium constant Keq can be derived by rearranging Equation 12.2. K eq = e−∆G °ʹ/RT = e−(−7100 J · mol = e2.87 = 17.6

−1

)/(8.3145 J · K−1 · mol−1)(298 K)

At 37°C, T = 310 K. ∆G = ∆G °ʹ + RT ln

[glucose-6-phosphate] [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.

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 processes because its reactions occur with a relatively large negative change in free energy. Adenosine triphosphate (ATP) contains two phosphoanhydride bonds (Fig. 12.12). 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 a phosphoryl group is transferred to water—in other words, hydrolysis of the phosphoanhydride bond, such as ATP + H2O → ADP + Pi This is a spontaneous reaction with a ΔG°′ value of –30 kJ · mol–1.

Free Energy Changes in Metabolic Reactions

NH2 phosphoanhydride bonds O⫺ O⫺ O⫺ ⫺O

␥

P O

O

P ␤ O

O

P ␣ O

O

CH2 H

H

FIGURE 12.12 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. A reaction in which one or two phosphoryl groups are transferred to another compound (a reaction in which a phosphoanhydride bond is cleaved) has a large negative value of ΔG°′.

N

N

N

N O H

H

HO OH Adenosine AMP ADP

Q How does hydrolysis of a phosphoanhydride bond affect the net charge of a nucleotide?

ATP The following example illustrates the role of ATP in a coupled reaction. Consider the phosphorylation of glucose by inorganic phosphate (HPO2− or Pi), a thermodynamically 4 unfavorable reaction (ΔG°′ = +13.8 kJ · mol–1):

H HO

CH2OH O H OH H H

H

H ⫹ Pi OH

HO

CH2OPO2⫺ 3 O H H OH H OH H

OH

Glucose

317

⫹ H 2O

OH

Glucose-6-phosphate

When this reaction is combined with the ATP hydrolysis reaction, the values of ΔG°′ for each reaction are added:

glucose + Pi → glucose-6-phosphate + H2O ATP + H2O → ADP + Pi

∆G °ʹ +13.8 kJ·mol−1 −30.5 kJ·mol−1

glucose + ATP → glucose-6-phosphate + ADP

−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’s 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). But 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 free 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

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CH APTER 1 2 Metabolism and Bioenergetics

ADP  Pi

Nutrient

catabolism

Product

anabolism

ATP

Waste product

Precursor

In this scheme, it appears that the “energy” of the catabolized nutrient is transferred to ATP, then the “energy” of ATP is transferred to another product in a biosynthetic reaction. However, free 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 delocalization 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. Terminal phosphoryl group of ATP

Inorganic phosphate (Pi )

O

O

O

P

ADP

O

O

O O 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

–45.6

Phosphocreatine

–43.1

ATP → ADP + Pi

–30.5

Glucose-1-phosphate

–20.9

PPi → 2 Pi

–19.2

Glucose-6-phosphate

–13.8

Glycerol-3-phosphate

–9.2

P O

P

OH

O

2

ADP

O

O

3

P

O  H

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.

Free 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.

Free Energy Changes in Metabolic Reactions

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: 

H2N



H2N

O O H2O

C

NH

P

N

CH3 O

Pi

C

NH2

N

CH3

CH2

CH2

COO

COO

Phosphocreatine

Creatine

Creatine has lower free 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 produce ATP (Box 12.C). 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 Because the reactants and products are energetically equivalent, ΔG°′ values for these reactions are near zero. Another class of compounds that can release a large amount of free energy upon hydrolysis 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

CoA

CH3

C

O



CoA

SH

Acetyl-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 cells, releases enough free 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

319

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CH APTER 1 2 Metabolism and Bioenergetics

Box 12.C Powering Human Muscles 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. Cells cannot stockpile ATP; its concentration remains remarkably stable (between 2 and 5 mM in most cells) under widely varying levels of demand. Without phosphocreatine, a muscle would exhaust its ATP supply before it could be replenished by other, slower processes. Different types of physical activity make different demands on a muscle’s ATP-generating mechanisms. A single burst of activity is powered by the available ATP. Activities lasting up to a few seconds require phosphocreatine to maintain the ATP supply. Phosphocreatine itself is limited, so continued muscle contraction must rely on ATP produced by catabolizing glucose (obtained from the muscle’s store of glycogen) via glycolysis. 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 anaerobically (without the participation of O2). To continue its activity, it must switch to aerobic (O2-dependent) metabolism and further oxidize glucose via the citric acid cycle. The muscle also catabolizes Anaerobic systems

ATP

fatty acids, whose products also enter the citric acid cycle. 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 anaerobic to aerobic metabolism after about a minute and a half. In world-class athletes, the breakpoint occurs at about 150 to 170 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 reddishbrown color of the mitochondria gives 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.

Q 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

[Figure adapted from McArdle, W. D., Katch, F. I., and Katch, V. L., Exercise Physiology (2nd ed.), p. 348, Lea & Febiger (1986).]

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 free energy change of 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

Free Energy Changes in Metabolic Reactions

driven by the free 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

Free energy

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.

G

dam 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 gotten 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 free energy and energy “currencies” used by cells. • Describe the reasons why cells control metabolic reactions that have large negative free energy changes.

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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

• 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 is related to the actual cellular concentrations of reactants and products.

• Series of reactions known as metabolic pathways break down and synthesize biological molecules. Several pathways make use of the same small molecule intermediates.

• A thermodynamically unfavorable reaction may proceed when it is coupled to a favorable process involving ATP, whose phosphoanhydride bonds release a large amount of free energy when cleaved.

• 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.

• 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 metabolic pathway intermediate metabolite glycolysis citric acid cycle oxidation reduction redox reaction

cofactor coenzyme oxidative phosphorylation flux transcriptomics transcriptome microarray (DNA chip) proteomics proteome metabolomics metabolome

essential compound 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 elegans, a nematode worm; f. The Thiothrix bacteria, which oxidize hydrogen sulfide; g. Cyanobacteria (erroneously termed “blue-green algae” in the past).

Problems

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 cleaves the glycosidic bond between the second and third residues. The enzyme cannot accommodate branched chains. a. What are the main products of amylase digestion? b. 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. 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? 7. Unlike the monosaccharides described in Problem 6, sugar alcohols such as sorbitol (see Solution 11.17c) are absorbed via passive diffusion. Why? What process occurs more rapidly, passive diffusion or passive transport? 8. 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? 9. 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? 10. Hydrolysis of proteins begins in the stomach, catalyzed by the hydrochloric acid secreted into the stomach by parietal cells. Draw the reaction that shows the hydrolysis of a peptide bond. 11. How does the low pH of the stomach affect protein structure in such a way that the proteins are prepared for hydrolytic digestion? 12. Like the serine proteases (see Section 6.4), pepsin is made 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 acids 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? 13. The cleavage of peptide bonds in the stomach is catalyzed both by hydrochloric acid (see Problem 10) and by the stomach enzyme pepsin (see Problem 12). Peptide bond cleavage 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. 14. 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

323

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? 15. 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. 16. 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? 17. 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 acid promotes emulsification of fats in the stomach; that is, the products are more easily incorporated into micelles. Explain why. 18. Most of the fatty acids produced in the reaction described in Problem 17 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. 19. 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. 20. Some cholesterol is converted back to cholesteryl esters in the epithelial cells lining the small intestine (the reverse of the reaction described in Problem 19). 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. 21. 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. 22. 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. 23. The phosphorolysis reaction that removes glucose residues from glycogen 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? 24. 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.

12.2

Metabolic Pathways

25. 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.

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CH APTER 1 2 Metabolism and Bioenergetics

Acetyl-CoA

GAP

Pyruvate

Glycolysis Citric acid cycle Fatty acid metabolism TAG synthesis Photosynthesis Transamination 26. The pyruvate → acetyl CoA reaction can proceed only in the direction indicated. Given this limitation, consult Figure 12.10 and tell which of the following transformations are possible: a. acetyl-CoA → glucose; b. acetyl-CoA → fatty acids; c. acetyl-CoA → alanine. 27. For each of the (unbalanced) reactions shown below, tell whether the reactant is being oxidized or reduced. a. A reaction from the catabolic glycolytic pathway

O

H

O

C

H

C

OH

H

CH2OPO2 3

HPO42

C

OPO32

C

OH

CH2OPO32

33. 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?

H C

CH2

C

C

H

O

R

CH2

SCoA

CH2

CH2

C

34. Would you expect vitamin A to be more easily absorbed from raw or from cooked carrots? Explain.

SCoA

35. Refer to Table 12.2 and identify the vitamin required to accomplish each of the following reactions:

O c. A reaction associated with the catabolic glycolytic pathway

a.

C

O

C

O

CH3

H

CH2OPO32 O OH H OH H H

C C

OH

H

OH

O

C

N

H C

b. COO

CH2OPO32 O H OH H

C

COO

O  ATP  HCO 3

O  ADP  Pi

C

CH2

CH3

COO c. COO

O

OH

C

H

3N

O 

CH3 H

C

OH

CH3

H

H C

N O

d. A reaction associated with the anabolic pentose phosphate pathway

OH

O

O

O

H

31. Vitamin B12 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. 32. 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.

b. A reaction from the fatty acid synthesis pathway

R

30. A body-building 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?

OH

28. For each of the reactions shown in Problem 27, identify the cofactor as NAD+, NADP+, NADH, or NADPH. 29. 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: − CH4 + SO2− 4 → _____ + HS + H2O

b. Identify the reaction component that undergoes oxidation. c. Identify the component that undergoes reduction.

CH CH2

H N 3

CH CH3

CH2 COO

O  CoA



C

O

CH2 CH2 COO

d. COO

C

COO

COO

COO

O SH

H3C

C

S

CoA  CO2

CH3 36. Why is niacin technically not a vitamin? 37. The microbial enzyme lactate racemase contains a nickel-based prosthetic group. a. To which common coenzyme is this prosthetic

Problems

group related? b. Which two amino acid side chains hold the group in place?

NH

O

O O

P

Ni2

S N H

N

O O

O H

H

H

OH

OH

47. Calculate ΔG for the A ⇌ B reaction described in Problem 41 when the concentrations of A and B are 0.9 mM and 0.1 mM, respectively. In which direction will the reaction proceed? 48. Calculate ΔG for the C ⇌ D reaction described in Problem 41 when the concentrations of C and D are 0.9 mM and 0.1 mM, respectively. In which direction will the reaction proceed?

N S

325

H

38. 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 S2– to SO42– or the conversion of SO42– to S2– ? 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+?

49. a. The ΔG°′ 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 ΔG°′ value is twice as large. b. Carry out the same exercise for a hypothetical reaction whose ΔG°′ value is –10 kJ · mol–1. 50. a. Use the standard free energies provided in Table 12.4 to calculate ΔG°′ for the isomerization of glucose-1-phosphate to glucose-6-phosphate. Is this value consistent with the value of ΔG°′ shown in Sample Calculation 12.2? Is this reaction spontaneous under standard conditions? b. Is the reaction 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? 51. Calculate ΔG for the hydrolysis of ATP under cellular conditions, where [ATP] = 3 mM, [ADP] = 1 mM, and [Pi] = 5 mM. 52. The standard free energy change for the reaction catalyzed by triose phosphate isomerase is 7.9 kJ · mol–1.

H

H

12.3 Free Energy Changes in Metabolic Reactions 39. 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? 40. 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 39? 41. Consider two reactions: A ⇌ B and C ⇌ D. Keq 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. 42. Calculate the ΔG°′ values for the reactions described in Problem 41. Assume a temperature of 37°C. 43. Consider the reaction E ⇌ F, Keq = 1. a. Without doing any calculations, what can you conclude about the ΔG°′ 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. 44. Refer to the hypothetical reaction described in Problem 43. Determine the direction 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?

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. b. Calculate ΔG 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? 53. An apple contains about 72 Calories. Express this quantity in terms of ATP equivalents (that is, how many ATP → ADP + Pi reactions?). 54. 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 53). 55. Construct a graph plotting free energy versus the reaction coordinate for the following reactions: a. glucose + Pi → glucose6-phosphate; b. ATP + H2O → ADP + Pi; c. the coupled reaction. 56. 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? 57. The ΔG°′ 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?

45. 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?

58. The ΔG°′ for the formation of UDP–glucose from glucose1-phosphate and UTP is about zero. Yet the production of UDP– glucose is highly favorable. What is the driving force for this reaction?

46. Calculate the ratio of the concentration of glucose-6-phosphate (G6P) to the concentration of glucose-1-phosphate (G1P) (see Problem 45) that gives a free energy change of –2.0 kJ · mol–1. Assume a temperature of 37°C.

59. a. The complete oxidation of glucose releases a considerable amount of energy. The ΔG°′ for the reaction shown is –2850 kJ · mol–1.

glucose-1-phosphate + UTP ⇌ UDP–glucose + PPi

C6H12O6 + 6 O2 ⟶ 6 CO2 + 6 H2O

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CH APTER 1 2 Metabolism and Bioenergetics

How many moles of ATP could be produced under standard conditions from the oxidation of one mole of glucose, assuming about 33% efficiency? b. The oxidation of palmitate, a 16-carbon saturated fatty acid, releases 9781 kJ · mol–1. C16H32O2 + 23 O2 →16 CO2 + 16 H2O How many moles of ATP could be produced under standard conditions from the oxidation of one mole of palmitate, assuming 33% efficiency? c. Calculate the number of ATP molecules produced per carbon for glucose and palmitate. Explain the reason for the difference. 60. 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. Considering this fact and your answer to part b, suggest an explanation that is consistent with these findings. 61. a. How many apples (see Solution 53) would be required to provide the amount of ATP calculated in Problem 60? b. How many large hot chocolate drinks (see Solution 54) would be required? 62. 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? 63. Citrate is isomerized to isocitrate in the citric acid cycle (Chapter 14). The reaction is catalyzed by the enzyme aconitase. The ΔG°′ of 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? 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?

COO

CH2 HO

C

COO

CH2

COO

CH2 aconitase

COO

Citrate

HC HO

CH

COO COO

Isocitrate

64. 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.

CH2OPO2 3 O phosphoglucose H H H isomerase OH H OH OH H

OH

Glucose-6-phosphate

CH2OPO2 3 O CH2OH HO H OH H OH

H

Fructose-6-phosphate

a. What is the ΔG°′ for this reaction? Would this reaction proceed in the direction written under standard conditions? b. What is the ΔG

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? 65. 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 was 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? 66. Another way to promote the formation of glucose-6-phosphate from glucose (see Problem 65) is to couple the phosphorylation of glucose to the hydrolysis of ATP as shown in Section 12.3. a. Calculate Keq 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 glucose6-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 65) or the coupling of this phosphorylation to ATP hydrolysis? Explain. 67. Fructose-6-phosphate is phosphorylated to fructose-1,6-bisphosphate in the third step of the glycolytic pathway (Chapter 14). The phosphorylation of fructose-6-phosphate is described by the following equation: fructose-6-phosphate + Pi ⇌ fructose-1,6-bisphosphate ΔG°′ = 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 ΔG°′ 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 part a and c. What is the likely path for the cellular synthesis of fructose-1,6-bisphosphate? 68. One can envision two mechanisms for coupling ATP hydrolysis to the phosphorylation of fructose-6-phosphate (F6P) to fructose-1,6-bisphosphate (F16BP) (see Problem 67), yielding the same overall reaction: Mechanism I: ATP is hydrolyzed as F6P is transformed to F16BP: F6P + Pi ⇌ F16BP ATP + H2O ⇌ ADP + Pi Mechanism II: ATP transfers its γ phosphate directly to F6P in one step, producing F16BP. F6P + ATP + H2O ⇌ F16BP + ADP

Problems

Choose one of the above mechanisms as the more biochemically feasible and provide a rationale for your choice. 69. Glyceraldehyde-3-phosphate (GAP) is eventually converted to 3-phosphoglycerate (3PG) in the glycolytic pathway.

H

H

O

C

O

C

OH

H

CH2OPO2 3

C

O

C

OH

CH2OPO2 3

Glyceraldehyde-3-phosphate (GAP)

3-Phosphoglycerate (3PG)

O

H

C

OPO2 3

C

OH

327

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 ΔG°′ 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 2 Pi. 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 ΔG°′. 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. 72. DNA containing broken phosphodiester bonds (“nicks”) can be repaired by the action of a ligase enzyme. Formation of a new phosphodiester bond in DNA requires the free energy of ATP, which is hydrolyzed to AMP:

CH2OPO2 3 1,3-Bisphosphoglycerate (1,3-BPG)

ATP  nicked bond

Consider these two scenarios: I. GAP is oxidized to 1,3-BPG (ΔG°′ = 6.7 kJ · mol–1), which is subsequently hydrolyzed to yield 3PG (ΔG°′ = –49.3 kJ · mol–1) II. GAP is oxidized to 1,3BPG, which then transfers its phosphate to ADP, yielding ATP (ΔG°′ = –18.8 kJ · mol–1). Write the overall equations for the two scenarios. Which is more likely to occur in the cell, and why? 70. 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:

ligase AMP  PPi  phosphodiester bond

The equilibrium constant expression for this reaction can be rearranged to define a constant, C, as follows:

Keq =

[phosphodiester bond][AMP][PPi] [nick] [ATP]

[AMP] [PPi] [nick] = [phosphodiester bond] Keq [ATP] C=

Mechanism I: glutamate + NH3 ⇌ glutamine

[PPi] Keq [ATP]

[nick] = C[AMP] [phosphodiester bond]

ATP + H2O ⇌ ADP + Pi Mechanism II: glutamate + ATP ⇌ γ-glutamylphosphate + ADP γ-glutamylphosphate + H2O + NH3 ⇌ glutamine + Pi Write the overall equation for the reaction for each mechanism. Is one mechanism more likely than the other? Or are both mechanisms equally feasible for the conversion of glutamate to glutamine? Explain. 71. Palmitate is activated in the cell by forming a thioester bond to coenzyme A. The ΔG°′ for the synthesis of palmitoyl-CoA from palmitate and coenzyme A is 31.5 kJ · mol–1.

O H 3C

(CH2)14

C

O  CoA O

H3C

(CH2)14

C

S

CoA  H2O

a. What is the ratio of products to reactants at equilibrium for the reaction? 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 ΔG°′ for the reaction. What is the ratio of products to reactants at equilibrium

a. The ratio of nicked bonds to phosphodiester bonds at various concentrations of AMP was determined. Using the data provided, construct a plot of [nick]/[phosphodiester bond] versus [AMP] and determine the value of C from the plot. [AMP] (mM)

[Nick]/[Phosphodiester bond]

10 20 30 40 50

4.01 × 10–5 5.47 × 10–5 8.67 × 10–5 9.30 × 10–5 1.13 × 10–4

b. Determine the value of Keq for the reaction in which the concentrations of PPi and ATP are 1.0 mM and 14 μM, respectively. c. What is the value of ΔG°′ for the reaction? d. What is the value of ΔG°′ for nicked bond → phosphodiester bond reaction? (Note: The ΔG°′ for the hydrolysis of ATP to AMP and PPi is –48.5 kJ · mol–1 under the conditions used in the experiment.) e. The ΔG°′ for the hydrolysis of a typical phosphomonoester bond is –13.8 kJ · mol–1. Compare the stability of the phosphodiester bond in DNA to the stability of a typical phosphomonoester bond.

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CH APTER 1 2 Metabolism and Bioenergetics

Selected Readings Falkowski, P. G., Fenchel, T., and Delong, E. F., The microbial engines that drive Earth’s biogeochemical cycles, Science 320, 1034–1038 (2008). [Discusses the diversity and interconnectedness of metabolic processes.] Hanson, R. W., The role of ATP in metabolism, Biochem. Ed. 17, 86–92 (1989). [Provides an excellent explanation of why ATP is an energy transducer rather than an energy store.] 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.]

Smolin, L. A. and Grosvenor, M. B, Nutrition: Science and Applications, Wiley (2013). [Includes health-related information on macromolecular nutrients and vitamins.] Wishart, D. S., Jewison, T., Guo, A. C., Wilson, M., Knox, C., et al., HMDB 3.0—the human metabolome database in 2013. Nuc. Acids Res. 41(D1), D801–807. doi: 10.1093/nar/gks1065 (2013). [Describes the human metabolome database, with over 40,000 entries. Available at www.hmdb.ca.]

CHAPTER 13

Dmytro Pylypenko/shutterstock

Glucose Metabolism

Weddell seal pups are born with huge brains—75% of their adult size, compared to 25% for a human newborn—presumably to help them quickly learn to navigate under ice. Along with this extraordinarily large brain comes a high demand for glucose, the brain’s primary fuel, which must be supplied by the mother seal’s milk.

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.

329

330

CH APTER 1 3 Glucose Metabolism

L EARNING OBJECTIVES Describe the substrates, products, and chemical reaction for each step of glycolysis. • Compare the energyconsuming and energygenerating phases of glycolysis. • List the flux-control points for the pathway. • Describe the metabolic uses of pyruvate.

13.1

Pathways dealing with carbohydrates, highlighted in Figure 13.1, are part of the larger metabolic scheme introduced in Figure 12.11. 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:

BIOPOLYMERS Polysaccharides

1

4

MONOMERS Monosaccharides +

NH4

NAD+

2

3

NAD+

NADH

NADH, QH2 2- and 3-Carbon INTERMEDIATES

NAD+, Q

citric acid cycle

photosynthesis

CO2 NADH, QH2 ADP

O2

oxidative phosphorylation ATP

H2O NAD+, Q

Glycolysis

1. Each step of the pathway is catalyzed by a distinct enzyme. 2. The free 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 free 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 free energy all at once. In the cell, glucose combustion requires many steps so that the cell can recover its free 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.

Reactions 1–5 are the energy-investment phase of glycolysis The first five reactions of glycolysis can be considered a preparatory phase for the second, energy-producing phase. In fact, the first phase requires the investment of free energy in the form of two ATP molecules.

FIGURE 13.1 Glucose metabolism 1. Hexokinase In the first step of glycolysis, the enzyme hexokinase transfers a phosin context. (1) The polysaccharide phoryl group from ATP to the C6 OH group of glucose to form glucose-6-phosphate: glycogen is degraded to glucose, which is then catabolized by the CH2OH CH2OPO32⫺ glycolytic pathway (2) to the threecarbon intermediate pyruvate. O O H H H H Gluconeogenesis (3) is the pathway H H for the synthesis of glucose from ⫹ ADP ⫹ ATP OH H OH H hexokinase smaller precursors. Glucose can then OH OH HO HO be reincorporated into glycogen (4). The conversion of glucose to ribose, H H OH OH a component of nucleotides, is not Glucose Glucose-6-phosphate shown in this diagram.

CH2OH O H OH H

Glucose H

ATP

1 hexokinase

ADP

HO

FIGURE 13.2 The reactions of glycolysis. The substrates, products, and enzymes corresponding to the 10 steps of the pathway are shown. Shading indicates the substrates (purple) and products (green) of the pathway as a whole.

H OH

H OH O3POCH2 O H H H OH H OH HO

⫺2

Glucose-6-phosphate phosphoglucose 2 isomerase

H ⫺2

Fructose-6-phosphate ATP

3 phosphofructokinase

ADP

⫺2

Fructose-1,6-bisphosphate

HO

HO

H

O

Dihydroxyacetone phosphate triose phosphate 5 isomerase

2 Pi ⫹ 2 NAD⫹ 2 NADH ⫹ 2 H⫹

glyceraldehyde-36 phosphate dehydrogenase

2 1,3-Bisphosphoglycerate 2 ADP 2 ATP

H

OH

C C

Energy Investment Phase

CH2OPO2⫺ 3

O3POCH2 O

4 aldolase

⫹

CH2OH

H

H

HO

HO

H

H

Glyceraldehyde3-phosphate

OH

O3POCH2 O H

Q Next to each reaction, write the term that describes the type of chemical change that occurs.

OH

H

CH2OPO2⫺ 3 C

OH

CH2OPO2⫺ 3

O H

C C

O

CH2OH

OPO2⫺ 3 OH

CH2OPO2⫺ 3

7

phosphoglycerate kinase O

2 3-Phosphoglycerate

H

C C

O⫺ OH

CH2OPO2⫺ 3

phosphoglycerate 8 mutase O

2 2-Phosphoglycerate

H

C C

O⫺

Energy Payoff Phase

OPO2⫺ 3

CH2OH

2 H2O

9 enolase O

C

O⫺

2 Phosphoenolpyruvate

C

2 ADP

CH2

2 ATP

10 pyruvate kinase

2 Pyruvate

O

C C

OPO2⫺ 3

O⫺ O

CH3

331

332

CH APTER 1 3 Glucose Metabolism

A kinase is an enzyme that transfers a phosphoryl group from ATP (or another nucleoside triphosphate) to 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 prevents glucose from backing out of glycolysis. 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: 2O

6 3POCH2 5

H 4

HO

H OH 3

H

O H

6 2O POCH 3 2

H 1

OH

2

5

phosphoglucose isomerase

1

O

H

CH2OH

HO

H

OH 4

3

HO

OH

Glucose-6-phosphate

2

H

Fructose-6-phosphate

Because fructose is a six-carbon ketose (Section 11.1), it forms a five-membered ring. 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-1,6-bisphosphate. 6 3POCH2

2O

5

H

H 4

HO

1

O

2O

CH2OH

HO

2

OH 3

 ATP

phosphofructokinase

6 3POCH2 5

H

H 4

H

HO

Fructose-6-phosphate

1

CH2OPO32

O HO

2

 ADP

OH 3

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 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 fructose6-phosphate 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.

Glycolysis

(a)

(b)

Glucose

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.

Glucose ATP

Fructose-6-phosphate

Fructose-2,6bisphosphate

Fructose-6-phosphate

Pi ADP

Fructose-1,6-bisphosphate Fructose-1,6-bisphosphate Phosphoenolpyruvate Pyruvate Pyruvate The most potent activator of phosphofructokinase in mammals is the compound fructose-2,6-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

2O

3POH2C

H

HO

H

OH

HO

ATP

CH2OH

ADP

O

2O POH C 3 2

phosphofructokinase-2

H

Fructose-6-phosphate

CH2OH

H

HO

HO

H

H

O

4. Aldolase Reaction 4 converts the hexose fructose-1,6-bisphosphate to two three-carbon molecules, each of which bears a phosphate group.

CH2OPO32

CH2OPO2 3 1 C

O

HO

C

H

H

C

OH

H

C

OH

2 3 4 5

CH2OPO32 6 Fructose1,6-bisphosphate

PO32

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.

3

C

2

HO aldolase

O

CH2 1

Dihydroxyacetone phosphate

 O

C

H

1

H

C

2

OH

CH2OPO32 3

Glyceraldehyde3-phosphate

333

334

CH APTER 1 3 Glucose Metabolism

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.)

NH2 NH2

(CH2)4 Lys



O

O C



O

O

CH2

C

CH2OPO2 3

CH2

Asp

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.

HO

O

C

H

H C

O

H2O

H

OH

Fructose-1,6bisphosphate

C

H2N

H



(CH2)4

CH2OPO2 3

HO

O

O C

CH2OPO2 3

O

CH2OH Dihydroxyacetone phosphate

CH2OPO2 3 C

(CH2)4



C

NH

C

H

(CH2)4

H

CH2



O

O C

2. The active-site Lys residue reacts

CH2

with the substrate’s carbonyl group to form a Schiff base.

H2O

4. The Asp residue gives up

its proton to the remainder of the substrate, re-forming a Schiff base.

CH2OPO2 3 

C

NH

C

H

H C

O

H

OH

HO

C

(CH2)4 

O

O H

C CH2

CH2OPO2 3

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.

CH2OPO2 3 C HO

C H

NH

(CH2)4

H O

O C

CH2 H H

C C

O OH

CH2OPO2 3 Glyceraldehyde-3-phosphate FIGURE 13.4 The aldolase reaction.

Q Does this reaction follow an ordered or ping pong mechanism (see Section 7.2)?

SEE ANIMATED PROCESS DIAGRAM The enzymatic mechanism of aldolase

Glycolysis

335

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

O

HC

C

HI

CH2

SH  ICH2COO

HC

Iodoacetate

NH

O 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-3-phosphate— 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

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; see Section 6.3). In addition, the catalytic power of triose phosphate isomerase depends on a protein loop that closes over the active site (Fig. 13.5). FIGURE 13.5 Conformational changes in yeast triose phosphate isomerase. (a) One loop of the protein, comprising residues 166–176, is highlighted in green. (b) When a substrate binds to the enzyme, the loop closes over the active site to stabilize the reaction’s transition state. 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. [Structure

(a)

(b)

of the enzyme alone (pdb 1YPI) determined by T. Alber, E. Lolis, and G. A. Petsko; structure of the enzyme with the analog (pdb 2YPI) determined by E. Lolis and G. A. Petsko.]

336

CH APTER 1 3 Glucose Metabolism

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.

Reactions 6–10 are the energy-payoff phase of glycolysis So far, the reactions of glycolysis have consumed 2 ATP, but this investment pays off in the second phase of glycolysis when 4 ATP are produced, for a net gain of 2 ATP. All of the reactions of the second phase involve three-carbon intermediates, but keep in mind that each glucose molecule that enters the pathway yields two of these three-carbon units. Some species convert glucose to glyceraldehyde-3-phosphate by different pathways than the one presented above. However, the second phase of glycolysis, which converts glyceraldehyde-3-phosphate to pyruvate, is 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

OH  NAD  Pi

2 3 CH2OPO3

H

glyceraldehyde-3-phosphate dehydrogenase

1C 2C

OPO2 3 OH  NADH  H

2 3 CH2OPO3

1,3-Bisphosphoglycerate

Glyceraldehyde3-phosphate

Unlike the kinases that catalyze Reactions 1 and 3, glyceraldehyde-3-phosphate dehydrogenase 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 (Chapter 15). An active-site cysteine residue participates in the glyceraldehyde-3-phosphate dehydrogenase reaction (Fig. 13.6). The enzyme is inhibited by arsenate (AsO43–), which competes with Pi (PO43–) for binding in the enzyme active site.

7. Phosphoglycerate Kinase The product of Reaction 6, 1,3-bisphosphoglycerate, is an acyl phosphate.

O R

C

O O

P

O

O An acyl phosphate

Glycolysis

337

NAD NAD S S

H

H

O 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

O

O

O



O S

C R

FIGURE 13.6

The glyceraldehyde-3-phosphate dehydrogenase reaction.

Q Identify the reactant that undergoes oxidation and the reactant that undergoes reduction.

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:

H

OH

NADH

R

O

P

NAD NADH

4. NADH exits and is replaced by another molecule of NAD . Pi enters the active site.

3. NAD  oxidizes the substrate, forming a thioester intermediate.

H C



C

NAD S

O

R

2. The SH group of the active-site Cys residue forms a covalent bond with glyceraldehyde-3-phosphate.

H

R H

O

O

R



OPO2 3

NAD

NAD H

C

1C 2C

OPO2 3 OH  ADP

2 3 CH2OPO3

1,3-Bisphosphoglycerate

O

phosphoglycerate kinase

H

1C 2C

O OH  ATP

2 3 CH2OPO3

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. 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.

SEE ANIMATED PROCESS DIAGRAM The enzymatic mechanism of GAPDH

338

CH APTER 1 3 Glucose Metabolism

8. Phosphoglycerate Mutase In the next reaction, 3-phosphoglycerate is converted to 2-phosphoglycerate:

O

O

O

O C1

C1 H

C2 OH

H

C3 OPO2 3

phosphoglycerate mutase

H

C2 OPO2 3

H

C3 OH

H

H

3-Phosphoglycerate

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 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

O

O

O

C

C

O

H

C

OH

H

C

OPO2 3

O

His

P

H

O

C

H C

H

O

O C

O

OPO2 His 3

H

C

OPO2 3

OPO2 3

H

C

OH

H

O

P

His

O

H

3-Phosphoglycerate

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

O C1 H

C2 OPO2 3

H

C3 OH

O

O C enolase

H 2-Phosphoglycerate

C H

OPO2  H2O 3

C H

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

O C

O

O

C

C OPO2  ADP 3

CH2

Phosphoenolpyruvate

pyruvate kinase

C

O  ATP

CH3

Pyruvate

Glycolysis

339

The reaction actually occurs in two parts. First, ADP attacks the phosphoryl group of phosphoenolpyruvate to form ATP and enolpyruvate:

O

O C C

O

O O

CH2

O

O P

O



O

O

ATP

P

O

AMP

C H

O

Phosphoenolpyruvate

C

ADP

OH

CH2 Enolpyruvate

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. ΔG°′ for this step is –46 kJ · mol–1, so ΔG°′ for the net reaction (hydrolysis of phosphoenolpyruvate 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, phosphofructokinase, 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. 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.

Glucose

1 2

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. [Data from Newsholme, E. A. and Start, C., Regulation in Metabolism, p. 97, Wiley (1973).]

O

O C C

O

O C

OH

CH2 Enolpyruvate

C

O

CH3 Pyruvate

340

CH APTER 1 3 Glucose Metabolism

The glycolytic pathway can be considered to be a sort of pipe, with an entrance for glucose (or glucose-6-phosphate) and an exit for pyruvate. The control points for the pathway are 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 second phase of glycolysis: Glyceraldehyde-3-phosphate is converted to pyruvate with the synthesis of 2 ATP (in Reactions 7 and 10). Since each molecule of glucose yields two molecules of glyceraldehyde-3-phosphate, the reactions of the second phase of glycolysis must be doubled, so the yield is 4 ATP. Two molecules of ATP are invested in phase 1, 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 glucose-6-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 metabolized 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 deposition. 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. Q 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

ATP CH2OH

O

HOCH2 H

HO

H

OH

HO

H

ADP

fructokinase

HOCH2 H H HO

O

CH2OPO32⫺ ⬅ HO HO OH H H

H

CH2OPO32⫺

CH2OPO32⫺

C

O

C

C

H

C

OH

C

OH

CH2OH Fructose

Fructose-1-phosphate

fructose-1phosphate aldolase

O

HO

CH2 ⫹ H O C

H

C

OH

CH2OH

Glyceraldehyde

Glycolysis Glucose

FIGURE 13.8 Fates of pyruvate. Pyruvate may be converted to a two-carbon acetyl group linked to the carrier coenzyme A. Acetyl-CoA may be further broken down by the citric acid cycle or used to synthesize fatty acids. In muscle, pyruvate is reduced to lactate to regenerate NAD+ for glycolysis. Yeast degrade pyruvate to ethanol and CO2. Pyruvate can also be carboxylated to produce the four-carbon oxaloacetate.

NAD+ NADH CO2

CO2 Pyruvate

Acetyl-CoA NADH NAD+

Lactate

Oxaloacetate NADH NAD+

Ethanol + CO2

Q Beside each arrow, write the names of the enzymes that catalyze the process.

Pyruvate is converted to other substances What happens to the pyruvate generated by the catabolism of glucose? It can be further broken down to acetyl-CoA or used to synthesize other compounds such as oxaloacetate. The fate of pyruvate depends on the cell type and the need for metabolic free energy and molecular building blocks. Some of the options are diagrammed 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

O

H

C C

H

O C

O

CH3

 N

O

O  H

lactate dehydrogenase

HO

C H  CH3

R Pyruvate

H

C NH2

Lactate

This reaction, sometimes called the eleventh step of glycolysis, allows the muscle to function anaerobically for a minute or so (see Box 12-C). The net reaction for anaerobic glucose catabolism is glucose + 2 ADP + 2 Pi → 2 lactate + 2 ATP 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, 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. 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) catabolizes the removal of pyruvate’s carboxylate group to produce acetaldehyde. Next, alcohol dehydrogenase reduces acetaldehyde to ethanol.

O C



N R

NADH

341

NAD

NH2

342

CH APTER 1 3 Glucose Metabolism

O

C

O

CO2

O

C

pyruvate decarboxylase

CH3 Pyruvate

O

C

 H NADH NAD

CH3

alcohol dehydrogenase

Acetaldehyde

OH 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. Although glycolysis is an oxidative pathway, its end product pyruvate is still a relatively reduced molecule. The further catabolism of pyruvate begins with its decarboxylation to form a two-carbon acetyl group linked to coenzyme A.

CoASH CO2

O CH3

C

COO

Pyruvate

pyruvate dehydrogenase

O CH3

C

SCoA

Acetyl-CoA

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 H

C

H

alcohol dehydrogenase

CH3 Ethanol

O

C

O

H

CH3 Acetaldehyde

acetaldehyde dehydrogenase

C CH3 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. Q 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?

Glycolysis

343

Standard Free Energy Changes The resulting acetyl-CoA is a substrate for the citric acid cycle (Chapter 14), TA B L E 13. 1 for Glucose Catabolism which converts the acetyl carbon atoms to CO2. The complete oxidation of the six glucose carbons to CO2 releases much more free energy than the conΔG°′ version of glucose to lactate (Table 13.1). Much of this energy is recovered CATABOLIC PROCESS (kJ · mol–1) in the synthesis of ATP by the reactions of the citric acid cycle and oxidative –196 C6H12O6 → 2 C3H5O3– + 2 H+ phosphorylation (Chapter 15), pathways that require the presence of molecu(glucose) (lactate) lar oxygen. Pyruvate is not always destined for catabolism. Its carbon atoms provide C6H12O6 + 6 O2 → 6 CO2 + 6 H2O –2850 the raw material for synthesizing a variety of molecules, including, in the (glucose) 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

C pyruvate carboxylase

O

 ADP  Pi

CH2

CH3

COO

Pyruvate

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:

O C H

Lys residue

NH

HN

C

C

H C H

H2C S

O CH2

CH2

Biotin

CH2 CH2

C

O NH

(CH2)4

C CH NH

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, 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.

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CH APTER 1 3 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. 

ATP



O

O

O C

HO

P

O

O 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

SEE ANIMATED PROCESS DIAGRAM The reaction mechanism of pyruvate carboxylase

O



O C

N

NH

O R S O

O C C

3. The enzyme abstracts a proton from pyruvate, forming a carbanion.

O

O

O

4. The carbanion attacks the carboxyl group attached to biotin, generating oxaloacetate.

C C

O Biotinyl-Lys



CH3

CH2

Pyruvate

O

O C C

O

CH2 C 

O

FIGURE 13.9 The pyruvate carboxylase reaction.

L EARNING 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.

O

Oxaloacetate

13.2

Gluconeogenesis

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 gluconeogenesis 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).

Gluconeogenesis

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, Glucose ATP

Pi

hexokinase

glucose-6-phosphatase

H2O

ADP

Glucose-6-phosphate phosphoglucose isomerase

Fructose-6-phosphate ATP Pi phosphofructokinase

fructose bisphosphatase

ADP glycolysis

H2O gluconeogenesis

Fructose-1,6-bisphosphate 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

GDP  CO2 phosphoenolpyruvate carboxykinase

GTP Oxaloacetate

ATP

ADP  Pi Pyruvate

pyruvate carboxylase

ATP



CO2

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.

Q Compare the ATP yield of glycolysis with the ATP consumption of gluconeogenesis.

345

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CH APTER 1 3 Glucose Metabolism

phosphoenolpyruvate carboxykinase catalyzes the decarboxylation of oxaloacetate to form phosphoenolpyruvate:

O

O C C

HCO3  ATP

ADP  Pi

O

O C C

O

pyruvate carboxylase

CH3

O

O C

O phosphoenolpyruvate carboxykinase

CH2 O

Pyruvate

GTP

CO2  GDP

C

O

PO2 3

CH2

C O

Oxaloacetate

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 phosphoenolpyruvate 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.) Two molecules of phosphoenolpyruvate are converted to one molecule of fructose-1,6bisphosphate 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. 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 catalyzed 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 carboxykinase, 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) carefully 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

Glycogen Synthesis and Degradation

347

and fructose-1,6-bisphosphate. We have already seen that fructose-2,6-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 FBPase

fructose-6-phosphate + ATP → fructose-1,6-bisphosphate + ADP fructose-1,6-bisphosphate + H2O → fructose-6-phosphate + Pi

net

ATP + H2O → ADP + Pi

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. LEARNING OBJECTIVES

13.3

Glycogen Synthesis and Degradation

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.

Compare the processes of glycogen synthesis and degradation. • Identify the substrates and products for each process. • Compare the free energy needs of each pathway. • List the metabolic fates of glucose-6-phosphate.

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CH APTER 1 3 Glucose Metabolism

Glycogen synthesis consumes the free 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: 2O

3POCH2

H HO

HOCH2 O H

H

OH

O

H

H

H OH

phosphoglucomutase

OH

HO

Glucose-6-phosphate

H

H OH

H

H

OH

OPO32

Glucose-1-phosphate

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

The reaction catalyzed by UDP–glucose pyrophosphorylase

O O

P

O

P

O

H HO

H

O

O

CH2OH O H OH H

O

O

O

P

O

OH2C

O

H

H

H

H OH

HO H

UTP

O

O

N

O

P O

OH

Glucose-1-phosphate UDP–glucose pyrophosphorylase

PPi

inorganic pyrophosphatase

2 Pi O

H HO

CH2OH O H OH H H

OH

HN H O

P O

O

O

O O

P O

UDP–glucose

N

O

OH2C H 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 phosphoanhydride bond makes the formation of UDP–glucose an exergonic, irreversible process— that is, PPi hydrolysis “drives” a reaction that would otherwise be near equilibrium. The hydrolysis of PPi by inorganic pyrophosphatase is a common strategy in biosynthetic reactions;

Glycogen Synthesis and Degradation

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. 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. The steps of glycogen synthesis can be summarized as follows:

UDP  UDP–Glucose glycogen synthase

Glycogen

UDP

UDP–glucose pyrophosphorylase pyrophosphatase glycogen synthase

glucose-1-phosphate + UTP ⇌ UDP–glucose + PPi PPi + H2O → 2 Pi UDP–glucose + glycogen (n residues) → glycogen (n +1 residues) + UDP

net

glucose-1-phosphate + glycogen + UTP + H2O → glycogen + UDP + 2 Pi

The energetic price for adding one glucose unit to glycogen is the cleavage of one phosphoanhydride bond of UTP. 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 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 can remove α(1 → 6)-linked residues by hydrolysis. In the liver, phosphoglucomutase converts glucose-1-phosphate to glucose-6-phosphate, which is then hydrolyzed by glucose-6-phosphatase to release free glucose.

H2O

Pi glycogen

glycogen phosphorylase

glucose-1-phosphate

phosphoglucomutase

glucose-6-phosphate

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 glycogenderived 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 regulated by a variety of mechanisms linked to hormonal signaling. Likewise, the activity of glycogen synthase is subject to hormonal control. 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 the advantage of breaking down glycogen by phosphorolysis rather than hydrolysis. Explain why only some tissues contain glucose-6-phosphatase.

Pi

glucose-6phosphatase

glucose

349

350

CH APTER 1 3 Glucose Metabolism

L EARNING OBJECTIVES

13.4 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.

The Pentose Phosphate Pathway

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+) and biosynthetic enzymes (which generally use NADP+). 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 CH2OPO32  O NADP NADPH H H OH H glucose-6-phosphate OH dehydrogenase H

H HO

CH2OPO32 O H OH H H

OH

Glucose-6-phosphate

O

OH

6-Phosphoglucono-lactone

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:

O C

CH2OPO32 H HO

O H OH H

H OH

H2O O

H

6-phosphogluconolactonase

O

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OPO32 6-Phosphoglucono-lactone

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:

The Pentose Phosphate Pathway

O C

O

H

C

OH

HO

C

H

H

C

OH

H

C

OH

NADP

CO2  NADPH

CH2OH

6-phosphogluconate dehydrogenase

C

O

H

C

OH

H

C

OH

CH2OPO32

CH2OPO2 3 Ribulose-5-phosphate

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:

CH2OH

O

C

O

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

OH

H

C

ribulose-5-phosphate isomerase

CH2OPO32

CH2OPO32

Ribulose-5-phosphate

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 ribose, it also provides a reducing agent (NADPH) required for the reduction of ribose to deoxyribose. Ribonucleotide reductase carries out the reduction of nucleoside diphosphates (NDPs):

O O

P O

O

O O

P

O

O

H2C

O

H NDP

base

H

H

OH

OH

H

O

P O

O O

P

O

O

H

base

O

H2C H

H

OH

H

dNDP

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

H

351

352

CH APTER 1 3 Glucose Metabolism

C5

C5

C5

C3

C7

C5

C6

C4

C5

C6

C6

C3

FIGURE 13.11 Rearrangements of the products of the pentose phosphate pathway. Three of the five-carbon products of the oxidative phase of the pentose phosphate pathway are converted to two fructose-6-phosphate and one glyceraldehyde-3-phosphate by reversible reactions involving the transfer of two- and three-carbon units. Each square represents a carbon atom in a monosaccharide. This pathway also allows ribose carbons to be used in glycolysis and gluconeogenesis.

Q Draw the structures of the carbohydrates corresponding to each shape.

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. A set of reversible reactions transform five-carbon ribulose units into six-carbon units (fructose-6-phosphate) and three-carbon units (glyceraldehyde-3-phosphate). These transformations are accomplished mainly by the enzymes transketolase and transaldolase, which transfer two- and three-carbon units among various intermediates to produce a set of sugars containing three, four, five, six, or seven carbons (the reaction catalyzed by transketolase was introduced in Section 7.2). Figure 13.11 is a schematic view of this process. 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 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 13.12. As you examine this diagram, keep in mind the following points, which also apply to the metabolic pathways we will encounter in subsequent chapters: 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 free 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 how the cell catabolizes excess ribose groups.

Clinical Connection: Disorders of Carbohydrate Metabolism 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

353

FIGURE 13.12 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.

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

Clinical Connection: Disorders of Carbohydrate Metabolism 13.5

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

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.

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CH APTER 1 3 Glucose Metabolism

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 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 galactose-free 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 The glycogen storage diseases are a set of inherited disorders of 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 TA BLE 1 3 .2 cramps. The incidence of glycogen storage diseases is estimated to be as TYPE ENZYME DEFICIENCY high as 1 in 20,000 births, although some disorders are not apparent until adulthood. Twelve types of glycogen storage diseases have been described, I Glucose-6-phosphatase and the defect in each is listed in Table 13.2. II α-1,4-Glucosidase A defect of glucose-6-phosphatase (type I glycogen storage disease) aff ects both gluconeogenesis and glycogenolysis, since the phosphatase III Amylo-1,6-glucosidase catalyzes the final step of gluconeogenesis and makes free glucose avail(debranching enzyme) able from glycogenolysis. The enlarged liver and hypoglycemia can lead IV Amylo-(1,4 → 1,6)-transglycosylase to a host of other symptoms, including irritability, lethargy, and, in severe (branching enzyme) cases, death. A related defect is the deficiency of the transport protein that imports glucose-6-phosphate into the endoplasmic reticulum, where the V Muscle glycogen phosphorylase phosphatase is located. VI Liver glycogen phosphorylase Type III glycogen storage disease results from a deficiency of the glycogen debranching enzyme. This condition accounts for about one-quarter VII Phosphofructokinase of all cases of glycogen storage disease and usually affects both liver and VIII, IX, X Phosphorylase kinase muscle. The symptoms include muscle weakness and liver enlargement due XI GLUT2 transporter to the accumulation of glycogen that cannot be efficiently broken down. The symptoms of type III glycogen storage disease often improve with age and 0 Glycogen synthase disappear by early adulthood.

Summary

355

The most common type of glycogen storage disease is type IX. In this disorder, the kinase that activates glycogen phosphorylase is defective. Symptoms range from severe to mild and may fade with time. The complexity of this disease reflects the fact that the phosphorylase kinase consists of four subunits, with isoforms that are differentially expressed in the liver and other tissues. In the past, glycogen storage diseases were diagnosed on the basis of symptoms, blood tests, and painful biopsies of liver or muscle to assess its 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 3.5).

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 free 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 phase (reactions catalyzed by hexokinase, phosphoglucose isomerase, phosphofructokinase, aldolase, and triose phosphate isomerase) requires the investment of two ATP. The irreversible reaction catalyzed by phosphofructokinase is the rate-determining step and the major control point for glycolysis. The second phase of the pathway (reactions catalyzed by glyceraldehyde3-phosphate dehydrogenase, phosphoglycerate kinase, phosphoglycerate mutase, enolase, and pyruvate kinase) generates 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 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.

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CH APTER 1 3 Glucose Metabolism

Key Terms near-equilibrium reaction rate-determining reaction tautomerization feed-forward activation

glycolysis kinase metabolically irreversible reaction

pentose phosphate pathway glycogen storage disease

fermentation gluconeogenesis futile cycle glycogenolysis

Bioinformatics Brief Bioinformatics Exercises 13.1 Hexokinase Structure and Ligand Binding 13.2 Glycolysis and the KEGG Database

Problems 13.1

O

Glycolysis

1. Identify which of the ten reactions of glycolysis are a. phosphorylations; b. isomerizations; c. oxidation–reductions; d. dehydrations; 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 burns 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 very little 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

With inhibitor

90 0.30

125 0.12

OH O

P

O P

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 as a drug to treat the disease? 7. a. What is the ratio of fructose-6-phosphate (F6P) to glucose-6phosphate (G6P) under standard conditions? b. Under cellular conditions? In which direction does the reaction proceed under cellular conditions? 8. 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? 9. ADP stimulates the activity of phosphofructokinase (PFK), yet it is a product of the reaction and not a reactant. Explain this apparently contradictory regulatory strategy.

Problems

11. 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 (see Problem 10) to explain what happens. 12. 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? 13. 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. 14. 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? 15. 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? 16. 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. 17. What is the ratio of glyceraldehyde-3-phosphate (GAP) to dihydroxyacetone phosphate (DHAP) in cells at 37°C under non-equilibrium 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? 18. 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

C

O

CH2OPO2 3 Phosphoglycohydroxamate 19. 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? 20. Arsenate, AsO3− 4 , acts as a phosphate analog and can replace phosphate in the GAPDH reaction. The product of this reaction is 1-arseno-3-phosphoglycerate, which is unstable and spontaneously hydrolyzes to form 3-phosphoglycerate, as shown. What is the effect of arsenate on cells undergoing glycolysis?

O CH HCOH

C

 AsO43

GAPDH

OAsO32

HCOH H2COPO32

H2COPO32 O C

O

NAD NADH  H

O OAsO32

H2O

AsO43

HCOH

C

O

HCOH

H2COPO32

H2COPO32

21. 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. 22. The thermophilic archaebacterium Thermoproteus tenax expresses two GAPDH enzymes; one that carries out the reaction shown in the text (although NADP+ is the reactant rather than NAD+) and a second GAPDH enzyme that catalyzes the reaction shown here.

Glyceraldehyde-3-phosphate  NAD

T. tenax 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 protein1)

10. The “T” and “R” nomenclature used to describe the low- and high-affinity forms of hemoglobin (see 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 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).

357

25 20 15 10 5 0

0

NADP

5

10 [NAD] (mM) Control

AMP

15

20 Glucose-1-P

23. 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? 24. Vanadate, VO3− 4 , inhibits GAPDH, not by acting as a phosphate analog, but by interacting with essential—SH groups on the

358

CH APTER 1 3 Glucose Metabolism

enzyme. What happens to cellular levels of phosphate, ATP, and 2,3bisphosphoglycerate (see pathway below) when red blood cells are incubated with vanadate?

Glyceraldehyde-3-phosphate GAPDH

1,3-Bisphosphoglycerate PGK

2,3-Bisphosphoglycerate

3-Phosphoglycerate

assuming 33% efficiency? Compare your answer to Solution 33. 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)? 35. Studies have shown that the halophilic organism Halococcus saccharolyticus degrades glucose via the Entner–Doudoroff pathway rather than by the glycolytic pathway presented in the text. A modified scheme for the Entner–Doudoroff pathway is shown here. 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.

Glucose

25. The mechanism of plant phosphoglycerate mutase is different from the mechanism of mammalian phosphoglycerate mutase presented in the text. 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). a. What is the fate of the [32P] label when [32P]-labeled 3PG is added to cultured hepatocytes? b. What is the fate of the label when the labeled 3PG is added to plant cells?

NADP NADPH Gluconate H2O

26. Several studies have shown that aluminum inhibits phosphofructokinase in liver cells. a. Compare the production of pyruvate by perfused livers in control and aluminum-treated rats using fructose as an energy source. b. What would the experimental results be if glucose were used instead of fructose?

2-Keto-3-deoxygluconate (KDG)

27. Investigators who wish to deplete ATP in cultured cells do so by adding iodoacetate to the culture medium. Why does the addition of iodoacetate successfully deplete intracellular ATP?

2-Keto-3-deoxy-6-phosphogluconate (KDPG)

ATP ADP

Pyruvate

28. The term “turbo design” has been used to describe pathways such as glycolysis that have one or more ATP-consuming steps followed by one or more ATP-producing steps with a net yield of ATP production for the pathway overall. Mathematical models have shown that “turbo” pathways have the risk of substrate-accelerated death unless there is a “guard at the gate,” that is, a mechanism for inhibiting an early step of the pathway. In yeast, hexokinase is inhibited by a complex mechanism mediated by trehalose-6-phosphate synthase (TPSI). Mutant yeast in which TPSI is defective (there is no “guard at the gate”) die if grown under conditions of high glucose concentration. Explain why. 29. a. Explain why alcohol consumption is associated with increased risk of developing hypothermia. b. Drinking a glass of water for each alcoholic drink is a popular hangover-prevention strategy. Explain how increased water consumption might relieve some of the negative effects of alcohol consumption. 30. 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 would change the effectiveness of therapeutic drugs. 31. 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? 32. When leavened bread is made, the bread dough is “punched” down and then put in a warm place to “rise” to increase its volume. Give a biochemical explanation for this observation. 33. 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 be theoretically produced by the catabolism of glucose to lactate (see Table 13.1), assuming 33% efficiency? 34. How many molecules of ATP could theoretically be produced (see Problem 33) by the catabolism of glucose to CO2 (see Table 13.1),

Glyceraldehyde-3-phosphate 1,3-BPG

Pyruvate 36. Trypanosomes living in the bloodstream obtain all their free 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 this other pathway necessary? b. Would the pathway be necessary if the trypanosome excreted lactate rather than pyruvate? c. Why would this pathway be a good target for antiparasitic drugs?

13.2

Gluconeogenesis

37. 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. 38. A physician diagnoses an infant patient with a pyruvate carboxylase deficiency in part by measuring the patient’s blood levels of lactate and pyruvate. How would the patient’s [lactate]/[pyruvate] ratio compare to the ratio in a normal infant?

Problems

39. In an experiment, the activity of the fructose-1,6-bisphosphatase enzyme is measured at increasing concentrations of fructose2,6-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 (%)

Glycogen Synthesis and Degradation

49. 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. 50. 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?

100 80

51. 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 is 50/1? c. What advantage does degradation by phosphorolysis have over a simple hydrolysis, which would produce glucose instead of glucose-1-phosphate?

60 40 20 0

13.3

359

0

1 2 3 4 [Fructose-2,6-bisphosphate] (μm) Without AMP

5

glycogen (n  1 residues)  G1P G   3.1 kJ  mol1

With AMP

40. 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. 41. 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? 42. 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 41? 43. 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 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. 44. 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 43), and the compound also stimulates the activity of pyruvate kinase. a. What is the effect of adding brazilin to hepatocytes in culture? b. Why would brazilin be an effective treatment for diabetes? 45. Metformin is a drug that decreases the expression of phosphoenolpyruvate carboxykinase. Explain why metformin would be helpful in treating diabetes. 46. 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? 47. 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)? 48. Draw a diagram that illustrates how alanine (see Problem 47) 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?

phosphorylase

glycogen (n residues)  Pi

52. The mechanism of the phosphoglucomutase enzyme is similar to that of the plant mutase described in Problem 25 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? Ser CH2

Ser

O O

P

O

CH2

Ser OH

O

CH2

O

O

H HO

CH2OPO2 3 O H H OH H OH H

OH

Glucose-6phosphate

P

O

O

H HO

CH2OPO2 3 O H H H OH H OPO2 HO 3 H

OH

Glucose-1,6bisphosphate

CH2OH O H OH H H

H OPO2 3

OH

Glucose-1phosphate

53. Trehalose, a disaccharide consisting of two glucose residues (see Problems 11.32 through 11.34), is one of the major sugars in 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 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 have shown 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? Fat body Glycogen Glycolysis

Hemolymph Glucose

Glucose Trehalose

Trehalose

54. 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

360

CH APTER 1 3 Glucose Metabolism

(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 22. 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

63. Red blood cells synthesize and degrade 2,3-bisphosphoglycerate (BPG) as a detour from the glycolytic pathway, as shown in Problem 24. 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 hexokinase-deficient and pyruvate kinase–deficient erythrocytes. Identify which curve corresponds to which enzyme deficiency.

55. 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?

100

Oxygen saturation (%)

90

56. 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. 57. Write a mechanism for the nonenzymatic hydrolysis of 6-phosphogluconolactone to 6-phosphogluconate.

59. Several studies have shown that the metabolite glucose1,6-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

60. Xylulose-5-phosphate (an intermediate of the pentose phosphate pathway) 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 61. Type VII glycogen storage disease is actually 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? 62. 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.

70 60

Normal erythrocytes

50 40 30 20 10 0

0

10

20

30 40 pO2 (torr)

50

60

64. a. What happens to the [ADP]/[ATP] and [NAD+]/[NADH] ratios in red blood cells with a pyruvate kinase deficiency (see Problem 63)? 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. 65. During even mild exertion, individuals with McArdle’s disease experience painful muscle cramps due to a genetic defect in glycogen phosphorylase, the enzyme that breaks down glycogen. Yet the muscles in these individuals contain normal amounts of glycogen. What did this observation tell researchers about the pathways for glycogen degradation and glycogen synthesis? 66. Patients with McArdle’s disease have normal liver glycogen content and structure. Identify the type of glycogen storage disease as listed in Table 13.2. 67. A patient with McArdle’s disease (see Problem 66) 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)

58. 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.

80

60

40

Control

20

0

Patient

0

5

10

15 20 Time (min)

25

30

68. Patients with von Gierke’s disease (type I glycogen storage disease) 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

Selected Readings

von Gierke’s disease. b. Some patients with von Gierke’s disease also have enlarged kidneys. Explain why. 69. a. Does a patient with McArdle’s disease (see Problems 65–67) suffer from hypoglycemia, hyperglycemia, or neither? b. Does a patient with von Gierke’s disease (see Problem 68) suffer from hypoglycemia, hyperglycemia, or neither? 70. Would a feeding of cornstarch administered at bedtime help relieve the symptoms of type 0 glycogen storage disease? Explain why or why not. 71. Reduced glutathione, a tripeptide containing a Cys residue, is found in red blood cells, where it reduces organic peroxides formed in cellular structures exposed to high concentrations of reactive oxygen.

361

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. Glutathione is regenerated as shown in the following reaction:

-Glu

Cys

Gly

S



S -Glu

Cys



NADPH

H

Gly

2 -Glu

Cys

Gly



NADP

SH

2 -Glu

Cys

Gly



R

O

OH

SH Reduced glutathione

-Glu

Cys

Gly

S



S -Glu

Cys

Organic peroxide

R

OH



H2O

Gly

Oxidized glutathione

Use this information to predict the physiological effects of a glucose6-phosphate dehydrogenase deficiency. 72. 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.

Selected Readings Brosnan, J. T., Comments on metabolic needs for glucose and the role of gluconeogenesis, Eur. J. Clin. Nutr. 53, S107–S111 (1999). [A very readable review that discusses possible reasons why carbohydrates are used universally as metabolic fuels, why glucose is stored as glycogen, and why the pentose phosphate pathway is important.]

Özen, H., Glycogen storage diseases: New perspectives, World J. Gastroenterology 13, 2541–2553 (2007). [Describes the symptoms, biochemistry, and treatment of the major forms of these diseases.]

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.]

Roach, P. J., Depaoli-Roach, A. A., Hurley, T. D., and Tagliabracci, V. S., Glycogen and its metabolism: Some new developments and old themes, Biochem. J. 441, 763–787 (2012). [Includes discussions of the hormone-mediated regulation of key enzymes of glycogen synthesis and breakdown.]

Patra, K. C. and Hay, N., The pentose phosphate pathway and cancer, Trends Biochem. Sci. 39, 347–354 (2014). [Reviews the reactions of the pentose phosphate pathway.]

Paul Starosta/Getty Images, Inc.

CHAPTER 14 The Citric Acid Cycle

Metabolism converts the stored energy of fuel molecules into the chemical energy of ATP, but some energy is always lost as heat, which is one way animals can keep warm. Until recently, birds and mammals were the only organisms known to adjust their metabolism to maintain high body temperatures. However, the Tegu lizard can also generate heat during its reproductive phase, boosting its nest temperature as much as 10°C higher than the ambient temperature.

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). • Pyruvate can be converted to lactate, acetyl-CoA, or oxaloacetate (Section 13.1).

L EARNING 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. 362

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

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 CO2 molecules are products of the citric acid cycle.

363

The Pyruvate Dehydrogenase Reaction

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 transported into the mitochondria. 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. [Structure of the A. vinelandii core (pdb 1EAA) determined by A. Mattevei and W. G. J. Hol. Structure of the B. stearothermophilus core (pdb 1B5S) determined by T. Izard, A. Aevarsson, M. D. Allen, A. H. Westphal, R. N. Perham, A. De Kok, and W. G. Hol.]

pyruvate + CoA + NAD+ → acetyl-CoA + CO2 + 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

1. In the first step, which is catalyzed by E1 (also called pyruvate dehydrogenase), 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:

N

O

C O





CH3

O

H

C O

O



N

C

C

O OH

S CH3

CO2

CH2

N

O

CH2



N C

CH2 S

NH2 H N Thiamine pyrophosphate

H 3C

N C

H

O

O

S

C C

P

CH3

CH3 

O

O

CH3 

P

S

C CH3

Hydroxyethyl–TPP

CH3 Pyruvate

2. The hydroxyethyl group is then transferred to E2 of the pyruvate dehydrogenase complex. The hydroxyethyl acceptor is a lipoamide prosthetic group (Fig. 14.3).

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.

364

CH APTER 1 4 The Citric Acid Cycle

S

The transfer reaction regenerates the TPP cofactor of E1 and oxidizes the hydroxyethyl group to an acetyl group:

CH2 CH2

S

CH3

CH2 CH2 CH2 CH2

O

S

C CH3

H

O

CH3 

N C

H

C

S

N

H

CH3

O

S

S

 C S

C

CH3

S

O HS

Lipoamide

(CH2)4 CH

HS

S

NH

NH



N C

H

CH2 C



Lipoic acid

CH3

Lys

C O

Lipoamide

3. Next, E2 transfers the acetyl group to coenzyme A, producing acetyl-CoA and leaving a reduced lipoamide group. Coenzyme A

CoA

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.

Acetyl-CoA

SH

FIGURE 14.3

O

CH3

O

CoA

C

S

C

CH3



S HS HS HS

Recall that acetyl-CoA is a thioester, a form of energy currency (see Section 12.3). Some of the free 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.

FAD

HS 

S

S

SH  HS

S

FAD

SH

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).

NAD FAD

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.

The Eight Reactions of the Citric Acid Cycle

365

NAD FAD SH

5

SH S

OH CO2

CH3

C TPP

1

C

4

S S

HS

Lipoamide 2

HS

O CH3

FAD

S

HydroxyethylTPP

NADH  H

O O

C O

TPP

CH3

C

3

O

S CH3

Pyruvate

HS

C

S

CoA

Acetyl-CoA

CoA

Acetyl-dihydrolipoamide

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 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.

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.

Q Without looking at the text, write the net equation for the reactions shown here. How many vitamin-derived cofactors are involved?

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.

The Eight Reactions of the Citric Acid Cycle 14.2

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. 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

LEARNING OBJECTIVE Describe the substrate, product, and type of chemical reaction for each step of the citric acid cycle.

366

CH APTER 1 4 The Citric Acid Cycle 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).

BIOPOLYMERS Proteins Nucleic acids Polysaccharides Triacylglycerols

MONOMERS Amino acids Nucleotides Monosaccharides Fatty acids +

NH4

NAD+

NAD+ NADH

NADH, QH2 2- and 3-Carbon INTERMEDIATES

NAD+, Q

citric acid cycle

photosynthesis

NADH, QH2 ADP oxidative phosphorylation ATP

H2O

acetyl-CoA + GDP + Pi + 3 NAD + + Q → 2 CO2 + CoA + GTP + 3 NADH + QH2 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 is summarized in Figure 14.6.

CO2

O2

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 prokaryotes 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

1. Citrate synthase adds an acetyl group to oxaloacetate 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:

NAD+, Q

COO

COO C

O

CH2

H2O HSCoA

O  CH3

C

SCoA

COO Oxaloacetate

citrate synthase

CH2 HO

C

COO

CH2 COO

Acetyl-CoA

Citrate

Citrate synthase, a dimer, undergoes a large conformational change upon substrate binding (Fig. 14.7). 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.

The Eight Reactions of the Citric Acid Cycle

O

C

C SCoA Acetyl-CoA

O

COO

CH2

CoASH

CH2 ⫺

FIGURE 14.6 Reactions of the citric acid cycle.

COO⫺

CH3

COO⫺

HO

C

367

COO

Q Identify the number of carbon atoms in each intermediate.

⫺

CH2 COO⫺ Citrate

1 citrate synthase

Oxaloacetate

COO⫺

NADH

CH2 NAD⫹ COO⫺ CH

OH

CH2

2 aconitase

8 malate dehydrogenase

HO

Malate

3 isocitrate dehydrogenase

COO⫺

7 fumarase

H2O

COO⫺

CH2

CH

CH2 C

HC COO⫺

⫺

COO

Fumarate COO QH2

C

Isocitrate

COO⫺

CH2 C

CH2 COO⫺ 5 succinyl-CoA Succinate synthetase

GTP ⫹ CoASH

H

COO⫺

NAD⫹

O

NADH ⫹ CO2

COO⫺ ␣-Ketoglutarate

CH2

⫺

6 succinate CH2 dehydrogenase

Q

COO⫺

CH

4 ␣-ketoglutarate dehydrogenase

O

SCoA Succinyl-CoA

NAD⫹ ⫹ CoASH

NADH ⫹ CO2

GDP ⫹ Pi

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) When oxaloacetate (red, mostly buried) binds, each subunit undergoes a conformational change that creates a binding site for acetyl-CoA (an acetyl-CoA analog is shown here in orange). This conformational change explains why oxaloacetate must bind to the enzyme before acetyl-CoA can bind. [Structure of chicken citrate synthase alone (pdb 5CSC) determined by D.-I. Liao, M. Karpusas, and S. J. Remington; structure of citrate synthase with oxaloacetate and carboxymethyl-CoA (pdb 5CTS) determined by M. Karpusas, B. Branchaud, and S. J. Remington.]

(a)

(b)

368

CH APTER 1 4 The Citric Acid Cycle His 274

N

N

N

N

H

H OH 

OOC

C

Asp 375 

O

C

COO

CH2

CH2

O

HO

COO

O

C

Citrate 1. Oxaloacetate and then acetyl-CoA bind to the enzyme.

4. Hydrolysis releases CoA and citrate.

HSCoA  H H2O

N

N



OOC

COO

H

H

C

O Acetyl-CoA

CH2 Oxaloacetate

N

N

O

H2C H



C

SCoA

O

O

C

O

OH 

OOC

Citryl-CoA

C

C

CH2

CH2

HO

COO

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.

H

N

SCoA

C

O

3. The enolate attacks oxaloacetate to produce a citryl-CoA intermediate.

N H

O 

OOC



O

C CH2

H2C

COO

C HO

FIGURE 14.8 The citrate synthase reaction.

SCoA O

C

2. Aconitase isomerizes citrate to isocitrate SEE ANIMATED PROCESS DIAGRAM

The second enzyme of the citric acid cycle catalyzes the reversible isomerization of citrate to isocitrate:

Mechanism of citrate synthase

COO CH2 HO

C

COO

COO H2O

CH2 C

COO

CH2

CH

COO

COO

Citrate

Aconitate

The enzyme is named after the reaction intermediate.

COO CH2

H2O H

C

COO

HO

C

H

COO Isocitrate

The Eight Reactions of the Citric Acid Cycle

369

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 (also known as the Krebs 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).

3. Isocitrate dehydrogenase releases the first CO2 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.

COO CH2 H HO O

C C C

CH2

O

C H

COO

H  NADH

NAD

H O

C C

O

O

C

FIGURE 14.9 Stereochemistry of aconitase. The three-point attachment of citrate to the enzyme allows only one carboxymethyl group (shown in green) to react.

COO CH2

CO2

O

C

H

O

O

O

C C

O

C

COO CH2

H H

O O

Isocitrate

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 CoASH CO2

CH2 C

O

COO -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.

370

CH APTER 1 4 The Citric Acid Cycle

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 are released 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

The thioester succinyl-CoA releases a large amount of free energy when it is hydrolyzed (ΔG°′ = –32.6 kJ · mol–1). 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 α-Ketoglutarate reversible. In fact, the enzyme is named for the reverse reaction. Succinyl-CoA synthetase in the mammalian citric acid cycle gener4 ates GTP, whereas the plant and bacterial enzymes generate ATP (recall that GTP is energetically equivalent to ATP). An exergonic Succinyl-CoA reaction coupled to the transfer of a phosphoryl group to a nucleoside FIGURE 14.10 Fates of carbon diphosphate is termed substrate-level phosphorylation to distinatoms in the citric acid cycle. guish it from oxidative phosphorylation (Section 15.4) and photophosphorylation (Section The two carbon atoms that are 16.2), which are more indirect ways of synthesizing ATP. lost as CO2 in the reactions How does succinyl-CoA synthetase couple thioester cleavage to the synthesis of a nuccatalyzed by isocitrate leoside triphosphate? The reaction is a series of phosphoryl-group transfers that involve an dehydrogenase (step 3) and active-site histidine residue (Fig. 14.11). The phospho-His reaction intermediate must move a α-ketoglutarate dehydrogenase large distance to shuttle the phosphoryl group between the succinyl group and the nucleoside (step 4) are not the same carbons diphosphate (Fig. 14.12). that entered the cycle as acetyl3

CoA (red). The acetyl carbons become part of oxaloacetate and are lost in subsequent rounds of the 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

FIGURE 14.11 The succinyl-

CoA synthetase reaction. Q What is the fate of the free CoA molecule?

COO O

CH2  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

COO CH2 CH2 C

O

SCoA

OPO2 3

Succinyl-CoA

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.

SEE ANIMATED PROCESS DIAGRAM The reaction catalyzed by succinyl-CoA synthetase

His residue

N H

COO CH2

N

CH2 N H GDP

GDP  Pi N

PO2 3

GTP

3. The phosphoryl group is then transferred to GDP to form GTP.

PO2 3

N Phospho-His

 H

COO Succinate

The Eight Reactions of the Citric Acid Cycle

371

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

H succinate dehydrogenase

C

OOC

COO Succinate

COO

C

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

FIGURE 14.12 Substrate binding in succinyl-CoA synthetase. Succinyl-CoA (represented by coenzyme A, red) binds to the enzyme, and its succinyl group is phosphorylated. The succinyl phosphate then transfers its phosphoryl group to the His 246 side chain (green). A protein loop containing the phospho-His side chain must undergo a large movement because the nucleoside diphosphate awaiting phosphorylation (ADP, orange) binds to a site about 35 Å away. [Structure of E. coli succinyl-CoA synthetase (pdb 1CQI) determined by M. A. Joyce, M. E. Fraser, M. N. G. James, W. A. Bridger, and W. T. Wolodko.]

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 CH HC

COO H2O

H

C

OH

fumarase

H

C

H

COO Fumarate

COO 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 reaction

372

CH APTER 1 4 The Citric Acid Cycle

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.

L EARNING 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.

Glucose

glycolysis

2 Pyruvate

pyruvate processing 2 Acetyl-CoA

citric acid cycle

Thermodynamics of the Citric Acid Cycle

14.3

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 QH2 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 5 ATP 2 NADH of 10 ATP equivalents. The energy yield of a molecule of glucose, which generates two acetyl units, can be calculated (Fig. 14.13). 2 ATP A muscle operating anaerobically produces only 2 ATP per glucose, but under aerobic conditions when the citric acid cycle is fully functional, each glucose molecule generates about 32 ATP equivalents. This 2 NADH 5 ATP general phenomenon is called the Pasteur effect, after Louis Pasteur, who first observed that the rate of glucose consumption by yeast cells 2 CO2 decreased dramatically when the cells were shifted from anaerobic to aerobic growth conditions.

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.

Q Trace the fate of the six glucose carbon atoms.

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.

Thermodynamics of the Citric Acid Cycle

Acetyl-CoA Oxaloacetate

FIGURE 14.14 Regulation of the citric acid cycle. Inhibition is represented by red symbols, activation by green symbols.

Citrate

Malate

373

Isocitrate ADP NADH Ca2⫹ ␣-Ketoglutarate

Fumarate

Ca2⫹ Succinate

Succinyl-CoA

The activity of isocitrate dehydrogenase is inhibited by its reaction product, NADH. NADH also inhibits α-ketoglutarate dehydrogenase and citrate synthase. Both dehydrogenases are activated by Ca2+ ions, which generally signify the need to generate cellular free 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).

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 preexisting biochemical reactions. Clues to its origins can be found by examining the metabolism

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). 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, although characteristically rapid, may be limited by the availability of oxygen and other 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 NADPHdependent 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. Q How would a fumarase deficiency affect the levels of pyruvate, fumarate, and malate?

374

CH APTER 1 4 The Citric Acid Cycle

Acetyl-CoA Citrate

Oxaloacetate

oxidation

Malate

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. The modern citric acid cycle may have evolved by connecting these pathways.

Pyruvate CO2 Acetyl-CoA

Citrate

Oxaloacetate

Malate

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 H2S. 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). It is easy to theorize that the evolution of an enzyme to interconvert α-ketoglutarate and succinate could have created 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 accommodated 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).

Isocitrate BE FORE GOING ON CO2 -Ketoglutarate

Fumarate Succinate

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.

L EARNING 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.

Anabolic and Catabolic Functions of the Citric Acid Cycle 14.4

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.

Anabolic and Catabolic Functions of the Citric Acid Cycle

Fatty acids, cholesterol

Glucose

FIGURE 14.17 Citric acid cycle intermediates as biosynthetic precursors.

Citrate

Oxaloacetate Pyruvate Malate

Isocitrate

Fumarate

-Ketoglutarate

Succinate

Amino acids, nucleotides

Succinyl-CoA Heme

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 α-ketoglutarate (sometimes called 2-oxoglutarate) can undergo reductive amination by glutamate dehydrogenase to produce the amino acid glutamate:

COO

NADH  H

NAD

CH2

CH2  NH 4

CH2 C

COO

O

COO -Ketoglutarate

glutamate dehydrogenase

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. ATP-citrate lyase then catalyzes the reaction ATP + citrate + CoA → ADP + Pi + oxaloacetate + acetyl-CoA The resulting acetyl-CoA is used for fatty acid and cholesterol synthesis, which take place in the cytosol. 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

375

376

CH APTER 1 4 The Citric Acid Cycle INNER MITOCHONDRIAL MEMBRANE

FIGURE 14.18 The citrate

transport system. Both citrate and pyruvate cross the inner mitochondrial membrane via specific transport proteins. This system allows carbon atoms from mitochondrial acetyl-CoA to be transferred to the cytosol for the synthesis of fatty acids and cholesterol. Q Can the same CO2 molecule released in the malic enzyme reaction be used in the pyruvate carboxylase reaction?

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

lyase reaction can be converted to malate by 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 and be converted back to oxaloacetate to complete the cycle shown in Figure 14.18. In plants, isocitrate is diverted from the citric acid cycle in a biosynthetic pathway known as the glyoxylate 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 + CO2 + ATP + H2O → oxaloacetate + ADP + Pi 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 oxaloacetate 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

Anabolic and Catabolic Functions of the Citric Acid Cycle

377

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



COO

CH

OH

COO Isocitrate

CH2 Fumarate

CH

isocitrate lyase

CH2

O

COO Succinate

C

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 gluconeogenesis. 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 isocitrate 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. Q Write the net equation for the glyoxylate cycle as shown here.

H COO

COO Glyoxylate

O CH3

C

SCoA

CH malate synthase

OH

CH2 COO

Oxaloacetate 

Malate

Acetyl-CoA

gluconeogenesis

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 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

Aspartate

COO 

C

O

CH3 Pyruvate

COO C

O

CH2 COO Oxaloacetate

COO 

 H3N

C H CH3

Alanine

378

CH APTER 1 4 The Citric Acid Cycle

pyruvate, amino acids

FIGURE 14.19 Anaplerotic

reactions of the citric acid cycle.

Oxaloacetate

Citrate

Malate

Isocitrate

Fumarate

-Ketoglutarate amino acids

amino acids Succinate

Succinyl-CoA

amino acids, odd-chain fatty acids 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 fourfold 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 C

O 

CH3

COO

CH2 CH2 H C

NH 3



H3N

COO

CH2

C H 

CH2

CH3

C

COO Pyruvate

Glutamate

O

COO Alanine

-Ketoglutarate

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.

Problems

379

Summary 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

In addition, succinyl-CoA synthetase yields one molecule of GTP or ATP. • 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.

• The eight reactions of the citric acid cycle function as a multistep catalyst to convert the two carbons of acetyl-CoA to 2 CO2.

14.4 Anabolic and Catabolic Functions of the Citric Acid Cycle

14.3

• 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.

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 phosphorylation.

Key Terms mitochondrial matrix multienzyme complex citric acid cycle

substrate-level phosphorylation Pasteur effect

carcinogenesis glyoxylate pathway

glyoxysome anaplerotic reaction

Bioinformatics Brief Bioinformatics Exercises 14.1 Viewing and Analyzing the Pyruvate Dehydrogenase Complex 14.2 The Citric Acid Cycle and the KEGG Database

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 dehydrogenase 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. Arsenite is toxic in part because it binds to sulfhydryl compounds such as dihydrolipoamide, as shown in the figure. What effect would the presence of arsenite have on the citric acid cycle? OH O

As OH

Arsenite

S O



 2 H2O

As S

HS

R

HS R Dihydrolipoamide

380

CH APTER 1 4 The Citric Acid Cycle 0.45

7. How is the activity of the pyruvate dehydrogenase complex affected by a. a high [NADH]/[NAD+] ratio or b. a high [acetyl-CoA]/ [CoASH] ratio? 8. The activity of the pyruvate dehydrogenase complex is also 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. 9. The PDH kinase and PDH phosphatase enzymes (see Problem 8) 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+? 10. Acetyl-CoA produced by the pyruvate dehydrogenase complex can enter the citric acid cycle or can be used to synthesize fatty acids (see Chapter 17). Hepatocytes in culture were incubated with fatty acids and the activity of PDH kinase (see Problem 8) was measured. Would you expect fatty acids to stimulate or inhibit the kinase? 11. 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 dehydrogenase deficiency administer thiamine as a first course of treatment. Explain why. 12. A second strategy to treat a pyruvate dehydrogenase deficiency disease (see Problem 11) involves administering dichloroacetate, a compound that inhibits pyruvate dehydrogenase kinase (see Problem  8). How might this strategy be effective?

14.2 The Eight Reactions of the Citric Acid Cycle 13. 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. 14. Site-directed mutagenesis techniques were used to synthesize a mutant citrate synthase enzyme in which the active site histidine was converted to an alanine. Why did the mutant citrate synthase enzyme exhibit decreased catalytic activity? 15. 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-acetonylCoA? Explain.

C

0.35

with I

0.25 no I 0.15 0.05 0

−1.0

−0.5

0

0.5

1.0

1/[acetyl-CoA] (μM−1)

16. 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.

OH CoA

S

CH2

C O

Carboxymethyl-CoA (transition state analog) 17. Kinetic studies with aconitase revealed that trans-aconitate is a competitive inhibitor of the enzyme if cis-aconitate is used as the substrate. But if citrate is used as the substrate, trans-aconitate is a noncompetitive inhibitor. Propose a hypothesis that explains this observation. 18. In a yeast mutant, the gene for aconitase is nonfunctional. What are the consequences for the cell, particularly with regard to energy production? 19. The ΔG°′ value for the isocitrate dehydrogenase reaction is –21 kJ · mol–1. What is Keq for this reaction? Assume T = 25°C. 20. The crystal structure of isocitrate dehydrogenase shows that there is 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? 21. 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. 22. Using the mechanism you drew for Problem 21, explain how succinyl phosphonate (below) inhibits α-ketoglutarate dehydrogenase.

O⫺ ⫺O

P

O

C

O

CH2 CH2 COO⫺ Succinyl phosphonate 23. Succinyl-CoA synthetase is also called succinate thiokinase. Why is the enzyme considered to be a kinase?

O H3C

1/v (min . nmol−1)

6. Using the pyruvate dehydrogenase complex reaction as a model, reconstruct the TPP-dependent yeast pyruvate decarboxylase reaction in alcoholic fermentation (see Box 13.B).

CH2

S

S-Acetonyl-CoA

CoA

24. The succinyl-CoA synthetase reaction is shown in Figure 14.11, but mechanistic details are not provided. The His residue shown in

Problems

25. Malonate is a competitive inhibitor of succinate dehydrogenase. What citric acid cycle intermediates accumulate if malonate is present in a preparation of isolated mitochondria? 26. 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? 27. The ΔG°′ 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? 28. 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. Wild type enzyme Vmax (μmol · min–1 · mg–1)

345

E315Q mutant enzyme 32

KM (mM)

0.21

0.25

kcat /KM (M–1 · s–1)

5.6 × 106

4.3 × 105

29. 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? 30. 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 labeled? b. If citrate becomes radioactively labeled, what can you conclude about the connection between glycosomal and mitochondrial pathways in the parasite?

14.3 Thermodynamics of the Citric Acid Cycle 31. 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. 32. Predict the effect of the following metabolites on the activity of citrate synthase: a. NADH; b. citrate; c. succinyl-CoA; and d. ATP. 33. Citrate competes with oxaloacetate for binding to citrate synthase. Isocitrate dehydrogenase is activated by Ca2+ ions, which are released when 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)? 34. 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.

35. 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. 36. The scientists who carried out the hyperoxia experiments described in Problem 35 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.) 37. 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 non-phosphorylated enzyme show no significant conformational differences. a. How does phosphorylation regulate isocitrate dehydrogenase activity? b. To confirm their hypothesis, 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? 38. 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 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? 150 120 [NAD] / [NADH]

the figure is located close to a Glu residue (not shown) that participates in catalysis. Mutating this Glu residue to Asp has 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?

381

90 mutant

60 30

wild-type 0

0

12

24

36

Time (h)

39. A patient with an α-ketoglutarate 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. 40. Succinyl-CoA inhibits both citrate synthase and α-ketoglutarate dehydrogenase. How is succinyl-CoA able to inhibit both enzymes? 41. 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.

382

CH APTER 1 4 The Citric Acid Cycle

42. Individuals with a mutation in the gene for the α subunit of succinyl-CoA synthetase (see Problem 41) 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 20s. Why is the prognosis for patients with the β subunit mutation better than for patients with the α subunit mutation? 43. Why would a deficiency of succinate dehydrogenase lead to a shortage of coenzyme A? 44. Individuals who are deficient in fumarase develop lactic acidosis. Explain why. 45. Malate dehydrogenase is more active in cells oxidizing glucose aerobically than in cells oxidizing glucose anaerobically. Explain why. 46. Acetyl-CoA acts as an allosteric activator of pyruvate carboxylase. S-acetonyl-CoA (see Problem 15) 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? 47. 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? 48. Certain microorganisms with an incomplete citric acid cycle decarboxylate α-ketoglutarate to produce succinate semialdehyde. A dehydrogenase then converts succinate semialdehyde to 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?

COO

COO

CH2

CH2

CH2

CH2

C

C

O

O

O

COO

Succinate

-Ketoglutarate

CO2

COO CH2 CH2 C

NADH NAD

O

H Succinate semialdehyde

14.4 Anabolic and Catabolic Functions of the Citric Acid Cycle 49. 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?

52. A bacterial mutant with low levels of isocitrate dehydrogenase is able to grow normally when the culture medium is supplemented with glutamate. Explain why. 53. 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. 54. 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. 55. 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. 56. The physician treats the patient described in Problem 55 by administering glutamine. Explain why glutamine supplements are effective in treating the disease. 57. Physicians often attempt to treat a pyruvate carboxylase deficiency by administering biotin. Explain why this strategy might be effective. 58. Patients with a pyruvate dehydrogenase deficiency and patients with a pyruvate carboxylase deficiency (see Problems 55–57) both have high blood levels of pyruvate and lactate. Explain why. 59. 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. 60. 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, isocitrate dehydrogenase is phosphorylated. a. Draw a diagram showing how acetate is metabolized in E. coli. b. When glucose is added to the culture, the phosphate group is removed from isocitrate dehydrogenase. How does flux through the metabolic pathways change in E. coli when glucose is the food source instead of acetate? 61. 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. 62. 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? 63. 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

50. 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.

Fumarate

51. Describe how the aspartate + pyruvate transamination reaction could function as an anaplerotic reaction for the citric acid cycle.

adenylosuccinate lyase

adenosine deaminase

NH 4

AMP

IMP Aspartate  GTP Adenylosuccinate

adenylosuccinate synthetase

GDP  Pi

Problems

64. 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. 65. 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. 66. 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 HO

C

HO

CH

COO

COO COO

Hydroxycitrate 67. The bacterium Helicobacter pylori colonizes the upper gastrointestinal tract in humans and causes chronic gastritis, ulcers, and possibly gastric cancer. Understanding the metabolism of this organism is essential in the development of new drugs to treat these diseases. The citric acid “cycle” in H. pylori is a noncyclic, branched pathway, as shown below. Succinate is produced in the “reductive branch,” whereas α-ketoglutarate is produced in the “oxidative branch.” The two branches are linked by the α-ketoglutarate oxidase reaction. a. Compare and contrast the citric acid cycle in H. pylori with the citric acid cycle in mammals. b. What enzymes might serve to regulate the citric acid cycle in H. pylori? c. What enzymes might be used as drug targets for persons suffering from H. pylori–induced gastritis or ulcers?

Glyoxylate  Acetyl-CoA malate synthase

Malate H 2O

fumarase

NAD NADH

aconitase

reductase

Succinate CO2, QH2

Acetyl-CoA

citrate synthase

Citrate

Fumarate QH2 fumarate Q

Oxaloacetate

malate dehydrogenase

Isocitrate -ketoglutarate oxidase

isocitrate dehydrogenase

Q

NADP

NADPH -Ketoglutarate

68. H. pylori (see Problem 67) uses amino acids and fatty acids present in the gastrointestinal tract as a source of biosynthetic intermediates.

383

a. Describe how H. pylori uses the acetyl-CoA derived from fatty acid breakdown to synthesize glucose and glutamate. b. Describe how H. pylori converts aspartate to glutamate. 69. 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? 70. 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. But 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? 71. 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

COO Phosphoenolpyruvate



HCO 3

PPC

CH2 C

O



Pi

COO Oxaloacetate

72. 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? 73. 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 α-ketoglutarate. b. Give the sequence of enzymes that can convert α-ketoglutarate to pyruvate. 74. 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.

384

CH APTER 1 4 The Citric Acid Cycle

Selected Readings Barry, M. J., Enzymes and symmetrical molecules, Trends Biochem. Sci. 22, 228–230 (1997). [Recounts how experiments and insight revealed that the symmetrical citrate molecule can react asymmetrically.] Brière, J.-J., Favier, J., Giminez-Roqueplo, A.-P., and Rustin, P., Tricarboxylic acid cycle dysfunction as a cause of human disease and tumor formation, Am. J. Physiol. Cell Physiol. 291, C1114–C1120 (2006). [Includes discussions of the metabolic role of the citric acid cycle and its location.] 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.]

Owen, O. E., Kalhan, S. C., and Hanson, R. W., The key role of anaplerosis and cataplerosis for citric acid cycle function, J. Biol. Chem. 277, 30409–30412 (2002). [Describes the addition (anaplerosis) and removal (cataplerosis) of citric acid cycle intermediates in different organ systems.] 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.]

Courtesy Kathryn Tosney

CHAPTER 15 Oxidative Phosphorylation

Loricifera, a group of tiny (< 1 mm) organisms that resemble armored jellyfish, live in marine sediments. Remarkably—for a multicellular organism—they are entirely anaerobic and lack mitochondria, the site of aerobic metabolism in all other animals.

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 free energy of the redox reactions is conserved in the formation of a transmembrane gradient of protons, another type of free 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

In the scheme introduced in Figure 12.11, 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

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. 385

386

CH APTER 1 5 Oxidative Phosphorylation 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 molecular oxygen. The free energy of this process is conserved in a manner that powers the synthesis of ATP from ADP + Pi.

BIOPOLYMERS Proteins Nucleic acids Polysaccharides Triacylglycerols

MONOMERS Amino acids Nucleotides Monosaccharides Fatty acids +

NH4

NAD+

NAD+

NADH

NADH, QH2 2- and 3-Carbon INTERMEDIATES

NAD+, Q

citric acid cycle

photosynthesis

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 free energy transformation. The flow of electrons from reduced compounds such as NADH and QH2 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: Aoxidized + Breduced ⇌ Areduced + Boxidized 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:

CO2 NADH, QH2 ADP

O2

oxidative phosphorylation ATP

H2O NAD+, Q

FADH2 (reduced)

+

Q

(oxidized)

⇌

FAD (oxidized)

+

QH2 (reduced)

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).

Reduction potential indicates a substance’s tendency to accept electrons 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. Using the example above, the half-reaction for ubiquinone (by convention, written as a reduction reaction) is Q + 2 H+ + 2e− ⇌ QH2 (the reverse reaction would describe an oxidation half-reaction). 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

The Thermodynamics of Oxidation–Reduction Reactions

25°C, the pH is 7.0, and all species are present at concentrations of 1 M). The greater the value of Ɛ°ʹ, 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°ʹ ) by the Nernst equation:

Ɛ = Ɛ°ʹ −

[Areduced ] RT ln nF [Aoxidized ]

[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

Ɛ = Ɛ°ʹ −

0.026 V [Areduced ] ln n [Aoxidized ]

[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°ʹ (Sample Calculation 15.1).

Standard Reduction Potentials of Some Biological Substances

TA B L E 15.1

HALF-REACTION 1 2

+

Ɛ °′ (V)

−

O2 + 2 H + 2 e ⇌ H2O

SO2− 4 NO3−

+

+2H + + 2 H+ +

0.815

−

2 e ⇌ SO2− 3 + H2O 2e − ⇌ NO2− + H2O 3+ −

0.48 0.42

Cytochrome a3 (Fe ) + e ⇌ cytochrome a3 (Fe2+)

0.385

Cytochrome a (Fe3+) + e− ⇌ cytochrome a (Fe2+)

0.29

3+

−

2+

Cytochrome c (Fe ) + e ⇌ cytochrome c (Fe )

0.235

Cytochrome c1 (Fe3+) + e− ⇌ cytochrome c1 (Fe2+)

0.22

3+

−

2+

Cytochrome b (Fe ) + e ⇌ cytochrome b (Fe )(mitochondrial)

0.077

Ubiquinone + 2 H+ + 2 e− ⇌ ubiquinol

0.045

Fumarate− + 2 H+ + 2 e− ⇌ succinate−

0.031

FAD + 2 H+ + 2 e− ⇌ FADH2 (in flavoproteins)

~ 0.

Oxaloacetate− + 2 H+ + 2 e− ⇌ malate−

− 0.166

Pyruvate− + 2 H+ + 2 e− ⇌ lactate−

− 0.185

+

−

Acetaldehyde + 2 H + 2 e ⇌ ethanol

− 0.197

S + 2 H+ + 2 e− ⇌ H2S

− 0.23

+

−

Lipoic acid + 2 H + 2 e ⇌ dihydrolipoic acid

− 0.29

NAD+ + H+ + 2 e− ⇌ NADH

− 0.315

+

+

−

NADP + H + 2 e ⇌ NADPH

− 0.320

Acetoacetate− + 2 H+ + 2 e− ⇌ 3-hydroxybutyrate−

− 0.346

Acetate− + 3 H+ + 2 e− ⇌ acetaldehyde + H2O

− 0.581

Source: 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).

387

388

CH APTER 1 5 Oxidative Phosphorylation

SAMP LE CA LCU LAT I O N 1 5 . 1 SEE SAMPLE CALCULATION VIDEOS

Problem Calculate the reduction potential of fumarate (Ɛ°ʹ = 0.031 V) at 25°C when [fumarate] = 40 μM and [succinate] = 200 μM.

Solution Use Equation 15.2. Fumarate is the oxidized compound and succinate is the reduced compound. Ɛ = Ɛ°ʹ −

0.026 V [Areduced ] ln n [Aoxidized ]

= 0.031 V −

0.026 V (2 × 10−4 ) ln 2 (4 × 10−5 )

= 0.031 V − 0.021 V = 0.010 V

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°ʹ for NAD+ (–0.315 V) is lower than Ɛ°ʹ 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–ubiquinone reaction, the net reaction is the ubiquinone reduction half-reaction (the half-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). Note that because the NAD+ half-reaction has been reversed to indicate oxidation, we have also reversed the sign of its Ɛ°ʹ value: NADH ⇌ NAD + + H + + 2 e − Q + 2 H + + 2 e − ⇌ QH2 net: NADH + Q + H + ⇌ NAD + + QH2

Ɛ°ʹ = +0.315 V Ɛ°ʹ = 0.045 V ∆Ɛ°ʹ = +0.360 V

When the two half-reactions are added, their reduction potentials are also added, yielding a ∆Ɛ°ʹ 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°ʹ, as shown above, is just a shortcut to simplify the task of calculating ∆Ɛ°ʹ. Another method for calculating ∆Ɛ°ʹ uses the following equation: ∆Ɛ°ʹ = Ɛ°ʹ(e− acceptor) − Ɛ°ʹ(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: ∆G °ʹ = −nF ∆Ɛ°ʹ or ∆G = −nF ∆Ɛ

[15.4]

The Thermodynamics of Oxidation–Reduction Reactions

389

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. NADH

Q Determine the total free energy change for the oxidation of NADH by O2.

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 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.

NAD+

Complex I

0

ΔG°ʹ = −69.5 kJ · mol−1

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

BEFORE GOING ON 0.6

• Explain why an oxidation–reduction reaction must include both an oxidant and a reductant. • When two reactants are mixed together, explain how you can predict which one will become reduced and which one will become oxidized. • Explain how adding the Ɛ°ʹ values for two half-reactions yields a value of ∆Ɛ°ʹ and ∆G°ʹ for an oxidation–reduction reaction. • Select the two half-reactions from Table 15.1 that would be most likely to form a freely reversible (near-equilibrium) redox reaction.

0.8

O2

H2O

S A MPLE CALCULATIO N 15.2 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°ʹ: malate→ oxaloacetate + 2 H + + e− NAD + + H + + 2e− → NADH net: malate + NAD + → oxaloacetate + NADH + H +

Ɛ°ʹ = +0.166 V Ɛ°ʹ = −0.315 V ∆Ɛ°ʹ = −0.149 V

Method 2 Identify the electron acceptor (NAD+) and electron donor (malate). Substitute their standard reduction potentials into Equation 15.3: ∆Ɛ°ʹ = Ɛ°ʹ(e− acceptor) − Ɛ°ʹ(e− donor) = −0.315 V − (−0.166 V) = −0.149 V

SEE SAMPLE CALCULATION VIDEOS

390

CH APTER 1 5 Oxidative Phosphorylation

Both Methods The ∆Ɛ°ʹ for the net reaction is –0.149 V. Use Equation 15.4 to calculate ∆G°ʹ: ∆G°ʹ = −nF ∆Ɛ°ʹ = −(2)(96,485 J·V−1 ·mol−1 )(−0.149 V) = +28,750 J·mol−1 = +28.8 kJ·mol−1 The reaction has a positive value of ∆G°ʹ 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.)

L EARNING 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.

15.2

Mitochondrial Electron Transport

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 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 O2, 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 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 Outer membrane architecture that encloses a space called the mitochondrial matrix. Because the Inner membrane inner mitochondrial membrane prevents the transmembrane movements of ions and small molecules (except via specific transport proteins), the composition of Intermembrane space the matrix differs from that of the space between the inner and outer membranes. Matrix In fact, the ionic composition of the intermembrane space is considered to be equivalent to that of the cytosol due to the presence of the porins in the outer mitochondrial membrane (Fig. 15.3). Mitochondria are customarily shown as kidney-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 FIGURE 15.3 Model of mitochondrial slices, reveals that mitochondria are highly variable structures. For example, the structure. The relatively impermeable cristae may be irregular and bulbous rather than planar and may make several inner mitochondrial membrane tubular connections with the rest of the inner mitochondrial membrane (Fig. encloses the protein-rich matrix. The 15.4b). Moreover, a cell may contain hundreds to thousands of discrete bacintermembrane space has an ionic teria-shaped mitochondria, or a single tubular organelle may take the form of an composition similar to that of the extended network with many branches and interconnections (Fig. 15.4c). Indicytosol because the outer mitochondrial vidual mitochondria can move around the cell and undergo fusion (joining) and membrane is permeable to substances with fission (separating). masses of less than about 10 kD.

Mitochondrial Electron Transport

(a)

(b)

FIGURE 15.4 Images of mitochondria. (a) Conventional electron micrograph showing cristae as a system of planar baffles. [K. Porter/Photo Researchers.] (b) Three-dimensional reconstruction of a mitochondrion by electron tomography, showing irregular tubular cristae. [Courtesy Carmen Mannella, Wadsworth Center, Albany, New York.] (c) Electron

(c)

micrograph of a mammalian fibroblast, showing a network of tubular mitochondria (labeled with a green fluorescent dye). The remainder of the cytosol is delineated by microtubules (labeled with a red fluorescent dye). [Courtesy Michael P. Yaffe. From Science 283, 1493–1497 (1991).]

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 (across one or both membranes) by special mechanisms. 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).

+ H3N

COO−

COO−

C H

C O

CH2

CH2

COO−

COO−

Aspartate

Oxaloacetate

NADH + H+

391

NAD+

SEE GUIDED TOUR Oxidative Phosphorylation

COO− HO C H

matrix malate dehydrogenase

CH2 COO−

Malate

MATRIX

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

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.

392

CH APTER 1 5 Oxidative Phosphorylation

ATP MATRIX

INTERMEMBRANE SPACE

ADP

(a)

H+

Pi

(b)

FIGURE 15.6 Mitochondrial transport systems. (a) The adenine nucleotide translocase binds either ATP or ADP and changes its conformation to release the nucleotide on the opposite side of the inner mitochondrial membrane. This transporter 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.

Q Does the activity of either of these transporters contribute to the mitochondrial membrane potential?

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 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). The protein complexes that carry out electron transport and ATP synthesis are oriented in the inner mitochondrial membrane so that they can bind the NADH, ADP, and Pi present in the matrix. Electron microscopy studies show that Complexes I, III, and IV form a “supercomplex,” which likely increases the efficiency of electron transfer between them.

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+ + QH2 Complex I is the largest of the electron transport proteins in the mitochondrial respiratory chain, with 44 different subunits and a total mass of about 980 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). Electron transport takes place in the peripheral arm, 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. 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,

Mitochondrial Electron Transport

393

(b)

(a) FIGURE 15.7 Structure of bacterial Complex I. (a) The 16

subunits 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” projects into the cytosol in bacteria (or mitochondrial matrix in eukaryotes). (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. [Structure of Thermus thermophilus Complex I (pdb 4HEA) determined by R. Baradaran, J. M. Berrisford, G. S. Minhas, and L. A. Sazanov.]

Q Identify the FMN group, the 4Fe – 4S clusters, and the 2Fe–2S clusters in part (b).

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

CH2OPO32⫺ HO HO HO

C C C

HO

H 2

H

H⫹,

2

e⫺

H3C

N

H

HO

C

H

HO

C

H

Cys Fe

Cys

Fe S

Cys

2Fe–2S Cys S

Fe

Fe Fe

Cys S

S S

Fe

CH2 H

CH2 H3C

C

S

Cys

CH2OPO2⫺ 3

H

Cys

N

O N

N O

Flavin mononucleotide (FMN)

H

H3C H 3C

N

N N

N H

O H

O

FMNH2

FIGURE 15.8 Flavin mononucleotide (FMN). This prosthetic group resembles flavin adenine 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.

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.

394

CH APTER 1 5 Oxidative Phosphorylation

and detailed analysis of the crystal structure 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+ transporters (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. 2.14 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 + QH2

NADH + H+

MATRIX

2e− Complex I INTERMEMBRANE SPACE

4 H+

FIGURE 15.10 Complex I function. As two electrons from the water-soluble NADH are transferred to the lipid-soluble ubiquinone, four protons are translocated from the matrix into the intermembrane space.

Succinate dehydrogenase is the only one of the citric acid cycle enzymes that is not soluble in the mitochondrial matrix; it is embedded in the inner membrane. Like the other respiratory complexes, it contains several redox centers, including an FAD group. Succinate dehydrogenase is also called Complex II of the mitochondrial respiratory chain. However, it is more like a tributary because it does not undertake proton translocation and therefore does not directly contribute the free energy of its oxidation–reduction reaction toward ATP synthesis. Nevertheless, it does feed reducing equivalents as ubiquinol into the electron transport chain (Fig. 15.11a). A major source of ubiquinol is fatty acid oxidation, another energy-generating catabolic pathway that takes place in the mitochondrial matrix. A membrane-bound fatty acyl-CoA dehydrogenase catalyzes the oxidation 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 fatty acid also produces NADH that is reoxidized by the mitochondrial electron transport chain, starting with Complex I. Electrons from cytosolic NADH can also enter the mitochondrial NAD+ ubiquinol pool through the actions of a cytosolic and a mitochondrial glycerol-3-phosphate dehydrogenase (Fig. 15.11c). This system, which shuttles electrons from NADH to ubiquinol, bypasses Complex I. Q QH2

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 are proteins with heme prosthetic groups. The name cytochrome literally 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

Mitochondrial Electron Transport R

R

H C H

Succinate

Fumarate

395

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

QH2

2e−

(b)

H2C OH

H2C OH

C O

HO C H

CH2OPO32−

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, which is one step of the fatty acid oxidation pathway, generates ubiquinol. R represents the hydrocarbon tail of the fatty acid. (c) In the glycerol-3-phosphate shuttle system, electrons from cytosolic NADH are used by a cytosolic glycerol-3-phosphate dehydrogenase to reduce dihydroxyacetone phosphate to glycerol-3-phosphate. The mitochondrial enzyme, embedded in the inner membrane, then reoxidizes the glycerol-3-phosphate, ultimately transferring the two electrons to the membrane ubiquinone pool. FIGURE 15.11

Q

CH2OPO32−

Dihydroxyacetone phosphate

Glycerol-3phosphate

cytosolic dehydrogenase

NADH + H+

NAD+

(c)

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: QH2 + 2 cytochrome c (Fe3+ ) ⇌ Q + 2 cytochrome c (Fe2+ ) + 2 H + Complex III itself contains two cytochromes (cytochrome b and cytochrome c1) that are integral membrane proteins. These two proteins, along with an iron–sulfur protein (also called the Rieske protein), form the functional core of Complex III (these same three subunits are the only ones that have homologs in the corresponding bacterial respiratory complex). Altogether, each monomer of Complex III is anchored in the membrane by 14 transmembrane α helices (Fig. 15.13). The flow of electrons through Complex III is complicated, in part because 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

CH2

CH

CH3

H3C

CH N

CH2

N Fe3⫹

N

N

H3C

CH3 CH2

CH2

CH2

CH2

COO⫺

COO⫺ Heme b

FIGURE 15.12 The heme group of a b cytochrome. The planar porphyrin ring surrounds a central Fe atom, shown here in its oxidized (Fe3+) state. The heme substituent groups that are colored blue differ in the a and c cytochromes (the heme group of hemoglobin and myoglobin has the b structure; see Section 5.1).

396

CH APTER 1 5 Oxidative Phosphorylation

MATRIX

INTERMEMBRANE SPACE

(a)

(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). [Structure (pdb 1BE3) determined by S. Iwata, J. W.

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 FIGURE 15.13

Lee, K. Okada, J. K. Lee, M. Iwata, S. Ramaswamy, and B. K. Jap.]

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

cyt c(Fe3+)

2 H+

2 H+

cyt c(Fe2+)

Round 2

QH2 . Q−

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.

cyt b 5 e− e− 4

Q QH2

3. The oxidized ubiquinone diffuses to another quinone-binding site, where it accepts the electron from cytochrome b, becoming a half-reduced semiquinone (.Q −).

2 H+ FIGURE 15.14

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+)

The Q cycle.

Q Write an equation for round 1 and for round 2.

Mitochondrial Electron Transport

397

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.

Q How does the proton-translocating mechanism of Complex III differ from the one in Complex I?

2 H+ MATRIX

Complex III

QH2 + Q 2 QH2

2e−

INTERMEMBRANE SPACE

4 H+ 2 cyt c(Fe3+)

2 cyt c(Fe2+)

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.

Complex IV oxidizes cytochrome c and reduces O2 Just as ubiquinone ferries electrons from Complex I and other enzymes to Complex III, cytochrome c ferries electrons between Complexes III and IV. 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 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 (Fe2+ ) + O2 + 4 H+ ⟶ 4 cytochrome c (Fe3+ ) + 2 H2O 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).

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. Cytochrome c transfers one electron at a time from Complex III to Complex IV. [Structure of tuna cytochrome c (pdb 5CYT) determined by T. Takano.]

MATRIX

INTERMEMBRANE SPACE

(a) FIGURE 15.17 Structure of cytochrome c oxidase. (a) The 13 subunits in each monomeric 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)

(b) from one half of the complex are shown in space-filling form. [Structure (pdb 2OCC) determined by T. Tsukihara and M. Yao.]

Q Locate the copper ion and heme iron of the binuclear center where O2 is reduced.

398

CH APTER 1 5 Oxidative Phosphorylation

Fe2⫹

Cu⫹

FIGURE 15.18 A proposed model for the cytochrome c oxidase reaction. Although the exact sequence of proton and electron transfers is not known, the reaction intermediates shown here are inferred from spectroscopic and other evidence. An enzyme tyrosine radical (not shown) plays a role in electron transfer.

O2 Fe2⫹

Cu⫹

O2 H⫹,e⫺ Fe4⫹

Cu2⫹

O2⫺

OH⫺

H⫹ H2O Fe4⫹

Cu2⫹

O2⫺ H⫹,e⫺ Fe3⫹

Cu2⫹

OH⫺ H⫹,2 e⫺ H2O Fe2⫹

Each 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. Note that the chemical reduction of O2 to H2O consumes four protons from the mitochondrial matrix. One possible sequence of reaction intermediates is shown in Figure 15.18. The incomplete reduction of O2 to H2O is believed to generate free radicals that can damage mitochondria (Box 15.A). Cytochrome c oxidase also relays four additional protons from the matrix to the intermembrane space (two protons for every pair of electrons). The protein complex 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). As the reduced cofactors NADH and ubiquinol deliver their electrons to the electron transport chain, they become oxidized. The cofactors rejoin the pool of oxidized 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. 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).

Cu⫹

Box 15.A Free Radicals and Aging The oxidizing power of molecular oxygen allows aerobic metabolism—a far more efficient strategy than anaerobic metabolism —but comes at a cost. Partial reduction of O2 by Complex IV, or possibly via side reactions carried out in Complexes I and III, can produce the superoxide free radical, · O2−. O2 + 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. Such reactivity means that a free radical, although extremely short-lived (the half-life of · O2− is 1 × 10−6 seconds), can chemically alter nearby molecules. Presumably, the most damage is felt by mitochondria, whose proteins, lipids, and DNA are all susceptible to oxidation as superoxide steals an electron. As the damage accumulates, the mitochondria become less efficient and eventually nonfunctional, at which point the cell self-destructs. According to the free radical theory of aging, oxidative damage mediated by · O2− and other free radicals is responsible for the degeneration of tissues that occurs with aging. Several lines of evidence support the link between free radicals and aging. First, the tissues of some individuals with progeria,

a form of accelerated aging, appear to produce higher than normal levels of oxygen free radicals. Second, cells of all kinds are equipped with antioxidant mechanisms, suggesting that these components perform an essential function. For example, the enzyme superoxide dismutase converts superoxide to a less toxic product, peroxide: 2 ·O−2 + 2 H+ → H2O2 + O2 Other cellular components, such as ascorbate (see Box 5.C) and α-tocopherol (see Box 8.B) may protect cells from oxidative damage by scavenging free radicals. Finally, animal experiments suggest that caloric restriction, which extends life spans, generates fewer free radicals by decreasing the availability of fuel molecules that undergo oxidative metabolism. Unfortunately, studies in humans have not yielded conclusive evidence that consuming particular antioxidants or decreasing fuel consumption diminishes the degeneration that normally accompanies aging. Q Free radicals have been identified as hormone-like signaling molecules in both animals and plants. How might this information alter the free radical theory of aging?

Chemiosmosis

1 2 MATRIX

399

O2 + 2 H+ H2O

Complex IV 2e −

INTERMEMBRANE SPACE

2 H+

2 cyt c(Fe3+) 2 cyt c(Fe2+)

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 O2 →H2O (the full reduction of O2 requires four electrons).

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.

15.3

LEARNING OBJECTIVES

Chemiosmosis

The electrons collected from metabolic fuels during their oxidation are now fully disposed of 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 (∆G°ʹ = +30.5 kJ · mol−1). Complex I: NADH → QH2 Complex III: QH2 → cytochrome c Complex IV: cytochrome c→ O2

∆G°ʹ = −69.5 kJ·mol−1 ∆G°ʹ = −36.7 kJ·mol−1 ∆G°ʹ = −112.0 kJ·mol−1

NADH → O2

∆G°ʹ = −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.

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.

400

CH APTER 1 5 Oxidative Phosphorylation

MATRIX +

+

+

+ +

I +

+ + ++ + + + ++ +++ + + + + + + +

Chemiosmosis links electron transport and oxidative phosphorylation +

+

+

III

INTERMEMBRANE SPACE

+

+

IV

+ ++++ + ++ + + + ++ + + + + + +

+

+

+

++ + + + +++++ + + + + + + + + + +

FIGURE 15.20 Generation of a proton 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.

Until the 1960s, the connection between respiratory electron transport (measured as O2 consumption) and ATP synthesis was a mystery. Credit for discovering the connection belongs primarily to Peter Mitchell, 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 protons represents a source of free energy, also called a protonmotive force, that can drive the activity of an ATP synthase. 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 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).

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. 15.20). This imbalance of protons, a nonequilibrium state, has an associated free energy (the force that would restore the system to equilibrium). The free 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 ∆G = RT ln

[H+]out [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 ∆G = ZF ∆ψ

[15.6]

where Z is the ion’s charge (+1 in this case) and Δψ 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): ∆G = RT ln

[H +] out + ZF ∆ψ [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.

ATP Synthase

401

S A MPLE CALCULATIO N 15.3 Problem

SEE SAMPLE CALCULATION VIDEOS

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 ∆G = 2.303 RT (pHin − pHout ) + ZF ∆ψ Substituting known values gives ∆G = 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

BEFORE GOING ON LEARNING OBJECTIVES • 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

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 F0, 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. 15.21). 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.

ADP + Pi

ATP synthase rotates as it translocates protons Not surprisingly, the overall structure of ATP synthase is conserved among different species. The F1 component consists of three α and three β subunits surrounding a central shaft. The membrane-embedded portion of ATP synthase includes an a subunit, two b subunits that extend upward to interact with the F1 component, and a ring of c subunits (Fig. 15.22). The exact number of c subunits varies with the source; bovine mitochondrial ATP synthase, for example, has 8 c subunits, while some bacterial enzymes have 15 c subunits.

ATP synthase function. As protons flow through the F0 component from the intermembrane space to the matrix, the F1 component catalyzes the synthesis of ATP from ADP + Pi. FIGURE 15.21

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.

ATP

F1 H+ MATRIX

F0

INTERMEMBRANE SPACE

402

CH APTER 1 5 Oxidative Phosphorylation

FIGURE 15.22 Structure of ATP synthase. (a) Model of the mammalian enzyme with individual subunits labeled. A spherical structure consisting of three α and three β subunits is connected via a central stalk (subunits γ, δ, and ɛ) to the membrane-embedded c ring. The a subunit is closely associated with the c ring, and a peripheral stalk containing several subunits (including b) links subunit a to the catalytic domain. (b) X-Ray structure of bovine ATP synthase at 3.5-Å resolution. Some subunits are not visible in the crystal structure. [Structure (pdb 2XND) determined by I. N. Watt, M. G. Montgomery, M. J. Runswick, A. G. W. Leslie, and J. E. Walker.]

␣

b

␣

␤

F1

MATRIX

c c

c

␥

⑀ c c ␦

c

c

F0

a

INTERMEMBRANE SPACE

c

(a)

(b)

In all species, proton transport through ATP synthase involves the 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. 15.23). When properly positioned at the a subunit, a c subunit can take up a proton from the intermembrane space. A slight rotation of the c ring brings another c subunit into position so that it can release its bound proton into the matrix. The favorable thermodynamics of proton translocation force the c ring to keep moving in one direction. Experiments show that depending on the 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 V-ATPases, in fact function as active transporters that use the free energy of the ATP hydrolysis reaction to drive ion movement across the membrane. Attached to the c ring and rotating along with it are the γ, δ, and ɛ subH+ MATRIX units (Fig. 15.22). The δ and ɛ subunits are relatively small, but the γ subunit consists of two long α helices arranged as a bent coiled coil that protrudes into c ␥ c c c the center of the globular F1 structure. The three α and three β subunits of F1 c c c c have similar tertiary structures and are arranged like the sections of an orange around the γ subunit (Fig. 15.24). Although all six subunits can bind adenine H+ nucleotides, only the β subunits have catalytic activity (nucleotide binding to H+ the α subunits may play a regulatory role). a A close examination of the F1 assembly reveals that the γ subunit interacts asymmetrically with the three pairs of αβ units. In fact, each αβ unit has a slightly different conformation, and model-building indicates that for steric INTERMEMBRANE SPACE reasons, the three units cannot simultaneously adopt the same conformation. H+ The three αβ pairs 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, FIGURE 15.23 Mechanism of proton transport since it is held in place by the peripheral arm that is anchored to the a subunit by ATP synthase. When a c subunit (pink) binds (see Fig. 15.22a). a proton from one side of the membrane, it moves For an ATP synthase containing 8 c subunits, the transmembrane moveaway from the a subunit (blue). Because the c ment of each proton could potentially turn the γ shaft by 45° (360° ÷ 8). Howsubunits form a ring, rotation brings another c ever, videomicroscopy experiments indicate that the γ subunit rotates in steps subunit toward the a subunit, where it releases its of 120°, interacting successively with each of the three αβ pairs in one full bound proton to the opposite side of the membrane. rotation of 360°. Electrostatic interactions between the γ and β subunits apparIn mammalian ATP synthase (shown here), one ently act as a catch that holds the γ subunit in place while translocation of two complete rotation of the c ring corresponds to the to three protons builds up strain. Translocation of the next proton causes the translocation of eight protons. The γ subunit in the γ subunit to suddenly snap into position at the next β subunit, a movement of center of the c ring rotates along with it.

ATP Synthase

403

120°. This mechanism accounts for the variation in the number of c subunits in ATP synthases from different sources. The c ring spins in small increments (24° to 45°, depending on the number of c subunits), but the γ subunit makes just three large shifts of 120°.

The binding change mechanism explains how ATP is made

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.

[Structure (pdb 1E79) determined by C. Gibbons, M. G. Montgomery, A. G. W. Leslie, and J. E. Walker.]

ADP + Pi

O

AD

L

T

i

P .P

ADP.Pi L O

ATP

The P:O ratio describes the stoichiometry of oxidative phosphorylation

T

AT P

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. Experiments with the isolated F1 component of ATP synthase show that in the absence of F0, 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).

FIGURE 15.24 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 structure shown in Figure 15.22b.

P AT

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, rotationdriven 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. 15.25):

ATP T L

O

AD

.Pi P

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 nonintegral values would be difficult to reconcile with most biochemical reactions, but

ATP

ATP

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. 15.24. 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. Because each of the three catalytic sites cycles through the three conformational states, ATP is released from one of the three β subunits with each 120° rotation of the γ subunit. FIGURE 15.25

T

L

i

AD P .P

P AT

O

404

CH APTER 1 5 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 F0 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.

Q Why would increasing the activity of UCP promote weight loss?

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. 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 F0 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 O2 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 depends on the rate of fuel catabolism In most metabolic pathways, control is exerted at highly exergonic (irreversible) steps. In oxidative phosphorylation, this step would be the reaction catalyzed by cytochrome c oxidase (Complex IV; see Fig. 15.2). However, there are no known effectors of cytochrome c oxidase activity. Apparently, the close coupling between generation of the proton gradient and ATP synthesis allows oxidative phosphorylation to be regulated primarily by the availability of reduced cofactors (NADH and QH2) produced by other metabolic processes.

Summary

405

Less important regulatory mechanisms may involve the availability of the substrates ADP and Pi (which depend on the activity of their respective transport proteins). Experiments with ATP synthase show that when ADP and Pi are absent, the β subunits cannot undergo the conformational changes required by the binding change mechanism. The γ subunit therefore remains immobile, and no protons are translocated through the c ring. This tight coupling between ATP synthesis and proton translocation prevents the waste of the free energy of the proton gradient. There is also evidence that mitochondria contain a regulatory protein that binds to ATP synthase to inhibit its rate of ATP hydrolysis. The inhibitor is sensitive to pH, so it does not bind to ATP synthase when the matrix pH is high (as it is when electron transport is occurring). However, if the matrix pH drops as a result of a momentary disruption of the proton gradient, the inhibitor binds to ATP synthase. This regulatory mechanism prevents ATP synthase from operating in reverse as an ATPase. The processes of electron transport and oxidative phosphorylation, like other metabolic pathways, maintain a steady state. 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 the c ring of ATP synthase. Similarly, the cell’s concentration of ATP remains more or less constant. 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 rates of ATP consumptions and synthesis may fluctuate dramatically, depending on the cell’s activity level, but they must be balanced. The constant cleavage and regeneration of ATP continues as long as the cell has the fuel and oxygen to support oxidative phosphorylation. 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.

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.

• 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. • 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.

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.

15.3

Chemiosmosis

• The chemiosmotic theory describes how proton translocation during mitochondrial electron transport generates an electrochemical gradient whose free energy drives ATP synthesis.

406

15.4

CH APTER 1 5 Oxidative Phosphorylation

ATP Synthase

• The energy of the proton gradient is tapped as protons spontaneously 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 Pi, 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.

Key Terms oxidative phosphorylation oxidizing agent (oxidant) reducing agent (reductant) half-reaction Ɛ°′ Ɛ Nernst equation

F cellular respiration electron transport chain mitochondrion mitochondrial matrix intermembrane space cristae

electron tomography redox center proton wire cytochrome Q cycle 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 Diversity of the Electron-Transport Chain and the KEGG Database

Problems 15.1 The Thermodynamics of Oxidation– Reduction Reactions

7. Calculate the standard free energy change for the reduction of pyruvate by NADH. Consult Table 15.1 for the relevant half-reactions. Is this reaction spontaneous under standard conditions?

1. Identify the oxidized and reduced forms from the following pairs: a. malate/oxaloacetate; b. pyruvate/lactate; c. NADH/NAD+; d. fumarate/succinate.

8. Calculate the standard free energy change for the reduction of oxygen by cytochrome a3. Consult Table 15.1 for the relevant half-reactions. Is this reaction spontaneous under standard conditions?

2. Identify the oxidized and reduced forms from the following pairs: a. Fe(CN)63−/Fe(CN)62−; b. H2O2/O2; c. FMN/FMNH2 (see Fig. 15.8); d. α-ketoglutarate/isocitrate. 3. Calculate the reduction potential of fumarate at 37°C when [fumarate] = 80 μM and [succinate] = 100 μM. 4. 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. 5. 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? 6. 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?

9. Calculate the standard free energy change for the oxidation of malate by ubiquinone. Is the reaction spontaneous under standard conditions? 10. In yeast, alcohol dehydrogenase reduces acetaldehyde to ethanol (Section 13.1). Calculate the free energy change for this reaction under standard conditions. 11. In one of the final steps of the pyruvate dehydrogenase reaction (see Section 14.1), E3 reoxidizes the lipoamide group of E2, then NAD+ reoxidizes E3. Calculate ΔG°′ for the electron transfer from dihydrolipoic acid to NAD+. 12. 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. 13. Acetaldehyde may be oxidized to acetate. Would NAD+ be an effective oxidizing agent? Explain.

Problems

14. Acetoacetate may be reduced to 3-hydroxybutyrate. What serves as a better reducing agent, NADH or FADH2? Explain. 15. For every two QH2 that enter the Q cycle, one is regenerated and the other passes its two electrons to two cytochrome c1 centers. The overall equation is QH2 + 2 cyt c1 (Fe3+ ) + 2 H + →Q + 2 cyt c1 (Fe2+ ) + 4 H + Calculate the free energy change associated with the Q cycle. 16. Why is succinate oxidized by FAD instead of by NAD+? 17. a. What is the ΔƐ value for the oxidation of ubiquinol by cytochrome 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 in part a. Assume T = 25°C. 18. a. What is the ΔƐ value for the oxidation of cytochrome c by the CuA redox center in Complex IV (see Problem 12) 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 in part a. 19. 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) cyt c1 (Fe3+) + e– → cyt c1 (Fe2+)

Ɛ°′ = 0.280 V Ɛ°′ = 0.215 V

20. If the ΔG value for the reaction described in Problem 19 is –6.0 kJ · mol−1, what is the value of ΔƐ? 21. Calculate the overall efficiency of oxidative phosphorylation, assuming standard conditions, by comparing the free energy potentially available from the oxidation of NADH by O2 and the free energy required to synthesize 2.5 ATP from 2.5 ADP. 22. Using the percent efficiency calculated in Problem 21, calculate the number of ATP generated by: a. Complex I (where NADH is oxidized by ubiquinone); b. Complex III (where ubiquinol is oxidized by cytochrome c); c. Complex IV (where cytochrome c is oxidized by molecular oxygen). 23. Calculate ΔƐ°′ and ΔG°′ for the succinate dehydrogenase (Complex II) reaction. 24. Refer to the solution for Problem 23. Does this reaction provide sufficient free energy to drive ATP synthesis under standard conditions? Explain.

15.2 Mitochondrial Electron Transport 25. 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. a. What happens to oxygen consumption when these inhibitors are added to a suspension of respiring mitochondria? b. What is the redox state of the electron carriers in the electron transport chain when each of the inhibitors is added separately to the mitochondrial suspension? 26. What is the effect of added succinate on rotenone-blocked, antimycin A–blocked, or cyanide-blocked mitochondria (see Problem 25)? In other words, can succinate help “bypass” the block? Explain. 27. 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 A–blocked,

407

or cyanide-blocked mitochondria described in Problem 25? b. Ascorbate (vitamin C) can donate a pair of electrons to cytochrome c. Can ascorbate act as a bypass for the rotenone-blocked, antimycin A–blocked, or cyanide-blocked mitochondria described in Problem 25? 28. 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? 29. If cyanide poisoning (see Problem 25) is diagnosed immediately, it can be treated by administering nitrites that can oxidize the Fe2+ in hemoglobin to Fe3+. Why is this treatment effective? 30. The effect of the drug fluoxetine (Prozac) 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 25–27). a. How do pyruvate, malate, and succinate serve as substrates for electron transport? b. What is the effect of fluoxetine on electron transport? [Fluoxetine] (mM)

Rate of electron transport pyruvate + succinate + ascorbate + malate rotenone TMPD

0

160

140

180

0.15

80

130

120

31. Complex I, succinate dehydrogenase, acyl-CoA dehydrogenase, 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? 32. What side chains would you expect to find as part of a proton wire in a proton-translocating membrane protein? 33. 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? 34. 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. 35. Cytochrome c is easily dissociated from isolated mitochondrial membrane preparations, but the isolation of cytochrome c1 requires the use of strong detergents. Explain why. 36. 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? 37. 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? 38. 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 have involved 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). 39. 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 myoglobinfacilitated oxygen diffusion describes myoglobin’s role in transporting

408

CH APTER 1 5 Oxidative Phosphorylation

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. 40. 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. 41. 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? 42. 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?

H+, such as Na+ and Cl−. a. Why was this thought to be important? b. Could ATP still be synthesized if the membrane were permeable to other ions? 53. 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? 54. How does transport of inorganic phosphate from the intermembrane space to the mitochondrial matrix affect the pH difference across the inner mitochondrial membrane? 55. Pioglitazone, a drug used to treat diabetes, causes some membraneembedded 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.

43. 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 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?

56. Metformin, another diabetes drug, suppresses the activity of mitochondrial glycerol-3-phosphate dehydrogenase. Predict the effect of metformin on electron transport and ATP production.

44. 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?

57. In experimental systems, the F0 component of ATP synthase can be reconstituted into a membrane. F0 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 protontranslocating activity of F0? Explain.

15.3

59. A bacterial ATP synthase has 10 c subunits, and a chloroplast ATP synthase has 14 c subunits. Would you expect the bacterium or the chloroplast to have a higher P:O ratio?

Chemiosmosis

45. What is the free energy change for generating the electrical imbalance of protons in neuroblastoma cells, where Δψ is 81 mV? 46. What is the free energy change for generating the electrical imbalance of protons in respiring mitochondria in culture, where Δψ is 150 mV? 47. 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. 48. 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. 49. What size 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. 50. What size 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. 51. 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. 52. Mitchell’s original chemiosmotic hypothesis relies on the impermeability of the inner mitochondrial membrane to ions other than

15.4

ATP Synthase

58. 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.

60. Experiments indicate that the c ring of ATP synthase spins at a rate of 6000 rpm. How many ATP molecules are generated each second? 61. In addition to its effects on electron transport, fluoxetine (see Problem 30) can also inhibit ATP synthase. Why might long-term use of fluoxetine be a concern? 62. 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. 63. 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 lactate or ethanol. Do organisms that can completely oxidize glucose have an advantage over organisms that cannot? 64. 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. 65. 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. The O2 that we breathe in is not directly converted

Problems

to the CO2 that we breathe out. Write a balanced equation for the complete combustion of glucose and oxygen. 66. 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. How much ATP is generated per mole of glucose when the glycerol-3-phosphate shuttle is used? 67. 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. 68. Consider the adenine nucleotide translocase and the Pi – H+ symport protein that import ADP and Pi, the substrates for oxidative phosphorylation, into the mitochondrion (see Fig. 15.6). a. How does the activity of the adenine nucleotide translocase 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 conclude about the thermodynamic force that drives the two transport systems? 69. 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 facilitates hexokinase binding to the outer mitochondrial membrane, where it then becomes closely associated with the adenine nucleotide translocase. Explain why this process benefits the cancer cell. 70. The compounds atractyloside and bongkrekic acid both bind tightly to and inhibit the adenine nucleotide translocase. How do these compounds affect ATP synthesis? Electron transport? 71. The compound tetramethyl-p-phenylenediamine (TMPD) donates a pair of electrons directly to Complex IV (see Problem 27a). What is the P:O ratio of this compound? 72. Ascorbate (vitamin C) can donate a pair of electrons to cytochrome c. What is the P:O ratio for ascorbate? 73. 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. 74. Experiments in the late 1970s attributed the Pasteur effect (see Problem 73) to the stimulation of hexokinase and phosphofructokinase under anaerobic conditions. Upon exposure to oxygen, the stimulation of these enzymes ceases. Why are these enzymes more active in the absence of oxygen? 75. 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? 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

NO2

Dinitrophenol

409

76. 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 ATPhydrolyzing activity of ATP synthase. 77. Oligomycin is an antibiotic that blocks proton transfer through the F0 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? d. What changes occur when dinitrophenol (see Problem 75) is then added to the suspension? 78. The compound carbonylcyanide-p-trifluoromethoxy phenylhydrazone (FCCP) is an uncoupler similar to DNP (see Problem 75). Describe how FCCP acts as an uncoupler.

N

H

C C

N

N N

OCF3

C FCCP

79. Dinitrophenol (see Problem 75) 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? 80. 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) of 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? 81. UCP1 is an uncoupling protein in brown fat (Box 15.B). Experiments using UCP1-knockout mice (animals missing the gene for UCP1) resulted in the discovery of a second uncoupling protein named UCP2. 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 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. But the body temperatures of the cold-exposed knockout mice decreased 10°C or more. Explain. 82. 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? 83. The gene that codes for an uncoupling protein (see Problem 82) in potatoes was isolated. The results of a Northern blot analysis (which detects mRNA) are shown below. What is your interpretation

410

CH APTER 1 5 Oxidative Phosphorylation

of these results? How does the mRNA level affect thermogenesis in the potato? 20°C Days:

0

1

Glutamate

glutamate transporter

4°C 2

3

0

1

INTERMEMBRANE SPACE

2

MATRIX

3

glutamate dehydrogenase α-Ketoglutarate Glutamate

84. 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. 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?

NAD+

NADH

85. 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? 86. 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 Boekema, E. J. and Braunm H.-P., Supramolecular structure of the mitochondrial oxidative phosphorylation system, J. Biol. Chem. 282, 1–4 (2007). [Presents evidence that Complexes I through IV form “supercomplexes.”] Boyer, P. D., Catalytic site forms and controls in ATP synthase catalysis, Biochim. Biophys. Acta 1458, 252–262 (2000). [Reviews the steps of ATP synthesis and hydrolysis, along with experimental evidence and alternative explanations; by the originator of the binding change mechanism.] Friedman, J. R. and Nunnari, J., Mitochondrial form and function, Nature 505, 335–343 (2014). [Descibes some attributes of mitochondria as semi-independent cellular organelles.]

Watt, I. N., Montgomery, M. G., Runswick, M. J., Leslie, A. G. W., and Walker, J. E., Bioenergetic cost of making an adenosine triphosphate molecule in animal mitochondria, Proc. Nat. Acad. Sci. 107, 16823–16827 (2010). [Includes details of ATP synthase structure.] Zickerman, V., Wirth, C., Nasiri, H., Siegmund, K., Schwalbe, H., Hunte, C., and Brandt, U., Mechanistic insight from the crystal structure of mitochondrial complex I, Science 347, 44–49 (2015). [Describes the structure of a eukaryotic Complex I and a possible mechanism for linking electron transport to proton pumping.]

CHAPTER 16

Mark Newman/Getty Images

Photosynthesis

The bright green color of the anemone Anthopleura xanthogrammica is due in part to the presence of symbiotic algae in its tentacles. The algae obtain nutrients from the anemone and in turn supply it with O2, a by-product of 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). • 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 free 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

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.

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.

411

412

CH APTER 1 6 Photosynthesis 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 biosynthetic reactions.

BIOPOLYMERS Proteins Nucleic acids Polysaccharides Triacylglycerols

MONOMERS Amino acids Nucleotides Monosaccharides Fatty acids +

NH4

NAD+

NAD+ NADH

NADH, QH2 2- and 3-Carbon INTERMEDIATES

NAD+, Q

citric acid cycle

photosynthesis

CO2 NADH, QH2 O2

ADP oxidative phosphorylation ATP

H2O NAD+, Q

These organisms absorb enough solar energy to undertake the energetically costly oxidation 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 ion-impermeable 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.

Pigments absorb light of different wavelengths 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: E=

hc λ

[16.1]

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. Inner membrane

FIGURE 16.2 The chloroplast.

(a) Electron micrograph of a chloroplast from tobacco. [Dr.

Stroma Thylakoid membrane

Outer membrane

Jeremy Burgess/Photo Researchers, Inc.]

Thylakoid lumen

(b) Model. The stacked thylakoid membranes are known as grana (singular, granum). Q Compare these images to the images of a mitochondrion in Figure 15.4.

(a)

(b)

Chloroplasts and Solar Energy

413

S A MPLE CALCULATIO N 16.1 Problem

SEE SAMPLE CALCULATION VIDEOS

Calculate the energy of a photon with a wavelength of 550 nm.

Solution hc λ (6.626 × 10 −34 J·s)(2.998 × 108 m · s −1 ) E= 550 × 10 −9 m E = 3.6 × 10 −19 J E=

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. The second most common 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 types of pigments absorb all the wavelengths of visible light (Fig. 16.4).

CH2

(a)

CH

CH3

H3C

CH2 N

CH3

N Mg

N

N

H3 C

CH3 CH2

H3C

H3C

H3C

H3C

O

CH3

O

CH2

C

C

O

O

CH3

O

Chlorophyll a

(b)

H3C CH3

H3C CH3

CH3

CH3

␤-Carotene

CH3

⫺OOC

(c)

CH3

COO⫺

CH3

CH2 CH2

CH3

CH3 CH CH3

CH2 CH2

CH3 CH3 CH2

H O

N H

C H

N H

C H

N H

Phycocyanin

C H

N H

O

H3C CH3

FIGURE 16.3 Some common chloroplast photoreceptors. (a) Chlorophyll a. In chlorophyll b, a methyl group (blue) is replaced by an aldehyde group. 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. (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.

414

CH APTER 1 6 Photosynthesis

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.

Chlorophyll a

Carotenoids

Absorbance

Q Use this diagram to explain the color of each type of pigment molecule.

Solar radiation

Chlorophyll b

Phycocyanin

400

500 Wavelength (nm)

600

700

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 a recipient molecule. In this process, called photooxidation, the excited molecule becomes oxidized and the acceptor molecule becomes reduced. Another electron-transfer reaction is required to restore the photooxidized molecule to its original reduced state. All of these energy-transferring processes occur in chloroplasts to some extent, but exciton transfer and photooxidation are the most important for photosynthesis.

Light-harvesting complexes transfer energy to the reaction center Chlorophyll excitation Chlorophyll*

Xox

X 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.

The primary reactions of photosynthesis occur at specific chlorophyll molecules called reaction centers. However, chloroplasts contain many more 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 protein microenvironment of each photoreceptor influences the wavelength (and therefore the energy) of

The Light Reactions

(a) 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 of chlorophyll molecules (yellow and green) and carotenoids (red). The pigments are all within a few angstroms of each other. (a) Side view. The extracellular side is at the top. (b) Top

(b)

415

(c)

view. (c) Top view showing only the chlorophyll molecules. The 18 chlorophyll molecules in the inner ring (green) overlap so that excitation energy may be delocalized over the entire ring. [Structure (pdb 1KZU) determined by R. J. Cogdell, A. A. Freer, N. W. Isaacs, A. M. Hawthornthwaite-Lawless, G. McDermott, M. Z. Papiz, and S. M. Prince.]

the photon it can absorb (just as 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

FIGURE 16.7 Function of lightharvesting 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.

The Light Reactions

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

416

CH APTER 1 6 Photosynthesis

L EARNING 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.

FIGURE 16.8 Structure of cyanobacterial Photosystem II. The proteins are shown as gray ribbons, and the various prosthetic groups and cofactors 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. [Structure of Photosystem II from Synechococcus elongatus (1S5L) determined by K. N. Ferreira, T. M. Iverson, K. Maghlaoui, J. Barber, and S. Iwata.]

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

FIGURE 16.9 Arrangement of prosthetic groups in Photosystem II. The green chlorophyll groups constitute the photooxidizable P680. The two “accessory” chlorophyll groups (yellow) do not undergo oxidation or reduction. An electron from P680 travels to one of the pheophytin groups (orange), which are essentially chlorophyll molecules without the central Mg2+ ion. Next, the electron is transferred to a tightly bound plastoquinone molecule (blue) and then to a loosely bound plastoquinone (not shown). An iron atom (red) may assist the final electron transfer. The lipid tails of the prosthetic groups are not shown.

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 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). Several dozen chlorophyll molecules in 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.

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. 16.9). Although the prosthetic groups in Photosystem II are arranged more or less symmetrically,

The Light Reactions

417

they do not all directly participate in electron transfer. The electron eventually reaches a plastoquinone molecule on the stromal side of Photosystem II. Plastoquinone (PQ) is similar to mammalian mitochondrial ubiquinone (see Section 12.2).

O H

H3C

CH3 H3C

[CH2

CH

C

CH2]n

H

O Plastoquinone

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

FIGURE 16.10 Structure of the Mn4CaO5 cluster. Atoms are colorcoded: 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. The Asp, Glu, and His side chains that hold the cluster in place are not shown. [Structure of PSII (pdb 3WU2) determined by Y. Umena, K. Kawakami, J.-R. Shen, and N. Kamiya.]

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 H2 O → O2 + 4 H + + 4 e − The electrons derived from H2O are used to restore photooxidized P680 to its reduced state. The catalyst for the water-splitting reaction is a cofactor with the composition Mn4CaO5 (Fig. 16.10). 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 Tyrosine radical

The oxidation of water is a thermodynamically demanding reaction because O2 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. 16.11). Recall that the lower the reduction potential, the higher the free 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 photooxidation events in Photosystem II are required to oxidize two H2O molecules and produce one O2 molecule. Figure 16.12 summarizes the functions of Photosystem II.

418

CH APTER 1 6 Photosynthesis −1.4

FIGURE 16.11 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.

−1.2 −1.0 P680*

−0.8 −0.6

Pheophytin

−0.4

Ɛ°′(V)

Q Will pheophytin be in the oxidized or reduced state when it is dark?

−0.2

PQA

0 PQB

+0.2 +0.4 +0.6 +0.8 +1.0 +1.2

H2O

.

Y

P680

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 flow of electrons among its redox groups, to the final transfer of electrons to a mobile electron carrier. The cytochrome b6 f complex contains eight subunits in each of its monomeric halves (Fig. 16.13). 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 chloroplast 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

4 H+ STROMA

Photosystem II 4e − LUMEN

2 PQH2 2 PQ

Oxygenevolving complex 2 H2O

4 H+ O2

FIGURE 16.12 Photosystem II function. For

every oxygen molecule evolved, two plastoquinone molecules are reduced.

FIGURE 16.13 Structure of cyanobacterial cytochrome b6 f. Each subunit of the dimeric complex is a different color. The prosthetic groups are not shown. [Structure (pdb 1VF5) determined by G. Kurisu, H. Zhang, J. L. Smith, and W. A. Cramer.]

Q Compare this structure to the functionally similar mitochondrial cytochrome bc1 (Complex III) in Figure 15.13.

The Light Reactions

419

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. 16.14). Plastocyanin functions as a one-electron carrier by cycling between the Cu+ and Cu2+ oxidation states. Like cytochrome c, plastocyanin is a peripheral 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. 16.15). The resulting pH gradient between the stroma and the lumen is a source of free energy that drives ATP synthesis, as described below.

A second photooxidation occurs at Photosystem I Photosystem I, like Photosystem II, is a large protein complex containing multiple pigment molecules. The Photosystem I in the cyanobacterium Synechococcus is a symmetric trimer with 12 proteins in each monomer (Fig. 16.16). Ninety-six chlorophyll molecules and 22 carotenoids operate as a built-in light-harvesting complex for Photosystem I. In the core of each monomer, 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 electron to achieve a low-energy oxidized state, P700+. The group is then reduced by accepting an electron donated by plastocyanin.

FIGURE 16.14 Plastocyanin. The redox-active copper ion (green) is coordinated by a Cys, a Met, and two His residues (yellow). [Structure of plastocyanin from poplar leaves (pdb 1PLC) determined by J. M. Guss and H. C. Freeman.]

4 H+ STROMA

P700*

e−

Cytochrome b6f

2 PQH2 + 2 PQ

4e−

4 PQH2 P700

P700+

LUMEN

8 H+ 4 Plastocyanin (Cu2+) Plastocyanin (Cu2+)

4 Plastocyanin (Cu+)

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

FIGURE 16.15 Cytochrome b6 f function. The stoichiometry shown for the cytochrome b6 f Q cycle reflects the four electrons released by the oxygenevolving complex of Photosystem II.

FIGURE 16.16 Structure of cyanobacterial Photosystem I. 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. Only one monomer of the trimeric complex is shown. The stroma is at the top. [Structure of the Synechococcus Photosystem I (pdb IJB0) determined by P. Jordan, P. Fromme, H. T. Witt, O. Klukas, W. Saenger, and N. Krauss.]

420

CH APTER 1 6 Photosynthesis

FIGURE 16.17 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.

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. 16.17). 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. 16.18). Reduced ferredoxin participates in two different electron transport pathways in the chloroplast, which are known as noncyclic and cyclic electron flow. In noncyclic 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. 16.19). The net result of noncyclic electron flow is therefore the transfer of electrons from water, through Photosystem II, cytochrome b6 f, Photosystem I, and then on to NADP+. 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. 16.20). The zigzag pattern is due to the two photooxidation events, which markedly decrease the reduction potentials of P680 and P700. 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. There, 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. 16.21). 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 O2 or NADPH). By varying the proportion of electrons that follow the noncyclic and cyclic pathways through Photosystem I, a photosynthetic cell can vary the proportions of ATP and NADPH produced by the light reactions.

2 NADPH

2 NADP+ + 2 H+

4 Ferredoxinox 4 Ferredoxinred STROMA

Photosystem I 4e−

LUMEN

4 Plastocyanin (Cu+)

FIGURE 16.18 Ferredoxin. The 2Fe–2S cluster is shown in orange. [Structure of ferredoxin from the cyanobacterium Anabaena (pdb 1CZP) determined by R. Morales, M. H. Charon, and M. Frey.]

4 Plastocyanin (Cu2+)

FIGURE 16.19 Noncyclic 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. Therefore, 2 NADPH are produced for every molecule of O2.

The Light Reactions −1.4

FIGURE 16.20 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). Excitation of P680 and P700 ensures that electrons follow a thermodynamically favorable pathway to groups with increasing reduction potential.

P700*

−1.2 −1.0 P680*

−0.8

Ferredoxin

−0.6 −0.4

Ɛ°′(V)

−0.2

NADP+

Plastoquinone

0 Cytochrome b6f

+0.2

P700

+0.6 +0.8

Q Compare the redox changes depicted here with those of the mitochondrial electron transport chain in Figure 15.2.

Plastocyanin

+0.4

421

H2O

+1.0

P680

+1.2

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 highly homologous to mitochondrial and bacterial ATP synthases. The CF1CF0 complex (“C” indicates chloroplast) consists of a proton-translocating integral membrane component (CF0) mechanically linked to a soluble CF1 component where ATP synthesis occurs by a binding change mechanism (as described in Fig. 15.25). The movement of protons from the thylakoid lumen to the stroma provides the free energy to drive ATP synthesis (Fig. 16.22). 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.

ADP + Pi

ATP

CF1 H+

Ferredoxinred 2

STROMA

H+ Ferredoxinox

STROMA

Cytochrome b6f

PQH2

CF0

Photosystem I

e−

e−

LUMEN

PQH2 LUMEN

Plastocyanin (Cu+) 2 H+ Plastocyanin (Cu2+)

FIGURE 16.21 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 16.22 Photophosphorylation. As protons traverse the CF0 component of chloroplast ATP synthase (following their concentration gradient from the lumen to the stroma), the CF1 component carries out ATP synthesis.

422

CH APTER 1 6 Photosynthesis

Assuming noncyclic 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. 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. • Discuss the yields of O2, NADPH, and ATP in cyclic and noncyclic 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.

L EARNING OBJECTIVES Describe the steps of carbon fixation by the Calvin cycle. • Distinguish the two activities of rubisco. • Summarize the function of most 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.

16.3

Carbon Fixation

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 atmospheric carbon dioxide in biologically useful organic molecules, a process called carbon fixation.

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. 16.23). This reaction itself does not require ATP or NADPH, but the 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 acetyl 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. Rubisco is easily the most abundant biological catalyst. 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. 16.24). In some archaebacteria, rubisco has ten 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.

Carbon Fixation

CH2OPO32⫺

1

O

C

2

H 3C

OH

H 4C

OH

CH2OPO32⫺

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.

⫺

O

H

H⫹

C C

O

C

OH

H 2. The enediolate intermediate nucleophilically attacks CO2 .

CH2OPO32⫺ CO2

CH2OPO32⫺

5

Ribulose-1,5-bisphosphate

Enediolate

CH2OPO32⫺ HO

H

CH2OPO32⫺ HO

C

H

COO⫺ COO⫺ H

C

H2O

OH

CH2OPO32⫺

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

C

COO⫺

C

O

C

OH

CH2OPO32⫺

H⫹

CH2OPO32⫺

⫹

423

3. H2O attacks C3.

CH2OPO32⫺

2 3-Phosphoglycerate FIGURE 16.23 The rubisco carboxylation reaction.

Q Before this reaction was understood, scientists believed that carbon fixation involved the reaction of CO2 with a two-carbon molecule. Explain.

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

⫺O

H2O

H C

OH ⫹ O2

H C

OH

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).

FIGURE 16.24 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. [Structure (pdb 1RCX) determined by T. C. Taylor and I. Anderson.]

424

CH APTER 1 6 Photosynthesis

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 condensation of bicarbonate (HCO3–) with phosphoenolpyruvate to yield oxaloacetate, which is then reduced to malate. 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 C4 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 global warming 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 the warming trend 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.

Q Operation of the C4 pathway consumes ATP. Using the pathway shown here, indicate where this occurs.

COO⫺

COO⫺ OPO32⫺

C

C

CH3

CH2

Pyruvate

Phosphoenolpyruvate

CO2

carbonic anhydrase

HCO⫺ 3 Pi

phosphoenolpyruvate carboxylase

COO⫺ C

O

malic enzyme

NADPH ⫹ CO2

Calvin cycle

NADP⫹

COO⫺ malate dehydrogenase

O

CH2 NADPH NADP⫹

COO⫺ Oxaloacetate

CHOH CH2 COO⫺ Malate

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 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. 16.25). 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. 16.25) is identical to the phosphoglycerate kinase reaction of glycolysis (see Section 13.1). Next, bisphosphoglycerate is reduced by the chloroplast enzyme glyceraldehyde3-phosphate dehydrogenase, which resembles the glycolytic enzyme but uses NADPH rather than NADH. The NADPH is the product of the light reactions of photosynthesis.

Carbon Fixation

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

H

C

Ribulose-1,5-bisphosphate

OH

CH2OPO2⫺ 3 2 3-Phosphoglycerate

3

ATP

phosphoglycerate kinase

4

H

O

glyceraldehyde-3-phosphate dehydrogenase

C H

C

OH

CH2OPO2⫺ 3

2 Glyceraldehyde-3-phosphate

OPO2⫺ 3

O C H

Pi ⫹ NADP⫹ NADPH

ADP

C

CH2OPO2⫺ 3 2 1,3-Bisphosphoglycerate

FIGURE 16.25 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 glyceraldehyde-3-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 glyceraldehyde3-phosphate to glucose consists of reactions that require no further input of ATP. Glyceraldehyde3-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 remains part of the Calvin cycle and enters a series of isomerization and group-transfer reactions that regenerate ribulose-5-phosphate. These interconversion reactions are similar to those of the pentose phosphate pathway (Section 13.4). We can represent them simply by showing how carbon atoms are shuffled among three- to seven-carbon sugars: C3 + C3 → C6 C3 + C6 → C4 + C5 C3 + C4 → C7 C3 + C7 → C5 + C5 net reaction: 5 C3 → 3 C5 Consequently, if the Calvin cycle starts with three five-carbon ribulose molecules, so that three CO2 molecules are fixed, the products are six three-carbon glyceraldehyde-3-phosphate molecules, five of which are recycled to form three ribulose molecules, leaving the sixth (representing the three fixed CO2) as its net product.

3 CO2

3 C5

6 C3

OH

3 C5

425

426

CH APTER 1 6 Photosynthesis

The net equation for the Calvin cycle, including the ATP and NADPH cofactors, is 3 CO2 + 9 ATP + 6 NADPH → glyceraldehyde-3-phosphate + 9 ADP + 8 Pi + 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 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 noncyclic 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 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 CO2 with the ɛ-amino group: (CH2 ) 4NH2 + CO2 ⇌ (CH2 ) 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 ATP

Adenosine

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.

Carbon Fixation

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. Glyceraldehyde3-phosphate or its isomer dihydroxyacetone phosphate is transported out of the chloroplast 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:

HOCH2 H HO

O H OH

H

H

OH

2⫺O

H UDP

O

3POH2C

H

⫹

CH2OH

HO

H

OH H

OH

UDP–Glucose

Fructose-6-phosphate

UDP

H HO

CH2OH O H OH H H

O

H HOCH2 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

HO

CH2OH OH

OH

H

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,

427

428

CH APTER 1 6 Photosynthesis

each containing approximately 36 cellulose polymers, and all embedded in an amorphous matrix of other polysaccharides (see Box 11.A; 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 become incorporated into other compounds such as monosaccharides. • Identify the source of the ribulose-1,5-bisphosphate used for carbon fixation by rubisco. • 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.

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 Photosystem 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 noncyclic electron flow Z-scheme cyclic electron flow photophosphorylation

dark reactions carbon fixation photorespiration C4 pathway Calvin cycle quantum yield

Problems

429

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. photophosphorylation; 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.

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)? 15. 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.

3. The thylakoid membrane contains some unusual lipids. One of these is galactosyl diacylglycerol; a β-galactose residue is attached to the first glycerol carbon. Draw the structure of galactosyl diacylglycerol.

16. 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?

4. Thylakoid membranes contain lipids with a high degree of unsaturation. What does this tell you about the character of the thylakoid membrane?

16.2

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. At what wavelength would a mole of photons have an energy of 250 kJ? 8. At what wavelength would a mole of photons have an energy of 400 kJ? 9. 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. 10. 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. 11. Red tides are caused by 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. 12. Some photosynthetic bacteria live in murky ponds where visible light does not penetrate easily. What wavelengths might the photosynthetic pigments in these organisms absorb? 13. Compare the structures of chlorophyll a (Fig. 16.3) and the reduced form of heme b (Fig. 15.12). 14. 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

The Light Reactions

17. 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? 18. 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? 19. 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? 20. When the antifungal agent myxothiazol is added to a suspension of chloroplasts, the QH2/Q ratio increases. Where does myxothiazol inhibit electron transfer? 21. 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? 22. 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?

430

CH APTER 1 6 Photosynthesis

Mutant

Characteristics

Ala → Cys Ala → Ser Ala → Ile

Similar to wild-type Photosynthesis impaired Photoautotrophic growth and photosynthesis impaired; resistant to herbicides Not photosynthetically competent

Ala → Arg

23. Use Equation 15.4 to calculate the free energy change for transforming one mole of P680 to P680*. 24. Determine the wavelength of the photons whose absorption would supply the free energy to transform one mole of P680 to P680* (see Problem 23). 25. 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 and Δψ is –50 mV. 26. Compare the free energy of proton translocation calculated in Problem 25 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? Which contributes a larger component of the free energy for each process, the pH difference or the membrane potential? 27. Calculate the standard free energy change for the oxidation of one molecule of water by NADP+. 28. Calculate the energy available in two photons of wavelength 600 nm. Compare this value with the standard free energy changes you calculated in Problem 27. Do two photons supply enough energy to drive the oxidation of one molecule of water by NADP+? 29. 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? 18

30. If radioactively labeled water (H2 O) is provided to a plant, where does the label appear? 31. Predict the effect of an uncoupler such as dinitrophenol (see Box 15.B) on production of a. ATP and b. NADPH by a chloroplast. 32. 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? 33. Does the quantum yield of photosynthesis increase or decrease for systems where a. the CF0 component of ATP synthase contains more c subunits or b. when the proportion of cyclic electron flow through Photosystem I increases? 34. Oligomycin inhibits the proton channel (F0) of the ATP synthase enzyme in mitochondria but does not inhibit CF0. 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.

16.3

Carbon Fixation

35. Defend or refute this statement: The “dark” reactions are so named because these reactions occur only at night. 36. 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?

37. 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? 38. As described in the text, 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? 39. A tiny acorn grows into a massive oak tree. Using what you know about photosynthesis, what accounts for the increase in mass? 40. The ΔG°′ value for the rubisco reaction is –35.1 kJ · mol–1 and the ΔG is –41.0 kJ · mol–1. What is the ratio of products to reactants under normal cellular conditions? Assume a temperature of 25°C. 41. 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. 42. Some plants synthesize the sugar 2-carboxyarabinitol-1-phosphate. This compound inhibits the activity of rubisco. a. What is the probable mechanism of action of the inhibitor? b. Why do plants synthesize the inhibitor at night and break it down during the day?

CH2OPO2⫺ 3 HO

C

COO⫺

H

C

OH

H

C

OH

CH2OH 2-Carboxyarabinitol1-phosphate 43. An “activating” CO2 reacts with a lysine side chain on rubisco to carboxylate it. 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? 44. Chloroplast phosphofructokinase (PFK) is inhibited by ATP and NADPH. How does this observation link the light reactions to the regulation of glycolysis? 45. Crabgrass, a C4 plant, remains green during a long spell of hot dry weather when C3 grasses turn brown. Explain this observation. 46. 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. 47. 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 dihydroxyacetone 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. Show how the same antiport could “transport” ATP and reduced cofactors from the chloroplast to the cytosol. 48. The sedoheptulose bisphosphatase (SBPase) enzyme in the Calvin cycle catalyzes the removal of a phosphate group from C1

Selected Readings

of sedoheptulose-1,7-bisphosphate (SBP) to produce sedoheptulose7-phosphate (S7P). The ΔG°′ for this reaction is –14.2 kJ · mol–1 and the ΔG 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. 49. 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

OPO32⫺ ⫹ HCO3⫺

COO⫺ PEPC

CH2 PEP

C

O ⫹ HPO42⫺

CH2 COO⫺ OAA

50. In germinating oil seeds, triacylglycerols are rapidly converted to sucrose and protein. What is the role of PEPC (see Problem 49) in this process? 51. The use of transgenic plants is increasing. These plants are constructed by inserting a new gene into the plant genome to give the plant a desirable characteristic, such as resistance to pesticides or frost, or to confer a nutritional benefit. Transformed plants contain the gene of interest, a promoter (to induce expression of the gene), and a terminator. The cauliflower mosaic virus (CaMV) is often used as a promoter in transgenic plants. A partial DNA sequence is shown below. Design a set of 18-bp primers (see Figure 3.18) that you could use to detect the presence of this promoter in a transgenic plant.

52. Novartis has 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 cell and ultimately resulting in the death of the organism. Is the Bt endotoxin toxic to humans? Explain why or why not. 53. The Monsanto Company has 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.

Shikimate ATP Shikimate-3-P PEP

EPSPS

5-Enolshikimate-3-P

Chorismate Phenylalanine

GTAGTGGGATTGTGCGTCATCCCTTACGTCAGT · · · (112 bases) · · · TCAACGATGGCCTTTCCTTTATCGCAATGATGGCATTTGTAGGAGC

431

Tyrosine

54. Is Roundup® (see Problem 53) toxic to humans? Explain why or why not.

Selected Readings Allen, J. F. and Martin, W., Out of thin air, Nature 445, 610–612 (2007). [A short discussion of the cyanobacterial shift to oxygenic photosynthesis billions of years ago.] Andersson, I., Catalysis and regulation in rubisco, J. Exp. Botany 59, 1555–1568 (2008). [Discusses the structure, chemistry, and regulation of rubisco.] Busch A. and Hippler M., The structure and function of eukaryotic photosystem I, Biochim. Biophys. Acta. 1807, 864–77 (2011).

[Discusses electron transfer between the light-harvesting groups and the core of Photosystem I.] Evans, J. R., Improving photosynthesis, Plant Physiol. 162, 1780–93 (2013). [Focuses on carbon fixation and how the process could be engineered for greater efficiency.] Suga, M., et al., Native structure of photosystem II at 1.95 Å resolution viewed by femtosecond X-ray pulses, Nature 517, 99–103 (2015). [Provide details on the fine structure of Photosystem II.]

© Nature Photographers Ltd/Alamy Stock Photo

CHAPTER 17 Lipid Metabolism

Although many bird species migrate long distances, some even from the Arctic to the Antarctic, the record for the longest non-stop flight belongs to the bar-tailed godwit, which migrates from Alaska to New Zealand (and back again) without refueling along the way. In addition to having a size and shape adapted for efficient flight, the godwit powers its long migration by catabolizing significant amounts of stored fat.

DO YOU REMEMBER? • Lipids are predominantly hydrophobic molecules that can be esterified but cannot form polymers (Section 8.1). • Metabolic 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). • Cells also use the free energy of other phosphorylated compounds, thioesters, reduced cofactors, and electrochemical gradients (Section 12.3).

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. But unlike other classes of biological molecules, lipids are insoluble in water, so we will begin by looking at how they travel between organs. L EARNING 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. 432

17.1

Lipid Transport

Approximately half of all deaths in the United States are linked to the vascular disease atherosclerosis (a term derived from the Greek athero, “paste,” and sclerosis, “hardness”). Atherosclerosis is a slow progressive disease that begins with the accumulation of lipids in the walls of large blood vessels. The trapped lipids initiate inflammation by triggering the production of chemical signals that attract white blood cells, particularly macrophages. These cells engorge themselves by taking up the accumulated lipids and continue to recruit more macrophages, 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 may undergo calcification, as occurs in bone formation. This accounts for the “hardening” of the

Lipid Transport

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). What is the source of the lipids that accumulate in vessel walls? They are deposited by 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). 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. 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. The LDL receptor itself is not degraded but cycles back to the cell surface, ready for another round of receptor-mediated endocytosis of an LDL particle. LDL

433

FIGURE 17.1 An atherosclerotic plaque in an artery. Note the thickening of the vessel wall. [James Cavallini/BSIP/Phototake.]

cholesterol LDL receptor

CELL

TA B L E 17.1

Characteristics of Lipoproteins % CHOLESTEROL AND % % CHOLESTERYL PROTEIN TRIACYLGLYCEROL ESTER

LIPOPROTEIN

DIAMETER (Å)

DENSITY (g · cm–3)

Chylomicrons

1000–5000

100 proteins (gray), three of the five snRNAs (green), and size the RNA and proteins that make up the spliceosome and a portion of an intron (orange) [Structure (3JB9) determined by C. Yan, J. that destroy intronic RNA and incorrectly spliced transcripts. Hang, R. Wan, M. Huang, C. Wong, and Y. Shi.] 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. 21.24). 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 (see Table 3.4).

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 life span of a given mRNA molecule is another regulated aspect of gene expression: mRNA molecules decay at different rates. In mammalian cells, mRNA life spans range from less than an hour to about 24 hours. 5′ splice site

5′

3′ splice site

branch point

AG GUAAGU

A

CAG G

3′

20–50 bases 5′ exon

intron

3′ exon

FIGURE 21.22 Consensus sequence at eukaryotic mRNA splice sites. Nucleotides shown in bold

are invariant.

570

CH APTER 2 1 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 21.23 mRNA splicing.

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 destruction of the entire message (Fig. 21.25). 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. 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. gene mRNA transcripts Striated muscle Smooth muscle Nonmuscle/ fibroblast Brain FIGURE 21.24 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. [After Breitbart, R. E., Andreadis, A., and Nadal-Ginard, B., Annu. Rev. Biochem. 56, 481 (1987).]

RNA Processing AAAAAAAAAAAAA 3′ Poly(A) tail

5′ Cap

571

FIGURE 21.25 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.

AAAAAA Deadenylase

Decapping enzyme

5′ exonuclease

3′ exonuclease

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. 21.26). 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

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.

3. RISC unwinds and degrades the passenger RNA strand.

mRNA

SEE ANIMATED PROCESS DIAGRAM FIGURE 21.26 RNA interference. The steps involving siRNA are shown. The miRNA pathway for inactivating mRNA is similar.

Mechanism of RNA interference

572

CH APTER 2 1 Transcription and RNA 45S RNA

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 18S 5.8S 28S locate complementary target mRNAs. Whereas an siRNA specifically seeks and destroys an mRNA that is perfectly complementary, an FIGURE 21.27 Eukaryotic rRNA processing. The initial miRNA can bind to a large number of target mRNAs—possibly transcript of about 13.7 kb has a sedimentation coefficient of hundreds—because it forms base pairs with a stretch of only 6 or 45S. Three smaller rRNA molecules (18S, 5.8S, and 28S) are 7 nucleotides. The captured mRNAs are unavailable for translation derived from it by the action of nucleases. and are susceptible to the standard mechanisms for RNA degradation diagrammed in Figure 21.25. 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. Several clinical trials are under way to test whether siRNAs can turn off the expression of viral genes in order to block 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 consumers who avoid traditional genetically modified foods that contain foreign genes (see Box 3.A).

5′

3′

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 region in the nucleus. The initial eukaryotic rRNA transcript is cleaved and trimmed by endo- and exonucleases to yield three rRNA molecules (Fig. 21.27). 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

O H

H

N

O

Ribose

N N

Ribose Uridine

O

N H

Pseudouridine (␺)

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

RNA Processing

NH2 H3C

⫹

NH N

N N

O

⫹

N N Ribose

3-Methylcytidine (m3C)

CH3

Ribose

NH2

O

O H (CH3)2N

FIGURE 21.28 Some modified nucleotides in tRNA molecules. The parent nucleotide is in black; the modification is shown in red.

N 6-Isopentenyladenosine (i 6A)

1-Methyladenosine (m1A)

H3 C

CH3

C

N

N Ribose

CH

N

N

N

CH2

H

N

N N

N

O Ribose

N 2,N 2-Dimethylguanosine (m22G)

N N

573

H H H H

Ribose Dihydrouridine (D)

rRNAs eventually combine with some 80 different ribosomal proteins to 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 100 or so known nucleotide modifications are shown in Figure 21.28. 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. N 6-Methyladenosine residues seem to favor mRNA degradation, while 5-hydroxymethylcytosine residues enhance mRNA translation. It is likely that RNA modifications provide binding sites for proteins or influence RNA secondary structure. Unlike DNA, whose conformational flexibility is considerably constrained by its doublestranded 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. 21.29). 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. 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

574

CH APTER 2 1 Transcription and RNA

O

A R

A N

R

N R

H N

O

N

N

H

G

N N

N N

N

O

N R

R

G

R N

O

R

N N

G FIGURE 21.29

O

N

N H

O

N

C

R

NH2

N R R

NH2

N H

N

H N H

H N

N H

N

N

H N

N

N

O

N N

H

H

A

H

O

N N

H

U

N N

N

H

N

N

H N

H

H

N

N

A

U

Some nonstandard base pairs. R represents the ribose–phosphate backbone.

true catalyst (Fig. 21.30). 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 repository 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 FIGURE 21.30 RNase P. In this model of the Thermotoga maritima enzyme, the 347-nucleotide RNA is gold and the product tRNA is red. The small protein component (117 amino acids) is green. [Structure (pdb 3Q1Q) determined by N. J. Reiter, A. Osterman, A. Torres-Larios, K. K. Swinger, T. Pan, and A. Mondragon.]

• 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.

Summary 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.

• Gene expression may be regulated by altering histones through acetylation, phosphorylation, and methylation, by methylating DNA, and by rearranging nucleosomes.

Problems

• 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. • Regulatory DNA sequences may affect transcription through binding proteins that interact with RNA polymerase via the Mediator complex. • The bacterial lac operon illustrates the regulation of transcription by a repressor protein.

21.2 RNA Polymerase • Eukaryotic RNA polymerase II transcribes protein-coding genes. It requires no primer and polymerizes ribonucleotides to generate an RNA chain that forms a short double helix with the template DNA. • The polymerase acts processively along the DNA template but reverses to allow the excision of a mispaired nucleotide.

575

• The elongation phase of transcription in eukaryotes is triggered by phosphorylation of the C-terminal domain of RNA polymerase II. • Transcription termination in prokaryotes involves destabilization of the DNA–RNA hybrid helix. In eukaryotes, transcription termination is linked to polymerase pausing and 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. • RNA interference is a pathway for inactivating mRNAs according to their ability to pair with a complementary siRNA or miRNA. • rRNA and tRNA transcripts are processed by nucleases and enzymes that modify particular bases. • The chemical and structural variability of RNA molecules makes it possible for some to function as enzymes.

Key Terms transcription gene mRNA rRNA tRNA operon RNA processing RNA polymerase ncRNA

histone code CpG island imprinting epigenetics promoter consensus sequence TATA box general transcription factor enhancer

activator silencer repressor cap poly(A) tail intron exon splicing spliceosome

small nuclear RNA (snRNA) ribozyme RNAi siRNA miRNA nucleolus small nucleolar RNA (snoRNA) RNA world

Bioinformatics Brief Bioinformatics Exercises 21.1 Viewing and Analyzing RNA Polymerase 21.2 RNA Polymerase, Transcription, and the KEGG Database 21.3 Viewing and Analyzing the lac Repressor

Problems 21.1 Initiating Transcription 1. Why does the genome contain so many more genes for rRNA than mRNA? 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? 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 also induce conformational changes in the DNA. Describe the purpose of these two modes of interaction with the DNA. 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. How do these modifications change the character of their respective side chains? 6. How do the modifications described in Problem 5 decrease the binding affinity between the histones and the DNA?

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CH APTER 2 1 Transcription and RNA

7. 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. 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. In addition to the modifications listed in Table 21.2, histones may also 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. The sulfonium ion’s methyl group is used in methyl-group transfer reactions. a. The demethylated S-adenosylmethionine is then hydrolyzed to produce adenosine and a nonstandard amino acid. Draw the structure of this amino acid. 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 residues in DNA. If a demethylase carries out a hydrolytic reaction to restore cytosine residues, what is the other reaction product?

H CH3

⫹

S

CH2

CH2

O H HO

COO⫺

17. 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? 18. 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

GAGCATATAAGGTGAGGTAGGATCAGTTGCTCCTCACATTT 19. The enzyme EZH2 is a histone lysine methyltransferase that is upregulated in various types of cancers. How does this upregulation affect lysine residues 4 and 27 of histone 3? How does this affect the transcription of genes associated with this histone? 20. 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? 21. 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

NH⫹ 3

CH2

H

C

16. Predict the effect of a mutation in one of the bases in either the –35 or the –10 region of the promoter.

Fully base paired One-base bulge Four-base bulge Eight-base bulge

Adenine H

Kd (nM) 315 0.52 0.0025 0.0013

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. 20.18). 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. a. The sense (coding strand) of 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 contains an AT-rich region and why is this region composed of A:T and not G:C base pairs? AAAATAAATGCTTGACTCTGTAGCG+1

GGAAGGCGTATTATCCAACACCC

b. Use the data in the table to calculate the ΔG°′ 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? 22. 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. 23. 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? 24. The genes of the lac operon are not expressed when the lac repressor binds to the operator. But 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.

Problems

27. 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 phenylGal to growth medium distinguish between wild-type bacterial cells and cells that have a mutation in the lacI gene? 28. 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

I

2000

Polymerase activity Absorbance

1500

1

II

1000

0.5

500

III

0 15

25

20

30 35 Fraction number

0 45

40

35. Radioactively labeled γ-[32P]GTP is added to a bacterial culture undergoing transcription. Is the resulting RNA labeled? If so, where? 36. 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?

21.2 RNA Polymerase

37. The C-terminal domain of RNA polymerase II projects away from the globular portion of the protein. Why?

29. RNA synthesis is much less accurate than DNA synthesis. Why does this not harm the cell?

38. In an experiment, RNA polymerase II was truncated so that its C-terminal domain (CTD) was missing. How would this affect cells?

30. 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.

39. The DNA sequence of a hypothetical E. coli terminator is shown below. N stands 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.

31. Explain why the adenosine derivative cordycepin inhibits RNA synthesis.

5′ · · · NNAAGCGCCGNNNNCCGGCGC T T T TT TNNN · · · 3′ 3′ · · · NN T T CGCGGC NNNNGGC CGCGAAAAAANNN · · · 5′

NH2 N

N HOCH2 H

N

N O H

H

H H

OH

Cordycepin

32. How does your answer to Problem 31 provide evidence to support the hypothesis that transcription occurs in the 5′→3′ direction, not the 3′→5′ direction? 33. The activity of RNA polymerase II is inhibited by the mushroom toxin α-amanitin (Kd = 10–8 M). In 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? 34. 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 33, the investigators noted that RNA polymerase I was fully active in the presence

40. Inosine triphosphate (ITP; Section 18.5) is added to a culture of bacteria, which use it in place of GTP. Inosine (I) forms base pairs with cytidine (C), and the I:C base pairs form two hydrogen bonds. a. Write the sequence of the mRNA transcript that is made using the top strand of the gene shown in Problem 39 as the coding strand. b. Draw the hairpin structure that would form in this RNA transcript. Compare the stability of this RNA hairpin with the RNA hairpin you drew in Problem 39b. How is termination of transcription affected by the ITP substitution? 41. Formation of an RNA hairpin cannot be the sole factor in the termination of transcription in prokaryotes. Why? 42. The addition of β,γ-imido nucleoside triphosphates to cells in culture has been shown to inhibit Rho-dependent termination. Explain why.

O ⫺O

P O⫺

O HN

P O⫺

O O

P

Base O

CH2 O

O⫺ H

␤,␥-Imido nucleoside triphosphate

H

H

OH

OH

H

43. 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

Absorbance at 280 nm

26. 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?

of 5 mM Mg2+ ions whereas polymerases II and III were only 50% active. How did these results support the conclusions that the three peaks constituted three different forms of RNA polymerase? Polymerase activity (units)

25. 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?

577

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CH APTER 2 1 Transcription and RNA

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? 44. 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 would be 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

Kd (nM)

FAD FMN Riboflavin

300 5 3000

45. 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, then resumes transcription, in effect retranscribing 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. 46. 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

47. 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? 48. The bacterial enzyme polynucleotide phosphorylase (PNPase) is a 3′→5′ exoribonuclease that degrades mRNA. a. The enzyme catalyzes a phosphorolysis reaction, as does glycogen phosphorylase (see Section 13.3), rather than hydrolysis. 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? 49. Name the template, substrates, and product for the following polymerases: a. DNA polymerase; b. human telomerase; c. RNA polymerase; d. poly(A) polymerase; e. tRNA CCA-adding enzyme. 50. 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. enhancers; h. 3′ CCA sequence; i. 5′ and 3′ splice sites; j. promoter; k. branch point. 51. Why are only mRNAs capped and polyadenylated? Why do these post-transcriptional modifications not take place on rRNA or tRNA? 52. 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? 53. 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? NH2

O

N

N

H2N

O

N⫹ O H2C H H

P O⫺

H H

OH

O

OH

O O

P

N

N

O

CH2 O

O⫺ H

H

H

O

OH

H

RNA

54. The poly(A) polymerase that modifies the 3′ end of mRNA molecules differs from other polymerases. a. The active sites of DNA and RNA polymerases are large enough to accommodate a

Selected Readings

double-stranded polynucleotide, but the active site of poly(A) polymerase is much narrower. Explain. b. Explain how the substrate specificity of poly(A) polymerase differs from that of a conventional RNA polymerase. 55. 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 cell-free system containing mRNA and RNases? 56. The only mRNA transcripts that lack poly(A) tails are those encoding histones. Why do these mRNA transcripts not require poly(A) tails? 57. ATP can be labeled with 32P at any one of its three phosphate groups, designated α, β, and γ (see Fig. 12.12). 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, or c. γ position? 58. 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. 59. 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? 60. Introns are removed co-transcriptionally rather than posttranscriptionally. Why is this a good cellular strategy? 61. A portion of the gene for the β chain of hemoglobin 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. · · · CCCTGGGCAGGTTGGTA · · · · · · · · TTTCCCACCCTTAGGCTGCT · · ·

579

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 activity studied. 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

k (min–1)

Kd (nM)

Wild-type U69 U69 → G69 U69 → C69 U69 deletion U69 + U70

0.26 0.0034 0.0056 0.0056 0.0054

1.7 73 3 7 181

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 structure. Use what you have learned about the primary, secondary, and tertiary structures of proteins to describe the effect of base substitution on the structure and activity of the ribozyme RNase P. 64. Explain why the vast majority of nucleic acids with catalytic activity are RNA rather than DNA. 65. 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 eliminated 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′

62. The hemoglobin β chain gene contains three exons, so two introns must be removed from the primary mRNA transcript (see Problem 61). What types of mutations would result in splicing errors? How would the mutations affect the protein translated from the improperly spliced mRNA? 63. 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

66. Some tRNA molecules include sulfur-containing nucleotides. Draw the structures of 4-thiouridine and 2-thiocytidine. 67. Biochemistry textbooks published a few decades ago often used the phrase “one gene, one protein.” Why is this phrase no longer accurate? 68. 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 Gilmour, D. S. and Fan, R., Derailing the locomotive: transcription termination, J. Biol. Chem. 283, 661–664 (2008). [Outlines the mechanisms for transcription termination in prokaryotes and eukaryotes.] Liu, X., Bushnell, D. A., and Kornberg, R. D., RNA polymerase II transcription: structure and mechanism, Biochim. Biophys. Acta 1829, 2–8 (2013). [Summarizes the essential components of the transcription initiation and elongation complexes.]

Roy, A. L. and Singer, D. S., Core promoters in transcription: old problems, new insights, Trends Biochem. Sci. 40, 165–171 (2015). [Points out that a majority of mammalian genes do not have recognizable promoter sequences.] Sashital, D. and Doudna, J. A., Structural insights into RNA interference, Curr. Opin. Struct. Biol. 20, 90–97 (2010). [Includes information on the structure of RISC.]

Murakami, K. S., Structural biology of bacterial RNA polymerase, Biomolecules 5, 848–864 (2015). [Summarizes what is known about the structure and function of the bacterial enzyme.]

Sharp, P. A., The discovery of split genes and RNA splicing, Trends Biochem. Sci. 30, 279–281 (2005). [A brief summary of splicing by one of its discoverers.]

Nilsen, T. W. and Graveley, B. R., Expansion of the eukaryotic proteome by alternative splicing, Nature 463, 457–463 (2010). [Reviews the advantages of alternative splicing and how it might be regulated.]

Washburn, R. S. and Gottesman, M. E., Regulation of transcription elongation and termination. Biomolecules 5, 1063–1078 (2015). [Focuses on the transcription in E. coli.]

TessarTheTegu/Shutterstock

CHAPTER 22 Protein Synthesis

In many insects that undergo metamorphosis, the larval stages are characterized by the accumulation of large amounts of storage proteins. During the pupal stage, these molecules are broken down to supply the amino acids needed to synthesize proteins for the adult body.

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). • rRNA and tRNA transcripts are modified to produce functional molecules (Section 21.3).

L EARNING 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. 580

In the decade that followed Watson and Crick’s 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

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 Crick’s work on the structure of DNA, he hypothesized that translation required “adaptor” molecules (subsequently identified as tRNA) that carried an amino acid and recognized genetic information in the form of nucleotides. The 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.

tRNA and the Genetic Code

The genetic code is redundant

581

Ile Ser Arg Thr 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 Pro Ser Arg Glu 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 deHis Leu Glu Ser letion 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 FIGURE 22.1 Reading frames. Even it maintains the proper reading frame for translation. Since a given nucleotide sequence with a nonoverlapping genetic code based in an mRNA molecule can potentially have three different reading frames (Fig. 22.1), the on nucleotide triplets, a given nucleotide sequence has three possible reading selection of the proper one depends on the precise identification of a translation start site. frames. An mRNA molecule can therefore The genetic code, shown in Table 22.1, is said to be redundant because several mRNA potentially specify three different amino codons may correspond to the same amino acid. In fact, most amino acids are specified by acid sequences. 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 appear to cluster; for example, U at the second codon 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. 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 a mammalian cell as many as 150 (this is an obvious example of redundancy in biological systems, since only 20 different amino acids TA B L E 22.1 FIRST POSITION (5ʹ END)

The Standard Genetic Code SECOND POSITION

U

C

A

G

THIRD POSITION (3ʹ 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

582

CH APTER 2 2 Protein Synthesis

are routinely incorporated into polypeptides). tRNAs that bear the same amino acid but have different anticodons are called isoacceptor tRNAs. 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 up to one-quarter are post-transcriptionally modified (the structures of some of these modified nucleotides are shown in Fig. 21.28). 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 (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.

SEE GUIDED TOUR Protein Synthesis

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. As expected, most AARSs 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 C D loop T ψ 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

(a)

C U C G

A

TψC loop

Anticodon Phe

FIGURE 22.2 Structure of yeast tRNA . (a) Secondary structure. 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, with the various structures colored as in part (a). The long

(b) 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. The anticodon and acceptor ends of the molecule are separated by about 75 Å. [Structure (pdb 4TRA) determined by E. Westhof, P. Dumas, and D. Moras.]

Q Why is it important that the bases in the anticodon loop point outward?

tRNA and the Genetic Code

583

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

CH2 O

O⫺

H

Adenine H

3⬘

2⬘

O

H

tRNA

H

H

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 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).

R

H

O

C

C

O⫺

⫹

FIGURE 22.3 The aminoacyl–tRNA synthetase reaction.

ATP

NH⫹ 3 Amino acid

PPi

R

H

O

C

C

O O

NH⫹ 3

P

O

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.

Ribose–Adenine

Q How many “high-energy” phosphoanhydride bonds break in this process? How many “high-energy” acylphosphate bonds form?

TA B L E 22. 2

O⫺

AMINO ACIDS

Aminoacyl–adenylate (aminoacyl–AMP)

tRNA AMP

R

H

O

C

C

O

tRNA

NH⫹ 3 Aminoacyl–tRNA

Classes of Aminoacyl– tRNA Synthetases

Class I

Arg Cys Gln Glu Ile

Leu Met Trp Tyr Val

Class II

Ala Asn Asp Gly His

Lys Pro Phe Ser Thr

2. The amino acid, which has been “activated” by its adenylation, reacts with tRNA to form an aminoacyl–tRNA and AMP.

584

CH APTER 2 2 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. 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). AARSs are modular proteins with a catalytic domain, where amino acid activation and transfer to a tRNA occur, as well as a domain that binds the tRNA anticodon or, in some cases, another part of the tRNA such as the variable loop. 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.

Some synthetases have proofreading activity Studies of individual AARS enzymes indicate that the amino acid binding site 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, by their ability to form hydrogen bonds with the protein. In some cases, the specificity of tRNA aminoacylation may be enhanced through proofreading by the AARS. For example, IleRS almost always produces Ile–tRNAIle and only rarely transfers valine (about once every 50,000 reactions) even though valine differs from isoleucine only by a single methylene group and should easily fit into the IleRS active site. The high fidelity of IleRS requires two active sites that participate in a “double-sieve” mechanism to prevent the synthesis of mischarged tRNAIle. 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. 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.

tRNA anticodons pair with mRNA codons During translation, tRNA molecules align with mRNA codons, base pairing in an antiparallel fashion, for example

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. [Structure of the E. coli complex (pdb 1QRT) determined by J. G. Arnez and T. A. Steitz.]

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.

tRNA anticodon

– C –G– I –

– C –G– I –

– C –G– I –

mRNA codon

– G – C –U–

–G–C–C–

–G–C–A–

As Francis Crick originally proposed in the wobble hypothesis, the third codon position and the 5′ anticodon position experience some flexibility, or wobble, in the

tRNA and the Genetic Code

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 less than 40 tRNAs (the reasons why mammalian cells contain over 150 tRNAs are not clear). Variations in tRNA anticodon sequences allow nonstandard amino acids to be occasionally incorporated into polypeptides at positions corresponding to stop codons (Box 22.A).

TA B L E 22. 3

585

Allowed Wobble Pairs at the Third Codon– Anticodon Position

5′ ANTICODON BASE

3′ CODON BASE

G U A, G U, C U, C, A

C A U G I

BEFORE GOING ON • Summarize the features of the genetic code. • 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 don’t require 61 different codons. • Identify some codons whose meanings would change following a single-nucleotide substitution. Identify some codons whose meanings would not change.

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 delivered to 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.

O

NH CH C

CH2

CH2

CH2

CH2

NH

CH3

C

N

O 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, fluorescent 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. Q What is the disadvantage for a cell to have a variant tRNA that can insert an amino acid at a stop codon?

586

CH APTER 2 2 Protein Synthesis

L EARNING OBJECTIVES Recognize the major features of the ribosome. • Explain the importance of ribosomal RNA. • Identify the three tRNA binding sites in the ribosome.

22.2

Ribosome Structure

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 mammalian cell may contain as many as 108 protein molecules. An analysis of the human proteome, a set of about 20,000 proteins encoded by the human genome, by mass spectrometry (Section 4.6) reveals that 10,000–12,000 proteins make up the core proteome in different types of human cells, with about 2350 proteins accounting for 75% of the total protein mass in every cell. These abundant “housekeeping” proteins include histones, ribosomal proteins, metabolic enzymes, and cytoskeletal proteins.

(a)

The ribosome is mostly RNA

(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. Note how the overall shape of the 30S subunit reflects the structure of the rRNA. [Structure (pdb 1J5E) determined by B. T. Wimberly, D. E. Brodersen, W. M. Clemons, Jr., R. J. Morgan-Warren, A. P. Carter, C. Vonrhein, T. Hartsch, and V. Ramakrishnan.]

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 bacterial cell may contain 20,000 ribosomes and a yeast cell, about 200,000. This accounts 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 twothirds of the mass of a ribosome is due to the rRNA; the remainder is due to dozens of different proteins (over 80 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.

TAB L E 22. 4

Ribosome Components RNA

POLYPEPTIDES

Small subunit (30S)

16S

21

Large subunit (50S)

23S, 5S

31

Small subunit (40S)

18S

33

Large subunit (60S)

28S, 5.8S, 5S

49

E. coli ribosome (70S)

Mammalian ribosome (80S)

Ribosome Structure

587

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 structures, along with the unique eukaryotic protein components, surround a core structure that is shared with the simpler bacterial ribosome (Fig. 22.7).

Three tRNAs bind to the ribosome 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 FIGURE 22.6 Structure of the 50S ribosomal subunit from aminoacyl ), which accommodates an incoming aminoacyl–tRNA; the P site (for peptidyl), which binds the tRNA with the growing Haloarcula marismortui. The 50S subunit is shown with rRNA polypeptide chain; and the E site (for exit), which transiently binds in gray and proteins in green. Most of the ribosomal proteins are not visible in this view. The protein-free central area forms the a deacylated tRNA after peptide bond formation. The anticodon interface with the 30S subunit. [Structure (pdb 1JJ2) determined by ends of the tRNAs extend into the 30S subunit to pair with mRNA D. J. Klein, T. M. Schmeing, P. B. Moore, and T. A. Steitz.] 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, 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, a large basket-shaped contraption that regulates molecular traffic into and out of the nucleus. Ribosomes in the cytoplasm then translate the mRNA into protein. Cryoelectron tomography, a method for reconstructing three-dimensional cellular structures by analyzing frozen slices by electron

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. [From M.

The exit tunnel

Yusupov, Science 334, 1524–1529 (2011). Reprinted with permission from AAAS. Courtesy Marat Yusupov.]

Q Which areas of the ribosome appear to be most conserved between bacteria and eukaryotes? Why?

588

CH APTER 2 2 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. [From M. Schmeing, Nature 461, 1234–1242 (2009). Reprinted by permission from Macmillan Publishers, Ltd. Photo Courtesy of M. Schmeing, McGill University.]

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). 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.

FIGURE 22.9 Cryoelectron 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. [Courtesy of Wolfgang Baumeister, Max Planck Institute of Biochemistry]

Translation

22.3

LEARNING OBJECTIVES

Translation

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 mRNA sequence 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. The initiation codon is recognized by an initiator tRNA that has been charged with methionine. This tRNA does not recognize other Met codons that occur elsewhere in the coding sequence of the mRNA. In bacteria, the methionine attached to the initiator tRNA is modified by the transfer of a formyl group from tetrahydrofolate (see Section 18.2). The resulting aminoacyl group is designated fMet, and the initiator tRNA is known as tRNAfMet:

CH3 S CH2 O HC

CH2 O NH

CH

C

O

tRNA fMet

N-Formylmethionine–tRNA Met f (fMet–tRNA fMet )

Because the amino group of fMet is derivatized, it cannot form a peptide bond. Consequently, fMet can be incorporated only at the N-terminus of a polypeptide. Later, the formyl group or the entire fMet residue may be removed. In eukaryotic and archaebacterial cells, the initiator tRNA, designated tRNAMet 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. fMet–tRNAfMet 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. 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

589

C A C U A G

rRNA FIGURE 22.10 Alignment of a Shine–Dalgarno sequence with 16S rRNA. A region near the 3′ end of the 16S rRNA molecule (shaded purple) is complementary to the Shine–Dalgarno sequence (blue) in an mRNA, about 10 nucleotides upstream of the initiation codon (green). This mRNA–rRNA interaction helps position the bacterial mRNA at the start of translation. Note that the Shine–Dalgarno sequence shown is a consensus sequence that varies slightly from gene to gene.

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.

590

CH APTER 2 2 Protein Synthesis

FIGURE 22.11 Summary of translation initiation in E. coli. Similar events occur during translation initiation in eukaryotes, when a 40S and a 60S subunit associate following the binding of Met–tRNAMet to an i initiation codon.

30S GTP 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′

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

SEE ANIMATED PROCESS DIAGRAM

AUG

The tRNA bearing the initial fMet is positioned in the P site of the ribosome.

Translation initiation in E. coli

poly(A) tail

stop codon

After the 30S–mRNA–fMet–tRNAfMet 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 fMet–tRNAfMet at the P site—to bind a second aminoacyl–tRNA in order to form the first peptide bond (Fig. 22.11). In eukaryotes, translation initiation requires 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 initiation require the helicase activity of the ribosome to remove secfactors ondary structure in the mRNA that would impede translacap-binding protein tion. The 40S subunit scans the mRNA in an ATP-dependent manner until it encounters the first AUG codon, which is eIF-2 typically 50 to 70 nucleotides downstream of the 5′ cap (Fig. Met 22.12). The initiation factor eIF2 (the e signifies eukaryotic) tRNA i 40S hydrolyzes its bound GTP and dissociates, and the 60S subAUG unit 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. 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 that is complementary to the anticodon of the initiator tRNA.

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–

Translation

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 isn’t 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 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. 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. 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 anti-codon 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 does not monitor 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.

(a)

(b)

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. [Courtesy Venki Ramakrishnan. From Science 292, 897–902 (2001). Reproduced with permission of AAAS.]

591

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). [Structure (pdb 1TTT) determined by P. Nissen, M. Kjeldgaard, S. Tharp, G. Polekhina, L. Reshetnikova, B. F. C. Clark, and J. Nyborg.]

592

CH APTER 2 2 Protein Synthesis Correct aminoacyl–tRNA

GTP

GDP

EF-Tu

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

3. If the tRNA anticodon and mRNA codon are mismatched, the aminoacyl–tRNA dissociates before EF-Tu hydrolyzes GTP.

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.

Ribosome ready for transpeptidation

GTP EF-Tu

FIGURE 22.15 Function of EF-Tu in translation elongation in E. coli.

Ribosome with a mismatched tRNA

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. 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, 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).

TAB 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

Translation P site

A site

P site

A site

NH R

CH

O

C

NH

NH

R

CH

R

CH

O

C

O

C

NH

NH2

NH

R

CH

R

CH

R

CH

O

C

O

C

O

C

O

O

tRNA

tRNA

tRNA

Aminoacyl–tRNA

Uncharged tRNA

Peptidyl–tRNA

OH

O tRNA Peptidyl–tRNA

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 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. 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, although it is not clear how events at the peptidyl transferase site in the large subunit are communicated to the mRNA decoding site in the small subunit. 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.

593

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.

Q Compare this reaction to the condensation reaction of two amino acids shown in Section 4.1.

594

CH APTER 2 2 Protein Synthesis

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.

H3C

NH2

CH3

N

N

N

tRNA

HOCH2

N

N O

H

H

H

OH

C

N

N O3POCH2 O H

H

O

OH

H

H

C

H HN

N

N

O

CH

CH2

OH

NH⫹ 3

Tyrosyl–tRNA

O

CH

CH2

OCH3

NH⫹ 3

O 2N

OH CH2OH

O

C

C

C

H

H

NH

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 antibiotic 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.

Puromycin The antibiotic chloramphenicol interacts with the active-site nucleotides, including the catalytically essential A2451, to prevent transpeptidation.

Q Explain why some of the side effects of antibiotic use in humans can be traced to impairment of mitochondrial function.

EF-G bears a striking resemblance to the EF-Tu–tRNA complex (Fig. 22.17), 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 would also bump the deacylated tRNA from the P site 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 unidirectionally. The E. coli ribosomal elongation cycle is shown in Figure 22.18. 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.

FIGURE 22.17 Structure of EF-G from T. thermophilus. [Structure (pdb 2BV3) determined by S. Hansson, R. Singh, A. T. Gudkov, A. Liljas, and D. T. Logan.]

Q 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).

Translation

GTP EF-Tu

GDP

P site

A site

595

FIGURE 22.18 The E. coli ribosomal elongation cycle.

EF-Tu

mRNA

delivery of aminoacyl–tRNA

SEE ANIMATED PROCESS DIAGRAM

GDP transpeptidation

Elongation cycle in E. coli ribosomes

EF-G–GDP

GTP

translocation

GTP

EF-G–GTP

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 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 position, 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. 22.19). The amide group of the Gln residue promotes transfer of the peptidyl group from the P-site tRNA to water, apparently by stabilizing the transition state FIGURE 22.19 Structure of RF-1. The RF-1 of this hydrolysis reaction. The product of the reaction is an untethered polypep- protein is shown as a purple ribbon with its tide that can exit the ribosome. At one time, release factors were believed to act by anticodon-binding Pro–Val–Thr (PVT) sequence mimicking tRNA molecules, as EF-G does. However, it now appears that release in red and its peptidyl transferase–binding Gly– factors undergo conformational changes upon binding to the ribosome, so they Gly–Gln (GGQ) sequence in orange. This image shows the structure of RF-1 as it exists when don’t operate straightforwardly as tRNA surrogates. In E. coli, an additional RF (RF-3), which binds GTP, promotes the bind- bound to a ribosome. [Structure of the T. thermophilus ribosome with RF-1 (pdb 3D5A) determined by M. ing of RF-1 or RF-2 to the ribosome. In eukaryotes, eRF3 performs this role. Laurberg, H. Asahara, A. Korostelev, J. Zhu, S. Trakhanov, Hydrolysis of the GTP bound to RF-3 (or eRF3) allows the release factors to and H. F. Noller.]

596

CH APTER 2 2 Protein Synthesis

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

GTP

RF-3–GTP

GDP

RF-3–GDP

FIGURE 22.20 Translation termination in E. coli.

SEE ANIMATED PROCESS DIAGRAM Translation termination in E. coli

dissociate (Fig. 22.20). This leaves the ribosome with a bound mRNA, an empty A site, and a deacylated tRNA in the P site. 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 tRNA. 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 in vivo A ribosome can extend a polypeptide chain by approximately 20 amino acids every second in bacteria and by about 4 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. 22.21). 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. (a)

FIGURE 22.21 A polysome. (a) In this electron micrograph, a single mRNA strand encoding silkworm fibroin is studded with ribosomes. Arrows indicate growing fibroin polypeptide chains. [Courtesy Oscar L. Miller, Jr. and Steven L. McKnight, University of Virginia.]

(b) In this cryoelectron microscopy–based reconstruction of a

(b)

polysome, ribosomes are blue and gold, and the emerging polypeptides are red and green. [Courtesy Wolfgang Baumeister and Julip Ortiz, from Cell 136, 261–271 (2009).]

Q Identify the polysomes in Fig. 22.9.

Post-Translational Events

597

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.

22.4

Post-Translational Events

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.

Chaperones promote protein folding

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.

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 only partly 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. 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, chaperones bind to exposed hydrophobic patches on the protein surface (recall that hydrophobic groups tend to aggregate, which could lead to nonnative protein structure or protein aggregation and precipitation). Many chaperones are ATPases that use the free energy of ATP hydrolysis to drive conformational changes that allow them to bind and release a polypeptide substrate while it assumes its native shape. Chaperones were 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. 22.22). FIGURE 22.22 Trigger factor bound to 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 a ribosome. The ribosome-binding portion patch and are thereby prevented from sticking to each other or to other cellular compo- of trigger factor is shown with its domains nents. Trigger factor may dissociate from the ribosome but remain associated with the in different colors. It binds to the 50S nascent (newly formed) polypeptide until another chaperone takes over. Eukaryotes ribosomal subunit where a polypeptide (magenta helix) emerges. PT represents the lack trigger factor, although they have other small heat-shock proteins that function peptidyl transferase active site. [Courtesy in the same manner to protect newly made proteins. These chaperones are extremely Nenad Ban, Eidgenossische Technische Hochschule abundant in cells, so there is at least one per ribosome. Honggerberg, Zurich.]

598

CH APTER 2 2 Protein Synthesis

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. 22.23). 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 FIGURE 22.23 The GroEL/GroES two GroEL rings of the chaperonin complex act in a reciprocating fashion to promote chaperonin complex. The two seventhe folding of two polypeptide chains in a safe environment (Fig. 22.24). Note that subunit GroEL rings, viewed from the 7 ATP are consumed for each 10-second protein-folding opportunity. If the released side, are colored red and yellow. A substrate has not yet achieved its native conformation, it may rebind to the chaperonin seven-subunit GroES complex (blue) caps complex. Only about 10% of bacterial proteins seem to require the GroEL/GroES the so-called cis GroEL ring. [Structure (pdb chaperonin complex, and most of these range in size from 10 to 55 kD (a protein larger 1AON) determined by Z. Xu, A. L. Horwich, and than about 70 kD probably could not fit inside the protein-folding chamber). ImmuP. B. Sigler.] nocytological 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 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.

1. A GroEL ring binds 7 ATP and an unfolded polypeptide, which associates with hydrophobic patches on the GroEL subunits.

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 FIGURE 22.24 The chaperonin

reaction cycle.

5. The cis ring releases its GroES cap, the 7 ADP, and the folded substrate polypeptide.

ATP

Post-Translational Events

The signal recognition particle targets some proteins for membrane translocation

599

signal sequence binding groove

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 a cell’s total proteins. In both prokaryotic and eukaryotic cells, most membrane and secretory proteins FIGURE 22.25 The SRP signal peptide are synthesized by ribosomes that dock with the plasma membrane (in prokaryotes) or binding domain. This model shows the endoplasmic reticulum (in eukaryotes). The protein that emerges from the ribosome is molecular surface of a portion of the E. coli inserted into or through the membrane co-translationally. The membrane translocation signal recognition particle. The protein is system is fundamentally similar in all cells and requires a ribonucleoprotein known as magenta with hydrophobic residues in yellow. Adjacent RNA phosphate groups the signal recognition particle (SRP). As in other ribonucleoproteins, including the ribosome and the spliceosome are red, and the rest of the RNA is dark (see Section 21.3), the RNA component of the SRP is highly conserved and is essen- blue. A signal peptide binds in the SRP tial for SRP function. The E. coli SRP consists of a single multidomain protein and groove, making hydrophobic as well as a 4.5S RNA. The mammalian SRP contains a larger RNA and six different proteins, electrostatic contacts with the protein and RNA. [From R. Batey, Science 287, 1232–1239 but its core is virtually identical to the bacterial SRP. The RNA component of the SRP (2000). Reprinted with permission from AAAS. includes several nonstandard base pairs, including G:A, G:G, and A:C, and interacts Courtesy Robert Batey.] 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 SEE ANIMATED amino acids preceded by at least one positively charged residue. For example, human proinPROCESS DIAGRAM sulin (the polypeptide precursor of the hormone insulin) has the following signal sequence: MALWMRLLPLLALLALWGPDPAAAFVN....

The secretory pathway

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-rich protein domain. The flexible side chains of the hydrophobic Met residues allow the pocket to accommodate helical signal peptides of variable sizes and shapes. In addition to the Met residues, the SRP binding pocket contains a segment of RNA, whose negatively charged backbone interacts electrostatically with the positively charged N-terminus and basic residue of the signal peptide (Fig. 22.25). 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 translation elongation, possibly via conformational changes in the ribosome resulting from SRP 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 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. 22.26). 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 FIGURE 22.26 Ribosome bound to one or more hydrophobic protein segments to move sideways into the lipid bilayer, Sec61. In this cryoEM-based image, a which explains how transmembrane proteins (Section 8.3) become incorporated into yeast ribosome (partially cut away) is the membrane. bound to Sec 61 (pink), which is The driving force for co-translational protein translocation is provided by the ribo- positioned at the end of the polypeptide some: the process of polypeptide synthesis simply pushes the chain through the chan- exit tunnel. Portions of the nascent nel. In situations where a completed polypeptide is inserted across or into a membrane polypeptide (NC, green) are visible in the post-translationally (after it has left the ribosome), the transport system—which may exit tunnel. [From R. Beckmann, Science 326, 1369–1373 (2009). Reprinted with permission from involve the Sec translocon or other machinery—relies on some sort of motor protein AAAS. Courtesy Roland Beckmann.]

600

CH APTER 2 2 Protein Synthesis Signal peptide Ribosome mRNA

1. The SRP binds to the signal peptide when it emerges from the ribosome. Polypeptide elongation halts.

SRP

ER LUMEN

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

3. Signal peptidase removes the signal peptide from the growing polypeptide. Signal peptidase

Polypeptide

4. The rest of the polypeptide is translocated into the ER lumen. From here it is transported by vesicles to the extracellular space.

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 22.27. 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 plasma membrane via vesicles. The proteins may undergo processing (described next) en route to their final destination.

Many proteins undergo covalent modification

Proteolysis is part of the maturation pathway of many proteins. For example, after it has entered the endoplasmic reticulum lumen and had its signal peptide removed and its cysteine side chains cross-linked as disulfides, the insulin precursor FIGURE 22.27 Membrane translocation of a eukaryotic secretory protein. undergoes proteolytic processing. The prohormone is cleaved at two sites to generate the mature hormone (Fig. 22.28). Over 200 other types of post-translational modification have been documented. 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, 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 CC chain terminal groups by amidation. Fatty acyl chains and other lipid groups are added to proteins to anchor them to membranes (SecP P G E L A L AQ L G G L G G G L E VAG AQ P N Q tion 8.3). The addition and removal of phosphoryl groups is a Q A K powerful mechanism for allosterically regulating cellular signaling E R components (Section 10.2) and metabolic pathways (Section 19.2). A chain R G S S Modification of a protein by covalently attaching another I R V E Q C C T S I C S LY Q L E N YC N C protein occurs during protein degradation, when ubiquitin is coA S S valently linked to a target protein (Section 12.1). A related K S S protein, called SUMO (for small ubiquitin-like modifier), is also P N F V N Q H L C G S H LV E A LY LV C G E R G F F Y T covalently attached to a target protein’s lysine side chains, but B chain rather than mark the protein for degradation, as ubiquitin does, SUMO is involved in various other processes, including protein FIGURE 22.28 Conversion of proinsulin to insulin. The transport into the nucleus. prohormone, with three disulfide bonds, is proteolyzed at two All the post-translational modifications mentioned above are bonds (indicated by arrows) to eliminate the C chain. The mature insulin hormone consists of the disulfide-linked A and B chains. catalyzed by specific enzymes, which act more or less reliably

Key Terms

601

depending on the nature of the modification and the cellular context. One consequence of post-translational processing therefore is that proteins may exhibit a great deal of variation beyond the sequence of amino acids that is specified by the genetic code.

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 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 threebase 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–tRNA 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 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. • 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.

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 wobble hypothesis

ribosome A site P site E site initiation factor (IF) G protein

elongation factor (EF) transpeptidation translocation release factor (RF) ribosome recycling factor (RRF) polysome

post-translational modification molecular chaperone signal recognition particle (SRP) signal peptide translocon glycosylation

602

CH APTER 2 2 Protein Synthesis

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. What is the source of these proteins 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 3′-GCTACTCGGAAAGTCGTGGCGT T AG T GAGG T T G C G C G C C A C G A A T C A C T C C A A C GC G C G G T G 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 cell-free 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. a. What polypeptides are synthesized from a nonrandom poly(UAUC) template? 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. The 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. 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? 16. 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? 17. Why doesn’t GlyRS need a proofreading domain? 18. 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

Problems

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. 19. Name the amino acids that are attached to the 3′ end of the E. coli tRNA molecules with the anticodon sequences a. GUG, b. GUU, or c. CGU. 20. 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? 21. 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? 22. Draw the “wobble” base pair that forms between inosine (I) and adenosine.

N

N

H

N Inosine 23. 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 tRNALeu cmo5 UAG ? 24. Some RNA transcripts are substrates for an adenosine deaminase. 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. 25. 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. 26. 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.

COO⫺ H

C

CH2

NH⫹ 3

CH2

CH2

nascent protein? Why is the tunnel lined with hydrophilic and not hydrophobic amino acids? 29. 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. 21.27). What is the advantage of this operon-like arrangement of rRNA genes? 30. 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? 31. 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. 32. What is the effect of adding EDTA, a chelating agent specific for divalent cations, to a bacterial cell extract carrying out protein synthesis?

O

N

603

CH3

Norleucine

Experimental results have shown that peptides containing norleucine in place of leucine were not produced unless the cells contained a mutant LeuRS. Explain these results.

22.2 Ribosome Structure 27. 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? 28. Newly synthesized proteins emerge from the ribosome via an exit tunnel (see Fig. 22.7), 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

33. 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? 34. 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. 35. 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? 36. Propose at least one hypothesis to explain why eukaryotic ribosomes are more complex than prokaryotic ribosomes.

22.3

Translation

37. Indicate the substrates, product, template, primer, enzyme, and cellular location for the following eukaryotic processes: a. replication; b. transcription; c. translation. 38. 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? 39. 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. What polypeptide was synthesized? What would the result be if the mRNA were read in the 3′ → 5′ direction? How does this directionality allow prokaryotes to begin translation before transcription is complete? 5ʹ-AAAA · · · AAAC-3ʹ 40. Mycobacteriophage genes occasionally begin with GUG or even more rarely, UUG (rather than AUG). Which amino acids correspond to these codons? 41. The translation initiation sequence for the ribosomal protein L10 is shown. Draw a diagram that shows how the Shine–Dalgarno

604

CH APTER 2 2 Protein Synthesis

sequence aligns with the appropriate sequence on the 16S rRNA. Identify the initiation codon.

enzyme, which cleaves the peptidyl group from the tRNA? What does this tell you about the ability of ribosomes to carry out protein synthesis?

5ʹ-CUACCAGGAGCAAAGCUAAUGGCUUUA-3ʹ

55. 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.

42. 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.

Peptidyl group

5ʹ-UAACUAGGAUGAAAUGCAUGUCUAAG · · ·

Ala Asp Lys Phe Pro

43. Why does eukaryotic translation initiation require proteins that recognize the 5′ cap and the poly(A) tail on the RNA?

45. What happens when colicin E3, which cleaves on the 5′ side of A1493 in the 16S rRNA, is added to a bacterial culture? 46. Explain why modification or mutagenesis of prokaryotic 16S rRNA at position 1492 or 1493 increases the error rate of translation. 47. 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 could prevent the incorporation of the wrong amino acid in a protein. Aminoacyl–tRNA Ala

Ala-tRNA Gln-tRNAAla Gln-tRNAGln Ala-tRNAGln

Dissociation constant (nM) 6.2 0.05 4.4 260

48. 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. 49. 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? 50. Predict the effect on protein synthesis if EF-Tu were able to recognize and form a complex with fMet−tRNAMet f . 51. Identify the peptide encoded by the DNA sequence shown below (the top strand is the coding strand): C GA T AA T G T C C GA C C AAG C GA T C T C G T AG C A GC T A T T A C AGG C T GG T T C G C T AGAG C A T C G T 52. 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. wild-type

ACACCATGGTGCATCTGACT

mutant

ACACCATGGTTGCATCTGAC

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? 53. 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? 54. 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

57 8 100 16 0.14

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? 56. 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. 57. 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 result. 58. 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 text.] 59. 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). 60. In 1961, Howard Dintzis carried out a series of experiments demonstrating that translation occurs in the N → C 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 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

44. 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?

Rate of peptidyl transfer (s–1)

80 60

Time of incubation: 4 minutes

40 20 0

0

20

40

60 80 100 Amino acid number

120

140

Problems

61. 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? 62. 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 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?

22.4

Post-Translational Events

63. In prokaryotes, translation can begin even before an mRNA transcript has been completely synthesized. Why is co-transcriptional translation not possible in eukaryotes? 64. 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? 65. In 1957, Christian Anfinsen carried out a denaturation experiment in vitro with ribonuclease, a pancreatic enzyme consisting of a single 124-amino-acid chain cross-linked by four disulfide bonds (see Problem 4.47). 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? 66. In another set of experiments, the lysine side chains on the surface of ribonuclease (see Problem 65) were covalently attached to an eightresidue 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? 67. Chaperones are located not just in the cytosol but in the mitochondria as well. What is the role of mitochondrial chaperones? 68. 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? 69. Multidomain proteins tend to fold better inside cagelike chaperonin structures (such as GroEL/GroES in E. coli) than with cytosolic chaperones. Explain why.

605

70. 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? 71. The disease β thalassemia results from a defect in a β globin gene. Heterozygotes (who have one normal and one abnormal gene) may develop mild anemia, but homozygotes (who have two defective β globin genes) 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 70). b. Explain why a mutation in an α globin gene would reduce the severity of anemia in β thalassemia. 72. Immature red blood cells (see Problem 70) 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? 73. 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 74. 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? 75. 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. 76. 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? 77. 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? 78. 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? 79. Polyglutamine diseases are neurodegenerative disorders caused by mutations in the DNA that produce triplet CAG repeats. Proteins translated from the mutated genes contain long stretches of glutamine residues that interfere with protein folding. Polyglutamine proteins

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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 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?

80. The protein c-Myc is a leucine zipper protein 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. 81. In the N-myristoylation process, myristic acid (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. 82. In the palmitoylation process, 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 Moore, P. B., The ribosome returned, J. Biol. 8, 8 (2009). [Reviews many aspects of translation, including the accuracy of decoding.] Neumann, H., Rewiring translation—genetic code expansion and its applications, FEBS Lett. 586, 2057–2064 (2012). [Describes how researchers introduce unusual amino acids into proteins, and what questions can be answered through these techniques.] Nirenberg, M., Historical review: deciphering the genetic code—a personal account, Trends Biochem. Sci. 29, 46–54 (2004). [Describes the experimental approaches used to elucidate the genetic code.] Ramakrishnan, V., The ribosome emerges from a black box, Cell 159, 979–984 (2014). [A short review of the ribosome’s structure and operation, by a leading ribosome researcher.]

Saibil, H., Chaperone machines for protein folding, unfolding, and disaggregation, Nat. Rev. Mol. Cell Biol. 14, 630–642 (2013). [Reviews the structures and mechanisms of a variety of molecular chaperones.] Schmeing, T. M. and Ramakrishnan, V., What recent ribosome structures have revealed about the mechanism of translation, Nature 461, 1234–1242 (2009). [Discusses initiation, elongation, release, and recycling, including the proteins that interact with the ribosome at each step.]

Glossary Numbers and Greek letters are alphabetized as if they were spelled out. Aerobic. Occurring in or requiring oxygen.

Anneal. To allow base pairing between complementary single polynucleotide strands so that double-stranded segments form.

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.

Affinity chromatography. A procedure in which a molecule is isolated by its ability to bind specifically to a second immobilized molecule.

A site. The ribosomal binding site that accommodates an aminoacyl–tRNA.

Agonist. A substance that binds to a receptor so as to evoke a cellular response.

Anomers. Sugars that differ only in the configuration around the carbonyl carbon that becomes chiral when the sugar cyclizes.

Abasic site. The deoxyribose residue remaining after the removal of a base from a DNA strand.

Aldose. A sugar whose carbonyl group is an aldehyde.

Antagonist. A substance that binds to a receptor but does not elicit a cellular response.

Alkalosis. A pathological condition in which the pH of the blood rises above its normal value of 7.4.

Antenna pigment. A molecule that transfers its absorbed energy to other pigment molecules and eventually to a photosynthetic reaction center.

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.

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.

Anticodon. The sequence of three nucleotides in a tRNA that recognizes an mRNA codon through complementary base pairing. Antiparallel. tions.

Running in opposite direc-

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.

Antiparallel β sheet. See β sheet.

α-amino acid.

Apoptosis. Programmed cell death that results from extracellular or intracellular signals and involves the activation of enzymes that selectively degrade cellular structures.

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.

Antiport. Transport that involves the simultaneous transmembrane movement of two molecules in opposite directions. Antisense strand.

See noncoding strand.

Aquaporin. A membrane protein that facilitates the transmembrane movement of water molecules.

α helix. A regular secondary structure of polypeptides, with 3.6 residues per right-handed turn and hydrogen bonds between each backbone CO group and the backbone NH group that is four residues further.

Archaea. One of the two major groups of prokaryotes.

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.

ATPase. An enzyme that catalyzes the hydrolysis of ATP to ADP + Pi.

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 oxygen.

independently

Atherosclerosis. A disease characterized by the formation of cholesterol-containing fibrous plaques in the walls of blood vessels.

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. Autophosphorylation. The phosphorylation of a kinase by another molecule of the same kinase. Axon. The extended portion of a neuron that conducts an action potential from the cell body to a synapse with a target cell.

of

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.

B-DNA. The standard double-helical DNA.

conformation

of

Backbone. The atoms that form the repeating linkages between successive residues of a polymeric molecule, exclusive of the side chains.

G-1

G-2

Glossary

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 solventexposed 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. 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-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. Carbanion. A compound that bears a negative charge on a carbon atom. Carbohydrate. A compound with the formula (CH2O)n , where n ≥ 3. Also called a saccharide. Carbon fi xation. 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 oxidized, 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. C 4 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.

Glossary

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.

CRISPR. Clustered regularly interspersed short palindromic repeats, short DNA segments involved in bacterial defense against bacteriophages.

Codon. The sequence of three nucleotides in DNA or RNA that specifies a single amino acid.

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.

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. 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). 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.

Cristae. The invaginations of the inner mitochondrial membrane. Cross-talk. The interactions of different signal transduction pathways through activation of the same signaling components. Cryoelectron microscopy. A variation of electron crystallography in which electron diffraction data from a molecular structure 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.

G-3

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.

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.

Discontinuous synthesis. A mechanism whereby the lagging strand of DNA is synthesized as a series of fragments that are later joined.

Dark reactions. The photosynthetic reactions in which NADPH and ATP produced by the light reactions are used to incorporate CO2 into carbohydrates.

Dissociation constant (Kd). The ratio of the products of the concentrations of the dissociated species to those of their parent compounds at equilibrium.

Deamination. The hydrolytic or oxidative removal of an amino group.

Disulfide bond. A covalent SS linkage, often between two Cys residues in a protein.

ΔG.

Divergent evolution. The accumulation of changes in species that share an ancestor.

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 oxygenbinding conformation.

DNA (Deoxyribonucleic acid). A polymer of deoxynucleotides whose sequence of bases encodes genetic information in all living cells. 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.

G-4 Ɛ. Ɛ °′.

Glossary

See reduction potential. 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.

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 crystallography. A technique for determining molecular structure by analyzing the pattern of diffraction of a beam from an electron microscope. See also cryoelectron microscopy. 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. Electronegativity. A measure of an atom’s affinity for electrons. Electrophile. A compound containing an electron-poor center. An electrophile (electron-lover) 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 receptor-mediated 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. Endothermic reaction. A reaction that absorbs heat from the surroundings so that its change in enthalpy (ΔH ) 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.

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 (ΔH) 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.

See Faraday constant.

F-actin. The polymerized form of the protein actin. See also G-actin. 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. 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.

Euchromatin. The transcriptionally active, relatively uncondensed chromatin in a eukaryotic cell.

Fibrous protein. A protein characterized by a stiff, elongated conformation, that tends to form fibers.

Eukarya. See eukaryote.

First-order reaction. A reaction whose rate is proportional to the concentration of a single reactant.

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 spontaneous process). Exocytosis. The fusion of an intracellular vesicle with the plasma membrane in order to release the contents of the vesicle outside the cell.

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.

See translocase.

Fluid mosaic model. A model of biological membranes in which integral membrane proteins float and diffuse laterally in a fluid lipid layer.

Glossary

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. 454 sequencing.

See pyrosequencing.

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. energy (ΔG‡). Free radical. electron.

See activation

A molecule with an unpaired

Futile cycle. Two opposing metabolic reactions that function together to provide a control point for regulating metabolic flux. 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 (GPCR). A transmembrane protein that binds an extracellular ligand and transmits the signal to the cell interior by interacting with an intracellular G protein. Gas constant (R). A thermodynamic constant equivalent to 8.3145 J · K–1 · mol–1.

Genome. The complete set of genetic instructions in an organism. Genome map. A reconstruction of an organism’s genome, based on DNA sequences and physical DNA features. 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.

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. GWAS.

See genome-wide association study.

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.

Glycan. See polysaccharide.

Haploid.

Glycerophospholipid. An amphipathic lipid in which two hydrocarbon chains and a polar phosphate derivative are attached to a glycerol backbone.

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.

Glycogen storage disease. An inherited defect in an enzyme or transporter that affects the formation, structure, or degradation of glycogen.

Gel filtration chromatography. See sizeexclusion chromatography.

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.

Glycolipid. A lipid to which carbohydrate is covalently attached.

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.

Gene therapy. The manipulation of genetic information in the cells of an individual in order to produce a therapeutic effect.

Glycosaminoglycan. An unbranched polysaccharide consisting of alternating residues of an amino sugar and a sugar acid.

General transcription factor. One of a set of eukaryotic proteins that are typically required for the synthesis of mRNAs. See also transcription factor.

Glycosidase. An enzyme that catalyzes the hydrolysis of glycosidic bonds.

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.

Glycosyltransferase. An enzyme that catalyzes the addition of a monosaccharide residue to a polysaccharide.

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.

Glycogenolysis. The enzymatic degradation of glycogen to glucose-1-phosphate.

Gene expression. The transformation by transcription and translation of the information contained in a gene to a functional RNA or protein product.

Glycosylation. The attachment of carbohydrate chains to a protein through N- or O-glycosidic linkages.

Gluconeogenesis. The synthesis of glucose from noncarbohydrate precursors.

Gated channel. A transmembrane channel that opens and closes in response to a signal such as changing voltage, ligand binding, or mechanical stress.

Gene. A unique sequence of nucleotides that encodes a polypeptide or RNA; it may include nontranscribed and nontranslated sequences that have regulatory functions.

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H. See enthalpy.

Having one set of chromosomes.

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.

Glycoside. A molecule containing a saccharide linked to another molecule by a glycosidic bond to the anomeric carbon.

Highly repetitive DNA. Clusters of short DNA sequences that are repeated side-by-side and are present at millions of copies in the human genome.

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.

Histone code. The correlation between patterns of covalent modification of histone proteins and the transcriptional activity of the associated DNA.

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Glossary

Histones. Highly conserved basic proteins that form a core to which DNA is bound in a nucleosome.

Imprinting. A heritable variation in the level of expression of a gene according to its parental origin.

Isoprenoid. A lipid constructed from fivecarbon units with an isoprene skeleton. Also called a terpenoid.

Homeostasis. The maintenance of constant internal conditions.

In vitro. In the laboratory (literally, in glass).

Isozymes. Different proteins that catalyze the same reaction.

Homo-. The same. In a homopolymer, all the subunits are identical.

Induced fit. An interaction between a protein and its ligand that induces a conformational change in the protein to enhance the protein’s interaction with the ligand.

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. cholesterol in the blood. Hyperglycemia. in the blood.

IF.

Elevated levels of

Elevated levels of glucose

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.

In vivo. In a living organism.

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. Intrinsic protein. protein.

See integral membrane

Intrinsically unstructured 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.

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. ketone.

A sugar whose carbonyl group is a

KI. See inhibition constant. Kinase. An enzyme that transfers a phosphoryl group between ATP and another molecule.

Ion exchange chromatography. A fractionation procedure in which charged molecules are selectively retained by a matrix bearing oppositely charged groups.

Kw. See ionization constant of water.

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.

Imino group. A portion of a molecule with the formula C NH.

Isoelectric point (pI). The pH at which a molecule has no net charge.

the

Ka. See acid dissociation constant.

Kinetics. The study of chemical reaction rates.

Isoacceptor tRNA. A tRNA that carries the same amino acid as another tRNA but has a different codon.

with

k. See rate constant.

Invariant residue. A residue in a protein that is the same in all evolutionarily related proteins.

formula

Imine. A molecule C NH.

K. See dissociation constant.

KM. See Michaelis constant. Kwashiorkor. A form of severe malnutrition resulting from inadequate protein intake; marked by abdominal swelling and reddish hair. See also marasmus. 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.

Glossary

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.

Matrix.

See mitochondrial matrix.

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.

Light reactions. The photosynthetic reactions in which light energy is absorbed and used to generate NADPH and ATP.

Messenger RNA (mRNA). A ribonucleic acid whose sequence is complementary to that of a protein-coding gene in DNA.

Lineweaver–Burk plot. A rearrangement of the Michaelis–Menten equation that permits the determination of KM and Vmax from a linear plot.

Metabolic acidosis. A low blood pH caused by the overproduction or retention of hydrogen ions.

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.

Metabolic alkalosis. A high blood pH caused by the excessive loss of hydrogen ions. Metabolic fuel. A molecule that can be oxidized to provide free energy for an organism.

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can therefore be used to identify active genes. Also called a DNA chip. Microbiome. 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.

See bilayer.

Metabolic pathway. A series of enzymecatalyzed reactions by which one substance is transformed into another.

Mismatch repair. A DNA repair pathway that removes and replaces mispaired nucleotides on a newly synthesized DNA strand.

Lipid-linked protein. A protein that is anchored to a biological membrane via a covalently attached lipid.

Metabolic syndrome. A set of symptoms related to obesity, including inulin resistance, atherosclerosis, and hypertension.

Mitochondrial matrix. The gel-like solution of enzymes, substrates, cofactors, and ions in the interior of the mitochondrion.

Lipolysis. The degradation of a triacylglycerol so as to release fatty acids.

Metabolically irreversible reaction. A reaction whose value of ΔG is large and negative so that the reaction cannot proceed in reverse.

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.

Lipid bilayer.

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.

Metabolism. The total of all degradative and biosynthetic cellular reactions. Metabolite. A reactant, intermediate, or product of a metabolic reaction.

Locus. The chromosomal location of a gene or other DNA feature.

Metabolome. The complete set of metabolites produced by a cell or tissue.

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).

Metabolomics. The study of all the metabolites produced by a cell or tissue.

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.

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. Mass spectrometry. A technique for identifying molecules by measuring the massto-charge ratios of gas-phase ions, such as peptide fragments.

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. 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 + k 2)/k 1; 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

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. Sequences of DNA that are present at hundreds of thousands of copies in the human genome. Molecular chaperone. A protein that binds to unfolded or misfolded proteins in order to promote their normal folding. Monogenic disease. A disease linked to a defect in a single gene. Monomer. A structural unit from which a polymer is built up. 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.

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Glossary

Mutation. A heritable alteration in an organism’s genetic material.

enzyme such that the apparent Vmax decreases but KM is not affected.

Oligopeptide. A polypeptide consisting of a few amino acid residues.

Myelin sheath. The multilayer coating of sphingomyelin-rich membranes that insulates a mammalian neuron.

Noncyclic 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.

Oligosaccharide. A polymeric carbohydrate containing a few monosaccharide residues. In glycoproteins, the groups are known as N-linked and O-linked oligosaccharides.

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. 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([A reduced ]/[A oxidized ]). 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 fi xation. The process by which atmospheric N2 is converted to a biologically useful form such as NH3. NMR spectroscopy. See nuclear magnetic resonance spectroscopy. Noncoding RNA (ncRNA). An RNA molecule 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

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 molecular structure of a protein or nucleic acid. Nuclease. acids.

An enzyme that degrades nucleic

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. The region of the eukaryotic nucleus where rRNA is processed and ribosomes are assembled. Nucleophile. A compound containing an electron-rich group. A nucleophile (nucleuslover) reacts with an electrophile (electronlover). 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.

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. Ordered mechanism. A multisubstrate reaction with a compulsory order of substrate binding to the enzyme. ORF.

See open reading frame.

Orientation effects. entation effects.

See Proximity and ori-

Orphan gene. A gene that appears to have no counterpart in the genome of another species. 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 reactants during the transition state and thereby lowers its energy. Oxyhemoglobin. Hemoglobin that contains bound oxygen or is in the oxygen-binding conformation. P site. The ribosomal binding site that accommodates a peptidyl–tRNA.

O-linked oligosaccharide. An oligosaccharide linked to the hydroxyl group of a protein Ser or Thr side chain.

Palindrome. A segment of DNA that has the same sequence on each strand when read in the 5ʹ → 3ʹ direction.

Okazaki fragments. The short segments of DNA formed in the discontinuous laggingstrand synthesis of DNA.

Parallel β sheet.

Oligonucleotide. A polynucleotide consisting of a few nucleotide residues.

Passive transport. The thermodynamically spontaneous protein-mediated transmembrane

See β sheet.

Partial oxygen pressure ( pO2). The concentration of gaseous O2 in units of torr.

Glossary

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. 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.

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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.

Polynucleotide. See nucleic acid.

Photoreceptor. or pigment.

Polyprotein. A polypeptide that undergoes proteolysis after its synthesis to yield several separate protein molecules.

A light-absorbing molecule,

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.

Polypeptide. A polymer consisting of amino acid residues linked in linear fashion by peptide bonds.

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.

Peroxisome. A eukaryotic organelle with specialized oxidative functions, including fatty acid degradation.

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.

p50. The ligand concentration (or pressure for a gaseous ligand) at which a binding protein such as hemoglobin is half-saturated with ligand.

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).

PPi. A pyrophosphoryl group: H3P2O6, H2P2O6–, HP2O62–, or P2O63–.

pH. A quantity used to express the acidity of a solution, equivalent to –log[H+].

Plasmid. A small circular DNA molecule that autonomously replicates and may be used as a vector for recombinant DNA.

Primase. The enzyme that synthesizes a segment of RNA to be extended by DNA polymerase during DNA replication.

(+) end. The end of a polymeric filament where growth is faster. See also (–) end.

Primer. An oligonucleotide that base pairs with a template polynucleotide strand and is extended through template-directed polymerization.

Phage.

See bacteriophage.

Pharmacokinetics. The behavior of a drug in the body, including its metabolism and excretion. Phosphatase. An enzyme that hydrolyzes phosphoryl ester groups.

P:O ratio. The ratio of the number of molecules of ATP synthesized from ADP + Pi to the number of atoms of oxygen reduced.

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).

Point mutation. The substitution of one base for another in DNA, arising from mispairing during DNA replication or from chemical alterations of existing bases.

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.

Polarity. Having an uneven distribution of charge.

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.

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.

pO2. See partial oxygen pressure.

Primary structure. The sequence of residues in a polymer.

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 modification. 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.

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Glossary

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 ATPdependent 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.

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.

Protofilament. One of the 13 linear polymers of tubulin subunits that forms a microtubule.

Rate equation. A mathematical expression for the time-dependent progress of a reaction as a function of reactant concentration.

Proton jumping. The rapid movement of a proton among hydrogen-bonded water molecules.

Rational drug design. The synthesis of more effective drugs based on detailed knowledge of the target molecule’s structure and function.

Proton wire. A series of hydrogen-bonded water molecules and protein groups that can relay protons from one site to another.

Reactant. One of the starting materials for a chemical reaction.

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. Purine. A derivative of the compound purine, such as the nucleotide base adenine or guanine. Pyrimidine. A derivative of the compound pyrimidine, such as the nucleotide base cytosine, uracil, or thymine. Pyrosequencing. A procedure for determining the sequence of nucleotides in DNA by detecting a flash of light generated by the addition of a new nucleotide to a growing DNA strand. Also known as 454 sequencing.

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 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.

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. Reading frame. The grouping of nucleotides in sets of three whose sequence corresponds to a polypeptide sequence. Real-time PCR. A variation of the polymerase chain reaction in which the level of gene expression can be measured. Also known as quantitative PCR (qPCR). 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 carries out the various activities of DNA replication. 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. 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. RF.

See release factor.

Ribonucleic acid.

See RNA.

Glossary

Ribosomal RNA (rRNA). The RNA molecules that provide structural support for the ribosome and catalyze peptide bond formation. Ribosome. The RNA-and-protein particle that synthesizes polypeptides under the direction of mRNA. Ribosome recycling factor (RRF). A protein that binds to a ribosome after protein synthesis to prepare it for another round of translation.

tration of one reactant or to the product of the concentrations of two reactants.

snoRNA.

Secondary active transport. Transmembrane transport of one substance that is driven by the free energy of an existing gradient of a second substance.

snRNA.

Secondary structure. The local spatial arrangement of a polymer’s backbone atoms without regard to the conformations of its substituent side chains.

G-11

See small nucleolar RNA.

SNP. See single-nucleotide polymorphism. 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.

Ribozyme. An RNA molecule that has catalytic activity.

Selection. A technique for distinguishing cells that contain a particular feature, such as resistance to an antibiotic.

RNA (Ribonucleic acid). A polymer of ribonucleotides, such as messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA).

Semiconservative replication. The mechanism of DNA duplication in which each new molecule contains one strand from the parent molecule and one newly synthesized strand.

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.

RNA interference (RNAi). A phenomenon in which short RNA segments direct the degradation of complementary mRNA, thereby inhibiting gene expression.

Sense strand.

Sphingomyelin.

Serine protease. A peptide-hydrolyzing enzyme that has a reactive Ser residue in its active site.

RNA polymerase.

Spliceosome. A complex of protein and snRNA that carries out the splicing of immature mRNA molecules.

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.

Splicing. The process by which introns are removed and exons are joined to produce a mature RNA transcript.

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.

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.

See coding strand.

Signal peptide. A short sequence in a membrane or secretory protein that binds 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 is transmitted 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.

Saturated fatty acid. A fatty acid that does not contain any double bonds in its hydrocarbon chain.

Single-nucleotide polymorphism (SNP). A nucleotide sequence variation in the genomes of two individuals from the same species.

Saturation. The state in which all of a macromolecule’s ligand-binding sites are occupied by ligand.

siRNA.

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. 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 cellsurface receptor. Second-order reaction. A reaction whose rate is proportional to the square of the concen-

See small interfering RNA.

Site-directed mutagenesis. A technique in which a cloned gene is mutated in a specific manner. 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 20- to 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.

See sphingolipid.

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. reaction.

A reactant in an enzymatic

G-12

Glossary

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 chemically 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. 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 state. One of two conformations of an allosteric protein; the other is the R state. 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 eukaryotic 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.

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. 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.

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.

Transcriptomics. The study of the genes that are transcribed in a certain cell type or at a certain time.

Triglyceride.

Transfer RNA (tRNA). The small L-shaped RNAs that deliver specific amino acids to ribosomes according to the sequence of a bound mRNA.

Triose.

Transgenic organism. An organism that stably expresses a foreign gene.

Trisaccharide. A carbohydrate consisting of three monosaccharides.

Transition mutation. A point mutation in which one purine (or pyrimidine) is replaced by another purine (or pyrimidine).

tRNA. See transfer RNA.

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.

Tumor suppressor gene. A gene whose loss or mutation may lead to cancer.

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 lipid from one bilayer leaflet to another. Also called a flippase. 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.

Thioester. A compound containing an ester linkage to a sulfur rather than an oxygen atom.

Translocon. The complex of membrane proteins that mediates the transmembrane movement of a polypeptide.

3ʹ end. The terminus of a polynucleotide whose C3ʹ is not esterified to another nucleotide residue.

Transmissible spongiform encephalopathy (TSE). A fatal neurodegenerative disease caused by infection with a prion.

See triacylglycerol.

Trimer. An assembly consisting of three monomeric units. A three-carbon sugar.

Triple helix. The right-handed helical structure formed by three left-handed helical polypeptide chains in collagen.

TSE. See transmissible spongiform encephalopathy.

Turnover number. See catalytic constant. 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.

Glossary

v. Velocity (rate) of a reaction. 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 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.

Vitamin. A metabolically required substance that cannot be synthesized by an animal and must therefore be obtained from the diet.

Y. See fractional saturation.

Vmax. Maximal velocity of an enzymatic reaction.

Z.

v0. Initial velocity of an enzymatic reaction. 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-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.

G-13

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 1. a. carboxylic acid; b. amine; c. ester; d. alcohol. 3.

Hydroxyl group

21. 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 much different from that of starch.

OH

Imino group N O

CH3

Ether linkage [From Li, S.-Y., Wang, X.-B., and Kong, L.-Y. Eur. J. Med. Chem. 71, 36–45 (2014).] 5. 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. 7. a. C and H plus some O; b. C, H, and O; c. C, H, O, and N plus small amounts of S. 9. You should measure the nitrogen content, since this would indicate the presence of protein (neither lipids nor carbohydrates contain appreciable amounts of nitrogen). 11. 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. 13.

O

Carbonyl

C H H HO

OH H

H

OH

H

OH

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.

Hydroxyl

CH2 OH

15. Uracil has a carbonyl functional group, whereas cytosine has an amino functional group. 17. 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 carboxylate group are ionized, which render the molecule “saltlike.” Glucose is also water soluble because its aldehyde group and many hydroxyl groups are able to form hydrogen bonds with water. 19. 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. On the other hand, proteins 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

23. 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. 25. 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. 27. The dissolution of ammonium nitrate in water is a highly endothermic process, as indicated by the positive value of ΔH. 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. 29. The dissolution of urea in water is an endothermic process and has a positive ΔH value. In order for the process to be spontaneous, the process must also have a positive ΔS 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. 31. 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. 33. 0 > –14.3 kJ · mol–1 – (273 + 25 K)(ΔS) 14.3 kJ · mol–1 > – (273 + 25 K)(ΔS) –48 J · K–1 · mol–1 > ΔS ΔS could be any positive value, or it could have a negative value smaller than – 48 J · K–1 · mol–1. 35. a. Entropy decreases when the antibody–protein complex binds because the value of ΔS is negative. b. ΔG = ΔH – TΔS ΔG = –87,900 J · mol–1 – (298 K)(–118 J · K–1 · mol–1) ΔG = –52.7 kJ · mol–1 The negative value of ΔG indicates that the complex forms spontaneously. 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., Allen, M. J., and Nall, B. T. Biochemistry 34, 5831–5838 (1995).] 37. a. The conversion of glucose to glucose-6-phosphate is not favorable because the ΔG value for the reaction is positive, indicating an endergonic process. b. If the two reactions are coupled, the overall reaction is the sum of the two individual reactions. The ΔG value is the sum of the ΔG values for the two individual reactions: S-1

S-2

Odd-Numbered Solutions

ATP + glucose → ADP + glucose-6-phosphate ΔG = –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 ΔG value of the coupled reaction is negative, which indicates that the reaction as written is favorable. 39. C (most oxidized), A, B (most reduced) 41. a. oxidized; b. oxidized; c. oxidized; d. reduced. 43. 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 ΔG) than the conversion of glucose carbons to CO2. Therefore, palmitate carbons provide more free energy than glucose carbons. 45. 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. 47. A

B C

Chapter 2

7. Identical hydrogen bonding patterns in the two molecules are shown as open arrows in Solution 6. 9. a. H < C < S < N < O < F b. 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 hydrogen’s, cannot. 11. a. van der Waals forces (dipole–dipole interactions); b. hydrogen bonding; c. van der Waals forces (London dispersion forces); d. ionic interactions. 13. 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. 15. 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. 17. The positively charged ammonium ion is surrounded by a shell of water molecules that are oriented so that the partially negatively charged oxygen atoms interact with the positive charge on the ammonium ion. Similarly, the negatively charged sulfate ion is hydrated with water molecules oriented so that the partially positively charged hydrogen atoms interact with the negative charge on the sulfate anion. (Not shown in the diagram is the fact that the ammonium ions outnumber the sulfate ions by a 2:1 ratio. Also note that the exact number of water molecules shown is unimportant.)

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°. 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. 5. The arrows point toward hydrogen acceptors and away from hydrogen donors.

O ⫹

H3N

CH

N H

CH2 COO⫺

CH C

CH3

O

Aspartame O H O

H2N

S

O N H

O NH2

O

Sulfanilamide

H N

N

N

Uric acid H

H H

O

O

H

H

O

H

O

H

H H

O H

H

O NH⫹ 4 O H

H

H

H SO42⫺

H

O

H

H O

O H

H

19. Structure A depicts a polar compound, while structure B depicts an ionic compound similar to a salt like sodium chloride. This is 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.

CH2 O

C

H

21. 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.

Odd-Numbered Solutions

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.

23. Polar and nonpolar regions of the detergents are indicated. nonpolar

CH3 H3C(H2C)15

nonpolar

N



CH3

CH3

OH CH3

CH3

polar

polar O

CH3

Hexadecyltrimethylammonium

O

Cholate HO

OH

25. Compounds A and D are amphiphilic, compound B is nonpolar, and compounds C and E are polar. 27. a. In the nonpolar solvent, AOT’s polar head group faces the interior of the micelle, and its nonpolar tails face the solvent. CH2CH3 Nonpolar tails

39. 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. But 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. 41.

−

O

H3C

(CH2)3

CH CH2

O

C

CH2 O

H3C

(CH2)3

CH CH2

O

C

CH S

CH2CH3

S-3

O

O

Polar head

O AOT

b. The protein, which contains numerous polar groups, interacts with the polar AOT groups in the micelle interior.

43. 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. 2− 2− 2− 3− 3− 2− 45. a. C2O2− 4 b. SO3 c. HPO4 d. CO3 e. AsO4 f. PO4 g. O2

(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.)

47. a. The final concentration of HNO3 is

Isooctane Water

Protein

pH = −log[ H + ]

29. a. The nonpolar core of the lipid bilayer helps prevent the passage of water since the polar water molecules cannot easily penetrate 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.

pH = −log(0.038) pH = 1.4 (0.015 L)(1.0 M) = 0.029 M 0.515 L 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.) b. The final concentration of KOH is

31. a. CO2 is nonpolar and would be able to cross a bilayer. b. Glucose is polar and would not be able to pass through a bilayer because the presence of the hydroxyl groups means glucose is highly hydrated and would not be able to pass through the nonpolar tails of the molecules forming the bilayer. c. DNP is nonpolar and would be able to cross a bilayer. d. Calcium ions are charged and are, like glucose, highly hydrated and would not be able to cross a lipid bilayer.

Kw = 1.0 × 10−14 = [ H + ] [ OH− ] 1.0 × 10−14 [ OH− ] 1.0 × 10−14 [ H+ ] = (0.029 M) [ H+ ] =

33. 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. 35. 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. 37. 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 ÷

[ H + ] = 3.4 × 10−13 M pH = −log[ H + ] pH = −log(3.4 × 10−13 ) pH = 12.5 49.

CH2 HO

C

COOH

COOH

CH2

COOH

Citric acid

H N

H

Piperidine

S-4

Odd-Numbered Solutions O 

H2N

H

CH

C

OH

HO

O

O

C

C

Finally, calculate the volume of glacial acetic acid needed: OH

CH2

O

CH2 CH2

NH

The addition of 3.3 mL to a 500-mL solution dilutes the solution by less than 1%, which doesn’t introduce significant error. H2CO3 →H + + HCO−3

57. a.

CH2

O

H2N H

N H

O

Barbituric acid

Lysine O O

0.057 moles = 0.0033 L, or 3.3 mL 17.4 mol · L−1

Oxalic acid

NH CH2

CH2

S

O

O

4-Morphine ethanesulfonic acid (MES)

HCO−3 →H + + CO2− 3 b. The pK of the first dissociation is closer to the pH; therefore the weak acid present in blood is H2CO3 and the conjugate base is HCO3–. c.

pH = pK + log

51. Calculate the final concentrations of the weak acid (H2PO−4 ) and + conjugate base (HPO2− 4 ). Note that K is a spectator ion.

7.40 = 6.35 + log

(0.025 L)(2.0 M) = 0.25 M 0.200 L (0.050 L)(2.0 M) = 0.50 M [ HPO2− 4 ] = 0.200 L [ H2PO−4 ] =

Next, substitute these values into the Henderson–Hasselbalch equation using the pK values in Table 2.4: [ A− ] pH = pK + log [ HA ]

1.05 = log 11.2 =

(0.01 L)(0.05 M) = 9.4 × 10−4 M 0.53 L (0.02 L)(0.02 M) = 7.5 × 10−4 M [ borate] = [ A− ] = 0.53 L [ A− ] pH = pK + log [ HA ] 7.5 × 10−4 = 9.24 + log 9.4 × 10−4 = 9.24 − 0.10 = 9.14

[ boric acid] = [ HA ] =

55. First, determine the ratio of [A−] to [HA]: [ A− ] pH = pK + log [ HA ] [ A− ] = pH − pK log [ HA ] [ A− ] = 10(pH−pK) [ HA ] Substitute the values for the desired pH (5.0) and the pK (4.76): [ A− ] = 10(5.0−4.76) = 100.24 = 1.74 [ HA ] Calculate the number of moles of acetate (A−) already present: (0.50 L)(0.20 mol · L−1) = 0.10 moles acetate Calculate the moles of acetic acid needed, based on the calculated ratio: [ A− ] = 1.74 [ HA ] 0.10 moles [ HA ] = 1.74 [ HA ] = 0.057 moles

24 × 10−3 M [H2CO3]

24 × 10−3 M [H2CO3]

59. a. CH3 C

O

COO

Pyruvate

pH = 6.82 + 0.30 53. The final volume is 500 mL + 10 mL + 20 mL = 0.53 L

24 × 10−3 M [H2CO3]

[H2CO3] = 2.1 × 10−3 M = 2.1 mM

pH = 6.82 + log (0.50 M)/(0.25 M) pH = 7.12

[HCO−3] [H2CO3]

b. Pyruvate predominates in the cell at pH 7.4. The pK values for carboxylic acid groups are typically in the 2–3 range; therefore, the carboxylate group will be unprotonated at physiological pH. 61. 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. 63. 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 2− 3− [HPO4 ]  [PO4 ]

14 12

[PO43−]

pK2 Midpoint two  2 [H2PO4 ]  [HPO4 ]

10 8 pH 6 4

pK1 Midpoint one  [H3PO4]  [H2PO4 ]

2 0 [H3PO4]

[HPO42]

[H2PO 4] 0.5

1.0 1.5 2.0 H+ ions dissociated

2.5

3.0

Odd-Numbered Solutions

H3PO4 →H + + H2PO−4

b.

needed is 0.005 mol/0.316 = 0.016 moles. The stock imidazolium chloride is 1 M, so the volume of imidazolium chloride to be added is

H2PO−4 →H + + HPO2− 4

0.016 mol = 0.016 L or16 mL 1.0 mol · L −1

+ 3− HPO2− 4 →H + PO4

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 H2PO4– and the weak base is HPO42–. d. The dissociation of the third proton has a pK of 12.38. Therefore, a buffer solution at pH 11 would consist of the weak acid 3− HPO2− 4 and its conjugate base, PO4 (supplied as the sodium salts Na2HPO4 and Na3PO4). 65. a. HO

(H2C)2



HN

N

(CH2)2

(H2C)2

N

N

(CH2)2

69. H + (aq) + HCO−3 (aq) ⇌ H2CO3 (aq) ⇌ H2O (l) + CO2 (aq) Failure to eliminate CO2 in the lungs would cause a buildup of CO2 (aq). This would shift the equilibrium of the above equations to the left. The increase in CO2 (aq) would lead to the increased production of carbonic acid, which would in turn dissociate to form additional hydrogen ions, causing acidosis. 71. 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.

SO 3

Weak acid (HA) HO

S-5

 SO 3 H

73. Ammonia and ammonium ions are in equilibrium, as represented by the following equation:

Conjugate base (A) b. The pK for HEPES is 7.55; therefore, its effective buffering range is 6.55–8.55. 0.10 mole 260.3 g × = 26 g c. 1.0 L × 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– 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− ] 0.10 mole − x = 2.82 = x [ HA ] 2.82x = 0.10 mol − x 3.82x = 0.10 mol x = 0.10 mol/3.82 = 0.0262 mol Calculate how much 6.0 M HCl to add: 0.0262 mol = 0.0044 L, or 4.4 mL 6.0 mol · L−1 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. 67.

[ A− ] pH = pK + log [ HA ] [ A− ] = pH − pK log [ HA ] − [A ] = 10(pH−pK) [ HA ] [ A− ] = 10(6.5−7.0) = 10−0.5 = 0.316 [ HA ]

Since the starting solution contains (0.5 L)(0.01 mol · L−1) = 0.005 mole of imidazole (A–), the amount of imidazolium chloride (HA)

NH4+ ⇌ H + + NH3 Carbonic acid and bicarbonate ions are in equilibrium, as represented by the following equation: H2CO3 ⇌ H + + HCO−3 Phosphate ions are in equilibrium, according to the following equation: H2PO−4 ⇌ H + + HPO2− 4 In metabolic acidosis, the concentration of protons increases, so the equilibrium shifts to form H2PO4–, carbonic acid, and ammonium ions. In order to bring the pH back to normal, the kidney excretes H2PO4– and ammonium ions, and bicarbonate ions are reabsorbed. The result is a decrease in the concentration of protons and an increase in blood pH. 75. 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 movement of HCO3– out of the cell through another exchange protein. 77. The cell-surface carbonic anhydrase can catalyze the conversion of H+ + HCO3– to CO2, which can then diffuse into the cell (the ionic H+ and HCO3– 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 wildtype 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.

S-6

Odd-Numbered Solutions

5. Thymine (5-methyl uracil) contains a methyl group attached to C5 of the pyrimidine ring of uracil.

via stacking interactions. In this way, contact with the aqueous solution is minimized, as described by the hydrophobic effect.

7.

27. a. The Tm is approximately 72°C. b. The melting curves are shown below.

NH2 H3C

N

1.4

O

5-Methylcytosine 9. The base 5-chlorouracil is a substitute for thymine (5-methyluracil). 11. [From Jordheim, L. P., Durantel, D., Zoulim, F., and Dumontet, C. Nat. Rev. Drug Discov. 12, 447–464 (2013).] NH2 N

N

Cl

H

H

H

OH

OH

13. a. A diphosphate bridge links the ribose groups in each dinucleotide. This linkage is a variation of the monophosphate bridge (phosphodiester linkage) in DNA and RNA. b. The adenosine group in CoA bears a phosphoryl group on C3ʹ. 15.

O O P

OH

O

N G

O CH2

5

N

O

N

NH NH2

3

H

N N

N

H

H H O

H

H

H

H 2

5 CH2

A N

O

OH

D. discoideum

O P

S. albus

1.2 Tm

Tm

1.1

30

50 70 Temperature (°C)

90

29. 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.

H

8-Chloroadenosine

O

1.3

1.0

N

N CH2OH O

Relative absorbance at 260 nm

N H

O

O

NH2

17. The organism 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

31. 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. 33. 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. 35. a. 3′-TGTGGTACCACGTAGACTGA-5′ b. 5′-ACACCAUGGUGCAUCUGACU-3′ 37. a. A poly-Phe polypeptide was produced. b. Poly A produces poly-Lys; poly C yields poly-Pro; and poly G yields poly-Gly. 39. 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. 41. a. First reading frame: AGG TCT TCA GGG AAT GCC TGG CGA GAG GGG AGC Arg - Ser - Ser - Gly - Asn - Ala - Trp - Arg - Glu - Gly - SerAGC TGG TAT CGC TGG GCC CAA AGG C Ser - Trp - Tyr - Arg - Trp - Ala - Gln - Arg

19. The total amount of purines (A + G) in DNA must equal the total amount of pyrimidines (C + T) because each base pair in the double-stranded DNA molecule consists of a purine and a pyrimidine. This is not true for RNA, which is single-stranded.

Second reading frame: A GGT CTT CAG GGA ATG CCT GGC GAG AGG GGA GCA - Gly - Leu - Gln - Gly - Met - Pro - Gly - Glu - Arg - Gly - Ala-

21. It is a G:C base pair.

GCT GGT ATC GCT GGG CCC AAA GGC Ala - Gly - Ile - Ala - Gly - Pro - Lys - Gly

23. 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. 25. 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

Third reading frame: AG GTC TTC AGG GAA TGC CTG GCG AGA GGG GAG CAG --Val - Phe - Arg - Glu - Cys - Leu - Ala - Arg - Gly - Glu - GlnCTG GTA TCG CTG GGC CCA AAG GC Leu - Val - Ser - Leu - Gly - Pro - Lysb. The second reading frame, which produces a protein in which every third amino acid is Gly, is the correct reading frame.

Odd-Numbered Solutions

CTCAGAGTTCACC ATG GGC TCC ATC GGT GCA GCA AGC Met Gly Ser Ile Gly Ala Ala Ser ATG GAA ··· 1104 bp ·· TTC TTT GGC AGA TGT GTT TCC Met Glu ···················· Phe Phe Gly Arg Cys Val Ser CCT TAA AAAGAA Pro * 47. 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 sequence of the normal protein in this region of the gene is ····Asn-Ile-Asp-Thr····. The amino acid sequence of the mutated CF gene (in which two bases, A and T, are missing), is ····Asn-Arg-Tyr····. This is a frameshift mutation and all of the amino acids past the deletion will differ from the amino acids in the normal protein. 49. 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.) 51. 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.)

57. 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).] 59. MspI, AsuI, EcoRI, PstI, SauI, and NotI generate sticky ends. AluI and EcoRV generate blunt ends. 61. AluI cleaves the sequence at two locations and EcoRI and NotI cleave the sequence at one location each. The other enzymes listed in Table 3.5 do not cleave this segment of the plasmid DNA. EcoRI

AluI

AluI

NotI

CGCGGATCCGAATTCGAGCTCCGTCGACAAGCTTGCGGCCGCACTCGAG

63. a. A HaeII digest of the linear 2743-bp pGEM-3Z plasmid (see Figure 3.16) would produce four bands, as shown below (1–323 produces a 323-bp band, 324–693 produces a 370-bp band, 694–2564 produces an 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.] Circular

45. First, identify the translation start site, the Met residue whose codon is AUG in the mRNA (see Table 3.3) or ATG in the DNA. Translation stops at the DNA sequence TAA, which corresponds to the stop codon UAA in the mRNA. Use Table 3.3 to decode the intervening codons, substituting U for T.

55. If a SNP occurs every 300 nucleotides or so, and if there are about 3 million kb (3 × 109 nucleotides) in the human 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]

Linear

43. The genetic code (shown in Table 3.3) is redundant. Since there are 64 different possibilities for 3-base codons and only 20 amino acids, most amino acids have more than one codon. If a mutation happens to occur in the third position (3ʹ end), the mutation might not alter the protein sequence. For example, GUU, GUC, GUA, and GUG all code for valine. A mutation in the third position of a valine codon would still result in the selection of valine and would have no effect on the amino acid sequence of the protein.

S-7

3000 bp 1500 bp 1000 bp

500 bp

53. a. The first reading frame is the longest ORF. First reading frame: TAT GGG ATG GCT GAG TAC AGC ACG TTG AAT GAG Tyr - Gly - Met - Ala - Glu - Tyr - Ser - Thr - Leu - Asn - GluGCG 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 - Gly - Trp - Leu - Ser - Thr - Ala - Arg -Stop - Met - Arg-

100 bp

65. 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-TGCCATTGCA-5ʹ 67. The primers are shown in red below.

CGA TGG CCG CTG GTG ATG Arg - Trp - Pro - Leu - Val - Met

5ʹ-ATGATTCGCCTCGGGGCTCCCCAGTCGCTGGTGCTGCT GACGCTGCTCGTCG-3ʹ ← 3ʹ-ACGACT GCGACGAGCAGC-5ʹ

Third reading frame:

5ʹ-ATGATTCGCCTCGGGGCT-3ʹ → 3ʹ-TACTAAGCGGAGCCCCGAGGGGTCAGCGACCACGACGACTGCGACGAGCAGC-5ʹ

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.

69. Primers with high GC content have high Tm values. If the annealing temperature is much lower than the melting temperature, improper base pairing may occur (see Figure 3.8) and the desired gene fragment may not be amplified. 71. To amplify the protein-coding 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

S-8

Odd-Numbered Solutions

b. D-Lysine may be lacking in nutritional value because the enzymes that are responsible for protein synthesis may not recognize it. As a result, D-lysine will not be incorporated into the proteins synthesized by the organism.

would each be nine bases long). Use Table 3.3 to find the codons that correspond to the first three residues: Met

Gly

Ser

AUG

GGU GGC GGA GGG

UCU UCC UCA UCG AGU AGC

COO H

-amino group

NH 3 chiral carbon

C

CH2 CH2

Using just the topmost set of codons, a possible DNA primer would therefore have 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. 3.18). 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

CH2 ε-amino group

CH2 NH 3 D-Lysine

3. a. His, Phe, Pro, Tyr, Trp; b. His, Phe, Tyr, Trp; c. His, Cys; d. Gly; e. Arg, Lys; f. Asp, Glu; g. Cys, Met. 5. From least soluble to most soluble: Val, Trp, Thr, Ser, Arg. You can use Table 4.3 as a guide, but you should also be able to do this type of problem without using the table. 7. a.

Again, considering just the topmost set of codons, a probable DNA 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.

Mg2 COO

H

C

CH2

CH2

COO

NH 3

b. In the body, the glutamate and the Mg2+ ion separate, and the amino group remains protonated (since the pH of body fluids, ∼7.4, is less than the pK value of the amino group, ∼9.0). Therefore, magnesium glutamate and monosodium glutamate yield the same form of glutamate.

73. ATTGTTCCCACAGACCG CGGCGAAGCATTGTTCC ACCGTGTTTCCGACCG TTGTTCCCACAGACCGTG

9. The Asp residue is more likely to form hydrogen bonds when protonated, as shown below. The carboxylic acid side chain can participate in three hydrogen bonds, whereas the carboxylate group can participate in only two. (Arrows point toward hydrogen bond acceptors and away from hydrogen bond donors.) [From Pace, C. N., Grimsley, G. R., and Scholtz, J. M. J. Biol. Chem. 284, 13285–13289 (2009).]

75. 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.

Chapter 4 1. a.

-amino group H

3N

O

COO C

C

O

H

H

CH2

chiral carbon O

CH2

C

CH2 CH2 NH 3 L-Lysine

O

11.

ε-amino group

N-terminus



H3N

peptide bond

peptide bond

-amino group

O CH

CH2 HN N

C

-carboxylate group

O H N

CH

C

H N

CH

COO

CH2

CH2

CH2

CH2

CH2

COO

CH2 NH 3

-amino group

C-terminus

-carboxylate group

Odd-Numbered Solutions

13.

O

O H

3N

C

CH

H N

CH2

C

O H N

CH2

C

O H N

CH

C

CH

C

H N

C

CH2

O H N

C

CH

O

CH2

CH2

CH2

CH2

O

O H N

S-9

CH2 S CH3 OH

OH

15. O H

3N

C

CH CH2

O

O H N

CH

C

H N

CH2 

COO

CH CH2 CH2

C

H N

H N

CH2 COO

CH

C

N-terminus Glu Tyr C-terminus Net charge

N-terminus 3 Asp C-terminus Net charge

CH2

CH2

NH3

NH 3

N-terminus His Lys Glu C-terminus Net charge

C

C

N H

H O

Charge +1 –1 0 –1 –1

H

29. 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. 31. 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. 33.

Q

Charge +1 –3 –1 –3

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

C

CH2

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. Group

H O

CH2

17. 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. Group

O

O

CH2 

4

CH2

OH

C

CH

25. a. tertiary; b. quaternary; c. primary; d. secondary. 27. H

O

O

Charge +1 0 +1 –1 –1 0

19. 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. 21. 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. 23. There are six possible sequences: HPR, HRP, PHR, PRH, RHP, and RPH.

M

R

W

E

K

R

W

H

G

R W

Y K

G V

L

W

The polar amino acid residues are shown in red and the nonpolar residues in blue. The polar residues are mainly on one side of the helix while the nonpolar residues are on the other side. Quite a few of the polar side chains are positively charged. [This is an example of an amphipathic helix. From Martoglio, B., Graf, R., and Dobberstein, B., EMBO J. 16, 6636–6645 (1997).] 35. The highlighted region, consisting of mainly nonpolar residues, is about the right length to span the membrane (see Problem 34). PPEEETGERVQLAHHFSEPEITLIIFGVMAGVIGTILLISYGIRRLIKKSPSDVKPLPSPDTD 37. Triose phosphate isomerase is an example of an α/β protein. 39. Cys, His, and Tyr are the most likely to be uncharged and buried in the hydrophobic core of the protein. Cys and Tyr are protonated and neutral, and His, with a near-neutral pK, is often unprotonated and uncharged. [From Pace, C. N., Grimsley, G. R., and Scholtz, J. M. J. Biol. Chem. 284, 13285–13289 (2009).]

S-10

Odd-Numbered Solutions

41. It’s 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. 43. There are many possible answers for this question. An example is shown for each.

a.

O H N

CH

C

b.

O H N

CH2

CH

C

Hydrogen bonding between Tyr and Ser

NH 3

CH

H

O

CH2

HN

O H

COO

CH2 O C

O H N

CH2

CH2 CH2 A Lys-Glu CH2 ion pair

c.

CH

C

CH

CH3

CH2 van der Waals CH3 forces between Leu and Val CH3 H 3C

CH O

HN

CH

C

CH2 O HN

CH

C

45. 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. 47. 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. 49. 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. 51. Proline does not fit well into the structure of the α helix, both because of its geometry (see Problem 31) and the absence of a peptide NH group to contribute to hydrogen bonding. This amino acid substitution would produce 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., Gaetani, M., Ortiz, V., Bhasin, N., Harper, S., Gallagher, P. G., Speicher, D. W., and Discher, D. E., Blood 109, 3538–3543 (2007).]

59. 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. 61. 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., Lei, M., and Huo, S. Protein Sci. 12,1222– 1231 (2003).] 63. 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) = 6.25. 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 65. 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 likely contains 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 likely contains an abundance of these residues. 67. At pH 7.0, the peptide likely has 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. 69. 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 kDa. Lane 3 shows TGF-β in the absence of 2-mercaptoethanol; here the intact protein has a molecular mass of 25 kDa. [From Assoian, R. K., Komoriya, A., Meyers, C. A., Miller, D. M., and Sporn, M. B., J. Biol. Chem. 258, 7155–7160 (1983)].

53. 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.)

57. 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.

100

63

Mass (kDa)

55. The positively charged Arg residues and the negatively charged Asp residues are likely to be found on the surface of the monomer. These residues likely 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., Misquitta, S., Blond, S. Y., Adams, E., and Colman, R. F., J. Biol. Chem. 283, 32800–32888 (2008).]

e1

Lan

40

25

16

10

e2

Lan

e3

Lan

Odd-Numbered Solutions

71. The amino-terminal residue is Ala. The carboxyl-terminal residue must be Met, since the dodecapeptide was not cleaved when CNBr was added. Chymotrypsin cleaves after Phe. Fragment II contains the Asp, so it appears in the sequence first, and Phe must be the cleavage site. Trypsin cleaves after Lys. Since fragment III contains Asp, Lys must be the cleavage site. Elastase cleaves after Gly, Val, and Ser. Val must occupy the second position followed by a Pro, and this Val was not cleaved. [Based on Anastasi, A., Montecucchi, P., Erspamer, V. and Visser, J., Experientia 33, 857–858 (1977).] Trypsin

5. 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: a.

Chymotrypsin

b.

Asp–Val–Pro–Lys–Ser–Asp–Gln–Phe–Val–Gly–Leu–Met Elastase

Elastase

73. Sequencing this particular peptide using chemical or enzymatic cleavage methods would be difficult. There is only one N-terminal Met, so CNBr would not cleave the peptide. Cleavage with trypsin would not occur because the peptide lacks Lys and Arg residues. Cleavage with elastase would yield single amino acids, so locating areas of overlap would be difficult. Cleavage with chymotrypsin would produce three fragments, but the lack of a second cleavage agent would make it difficult to put the peptide fragments in the correct order. [From Burstein, I. and Schechter, Y., Biochem. 17, 2392–2400 (1978).] 75. Leu and Ile are isomers and have the same mass; therefore, mass spectrometry cannot distinguish them. 77.

Phe

Val H3C



Ala

O

CH3

O

H N

H3N N H

Asp

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.

pO2 p50 + pO2 p50 × Y pO2 = 1−Y 2.8 torr × 0.25 pO2 = (1 − 0.25) pO2 = 0.93 torr Y=

pO2 p50 + pO2 p50 × Y pO2 = 1−Y 2.8 torr × 0.90 pO2 = (1 − 0.90) pO2 = 25 torr Y=

7. In the arteries, nearly all the hemoglobin is oxygenated and therefore takes on the color of the Fe(II), in which the sixth coordination site is occupied by O2. Blood that has passed through the capillaries and given up some of its oxygen contains a mixture of oxy- and deoxyhemoglobin. Deoxyhemoglobin, in which the Fe(II) has only five ligands, imparts a bluish tinge to venous blood. 9. 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., Bonaventura, J., Wittenberg, J. B., and Block, B. A., Am. J. Physiol. Regul. Integr. Comp. Physiol. 280, R1123–R1133 (2001).] 11. 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 E7

His F8

Human SEDLKKHGATVLTALGGILKKKGH H EAEIKPLAQSHA Harbor seal SEDLRKHGKTVLTALGGILKKKGH H DAELKPLAQSHA

13. Use Equation 5.4 to calculate the fractional saturation (Y ) for hyperbolic binding, letting K = p50 = 26 torr:

Chapter 5 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. 3. Myoglobin facilitates O2 diffusion in the cell by acting as a “molecular bucket brigade,” accepting O2 delivered to the cell by hemoglobin and then transferring it to mitochondrial proteins. Myoglobin is 50% saturated with O2 when the intracellular pO2 is equal to its p50 value, so it functions most effectively under these conditions. At an intracellular pO2 greater than p50, O2 remains bound to myoglobin and is not transferred; at pO2 less than p50, myoglobin doesn’t bind sufficient O2. In either case, the transfer of O2 from hemoglobin to mitochondrial proteins is compromised.

S-11

Y=

pO2 p50 + pO2

30 torr = 0.54 26 torr + 30 torr 100 torr = 0.79 At 100 torr, Y = 26 torr + 100 torr

At 30 torr, Y =

Therefore, if hemoglobin exhibited hyperbolic oxygen-binding behavior, 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).

S-12

Odd-Numbered Solutions

15. Y = Y=

( pO2 ) n ( p50 ) n + ( pO2 ) n

causes the third equation to shift to the right. This depletes CO2(aq), which causes the second equation to shift right, thus depleting carbonic acid. This causes the first equation to shift right as hydrogen ions and HCO3‒ combine to form more carbonic acid. Depletion of hydrogen ions results in a basic pH.

(25 torr) 3 (40 torr) 3 + (25 torr) 3

Y = 0.20

H + (aq) + HCO3− (aq) ⇌ H2CO3 (aq)

( pO2 ) n Y= ( p50 ) n + ( pO2 ) n Y=

H2CO3 (aq) ⇌ CO2 (aq) + H2O(1)

(120 torr) 3 (40 torr) 3 + (120 torr) 3

Y = 0.96 17. 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. 19. The increased O2 release is the result of the Bohr effect. The increase in [H+] promotes the shift from the oxy to the deoxy conformation of hemoglobin. The decrease in oxygen affinity improves oxygen delivery to the muscle, where it is needed. ‒ 3

21. a. HCO 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 HCO3‒. b. Both curves are sigmoidal. According to the investigators, the p50 for crocodile hemoglobin is 6.8 torr in the absence of HCO3‒ and 44 torr in its presence. The higher p50 value in the presence of HCO3‒ indicates that the crocodile hemoglobin has a lower affinity for oxygen.



w/o HCO3 Y



w/ HCO3

pO2

c. The positively charged Lys side chain forms an ion pair with the negatively charged HCO3‒. The phenolate oxygens on the two tyrosine side chains act as hydrogen bond acceptors for the bicarbonate hydrogen. [From Komiyama, N. H., Miyazaki, G., Tame, J., and Nagai, K., Nature 373, 244–246 (1995).]

CO2(aq) ⇌ CO2(g) 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. 29. 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. 31. 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 would be less sensitive to pH changes than normal hemoglobin. [From Perutz, M. F., Fermi, G., and Shih, T.-B., Proc. Natl. Acad. Sci. USA 81, 4781–4784, (1984).] 33. 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., Winkler, K., Louis, J., Rea, C., Blume, K., and Beutler, E., Blood 58, 813–817 (1981).] 35. 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.

O

O

23. Negatively charged Glu side chains on the surface of the oxygenated lamprey hemoglobin monomer resist association due to the charge–charge repulsion. But 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., Maillett, D. H., Knapp, J., Olson, J. S., and Riggs, A. F., J. Biol. Chem. 275, 13517–13528 (2000).]

b.

25. 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 barheaded goose hemoglobin can more easily bind oxygen in order to deliver it to the tissues. [From Jessen, T.-H., Weber, R. E., Fermi, G., Tame, J., and Braunitzer, G., Proc. Natl. Acad. Sci. USA 88, 6519–6522 (1991).]

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 oxygenated form of hemoglobin and an increase in its oxygen affinity. [From Bonaventura, J., Bonaventura, C., Sullivan, B., Ferruzzi, G., McCurdy, P. R., Fox, J., and Moo-Penn, W. F., J. Biol. Chem. 251, 7563–7571 (1976).]

27. a. A person hyperventilates in order to obtain more O2, and the blood pH increases as a result, as shown by the equations below. Excessive removal of CO2(g) from the lungs during hyperventilation

HN

C

CH

NH 4

CH2 C

HN

C

CH CH2

O

NH2

H2O

C

O

O

37. 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

Odd-Numbered Solutions

readily. The substitution is quite near His 146, which binds protons (see Problem 20), and may decrease this side chain’s 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. 39. Both types of molecules are proteins and consist of polymers of amino acids. Both contain elements of secondary structure. But 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. 41. 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. 43. 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. 45. 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. 47. 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. 49. 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.

S-13

4 in both sequences are hydrophobic, Trp and Tyr are much larger 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. 5.27). 57. 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. 59. 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 quaternary structure referring to the association of collagen molecules. 61.

O

R

C

H N

C

N

N

C

H

O

O

R

C

C

C

C

N

O

R

H

63.

H

O NH

CH

2

C 1

CH2

3

CH2

4

CH

5

OH

CH2

6

NH 3

53. As shown in Figure 5.24, 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., Jordan, M. A., Chu, K. C., and Wilson, L., J. Biol. Chem. 271, 29807–29812 (1996).]

65. 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., Yiannias, J. A., Windgassen, E. B., and Gornet, M. K., Int. J. Dermatol. 45, 909–913 (2006).]

55. 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

67. 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

51. Colchicine, which promotes microtubule depolymerization, inhibits the mobility of the neutrophils because cell mobility results from polymerization and depolymerization of microtubules.

S-14

Odd-Numbered Solutions

cells themselves and thus facilitates the preparation of cells for culturing. [Source: Worthington Biochemical Corporation.] 69. 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).] 71. 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 doesn’t 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., Peyton, D. H., and Bächinger, H. P., FEBS Lett. 473, 237–240 (2000).] 73. Because collagen has such an unusual amino acid composition (nearly two-thirds of the protein’s residues are Gly, Pro, or Pro derivatives), it contains relatively fewer 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. 75. Since individuals with severe cases of osteogenesis imperfecta do not survive to reproductive age, their particular genetic defect is not passed on. Hence, most cases arise from new mutations. 77. 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. 79. 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. 81. 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. 83. 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.

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., Freed, J. D., and Cowan, J. A., J. Am. Chem. Soc. 122, 8814–8824 (2000).] 3.57 h −1 = 9.9 × 107 3.6 × 10 −8 h −1 5. For adenosine deaminase: 370 s −1 = 2.1 × 1012 1.8 × 10 −10 s −1 For triose phosphate isomerase: 4300 s −1 = 1.0 × 109 4.3 × 10 −6 s −1 The rate of the uncatalyzed reaction is slower for the adenosine deaminase reaction than for the triose phosphate isomerase reaction. But 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. 7. 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

H N

N

3N

O

O

O

O

H N N H

O

O

9. 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. 11. a.

b.

COO

COO

H C

H

succinate dehydrogenase CH

H C

H

CH

COO

COO

Succinate

Fumarate

COO H C H H C

OH

COO

Malate

COO

malate dehydrogenase

H C H

C

O

COO

Oxaloacetate

c. Both succinate dehydrogenase and malate dehydrogenase are oxidoreductases.

Odd-Numbered Solutions

13. NH 3 OOC

H2C

COO

C H

O OOC



C

CH2

CH2

COO

α–Ketoglutarate

Aspartate

aspartate aminotransferase NH 3

O OOC

H2C

C

COO



OOC

Oxaloacetate

C H

CH2

CH2

COO

Glutamate

15. a. Reaction 4, b. Reaction 1, c. Reaction 3, d. Reaction 2. 17. a. 2 H2O2 ⇌ O2 + 2 H2O b. 2 glutathione + H2O2 ⇌ 2 glutathione disulfide + 2 H2O

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. 29. His 57 abstracts a proton from Ser 195, thus rendering the serine oxygen a better nucleophile. 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. 31.

Fast



F

P

CH2OH

CH3 CH CH3 O

CH3

Sarin

HF CH2

Slow

Modified enzyme

O

H3C

CH

H3C

G

O

Acetylcholinesterase

19. 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. 21. a.

S-15

O

P

O

CH3

G

33. His residues are often involved in proton transfer. A carboxymethylated His would be unable to donate or accept protons. Reaction coordinate One-step

b.

O

Reaction coordinate Two-step

HN

CH CH2

G

G

O

C

HN HBr

CH2

 BrCH2COO N

HN

N

N

c.

CH2COO

Reaction coordinate

Reaction coordinate

Positive free energy change

Negative free energy change

[From Shapiro, R., Weremowicz, S., Riordan, J. F., and Vallee, B., Proc. Natl. Acad. Sci USA 84, 8783–8787 (1987).]

Reaction coordinate

35. 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 are not directly involved in catalysis and/or because they are shielded in some way that prevents them from reacting with NEM.

G

G

Reaction coordinate Initial slow step

d.

C

CH

G

Initial fast step G

Reaction coordinate

Reaction coordinate

23. Yes. An enzyme decreases the activation energy barrier for both the forward and the reverse directions of a reaction. 25. 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. 27. 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

37. 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. 39. [From Kong, L., Shaw, N., Yan, L., Lou, Z., and Rao, Z., J. Biol. Chem. 290, 7160–7168 (2015).] Asp CH2 C

O

His

Cys

CH2

CH2

O

S HN

NH R C

H N

C

O

RN

41. a. Glu 35 has a pK of 5.9 and Asp 52 has a pK of 4.5. b. Lysozyme is inactive at pH 2.0 because both the Glu and the Asp are protonated. The Asp is no longer negatively charged and cannot nucleophilically attack the carbocation intermediate.

S-16

Odd-Numbered Solutions

b. The reaction coordinate diagram will look like the one in Figure 6.7, since this is a two-step reaction. Each step has a characteristic activation energy. The acetylated chymotrypsin is the intermediate. c. Yes, chymotrypsin and trypsin use the same catalytic mechanism, so trypsin can act as an esterase as well as a protease, although the reaction is not as rapid. [From Stewart, J. A. and Ouellet, L., Can. J. Chem. 37, 751–759 (1952).]

Lysozyme is inactive at pH 8.0 because both the Glu and the Asp are unprotonated. The Glu would be unable to donate a hydrogen to cleave the bond between the sugar residues. c. Asp 52 Glu 35

COO

Glu 35

OOC

COO



O Asp 52 O

45. a. Nucleophile

Acid catalyst

Enzyme CH 2

HO

RN

Peptide

H N

C

CH2

NH O

HN

RC

CH2

S C

NH NH

RC

RN

O

N H

First tetrahedral intermediate

H

O

H

CH2

S

H2O

Acid catalyst

Glu 35

H

O Asp 52

COO

H

O

N

RC

Amine product Acid catalyst CH2

CH2

S O

Glu 35

Asp 52

COOH

OH

CH2

N

OH

CH2

S

NH

C

O

H

RN

OOC

Base catalyst

NH

C RN

H O

Acyl–enzyme intermediate H

Second tetrahedral intermediate

N

Water

Carboxylate product O

43. a. In the first part of the reaction, the ester bond is cleaved and the chymotrypsin is acetylated. The p-nitrophenolate ion is quickly released, which accounts for the rapid increase in absorbance seen at 410 nm. The enzyme must be regenerated before a second round of catalysis can begin, which requires a deacetylation step. This step is much slower than the first step. Once the acetate is released, the enzyme is regenerated and another molecule of substrate can bind and react. Thus a steady state is reached and the absorbance increases at a uniform rate until the substrate is depleted. Chymotrypsin CH2

RN

Nucleophile

C

O

O

b. The mechanism employs acid–base catalysis as well as covalent catalysis. c. The reaction coordinate diagram would look like the one in Solution 22, since the mechanism is similar to that of chymotrypsin, employing two relatively unstable tetrahedral intermediates and a stable acylated intermediate. d. Bromelain acts as a meat tenderizer because it is a protease and can hydrolyze the peptide bonds in the structural proteins of the meat. This makes the meat easier to chew and digest. e. Cys has a pK of ~3 since it must be unprotonated to act as a nucleophile, and His has a pK of ∼8 since it must be protonated to be active; see part a.

NO2  H

CH2

O

C

47.

H OH

O (CH2)2

O

NH HN

NO2

p-Nitrophenolate (yellow)

CH2

S

p-Nitrophenylacetate

fast

 H O

Acid catalyst

Enzyme CH 2

O

OH  H3C

C

C

OH

RN

C

N RC

O

Glu 224

CH3

Zn2 His His Glu

Modified chymotrypsin H2O slow O

Chymotrypsin CH2

OH  CH3COO  H

(CH2)2

C

RN O

C

Glu 224

O

Zn2 His His Glu

OH H 2N RC

Odd-Numbered Solutions

[From Li, L., Binz, T., Niemann, H., Singh, B. R., Biochemistry 39, 2399–2405 (2000).] 49. a. The deamidation reaction for asparagine is shown. The deamidation reaction for glutamine is similar. O CH

NH

b.

O  H2O

C

CH

NH

CH2

C

C

O

O

NH2

O

Asparagine

Aspartate

CH2

CH2

CH2

CH2

C

1

O

C

A

H



C

CH2

CH2

HA

CH2

2 HO

O

C

O

NH 3

NH 3

NH2

NH 4

A O H

H

CH2

4

CH2

NH 4

C

3

CH2 CH2 NH3

O

O

C

O

O

H

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. Amino terminal Asn residues are not deamidated because a four-membered ring, which is unstable, would result. [From Wright, H. T., Crit. Rev. Biochem. Mol. Biol. 26, 1–52 (1991).] O NH2

CH

C

NH

NH

CH2 O

O

CH2 C

C

NH

CH2 CH2  NH3  HA

O

NH2

C

CH

H

A

51. 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. 53. 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. 55. 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. 57. 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. 59. Because enzymes bind more tightly to their transition states than to their substrates, a drug designed to treat a disease by inhibiting a

S-17

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. 61. 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. 63. a. Trypsin cleaves peptide bonds on the carboxyl side of Lys and Arg residues, which are positively charged at physiological pH. These residues fit into the specificity pocket and interact electrostatically with Asp 189. b. A mutant trypsin with a positively charged Lys residue in its specificity pocket would no longer prefer basic side chains because the like charges would repel one another. The mutant trypsin might instead prefer to cleave peptide bonds on the carboxyl side of negatively charged residues such as Glu and Asp, whose side chains could interact electrostatically with the positively charged Lys residue. c. If the substrate specificity pocket does not include a positively charged Lys residue, then there would be no reason to expect the mutant enzyme to prefer substrates with acidic side chains. Instead, the mutant enzyme would be more likely to prefer substrates with nonpolar side chains such as Leu or Ile. [From Graf, L., Craik, C. S., Patthy, A., Roczniak, S., Fletterick, R. J., and Rutter, W. J. Biochemistry 26, 2616–2623 (1987).] 65. During chymotrypsin activation, chymotrypsin cleaves other chymotrypsin molecules at a Leu, a Tyr, and an Asn residue. Only one of these (Tyr) fits the standard description of chymotrypsin’s specificity. Clearly, chymotrypsin has wider substrate specificity, probably determined in part by the identities of residues near the scissile bond. 67. No, the compound shown in Problem 8 would not be hydrolyzed by chymotrypsin. The side chain on the carboxyl side of the amide bond is an arginine side chain, which would not fit into chymotrypsin’s specificity pocket. 69. a. Persistent activation of trypsinogen to trypsin also results in the activation of chymotrypsinogen to chymotrypsin (see Solution 66) 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., Ohmuraya, M., and Baba, H., Postgrad. Med. J. 82, 775–778 (2006).] 71. A protease with extremely narrow substrate specificity (that is, a protease with a single target) would pose no threat to nearby proteins because these proteins would not be recognized as substrates for hydrolysis. 73. 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. 75. 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. O

77. CH2

CH2

C

NH2  NH3 CH2

Factor XIIIa

CH2

CH2

CH2

NH4

O CH2

CH2

C

NH

CH2

CH2

CH2

CH2

S-18

Odd-Numbered Solutions

79. 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. 81. 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.

17. v = k[trehalose] v = (3.3 × 10 −15 s −1 )(0.050 M) v = 1.7 × 10 −16 M·s −1 19. Using Equation 7.2 and the k value provided in Problem 16: v = k[sucrose] v [sucrose] = k 5.0 × 10 −3 M·s −1 [sucrose] = 1.0 × 104 s −1 [sucrose] = 5.0 × 10 −7 M = 0.50 μM

Chapter 7 1. The hyperbolic shape of the velocity versus substrate 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). 3. v = −

21. a. v = k[ Enzyme] [ Pi ] b. Using Equation 7.3: v = k[Enzyme] [Pi] v = (3.9 × 106 M −1 ·s −1 )(15 × 10 −12 M)(50 × 10 −3 M) v = 2.9 × 10 −6 M·s −1

d[S] dt

0.025 M v=− 440 y × 365 d·y −1 × 24 h·d −1 × 3600 s ·h −1

[From Meadow, N. D., Savtchenko, R. S., Nezami, A., and Roseman, S., J. Biol. Chem. 280, 41872–41880 (2005).] zero order

23.

v = −1.8 × 10 −12 M·s −1 5. v = − v=−

d[S] dt

v0

0.025 M 6 × 106 y × 365 d·y −1 × 24 h·d −1 × 3600 s ·h −1

first order

v = −1.3 × 10 −16 M·s −1 d[P] 25 × 10 −6 M = v= dt 50 d × 24 h·d −1 × 3600 s ·h −1 v = 5.8 × 10 −12 M·s −1

7.

[From Bryant, R. A. R. and Hansen, D. E., J. Am. Chem. Soc. 118, 5498–5499 (1996).] d[S] dt 0.065 M v=− 60 s v = − 1.1 × 10 −3 M·s −1

11. v = −

A→B+C A+B→C 2A→B 2A→B+C Units of k

v0 =

Vmax [S] KM + [S] (65 μmol·min −1 )(1.0 μM) (0.135 μM) + (1.0 μM)

27. Rearrange Equation 7.21 to solve for KM, then substitute the values for v0, substrate concentration and Vmax: Vmax [S] v0 = KM + [S] [S](Vmax − v0 ) KM = v0

Molecularity

Rate equation

Unimolecular Bimolecular Bimolecular Bimolecular

v = k[A] v = k[A][B] v = k[A]2 v = k[A]2

[From Doğru, Y. Z. and Erat, M., Food Res. Int. 49, 411–415 (2012).]

Order

29. Rearrange Equation 7.21 to solve for substrate concentration, then substitute the values for v0, KM and Vmax:

Reaction velocity proportional to . . .

s–1 [A] First M–1 · s–1 [A] and [B] Second M–1 · s–1 [A] squared Second M–1 · s–1 [A] squared Second 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. 15. v = k [sucrose] v = (5.0 × 10 −11 s −1 )(0.050 M) v = 2.5 × 10 −12 M·s −1

25. v0 =

v0 = 57 μmol·min −1

9. (5.8 × 10 −12 M·s −1 )(4.7 × 1011 ) = 2.7 M ·s −1

13. a. Reaction

[S]

KM =

(10 mM)(0.36 − 0.23 U ·min −1 ·mL −1 ) 0.23 U ·min −1 ·mL −1

KM = 5.6 mM

v0 =

Vmax [S] KM + [S]

[S] =

v0 KM Vmax − v0

[S] =

(0.3 μmol·min −1 ·mL −1 )(0.6 mM)

(1.1 − 0.3 μmol·min −1 ·mL −1 ) [S] = 0.2 mM

[From Botman, D., Tigchelaar, W., and Van Noorden, C. J. F., J. Histochem. Cytochem. 62, 813–826 (2014).] 31. The Vmax is approximately 30 μM · s–1 and the KM is approximately 5 μM.

Odd-Numbered Solutions

33. a. When v0 = 0.75Vmax, Vmax [S] 0.75Vmax = [S] + KM

of the substrate AAPF with the enzyme, and this ratio is larger in the mutant enzyme than in the wild-type, as shown below: Wild-type:

Vmax cancels out on both sides. [S] 0.75 = [S] + KM 0.75( [S] + KM ) = [S] 0.75KM = 0.25[S] 3KM = [S] Thus, the substrate concentration is three times as high as the KM. b. When v0 = 0.9Vmax, 0.9Vmax =

Vmax [S] [S] + KM

Vmax cancels out on both sides. [S] [S] + KM 0.9( [S] + KM ) = [S] 0.9KM = 0.1[S] 9KM = [S] 0.9 =

Thus, the substrate concentration is nine times as high as the KM. 35. 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 has precipitated out of solution during the reaction.

kcat 21 s −1 = = 1.1 × 104 M −1 · s −1 KM 1.9 × 10 −3 M Mutant: kcat 120 s −1 = = 6.0 × 104 M −1 · s −1 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. 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 would remove 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., Morinaga, Y., Ikemura, H., and Inouye, M., J. Biol. Chem. 263, 19592–19596 (1988).] 45. The Vmax can be calculated by taking the reciprocal of the y intercept:

37. 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” on the enzyme (see Fig. 6.17) 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. 39. a.

kcat =

Vmax [E]T

b m 4.41 × 10 −4 μM −1 ·h 0.26 μM −1 ·h·μM

x int = − 1.70 × 10 −3 μM −1 1 KM = − x int 1 KM = − −1.70 × 10 −3 μM −1 KM = 590 μM 47. 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 (s . mM1)

43. 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 wildtype enzyme, which means that the mutant enzyme has a greater catalytic efficiency and a higher turnover rate of substrate converted to product per minute. The kcat /KM ratio reflects the specific reactivity

1 4.41 × 10 −4 μM −1 ·h

x int = −

kcat 4.0 s −1 = = 2.9 × 104 s −1 ·M −1 KM 1.4 × 10 −4 M

b. Use the kcat for catechol from Solution 40b and the KM given in the problem: kcat 0.33 s−1 = = 4.2 × 102 s −1 ·M −1 KM 7.9 × 10 −4 M

Vmax =

x int = −

4.0 × 10 −7 M·s −1 1.0 × 10 −7 M = 4.0 s −1

41. a. Use the kcat for dopamine from Solution 40a and the KM from Solution 27: kcat 0.30 s−1 = = 5.4 × 101 s −1 ·M −1 KM 5.6 × 10 −3 M

1 y int

The KM can be determined by first calculating the x intercept and then taking its reciprocal:

The kcat is the turnover number, which is the number of catalytic cycles per unit time. Each molecule of the enzyme therefore undergoes 4 catalytic cycles per second. b.

Vmax =

Vmax = 2.27 × 103 μM ·h −1

kcat = kcat

S-19

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

Odd-Numbered Solutions

51. a. competitive; b. uncompetitive; c. noncompetitive; d. mixed; e. mixed. 53. a. Since the structures are similar (both have choline groups), 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. 55. Calculate α using Equation 7.29, then use α to compare the KM values with and without inhibitor: [I] KI 4 μM α=1+ 2 μM α=3 K app M α= KM K app M 3= 10 μM K app M = 30 μM α=1+

57. a. NADPH is structurally similar to NADP+ and is likely to be a competitive inhibitor. b. The Vmax is the same in the presence and absence of the inhibitor since inhibition can be overcome at high substrate concentrations. The KM increases because a higher concentration of substrate is 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 NADP+ as a cofactor. The differences in Vmax are not as great. [From Hansen, T., Schicting, B., and Schonheit, P., FEMS Microbiol. Lett. 216, 249–253 (2002).] K app M KM 40 μM α= =4 10 μM [I] α = 1+ KI 30 μM 4 = 1+ KI KI = 10 μM

0.40 1/v0 (nmol1 . min)

49. 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. a. Lineweaver–Burk plots are shown below. The KM is calculated from the x intercept, the Vmax from the y intercept (see Solution 45).

0.20 y  1.899x  0.035

0.10 0

0.02 0.04 0.06 1/[S] ( M1)

0.08

0.10

Without p6* With p6*

Without p6*

With p6*

–0.0184 54 0.035 28.6

–0.011 90 0.038 26.3

–1

x intercept (μM ) KM (μM) y intercept (min · nmol–1) Vmax (nmol · min–1)

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 has increased nearly twofold, indicating that p6* is competing with the substrate for binding to the active site of the enzyme. app KM KM 90 μM = 1.67 α= 54 μM [I] α=1+ KI [I] KI = α−1 10 μM KI = 1.67 − 1 KI = 15 μM

c. α =

[From Paulus, C., Hellebrand, S., Tessmer, U., Wolf, H., Kräusslich, H.-G., and Wagner, R., J. Biol. Chem. 274, 21539–21543 (1999).] 63. a. The Lineweaver–Burk plot is shown. The KM is calculated from the x intercept, the Vmax from the y intercept (see Solution 45). 1.50

59. α =

[From Gross, R. W. and Sobel, B. E., J. Biol. Chem. 258, 5221– 5226 (1983).]

y  3.434x  0.038

0.30

0.02

1/v0 (s . nM1)

S-20

y  9.137x  0.035

1.00

0.50 y  0.912x  0.035

0.05

0

0.05

0.10

0.15

1/[FDP] ( M1) Without vanadate With vanadate

–1

x intercept (μM ) KM (μM) y intercept (s · nM–1) Vmax (nM · s–1)

Without vanadate

With vanadate

–0.038 26 0.035 28.6

–0.0038 260 0.035 28.6

S-21

Odd-Numbered Solutions

b. The inhibitor is a competitive inhibitor. The Vmax is the same in the presence and absence of the inhibitor, but the KM has increased tenfold, indicating that the vanadate is competing with the substrate for binding to the active site of the enzyme. 65. The compound is a transition state analog (it mimics the planar transition state of the reaction) and therefore acts as a competitive inhibitor. 67. The structure of coformycin structurally resembles the proposed transition state for adenosine deaminase and this supports the proposed structure. However, 1,6-dihydroinosine has a KI of 1.5 × 10–13 M whereas coformycin’s KI is about 0.25 μM; thus 1,6-dihydroinosine more closely resembles the transition state than does coformycin. 69. a. The Lineweaver–Burk plot is shown below. The KM is calculated from the x intercept, the Vmax from the y intercept (see Solution 45). 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. 1/v0 (pmol1 . min)

1400 1200 1000

0.2

800 y  795x  148

400 200 0

0.2 0.4 0.6 0.8 1/[Pregnenolone] ( M1)

1.0

No inhibitor With inhibitor

Without inhibitor With inhibitor –1

x intercept (μM ) KM (μM) y intercept (pmol–1 · min) Vmax (pmol · min–1)

–0.186 5.4 148 6.8 × 10–3

77. 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. 79. 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.

y  1066x  200

600

75. 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 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.

–0.188 5.3 200 5.0 × 10–3

81. 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. 83. Because cytochrome P450 enzymes can modify warfarin to hasten its excretion, it is helpful to know which P450 enzymes 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.

Chapter 8 1. a.

COO

(CH2)12

H3C

Myristate (14:0) b.

CH

CH

(CH2)5

H3C

(CH2)7

COO

Palmitoleate (16:1n-7) b. Troglitazone inhibits the hydroxylase competitively (see Solution 62) 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., Auchus, J., and Miller, W. L., J. Biol. Chem. 276, 16767–16771 (2001).] 71. The KM is calculated from the x intercept, the Vmax from the y intercept (see Solution 45). 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 does not compete with the substrate for binding to the enzyme. [From Kubo, I., Chen, Q.-X., and Nihei, K.-I., Food Chem. 81, 241–247 (2003).]

x intercept (mM–1) KM (mM) y intercept (min · OD–1) Vmax (OD · min–1)

Without inhibitor

With inhibitor

–0.99 1.01 1.51 0.66

–2.70 0.37 4.27 0.23

73. It is difficult to envision 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.

c. H3C

CH

(CH

CH2

CH2)3

COO

(CH2)6

-Linolenate (18:3n-3) d.

CH

CH

(CH2)7

H3C

(CH2)13

COO

Nervonate (24:1n -9) 3. H3C

(CH2)4

CH

CH

CH2

CH

CH

(CH2)4

CH

CH

(CH2)3

COO

Sciadonate (all-cis- 5,11,14-eicosatrienoate)

[From Sayanova, O., Haslam, R., Venegas Caleron, M., and Napier, J. A., Plant Physiol. 144, 455–467 (2007).] 5. a. H3C (CH2)13 CH

CH

(CH2)2

CH

CH

(CH2)3

COO

cis, cis- 5,9-Tetracosodienoate

b. H3C (CH2)4

(CH

CH

CH2)2 (CH2)2

(CH2)3 CH

CH

CH

all cis- 5,9,15,18-Tetracosotetraenoate H

7.

H3C

(CH2)7

C

C

(CH2)7

COOH

H

Elaidic acid (trans- 9-octadecenoic acid)

CH

(CH2)3

COO

S-22

Odd-Numbered Solutions

9.

25. Both DNA and phospholipids have exposed phosphate groups that are recognized by the antibodies.

O CH2

O H 3C

(H2C)14

C

O

CH3

27. 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.

O

CH

O

(CH2)14

C

O

CH2

(CH2)14

C

CH3

Tripalmitin (CH2)14

C

OH

CH2

O

11. a. H3C

O

O 

H

C

2

O

C

(CH2)14

CH3

OH

CH2

b. The monoacylglycerol and fatty acid products are both amphipathic molecules, with a polar head and one nonpolar tail. These molecules form micelles as a consequence of the hydrophobic effect (see Section 2.2). O CH2

13. HO

C

O

C

(CH2)7

H

CH3

C

C

(CH2)7

CH3

H

CH2

15. All except phosphatidylcholine have hydrogen-bonding head groups. 17. 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. 19. The polar head group is indicated in blue; the two nonpolar tails in red. R1 R2 O

C

C

O

O

H2C

O

CH

O

HO H

P

O

O

H

P

O

C

C

H

OH

CH3 O

(CH2)2

N

CH3

CH3

O CH2

CH

O

OH

H HO OH H

O O

O

DPPC

(CH2)14 (CH2)14 CH3

CH3

CH

CH

CH

NH

CH

C

O

(H2C)12 (CH2)26 CH3 CH2 O

43. 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. 45. 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. 47. 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.

23. Glucose O CH2 HO

37. 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. 41. 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.

CH2

H

21.

35. 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 and which may not always be available.

39. b is polar, d is nonpolar, and a, c, and e are amphipathic.

O

O

31. Vitamin A, and the compound from which it is derived, β-carotene, are lipid-soluble molecules. The vegetables in a typical salad do not contain large amounts of lipid. The addition of the lipid-rich avocado provides a means to solubilize the β-carotene and thus increase its absorption. [From Unlu, N. Z., Bohn, T., Clinton, S. K., and Schwartz, S. J., J. Nutr. 135, 431–436 (2005).] 33. Oxidation of retinal (an aldehyde) yields retinoic acid (a carboxylic acid).

OH

[From Chao-Mei, Y., Curtis, J. M., Wright, J. L. C., Ayer, S. W., and Fathi-Afshar, Z. R., Can. J. Chem. 74, 730–735 (1996).]

H2C

29. The spicy ingredient in the food is a powder made from peppers that contains the hydrophobic compound capsaicin. 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.

C O

(CH2)6(CH2CH

CH)2(CH2)4CH3

49. 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.

Odd-Numbered Solutions

51. The lipids from the meat of the reindeer slaughtered in February contained fewer unsaturated acyl chains than lipids from healthy reindeer. 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).] 53. 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 van der Waals 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). 55. 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. 57. No. Higher temperatures increase fatty acid fluidity. To counter the effect of temperature, the plants make relatively more fatty acids with higher melting points. Dienoic acids have higher melting points than trienoic acids because they are more saturated. Therefore, the plants convert fewer dienoic acids into trienoic acids. 59. 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-facing leaflet, this side of the membrane is more negatively charged than the other side. 61. 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 tails as wavy lines. 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.

63. a.

O

HN

O

CH

b.

C

O HN

CH

CH2

CH2

O

O

C (CH2)7 CH

S-23

O

C

C (CH2)6 CH3

CH CH2)5 CH3

65. The membrane-spanning segment is a stretch of 19 residues (highlighted) that are all uncharged and mostly hydrophobic. [Source: UniProt] DAAYIQLIYPVTNFQKHMIGICVTLTVIIVCSVFIYKIFKIDIVLWYRDS 67. 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. 69. 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).] 71. 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 proteins that serve as transporters. Proteins that transport water, known as aquaporins, have been identified. [From Kleinzeller, A., News Physiol. Sci. 12, 49–54 (1997).] 73. After fusion, the green and red markers were segregated because they represent cell-surface proteins derived from 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.

Chapter 9 1. a. Use Equation 9.2 and substitute the values for [Na+]in and [Na+]out: ∆ψ = 0.058 V log SDS

[Na +]in [Na +]out

10 × 10−3 M 100 × 10−3 M ∆ψ = −0.058 V = −58 mV

∆ψ = 0.058 V log

S-24

b.

Odd-Numbered Solutions

∆ψ = 0.058 V log

[Na +]in [Na +]out

9. Use Equation 9.4 and let Z = 2 and T = 310 K: ∆G = RT ln

40 × 10−3 M 25 × 10−3 M ∆ψ = 0.012 V = 12 mV

∆ψ = 0.058 V log

[Na +]in ∆ψ = 0.058 V log [Na +]out [Na +]in +0.050 V = 0.058 V log [Na+] out [Na +] in 0.86 = log [Na + ] out [Na +] in 100.86 = [Na +] out [Na +] in 7.3 = 1 [Na +] out

+ ZF∆ψ

0.1 × 10−6 M 2 × 10−3 M −3 −1 −1 + (2)(96,485 × 10 kJ·V ·mol )(−0.05 V)

∆G = −25.5 kJ·mol−1 − 9.6 kJ·mol−1 ∆G = −35.1 kJ·mol−1 Ca2+ ions move spontaneously from the outside of the cell to the cytosol. 11. Use Equation 9.4 and let Z = 1 and T = 293 K: ∆G = RT ln

[K +]in + ZF∆ψ [K +]out

25 × 10−3 M 100 × 10−3 M −3 −1 −1 + (1)(96,485 × 10 kJ·V ·mol )(+ 0.05 V)

∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(293 K) ln

When a nerve cell is depolarized, sodium ions enter the cell; therefore the [Na+]in /[Na+]out ratio is more than 100-fold greater in the depolarized cell than in the resting cell. 5. Use Equation 9.4 and substitute the value of the ratio calculated in Solution 3: [Na +]in + ZF∆ψ [Na +]out

7.3 ∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(310 K) ln 1 + (1)(96,485 × 10−3 kJ·V−1 ·mol−1 )(+0.050 V) ∆G = 5.1 kJ·mol−1 + 4.8 kJ·mol−1 ∆G = 9.9 kJ·mol−1 The free energy change for this process is positive, so there is no longer a driving force to move additional Na+ ions into the cell. 7. The extracellular Na+ concentration is 150 mM and the intracellular concentration is about 12 mM, whereas the extracellular K+ concentration is about 4 mM and the intracellular concentration is 140 mM. Use Equation 9.4 and let Z = 1 for both ions and T = 293 K: ∆G = RT ln

[Ca2+]out

∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(310 K) ln

3. Use Equation 9.2 and substitute the values for [Na+]in and [Na+]out:

∆G = RT ln

[Ca2+]in

[Na +]in + ZF∆ψ [Na +]out

12 × 10−3 M 150 × 10−3 M −3 −1 −1 + (1)(96,485 × 10 kJ·V ·mol )(−0.070 V)

∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(293 K) ln

∆G = −6.15 kJ·mol−1 − 6.75 kJ·mol−1 ∆G = −12.90 kJ·mol−1 [K +]in ∆G = RT ln + ZF∆ψ [K +]out

∆G = −3.3 kJ·mol−1 + 4.8 kJ·mol−1 ∆G = 1.4 kJ·mol−1 This process is not spontaneous. 13. 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: ∆ψ −70 mV = 310 K 313 K ∆ψ = −70.7 mV The difference in membrane potential at the higher temperature would not significantly affect the neuron’s activity. b. It is more likely that an increased temperature would increase the fluidity of cell membranes. 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. 15. Use Equation 9.3 and substitute the values for extracellular and cytosolic glucose: [glucose]in a. ∆G = RT ln [glucose]out 0.5 × 10−3 M ∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(293 K) ln 5 × 10−3 M −1 ∆G = −5.6 kJ·mol b. ∆G = RT ln

[glucose]in [glucose]out

∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(293 K) ln

5 × 10−3 M 0.5 × 10−3 M

∆G = 5.6 kJ·mol−1 17. Use Equation 9.3:

140 × 10−3 M 4 × 10−3 M −3 −1 −1 + (1)(96,485 × 10 kJ·V ·mol )(−0.070 V) ∆G = 8.66 kJ·mol−1 − 6.75 kJ·mol−1 ∆G = 1.91 kJ·mol−1

∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(293 K) ln

+

The free energy change for the movement of Na ions into the cell is negative, indicating that movement of these ions in this direction occurs passively. In contrast, the free energy change for the movement of K+ ions into the cell is positive, indicating that K+ ions passively move from the inside of the cell to the outside. An active transport process is required to transport K+ ions into the cell.

a. ∆G = RT ln

[Glucose]in [Glucose]out

∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(310 K) ln

0.5 × 10−3 M 15 × 10−3 M

∆G = −8.8 kJ·mol−1 [Glucose]in b. ∆G = RT ln [Glucose]out ∆G = (8.3145 × 10−3 kJ·K−1 ·mol−1 )(310 K) ln ∆G = −5.4 kJ·mol−1

0.5 × 10−3 M 4 × 10−3 M

Odd-Numbered Solutions

S-25

19. The less polar a substance, the faster it can diffuse through the lipid bilayer. From slowest to fastest: C, A, B.

transport. This experiment also shows that the glucose transporter is asymmetrically arranged in the erythrocyte membrane.

21. 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.

35. As the glutamate (charge –1) enters the cell, 4 positive charges also enter (3 Na+, 1 H+) for a total of 3 positive charges. Since 1 K+ exits the cell at the same time, a total of 2 positive charges are added to the cell for each glutamate transported inside.

23. 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).] 25. 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. 27. 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. 29. The hydroxyl and amido groups act as proton donors to coordinate the negatively charged chloride ion. Cations could not interact with the protons and so would be excluded.

Glucose flux (mM . cm . s−1 × 106)

31. a. A transport protein, like an enzyme, carries out a chemical reaction (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). 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 · s–1 and the KM is about 0.5 mM.

37. 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 HCO3– ions. The HCO3– ions are transported out of the cell by Band 3 in exchange for Cl– ions, which enter the cell. The HCO3– 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. 39. Lactate and protons are transported in symport (see Fig. 9.14), along a concentration gradient. This explains why the transport of lactate is greater when the pH of the buffer is low and has a high concentration of hydrogen ions. [From Elliott, J. L., Saliba, K. J., and Kirk, K., Biochem. J. 355, 733–739 (2001).] 41. a. The maximal velocity is estimated to be between 6 and 7 pmol choline · mg–1 · mL–1. The KM is the substrate concentration at half maximal velocity and is about 40 mM. b. The choline transporter responds to increased choline concentrations by increasing its rate of transport so that efficient uptake occurs over a range of choline concentrations near the KM. At low concentrations of choline (10 μM), the transporter operates at 20% of its maximal velocity, whereas at high concentrations of choline (80 μM), the transporter operates at nearly 100% of its maximal velocity. The KM value of 40 μM is between the low and high physiological concentrations of choline. c. It is possible that the choline transporter cotransports hydrogen ions and choline. A hydrogen ion might be exported when choline is imported. This is an example of antiport transport. d. TEA is structurally similar to choline and acts as a competitive inhibitor. TEA might bind to the choline transporter, preventing choline from binding. In this manner, TEA is brought into the cell and choline transport is inhibited. [From Sinclair, C. J., Chi, K. D., Subramanian, V., Ward, K. L., and Green, R. M., J. Lipid Res. 41, 1841–1847 (2000).] 43.

O H N

CH

C

CH2 C

Vmax

O

O

1.0

O

P

O

O

Aspartyl phosphate 0.5

KM 0

0

2

6 4 [Glucose] (mM)

8

10

33. 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. But extracellular exposure of the ghost transporter to trypsin has no effect, so there is no trypsin-sensitive extracellular domain that is essential for

45. 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 was 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 44). 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., Geindre, M., and Deshusses, J., Biochem. J. 291, 297–301 (1993.)] 47. 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

Odd-Numbered Solutions

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. 49. The ABC transporters bind ATP, then undergo a conformational change as the ATP is hydrolyzed and Pi is released, leaving ADP. Vanadate, a phosphate analog, might 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. 51. 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. 53. 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. 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. Peptide

Peptide

H+

H+

H+

H+

61. 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). 63. 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. 65. 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 cone-shaped, thereby increasing bilayer curvature, which is a necessary step in the formation of a new vesicle by budding. 67. Two lipid bilayers separate the damaged organelle from the rest of the cell.

Na+

69. Receptor-mediated endocytosis reaches a maximum when all receptor sites are occupied, whereas the rate of pinocytosis is proportional to 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+

ATP

3 Na+

2 K+

55. Acetylcholinesterase inhibitors prevent the enzyme from breaking down acetylcholine. This increases the concentration of acetylcholine in the synaptic cleft and increases the chances that acetylcholine 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).] 57.

59. 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.

Concentration of substance internalized

S-26

Receptor-mediated endocytosis

Pinocytosis

CH(CH3)2



F

P

CH2OH

O

DIPF

O CH(CH3)2 HF

O O

P

Chapter 10 1. Signal 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:

CH2

(H3C)2HC

Time

O

Acetylcholinesterase

Modified enzyme

Kd =

O

O CH(CH3)2

Kd =

[R] [L] [R ·L] (2.5 × 10 −3 M) (125 × 10 −6 M)

22.5 × 10 −3 M Kd = 1.4 × 10−5 M = 14 μM

S-27

Odd-Numbered Solutions

5. Use Equation 10.1 and solve for [R · L]:

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. [From Jefferson, J. R., Harmon, J. T., and Jamieson, G. A., Blood 71, 110–116 (1988).]

[R] [L] Kd = [R ·L] [R] [L] [R ·L] = Kd (5 × 10 −3 M) (18 × 10 −3 M) [R ·L] = 3 × 10 −3 M −3 [R ·L] = 30 × 10 M = 30 mM 7. Let [R · L] = x and [R] = 0.010 – x Kd =

[R][L] [R ·L]

17. The Kd is obtained from the x-intercept in the double-reciprocal plot.

[R][L] [R ·L] = x = Kd (0.010 − x)(2.5 × 10

1/[R . L] ( M1)

x=

10 −3

M)

−3

1.5 × 10 M

(0.000025 − 0.0025x) 0.0015 0.0015x = 0.000025 − 0.0025x 0.0040x = 0.000025 x = 0.00625 = 6.25 mM = [R ·L] x=

8 6 4

1/Kd

2

The percentage of receptors occupied by ligand is 6.25 mM/10 mM or 62.5%. 4

9. The Kd estimated from the curve is about 0.1 mM.

3

2

1

0

1

2

3

4

1/[L] ( M1)

1.0

1 Kd 1 −2 μM−1 = − Kd

[R. L]/[R]T

x-int = −

0.6

Kd = 0.5 μM Kd 0.2 0.1

0.2

0.3

0.4

0.5

0.6

[L] (mM)

11.

Kd = [R]T = [R] = Kd = Kd [R ·L] =

[R] [L] [R ·L] [R] + [R ·L] [R]T − [R ·L] ( [R]T − [R ·L]) [L] [R ·L] [R]T [L] − [R ·L] [L]

0.7

19. 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. 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. 21. 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. 23. If receptors are removed from the cell surface, the ligand cannot bind and an intracellular response cannot occur.

(Kd [R ·L]) + ( [R ·L] [L]) = [R]T [L] [R ·L](Kd + [L]) = [R]T [L] [L] [R ·L] = [R]T Kd + [L] 13.

[L] [R ·L] = [R]T [L] + Kd [L] 100 = 1000 [L] + 1.0 × 10−10 M 0.10( [L] + 1.0 × 10−10 M) = [L] 0.10[L] + 1.0 × 10−11 M = [L] 1.0 × 10−11 M = 0.9[L] [L] = 1.11 × 10−11 M

O

25. a.

H N

CH

C

CH2 S C

(CH2)14

CH3

O

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.

S-28

Odd-Numbered Solutions

27. 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. 29. 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. 31. 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. 33. 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 increased cAMP concentration leads to the loss of water and electrolytes from the cells and results in diarrhea that can be fatal. 35. GTPγS can bind to a G protein, but since it cannot be hydrolyzed, the G protein is in a persistently active state. If GTPγS binds to a stimulatory G protein, adenylate cyclase is continually active, which increases the cellular cAMP concentration. If GTPγS binds to an inhibitory G protein (see Problem 34), adenylate cyclase is continually inhibited, and the cellular cAMP concentration decreases. 37. 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. 39. 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. 41. 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. 43. 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. 45. 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 36). It’s 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. 47. If NO synthase is missing, the signaling pathway described in Problem 45 cannot be completed. NO cannot be synthesized, and subsequent steps, including the production of the second messenger 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. 49. 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 Problem 45, 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. 51. 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, it is not available to activate any other Ca2+-sensitive proteins that might be involved in normal cell signaling. 53. Since the growth factor stimulates kinase activity, the H2O2 second messenger is likely to produce similar responses, so it must inactivate the phosphatases. 55. 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 transporters 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. 57. 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. 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. 59. As noted in Problem 39, 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. 61. 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

Odd-Numbered Solutions

cell death). Thus inhibition of PI3K results in the death of the cancer cell and is an effective treatment.

75.

High osmolarity (salt or glucose)

63. 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 get phosphorylated. [From Nallagatla, S. R., Toroney, R., and Bevilacqua, P. C., Curr. Opin. Struct. Biol. 21, 119–127 (2011).]

Ras

69. 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 59), which leads to the phosphorylation of transcription factors that promote the expression of proteins involved in the cell cycle, ultimately leading to cell survival. 71. a.

O Enzyme

C

OH COCH3

 CH2OH

MEK PKA MAP kinase

C O

Activation of glycolysis Accumulation of glycerol

Chapter 11 1. a. aldotetrose; b. ketopentose; c. aldohexose; d. ketopentose. 3. Coenzyme A, NAD, and FAD all contain ribose residues. 5. Sugars b and d are epimers. 7.

CHO HO

C

HO

C

H

H

C

OH

H

C

OH

CH3



H

CH2OH

O C

CH2O

Pi

PFK2

O Acetylsalicylic acid

Acetylated enzyme

Raf

cAMP

65. 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. 67. Arachidonic acid is the substrate for the production of prostaglandins, many of which have inflammatory properties. Stimulating 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.

S-29

OH

9. a.

OH 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. But it’s possible that the acetylation alters the structure of the active site such that the arachidonate 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). 73. Phospholipase A2 catalyzes the release of arachidonate from membrane phospholipids. Blocking this reaction would prevent the COX-catalyzed conversion of arachidonate to proinflammatory prostaglandins.

b.

CH2OH



C

O

HO

C

H

HO

C

H

H

C

OH

CH2OH

CH2OH C

O

H

C

OH

H

C

OH

HO

C

H

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. 11. HO H

CH2OH O H OH H H

H OH

OH

-D-Galactose

HO H

CH2OH O H OH H H

OH H

OH

-D-Galactose

S-30

Odd-Numbered Solutions

13. A five-membered ring results. HOCH2

O

H

27. a.

HOCH2

H H

H

H

OH OH

O

OH

OH

H OH

-D-Ribose

H

H

H

HO

OH

H

-D-Ribose

COO

b.

H

C

OH

HO

C

H

H

C

OH

H

C

OH

b.

O H OH

H

H

OH

CH3I H3CO

OH

H3CO

O

H

OH

CH2OCH3 O H H OCH3 H OCH3 H

OH

CHO

C

HO

C

H

H

C

OH

H

C

OH

OH

31. H

CH2OH

HO

Sorbitol COO

19.

O

H

C

H

H

C

OH

H

C

OH

b.

CHO H

C

OH

HO

C

H

HO

C

H

H

C

OH

COOH OH

O H OH

H

H

H

HO

H H

35. H

CH2OPO32 C

O

HO

C

H

H

C

OH

H

C

OH

H

C

OH

CH3

C

CH2OH O H OH H

CH2OH O H OH H

H O

H

OH

OH H

OH

H

H

HO

OH

H

O

37.

R H

H

H H

OH

OH

CH2OH O H OH H

HO

H

COOH OH

H

N-Acetylneuraminic acid

OH H

O H OH

H

H C

OH

RH C

OH

CH2OH

H

OH

H O

OH

H

HO

H

OH

CH2

H

O

CH2OH O H OH H

OH

CH2OH O H OH H H

H

H

C

CH2OCH3

OH

OH

CH2OPO32 H N

H

2,3,4,6-Tetra-Omethyl-D-glucose

H

H

OH

Galacturonate

O

OCH3

OH

COOH

25.

H

C

33. Trehalase digestion produces glucose, which exists in solution as a mixture of the α and β anomers.

CH2OH

23.

C

H

Cellobiose is a reducing sugar. The anomeric carbon of the glucose on the right side is free to reverse the cyclization reaction to re-form the aldehyde functional group, which can be reduced.

KDG 21. a.

H3CO

OCH3

CH2OH O H OH H H

C

OCH3

29. Lactose is a reducing sugar because it has a free anomeric carbon (C1 of the glucose residue). Sucrose is not a reducing sugar because the anomeric carbons of both glucose and fructose are involved in the glycosidic bond.

CH2OH H

C

CH3OH

CH2OH

c.

H

H, H2O

1,2,3,4,6-Penta-Omethyl-D-glucose

Gluconate

OCH3

1,2,3,4,6-Penta-Omethyl-D-glucose

CH2OCH3 O H H OCH3 H OCH3

H

CH2OH H

H

H

-D-Glucose

15. 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. 17. a.

CH2OH O H OH H

OH CH2OH O H H OH

H OH

H

OH

39. In order for sorbitol to be catabolized to yield energy, it would need to enter the same catabolic pathway as glucose. The enzymes that catalyze glucose catabolism are specific for glucose and do not bind to sorbitol, so the sugar alcohol is not metabolized and passes

Odd-Numbered Solutions

through the body undigested. In this manner, sorbitol contributes no calories to the food containing it, but it is nearly identical to its parent monosaccharide in taste and sweetness.

59.

H O

HO

H

CH2OH O H H

H O

HO

OH

H

H

H

CH2OH O H H

HO

OH

CH2 O

OH

O

H

HO

H

O

49. Pectin is a highly hydrated polysaccharide, so it thickens the fruit preparation and helps turn it into a gel. 51. The cellulose-based plant cell wall is strong and rigid, but it must be remodeled as the plant cell grows. The cell uses cellulase to weaken the cell wall so that the cell can expand. 53. Four. Both starch and glycogen contain glucose residues linked by α(1→4) glycosidic bonds and α(1→6) glycosidic bonds at branch points. 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. HOCH2

O O

HOCH2 O

O

H

H

H H

H H

OH

P

O

O

O

P O

N

O

CH2 H

H

H

O

H OH

OH

63. 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.

47. 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 roughage, or fiber, to the diet, but these foods neither provide nor cost the body much in the way of energy.

H

O

O

O

O

CH3

H

HO

HN H

NH

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). Nondigestible carbohydrates are often classified by food manufacturers as “fiber”; thus inulin extracted from chicory root is often added to processed foods to boost their fiber content.

H

O

C O

CH2OH O H OH H H

HO

OH

CH

61.

HO

CH2OH O

C H

[From Schirm, M., Schoenhofen, I. C., Logan, S. M., Waldron, K. C., and Thibault, P., Anal. Chem. 77, 7774–7782 (2005).]

H

O

NH

R  H (Ser) R  CH3 (Thr)

C

CH2

R

CH3

H

CH2OH

55.

NH

H

45. a.

O

H

C

43. Starch, glycogen, cellulose, and chitin are homopolymers. Peptidoglycan and chondroitin sulfate are heteropolymers.

OH

H

H

OH H

H

O H OH

HO

CH2OH O H OH H

H

CH2OH H

41.

S-31

H

HO H

OH

57. The N-linked saccharide is N-acetylglucosamine, and the bond has the β configuration. The O-linked oligosaccharide is N-acetylgalactosamine, and the bond has the α configuration.

65. a. The monosaccharide is N-acetylglucosamine. b. The amide bond forms between Ala and the carboxylate group on the C4 substituent in the disaccharide. The arrow indicates the linkage site. CH2OH H

CH2OH

O H OH

H O

H

O H

H

NH

C O

O

H H

H H

CH3 O

CH3CHCOO

NH

C O

CH3 n

67. 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).]

Chapter 12 1. a. chemoautotroph; b. photoautotroph; c. chemoautotroph; d. heterotroph; e. heterotroph; f. chemoautotroph; g. photoautotroph. 3. The pH of the stomach is ∼2. 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

S-32

Odd-Numbered Solutions

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 completely hydrolyze starch to its component monosaccharides, since only monosaccharides can be absorbed. 7. Sugar alcohols are not present in abundance naturally, which explains the absence of transporters for these molecules. Passive diffusion is less effective than passive transport (as shown in Problem 9.21). 9. Because the products of nucleic acid digestion are relatively large, charged nucleotides, a transport protein is required to facilitate their movement across the cell membrane. The transport protein most likely uses the free energy of a Na+ gradient (active transport), as is the case for intestinal monosaccharide and amino acid transporters. 11. The low pH denatures the protein, unfolding it so peptide bonds are more accessible to proteolytic digestion by stomach enzymes. 13. 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. 15. 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. Na+

Amino acid

Na+

Amino acid

ATP

Amino acid transporter +

Na

K

+

17. a. O

C

O

H2C

O R1

O

CH O

C

Fatty acid

Cholesterol

HO

21. 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, and so on. 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. 23. The phosphorylated glucose molecule is not recognized by the glucose transporter. Removal of the phosphate group allows the glucose to more easily leave the cell. 25. Acetyl-CoA

R3

Triacylglycerol

O

Gastric lipase H2O

R2

C

C

H2C O

O

O

C

R1

CH

H2C

O

R3

Citric acid cycle

✔

Fatty acid metabolism

✔

GAP

Pyruvate

✔

✔

TAG synthesis

✔

Photosynthesis

✔ ✔

29. a. CH4 + SO42 – → HCO3– + HS– + H2O; b. CH4 is oxidized to HCO3–; c. SO42– is reduced to HS –.

Na,K-ATPase Amino acid

R2

Cholesteryl esterase O

27. a. oxidized; b. reduced; c. reduced; d. oxidized.

ADP + Pi

C

C

Transamination

K+

O

(H2C)17

Glycolysis

INTESTINAL CELL

H2C

H3C

Na-amino acid transporter

INTESTINAL SPACE

O

O

OH

Diacylglycerol

31. 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. 33. a.

O H N

19.

C

CH2

Fatty acid

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.

CH

OOC

CH

COO

-Carboxyglutamate 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. 35. a. vitamin C; b. biotin; c. pyridoxine; d. pantothenic acid.

O H3C

(H2C)17

C

O

Cholesteryl ester

37. a. The prosthetic group is related to NAD+, as it contains a nicotinamide–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:

Odd-Numbered Solutions

Keq = e − ∆G°ʹ/RT

NH N Ni2

S

phosphate

O

O O

P

O

His

Keq = e −20,000 J·mol

H

Small changes in ΔG°′ result in large changes in Keq. Doubling the ΔG°′ value (a positive, unfavorable value) leads to a 60-fold decrease in Keq.

N nicotinamide

Keq = e − ∆G°ʹ/RT

b.

H

Keq = e −(−10,000 J·mol

H

H OH

/(8.3145 J·K − 1 ·mol − 1)(298 K)

Lys

O

O

−1

Keq = e −8.07 = 0.00031

S N H

ribose

S-33

−1

)/(8.3145 J·K − 1 ·mol − 1)(298 K)

−1

)/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e4.04 = 57

OH

Keq = e − ∆G°ʹ/RT

39. Use Equation 12.2:

Keq = e −(−20,000 J·mol

a. ∆G°ʹ = −RT ln Keq

Keq = e8.07 = 3200

∆G°ʹ = −(8.3145 J·K −1 ·mol −1 )(298 K)ln 0.25 ∆G°ʹ = 3400 J·mol−1 = 3.4 kJ·mol −1 b. ∆G°ʹ = −RT ln Keq

The same conclusion can be made: Small changes in ΔG°′ lead to large changes in Keq. Doubling a (favorable) ΔG°′ results in a Keq value that is nearly 60 times as large.

∆G°ʹ = −(8.3145 J·K −1 ·mol −1 )(310 K)ln 0.25 ∆G°ʹ = 3600 J·mol −1 = 3.6 kJ·mol −1 c. The ΔG°′ values for both reactions are positive, so the reaction is not spontaneous at either temperature. 41. Because Keq is the ratio of the product concentration to the reactant concentration at equilibrium, the reaction with the larger Keq 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. 43. a. Since Keq = 1, ln Keq = 0 and ΔG°′ 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. 45. Use Equation 12.3: ∆G = ∆G°ʹ + RT ln

[G6P] [G1P]

51. The complete reaction is ATP + H2O → ADP + Pi. Use Equation 12.3 and the value of ΔG°′ from Table 12.4. The concentration of water is assumed to be equal to 1. [ADP] [Pi] ∆G = ∆G°ʹ + RT ln [G1P] ∆G = −30,500 J·mol −1 + (8.3145 J·K −1 ·mol −1 )(310 K) (1 × 10 −3 M)(5 × 10 −3 M ln( ) 3 × 10 −3 M ∆G = −30,500 J · mol −1 + (−16,500 J · mol −1 ) = −47 kJ · mol −1 53. First, convert Calories to joules: 72,000 cal × 4.184 J/cal = 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 9.8 moles of ATP. 55. a

∆G = −7100 J · mol −1 + (8.3145 J · K −1 · mol −1 ) (310 K)

G

−3

20 × 10 M ln( 5 × 10 −3 M )

c

∆G = −7100 J · mol −1 + 3600 J · mol −1

b

∆G = −3500 J · mol −1 = −3.5 kJ · mol −1 47. Use Equation 12.3 and the ΔG°′ for the reaction calculated in Solution 42: [B] [A] ∆G = −5900 J · mol −1 + (8.3145 J · K −1 · mol −1 )(310 K) ∆G = ∆G°ʹ + RT ln

ln(

0.1 × 10 −3 M 0.9 × 10 −3 M )

∆G = −11, 600 J · mol −1 = −11.6 kJ · mol −1 The reaction will proceed as written, with A converted to B until the ratio of [B]/[A] = 10/1.

Reaction coordinate

57. 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 ΔG° 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 serve to 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 ΔG°′ value that is more negative. 59. a. The synthesis of ATP from ADP requires 30.5 kJ · mol–1 of energy:

49. a. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): Keq = e

−4.04

−1

/(8.3145 J·K − 1 ·mol − 1)(298 K)

= 0.018

∆G°ʹ = +30.5 kJ · mol −1

2850 kJ·mol −1 × 0.33 = 30.8 ATP 30.5 kJ·mol −1

− ∆G°ʹ/RT

Keq = e −10,000 J·mol Keq = e

ADP + Pi → ATP + H2O

b.

9781 kJ·mol −1 × 0.33 = 106 ATP 30.5 kJ·mol −1

S-34

Odd-Numbered Solutions

c. For glucose, 30.8 ATP/6 carbons = 5.1 ATP/carbon. For palmitate, 106 ATP/16 carbons = 6.6 ATP/carbon. Most of the carbon atoms of fatty acids are fully reduced —CH2— groups. Most of the carbon atoms of glucose have hydroxyl groups attached to them (—CHOH—) and are therefore already partly oxidized. Consequently, more free energy is available from a carbon in a triacylglycerol than from a carbon in a glycogen molecule. 61. a. The apple provides 9.8 moles of ATP (see Solution 53). The moderately active female described in Problem 60 requires 100 moles ATP daily. Therefore 100 mol ATP/9.8 mol · apple–1 = 10 apples. Keeping the 33% efficiency in mind, 10/0.33 = 30 apples would be required. b. The hot chocolate drink provides 104 moles of ATP (see Solution 54). Therefore 100 mol ATP/104 mol · hot chocolate drink–1 = 1 drink. Keeping the 33% efficiency in mind, 1/0.33 = 3 drinks would be required. In other words, one of these drinks provides one-third of the daily energy needs for the moderately active female.

is not a feasible route to the production of this compound for the glycolytic pathway. d.

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. 67. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): a.

/(8.3145 J · K − 1 · mol − 1)(298 K)

Keq = e −2.02 = 0.13 b. Since [isocitrate] = 0.133 [citrate] [isocitrate] = 0.133 [citrate] The total concentration of isocitrate and citrate is 2 M, so Keq =

[isocitrate] = 2 M − [citrate] Combining the two equations gives

F6P + ATP + H2O ⇌ F16BP + ADP c.

−1

/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e −6.94 = 9.7 × 10 −4 [F16BP] [ADP] Keq = [F6P] [ATP] [F16BP](1 × 10 −3 M) 9.7 × 10 −4 = [F6P](3 × 10 −3 M) [F16BP] 2.9 × 10 −3 = [F6P] 1 d. The conversion of fructose-6-phosphate to fructose-1,6bisphosphate is unfavorable. The ratio of products to reactants at equilibrium is 2.2 × 10–11 under standard conditions. But 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 fructose6-phosphate increases to 2.9 × 10–3, a change of eight orders of magnitude.

/(8.3145 J·K − 1 ·mol − 1)(298 K)

c. Under the given conditions, the reaction would produce only 9.5 × 10–8 M glucose-6-phosphate from 5 mM glucose and thus

∆G°ʹ = 17.2 kJ·mol −1

Keq = e − ∆G°ʹ/RT Keq = e −17,200 J·mol

Keq = e − ∆G°ʹ/RT

b. Use the equilibrium constant expression and the Keq calculated in part a to solve for the equilibrium concentration of G6P: [G6P] Keq = [glucose] [Pi] [G6P] 0.0038 = −3 (5 × 10 M)(5 × 10 −3 M) [G6P] = 9.5 × 10 −8 M

∆G°ʹ = −30.5 kJ·mol −1

ATP + H2O ⇌ ADP + Pi

65. 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°′ for the hydrolysis of glucose-6-phosphate (G6P) found in Table 12.4:

Keq = e −5.57 = 0.0038

∆G°ʹ = 47.7 kJ·mol −1

F6P + Pi ⇌ F16BP

c. The preferred direction under standard conditions is toward the formation of citrate. d. The reaction occurs in the direction of isocitrate synthesis because standard conditions do not exist in the cell. Also, the reaction is the second step of an eight-step pathway, so isocitrate is removed as soon as it is produced in order to serve as the reactant for the next step of the pathway.

−1

/(8.3145 J·K − 1 ·mol − 1)(298 K)

b.

0.133[citrate] = 2 M − [citrate] 1.133[citrate] = 2 M [citrate] = 1.77 M [isocitrate] = 2 M − 1.77 M = 0.23 M

Keq = e −13,800 J·mol

−1

Keq = e −19.2 = 4.4 × 10 −9 [F16BP] Keq = [F6P] [Pi] [F16BP] 4.4 × 10 −9 = [F6P](5 × 10 −3 M) [F16BP] 2.2 × 10 −11 = [F6P] 1

Keq = e − ∆G°ʹ/RT −1

Keq = e − ∆G°ʹ/RT Keq = e −47,700 J·mol

63. a. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): Keq = e −5000 J · mol

[G6P] [glucose] [Pi] 250 × 10 −6 M 0.0038 = [glucose](5 × 10 −3 M) [glucose] = 13 M Keq =

69. GAP ⇌ 1,3BPG I. 1,3BPG + H2O ⇌ 3PG + Pi GAP + H2O ⇌ 3PG + Pi

ΔG°′ = 6.7 kJ · mol–1 ΔG°′ = – 49.3 kJ · mol–1

GAP ⇌ 1,3BPG 1,3BPG + ADP ⇌ 3PG + ATP II. GAP + ADP ⇌ 3PG + ATP

ΔG°′ = 6.7 kJ · mol–1 ΔG°′ = – 18.8 kJ · mol–1

ΔG°′ = – 42.6 kJ · mol–1

ΔG°′ = – 12.1 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.

Odd-Numbered Solutions

71. a. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): Keq = e − ∆G°ʹ/RT Keq = e −31,500 J·mol Keq = e

−12.7

−1

/(8.3145 J·K − 1 ·mol − 1)(298 K)

= 3.0 × 10

−6

[palmitoyl-CoA] Keq = [palmitate] [CoA] [palmitoyl-CoA] 3.0 × 10 −6 = 1 [palmitate] [CoA] The ratio of products to reactants is 3.0 × 10–6:1. The reaction is not favorable. b. Coupling the synthesis of palmitoyl-CoA with ATP hydrolysis to ADP (shown in Table 12.4) produces a standard free energy change of 1.0 kJ · mol–1 for the coupled process [31.5 kJ · mol–1 + (–30.5 kJ · mol–1) = 1.0 kJ · mol–1]. palmitate + CoA + ATP → palmitoyl-CoA + ADP + Pi Keq = e − ∆G°ʹ/RT Keq = e −1000 J·mol

−1

/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e −0.40 = 0.67 0.67 [palmitoyl-CoA] [ADP] [Pi] = 1 [palmitate] [CoA] [ATP] Coupling the synthesis of palmitoyl-CoA with the hydrolysis of ATP to ADP improves the [product]/[reactant] ratio, but the formation of products is still not favored. c. Coupling the synthesis of palmitoyl-CoA with ATP hydrolysis to AMP (shown in Table 12.4) produces a standard free energy change of –14.1 kJ · mol–1 for the coupled process [31.5 kJ · mol–1 + (–45.6 kJ · mol–1) = –14.1 kJ · mol–1]. palmitate + CoA + ATP → palmitoyl-CoA + AMP + PPi Keq = e − ∆G°ʹ/RT Keq = e −(−14,100 J·mol

−1

)/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e5.7 = 296 296 [palmitoyl-CoA] [AMP] [PPi] = 1 [palmitate] [CoA] [ATP] Coupling the synthesis of palmitoyl-CoA with the hydrolysis of ATP to AMP improves the [product]/[reactant] ratio considerably. The formation of products is now favored. d. Coupling the synthesis of palmitoyl-CoA with ATP hydrolysis to AMP and PPi followed by PPi hydrolysis (shown in Table 12.4) produces a standard free energy of –33.3 kJ · mol–1 for the coupled process [–14.1 kJ · mol–1 + (–19.2 kJ · mol–1) = –33.3 kJ · mol–1]. palmitate + CoA + ATP + H2O → palmitoyl-CoA + AMP + 2 Pi Keq = e − ∆G°ʹ/RT Keq = e −(−33,300 J·mol

−1

)/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e13.4 = 6.9 × 105 6.9 × 105 [palmitoyl-CoA] [AMP] [Pi]2 = 1 [palmitate] [CoA] [ATP] Coupling the activation of palmitate to palmitoyl-CoA with the hydrolysis of ATP to AMP, with subsequent hydrolysis of pyrophosphate, is a thermodynamically effective means of accomplishing the reaction. Coupling the reaction with hydrolysis of ATP to ADP is not effective.

S-35

Chapter 13 1. a. Reactions 1, 3, 7, and 10; b. Reactions 2, 5, and 8; c. Reaction 6; d. Reaction 9; e. Reaction 4. 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 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., Weber, I. T., Harrison, R. W., and Bell, G. I., J. Biol. Chem. 269, 21925–21928 (1994). 7. a. Rearrange Equation 12.2 to solve for Keq, as shown in Sample Calculation 12.2, then use the equilibrium constant expression to determine the [F6P]/[G6P] ratio: Keq = e − ∆G°ʹ/RT Keq = e −2200 J·mol

−1

/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e −0.88 [F6P] 0.41 Keq = = 1 [G6P] b. Rearrange Equation 12.3 to solve for the [F6P]/[G6P] ratio: [F6P] ∆G = ∆G°ʹ + RT ln [G6P] [F6P] e(∆G− ∆G°ʹ)/RT = [G6P] [F6P] (−1400 J·mol − 1 −2200 J·mol − 1)/(8.3145 J·K − 1 ·mol − 1)(310 K) = e [G6P] [F6P] 0.25 = e −1.4 = [G6P] 1 The reaction will proceed in the forward direction. 9. 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. 11. 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 ½Vmax. PEP stabilizes the T form of PFK (see Solution 10b). 13. Glycerol can serve as an energy source because it can be converted 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 glycolytic pathway by first being converted to glucose-6-phosphate, then fructose-6-phosphate. The next step, conversion to fructose-1,6-bisphosphate, requires phosphofructokinase. Thus, glycerol is a suitable substrate for this mutant, but glucose is not.

S-36

Odd-Numbered Solutions

15. 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 is cleaved and the first substrate is released. 17. Rearrange Equation 12.3 to solve for the [GAP]/[DHAP] ratio: [GAP] [DHAP] [GAP] = [DHAP]

∆G = ∆G°ʹ + RT ln e(∆G− ∆G°ʹ )/RT e(4400 J·mol

−1

−7900 J·mol − 1)/(8.3145 J·K − 1 ·mol − 1)(310 K)

produce. If glycolysis cannot be completed, the “investment” stage consumes 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. 29. a. Ethanol stimulates vasodilation, a relaxation of the blood vessels that widens the vessel and increases blood flow. This allows heat to escape more readily from the body, leading to hypothermia (a decrease in the core body temperature). b. When ethanol is consumed, the hypothalamus is less able to regulate osmotic pressure and as a result, the kidneys excrete water at a greater rate. Drinking water can replace some of this water and may alleviate symptoms of hangover caused by excessive alcohol consumption. 31. a. Methanol reacts with alcohol dehydrogenase (as ethanol does) to produce formaldehyde (Box 13.B).

=

[GAP] [DHAP]

[GAP] 0.26 = e −1.36 = [DHAP] 1 The ratio of [GAP] to [DHAP] is 0.26:1, which seems to indicate that 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. 19. 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). b. The structure of GAPDH in cancer cells is probably different from the structure of GAPDH in normal cells. The structure of the active site in GAPDH from cancer cells might be altered in such a way that the binding of methylglyoxal is permitted, which then precludes the binding of the substrate. Or 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., Basu, N., and Ray, S., Mol. Cell. Biochem. 177, 21–26 (1997).] 21. 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. 23. Phosphoglycerate kinase catalyzes the conversion of 1,3bisphosphoglycerate 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. 25. 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 position 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 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. 27. Iodoacetate inactivates aldolase, which leads to the accumulation of fructose-1,6-bisphosphate, a metabolite that “costs” 2 ATP to

NAD H3 C

NADH, H

OH

Methanol

O H

alcohol dehydrogenase

C

H

Formaldehyde

b. Administering ethanol is a good antidote because ethanol will compete with methanol for binding to alcohol dehydrogenase and will produce the less harmful acetaldehyde. This allows time for methanol to be eliminated from the system. [From Cooper, J. A. and Kini, M., Biochemical Pharmacology 11, 405–416 (1962).] 33. The glucose–lactate pathway releases 196 kJ · mol–1 of free energy, enough theoretically to drive the synthesis of (196/30.5) × 0.33, or about 2 ATP. 35. 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., Selig, M., Xavier, K. B., Santos, H., and Schönheit, P., Arch. Microbiol. 175, 52–61 (2001).] 37. 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. 39. 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). 41. 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 glucose-6-phosphatase. [In fact, insulin has been shown to suppress the transcription of PEPCK and glucose-6-phosphatase.] 43. 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

Odd-Numbered Solutions

45. 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 44, which helps decrease the level of circulating glucose in patients with diabetes. 47. Deamination of alanine produces pyruvate, a gluconeogenic substrate, and deamination of aspartate produces oxaloacetate, an intermediate of gluconeogenesis. 49. The starch in the grains must be converted to glucose because the yeast use glucose as a substrate for fermentation. 51. 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 to 1. Keq = e − ∆G°ʹ/RT Keq = e −(3100 J·mol

−1

)/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e −1.25 Keq =

0.29 [glycogen (n − 1 residues)] [G1P] = 1 [glycogen (n residues)] [Pi]

0.29 1[G1P] = 1 1[Pi] [Pi] 3.5 = 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: ∆G = ∆G°ʹ + RT ln

[glycogen (n − 1)residues] [G1P] [glycogen (n residues)] [Pi]

= 3100 J·mol −1 + (8.3145 J·K −1 ·mol −1 × 310 K)ln

(1)(1) (1)(50)

= 3100 J·mol −1 + (−10,100 J·mol −1 ) = −7.0 kJ·mol −1 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 hexokinase step and saves a molecule of ATP. Hydrolysis, which produces glucose, would require expenditure of an ATP to phosphorylate glucose to glucose-6-phosphate. 53. 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., Clark, C. P., Gondim, K. C., and Wells, M. A., Insect Biochem. Mol. Biol. 31, 165–170 (2001).] 55. 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 also enter the pentose phosphate pathway.

CH2OPO 32

57. H HO

CH2OPO 2 3

O H

H

OH

O HO

H

O C

O

H

H OH

O

H

C

OH

HO

C

H

H

C

OH

H

C

OH

H OH

H

H

OH

O O H H

H

CH2OPO 32 H O H OH H

O H

H HO

H

O O H

OH

CH2OPO32

59. G16BP inhibits hexokinase but stimulates 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).] 61. The low PFK activity means that glycolysis cannot generate a sufficient amount of the 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, releasing myoglobin into the blood, which subsequently appears in the urine. 63. Hexokinase-deficient erythrocytes have low levels of all glycolytic intermediates, since hexokinase catalyzes the first 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 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. 100 90

Hexokinase deficient

80 Oxygen saturation (%)

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.

S-37

70 60 Normal erythrocytes

50 40 30

Pyruvate kinase deficient

20 10 0

0

10

20

30

40

pO2 (torr)

50

60

S-38

Odd-Numbered Solutions

65. This observation revealed that the pathways for glycogen degradation and synthesis must be different, since a defect in the degradative pathway has no effect on the synthetic pathway. 67. 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 gluconeogenesis. 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., Wyngaarden, J. B., and Fredrickson, D. S., The Metabolic Basis of Inherited Disease, pp. 151–153, McGraw-Hill, New York (1978).] 69. a. Blood glucose concentrations are regulated by pancreatic hormones acting on the liver to stimulate glycogen synthesis or degradation, whatever is appropriate. Since the patient’s liver enzymes appear to function normally (see Problem 67), the blood glucose concentration is properly regulated and the patient is neither hypo- nor hyperglycemic. b. As described in Solution 68, patients with von Gierke’s disease cannot convert glucose-6-phosphate to glucose in the absence of glucose-6-phosphatase, thus these patients are hypoglycemic.

large doses of thiamine might be a successful treatment option if the E1 mutation weakens the interactions between the protein and thiamine. 13. 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. 15. a.

O Br

CH2

C



CH3

CoA

SH O

HBr H3C

C

CH2

S

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 of the inhibitor. The KM increases in the presence of the inhibitor, indicating that the enzyme’s affinity for its substrate decreases in the presence of the inhibitor. The inhibitor competes with acetyl-CoA for binding to the citrate synthase active site. S-acetonyl-CoA can do this because its structure resembles that of acetyl-CoA.

71. The pentose phosphate pathway in the red blood cell generates NADPH, which is used to regenerate oxidized glutathione. Glucose-6-phosphate dehydrogenase catalyzes the first step of the oxidative branch of the pathway. Its deficiency results in a decreased output of NADPH from the pathway. As a result, glutathione remains in the oxidized form and cannot fulfill its roles of decreasing the concentrations of organic peroxides, maintaining red blood cell shape, and keeping the iron ion of hemoglobin in the +2 form. Hemolytic anemia is the likely result.

17. 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. But 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).]

Chapter 14

19. See Sample Calculation 12.2. Keq can be calculated by rearranging Equation 12.2:

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.

Keq = e −∆G°ʹ/RT

3. The purpose of steps 4 and 5 is to regenerate the enzyme. In step 3, the product acetyl-CoA is released, but the lipoamide prosthetic group of E2 is reduced. 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.

Keq = 4.8 × 103

5. Arsenite reacts with the reduced lipoamide group on E2 of the pyruvate dehydrogenase complex to form a compound with the structure shown in the figure. The enzyme cannot be regenerated and can no longer catalyze the conversion of acetyl-CoA to pyruvate. The α-ketoglutarate dehydrogenase complex has a lipoamide group on its E2 subunits and will be inhibited as well. The entire citric acid cycle cannot function, glucose cannot be oxidized aerobically, and respiration comes to a halt, which explains why these compounds are so toxic. 7. In both cases, the activity of the pyruvate dehydrogenase complex decreases, as both NADH and acetyl-CoA are products of the reaction. Rising concentrations of NADH and acetyl-CoA decrease pyruvate dehydrogenase activity by competing with NAD+ and CoASH for binding sites on the enzyme. 9. Ca2+ inhibits PDH kinase and activates PDH phosphatase. In this way, the pyruvate dehydrogenase complex is active and can funnel substrates from glycolysis into the citric acid cycle to provide ATP needed by myosin (see Section 5.4) for muscle contraction. 11. The E1 subunit of the pyruvate dehydrogenase complex requires TPP, the phosphorylated form of thiamine, as a cofactor. Administering

Keq = e −(−21,000 J·mol

−1

) /(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e8.5

21. 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. COO

COO CO2

CH2

CH2

CH2 C

CH2

O

HO

COO

C TPP

-Ketoglutarate Step 2. The succinyl group is then transferred to the lipoamide prosthetic group of E2 of the α-ketoglutarate dehydrogenase complex. COO

COO TPP

CH2

HO

CH2

CH2

CH2

C

C

S

O

S

TPP S R

Lipoamide

HS R

Odd-Numbered Solutions

Step 3. The succinyl group is transferred to coenzyme A, and the lipoamide group is reduced. COO

HS

reduced lipoamide HS

CH2 CH2 C

COO CH2 R

CH2 C

O

S

Coenzyme A

O

CoA

Succinyl-CoA HS R

Steps 4 and 5. The last two steps are the same as for the pyruvate 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. 23. 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). 25. 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. 27. Rearrange Equation 12.3 to solve for the [malate]/[fumarate] ratio: ∆G = ∆G°ʹ + RT ln

[malate] [fumarate]

[malate] [fumarate] [malate] −1 −1 −1 −1 e(0 J·mol +3400 J·mol )/(8.3145 J·K ·mol )(310 K) = [fumarate] [malate] 3.7 = e1.32 = [fumarate] 1 e(∆G−∆G°ʹ )/RT =

The ratio of malate to fumarate is 3.7 to 1, indicating that the reaction proceeds in the direction of formation of malate. This is not a control point for the citric acid cycle because the ΔG is close to zero, indicating it is a near-equilibrium reaction. 29. a. The isotopic label on C4 of oxaloacetate is released as 14 CO2 in the α-ketoglutarate dehydrogenase reaction. b. The isotopic label on C1 of acetyl-CoA is scrambled at the succinyl-CoA synthetase step. Because succinate is symmetrical, C1 and C4 are chemically equivalent, so in a population of molecules, both C1 and C4 would appear to be labeled (half the label would appear at C1 and half at C4). Consequently, one round of the citric acid cycle would yield oxaloacetate with half the labeled carbon at C1 and half at C4. Both of these labeled carbons would be lost as 14CO2 in a second round of the citric acid cycle. 31. 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. 33. During the resting state, citric acid cycle activity is low as a result of the low activities of the two enzymes described. Any 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

S-39

of its inhibitor, so the activity of citrate synthase also increases. 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. 35. 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 will be decreased. b. If the citric acid cycle and mitochondrial respiration are not functioning, the cell turns to glycolysis to produce the ATP required 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 during hyperoxia. [From Allen, C. B., Guo, X. L., and White, C. W., Am. J. Physiol. 274 (3 Pt. 1), L320–L329 (1998).] 37. 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. Construction of the mutant supports this hypothesis. The introduction of the negatively charged Asp residue in place of the Ser residue similarly introduces a negative charge to the active site and prevents isocitrate binding in the same manner. [From Dean, A. M., Lee, M. H. I., and Koshland, D. E., J. Biol. Chem. 264, 20482–20486 (1989).] 39. Because of the enzyme deficiency, the citric acid cycle cannot be completed; in the absence of a functional citric acid cycle, glucose must be oxidized anaerobically. Lactate is a product of anaerobic oxidation of glucose; thus lactate levels are elevated in the patients. The pyruvate → lactate transformation is reversible, so if lactate levels rise due to increased glycolytic 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).] 41. Succinyl-CoA synthetase catalyzes the only substrate-level phosphorylation reaction in the citric acid cycle. The enzyme with the ADP-specific β subunit could produce ATP in the brain and muscle to meet the energy needs of these tissues. The enzyme with the GDP-specific β subunit could produce GTP needed by phosphoenolpyruvate carboxykinase for gluconeogenesis in the liver and kidney. 43. 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. 45. When cells obtain energy in the absence of oxygen, glucose is oxidized to pyruvate, which is subsequently reduced to lactate with concomitant regeneration of NAD+. The citric acid cycle is not active, and the cells respond by down-regulating the activity of the citric acid cycle enzyme malate dehydrogenase. In the presence of oxygen, pyruvate is converted to acetyl-CoA, which enters the citric acid cycle so that ATP can be produced by oxidative phosphorylation. Under aerobic conditions, malate dehydrogenase is absolutely essential for the regeneration of oxaloacetate as part of the citric acid cycle; therefore, activity levels are much higher. 47. The phosphofructokinase reaction is the major rate-control point for the pathway of glycolysis. Inhibiting phosphofructokinase slows the glycolytic pathway, so the production of pyruvate and then acetylCoA can be decreased when the citric acid cycle is operating at maximum capacity and the citrate concentration is high.

S-40

Odd-Numbered Solutions

49. 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. 51. 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. 53. 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, glyceraldehyde-3phosphate, tryptophan, phenylalanine, and pentadecanoate can serve as gluconeogenic substrates because at least one of their breakdown products can be converted to oxaloacetate. Palmitate and 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. 55. 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. 57. 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. 59. 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 acceptor in the electron transport chain that reoxidizes NADH and QH2. 61. 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. 63. 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 adenosine 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). 65. 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. But a varied diet provides ample B vitamins, so supplementation isn’t necessary, and B vitamins are water-soluble so vitamins ingested in excess of what is needed are excreted. 67. a. The bacterial pathway is exclusively biosynthetic; acetyl-CoA carbons are not oxidized to CO2. Isocitrate dehydrogenase is NADP+-dependent rather than NAD+-dependent. H. pylori lacks α-ketoglutarate dehydrogenase and instead has α-ketoglutarate oxidase. The enzyme succinyl-CoA synthetase is missing in H. pylori. This enzyme catalyzes the only substrate-level phosphorylation reaction in the citric acid cycle; therefore, no GTP is produced in the citric acid cycle of this organism. Succinate dehydrogenase is missing. Fumarate reductase is present. A glyoxylate cycle enzyme, malate synthase, is part of this “cycle.” b. Citrate synthase, isocitrate dehydrogenase, and α-ketoglutarate oxidase may serve as regulatory control points since these enzymes catalyze irreversible reactions and may be subject to regulation by allosteric modulators. c. Enzymes unique to H. pylori would be good therapeutic targets: α-ketoglutarate oxidase, fumarate reductase, and malate synthase. [From Pitson, S. M., Mendz, G. L., Srinivasan, S., and Hazell, S. L., Eur. J. Biochem. 260, 258–267 (1999).] 69. Enzymes of the glyoxylate pathway, particularly malate dehydrogenase 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. 71. 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. 73. a. 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

Odd-Numbered Solutions

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.

S-41

Dihydrolipoic acid→lipoic acid + 2 H + + 2 e −

Ɛ°ʹ = 0.29 V

NAD + + H + + 2 e − →NADH

Ɛ°ʹ = −0.315 V

Dihydrolipoic acid + NAD + →lipoic acid + NADH + H + ∆Ɛ°ʹ = −0.025 V Use Equation 15.4 to calculate ∆G°ʹ for this reaction:

Chapter 15 1. a. Oxaloacetate is oxidized; malate is reduced; b. pyruvate is oxidized; lactate is reduced; c. NAD+ is oxidized; NADH is reduced; d. fumarate is oxidized; succinate is reduced (see Table 15.1). 3. Fumarate is the oxidized compound; succinate is the reduced compound; Ɛ°ʹ = 0.031 V (Table 15.1). Use Equation 15.1: Ɛ = Ɛ°ʹ −

[succinate] RT ln nF [fumarate]

Ɛ = 0.031 V −

(8.3145 J·K

−1

∆G°ʹ = −nF ∆Ɛ°ʹ ∆G°ʹ = −(2)(96,485 J·V −1 ·mol −1 )(−0.025 V) ∆G°ʹ = 4800 J·mol −1 = 4.8 kJ·mol −1 13. Consult Table 15.1 for the relevant half-reactions involving acetaldehyde and NAD+. Reverse the acetaldehyde half-reaction and the sign of its Ɛ°ʹ value to indicate oxidation; then combine the half-reactions and their Ɛ°ʹ values. acetaldehyde + H2O →acetate + 3 H + + 2 e −

−1

·mol )(310 K)

(2)(96,485 J·V −1 ·mol −1 ) Ɛ = 0.031 V − (0.0134 V × 0.223) = 0.028 V

ln

(100 × 10

−6

M)

(80 × 10 −6 M)

5. a. Rearrange Equation 15.2 to solve for Ɛ°ʹ, then enter the values given in the problem: [Areduced] 0.026 V ln n [Aoxidized] [Areduced] 0.026 V Ɛ°ʹ = Ɛ + ln n [Aoxidized]

Ɛ = Ɛ°ʹ −

b. A possible identity for substance A is nitrate. 7. Reverse the NADH half-reaction and the sign of its Ɛ°ʹ to indicate oxidation; then combine the half-reactions and their Ɛ°ʹ values. pyruvate + 2 H + + 2 e − →lactate

Ɛ°ʹ = −0.185 V

NADH →NAD + + H + + 2 e −

Ɛ°ʹ = 0.315 V ∆Ɛ°ʹ =

0.130 V

Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF∆Ɛ°ʹ ∆G°ʹ = −(2)(96,485 J·V −1 ·mol −1 )(0.130 V) ∆G°ʹ = −25,100 J·mol −1 = −25.1 kJ·mol −1 The reduction of pyruvate by NADH (Section 13.1) is spontaneous under standard conditions. 9. Reverse the malate half-reaction and the sign of its Ɛ°ʹ to indicate oxidation; then combine the half-reactions and their Ɛ°ʹ values. (Note that ubiquinone is Q and ubiquinol is QH2.)

Q + malate→QH2 + oxaloacetate

Ɛ°ʹ = 0.581 V

−

NAD + H + 2 e →NADH

Ɛ°ʹ = −0.315 V

acetaldehyde + H2O + NAD + →NADH + acetate + 2 H + ∆Ɛ°ʹ = 0.266 V Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF ∆Ɛ°ʹ ∆G°ʹ = −(2)(96,485 J·V −1 ·mol −1 )(0.266 V) ∆G°ʹ = −51,300 J·mol −1 = −51.3 kJ·mol −1

15. 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°ʹ value to indicate oxidation; multiply the coefficients in the cytochrome c1 equation by 2 so that the number of electrons transferred is equal; then combine the half-reactions and their Ɛ°ʹ values.

(5 × 10 −6 M) 0.026 V ln 2 (200 × 10 −6 M) Ɛ°ʹ = 0.47 V + (−0.05 V) = 0.42 V

Q + 2 H + + 2 e − →QH2 malate→oxaloacetate + 2H + + 2 e −

+

The oxidation of acetaldehyde by NAD+ is spontaneous, as shown by the negative ∆G°ʹ value, so yes, NAD+ is an effective oxidizing agent.

Ɛ°ʹ = 0.47 V +

NADH + pyruvate + H + →NAD + + lactate

+

Ɛ°ʹ = 0.045 V Ɛ°ʹ = 0.166 V ∆ Ɛ°ʹ = 0.211 V

Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF∆Ɛ°ʹ ∆G°ʹ = −(2)(96,485 J·V −1 ·mol −1 )(0.211 V) ∆G°ʹ = −40,700 J·mol −1 = −40.7 kJ·mol −1 The reduction of oxidation of malate by ubiquinone is spontaneous under standard conditions. 11. The relevant reactions and their Ɛ°ʹ values are obtained from Table 15.1:

QH2 →Q + 2 H + + 2 e − 3+

−

Ɛ°ʹ = −0.045 V 2+

2 cyt c1 (Fe ) + 2 e →2 cyt c1 (Fe ) 3+

Ɛ°ʹ = 0.220 V 2+

QH2 + 2 cyt c1 (Fe ) →Q + 2 cyt c1 (Fe )

∆Ɛ°ʹ = 0.175 V

Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF ∆Ɛ°ʹ ∆G°ʹ = −(2)(96,485 J·V −1 ·mol −1 )(0.175 V) ∆G°ʹ = −33,800 J·mol −1 = −33.8 kJ·mol −1 The reaction is spontaneous under standard conditions. 17. a. Use Equation 15.2 to determine the Ɛ values for these two half-reactions. QH2 →Q + 2 H + + 2 e − 0.026 V [QH2] Ɛ = Ɛ°ʹ − ln n [Q] 0.026 V ln 10 Ɛ = −0.045 V − 2 Ɛ = −0.075 V

Ɛ°ʹ = −0.045 V

2 cyt c (Fe3+ ) + 2 e − →2 cyt c (Fe2+ )

Ɛ°ʹ = 0.235 V

0.026 V [cyt c (Fe2+ )] Ɛ = Ɛ°ʹ − ln n [cyt c (Fe3+ )] 0.026 V 1 ln Ɛ = 0.235 V − 2 5 Ɛ = 0.256 V QH2 →Q + 2 H + + 2 e −

Ɛ = −0.075 V

2 cyt c (Fe3+ ) + 2 e − →2 cyt c (Fe2+ )

Ɛ = 0.256 V

QH2 + 2 cyt c (Fe3+ ) →Q + 2 H + + 2 cyt c (Fe2+ ) ∆Ɛ = 0.181 V

S-42

Odd-Numbered Solutions

b. Use Equation 15.4 to calculate ΔG for this reaction: ∆G = −nF∆Ɛ ∆G = −(2)(96,485 J·V −1 ·mol −1 )(0.181 V) ∆G = −34,900 J·mol −1 = −34.9 kJ·mol −1 19. Reverse the half-reaction for the iron–sulfur protein to indicate that it is being oxidized. Add the two half-reactions to obtain the ΔƐ°′ for the reaction. FeS (red) →FeS (ox) + e − 3+

−

Ɛ°ʹ = −0.280 V 2+

cyt c1 (Fe ) + e →cyt c1 (Fe ) 3+

Ɛ°ʹ = 0.215 V 2+

FeS (red) + cyt c1 (Fe ) →FeS (ox) + cyt c1 (Fe ) ∆ Ɛ°ʹ = −0.065 V Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF ∆Ɛ°ʹ ∆G°ʹ = −(1)(96,485 J·V −1 ·mol −1 )(−0.065 V) ∆G°ʹ = 6300 J·mol −1 = 6.3 kJ·mol −1 The positive ∆G°ʹ indicates that the electron transfer is unfavorable under standard conditions. However, cellular conditions are not necessarily standard conditions, and the ΔG 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 ∆G°ʹ indicates. 21. Consult Table 15.1 for the relevant half-reactions involving O2 and NADH. Reverse the NADH half-reaction and the sign of its Ɛ°ʹ value to indicate oxidation; then combine the half reactions and their Ɛ°ʹ values. 1 2

O2 + 2 H + + 2 e − →H2O +

+

Ɛ°ʹ = 0.815 V −

Ɛ°ʹ = 0.315 V

+

∆ Ɛ°ʹ = 1.130 V

NADH →NAD + H + 2 e 1 2

+

NADH + O2 + H →NAD + H2O

Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF∆Ɛ°ʹ ∆G°ʹ = −2(96,485 J·V −1 ·mol −1 )(1.130 V) ∆G°ʹ = −218,000 J·mol −1 = −218 kJ·mol −1 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%. 23. The relevant reactions and their Ɛ°ʹ values are obtained from Table 15.1: succinate → fumarate + 2 H + + 2 e −

Ɛ°ʹ = −0.031 V

Q + 2 H + + 2 e− →QH2

Ɛ°ʹ = 0.045 V

succinate + Q →fumarate + QH2

∆ Ɛ°ʹ = 0.014 V

Use Equation 15.4 to calculate ∆G°ʹ for this reaction: ∆G°ʹ = −nF ∆Ɛ°ʹ ∆G°ʹ = −2(96,485 J·V −1 ·mol −1 )(0.014 V) ∆G°ʹ = −2,700 J·mol −1 = −2.7 kJ·mol −1 25. 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 the oxygen, the final electron acceptor. b. In rotenone- or amytal-blocked mitochondria, NADH and Complex I redox centers are reduced while components from ubiquinone on are oxidized. In antimycin A–blocked mitochondria, NADH, Complex I

redox centers, ubiquinol, and Complex III redox centers are reduced while cytochrome c and Complex IV redox centers are oxidized. In cyanide-blocked mitochondria, all of the electron transport components are reduced and only oxygen remains oxidized. 27. a. Adding tetramethyl-p-phenylenediamine to rotenone-blocked and antimycin A–blocked mitochondria effectively bypasses the block as the compound donates its electrons to Complex IV and electron transport resumes. Adding tetramethyl-p-phenylenediamine is not an effective bypass for cyanide-blocked mitochondria because cyanide inhibits electron transport in Complex IV. b. Similarly, ascorbate, which donates its electrons to cytochrome c and then to Complex IV, can act as an effective bypass for antimycin A–blocked mitochondria but not cyanide-blocked mitochondria. 29. Cyanide binds to the Fe2+ in the Fe–Cu center of cytochrome a3 (see Problem 25). 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)

CN

cytochrome a3 (Fe3 Cu) HbO2 (Fe2)

31. All 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. 33. Like the lipids that compose 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 phospholipids that compose the cell membrane. Coenzyme Q literally dissolves in the membrane, which facilitates rapid diffusion. 35. 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 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. 37. 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. 39. 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. 41. Myoglobin functions in muscle cells to facilitate oxygen diffusion throughout the cell (see Problem 39) and possibly assumes this same

Odd-Numbered Solutions

S-43

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. 55. 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., Solís-Muñoz, P., Fernández-Moreira, D., MuñozYagüe, T., and Solís-Herruzo, J. A., BMC Biology 11, 88 (2013).] 57. F0 acts as a proton channel as the c ring rotates, feeding protons through the a subunit (see Fig. 15.23). 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. 59. Since it has 10 c subunits, the bacterial 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. ∆G = 2.303 RT (pHin − pHout ) + ZF∆ψ 61. By decreasing both the rate of electron transport (see Solution ∆G = 2.303(8.3145 J·K −1 ·mol −1 )(310 K)(7.6 − 7.2) 30) and ATP synthesis, fluoxetine decreases the rate of ATP production in the brain. The brain relies on a constant source of ATP + (1)(96,485 J·V −1 ·mol −1 )(0.200 V) for proper function, so decreased ATP production could lead to an −1 −1 −1 ∆G = 2400 J·mol + 19,300 J·mol = 21.7 kJ·mol impairment of brain function. 49. Use the rearranged form of Equation 15.7 shown in Sample 63. A total of 32 ATP are obtained from the exergonic oxidation of Calculation 15.3 and solve for (pHin – pHout); then enter the values glucose under aerobic conditions (see Fig. 14.13), whereas only 2 ATP given in the problem: are produced when glucose is oxidized in the absence of oxygen to ∆G = 2.303 RT (pHin − pHout ) + ZF∆ψ lactate or ethanol (Section 13.1). Organisms that can oxidize glucose in the presence of oxygen have an advantage over anaerobic organisms ∆G − ZF∆ψ pHin − pHout = because these organisms can extract more energy per glucose. This 2.303RT may have been important in evolution. −1 −1 −1 (30,500 J·mol ) − (1)(96,485 J·V ·mol )(0.170 V) pHin − pHout = 65. a. Substrate-level phosphorylation is catalyzed by phosphogly2.303(8.3145 J·K −1 ·mol −1 )(298 K) cerate kinase and pyruvate kinase in glycolysis, and by succinyl-CoA 14,100 J·mol −1 synthetase in the citric acid cycle. b. C6H12O6 + 6 O2 → 6 CO2 + pHin − pHout = = 2.5 5700 J·mol −1 6 H2O. 51. a. The pH of the intermembrane space is lower than the pH of 67. Organic compounds are oxidized by oxygen through the activity of the mitochondrial matrix because protons are pumped out of the mathe electron transport complexes of mitochondria. This activity genertrix, across the inner membrane, and into the intermembrane space. ates the proton gradient that serves as the energy reservoir to drive the The increase in concentration of protons in the intermembrane space addition of Pi to ADP to synthesize ATP. In the absence of the ADP decreases the pH; the deficit of protons in the matrix results in an reactant, ATP synthase is unable to synthesize ATP. As a result of the increase in pH. b. Detergents disrupt membranes. An intact inner mitight coupling between oxidative phosphorylation and electron transtochondrial membrane is required for oxidative phosphorylation to port, electron transport and oxygen consumption also come to a halt. take place. Without an intact membrane, an electrochemical gradient, 69. Locating hexokinase II on the outer mitochondrial membrane is which is the energy reservoir that drives ATP synthesis, cannot be strategic because ATP produced by oxidative phosphorylation is exestablished, and ATP synthesis does not occur. ported from the mitochondrial matrix to the cytosol via the translocase. Hexokinase II is poised to capture one of its reactants—ATP—so the 53. a. K+ phosphorylation of glucose to glucose-6-phosphate (G6P) is especially rapid. This allows the tumor cell to funnel G6P into cellular pathways Nigericin needed to sustain the cell, such as glycolysis and the pentose phosphate pathway. Akt (protein kinase B; see Problem 10.43) also promotes cell growth and survival in other ways. [From Mathupala, S. P., Ko, Y. H., and Pedersen, P. L., Oncogene 25, 4777–4786 (2006).] + K H+ 71. The donation of a pair of electrons to Complex IV results in the Valinomycin synthesis of about 1.3 ATP per atom of oxygen (½ O2; see Solution 22c). MATRIX Therefore, the P:O ratio of this compound is 1.3. 73. a. Aerobic oxidation of glucose yields 32 ATP per glucose INTERMEMBRANE SPACE whereas alcoholic fermentation of glucose by the yeast yields only 2 CYTOSOL

role in tumor cells. The resulting increase in oxygen concentration may allow the tumor cells to oxidize a higher percentage of the available glucose aerobically and thus obtain more ATP per molecule of glucose than if glucose were oxidized anaerobically. [From Kristiansen, G., et al., J. Biol. Chem., 286, 43417–43428 (2011).] 43. 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. These results suggest that even though oxidative damage to mitochondria increases with age (see Box 15.A), exercise can help maintain or increase mitochondrial function. [From Menshikova, E. V., Ritov, V. B., Fairfull, L., Ferrell, R. E., Kelley, D. E., and Goodpaster, B. H., J. Gerontol. A-Biol. 61, 534–540 (2006)]. 45. The free energy change for generating the electrical imbalance is calculated using Equation 15.6: ∆G = ZF∆ψ ∆G = (1)(96,485 J·V −1 ·mol −1 )(0.081 V) ∆G = 7800 J·mol −1 = 7.8 kJ·mol −1 47. Use the rearrangement of Equation 15.7 as shown in Sample Calculation 15.3:

S-44

Odd-Numbered Solutions

ATP per glucose (see Solution 63). 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 consumption of glucose decreases when the cells are exposed to oxygen because fewer glucose molecules must be oxidized to yield the same amount of ATP. b. Both ratios will initially increase, as the citric acid cycle (which does not operate under anaerobic conditions) produces more NADH equivalents for electron transport. The [ATP]/[ADP] 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. 75. 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

NO2

H H

NO2 O

O

O



N

O

N

O

O N

O

O

N

O

O

O N

O

O

N

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. 77. a. ATP synthesis decreases dramatically in the presence of oligomycin, since proton transport is required to rotate the γ subunit of ATP synthase and trigger the sequential conformational changes of the β subunits that catalyze the phosphorylation of ADP to ATP. b. Since oxidative phosphorylation and electron transport are coupled, a decrease in the rate of oxidative phosphorylation also affects the rate of electron transport. If ATP synthesis is not occurring, the proton gradient is not “discharged” and the rate of electron transport decreases. c. A decrease in the rate of electron transport also decreases the rate of oxygen consumption. d. If dinitrophenol is added, the proton gradient is dissipated, or “discharged,” but not in a way that leads to ATP synthesis. Therefore, ATP synthesis still does not occur, but electron transport and oxygen consumption resume, with the free energy of the process released as heat.

79. 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. 81. 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 the normal mice, the cold temperature stimulated the uncoupling protein. As a result, the energy of electron transport was dissipated as heat rather than used to synthesize ATP. This helped the mice maintain normal body temperature. But the UCP1-knockout mice lacked the uncoupling protein and were unable to uncouple oxidation from phosphorylation. Thus, they could not generate “extra” heat and their body temperatures decreased as a result. [From Enerbäck, S., Jacobsson, A., Simpson, E. M., Guerra, C., Yamashita, H., Harper, M.-E., and Kozak, L. P., Nature 387, 90–94 (1997).] 83. 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 in the rate of mRNA degradation would produce the same results). The increased amount of mRNA likely results in an increase in 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., Klein, M., Riesmeier, J. W., Müller-Röber, B., Fleury, C., Bouillaud, F., and Ricquier, D., Nature 389, 135–136 (1997).] 85. 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 lifespan by favoring maintenance over growth, which delays aging (see Box 15.A). 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.

Odd-Numbered Solutions

3. HO H

CH2OH O H OH H H

OH

O H

CH2 HC

O

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 E E E

hc = λ (6.626 × 10 −34 J·s)(2.998 × 108 m·s −1 ) = 680 × 10 −9 m −19 = 2.9 × 10 J = (2.9 × 10 −19 J·photon −1 ) × (6.022 × 1023 photons ·mol −1 )

E = 1.8 × 105 J·mol −1 = 180 kJ·mol −1 7. Multiply Planck’s law (Equation 16.1) by Avogadro’s number (N), then solve for λ: hc ×N λ hcN λ= E (6.626 × 10 −34 J· s)(2.998 × 108 m·s −1 )(6.022 × 1023 mol −1 ) λ= 250 × 103 J·mol −1

E=

λ = 4.8 × 10 −7 m = 480 nm

S-45

21. 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 membrane, which facilitates rapid diffusion. 23. The difference in reduction potential between P680* and P680 is –0.8 V – 1.15 V = –1.95 V. ∆G°ʹ = −nF∆Ɛ°ʹ ∆G°ʹ = −(1)(96,485 J·V −1 ·mol −1 )(−1.95 V) ∆G°ʹ = 188,000 J·mol −1 = 188 kJ·mol −1 25. Use Equation 15.7, as applied in Sample Calculation 15.3. The matrix and the stroma are both in. ∆G = 2.303 RT (pHin − pHout ) + ZF∆ψ ∆G = 2.303 (8.3145 J·K −1 ·mol −1 )(298 K)(3.5) + (1)(96,485 J·V −1 ·mol −1 )(−0.05 V) ∆G = 20,000 J·mol −1 − 4800 J·mol −1 = 15.2 kJ·mol −1 27. Consult Table 15.1 for the reduction potentials of the relevant half-reactions, reversing the sign for the water oxidation half-reaction. H2O → 12 O2 + 2 H + + 2 e − +

+

Ɛ°ʹ = −0.815 V

−

NADP + H + 2 e → NADPH +

H2O + NADP →

1 2 O2

+ NADPH + H

Ɛ°ʹ = −0.320 V +

∆Ɛ°ʹ = −1.135 V

Use Equation 15.4 to calculate ∆G°ʹ : ∆G°ʹ = −nF∆Ɛ°ʹ ∆G°ʹ = −(2)(96,485 J·V −1 ·mol −1 )(−1.135 V) ∆G°ʹ = 219,000 J·mol −1

9. If the synthesis of each ATP requires 30.5 kJ · mol–1, then 5.9 mol ATP (180/30.5) could be synthesized.

Divide by Avogadro’s number to obtain the free energy change per molecule:

11. 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.

219,000 J·mol −1 = 3.6 × 10 −19 J·molecule −1 6.022 × 1023 molecules ·mol −1

13.

A

D

B N

N

N

N C

The central metal ion in chlorophyll a is Mg2+, whereas in heme b 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, branched-chain alcohol. 15. 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. 17. The order of action is water–plastoquinone oxidoreductase (Photosystem II), then plastoquinone–plastocyanin oxidoreductase (cytochrome b6 f ), then plastocyanin–ferredoxin oxidoreductase (Photosystem I). 19. 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.

29. The final electron acceptor in photosynthesis is NADP+. The final electron acceptor in mitochondrial electron transport is oxygen. 31. 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. 33. a. More c subunits means 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. 35. 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. 37. 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.

S-46

Odd-Numbered Solutions

39. The increase in mass comes from carbon dioxide. 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. 41. 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. 43. 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 the 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 carboxylation reaction that activates rubisco. 45. 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. But 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. 47.

NADP+ NADPH Glyceraldehyde 1,3-Bisphospho3-phosphate glycerate ADP ATP 3-Phosphoglycerate

Dihydroxyacetone phosphate

Chapter 17 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 have the highest density. 3. Cholesteryl esters and triacylglycerols are nonpolar; cholesterol and phospholipids are amphipathic. 5. 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. 7. a. HDL remove excess cholesterol from tissues and transport it back to the liver. This prevents the accumulation of cholesterol in vessel walls that leads to atherosclerosis. b. HDL level alone does not indicate the risk of developing atherosclerosis, since the level of LDL, the activity of the LDL receptor, and other factors such as smoking or vessel wall injuries resulting from infection can all influence the likelihood of developing the disease. 9. 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 β and γ subunits dissociate. 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. 11. a. fatty acid + CoA + ATP→acyl-CoA + AMP + PPi ∆G°ʹ = 0 kJ·mol −1 PPi + H2O → 2 Pi ∆G°ʹ = −19.2 kJ·mol −1 fatty acid + CoA + ATP + H2O →acyl-CoA + AMP + 2 Pi ∆G°ʹ = −19.2 kJ·mol −1

STROMA

CYTOSOL

Dihydroxyacetone phosphate

3-Phosphoglycerate ATP

Glyceraldehyde 3-phosphate

NAD+

ADP 1,3-Bisphosphoglycerate NADH

49. 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. 51. The primers should have the following sequences: GTAGTGGGATTGTGCGTC and GCTCCTACAAATGCCATC. 53. 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 this inhibitor 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. The equilibrium constant can be determined by rearranging Equation 12.2 (see Sample Calculation 12.2): Keq = e − ∆G°ʹ/RT Keq = e −(−19,200 J·mol

−1

)/(8.3145 J·K − 1 ·mol − 1)(298 K)

Keq = e7.75 = 2.3 × 103 13. 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. 15. 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 must rely 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. 17. Medium-chain acyl-CoA (4 –12 carbons) accumulates in individuals with MCAD deficiency, since the conversion of fatty acyl-CoA to enoyl-CoA is blocked. Acylcarnitine esters would also accumulate. 19. 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

Odd-Numbered Solutions

(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.  Oxidation Fatty acyl-CoA

Citric acid cycle Succinate

Enoyl-CoA

Fumarate

Hydroxyacyl-CoA

Malate

Ketoacyl-CoA

Oxaloacetate

21. a. Benzoate was produced when the dogs were fed phenylpropionate. b. Phenylacetate was produced when the dogs were fed phenylbutyrate.

 COO

 CH2

CH2

Phenylpropionate

O COO



H 3C

C

CoA

S

Acetyl-CoA Benzoate

CH2

 CH2

CH2

 COO

Phenylbutyrate

O CH2

COO



H3C

C

S

CoA

Acetyl-CoA

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, 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 fatty acyl-CoA. This gives a final total of 112.5 ATP. Note that a C17 fatty acid yields more ATP than palmitate (106 ATP) and less than oleate (118.5 ATP). 27. The patient could be treated with injections of vitamin B12 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. 29. 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. 31. 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. 33. Phase I

Phenylacetate 23. a. Palmitate goes through seven cycles of β oxidation. The first six cycles each produce 1 QH2, 1 NADH, and 1 acetyl-CoA. The seventh cycle produces 1 QH2, 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 acetyl-CoA 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 gives 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. 25. A C17 fatty acid goes through seven cycles of β oxidation. The first six cycles each produce 1 QH2, 1 NADH, and 1 acetyl-CoA. The seventh cycle produces 1 QH2, 1 NADH, 1 acetyl-CoA, and 1 propionyl-CoA. Each QH2 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 QH2 × 1.5 = 1.5 ATP; 3 NADH × 2.5 = 7.5 ATP;

S-47

O Adenosine

O

P

O

P

O

O O

O

O

P

O

O



O ADP C OH

ATP

O HO

P O

O

Pi

O C

biotinyl-enzyme



O

Carboxyphosphate

O

O

C O



NH

HN

O S

(CH2)4

C

Biotinyl-enzyme

NH

(CH2)4

E

S-48

Odd-Numbered Solutions

synthetic pathway are separate proteins. Triclosan actually inhibits the bacterial enoyl-ACP reductase. The enzymes of the mammalian multifunctional enzyme must be arranged in such a way as to preclude the binding of triclosan to the active site of the enoyl-ACP reductase. 45. In mammals, fatty acid desaturation requires O2 for dehydrogenation. Because bacteria use a different enzyme to introduce a double bond, inhibition of that enzyme could block bacterial growth without harming the mammalian host.

O

O C

NH

N

O

O (CH2)4

S

C

NH

E

(CH2)4

Carboxybiotinyl-enzyme Phase II Acetyl-CoA

O

CoAS

Acetyl-CoA enolate

O

CoAS

Malonyl-CoA

C

C

CH2 H

CH2

CH2 C

O

O

C

B:

O 

 

O

O C

N

NH

 O

BH N

NH

O HN

HO H3C

C O

47.

O

CoAS

NH

(CH2)46

CH β

COO− CH α

CH3

49. 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. Oleate and palmitoleate, with a double bond at the 9,10 position, are not essential fatty acids. Linoleate has a second double bond at the 12,13 position and therefore is an essential fatty acid. α-Linolenate has double bonds at positions 9,10, 12,13, and 15,16 and is also essential.

O

O C

Carboxybiotinyl-enzyme

(CH2)25

Biotinyl-enzyme

(CH2)7 O

35. 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. 37. Acetyl-CoA carboxylase generates malonyl-CoA, the substrate for fatty acid synthase. Mice without acetyl-CoA carboxylase are unable to synthesize fatty acids and hence store less fat in their bodies. Malonyl-CoA also inhibits carnitine acyltransferase. In the absence of acetyl-CoA carboxylase to produce this inhibitor, transport of fatty acids into mitochondria cannot be regulated, and mitochondrial β oxidation proceeds continuously. 39. The citrate transporter (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. 41. 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. 43. Mammalian fatty acid synthase differs structurally from bacterial fatty acid synthase; thus, triclosan can inhibit 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

O

C

CH O

(CH2)7

CH CH2

CH

CH

O

CH

CH

C

CH2

CH2

CH

CH

CH

CH

CH

O C

O

(CH2)7

CH

O

(CH2)7 CH CH

(CH2)7

(CH2)4

CH2

(CH2)5

CH3

CH3

CH3

CH3

Oleate

Linoleate

-Linolenate

Palmitoleate

51. 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 malonylCoA. The decarboxylation reaction is accompanied by hydrolysis of a thioester bond, which releases a similar amount of free energy as phosphoanhydride bond hydrolysis. 53. 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. 55. The synthesis of the ketone body acetoacetate does not require the input of free energy (the thioester bonds of 2 acetyl-CoA are

Odd-Numbered Solutions

hydrolyzed; see Fig. 17.16). Conversion of acetoacetate to 3hydroxybutyrate 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. 57. 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. 59. a. Because a thioesterase can reverse the inhibition, the modification must involve the acylation of a cysteine side chain, as shown here. 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 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). 61. 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.) 63. 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. 65. 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. 67. The first enzyme in the pathway shown in Figure 17.19 is choline kinase, which catalyzes the ATP-dependent phosphorylation of choline to produce phosphocholine, which is needed for phosphatidylcholine 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. 69. 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

S-49

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. 71. a. Fumonisin inhibits ceramide synthase. The concentration of the final product, ceramide, decreased, while other lipid synthetic pathways were not affected. The first enzyme in the pathway, serine palmitoyl transferase, was not the target of fumonisin B1 because there was no significant decrease in its product, 3-ketosphinganine. The second enzyme in the pathway, 3-ketosphinganine reductase, was also not a target because if it were, the substrate of this reaction would have accumulated in the presence of fumonisin. Accumulation of sphinganine indicates that ceramide synthase is inhibited because sphinganine cannot be converted to dihydroceramide and instead accumulates. 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., Norred, W. P., Bacon, C. W., Riley, R. T., and Merrill, A. H., J. Biol. Chem. 266, 14486 –14490 (1991).] 73. 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.

Chapter 18 1. The ATP-induced conformational change must decrease Ɛ°′ 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 could not reduce N2. 3. Leghemoglobin, like myoglobin, is an O2-binding protein. Its presence decreases the concentration of free O2 that would otherwise inactivate the bacterial nitrogenase. 5. a.

N2

* *

NH 4

* NO 2

NO 3

*

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. 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. 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 interfere with substrate binding.

S-50 9.

Odd-Numbered Solutions

O

a. H

3N

O

C

CH2

COO

Glu

C

-Ketoglutarate O

b. H

3N

b. α-ketoglutarate + NH +4 + NAD(P)H →Glu + H2O + NAD(P)+ Glu + α-keto acid→α-ketoglutarate + amino acid

CH2

α-keto acid + NAD(P)H + NH +4 →NAD(P) + + amino acid + H2O

CH2

CH2

19. In the glutamine synthetase reaction, ATP donates a phosphoryl group to glutamate, which 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 instead of ADP and phosphate as in the glutamine synthetase reaction.

NH NH 2

NH 2

C

NH2 O

c.

COO

Glu O

C

CH

C

-Ketoglutarate

CH2

O

21. a. 3-Phosphoglycerate is derived from glycolysis, illustrating the importance of glycolytic intermediates in amino acid biosynthesis. b. The second reaction of the serine biosynthetic pathway is catalyzed by a transaminase.

CH2

OH

OH O

d. 3N

O

CH2

NH2

H

NH4+ + ATP + NADPH + α-keto acid→ ADP + Pi + NADP + + amino acid

CH2

C

3N

Glu + α-keto acid→α-ketoglutarate + amino acid

C

-Ketoglutarate

NH

H

Gln + α-ketoglutarate + NADPH →2 Glu + NADP +

COO

O

CH2

Glu + ATP + NH +4 →Gln + Pi + ADP

O

H

Glu

C

CH

17. a.



C

CH

COO

Glu O

C

-Ketoglutarate

CH2

23. 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.

O

CH2

25. Pyruvate

11. 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.

3-Phosphoglycerate

COO

O H

3N

CH

C

CH2 COO

Aspartate

O

C 

CH2

C

AST

CH2 COO

-Ketoglutarate

O

CH2

Ser

Gly

Oxaloacetate -Ketoglutarate

transamination

Glu

Gln synthetase

transamination

Gln

H



COO

Oxaloacetate

3N

CH

C

O

Asp

Asn synthetase

O

COO

O

Ala

Cys

13. a. leucine; b. methionine; c. tyrosine; d. valine. 15. a. Both reactions are reversible. The ALT reaction is shown in Section 18.2; the AST reaction is shown below.

transamination

Asn

Glu

Arg

CH2 CH2

Pro

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, can react 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 text). Pyruvate can then enter gluconeogenesis. [From Botros, M. and Sikaris, K. A., Clin. Biochem. Rev. 34, 117–130 (2013).]

27. Cysteine is the source of taurine. The cysteine sulfhydryl group is oxidized to a sulfonic acid, and the amino acid is decarboxylated. 29. 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. 31. a. Tyrosine is not an essential amino acid because it can be synthesized from phenylalanine. As long as the diet contains sufficient amounts of phenylalanine, tyrosine is not essential. b. Phenylalanine hydroxylase catalyzes the transformation of phenylalanine to tyrosine, so if this enzyme were not functional, tyrosine would become essential and would need to be obtained in the diet.

Odd-Numbered Solutions

33. a. The values of KM and Vmax are shown in the table. Thr only Thr + Ile –1

–1

Vmax (μmol · mg · min ) KM (mM)

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 KM indicate that isoleucine, the end product of the pathway, acts as a negative allosteric inhibitor of the enzyme that catalyzes an early, committed step of its own synthesis. The velocity versus substrate concentration curve obtained for threonine 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 threonine 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).] 35. 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 gluconeogenesis. Acetyl-CoA can enter the citric acid cycle, be converted to acetoacetate, or be used for fatty acid synthesis. Acetoacetate is a ketone body and can be converted to acetyl-CoA for the citric acid cycle or fatty acid synthesis. 37. 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. 39. Threonine catabolism yields glycine and acetyl-CoA. 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 methylenetetrahydrofolate 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. [From Wang, J., Alexander, P., Wu, L., Hammer, R., Cleaver, O., and McKnight, S. L., Science 325, 435–439 (2009).] 41. 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. 43. a. Acetyl-CoA can enter the citric acid cycle if sufficient oxaloacetate is available; if not, excess acetyl-CoA is converted to ketone bodies. Leucine differs from isoleucine in that leucine is exclusively ketogenic, generating a ketone body and a ketone body precursor upon degradation. Isoleucine produces propionyl-CoA along with

S-51

acetyl-CoA; the former can be converted to succinyl-CoA (see Solution 41) and then to glucose. Thus, isoleucine is glucogenic as well as 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 Figure 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. 45. 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 to the active tetrameric 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 must lead to phosphorylation of the enzyme. d. Phenylalanine hydroxylase is most active when the glucagon concentration is high, corresponding to the fasting state. Under these circumstances, phenylalanine can be degraded to produce acetoacetate (a ketone body) and fumarate (which can be converted to glucose); both compounds provide necessary resources in the fasting state. 47. 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. 49. 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 all generate free ammonia. 51. 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. 53. 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. 55. 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 will enter the urea cycle, and the presence of arginine under these conditions ensures that sufficient ornithine will be available to react with the carbamoyl phosphate. 57. a. Acetylation of the lysine side chain removes the positive charge on the ε-amino group, as shown below. b. OTC catalyzes the condensation of carbamoyl phosphate with ornithine to form citrulline. The phosphate group on carbamoyl phosphate and the

S-52

Odd-Numbered Solutions

carboxylate group on ornithine are both negatively charged; one of these groups may form an ion pair with the positively charged ε-amino group. If the lysine is acetylated, the positive charge is removed and the ion pair cannot form. c. The glutamine side chain has an amide functional group that structurally resembles the acetylated Lys side chain. If the hypothesis in part b is correct, the mutant enzyme should have lower affinity for the substrate (this is in fact the case). [From Xiong, Y. and Guan, K.-L., J. Cell. Biol. 198, 155–164 (2012).] O HN

CH

C

(CH2)4 NH C

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 salvage pathway. Only hybridomas that result from the fusion of a lymphocyte (which can carry out the salvage pathway) and a myeloma cell (which can divide in culture indefinitely) will survive in HAT medium. 75. 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). 77. 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.

O

CH3

N-acetyl-Lys 59. 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. 61. a. A urea cycle enzyme deficiency decreases the rate at which nitrogen can be eliminated as urea. Since the sources of nitrogen for urea synthesis include free ammonia, low urea cycle activity may lead to high levels of ammonia in the body. b. A low-protein diet might reduce the amount of nitrogen to be excreted. 63. 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. 65. Glutaminase: glutamine→glutamate + NH +4 Glutamate dehydrogenase: glutamate + NAD(P) + →α-ketoglutarate + NH +4 + NAD(P)H net: glutamine + NAD(P)+ →α-ketoglutarate + 2 NH +4 + NAD(P)H 67. a. H. pylori urease converts urea to NH3 and CO2. The ammonia has a pK of 9.25, so it combines with protons to produce NH +4 . The resulting decrease in hydrogen ion concentration helps the bacteria maintain an intracellular pH higher than the environmental pH. b. Urease on the cell surface increases the pH of the fluid surrounding the cell, creating a more hospitable microenvironment for bacterial growth. 69. 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. 71. Inhibiting HGPRT would block 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 relies on salvage reactions using the host cell’s hypoxanthine. 73. Lymphocytes that have not fused with a myeloma cell are unable to use the de novo synthetic pathway because it is blocked. These cells are still able to use the HGPRT salvage pathway, but the cells

Chapter 19 1. The two main metabolites at the “crossroads” are pyruvate and acetyl-CoA. Pyruvate is the main product of glycolysis. It can be converted to acetyl-CoA by pyruvate dehydrogenase. Pyruvate is produced from a transamination reaction involving alanine. Pyruvate can be carboxylated to oxaloacetate for gluconeogenesis. Acetyl-CoA is a product of fatty acid degradation and one of the reactants in the citric acid cycle. Acetyl-CoA is a product of the degradation of ketogenic amino acids. Acetyl-CoA can be used to synthesize fatty acids and ketone bodies. 3. 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 in order to maximize ATP production. 5. 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). 7. 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 can also serve as the source of the NADPH required for fatty acid biosynthesis, via the pentose phosphate pathway (see Section 13.4). 9. 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. 11. 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. 13. 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

Odd-Numbered Solutions

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 why the patient suffers from lactic acidosis. Decreased oxaloacetate levels decrease 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 to cause ketosis. c. Acetyl-CoA stimulates pyruvate carboxylase activity (see Section 14.4). Adding acetyl-CoA would allow the investigators to determine whether there was a slight amount of pyruvate carboxylase activity that could be detected by adding this activator. 15. 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., Ogier, H., Marsac, C., Munnich, A., Charpentier, C., and Saudubray, J. M., Pediatrics 68, 914 (1981).] 17. Fatty acid oxidation must be a major source of metabolic free energy during metamorphosis. 19. The KM for hexokinase is about 0.1 mM, which is lower than the fasting blood glucose concentration. 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. 21. 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 additional components of the signaling pathway. These interactions could not occur if a tyrosine phosphatase removed the phosphoryl groups attached to the Tyr residues. 23. 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. 25. 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. 27. a. AMPK increases the expression of GLUT4, which increases ATP production via glucose catabolism. b. AMPK decreases expression of the gluconeogenic enzyme glucose-6-phosphatase, since gluconeogenesis consumes cellular ATP. 29. a. inhibit; b. stimulate. 31. 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. 33. 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 phosphorylase

S-53

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. A deficiency in the muscle glycogen phosphorylase enzyme (Type V) is the most likely explanation. 35. Phosphorylase kinase is regulated by phosphorylation, which causes a conformational change that activates the enzyme. The enzyme is phosphorylated by protein kinase A (Figure 19.11), so the activity of the enzyme therefore depends somewhat on the G protein–coupled receptor pathway activated by either glucagon or epinephrine. But 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. 37. These two enzymes are part of the gluconeogenic pathway. Their concentrations increase when dietary fuels are not available so that the liver can supply other tissues with newly synthesized glucose. 39. If 3000 g of fat are utilized at a rate of 75 g/day, the fast can last for 40 days before death occurs. 41. After a few days of a diet low in carbohydrate, glycogen stores are depleted and the liver converts fatty acids to ketone bodies to be used as fuel for muscle and other tissues. Acetone is produced from the nonenzymatic decarboxylation of the ketone body acetoacetate. The relatively nonpolar acetone passes from blood capillaries into the lung alveoli, and its smell can be detected in exhaled breath. 43. 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; the result is that cellular cAMP concentration decreases. 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. 45. Phosphorylation and inactivation of GSK3 indirectly results in the activation of glycogen synthase, as described in Solution 25. Promoting the storage of glucose as glycogen in muscle and liver reverses the symptoms of insulin resistance. 47. a. The cytochrome c content is high because of the large number of mitochondria that allow brown adipose tissue to oxidize metabolic fuels aerobically, funneling the reduced coenzymes through the electron transport chain. Uncoupling oxidative phosphorylation permits the energy of electron transport to be dissipated as heat (see Box 15.B). b. When subjects were exposed to cold, uptake of the labeled glucose into brown adipose tissue increased to provide reduced coenzymes for electron transport and thermogenesis. 49. 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.

S-54

Odd-Numbered Solutions

51. 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 50, 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 have to be targeted in such a way that it would be delivered to the mitochondrial matrix, where ACC2 is located. [From Abu-Elheiga, L., Matzuk, M. M., Abo-Hashema, K. A. H., and Wakhil, S. J., Science 291, 2613–2616 (2001).] 53. The drugs can activate the intracellular tyrosine kinase domains of the insulin receptor, bypassing the need for insulin to bind to the receptor. 55. a. In the absence of insulin, the cells of the diabetic are not able to take up glucose, which results in hyperglycemia following a meal. If glucagon is also present, it stimulates the liver to break down glycogen, leading to the release of glucose from the liver. The lack of insulin and the excess glucagon both contribute to the hyperglycemia experienced by the diabetic. b. Eliminating the Asp at position 9 results in an analog with decreased affinity for the receptor and with little biological activity, indicating that the Asp plays a role in both binding and signal transduction. Replacing the Asp with a positively charged Lys decreases the binding affinity by about half but completely eliminates the biological activity. The Asp evidently plays an important role in binding, but conservation of the negative charge does not seem to be critical, since a positive charge does not abolish binding. Hence some other aspect of the Asp side chain structure is important for binding. c. Removing the positive charge at position 12 greatly decreases binding affinity. But once the Lys12 → Glu12 analog binds, it is capable of eliciting a biological response. The addition of a negative charge at position 12 virtually abolishes binding, so it’s possible that the positively charged group at position 12 forms an ion pair with a negatively charged side chain on the glucagon receptor. d. The des-His1 glucagon has both decreased binding affinity and biological activity, indicating that the histidine at position 1 is important for binding but plays a greater role in signal transduction. The des-His1des-Asp9 analog does not bind well (only 7% of the control) and has no biological activity. Interestingly, the des-His1-Lys9 derivative binds well (70%) but has no biological activity. This indicates that the replacement of aspartate by lysine at position 9 preserves characteristics that are important for binding. However, once bound, the analog does not trigger signal transduction. e. The des-His1-Lys9 is the best antagonist because it binds to the receptor with 70% of the affinity of the native hormone but has no biological activity. In this derivative, the His residue important for activity has been deleted and the Asp9 residue has been changed to Lys. The Lys residue at position 12, which is critical for binding, is retained. [From Unson, C. G., et al., J. Biol. Chem. 266, 2763–2766 (1991); and Unson, C. G., et al., J. Biol. Chem. 273, 10308–10312 (1998).] 57. 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 would increase the concentration of fructose-2,6bisphosphate and would therefore stimulate glycolysis and inhibit

gluconeogenesis. The increase in glucose utilization and decrease in glucose production would lower the level of glucose in the blood in the diabetic patient. [From Hardie, D. G., Hawley, S. A., and Scott, J. W., J. Physiol. 574, 7–15 (2006).] 59. a.

H HO

CH2OH O H OH H H

ADP H OH

H

2-Deoxy-D-glucose

ATP

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. 61. 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 glutamine can be deaminated to form glutamate, which is the precursor for the synthesis of a number of 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. 63. 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., Locasale, J. W., Swanson, K. D., Sharf, H., Heffron, G. J., AmadorNoguez, D., Cristofk, H. R., Wagner, G., Rabinowitz, J. D., Asara, J. M., and Cantley, L. C., Science 329, 1492–1499 (2010).] 65. 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. 67. 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. 69. 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.

Chapter 20 1. Parental 15N-labeled DNA strands are shown in black, and newly synthesized 14N DNA strands are shown in gray. The original

S-55

Odd-Numbered Solutions 15

N-labeled parental DNA strands persist throughout succeeding generations, but their proportion of the total DNA decreases as new DNA is synthesized.

O H N

C

CH CH2 CH2

15N

N-acetyl lysine

CH2

DNA (heavy)

CH2 HN

C

CH3

O

+

First generation

Second generation

+

3 4

Third generation

Fourth generation

7 8

Hybrid DNA

+

+

+

+

14N

DNA (light)

9. The nucleotide lacks the free 3ʹ hydroxyl group that serves as the attacking nucleophile for an incoming dNTP. Removing the azidomethyl group at the end of each reaction cycle generates a 3ʹ OH group that can support nucleotide addition in the next reaction cycle. 11. PPi is the product of the polymerization reaction catalyzed by DNA polymerase. This reaction also requires a template DNA strand and a primer with a free 3ʹ end. a. There is no primer strand, so no PPi is produced. b. There is no primer strand, so no PPi is produced. c. PPi is produced. d. No PPi is produced because there is no 3ʹ end that can be extended. e. PPi is produced. f. PPi is produced. 13. 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). 15. 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 will increase accuracy.

1 4

1 8

3. a. Yes. By moving along a single DNA strand, the helicase can act 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 up to five 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., Narayan, M., and Patel, S. S., J. Mol. Biol. 321, 807–819 (2002).] 5. a. DNA replication (and hence bacterial growth) halts immediately at the nonpermissive temperature because the DNA cannot be unwound ahead of the replication fork in the absence of the helicase. b. Bacterial growth slows and then stops because the role of DnaA is to locate the replication origin (see Problem 2). When the temperature shifts to the nonpermissive temperature, DNA replication already underway is not affected (those cells can complete cell division), but another round of replication cannot begin in the absence of functioning DnaA. 7. Acetic anhydride reacts with the ε-amino group of lysine side chains to produce N-acetyl lysine. This abolishes the positive charge that the SSB protein requires to bind to negatively charged DNA:

17. First, the cell contains roughly equal concentrations of the four deoxynucleotide substrates for DNA synthesis; this minimizes the chance for an overabundant dNTP to take the place of another or for the wrong dNTP to take the place of a scarce dNTP. Second, DNA polymerase requires accurate pairing between the template base and the incoming base. Third, the 3ʹ→5ʹ exonuclease proofreads the newly formed base pair. Fourth, the removal of the RNA primer and some of the adjacent DNA helps minimize errors introduced by primase and by the DNA polymerase at the 5ʹ end of a new DNA segment. Finally, DNA repair mechanisms can excise mispaired or damaged nucleotides. 19. The RNase most likely detects the geometry of the polynucleotide chain, which shifts from the wide and shallow A-form where RNA is present to the narrower and steeper B-form of DNA. 21.

O E

Lys

NH2N R

O

O

P

O



O

O NAD

1



NMN

O E

Lys



NH2

P O

O

R

P

A

O

R

A

S-56

Odd-Numbered Solutions

NH2

31.

O

H

P

OH

2

O

N

O

O

O N

O N

O

E

O

Lys

N

NH2

P OH O O O P  O R A O

O

H

b.

H N

O

O

23. a. DNA polymerase; b. reverse transcriptase or telomerase; c. primase or RNA polymerase.

N

25. a.

N

H

N

N

H H

N

N H

H H

N

N H

b.

O

O

N

N H H

N H H

N

N

N

H N

N

H

N O

N

Cytosine

c. An A:T base pair is converted to a C:G base pair.

N

O

N

Hypoxanthine

N

H

O

H N

N

Hypoxanthine

P

N

Adenine

NH

N

O

N

H

O

33. a.

AMP

N

N

8-Oxoguanine

N

H

N

H

35. The structures of the intercalating agents resemble A:T and G:C base pairs, which explains why they are able to 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). 37. Bromouracil causes an A:T to G:C transition.

T T G G G A T

A

G G

A G G G T

O

T G

Br

N

O

H

T

G

N

N

H

3

O

N

N

H

O Br

NH

N

H N

N

N

G

O

G

N H

A

5-Bromouracil (keto tautomer)

T

N H

O

5-Bromouracil (enol tautomer)

T

H

39. a. 27. 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. 29. a. transversion; b. transversion; c. transversion; d. transition; e. transversion; f. transition.

N H

N

N N N O Cytosine

H

N

N H

2-Aminopurine

N

H

N

H Guanine

Odd-Numbered Solutions

b.

DNA synthesis. This may result in the loss of control of the cell cycle, converting a non-cancerous cell to a cancerous cell. [From Giacinti, C. and Giordano, A., Oncogene 25, 5220–5227 (2006).]

H N H N

H

N O

H



N

N N

N

N

H

N H N H O

57. 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. 59. 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.

H

N

S-57

N

N H

N

N

N

H [From Sowers, L. C., Boulard, Y., and Fazakerley, G. V., Biochemistry 29, 7613–7620 (2000).] 41. During the course of the reaction, a methyl group from the O6-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. 43. Most likely the thymine–thymine dimer, since this lesion forms upon exposure of the DNA to ultraviolet light. 45. All of these deaminations produce bases that are foreign to DNA; therefore, they can be quickly spotted and repaired before DNA has replicated and the damage is passed on to the next generation. 47. 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. 49. a. The human genome contains 3.083 × 109 bp, which means that every cell contains twice this much DNA to be replicated. An error rate of 1/22,000 would generate (2)(3.083 × 109)/(22,000) = 138,100 errors. b. Reducing the error rate 100-fold would reduce the number of errors to about 1381. 51. 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. 53. Without functional DNA repair enzymes, additional mutations may arise in genes that are involved regulating cell growth. In the absence of proper growth controls, cells may begin to proliferate at an accelerated rate. 55. Normally, the Rb protein acts as a tumor suppressor by preventing the cell from synthesizing DNA, a prerequisite 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

61. 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 made, so the cell relies less on aerobic respiration as a source of ATP and relies more on glycolysis. 63. 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. 65. 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. 67. a. In all of the variants, a neutral or negatively charged amino acid is replaced by a positively charged amino acid (Lys or Arg). It’s reasonable to hypothesize that an enzyme with an increased number of positive charges might bind more tightly to the negatively charged DNA backbone. b. Intact, double-stranded DNA has a lower absorbance at 260 nm than does single-stranded DNA (see Fig. 3.7). An increase in absorbance at 260 nm over time is a useful measurement of the catalytic activity of DNase, since the products of the reaction are short, single-stranded oligonucleotides. c. All of the variants have lower KM values and higher Vmax values than the wild-type DNase, indicating that the variants catalyze the reaction more efficiently than the wild-type enzyme. The lower KM values indicate tighter binding, possibly due to the formation of ion pairs between the variant enzymes and the DNA. The replacement of three amino acids yields a variant enzyme in which the KM and Vmax values are optimized. d. The plasmid DNA normally exists in a supercoiled circle, as shown in the control lane. The wild-type DNase nicks the DNA on one strand to convert the plasmid to a relaxed circular DNA. The E13R/N74K/ T205K (+3) mutant cleaves supercoiled DNA better than the E13R/ T14K/N74K/T205K (+4) mutant. Both mutants produce linear DNA, whereas the wild-type DNase does not. This indicates that the variants can cut both strands, whereas the wild-type enzyme cuts only one strand.

S-58

Odd-Numbered Solutions

69. 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.

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 Section 18.2).

71. 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. 73. The high degree of sequence conservation from cows to peas indicates that the sequence of the histone H4 protein is so vital to its function that amino acid substitutions, especially those that are nonconservative in nature, would disrupt the function of the protein and thus cannot be tolerated. 75. a. About 146 bp of DNA are wrapped around a histone octamer to form a nucleosome. The nucleosomes are separated by about 20–40 bp of DNA. The “linker” DNA is more susceptible to hydrolysis, so brief treatment with the nuclease yields DNA fragments about 200 bp long. b. If the nuclease treatment is prolonged, the entire stretch of “linker” DNA is digested, leaving only the 146 bp segment of DNA that is wrapped around the histone octamer and thus protected from digestion.

Chapter 21 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 groups. Electrostatic interactions involve the ionic phosphate groups of the DNA backbone and are therefore sequence independent. 5. a. b. c. CH2

CH2

CH2

CH2

O O

CH2

N

P

O

N O

O

CH2

O

O

NH C

P

O

CH3

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. 7.

O NH

CH

C

O NH

CH

C

CH2

CH2

CH2

CH2

CH2

CH2

CH2

NH

NH 2

C

CH3

NH2



N H

CH3

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

COO⫺ ⫹

H3N

C

H

CH2 CH2 SH

b. The other product is methanol, CH3OH. 13. a.

O H

O

NH2

C

N N

HO

C

N N

O

R

5-Formylcytosine

NH2

O

R

5-Carboxylcytosine

b. A DNA glycosylase recognizes the oxidized base and an endonuclease then hydrolyzes the phosphodiester backbone. DNA polymerase then fills in the gap, and ligase seals the nick (see Fig. 20.18). 15. 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. AAAATAAATGCTTGACTCTGTAGCG+1

GGAAGGCGTATTATCCAACACCC 17. Affinity chromatography takes advantage of the ability of the desired protein to bind to a specific ligand. 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).] 19. Upregulation of the methylase increases the extent of methylation of Lys 4 (K4) and Lys 27 (K27) in histone 3. Since H3K4me3associated genes normally tend to be in transcriptionally active chromatin (see Table 21.2), these genes are hyperactivated in cancer cells. The opposite occurs with H3K27me3-associated genes, which are normally transcriptionally inactive and are hypersilenced in cancer cells. [From Stark, G. R., Wang, Y., and Lu, T., Cell Res., 21, 375–380 (2011).] 21. 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: ∆G°ʹ = −RT ln (1/Kd ) ∆G°ʹ = −(8.3145 J·K −1 ·mol −1 )(298 K) ln (1/315 × 10 −9 ) ∆G°ʹ = −37,000 J·mol−1 = −37 kJ·mol −1

Odd-Numbered Solutions

For the eight-base bulge, ∆G°ʹ = −RT ln (1/Kd ) ∆G°ʹ = −(8.3145 J·K −1 ·mol −1 )(298 K) ln (1/0.0013 × 10 −9 ) ∆G°ʹ = −68,000 J·mol−1 = −68 kJ·mol −1 Polymerase binding to the eight-base bulge DNA (a mimic of melted 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 ΔG°′ values for polymerase binding to double-stranded DNA and to the eight base bulge is – 68 – (–37) = –31 kJ · mol–1. This value estimates 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).] 23. 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. 25. 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. 27. 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 cleave 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 phenyl-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. 29. 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. 31. 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. 33. 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. 35. The 5′ end of any prokaryotic RNAs ending in G will be labeled. The 5′ ends of RNA transcripts have 5′ triphosphate groups containing the labeled γ-phosphate. The phosphodiester bonds in the RNA will not be labeled because these phosphate groups come from the α-phosphates of the nucleoside triphosphates (the γ-phosphates are released as pyrophosphate). *pppG + pppN ⇌ *pppGpN + PPi 37. The C-terminal domain (CTD) is phosphorylated on multiple serine residues when RNA polymerase transitions to elongation mode.

S-59

The many negatively charged phosphate groups cause charge–charge repulsions that push this domain away from the globular domain of the RNA polymerase as well as away from the negatively charged DNA. 39. a. 5′ . . . NNAAGCGCCGNNNNCCGGCGCUUUUUUNNN . . . 3′ b.

N N G C C G C G A A 5․․․NN

N

N C C G G C G C U U UUUUNNN․․․3

41. 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. 43. a.

Product absent Repressor RNA polymerase

transcription

DNA

Operator Operon

Product present

RNA polymerase no transcription Product

Repressor

b.

Product absent mRNA translation DNA

Operon Product present Product RNA-binding protein mRNA terminate transcription or block translation DNA

S-60

Odd-Numbered Solutions

c. If no protein were involved, the operon’s product would have to interact directly with the mRNA. 45. 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.

G A C G A C G A C CAG

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. 59. 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 lariat-shaped intermediate (see Fig. 21.23) requires a segment of RNA long enough to curl back on itself without strain. 61.

5′ splice site ↓ . . . CCCTGGGCAGGTTGGTA . . .

C

3′ splice site ↓ . . . TTTCCCACCCTTAGGCTGCT . . .

A G C A G C A G CAG

[From Fabre, E., Dujon, B., and Richard, G.-F., Nuc. Acids Res. 30, 3540–3547 (2002).] 47. 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). 49. 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

51. 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 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. 53. a. The cap is an NAD+ group (see Fig. 3.2). b. If the capping enzyme preferentially uses NAD+ over NADH, or if the capping group was accessible to redox enzymes that used NAD+ or NADH as a cofactor, the oxidation state of the cap could reflect the cell’s overall redox balance, and the capped RNA might interact differently with proteins that recognize an NAD+ cap than with proteins that recognize an NADH cap. c. Because the nicotinamide nucleotide is linked by a 5′–5′ bond, it is probably resistant to most cellular nucleases. 55. 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. 57. 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

63. a. Mutating U69 to G or C dramatically decreases activity as shown by the nearly 50-fold decrease in the value of 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 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 structure in proteins refers to the overall three-dimensional shape of the macromolecule; similarly for the ribozyme, the tertiary structure refers to the threedimensional shape of the molecule. In this study, changing the identity of a base (or adding a base) altered the primary structure of the RNase P in a way that changed the secondary and tertiary structure of the ribozyme. These changes affected substrate binding and catalysis to different extents. [From Kaye, N. M., Zahler, N. H., Christian, E. L., and Harris, M. E., J. Mol. Biol. 324, 429–442 (2002).] 65. 5′ . . . G G A G U A C C C U G A U G A G A U C . . . 3′ 3′ . . . C C U C A U G G G A C U A C U C U A G . . . 5′ [From Takei, Y., Kadomatsu, K., Yuzawa, Y., et al., Cancer Res. 64, 3365–3370 (2004).] 67. The mRNA from a gene may be alternatively spliced to yield several different types of proteins. This increases the diversity of the proteins produced by the cell without a correspondingly large number of genes.

Chapter 22 1. A hypothetical quadruplet code would have 44, or 256, possible combinations. 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. 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.

Odd-Numbered Solutions

7. 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. Two of the six reading frames do not contain stop codons and are therefore open reading frames. R * A FQ H R L V R L R A T D E P FS T A* *GCA P M S L SA P L S E V A R H R GAQP H * A V L K G S S V A RNLT K R C * K A H WRA T S L S G AE 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 tetrapeptide 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 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.

S-61

assist the 23S rRNA in forming the necessary three-dimensional 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. 35. 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. 21.19). 37. Process

Replication

Substrates dATP, dCTP, dGTP, dTTP

Product

Transcription

Translation

ATP, CTP, GTP, UTP

20 different amino acids, each linked to a tRNA

two identical a single-stranded a polypeptide double helices RNA molecule chain

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

Primer

RNA primer

no primer needed

19. a. His; b. Asn; c. Thr [Source: tRNAdb, http://trna.bioinf.unileipzig.de/DataOutput/].

Met attached to an initiator tRNA

Enzyme

21. 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 important for specifying an amino acid (see Table 22.1), wobble at the third position may not affect translation.

DNA polymerase

RNA polymerase ribosomal peptidyl transferase (rRNA)

Cellular nucleus location

15. Gly and Ala; Val and Leu; Ser and Thr; Asn and Gln; Asp and Glu. 17. 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.

23. CUG, CUA, and CUU. [From Sørensen, M. A., et al., J. Mol. Biol. 354, 16 –24 (2005).] 25. 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 was 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.

nucleus

cytosol

39. The polypeptide would contain all lysine residue with asparagine at the C-terminus. If the mRNA were read from 3′ → 5′, the polypeptide would consist of an N-terminal glutamine followed by a series of lysine residues. 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. 41. 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′

27. 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.

43. 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.

29. 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. 31. Ribosomal inactivating proteins catalyze the removal of adenine residues from ribosomal RNA. This is analogous to removing a side chain from an amino acid residue in a protein. Like proteins, ribosomal RNAs have specific residues that are essential to their function; removing these residues causes loss of activity.

45. 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.

33. The peptidyl transferase activity lies entirely within the 23S rRNA; that is, 23S rRNA is a ribozyme. The proteins might be necessary to

47. The correctly charged tRNAs (Ala–tRNAAla and Gln–tRNAGln) bind to EF-Tu with approximately the same affinity, so they are delivered

S-62

Odd-Numbered Solutions

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–tRNAAla 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. 49. 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. 51. 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. 53. 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 GTP-hydrolyzing proteins are indicated in parentheses): Aminoacylation (AARS) Translation initiation (IF-2) Positioning of each aminoacyl–tRNA (EF-Tu) Translocation after each transpeptidation (EF-G) Termination (RF-3)

2 × 20 ATP 1 GTP 19 GTP 19 GTP 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. 55. 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., Brenner, S., Beringer, M., and Rodnina, M. V., J. Biol. Chem. 283, 32229–32235 (2008).] 57. 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 nucleophilic the amino group (the less likely it is to be protonated). 59. a. transcription; b. transcription; c. translation; d. transcription; e. transcription; f. translation; g. transcription. 61. 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 aminoacylation. This would minimize the ability of the mutated tRNA to insert the amino acid at a position corresponding to a stop codon. 63. 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. 65. 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. 67. 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. 69. 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. 71. 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. 73. The basic residue is highlighted in gray; the hydrophobic core is underlined. MKWVTFISLLLLFSSAYSRGV 75. 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. 77. 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.

Odd-Numbered Solutions

79. a.

b. O

O H N

O

CH

C

H N

CH

C

CH2

CH2

O

CH2

P O⫺

O⫺

Polyglutamine protein

CH2 CH2 HN

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., Palazzolo, I., and Poletti, A., Hum. Mol. Gen. 18, R40–R47 (2009).] O

81. C

O

Ubiquitin

S-63

H3C

(H2C)12

C O

H N

CH H

C

Index Page references followed by T indicate tables. Page references followed by F indicate figures.

A A, see Adenine; Alanine AARSs, see Aminoacyl-tRNA synthetases Abasic sites, 532 ABC transporters, 247 Ab initio methods, 66 ABO blood group system, 293 ACC, see Acetyl-CoA carboxylase ACES (N-[2-Acetamido]-2aminoethanesulfonic acid), 37T Acetaldehyde: ethanol from, 163–164 in fermentation, 341, 342 oxidation state of carbon in, 13T standard reduction potential of, 387T Acetaldehyde dehydrogenase, 342 N-(2-Acetamido)-2-aminoethanesulfonic acid (ACES), 37T Acetaminophen, 205, 276 Acetate: from acetylcholine, 199, 200, 250 in alcohol metabolism, 342 standard reduction potential of, 387T Acetic acid: laboratory synthesis of, 15 oxidation state of carbon in, 13T pK of, 36, 37T titration of, 40–41, 41F Acetimidoquinone, 205 Acetoacetate: decarboxylation of, 161–163 in ketogenesis, 450, 451F in ketone body catabolism, 452F standard reduction potential of, 387T Acetoacetyl-ACP, 446F, 447 Acetoacetyl-CoA: in cholesterol synthesis, 455F in ketogenesis, 451F in ketone body catabolism, 452F Acetone: from acetoacetate, 161–163 from ketone bodies, 450, 451F oxidation state of carbon in, 13T Acetylcholine: hydrolysis of, 199, 200 at nerve–muscle synapse, 248–250, 249F Acetylcholine receptors, 255, 264 Acetylcholinesterase, 258 at nerve–muscle synapse, 250 noncompetitive inhibition of, 199, 200 Acetyl-CoA: in amino acid catabolism, 478, 479 in anaplerotic reactions, 376 in β oxidation, 436–439, 438F, 441F as biosynthetic precursor, 375 cholesterol from, 454–457, 455F in citric acid cycle, 365, 366, 368F in fatty acid synthesis, 444, 445, 446F in glyoxylate pathway, 377 and ketone bodies, 450–452, 451F, 452F in lipid metabolism, 457 as metabolic intermediate, 307, 307F from pyruvate, 341F, 342, 343, 363–365, 365F standard free energy of hydrolysis for, 319 in urea cycle, 482, 483 Acetyl-CoA carboxylase (ACC), 516 in fatty acid synthesis, 444–445 regulation of, 449, 449F Acetyl-Cys, 447 Acetyl-dihydrolipoamide, 365F

Acetylene, 13T N-Acetylglucosamine, 294 N-Acetylglutamate, 482–483 Acetyl group, in citrate synthase reaction, 366–368 Acetyl-Lys, 553 Acetylsalicylic acid (aspirin), 50, 275, 276, 281 O-Acetylserine, 472 Acids: and buffers, 42 defined, 34, 35 pK values of, 36–40, 37T Acid–base catalysis, 160–161, 162F Acid–base chemistry, 33–45 of buffers, 40–42 in human body, 42–45 hydrogen and hydroxide ions in, 33–34 pH in, 34–40 pK value in, 36–40 Acid catalysis, 161 Acid dissociation constant (Ka), 36 Acidic solutions, 34 Acidification, ocean, 35 Acidosis, 44, 45 Aconitase, 368–369, 369F Aconitate, 368 ACP (acyl carrier protein), 443–444, 444F Acquired immune deficiency syndrome (AIDS), 198. See also Human immunodeficiency virus (HIV) Actin: monomeric, 130, 130F in myosin–actin reaction cycle, 142, 143, 143F as real-time PCR benchmark, 74 F-Actin, 131 G-Actin, 131 Actin filaments (microfilaments), 130–132 abundance of, 130–131 assembly of, 131, 131F distribution of, in cell, 130F treadmilling by, 131–132, 132F Action potential, 237, 237F Activation: in allosteric enzyme regulation, 200–203 auto-, 171 of chymotrypsin, 171–172, 172F of fatty acids, 435–436 of fatty acid synthesis, 449 feed-forward, 339 of G proteins, 265–266 of insulin receptor, 272, 272F of Photosystem II, 416–417 of protein kinase A, 267 Activation energy (ΔG‡), 158–160 Activators, 558, 560 Active chromatin, 554T Active site, 168–169 Active transport, 245–248 by ABC transporters, 247 by Na,K-ATPase, 245–247 secondary, 247–248 Active transporters, 240, 245–248 Acute lung injury, 50 Acyladenylate, 436 Acyl-carnitine, 437F Acyl carrier protein (ACP), 443–444, 444F Acyl-CoA: in β oxidation, 436–438, 442

in carnitine shuttle system, 437F in fatty acid activation, 436 triacylglycerols and phospholipids from, 452–454 Acyl-CoA:cholesterol acyltransferase, 457 Acyl-CoA dehydrogenase: in β oxidation, 438, 438F, 440 and ubiquinol, 394, 395F Acyl-CoA oxidase, 442 Acyl-CoA synthetase, 435, 436, 453 Acylcovir, 204 Acyl–enzyme intermediate, 165F Acyl groups, 4T, 217 Acyl homoserine lactones, 262 Acyl phosphates, 336 Acyltransferase, 453, 453F Adenine (A), 54T and Chargaff ’s rules, 56–57 as complement of T (or U), 59 and DNA structure, 56–57 in genetic code, 63T, 581T and melting temperature, 59 as purine base, 53 salvage pathways for, 489 wobble pairing with, 585F Adenosine, 54T, 198, 261, 262 Adenosine deaminase, 156T, 181, 212 Adenosine diphosphate (ADP), 54T and actin, 131 in activation of phosphofructokinase, 203 in Calvin cycle, 426 in gluconeogenesis, 345F, 346 in glycolysis, 330, 331F, 338–339 and kinesin, 149F mitochondrial transport of, 392, 392F and myosin, 147F and rate of oxidative phosphorylation, 405 Adenosine-5′-(β,γ-methylene) triphosphate (AMPPCP), 258 Adenosine monophosphate (AMP), 54, 54T in DNA ligation, 527 in fatty acid activation, 435, 436 from phosphorylation of cAMP, 268 in purine synthesis, 485–486, 486F Adenosine receptor, 261–263 Adenosine triphosphate (ATP), 54T. See also ATP hydrolysis; ATP synthesis about, 5–6 from β oxidation, 439 in Calvin cycle, 426 in DNA ligation, 527 in fatty acid activation, 435, 436 in fatty acid synthesis, 443, 447 in gluconeogenesis, 345F, 346 from glucose, 372, 372F in glycogenolysis, 349 in glycolysis, 330, 331F, 336, 338–339 and hexokinase, 169 in histidine synthesis, 474 mitochondrial transport of, 392, 392F in nitrogen fixation, 465 in nucleotide synthesis, 485, 487 powering muscle with, 320 resonance stabilization of phosphoryl group in, 318 structure of, 316, 317F in tRNA aminoacylation, 582–584 in urea cycle, 482, 483F S-Adenosylmethionine, 576

Adenylate, see Adenosine monophosphate (AMP) Adenylate cyclase, 266 Adenylosuccinate, 486F Adipocytes, 304, 304F Adiponectin, 505T, 506 Adipose tissue: AMPK effects on, 506, 506T brown and white, 508–509, 509F hormones produced by, 505–506, 505T insulin action on, 503, 503T metabolic roles of, 498, 499, 499F polymer storage in, 304 A-DNA, 59, 564 ADP, see Adenosine diphosphate ADP–glucose, 426–427 Adrenaline, see Epinephrine α-Adrenergic receptors, 269 β2-Adrenergic receptors, 265, 265F, 269 Adrenoleukodystrophy (ALD), 78T, 82 Adult onset diabetes, 509–511 Aerobic glycolysis, 512 Aerobic muscle operation, 320 Aerobic organisms, 16 Affinity, of receptors, 261–263 Affinity chromatography, 109, 278 Aging, free radicals and, 398 Agonists, 261 Agre, P., 243 AICAR (5-aminoimidazole-4carboxamide ribonucleoside), 517 AIDS (acquired immune deficiency syndrome), 198. See also Human immunodeficiency virus (HIV) Airways, proteoglycans on surface of, 295F Alanine (Ala, A): abbreviations for, 87 chiral isomers of, 88 genetic code for, 63T, 581T as glucogenic amino acid, 477T in glucose–alanine cycle, 501, 501F as hydrophobic amino acid, 87F, 88, 99T as metabolic intermediate, 307, 307F as nonessential amino acid, 470T percent occurrence of, 91F representations of, 3, 5F synthesis of, 470 in transamination, 377, 378, 467, 477 β-Alanine, 490 Alanine aminotransferase, 157, 378 Alanine transaminase (ALT) tests, 177, 469 AlaRS, 602 Albuterol, 206T Alcaligenes, 93T, 104F Alcaptonuria, 480 Alcohol(s). See also Ethanol bonding of water with, 27 as compound, 4T Alcohol dehydrogenase: in alcohol metabolism, 342 in fermentation, 342, 343 metal ion catalysis with, 163–164 ALD (adrenoleukodystrophy), 78T, 82 Aldehydes, 4T Aldolase: in gluconeogenesis, 345F in glycolysis, 331F, 333–335, 334F Aldose, 284 Aldose reductase, 510 Algal blooms, 407, 429 Alkalosis, 44–45

I-1

I-2 Index Alleles, 74 Allolactose, 561 Allosteric enzymes, 194 Allosteric proteins, 126 Allosteric regulation, 200–203 α-Amino acids, 87 α Anomer, 285–286, 285F α Carbon (Cα), 87, 88 α-chain, hemoglobin: genetic variants in, 127–129, 129T myoglobin vs., 121–123, 122F α Helix: of DNA-binding proteins, 558 in lipid bilayers, 226, 226F proteins with, 95, 95F, 98 α subunit (ATP synthase), 402 α (degree of inhibition), 196 ALS (amyotrophic lateral sclerosis), 408 Alternative splicing, 569, 570F ALT (alanine transaminase) tests, 177, 469 AluI, 69T Aluminum ions, 199, 200 Alzheimer’s disease, 105–106, 105F Amides, 4T Amido group, 4T Amines, 4T, 27 Amino acids, 86–94. See also Proteins; specific amino acids about, 3 and aminoacyl–tRNA synthetases, 583F aromatic, 472–474 branched-chain, 472 chemical properties of, 87–89 from citric acid cycle intermediates, 375 defined, 86 essential, 312T, 469, 474 genetic code for, 581, 581T glucogenic, 477–480, 477T ketogenic, 477, 477T, 479–480 and mass spectrometry, 109–110 mobilization of, 305 in myoglobin, 122, 122F nonessential, 469–471 from nucleotide degradation, 489–490 and nucleotide sequences, 62–63, 63T peptide bonds linking, 90–93 polymerization of, 6, 7 and primary structure of proteins, 93–94 signaling molecules from, 475–476 storage of, 304 sulfur-containing, 472 transamination of, 467–469, 468F variants of, in translation, 585 D Amino acids, 88 L Amino acids, 88 Amino acid biosynthesis, 469–476 for aromatic amino acids, 472–474 for branched-chain amino acids, 472 cellular location of, 498F glycolysis and, 512 metabolites in, 470–471 signaling molecule precursors from, 475–476 for sulfur-containing amino acids, 472 Amino acid catabolism, 476–480 cellular location of, 498F diseases related to, 480 of glucogenic vs. ketogenic amino acids, 477–480, 477T Amino acid–PLP Schiff base, 468F Amino acid residues, 90 Aminoacyl–adenylate, 583F Aminoacylation, tRNA, 582–584, 583F Aminoacyl site, see A site Aminoacyl–tRNA: delivery to ribosome of, 591–592 in peptidyl transferase reaction, 593F structure of, 583, 583F

Aminoacyl–tRNA synthetases (AARSs), 582–584, 583F classes of, 583, 583T proofreading by, 584 for variant amino acids, 585 γ-Aminobutyrate (GABA), 475 Aminoglycoside, 176 Amino groups, 4T, 86 hydrogen bonding with, 95 ionic interactions for, 26 in peptide bonds, 90 pK of, 40 transamination of, 467–469 5-Aminoimidazole-4-carboxamide ribonucleoside (AICAR), 517 β-Aminoisobutyrate, 490 α-Amino-β-ketobutyrate, 478 2-Aminopurine, 547 Amino sugars, 287F Aminotransferase, 467–469 Ammonia: assimilation of, 466–467 diffusion of, 256 neurotoxicity of, 481 in nitrogen fixation, 465–466 pK of, 36 Ammonium ions: in acid–base balance of humans, 44, 45 pK of, 36, 37T Ammonium sulfate, 47 Amoeba, 1 Amoxicillin, 206T AMP, see Adenosine monophosphate AMP-dependent protein kinase (AMPK), 506, 506F, 506T Amphipathic molecules, 31, 218 Amphiphilic molecules, 31, 31F Amphotericin B, 242 AMPK, see AMP-dependent protein kinase AMPPCP (Adenosine-5′-[β,γmethylene] triphosphate), 258 ampR gene, 70 amu (atomic mass units), 7 Amylase, 302, 302F, 323 Amylin, 505T Amylo-1,6-glucosidase, 354T Amyloid deposits, 105–106, 106F Amyloid-β, 105–106 Amylose, 289, 289F Amylo-(1,4 → 1,6)-transglycosylase, 354T Amyotrophic lateral sclerosis (ALS), 408 Anabolism: and catabolism, 302F in citric acid cycle, 374–378 defined, 301 Anaerobic muscle operation, 320 Anaerobic organisms, 16 Anandamide, 281 Anaplerotic reactions, 376–378, 378F Anemia, 123, 313T, 460 Anfinsen, C., 116, 605 Angiogenin, 179 Angstroms, 7 Animals. See also specific species digestion of cellulose by, 291 genome size and gene number of, 64T sweating by, 32 transgenic, 72 Animal cells: membrane potentials of, 237 Na+ and K+ in, 235, 236F Anionic groups, 15 Annealing of DNA, 60 Anomers, 285–286 α Anomer, 285–286, 285F β Anomer, 285–286, 285F Anomeric carbon, 286 Anopheles mosquito, 52

Antagonists, 263 Antenna pigments, 414 Anthopleura xanthogrammica, 411 Antibiotics, 70, 594 Anticodons. See also tRNA AARS interactions with, 582–583 codon pairing with, 584–585 defined, 581 ribosomal sensor for, 591, 591F wobble pairing of, 585F Antidepressants, 250 Antidiabetic drugs, 510T Antimycin A, 430 Antiparallel β sheets, 95, 96F Antiparallel DNA strands, replication of, 522–523 Antiparallel polynucleotide strands, 57 Antiporters, 245, 245F Antithrombin, 174, 175, 175F AOT [bis-(2-ethylhexyl)sulfosuccinate], 48 AP endonuclease, 534–535, 534F Aplysia, 464 ApoB, 82 Apolipoprotein B-100, 463 Apoptosis, 538 AP site, 532 Apurinic site, 532 Apyrimidinic site, 532 Aquaporins, 243–244, 243F, 244F Aquaporin 1, 243–244 Aqueous chemistry, 24–45. See also Water acid–base balance in humans, 42–45 acid–base chemistry, 33–45 buffers, 40–42 hydrogen bonding, 24–29 hydrophobic effect, 29–32 Arabidopsis thaliana, 322 genome size and gene number, 64T as model organism, 65F noncoding DNA segments for, 66 Arachidic acid, 217T Arachidonate: in eicosanoid synthesis, 275, 275F synthesis of, 448, 448F Arachidonic acid, 217T Ara h8 protein, 117 Archaea. See also specific species evolution of, 16, 18F genome size and gene number of, 64T membrane lipids of, 218–220, 220F rubisco in, 422 Archaeal lipids, 231 Archaeoglobus fulgidus, 64T Arginase, 482, 483F Arginine (Arg, R): abbreviations for, 87 catabolism of, 478 as charged amino acid, 87F, 89 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T nitric oxide from, 476 as nonessential amino acid, 469, 470T percent occurrence of, 91F pK value of groups in, 91T synthesis of, 470 in urea cycle, 482, 483F Argininosuccinase, 482, 483F Argininosuccinate, 483F Argininosuccinate synthetase, 482, 483F ArgRS, 584 Aromatic amino acids, 472–474 Arrestin, 269, 269F Arsenate, 357 Arsenite, 379 Arteries, plaques in, 432–433, 433F Artificial selection, 17 Ascorbate, 137, 313T, 407 Ascorbic acid, 313T A site (aminoacyl site): defined, 587

in elongation, 590, 591 and peptidyl transferase reaction, 593, 594 Asn, see Asparagine AsnRS, 584 Asp, see Aspartate Asparaginase, 478 Asparagine (Asn, N), 3 abbreviations for, 87 catabolism of, 478 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T as nonessential amino acid, 470T percent occurrence of, 91F as polar amino acid, 87F, 89 synthesis of, 470 Asparagine synthetase, 470, 492 Aspartame, 114 Aspartate (Asp, D): abbreviations for, 87 as acid–base catalyst, 162F catabolism of, 477, 478 as charged amino acid, 87F, 89 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T in malate–aspartate shuttle system, 391, 391F as nonessential amino acid, 470T percent occurrence of, 91F pK value of groups in, 91T in synthesis reactions, 470, 473 in transamination, 377 in urea cycle, 482 and zinc fingers, 101 Aspartate aminotransferase, 209 Aspartate aminotransferase (AST) tests, 177, 469 Aspartate transcarbamoylase, 213 Aspartyl aminopeptidase, 181–182 Aspartyl proteases, 180 Aspergillus nidulans, 360 Aspirin (acetylsalicylic acid), 50, 275, 276, 281 Assimilation, nitrogen, 466–469 Association studies, genome-wide, 68 AST (aspartate aminotransferase) tests, 177, 469 a subunit (ATP synthase), 402 AsuI, 69T Atherosclerosis, 432–434, 433F Atkins diet, 516 ATM protein kinase, 539 Atomic mass units (amu), 7 Atorvastatin, 197, 206T, 456F ATP, see Adenosine triphosphate ATPases, 245–247 ATP-citrate lyase, 375, 376, 444F ATP hydrolysis: and actin, 130–131, 130F coupling of reactions to, 316–318 in fatty acid oxidation, 436 free energy change for, 318T in kinesin reaction cycle, 145–146, 145F in myosin–actin reaction, 142–143, 143F in Na,K-ATPase reaction, 246 ATP synthase, 401–405 binding change mechanism for, 403 function of, 401, 401F P:O ratio for, 403, 404 in proton translocation, 401–403 rate of fuel catabolism by, 404–405 structure of, 401, 402F and uncoupling agents, 404 ATP synthesis. See also Oxidative phosphorylation in photosynthesis, 420–422 P:O ratio for, 404 uncoupling of, 404

In dex Autoactivation, 171 Autoimmune diseases, 231 Autophagy, 258–259 Autophosphorylation, 271–273 Auxin, 261T Avandia (rosiglitazone), 205, 510, 510T Avery, O., 80 Axons, 237, 238F 3′-Azido-2′,3′-dideoxythymidine (Zidovudine, AZT), 530 Azotobacter vinelandii, 363F

B Bacillus amyloliquefaciens, 170F Bacillus anthracis, 280 Bacillus stearothermophilus: phosphofructokinase of, 201, 202F, 357 pyruvate dehydrogenase complex of, 363, 363F Bacillus thuringiensis, 431 Backbone, polypeptide, 91, 94–95 Bacteria. See also specific species amino acid synthesis in, 472–474 aminoacyl–tRNA synthetases of, 584 biofilms of, 291 cell walls of, 295 Complex I of, 392, 393F DNA gyrase of, 543 evolution of, 16, 18F genome size and gene number of, 64T in microbiome, 500 mismatch repair system of, 533 origin of replication for, 520 phosphofructokinase regulation in, 333F pore formation as defense for, 242 pyruvate dehydrogenase in, 363 quorum sensing in, 262 recombinant, 70, 71F ribosome recycling factor of, 596 ribosomes of, 586, 587, 588F rubisco in, 422 trigger factors of, 597 tRNA processing in, 573 Bacteriophages, 69, 603 Bacteriophage λ, 70, 70F, 558F Bacteriophage T7, 546, 576 Bacteriorhodopsin, 226F Baker’s yeast, see Yeast Ball-and-stick model, 5F Banting, F., 509, 509F Bar-headed goose, 149 Bar-tailed godwit, 432 Bases: in buffers, 40–42 conjugate, 36, 40, 41 defined, 34, 35 nucleotide, 53, 54T, 59–60 Schiff, 162, 163, 467, 468F strong, 40, 41 Base catalysis, 161 Base excision repair, 533–535, 534F Base pairs (bp): axial view of, 59, 60F nonstandard, in RNA, 573, 574F and structure of DNA, 57–59 as unit for DNA segments, 58 Basic solutions, 34 Bassham, J., 424 Bcr-Abl kinase, 273 B-DNA, 59, 564 Beer, 359 Beeswax, 220 Behenic acid, 217T Benedict’s solution, 297 Benson, A., 424 Benzene, 29T 6-O-Benzyl-D-galactose, 256 Beriberi, 312, 313, 313T, 379 Best, C., 509, 509F β Anomer, 285–286, 285F

β Barrels: formation of, 226–227, 227F of porins, 240–241 β Cell exhaustion, 511 β-chain, of hemoglobin, 579 genetic variants of, 127–129, 129T myoglobin vs., 121–123, 122F β3-GalT enzyme, 210 β Oxidation, 436–439 cellular location of, 498F fatty acid synthesis vs., 443–444 of odd-chain fatty acids, 440–442 in peroxisomes, 442–443 reactions of, 436–439, 438F of unsaturated fatty acids, 439–440 β Sheets: antiparallel, 95, 96F β barrel from, 226–227 of DNA binding proteins, 559 parallel, 95, 96F of proteins, 95–96, 98 β subunit (ATP synthase), 402, 403 Bicarbonate: in blood buffering, 43, 43F in C4 pathway, 424 in oceans, 35 production of, 44, 44F reabsorption of, 43, 43F Biguanides, 510T Bilayers, see Lipid bilayers Bile acids, 457 Bimolecular reactions, 186, 191 Binding change mechanism, 403, 403F Biochemical standard conditions, 315, 315T Biochemistry: themes of, 1–2 units in, 7 Bioconversion process, 290 Biocytin, 313T Biofilms, 291, 291F Biofuels, 290 Bioinformatics, 1 Biological fluids, pH of, 34, 34T Biological markers, 293–294 Biological membranes, see Lipid bilayers; Membranes Biological molecules, 3–9 in cells, 3, 5–6 elements found in, 3, 3F laboratory synthesis of, 14–15, 15F in polymers, 6–9 relative strength of bonds in, 26, 26F in solution, 28, 29 solvated nature in vivo, 28 Biological polymers (biopolymers). See also Carbohydrates; Nucleic acids; Polysaccharides; Proteins digestion of, 302, 302F functions of, 9T types, 6–9 Biosynthesis of amino acids, see Amino acid biosynthesis Biosynthetic precursors: citric acid cycle intermediates as, 375–376, 375F from metabolism in cancer cells, 512 for signaling molecules, 475–476 Biotin, 313T, 343 1,3-Bisphosphoglycerate: in Calvin cycle, 424, 425F free energy change for hydrolysis of, 318T in gluconeogenesis, 345F in glycolysis, 331F, 336–337, 337F 2,3-Bisphosphoglycerate: and carbohydrate metabolism disorders, 354 and oxygen binding by hemoglobin, 127, 127F Bis-(2-ethylhexyl)sulfosuccinate (AOT), 48

Bisubstrate reactions, 193 Blackburn, E., 529 Black smokers, 15, 15F Blindness, 313T Blood. See also Red blood cells ABO group, 293 buffering for, 42–44, 50 coagulation of, 173–175 mass spectrometry of, 110 pH of, 34, 34T Blood clots, 173, 173F Blue-green algae, see Cyanobacteria Blunt ends, 69 Body temperature, 362 Bohr effect, 126, 126F Bone, collagen in, 137, 139–140 Bone alkaline phosphatase, 213 Bonito, 148 Bordetella pertussis, 279 Boric acid, 37T Botox, 258 Botulinum toxin, 323 Bovine ATP synthase, 402F Bovine chymotrypsinogen, 170T Bovine insulin, 93T Bovine pancreatic trypsin inhibitor, 172–173, 172F, 179F Bovine papillomavirus E5, 115 Bovine trypsin, 170T Boyer, P, 403 bp, see Base pairs Brain: effect of Alzheimer’s disease on, 105F rat, metabolite profile of, 311 Brain glutaminase, 208 Branched-chain amino acids, 472 Branching proteins, 132 Brazilin, 359 BRCA1, 539 BRCA2, 539 Breast cancer, 539 Bromelain, 180 Bromodomains, 554, 554F 5-Bromouracil, 547 Brown adipose tissue, 508–509, 509F Brown fat, 404 Brown microalgae, 283, 299 Buffers, 40–42 Buffering capacity, 41 1-Butanol, 29T Butenoyl-ACP, 446F Butyryl-ACP, 446F

C C, see Cysteine; Cytosine C4 pathway, 423, 424 Ca2+ (calcium ions), 270, 505 Ca2+ (calcium) channels, 248, 249F Cachexia, 513 cADPR (cyclic ADP-ribose), 81 Caenorhabditis elegans, 64T, 65, 65F Caffeine, 263 cal (calorie), 315 Calcineurin, 103F, 279 Calcium carbonate, 35 Calcium (Ca2+) channels, 248, 249F Calcium ions (Ca2+), 270, 505 Calmodulin, 270 Calorie (cal), 315 Cα (α carbon), 87, 88 Calvin, M., 424, 430 Calvin cycle, 424–428 initial reactions of, 424, 425F sucrose and starch from, 426–428 sugar molecules in, 424–426 cAMP (cyclic AMP): activation of protein kinase A by, 267 G protein and generation of, 266 phosphorylation of, 268 cAMP phosphodiesterase, 268

I-3

Cancer. See also DNA repair and cell signaling, 273 and DNA repair, 539–540 as genetic disease, 538–540 metabolism of cancer cells, 511–513, 513F Cancer Genome Atlas Project, 539 Cannabis sativa, 260 3′ Cap, 567, 567F Capping proteins, 132 Capsaicin, 222 Carbamate, 482F Carbamoyl phosphate, 482, 482F Carbamoyl phosphate synthetase: composition of, 93T in urea cycle, 481–483, 482F Carbanions, 161, 468F Carbohydrates, 283–296. See also Fructose; Glucose; Monosaccharides and ABO blood group system, 293 about, 5 from carbon dioxide, 308 covalent modifications of, 102F defined, 283 glycoproteins, 291–295 and metabolic disorders, 353–355 as metabolic fuels, 508T polysaccharides, 287–291 Carbon: α, 87, 88 anomeric, 286 in biological molecules, 3F in citric acid cycle, 370, 370F electronegativity of, 26T oxidation states of, 13, 13T reduction/oxidation of compounds containing, 13, 13F Carbonate ions, 35 Carbon dioxide. See also Carbon fixation in Calvin cycle, 426 in citric acid cycle, 365–366, 369–370 and ocean acidification, 35 in oxidation–reduction reactions, 308 oxidation state of carbon in, 13T in pyruvate dehydrogenase reaction, 362 from respiration, 256 Carbon fixation, 422–426 C4 pathway in, 424 Calvin cycle for, 424–428 defined, 422 in modern cells, 16 regulation of, 426 rubisco as catalyst for, 422–423 sucrose and starch from, 426–428 Carbonic acid, 35, 43 Carbonic anhydrase: in blood buffering, 43 and C4 pathway, 424 hydration of carbon dioxide by, 180 kcat of, 190T rate enhancement by, 156T Carbon monoxide, 13T, 148 Carbon monoxide poisoning, 124 Carbon skeletons, amino acid, 359 Carbonyl groups, 4T, 27 2-Carboxyarabinitol-1-phosphate, 430 Carboxyhemoglobin, 124 Carboxylates (carboxylic acids), 4T, 40 Carboxylate groups: in amino acids, 86 in carboxylic acids, 4T ionic interactions with, 26 in peptide bonds, 90 pK of, 40 Carboxyl groups, 4T Carboxylic acids, 4T, 40 Carboxymethyl-CoA, 380 Carboxymethyl (CM) groups, 108 Carboxyphosphate, 482F

I-4 Index Carcinogens, 533 Carcinogenesis, 373, 538–539 Carnitine acyltransferase, 437F, 449 Carnitine deficiency, 460 Carnitine shuttle system, 436, 437F, 449F β-Carotene, 72, 413F Carsonella ruddii, 83 Cas9 nuclease, 77 Catabolism. See also Glycolysis of amino acids, 476–480 and anabolism, 302F in citric acid cycle, 374–378 cofactors from, 309 defined, 301 of ketone bodies, 452F of propionyl-CoA, 440, 441F rate of oxidative phosphorylation and, 404–405 of sugars other than glucose, 340 Catalysis: acid, 161 acid–base, 160–161, 162F base, 161 in citric acid cycle, 372 covalent, 161–163, 163F electrostatic, 169 metal ion, 163–164 Catalysts. See also Enzymes amino acid side chains as, 161, 162F for carbon fixation, 422–423 defined, 155 effect on reaction of, 159–160, 160F enzymes as, 13 for fatty acid synthesis, 444–447 fumarase as, 371 for peptide bond formation, 593–595 properties of, 166–169 protein groups as, 163F RNase P as, 573, 574 succinyl-CoA synthetase as, 370 Catalytic constant (kcat): defined, 190 experimental determination of, 191–192 and kcat/KM ratio, 190–191 of selected enzymes, 190T Catalytic converters, 155 Catalytic efficiency, 190–191 Catalytic mechanisms, 158–166 acid–base catalysis, 160–161 and activation energy, 158–160 of chymotrypsin, 164–166 covalent catalysis, 161–163 metal ion catalysis, 163–164 Catalytic perfection, 191 Catalytic triad, 164–166, 164F Cataracts, 510, 510F Catechol, 475 Catecholamines, 475, 504, 505 CATH system, 98 Cationic groups, 15 Cattle, timing of slaughter, 151. See also Cows CCR5 gene, 78 cDNA (complementary DNA), 530 CDP (cytidine diphosphate), 54T CDP–choline, 454F CDP–diacylglycerol, 455F CDP–ethanolamine, 454F Celebrex, 276 Cells, 2F. See also specific types absorption of digestion products by, 302–303 biological molecules in, 3, 5–6 cryoelectron tomography of, 587, 588, 588F immortality for, 531 membranes of, 215 metabolic pathways of, 310 origins of modern, 16–18 replication of, 14 Cell crawling, 132, 132F

Cell division, 134F Cell-free translation systems, 605 Cell membranes, see Membranes Cellular respiration, 390, 404 Cellulose: in biofuels, 290 as glucose polymer, 8, 9F structural support from, 289–291 structure of, 289, 290, 290F Cell walls, bacterial, 295 Central dogma of molecular biology, 61–64. See also Protein synthesis and decoding of DNA, 62–63 and disease caused by mutations, 63–64 Centromeres, 528 Ceramide, 270 Ceramide-1-phosphate, 281 Cerebrosides, 219F Cerulenin, 450 CF, see Cystic fibrosis CF1CF0 complex, 421, 421F CFTR gene, 602, 605 cGAMP (cyclic guanosine monophosphate-adenosine monophosphate), 81 Channeling, 474 Chaperones, 293, 597–598 Chaperonins, 598, 598F Chargaff, E., 53, 56 Chargaff’s rules, 56–57 Charged side chains, amino acid, 87F, 89 Chase, M., 80 Chemical labeling, 164 Chemical methylating agents, 548 Chemical reactions. See also Rate of reaction; specific reactions anaplerotic, 376–378, 378F bimolecular, 186, 191 coupled, 12–13, 13F effect of catalysts on, 159–160, 160F endergonic/endothermic, 11–12 enzyme-catalyzed, 187–189, 187F exergonic/exothermic, 11–12 first-order, 186 irreversible, 332 multistep, 193–194 multisubstrate, 193 near-equilibrium, 332 nonspontaneous, 11–12 second-order, 186, 191 spontaneous, 11–12 unimolecular, 186 in vitro vs. in vivo, 12 Chemiosmosis, 399–401 electron transport and oxidative phosphorylation in, 400 and photosynthetic light reactions, 421–422 proton gradient in, 400–401 Chemiosmotic theory, 400 Chemoautotrophs, 301 Chesapeake hemoglobin variant, 129T Chimpanzee genome, 83 Chinese restaurant syndrome, 90 Chirality, 88, 284–285 Chitin, 291 Chlamydomonas reinhardtii, 429 Chloramphenicol, 594 Chlorine ions, 32F 8-Chloroadenosine, 81 Chlorophyll a, 413F Chloroplasts, 411–415 defined, 412 evolution of, 17, 18 light-harvesting complexes in, 414–415 pigments in, 412–414 structure of, 412, 412F Cholate, 48, 457

Cholecystokinins, 505T Cholesterol, 6 in bilayers, 224 conversion of squalene to, 455, 456F function of, 219, 220 as hormone precursor, 274 lipoprotein handling of, 433, 434 percentage of, in lipoproteins, 433T synthesis of, 454–457, 455F Cholesteryl esters, 457 Cholesteryl stearate, 303 Choline: from acetylcholine, 199, 200, 250 as essential substance, 312T in phospholipid synthesis, 454F Choline transport proteins, 256–257 Chondroitin sulfate, 294F Chorismate, 472, 473 Chorismate mutase, 156T Chorispora bungeana, 233 Chromatin: remodeling of, 554, 554F transcriptionally active vs. silent, 554T Chromatography, 107–109, 278 Chromium, 407 Chromosomes: defined, 52 DNA packaging in, 544 of eukaryotes, 528, 528F extension of, by telomerase, 529, 531 size of, 519 Chylomicrons: characteristics of, 433, 433T cholesterol and cholesteryl esters in, 323 in lipoprotein function, 434F Chymotrypsin, 182 activation of, 171–172, 172F in amino acid sequencing, 110 catalytic mechanism of, 164–166 cleavage site for, 109T evolution of, 170 function of, 155 irreversible inhibition of, 195 and protease inhibitors, 172–173 rate enhancement by, 156T ribbon model of, 155F structure of, 155, 170 substrate specificity of, 156, 170–171, 171F α Chymotrypsin, 171, 172F δ Chymotrypsin, 171, 172F π Chymotrypsin, 171, 172F Chymotrypsinogen: activation of, 171–172, 172F percent sequence identity for, 170T Ciprofloxacin, 543 Circularization, of mRNA, 590, 590F Citrate, 2F, 326 as biosynthetic precursor, 375 in citric acid cycle, 366–369, 367F, 368F in fatty acid synthesis, 444, 444F transport system for, 376F Citrate synthase, 2F in citrate transport system, 444F in citric acid cycle, 366–368, 367F, 368F in fatty acid synthesis, 444 regulation by, 372 Citrate transport system, 444, 444F, 449F Citric acid cycle, 362–379 aconitase in, 367F, 368–369 anabolic and catabolic functions of, 374–378 and cancer cell metabolism, 512 cellular location of, 498F citrate synthase in, 366–368, 367F, 368F

context for reactions of, 365–366, 366F defined, 308, 365 fumarase in, 367F, 371 isocitrate dehydrogenase in, 367F, 369 α-ketoglutarate dehydrogenase in, 367F, 369–370 malate dehydrogenase in, 367F, 371–372 and pyruvate dehydrogenase reaction, 362–365 reactions of, 365–372, 367F succinate dehydrogenase in, 367F, 370, 371 succinyl-CoA synthetase in, 367F, 370 thermodynamics of, 372–374 Citrulline, 476, 482, 483F Citryl-CoA, 368F Clamp: DNA polymerase, 524, 524F RNA polymerase, 564 Class I aminoacyl–tRNA synthetases, 583, 583T Class I enzymes, 487, 496 Class II aminoacyl–tRNA synthetases, 583, 583T Classic hemophilia, 182 Clathrin, 252, 252F, 253 Cleavage sites, restriction endonuclease, 69, 69T Clinical trials, 205 Clones, 70–71, 71F Closed conformation of DNA polymerase, 525, 525F Clostridium botulinum, 180 Clostridium butyricum, 60T Clostridium perfrigens, 151 Clostridium tetani, 258 Clotrimazole, 280 Clover, 465F Cloverleaf secondary structure of tRNA, 582, 582F Clustered regularly interspersed short palindromic repeats (CRISPRs), 77 CM (carboxymethyl) groups, 108 CMP (cytidine monophosphate), 54T c-Myc protein, 606 Coagulation, blood, 173–175 Coagulation cascade, 174, 174F Coagulation factors, 71T, 174, 174T Coat proteins, 252, 253 Cobalamin, 313T, 314, 442 Cobalamin coenzymes, 313T, 442, 442F Cockayne syndrome, 535 Coding segments of genome, 66, 66F Coding strand, 62 Codons, 62. See also mRNA defined, 580 pairing of anticodons with, 584–585 and redundancy in genetic code, 581 ribosomal sensor for pairing of, 591, 591F wobble pairing of, 585F Coenzymes (cofactors), 160, 160F, 309 cobalamin, 313T, 442, 442F as energy currency, 319 FeMo, 465, 465F flavin, 313T nicotinamide, 313T Coenzyme A, 55F, 313T. See also Acetyl-CoA and acyl-carrier proteins, 444F in citric acid cycle, 366, 368F in fatty acid activation, 435–436 functions of, 54 in pyruvate dehydrogenase reaction, 364, 365F Coenzyme Q, see Ubiquinol; Ubiquinone Cofactors, see Coenzymes

In dex Coiled coils, in keratin, 135, 135F Colchicine, 134, 150 Collagen: cross-linking of, 139 fibers of, 137, 137F genetic defects in, 139–140 and glycosaminoglycans, 294 structure of, 136–139, 138F Collagenase, 151 Collagen type I, 140 Collagen type II, 140 Collagen type III, 140 Colony stimulating factor, 71T Competitive inhibition, 195–197, 195F, 200 Complement, 16, 242 Complementarity, of nucleotide bases, 59 Complementary DNA (cDNA), 530 Complementary strand, 74–76 Complex I, 392–394 bacterial, 392, 393F free energy from, 399 function of, 394F in mitochondrial electron transport, 389F Complex II, 394, 395F. See also Succinate dehydrogenase Complex III, 394–397 free energy from, 399 function of, 397F mammalian, 396F in mitochondrial electron transport, 389F Complex IV, 397–399 free energy from, 399 function of, 399F in mitochondrial electron transport, 389F model for reaction of, 398, 398F and rate of oxidative phosphorylation, 404 structure of, 397, 397F Complex V, 401. See also ATP synthase Compounds, dissolution of, 28–29 Concentration. See also Substrate concentration at biochemical standard state, 315, 315T in enzyme-catalyzed reactions, 187–188, 187F reactant, 314–316, 315T Condensation reaction, 90 Conformation(s), 7 in citrate synthase, 367F of Complex I, 394 of DNA polymerase, 525, 525F of gated channels, 242–243 in G protein signaling pathway, 265 of hemoglobin, 124–126, 125F of Na,K-ATPase, 246–247 proteins with multiple, 102–103, 103F of RecA and DNA, 536, 537F of RNA polymerase, 564–565 of transport proteins, 244–245 of triose phosphate isomerase, 335F Conjugate base, 36, 40, 41 Conotoxins, 235 Consensus sequences: at mRNA splice sites, 568, 569F of promoters, 556, 556F Conservative replication of DNA, 520, 520F Conservative substitution, 123 Constitutive proteins, 537–538 Contraction, muscle, 142, 142F Conus snails, 235 Convergent evolution, 170 Cooperative binding: kinetics of reactions with, 194, 194F of oxygen, 123–126 Copper ions, 397F

CoQ10 supplement, 324. See also Ubiquinone Coral, 35 Cordycepin, 577 Corey, R., 80, 95 Cori, C., 499 Cori, G., 499 Cori cycle, 499, 500, 500F Corn (maize): genetically modified, 72 genome size and gene number, 64T, 66 Cortisol, 261T, 274 Cotton, genetically modified, 72 Coupled reactions: free energy change in, 12–13, 13F metabolic, 316–318 Covalent bonds, 26, 26F Covalent catalysis, 161–163, 163F Covalent modification: and action of insulin, 504 of DNA, 555 as post-translational events, 600–601 of proteins, 102F Cows. See also entries beginning Bovine histones of peas and, 550 lactation in, 515 timing of slaughter for cattle, 151 Cowtown hemoglobin variant, 149 COX (cyclooxygenase) inhibitors, 276 CpG islands, 555 Crabgrass, 430 C-reactive protein, 93T Creatine, 319 Creatine kinase, 177 Creutzfeldt-Jakob disease, 106 Crick, F., 56, 57, 61, 82, 519, 580, 584 CRISPR-Cas9 system: gene editing with, 77–79, 77F and recombination, 536 CRISPRs (clustered regularly interspersed short palindromic repeats), 77 Cristae, 390, 391F Crocodiles, 149 Crohn’s disease, 68 Cross-linking of collagen, 139 Cross-talk, 270 Cryoelectron crystallography, 112 Cryoelectron tomography, 29F, 587, 588, 588F γβ-Crystallin, 98F c Subunit (ATP synthase), 402, 403 C-terminus, 90, 91T CTP, see Cytidine triphosphate CTP synthase, 487 Cuphea plant, 230, 232 Curved arrow convention, 162 C-value paradox, 83 Cyanide poisoning, 407 Cyanobacteria, 322, 492 cytochrome b6 f of, 418, 418F photosynthesis in, 411–412 Photosystem II of, 416, 416F Photosystem I of, 419, 419F Cyanogen bromide, 109 Cyanuric acid, 47 Cyclic ADP-ribose (cADPR), 81 Cyclic AMP, see cAMP Cyclic electron flow, 420, 421F Cyclic guanosine monophosphateadenosine monophosphate (cGAMP), 81 Cyclization, in carbohydrates, 285–286 Cyclooxygenase (COX) inhibitors, 276 Cyclophilin, 190T Cyclosporine A, 279 Cymbalta (duloxetine), 206T Cysteine (Cys, C), 3 abbreviations for, 87 as acid–base catalyst, 162F

catabolism of, 478 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T as nonessential amino acid, 469, 470T percent occurrence of, 91F pK value of groups in, 91T as polar amino acid, 87F, 89 synthesis of, 472 and zinc fingers, 101 Cysteine proteases, 179 Cystic fibrosis (CF), 82, 117, 602 Cytidine, 54T Cytidine deaminase, 190T, 212 Cytidine diphosphate (CDP), 54T Cytidine monophosphate (CMP), 54T Cytidine triphosphate (CTP): nucleic acids in, 54, 54T in pyrimidine synthesis, 486–487 Cytidylate, 54T Cytochromes, 394–399 Cytochrome a, 387T Cytochrome a3, 387T Cytochrome b: in Complex III, 395, 396F heme group of, 395F in Q cycle, 396F standard reduction potential of, 387T Cytochrome b6, 418 Cytochrome b6 f, 418–419, 418F Cytochrome bc1, see Complex III Cytochrome c, 233, 397F electron transfer from ubiquinol to, 394–397 free energy change for reaction of, 389F oxidation of, 397–399 in Q cycle, 396F standard reduction potential of, 387T Cytochrome c1: in Complex III, 395, 396F standard reduction potential of, 387T Cytochrome c oxidase, see Complex IV Cytochrome f, 419 Cytochrome P450, 204–205 Cytokinesis, 143 Cytosine (C), 54T and Chargaff’s rules, 56–57 as complement of guanine, 59 deamination of, 532 and DNA structure, 56–57 in genetic code, 63T, 581T hydrogen bonding of, 27 and melting temperature of DNA, 59, 60, 60T as pyrimidine base, 53 wobble pairing with, 585F Cytoskeletal proteins, 130, 130F. See also specific types Cytoskeleton, 130

D D, see Aspartate Daltons (D), 7 D amino acids, 88 dAMP (deoxyadenosine monophosphate), 54 Danielli, J., 234 Dark reactions, 422, 426. See also Carbon fixation Darwin, C., 14 dATP, 547 Davson, H., 234 DCCD (dicyclohexylcarbodiimide), 409 DCMU (3-[3,4-Dichlorophenyl]-1,1dimethylurea), 429 dCTP (deoxycytidine triphosphate), 54 ddC (2′,3′-Dideoxycytidine, Zalcitabine), 530 DEAE (diethylaminoethyl) groups, 108 Deafness, 144 Deamination, 477, 532

I-5

Deep-sea hydrothermal vent worm, 151 Degree of inhibition (α), 196 Demospongic fatty acids, 230 Denaturation: of DNA, 59–60 of proteins, 101 Denitrification, 465 Deoxyadenosine monophosphate (dAMP), 54 Deoxycytidine triphosphate (dCTP), 54 2-Deoxy-D-glucose, 517 Deoxyguanosine diphosphate (dGDP), 54 Deoxyhemoglobin, 124–129, 125F Deoxynucleoside triphosphate (dNTP), 522F Deoxynucleotides, 54, 54T Deoxyribonuclease (DNase), 546, 549–550 Deoxyribonucleic acid, see DNA Deoxyribonucleotides, 487–488 2′-Deoxyribose, 53, 287 Deoxythymidylate (dTMP), 195, 489 Deoxyuridylate (dUMP), 195, 489 Desaturases, 448 Desensitization, receptor, 264, 269 d5SICS, 82 dGDP (deoxyguanosine diphosphate), 54 dGTP, 547 DHA (4,7,10,13,16,19-docosahexaenoic acid), 216, 217T DHF (dihydrofolate reductase), 46–47 Diabetes, 509–511 drugs for treatment of, 510T hyperglycemia with, 509–510 ketone bodies in, 51 and metabolic syndrome, 511 Diabetes mellitus, 305, 502, 509 Diacylglycerol: in phosphoinositide signaling pathway, 269 in phospholipid synthesis, 454F in triacylglycerol synthesis, 453F Diazotrophs, 465 DIC (disseminated intravascular coagulation), 182 Dicer (ribonuclease), 571, 572 3-(3,4-Dichlorophenyl)-1,1dimethylurea (DCMU), 429 Dictyostelium discoideum, 29F, 60T Dicyclohexylcarbodiimide (DCCD), 409 2′,3′-Dideoxycytidine (Zalcitabine, ddC), 530 Dielectric constants, 28, 29T Diet: and atherosclerosis, 433 and heart disease, 448 Dietary guidelines, 303 Diethylaminoethyl (DEAE) groups, 108 Diffraction pattern, 111 Diffusion: of ammonia, 256 lateral, 224–225 myoglobin-facilitated, 407 through lipid bilayers, 31–32, 32F transverse, 224 Diffusion-controlled limit, 191 Digestion, products of, 302–303 Dihydrobiopterin, 479, 480 Dihydrofolate, 489 Dihydrofolate reductase (DHF), 46–47 1,6-Dihydroinosine, 198 Dihydrolipoic acid, 387T Dihydrouridine, 573F Dihydroxyacetone, 283, 284 Dihydroxyacetone phosphate, 395F from fructose metabolism, 340 in gluconeogenesis, 345F in glycolysis, 331F, 333–336, 334F in sucrose synthesis, 427

I-6 Index Dihydroxyacetone phosphate (continued) in triacylglycerol synthesis, 452, 453 in triose phosphate isomerase reaction, 184, 197 Diisopropylphosphofluoridate (DIPF), 164, 195 Dimethyldecylphosphine oxide, 48 N 2,N 2-Dimethylguanosine, 573F Dinitrophenol (DNP), 409 Dintzis, H., 604 Dipalmitoylphosphatidylcholine, 231 Dipeptides, in chymotrypsin activation, 172, 172F DIPF (diisopropylphosphofluoridate), 164, 195 Diphosphoric acid esters, 4T Diphosphoryl group, 4T Diploid state, 65 Dipole–dipole interactions, 27 Disaccharides, 283, 288 Discontinuous DNA synthesis, 523 Disease(s). See also specific diseases of amino acid metabolism, 480 biochemical causes of, 2 gene therapy for, 78, 78F mutations as cause of, 63–64 protein misfolding as cause of, 105–106 RNA interference for treatment of, 572 single-nucleotide polymorphisms as cause of, 68 Disorders of fuel metabolism, 507–511 carbohydrate-related disorders, 353–355 diabetes, 509–511 metabolic syndrome, 511 obesity, 508–509, 511 starvation, 507–508 Disseminated intravascular coagulation (DIC), 182 Dissociation constant (Kd), 261 Disulfide bonds: between amino acids, 89 and protein stability, 101, 101F Divergent evolution, 170 DNA (deoxyribonucleic acid), 2F, 52–79. See also Genes; Protein synthesis; Replication altering, 76–79 amplification of, 71–74 in central dogma of molecular biology, 61–64 complementary, 530 covalent modification of, 555 decoding of, 62–63 denaturation/renaturation of, 59–61 double helix structure of, 56–59 double-stranded, 521–522, 535–538, 537F in genomics, 64–68 in glucocorticoid–DNA complex, 275F hybridization of mRNA and, 567–568, 568F hydrogen bonding in, 27 information contained in, 52 laboratory manipulations of, 68–79 melting curve for, 59, 60, 60F melting temperature of, 59–60, 60T model of, 57–58, 58F and nucleic acid structures, 56–61 nucleotides of, 52–55 palindromic, 69 proofreading of, 525 recombinant, 69–71, 70F in RNA–DNA hybrid double helix, 59, 59F and RNA polymerase, 563, 563F sequencing of, 74–76 single-stranded, 521–522, 536 A-DNA, 59

B-DNA, 59 DNA-binding domains (replication protein A), 521, 521F DNA binding proteins, 558–559 DNA breathing, 550 DNA chips, 311 DNA fingerprinting, 74 DNA glycosylase, 533–534, 534F DNA gyrase, 543 DNA helicases, 521–522, 521F DnaK, 598 DNA ligase, 70, 527–528 dNaM, 82 DNA methyltransferase, 555, 576 DNA packaging, 540–544 effects of, on transcription, 553–555 negative supercoiling for, 541 in nucleosomes, 543–544 topoisomerases in, 541–543 DNA photolyase, 533 DNA polymerases, 86F, 522–525 in base excision repair, 535 defined, 520 in DNA sequencing, 74–76, 75F, 76F errors made by, 531–532, 535 in factory model of replication, 520 mechanism of, 522–525, 522F in nucleotide excision repair, 535 in polymerase chain reaction, 71, 72 problems faced by, 522–523 proofreading by, 525 structure of, 523–524 at telomeres, 528 DNA polymerase I, 524–526, 524F DNA polymerase III, 548 DNA polymerase V, 548 DNA polymerase α, 525 DNA polymerase-associated clamps, 524, 524F DNA polymerase δ, 548 DNA polymerase ε, 525 DNA polymerase η, 535 DNA polymerase μ, 536 DNA repair, 533–538 base excision repair, 533–535 and cancer, 539–540 and DNA damage, 531–533 by enzymes, 533 mismatch repair, 533 nonhomologous end-joining, 535–536 nucleotide excision repair, 535 recombination, 536–538 DNA replication, see Replication DNA–RNA hybrid double helix, 59, 59F DNase (deoxyribonuclease), 546, 549–550 DNP (dinitrophenol), 409 dNTP (deoxynucleoside triphosphate), 522F 4,7,10,13,16,19-Docosahexaenoic acid (DHA), 216, 217T Dodder vine, 551 Dodecane, 29 Domains, 98 bromo-, 554, 554F DNA-binding, 521, 521F Dopamine, 475 Double helix structure: of DNA, 56–59 RNA–DNA hybrid with, 59, 59F Double-stranded DNA: nonhomologous end-joining to repair, 535–536 recombination to repair, 536–538, 537F single-stranded DNA from, 521–522 Down syndrome, 116 Doxorubicin, 549 Drosophila melanogaster, 64T, 67, 68F Drugs: antidiabetic, 510T commonly prescribed, 206T

development of, 204–206 fluorine in, 28 and microtubule dynamics, 134–135 pharmacokinetics of, 204 Drug design, rational, 204–205 Drug resistance, 247 D sugars, 284–285 dTMP (deoxythymidylate), 195, 489 Duloxetine, 206T dUMP (deoxyuridylate), 195, 489

E E, see Glutamate E. coli, see Escherichia coli E1, 363, 363F, 364 E2, 363–364, 363F E3, 363, 363F, 364 Eastern skunk cabbage, 409 Ebola virus, 72 EBS (epidermolysis bullosa simplex), 136 EC (Enzyme Commission), 157 EcoRI, 69, 69T, 116 EcoRV, 69, 69T, 116 Edema factor, 280 Edidin, Michael, 234 Edman degradation, 109 Edmonson, A. B., 115 Edrophonium, 210 EDTA, 603 eEF1α, 592T eEF2, 592T EFs (elongation factors), 591–593, 592T Effectors, negative vs. positive, 203 EF-G: function of, 592T in peptidyl transferase reaction, 593, 594, 594F and ribosome recycling factor, 596 EF-Tu: and EF-G, 594 function of, 592, 592F, 592T tRNA complex with, 591, 591F Ehlers–Danlos syndrome, 140 EI (enzyme–inhibitor) complex, 196 Eicosanoids, 275, 275F 5,8,11,14,17-Eicosapentaenoic acid (EPA), 216, 217T eIF2, 590, 592T 80S ribosome, 586F Elastase, 109T percent sequence identity for, 170T specificity pocket of, 170, 171, 171F structure of, 170, 170F substrate specificity of, 170 Electrochemical gradients, 320, 400–401 Electrons: in oxidation–reduction reactions, 308–309, 386 tendency to accept, 386–388 Electron crystallography, 112 Electronegativity, 26, 26T, 28 Electron flow: cyclic, 420, 421F in cytochrome b6 f, 419 noncyclic, 420, 420F in Photosystem II, 418F and reduction potential, 388 Electron tomography, 390, 391F Electron transport chain, 255, 389F, 390–399 Complex I in, 392–394 Complex III in, 394–397 Complex IV in, 397–399 and free radicals, 398 and mitochondrial membrane compartments, 390–392 and oxidative phosphorylation, 400 and ubiquinol from oxidation reactions, 394 Electrophiles, 163

Electrophoresis, 70 Electrostatic catalysis, 169 Electrostatic forces, 26–28 Elements, in biological molecules, 3, 3F Elongases, 447, 448 Elongation: in transcription, 564–565, 565F in translation, 590–592 Elongation cycle, E. coli, 594, 595F Elongation factors (EFs), 591–593, 592T Embryonic stem cells, 494 Enalapril, 214 Enantiomers, 284–285 (−) End: of actin filaments, 131 of microtubules, 134, 145 (+) End: of actin filaments, 131 of microtubules, 134, 145 Endergonic reactions, 11–12 End-joining, nonhomologous, 535–536, 536F Endocytosis, 252, 252F, 253 Endonucleases, 59 AP, 534–535, 534F restriction, 69, 69T Endopeptidases, 90 Endothelin, 7, 7F Endothermic reactions, 11 Enediolate intermediate, 197, 423F Energy. See also Free energy (G) activation, 158–160 in citric acid cycle, 372 during glycolysis, 330–340 and metabolism, 10–14 in organisms, 3 in photoexcited molecules, 414, 414F of photons, 412 Enhancers, 558–560, 559F Enolase: in gluconeogenesis, 345F in glycolysis, 331F, 338 Enolpyruvate, 339 Enoyl-ACP reductase, 446F, 450 Enoyl-CoA, 438F, 442, 443 Enoyl-CoA hydratase, 438F, 439, 440 Enoyl-CoA isomerase, 439 Enthalpy (H): and free energy, 10–11 and solvation, 29, 30 Enthalpy change (ΔH), 11 Entner–Doudoroff pathway, 358 Entropy (S): in free energy, 10–11 illustration of, 10F in living organisms, 14 Entropy change (ΔS), 11, 30 Environment, cancer and, 538 Enzymes, 154–182. See also Proteins; Ribozymes; specific enzymes allosteric, 194 in blood coagulation, 173–175 as catalysts, 13 catalytic mechanisms of, 158–166 chymotrypsin as model for, 169–173 Class I, 487, 496 classification of, 157T defined, 155 features of, 154–157 in gluconeogenesis, 345–346 glycogen storage diseases caused by, 354–355, 354T glycolytic, 345–346, 353–354 multienzyme complexes, 365 multifunctional, 445 mutations in, 373 nomenclature for, 157–158 in pyruvate dehydrogenase complex, 363 regulating activity of, 203, 203F. See also Enzyme inhibition

In dex repair, 533 saturated, 185 unique properties of, 166–169 Enzyme-catalyzed reactions: concentration changes in, 187, 187F rate equations for, 187–189. See also Michaelis–Menten equation Enzyme Commission (EC), 157 Enzyme inhibition, 194–207 and allosteric regulation, 200–203 competitive, 195–197, 195F by drugs, 204–206 factors influencing, 203 irreversible, 195 mixed, 199 noncompetitive, 199, 200F by transition state analogs, 197–199 uncompetitive, 200 Enzyme–inhibitor (EI) complex, 196 Enzyme kinetics, 183–194 about, 183–185 defined, 184 Michaelis–Menten equation for, 186–192 for non-Michaelis–Menten enzymes, 192–194 Enzyme–PLP Schiff base, 467, 468F Enzyme–substrate (ES) complex, 184–185 EPA (5,8,11,14,17-eicosapentaenoic acid), 216, 217T Epidermolysis bullosa simplex (EBS), 136 Epigenetic modifications, 555 Epimers, 285 Epinephrine, 278, 459 extracellular signals from, 261T and glucagon action, 504–505, 505F as ligand for β-adrenergic receptor, 265 in phosphoinositide signaling pathway, 269 from tyrosine, 178, 475 EPO (erythropoietin), 71T, 261T ℰ (reduction potential): free energy change from, 388–390 in Photosystem II, 417, 418F standard, 386–387, 387T and tendency to accept electrons, 386–388 ℰº′ (standard reduction potential), 386–387, 387T Equilibrium, 11 Equilibrium constant (Keq), 314 Eranthis seed oil, 230 Erasers, 554 eRF1, 592T, 595 eRF3, 595, 596 Ergosterol, 242, 462 Erythrocyte ghosts, 256 Erythromycin, 594 Erythropoietin (EPO), 71T, 261T Erythrose-4-phosphate, 193, 194, 473 Escherichia coli (E. coli): carbamoyl phosphate synthetase of, 93T citric acid cycle of, 374 cytoplasmic volume and growth medium for, 49 DNA damage in, 548 DnaK in, 598 DNA ligase of, 546 DNA polymerase-associated clamps of, 524, 524F DNA polymerases of, 86F, 523–526, 524F EcoRI from, 69, 69T EcoRV from, 69, 69T EF-Tu in, 591 elongation cycle in, 594, 595F evolution of, 17F fumarase of, 104F

generation time for, 17 genome of, 64T, 67 glutamine synthetase of, 466, 467F, 492 GroEL/GroES complex of, 598 isocitrate dehydrogenase in, 382 lac operon of, 560–561 maltoporin of, 86F as model organism, 65F nitrogenous bases used by, 81 OmpF protein of, 240–241, 240F peptidoglycan of, 295, 295F promoters for, 556, 556F, 576 RecA of, 536, 537F release factors of, 595 replication in, 527–528, 527F ribonuclease H1 from, 179 ribosome components of, 586T RNA polymerase of, 556 signal recognition particle of, 599 single-strand binding protein of, 521, 546 transcription termination in, 565 translation initiation in, 589, 590, 590F translation termination in, 595–596, 596F transpeptidation in, 593 tRNA in, 603 ES (enzyme–substrate) complex, 184–185 E site (exit site), 587, 591 Esomeprazole, 206T Essential amino acids, 312T, 469–474 defined, 469 nonessential vs., 470T synthesis of, 471–474 Essential fatty acids, 312T, 448–449 Essential substances, 312, 312T Esters, 4T Ester linkage, 4T Ethane, 13T Ethanol: from acetaldehyde, 163–164 dielectric constant for, 29T metabolism of, 342 oxidation state of carbon in, 13T from pyruvate, 341–342, 341F standard reduction potential of, 387T Ethanolamine, 453, 454F Ethanolamine head group (lipids), 220F Ethene, 13T Ethers, 4T Ether linkage, 4T Ethidium bromide, 547 Ethyl alcohol, see Ethanol N-Ethylmaleimide, 179 Eubacteria, see Bacteria Euchromatin, 544 Eukarya, 16–18, 18F Eukaryotes: cell nucleus of, 544, 544F chromosomes of, 528, 528F circularization of mRNA for, 590, 590F DNA-binding proteins of, 558–559 DNA ligation for, 527 DNA packaging for, 543–544 evolution of, 16–18, 17F, 18F helicase of, 521 membrane translocation for, 600F mRNA processing in, 567–569 oligosaccharides of, 292 origins of replication in, 520 possible origin of, 18, 18F promoter elements for, 556, 556F pyruvate dehydrogenase in, 363 ribosomal subunits of, 587, 587F ribosomes of, 586–588, 586T RNA polymerases of, 556–557, 562 rRNA processing in, 572, 572F signal recognition particle of, 599–600

splicing reactions for, 567–569 telomeres of, 528–529 transcription elongation in, 564–565 transcription factors for, 557 transcription termination in, 566 translation factors of, 592T translation initiation in, 590 Evolution: of citric acid cycle, 373–374 divergent vs. convergent, 170 of globins, 123, 123F of myoglobin, 121–123 of organisms, 2 and origin of life, 14–17 process of, 17 of serine proteases, 170 Evolutionary tree, 17, 18F Exciton transfer, 414, 415 Exercise, 32 Exergonic reactions, 11–12 Exit site (E site), 587, 591 Exocytosis, 248–249, 252 Exons, 568 Exonucleases: defined, 59 3′→5′ , 525, 526 in transcription termination, 566 Exopeptidases, 90 Exosomes, 253 Exothermic reactions, 11 Explosives, 110 Expression, see Gene expression Extracellular fluid, 32, 32F Extracellular matrix, 136–137 Extracellular signals, 261, 261T, 264 Extrinsic membrane proteins, 225F, 226 EZH2 enzyme, 576

F F, see Phenylalanine ℱ (Faraday constant), 236, 387 F-actin, 131 Factor VII, 174, 174T Factor IX, 174, 174T, 175, 182 Factor Xa, 174 Factor XIa, 174, 174T Factory model of replication, 520 FAD, see Flavin adenine dinucleotide FADH2 (flavin adenine dinucleotide, reduced form), 371, 387T Faraday constant (ℱ), 236, 387 FAS (fatty acid synthase) inhibitor, 516 Fat(s). See also Adipose tissue dietary, 433, 448 subcutaneous vs. visceral, 511 F-ATP synthase, 401 Fat toxicity, 511 Fatty acids: defined, 216 essential, 312T, 448–449 free, 435 hydrocarbon chains of, 221 insulin and uptake of, 503, 510 metabolic control mechanisms of, 449, 449F as metabolic fuels, 507, 508T mobilization of, 305 names and structures of, 217T odd-chain, 440–442 omega-3, 216 omega-6, 216 saturated, 216, 217T storage of, 304 ubiquinol from, 394 unsaturated, 216, 217T, 439–440 Fatty acid oxidation, 435–443 activation of fatty acids for, 435–436 β oxidation, 436–439 of odd-chain fatty acids, 440–442 in peroxisomes, 442–443 synthesis of fatty acids vs., 443 for unsaturated fatty acids, 439–440

I-7

Fatty acid synthase: mammalian, 445F reactions catalyzed by, 445–447 regulation of, 449, 449F Fatty acid synthase (FAS) inhibitor, 516 Fatty acid synthesis, 443–452, 446F acetyl-CoA carboxylase in, 444–445 activation and inhibition of, 449–450 cellular location of, 498F elongation and desaturation processes in, 447–449 fatty acid synthase in, 445–447 ketone bodies in, 450–452 Fatty acyl-CoA, 438F Fatty acyl groups, 435 Favorable reactions, 316–318 F0 component (ATP synthase), 401, 401F F1 component (ATP synthase), 401–403, 401F, 403F FDG ([18F]fluorodeoxyglucose), 356 FDNP (1-fluoro-2,4-dinitrophenol), 179 Feedback inhibitors, 201–203 Feed-forward activation, 339 FeMo cofactor, 465, 465F Fermentation, 341 Ferredoxin, 420, 420F, 426 Ferredoxin–NADP+ reductase, 420 Fe–S proteins, see Iron–sulfur proteins Fetal hemoglobin, 127, 149 Fever, 255 Fiber, 299 Fibrin, 173 Fibrinogen, 173, 174T Fibroblasts, 391F Fibrous proteins, 97. See also Motor proteins 50S subunit, 586, 586T, 587, 587F Fireflies, luciferase enzyme of, 154 First law of thermodynamics, 10 First-order reactions, 186 Fischer, E., 166, 207 Fischer projections, 285–286 Fish. See also specific species actin microfilament dynamics for, 132F hemoglobin in, 149 FISH (fluorescence in situ hybridization), 81 Fish oil, 230 5′ End, 56 5′ Capping, 567, 567F, 570 Fixation: carbon, see Carbon fixation nitrogen, 464–466 Flavin adenine dinucleotide (FAD), 578 in citric acid cycle, 371 standard reduction potential of, 387T structure and function of, 54, 55F Flavin adenine dinucleotide, reduced form (FADH2), 371, 387T Flavin coenzymes, 313T Flavin mononucleotide (FMN), 392, 393, 393F Flavin mononucleotide, reduced form (FMNH2), 393F Flavodoxin, 98F Flip-flop, in lipid bilayers, 224 Flippases, 224 Fluid mosaic model, 228–229, 228F Fluorescence, 414 Fluorescence in situ hybridization (FISH), 81 Fluorescence photobleaching recovery studies, 234 Fluorine: electronegativity of, 26T in pharmaceuticals, 28 [18F]Fluorodeoxyglucose (FDG), 356 1-Fluoro-2,4-dinitrophenol (FDNP), 179 5-Fluorouracil, 195, 496

I-8 Index Fluoxetine: effect of, on rat brain, 407 fluorine in, 28 and serotonin transport, 250, 475 and signaling by serotonin receptor, 258 Flurofamide, 181 Flu virus, 22 Flux, 310, 321 FMN, see Flavin mononucleotide FMNH2 (flavin mononucleotide, reduced form), 393F Folate, 471, 472 Folding of proteins, see Protein folding Folic acid, 313T Food: and dietary guidelines, 303 digestion of, 302–303 genetically modified, 72 RNA interference for, 572 Food and Drug Administration, 205 Food groups, 303 Formaldehyde, 13T, 14 Formamide, 29T Formic acid, 13T, 37T N-Formylmethionine–tRNAfMet, 589, 590 40S subunit, 586, 586T, 587F 454 sequencing, 75, 75F Fractional saturation (Y), 121 Franklin, R., 57 Free energy (G): from β oxidation, 438–439 from chemiosmosis, 421–422 and entropy/enthalpy, 10–11 in metabolic reactions, 318–321 Free energy change, see ΔG Free energy change of reaction (ΔGreaction), 159, 159F. See also Standard free energy change of reaction (ΔGº′) Free energy of activation (ΔG‡), 158–160 Free radicals, 398 Fructose: catabolism of, 340 structure of, 284 in sucrose, 288 Fructose bisphosphatase, 345F, 346–347 Fructose-1,6-bisphosphate: allosteric regulation of, 201–203 in gluconeogenesis, 345F, 346–347 in glycolysis, 331F, 333–335, 334F Fructose-2,6-bisphosphate, 333, 333F Fructose intolerance, 354, 360 Fructose-1-phosphate, 340, 354 Fructose-6-phosphate, 326 in gluconeogenesis, 345F, 346–347 in glycolysis, 331F, 332–333 and pentose phosphate pathway, 352 in phosphofructokinase reaction, 201–203 structure of, 287 in sucrose synthesis, 427 in transketolase reaction, 193, 194 Fructose-1-phosphate aldolase, 360 Fruit flies, see Drosophila melanogaster F-type ATPases, 247 Fuel metabolism integration, 498–501 and intestinal microbiome, 500 metabolites in, 499–501 and organ specialization, 498–499 Fuels, metabolic, see Metabolic fuels Fuel storage: and insulin, 503–504 in polysaccharides, 288–289 Fumarase: in β oxidation, 441F in citric acid cycle, 367F, 371 mutations in, 373 quaternary structure, 104F

Fumarate: in β oxidation, 441F in citric acid cycle, 367F, 370, 371 standard reduction potential of, 387T from succinate, 195–196 in urea cycle, 482, 483F Fumarate hydratase, see Fumarase Fumonisins, 462 Functional groups. See also specific Functional groups acid and base properties of, 40 hydrogen bonding in, 26, 27 types of, 4, 4T Fungi: genome size and gene number, 64T in microbiome, 500 Funk, C., 312 Fusion proteins, 114 Futile cycles, 347

G G, see Glycine; Guanine G, see Free energy ΔG‡ (free energy of activation), 158–160 ΔG (free energy change), 11–13 in coupled reactions, 12–13, 13F and energy sources for metabolic reactions, 318–321 for glycogen synthesis, 348–349 for glycolysis reactions, 332, 339, 339F in human muscles, 320 and hydrophobic effect, 30 in metabolic reactions, 314–321 for proton gradient, 400 and reactant concentration, 314–316 and reaction coupling, 316–318 and reduction potential, 388–390 regulation of reactions with large, 321 for spontaneous processes, 11–12 for transmembrane ion movement, 237, 239 ΔGº′, see Standard free energy change of reaction ΔGreaction (free energy change of reaction), 159, 159F G16BP (glucose-1,6-bisphosphate), 360 GABA (γ-aminobutyrate), 475 G-actin, 131 Galactose: in carbohydrate metabolism disorders, 354 catabolism of, 340 from lactose, 560 in lactose, 288 D-Galactose, 285 β-Galactosidase, 70, 560, 561 Galactosyl transferase, 210 Galacturonate, 290 γ subunit (ATP synthase), 402, 403 Gangliosides, 219F Garrod, A., 480 Gas constant (R), 236 Gastric juice, 34T Gastric lipase, 323 Gastric ulcers, 495 Gated ion channels, 242–243 GCN4 transcription factor, 559F GDP, see Guanosine diphosphate GEFs (guanine nucleotide exchange factors), 269, 280 Gelatin, 152 Gel electrophoresis, SDS-PAGE, 109, 109F Gel filtration chromatography, 107 GenBank, 64 Genes. See also DNA as cause of disease, 63–65 in central dogma of molecular biology, 61–62

defined, 553–554 functional classifications of, 67, 68F homologous, 67 identifying, 66–67 Mendel’s experiments with, 52 number of, 64T, 65–66 orphan, 67 removing introns from, 567–569 Gene expression. See also Transcription and central dogma, 61–62 components for, 580 by operons, 560–561 Gene knockouts, 78 General transcription factors, 557–558 Gene therapy, 78, 78T Genetically modified organisms, 72, 78 Genetic code, 63T, 581T for amino acid variants, 585 and central dogma, 62 redundancy of, 581 and tRNA, 582–585 Genetic defects, DNA testing for, 76 Genetic Testing Registry, 64 Gene transfer, horizontal, 67 Genome(s). See also Human genome chimpanzee, 83 defined, 62 of model organisms, 65F number of genes and size of, 64T Genome maps, 67, 67F Genome sequencing, 66–67 Genome-wide association studies (GWAS), 68 Genomics, 64–68 biological functions in, 67–68 organism complexity and gene number in, 65–66 sequence comparisons in, 66–67 Gentobiose, 298 Geraniol, 220, 222 Ghrelin, 505T, 506 Giant barrel sponge, 183 Gleevec (imatinib), 273 Glipizide, 510T Gln, see Glutamine GlnRS, 584, 584F Globins, 121–123, 123F, 605 Globular proteins, 97, 151 Glu, see Glutamate Glucagon, 494 in mammalian fuel metabolism, 504–505, 505F structure of, 504, 504F Glucocerebroside, 231 Glucocorticoids, 274 Glucocorticoid receptors, 275F Glucogenic amino acids, 477–480, 477T Glucokinase, 502–503, 502F Gluconeogenesis, 344–347 cellular location of, 498F and Cori cycle, 500 defined, 344 enzymes in, 345–346 and glucose–alanine cycle, 501 regulation of, 346–347 summary of reactions, 345F Gluconeogenic enzymes, 345–346 Glucophage (metformin), 510T Glucosamine, 287, 287F Glucose, 5, 284. See also Glycogen; Glycolysis aqueous solubility of, 28 ATP yield from, 372, 372F in cancer cells, 512 central role of, 329 chair conformation of, 286 in diabetes, 509–510 Haworth and Fischer projections of, 285, 285F and hexokinase, 169 as hydrophilic substance, 30 and insulin release, 502–503

from lactose, 560 in lactose, 288 phosphorylation of, 317 polymers of, 8, 9F reaction of methanol with, 286, 286F in starvation, 507–508 storage of, 304 in sucrose, 288 universality of, 14 D-Glucose, 284–285 Glucose–alanine cycle, 501, 501F Glucose-1,6-bisphosphate (G16BP), 360 Glucose metabolism, 329–355 context for reactions of, 330F and disorders of carbohydrate metabolism, 353–355 gluconeogenesis in, 344–347 glycogen synthesis and degradation, 347–349 glycolysis in, 330–344 pentose phosphate pathway in, 350–352 summary of reactions in, 352, 353F Glucose-6-phosphatase: deficiency in, 354, 354T in gluconeogenesis, 345F, 346 in glycogenolysis, 349 Glucose-1-phosphate: free energy change for hydrolysis of, 318T in glycogenolysis, 349 in glycogen synthesis, 348 in mobilization of metabolic fuels, 305, 305F in starch synthesis, 426 in sucrose synthesis, 427 Glucose-6-phosphate, 514 free energy change for hydrolysis of, 318T in gluconeogenesis, 345F in glycogenolysis, 349 in glycogen synthesis, 348 in glycolysis, 330, 331F, 332 in pentose phosphate pathway, 350 from phosphorylation of glucose, 317 in starch synthesis, 426 in sucrose synthesis, 427 Glucose-6-phosphate dehydrogenase, 211 deficiency of, 354 in pentose phosphate pathway, 350 Glucose transporters: of intestinal cells, 248, 248F of red blood cells, 244F α-1,4-Glucosidase, 354T Glucotrol (glipizide), 510T Glucuronate, 287F GluRS, 584 GLUT1 transporter, 256 GLUT2 transporter, 354T GLUT4 transporter, 503, 503F Glutamate (Glu, E): abbreviations for, 87 as acid–base catalyst, 162F in alanine aminotransferase reaction, 378 in ammonia assimilation, 466, 467 catabolism of, 477, 478 as charged amino acid, 87F, 89 from citric acid cycle intermediates, 375 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T as mitochondrial respiration substrate, 410 nervous system signaling by, 90 as neurotransmitter, 256 as nonessential amino acid, 470T percent occurrence of, 91F pK value of groups in, 91T in synthesis reactions, 470, 474

In dex transamination of, 467 and urea cycle, 481–483 and zinc fingers, 101 Glutamate dehydrogenase: in cancer cells, 512, 513 and nitrogen for urea cycle, 481, 482 reductive amination by, 375 Glutamate synthase, 466–467 Glutamine (Gln, Q): abbreviations for, 87 in ammonia assimilation, 466 in cancer cells, 512–513 catabolism of, 478, 481 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T metabolism of, 44 as nonessential amino acid, 470T percent occurrence of, 91F as polar amino acid, 87F, 89 in synthesis reactions, 470, 474 Glutamine synthetase, 492 in ammonia assimilation, 466–467 composition, 93T structure of, 467F γ-Glutamylcysteine synthetase, 210 Glutaredoxin, 112F Glutathione, 114, 361 Glutathione transferase, 116 GLUT proteins, 244, 245F GLUT1, 256 GLUT2, 354T GLUT4, 503, 503F Gly, see Glycine Glycans, 287–288 Glyceraldehyde, 283, 284, 340 D-Glyceraldehyde, 284 L-Glyceraldehyde, 284 Glyceraldehyde-3-phosphate, 287 in Calvin cycle, 425, 425F, 426 in gluconeogenesis, 345F in glycolysis, 331F, 333–336, 334F, 337F as metabolic intermediate, 307F and pentose phosphate pathway, 352 in starch synthesis, 426 in sucrose synthesis, 427 in transketolase reaction, 193, 194 in triose phosphate isomerase reaction, 184, 197 Glyceraldehyde-3-phosphate dehydrogenase, 357 in Calvin cycle, 424, 425F in gluconeogenesis, 345F, 346 in glycolysis, 331F, 336, 337F as two-domain protein, 99F Glycerol, 217, 435, 459 Glycerol-3-phosphate: free energy change for hydrolysis of, 318T from fructose metabolism, 340 in triacylglycerol synthesis, 452, 453F Glycerol-3-phosphate dehydrogenase: in oxidative phosphorylation, 394, 395F in triacylglycerol synthesis, 452 Glycerophospholipids: defined, 217–219 synthesis of, 453–454 Glycinamide, 37T Glycine (Gly, G): abbreviations for, 87 catabolism of, 478–479 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T as nonessential amino acid, 470T percent occurrence of, 91F as polar amino acid, 87F, 89 and structure of collagen, 137, 138 synthesis of, 471

Glycine cleavage system, 478 Glycine phosphonate, 473 Glycogen. See also Glucose degradation of, 349 fuel storage in, 288–289 and insulin, 503 phosphorolysis of, 304–305, 305F and starvation, 503 structure of, 304, 304F synthesis of, 347–349 Glycogen debranching enzyme, 354, 354T Glycogenolysis, 349 Glycogen phosphorylase: in glycogenolysis, 349 glycogen storage diseases involving, 354T, 355 and insulin, 503 and protein kinase A, 267 reciprocal regulation of glycogen synthase and, 504, 504F Glycogen storage diseases, 354–355, 354T, 515 Glycogen synthase: glycogen storage diseases involving, 354T in glycogen synthesis, 349 and insulin, 503 reciprocal regulation of glycogen phosphorylase and, 504, 504F Glycogen synthase kinase 3, 515 Glycolipids, 218 Glycolysis, 330–344 and alcohol metabolism, 342 in cancer cells, 512 catabolism of other sugars vs., 340 cellular location of, 498F and Cori cycle, 500 defined, 330 energy-investment phase of, 330–336 energy-payoff phase of, 336–340 and gluconeogenesis, 345F intermediates of, 307, 307F powering muscles with, 320 pyruvate’s fate after, 341–344 standard free energy changes for, 342–343, 343F summary of reactions, 331F Glycolytic enzymes: deficiencies in, 353–354 in gluconeogenesis, 345–346 Glycomics, 287 Glycophorin, 115 Glycoproteins, 291–295 in ABO blood group system, 293 defined, 229 oligosaccharides, 292–294 peptidoglycan, 295 proteoglycans, 294–295 Glycosaminoglycans, 294–295 Glycosidases, 292 Glycosides, 286 Glycosidic bonds: defined, 8, 286 in disaccharides, 288–289 Glycosylation, 294, 600 Glycosylphosphatidylinositol, 227F, 228 Glycosylphosphatidylinositol (GPI) group, 233 Glycosyltransferases, 292 Glyoxylate, 377 Glyoxylate cycle, 376, 377 Glyoxysome, 377 Glyphosate, 473 GMP, see Guanosine monophosphate Golden Rice, 72 Gout, 134, 150 GPCRs, see G protein-coupled receptors GPI (glycosylphosphatidylinositol) group, 233 G proteins, 265–270, 267F activation of, 265–266

and activation of protein kinase A, 267 and adenylate cyclase, 266 and calmodulin, 270 defined, 263, 590 in GPCR–G protein complex, 265–266, 266F in phosphoinositide pathway, 269–270 switching off signals from, 267–269 G protein–coupled receptors (GPCRs): activation of G protein by, 265–266 defined, 263 in GPCR–G protein complex, 265–266, 266F signal transduction pathway for, 263, 263F transmembrane helices of, 265 G quartet, 547 Gradients: free energy associated with, 320 proton, 400–401, 400F in secondary active transport, 247–248 Great Lakes hemoglobin variant, 149 Green fluorescent protein, 114 Grelag goose, 149 Griffith, F., 80 GroEL/GroES complex, 598, 598F Growth factor receptors, 272 Growth hormone, 71T, 100F, 261T GTP, see Guanosine triphosphate GTPγS, 279 Guanine (G), 54T and Chargaff ’s rules, 56–57 as complement of C, 59 and DNA structure, 56–57 in genetic code, 63T, 581T hydrogen bonding by, 27 and melting point of DNA, 60, 60T and melting temperature, 59, 60T oxidative damage to, 532 as purine base, 53 wobble pairing with, 585F Guanine nucleotide exchange factors (GEFs), 269, 280 Guanosine, 54T Guanosine diphosphate (GDP), 54, 54T in citric acid cycle, 366 in gluconeogenesis, 345F in G protein signaling pathway, 266 tubulin binding of, 133 Guanosine monophosphate (GMP): nucleoside in, 54T in purine synthesis, 485–486, 486F Guanosine triphosphate (GTP), 54T in β oxidation, 439 in citric acid cycle, 366, 370, 370F, 372 in gluconeogenesis, 345F, 346 in GMP synthesis, 485 in G protein signaling pathway, 266 in protein elongation, 592 in transpeptidation, 594 tubulin binding of, 133 Guanylate, 54T Guanylyl-(αβ)-methylenediphosphonate (GMP · CPP), 150 Guide RNA, in CRISPR-Cas9 system, 77, 77F GWAS (genome-wide association studies), 68

H H, see Histidine H, see Enthalpy H+, see Hydrogen ion ΔH (enthalpy change), 11 H3K4me3 modification, 554T H3K9ac modification, 554T H3K9me2 modification, 554T H3K27me3 modification, 554T H4K16ac chromatin, 554T Haemophilus influenzae, 64T, 65

I-9

Hair: keratins in, 136, 136F straight and curly, 151 Half-reactions, 386 Haloarcula marismortui, 587F Halococcus saccharolyticus, 358 Hangover, 342 Haploid state, 65 Haptocorrin, 324 Hard keratins, 151 Hartnup disease, 324 HAT (histone acetyltransferase), 576 Haworth projections, 285–286 HDL, see High-density lipoproteins Head groups: ethanolamine, 220F polar, 217–219 Heart disease, 448 Heat, rate of reaction and, 155 Heat exhaustion, 32 Heat-shock protein (Hsp), 116, 597 HeLa cell, 588F Helical wheel, 115 Helicases, 521–522, 521F Helicobacter pylori, 383, 495 Helix–turn–helix (HTH) motif, 558 Heme: in Complex IV, 397F in cytochromes, 394, 395F in myoglobin, 120, 121F Hemicellulose, 290, 299 Hemoglobin: affinity for carbon monoxide of, 124 α and β chains of, 121, 122, 122F amino acid sequence for, 122, 122F composition of, 93T conformational shift in, 124–126, 125F cooperative binding by, 194 evolution of, 121–123 fetal, 127, 149 genetic variations in, 127–129, 129T, 149–150 H+ ion and BPG regulation of, 126–127 levels of structure for, 94F oxygen binding to, 123–127 quaternary structure of, 104, 104F, 105 in sickle cell disease, 63 subunit interactions in, 125, 125F Hemoglobin C, 129 Hemoglobin S, 128–129, 128F Hemophilia, 78T, 182 Henderson–Hasselbalch equation, 38–40 Hen egg white lysozyme, 101F Henseleit, K., 481 Hepatitis, 73 Hepatocytes, 2F HEPES (N-2-hydroxyethylpiperazineN′-2-ethanesulfonic acid), 37T Herbicides, 473 Herceptin (trastuzumab), 273 Hereditary pancreatitis, 182 Hershey, A., 80 hetero- (prefix), 104 Heterochromatin, 544 Heterotrophs, 301–302 Hexacoordinate Fe(II), 148 Hexadecyltrimethylammonium, 48 Hexokinase, 181 and glucokinase, 502, 502F in glycolysis, 330, 331F, 332, 339 induced fit in, 169, 169F Hexokinase II, 409 Hexoses, 284 HIFs (hypoxia-inducible factors), 373 High altitude acclimatization, 149 High-density lipoproteins (HDL): function of, 433, 434, 434F properties of, 433T

I-10 Index Highly repetitive sequences, in DNA, 66 High osmolarity glycerol (HOG) pathway, 281 High-performance liquid chromatography (HPLC), 109 Hill coefficient, 148 HindIII, 70F Hirudin, 182 Hirudo medicinalis, 182 Histidine (His, H): abbreviations for, 87 as acid–base catalyst, 162F catabolism of, 484 as essential amino acid, 312T, 470T genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T percent occurrence of, 91F pK value of ionizable groups in, 91T as polar amino acid, 87F, 89 synthesis of, 474 and zinc fingers, 101 Histones: amino acids in, 114 defined, 543 modifications to, 553–554, 553F, 554T Histone acetyltransferase (HAT), 576 Histone code, 553, 554 Histone methyltransferase (HMT), 576 HIV, see Human immunodeficiency virus HIV protease, 181, 198–199, 198F, 211 HIV reverse transcriptase, 93T, 530 HMG-CoA: in cholesterol synthesis, 454, 455, 455F in ketogenesis, 451F and statins, 455, 456, 456F HMG-CoA lyase, 451F HMG-CoA reductase, 455, 455F HMG-CoA synthase, 451F, 455F HMT (histone methyltransferase), 576 H3O+ (hydronium ion), 33, 35 HOG (high osmolarity glycerol) pathway, 281 Homeostasis, 2 homo- (prefix), 104 Homoarginine, 213 Homocysteine, 472 Homocystinuria, 472 Homodimer, 104 Homogentisate, 480 Homogentisate dioxygenase, 480 Homologous genes, 67 Homologous proteins, 123 Homologous recombination, 536, 537F Homo sapiens, see Humans Homotetramer, 104 Homotrimer, 104 Horizontal gene transfer, 67 Hormones, 501–507 and AMP-dependent protein kinases, 506 defined, 261, 501 epinephrine, 504–505 glucagon, 504–505 growth, 71T, 100F, 261T insulin, 502–504 lipid, 274–275, 274F peptide, 234 regulation of fuel metabolism by, 505–506, 505T Hormone response elements, 274 Hormone-sensitive lipase, 505 Hot packs, 21 HPLC (high-performance liquid chromatography), 109 Hsp (heat-shock protein), 116, 597 Hsp90, 605 HTH (helix–turn–helix) motif, 558

Humans, 2F. See also Cancer; Mammals; Mammalian fuel metabolism acid–base balance in, 42–45 chromosomes of, 53F C-reactive protein of, 93T cryoelectron tomography of cell, 588F diet of, 302, 303 DNA melting temperature for, 60T DNA molecules of, 519 DNA polymerase-associated clamp of, 524F DNA replication in, 525 endothelin of, 7, 7F essential and nonessential amino acids for, 469–474, 470T essential substances for, 312, 312T gene functions of, 67, 68F genomic analysis for, 67 hair of, 136, 136F hemoglobin of, 104F insulin of, 502F intestinal microbiome of, 500 ionic composition of intracellular vs. extracellular fluid for, 32F myoglobin of, 93T number of genes for, 64T, 65 pH of blood of, 34 powering muscles of, 320 preinitiation complex of, 557F proinsulin of, 599 protein-coding sequences for, 552 proteome size of, 586 skin of, 135F sweating by, 32 telomeric DNA of, 529, 531 Type I topoisomerase of, 542, 542F vitamins for, 312–314 water in make up of, 24 Human genome: chimpanzee vs., 83 map of, 67, 67F noncoding segments in, 66, 66F repetitive sequences in, 66 size of, 64T, 65 Human immunodeficiency virus (HIV), 530 gene therapy for, 78 PCR test for, 72, 73 treatment for, 198–199 Human Microbiome Project, 500 Humpback whale, 119 Huntington’s disease, 78 Hybridization, DNA and mRNA, 567–568, 568F Hydrated ions, 28 Hydration (hydration reaction): fumarase as catalyst for, 371 with nonpolar molecules, 30, 30F Hydrocarbon chains (fatty acid), 216–217 Hydrochloric acid, 35, 40, 41 Hydrocodone, 206T Hydrogen, 26T Hydrogen bonding: and characteristics of water, 24–26 in collagen, 139 electrostatic forces in, 26–28 low-barrier, 168, 168F and protein stability, 100 relative strength of, 26, 26F and secondary structures of proteins, 95 and water as solvent, 28–29 Hydrogen cyanide, 14 Hydrogen ion (H+): in acid–base chemistry, 33–40 and hydroxide ion, 34–35, 34F and oxygen-binding by hemoglobin, 126 and pH, 34, 34F, 35 Hydrogenobacter, 322

Hydrogen peroxide, 280, 443 Hydrogen sulfide, 387T Hydrolases, 157T Hydrolysis, 90. See also ATP hydrolysis of acetylcholine, 199, 200 of peptide bonds, 164–166 phosphate, 318–319, 318T rate of, 154–155 Hydronium ion (H3O+), 33, 35 Hydrophilic interactions, 31 Hydrophilic substances, 29 Hydrophilic surfaces (proteins), 98–99 Hydrophobic collapse, 102 Hydrophobic cores: of lipid bilayers, 31–32 of proteins, 98–99 Hydrophobic effect: and properties of water, 29–32 in protein folding, 100, 100F in quaternary structure, 104 Hydrophobic polypeptide segments, 597 Hydrophobic side chains, 87F, 88 Hydrophobic substances, 29, 215 Hydrothermal vents, 15F Hydrothermal vent worms, 151 Hydroxide ion (OH−): in acid–base chemistry, 33–40 hydrogen ion and, 33–34, 34F and pH of solution, 35 3-Hydroxyacyl-ACP, 447 3-Hydroxyacyl-ACP dehydrogenase, 446F 3-Hydroxyacyl-CoA, 438F 3-Hydroxyacyl-CoA dehydrogenase, 438F Hydroxyapatite, 140 N-Hydroxyarginine, 476 3-Hydroxybutyrate: in ketogenesis, 450, 451F in ketone body catabolism, 452F standard reduction potential of, 387T 3-Hydroxybutyrate dehydrogenase, 451F, 452F Hydroxybutyryl-ACP, 446F Hydroxycitrate, 382 N-2-Hydroxyethylpiperazine-N′-2ethanesulfonic acid (HEPES), 37T Hydroxyethyl-TPP, 363, 365F 2-Hydroxyglutarate, 373 Hydroxyl groups, 4T, 89, 90 p-Hydroxyphenylpyruvate, 480 p-Hydroxyphenylpyruvate dioxygenase, 480 Hydroxyproline, 137 Hydroxyurea, 149 Hypercholesterolemia, 434 Hyperglycemia, 509–510 Hyperoxia, 381 Hyperthyroidism, 516 Hypothalamus, 506, 506T Hypoxanthine, 81 Hypoxia-inducible factors (HIFs), 373

I I, see Isoleucine Ibuprofen, 276 Ice, 25, 25F IDLs (intermediate-density lipoproteins), 433T IFs (initiation factors), 589–590 IF-1, 589 IF-2, 589, 590, 592T IF-3, 589 Ile, see Isoleucine IleRS, 584, 602 Illumina sequencing method, 75–76, 76F, 546 Imatinib (Gleevec), 273 Imidazole, 181

Imidazolium ion, 37T, 89 β,γ-Imido nucleoside triphosphate, 577 Imines, 4T, 162, 163 Imino group, 4T, 137 Immortal cells, 531 Immunoglobulin, 93T IMP, 485–486, 486F Imprinting, 555 Inborn errors of metabolism, 480 Incretins, 505T, 506 Indole, 473, 474 Indole-3-glycerol phosphate, 473 Induced fit, 168, 169 Ingram, V., 128 Inhibition, enzyme, see Enzyme inhibition Inhibition constant (KI), 196–197 Inhibitors of fatty acid synthesis, 449–450 Initial velocity (v0), 188 Initiation factors (IFs), 589–590 Initiation of transcription, 552–561 covalent modification of DNA for, 555 and definition of gene, 552–553 DNA-binding proteins at, 558–559 and DNA packaging, 553–555 enhancers and silencers at, 558–560 for prokaryotic operons, 560–561 at promoters, 555–560 transcription factors at, 557 Initiation of translation, 589–590 Initiator tRNA, 589–590 Inorganic phosphate, see Pi Inorganic pyrophosphate (PPi): in DNA ligation, 527 in fatty acid activation, 436 free energy change for hydrolysis of, 318T in glycogen synthesis, 348–349 in pyrosequencing, 75 Inosine, 198 Inosine triphosphate, 577 Inositol, 455F Inositol trisphosphate, 269 Instant cold packs, 21 Insulin, 116, 359, 494 composition of, 93T covalent modification of, 600, 600F and diabetes, 509–510 fuel use and storage in response to, 503–504 in insulin receptor, 271 metabolic effects of, 503T as recombinant protein, 71T release of, 502–503 and starvation, 507 structure of, 86F, 502, 502F Insulin-dependent diabetes, 509 Insulin receptor: activation of, 272, 272F insulin binding by, 271 and release of insulin, 503 structure of, 271F Insulin resistance, 509–511, 516 Integral membrane proteins, 225–226, 225F Interconversion reactions: in Calvin cycle, 425 in pentose phosphate pathway, 352, 352F Interleukin-1 receptors, 233 Intermediates, 307–308. See also specific types of citric acid cycle, 375–378 defined, 306 of glycolysis, 307, 307F roles of, 307F Intermediate-density lipoproteins (IDLs), 433T Intermediate filaments: defined, 130

In dex distribution of, in single cell, 130F keratin as, 135–136 model of, 136–137, 137F Intermembrane space, 390 International Union of Biochemistry and Molecular Biology, 157 Intestine: glucose transport into cells of, 247, 248F microbiome of, 500 products of digestion in, 302, 303 Intracellular fluid, 32, 32F Intrinsically unstructured proteins, 103, 103F Intrinsic factor, 460 Intrinsic membrane proteins, 225–226, 225F Introns, 567–569 Inulin, 299 Invariant residues, 123, 148 in vitro reactions, 12 in vivo reactions, 12 Iodoacetate, 179, 335 Ions, in intra- and extracellular fluids, 32, 32F Ion channels. See also Potassium (K+) channels calcium, 248, 249F gated, 242–243 selectivity of, 241 sodium, 237 Ion exchange chromatography, 107, 108, 108F Ionic interactions, 26, 26F Ionization constant of water (Kw), 34 Ionization of acids, 36–37 Ionizing radiation, 532–533 Ion movement: and conformation of Na,K-ATPase, 246–247 and membrane potential, 237, 238 by membrane proteins, 237, 239–240 Ion pairs, 99–101, 100F Iron deficiency, 72 Iron-molybdenum complex (FeMo), 465, 465F Iron–sulfur proteins: in Complex I, 393, 393F in Complex III, 395, 396F in Q cycle, 396F Irregular secondary structure, 96, 97 Irreversible inhibitors, 195 Irreversible reactions, 332 Isoacceptor tRNAs, 582 Isocitrate, 367F, 368–369, 377 Isocitrate dehydrogenase, 382 mutations in, 373 reactions of, 367F, 369 regulation of citric acid cycle by, 372, 373 Isocitrate lyase, 377 Isoelectric point (pI), 108 Isoleucine (Ile, I): abbreviations for, 87 catabolism of, 479 as essential amino acid, 312T, 470T genetic code for, 63T, 581T as glucogenic/ketogenic amino acid, 477T as hydrophobic amino acid, 87F, 88 hydrophobicity of, 99T percent occurrence of, 91F synthesis of, 472 Isomaltase, 323 Isomerases, 157T Isomerization reactions: in Calvin cycle, 425 in citric acid cycle, 368–369 in pentose phosphate pathway, 351, 352F

in unsaturated fatty acid degradation, 440 Isoniazid, 450 6 N -Isopentenyladenosine, 573F Isopentenyl pyrophosphate, 454 Isoprene, 220 Isoprenoids, 220, 455 Isozymes, 158

J Jasmonate, 275 Joule (J), 315 Juvenile onset diabetes, 509

K K, see Lysine K+, see Potassium ions Ka (acid dissociation constant), 36 kcat, see Catalytic constant Kd (dissociation constant), 261 Keq (equilibrium constant), 314 KI (inhibition constant), 196–197 KM, see Michaelis constant Kw (ionization constant of water), 34 Kansas hemoglobin variant, 129T Kassinin, 117 kb (kilobase pairs), 58–59 K+ channels, see Potassium (K+) channels Kendrew, J., 97, 111 Keratan sulfate, 300 Keratins, 135–136 Ketimine, 468F α-Keto acids, 467, 468F 3-Ketoacyl-ACP reductase, 446F 3-Ketoacyl-ACP transferase, 446F Ketoacyl-CoA, 438F 3-Ketoacyl-CoA transferase, 452F α-Ketobutyrate, 472 Keto-enol tautomerization, 161 Ketogenesis, 450, 451F, 498F Ketogenic amino acids, 477, 477T, 479–480 α-Ketoglutarate: in amino acid catabolism, 477, 478 amino acid synthesis from, 470 in ammonia assimilation, 466, 467 in anaplerotic reactions, 378 as biosynthetic precursor, 375, 375F in citric acid cycle, 367F, 369–370 in transamination, 467, 469 in urea cycle, 481 α-Ketoglutarate dehydrogenase: reactions of, 367F, 369–370 regulation of citric acid cycle by, 372 Ketones, 4T Ketone bodies: catabolism of, 452F in diabetes, 510 in fatty acid synthesis, 450–452 in starvation, 508 Ketose, 284 Ketosteroid isomerase, 190T Khorana, H. G., 82, 602 Kidneys: in acid–base balance, 43–44 antiporter proteins in, 258 in glutamine catabolism, 481 ions in, 51 metabolic roles of, 499, 499F responses, to acidosis, 45 Kilobase pairs (kb), 58–59 Kinases, 264, 332. See also specific enzymes and types Kinesin, 143–146 processivity of, 146 reaction cycle of, 145–146, 145F structure of, 143–146, 144F Kinetics, 184. See also Enzyme kinetics kcat/KM ratio, 190–191 Klenow fragment, 86F

Knoop, F., 460 Koala, 497 Koshland, D., 168, 181 Krebs, H., 369, 481 Krebs cycle, see Citric acid cycle Ku protein, 536, 536F Kwashiorkor, 507

L L, see Leucine lac operon, 560–561, 576–577 lac repressor, 561, 561F Lactate, 517 and Cori cycle, 500 from pyruvate, 341, 341F standard reduction potential of, 387T Lactate dehydrogenase, 341 Lactate racemase, 324 Lactobacillus, 233 Lactococcus, 258 Lactose, 288, 340, 560 Lactose intolerance, 298 Lactose permease, 561 lacY gene, 576 lacZ gene, 70 Lagging strand: defined, 523 RNase and ligase for, 525–528 Lambert–Eaton syndrome, 258 L amino acids, 88 Lamprey, 149 Large intervening noncoding RNA (lincRNA), 552T Larval stages, 580 Lateral diffusion, 224–225 Lauric acid, 217T LDL, see Low-density lipoproteins Leading strand, 523 Leber’s congenital amaurosis, 78T Le Châtelier’s principle, 42 Leghemoglobin, 492 Leptin, 516 appetite suppression by, 516 and obesity, 508 regulation of fuel metabolism by, 505, 505T Lesch–Nyhan syndrome, 496 Lethal factor, 280 Leucine (Leu, L): abbreviations for, 87 catabolism of, 479 as essential amino acid, 312T, 470T genetic code for, 63T, 581T as hydrophobic amino acid, 87F, 88 hydrophobicity of, 99T as ketogenic amino acid, 477T percent occurrence of, 91F synthesis of, 472 Leucine zipper motifs, 559 Leukemia, 273 Levothyroxine, 206T Lewis dot structures, 162 Life: origin and evolution of, 14–18 thermodynamic possibility of, 12–14 Ligands: of allosteric proteins, 126 binding of receptors to, 261–263, 261F defined, 260 Ligases: reactions catalyzed by, 157T in replication, 527–528 Light: absorption of, 412–414, 414F availability of, 426 Light-harvesting complexes, 414–415, 415F Light reactions (photosynthesis), 415–422 and carbon fixation, 426 and chemiosmosis, 421–422 cytochrome b6 f in, 418

I-11

defined, 415 in Photosystem I, 419–421 in Photosystem II, 416–419 Lignin, 291 Lignoceric acid, 217T Limonene, 220, 222 lincRNA (Large intervening noncoding RNA), 552T Lineweaver–Burk plot: for competitive inhibition, 196, 196F KM and Vmax from, 191–192, 191F Linkages, 4T Linoleate: in arachidonate synthesis, 448F degradation of, 439–440 as essential fatty acid, 312T, 448–449 hydrocarbon chain of, 216 Linoleic acid, 217T Linolenate: in arachidonate synthesis, 448F as essential fatty acid, 312T, 448–449 α-Linolenic acid, 216, 217T γ-Linolenic acid, 217T Lipases. See also Phospholipases digestion by, 302, 302F gastric, 323 hormone-sensitive, 505 lipoprotein, 435 pancreatic, 230 Lipids, 215–222. See also Glycerophospholipids; Triacylglycerols about, 6 characteristics of, 432 defined, 215 fatty acids as, 216–217 physiological functions of, 219–222 with polar head groups, 217–219 Lipid bilayers, 31F, 222–225, 222F. See also Membranes α helices and, 226 asymmetry of, 224–225 defined, 222 diffusion and, 31–32, 32F fluid structure of, 223–224 hydrophobic core of, 31–32 membrane proteins in, 225–226 simulation of, 223F Lipid flippases, 247 Lipid hormones, 274–275, 274F Lipid-linked proteins, 225F, 227–228, 227F Lipid metabolism, 432–459 and cholesterol synthesis, 454–457 context for reactions of, 435, 435F and diabetes, 510 fatty acid oxidation in, 435–443 and fatty acid synthesis, 443–452 and lipid transport, 432–434 summary of, 457–458 and triacylglycerol/phospholipid synthesis, 452–454 Lipid transport, 432–434 Lipitor (atorvastatin), 206T, 456F Lipoamide: and lipoic acid, 313T and pyruvate dehydrogenase, 363–364, 365F structure of, 364F Lipoic acid, 313T, 387T Lipolysis, 504–505 Lipoproteins, 234, 433–434 characteristics of, 433, 433T defined, 303 function of, 433–434, 434F structure of, 433F Lipoprotein lipase, 435 Lisdexamfetamine, 206T Lisinopril, 206 Listeria monocytogenes, 299

I-12 Index Liver, 2F amino acid catabolism in, 476–477 AMPK effects on, 506F Cori cycle in, 499, 500, 500F glucose–alanine cycle in, 501, 501F and glycogen storage diseases, 354–355 insulin action on, 503T metabolic roles of, 498, 499F products of digestion in, 303 during starvation/malnutrition, 507–508 tumors in, 538F Liver glycogen phosphorylase, 354T LmrA transporter, 258 Lock-and-key model, 166–168 Locus, of DNA, 74 Loligo pealeii, 231 London dispersion forces, 27 Loops, polypeptide, 97 Loricifera, 385 Lovastatin, 456F Low-barrier hydrogen bonds, 168, 168F Low-density lipoproteins (LDL), 259 function of, 433–434, 434F and heart disease, 448 properties of, 433T L sugars, 284–285 Luciferase, 75, 75F, 154 Lung injury, acute, 50 Lyases, 157T Lymphotactin, 103F Lysine (Lys, K): abbreviations for, 87 as acid–base catalyst, 162F as charged amino acid, 87F, 89 and cross-linking of collagen, 139 as essential amino acid, 312T, 470T genetic code for, 63T, 581T hydrophobicity of, 99T as ketogenic amino acid, 477T percent occurrence of, 91F pK value of groups in, 91T Lysosomes, 252, 293, 305 Lysozyme, 101F, 179, 299 Lysyl hydroxylase, 151

M M, see Methionine M (moles per liter), 7 McArdle’s disease, 360 McCarty, M., 80 Mackerel, 148 MacLeod, C., 80 Macromolecules, 2, 6 Macronutrients, 303 Mad cow disease, 106 Magnesium ions: and DNA polymerase, 524 and enolase reaction, 338 in rubisco/Calvin cycle enzymes, 426 Maize, see Corn Major groove, of DNA, 58, 58F Malaria, 128–129 Malate: in β oxidation, 441F as biosynthetic precursor, 375F, 376 in C4 pathway, 424 in citric acid cycle, 367F, 371–372 in glyoxylate pathway, 377 standard reduction potential of, 387T Malate–aspartate shuttle system, 391, 391F Malate dehydrogenase, 177 in C4 pathway, 424 in citrate transport system, 376 in citric acid cycle, 367F, 371–372 in malate–aspartate shuttle system, 391F Malate synthase, 377 Malathion, 258 4-Maleylacetoacetate, 480

Malic enzyme: in β oxidation, 441F in C4 pathway, 424 in citric acid cycle, 376 Malonate, 196, 381 Malonyl-ACP, 446F, 447, 450 Malonyl-CoA, 445, 446F, 449 Maltoporin, 86F Mammals. See also specific types ATP synthase of, 402F Complex III of, 396F DNA methyltransferases of, 555 essential fatty acids for, 448–449 fatty acid synthase of, 445F fatty acid synthesis in, 443, 444 folate requirement for, 471 as heterotrophs, 302 phosphofructokinase regulation in, 333, 333F pore formation as defense of, 242 repair enzymes of, 533 ribosome components of, 586T RNA polymerase of, 562–563, 562F signal recognition particle of, 599 snoRNAs of, 572, 573 as transgenic organisms, 72 Mammalian fuel metabolism, 497–514. See also Oxidative phosphorylation in cancer cells, 511–513 disorders of, 507–511 hormonal control of, 501–507 integration of, 498–501 pathways of, see Metabolic pathways Manduca sexta, 359 Manganese cluster (Photosystem II), 417, 417F MAP kinase pathway, 280 Maple syrup urine disease, 480, 494 Marasmus, 507 Mass action ratio, 315–316 Mass spectrometry, 109–110, 111F Matrix: extracellular, 136–137 mitochondrial, 363, 390 Maximum reaction velocity, see Vmax MDMA (3,4-methylenedioxymethylamphetamine), 258 Mechanosensitive channels, 243, 243F Mediator complex, 558, 560, 560F, 565 Melamine, 47 Melatonin, 177, 475 Melittin, 234 Melting curve, DNA, 59, 60, 60F Melting point, lipid bilayer, 223–224 Melting temperature (Tm), of DNA, 59–60, 60T Membranes, 222–229. See also Lipid bilayers fluid mosaic model for, 228–229 mitochondrial, 390–392 oxidation–reduction reactions at, 309 proteins anchored in, 227–228 proteins that span, 226–227 Membrane fusion, 248–253, 252F and antidepressants, 250 and exosomes, 253 at nerve–muscle synapses, 248–250, 249F SNAREs in, 251–252 via endocytosis, 252, 253 Membrane potential (Δψ): defined, 236 ion movements and, 237, 238 for proton gradients, 400 Membrane proteins, 225–228, 225F with α helix structure, 226, 226F with β barrel structure, 226–227, 227F glycoproteins as, 229 integral, 225–226, 225F limitations on mobility of, 228, 229, 229F

lipid-linked, 225F, 227–228, 227F mediation of ion movement by, 237, 239–240 membrane translocation for, 599–600 Membrane translocation, 599–600, 600F Membrane transport, 235–254 active transport, 245–248 and membrane fusion, 248–253 passive transport, 240–245 thermodynamics of, 235–240 Mendel, G., 52 Menten, M., 189 Meselson, M., 82, 519–520 Messenger RNA, see mRNA Met, see Methionine Metabolic acidosis, 44 Metabolic alkalosis, 44–45 Metabolically irreversible reactions, 332 Metabolic fuels, 301–306. See also Glucose metabolism; Mammalian fuel metabolism; Oxidative phosphorylation amino acids as, 476–477 defined, 304 and dietary guidelines, 303 and insulin, 503–504 mobilization of, 304–306 monomers as, 304 products of digestion as, 302–303 sources of, 508, 508T Metabolic pathways, 306–314 common intermediates of, 307–308 complexity of, 310–312 human, vitamins in, 312–314 locations of, 498, 498F and metabolomics vs. transcriptomics/ proteomics, 311 oxidation–reduction reactions in, 308–309 Metabolic reactions, 301–322 free energy changes during, 314–321 fuels for, 301–306 in glucose metabolism, 352, 353F glucose metabolism in context of, 330F pathways for, 306–314 Metabolic syndrome, 511 Metabolism. See also Citric acid cycle; specific types, e.g.: Glucose metabolism of alcohol, 342 defined, 301 and energy, 10–14 features of, 310, 310F genomic analysis of, 67 microbiome’s contribution to, 500 nucleotide functions in, 54 vitamins in, 312–314 Metabolites: common amino acid, 470–471 defined, 306 in organs, 499–501 Metabolome, 311 Metabolomics, 311 Metal ion catalysis, 163–164 Metamorphosis, 580 Metformin, 359, 408, 510, 510T Methane, 13T, 308 Methane hydroxylase, 104F Methanococcus jannaschii, 64T Methanol, 358 dielectric constant for, 29T reaction with glucose of, 286, 286F Methionine (Met, M): abbreviations for, 87 as essential amino acid, 312T, 470T genetic code for, 63T, 581, 581T as glucogenic amino acid, 477T as hydrophobic amino acid, 87F, 88, 99, 99T initiation of translation at, 589 percent occurrence of, 91F

in SRP binding, 599 synthesis of, 472 Methionine synthase, 472 Methotrexate (MTX), 46–47, 489 1-Methyladenosine, 573F N 6-Methyladenosine, 573 Methylammonium ion, 37T Methylation: chemical agents for, 548 of DNA, 555, 572 and thymidine nucleotides, 488–489 3-Methylcytidine, 573F 5-Methylcytosine, 555, 576 3,4-Methylenedioxymethylamphetamine (MDMA), 258 N 5,N 10-Methylenetetrahydrofolate, 471F, 489 Methyl groups, 27 6 O -Methylguanine, 533 3-Methylhistidine, 494 (R)-Methylmalonyl-CoA, 441F (S)-Methylmalonyl-CoA, 441F Methylmalonyl-CoA mutase, 440, 441F, 442 Methylmalonyl-CoA racemase, 441F Methyltransferases, 533, 555, 576 Mevacor (lovastatin), 456F Mevalonate, 454–456, 455F Mice: DNA methyltransferase of, 576 embryonic stem cells in, 494 immunoglobulin of, 93T obesity in, 508F P-glycoprotein of, 247F PrP gene in, 116 Zif268 transcription factor of, 559F Micelles, 31, 31F Michaelis, L., 189 Michaelis constant (KM), 189–192 defined, 188 experimental determination of, 191–192 graphical determination of, 189–190, 190F and kcat/KM ratio, 190–191 Michaelis–Menten equation, 186–192 for competitive inhibition, 196 defined, 189 enzymes that do not follow, 192–194 kcat in, 190 kcat/KM in, 190–191 KM in, 189–192 and rate equations, 186–189 Vmax determination for, 191–192 Microarrays, 311 Microbes, metabolic interdependence of, 310–311 Microbiome, 500 Microenvironment, 91, 168–169 Microfilaments, see Actin filaments Micro RNA, see miRNA Microtubules, 130 assembly of, 133, 133F depolymerizing, 134, 134F distribution of, in single cell, 130F of dividing cells, 134F drugs affecting, 134–135 and kinesin, 143–146 tubulin in, 132–134 Microvesicles, 253 Miescher, F., 52 Milledgeville hemoglobin variant, 129T, 149 Minerval, 230 Minor grooves, of DNA, 58, 58F Minoxidil, 151 (–) End: of actin filaments, 131 of microtubules, 134, 145 miRNA (micro RNA), 552T, 570 Mismatch repair, 533, 533F Mitchell, P., 400

In dex Mitochondria, 2F. See also Membranes chloroplasts vs., 412, 421 defined, 390 electron transport in, 389F, 390–399 evolution of, 17–18 images of, 390, 391F membranes of, 390–392 regulation of oxidative phosphorylation by, 405 structure of, 390, 390F transport systems of, 392, 392F Mitochondrial matrix, 363, 390 Mixed inhibition, 199, 200F Mn4CaO5 cluster (Photosystem II), 417, 417F Mobilization, of metabolic fuels, 304–306 Model organisms for genetic studies, 65F Moderately repetitive sequences, in DNA, 66 Molecular biology, central dogma of, 61–64 Molecular chaperones, 102, 597–598 Molecules, 2F Moles per liter (M), 7 Mollusks, hemoglobin from, 149 Monogenic diseases, 63 Monomers, 6, 7, 304 Monosaccharides, 283–287, 323 α and β anomers of, 285–286 and chirality of carbohydrates, 284–285 defined, 5, 283 derivatives of, 286–287, 287F from pentose phosphate pathway, 351–352 polymerization, 6, 7 in polysaccharides, 287 Monosodium glutamate (MSG), 90 Morpho butterfly, 515 Motor proteins, 141–146 kinesin, 143–146 myosin, 141–143 mRNA (messenger RNA): and binding of tRNA to ribosome, 587 circularization of, 590, 590F consensus sequence at splice sites in, 568, 569F and decoding of DNA, 62 defined, 552 5′ cap of, 567, 567F hybridization of DNA and, 567–568, 568F pairing with tRNA by, 584–585 poly(A) tail of, 567, 568F processing of, 567–572 splicing of, 567–569, 570F in transcriptomics, 311 in transpeptidation, 593 turnover for, 569–570, 571F MscL protein, 243F MscS protein, 243F MSG (monosodium glutamate), 90 MspI, 69T MTX (methotrexate), 46–47, 489 Mucins, 294, 295 Mulder, G. J., 52 Mullis, K., 71 Multienzyme complexes, 365 Multifunctional enzymes, 445 Multiple-hit hypothesis for carcinogenesis, 538–539 Multistep reactions, 193–194 Multisubstrate reactions, 193 Muscles: AMPK effects on, 506, 506T Cori cycle in, 499, 500, 500F effects of glycogen storage diseases on, 354–355 events at nerve–muscle synapses, 248–250, 249F

glucose–alanine cycle in, 501, 501F insulin action on, 503, 503T metabolic roles of, 499, 499F myosin in, 141–142, 141F powering, 320 Muscle contraction, 142, 142F Muscle glycogen phosphorylase, 354T Muscular dystrophy, 152 Mutagens, 533 Mutation(s), 17 in citric acid cycle enzymes, 373 defined, 531 disease caused by, 63–64 genetic engineering of, 76–79 in hemoglobin, 123, 127–129 in myosin, 144 nonsense suppressor, 605 point, 532 temperature-sensitive, 546 transition, 532 transversion, 532 and tumor growth, 539 Mutation hotspots, 548 MutS, 533, 533F Myasthenia gravis, 258 Mycobacteriophage, 603 Mycobacterium tuberculosis, 383, 450, 461 Mycoplasma genitalium, 64T, 65 Myelin sheath, 237 Myoglobin, 116, 120–123 α helix of, 95F amino acid sequence, 122, 122F composition of, 93T evolution of, 121–123 hydrophobic and hydrophilic residues in, 99, 99F ion pairs in, 100F oxygen binding to, 120–121 structure of, 120–122, 120F, 122F X-ray crystallography, 97 Myoglobin-facilitated oxygen diffusion, 407 Myosin, 141–143 head and neck region of, 141, 141F lever mechanism of, 142–143 mutations in, 144 structure of, 141–142, 141F Myosin V, 152 Myosin VIIa, 144 Myosin–actin reaction cycle, 142, 143, 143F Myristic acid, 217T Myristoylation, 227, 227F N-Myristoylation process, 606 Myxothiazol, 407, 429

N N, see Asparagine Na+, see Sodium ions Na+ (sodium) channels, 237 NAD+, see Nicotinamide adenine dinucleotide NADH: from β oxidation, 438–439 in citric acid cycle, 366, 369, 372 electron transfer to ubiquinone from, 392–394 in electron transport chain, 390 as energy currency, 385 free energy change for reaction of, 388, 389F in gluconeogenesis, 345F, 346 and glycerol-3-phosphate dehydrogenase, 394 in glycolysis, 330, 331F, 336, 337F, 341 in mitochondria, 391 oxidation of, 398 standard reduction potential of, 387T in urea cycle, 482 NADH dehydrogenase, see Complex I

NADH:ubiquinone oxidoreductase, see Complex I NADP+, see Nicotinamide adenine dinucleotide phosphate NADPH: in Calvin cycle, 424, 426 in fatty acid synthesis, 443, 447 from pentose phosphate pathway, 350–352 in photosynthesis, 420 standard reduction potential of, 387T Na,K-ATPase, 245–248 conformational changes by, 245–247 and glucose transport in intestinal cells, 247, 248F reaction cycle of, 246, 246F structure of, 246–247, 247F Nanometers (nm), 7 Native structure, protein, 101 Natural selection, 16 ncRNA (noncoding RNA): defined, 67, 552 functions of, 553 types of, 552T NDPs (nucleoside diphosphates), 351 Near-equilibrium reactions, 332 Negative effectors, 203 Negative supercoiling, 541 Nernst equation, 387 Nerves, see Neurons Net charge per ion (Z), 236, 400 Neuraminidase, 212 Neurons: action potentials of, 237, 237F events at nerve–muscle synapse, 248–250, 249F kinesin and, 146, 146F membrane-related events of, 235 propagation of nerve impulses by, 238F voltage-gated K+ channels of, 242, 243 Neuropeptide Y, 505T Neurotransmitters. See also specific types from amino acids, 475–476 defined, 248 in nerve–muscle synapse, 248–250 Neutral solutions, 34 Nevirapine, 530 Newsholme, E., 347 Nexium (esomeprazole), 206T NFAT (nuclear factor of activated T cells), 279 Niacin, 313–314, 313T Nicholson, G., 228 Nick, in DNA, 526 Nick translation, 527 Nicotinamide, 313T Nicotinamide adenine dinucleotide (NAD+), 54, 55F in citric acid cycle, 366, 369 in DNA ligation, 527 as electron shuttle, 398 in gluconeogenesis, 345F in glycolysis, 330, 331F, 336, 337F, 341 in oxidation–reduction reactions, 309 in pyruvate dehydrogenase complex, 364 standard reduction potential of, 387T Nicotinamide adenine dinucleotide phosphate (NADP+): in Calvin cycle, 426 in oxidation–reduction reactions, 309 in pentose phosphate pathway, 351 in photosynthesis, 420 standard reduction potential of, 387T Nicotinamide coenzymes, 313T Nicotinamide mononucleotide, 527 Nigericin, 408 Nirenberg, M., 82, 602

I-13

Nitrate, 387T, 465 Nitrate reductase: composition of, 93T in nitrogen cycle, 465, 466F Nitric oxide, 261T, 279, 476 Nitric oxide synthase, 476 Nitrification, 465 Nitrite, 465 Nitrite reductase, 104F Nitrogen. See also Urea cycle assimilation of, 466–469 electronegativity of, 26T function of, 464 for urea cycle, 481–482 Nitrogenase, 465–466, 465F, 466F Nitrogen cycle, 465, 466F Nitrogen fixation, 464–466 Nitrogen-fixing organisms, 465 Nitrogen metabolism, 464–491 amino acid biosynthesis in, 469–476 amino acid catabolism in, 476–480 context for reactions in, 469, 470F nitrogen fixation and assimilation in, 465–469 and nucleotide metabolism, 485–491 urea cycle in, 480–484 Nitroglycerin, 280, 476 p-Nitrophenol, 37T p-Nitrophenylacetate, 156–157, 180 N-Linked oligosaccharides: processing of, 292, 292F structure of, 292–293, 293F nm (nanometers), 7 NMR (nuclear magnetic resonance) spectroscopy, 112 Noncoding RNA, see ncRNA Noncoding segments of genome, 66, 66F Noncoding strand, 62 Noncompetitive inhibition, 199, 200F Noncyclic electron flow, 420, 420F Nonessential amino acids, 469–471 defined, 469 essential vs., 470T synthesis of, 470–471 Nonhomologous end-joining, 535–536, 536F Non-insulin-dependent diabetes, 509–511 Nonketotic hyperglycinemia, 494 Nonpolar molecules: aggregation of, in water, 30, 30F diffusion through vesicles for, 31 electrostatic interactions of, 27 hydration of, 30, 30F Nonreducing sugars, 286 Nonsense suppressor mutations, 605 Nonspontaneous reactions, 11–12 Norepinephrine (noradrenaline), 265, 475, 504 NotI, 69T N-terminus, 90, 91T, 109 Nuclear factor of activated T cells (NFAT), 279 Nuclear magnetic resonance (NMR) spectroscopy, 112, 112F Nucleases, 59. See also specific nucleases Nucleic acids. See also DNA; RNA about, 7–8, 8F defined, 52 denaturation and renaturation of, 59–61 in double helix, 56–59 functions of, 9T in nucleotides, 53–54 single-stranded, 59 structures of, 56–61 universality of, 14 Nucleolus, 572 Nucleophiles, 163 Nucleosides, 53–54, 54T

I-14 Index Nucleoside triphosphates, 319 Nucleosomes, 543–544, 543F in chromatin remodeling, 554–555, 554F DNA packaging in, 553 Nucleosome sliding, 555F Nucleotides, 52–55, 54T about, 5–6 cellular functions of, 54–55 defined, 52 digestion products in synthesis of, 304 evolutionary tree based on, 18F functions of, 54–55, 55F identifying, in DNA sequencing, 74, 75 nucleic acids in, 53–54 polymerization of, 6, 7 in rRNA and tRNA processing, 572–573 sequences of, and amino acids, 62–63, 63T Nucleotide bases: and DNA melting temperature, 59–60 nucleic acids in, 53, 54T Nucleotide diphosphates (NDPs), 351 Nucleotide excision repair, 535–536, 535F Nucleotide metabolism, 485–491 methylation in, 488–489 nucleotide degradation products, 489–490 purine synthesis in, 485–486 pyrimidine synthesis in, 486–487 ribonucleotide reductase in, 487–488 Nucleotide synthesis: cellular location of, 498F of purines, 485–486 of pyrimidines, 486–487 Nucleus, cell, 544, 544F

O Obesity, 508F causes of, 508–509 and metabolic syndrome, 511 Ocean acidification, 35 n-Octylglucoside, 48 Odd-chain fatty acids, 440–442 Ogston, A., 369 Ohio hemoglobin variant, 150 Okazaki fragments, 523, 526 Olavius algarvensis, 301 Oleate, 216, 439 Oleic acid, 217T Olestra, 232 Oligomycin, 409, 430 Oligonucleotides, 59 Oligonucleotide probes, 60 Oligopeptides, 92 Oligosaccharides, 292–294 in ABO blood group system, 293 as biological markers, 293–294 defined, 292 O- vs. N-linked, 292–293 polysaccharides vs., 287–288 O-Linked oligosaccharides, 292–293 Omega-3 fatty acids, 216 Omega-6 fatty acids, 216 OMIM (Online Mendelian Inheritance in Man), 64 OmpF protein, 240–241, 240F Oncogenes, 273 Online Mendelian Inheritance in Man (OMIM), 64 Open conformation of DNA polymerase, 525, 525F Open reading frames (ORFs), 66, 82 Operons: defined, 552 gene expression by, 560–561 lac, 560–561, 576–577 Oral rehydration therapy (ORT), 323 Ordered mechanism, 193

ORFs (open reading frames), 66, 82 Organs, 2F. See also specific organs metabolites in, 499–501 specialization of, 498–499 Organelles, 2F, 31 Organisms: aerobic vs. anaerobic, 16 entropy in, 14 genetically modified, 72, 78 levels of organization in, 2, 2F metabolic interdependence of, 310–311 model, 65F nitrogen-fixing, 465 photosynthetic, 13, 16, 411 replication of, 14 transgenic, 72 Ornithine, 483F Ornithine transcarbamoylase, 482, 483F Orotidine-5′-monophosphate decarboxylase, 156T Orphan genes, 67 ORT (oral rehydration therapy), 323 Oryza sativa, see Rice Oseltamivir, 212 Osmosis, 48, 243 Osteogenesis imperfecta, 140, 152 Ouabain, 257, 514 Ovalbumin, 568F Overton, C., 234 Oxaloacetate: in amino acid catabolism, 477, 478 from anaplerotic reactions, 376, 377 as biosynthetic precursor, 375, 375F, 376 in C4 pathway, 424 citrate from, 366–368, 367F, 368F in citrate transport system, 444, 444F in gluconeogenesis, 345F, 346 in malate–aspartate shuttle system, 391F from malate dehydrogenase reaction, 367F, 371–372 as metabolic intermediate, 307, 307F from pyruvate, 341F, 343, 344F standard reduction potential of, 387T in transamination reactions, 469, 470 Oxalobacter formigenes, 256 Oxazolidinones, 605 Oxidants, 386 Oxidation (oxidation reactions), 13, 13F. See also Fatty acid oxidation β, 436–443 of cytochrome c, 397–399 defined, 308 of pentose phosphate pathway, 350–351 photooxidation, 414, 417–421 ubiquinol from, 394, 395F of water, 417–418 Oxidation–reduction (redox) reactions: in metabolic pathways, 308–309 and Photosystem II, 416–417 thermodynamics of, 385–390 Oxidation state, of carbon, 13, 13T Oxidative decarboxylation reactions, 369–370 Oxidative phosphorylation, 385–406. See also Mitochondria ATP synthase in, 401–405 cellular location of, 498F in chemiosmosis, 399–401 and citric acid cycle, 372 context for reactions of, 385–386, 386F defined, 309, 385 and mitochondrial electron transport, 390–400 powering muscles with, 320 and redox reaction thermodynamics, 385–390 Oxidizing agents, 386

Oxidoreductases, 157T 8-Oxoguanine, 532 Oxyanion hole, 167, 167F Oxygen: and citric acid cycle, 372 concentration of, 120–121, 121F cooperative binding of, 123–126 diffusion through bilayers of, 31 DNA damage by, 531–532 electronegativity of, 26T in electron transport chain, 390 free energy change for reaction of, 389F H+ and BPG and binding of, 126–127 and nitrogen fixation, 465 partial pressure of, 121 reduction of, 397–399 Oxygen-binding curves: for hemoglobin, 123–124, 124F for myoglobin, 121, 122F Oxygen-binding proteins, 120–139 hemoglobin, 121–129 myoglobin, 120–123 Oxygen-evolving complex (Photosystem II), 417–418 Oxygen pressure at 50% saturation ( p50), 121 Oxyhemoglobin, 125–129, 125F

P P, see Proline p50 (oxygen pressure at 50% saturation), 121 Pi (inorganic phosphate), 131 in citric acid cycle, 366 in fatty acid oxidation, 436 in gluconeogenesis, 345F, 346 in glycolysis, 330, 331F, 336, 337F mitochondrial transport of, 392, 392F phosphorylation of glucose by, 317 and rate of oxidative phosphorylation, 405 resonance stabilization of, 318 p53 tumor suppressor, 539–540, 540F P56, 216 P680, 416, 417 P700, 419–420 Packaging, DNA, see DNA packaging Paclitaxel, 134–135, 150 PAH, see Phenylalanine hydroxylase Palindromic DNA, 69 Palmitate, 327 as amphiphilic molecule, 31 covalent modifications of, 102F in fatty acid synthesis, 446F, 447 hydrocarbon chains of, 216 structure of, 6 Palmitic acid, 217T Palmitoleic acid, 217T Palmitoyl-ACP, 446F Palmitoylation, 227, 227F, 606 Palmitoyl thioesterase, 446F Pancreatic amylase, 21 Pancreatic β cells, 502, 511 Pancreatic islet cells, 502, 502F Pancreatic juice, 34T Pancreatic lipase, 230 Pandinin 2 peptide, 115 Pantothenic acid, 313T Parallel β sheets, 95, 96F Paramecium, 17F Parathion, 178–179, 258 Parkinson’s disease, 106 Partial pressure of oxygen ( pO2), 121 Passive transport, 240–245 in aquaporins, 243–244 and conformations of transport proteins, 244–245 and defensive pore-forming, 242 in gated ion channels, 242–243 in ion channels, 242 in porins, 240–241

Passive transporters, 237, 239–245 Pasteur, L., 341, 372 Pasteur effect, 372 Pauling, L., 80, 95, 128, 166 PCR (polymerase chain reaction), 71–74, 73F PCSK9 protein, 434 PDH, see Pyruvate dehydrogenase PDK (pyruvate dehydrogenase kinase), 380 Peanut oil, 233 Peas, histones of, 550 Pectin, 290 Pellagra, 313–314, 313T Pentacoordinate Fe(II), 148 Pentoses, 284 Pentose phosphate pathway, 350–352 cellular location of, 498F isomerization and interconversion reactions of, 351–352, 352F oxidative reactions of, 350–351 PEPC, see Phosphoenolpyruvate carboxylase PEPT1 transporter, 258 Peptides, 92, 599–600 Peptide bonds, 7 between amino acids, 90–93 formation of, 593–595 hydrolysis of, 164–166 and secondary structure of proteins, 94–97 Peptide hormones, 234 Peptidoglycan, 295, 295F Peptidyl site, see P site Peptidyl transferase, 593–595, 593F Peptidyl–tRNA, 593F Peripheral membrane proteins, 225F, 226 Permeability coefficient, 255 Pernicious anemia, 460 Peroxisomes, 442–443, 442F Perutz, M., 124 PET scan, see Positron emission tomography scan pGEM-3Z, 70, 71F, 84 P-glycoprotein, 247, 247F pH: altering, 34–35 of biological fluids, 34, 34T defined, 34 and Henderson–Hasselbalch equation, 38–40 and hydrogen ions, 34, 34F of oceans, 35 physiological, 34 and pK, 37–40 standard state, 315, 315T Phagocytes, 383 Phalloidin, 150 Pharmaceuticals, see Drugs Pharmacokinetics, 204 Phenol, pK, 37T Phenylalanine (Phe, F): abbreviations for, 87 catabolism of, 479, 480 as essential amino acid, 312T, 470T genetic code for, 63T, 581T as glucogenic/ketogenic amino acid, 477T as hydrophobic amino acid, 87F, 88F, 99, 99T percent occurrence of, 91F synthesis of, 473 Phenylalanine hydroxylase (PAH), 208 in phenylalanine degradation, 479 and PKU, 480 in tyrosine synthesis, 473 Phenyl-β-D-galactose, 577 Phenylketonuria (PKU), 480, 494 Phosphoethanolamine methyltransferase, 462 Phorbol esters, 279

In dex Phosphatases, 268, 274, 453F. See also specific enzymes Phosphate. See also Pi (inorganic phosphate) hydrolysis of, 318–319, 318T laboratory synthesis of, 14 pK of, 37T Phosphate-buffered saline, 42 Phosphatidate, 453F, 455F Phosphatidylcholine, 218, 453, 454F Phosphatidylethanolamine, 218 phosphatidylserine from, 453 synthesis of, 453, 454F Phosphatidylglycerol, 218 Phosphatidylinositol, 231, 258, 454, 455F Phosphatidylinositol bisphosphate, 269 Phosphatidylserine, 218, 453 Phosphoanhydride bonds, 4T, 318 Phosphocholine, 454F Phosphocreatine, 318T, 319, 320 Phosphodiesters, 4T Phosphodiester bonds, 7–8, 56 Phosphoenolpyruvate, 21 in amino acid synthesis, 473 in C4 pathway, 424 conformational change for, 201, 202, 202F free energy change for hydrolysis of, 318T in gluconeogenesis, 345F, 346 in glycolysis, 331F, 338–339 and phosphofructokinase, 201–203, 202F, 332, 333F Phosphoenolpyruvate carboxykinase, 345F, 346 Phosphoenolpyruvate carboxylase (PEPC, PPC), 383, 424, 431 Phosphoester linkage, 4T Phosphoethanolamine, 454F Phosphofructokinase, 21, 461 allosteric regulation of, 201–203 conformational change in, 201, 202, 202F deficiency in, 354T and gluconeogenesis, 347 in glycolysis, 331–333, 333F and phosphoenolpyruvate, 201–203, 202F, 332, 333F structure of, 202F Phosphofructokinase-2, 333 Phosphoglucomutase: in glycogenolysis, 349 in glycogen synthesis, 348 in starch synthesis, 426 6-Phosphogluconate, 350–351 6-Phosphogluconate dehydrogenase, 350–351 6-Phosphogluconolactonase, 350 6-Phosphoglucono-δ-lactone, 350 Phosphoglucose isomerase: in gluconeogenesis, 345F in glycolysis, 331F, 332 2-Phosphoglycerate: in gluconeogenesis, 345F in glycolysis, 331F, 338 3-Phosphoglycerate: in Calvin cycle, 424, 425F in gluconeogenesis, 345F in glycolysis, 331F, 337–338 and rubisco, 423, 423F serine from, 471 Phosphoglycerate kinase, 86F in Calvin cycle, 425F in gluconeogenesis, 345F in glycolysis, 331F, 336, 337 Phosphoglycerate mutase: in gluconeogenesis, 345F in glycolysis, 331F, 338 Phosphoglycohydroxamate, 197–198 2-Phosphoglycolate, 423 Phospho-His intermediate, 370F

3-Phosphohydroxypyruvate, 471 Phosphoinositide signaling pathway, 269–270 Phospholipases, 218, 218F Phospholipase A1, 232 Phospholipase A2, 275 Phospholipase C, 269, 270 Phospholipids, synthesis of, 453–454 5-Phosphoribosyl pyrophosphate (PRPP), 474, 485 Phosphoribulokinase, 425F Phosphoric acid, 37T, 50 Phosphoric acid esters, 4T Phosphorolysis, 304, 305, 305F Phosphorylase kinase, 516 composition of, 93T deficiency in, 354T and protein kinase A, 505 Phosphorylation: autophosphorylation, 271–273 of protein kinase A, 267 substrate-level, 370 Phosphoryl group, 4T covalent modifications of, 102F resonance stabilization of, 318 transfer of, 318–319, 318T 3-Phosphoserine, 471 Photoautotrophs, 301 Photoexcited molecules, 414, 414F Photons, 412 Photooxidation: defined, 414 in Photosystem I, 419–421 in Photosystem II, 417–418 Photophosphorylation, 421, 421F Photoreceptors, 413–414, 413F Photorespiration, 423 Photosynthesis, 411–428. See also Chloroplasts carbon fixation in, 422–428 chloroplasts, 411–415 context for reactions in, 412F light reactions of, 415–422 thermodynamics of, 13 Photosynthetic organisms, 13, 16, 411 Photosystem I, 419–421 linking of Photosystem II and, 418–419 photooxidation reaction in, 419–421 Photosystem II, 416–419 function of, 418F as light-activated oxidation–reduction enzyme, 416–417 linking of Photosystem I and, 418–419 oxidation of water by, 417–418 Phycocyanin, 413, 413F Phylloquinone, 221, 313T Physiological pH, 34, 89 Phytanate, 443 Phytol, 233 pI (isoelectric point), 108 Pig, insulin from, 86F Pigments, 412–414, 414F Ping pong mechanism, 193 Pinocytosis, 252 Pioglitazone, 408 pK: defined, 36–37 of ionizable groups in polypeptides, 91T and pH, 37–40 of selected acids, 37T PKA, see Protein kinase A PKR protein kinase, 281 PKU (phenylketonuria), 480, 494 Planck’s law, 412 Plants. See also specific species aromatic amino acid synthesis in, 472–474 genome size and gene number of selected, 64T

rubisco in, 422 transgenic, 72 Plaques, atherosclerotic, 432–433, 433F Plasma, coagulation factors in, 174, 174T Plasmalogens, 232 Plasma membranes, 251–252. See also Lipid bilayers Plasmids, 70 Plasmodium falciparum, 128, 129, 149 Plasmodium vivax, 256 Plastocyanin, 86F, 419, 419F Plastoquinol, 417 Plastoquinone, 417, 429 Plotosus japonicus, 24 PLP, see Pyridoxal-5′-phosphate (+) End: of actin filaments, 131 of microtubules, 134, 145 Pneumococcus, 80 PNPase (polynucleotide phosphorylase), 578 pO2 (partial pressure of oxygen), 121 Point mutation, 532 Polar head groups, lipids with, 217–219 Polar molecules, 25 bilayers and diffusion of, 31, 31F electrostatic interactions of, 27 Polar side chains, amino acid, 87F, 88–89 Poly(A)-binding protein, 567, 568F Polyacrylamide gel electrophoresis, 109, 109F Polyadenylate tail, see Poly(A) tail Poly(A) polymerase, 578–579 Poly(A) tail, 567, 568F, 570 Polygenic diseases, 63 Polyglutamine diseases, 605–606 Polymers. See also Biological polymers and monomers, 6, 7 nucleic acids as, 53–54 storage of, in body, 304 Polymerases, 59. See also DNA polymerases; RNA polymerases Polymerase chain reaction (PCR), 71–74, 73F Polymerization, 7 of actin, 131 in laboratory synthesis, 15–16 and storage of monomers in body, 304 Polynucleotides, 7–8, 56 Polynucleotide phosphorylase (PNPase), 578 Polypeptides, 86. See also Proteins; Protein synthesis chromatography and properties of, 107–109 defined, 7 laboratory synthesis of, 15 Polypeptide backbone, 91, 94–95 Polyphenol oxidase, 208 Polyproteins, 198 Polyprotic acid, 36, 37 Polysaccharides, 287–291 about, 8–9 in biofilms, 291 in cellulosic biofuels, 290 defined, 283 disaccharides, 288 fuel storage in, 288–289 functions of, 9T structural support from, 289–291 Polysomes, 596, 596F Poplar, 86F P:O ratio, 403, 404 Porcine elastase, 170T Pores: of aquaporins, 244, 244F defensive formation of, 242 Porins: aquaporins, 243–244, 243F as β barrel proteins, 240–241

I-15

Positive effectors, 203 Positron emission tomography (PET) scan, 512, 512F, 517 Post-translational events, 597–601 covalent modification as, 600–601 membrane translocation as, 599–600 protein folding as, 597–598 Post-translational processing, 102, 597 Potassium (K+) channels: gated, 241–243, 242F in nerve stimulation, 237 selectivity filters of, 241, 241F of Streptomyces lividans, 241, 241F Potassium ions (K+): in animal cells, 235, 236F in intra- and extracellular fluids, 32F and Na,K-ATPase, 246 PPi, see Inorganic pyrophosphate (PPi) PP1 (protein phosphatase 1), 213 PPC, see Phosphoenolpyruvate carboxylase Prebiotic world, 14–16 Prednisone, 277, 281 Preinitiation complex, 557, 557F Prenylation, 227F, 228 Presbyterian hemoglobin variant, 150 Pressure, standard state, 315, 315T Primary structure of proteins, 93, 94F Primase, 522, 523, 526 Primers: defined, 522 excision of, 526, 526F for PCR, 72 Prions, 106 Pro, see Proline Probes, oligonucleotide, 60 Processing: post-translational, 102, 597 of RNA, see RNA processing Processivity: of DNA polymerases, 524 of kinesin, 146 of RNA polymerase, 564 Product inhibition, 197 Products: of digestion, 302–303 free energy of reactants and, 159, 159F Proelastase, 176 Programmed cell death, see Apoptosis Proinsulin, 599, 600F Prokaryotes: biofilms of, 291 DNA ligation for, 527 evolution of, 16, 17F, 18F gene number and complexity of, 66 operons of, 552, 560–561 ribosomes of, 586, 586T transcription elongation in, 564 transcription termination in, 565, 566F translation factors of, 592T Proline (Pro, P), 115 abbreviations for, 87 catabolism of, 484 genetic code for, 63T, 581T as glucogenic amino acid, 477T as helix disrupter, 115 as hydrophobic amino acid, 87F, 88, 99T and hydroxyproline, 137 as nonessential amino acid, 470T percent occurrence of, 91F synthesis of, 470 Proline racemase, 212 Promoters, 554–560 defined, 555 E. coli, 556, 556F and enhancers/silencers, 558–560 initiation of transcription at, 555–557 transcription factors and, 557

I-16 Index Proofreading: by aminoacyl-tRNA synthetases, 584 by DNA polymerases, 525, 525F by RNA polymerases, 564 1-Propanol, 29T Propionyl-CoA, 440–442, 441F Propionyl-CoA carboxylase, 440, 441F Prostaglandin H2, 275F Prosthetic groups, 120 of Photosystem I, 420, 420F of Photosystem II, 416F Protamines, 550 Proteases. See also Serine proteases aspartyl, 180 cysteine, 179 digestion by, 302, 302F HIV, 181, 198–199, 198F, 211 in mass spectrometry, 109, 109T and protease inhibitors, 172–173 Protease inhibitors, 172–173 Proteasomes: protein degradation by, 305–306, 306F structure of, 305, 305F Proteins. See also Membrane proteins; specific types and proteins about, 7 allosteric, 126 branching, 132 capping, 132 composition of selected, 93T constitutive, 537–538 defined, 86 degradation of, 305–306, 306F functions of, 9T as metabolic fuels, 508T percentage of, in lipoproteins, 433T roles of, 85 Protein folding: diseases caused by, 105–106, 105F hydrophobic effect in, 100, 100F as post-translational event, 597–598 and secondary structures, 101–102 Protein folding funnel, 101–102, 102F Protein groups, as catalysts, 163F Protein kinase A (PKA): activation of, 267 in fuel mobilization, 505 inactive form of, 267F structure of, 268F Protein kinase B, 279 Protein kinase C, 269–270 Protein kinase G, 279 Protein phosphatase 1 (PP1), 213 Protein structure, 85–113 and amino acids, 86–94 analyzing, 107–112 and disease related to protein misfolding, 105–106 primary, 93, 94F quaternary, 93, 94F, 104–105 secondary, 93–97, 94F tertiary, 93, 94F, 97–103 Protein synthesis, 580–601. See also Transcription; Translation antibiotic inhibitors of, 594 post-translational events in, 597–601 and proteomics, 311 and ribosome structure, 586–588 in translation, 589–597 tRNA in, 582–585 Proteoglycans, 294–295 Proteolysis, 171–172, 600 Proteomes, 311 Proteomics, 311 Prothrombin, 174T Protofilaments, 133 Protons, translocation of, 401–403, 402F Proton gradients, 400–401, 400F Proton jumping, 33, 33F Protonmotive force, 400

Proton wires, 394, 398 Providence hemoglobin variant, 129T, 150 Proximity and orientation effects, 168, 168F Prozac (fluoxetine): effect of, on rat brain, 407 fluorine in, 28 and serotonin transport, 250, 475 and signaling by serotonin receptor, 258 PrP gene, 116 PRPP (5-phosphoribosyl pyrophosphate), 474, 485 PrPSc, 106 Pseudomonas aeruginosa, 255, 291F Pseudouridine, 572 D-Psicose, 297 P site (peptidyl site): defined, 587 in initiation and elongation, 589–591 in peptidyl transferase reaction, 593, 594 Δψ, see Membrane potential PstI, 69T PTP1B enzyme, 212, 516 P-type ATPases, 247 Pupal stages, 580 Purine, 53 Purine nucleotides, 53 degradation of, 489–490 synthesis of, 485–486 Purine nucleotide cycle, 382 Puromycin, 594 Purple nonsulfur bacteria, 323 Purple photosynthetic bacteria, 414 Pyl (pyrrolysine), 585 Pyridoxal-5′-phosphate (PLP): as coenzyme, 313T transamination catalyzed by, 467–469, 468F Pyridoxine, 467 Pyrimidine, 53 Pyrimidine nucleotides, 53 degradation of, 490 synthesis of, 486–487 Pyrococcus, rubredoxin, 93T Pyrococcus furiosus, 72 Pyrophosphate, see Inorganic pyrophosphate (PPi) Pyrophosphoryl group, 4T Pyrosequencing, 75, 75F Pyrrolysine (Pyl), 585 Pyruvate: acetyl-CoA from, 363–365, 365F in amino acid catabolism, 477, 478 amino acid synthesis from, 470 in anaplerotic reactions, 376–378 in β oxidation, 441F in C4 pathway, 424 conversion of, after glycolysis, 341–344, 341F in gluconeogenesis, 345–346, 345F in glycolysis, 330, 331F, 338–339 from malate, 376 as metabolic intermediate, 307, 307F standard reduction potential of, 387T in transamination reaction, 467 Pyruvate carboxylase: deficiency in, 358, 515 in fermentation, 341, 342 in gluconeogenesis, 345, 345F, 346 and oxaloacetate from pyruvate, 343, 344F Pyruvate decarboxylase, 93T, 157, 376 Pyruvate dehydrogenase (PDH), 362–365 in catabolism of propionyl-CoA, 441F E1, E2, and E3 enzymes of, 363 PDH kinase as regulator of, 514 pyruvate to acetyl-CoA conversion by, 342, 363–365, 365F

Pyruvate dehydrogenase kinase (PDK), 380 Pyruvate kinase, 214 deficiencies in, 353–354 in glycolysis, 331F, 338–339 PYY3-36 oligopeptide, 505T, 506

Q Q, see Glutamine; Ubiquinone Q cycle, 395, 396F QH· (ubisemiquinone), 310 QH2, see Ubiquinol Quantitative PCR (qPCR), 73, 74 Quantum yield, of photosynthesis, 426 Quaternary structure of proteins, 104–105 defined, 93, 94F hemoglobin, 104, 105 Quorum sensing, 262

R R, see Arginine R (gas constant), 236 R. oligosporus, 298 Raffinose, 298 Rafts (membrane), 224, 234 Ranaspumins, 85 Randle hypothesis, 461 Random mechanism, 193 Ras signaling pathway, 272, 273F, 280, 549 Rat, metabolite profile of brain, 311 Rate constant (k), 186, 187 Rate-determining reactions, 333 Rate equations, 186–189 Rate of reaction, 11. See also Enzyme kinetics; v (reaction velocity) activation energy and, 158, 159 enzymes and, 155–156, 156T with large free energy changes, 321 for oxidative phosphorylation, 404–405 Rational drug design, 204–205 Rb protein, 549 R conformational state, of hemoglobin, 125 Reactants, free energy of, 159, 159F Reactant concentration: and free energy change, 314–316 and reaction rate, 155 standard state, 315, 315T Reaction centers, 414–415 Reaction coordinate, 158–159, 159F Reaction mechanisms: catalytic, see Catalytic mechanisms depicting, 162 Reaction specificity, of enzymes, 156 Reaction velocity, see v Reading frames: open, 66, 82 for translation, 581, 581F Real-time PCR, 73, 74 RecA protein, 536, 537F Receptors. See also specific receptors binding of ligands to, 261–263, 261F defined, 260 desensitization of, 264, 269 types of, 263–264 Receptor-mediated endocytosis, 252 Receptor tyrosine kinases, 270–273 autophosphorylation of, 271–273 binding of insulin to, 271 defined, 264 signal transduction pathway for, 263F, 264 Recognition sequences, restriction endonuclease, 69, 69T Recombinant DNA, 69–71, 70F Recombination, DNA repair by, 536–538

Red blood cells: carbohydrate metabolism disorders involving, 353–354 globin synthesis in immature, 605 glucose transporters of, 244F normal vs. sickled, 128, 128F protein pumps in, 49 Redox centers, 392, 393F Redox reactions, see Oxidation–reduction reactions Red tide, 429 Reducing agents, 386 Reducing sugars, 286 Reductants, 386 Reduction (reduction reaction), 13, 13F defined, 308 oxygen in, 397–399 in unsaturated fatty acid degradation, 440 Reduction potential, see ℰ Reductive biosynthetic pathway, 374, 374F Refsum’s disease, 460 Regular secondary structure, 95–96 Regulation: allosteric, 200–203 of carbon fixation, 426 of citric acid cycle, 372–373, 373F of enzymes, 200–203 of gluconeogenesis, 346–347 of glycolysis, 339, 340 of hemoglobin, 126–127 of metabolism, 310, 321, 435. See also Mammalian fuel metabolism of phosphofructokinase, 332–333, 333F of pyruvate dehydrogenase reaction, 365 Reindeer meat, 233 Release factors (RFs), 595–596 Relenza (zanamivir), 212 Renaturation: of DNA, 60–61, 61F of proteins, 101 Renin, 180 Replication, 519–531. See also DNA polymerases in cancer, 538–540 and central dogma, 61, 62F conservative, 520, 520F defined, 519 DNA polymerase in, 522–525 factory model of, 520 helicases in, 521–522 of lagging strand, 525–528 of living organisms, 14 machinery of, 519–528 model of, 523F packaging after, 540–544 of primitive RNA, 15–16, 16F semiconservative, 520, 520F separation of transcription and, 562, 562F at telomeres, 528–531 transcription vs., 551–552 Replication-coupling assembly factor, 544 Replication foci, 520F Replication fork, 520 Replication protein A, 521, 521F Replisomes, 527 Repressors, 560, 561, 561F Residues: amino acid, 90 defined, 7 invariant, 123, 148 nucleotide, 56 protein, 123 variable, 123 Resistin, 505T, 506 Resonance stabilization, 318 Respiration, cellular, 390, 404 Respiratory acidosis, 45

In dex Respiratory alkalosis, 45 Restriction digests, 70, 70F Restriction endonucleases, 69, 69T Retina, of eye, 257 Retinoate (retinoic acid), 274, 274F Retinol, 221, 313T Reverse osmosis, 48 Reverse transcriptase, 93T, 529, 530 RFs (release factors), 595–596 RF-1, 592T, 595, 595F RF-2, 592T RF-3, 595, 596 R group, 86 Rhamnose, 290 Rhodopseudomonas acidophila, 415F Riboflavin, 313T Ribonucleases (RNases): A, 179 denaturation experiments with, 116, 605 H, 526, 546 H1, 179 P, 573, 574, 574F, 579 in replication of lagging strand, 525–527 Ribonucleic acid, see RNA Ribonucleotide reductase, 487–488, 488F Ribonucleotides, 487–488 Ribose, 53, 284, 287 Ribose-1-phosphate, 490 Ribose-5-phosphate: in pentose phosphate pathway, 350–352 in purine nucleotide synthesis, 485, 486F source of, 304 Ribose phosphate pyrophosphokinase, 485 Ribosomal inactivating proteins (RIPs), 603 Ribosomal RNA, see rRNA Ribosomes: bacterial, 587, 588F components of, 586, 586T decoding of RNA in, 62 defined, 586 delivery of tRNA to, 590–592 and membrane translocation, 599–600 RNA in, 586–587 structure of, 586–588, 586F tRNA binding to, 587–588 Ribosome recycling factor (RRF), 596 Ribozymes, 155, 178, 569, 573 Ribulose-1,5-bisphosphate: in Calvin cycle, 424, 425F and rubisco, 423, 423F Ribulose bisphosphate carboxylase, 93T Ribulose-5-phosphate: in Calvin cycle, 424, 425, 425F in pentose phosphate pathway, 351, 352 Ribulose-5-phosphate isomerase, 351 Rice: genetically modified, 72 genome size and gene number for, 64T, 66 Rickets, 313T Rifampicin, 577 Rigor mortis, 152 RIPs (ribosomal inactivating proteins), 603 RISC, see RNA-induced silencing complex Ritonavir, 199 [R · L]:[R]T ratio, 278 RNA (ribonucleic acid). See also specific types, e.g.: mRNA (messenger RNA) backtracking by, 564, 564F ball-and-stick model for, 8F defined, 53

genes encoding, 63 guide, in CRISPR-Cas9 system, 77, 77F hydrogen bonding in, 27 noncoding, 67 nonstandard base pairs in, 573, 574F in ribosome, 586–587 and RNA polymerase, 563F secondary structure of, 573–574 self-replication of, 15–16, 16F structure of, 59 RNA–DNA hybrid double helix, 59, 59F RNAi (RNA interference), 82, 570–572, 571F RNA-induced silencing complex (RISC), 571, 571F, 572 RNA polymerases, 562–566 backtracking RNA in, 564, 564F in Cockayne syndrome, 535 defined, 552 with DNA and RNA, 563, 563F function of, 522 and Mediator complex, 560, 560F as processive enzyme, 564 structure and mechanism of, 562–564 in transcription elongation, 564–565 in transcription initiation, 556, 557 in transcription termination, 565–566 RNA polymerase I, 562 RNA polymerase II, 562, 562F RNA polymerase III, 562 RNA processing, 567–574 defined, 552 5′ caps and poly(A) tails in, 567 and mRNA turnover, 569–570 nucleotides in, 572–573 and RNA interference, 570–572 and secondary structure of RNA, 573–574 splicing of introns, 567–569 RNases, see Ribonucleases RNase A, 179 RNase H, 526, 546 RNase P, 573, 574, 574F, 579 RNA world, 574 Rofecoxib, 205 Root nodules, 465F Rosiglitazone, 205, 510, 510T Roundup (glyphosate), 473 “Roundup Ready” soybeans, 431 RRF (ribosome recycling factor), 596 rRNA (ribosomal RNA): decoding of DNA by, 62 defined, 552 as noncoding RNA, 552T processing of, 572, 572F, 573 in ribosome function, 586–587 Rubisco, 422–424, 423F, 425F Rubredoxin, 93T Rudder (protein loop), 564

S S, see Serine S, see Entropy ΔS (entropy change), 11, 30 Saccharomyces cerevisiae, see Yeast Saline, phosphate-buffered, 42 Saliva, pH of, 34T Salmonella, 93T Salvage pathways, 489–490 SAMDH1 enzyme, 547 Sandwich model, 234 Saquinavir, 199 Sarin, 178 Saturated enzymes, 185 Saturated fatty acids, 216, 217T Saturation, of myoglobin with oxygen, 121 SauI, 69T SBPase (sedoheptulose bisphosphatase), 430–431

SCARMD (severe childhood autosomal recessive muscular dystrophy), 115 Scatchard plot, 278 Schiff bases, 162, 163, 467, 468F Schiffer, M., 115 Sciadonate, 230 SCID (severe combined immunodeficiency), 78, 78T Scissile bonds, 164 Scrapie, 106 Scurvy, 137, 312, 313T SDS, see Sodium dodecyl sulfate SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis), 109, 109F Sec (selenocysteine), 585 Sec61, 599, 599F Secondary active transport, 247–248 Secondary structure of proteins, 94–97 α helix, 95 β sheet, 95–96 defined, 93 of hemoglobin, 94F irregular, 96, 97 and protein folding, 101–102 Secondary structure of RNA, 573–574 Second law of thermodynamics, 12 Second messengers: cyclic AMP as, 266 defined, 263 in phosphoinositide signaling pathway, 269–270 Second-order reactions, 186, 191 β-Secretase, 105 γ-Secretase, 105 Secretory proteins, 599–600, 600F SecY, 599 Sedoheptulose bisphosphatase (SBPase), 430–431 Selection, 70 Selective serotonin reuptake inhibitors (SSRIs), 250 Selectivity filters, ion channel, 241, 241F Selenocysteine (Sec), 585 Semiconservative replication of DNA, 520, 520F Semiquinone, 396F Serine (Ser, S): abbreviations for, 87 catabolism of, 478 genetic code for, 63T, 581T as glucogenic amino acid, 477T hydrophobicity of, 99T as nonessential amino acid, 470T percent occurrence of, 91F in phosphatidylserine synthesis, 453 as polar amino acid, 87F, 88–89 in synthesis reactions, 471, 472 Serine hydroxymethyltransferase, 471, 492, 493 Serine proteases: catalytic mechanism of, 165F, 166 defined, 164 evolution of, 170 percent sequence identity among, 170T specificity pockets of, 170, 171F Serotonin, 250, 258, 475 Ser/Thr kinases, 267 Sertraline, 250 Serum glutamate-oxaloacetate transaminase (SGOT), 469 Serum glutamate-pyruvate transaminase (SGPT), 469 Set-point, body weight, 508 70S ribosome, 586, 586T Severe childhood autosomal recessive muscular dystrophy (SCARMD), 115 Severe combined immunodeficiency (SCID), 78, 78T

I-17

Severing proteins, 132 SGOT (serum glutamate-oxaloacetate transaminase), 469 SGPT (serum glutamate-pyruvate transaminase), 469 Shewanella bacteria, 208 Shine–Dalgarno sequence, 589, 589F Short interfering RNAs, see Small interfering RNAs Shotgun DNA sequencing, 76 Sickle-cell disease, 63, 128, 129 Sickle-cell hemoglobin (hemoglobin S), 128–129, 128F Side chains (amino acid), 88–89 as catalysts, 161, 162F charged, 87F, 89 hydrophobic, 87F, 88 polar, 87F, 88–89 σ Factor (RNA polymerase), 556 Signaling, effects of, 264 Signaling molecules, 475–476 Signaling pathways, 260–277 and aspirin, 276 and cancer, 273 features of, 260–264 of G proteins, 265–270 of lipid hormones, 274–275 of receptor tyrosine kinases, 270–273 switching off, 264, 267–269 Signal peptidase, 600 Signal peptides, 599–600 Signal recognition particle (SRP), 599–600, 599F Signal transduction, 260, 263F Silencers, 560 Silent chromatin, 554T Simian virus 40 (SV40), 546 Singapore hemoglobin variant, 129T Singer, S. J., 228 Single-nucleotide polymorphisms (SNPs), 67–68 Single-strand binding protein (SSB), 521, 522 Single-stranded DNA: conversion of double-stranded DNA to, 521–522 in recombination, 536 Single-stranded nucleic acids, 59 siRNAs (small interfering RNAs), 552T, 570 Site-directed mutagenesis, 77 16S rRNA, 586F, 589, 589F 60S subunit, 586, 586T, 587F Size-exclusion chromatography, 107, 107F Skin, keratins in, 135, 135F Small interfering RNAs (siRNAs), 552T, 570 Small intestine, amino acid catabolism in, 476–477 Small nuclear RNAs (snRNAs), 552T, 568 Small nucleolar RNA molecules (snoRNAs), 552T, 572 Small ubiquitin-like modifier protein (SUMO), 600 SNAREs (soluble N-ethylmaleimidesensitive-factor attachment protein receptor), 251–252, 251F snoRNAs (small nucleolar RNA molecules), 552T, 572 SNPs (single-nucleotide polymorphisms), 67–68 snRNAs (small nuclear RNAs), 552T, 568 Soap, 231 Sodium (Na+) channels, 237 Sodium chloride, 28 Sodium dodecyl sulfate (SDS), 48, 109, 116 Sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE), 109, 109F

I-18 Index Sodium hydroxide, 35, 40, 41 Sodium ions (Na+): distribution of in animal cells, 235, 236F and glucose transport, 247 in intra- and extracellular fluids, 32F and Na,K-ATPase, 246 Solar energy, in chloroplasts, 411–415 Soluble N-ethylmaleimide-sensitivefactor attachment protein receptor (SNAREs), 251–252, 251F Solute, 28 Solvated ions, 28 Solvents, 28–29, 29F Sophorose, 298 Sorbitol, 298, 510 Soybeans, 72, 431 Sp1 protein, 576 Space-filling model, 5F Specificity pockets, 170, 171, 171F Spectrin, 116 Sphingolipids, 218, 219F, 270 Sphingomyelins, 218, 219F, 270 Sphingosine, 219F, 270 Sphingosine-1-phosphate, 281 Spinach, 93T, 423F Spliceosomes, 568–569, 569F Splicing: alternative, 569, 570F of mRNA, 567–570, 570F in RNA processing, 567–570 Spontaneous reactions, 11–12 Sports drinks, 32 SQDG (sulfoquinovosyldiacylglycerol), 231 Squalene, 455, 456F SRP (signal recognition particle), 599–600, 599F SSB (single-strand binding protein), 521, 522 SSRIs (selective serotonin reuptake inhibitors), 250 Stacking interactions, 59 Stahl, F., 82, 519–520 Standard conditions (standard state), 315, 315T Standard free energy change of reaction (ΔGº′): calculating, 314–316 for citric acid cycle, 366, 370, 371 for glucose catabolism, 342–343, 343T for glycolysis, 335–337, 339 for phosphate hydrolysis, 318–319, 318T Standard reduction potential (ℰ º′), 386–387, 387T Standard state, 315, 315T Staphylococcal nuclease, 114–115, 156T, 190T Starch: from Calvin cycle products, 426–428 fuel storage in, 288–289 as glucose polymer, 8, 9F Starch synthase, 426–427 Start codon, 66 Starvation, 507–508 Statins, 455, 456, 456F Steady state, 187 Stearate, 216 Stearic acid, 217T Stereochemistry, of aconitase, 369, 369F Stereocilia, 144 Stereoisomerism, of carbohydrates, 284–285 Steroids, 274 Sticky ends, 69 Stoichiometry, of oxidative phosphorylation, 403, 404 Stomata, 424 Stonewashed cotton clothing, 299

Stop codons: in genetic code, 62, 581, 581T and open reading frame, 66 termination of translation at, 595 Streptavidin, 111F Streptomyces albus, 60T Streptomyces lividans, 241, 241F Streptomycin, 594 Stroma, 412 Strong bases, 40, 41 Structural formula, 5F Structural proteins, 130–140 actin filaments, 131–132 collagen, 136–140 keratin, 135–136 microtubules, 132–135 tubulin, 132–134 Subcutaneous fat, 511 Substitution, conservative, 123 Substrates: binding of, to succinyl-CoA synthetase, 370, 371F cooperative binding of, 194, 194F for enzymes, 156 kinetics of multisubstrate reactions, 193 suicide, 195 Substrate concentration: at half-maximal velocity, see Michaelis constant (KM) on Lineweaver–Burk plots, 191–192 reaction velocity and, 184–185, 185F Substrate-level phosphorylation, 370 Substrate specificity, 170–171 Subtilisin, 170, 170F, 209 Subunits, 104–105 Subunit interactions, in hemoglobin, 125, 125F Succinate, 177, 195–196 in β oxidation, 441F in citric acid cycle, 367F, 370F, 371 in glyoxylate pathway, 377 in ketone body catabolism, 452F pK of, 37T standard reduction potential of, 387T Succinate dehydrogenase: in β oxidation, 441F in citric acid cycle, 367F, 370, 371 competitive inhibition of, 195–196 as Complex II, 394 electron transfer by, 386 Succinic acid, pK, 37T Succinyl-CoA: in amino acid catabolism, 479 from anaplerotic reactions, 377 in β oxidation, 441F as biosynthetic precursor, 375, 375F in citric acid cycle, 367F, 369–370, 370F in ketone body catabolism, 452F Succinyl-CoA synthetase: in β oxidation, 441F in citric acid cycle, 370, 370F substrate binding in, 371F Succinyl phosphate, 370F Sucrose: from Calvin cycle products, 426, 427 catabolism of, 340 structure of, 288 Sucrose-6-phosphate, 427 Sugars, 5. See also Carbohydrates in Calvin cycle, 424–426 catabolism of, 340. See also Glycolysis L and D, 284–285 reducing vs. nonreducing, 286 Sugar–phosphate backbones, 57, 58 Suicide substrates, 195 Sulfate, 387T Sulfhydryl group, 4T Sulfonamides, 493 Sulfonylureas, 510T

Sulfoquinovosyldiacylglycerol (SQDG), 231 Sulfur-containing amino acids, 472 Sulfurylase, 75, 75F SUMO (small ubiquitin-like modifier protein), 600 Supercoiling, of DNA, 541–543, 541F Superoxide dismutase, 398 Superoxide free radical, 398 Surface tension, 25–26, 26F SV40 (simian virus 40), 546 Sweat, 32 Symporters, 245, 245F Synapses, nerve–muscle, 248–250, 249F Synaptic vesicles, 248 Synechococcus species, 419, 419F Synechocystis PCC6803, 64T, 65 Synthesis: of amino acids, see Amino acid biosynthesis of cholesterol, 454–457 citric acid cycle as pathway for, 373–374 discontinuous, 523 of eicosanoids, 275, 275F of fatty acids, 443–452 of glycogen, 347–349 of nucleotides, 304, 485–489 of phospholipids, 453–454 of polypeptides, 14, 15 of protein, see Protein synthesis of starch and sucrose, 426–428 of triacylglycerols, 452–453, 453F Synthroid (levothyroxine), 206T α-Synuclein, 106 Syracuse hemoglobin variant, 129T Szent-Györgyi, A., 372

T T, see Threonine; Thymine Tm (melting temperature), of DNA, 59–60, 60T T3 (triiodothyronine), 274, 274F T4 (thyroxine), 274, 274F Tachystatin, 98F TAFs (TBP-associated factors), 576 TAF1, 557 Tagatose, 297 Tamiflu (oseltamivir), 212 Tardigrades, 519 TATA-binding protein (TBP), 557, 557F TATA box, 556, 557 Tau protein, 105 Taurine, 493 Tautomers, 161 Tautomerization, 339 TBP (TATA-binding protein), 557, 557F TBP-associated factors (TAFs), 576 T conformational state, of hemoglobin, 125, 126 TDP (thymidine diphosphate), 54T Tegu lizard, 362 Telomerase, 528–531, 529F and cell immortality, 531 chromosome extension by, 529, 531 Telomeres, 528–531 and cell immortality, 531 DNA synthesis at, 529, 529F extension of chromosomes at, 529, 531 HIV reverse transcriptase at, 530 Temperature, 10 body, 362 melting, 59–60, 60T and rate of reaction, 155 and spontaneity, 12 standard state, 315, 315T Temperature-sensitive mutations, 546 Tendons, collagen in, 137

Termination: of transcription, 565–566, 566F of translation, 595–596 Ter sequences, 525 Tertiary structure of proteins, 97–103 defined, 93 of hemoglobin, 94F hydrophobic cores, 98–99 and protein stability, 99–101 for proteins with multiple conformations, 102–103 and secondary structures, 101–102 Tertiary structure of tRNA, 582, 582F Tetrahedral intermediate, 165F Tetrahydrobiopterin, 479, 480 Tetrahydrocannabinol (THC), 260, 281 Tetrahydrofolate, 313T, 471, 471F Tetramethyl-p-phenylenediamine (TMPD), 407 Tetroses, 284 TFIIB, 557, 564 TFIID, 557 TFIIE, 557 TFIIF, 557 TFIIH, 557 TGF-β (transforming growth factor β), 117 Thalassemias, 129 β-Thalassemia, 78T, 605 Thalidomide, 87 THC (tetrahydrocannabinol), 260, 281 Thermodynamics, 10–14 of citric acid cycle, 372–374 first law of, 10 of membrane transport, 235–240 of oxidation–reduction reactions, 385–390 second law of, 12 Thermogenesis, 404 Thermolysin, 109T Thermoproteus tenax, 357, 359–360 Thermotoga maritima, 574F Thermus aquaticus, 72, 525F Thermus thermophilus, 586, 586F, 594F Thiamine, 312, 313, 313T Thiamine pyrophosphate (TPP): in pyruvate dehydrogenase reaction, 363, 364, 365F structure of, 363F and vitamins, 313T Thiazolidinediones, 510, 510T Thick filaments (myosin), 142, 142F Thin filaments (actin), 142, 142F Thioesters, 319 Thiogalactoside transacetylase, 561 Thiols, 4T Thiolase: in β oxidation, 438F in cholesterol synthesis, 455F in ketogenesis, 451F in ketone body catabolism, 452F Thiolate anion, 89 Thiol groups, 89 Thiophenol, pK, 37T Thioredoxin, 430, 487 Thiotrix bacteria, 322 30S subunit, 586, 586T, 589 Thr, see Threonine 3′ end, 56 3′→5′ Exonuclease, 525, 526 3′ Polyadenylation, 567 3′ Cap, 567, 567F Threonine (Thr, T): abbreviations for, 87 catabolism of, 478 as essential amino acid, 312T, 470T genetic code for, 63T, 581T as glucogenic/ketogenic amino acid, 477T hydrophobicity of, 99T percent occurrence of, 91F as polar amino acid, 87F, 89 Threonine deaminase, 493

In dex Thrombin, 173–174, 174F Thromboxane, 261T, 275, 275F, 281 Thylakoid, 412 Thylakoid membrane, 429 Thymidine, 54T Thymidine diphosphate (TDP), 54T Thymidine monophosphate (TMP), 54T Thymidine nucleotides, synthesis of, 488–489 Thymidine triphosphate (TTP), 54T Thymidylate, 54T Thymidylate synthase, 195, 489 Thymine (T), 54T and Chargaff’s rules, 56–57 as complement of A, 59 and DNA structure, 56–57 and melting temperature, 59 as pyrimidine base, 53 UV damage to, 532 Thymine dimers, 532, 533 Thyroxine (T4), 274, 274F Tissue plasminogen activator, 71T Titration, 40–41, 41F T-loops, in DNA, 531, 531F TMP (thymidine monophosphate), 54T TMPD (tetramethyl-pphenylenediamine), 407 TNFα (tumor necrosis factor α), 511 Tobacco, 412F Tocopherol, 313T α-Tocopherol, 221 Tofu, 181 Topoisomerases, 521, 541–543 Tosyl-phenylalanine chloromethyl ketone (TPCK), 178 Toxoplasma gondii, 215 TPP, see Thiamine pyrophosphate Trace elements, 3 Transacylase, 446F Transaminases, 467–469, 478, 480 Transamination, 467–469 as anaplerotic reaction, 377–378 and glucose–alanine cycle, 501 Transcription, 61, 551–575 defined, 552 DNA packaging and, 553–555 elongation in, 564–565 initiation of, 552–561 replication vs., 551–552 RNA polymerase in, 562–566 RNA processing for, 567–574 spatial separation of replication and, 562, 562F termination of, 565–566 Transcription factors: autophosphorylation of, 272 defined, 558 general, 557–558 zinc fingers of, 103F Transcriptome, 311 Transcriptomics, 311 Transesterification, 568–569 Trans fatty acids, 230, 448 Transferases, 157T. See also specific enzymes Transfer RNA, see tRNA Transforming growth factor β (TGF-β), 117 Transgenic organisms, 72 Transgenic plants, 431 Transition mutations, 532 Transition state analogs, 197–198 Transition states: in catalytic mechanisms, 158, 159 for serine protease reactions, 165F stabilization of, 166–168, 167F Transketolase, 193–194 Translation, 61, 589–597 antibiotics in, 594 defined, 580 elongation in, 590–592 initiation of, 589–590

peptide bond formation in, 593–595 termination of, 595–596 in vivo efficiency of, 596 Translation factors, 592T Translocases, 224 Translocation: defined, 595 membrane, 599–600 of protons, 401–403, 402F Translocons, 599 Transmembrane helices (GPCRs), 265 Transmembrane proteins: with α helix structure, 226 with β barrel structure, 226–227 7-Transmembrane (7TM) receptors, see G protein–coupled receptors (GPCRs) Transmissible spongiform encephalopathies (TSEs), 106 Transpeptidation, 593–595 Transport. See also Electron transport chain; Membrane transport citrate transport system, 444, 444F lipid transport, 432–434 Transporters (transport proteins): ABC, 247 active, 240, 245–248 choline, 256–257 conformations of, 244–245 glucose, 244F, 248, 248F GLUT, 256, 354T, 503, 503F LmrA, 258 passive, 237, 239–245 PEPT1, 258 types of, 245, 245F Transporter Classification Database, 237 Transposable elements, 66 Transthyretin, 117 Transverse diffusion, 224 Transversions (transversion mutations), 532 Trastuzumab (Herceptin), 273 Treadmilling: by actin filaments, 131–132, 132F by microtubules, 134 Tree frogs, 85 Trehalase, 298 Trehalose, 359 Triacylglycerols: defined, 217 fatty acid oxidation for, 435 in lipoproteins, 433T mobilization of, 305 storage of, 304 synthesis of, 452–453, 453F TRiC chaperonin complex, 598 Triclosan, 450 Trienoic acids, 233 Trifluoroacetic acid, 37T Trigger factor, 597–598, 597F Triglycerides, 217. See also Triacylglycerols Triiodothyronine (T3), 274, 274F Trioses, 284 Triose phosphate isomerase, 97, 98F, 156T in gluconeogenesis, 345F in glycolysis, 331F, 335–336, 335F kcat of, 190T reaction kinetics for, 184, 184F transition state analogs for, 197–198 Triple helix, collagen as, 136–139, 138F Trisaccharides, 283 Tris(hydroxymethyl)-aminomethane (Tris), 37T tRNA (transfer RNA), 580–585 aminoacyl–, 583, 583F aminoacylation of, 582–584 binding of, to ribosome, 587–589 and central dogma of molecular biology, 62

defined, 552 EF-Tu complex with, 591, 591F initiator, 589–590 isoacceptor, 582 modified nucleotides in, 573, 573F as noncoding RNA, 552T pairing of mRNA with, 584–585 processing of, 572–573 and proofreading by synthetases, 584 and redundancy of genetic code, 581 structure of, 59F, 581–582 in translation, 589–597 tRNAAla, 584 tRNAGln, 584, 584F tRNAPhe, 582F Troglitazone, 212, 213 Trp, see Tryptophan Trypanosoma brucei, 210, 257, 381 Trypanosoma cruzi, 356 Trypanosomes, 358 Trypsin: in amino acid sequencing, 110 with bovine pancreatic trypsin inhibitor, 179F cleavage site for, 109, 109T percent sequence identity for, 170T specificity pocket, 170, 171F structure of, 170, 170F substrate specificity of, 170 and trypsin inhibitor, 172–173, 172F Trypsin inhibitor, 172–173, 172F Tryptophan (Trp, W), 177 abbreviations for, 87 catabolism of, 479 as essential amino acid, 312T, 470T genetic code for, 63T, 581, 581T as glucogenic/ketogenic amino acid, 477T as hydrophobic amino acid, 87F, 88, 99T serotonin from, 475 synthesis of, 473–474 Tryptophan synthase, 473, 474, 474F TSEs (transmissible spongiform encephalopathies), 106 TTP (thymidine triphosphate), 54T Tubulin: drug interactions with, 134–135 and kinesin, 145–146 microtubules of, 132–134 α-Tubulin, 133–134, 133F β-Tubulin, 133–134, 133F Tubulin dimer, 133, 133F Tumor necrosis factor α (TNFα), 511 Tumors, 538–539, 538F. See also Cancer Tumor suppressor genes, 539–540 Tuna, 148 Turbo design pathways, 358 Turnover, mRNA, 569–570, 571F Turnover number, 190 Twists, DNA, 541, 542 Two-dimensional electrophoresis, 603 Type 1 diabetes, 509 Type I topoisomerases, 541–542, 541F, 542F Type 2 diabetes, 68, 359, 509–511 Type II topoisomerases, 542–543, 542F Type AB blood, 293 Type A blood, 293 Type B blood, 293 Type O blood, 293 Tyrosinase, 213 Tyrosine (Tyr, Y), 178 abbreviations for, 87 as acid–base catalyst, 162F catabolism of, 479, 480 catecholamines derived from, 475 as essential amino acid, 469 genetic code for, 63T, 581T as glucogenic/ketogenic amino acid, 477T

I-19

hydrophobicity of, 99T as nonessential amino acid, 470T percent occurrence of, 91F pK value of ionizable groups in, 91T as polar amino acid, 87F, 89 synthesis of, 473 Tyrosine kinases, 264, 270–273 Tyrosine radical, 417 Tyrosyl–tRNA, 594 TyrRS, 584

U U, see Uracil Ubiquinol (QH2, Coenzyme Q reduced form): from β oxidation, 438–439 in citric acid cycle, 366, 370–372 electron transfer to cytochrome c from, 394–397 as energy currency, 385 in mitochondria, 391 oxidation of, 398 from oxidation reactions, 394, 395F in Q cycle, 396F from reduction of ubiquinone, 309 standard reduction potential of, 387T Ubiquinol:cytochrome c oxidoreductase, see Complex III Ubiquinone, 2F Ubiquinone (Q, coenzyme Q). See also Complex I in citric acid cycle, 366 in Complex III net reaction, 395 as electron shuttle, 398 electron transfer from NADH to, 392–394 free energy change for reaction of, 388, 389F half-reaction for, 386 as isoprenoid, 220 reduction of, 309 standard reduction potential of, 387T Ubiquitin, 117, 305–306, 306F Ubisemiquinone (QH·), 310 Ucephan, 495 UCP1, 409 UDP (uridine diphosphate), 54T UDP–glucose, 348 in cellulose synthesis, 427–428 in sucrose synthesis, 427 UDP–glucose pyrophosphorylase, 348, 349 Ultraviolet light, 532, 535 UMP (uridine monophosphate), 54T, 486–487 Uncompetitive inhibitors, 200, 201F Uncoupling agents (uncouplers), 404 Unfavorable reactions, coupling of, 316–318 Unimolecular reactions, 186 Uniporters, 245, 245F Units, in biochemistry, 7 U.S. Department of Agriculture, 303 Unsaturated fatty acids: defined, 216 degradation of, 439–440 names and structures of, 217T Uracil (U), 53, 54T as complement of A, 59 from cytosine deamination, 532 degradation of, 490 in genetic code, 63T, 581T wobble pairing with, 585F Uracil-DNA glycosylase, 533–534, 534F Urate, 490 Urea: and aquaporins, 243 dissolution in water of, 21 excretion of nitrogen as, 481, 484 protein denaturation with, 101 in urea cycle, 483F as waste product, 20

I-20 Index Urea cycle, 480–484 cellular location of, 498F defined, 481 and glucose–alanine cycle, 501 glutamate in, 481–482 reactions in, 482–484, 483F reactions related to, 484F Urease, 484 β-Ureidoisobutyrate, 490 β-Ureidopropionate, 490 Uric acid, 484, 490 Uridine, 54T, 490, 572 Uridine diphosphate (UDP), 54T Uridine monophosphate (UMP), 54T, 486–487 Uridine triphosphate (UTP), 54T in glycogen synthesis, 348–349 in pyrimidine synthesis, 486–487 Uridylate, 54T, 486–487 Urinalysis, 110 Urine, pH of, 34T, 49 Usher syndrome, 144 UTP, see Uridine triphosphate

V V, see Valine v0 (initial reaction velocity), 188 Vmax (maximum reaction velocity): defined, 188 experimental determination of, 191–192 inhibitors affecting, 199–200 v (reaction velocity): and competitive inhibitors, 196, 196F defined, 184 initial, 188 and KM, 189–190 on Lineweaver–Burk plots, 191–192 in Michaelis–Menten equation, 187, 188 and mixed inhibitors, 200, 200F in rate equations, 186 substrate concentration and, 184–185 Valine (Val, V): abbreviations for, 87 degradation of, 479, 479F as essential amino acid, 312T, 470T genetic code for, 63T, 581T as glucogenic amino acid, 477T

as hydrophobic amino acid, 87F, 88, 99T percent occurrence of, 91F synthesis of, 472 Vanadate, 258, 357 Van der Waals interactions, 26F, 27, 27F Van der Waals radii, 26 Variable residues, 123 Vascular endothelial growth factor (VEGF), 579 Vasoinhibins, 179 Vasotec (enalapril), 214 Vectors, cloning, 70–71, 71F Vegetable oil, 29, 30 VEGF (vascular endothelial growth factor), 579 Velocity of reaction, see v (reaction velocity) Ventolin (albuterol), 206T Very low-density lipoproteins (VLDL): function of, 433, 434F properties of, 433T Vesicles: of kinesin, 144, 145 and lipid bilayers, 31 microvesicles, 253 and SNARES, 251–252 synaptic, 248 Viagra, 280 Vibrio cholerae, 279 Vinblastine, 150 Vioxx, 205, 276 Viruses, RNAi treatment of, 572. See also specific viruses Visceral fat, 511 Visible light, 414F Vitamins, 221, 312–314 in diet, 54 lipid-soluble, 221 roles of, 313T Vitamin A, 72, 221, 313T Vitamin B1, 313T Vitamin B2, 313T Vitamin B3, 313T Vitamin B5, 313T Vitamin B6, 313T, 467 Vitamin B7, 313T Vitamin B12, 313T, 324

Vitamin C, 137, 313T Vitamin D, 221, 313T Vitamin D2, 221 Vitamin D3, 221 Vitamin E, 313T Vitamin K, 221, 313T VLDL, see Very low-density lipoproteins Voltage-gated channels: calcium, 248, 249F potassium, 241, 242, 242F von Gierke’s disease, 360–361 V-type ATPases, 247 Vyvanse (lisdexamfetamine), 206T

W W, see Tryptophan Warburg, O., 512, 549 Warburg effect, 512 Warfarin, 214 Water: and acid–base chemistry, 33–44 and aquaporins, 243–244 characteristics of, 24–26 concentration of, at standard state, 315, 315T dielectric constant for, 29T electronic structure of, 25, 25F in human sweat, 32 hydrogen bonding in, 24–29 oxidation of, in photosynthesis, 417–418 as requirement for life, 24 solvent properties of, 28–29 standard reduction potential of, 387T Water strider, 26F Watson, J., 56, 57, 61, 82, 519 Wavelength, of absorbed light, 412–414 Weak acids, 36, 40, 41 Weddell seals, 329 West Nile virus, 73 White adipose tissue, 508, 509, 509F Wiskott-Aldrich syndrome, 78T Wobble hypothesis, 584–585, 585T Writhes, DNA, 541, 542

X Xanthosine monophosphate (XMP), 486F Xenopus, 103F

Xeroderma pigmentosum, 535 XMP (xanthosine monophosphate), 486F X-ray crystallography, 110–112, 112F Xylitol, 287, 287F Xylose, 181, 290 Xylose-5-phosphate, 360 Xylulose-5-phosphate, 193, 194

Y Y, see Tyrosine Y (fractional saturation), 121 Yeast, 322 codon usage in, 602 enzyme kinetics studies in, 184 enzyme studies in, 160 GCN4 transcription factor of, 559F gene functions in, 67 genome of, 64T HOG pathway in, 281 as model organism, 65F phosphoglycerate kinase, 86F, 93T proteasome core of, 305F pyruvate decarboxylase, 93T spliceosome of, 569F transcription factor GAL4, 101F triose phosphate isomerase, 335F tRNAPhe, 582F Yersinia pestis, 281 Yield, quantum, 426

Z Z (net charge per ion), 236, 400 Zalcitabine (2′,3′-dideoxycytidine, ddC), 530 Zanamivir, 212 Zea mays, see Corn Zebrafish, 114 Zidovudine (3′azido-2,′3′dideoxythymidine, AZT), 530 Zif268 transcription factor, 559F Zinc fingers, 101, 101F, 558–559, 559F Zoloft (sertraline), 250 Zovirax (acylcovir), 204 Z-scheme, 420, 421F Zymogens, 171

Common Functional Groups and Linkages in Biochemistry Structurea

Compound Name

Amineb

RNH2 or R2NH or R3N or

Functional Group

RNH⫹ 3 R2NH⫹ 2 R3NH⫹

N

⫹

or

N

(amino group)

Alcohol

ROH

OH (hydroxyl group)

Thiol

RSH

SH (sulfhydryl group)

Ether

ROR

O

Aldehyde

Ketone

R

R

O

O

C H

C

O

O

C

R

C

O Carboxylic acidb (Carboxylate)

O (carbonyl group), R

C O

(carbonyl group), R

C

R

C O

OH or

C O

OH (carboxyl group) or

R

C

O⫺

C

O⫺ (carboxylate group)

R

C

(acyl group)

(acyl group)

O

O

O Ester

(ether linkage)

C

OR

O

(ester linkage)

C N

(amido group)

O Amide

Imineb

R

C O

NH2

R

C O

NHR

R

C

NR2

R R

NH or NR or

O

R R

NH⫹ 2 NHR⫹

C

O Phosphoric acid esterb

R

O

P

O

P

OH or

O

O

P

O⫺

O

P O⫺

a

P

O

P

O OH or

P

O O

P

O OH or

P O⫺

P

O O

OH

OH O O

O

P

P

O

O

(phosphoanhydride linkage)

OH O

O O⫺

O⫺ (phosphoryl group, Pi )

O⫺

P

O OH or

P

O O

P

O⫺

O⫺ OH O⫺ OH (diphosphoryl group, pyrophosphoryl group, PPi )

R represents any carbon-containing group. In a molecule with more than one R group, the groups may be the same or different. Under physiological conditions, these groups are ionized and hence bear a positive or negative charge.

b

H (imino group)

(phosphoester linkage)

OH

OH O R

⫹

N

OH

O R

C

O

O⫺

Diphosphoric acid esterb

or

O

OH O R

N

USEFUL CONSTANTS Avogadro’s number Gas constant (R) Faraday (F ) Kelvin (K)

6.02 × 1023 molecules · mol–1 8.314 J · K –1 · mol–1 96,485 J · V –1 · mol–1 °C + 273

KEY EQUATIONS Henderson–Hasselbalch equation pH = pK + log

[A− ] [HA]

Michaelis–Menten equation Vmax [S] v0 = KM + [S] Lineweaver–Burk equation KM 1 1 1 + = v0 ( V max ) [S] V max

Nernst equation Ɛ = Ɛ°ʹ −

[A reduced ] [A reduced ] 0.026 V RT ln ln or Ɛ = Ɛ°ʹ − n nF [A oxidized ] [A oxidized ]

Thermodynamics equations ∆G = ∆H − T∆S ∆G °ʹ = −RT ln K eq ∆G = ∆G °ʹ + RT ln ∆G°ʹ = −nF∆Ɛ°ʹ

[C] [D] [A] [B]

NUCLEIC ACID BASES, NUCLEOSIDES, AND NUCLEOTIDES Base Formula

Base (X = H)

Nucleoside (X = ribose or deoxyribose)

Nucleotide (X = ribose phosphate or deoxyribose phosphate)

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)

NH2 N

N

N

N

X O H

N

N N

H2N

N X

NH2 N O

N X O

H

CH3

N

O

N X O

H O

N N X

The Standard Genetic Code First Position (5′ end)

Third Position (3′ end)

Second Position

U

C

A

G

U

UUU UUC UUA UUG

Phe Phe Leu Leu

UCU UCC UCA UCG

Ser Ser Ser Ser

UAU UAC UAA UAG

Tyr Tyr Stop Stop

UGU UGC UGA UGG

Cys Cys Stop Trp

U C A G

C

CUU CUC CUA CUG

Leu Leu Leu Leu

CCU CCC CCA CCG

Pro Pro Pro Pro

CAU CAC CAA CAG

His His Gln Gln

CGU CGC CGA CGG

Arg Arg Arg Arg

U C A G

A

AUU AUC AUA AUG

Ile Ile Ile Met

ACU ACC ACA ACG

Thr Thr Thr Thr

AAU AAC AAA AAG

Asn Asn Lys Lys

AGU Ser AGC Ser AGA Arg AGG Arg

U C A G

G

GUU GUC GUA GUG

Val Val Val Val

GCU GCC GCA GCG

Ala Ala Ala Ala

GAU GAC GAA GAG

Asp Asp Glu Glu

GGU GGC GGA GGG

U C A G

Gly Gly Gly Gly