GATE 2022 - Biotechnology - 22 Years Chapter wise Solved Papers (2000-2021)
 9789391061159

Table of contents :
Cover
Title
Copyrights
Contents
Preface
About GATE
GATE Syllabus
Verbal Ability
Chapter 1 : English Grammar
Chapter 2 : Sentence Completion
Chapter 3 : Synonyms
Chapter 4 : Antonyms
Chapter 5 : Reasoning Ability
Numerical Ability
Chapter 1 : Numbers and Algebra
Chapter 2 : Percentage and Its Applications
Chapter 3 : Time and Work
Chapter 4 : Ratio, Proportion and Mixtures
Chapter 5 : Permutations and Combinations & Probability
Chapter 6 : Miscellaneous
Engineering Mathematics
Chapter 1 : Linear Algebra
Chapter 2 : Calculus
Chapter 3 : Differential Equations
Chapter 4 : Probability and Statistics
Technical Section
Chapter 1 : Biomolecules : Constituents of Life
Chapter 2 : Bioenergetics and Metabolism
Chapter 3 : Biophysical and Biochemical Techniques
Chapter 4 : Cell Biology
Chapter 5 : Microbiology
Chapter 6 : Immunology
Chapter 7 : Genetics - I : Classical and Population Genetics
Chapter 8 : Genetics - II : Molecular Genetics
Chapter 9 : Advanced Biotechnology
Chapter 10 : Enzyme Science and Technology
Chapter 11 : Biochemical Engineering and Bioprocess Technology
Chapter 12 : Bioinformatics
SOLVED PAPER - 2017
SOLVED PAPER 2018
SOLVED PAPER - 2019
SOLVED PAPER - 2020
SOLVED PAPER - 2021

Citation preview

Title

: GATE 2022 - Biotechnology - 22 Years Chapter wise Solved Papers (2000-2021)

Language

: English

Author’s Name : Dr. Prabhanshu Kumar, Dr. Pawan Kr. Maurya, Er. Preeti T. Kumar Copyright ©

: 2021 Dr. Prabhanshu Kumar, Dr. Pawan Kr. Maurya, Er. Preeti T. Kumar

No part of this book may be reproduced in a retrieval system or transmitted, in any form or by any means, electronics, mechanical, photocopying, recording, scanning and or without the written permission of the Author/Publisher.

Published by : Career Launcher Infrastructure (P) Ltd. A-45, Mohan Cooperative Industrial Area, Near Mohan Estate Metro Station, New Delhi - 110044 Marketed by : G.K. Publications (P) Ltd. Plot No. 9A, Sector-27A, Mathura Road, Faridabad, Haryana-121003 ISBN : 978-93-91061-15-9 Printer’s Details : Made in India, New Delhi. For product information : Visit www.gkpublications.com or email to [email protected]

 Preface  About GATE  GATE Syllabus

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S olved Papers (Chapter-Wise) Verbal Ability 1. English Grammar MCQ Type Questions

1.1

MCQ Type Questions

2.1

Numerical Type Questions

2.1



Answers

1.4



Explanations

1.4

2. Sentence Completion MCQ Type Questions

2.1 - 2.4 2.1



Answers

2.4



Explanations

2.4

3. Synonyms MCQ Type Questions

3.1

Answers

3.2



Explanations

3.2

MCQ Type Questions

Answers

2.2



Explanations

2.2

3.1 - 3.4

MCQ Type Questions

3.1

Numerical Type Questions

3.2



Answers

3.2



Explanations

3.3

4. Ratio, Proportion and Mixtures 4.1 - 4.2

4.1 - 4.2

MCQ Type Questions

4.1

4.1



Answers

4.1



Explanations

4.2



Answers

4.1



Explanations

4.2

5. Reasoning Ability



3. Time and Work

3.1 - 3.2



4. Antonyms

2. Percentage and Its Applications 2.1 - 2.2

1.1 - 1.4

5. Permutations and Combinations &

5.1 - 5.14

MCQ Type Questions

5.1

Numerical Type Questions

5.9

Probability

5.1 - 5.4

MCQ Type Questions

5.1 5.2



Answers

5.10

Numerical Type Questions



Explanations

5.10



Answers

5.2



Explanations

5.3

6. Miscellaneous

6.1 - 6.6

Numerical Ability 1. Numbers and Algebra

1.1 - 1.12

MCQ Type Questions

1.1

Numerical Type Questions

1.4

MCQ Type Questions

6.1

Numerical Type Questions

6.3



Answers

1.6



Answers

6.4



Explanations

1.7



Explanations

6.4

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5. Microbiology

E ngineering Mathematics 1. Linear Algebra

1.1 – 1.3

5.1 – 5.6



Answers

5.3



Explanations

5.4

6. Immunology

6.1 – 6.12



Answers

1.2



Explanations

1.2



Answers

6.6

2.1 – 2.3



Explanations

6.7

2. Calculus –

Answers

2.2



Explanations

2.2

3. Differential Equations

3.1 – 3.3



Answers

3.1



Explanations

3.2

4. Probability and Statistics

7. Genetics - I : Classical and Population Genetics 7.1 – 7.5

Answers

4.1



Explanations

4.1

1.1 – 1.4



Answers

1.2



Explanations

1.3

2. Bioenergetics and Metabolism Answers

2.4



Explanations

2.5

3. Biophysical and Biochemical Techniques – –

Answers Explanations

4. Cell Biology

3.4

4.1 – 4.6 4.4



Explanations

4.4

7.3



Answers

8.7



Explanations

8.8

9.1 – 9.34



Answers

9.18



Explanations

9.18



Answers

10.9



Explanations

10.9



Answers

11.10



Explanations

11.11

12. Bioinformatics

3.4

Answers

Explanations

11. Biochemical Engineering and Bioprocess Technology 11. 1 – 11.20

3.1 – 3.6





10. Enzyme Science and Technology 10.1 – 10.14

2.1 – 2.8



7.3

9. Advanced Biotechnology

Technical Section 1. Biomolecules : Constituents of Life

Answers

8. Genetics - II : Molecular Genetics 8.1 – 8.12

4.1 – 4.2





Answers

12.3



Explanations

12.3

Solved Solved Solved Solved Solved

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12.1 – 12.6



Paper Paper Paper Paper Paper

2017 2018 2019 2020 2021

1 - 16 1 - 16 1 - 14 1 - 14 1 - 16

IIT Institutes

Indian Institute of Technology, Roorkee

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Whether you want to work with esteemed PSUs including the Maharatna and Navratna companies or do your Masters from IITs and IISc, GATE is your entry to career and academic avenues post your graduation. Thousands of students write the GATE paper annually. The level of competition is fierce, owing to the increasing competition every year for a limited number of seats. If you are a serious aspirant, it is advisable to start preparing for GATE from the second year of your Graduation with the right books. GK Publications has been the guiding force for GATE aspirants for the last 3 decades. Better known as the ‘‘Publisher of Choice'' to students preparing for GATE and other technical test prep examinations in the country, we published the first set of books in 1994 when GATE exam, both objective and conventional, was conducted in the paper and pencil environment and used as a checkpoint for entry to postgraduate courses in IITs and IISCs. At that time, students had little access to technology and relied mainly on instructor-led learning followed by practice with books available for these examinations. A lot has changed since then! Today, GATE is conducted in an online-only mode with multiple choice and numerical based questions.The exam pattern too has evolved over the years. Instead of theory, the questions today test a student's basic understanding and his capacity to put fundamentals to practical applications.The exam only rewards those with a concrete foundational clarity of basics. We, at GKP, have also embraced change. Today, our preparatory manuals are not mere books but torchbearers in your GATE preparation journey. We understand that tackling the syllabus one chapter at a time is not only easy but also allows students to cope with the looming exam pressure. A major game changer is the habit to practice and solve previous year questions and this is why, our GATE 2021 Chapter-wise Solved Papers of Biotechnology is your best bet to be GATE READY by 2021! The book carries detailed analysis of previous years' solved papers (all sets) so that you are well-versed with the latest exam pattern and how it has evolved over time. Both MCQs and Numerical have been explained in a lucid and detailed fashion, so that you take your preparation forward - one question at a time. With each question, its marks' weightage has also been given so that you can gauge the level of question and make judicious use of your time. Today's students have easy access to technology and the concept of a monologue within the classroom has changed to dialogue where students come prepared with concepts and then discuss topics. We have launched a web portal and mobile app to help you take your preparation on the go - so that you learn, revise and practice mock tests online. The mobile app provides access to video lectures, short tests and provides regular updates about the exam. The web portal in addition to what is available on the app provides full-length mock tests to mimic the actual exam to help you gauge your level of preparedness. The combination of practice content in print, video lectures, and short and full-length tests on mobile and web make this product complete courseware for GATE preparation. We also know that improvement is a never-ending process and hence we welcome your suggestions and feedback or spelling and technical errors if any. Please write to us at [email protected] We hope that our small effort will help you prepare well for the examination. We wish you all the best! Team GKP

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About GATE The Graduate Aptitude Test in Engineering (GATE) conducted by IISc and IITs has emerged as one of the bench mark tests for engineering and science aptitude in facilitating admissions for higher education (M.Tech./Ph.D.) in IITs, IISc and various other Institutes/Universities/Laboratories in India. With the standard and high quality of the GATE examination in 27 disciplines of engineering and science, Humanities and Social Sciences subjects, it identifies the candidate's understanding of a subject and aptitude and eligibility for higher studies. During the last few years, GATE score is also being used as one of the criteria for recruitment in Government Organizations such as Cabinet Secretariat, and National/State Public Sector Undertakings in India. Because of the importance of the GATE examination, the number of candidates taking up GATE exams has increased tremendously. GATE exams are conducted by the IITs and IISc as a computer based test having multiple choice questions and numerical answer type questions. The questions are mostly fundamental, concept based and thought provoking. From 2017 onwards GATE Exam is being held in Bangladesh, Ethiopia, Nepal, Singapore, Sri Lanka and United Arab Emirates. An Institute with various nationalities in its campus widens the horizons of an academic environment. A foreign student brings with him/her a great diversity, culture and wisdom to share. Many GATE qualified candidates are paid scholarships/assistantship, especially funded by Ministry of Human Resources Development, Government of India and by other Ministries. IIT, Kharagpur is the Organizing Institute for GATE 2022.

Why GATE? Admission to Post Graduate and Doctoral Programmes Admission to postgraduate programmes with MHRD and some other government scholarships/ assistantships in engineering colleges/institutes is open to those who qualify through GATE. GATE qualified candidates with Bachelor’s degree in Engineering/Technology/Architecture or Master’s degree in any branch of Science/Mathematics/Statistics/Computer Applications are eligible for admission to Master/Doctoral programmes in Engineering/Technology/Architecture as well as for Doctoral programmes in relevant branches of Science with MHRD or other government scholarships/ assistantships. Candidates with Master’s degree in Engineering/Technology/Architecture may seek admission to relevant Ph.D programmes with scholarship/assistantship without appearing in the GATE examination.

Financial Assistance A valid GATE score is essential for obtaining financial assistance during Master’s programs and direct Doctoral programs in Engineering/Technology/Architecture, and Doctoral programs in relevant branches of Science in Institutes supported by the MHRD or other Government agencies. As per the directives of the MHRD, the following procedure is to be adopted for admission to the post-graduate programs (Master’s and Doctoral) with MHRD scholarship/assistantship. Depending upon the norms adopted by a specific institute or department of the Institute, a candidate may be admitted directly into a course based on his/her performance in GATE only or based on his/her performance in GATE and an admission test/interview conducted by the department to which he/she has applied and/or the

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candidate’s academic record. If the candidate is to be selected through test/interview for post-graduate programs, a minimum of 70% weightage will be given to the performance in GATE and the remaining 30% weightage will be given to the candidate’s performance in test/interview and/or academic record, as per MHRD guidelines. The admitting institutes could however prescribe a minimum passing percentage of marks in the test/interview. Some colleges/institutes specify GATE qualification as the mandatory requirement even for admission without MHRD scholarship/assistantship. To avail of the financial assistance (scholarship), the candidate must first secure admission to a program in these Institutes, by a procedure that could vary from institute to institute. Qualification in GATE is also a minimum requirement to apply for various fellowships awarded by many Government organizations. Candidates are advised to seek complete details of admission procedures and availability of MHRD scholarship/assistantship from the concerned admitting institution. The criteria for postgraduate admission with scholarship/assistantship could be different for different institutions. The management of the post-graduate scholarship/assistantship is also the responsibility of the admitting institution. Similarly, reservation of seats under different categories is as per the policies and norms prevailing at the admitting institution and Government of India rules. GATE offices will usually not entertain any enquiry about admission, reservation of seats and/or award of scholarship/ assistantship.

PSU Recruitments As many as 37 PSUs are using GATE score for recruitment. It is likely that more number of PSUs may start doing so by next year. Below is the list of PSUs: MDL, BPCL, GAIL, NLC LTD, CEL, Indian Oil, HPCL, NBPC, NECC, BHEL, WBSEDCL, NTPC, ONGC, Oil India, Power Grid, Cabinet Secretariat, Govt. of India, BAARC, NFL, IPR, PSPCL, PSTCL, DRDO, OPGC Ltd., THDC India Ltd., BBNL, RITES, IRCON, GHECL, NHAI, KRIBHCO, Mumbai Railway Vikas Corporation Ltd. (MRVC Ltd.), National Textile Corporation, Coal India Ltd., BNPM, AAI, NALCO, EdCIL India. Important :

1. Admissions in IITs/IISc or other Institutes for M.Tech./Ph.D. through GATE scores shall be advertised separately by the Institutes and GATE does not take the responsibility of admissions. 2. Cabinet Secretariat has decided to recruit officers for the post of Senior Field Officer (Tele) (From GATE papers of EC, CS, PH), Senior Research Officer (Crypto) (From GATE papers of EC, CS, MA), Senior Research Officer (S&T) (From GATE papers EC, CS, CY, PH, AE, BT) in the Telecommunication Cadre, Cryptographic Cadre and Science & Technology Unit respectively of Cabinet Secretariat. The details of the scheme of recruitment shall be published in National Newspaper/Employment News by the concerned authority. 3. Some PSUs in India have expressed their interest to utilize GATE scores for their recruitment purpose. The Organizations who intend to utilize GATE scores shall make separate advertisement for this purpose in Newspapers etc.

Who Can Appear for GATE? Eligibility for GATE Before starting the application process, the candidate must ensure that he/she meets the eligibility criteria of GATE given in Table. (x)

Eligibility Criteria for GATE 2022 Degree/Program

B.E. / B.Tech. / B.Pharm.

B. Arch.

B.Sc. (Research) / B.S.

Pharm. D. (after 10+2)

M.B.B.S.

M. Sc. / M.A. / MCA or equivalent

Int. M.E./ M.Tech. (Post-B.Sc.) Int. M.E./ M.Tech. or Dual Degree (after Diploma or 10+2)

B.Sc. / B.A. / B.Com.

Int. M.Sc. / Int. B.S. / M.S.

Professional Society Examinations (equivalent to B.E. / B.Tech. / B.Arch.)

Description of Eligible Candidates

Expected Year of Completion

Currently in the 3rd year or higher grade or already completed

2022

Bachelor’s degree of Architecture (5- year course) / Naval Architecture (4- year course) / Planning (4- year course)

Currently in the 3rd year or higher grade or already completed

2023 (for 5-year program), 2022 (for 4-year program)

Bachelor’s degree in Science (Post-Diploma/4 years after 10+2)

Currently in the 3rd year or higher grade or already completed

2022

Qualifying Degree/Examination

Bachelor’s degree in Engineering / Technology (4 years after 10+2 or 3 years after B.Sc. / Diploma in Engineering / Technology)

6 years degree program, consisting of internship or residency training, during third year onwards Degree holders of M.B.B.S. and those who are in the 5th/6th/7th semester or higher semester of such programme.

Currently in the 3rd/4th/5th/6th year or already completed

2024

5th, 6th, 7th or higher semester or already completed

2022

Currently in the first year or higher or already Completed

2022

Post-B.Sc Integrated Master’s degree programs in Engineering/ Technology (4-year program)

Currently in the 1st / 2nd/3rd/4th year or already completed

2024

Integrated Master’s degree program or Dual Degree program in Engineering/Technology (5-year program)

Currently in the 3rd /4th/5th year or alreadycompleted

2023

Master’s degree in any branch of Arts/Science/Mathematics/Statistics/ Computer Applications or equivalent

Bachelor degree in any branch of Science / Arts / Commerce (3 years program)

Integrated M.Sc. or 5-year integrated B.S.-M.S. program

B.E./B.Tech./B.Arch. equivalent examinations of Professional Societies, recognized by MoE/UPSC/AICTE (e.g. AMIE by Institution of EngineersIndia, AMICE by the Institute of Civil Engineers-India and so on)

Currently in the 3rd year or already completed Currently in the 3rd year or higher or already completed

Completed Section A or equivalent of such professional courses

2021

2022

NA

In case a candidate has passed one of the qualifying examinations as mentioned above in 2020 or earlier, the candidate has to submit the degree certificate / provisional certificate / course completion certificate / professional certificate / membership certificate issued by the society or institute. In case, the candidate is expected to complete one of the qualifying criteria in 2022 or later as mentioned above, he/she has to submit a certificate from Principal or a copy of marks card for section A of AMIE. (xi)

Certificate From Principal Candidates who have to submit a certificate from their college Principal have to obtain one from his/ her institution beforehand and upload the same during the online submission of the application form.

Candidates With Backlogs Candidates, who have appeared in the final semester/year exam in 2022, but with a backlog (arrears/ failed subjects) in any of the papers in their qualifying degree should upload a copy of any one of the mark sheets of the final year, OR obtain a declaration from their Principal along with the signature and seal beforehand and upload the same during the online submission of the application form.

GATE Structure Structure of GATE GATE 2022 will be conducted on 27 subjects (papers). Table below shows the list of papers and paper codes for GATE 2022. A candidate is allowed to appear in ANY ONE or UP TO TWO papers of the GATE examination. However, note that the combination of TWO papers in which a candidate can appear MUST be from the pre-defined list as given in Table. Also note that for a paper running in multiple sessions, a candidate will be mapped to appear for the examination in one of the sessions ONLY. List of GATE Papers and Corresponding Codes GATE Paper

Code

GATE Paper

Code

Aerospace Engineering

AE

Instrumentation Engineering

IN

Agricultural Engineering

AG

Mathematics

MA

Architecture and Planning

AR

Mechanical Engineering

ME

Bio-medical Engineering

BM

Mining Engineering

MN

Biotechnology

BT

Metallurgical Engineering

MT

Civil Engineering

CE

Petroleum Engineering

PE

Chemical Engineering

CH

Physics

PH

Computer Science and Information Technology

CS

Production and Industrial Engineering

PI

Chemistry

CY

Statistics

ST

Electronics and Communication Engineering

EC

Textile Engineering and Fibre Science

TF

Electrical Engineering

EE

Engineering Sciences

XE*

Environmental Science & Engineering

ES

Humanities & Social Sciences

XH**

Ecology and Evolution

EY

Life Sciences

XL***

Geology and Geophysics

GG

Note: Environmental Science and Engineering (ES) and Humanities and Social Sciences (XH) are two new papers introduced in GATE 2022. (xii)

*XE Paper Sections Engineering Mathematics (Compulsory) (15 marks)

Cod e A

**XH Paper Sections Reasoning and Comprehension (Compulsory)

Code

B1

***XL Paper Sections

Code

Chemistry (Compulsory)

P

(25 marks)

(25 marks)

Any TWO optional Sections (2x35 = 70 marks)

Any ONE optional Section (60 marks)

Any TWO optional Sections (2x30 = 60 marks)

B

Economics

C1

Biochemistry

Q

Materials Science

C

English

C2

Botany

R

Solid Mechanics

D

Linguistics

C3

Microbiology

S

Thermodynamics

E

Philosophy

C4

Zoology

T

Polymer Science and Engineering

F

Psychology

C5

Food Technology

U

Food Technology

G

Sociology

C6

Atmospheric and

H

Fluid Mechanics

Oceanic Sciences

*XE (Engineering Sciences), **XH (Humanities & Social Sciences), ***XL (Life Sciences), papers are of general nature and will be comprised of Sections listed in the above table Note: Each subject/paper is of total 100 marks. General Aptitude (GA) section of 15 marks is common for all papers. Hence remaining 85 marks are for the respective subject/paper code. Combination of Two Papers Allowed to Appear in GATE 2022 (subject to availability of infrastructure and schedule) Code of The First (Primary) Paper AE AG AR BM BT CE CH CS CY EC EE ES EY GG IN MA ME MN MT PE PH PI ST TF XE XH XL

Codes of Papers Allowed as The Second Paper XE ES CE BT / XL BM / XL AR / ES CY / PE / XE MA CH / XL IN / PH IN AG / CE XL MN / PE / PH EC / EE / PH CS / PH / ST XE GG / XE PH / XE CH / GG / XE EC / GG / IN / MA / MT / ST XE MA / PH XE AE / CH / ME / MN / MT / PE / PI / TF ---BM / BT / CY / EY

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General Aptitude Questions All the papers will have a few questions that test the General Aptitude (Language and Analytical Skills), apart from the core subject of the paper.

Duration and Examination Type All the papers of the GATE 2022 examination will be for 3 hours duration and they consist of 65 questions for a total of 100 marks. Since the examination is an ONLINE computer based test (CBT), at the end of the stipulated time (3-hours), the computer screen will automatically close the examination inhibiting any further action. Candidates will be permitted to occupy their allotted seats 40 minutes before the scheduled start of the examination. Candidates can login and start reading the instructions 20 minutes before the start of examination. The late login time (if any) recorded by the computer system MUST NOT be beyond 30 minutes from the actual starting time of the examination. Under NO circumstances, will a candidate be permitted to login after 30 minutes from the actual examination starting time. Candidates will NOT be permitted to leave the examination hall before the end of the examination.

Pattern of Question Papers GATE 2022 may contain questions of THREE different types in all the papers: (i) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each, in all the papers and sections. These questions are objective in nature, and each will have choice of four answers, out of which ONLY ONE choice is correct. Negative Marking for Wrong Answers: For a wrong answer chosen in a MCQ, there will be negative marking. For 1-mark MCQ, 1/3 mark will be deducted for a wrong answer. Likewise, for 2-mark MCQ, 2/3 mark will be deducted for a wrong answer. (ii) Multiple Select Questions (MSQ) carrying 1 or 2 marks each in all the papers and sections. These questions are objective in nature, and each will have choice of four answers, out of which ONE or MORE than ONE choice(s) is / are correct. Note: There is NO negative marking for a wrong answer in MSQ questions. However, there is NO partial credit for choosing partially correct combinations of choices or any single wrong choice. (iii) Numerical Answer Type (NAT) Questions carrying 1 or 2 marks each in most of the papers and sections. For these questions, the answer is a signed real number, which needs to be entered by the candidate using the virtual numeric keypad on the monitor (keyboard of the computer will be disabled). No choices will be shown for these types of questions. The answer can be a number such as 10 or –10 (an integer only). The answer may be in decimals as well, for example, 10.1 (one decimal) or 10.01 (two decimals) or –10.001 (three decimals). These questions will be mentioned with, up to which decimal places, the candidates need to present the answer. Also, for some NAT type problems an appropriate range will be considered while evaluating these questions so that the candidate is not unduly penalized due to the usual round-off errors. Candidates are advised to do the rounding off at the end of the calculation (not in between steps). Wherever required and possible, it is better to give NAT answer up to a maximum of three decimal places. Note: There is NO negative marking for a wrong answer in NAT questions. Also, there is NO partial credit in NAT questions.

Marking Scheme – Distribution of Marks and Questions General Aptitude (GA) Questions In all papers, GA questions carry a total of 15 marks. The GA section includes 5 questions carrying 1mark each (sub-total 5 marks) and 5 questions carrying 2-marks each (sub-total 10 marks). (xiv)

Question Papers other than GG, XE, XH and XL These papers would contain 25 questions carrying 1-mark each (sub-total 25 marks) and 30 questions carrying 2-marks each (sub-total 60 marks) consisting of some MCQ type questions, while the remaining may be MSQ and / or NAT questions.

GG (Geology and Geophysics) Paper Apart from the General Aptitude (GA) section, the GG question paper consists of two parts: Part A and Part B. Part A is compulsory for all the candidates. Part B contains two sections: Section 1 (Geology) and Section 2 (Geophysics). Candidates will have to attempt questions in Part A and questions in either Section 1 or Section 2 of Part B. The choice of Section 1 OR Section 2 of Part B has to be made at the time of online registration in GOAPS. At the examination hall, candidate cannot request for change of section. Part A consists of 25 questions carrying 1-mark each (sub-total 25 marks and some of these will be MSQ and/or numerical answer type questions while remaining questions will be MCQ type). Either section of Part B (Section 1 and Section 2) consists of 30 questions carrying 2-marks each (sub-total 60 marks and some of these will be MSQ and/or numerical answer type questions while remaining questions will be MCQ type).

XE Paper (Engineering Sciences) A candidate appearing in the XE paper has to answer the following: • GA – General Aptitude carrying a total of 15 marks. • Section A – Engineering Mathematics (Compulsory): This section contains 11 questions carrying a total of 15 marks: 7 questions carrying 1-mark each (sub-total 7 marks), and 4 questions carrying 2-marks each (sub-total 8 marks). Some questions will be of numerical answer type while remaining questions will be MCQ type. • Any two of XE Sections B to H: The choice of two sections from B to H can be made during the examination after viewing the questions. Only TWO optional sections can be answered at a time. A candidate wishing to change midway of the examination to another optional section must first choose to deselect one of the previously chosen optional sections (B to H). Each of the optional sections of the XE paper (Sections B through H) contains 22 questions carrying a total of 35 marks: 9 questions carrying 1-mark each (sub-total 9 marks) and 13 questions carrying 2-marks each (subtotal 26 marks). Some questions will be of MSQ and/or numerical answer type while remaining questions will be MCQ type.

XH Paper (Humanities and Social Sciences) A candidate appearing in the XH paper has to answer the following: • GA – General Aptitude carrying a total of 15 marks. • Section B1 – Reasoning and Comprehension (Compulsory): This section contains 15 questions carrying a total of 25 marks: 5 questions carrying 1-mark each (sub-total 5 marks) and 10 questions carrying 2-marks each (sub-total 20 marks). Some questions will be of MSQ and/or numerical answer type while remaining questions will be MCQ type. • Any ONE of XH Sections C1 to C6: The ONE choice of section from C1 to C6 has to be made at the time of online registration in GOAPS. At the examination hall, candidate cannot request for change of section. Each of the optional sections of the XH paper (Sections C1 through C6) contains 40 questions carrying a total of 60 marks: 20 questions carrying 1-mark each (sub-total 20 marks) and 20 questions carrying 2-marks each (sub-total 40 marks). Some questions will be of MSQ and/or numerical answer type while remaining questions will be MCQ type. (xv)

XL Paper (Life Sciences) A candidate appearing in the XL paper has to answer the following: • GA – General Aptitude carrying a total of 15 marks. • Section P–Chemistry (Compulsory): This section contains 15 questions carrying a total of 25 marks: 5 questions carrying 1-mark each (sub-total 5 marks) and 10 questions carrying 2-marks each (sub-total 20 marks). Some questions will be of MSQ and/or numerical answer type while remaining questions will be MCQ type. • Any two of XL Sections Q to U: The choice of two sections from Q to U can be made during the examination after viewing the questions. Only TWO optional sections can be answered at a time. A candidate wishing to change midway of the examination to another optional section must first choose to deselect one of the previously chosen optional sections (Q to U). Each of the optional sections of the XL paper (Sections Q through U) contains 20 questions carrying a total of 30 marks: 10 questions carrying 1-mark each (sub-total 10 marks) and 10 questions carrying 2-marks each (sub-total 20 marks). Some questions will be of MSQ and/or numerical answer type while remaining questions will be MCQ type.

GATE Score After the evaluation of the answers, the actual (raw) marks obtained by a candidate will be considered for computing the GATE Score. For multi-session papers (subjects), raw marks obtained by the candidates in different sessions will be converted to Normalized marks for that particular subject. Thus, raw marks (for single session papers) or normalized marks (for multi-session papers) will be used for computing the GATE Score, based on the qualifying marks.

Calculation of Normalized Marks for Multi-Session Papers In GATE 2022 examination, some papers may be conducted in multi-sessions. Hence, for these papers, a suitable normalization is applied to take into account any variation in the difficulty levels of the question papers across different sessions. The normalization is done based on the fundamental assumption that "in all multi-session GATE papers, the distribution of abilities of candidates is the same across all the sessions". This assumption is justified since the number of candidates appearing in multi-session papers in GATE 2022 is large and the procedure for allocation of session to candidates is random. Further, it is also ensured that for the same multi-session paper, the number of candidates allotted in each session is of the same order of magnitude. Based on the above, and considering various normalization methods, the committee arrived at the following formula for calculating the normalized marks for the multi-session papers. Nsormalization mark of j th candidate in the i th session  Mij is given by g g   Mt – Mq ( M – M )  M g M ij ij iq q M ti – Miq

where Mij : is the actual marks obtained by the j th candidate in ith session

Mtg : is the average marks of the top 0.1% of the candidates considering all sessions Mqg : is the sum of mean and standard deviation marks of the candidates in the paper considering all sessions

Mti : is the average marks of the top 0.1% of the candidates in the ith session Miq : is the sum of the mean marks and standard deviation of the ith session

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Calculation of GATE Score for All Papers For all papers for which there is only one session, actual marks obtained by the candidates will be used for calculating the GATE 2022 Score. For papers in multi-sessions, normalized marks will be calculated corresponding to the raw marks obtained by a candidate and the GATE 2022 Score will be calculated based on the normalized marks. The GATE 2022 score will be computed using the formula given below. GATE Score = Sq  (St – Sq )

( M – Mq ) ( Mt – Mq )

where M : marks obtained by the candidate (actual marks for single session papers and normalized marks for multi-session papers) Mq : is the qualifying marks for general category candidate in the paper

Mt : is the mean of marks of top 0.1% or top 10 (whichever is larger) of the candidates who appeared in the paper (in case of multi-session papers including all sessions) Sq : 350, is the score assigned to Mq St : 900, is the score assigned to Mt In the GATE 2022 the qualifying marks (Mq) for general category student in each subject will be 25 marks (out of 100) or    , whichever is larger. Here  is the mean and  is the standard deviation of marks of all the candidates who appeared in the paper. After the declaration of results, GATE Scorecards can be downloaded by the GATE qualified candidates ONLY. The GATE 2022 Committee has the authority to decide the qualifying mark/score for each GATE paper. In case of any claim or dispute with respect to GATE 2022 examination or score, the Courts and Tribunals in Mumbai alone will have the exclusive jurisdiction to entertain and settle them.

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GATE Syllabus General Aptitude Verbal Aptitude Basic English Grammar: tenses, articles, adjectives, prepositions, conjunctions, verb-noun agreement, and other parts of speech. Basic Vocabulary: words, idioms, and phrases in context, Reading and comprehension, Narrative sequencing.

Quantitative Aptitude Data Interpretation: data graphs (bar graphs, pie charts, and other graphs representing data), 2-and 3-dimensional plots, maps, and tables. Numerical Computation and Estimation: ratios, percentages, powers, exponents and logarithms, permutations and combinations, and series, Mensuration and geometry, Elementary statistics and probability.

Analytical Aptitude Logic: Deduction and induction, Analogy, Numerical relations and reasoning.

Spatial Aptitude Transformation of shapes: translation, rotation, scaling, mirroring, assembling, and grouping Paper folding, cutting, and patterns in 2 and 3 dimensions.

Section 1: Engineering Mathematics Linear Algebra: Matrices and determinants; Systems of linear equations; Eigen values and Eigen vectors. Calculus: Limits, continuity and differentiability; Partial derivatives, maxima and minima; Sequences and series; Test for convergence. Differential Equations: Linear and nonlinear first order ODEs, higher order ODEs with constant coefficients; Cauchy’s and Euler’s equations; Laplace transforms. Probability and Statistics: Mean, median, mode and standard deviation; Random variables; Poisson, normal and binomial distributions; Correlation and regression analysis. Numerical Methods: Solution of linear and nonlinear algebraic equations; Integration by trapezoidal and Simpson’s rule; Single step method for differential equations.

Section 2: General Biology Biochemistry: Biomolecules - structure and function; Biological membranes - structure, membrane channels and pumps, molecular motors, action potential and transport processes; Basic concepts and regulation of metabolism of carbohydrates, lipids, amino acids and nucleic acids; Photosynthesis, respiration and electron transport chain. Enzymes - Classification, catalytic and regulatory strategies; Enzyme kinetics - MichaelisMenten equation; Enzyme inhibition - competitive, non-competitive and uncompetitive inhibition. Microbiology: Bacterial classification and diversity; Microbial Ecology - microbes in marine, freshwater and terrestrial ecosystems; Microbial interactions; Viruses - structure and classification; Methods in microbiology; Microbial growth and nutrition; Nitrogen fixation; Microbial diseases and host-pathogen interactions; Antibiotics and antimicrobial resistance. Immunology: Innate and adaptive immunity, humoral and cell mediated immunity; Antibody structure and function; Molecular basis of antibody diversity; T cell and B cell development; Antigen-antibody reaction; Complement; Primary and secondary lymphoid organs; Major histocompatibility complex (MHC); Antigen processing and presentation; Polyclonal and monoclonal antibody; Regulation of immune response; Immune tolerance; Hypersensitivity; Autoimmunity; Graft versus host reaction; Immunization and vaccines.

Section 3: Genetics, Cellular and Molecular Biology Genetics and Evolutionary Biology: Mendelian inheritance; Gene interaction; Complementation; Linkage, recombination and chromosome mapping; Extra chromosomal inheritance; Microbial genetics - transformation, transduction and conjugation; Horizontal gene transfer and transposable elements; Chromosomal variation; Genetic disorders; Population genetics; Epigenetics; Selection and inheritance; Adaptive and neutral evolution; Genetic drift; Species and speciation. Cell Biology: Prokaryotic and eukaryotic cell structure; Cell cycle and cell growth control; Cell-cell communication; Cell signaling and signal transduction; Post-translational modifications; Protein trafficking; Cell death and autophagy; Extra-cellular matrix. (xix)

Molecular Biology: Molecular structure of genes and chromosomes; Mutations and mutagenesis; Regulation of gene expression; Nucleic acid - replication, transcription, splicing, translation and their regulatory mechanisms; Non-coding and micro RNA; RNA interference; DNA damage and repair.

Section 4: Fundamentals of Biological Engineering Engineering Principles Applied to Biological Systems: Material and energy balances for reactive and non-reactive systems; Recycle, bypass and purge processes; Stoichiometry of growth and product formation; Degree of reduction, electron balance, theoretical oxygen demand. Classical thermodynamics and Bioenergetics: Laws of thermodynamics; Solution thermodynamics; Phase equilibria, reaction equilibria; Ligand binding; Membrane potential; Energetics of metabolic pathways, oxidation and reduction reactions. Transport Processes: Newtonian and non-Newtonian fluids, fluid flow - laminar and turbulent; Mixing in bioreactors, mixing time; Molecular diffusion and film theory; Oxygen transfer and uptake in bioreactor, kLa and its measurement; Conductive and convective heat transfer, LMTD, overall heat transfer coefficient; Heat exchangers.

Section 5: Bioprocess Engineering and Process Biotechnology Bioreaction engineering: Rate law, zero and first order kinetics; Ideal reactors - batch, mixed flow and plug flow; Enzyme immobilization, diffusion effects - Thiele modulus, effectiveness factor, Damkoehler number; Kinetics of cell growth, substrate utilization and product formation; Structured and unstructured models; Batch, fed-batch and continuous processes; Microbial and enzyme reactors; Optimization and scale up. Upstream and Downstream Processing: Media formulation and optimization; Sterilization of air and media; Filtration - membrane filtration, ultrafiltration; Centrifugation - high speed and ultra; Cell disruption; Principles of chromatography - ion exchange, gel filtration, hydrophobic interaction, affinity, GC, HPLC and FPLC; Extraction, adsorption and drying. Instrumentation and Process Control: Pressure, temperature and flow measurement devices; Valves; First order and second order systems; Feedback and feed forward control; Types of controllers - proportional, derivative and integral control, tuning of controllers.

Section 6: Plant, Animal and Microbial Biotechnology Plants: Totipotency; Regeneration of plants; Plant growth regulators and elicitors; Tissue culture and cell suspension culture system - methodology, kinetics of growth and nutrient optimization; Production of secondary metabolites; Hairy root culture; Plant products of industrial importance; Artificial seeds; Somaclonal variation; Protoplast, protoplast fusion - somatic hybrid and cybrid; Transgenic plants - direct and indirect methods of gene transfer techniques; Selection marker and reporter gene; Plastid transformation. Animals: Culture media composition and growth conditions; Animal cell and tissue preservation; Anchorage and non-anchorage dependent cell culture; Kinetics of cell growth; Micro & macro-carrier culture; Hybridoma technology; Stem cell technology; Animal cloning; Transgenic animals; Knock-out and knock-in animals. Microbes: Production of biomass and primary/secondary metabolites - Biofuels, bioplastics, industrial enzymes, antibiotics; Large scale production and purification of recombinant proteins and metabolites; Clinical-, food- and industrial- microbiology; Screening strategies for new products.

Section 7: Recombinant DNA Technology and Other Tools in Biotechnology Recombinant DNA technology: Restriction and modification enzymes; Vectors - plasmids, bacteriophage and other viral vectors, cosmids, Ti plasmid, bacterial and yeast artificial chromosomes; Expression vectors; cDNA and genomic DNA library; Gene isolation and cloning, strategies for production of recombinant proteins; Transposons and gene targeting. Molecular tools: Polymerase chain reaction; DNA/RNA labelling and sequencing; Southern and northern blotting; In-situ hybridization; DNA fingerprinting, RAPD, RFLP; Site-directed mutagenesis; Gene transfer technologies; CRISPR-Cas; Biosensing and biosensors. Analytical tools: Principles of microscopy - light, electron, fluorescent and confocal; Principles of spectroscopy - UV, visible, CD, IR, fluorescence, FT-IR, MS, NMR; Electrophoresis; Micro-arrays; Enzymatic assays; Immunoassays - ELISA, RIA, immunohistochemistry; immunoblotting; Flow cytometry; Whole genome and ChIP sequencing. Computational tools: Bioinformatics resources and search tools; Sequence and structure databases; Sequence analysis - sequence file formats, scoring matrices, alignment, phylogeny; Genomics, proteomics, metabolomics; Gene prediction; Functional annotation; Secondary structure and 3D structure prediction; Knowledge discovery in biochemical databases; Metagenomics; Metabolic engineering and systems biology.

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Verbal Ability

English Grammar

1.1

1

English Grammar

C HAPTER MCQ TYPE QUESTIONS 2016 1. Computers were invented for performing only high-end useful computations. However, it is no understatement that they have taken over our world today. The internet, for example, is ubiquitous. Many believe that the internet itself is an unintended consequence of the original invention. With the advent of mobile computing on our phones, a whole new dimension is now enabled. One is left wondering if all these developments are good or, more importantly, required. Which of the statement(s) below is/are logically valid and can be inferred from the above paragraph? (i) The author believes that computers are not good for us. (ii) Mobile computers and the internet are both intended inventions (a) (i) only

(b) (ii) only

(c) both (i) and (ii)

(d) neither (i) nor (ii)

2. All hill-stations have a lake. Ooty has two lakes. Which of the statement(s) below is/are logically valid and can be inferred from the above sentences?

5. The policeman asked the victim of a theft, “What did you ?” (a) loose

(b) lose

(c) loss

(d) louse

6. In a world filled with uncertainty, he was glad to have many good friends. He had always assisted them in times of need and was confident that they would reciprocate. However, the events of the last week proved him wrong. Which of the following inference(s) is/are logically valid and can be inferred from the above passage? (i) His friends were always asking him to help them. (ii) He felt that when in need of help, his friends would let him down. (iii)He was sure that his friends would help him when in need. (iv) His friends did not help him last week. (a) (i) and (ii)

(b) (iii) and (iv)

(c) (iii) only

(d) (iv) only

7. An apple costs Rs. 10. An onion costs Rs. 8. Select the most suitable sentence with respect to grammar and usage. (a) The price of an apple is greater than an onion. (b) The price of an apple is more than onion.

(i) Ooty is not a hill-station.

(c) The price of an apple is greater than that of an onion.

(ii) No hill-station can have more than one lake.

(d) Apples are more costlier than onions.

(a) (i) only

(b) (ii) only

(c) both (i) and (ii)

(d) neither (i) nor (ii)

3. Identify the correct spelling out of the given options : (a) Managable

(b) Manageable

(c) Mangaeble

(d) Managible

4. Choose the statement(s) where the underlined word is used correctly : (i) A prone is a dried plum. (ii) He was lying prone on the floor. (iii) People who eat a lot of fat are prone to heart disease.

8. He turned a deaf ear to my request. What does the underlined phrasal verb mean? (a) ignored

(b) appreciated

(c) twisted

(d) returned

2015 9. Choose the statement where underlined word is used correctly (a) The minister insured the victims that everything would be all right. (b) He ensured that the company will not have to bear any loss.

(a) (i) and (iii) only

(b) (iii) only

(c) The actor got himself ensured against any accident.

(c) (i) and (ii) only

(d) (ii) and (iii) only

(d) The teacher insured students of good results

1.2

English Grammar

10. The following question presents a sentence, part of which is underlined. Beneath the sentence you find four ways of phrasing the underline part. Following the requirements of the standard written English, select the answer that produces the most effective sentence. Tuberculosis, together with its effects, ranks one of the leading causes of death in India. (a) ranks as one of the leading causes of death (b) rank as one of the leading causes of death (c) has the rank of one of the leading causes of death (d) are one of the leading causes of death 11. Choose the statement where underlined word is used correctly.

16. In the following sentence certain parts are underlined and marked P, Q and R. One of the parts may contain certain error or may not be acceptable in standard written communication. Select the part containing an error. Choose D as your Answer: if there is no error. The student corrected all the errors that P the instructor marked on the answer book R Q (a) P (b) Q (c) R

(d) No Error

17. Choose the statement where underlined word is used correctly. (a) The industrialist had a personnel jet.

(a) When the teacher eludes to different authors, he is being elusive

(b) I write my experience in my personnel diary.

(b) When the thief keeps eluding the police, he is being elusive

(d) Being religious is a personnel aspect.

(c) Matters that are difficult to understand, identify or remember are allusive (d) Mirages can be allusive, but a better way to express them is illusory 12. Fill in the blank with the correct idiom/phrase. That boy from the town was a __________ in the sleepy village. (a) Dog out of herd

(b) Sheep from the heap

(c) Fish out of water

(d) Bird from the flock

13. Select the appropriate option in place of underlined part of the sentence. Increased productivity necessary reflects greater efforts made by the employees. (a) Increase in productivity necessary (b) Increase productivity is necessary (c) Increase in productivity necessarily (d) No improvement required 14. Ram and Shyam shared a secret and promised to each other that it would remain between them. Ram expressed himself in one of the following ways as given in the choices below. Identify the correct way as per standard English. (a) It would remain between you and me (b) It would remain between me and you

(c) All personnel are being given the day off. 18. Out of the following four sentences, select the most suitable sentence with respect to grammer and usage: (a) Since the report lacked needed information, it was of no use to them. (b) The report was useless to them because there were no needed information in it. (c) Since the report did not contain the needed information, it was not real useful to them. (d) Since the report lacked needed information, it would not had been useful to them. 19. Select the alternative meaning of the underlined part of the sentence. The chain snatchers took to their heels when the police party arrived. (a) took shelter in a thick jungle (b) open indiscriminate fire (c) took to flight (d) unconditionally surrendered

2014 20. Which of the following options is the closest in meaning to the word underlined in the sentence below?

(c) It would remain between you and I

In a democracy, everybody has the freedom to disagree with the government.

(d) It would remain with me

(a) dissent

(b) descent

(c) decent

(d) decadent

15. What is the adverb for the given word below? Misogynous

21. ‘Advice' is .

(a) Misogynousness

(b) Misogynity

(a) a verb

(b) a noun

(c) Misogynously

(d) Misogynous

(c) an adjective

(d) both a verb and a noun

English Grammar

1.3

22. Which of the options given below best completes the following sentence?

28. Choose the grammatically CORRECT sentence: (a) Two and two add four.

She will feel much better if she .

(b) Two and two become four.

(a) will get some rest

(c) Two and two are four.

(b) gets some rest

(d) Two and two make four.

(c) will be getting some rest 29.

(d) is getting some rest 23. While trying to collect an envelope I

should learn mechanics, II

from under the table, Mr. X fell down and II III

mathematics and III

was losing consciousness. IV Which one of the above underlined parts of the sentence is NOT appropriate?

how to do computation. IV Which of the above underlined parts of the sentence is not appropriate?

(a) I

(b) II

(a) I

(b) II

(c) III

(d) IV

(c) III

(d) IV

24. Which of the following options is the closest in meaning to the phrase underlined in the sentence below? It is fascinating to see life forms cope with varied environmental conditions.

2012 30. Choose the most appropriate alternative from the options given below to complete the following sentence :

(a) Adopt to

(b) Adapt to

Suresh's dog is the one  was hurt in the stampede.

(c) Adept in

(d) Accept with

(a) that

(b) which

(c) who

(d) whom

25. In a press meet on the recent scam, the minister said, “The buck stops here”. What did the minister convey by the statement? (a) He wants all the money (b) He will return the money (c) He will assume final responsibility (d) He will resist all enquiries

2013 26.

All engineering students I

The professor ordered to the students to go I II III out of the class. IV Which of the above underlined parts of the sentence is grammatically incorrect? (a) I

(b) II

(c) III

(d) IV

27. Choose the grammatically INCORRECT sentence:

31. Choose the grammatically INCORRECT sentence : (a) They gave us the money back less the service charges of Three Hundred rupees. (b) This country's expenditure is not less than that of Bangladesh. (c) The committee initially asked for a funding of Fifty Lakh rupeesr, but later settled for a lesser sum. (d) This country's expenditure on educational reforms is very less. 32. One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT ? I requested that he should be given the driving test today instead of tomorrow. (a) requested that

(a) He is of Asian origin.

(b) should be given

(b) They belonged to Africa.

(c) the driving test

(c) She is an European.

(d) instead of tomorrow

(d) They migrated from India to Australia.

1.4

English Grammar

ANSWERS MCQ Type Questions 1. (d)

2. (d)

3. (b)

4. (d)

5. (b)

6. (b)

7. (c)

8. (a)

9. (b)

10. (a)

11. (b)

12. (c)

13. (c)

14. (a)

15. (c)

16. (b)

17. (c)

18. (a)

19. (c)

20. (a)

21. (b)

22. (b)

23. (d)

24. (b)

25. (c)

26. (b)

27. (c)

28. (d)

29. (d)

30. (a)

31. (d)

32. (b)

EXPLANATIONS MCQ TYPE QUESTIONS 4. ‘lying prone’ means lie down flat. ‘Prone to’ means vulnerable to. 5. Lose (verb) 9. Insured–the person, group, or organization whose life or property is covered by an insurance policy. Ensured–to secure or guarantee

11. Elusive: Difficult to answer. 12. From the statement, it appears that boy found it tough to adapt to a very different situation. 16. The is not required in ‘Q’. 30. The correct answer is ‘that’. In the given sentence ‘was hurt in the stampede’ determines which dog belongs to Suresh.

„„

Sentence Completion

2.1

2

Sentence Completion

C HAPTER MCQ TYPE QUESTIONS 2016

1. The man who is now Municipal Commissioner worked as ____________________. (a) the security guard at a university (b) a security guard at the university (c) a security guard at university (d) the security guard at the university 2. Nobody knows how the Indian cricket team is going to cope with the difficult and seamer-friendly wickets in Australia. Choose the option which is closest in meaning to the underlined phrase in the above sentence. (a) put up with

(b) put in with

(c) put down to

(d) put up against

3. The chairman requested the aggrieved shareholders to _________ him. (a) bare with

(b) bore with

(c) bear with

(d) bare

4. The unruly crowd demanded that the accused be _____________ without trial. (a) hanged

(b) hanging

(c) hankering

(d) hung

5. Which of the following is CORRECT with respect to grammar and usage? Mount Everest is ____________. (a) the highest peak in the world (b) highest peak in the world (c) one of highest peak in the world (d) one of the highest peak in the world 6. Despite the new medicine’s ______________ in treating diabetes, it is not ______________widely. (a) effectiveness --- prescribed (b) availability --- used (c) prescription --- available (d) acceptance --- proscribed 7. The students _____ the teacher on teachers’ day for twenty years of dedicated teaching. (a) facilitated

(b) felicitated

(c) fantasized

(d) facillitated

8. After India’s cricket world cup victory in 1985, Shrotria who was playing both tennis and cricket till then, decided to concentrate only on cricket. And the rest is history. What does the underlined phrase mean in this context? (a) history will rest in peace (b) rest is recorded in history books (c) rest is well known (d) rest is archaic 9. If I were you, I __________ that laptop. It’s much too expensive. (a) won’t buy (b) shan’t buy (c) wouldn’t buy (d) would buy 10. Choose the most appropriate set of words from the options given below to complete the following sentence. _____ _________ is a will, _________ is a way. (a) Wear, there, their (b) Were, their, there (c) Where, there, there (d) Where, their, their

2015 11. Choose the most appropriate word from the options given below to complete the following sentence. The official answered _________ that the complaints of the citizen would be looked into. (a) respectably (b) respectfully (c) reputably (d) respectively 12. Choose the most appropriate word from the options given below to complete the following sentence. The principal presented the chief guest with a , as token of appreciation. (a) momento (b) memento (c) momentum (d) moment 13. Choose the appropriate word/phrase, out of the four options given below, to complete the following sentence : Frogs . (a) Croak (b) Roar (c) Hiss (d) Patter 14. Extreme focus on syllabus and studying for tests has become such a dominant concern of Indian students that they close their minds to anything  to the requirements of the exam (a) related (b) extraneous (c) outside (d) useful

2.2

Sentence Completion

15. The Tamil version of  John Abraham-starrer Madras café  cleared by the Censor Board with no cuts last week but the film’s distributors  no takers among the exhibitors for a release in Tamil Nadu this Friday. (a) Mr., was, found, on (b) a, was, found, at (c) the, was, found, on (d) a, being, find at 16. Choose the appropriate word/phase, out of the four options given below, to complete the following sentence: Apparent lifelessness ______ dormant life. (a) harbours (b) lead to (c) supports (d) affects 17. Choose the correct verb to fill in the blank below : Let us _____________. (a) Introvert (b) Alternate (c) Atheist (d) Altruist 18. Choose the most appropriate word from the options given below to complete the following sentence : If the Athlete had wanted to come first in the race, he ___________ several hours everyday (a) should practice (b) should have practiced (c) practiced (d) should be practicing 19. Choose the appropriate word/phrase, out of the four options given below, to complete the following sentence: Dhoni, as well as the other team members of Indian team, _____ present on the occasion. (a) were (b) was (c) has (d) have 20. We ________ our friend’s birthday and we _______ how to make it up to him. (a) completely forgot — don’t just known (b) forget completely — don’t just know (c) completely forget —just don’t know (d) forgot completely —just don’t know 21. Which of the following options is the closest in meaning to the sentence below? She enjoyed herself immensely at the party. (a) She had a terrible time at the party (b) She had a horrible time at the party (c) She had a terrific time at the party (d) She had a terrifying time at the party 22. Didn’t you buy _______ when you went shopping? (a) any paper (b) much paper (c) no paper (d) a few paper

2014 23. Choose the most appropriate word from the options given below to complete the following sentence: A person suffering from Alzheimer’s disease _________ short-term memory loss. (a) experienced (b) has experienced (c) is experiencing (d) experiences 24. Choose the most appropriate word from the options given below to complete the following sentence: _________ is the key to their happiness; they are satisfied with what they have. (a) Contentment (b) Ambition (c) Perseverance (d) Hunger 25. Choose the most appropriate word from the options given below to complete the following sentence: One of his biggest _____ was his ability to forgive. (a) vice (b) virtues (c) choices (d) strength 26. After the discussion, Tom said to me, ‘Please revert!’ He expects me to . (a) retract (b) get back to him (c) move in reverse (d) retreat 27. The value of one U.S. Dollar is 65 Indian Rupees today, compared to 60 last year. The Indian Rupee has . (a) Depressed (b) Depreciated (c) Appreciated (d) Stabilized 28. Choose the most appropriate word from the options given below to complete the following sentence: Communication and interpersonal skills are  important in their own ways. (a) each (b) both (c) all (d) either 29. Choose the most appropriate pair of words from the options given below to complete the following sentence: She could not  the thought of  the election to her bitter rival. (a) bear, loosing (b) bare, loosing (c) bear, losing (d) bare, losing 30. Choose the most appropriate phrase from the options given below to complete the following sentence: The aircraft _________ take off as soon as its flight plan was filed. (a) is allowed to (b) will be allowed to (c) was allowed to (d) has been allowed to

Sentence Completion

2.3

31. Choose the most appropriate word from the options given below to complete the following sentence: Many ancient cultures attributed disease to supernatural causes. However, modern science has largely helped ______ such notions. (a) impel (b) dispel (c) propel (d) repel

40. Choose the most appropriate alternative from the options given below to complete the following sentence : If the tired soldier wanted to lie down, he  the mattress out on the balcony. (a) should take (b) shall take (c) should have taken (d) will have taken

32. If she __________ how to calibrate the instrument, she ________ done the experiment. (a) knows, will have (b) knew, had (c) had known, could have (d) should have known, would have 33. Who _______ was coming to see us this evening ?

41. Choose the most appropriate word from the options given below to complete the following sentence : Given the seriousness of the situation that he had to face, his  was impressive. (a) beggary (b) nomenclature (c) jealousy (d) nonchalance

(a) you said (b) did you say (c) did you say that (d) had you said 34. Choose the most appropriate word from the options given below to complete the following sentence: He could not understand the judges awarding her the first prize, because he thought that her performance was quite ___________________. (a) Superb (b) Medium (c) Mediocre (d) Exhilarating

2013 35. Friendship, no matter how ________ it is, has its limitations. (a) cordial (b) intimate (c) secret (d) pleasant 36. Were you a bird, you_____ in the sky. (a) would fly (b) shall fly (c) should fly (d) shall have flown 37. Complete the sentence: Dare __________ mistakes. (a) commit (b) to commit (c) committed (d) committing 38. The Headmaster  to speak to you. Which of the following options is incorrect to complete the above sentence? (a) is wanting (b) wants (c) want (d) was wanting

2012 39. Choose the most appropriate alternative from the options given below to complete the following sentence : Despite several  the mission succeeded in its attempt to resolve the conflict. (a) attempts (b) setbacks (c) meetings (d) delegations

2011 42. Choose the most appropriate word from the options given below to complete the following sentence: If you are trying to make a strong impression on your audience, you cannot do so by being understated, tentative or _______ . (a) hyperbolic (b) restrained (c) argumentative (d) indifferent 43. Choose the most appropriate word(s) from the options given below to complete the following sentence. I contemplated ______ Singapore for my vacation but decided against it. (a) to visit (b) having to visit (c) visiting (d) for a visit 44. Choose the most appropriate word from the options given below to complete the following sentence: Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which_______treatments are unsatisfactory. (a) similar (b) most (c) uncommon (d) available 45. Choose the most appropriate word from the options given below to complete the following sentence: It was her view that the country's problems had been_________by foreign technocrats, so that to invite them to come back would be counterproductive. (a) identified (b) ascertained (c) exacerbated (d) analysed

2.4

Sentence Completion

2010 46. Choose the most appropriate word from the options given below to complete the following sentence: If we manage to _________ our natural resources, we would leave a better planet for our children. (a) uphold (b) restrain (c) cherish (d) conserve

47. Choose the most appropriate word from the options given below to complete the following sentence: His rather casual remarks on politics ________his lack of seriousness about the subject. (a) masked (b) belied (c) betrayed (d) suppressed

ANSWERS MCQ Type Questions 1. (b)

2. (a)

3. (c)

4. (a)

5. (a)

6. (a)

7. (b)

8. (c)

9. (c)

10. (c)

11. (b)

12. (b)

13. (a)

14. (b)

15. (c)

16. (a)

17. (b)

18. (b)

19. (b)

20. (c)

21. (c)

22. (a)

23. (d)

24. (a)

25. (b)

26. (b)

27. (b)

28. (b)

29. (c)

30. (c)

31. (b)

32. (c)

33. (b)

34. (c)

35. (b)

36. (a)

37. (b)

38. (c)

39. (b)

40. (a)

41. (d)

42. (b)

43. (c)

44. (d)

45. (c)

46. (a)

47. (c)

EXPLANATIONS MCQ TYPE QUESTIONS 2. From the given underlined phrase copewith means put up with. 4. ‘hanged’ means death by hanging ‘hung’ is used only with things and not with people. 6. Here ‘effectiveness’ is noun and ‘prescribed’ is verb. So these words are apt and befitting with the given word ‘medicine’. 12. The principal presented the chief guest with a memento, as token of appreciation. 13. Frogs make ‘croak’ sound. 14. Extraneous -irrelevant or unrelated to the subject being dealt with. 15. John-Abraham starrer Madras Cafe talks about the movie not the person, so Mr. is ruled out. ‘Find no takers’ is not the correct phrase. At this Friday is incorrect. So, option (c) is correct. 16. Apparent: looks like; dormant: hidden Harbour: give shelter; Effect (verb): results in 18. For condition regarding something which already happened, should have practiced is the correct choice. 23. The correct word to fill in the blank is experiences. This is because the sentence is in present tense and the sense is in present indefinite.

24. Contentment is the right word which is appropriate to the meaning of the sentence. Contentment is the sign of true happiness as it gives a sense of fulfillment. 39. The word despite tells us that mission succeeded even though there were problems. Hence the correct answer is Setbacks. 42. Tone of the sentence clearly indicates a word that is similar to understated is needed for the blank. Alternatively, word should be antonym of strong (fail to make strong impression). Therefore, best choice is restrained which means controlled / reserved / timid. 43. Contemplate is a transitive verb and hence is followed by a gerund Hence the correct usage of contemplate is verb + ing form. 44. The context seeks to take a deviation only when existing / present / current / alternative treatments are unsatisfactory. So word for the blank should be a close synonym of existing / present / current / alternative. Available is the closest of all. 45. Clues: foreign technocrats did something negatively to the problems – so it is counterproductive to invite them. All other options are non-negative. Best choice is exacerbated which means aggravated or worsened. 47. Betrayed, means ‘showed’ or revealed.

„„

Synonyms

3.1

3

Synonyms

C HAPTER MCQ TYPE QUESTIONS 2016 1. The Buddha said, “Holding on to anger is like grasping a hot coal with the intent of throwing it at someone else; you are the one who gets burnt.” Select the word below which is closest in meaning to the word underlined above. (a) burning

(b) igniting

(c) clutching

(d) flinging

2015 2. The word similar in meaning to ‘dreary’ is (a) cheerful

(b) dreamy

(c) hard

(d) dismal

3. Choose the word most similar in meaning to the given word : Educe (a) Exert

(b) Educate

(c) Extract

(d) Extend

4. Choose the most suitable one word substitute for the following expression Connotation of a road or way (a) Pertinacious

(b) Viaticum

(c) Clandestine

(d) Ravenous

5. Choose the word most similar in meaning to the given word:

2014 8. A student is required to demonstrate a high level of comprehension of the subject, especially in the social sciences. The word closest in meaning to comprehension is (a) understanding (b) meaning (c) concentration (d) stability 9. While receiving the award, the scientist said, "I feel vindicated". Which of the following is closest in meaning to the word ‘vindicated’? (a) punished (b) substantiated (c) appreciated (d) chastened 10. Choose the word that is opposite in meaning to the word “coherent”. (a) sticky (b) well-connected (c) rambling (d) friendly 11. Match the columns: Column-1 Column-2 1. eradicate P. misrepresent 2. distort Q. soak completely 3. saturate R. use 4. utilize S. destroy utterly (a) 1 : S, 2 : P, 3 : Q, 4 : R (b) 1 : P, 2 : Q, 3 : R, 4 : S (c) 1 : Q, 2 : R, 3 : S, 4 : P (d) 1 : S, 2 : P, 3 : R, 4 : Q

2013

Awkward (a) Inept

(b) Graceful

(c) Suitable

(d) Dreadful

6. A generic term that includes various items of clothing such as a skirt, a pair of trousers and a shirt as (a) fabric (b) textile (c) fibre (d) apparel 7. Which of the following combinations is incorrect? (a) Acquiescence – Submission (b) Wheedle – Roundabout (c) Flippancy – Lightness (d) Profligate – Extravagant

12. Which of the following options is the closest in meaning to the word given below: Primeval (a) Modern (b) Historic (c) Primitive (d) Antique 13. Which one of the following options is the closest in meaning to the word given below? Nadir (a) Highest (b) Lowest (c) Medium (d) Integration 14. They were requested not to quarrel with others. Which one of the following options is the closest in meaning to the word quarrel? (a) make out

(b) call out

(c) dig out

(d) fall out

3.2

Synonyms

2012

2011

15. Which one of the following options is the closest in meaning to the word given below?

17. Which of the following options is the closest in the meaning to the word below :

Mitigate

Inexplicable

(a) Diminish

(b) Divulge

(a) Incomprehensible (b) Indelible

(c) Dedicate

(d) Denote

(c) Inextricable

16. Which one of the following options is the closest in meaning to the word given below ? Latitude

(d) Infallible

2010 18. Which of the following options is the closest in meaning to the word below:

(a) Eligibility

(b) Freedom

Circuitous

(c) Coercion

(d) Meticulousness

(a) cyclic

(b) indirect

(c) confusing

(d) crooked

ANSWERS MCQ Type Questions 1. (c)

2. (d)

3. (c)

4. (b)

5. (a)

6. (d)

7. (b)

8. (a)

11. (a)

12. (c)

13. (b)

14. (b)

15. (a)

16. (b)

17. (a)

18. (b)

9. (b)

10. (c)

EXPLANATIONS MCQ TYPE QUESTIONS 1. The meaning of underlined word grasping means clutching (or holding something tightly). 2. dreary - depressingly dull and bleak or repetitive. 3. The word similar in meaning to Educe is Extract. 13. Given: A word “Nadir” To find: Closest meaning word to ‘Nadir’ in the given options. Analysis: The dictionary meaning of ‘Nadir’ is ‘Lowest’ which is potion (b) Source: Google Given: A sequence 44, 42, 40,...... To find: Maximum sum of the sequence Analysis: To find the maximum sum, we have to ignore all the negative terms (they will only decrease the sum). Hence, we terminate the sequence at 0. Now, the sequence is 44, 42,......,2, 0.

This sequence is an Arithmetic progression with common difference –2. Now, the problem reduces to finding the sum of this A.P. To find the sum of A.P., we have to find the number of terms in A.P. Let n be the number of terms a be the first term l be the last term d be the common difference. n=? a = 44, l = 0, d = – 2 Now, l = a + (n – 1)d 15. ‘Mitigate’ means to make something less harmful, serious, intense etc. Hence correct answer is ‘Diminis’ 17. Inexplicable means not explicable; that cannot be explained, understood, or accounted for. So the best synonym here is incomprehensible. 18. Circuitous means round about or not direct So circuitous : indirect.

„„

Antonyms

4.1

4

Antonyms

C HAPTER MCQ TYPE QUESTIONS 2015 1. Which word is not a synonym for the word vernacular? (a) regional (b) indigeneous (c) indigent (d) colloquial

2011 2. Choose the word from the options given below that is most nearly opposite in meaning to the given word: Amalgamate (a) merge (b) split (c) collect (d) separate 3. Choose the word from the options given below that is most nearly opposite in meaning to the given word : Frequency (a) periodicity (b) rarity (c) gradualness (d) persistency

ANSWERS MCQ Type Questions 1. (c)

2. (b)

3. (b)

4.2

Antonyms

EXPLANATIONS MCQ TYPE QUESTIONS 1. Vernacular - expressed or written in the native language of a place Indigent - deficient in what is requisite 2. Amalgamate means combine or unite to form one organization or structure. So the best option here

is split. Separate on the other hand, although a close synonym, it is too general to be the best antonym in the given question while Merge is the synonym; Collect is not related. 3. Best antonym is rarity which means shortage or scarcity.

„„

Reasoning Ability

5.1

5

Reasoning Ability

C HAPTER MCQ TYPE QUESTIONS 2016 1. Find the odd one in the following group of words. mock, deride, praise, jeer (a) mock

(b) deride

(c) praise

(d) jeer

2. Pick the odd one from the following options.

6. Sourya committee had proposed the establishment of Sourya Institutes of Technology (SITs) in line with Indian Institutes of Technology (IITs) to cater to the technological and industrial needs of a developing country. Which of the following can be logically inferred from the above sentence? Based on the proposal,

(a) CADBE

(b) JHKIL

(i) In the initial years, SIT students will get degrees from IIT.

(c) XVYWZ

(d) ONPMQ

(ii) SITs will have a distinct national objective.

3. Pick the odd one out in the following : 13, 23, 33, 43, 53 (a) 23

(b) 33

(c) 43

(d) 53

4. R2D2 is a robot. R2D2 can repair aeroplanes. No other robot can repair aeroplanes. Which of the following can be logically inferred from the above statements? (a) R2D2 is a robot which can only repair aeroplanes. (b) R2D2 is the only robot which can repair aeroplanes. (c) R2D2 is a robot which can repair only aeroplanes. (d) Only R2D2 is a robot. 5. A poll of students appearing for masters in engineering indicated that 60% of the students believed that mechanical engineering is a profession unsuitable for women. A research study on women with masters or higher degrees in mechanical engineering found that 99% of such women were successful in their professions. Which of the following can be logically inferred from the above paragraph? (a) Many students have misconceptions regarding various engineering disciplines. (b) Men with advanced degrees in mechanical engineering believe women are well suited to be mechanical engineers. (c) Mechanical engineering is a profession well suited for women with masters or higher degrees in mechanical engineering. (d) The number of women pursuing higher degrees in mechanical engineering is small.

(iii) SIT like institutions can only be established in consulation with IIT. (iv) SITs will serve technological needs of a developing country. (a) (iii) and (iv) only

(b) (i) and (iv) only

(c) (ii) and (iv) only

(d) (ii) and (iii) only

7. Fact : If it rains, then the field is wet. Read the following statements: (i) It rains

(ii) The field is not wet

(iii) The field is wet

(iv) It did not rain

Which one of the options given below is NOT logically possible, based on the given fact? (a) If (iii), then (iv)

(b) If (i), then (iii)

(c) If (i), then (ii)

(d) If (ii), then (iv)

8. Students taking an exam are divided into two groups, P and Q such that each group has the same number of students. The performance of each of the students in a test was evaluated out of 200 marks. It was observed that the mean of group P was 105, while that of group Q was 85. The standard deviation of group P was 25, while that of group Q was 5. Assuming that the marks were distributed on a normal distribution, which of the following statements will have the highest probability of being TRUE? (a) No student in group Q scored less marks than any student in group P. (b) No student in group P scored less marks than any student in group Q. (c) Most students of group Q scored marks in a narrower range than students in group P. (d) The median of the marks of group P is 100.

5.2

Reasoning Ability

9. A smart city integrates all modes of transport, uses clean energy and promotes sustainable use of resources. It also uses technology to ensure safety and security of the city, something which critics argue, will lead to a surveillance state. Which of the following can be logically inferred from the above paragraph? (i) All smart cities encourage the formation of surveillance states. (ii) Surveillance is an integral part of a smart city. (iii) Sustainability and surveillance go hand in hand in a smart city. (iv) There is a perception that smart cities promote surveillance. (a) (i) and (iv) only

(b) (ii) and (iii) only

(c) (iv) only

(d) (i) only

10. Find the missing sequence in the letter series.

(c) Mr. X is taller than Mr. Y (d) Mr. X is lengthier than Mr. Y 14. Social science disciplines were in existence in an amorphous form until the colonial period when they were institutionalized. In varying degrees, they were intended to further the colonial interest. In the time of globalization and the economic rise of postcolonial countries like India, conventional ways of knowledge production have become obsolete. Which of the following can be logically inferred from the above statements? I. Social science disciplines have become obsolete. II. Social science disciplines had a pre-colonial origin. III. Social science disciplines always promote colonialism.

(a) SUWY

(b) TUVW

IV. Social science must maintain disciplinary boundaries.

(c) TVXZ

(d) TWXZ

(a) II only

(b) I and III only

(c) II and IV only

(d) III and IV only

B, FH, LNP, _ _ _ _.

11. Michael lives 10 km away from where I live. Ahmed lives 5 km away and Susan lives 7 km away from where I live. Arun is farther away than Ahmed but closer than Susan from where I live. From the information provided here, what is one possible distance (in km) at which I live from Arun’s place? (a) 3.00

(b) 4.99

(c) 6.02

(d) 7.01

12. Leela is older than her cousin Pavithra. Pavithra’s brother Shiva is older than Leela. When Pavithra and Shiva are visiting Leela, all three like to play chess. Pavithra wins more often than Leela does. Which one of the following statements must be TRUE based on the above? (a) When Shiva plays chess with Leela and Pavithra, he often loses.

15. Two and a quarter hours back, when seen in a mirror, the reflection of a wall clock without number markings seemed to show 1 : 30. What is the actual current time shown by the clock? (a) 8:15

(b) 11:15

(c) 12:15

(d) 12:45

16. M and N start from the same location. M travels 10 km East and then 10 km North-East. N travels 5 km South and then 4 km South-East. What is the shortest distance (in km) between M and N at the end of their travel? (a) 18.60

(b) 22.50

(c) 20.61

(d) 25.00

17. M has a son Q and a daughter R. He has no other children. E is the mother of P and daughter-inlaw of M. How is P related to M?

(b) Leela is the oldest of the three.

(a) P is the son-in-law of M.

(c) Shiva is a better chess player than Pavithra.

(b) P is the grandchild of M.

(d) Pavithra is the youngest of the three. 13. Based on the given statements, select the appropriate option with respect to grammar and usage. Statements : I. The height of Mr. X is 6 feet. II. The height of Mr. Y is 5 feet. (a) Mr. X is longer than Mr. Y. (b) Mr. X is more elongated than Mr. Y

(c) P is the daughter-in-law of M. (d) P is the grandfather of M. 18. The number that least fits this set: (324, 441, 97 and 64) is ________. (a) 324 (b) 441 (c) 97 (d) 64

Reasoning Ability

5.3

19. The overwhelming number of people infected with rabies in India has been flagged by the World Health Organization as a source of concern. It is estimated that inoculating 70% of pets and stray dogs against rabies can lead to a significant reduction in the number of people infected with rabies. Which of the following can be logically inferred from the above sentences ? (a) The number of people in India infected with rabies is high (b) The number of people in other parts of the world who are infected with rabies is low (c) Rabies can be eradicated in India by vaccinating 70% of stray dogs (d) Stray dogs are the main source of rabies worldwide 20. A flat is shared by four first year undergraduate students. They agreed to allow the oldest of them to enjoy some extra space in the flat. Manu is two months older than Sravan, who is three months younger than Trideep. Pavan is one month older than Sravan. Who should occupy the extra space in the flat ? (a) Manu

(b) Sravan

(c) Trideep

(d) Pavan

21. Today, we consider Ashoka as a great ruler because of the copious evidence he left behind in the form of stone carved edicts. Historians tend to correlate greatness of a king at his time with the availability of evidence today. Which of the following can be logically inferred from the above sentences? (a) Emperors who do not leave significant sculpted evidence are completely forgotten. (b) Ashoka produced stone carved edicts to ensure that later historians will respect him. (c) Statues of kings are a reminder of their greatness. (d) A king’s greatness, as we know him today, is interpreted by historians. 22. Fact 1 : Humans are mammals. Fact 2 : Some humans are engineers. Fact 3 : Engineers build houses. If the above statements are facts, which of the following can be logically inferred ? I. All mammals build houses. II. Engineers are mammals. III. Some humans are not engineers. (a) II only

(b) III only

(c) I, II and III

(d) I only

2015 23. Mr. Vivek walks 6 meters North-East, then turns and walks 6 meters South- East, both at 60 degrees to East. He further moves 2 meters South and 4 meters West. What is the straight distance in meters between the point he started from and the point he finally reached? (a) 2 2

(b) 2

(d) 1 / 2 (c) 2 24. There are 16 teachers who can teach Thermodynamics (TD), 11 who can teach Electrical Sciences (ES), and 5 who can teach both TD and Engineering Mechanics (EM). There are a total of 40 teachers, 6 cannot teach any of the three subjects, i.e. EM, ES or TD. 6 can teach only ES. 4 can teach all three subjects, i.e. EM, ES and TD. 4 can teach ES and TD. How many can teach both ES and EM but not TD? (a) 1 (b) 2 (c) 3 (d) 4 25. The given question is followed by two statements: select the most appropriate option that solves the question Capacity of a solution tank A is 70% of the capacity of tank B. How many gallons of solution are in tank A and tank B? Statements: I. Tank A is 80% full and tank B is 40% full II. Tank A if full contains 14,000 gallons of solution (a) Statement I alone is sufficient (b) Statement II alone is sufficient (c) Either statement I or II alone is sufficient (d) Both the statements I and II together are sufficient 26. Operators and  are defined by ab ab a b= ; a b= ; a  b = ab. ab ab Find the value (66 6) (66 6) (a) – 2 (b) – 1 (c) 1 (d) 2 27. Humpty Dumpty sits on a wall every day while having lunch. The wall sometimes breaks. A person sitting on the wall falls if the wall breaks. Which one of the statements below is logically valid and can be inferred from the above sentences? (a) Humpty Dumpty always falls while having lunch (b) Humpty Dumpty does not fall sometimes while having lunch (c) Humpty Dumpty never falls during dinner (d) When Humpty Dumpty does not sit on the wall, the wall does not break

5.4

Reasoning Ability

28. If ROAD is written as URDG, then SWAN should be written as : (a) VXDQ

(b) VZDQ

(c) VZDP

(d) UXDQ

29. Select the pair that best expresses a relationship similar to that expressed in the pair : Children : Pediatrician (a) Adult : Orthopaedist (b) Females : Gynaecologist (c) Kidney : Nephrologist (d) Skin : Dermatologist 30. The head of a newly formed government desires to appoint five of the six selected members P,Q,R,S,T and U to portfolios of Home, Power, Defense, Telecom, and Finance. U does not want any portfolio if S gets one of the five. R wants either Home or Finance or no portfolio. Q says that if S gets either Power or Telecom, then she must get the other one. T insists on a portfolio if P gets one. Which is the valid distribution of portfolios? (a) P-Home, Q-Power, R-Defense, S-Telecom, T-Finance (b) R-Home, S-Power, P-Defense, Q-Telecom, T-Finance (c) P-Home, Q-Power, T-Defense, S-Telecom, U-Finance (d) Q-Home, U-Power, T-Defense, R-Telecom, P-Finance 31. Most experts feel that in spite of possessing all the technical skills required to be a batsman of the highest order, he is unlikely to be so due to lack of requisite temperament. He was guilty of throwing away his wicket several times after working hard to lay a strong foundation. His critics pointed out that until he addressed this problem success at the highest level will continue to elude him. Which of the statement (s) below is/are logically valid and can be inferred from the above passage? (i) He was already a successful batsman at the highest level (ii) He has to improve his temperament in order to become a great batsman (iii) He failed to make many of his good starts count (iv) Improving his technical skills will guarantee success

32. Alexander turned his attention towards India, since he had conquered Persia. Which one of the statements below is logically valid and can inferred from the above sentence? (a) Alexander would not have turned his attention towards India had he not conquered Persia. (b) Alexander was not ready to rest on his laurels, and wanted to march to India (c) Alexander was completely in control of his army and could command it to move towards India (d) Since Alexander’s kingdom extended to Indian borders after the conquest of Persia, he was keen to move further 33. Tanya is older than Enc. Cliff is older than Tanya. Eric is older than Cliff. If the first two statements are true, then the third statement is (a) True

(b) False

(c) Uncertain

(d) Data insufficient

34. Given below are two statements followed by two conclusions. Assuming these statements to be true, decide which one logically follows: Statements: I. No manager is a leader. II. All leaders are executive. Conclusions: I. No manager is a executive. II. All executive is a manager. (a) Only conclusion I follows. (b) Only conclusion II follows. (c) Neither conclusion I nor II follows. (d) Both conclusion I and II follow. 35. Find the missing sequence in the letter series below : A, CD, GHI, _?___, UVWXY (a) LMN

(b) MNO

(c) MNOP

(d) NOPQ

36. Ms. X will be in Bagdogra from 1/5/14 to 20/5/14 and from 22/5/14 to 31/5/14. On the morning of 21/5/14, she will reach Kochi via Mumbai. Which one of the statements below is logically valid and can be inferred from the above sentences? (a) Ms. X will be in Kochi for only one day only in May (b) Ms. X will be in Kochi for only one day in May

(a) (iii) and (iv)

(b) (ii) and (iii)

(c) Ms. X will be only in Kochi for one day in May

(c) (i), (ii) and (iii)

(d) (ii) only

(d) Only Ms. X will be in Kochi for one day in May

Reasoning Ability

37. Given below are two statements followed by two conclusions. Assuming these statements to be true, decide which one logically follows. Statement: I. All film stars are playback singers. II. All film directors are film stars. Conclusions: I. All film directors are playback singers. II. Some film stars are film directors. (a) Only conclusion I follows (b) Only conclusion II follows (c) Neither conclusion I nor II follows (d) Both conclusions I and II follow 38. Lamenting the gradual sidelining of the arts in school curricula, a group of prominent artists wrote to the Chief Minister last year, asking him to allocate more funds to support arts education in schools. However, no such increase has been announced in this year’s Budget. The artists expressed their deep anguish at their request not being approved, but many of them remain optimistic about funding in the future. Which of the statement(s) below is/are logically valid and can be inferred from the above statements? i. The artists expected funding for the arts to increase this year. ii. The Chief Minister was receptive to the idea of increasing funding for the arts. iii. The Chief Minister is a prominent artist. iv. Schools are giving less importance to arts education nowadays. (a) iii and iv (b) i and iv (c) i, ii and iv (d) i and iii 39. Four branches of a company are located at M, N, O, and P. M is north of N at a distance of 4 km; P is south of O at a distance of 2 km; N is southeast of O by 1 km. What is the distance between M and P in km? (a) 5.34 (b) 6.74 (c) 28.5 (d) 45.49 40. The given statement is followed by some courses of action. Assuming the statement to be true, decide the correct option. Statement: There has been a significant drop in the water level in the lakes supplying water to the city. Course of action: I. The water supply authority should impose a partial cut in supply to tackle the situation. II. The government should appeal to all the residents through mass media for minimal use of water.

5.5

III. The government should ban the water supply in lower areas. (a) Statements I and II follow. (b) Statements I and III follow. (c) Statements II and III follow. (d) All statements follow.

2014 41. In a group of four children, Som is younger to Riaz. Shiv is elder to Ansu. Ansu is youngest in the group. Which of the following statements is/are required to find the eldest child in the group? Statements 1. Shiv is younger to Riaz. 2. Shiv is elder to Som. (a) Statement 1 by itself determines the eldest child. (b) Statement 2 by itself determines the eldest child. (c) Statements 1 and 2 are both required to determine the eldest child. (d) Statements 1 and 2 are not sufficient to determine the eldest child. 42. X is 1 km northeast of Y. Y is 1 km southeast of Z. W is 1 km west of Z. P is 1 km south of W. Q is 1 km east of P. What is the distance between X and Q in km? (a) 1 (c)

(b)

3

2

(d) 2

43. Rajan was not happy that Sajan decided to do the project on his own. On observing his unhappiness, Sajan explained to Rajan that he preferred to work independently. Which one of the statements below is logically valid and can be inferred from the above sentences? (a) Rajan has decided to work only in a group. (b) Rajan and Sajan were formed into a group against their wishes. (c) Sajan had decided to give in to Rajan's request to work with him. (d) Rajan had believed that Sajan and he would be working together. 44. Find the odd one in the following group: ALRVX, EPVZB, ITZDF, OYEIK (a) ALRVX (b) EPVZB (c) ITZDF (d) OYEIK

5.6

Reasoning Ability

51. The multi-level hierarchical pie chart shows the population of animals in a reserve forest. The correct conclusions from this information are:

M am m al

In se ct

Honey -bee

Tige r

Beetle

Anuj Bhola Chandan Dilip Eswar Faisal (b) 2

6

5

1

3

4

(c) 4

2

6

3

1

5

(d) 2

4

6

1

3

5

46. Find the next term in the sequence: 13M, 17Q, 19S,  (a) 21W

(b) 21V

(c) 23W

(d) 23V

47. If ‘KCLFTSB’ stands for ‘best of luck’ and ‘SHSWDG’ stands for ‘good wishes’, which of the following indicates ‘ace the exam’?

Bird wk Ha

(i) Butterflies are birds (ii) There are more tigers in this forest than red ants (iii)All reptiles in this forest are either snakes or crocodiles (iv) Elephants are the largest mammals in this forest

(a) MCHTX

(b) MXHTC

(a) (i) and (ii) only

(c) XMHCT

(d) XMHTC

(b) (i), (ii), (iii) and (iv)

48. “India is a country of rich heritage and cultural diversity.” Which one of the following facts best supports the claim made in the above sentence? (a) India is a union of 28 states and 7 union territories. (b) India has a population of over 1.1 billion. (c) India is home to 22 official languages and thousands of dialects. (d) The Indian cricket team draws players from over ten states. 49. In which of the following options will the expression P < M be definitely true? (a) M < R > P > S

(b) M > S < P < F

(c) Q < M < F = P

(d) P = A < R < M

50. Find the next term in the sequence: 7G, 11K, 13M,  (a) 15Q (b) 17Q (c) 15P (d) 17P

sn ak e ile

4

ul Bulb

3

Leo par d

od oc Cr

1

ngo Dro

5

Butterfly

2

h ot M

(a) 6

t an ph Ele

R -a ed nt

Reptile

45. Anuj, Bhola, Chandan, Dilip, Eswar and Faisal live on different floors in a six-storeyed building (the ground floor is numbered 1, the floor above it 2, and so on). Anuj lives on an even-numbered floor. Bhola does not live on an odd numbered floor. Chandan does not live on any of the floors below Faisal's floor. Dilip does not live on floor number 2. Eswar does not live on a floor immediately above or immediately below Bhola. Faisal lives three floors above Dilip. Which of the following floor-person combinations is correct?

(c) (i), (iii) and (iv) only (d) (i), (ii) and (iii) only 52. Find the odd one in the following group Q,W,Z,B

B,H,K,M

W,C,G,J M,S,V,X

(a) Q, W, Z, B

(b) B, H, K, M

(c) W, C, G, J

(d) M, S, V, X

53. Lights of four colors (red, blue, green, yellow) are hung on a ladder. On every step of the ladder there are two lights. If one of the lights is red, the other light on that step will always be blue. If one of the lights on a step is green, the other light on that step will always be yellow. Which of the following statements is not necessarily correct? (a) The number of red lights is equal to the number of blue lights (b) The number of green lights is equal to the number of yellow lights (c) The sum of the red and green lights is equal to the sum of the yellow and blue lights (d) The sum of the red and blue lights is equal to the sum of the green and yellow lights

Reasoning Ability

54. Read the statements: All women are entrepreneurs. Some women are doctors. Which of the following conclusions can be logically inferred from the above statements? (a) All women are doctors (b) All doctors are entrepreneurs (c) All entrepreneurs are women (d) Some entrepreneurs are doctors 55. Find the odd one from the following group: (a) W,E,K,O (b) I,Q,W,A (c) F,N,T,X, (d) N,V,B,D 56. For submitting tax returns, all resident males with annual income below ` 10 lakh should fill up Form P and all resident females with income below ` 8 lakh should fill up Form Q. All people with incomes above ` 10 lakh should fill up Form R, except non residents with income above ` 15 lakhs, who should fill up Form S. All others should fill Form T. An example of a person who should fill Form T is (a) a resident male with annual income ` 9 lakh (b) a resident female with annual income ` 9 lakh (c) a non-resident male with annual income ` 16 lakh (d) a non-resident female with annual income ` 16 lakh 57. Which number does not belong in the series below? 2, 5, 10, 17, 26, 37, 50, 64 (a) 17 (b) 37 (c) 64 (d) 26 58. A dance programme is scheduled for 10.00 a.m. Some students are participating in the programme and they need to come an hour earlier than the start of the event. These students should be accompanied by a parent. Other students and parents should come in time for the programme. The instruction you think that is appropriate for this is (a) Students should come at 9.00 a.m. and parents should come at 10.00 a.m. (b) Participating students should come at 9.00 a.m. accompanied by a parent, and other parents and students should come by 10.00 a.m. (c) Students who are not participating should come by 10.00 a.m. and they should not bring their parents. Participating students should come at 9.00 a.m. (d) Participating students should come before 9.00 a.m. Parents who accompany them should come at 9.00 a.m. All others should come at 10.00 a.m.

5.7

59. The number of people diagnosed with dengue fever (contracted from the bite of a mosquito) in north India is twice the number diagnosed last year. Municipal authorities have concluded that measures to control the mosquito population have failed in this region. Which one of the following statements, if true, does not contradict this conclusion? (a) A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent (b) More cases of dengue are now reported because of an increase in the Municipal Office’s administrative efficiency (c) Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test (d) The number of people with malarial fever (also contracted from mosquito bites) has increased this year 60. Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely? (a) Strategies are now available for eliminating psychiatric illnesses (b) Certain psychiatric illnesses have a genetic basis (c) All human diseases can be traced back to genes and how they are expressed (d) In the future, genetics will become the only relevant field for identifying psychiatric illnesses

2013 61. Select the pair that best expresses a relationship similar to that expressed in the pair: Medicine: Health (a) Science : Experiment (b) Wealth : Peace (c) Education : Knowledge (d) Money : Happiness 62. Abhishek is elder to Savar. Savar is younger to Anshul. Which of the given conclusions is logically valid and is inferred from the above statements? (a) Abhishek is elder to Anshul (b) Anshul is elder to Abhishek (c) Abhishek and Anshul are of the same age (d) No conclusion follows

5.8

Reasoning Ability

63. Complete the sentence: Universalism is to particularism as diffuseness is to _____

67. Select the pair that best expresses a relationship similar to that expressed in the pair : water : pipe : :

(a) specificity

(a) cart : road

(b) neutrality

(b) electricity : wire

(c) generality

(c) sea : beach

(d) adaptation

(d) music: instrument

64. After several defeats in wars, Robert Bruce went in exile and wanted to commit suicide. Just before committing suicide, he came across a spider attempting tirelessly to have its net. Time and again, the spider failed but that did not deter it to retrain from making attempts. Such attempts by the spider made Bruce curious. Thus, Bruce started observing the near-impossible goal of the spider to have the net. Ultimately, the spider succeeded in having its net despite several failures. Such act of the spider encouraged Bruce not to commit suicide. And then, Bruce went back again and won many a battle, and the rest is history. Which one of the following assertions is best supported by the above information? (a) Failure is the pillar of success. (b) Honesty is the best policy. (c) Life begins and ends with adventures. (d) No adversity justifies giving up hope. 65. Statement: You can always give me a ring whenever you need. Which one of the following is the best inference from the above statement? (a) Because I have a nice caller tune. (b) Because I have a better telephone facility. (c) Because a friend in need is a friend indeed. (d) Because you need not pay towards the telephone bills when you give me a ring. 66. Statement: There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists, and so on. Which one of the following is the best inference from the above statement? (a) The emergence of nationalism in colonial India led to our Independence. (b) Nationalism in India emerged in the context of colonialism. (c) Nationalism in India is homogeneous. (d) Nationalism in India is heterogeneous.

68. All professors are researchers Some scientists are professors Which of the given conclusions is logically valid and is inferred from the above arguments : (a) All scientists are researchers (b) All professors are scientists (c) Some researchers are scientists (d) No conclusion follows

2012 69. Wanted Temporary, Part-time persons for the post of Field Interviewer to conduct personal interviews to collect and collate economic data. Requirements: High School-pass, must be available for Day, Evening and Saturday work. Transportation paid, expenses reimbursed. Which one of the following is the best inference from the above advertisement? (a) Gender-discriminatory (b) Xenophobic (c) Not designed to make the post attractive (d) Not gender-discriminatory 70. Given the sequence of terms, AD CG FK JP, the next term is (a) OV

(b) OW

(c) PV

(d) PW

2011 71. A transporter receives the same number of orders each day. Currently, he has some pending orders (backlog) to be shipped. If he uses 7 trucks, then at the end of the 4th day he can clear all the orders. Alternatively, if he uses only 3 trucks, then all the orders are cleared at the end of the 10th day. What is the minimum number of trucks required so that there will be no pending order at the end of the 5th day? (a) 4 (b) 5 (c) 6 (d) 7

Reasoning Ability

72. Few school curricula include a unit on how to deal with bereavement and grief, and yet all students at some point in their lives suffer from losses through death and parting. Based on the above passage which topic would not be included in a unit on bereavement? (a) How to write a letter of condolence (b) What emotional stages are passed through in the healing process (c) What the leading causes of death are (d) How to give support to a grieving friend 73. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair : Gladiator : Arena (a) dancer : stage (b) commuter : train (c) teacher : classroom (d) lawyer : courtroom 74. The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood. Serums to fight with diphtheria and tetanus were developed this way. It can be inferred from the passage, that horses were (a) given immunity of diseases (b) generally quite immune to diseases (c) given medicines to fight toxins (d) given diphtheria and tetanus serums

5.9

III. There are no twins. In what order were they born (oldest first)? (a) HSIG (b) SGHI (c) IGSH (d) IHSG 77. Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause. Which of the following statements best sums up the meaning of the above passage: (a) Modern warfare has resulted in civil strife (b) Chemical agents are useful in modern warfare (c) Use of chemical agents in warfare would be undesirable (d) People in military establishments like to use chemical agents in war 78. If 137 + 276 = 435 how much is 731 + 672? (a) 534

(b) 1403

(c) 1623

(d) 1513

NUMERICAL TYPE QUESTIONS 2015 1. Fill in the missing value

2010 75. The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed : Worker (a) fallow : land (b) unaware : sleeper (c) wit : jester (d) renovated : house 76. Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters). All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts: I. Hari’s age + Gita age > Irfan’s age + Saira’s age II. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest.

7 1

9

6

5

4

4

7

2

1

2

8

1

2

4

1

5

2

3

?

3

1

3

2014 2. The next term in the series 81, 54, 36, 24,.... is  3. Fill in the missing number in the series. 2

3

6

15

 157.5

630

4. What is the next number in the series? 12 35 81 173 357 ____

5.10

Reasoning Ability

ANSWERS MCQ Type Questions 1. (c)

2. (d)

3. (b)

4. (b)

5. (a)

6. (c)

7. (c)

8. (c)

9. (b)

10. (c)

11. (c)

12. (d)

13. (c)

14. (a)

15. (d)

16. (c)

17. (b)

18. (c)

19. (a)

20. (c)

21. (d)

22. (b)

23. (a)

24. (a)

25. (d)

26. (c)

27. (b)

28. (b)

29. (b)

30. (b)

31. (b)

32. (a)

33. (b)

34. (c)

35. (c)

36. (b)

37. (d)

38. (b)

39. (a)

40. (a)

41. (a)

42. (c)

43. (d)

44. (d)

45. (b)

46. (c)

47. (b)

48. (c)

49. (d)

50. (b)

51. (d)

52. (c)

53. (d)

54. (d)

55. (d)

56. (b)

57. (c)

58. (b)

59. (d)

60. (b)

61. (c)

62. (d)

63. (a)

64. (d)

65. (c)

66. (d)

67. (b)

68. (c)

69. (d)

70. (a)

71. (c)

72. (c)

73. (d)

74. (b)

75. (d)

76. (b)

77. (c)

78. (c)

Numerical Type Questions 1. (3)

2. (16)

3. (45)

4. (725 to 725)

EXPLANATIONS MCQ TYPE QUESTIONS

Distribution of P:

1. Given words ‘mock, deride and jeer’ are synonyms which means mockery. So, the given word praise is odd one. 2. C A D B E , J H K I L +1

+1

+1

+1

+1

+1

Group ‘Q’

X V Y W Z, O N P M Q +1

+1

+1

+1

+1

–1

So, option (d) is odd one from the given group. 3. The given number 33 is odd one out, because the remaining numbers are prime number.

Mean (2) = 85 Standard deviation (2) = 5 Pr ( –   x   + )  0.6827  68% within one standard deviation

2 – 2 = 85 – 5 = 80 2 + 2 = 85 + 5 = 90

7. Statements i and ii are not logically possible based on the given fact.

 Range of Q in one standard deviation is 80 to 90

8. Group ‘P’

Distribution of ‘Q’

Mean () = 105 Standard deviation (1) = 25 Pr ( –   x   + )  0.6827  68% within one standard deviation



1 – 1 = 105 – 25 = 80

68% within one standard deviation of Q is narrower

1 + 1 = 105 + 25 = 130

 68% within one standard deviation of Q means most students of group Q.

range = 80 to 130

 Most students of group ‘Q’ scored marks in a narrower range than students in group ‘P’

Reasoning Ability

5.11

10. The following letter series is in the order of even letters series

12

11

15.

1 2

10

11. From given data, the following diagram can be possible. 7 km

S

6.02 m 5 km

I

A Ah

M

Mirror image of 1 : 20 is 10 : 30

10 km

10 : 30 was the time two and quarter hour back so time now will be 12 : 45

Here S  Susan 16.

A  Arun

M

Ah  Ahmed M  Michael

10

(Starting point)  10 km O

5 2 km

5 2 km

5 km

From the above diagram, Arun lives farthest away than Ahmed means more than 5 km but closer than Susan means less than 7 km, from the given alternatives.

km

5 km 2 2 km

So, only option ‘c’ is possible.

4k

12. From the given question two statements will be followed.

2 2 km

m

O

N

From the given figure

For statment I

(OM)2  (ON)2

MN =

Arrange the given data according to their ages Shiva  Age Leela  Pavithra

OM = 5 2  5  2 2  5  7 2 ON = 10  5 2  2 2  10  3 2

For statement II

(5  7 2)2  (10  3 2)2

MN =

Arrange the given data according to their winning Pavithra

 20.61 km

Leela

So, from statement I and II, it is clear that only option (d) is possible (i.e., statement I). 13.

Height

25  98  70 2  100  18  60 2)

=

(+ or –) M

17.

Daughter R

Height 6 Feet 5 Feet

Q

Son Dau ght er-

(+)

in-la w

(+) Male Husband Female Grand (–) Child (+ or –) P E Mother

So, from the above relation diagram it is clear that P is the grandchild of M. 20. Manu age = sravan age + 2 months Manu age = Trideep age – 3 months

X

(Statement I)

Y

(Statement II)

Hence from the given figure Mr. X is taller than Mr. Y by 1 foot. 14. Social science disciplines had a pre-colonial origin.

Pavan age = Sravan’s age + 1 month From the above statement Trideep age > Man > Pavan > Sravan Hence, Trideep can occupy the extra space in the flat.

5.12

Reasoning Ability

TD

24.

Therefore concluding diagram can be

ES

E

4

1

L

6

0

11

E

1

or

E

E

L

or

M

M

L

35.

11 EM 25. Statement-I can be used to solve the question if capacity of both tanks is already known Statement-II can be used if it is known what quantity of each tank is full/empty. Therefore, by using both statements

36. Second sentence says that Ms. X reaches Kochi on 21/05/2014. Also she has to be in Bagdogora on 22/05/2014.

 She stays in Kochi for only one day in may.

Let capacity of tank B is x

0 1

70 x = 14000 100



4

80  14000 = 11200 gallons 100

2

N

S

P

41. Riaz > Som

...(i)

Shiv > Ansu

...(ii)

Ansu is youngest in the group

40  20000 = 8000 gallons 100

 From equations (i) and (ii)

 Total solution = 11200 + 8000 = 19200 gallons

 Riaz > Som > Shiv > Ansu

Solution in tank B =

28.

N 1km

W

Z w

X E

s 1km

1 km

66  6 72 6   66 6 = 66  6 60 5 (66 6)  (66 6) =

M

42.

66  6 60 5   26. 66 6 = 66  6 72 6

E

W

45°

x = 20000 gallons

Solution in tank A =

N

M

39.

1km

y W

P

1 km

Q

C E

5 6  1 6 5

R + 3 = U, O + 3 = R,

s

Here

XE = 1 km

A + 3 = D, D + 3 = G; S + 3 = V, W + 3 = Z,

QE =

A + 3 = D, N + 3 = Q 29. Community of people : Doctor 30. Since U does not want any portfolio, (c) and (d) are ruled out. R wants Home, or Finance or No portfolio, (a) is not valid. Hence option (b) is correct. 34. S + 1: M

S+2: L

L

E

2  2 2

 using Pythagoras theorem, XQ = (QE) 2  (XE) 2 =

2  1 = 3 km

 Statement I itself determines the eldest child which statement 2 is incorrect as Ansu is youngest in the group is not satisfied by statement 2 Hence option (a) is correct

Reasoning Ability

5.13

44. ALRVX  only one vowel

71. Let each truck carry 100 units. 2800 = 4n + e where n = normal 3000 = 10n + e where e = excess/pending

EPVZB  only one vowel ITZDF  only one vowel OYEIK  three vowels 45. (a) Anuj: Even numbered floor (2, 4, 6) (b) Bhola: Even numbered floor (2, 4, 6)



(c) Chandan lives on the floor above that of Faisal. (d) Dilip: not on 2nd floor.

n=

100 , 3

5 days  500 x =

(e) Eswar: does not live immediately above or immediately below Bhola



From the options its clear, that only option (b) satisfies condition (e).



So, correct Ans is (b).

500x =

e=

8000 3

8000 5100 + 3 3 8500 = 17 3 x >5

Minimum possible = 6

47. KCLFTSB: BST-Best, F-Of, LCK-Luck (Reverse order) SHSWDG: GD-Good, WSHS-Wishes (Reverse order) Similarly “ace the Exam’- C-Ace, T-The, XM-Exam 48. Diversity is shown in terms of difference language. 51. It is not mentioned that elephant is the largest animal. 52. Q

M

72. The given passage clearly deals with how to deal with bereavement and grief and so after the tragedy occurs and not about precautions. Therefore, irrespective of the causes of death, a school student rarely gets into details of causes—which is beyond the scope of the context. Rest all are important in dealing with grief. 73. Given relationship is worker: workplace. A gladiator is (i) a person, usually a professional combatant trained to entertain the public by engaging in mortal combat with another person or a wild. (ii) A person engaged in a controversy or debate, especially in public.

55. W, E, K, O,

I, Q, W, A, F, N, T, X, N, V, B, D,

+8 +6 +4

+8 +6 +4

+8 +6 +4

–2 –2

–2 –2

–2 –2

+8 +6 +2

Hence the odd one from the following group is N, V, B, D. 56. By reading the instructions, it is clear that person who fills form T is a resident female with annual income of ` 9 Lakh. 59. Among given option, option (d) i.e. the number of people with malarial fever has increased this year do not contradic the conclusion.

+2

70.

AD

, +3

+3 CG

, +4

+4 PK

, +5

+5 JP

, +6

OV

74. From the passage it cannot be inferred that horses are given immunity as in option (a), since the aim is to develop medicine and in turn immunize humans. Option (b) is correct since it is given that horses develop immunity after some time. Refer "until their blood built up immunities". Even option(c) is invalid since medicine is not built till immunity is developed in the horses. Option (d) is incorrect since specific examples are cited to illustrate and this cannot capture the essence. 76.

H + G > I + S ...(1) and G–S=1 ...(2) G is not oldest, S is not youngest  H+1>I Irfan older than Hari Gita older than Sarita From given option SGHI

5.14

Reasoning Ability

77. Use of chemical agents in warfare would be undesirable

2. 81 – 54 = 27; 27

2  18 3

276  010 111 110 100 011 101  435

54 – 36 = 18; 18 

2 =12 3

36 – 24 = 12; 12 

2 =8 3

Octal number system



78. 137  001 011 111



731  111 011 001 672  110 111 110 100 010 011

Hence

1623

NUMERICAL TYPE QUESTIONS 1. Middle number is the average of number on both sides

 Average of 3 and 3 is

33 6 = =3 2 2

3.

2

24 – 8 = 16 3

1.5

6 2

15 2.5

45 3

157.5 3.5

630 4

2nd number is in increasing order as shown 1st number above

4. The given series is 12, 35, 81, 173, 357, ....... The given series follows the following pattern 12  2 + 11 = 35 35  2 + 11 = 81 81  2 + 11 = 173 173  2 + 11 = 357 357  2 + 11 = 725

„„

Numerical Ability

Numbers and Algebra

1.1

1

Numbers and Algebra

CHAPTER MCQ TYPE QUESTIONS

P

2016 1. In a quadratic function, the value of the product of the roots (,  ) is 4. Find the value of

7 6.5

(b) 4n

(c) 22n–1

(d) 4n–1

2. Among 150 faculty members in an institute, 55 are connected with each other through Facebook® and 85 are connected through WhatsApp ® . 30 faculty members do not have Facebook® or WhatsApp ® accounts. The number of faculty members connected only through Facebook ® accounts is ______________. (a) 35 (b) 45 (c) 65 (d) 90 4y 3. If 9 y  6  3, then y2  is ________. 3 1 (a) 0 (b)  3 1 (c)  (d) Undefined 3

1 1 1 4. If q  and r  b  and s c  , the value of r q s abc is . a

(a) (rqs)–1

(b) 0

(c) 1

(d) r+q+s

5. The sum of the digits of a two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. (a) 39 (b) 57 (c) 66 (d) 93 6. Two finance companies, P and Q, declared fixed annual rates of interest on the amounts invested with them. The rates of interest offered by these companies may differ from year to year. Year-wise annual rates of interest offered by these companies are shown by the line graph provided below.

9 7.5

8

8 6

6.5

4

 n  n   n   n

(a) n4

10

9.5 8

9 8

Q

2000

2001

2002

2003

2004

2005

2006

If the amounts invested in the companies, P and Q, in 2006 are in the ratio 8 : 9, then the amounts received after one year as interests from companies P and Q would be in the ratio: (a) 2 : 3 (b) 3 : 4 (c) 6 : 7

(d) 4 : 3

2015 7. Read the following table giving sales data of five types of batteries for years 2006 to 2012

Year Type Type Type Type Type I

I

I

I

I

2006

75

144

114

102

108

2007 2008

90 96

126 114

102 75

84 105

126 135

2009

105

90

150

90

75

2010

90

75

135

75

90

2011 2012

105 115

60 85

165 160

45 100

120 145

Out of the following, which type of battery achieved highest growth between the years 2006 and 2012? (a) Type V (b) Type III (c) Type II (d) Type I 8. If a2 + b2 + c2 = 1, then ab + bc + ac lies in the interval

 2 (a) 1,   3  1  (b)  ,1 2 

 1 (c)  1,   2 (d) [2,–4]

1.2

Numbers and Algebra

Q. No. Marks

Answered Answered

Not

1

2

Correctly 21

Wrongly 17

Attempted 6

2 3

3 1

15 11

27 29

2 4

4 5

2 5

23 31

18 12

3 1

What is the average of the marks obtained by the class in the examination? (a) 2.290 (b) 2.970 (c) 6.795 (d) 8.795

Exports Item 1

Item 6

Item 6

11%

16%

19%

Item 2

Item 5

12%

Item 5

20%

20%

Item 3

Item 4

Item 1

Item 2

20%

12%

19%

22%

6%

Item 3 23%

(a) Item 2 (b) Item 3 (c) Item 6 (d) Item 5 13. The statistics of runs scored in a series by four batsmen are provided in the following table. Who is the most consistent batsman of these four? Batsman K L M N

2014 11. The monthly rainfall chart on 50 years of rainfall in Agra is shown in the following figure. Which of the following are true? (k percentile is the value such that k percent of the data fall below that value)

Revenues

4

(a) I only (b) I and II (c) II and III (d) I and III 10. The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking.

12. The total exports and revenues from the exports of a country are given in the two charts shown below. The pie chart for exports shows the quantity of each item exported as a percentage of the total quantity of exports. The pie chart for the revenues shows the percentage of the total revenue generated through export of each item. The total quantity of exports of all the items is 500 thousand tonnes and the total revenues are 250 crore rupees. Which item among the following has generated the maximum revenue per kg?

Ite m

9. If the list of letters, P, R, S, T, U is an arithmetic sequence, which of the following are also in arithmetic sequence? I. 2P, 2R, 2S, 2T, 2U II. P–3, R–3, S–3, T–3, U–3 III. P2, R2, S2, T2, U2

Average 31.2 46.0 54.4 17.9

Standard deviation 5.21 6.35 6.22 5.90

(a) K (b) L (c) M (d) N 14. The exports and imports (in crores of ` ) of a country from 2000 to 2007 are given in the following bar chart. If the trade deficit is defined as excess of imports over exports, in which year is the trade deficit 1/5th of the exports? 120 Exports

110

Imports

100 90

(i) On average, it rains more in July than in December (ii) Every year, the amount of rainfall in August is more than that in January (iii) July rainfall can be estimated with better confidence than February rainfall (iv) In August, there is at least 500 mm of rainfall (a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (iii) and (iv)

80 70 60 50 40 30 20 10 0 2000

(a) 2005 (b) 2004 (c) 2007 (d) 2006

2001

2002

2003

2004

2005

2006

2007

Numbers and Algebra

1.3

15. The table below has question-wise data on the performance of students in an examination. The marks for each question are also listed. There is no negative or partial marking in the examination. Q.No. Marks Answered 1

Answered

Not

Correctly

Wrongly

Attempted

21

17

6

2

21. Following table provides figures (in rupees) on annual expenditure of a firm for two years - 2010 and 2011.

2

3

15

27

2

3

2

23

18

3

What is the average of the marks obtained by the class in the examination?

Category

2010

2011

Raw material

5200

6240

Power & fuel

7000

9450

Salary & wages

9000

12600

Plant & machinery Advertising

20000

25000

15000

19500

Research & Development

22000

26400

In 2011, which of the following two categories have registered increase by same percentage? (a) Raw material and Salary & wages

(a) 1.34

(b) 1.74

(b) Salary & wages and Advertising

(c) 3.02

(d) 3.91

(c) Power & fuel and Advertising

Ratio of male to female students

16. The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students in 2011 and 2012 is equal, what is the ratio of male students in 2012 to male students in 2011? 3.5 3 2.5 2 1.5 1 0.5 0

2008

2009

2010

2011

2012

(a) 1 : 1

(b) 2 : 1

(c) 1.5 : 1

(d) 2.5 : 1

17. What is the average of all multiples of 10 from 2 to 198? (a) 90 (b) 100 (c) 110 (d) 120

(d) Raw material and Research & Development 22. What will be the maximum sum of 44, 42, 40,... ? (a) 502

(b) 504

(c) 506 (d) 500 23. The current erection cost of a structure is ` 13,200. If the labour wages per day increase by 1/5 of the current wages and the working hours decrease by 1/24 of the current period, then the new cost of erection in ` is (a) 16,500 (b) 15,180 (c) 11,000 (d) 10,120 24. Find the sum of the expression 1 1 1 1   +.....+ 1 2 2 3 3 4 80  81 (a) 7 (b) 8 (c) 9 (d) 10

2013

25. In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41°C and of Tuesday to Thursday was 43°C. If the temperature on Thursday was 15% higher than that of Monday, then the temperature in °C on Thursday was (a) 40 (b) 43 (c) 46 (d) 49 26. The set of values of p for which the roots of the equation 3x2 + 2x + p(p – 1) = 0 are of opposite sign is (a) (–, 0) (b) (0, 1) (c) (1, ) (d) (0, )

20. A number is as much greater than 75 as it is smaller than 117. The number is

27. Find the sum to n terms of the series 10 + 84 + 734 + …

18. The value of

12  12  12  ... is

(a) 3.464 (b) 3.932 (c) 4.000 (d) 4.444 2 19. If x is real and |x – 2x + 3| = 11, then possible values of |– x3 + x2 – x| include (a) 2, 4 (b) 2, 14 (c) 4, 52 (d) 14, 52

(a) 91 (b) 93

(a)

(c) 89 (d) 96

(c)



9 9n  1



10

9 9 1 n

8

 1

n

(b) (d)



9 9n  1



8

9 9 1 n

8

 1

n

2

1.4

Numbers and Algebra

28. If 3  X  5 and 8  Y  11 then which of the following options is TRUE? (a)

3 X 8   5 Y 5

(b)

3 X 5   11 Y 8

3 X 8 3 X 8     (d) 11 Y 5 5 Y 11 29. Following table gives data on tourists from different countries visiting India in the year 2011. (c)

Country USA England Germany Italy Japan Australia France

Number of Tourists 2000 3500 1200 1100 2400 2300 1000

Which two countries contributed to the one third of the total number of tourists who visited India in 2011? (a) USA and Japan (b) USA and Australia (c) England and France (d) Japan and Australia

2012 30. If (1.001)1259 = 3.52 and (1.001)2062 = 7.85, then (1.001)3321 = (a) 2.23 (b) 4.33 (c) 11.37 (d) 27.64 31. Raju has 14 currency notes in his pocket consisting of only ` 20 notes and ` 10 notes. The total money value of the notes is ` 230. The number of ` 10 notes that Raju has is (a) 5 (b) 6 (c) 9 (d) 10 32. The data given in the following table summarizes the monthly budget of an average household. Category

Amount (Rs.)

Food

4000

Clothing

1200

Rent

2000

Savings

1500

Other expenses

1800

The approximate percentage of the monthly budget NOT spent on savings is (a) 10% (b) 14% (c) 81%

(d) 86%

2011 33. P, Q, R and S are four types of dangerous microbes recently found in a human habitat. The area of each circle with its diameter printed in brackets represents the growth of a single microbe surviving human immunity system within 24 hours of entering the body. The danger to human beings varies proportionately with the toxicity, potency and growth attributed to a microbe shown in the figure below :

A pharmaceutical company is contemplating the development of a vaccine against the most dangerous microbe. Which microbe should the company target in its first attempt? (a) P (b) Q (c) R (d) S 34. The sum of n terms of the series 4 + 44 + 444 + ... is (a) (4/81) [10n+1 – 9n –1] (b) (4/81) [10n–1 – 9n –1] (c) (4/81) [10n + 1 – 9n –10] (d) (4/81) [10n – 9n – 10] 35. Given that f(y) = |y|/y, and q is any non-zero real number, the value of |f(q) – f(–q)| is (a) 0 (c) 1

(b) –1 (d) 2

NUMERICAL TYPE QUESTIONS 2016 1. The following graph represents the installed capacity for cement production (in tonnes) and the actual production (in tonnes) of nine cement plants of a cement company. Capacity utilization of a plant is defined as ratio of actual production of cement to installed capacity. A plant with installed capacity of at least 200 tonnes is called a large plant and a plant with lesser capacity is called a small plant. The difference between total production of large plants and small plants, in tonnes is ________.

Numbers and Algebra

Capacity/production(tonnes)

300

1.5

Installed Capacity

Actual Production

250

250 220 200 150

200 190 180 190 160 160 150 160

200 190 150

120

100

230

140

100

120

50 0

1

2

3

4 5 Plant Number

6

7

8

9

2. The numeral in the units position of 211870 + 146127  3424 is _________.

2015 3. The exports and imports (in crores of Rs.) of a country from the year 2000 to 2007 are given in the following bar chart. In which year is the combined percentage increase in imports and exports the highest ?

6. The smallest angle of a triangle is equal to two thirds of the smallest angle of a quadrilateral. The ratio between the angles of the quadrilateral is 3 : 4 : 5 : 6. The largest angle of the triangle is twice its smallest angle. What is the sum, in degrees, of the second largest angle of the triangle and the largest angle of the quadrilateral? 7. In a sequence of 12 consecutive odd numbers, the sum of the first 5 numbers is 425. What is the sum of the last 5 numbers in the sequence? 8. A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is ` 4,50,000. In 2013, the raw material expenses increased by 30% and all other expenses increased by 20%. What is the percentage increase in total cost for the company in 2013? rtation Transpo 10%

Labour 15%

Plant and Machinery 30%

Raw Material 20% Energy 25%

4. The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5 : 4. There are 40 males in Electrical Engineering. What is the difference between numbers of female students in the Civil department and the female students in the Mechanical department?

Electrical 20%

Transport- % 10 ation

Computer Science 20%

Mechanical 10%

9. A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the share of raw material, labour, energy, plant & machinery, and transportation costs in the total manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is ` 4,50,000. In 2013, the raw material expenses increased by 30% and all other expenses increased by 20%. If the company registered a profit of ` 10 lakhs in 2012, at what price (in `) was each air purifier sold?

Civile 30%

2014 5. A foundry has a fixed daily cost of ` 50,000 whenever it operates and variable cost of ` 800Q, where Q is the daily production in tonnes. What is the cost of production in ` per tonne for a daily production of 100 tonnes?

Labour 15%

Plant and Machinery 30%

Raw Material 20% Energy 25%

1.6

Numbers and Algebra

10. The sum of eight consecutive odd numbers is 656. The average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number?

Ratio of male to female students

11. The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students doubled in 2009, by what percent did the number of male students increase in 2009?

2

2

12. If (z + 1/z) = 98, compute (z + 1/z2) 13. In a survey, 300 respondents were asked whether they own a vehicle or not. If yes, they were further asked to mention whether they own a car or scooter or both. Their responses are tabulated below. What percent of respondents do not own a scooter?

Men

Women

Car

40

34

Scooter

30

20

3

Both

60

46

2.5

Do not own vehicle

20

50

Own vehicle

3.5

2

14. Round–trip tickets to a tourist destination are eligible for a discount of 10% on the total fare. In addition, groups of 4 or more get a discount of 5% on the total fare. If the one way single person fare is ` 100, a group of 5 tourists purchasing round–trip tickets will be charged ` ________.

1.5 1 0.5 0 2008

2009

2010

2011

2012

ANSWERS MCQ Type Questions 1. (b)

2. (a)

3. (c)

4. (c)

5. (a)

6. (d)

7. (d)

8. (b)

9. (b)

10. (c)

11. (b)

12. (d)

13. (a)

14. (d)

15. (c)

16. (c)

17. (b)

18. (c)

19. (d)

20. (d)

21. (d)

22. (c)

23. (a)

24. (b)

25. (c)

26. (b)

27. (d)

28. (b)

29. (c)

30. (d)

31. (a)

32. (d)

33. (d)

34. (c)

35. (d)

Numerical Type Questions 1. (120) 9. (20,000)

2. (7)

3. (2006) 10. (163)

4. (32)

5. (1300 to 1300)

11. (140 to 140)

12. (96)

6. (180 to 180) 13. (48)

14. (850)

7. (495)

8. (22)

Numbers and Algebra

1.7

EXPLANATIONS put

MCQ TYPE QUESTIONS

= 1

we get

 n  n  n  n  n  n     n  n 1. n n 1 1   n   n      n n  n  n  n  n is 4n.   n   n

2. From question Total number of faculty members = 150 The number of faculty members having facebook account = (FB) = 55 The number of faculty members having whatsapp = (W) = 85

The number of faculty members having both the accounts = (FB + W) – 120 = (55 + 85) – 120 = 20

 The number of faculty members connected only through Facebook accounts = 55 – 20 = 35

y =



qa = r r–b =

The number of faculty members do not have face book or Whats App accounts = 30 The number of faculty members having any account = 150 – 30 = 120

1 4 =  3 3

1 3 3 1 1 4  we get, =  9 3 9 9 1 . So, the required value is 3 1 4. Given q–a = r 1 1  = a r q and also put

n   n = (   )n = (4)n So, the value of

y = 1,

 

1 s

1 s r s = rb 1

b

=

s–c =



...(i)

1 sc



=

...(ii)

1 q 1 q

sc = q

...(iii)

From equation (i), qa = r (sc)a = r S

So the number of faculty members connected only through facebook accounts is 35. 3. Given,

(9y – 6) = 3 Or

– (9y – 6) = 3 9y – 6 = – 3

From equation (i), 9y = 9



y =1

Or from equation (ii),

 Now

9y = 3 1 y = 3

4y y = 3 2

...(ii)

=r 1

s ...(i)

(from eq. (iii))

s ac =  s  b

9y  6 = 3

It has two values either

ac



abc

(from eq. (ii)) 1

=s=s

abc = 1

So the value of abc is 1 5. Let the two digit number be xy Now sum of the digit of a two digit number is 12



x + y = 12

...(i)

Now according to question, (10y + x) – (10x + y) = 54



9y – 9x = 54



y–x =6

Now solving equation (i) and (ii) we get, x = 3 and y = 9 Hence the original number is 39.

 Option (a) is correct.

...(ii)

1.8

Numbers and Algebra

6. Amount invested by Company P in 2006 = ` 8x and amount invested by Company Q in 2006 = ` 9x Interest from Company P

` 8x  6 100 = ` 9x  4 100 4 3  Option (d) is correct. =

7. Type-I achieved a growth of 53% in the period which is higher than any other type of battery 10. Total number of students in class = 44 Average of marks, =

(2  21)  (3  15)  (1  11)  (2  23)  (5  31) 42

=

42  45  11  46  155 299  6.795 = 44 44

 Option (c) is correct. 11. In the question the monthly average rainfall chart for 50 years has been given. Let us check the options. (i) On average, it rains more in July than in December  correct. (ii) Every year, the amount of rainfall in August is more than that in January.

 may not be correct because average rainfall is given in the question. (iii)July rainfall can be estimated with better confidence than February rainfall.

 From chart it is clear the gap between 5 percentile and 95 percentile from average is higher in February than that in July  correct. (iv) In August at least 500 mm rainfall

 May not be correct, because its 50 year average. So correct option (b) (i) and (iii). 12. Item : 2

20  250  107 100 20  500  103 100

Item : 3

23  250  107 19  500  103 1.2 = Item 3 Item : 6 19  1.18  Item 6 16 Item : 5 20 5   1.6  1.6  Item 5 12 3 14. For 2005,

trade defict = (90 – 70) crores = 20 crores

1 5

Now

th

1 (70) crores 5 = 14 crores  20 crores.

of export =

Hence option (a) is wrong. For 2004, Trade defict =(80 – 70) crores = 10 crores Now

1 5

th

1 (70) crores 5 = 14 crores  20 crores.

of export =

Hence option (b) is wrong. For 2007 Trade defict = (120 – 110) crores = 10 crores Now

1 5

th

1 (110) crores 5 = 22 crores  10 crores.

of export =

Hence option (c) is wrong. For 2006, Trade defict = (120 – 100) crores = 20 crores Now

1 5

th

1 (100) crores 5 = 20 crores = Trade deficts

of export =

Hence option (d) is correct. 15. Total question 44  2 = 88 44  3 = 132 144 88 = 132 308 Total marks obtained = (212) + (153) + (232) = 133 Total Number of students = 44

0.5  104  5  103 1  Item 2 Average =

133 = 3.02 44

Numbers and Algebra

1.9

16. Take number of female students in 2011 = 100  Number of male in 2011 = 100 No. of female in 2012 = 100 No. of male in 2012 = 150 Ratio =

b2 – 4ac = 4 –4(1)(–8) = 36 > 0 Hence roots are real. and for x2 –2x + 14 = 0 b2 – 4ac = 4 –4(1)(14) < 0 Hence roots are imaginary  The value of x is the root of x2 –2x –8 = 0

150 100

17. All multiples of 10 from 2 to 198 = 10, 20,............190 Now

Now

x=

an = a + (n – 1)d 190 = 10 + (n – 1)  10

=

2  36 2

=

26 2

180 = (n – 1)  10 18 = n – 1



n = 19

Hence total number of multiples are 19. Now

Sn =

n [2a  (n  1) d] 2

=

19 [20  (18)(10)] 2

=

19 [20  180] = 1900 2

2  4  4(1)(8) 2

 x = 4, –2. Now for x= 4, –x3 + x2 – x = – 64 + 16 – 4 = 52. and for x = –2 –x3 + x2 – x = + 8 + 4 + 2 = 14 Hence possible values of –x3 + x2 – x = 14, 52 22. 0 =44 + (n – 1) (– 2) n – 1 = 22 n = 23

Now average of all multiples of 10 from 2 to198 =

 18.

=

10  20  ...  190 19 1900 =100 19

12  12  12  ...

For a particular type of question i.e.

a  a  a  ... we find a two consecutive

number whose product is a. The greater number among the two consecutive number is the answer. Here a = 12 i.e.

Sum

=

n ( a  l) 2

=

23 (44  0) 2

= 23  22 = 506 Hence (c) is the answer 23. Let ‘d’ be the # of days required Let ‘h’ be the daily working hours.  dh can be taken as a measure of the total work be done in order to erect the structure. Now working hours per day h =  # of days

d =

12 = 3 4 greater number

Hence 12  12  12  ... = 4 19. Here it is given that x is real x2 –2x + 3 = 11 Now  x2 – 2x + 3 = 11 or x2 –2x + 3 = –11 For x2 –2x + 3 = 11  x2 –2x –8 = 0

23 h 24

dh dh 24   d h  23  23 h    24 

Let

c = daily wages earlier



c=

13, 200 d

Now daily wages c =

6 c 5

 Total new cost = c  d

=

6 13200 24   d  16,500 5 d 23

1.10

Numbers and Algebra

24. Given: The expression S=

1



1 2

1 2 3

31. Let number of ` 20 notes be x and ` 10 notes be y

 ...... 

1



80  81

and

To find: Sum of the expression. Analysis: Each term of the expression is having irrational denominator. One can also see, that both the elements in the denominator of a term differ by only 1. This gives an idea that on rationalising the denominator, each denominator will become 1, so the sum will be easier than earlier. Rationalising the denominator, S=

2 1

3 2



 2   1  3   2

=

2

2

2

2

 ... 



81  80 81

  2

80



1 &

x = 9 and y = 5 Hence numbers of 10 rupee notes are 5. 32. Total budget = 10,500 Expenditure other than savings = 9000 Hence approximate percentage of monthly budget

2

9000 = 86% 10500

33. By observation of the table, we can say S

P

81

Occurs twice, once as positive and once a negative. So, after cancelling out, we are left with the following:

...(2)

Solving equations (1) and (2), we have

2 1 3 2 81  80   ....  1 32 81  80

Each term except

...(1)

x + y = 14

=

= 2  1  3  2  ....  81  80 Now, one can see that the terms will cancel out.

25.

20x + 10y = 230

Q

R

S

Requirement

800

600

300

200

Potency

0.4

0.5

0.4

0.8

34. Let S = 4 (1 + 11 + 111 + ...) =

4 (9 + 99 + 999 +...) 9

S = 81  1 = 9 – 1 = 8 So, the answer is (b).

=

4 {(10 – 1) + (102 – 1) + (103 – 1) + ......... } 9

Mon  Tues  Wed.  41 3

=

4 {(10 + 102 + ....+ 10n) – n} 9

=

 4  (10 n  1)  n 10 9 9 

=

4 {10n+1 – 9n – 10} 81

Mon + Tues + Wed. = 123

...(1)

Tues  Wed  Thurs.  43º 3

Tue + Wed + Thu. = 129° ...(2) (2) – (1) Tues + Wed + Thu – (Mon + Tues. + Wed.) = 129 – 123 = 6° 115 x  x  6º Thu. – Mon. = 6°  100

Thus. = Mon  Mon = x Thurs =

 30. Let



115 15 x = = 6° 100 100

x = 40°

f(y) =



f(q) =

f(–q) =

115 x 100

Thurs =

35. Given :

115  40  46º 100

1.001 = x x1259 = 3.52, and x2062 = 7.85 x3321 = x1259 . x2062 = 3.52  7.85 = 27.64

=

 f(q) – f(q) =

=

y y q q

q q – q q q q



2q q

q q

=2

Numbers and Algebra

1.11

NUMERICAL TYPE QUESTIONS 1. From the given graph large plants which are having installed capacity of at least 200 tonnes are 1, 4, 8 and 9. So, the total production of large plants

5. Fixed cost = ` 50,000 Variable cost = ` 800Q Q = daily production in tones For 100 tonnes of production daily, total cost of production = 50,000 + 800  100 = 130,000 So, cost of production per tonne of daily production

= 160 + 190 + 230 + 190 =

= 770 Now the remaining plants with installed capacity is less then 200 tonnes are 2, 3, 5, 6 and 7. So, total production of small plants = 150 + 160 + 120 + 100 + 120

130,000 = ` 1300. 100

6. Let the angles of quadrilateral are 3x, 4x, 5x, 6x So, 3x + 4x + 5x + 6x = 360 x = 20 Smallest angle of quadrilateral = 3  20 = 60

= 650 Then difference = 770 – 650 = 120 870

2. Unit digit of 211

+ 146127  3424 is 1 + 6  1 = 7

3. Increase in exports in 2006 =

100  70 = 42.8% 70

Increase in imports in 2006 =

120  90 = 33.3% 90

which is more than any other year. 4. According to question, There are 40 males in Electrical Engg.  Total number of females in Electrical Engg. = 72 – 40 = 32  Total number of students in Electrical Engg. = 72 20% = 72  30% = Total number of students in Civil Engg.  Total number of students in Civil Engg. 30  72 = 36 × 3 = 108 = 20

 Number of female students in Civil Engg. 4  108 = 4 × 12 = 48 9 Now, total number of students in Mechanical Engg.

=

10  72 = 36 20  Number of female students in Mechanical Engg.

=

=

4  36 = 16 9

 Difference between female students in Civil and Mechanical departments = 48 – 16 = 32

Smallest angle of triangle =

2  60  40 3

Largest angle of triangle = 2  40 = 60 Three angles of triangle are 40, 60, 80 Largest angle of quadrilateral is 120 Sum (2nd largest angle of triangle + largest angle of quadrilateral) = 60 + 120 =180. 7. 8 th observation is 7  2 = 14 more than 1 st observation 9th observation is 14 more than 2nd observation 10th observation is 14 more than 3rd observation 11th observation is 14 more than 4th observation 12th observation is 14 more than 5th observation Total 14  5 =70 Sum of the first five numbers = 425 Sum of last five numbers = 495 8. Transport (10%)

2012 300,000

2013 360,000

Labour (15%) Raw material (20%)

450,000 750,000

540,000 780,000

Energy (25%) 750,000 Plant and Machinery (30%) 900,000 Total

900,000 1,080,000

3, 000, 000 3, 660, 000

Percentage increase in total cost = 22% 9. Total expenditure =

15 x = 4,50,000 100

x = 3 106 Profit = 10 lakhs So, total selling price = 40,00,000 Total purifies = 200 S.P. of each purifier = (1)/(2) = 20,000 10. Eight consecutive odd number = 656 a – 6, a – 1, a – 2, a, a + 2, a + 4, a + 6 a + 8 = 656 a = 81

...(1) ...(2)

1.12

Numbers and Algebra

Smallest m = 75 Average consecutive even numbers



...(1)

2

a2 a a2 a4 = 87 4

 a = 86 Second largest number = 88

...(2)

Adding equation (1) and (2), 75 + 88 = 163 11. Ratio of male to female student in 2008 = 5 : 2 In 2009, let the male student be 5x and female student be 2x Now if number of female students double in 2009



Number of female student = 4x



Number of male student = 12x Gender

2008

2009

Male

5x

12x

Female

2x

4x

% of male student increase in 2009  12 x  5 x   =    100  % 5 x    

7  =   100  % 5   = 140%

12. Expanding

1   z  z  = 98   1 1 2  z  2  2( z)   = 98 z  z 1  z2 + 2 = 96 z 13. Total respondents = 300 Now men respondents who do not have scooter = 40 + 20 = 60 And women respondents who do not have scooter = 34 + 50 = 84

 Total respondents who do not have scooter = 60 + 84 = 144

 144   100  % = 48%  Required percent =   300  14. One way single person fare = ` 100

 Two way fare for single person = ` 200  For 5 persons two way fare = ` 1000 Now, total discount = (10 + 5)% = 15%

 15   1000  = ` 150 Discount amount = `   100  Amount to be paid = `(1000 – 150) = ` 850

„„

Percentage and Its Applications

2

Percentage and Its Applications

C HAPTER MCQ TYPE QUESTIONS 2016 1. In a huge pile of apples and oranges, both ripe and unripe mixed together, 15% are unripe fruits. Of the unripe fruits, 45% are apples. Of the ripe ones, 66% are oranges. If the pile contains a total of 5692000 fruits, how many of them are apples? (a) 2029198 (b) 2467482 (c) 2789080 (d) 3577422 2. A person moving through a tuberculosis prone zone has a 50% probability of becoming infected. However, only 30% of infected people develop the disease. What percentage of people moving through a tuberculosis prone zone remains infected but does not show symptoms of disease? (a) 15 (b) 33 (c) 35 (d) 37 3. (x % of y) + (y % of x) is equivalent to _______. (a) 2 % of xy (b) 2 % of (xy/100) (c) xy % of 100

2.1

(d) 100 % of xy

2014 4. The population of a new city is 5 million and is growing at 20% annually. How many years would it take to double at this growth rate? (a) 3 – 4 years (b) 4 – 5 years (c) 5 – 6 years (d) 6 – 7 years 5. One percent of the people of country X are taller than 6 ft. Two percent of the people of country Y are taller than 6 ft. There are thrice as many people in country X as in country Y. Taking both countries together, what is the percentage of people taller than 6 ft? (a) 3.0 (b) 2.5 (c) 1.5 (d) 1.25 6. Industrial consumption of power doubled from 2000-2001 to 2010-2011. Find the annual rate of increase in percent assuming it to be uniform over the years. (a) 5.6 (b) 7.2 (c) 10.0 (d) 12.2 7. The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For international comparison, the GDP is compared in US Dollars (USD) after conversion based on the market

exchange rate. During the period 2012-2013 the exchange rate for the USD increased from ` 50/ USD to ` 60/ USD. India’s GDP in USD during the period 2012-2013 (a) increased by 5% (b) decreased by 13% (c) decreased by 20% (d) decreased by 11%

2013 8. A firm is selling its product at ` 60 per unit. The total cost of production is ` 100 and firm is earning total profit of ` 500. Later, the total cost increased by 30%. By what percentage the price should be increased to maintained the same profit level. (a) 5 (b) 10 (c) 15 (d) 30

2012 9. An automobile plant contracted to buy shock absorbers from two suppliers X and Y. X supplies 60% and Y supplies 40% of the shock absorbers. All shock absorbers are subjected to a quality test. The ones that pass the quality test are considered reliable. Of X's shock absorbers, 96% are reliable. Of Y's shock absorbers, 72% are reliable. The probability that a randomly chosen shock absorber, which is found to be reliable, is made by Y is (a) 0.288 (b) 0.334 (c) 0.667 (d) 0.720

2011 10. There are two candidates P and Q in an election. During the campaign, 40% of the voters promised to vote for P, and rest for Q. However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q. 25% of the voters went back on their promise to vote for Q and instead voted for P. Suppose, P lost by 2 votes, then what was the total number of voters? (a) 100 (b) 110 (c) 90

(d) 95

NUMERICAL TYPE QUESTION 2015 1. From a circular sheet of paper of radius 30 cm, a sector of 10% area is removed. If the remaining part is used to make a conical surface, then the ratio of the radius and height of the cone is ____.

2.2

Percentage and Its Applications

ANSWERS MCQ Type Questions 1. (a)

2. (c)

3. (a)

4. (a)

5. (d)

6. (b)

7. (d)

8. (a)

9. (b)

10. (a)

Numerical Type Question 1. (2.06)

EXPLANATIONS MCQ TYPE QUESTIONS 1. Given,

5. Let number of people in country y = 100 So, number of people in country x = 300 Total number of people taller than 6ft in both the

Total no. of fruits = 5692000

1 2  100  5 100 100 % of people taller than 6ft in both the countries

countries = 300 

Unripe type of apples = 45% of 15% of 5692000

45 15   5692000 100 100 = 384210 =

Ripe type of apples = (100 – 66)%  (100 – 15)%

5  100 = 1.25%. 400 7. Per ` 100 final value ` 107

=

 5692000

34 85   5692000 100 100 = 1644988

100 107 Dollars final value 50 60 for 100 dollars?

 per

=

 Total no. of apples = 384210 + 1644988 = 2029198 2. The percentage of people moving through a tuberculosis prone zone remains infected but does not show symptoms of disease. = (100 – 30)%  50% =

70 50 35    35% 100 100 100

3. (x% of y) + (y% of x)

100  50 107  = 89.16 100 60 Descreased by 11% 9. X Y Supply 60% 40% Reliable 96% 72% Overall 0.576 0.288

=

 10.

P 40% –6% +15%

2 xy x y  y x = = 2% of xy 100 100 100  Option (a) is correct. 4.

r   A = P 1    100 

0.288 = 0.334 0.576  0.288 Q 60% +6% –15%

t

20   10 million = 5 million 1   100   1  2 = 1   5 

P(x) =

t

t

t

6 6 2 =   = log 2 = t log   5 5 0.301 = t 0.0791 0.301 t= = 3.8 years 0.0791  Hence around 3–4 years would it take to double the growth rate



49% 2% = 2



51% 100% = 100

NUMERICAL TYPE QUESTION 1. 90% of area of sheet = Cross sectional area of cone  0.9   30  30 =  r1  30 ( Slant height of cone (l) = 30)  27 cm = r1

 Height of the cone =

302  272 = 13.08 cm Hence ratio of radius and height is 2.06. „„

Time and Work

3.1

3

Time and Work

CHAPTER MCQ TYPE QUESTIONS 2016 1. P, Q, R and S are working on a project. Q can finish the task in 25 days, working alone for 12 hours a day. R can finish the task in 50 days, working alone for 12 hours per day. Q worked 12 hours a day but took sick leave in the beginning for two days. R worked 18 hours a day on all days. What is the ratio of work done by Q and R after 7 days from the start of the project? (a) 10:11

(b) 11:10

(c) 20:21

(d) 21:20

2. S, M, E and F are working in shifts in a team to finish a project. M works with twice the efficiency of others but for half as many days as E worked. S and M have 6 hour shifts in a day whereas E and F have 12 hours shifts. What is the ratio of contribution of M to contribution of E in the project? (a) 1 : 1 (b) 1 : 2 (c) 1 : 4 (d) 2 : 1 3. It takes 10 s and 15 s, respectively, for two trains travelling at different constant speeds to completely pass a telegraph post. The length of the first train is 120 m and that of the second train is 150 m. The magnitude of the difference in the speeds of the two trains (in m/s) is ______. (a) 2.0 (b) 10.0 (c) 12.0 (d) 22.0 4. The velocity V of a vehicle along a straight line is measured in m/s and plotted as shown with respect to time in seconds. At the end of the 7 seconds, how much will the odometer reading increase by (in m) ?

5. Ananth takes 6 hours and Bharath takes 4 hours to read a book. Both started reading copies of the book at the same time. After how many hours is the number of pages to be read by Ananth, twice that to be read by Bharath? Assume Ananth and Bharath read all the pages with constant pace. (a) 1

(b) 2

(c) 3

(d) 4

2015 6. An electric bus has onboard instruments that report the total electricity consumed since the start of the trip as well as the total distance covered. During a single day of operation, the bus travels on stretches M, N, O and P, in that order. The cumulative distances travelled and the corresponding electricity consumption are shown in the table below.

Stretch Comulative Electricity distance(km) used (kWh) M N

20 45

12 25

O P

75 100

45 57

The stretch where the electricity consumption per km is minimum is (a) M (b) N (c) O (d) P

2014 7. It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes? (a) 4 times the draining rate (b) 3 times the draining rate (c) 2.5 times the draining rate

(a) 0

(b) 3

(c) 4

(d) 5

(d) 2 times the draining rate

3.2

Time and Work

2013 8. A tourist covers half of his journey by train at 60 km/h, half of the remainder by bus at 30 km/h and the rest by cycle at 10 km/h. The average speed of the tourist in km/h during his entire journey is (a) 36 (b) 30 (c) 24 (d) 18 9. A car travels 8 km in the first quarter of an hour, 6 km in the second quarter and 16 km in the third quarter. The average speed of the car in km per hour over the entire journey is (a) 30 (b) 36 (c) 40 (d) 24

2011 10. The fuel consumed by a motorcycle during a journey while traveling at various speeds is indicated in the graph below.

From the given data, we can conclude that the fuel consumed per kilometre was least during the lap (a) P (b) Q (c) R (d) S

2010 11. 5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall? (a) 20 days (b) 18 days (c) 16 days (d) 15 days

NUMERICAL TYPE QUESTIONS 2015 1. A tiger is 50 leaps of its own behind a deer. The tiger takes 5 leaps per minute to the deer’s 4. If the tiger and the deer cover 8 metre and 5 metre per leap respectively, what distance in meters will the tiger have a run before it catches the deer?

2014 The distances covered during four laps of the journey are listed in the table below

Lap

Distance (kilometers)

Average speed (kilometers per hour)

P

15

15

Q

75

45

R

40

75

S

10

10

2. A man can row at 8 km per hour in still water. If it takes him thrice as long to row upstream, as to row downstream, then find the stream velocity in km per hour. 3. A train that is 280 metres long, travelling at a uniform speed, crosses a platform in 60 seconds and passes a man standing on the platform in 20 seconds. What is the length of the platform in metres?

ANSWERS MCQ Type Questions 1. (c)

2. (b)

3. (a)

4. (d)

11. (d) Numerical Type Questions 1. (800)

2. (4)

3. (560 to 560)

5. (c)

6. (b)

7. (a)

8. (c)

9. (c)

10. (a)

Time and Work

3.3

EXPLANATIONS 4th sec  triangle

MCQ TYPE QUESTIONS

1 1 2  1 2 5th sec  straight line

1. Q can finish 1 day’s 1 hour’s work

=

1 1  25  12 300 R can finish 1 day’s 1 hour’s work =

=0 6 sec  triangle th

1 = 50  12 =

1 1 11  2 2 7th sec  triangle =

1 600

1 1 11  2 2 Total Odometer reading at 7 seconds

Now Q working hours

=

= (7 – 2)  12 = 60 hr

1 1 1 1 =  1 1 1  0   m 2 2 2 2 =5m

 R working hours = 7  18 = 126 After 7 days, the ratio of work done by Q & R is Q :R

60 126 : 300 600 Q : R = 120 : 126 = 20 : 21 2. M is twice as efficient as E but worked for half as many days. So in this case they will do equal work if their shifts had same timings. But M’s shift is for 6 hours, while E’s shift for 12 hours. Hence, E will do twice the work as M. Ratio of contribution of M : E in work is 1 : 2. 3. Here the speed of first train (S1) =

120  12 m/s 10

150  10 m/s and the speed of second train (S2) = 15  The magnitude of difference in the speeds of two trains = (S1 – S2) = (12 – 10) m/s = 2 m/s 4. From the given figure. The odometer reading increases from starting point to end point. Odometer reading = Area of the given diagram Area of the velocity and time graph per second 1st sec  triangle 1 1 = 11  2 2 nd 2 sec  square =11=1 3rd sec  square + triangle = 11 

1 1 11 = 1 2 2

So, the odometer reading is increased by 5 m.

12 = 0.6 20 25 = 0.55 N 45 45 O = 0.6 75 57 P = 0.57 100 7. Vhalf = 30(s) drawing rate = s

6. For

M

Total volume = 60 S tank (s1)(10) – (s)10 = 30s s1(s) – s = 3s s1 = 4s s1 = 4 drawing rate 8. Given: A tourist covers half the journey at 60 km/h, one fourth at 30 km/h and the remaining one fourth at 10 km/h To find: Avearge speed of the whole journey. Analysis: One approach to this is to find individual time consumed in each of the three parts and use that value to get total time, and find average speed. So, Lets assume total length of journey = x km  First phase at 60 km/h Time t1 =

x 1 x Distance = = h 2 60 120 Speed

3.4

Time and Work

Second phase at 30 km/h Distance = Time t2 =

So part of work done in 1 day by 2 skilled, 6 semiskilled and 5 unskilled 2 6 5 1   = = 100 200 300 15 So work done by given workers in days = 15

x km 4 x x = 4.30 120

Third phase at 10 km/h

NUMERICAL TYPE QUESTIONS

x Distance = km 4

Time t3 = 

Total time =

x x h = 4.10 40

x 1 1 x x x    =    1 40  3 3 120 120 40 

x (1  1  3) 5x = = 40 3 120

 Average speed =

=

Total Distance Total time x  120 = 24 km/h 5x

Hence, the answer is (c) 9. Average Speed = Total Distance =

Total distance total time taken

8  6  16 30  60  15  15  15 45

Fuel consumption P

60 km/l

Q

90 km/l

R

75 km/l

S

30 km/l

Let at time ‘t’ the tiger catches the deer.  Distance travelled by deer + initial distance between them 50 × 8  400m = distance covered by tiger.  40 × t = 400 + 20t 400 = 20 min  t 200  total distance  400 + 20 × t = 800 m 2. Speed of man = 8;

= 40km/h. 10.

1. Tiger – 1 leap  8 meter Speed = 5 leap/hr = 40m/min Deer  1 leap = 5 meter speed = 4hr = 20m/min

Actual

15 1  l 60 4 75 5  l 90 6 40 8  l 75 15 10 1  l 30 3

11. 5 skilled workers build wall in 20 days 1 skilled worker build wall in 20  5 days Hence in 1 day, part of work done by skilled work 1 = 100 Similarly in 1 day part of work done by semiskilled workers 1 = 25  8 and in 1 day part of work done by un-skilled worker 1 = 30  10

Left distance = d

d 8 Upstream: Speed of stream = s

Time taken =

 speed upstream = S’ = (8 – s)

 d  t =   8s Downstream: Given speed downstream = t =

 3t = t 

3d = 8s

d 8s

d 8s

d 3d = 8s 8s  s = 4 km/hr 3. Let the length of platform = x.m 

and train length = 280 m (given) According to Question x  280 280 = 60 20



x = 560 m.

„„

Ratio, Proportion and Mixtures

4

4.1

Ratio, Proportion and Mixtures

C HAPTER

MCQ TYPE QUESTIONS 2015 1. A cube of side 3 units is formed using a set of smaller cubes of side 1 unit. Find the proportion of the number of faces of the smaller cubes visible to those which are NOT visible. (a) 1 : 4 (b) 1 : 3 (c) 1 : 2 (d) 2 : 3

2011 2. A container originally contains 10 litres of pure spirit. From this container 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this process is repeated one more time. How much spirit is now left in the container? (a) 7.58 litres (b) 7.84 litres (c) 7 litres (d) 7.29 litres

ANSWERS MCQ Type Questions 1. (c)

2. (d)

4.2

Ratio, Proportion and Mixtures

EXPLANATIONS MCQ TYPE QUESTIONS 1.

Number of faces per cube = 6 Total number of cubes = 9  3 = 27  Total number of faces = 27  6 = 162  Total number of non visible faces = 162 – 54 = 108



Number of visible faces 54 1 = = Number of non visible faces 108 2 3

3

729  10–1  9 2. 10  = 10   =  10   10  1000 

729  1 = 7.29 litres 1000

„„

Permutations and Combinations & Probability

5

CHAPTER

5.1

Permutations and Combinations & Probability

MCQ TYPE QUESTIONS 2016 1. Shaquille O’ Neal is a 60% career free throw shooter, meaning that he successfully makes 60 free throws out of 100 attempts on average. What is the probability that he will successfully make exactly 6 free throws in 10 attempts? (a) 0.2508

(b) 0.2816

(c) 0.2934

(d) 0.6000

2015 2. Four cards are randomly selected from a pack of 52 cards. If the first two cards are kings, what is the probability that the third card is a king? (a) 4/52

(b) 2/50

(c) 1/52  (1/52)

(d) 1/52  (1/52)  (1/50)

3. Five teams have to compete in a league, with every team playing every other team exactly once, before going to the next round. How many matches will have to be held complete the league round of matches? (a) 20

(b) 10

(c) 8

(d) 5

4. Right triangle PQR is to be constructed in the xy – plane so that the right angle is at P and line PR is parallel to the-axis. The x and y coordinates of P, Q, and R are to be integers that satisfy the inequalities: –4  x  5 and 6  y  16. How many different triangles could be constructed with these properties? (a) 110

(b) 1,100

(c) 9,900

(d) 10,000

5. A coin is tossed thrice. Let X be the event that head occurs in each of the first two tosses. Let Y be the event that a tail occurs on the third toss. Let Z be the event that two tails occurs in three tosses. Based on the above information, which one of the following statements is TRUE? (a) X and Y are not independent (b) Y and Z are dependent (c) Y and Z are independent (d) X and Z independent

6. Ram and Ramesh appeared in an interview for two vacancies in the same department. The probability of Ram’s selection is 1/6 and that of Ramesh is 1/8. What is the probability that only one of them will be selected? 47 1 (b) (a) 48 4 13 35 (c) (d) 48 48 7. Given set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is probability that the sum of the two numbers equals 16 ? (a) 0.20

(b) 0.25

(c) 0.30 (d) 0.33 8. The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p.c: 27 13 I. p + m + c = II. p + m + c = 20 20 1 III. (p)  (m)  (c) = 10 (a) Only relation I is true. (b) Only relation II is true. (c) Relations II and III are true. (d) Relations I and III are true.

2014 9. A five digit number is formed using the digits 1,3,5,7 and 9 without repeating any of them. What is the sum of all such possible five digit numbers? (a) 6666660

(b) 6666600

(c) 6666666 (d) 6666606 10. You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is (a) 1/4 (b) 1/3 (c) 1/2 (d) 2/3

5.2

Permutations and Combinations & Probability

2013

2010

11. Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7? (a) 13/90

(b) 12/90

15. Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4 digit numbers greater than 3000 can be formed? (a) 50 (b) 51 (c) 52 (d) 54

(c) 78/90

(d) 77/90

NUMERICAL TYPE QUESTIONS

12. What is the chance that a leap year, selected at random, will contain 53 Saturdays? (a) 2/7

(b) 3/7

(c) 1/7

(d) 5/7

13. In a factory, two machines M1 and M2 manufacture 60% and 40% of the autocomponents respectively. Out of the total production, 2% of M1 and 3% of M2 are found to be defective. If a randomly drawn autocomponent from the combined lot is found defective, what is the probability that it was manufactured by M2? (a) 0.35

(b) 0.45

(c) 0.5

(d) 0.4

2012 14. A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that day is 1 1 (a) (b) 4 16 7 9 (c) (d) 16 16

2015 1. How many four digit numbers can be formed with the 10 digits 0, 1, 2, ..... 9 if no number can start with 0 and if repetitions are not allowed?

2014 2. In any given year, the probability of an earthquake greater than Magnitude 6 occurring in the Garhwal Himalayas is 0.04. The average time between successive occurrences of such earthquakes is _____ years. 3. 10% of the population in a town is HIV+. A new diagnostic kit for HIV detection is available; this kit correctly identifies HIV+ individuals 95% of the time, and HIV– individuals 89% of the time. A particular patient is tested using this kit and is found to be positive. The probability that the individual is actually positive is ______ 4. A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs. The probability that the current batch is accepted is 

ANSWERS MCQ Type Questions 1. (a)

2. (b)

3. (b)

4. (c)

5. (d)

11. (d)

12. (a)

13. (c)

14. (c)

15. (b)

6. (b)

Numerical Type Questions 1. 4536

2. (25 to 25)

3. (0.47 to 0.48)

4. (0.8145)

7. (a)

8. (d)

9. (b)

10. (b)

Permutations and Combinations & Probability

5.3

EXPLANATIONS MCQ TYPE QUESTIONS 1. The question clearly explains that shaquille makes 60 free throws out of 100

60  0.6 Hence, Probability of free throw = 100 And probability of NOT free throw = 1 – 0.6 = 0.4 So required probability of exactly 6 throws in 10 attempts will be given by 10

C6 0.66  0.44 = 0.2508

2. There are 4 kings in a pack of 52 cards. If 2 cards are selected and both are kings, remaining cards will be 50 out of which 2 will be kings.

9. The digit in unit place is selected in 4! Ways The digit in tens place is selected in 4! Ways The digit in hundreds place is selected in 4! Ways The digit in thousands place is selected in 4! Ways The digit in ten thousands place is selected in 4! Ways Sum of all values for 1 4!  1  (100 + 101 + 102 + 103 +104) = 4!  11111  1 Similarly for ‘3’ 4!  (11111)  3 Similarly for ‘5’ 4!  (11111) 5 Similarly for ‘7’ 4!  (11111)  7 Similarly for ‘9’ 4!  (11111)  9

 sum of all such numbers 4! (11111) (1+ 3 + 5 + 7 + 9) = 24  (11111)  25

3. For a match to be played, we need 2 teams

= 6666600

No. of matches = No. of ways of selections 2 teams out of 5 = 5 C2 = 10 5.

x = {HHT, HHH} y depends on x z = {TTH, TTT}

 ‘d’ is the correct choice. 6. P (Ram) = 1/6; p(Ramesh) = 1/8 p(only at) = p (Ram) × p (not ramesh) + p(Ramesh) × p(n0 × Ram) 7 1 5 = 1    6 8 8 6 12 1   4 40 7. Here the given sets are A = {2, 3, 4, 5} and B = {11, 12, 13, 14, 15} desired outcome = {(2, 14), (3, 13), (4, 12), (5, 11)}  n(E) = 4 and total outcome = 4  5 = 20  n(s) = 20  Probability of an event =

n(E) n(s)

=

4 1 = = 0.2 20 5

10. The probability that the other face is tail is

1 . 3

11. Given: All the 2-digit numbers To find: Probability of selecting a number at random not divisible by 7 Analysis: Total number of 2-digit numbers 10, 11, ....., 99  90 numbers Unfavourable numbers i.e. numbers divisible by 7 : 14, 21, ......., 98 Number of unfavourable numbers = 13 This can be hand counted or we can use the following properties: Last term= First time + (no. – 1) Diff 98 = 14 + (no. – 1) 7  no. = 13  Probability =

Total – unfavourable Total

=

90  13 = 77/90 90

Hence, the solution is (d) 12. at a leap year 52 weeks, and 2 extra day they are (Mon, Tues) (Tues, Wed) (Wed Thu.) (Thus. Fri) (Fri. Sat) (Sat. Sun) (Sun Mond.) n(s) = 7 n(E) = 2 P(E) =

2 7

5.4

Permutations and Combinations & Probability

14.

NUMERICAL TYPE QUESTIONS 1. In thousands place, 9 digits except 0 can be placed In hundreds place, 9 digits can be placed (including 0, excluding the one used in thousands place) In tens place, 8 digits can be placed (excluding the ones used in thousands and hundreds place)

OB is the line when both A and B arrive at same time. Total sample sapce = 60  60 = 3600 Favourable cases = Area of OABC – 2(Area of SRC)

1  = 3600 – 2    45  45  2  = 1575

1575 7  Required probability = = 3600 16 15. 2, 2, 3, 3, 4, 4, 4, 4

x > 3000

In ones place, 7 digits can be placed (excluding the one used in thousands, hundreds and tens place) Total number of combinations = 9  9  8  7 = 4536 2. Given, in one year, probability of an earthquake > 6 magnitude = 0.04 So, average time between successive earthquakes = 1/0.04 = 25 years 3. Given, 10% of the population is HIV + , so probability = 0.1 for HIV+ and 0.9 for HIV– The diagnostic kit identifies  0.95 HIV+ and 0.89 HIV– correctly

322 — 3, 4

344 — 2, 3, 4

323 — 2, 3, 4

422 — 3, 4

324 — 2, 3, 4

423 — 2, 3, 4

332 — 2, 3, 4

424 — 2, 3, 4

333 — 2, 4

432 — 2, 3, 4

95 100

334 — 2, 3, 4

433 — 2, 3, 4

 Probabilities that none of the bulbs is

342 — 2, 3, 4

434 — 2, 3, 4

343 — 2, 3, 4

442 — 2, 3, 4 443 — 2, 3, 4 444 — 2, 3, 4

So, the required probability that the person is actually HIV+ = 0.95/2 = 0.475 4. Probability for one bulb to be non-defective is

4

95  defectives   = 0.8145 100  

„„

Miscellaneous

6.1

6

Miscellaneous

CHAPTER MCQ TYPE QUESTIONS 2016 1. In a 2  4 rectangle grid shown below, each cell is a rectangle. How many rectangles can be observed in the grid?

6. Which of the following curves represents the

  sin x     function y = ln  e   for x < 2 ?     Here, x represents the abscissa and y represents the ordinate. (a)

(a) 21

(b) 27

(c) 30

(d) 36

2.

(b)

Choose the correct expression for f(x) given in the graph. (a) f(x) = 1 – |x – 1|

(c)

(b) f(x) = 1 + |x – 1|

(c) f(x) = 2 – |x – 1| (d) f(x) = 2 + |x – 1| 3. The volume of a sphere of diameter 1 unit is ________ than the volume of a cube of side 1 unit. (a) least (b) less (c) lesser (d) low 4. A window is made up of a square portion and an equilateral triangle portion above it. The base of the triangular portion coincides with the upper side of the square. If the perimeter of the window is 6 m, the area of the window in m2 is . (a) 1.43 (b) 2.06 (c) 2.68 (d) 2.88 5. The binary operation is defined as a b = ab + (a + b), where a and b are any two real numbers. The value of the identity element of this operation, defined as the number x such that a x = a, for any a, is .

(d)

7. Given (9 inches)1/2 = (0.25 yards)1/2, which one of the following statements is TRUE? (a) 3 inches = 0.5 yards (b) 9 inches = 1.5 yards

(a) 0

(b) 1

(c) 9 inches = 0.25 yards

(c) 2

(d) 10

(d) 81 inches = 0.0625 yards

6.2

Miscellaneous

8. The Venn diagram shows the preference of the student population for leisure activities. Read books

Watch TV 12

13

19

7 44

17 15

Play sports

From the data given, the number of students who like to read books or play sports is ____. (a) 44

(b) 51

(c) 79

(d) 108

9. A wire of length 340 mm is to be cut into two parts. One of the parts is to be made into a square and the other into a rectangle where sides are in the ratio of 1:2. What is the length of the side of the square (in mm) such that the combined area of the square and the rectangle is a MINIMUM? (a) 30

(b) 40

(c) 120

(d) 180

10. Find the area bounded by the lines 3x + 2y = 14, 2x – 3y = 5 in the first quadrant (a) 14.95

(b) 15.25

(c) 15.70

(d) 20.35

11. A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x = 0.1 and has a slope of –0.02. What is the value of y at x = 5 from the fit?

17. Based on the given statements, select the most appropriate option to solve the given question. What will be the total weight of 10 poles each of same weight? Statements: (I) One fourth of the weight of a pole is 5 kg (II) The total weight of these poles is 160 kg more than the total weight of two poles. (a) Statement II alone is not sufficient (b) Statement II alone is not sufficient (c) Either I or II alone is sufficient (d) Both statements I and II together are not sufficient. 18. Consider a function f(x) = 1 – x on –1 x 1. The value of x at which the function attains a maximum, and the maximum value of function are: (a) 0, –1 (b) –1, 0 (c) 0, 1 (d) –1, 2 19. In a triangle PQR, PS is the angle bisector of QPR and QPS = 60. What is the length of PS? (a)

q  r

P

qr qr (b)  q  r 

(c) (d)

q

2

 r2

q  r

r

q

 s

2

Q

R

(i) ln x > ln y

(ii) ex > ey

qr p 20. Based on the given statements, select the most appropriate option to solve the given question. If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building? Statements: I. Each step is ¾ foot high. II. Each step is 1 foot wide. (a) Statement I alone is sufficient, but statement II alone is not sufficient. (b) Statement II alone is sufficient, but statementI alone is not sufficient. (c) Both statements together are sufficient, but neither statement alone is sufficient. (d) Statement I and II together are not sufficient.

(iii) yx > xy

(iv) cos x > cos y

2014

(a) –0.030 (b) –0.014

(c) 0.014

(d) 0.030

12. A square pyramid has a base perimeter x, and the slant height is half of the perimeter. What is the lateral surface area of the pyramid? (a) x2

(b) 0.75 x2 (c) 0.50 x2 (d) 0.25 x2

2015 1 5 13. If log x     , then the value of x is 3 7

125 25 49 343 (b) (c)  (d)  343 49 25 125 14. A function f(x) is linear and has a value of 29 at x = – 2 and 39 at x = 3. Find its value at x = 5. (a)

(a) 59

(b) 45

(c) 43

(d) 35

15. If x > y > I, which of the following must be true?

(a) i and ii

(b) i and iii

(c) iii and iv

(d) ii and iv

16. log tan1+ log tan2 +------+ log tan89 is ____. (a) 1

(b) 1

2

(c) 0

(d) –1

21. If y = 5x2 + 3, then the tangent at x = 0, y = 3 (a) passes through x = 0, y = 0 (b) has a slope of +1 (c) is parallel to the x-axis (d) has a slope of –1

Miscellaneous

6.3

22. Let f(x, y) = xnym = P. If x is doubled and y is halved, the new value of f is (a) 2n–m P (b) 2m–n P (c) 2(n – m) P (d) 2(m – n) P 23. A regular die has six sides with numbers 1 to 6 marked on its sides. If a very large number of throws show the following frequencies of occurrence: 1  0.167; 2  0.167; 3  0.152; 4  0.166; 5  0.168; 6  0.180. We call this die (a) irregular (b) biased (c) Gaussian (d) insufficient 24. Consider the equation : (7526)8 – (Y)8 = (4364)8, where (X)N stands for X to the base N. Find Y. (a) 1634 (b) 1737 (c) 3142 (d) 3162 25. At what time between 6 a.m. and 7 a.m. will the minute hand and hour hand of a clock make an angle closest to 60°? (a) 6 : 22 a.m. (b) 6 : 27 a.m. (c) 6 : 38 a.m. (d) 6 : 45 a.m. 26. The roots of ax2 + bx + c = 0 are real and positive a, b and c are real. Then ax2 + b|x| + c = 0 has (a) No roots (b) 2 real roots (c) 3 real roots (d) 4 real roots

2012 27. The cost function for a product in a firm is given by 5q2, where q is the amount of production. The firm can sell the product at a market price of ` 50 per unit. The number of units to be produced by the firm such that the profit is maximized is (a) 5

(b) 10

(c) 15

(d) 25

28. Which of the following assertions are CORRECT ? P : Adding 7 to each entry in a list adds 7 to the mean of the list Q : Adding 7 to each entry in a list adds 7 to the standard deviation of the list R : Doubling each entry in a list doubles the mean of the list S : Doubling each entry in a list leaves the standard deviation of the list unchanged (a) P, Q (b) Q, R (c) P, R (d) R, S 29. A political party orders an arch for the entrance to the ground in which the annual convention is being held. The profile of the arch follows the equation y = 2x – 0.1x2 where y is the height of the arch in meters. The maximum possible height of the arch is (a) 8 meters (b) 10 meters (c) 12 meters

(d) 14 meters

30. There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum

number of weighings required to identify the heavier bag is (a) 2 (b) 3 (c) 4 (d) 8

2011 31. If log (P) = (1/2) Log (Q) = (1/3) Log (R), then which of the following options is TRUE? (a) P2 = Q3 R2 (b) Q2 = PR (c) Q2 = R3 P (d) R = P2 Q2 32. The variable cost (V) of manufacturing a product varies according to the equation V = 4q, where q is the quanity produced. The fixed cost (F) of production of same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V + F)? (a) 5 (b) 4 (c) 7 (d) 6 33. Three friends R, S and T shared toffee from a bowl. R took 1/3rd of the toffees, but returned four to the bowl. S took 1/4th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally there in the bowl? (a) 38 (b) 31 (c) 48 (d) 41

NUMERICAL TYPE QUESTIONS 2015 1. In the given figure angle Q is a right angle, PS : QS = 3:1, RT : QT = 5:2 and PU : UR = 1:1. If area of triangle QTS is 20 cm2, then the area of triangle PQR in cm2 is ______.

2. If p, q, r, s are distinct integers such that: f (p, q, r, s) = max (p, q, r, s) g (p, q, r, s) = min (p, q, r, s) h (p, q, r, s) = remainder of (p  q)/(r  s) if (p  q) > (r  s) or remainder of (r  s)/(p  q) if (r  s) > (p  q) Also a function fgh (p, q, r, s) = f(p, q, r, s)  g(p, q, r, s)  h(p, q, r, s) Also the same operations are valid with two variable function of the form f(p, q). What is the value of fg (h(2, 5, 7, 3), 4, 6, 8

2014 3. When a point inside of a tetrahedron (a solid with four triangular surfaces) is connected by straight lines to its corners, how many (new) internal planes are created with these lines? _________.

6.4

Miscellaneous

ANSWERS MCQ Type Questions 1. (c) 2. (c) 3. (c) 11. (a) 12. (d) 13. (a) 21. (c) 22. (a) 23. (b) 31. (b) 32. (a) 33. (c) Numerical Type Questions 1. 280 2. (8) 3. (6)

4. (b) 14. (c) 24. (c)

5. (a) 15. (a) 25. (a)

6. (c) 16. (c) 26. (d)

7. (c) 17. (c) 27. (a)

8. (d) 18. (c) 28. (c)

9. (b) 19. (b) 29. (b)

10. (b) 20. (a) 30. (a)

EXPLANATIONS MCQ TYPE QUESTIONS 1. In the given 2  4 rectangle grid, the following type of rectangles are present. One figured rectangles = 8 Two figured rectangles = 10 Three figured rectangles = 4 Four figured rectangles = 5 Six figured rectangles = 2 Eight figured rectangles = 1  Total number of rectangles = 8 + 10 + 4 + 5 + 2 + 1 = 30 So the number of rectangles observed in the given grid is 30. 2. From the given graph function f(x) must be equals to zero at x = 3. From Option A : f(x) = 1 – |x – 1| at x =3 f(x) = 1 – |3 – 1| = 1 – 2 = – 1 So, it is false From Option B : f(x) = 1 + |x – 1| at x =3 f(x) = 1 + |3 – 1| = 1 + 2 = 3 So, it is false From Option C : f(x) = 2 – |x – 1| at x =3 f(x) = 2 – |3 – 1| = 2 – 2 = 0 So, it is true. From Option D : f(x) = 2 + |x – 1| at x =3 f(x) = 2 – |3 – 1| = 2 + 2 = 4 So, it is false. Hence the correct expression for f(x) given in the graph is 2 – |x – 1|. 3. ‘Lesser than’ is apt because the sentence should be in comparative degree.

4. The possible window will be as follows :

Perimeter of window = 5x = 6m x = 1.2 m So area of window = Area of triangle + Area of square

3 2 x  x2 4 3 (1.2)2  (1.2)2 = 2.06 m2 = 4 5. The binary operation is defined  a b = ab + (a + b) a x=a  From the equation ‘b’ is the variable Option A: x = 0 a o = a  0 + (a + 0) = 0 + a = a Option B: x = 1 a 1 = a  1 + (a + 1) = a + a + 1 = 2a + 1 Option C: x = 2 a 2  a  2 + (a + 2) = 2a + a + 2 = 3a+ 2 Option D: x = 10 a 10  a  10 + (a + 10) = 10a + a + 10 = 11a + 10  Option ‘A’ only True. 7. (9 inches)1/2 = (0.25 yards)1/2, Solving we get 9 inch = 0.25 yards (since 1 inch = 0.028 yard) 8. Given Venn diagram is =

Read books

Watch TV 12

13

7 44

17 15

Play sports

Miscellaneous

6.5

The number of students who like to read books or play sports = 13 + 12 + 44 + 7 + 15 + 17 = 108  14  A =  ,0   3 

10.

B = [0, 7], 5  C =  ,0  2   5  D = 0,   3 

E = [4, 1],

18. The given function is f (x) = 1 – x on –1  x  1 In order to find maximum and minimum value of function in interval [–1, 1] At x = –1, f (–1) = 1– –1 = 1–1 = 0 At x = –0.5, f (–0.5) = 1– –0.5 =1–0.5 = 0.5 At x = 0, f (0) = 1– 0 = 1 At x = 0.5, f (0.5) = 1– 0.5 = 0.5 At x = 1, f (l) = 1– 1 = 1–1 = 0  Hence, maximum value occurs at x = 0 and its maximum value is 1.  Option (c) is correct. 21. y = 5x2 + 3,

F = [0, 1]

 dy  = 10  0 = 0 Slope of tangent =    dx  x  0, y  3 Slope = 0  tangent is parallel to x-axis.

Required area = Area of  BFE + Area of quadrilateral FEOC

1 1  4  6   (4  2.5)  1 2 2 1 = 2  6   6.5 2 = 12 + 3.25 = 15.25 sq. units So, the area bounded by the given lines is 15.25 sq. units. 11. The equation of a straight line is y = mx + c m = slope = – 0.02 set (log x, y) Now, the line intercepts the abscissa at log x = 0.1 So, at abscissa y = 0  y = mx + c 0 = – 0.02  0.1 + c  c = 0.002 y = mx + c y = – 0.02  1og x + c At x =5  y = – 0.02  log 5 + 0.002 = – 0.030 So, the value of y is – 0.030. =

3

7 5 7 343 13. = x–1/3  = x1/3    = x x = 7 125 5 5 14. f(x) = 2x + 33 = 2 (5) + 33 = 43 15. For whole numbers, greater the value greater will be its log. Same logic for power of e. 16. log tan 1 + log tan 89 = log(tan 1 tan 89) = log(tan 1 cot 1) = log 1 = 0 Using the same logic total sum is ‘0’.

dy = 10x dx

m

1 P  = 2n X n   y m 2

22.

= 2n–m XnYm = 2n–m P 23. For a very large number of throws, the frequency should be same for unbiased throw. As it not same, then the die is baised. 24. (7526)8 – (y)8 = (4364)8  y8 = (7526)8 – (4364)8

7 4

4 5 3

(8 + 2=10)

2 6

6 4

3 1 4 2 When we have base 8, we borrow 8 instead of 10 as done in normal subtraction. 12 25. 11 1 10

2

9

3 4

8

5 6 From option At 6 a.m. Both hand – 180 aparts, At 6 : 20 a.m. – both hand apart  180 + 10 – 120 = 70 At 6 : 22 a.m. – both hand apart  180 + 12 – 132 = 60 (closest) 7

6.6

Miscellaneous 2

26. ax + bx + c = 0 Roots are real and positive  b2 – 4ac > 0 By the above condition we get two positive real roots Now, ax2 + b|x| + c = 0 This can be written as; ax2 + bx + c  = b2 – 4ac > 0 Now for ax2 – bx + c (–b)2 – 4ac  b2 – 4ac 2 Again b – 4ac > 0 So again we get real roots Thus we have 4 real roots of ax2 + b|x| + c = 0 27. P = 50q – 5q2

dp d2 p = 50 – 10q; C2, then C1 If A1 = A2, then A3 If C1 < C2, then C2 If A1 > A2, then A1 If A1 < A2, then A2 31.

log P = =

1 log Q 2 1 log (R) = k 3

 P = bk, Q = b2k, R = b3k Hence, Q2 = b4k = b3k bk = PR

32. Checking with all options in formula: (4q+100/q) i.e. (V+F). Option (a) gives the minimum cost. 33. Let total number of toffees in bowl be x 1 of toffees and returned 4 to the bowl 3 1  Number of toffees with R = x – 4 3 2 x+4 Remaining of toffees in bowl = 3 12  Number of toffees with S =  x  4  – 3 43  32  Remaining toffees in bowl =  x  4   4 43   132  Number of toffees with T =   x  4   4   2 2 43   Remaining toffees in bowl  1 3  2  =   x  4  4  + 2  2 4  3 

R took

 1 3  2    x  4  4  + 2 = 17 2 4  3  32     x  4 = 27 4 3

Given :



x = 48

NUMERICAL TYPE QUESTIONS 1. Let area of triangle PQR be ‘A’

1 SQ 1 =  4 PQ 1  3 2 QT 2 =  7 QR 2  5  Area of QTS = 1  SQ  QT 2 =

1 1  2    PQ    QR  2 4  7 

=

1 2 1      PQ  QR  4 7 2 

=

1  Area of PQR 14

1 A 14  A = 14 20 = 280 cm2 2. fg(h(2, 5, 7, 3), 4, 6, 8) = fg(1, 4, 6, 8) = f(1, 4, 6, 8)  g(1, 4, 6, 8) =81=8 Given 20 cm2 =

„„

Engineering Mathematics

Linear Algebra

1

1.1

Linear Algebra

C HAPTER 2016 1. The positive Eigen value of the following matrix is_________. 2 1  5 2    2. The value of determinant A given below is __________ .

 5 16 81    0 2 2 A=   0 0 16   

7. The solution for the following set of equations is, 5x + 4y + 10z = 13 x + 3y + z = 7 4x – 2y + z = 0 (a) x = 2, y = 1, z = 1 (b) x = 1, y = 2, z = 0 (c) x = 1, y = 0, z = 2 (d) x = 0, y = 1, z = 2 8. The solution to the following set of equations is 2x + 3y = 4

2015

4x + 6y = 0

3. 2x1 + x2 = 3 5x1 + bx2 = 7.5

(a) x = 0, y = 0

The system of linear equations in two variables shown above will have infinite solutions, if and only if, b is equal to _____.

(c) 4x = – 6y

 4 2 4. If A = 1 3 , then A2 + 3A will be   30 20  (a)   10 20 

28 10  (b)    4 18 

31 13  (c)    7 21

20 10  (d)    5 15 

5. What are the eigenvalues of the following matrix?  1 1  2 4   

(a) 2 and 3 (b) –2 and 3 (c) 2 and –3 (d) –2 and –3

2014 1 4  6. The eigenvalues of A = 2 3 are   (a) 2 ± i

(b) x = 2, y = 0 (d) No solution

2013 1 1  2 1  3 0  9. If P = 2 2 ,Q= 2 2 and R= 1 3 which one      

of the following statements is TRUE? (a) PQ = PR (b) QR= RP (c) QP = RP (d) PQ = QR 10 4  10. One of the eigen values of P = 18 12 is  

(a) 2

(b) 4

(c) 6

(d) 8

11. The solution of the following set of equations is x + 2y + 3z = 20 7x + 3y + z = 13 x + 6y + 2z = 0 (a) x = – 2, y = 2, z = 8

(b) –1, –2

(b) x = 2, y = –3, z = 8

(c) –1 ± 2i

(c) x = 2, y = 3, z = –8

(d) non-existent

(d) x = 8, y = 2, z = –3

Linear Algebra

1.2

2012

2011

12. What is the rank of the following matrix?

(a) 0

 5 3 1     6 2 4   14 10 0    (b) 1

(c) 2

(d) 3

13. Value of the determinant mentioned below is 1 0 1 0 4 7 0 2 1 1 1 1 2 0 2 1

(a) 24

(b) –30

(c) –24

(d) –10

ANSWERS 1. (3.0 : 3.0) 2. (160.0 : 160.0)

3. (2.5)

4. (a)

5. (a)

8. (d)

11. (b)

12. (c)

13. (a)

9. (a)

10. (c)

6. (c)

7. (b)

EXPLANATIONS 1.

2 1  A= ;I 5 2   |A – I| = 0

1 0  0 1  ; I   

 0  0    

1  2   A – I =  2     5 = – (2 – ) (2 + ) – 5 = –(4 – 2) – 5 = 2 – 4 – 5 = 2 – 9  |A – I| = 0 = 2 – 9 = 0 2 = 9   = ± 3 & +ve Eigen Value = + 3 2. Determinent A.



 5 16 81  A =  0 2 2   0 0 16    = 5(32 – 0) – 16 (0 – 0) + 81 (0 – 0) = 160. 3. The two equations gives the same line then we have infinite solution. 4.

4 2   4 2  A2 =    1 3   1 3   4  4  2  1 4  2  2 3  =   1  4  3  1 1  2  3  3  18 14  =  ,  7 11 12 6  3A =    3 9 18  12 14  6  A2 + 3A =    7  3 11  9 

30 20  =   10 20 

1   1 1   0  1 –  5. A –  = 0    – 0   =  –2 4 –   –2 4        (1 –) (4 –) – (–2) (1) = 0  2 – 5 + 6 = 0  2 – 3 – 2 + 6 = 0  ( – 2) (– 3) = 0

  = 2,  = 3 6. For Eigen value A –I = 0 1 A=  2  1 Now A –I =  2

Here

Now      

4  3  4  4    0  1    = 3    3   0    2 A – I = 0

4  1   =0  2 3     – (1 – ) (3 + ) + 8 = 0 (1 – ) (3 + ) – 8 = 0 3 – 2 –2 – 8 = 0 2 + 2+ 5 = 0

= =

2  4  4(1)(5) 2

2  4  20 2

2  4 i 2  16i2 = 2 2 = –1  2i =

1 4  Hence the Eigen value of A =   is  = –1  2i  2 3 

Linear Algebra

7.

8.

5x + 4y + 10z = 13 (i) x + 3y + z = 7 (ii) 4x – 2y + z = 0 (iii) From option x = 2, y = 1 and z = 1, doesnot satisfy equation (i) Hence option (i) is wrong From option (b) x = 1, y = 2, z = 0 For equation (i) 5x + 4y + 10z = 5(1) + 4 (2) + 0 = 13 Hence equation (i) is satisfied For equation (ii) x + 3y + z = (1) + 3 (2) + 0 = 7 Hence equation (ii) is satisfied For equation (iii) 4x – 2y + z = 4(1) – 2 (2) + 0 = 0 Hence all the three equations are satisfied for the given value of x, y and z Hence option (b) is correct. 2x + 3y = 4

(i)

4x + 6y = 0

(ii)

From option (a) Equation (i) is not satisfied From option (b) Equation (ii) is not satisfied From option (c) Equation (i) is not satisfied Also the given lines are parallel and hence they donot intersect Hence the given equation has no solution. 1 1  2 1  3 0  9. Given, P =  Q=  and R =     2 2  2 2  1 3  4 3  1 1   2 1  PQ=    =    8 6  2 2  2 2  1 1   3 0   4 3  P R =    =  2 2  1 3  8 6  Hence PQ =PR

1.3

10 – 4  10. Given, P =   18 –12  For a 2  2 matrix, the Eigen values are:

λ = ½ {(a11 + a22) +

4 a12 a21   a11 – a22  2 }

Therefore,  for the given matrix



=12 (10 – 12)  4 

 – 4)

 18 + 10 + 12  2 



= ½ {(– 2)  –288 + 484 } = ½ {(– 2) + 14} = 12/2 = 6 11. Given, x + 2y + 3z = 20 7x + 3y + z = 13 and x + 6y + 2z = 0 Dividing equation (iii) by 2 x  + 3y + z = 0 2 Equation (ii)

...(i) ...(ii) ...(iii)

 7x + 3y + z = 13 (ii) – (iii)

13 x = 13 2 Or x=2 Putting x = 2 in equation (i)  3  6y + 9z = 54 Putting x = 2 in equation (iii)  6y + 2z = – 2 Equation (iv) – (v)  0 + 7z = 56 or z = 8 Putting x = 2 and z = 8 in (iii) we have 2 + 6y + 16 = 0  6y = –18  y = – 3 

...(iv) ...(v)

x =2; y = – 3; and z = 8 12. A = 5(0 + 40) – 3(0 + 56) –1(60 – 28) = 200 – 200 = 0 But

5 3 6 2

0

Hence rank is 2 13. The value of determinant is 24.

„„

Calculus

2

Calculus

CHAPTER

6. The graph of the function

2016 1. The value of the integral 0.9

dx

 (1  x)(2  x) 0

F(x) =

x for 0 < x <  is k1 x 2  k2 x  1

is ____________.

(a)

2015

(b)

F(x)

F(x)

n

u  2. The limit of the function  1   as n  is  n 1/k10.5

(a) ln x (b) ln

2.1

1 x

(c)

(k1/k2)0.5

x

(d)

F(x)

F(x)

(c) e–x (d) ex

2014

(k1/k2)0.5

3. The limit of the function e

–2t

4. If y = xx, then

0.5

k2

sin (t) as t   , is ___

2013

dy is dx

7. If u = log (ex + ey), then

(a) xx(x – 1) (b) xx–1

(a) ex + ey

(c) xx(1 + log x)

(b) ex – ey

(d) ex(1 + log x) 5. Which of the following statements is true for the series given below? sn = 1 

x

1 2



1 3



(a) sn converges to log (n)

1 4

 ... 

1 n

(c)

1 e  ey x

(d) 1

x tan 8. Evaluate lim x 

1 x

(b) sn converges to n

(a) 

(b) 1

(c) sn converges to exp(n)

(c) 0

(d) -1

(d) sn diverges

y u  = x y

x

2.2

Calculus

ANSWERS 1. (1.65 : 1.75)

3. (0 to 0)

2. (d)

4. (c)

5. (d)

6. (a)

7. (d)

8. (b)

EXPLANATIONS n

x  2.  1   as n  n 



n  n – 1 x2 n x  1   ........  x +   n n  2i n2  n

As ncoefficient in n = 1 =1+x+

x 2 x3   ...... = ex 2! 3!

lim e2 t sin t = lim sin t t  t  l2 t

3.

sin t 1  2t  = lim   lim 2  t  t  t l t 2

sin t    0 = 0  lim t  t   4.

y=x

5.

1 dy x =  log x y dx x



dy = y(1 + log x) dx



dy = xx(1 + log x) [ y = xx] dx Sn = 1  1  1  1  .... 1 2 3 4 n

The given series is divergent 6. The graph of the function f(x) =

x k1 x  k2 x  1 2

For O < x < 

f(x)

x

Now taking log on both the side – log y = log xx 

log y = x log x

Now differentiating w.r.t to x

d d 1 dy log x  log x x = x dx dx y dx

(¼) 0.5

x

Calculus x

y

7. Given, u = log (e + e ) Now,

=

Let x =1/n; So for x α, n 0

 

 log e x  e y

=



x



=

8. Lim(x tan1/x), x  α

u u + y x

a ex  e y

=

 e

x e  ey x

ex x

 ey

ex  e y

e

x

 ey

y

 ex  e y +



 

 log e x  e y

+





+

 e 

2.3

=1

e

y x

ey x

 ey

 ey









Now, lim(1/n tan n) = lim(Sin n/n Cos n) = lim{(Sin n)/n}/lim(Cos n) We know lim(Sin n)/n, n0 = 1 and lim(Cos n), n  0 = 1 since Cos 0 = 1 Therefore, lim{(Sin n)/n}/lim(Cos n) = 1

„„

Differential Equations

3

Differential Equations

CHAPTER 2016

 x2  exp    2Dt  2Dt  c0

c( x, t) 

1. The Laplace transform F(s) of the function f(t) = cos (at), where a is constant, is __________. (a)

where D and c0 are assumed to be constant. Which of the following is correct?

s2 s2  a 2

a (b) 2 s  a2

(a)

c 2 c D 2 t x

(c)

s s  a2

(b)

c D  2 c  t 2 x2

(d)

s s  a2

(c)

2 c 2 c  D t 2 x2

(d)

2 c D 2 c  2 x2 t2

2

2

d2 y  y  0. The initial conditions for this second dx 2 order homogeneous differential equation are y(0) dy  3 at x = 0 = 1 and dx The value of y when x = 2 is ___________. 3. Consider the equation 2.

V=

aS bS

S2 c

Given a = 4, b = 1 and c = 9, the positive value of S at which V is maximum, will be _____.

2014 4. The concentration profile of a chemical at a location x and time t, denoted by c(x, t), changes as per the following equation,

2013 5. The Laplace transform of f(t) = 2t + 6 is 1 2  s s2 6 2 (c)  2 s s

3 6  s s2 6 2 (d)   2 s s

(b)

(a)

dy + y cot x = cosec x is dx (a) y = (c + x) cot x (b) y = (c + x) cosec x (c) y = (c + x) cosec x cot x

6. The solution to

(d) y = (c + x)

cosec x cot x

ANSWERS 1. (c)

3.1

2. (14.55 : 14.75)

3. (3.0 : 3.0)

4. (b)

5. (c)

6. (b)

Differential Equations

3.2

EXPLANATIONS 1. The Laplace Transfer 

F(s) = here

e

 st

Now, c1 & c2 find by given conditions.

F(t) dt



0

F(t) = cos (at) 



F(s) =

e

 st

3 = c 1 e 0 – c2 e 0 = c1 – c 2



c 1 – c2 = 3 c1 = 2, c2 = – 1



A B  (1  x) (2  x)

y = 2e x  e  x

Now, value of y when x = 2 is y = 2e2 – e–2 = 14.643 ...(i)

3.

V=

To find A multiply equation (i) with (1 – x) & put (1 – x) = 0.



To find B multiply equation (i) with (2 – x) & put (2 – x) = 0.





0

 V=

dx = (1  x)(2  x)

0.9



1

1

0.9

0

dx

0.9

dx

 1



0

0

0.9

0

0.9

1    log 2  x   1 0

1 1    log ax  b  c    ax  b a   =   log 0.1  log1    log1.1  log 2 = [2.303 + 0] – [– 0.0953 + 0.693] = 1.705. 2

d y  y  0, has the solution of type dx 2 m x m x y = c1 e 1 + c 2 e 2

Now, m1 & m2 find by m2 –1 = 0, m = ± 1, m1 = 1, m2 = – 1



36S , S  9S  9 2



  (1  x) dx  (2  x) dx 

 (1  x)   (2  x)   1 log 1  x 

y = c1ex + c2e–x

c

to become Vmax,

dV d2 V 0& 0 dS dS2 Now,

2.

b  S  S2

B= –1 0.9

=

aS

a  4, b  1,c  9.

A= 1



...(ii)

 From equation (i) & (ii)

1 A B  = (1  x)(2  x) (1  x) (2  x)



...(i)

dy = 3 at x = 0 dx

cos at dt

0

s F(s) = 2 s  a2 dx

c1 + c 2 = 1

For

Solving by integration by parts

 (1  x)(2  x) =

0 o y(0) = 1 = c1 e  c 2 e  c1  c 2

For

=

dV =0 dS

(S2  9S  9)(36)  36S(2S  9) 36S2  324  2 2 2 (S  9s  9) (S  9S  9)2

i.e. S2 + 9S + 9 – 2S2 – 9S = 0



– S2 + 9 = 0



S = ± 3.

Now,

d 2 V (S2  9S  9)2 (72S)  (36S2  324)(2)(2S  9) = (S2  95  9)4 dS2 for

S = + 3,

d2 V d 2S  0.1077  0, for S  3, 2  2.667  0 2 dS dS So, V is maximum when S = + 3.

Differential Equations

4. Given, concentration profile of a chemical, c (x, t) =

c0 2Dt

e

5. Given f(t) = 2t + 6 Applying the Laplace transform, we have

 x2     2Dt 

L{f(t)} = L(2t) + L(6) L(2t) = 2/s2

Now, change in concentration w.r.t. time is given by

  t 1  2 

c0 c t = t 2D

3.3

L(6) = 6/s Therefore the Laplace transform is 6/s + 2/s2 6. dy/dx + y Cotx = Cosec x

  e t  

 x2     2Dt 

   

To find the solution of the differential equation

 dy/dx = Cosec x – y Cotx  dy = Cosecx.dx – yCotx.dx 

x2 

c0 t    x 2    2Dt  c    e = 1  2 t 2D  t  2Dt  



 dy =   Cosecx – yCotx  .dx

 y = log(Cosecx – Cotx) – y log(Sinx) + C  y = log(1 – Cosx) – logSinx – y log(Sinx) + C

c  ^ 3c = D 2 t t ^ 2

 y = (x + c) Cosecx

„„

Probability and Statistics

4

4.1

Probability and Statistics

C HAPTER 2016

2012

1. Runs scored by a batsman in five one-day matches are 55, 75, 67, 88 and 15. The standard deviation is __________.

2015 2. Plasmid DNA (0.5 g) containing an ampicillin resistance marker was added to 200 l of competent cells. The transformed competent cells were diluted 10,000 times, out of which 50 l was plated on agar plates containing ampicillin. A total of 35 colonies were obtained. The transformation efficiency is _____ 106 cfu. g–1.

2014 3. If an unbiased coin is tossed 10 times, the probability that all outcomes are same will be ___  10–5

4. Consider the data set 14, 18, 14, 14, 10, 29, 33, 31, 25. If you add 20 to each of the values, then (a) both mean and variance change (b) both mean and variance are unchanged (c) the rnean is unchanged, variance changes (d) the mean changes, the variance is unchanged

2010 Statement for Linked Answer Questions (5 – 6): During bioconversion of sucrose to citric acid by Aspergillus niger final samples of 6 batches of fermentation broth were analyzed for citric acid content. The results (in g/L) were found to be 47.3, 52.2, 49.2, 52.4, 49.1 and 46.3. 5. The mean value of acid concentration will be (a) 49.4 (b) 51.0 (c) 48.2 (d) 50.8 6. The standard deviation for the above results is (a) 2.49 (b) 3.0 (c) 1.84 (d) 5.91

ANSWERS 1. (24.5 : 28.5)

3. (191 to 199)

2. (2.8)

4. (d)

5.(a)

6.(a)

EXPLANATIONS 1.

SD or  =

N

Variance =

2. DNA = 0.5 g

Variance

 (x – x ) 1

i=1

N

N = No. of samples = 5

x = average of run scored =

 =

 x / N  300 / 5  60 i

(55  60)2  (75  60)2  (67  60)2  (88  60)2  (15  60)2 5

25  225  49  784  2025   = 5  = 621.6  24.93

50 = 2.5  10–5 200  10000 Number of colonies TE = Dilution  Amount of DNA 35 = = 2.8  106 2.5 10 5  0.5

Dilution =

2

3108 5

3. An unbiased coin is tossed 10 times Hence probability of all outcomes to see same is 199  10–5 (approx) 4. If each observation is either increased, decreased, divided or multiplied by a constant, then the mean so obtained also increases, decreases, gets multiplied or gets divided respectively by the same constant.

4.2

Probability and Statistics

There will be no change in variance, if each observation is increased, decreased, divided or multiplied by a constant. 5. We have, for actual mean, in an individual series Mean, y = ( ∑ y)/n where, y = value of samples, n = number of samples Therefore,

∑ y = 47.3 + 52.2 + 49.2 + 52.4 + 49.1 + 46.3 = 296.5 Given, n = 6 Therefore, = ( ∑ y)/n = 296.5/6 = 49.42 Therefore, the mean value of acid concentration is 49.42 g/L

6. We know, for ungrouped data Standard deviation,σ = Where, y = value of samples, n = number of samples Given, y = ( ∑ y)/n = 49.4 Y

Y2

47.3 52.2 49.2 52.4 49.1 46.3

2237.29 2724.84 2420.64 2745.76 2410.81 2143.69

∑ y = 296.5

∑ y2 = 14683.036

Therefore, σ = =

14683.03 − (49.4 * 49.4) 6 2447.172 − 2440.36

=

6.182 = 2.49 „„

Technical Section

1

Biomolecules : Constituents of Life

C HAPTER 2016

2014

1. Disaccharide molecules that contain  (1  4) glycosidic linkage are

6. The 4-amino or 4-keto group of pyrimidine bases is located in the

(a) sucrose and maltose

(a) major groove of the double stranded DNA

(b) sucrose and isomaltose

(b) minor groove of the double stranded DNA

(c) maltose and isomaltose

(c) minor groove of the B form DNA but not the A form DNA

(d) lactose and cellobiose

2015 2. Which one of the following amino acids has the highest probability to be found on the surface of a typical globular protein in aqueous environment? (a) Ala

(b) Val

(c) Arg

(d) Ile

3. Levinthal's paradox is related to (a) protein secretion

(d) major groove of the B form DNA but not the A form DNA 7. Amino acid residue which is most likely to be found in the interior of water-soluble globular proteins is (a) Threonine

(b) Aspartic acid

(c) Valine

(d) Histidine

8. If a plant is shifted to cold temperature, which of the following changes would take place in its membrane?

(b) protein degradation (c) protein folding (d) protein trafficking 4. How many different protein sequences of 100 residues can be generated using 20 standard amino acids ? (a) 10020

(b) 100  20

(c) 20100

(d) 100!  20!

5. Match the reagents in Group I with their preferred cleavage sites in Group II. Group I P. Cyanogen bromide Q. o-Iodosobenzoate R. Hydroxylamine S. 2-Nitro-5-thiocyanobenzoate Group II

(a) Ratio of unsaturated to saturated fatty acids would increase (b) Ratio of unsaturated to saturated fatty acids would decrease (c) Absolute amount of both fatty acids would increase keeping the ratio same (d) Absolute amount of both fatty acids would remain unchanged 9. Which of the following statements with respect to the orientation of the nitrogenous bases to the pentose sugars, and the puckering of the sugar, is correct? (a) Anti, and 2’-endo in A form DNA (b) Anti, and 2’-endo in B form DNA

1. Carboxyl side of methionine

(c) Syn, and 3’-endo in A form DNA

2. Amino side of methionine

(d) Syn, and 3’ -endo in B form DNA

3. Carboxyl side of tryptophan

2013

4. Amino side of cysteine

10. Which one of the following aminoacids in proteins does NOT undergo phosphorylation?

5. Asparagine-glycine bonds (a) P-1, Q-3, R-5, S-4

(a) Ser

(b) P-2, Q-3, R-1, S-4

(b) Thr

(c) P-1, Q-2, R-5, S-4

(c) Pro

(d) P-4, Q-2, R-5, S-3

(d) Tyr

1.2 Biomolecules : Constituents of Life

2012

(a) P-2, Q-4, R-1, S-3

11. The molarity of water in a water : ethanol mixture (15 : 85, v/v) is approximately

(b) P-4, Q-1, R-2, S-3

(a) 0.85

(c) P-3, Q-1, R-2, S-4 (d) P-4, Q-2, R-1, S-3

(b) 5.55

Common Data for Question (14 - 15) :

(c) 8.5

The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right handed helix.

(d) 55.5

2011 12. The protein in eukaryotes which is subjected to degradation undergoes (a) phosphorylation

14. The number of h-helical turns permeating the membrane is (a) 5.6 turns (b) 3.5 turns

(b) carboxylation

(c) 6.5 turns

(c) ubiquitation

(d) 5.0 turns

(d) methylation

15. The number of amino acid residues present in the protein is

2010 13. Match the chemicals in Group I with the possible type/class in Group II Group I

Group II

P. Picloram

1. Vitamin

Q. Zeatin

2. Auxin

R. Thiamine

3. Amino Acid

S. Glutamine

4. Cytokinin

(a) 15 (b) 18 (c) 17 (d) 20

ANSWERS 1. (d)

2. (c)

3. (c)

4. (c)

5. (a)

11. (c)

12. (c)

13. (a)

14. (a)

15. (d)

6. (a)

7. (c)

8. (a)

9. (b)

10. (c)

Biomolecules : Constituents of Life 1.3

EXPLANATIONS 1. Glycosidic bond or glycosidic linkage is a type of covalent bond that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate. Lactose (  -Dgalactopyranosyl (1 4) D-glucopyranose) and cellobiose (  -D-glucopyranosyl (1  4) D-glucopyranose) are disaccharides whose monomeric units are connected by  (1  4) glycosidic bonds. 2. Arginine, lysine aspartic, glutamic and histidine can present in aqueous environment. 3. Finding the native folded state of protein. 100

4. 20 different protein sequences of 100 residues can be generated using 20 standard amino acids. 5.

Group I

Group II

P. Cyanogen bromide 1. Carboxyl side of methionine Q. o-Iodosobenzoate

3. Carboxyl side of tryptophan

R. Hydroxylamine

5. Asparagine-glycine bonds

S. 2-Nitro-5-thiocy

4. Amino side of cysteine

anobenzoate 6. The pyrimidine bases, Thymine and Cytosine in case of DNA have a 4-amino or a 4-keto group present with the amide ring. These 4-amino or 4keto groups are present in the major groove of the DNA molecule. 7. Globular proteins have no systematic structures. There may be single chains, two or more chains which interact in the usual ways or there may be portions of the chains with helical structures, pleated structures, or completely random structures. Common globular proteins include egg albumin, hemoglobin, myoglobin, insulin, serum globulins in blood, and many enzymes. Watersoluble globular proteins have hydrophobic amino acids present in the interior of the protein structure forming hydrophobic bonds. So, valine is the amino acid residue most likely to be present on the interior. 8. When a plant body is shifted from a warm climate to a cold climate, the number of long chain unsaturated fatty acids present in the plant would increase to preserve and store body heat as well

as reserved energy. So, ratio of unsaturated to saturated fatty acids would increase considerably in that plant. 9. Nitrogenous bases are always arranged in "anti" positions to their corresponding pentose sugar moieties. In B-form of DNA, sugar puckering occurs predominantly in the C2 position of the pentose sugar. The sugar pucker determines the shape of the a-helix, whether the helix will exist in the A-form or in the B-form. So, in C2-endo puckering occurs in B-form of DNA. 10. Proline is unique in that it is the only amino acid where the side chain is connected to the protein backbone twice, forming a five-membered nitrogen-containing ring. Proline is unable to occupy many of the main chain conformations easily adopted by all other amino acids. Proline can often be found in very tight turns in protein structures (i.e. where the polypeptide chain must change direction). It can also function to introduce kinks into alpha helices, since it is unable to adopt a normal helical conformation. Proline plays important roles in molecular recognition, particularly in intracellular signalling. The Proline side chain is very non-reactive and it does not undergo phosphorylation. 11. 18gm water in 100ml ethanol is 10M. Hence 15gm in 85ml is approximately- 8.5. 12. Ubiquitination is an enzymatic, protein posttranslational modification (PTM) process in which the carboxylic acid of the terminal glycine from the di-glycine motif in the activated ubiquitin forms an amide bond to the epsilon amine of the lysine in the modified protein. Ubiquitination, also known as ubiquitylation, regulates degradation of cellular proteins by the ubiquitinproteasome system, controlling a protein’s halflife and expression levels. This process involves the sequential action of ubiquitin-activating enzymes (E1), ubiquitin-conjugating enzymes (E2), and ubiquitin ligases (E3). The reaction catalyzed by each enzyme transfers a covalent bond with ubiquitin from one enzyme to the next and finally to a target protein. 13. Picloram is an auxin type of hormone absorbed by root. Zeatin is a member of the plant growth hormone family, cytokinins. Thiamine is a water-

1.4 Biomolecules : Constituents of Life

soluble vitamin of the B complex. Glutamine is one acids encoded by the standard genetic code CAA,CAG. 14. Given,

15. We know, the right handed α-helix has 3.6 residues per turn in the helix. But number of helix turns = 5.6

Width of lipid bilayer membrane= 30 Å

Therefore, No. of amino acids present in the protein = 5.6 * 3.6 = 20

We know, each turn of right handed α-helix has a pitch of = 5.1 Å

Therefore, No. of amino acid residues present in the protein is 20.

Thus, total turns = 30 / 5.1 = 5.6 turns.

„„

2

Bioenergetics and Metabolism

C HAPTER 2016

1. The equilibrium potential of a biological membrane for Na + is 55 mV at 37  C. Concentration of Na+ inside the cell is 20 mM. Assuming the membrane is permeable to Na+ only, the Na + concentration outside the membrane will be _________ mM. (Faraday constant: 23062 cal.V –1 .mol –1 , Gas constant: 1.98 cal.mol–1.K–1)

2015 2. The standard free energy change (G) for ATP hydrolysis is –30 kJ.mole –1 . The in vivo concentrations of ATP, ADP and Pi in E. coli are 7.90, 1.04 and 7.90 mM, respectively. When E. coli cells are cultured at 37  C, the free energy change (G) for ATP hydrolysis in vivo is ______ kJ. mole–1. 3. Match the compounds in Group I with the correct entries in Group II. Group I P. Cyanide

(a) a mixture of -D-Glucose and  -D-Glucose (b) a mixture of D-Glucose and L-Glucose (c) -D-Glucose only (d)  -D-Glucose only 6. The reactions leading to the formation of amino acids from the TCA cycle intermediates are (a) carboxylation

(b) isomerization

(c) transamination

(d) decarboxylation

7. Match the following photoreceptors with their prosthetic groups and spectral specificity Photoreceptor

1. Chromobilin

Q. Cryptochrome

2. FAD

R. Phytochrome

3. FMN

Absorption (nm) A. 400-500 B. 600-800

1. K+ ionophore

C. 500-600

R. Valinomycin 3. F1 subunit of ATP synthase 4. Cytochrome oxidase 5. Adenine nucleotide translocase (a) P-5, Q-2, R-3, S-1 (b) P-5, Q-2, R-1, S-3 (c) P-4, Q-2, R-1, S-3 (d) P-4, Q-5, R-3, S-1

2014 4. The most plausible explanation for a sudden increase of the respiratory quotient (RQ) of a microbial culture is that (a) cells are dying (b) yield of biomass is increasing

Moiety that absorbs light

P. Phototropin

Group II

Q. Antimycin A 2. Electron transfer from cytochrome b to cytochrome c1 S. Aurovertin

5. The product(s) resulting from the hydrolysis of maltose is/are

P

Q

R

(a) 3-A

1-A

1-B

(b) 1-B,

1-A

3-B

(c) 3-A

1-A

1-C

(d) 2-C

1-C

1-A

2013 8. Match the entries in Group I with the enzymes in Group II. Group I +

Group II

P. NAD

1. Glutathione peroxidase

Q. Selenium

2. Nitrogenase

R. Pyridoxal

3. Lactate dehydrogenase phosphate

S. Molybdenum

4. Glycogen phosphorylase

(a) P-3, Q-2, R-4, S-1 (b) P-4, Q-1, R-3, S-2

(c) the fermentation rate is increasing relative to respiration rate

(c) P-3, Q-1, R-4, S-2

(d) the maintenance rate is decreasing

(d) P-3, Q-4, R-2, S-1

2.2 Bioenergetics and Metabolism

Codes :

Common Data Questions Common Data for Questions 9 and 10 :

P

Q

R

S

A solution was prepared by dissolving 100 mg of protein X in 100 ml of water.

(a)

3

2

4

1

(b)

2

4

3

1

Molecular weight of protein X is 15,000 Da; Avogadro’s number = 6.022  1023.

(c)

4

3

1

2

(d)

3

4

2

1

9. Calculate the molarity ( M) of the resulting solution. (a) 66.6

(b) 6.6

(c) 0.67

(d) 0.067

10. The number of moleculespresent in this solution is (a) 40.15 1019

(b) 6.023 1019

(c) 4.015 1019

(d) 0.08 1019

2012

14. Match the vitamins in Group I with the processes/ reactions in Group II. Group I

Group II

P. Pantothenic acid

1. Electron transport

Q. Vitamin B2

2. Transfer of 1-C units

R. Vitamin B6

3. Decarboxylation

S. Folic acid

4. Fatty acid metabolism 5. Hydrolysis

Codes :

11. During photorespiration under low CO2 and high O 2 levels, O 2 reacts with ribulose 1, 5 - bisphosphate to yield (a) one molecule each of 3-phosp-hoglycerate and 2-phosphoglycolate (b) two molecules of 3-phosphoglycerate (c) two molecules of 2-phosphoglycolate (d) one molecule each of 3-phosp- hoglycerate and glyoxylate 12. Match the entries in Group I with the process parameters in Group II. Group I

Group II

P

Q

R

S

(a)

5

2

4

1

(b)

4

1

3

2

(c)

4

2

1

3

(d)

2

1

3

5

Common Data Questions Directions (Q. 15– 16) : In a muscle, the extracellular and intracellular concen-trations of Na+ are 150 mM and 12 mM and those of K+ are 2.7 mM and 140 mM respectively. Assume that the temperature is 25C and that the membnine potential is – 60 mV, with the interior more negatively charged than the exterior.

P. Clark electrode

1. Liquid level

(R = 8.314 J mol–1 K–1; F = 96.45 kJ mol–1 V–1)

Q. Redox probe

2. Dissolved oxygen

15. The free energy change for the transport of three Na+ out of the cell is

concentration R. Load cell

3. Vessel pressure

S. Diaphragm gauge

4. pH (anaerobic process)

Codes :

(a) + 1.5 kJ/mol (b) + 17.4 kJ/mol (c) + 18.9 kJ/mol (d) + 36.3 kJ/mol

P

Q

R

S

(a)

2

1

3

4

(b)

4

2

3

1

(a) +8.0 kJ/mol

(c)

2

4

1

3

(b) +11.6 kJ/mol

(d)

2

1

4

3

(c) +19.6 kJ/mol

16. The free energy change for the transport of two K+ into the cell is

13. Match the high energy compounds in Group I with the biosynthetic pathways for the molecules in Group II. Group I

Group II

(d) +31.2 kJ/mol

2011 17. An alternative to glycolysis pathway is

P. GTP

1. Fatty acid

(a) glyoxylate pathway

Q. UTP

2. Phospholipid

(b) pentose phosphate pathway

R. CTP

3. Protein

(c) citric acid cycle

S. Acyl coenzyme A

4. Peptidoglycan

(d) gluconeogenesis

Bioenergetics and Metabolism 2.3

18. In N-linked glycosylation, the oligosaccharide chain is attached to protein by

22. Oxidation reduction reactions with positive standard redox potential (E0) have

(a) asparagine

(b) arginine

(a) positive G0

(c) serine

(d) threonine

(b) negative G0

19. Match the products in Group I with their respective organisms in Group II. Group I

(c) positive E’ (d) negative E’ Linked Answer Questions

P. Glycerol

Statement for Linked Answer Questions (23 and 24) :

Q. Glutamic acid R. Curdlan S. Amphotericin B Group II 1. Corynebacterium glutamicum

The standard redox potential values for two halfreactions are given below. The value for Faraday’s constant is 96.48 kJ V–1 mol–1 and Gas constant R is 8.31 JK–1 mol–1. NAD+ + H+ + 2e–

2. Alcaligenes faecalis

+

FAD + 2H + 2e

3. Dunaliella salina 4. Streptomyces nodosus



 NADH

– 0.315 V

 FADH2

– 0.219 V

23. The Gfor the oxidation of NADH by FAD is

P

Q

R

S

(a) – 9.25 kJ mol–1

(a)

2

1

3

4

(b) – 103.04 kJ mol–1

(b)

4

2

1

3

(c) + 51.52 kJ mol–1

(c)

3

1

2

4

(d) – 18.5 kJ mol–1

(d)

2

1

4

3

20. Match items in Group I with Group II. Group I

24. The value of  G  , given K eq as 1.7, at 23  C will be (a) – 17.19 kJ mol–1 (b) – 19.8 kJ mol–1

P. Glycolytic pathway Q. Eukaryotic oxidative metabolism R. Glyoxylate cycle

(c) + 52.82 kJ mol–1 (d) – 117.07 kJ mol–1

S. Calvin cycle

2007

Group II

25. A protein binds to phosphocellulose column at pH 7.0 and elutes at pH 8.0. If the protein has to be further purified on a DEAE Sephacel column, the binding buffer should have a pH of

1. Chloroplast 2. Glyoxysomes 3. Mitochondria 4. Cytosol

(a) 5

(b) 6

(c) 7

(d) 8

P

Q

R

S

(a)

1

2

3

4

(b)

2

3

4

1

(c)

4

3

2

1

P. Catabolic product

1. Griseofulvin

(d)

3

4

1

2

Q. Bioconversion

2. Bakers yeast

26. Match items in group 1 with correct examples from those in group 2 Group 1

Group 2

2010

R. Biosynthetic product 3. 6-Aminopenici-

21. During lactic acid fermentation, net yield of ATP and NADH per mole of glucose is

S. Cell mass

llanic acid

(a) 2 ATP and 2 NADH

(a) P-4, Q-3, R-2, S-l

(b) 2 ATP and 0 NADH

(b) P-3, Q-4, R-l, S-2

(c) 4 ATP and 2 NADH

(c) P-4, Q-3, R-l, S-2

(d) 4 ATP and 0 NADH

(d) P-l, Q-4, R-3, S-2

4. Ethanol

2.4 Bioenergetics and Metabolism

27. The respiratory coefficient for the reaction aCHmOn + bO2 + cNH3  dCHaON+ eH2O + fCO2 is defined as (a) f/a

2004 29. The maximum reaction velocity (V m) for an enzyme catalyzed reaction was experimentally measured at two different temperatures and following results obtained :

(b) e/b

Temperature, C

(c) b/f

Vm mmol m

(d) f/b

–3

27

37

2.25

4.50

The energy of activation for the reaction is

2005

(a) 12834 cal mol–1

28. What product will result from complete hydrolysis of soluble dextran ?

(b) 25668 cal mol–1 (c) 6417 cal mol–1 (d) 19251 cal mol–1

(a) Sucrose only (b) Fructose only

2002

(c) Glucose and fructose only

30. Energy capture efficiency of the aerobic cells using glucose as a substrate is

(d) Glucose only

(a) 50%

(b) 40%

(c) 30%

(d) 20%

ANSWERS 1. (147.0 : 170.0)

2. (–48 to – 46)

3. (c)

4. (c)

5. (a)

6. (c)

7. (a)

10. (Marks to All)

11. (a)

12. (c)

13. (d)

14. (b)

15. (d)

21. (b)

22. (b)

23. (d)

24. (a)

25. (c)

8. (c)

9. (a)

16. (d)

17. (b)

18. (a)

19. (c)

20. (c)

26. (c)

27. (d)

28. (d)

29. (a)

30. (c)

Bioenergetics and Metabolism 2.5

EXPLANATIONS Eeq.Na  =

1.

starch. It is a reducing sugar. In the hydrolysis of any di- or polysaccharide, a water molecule helps to break the acetal bond. The acetal bond is broken; the H from the water is added to the oxygen glucose. The -OH is then added to the carbon on the other glucose. In this case the diastase enzyme is very specific and cuts the starch chain in two units of glucose which is maltose.

 Na    RT 0  ln  ZF  Na   i 

Given Eeq.Na  = 55 mV = 55  10–3 V R = 1.98 cal mol–1 K–1 T = 273 + 37 = 310 K F = 23062 cal V–1 mol–1

Na   = 20 mM. i Z= 1

 55  10–3 = 

 Na   1.98  310 0 ln   20 1  23062 

   

Na   = 20  7.897 = 157.94 mM 0

2. ATP + H2O  ADP + Pi

G = – 30 KJ/mol  = 37C G = G + RT ln

 ADP Pi   ATP 

= –30 + 8.314  10–3  (37 + 273) ln

(1.04 10 3  7.90  10 3 ) (7.90  10 3 )

4. High density culture is dependent upon high levels of available dissolved oxygen at times of maximum growth (using cascades of process parameter such as stirrer speed, gas flow and oxygen supplementation) or control of feed to match the quantity of substrate which is already present in the bulk culture. In a microbial culture, microbial metabolism is controlled on the basis of respiratory quotient (R.Q.). If, however, there occurs a sudden increase in the R.Q. of the microbial culture, it implies that rate of fermentation is increasing more than that of the rate of respiration.

6. Transamination refers to the transfer of an amine group from one molecule to another. This reaction is catalyzed by a family of enzymes called transaminases. Actually, the transamination reaction results in the exchange of an amine group on one acid with a ketone group on another acid. It is analogous to a double replacement reaction. The most usual and major keto acid involved with transamination reactions is alpha-ketoglutaric acid, an intermediate in the citric acid cycle. A specific example is the transamination of alanine to make pyruvic acid and glutamic acid. 7. Phototropin have flavin mononucleotide (FAD) molecules present in them which absorb light in the range of 400 to 500 nm of wavelength. Cryptochrome are comprised of flavin dinucleotide moieties which give light absorbance at the same range of 400-500 nm because of the same flavin molecule. Phytochrome has chromobilin present in it which is excited and shows absorbance within the range of 600-800 nm of wavelength.

5. Maltose or malt sugar is the least common disaccharide in nature. It is present in germinating grain, in a small proportion in corn syrup, and forms on the partial hydrolysis of

8. Lactate dehydrogenase (NAD+ oxidoreductase) is a ubiquitous enzyme among vertebrates and carries out the reaction: NAD+ + lactate NADH + Pyruvate. Glycogen Phosphorylase

= –30 – 17.70 = – 47.70 KJ/mol 3.

Group I

Group II

P. Cyanide

4. Cytochrome oxidase

Q. Antimycin A

2. Electron transfer from cytochrome b to cytochrome c1

R. Valinomycin

1. K+ ionophore

S. Aurovertin

3. F1 subunit of ATP synthase

2.6 Bioenergetics and Metabolism

catalyzes phosphorolytic cleavage of  (1, 4) glycosidic linkages of glycogen, releasing glucose1-phosphate as the reaction product. Pyridoxal phosphate (PLP), a derivative of vitamin B6, serves as prosthetic group for Glycogen Phosphorylase. Nitrogenase is a two-protein complex. One component, called Nitrogenase Reductase (NR) is an iron-containing protein that accepts electrons from ferredoxin, a strong reductant, and then delivers them to the other component, called Nitrogenase, or IronMolybdenum protein. There are several proteins in mammalian cells that can metabolize hydrogen peroxide and lipid hydroperoxides. These proteins include four selenium-containing glutathione peroxidases that are found in different cell fractions and tissues of the body. These proteins are called glutathione peroxidases. 9. Given, Molecular weight of the Protein X, m = 15,000 Da = 15,000 g/mol = 15 mg/mol Weight of the Protein in solution, W = 100 mg Volume of the solution, V = 100 mL Therefore, molarity of the solution = (W/m)  (1000/V) M = (100/15)  (1000/100) = 66.67 M 10. Number of moles present in the solution of protein = (100/15) = 6.667 moles Number of molecules present in the solution = no. of moles  Avogadro’s number = 6.667  6.022  1022 = 40.146  1022 molecules 11. During photo respiration under low CO 2 and high O 2 reacts with ribulose– 1, 5 bisphosphate to yield-one molecule each of 3-phosphoglycerate and 2-phosphoglycolate. 12. The Clark electrode is an electrode that measures oxygen on a catalytic platinum surface using the net reaction: O2 + 4e– + 2H2O  4OH– These are also known as dissolved oxygen electrodes. Redox probe also known as a pH probe is used for the measurement of pH. A load cell is a transducer that is used to create an electrical signal whose magnitude is directly proportional to the force being measured. It can be used to measure the liquid levels. To measure level, the load cell must be incorporated into the vessel's support structure.

The Diaphragm Pressure Gage uses the elastic deformation of a diaphragm (i.e. membrane) instead of a liquid level to measure the difference between an unknown pressure and a reference pressure. 13. Guanosine-5–triphosphate (GTP) is used as a source of energy for protein synthesis and gluconeogenesis. GTP is essential to signal transduction, in particular with G-proteins in second-messenger mechanisms where it is converted to guanosine diphosphate (GDP) through the action of GTPases. Peptidoglycan, also known as murein, is a polymer consisting of sugars and amino acids that form a mesh-like layer outside the plasma membrane of all bacteria (except Mycoplasma), forming the cell wall. Uridine triphosphate (UTP), which is a pyrimidine nucleotide, has the ability to act as an energy source for the synthesis of peptidoglycan. The phosphatidic acid is used to synthesize additional phospholipids utilizing cytidine triphosphate (CTP) as an energy source and creating a CDP-diacylglycerol molecule. Fatty-acyl-CoA Synthase, or more commonly known as yeast fatty acid synthase, is an enzyme complex responsible for fatty acid biosynthesis 14. Pantothenic acid, also called pantothenate or vitamin B 5 , is a water-soluble vitamin. Pantothenic acid is used in the synthesis of coenzyme A (CoA). CoA is also important in the biosynthesis of many important compounds such as fatty acids, cholesterol, and acetylcholine. Riboflavin is a water-soluble B vitamin, also known as vitamin B2. In the body, riboflavin is primarily found as an integral component of the coenzymes, flavin adenine dinucleotide (FAD) and flavin mononucleotide (FMN). FAD is part of the electron transport (respiratory) chain, which is central to energy production. Vitamin B6 is a water-soluble vitamin and is part of the vitamin B complex group. Several forms of the vitamin are known, but pyridoxal phosphate (PLP) is the active form and is a cofactor in many reactions of amino acid metabolism, including transamination, deamination, and decarboxylation. Folic acid is a B vitamin. It helps the body make healthy new cells. Transfer of 1-C units is important for many biosynthetic reactions. To carry out the transfer of 1-carbon units, NADPH must reduce folic acid two times in the cell.

Bioenergetics and Metabolism 2.7

17. The pentose phosphate pathway (PPP) is a process of glucose turnover that produce NaDPH as reducing equivalents and pentose as essential part of nucleotides. There are two different phases in a pathway. One is irreversible oxidase phase in which glucose-6-phosphate is converted to ribulose-6-phosphate by oxidative decarboxylation and NaDPH is generated. The other is reversible non- oxidative phase in which phosphorylated sugars are interconverted to generate xylulose 5p, ribulose-5p and ribose-5p. Phosphoribosyl pyrophosphate formed from ribose-5p is an activated compound used in the biosynthesis of histidine and purine/pyrimidine nucleotides. This pathway main purpose is to regenerate NaDP+ through an oxidation/reduction reaction. NaDPH is used for reductive reaction in anabolism, especially in fatty acid synthesis. 18. All N-linked glycans are derived from a common 14-sugar oligosaccharide synthesized in the ER, which is attached co-translationally to a protein while this is being translated inside the reticulum. The process of the synthesis of this glycan, known as Synthesis of the N-glycan precursor or LLO, constitutes one of the most conserved pathways in eukaryotes, and has been also observed in some eubacteria. The attachment usually happens on an asparagine residue within the consensus sequence asparagine-X-threonine by an complex called oligosaccharyl transferase(OST). After being attached to an unfolded protein, the glycan is used as a label molecule in the folding process (also known as Calnexin/Calreticulin cycle). 19. Glycerol is produced by Dunaliella salina. Glutamic acid is produced by corynebacterium glutamicum. Curdlan is synthesised by Alcaligenes faecalis . AmphotericinB is apolyene antifungal drug, often used intravenously for systemic fungal infections. It was originally extracted from Streptomyces nodosus, a filamentous bacterium. 20. In Plants, Cholroplast is the host for Glycolytic pathway. Oxidative oxidation in eukaryotes takes place at the Glyoxysomes. Mitochondria is the house of glyoxalate cycle in eukaryotes while cytosol host the Calvin cycle. 21. Lactic acid fermentation is caused by some fungi and bacteria. The most important lactic acid producing bacteria is lactobacillus. It is used to

convert lactose lactic acid in yogurt production. In lactic acid fermentation, NADH is oxidized into 2 molecules of NAD+ with a net gain of 2ATP and 2lactate. Pyruvate + 2NADH Æ Lactic Acid + 2NAD+ + 2ATP. 22. Redox potential E0 can be calculated by the ratio between Gibbs free energy and product of Faradays Number with number of participating moles. It implies that the value of Δ Go will be negative as Faraday’s constant is a positive number. 23. We have, from Gibbs free energy equation:-

ΔG = –nFΔE Where, ΔG = change in Gibbs free energy, n = number of electrons per mole product, F = Faraday’s constant,

ΔE = change in electrode potential of the reaction. Now, ΔGOXIDATION= –nFΔ(EREDUCTION – EOXIDATION) Given, from equation: n = 2 (2e– are exchanged) EOXIDATION = –0.315 V [Since NADH is oxidized to NAD+ by loss of 2e–] EREDUCTION = –0.219 V [Since FAD is reduced to FADH2 by accepting 2e–]

Therefore, ΔGOXIDATION= – 2 * 96.48 * {(–0.219) – (–0.315)} =– 18.524 kJ/mol

Therefore ΔG for oxidation of NADH by FAD is: –18.524 kJ/mol 24. We know, ΔG’ = ΔG0 + RT log Keq

Where, ΔG’= change in Gibbs free energy\ ΔG0 = standard change in Gibbs free energy R = Universal Gas constant T = absolute temperature Keq = equilibrium constant Given,

ΔG0 = –18.5 kJ/mol, R= 8.31 J/K/mol = 8.31 * 10–3 kJ/K/mol, Keq = 1.7, T = 230C = (273 + 23) K = 296K

Therefore, ΔG’ = –18.5 + [8.31 * 10–3 *296 * log(1.7)] = –17.19 kJ/mol

Therefore the value of ΔG’ is –17.19 kJ/mol

2.8 Bioenergetics and Metabolism

25. If the protein is most stable at pH's above its pI, then an anion-exchange gel matrix should be used. If it is most stable below its pI, then a cation-exchange gel matrix should be used. Here the protein eluted at pH 8 so it would take pH7 to bind in DEAE column. 26. Bioconversed product is 6 Amiono penicillinic Acid, Grisseofuvin is biosynthetic product, Bakers yeast have definite cell mass ie Biomass and Ethanol is a catabolic product. 27. The respiratory quotient (RQ) is calculated from the ratio RQ = CO2 eliminated / O2 consumed where the term "eliminated" refers to carbon dioxide (CO2) removed ("eliminated") from the body. So, RQ for the equation is f/b. 28. Complete hydrolysis of dextran yield solely β-dglucose i.e., hydrolysis is accompanied by inversion at carbon-1 in such a way that new reducing ends are released only in the β configuration.

29. log

K2 Ea  T2  T1  = K 1 2.303R  T1 T2  log(2) = Ea =

E a (10) 2.303  1.986(300  310)

0.3010  425359.5 10

= 12803.3 cal mol–1 30. Complete aerobic metabolism of glucose produces water and carbon dioxide as products. C6H12O6 + 6 O2  6 CO2 + 6 H2O Energy is released in this process. The overall “G of glucose breakdown is -720 kcal/mole. Normally about 32% of the energy is released is captured through the formation of ATP the remainder is released as heat.

„„

3

Biophysical and Biochemical Techniques

C HAPTER 2016

1. Cesium chloride density gradient centrifugation is commonly used for the separation of DNA molecules. The buoyant density, , of a double stranded Cs +DNA is given by the equation  = 1.66 + 0.098XG+C where XG+C denotes (a) total number of G and C (b) mole fraction of G + C (c) number of GC repeats (d) ratio of G+C to A+T content 2. Which one of the following techniques relies on the spin angular momentum of a photon?

where H is HETP (m) and u is linear liquid velocity of mobile phase (m.s–1). The values of A, B and C are 3  10–8 m2.s–1, 3 s and 6  10–5 m, respectively. The number of theoretical plates based on minimum HETP for a column of 66 cm length will be ___________.

2014 6. If the dissociation constant for solute-adsorbent binding is KD, the retention time of the solute in a chromatography column (a) increases with increasing KD (b) decreases with increasing KD

(a) CD spectroscopy

(c) passes through minimum with increasing KD

(b) Fluorescence spectroscopy

(d) is independent of KD

(c) IR spectroscopy (d) Raman spectroscopy 3. In dead-end filtration, rate of filtration is

2013 7. Match the entries in the Group I with the elution conditions in Group II.

(a) directly proportional to the square root of pressure drop across the filter medium

Group I

(b) inversely proportional to the pressure drop across the filter medium

Q. Hydrophobic chromatography

(c) inversely proportional to the viscosity of the solution

S. Chromatofocusing

(d) inversely proportional to the square of viscosity of the solution 4. The molar extinction coefficients of Trp and Tyr at 280 nm are 5690 and 1280 M –1 .cm –1 , respectively. The polypeptide chain of yeast alcohol dehydrogenase (37 kDa) contains 5 Trp and 14 Tyr residues. The absorbance at 280 nm of a 0.32 mg.mL –1 solution of yeast alcohol dehydrogenase measured in a cuvette of 1 cm pathlength will be _________. (Assume that the molar extinction coefficient values for Trp and Tyr apply to these amino acids in the yeast alcohol dehydrogenase). 5. A protein is to be purified using ion-exchange column chromatography. The relationship between HETP (Height Equivalent to Theoretical Plate) and the linear liquid velocity of mobile phase is given by:

A  Bu  C H= u

P. Ion-exchange R. Gel filtration column chromatography Group II 1. Isocratic solvent chromatography 2. Ampholytes 3. Increasing gradient of salt 4. Decreasing gradient of polarity (a) P-4,Q-1,R-2,S-3 (b) P-4,Q-3,R-1,S-2 (c) P-3,Q-4,R-1,S-2 (d) P-3,Q-4,R-2,S-1

2012 8. In transmission electron microscopy, electron opacity is greatly enhanced by treating the specimen with (a) ferrous ammonium sulfate (b) uranium acetate (c) sodium chloride (d) basic fuchsin

3.2 Biophysical and Biochemical Techniques

9. The helix content of a protein can be determined using (a) an infrared spectrometer

2009 12. Match the properties in Group 1 with the downstream operation in Group 2:

(b) a fluorescence spectrometer

Group 1

(c) a circular dichroism spectrometer

P. Size

1. Extraction

(d) a UV-Visible spectrophotometer

Q. Density

2. Distillation

R. Volatility

3. Filtration

S. Solubility

4. Sedimentation

10. Match the entries in Group I with the methods of sterilization in Group II.

Group 2

Group I

(a) P-3, Q-4, R-2, S-1

P. Serum

(b) P-4, Q-1, R-2, S-3

Q. Luria broth fïltration

(c) P-4, Q-3, R-1, S-2

R. Polypropylene tubes

(d) P-3, Q-2, R-4, S-1

S. Biological safety cabinets

2008

Group II

13. Match the coefficients in group 1 with their corresponding downstream processing steps given in group 2

1. Autoclave 2. Membrane

Group 1

3. UV irradiation 4. Gamma irradiation 5. Dry heat

P. Sedimentation coefficient Q. Partition coefficient R. Rejection coefficient

Codes :

S. Activity coefficient

P

Q

R

S

Group 2

(a) 5

3

1

4

1. Aqueous two phase extraction

(b) 1

4

5

3

2. Ultrafiltration

(c) 2

1

4

3

3. Dialysis

(d) 4

1

3

5

4. Centrifugation

2011

P

Q

R

S

(a) 3

1

4

2

Group I

(b) 2

1

4

3

P. Circular dichroism

(c) 4

3

1

2

Q. X-ray crystallography

(d) 4

1

2

3

11. Match the items in Group I with Group II.

R. Freeze-drying

2007

S. Ultracentrifugation

14. Match each parameter in group 1 with the appropriate measuring device in group 2

Group II

Group 1

1. Concentration 2. Sedimentation coefficient 3. Secondary structure determination 4. Tertiary structure determination

P. Pressure Q. Foam R. Turbidity S. Flow rate

P

Q

R

S

(a) 4

1

2

3

(b) 1

4

3

2

(c) 2

3

4

1

3. Diaphragm gauge

(d) 3

4

1

2

4. Rubber sheathed electrode

Group 2 1. Photometer 2. Rotameter

Biophysical and Biochemical Techniques 3.3

(a) P-3, Q-4, R-l, S-2 (b) P-l, Q-3, R-2, S-4

17. Cibacren Blue dye affinity chromatography can be used for affinity purification of

(c) P-4, Q-l, R-2, S-3

(a) NADPH dehydrogenase

(d) P-l, Q-2, R-3, S-4

(b) glucoamylase

2006

(c) subtilisin

15. When electroporation is used for introducing DNA into mammalian cells

(d) caspase

(P) a carrier for DNA is not required (Q) the lipid bilayer (membrane) interacts with an electric pulse to generate permeation sites (R) the viability of the cells becomes approximately zero percent (S) the first step involves absorption of DNA on the cell membrane (a) P, Q (b) Q, R

2002 18. Mechanism of separation of contaminants present in air by fibrous media are (a) Interception (b) Inertial impaction (c) Diffusion (d) All of these

2001 19. The limiting factor in the production of large quantities of ethanol as bio-fuel is

(c) P, S

(a) Lack of a balanced medium

(d) Q, S

(b) Ethanol toxicity to cells

2005 16. Ultrafiltration process can not be used for (a) fractionation of proteins (b) desalting (c) harvesting of cells (d) selective removal of solvents

(c) Expensive downstream processing steps (d) Only (b) and (c) of the above

3.4 Biophysical and Biochemical Techniques

ANSWERS 1. (b)

2. (a)

3. (c)

8. (b)

9. (c)

10. (c)

18. (d)

19. (d)

4. (0.37 : 0.43) 11. (d)

12. (a)

5. (1000.0 : 1000.0) 13. (d)

14. (a)

15. (a)

6. (b)

7. (c)

16. (c)

17. (a)

EXPLANATIONS 1. The density of DNA (g/mol or g/cm 3 ) as a function of its G : C content is given by following equation.

5. (i)

XG+C =

  1.66 0.098

u=

% XG + C

2. Circular dichroism (CD) is dichroism involving circularly polarized light, i.e., the differential absorption of left- and right-handed light. Lefthand circular (LHC) and right-hand circular (RHC) polarized light represent two possible spin angular momentum states for a photon, and so circular dichroism is also referred to as dichroism for spin angular momentum. 3. In dead end filtration, rate of filtration is inversely proportional to the viscosity of the solution. 4. Expected molar extinction sufficient of polypeptide chain E280nm = (5  5690) + (14  1280) = 46370 M–1 cm–1 Now, conc.of dehydrogenase C=

0.32  10 3 37

= 8.648  106



A/E = C  A = EC = 46370  8.648  10–6 A = 0.401

A

4 B  10 m / sec.

A = 3  10–8 m2/sec

as G + C is the mole fraction of G + C in DNA. So it can be expressed in terms of mole % G + C as falls

  1.66  100. = 0.098

A  Bu  C u

The minimum HETP occurs when

 = 1.66 + 0.098 XG + C 

H=

B = 3 sec C = 6  10–5 m (ii)  Now, (iii)

H = 6.6  10–4 = 6.6  10–2 cm H=

L , N = No. of theoretical plates N

L = Length of the column



H= =

L N 66cm  1000 6.6  10 2 cm

6. Chromatography involves a sample (or sample extract) being dissolved in a mobile phase (which may be a gas, a liquid or a supercritical fluid). The mobile phase is then forced through an immobile, immiscible stationary phase. The distribution of analytes between phases can often be described quite simply. An analyte is in equilibrium between the two phases; Amobile  Astationary The time between sample injection and an analyte peak reaching a detector at the end of the column is termed the retention time (tR). The dissociation constant for solute-adsorbent binding is KD. So, KD is indirectly proportional to the retention time of the column. So, retention time (tR) of a column decreases with increasing KD value.

Biophysical and Biochemical Techniques 3.5

7. In Ion exchange chromatography, be it cation based or anion based resin used, the elution of the substrate bound to the resins is done through increasing concentrations of salt in a stepwise gradient form of elution. For Hydrophobic column chromatography, ligands bound to the column through various interactions can be eluted by decreasing the level of polarity in the solution. In Gel filtration chromatography, molecules elute based on their different molecular weights through the process of sieving from the column by the use of any kind of isocratic solution of particular pH. In Chromatofocusing or also known as Isoelectric focussing, samples of proteins are precipitated at the isoelectric point using ampholytes of different charges. 8. In transmission electron microscope, to increase the electron opacity samples are impregnated with heavy metals such as uranyl acetate. Uranyl acetate is a water-soluble uranium compound and is often used as stains in electron microscopy. 9. The helix content of a protein can be determined using – circular dichromism spectroscopy. CD spectrum of unknown protein = A * [% alphahelical] + B * [% beta-sheet] + C * [% random coil] 10. Membrane filtration traps contaminants larger than the pore size on the surface of the membrane. If contaminants are smaller than the desired particle, decrease the membrane pore size and trap the product while passing the contaminants through the membrane. Heat labile fluids such as serum are sterilized by membrane filtration. A widely used method for heat sterilization is the autoclave. It can be used to sterilize bacterial cultures like Luria Broth. Gamma irradiation is a physical means of sterilisation or decontamination where products are exposed to gamma rays. It can be used to sterilize polypropylene tubes. In UV irradiation, Ultraviolet lamps are used to sterilize workspaces and tools used in biology laboratories and medical facilities. It can be used to sterile Biological safety cabinets. 11. Protein secondary structure can be determined by the Circular Dichromism where as XRD can be used for the Tertiary structure prediction of the protein. Freeze drying is an unit operation

for the concentration increase in a solution of limited solute. Ultra centrifugation is the unit operation for sedimentation. 12. All the properties in column A are phenomena unit for downstream process in column B. 13. Svedbarg Unit stands for sedimentation coefficient for centrifugation process. Partition coefficient for two layer aqueous extraction unit. Rejection coefficient is the unit for ultra filtration. Dialysis unit is activity coefficient. 14.  The Diaphragm Pressure Gage uses the elastic deformation of a diaphragm (i.e. membrane) instead of a liquid level to measure the difference between an unknown pressure and a reference pressure.

 In an antifoam system; if the foam contacts a rubber sheathed electrode, an electric unit actuates a solenoid valve to allow the passage o f a ste rile antif o am ag e n t in to th e fermentor.  A final widely used method for the determination of cell number is a turbidometric measurement or light scattering. This technique depends on the fact that as the number of cells in a solution increases, the solution becomes increasingly turbid (cloudy). The solution looks turbid because light passing through it is scattered by the microorganisms present and the turbidity is proportional to the number of microorganisms in the solution. The turbidity of a culture can be measured using a photometer or a spectrophotometer. Photometers, such as the Klett-Summerson device, use a red, green or blue filter providing a broad spectrum of light.  A rotameter is a device that measures the flow rate of liquid or gas in a closed tube. 15. Electroporation, or electropermeabilization, is a significant increase in the electrical conductivity and permeability of the cell plasma membrane caused by an externally applied electrical field. It is usually used in molecular biology as a way of introducing some substance into a cell, such as loading it with a molecular probe, a drug that can change the cell’s function, or a piece of coding DNA. 16. Ultrafiltration (UF) is a variety of membrane filtration in which hydrostatic pressure forces a

3.6 Biophysical and Biochemical Techniques

liquid against a semipermeable membrane. Suspended solids and solutes of high molecular weight are retained, while water and low molecular weight solutes pass through the membrane. This separation process is used in industry and research for purifying and concentrating macromolecular (10 3 - 106 Da) solutions, especially protein solutions. Ultrafiltration is not fundamentally different from microfiltration, nanofiltration or gas separation, except in terms of the size of the molecules it retains. Ultrafiltration is applied in cross-flow or dead-end mode and separation in ultrafiltration undergoes concentration polarization.

17. Cibacren blue dye (ligand) used in affinity chromatography, due to presence of sulfonated polyaromatic compounds which bind with considerable specificity and significant affinity to nucleotide-dependent enzymes and to a series of other proteins is used for glucoamylse purification. 18. The fibrous filter media are known to capture the airborne particles of different sizes due to Brownian diffusion, direct interception, inertial impaction, gravity settling, and particle sieving mechanisms. 19. For ethanol commercial utilization we need high purity which leads to heavy downstream processing cost along with high percentage of ethanol cause dehydration to the microbial inoculums leading to toxicity to cell.

„„

4

Cell Biology

C HAPTER 1. In a typical mitotic cell division cycle in eukaryotes, M phase occurs immediately after the (a) G0 phase (b) S phase (c) G1 phase (d) G2 phase 2. ATP biosynthesis takes place utilizing the H+ gradient in mitochondria and chloroplasts. Identify the correct sites of H+ gradient formation. (a) Across the outer membrane of mitochondria and across the inner membrane of chloroplast (b) Across the inner membrane of mitochondria and across the thylakoid membrane of chloroplast (c) Within the matrix of mitochondria and across the inner membrane of chloroplast

6. Choose the CORRECT sequence of steps involved in cytoplast production. (a) Digestion of cell wall  protoplast viability  cybrid formation  osmotic stabilizer (b) Osmotic stabilizer  digestion of cell wall  protoplast viability  cybrid formation (c) Protoplast viability  osmotic stabilizer  digestion of cell wall  cybrid formation (d) Osmotic stabilizer  digestion of cell wall  cybrid formation  protoplast viability

2014 7. If protoplasts are placed in distilled water, they swell and burst as a result of endosmosis. The plot representing the kinetics of burst is (a) Number of intact protoplast

2016

(d) Within the matrix of mitochondria and within the stroma of chloroplast 3. Which one of the following statements regarding G proteins is INCORRECT?

Time

(a) GDP is bound to G protein in the resting stage (b) Number of intact protoplast

(b) GTP bound a subunit cannot reassemble with  dimer (c) All G proteins are trimeric (d) Activation of G protein may result in activation or inhibition of the target enzymes

2015

Time

Number of intact protoplast

(c)

(a) Phosphatidylserine is presented on the outer cell surface (b) Cytochrome c is released from mitochondria (c) Mitochondrial membrane potential does not change (d) Annexin-V binds to the cell surface 5. The cytokinetic organelle in plant cells is (a) centriole (b) phragmoplast (c) proplastid (d) chromoplastid

Time

(d) Number of intact protoplast

4. Which one of the following is INCORRECT about a typical apoptotic cell ?

Time

4.2 Cell Biology

2012 8. Choose the correct signal transduction pathway. (a) Hormone  7 TM receptor  G protein  cAMP  PKA (b) Hormone  G protein  7 TM receptor  cAMP  PKA (c) Hormone  7 TM receptor  G protein  PKA  cAMP (d) Hormone  7 TM receptor  cAMP  G protein  PKA 9. Match the entries in Group I with those in Group II. Group I

12. A cell in G1 of interphase has 12 chromosomes. How many chromatids will be found per cell during metaphase II of meiosis? (a) 6

(b) 12

(c) 18

(d) 24

2010 13. Ames test is used to determine (a) the mutagenicity of a chemical (b) carcinogenicity of a chemical (c) both mutagenicity and carcinogenicity of a chemical (d) toxicity of a chemical 14. The formation of peptide cross-links between adjacent glycan chains in cell synthesis is called

P. MTT Q. Annexin V

(a) Transglycosylation

R. Methotrexate

(b) Autoglycosylation

S. Taxo

(c) Autopeptidation

Group II 1. Dihydrofolate reductase 2. Succinate ehydrogenase 3. Microtubules

(d) Transpeptidation 15. Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion : Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function.

4. Phosphatidylserine Codes : P

Q

R

S

(a) 3

1

4

2

(b) 2

4

1

3

(c) 2

3

4

1

(d) 4

2

1

3

10. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: Plants convert fatty acids into glucose. Reason: Plants have peroxisomes. (a) Both (A) and (R) are true but (R) is not the correct reason for (A)

Reason : cms can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non cms plant. (a) both (a) and (r) are true and (r) is the correct reason for (a) (b) both (a) and (r) are true and (r) is not the correct reason for (a) (c) (a) is false but (r) is true (d) (a) is true but (r) is false

2009 16. Caspases are involved in the process of

(b) Both (A) and (R) are true and (R) is the correct reason for (A)

(a) DNA replication

(c) (A) is true but (R) is false

(c) Antibody synthesis

(d) (A) is false but (R) is true

(d) Aposptosis

(b) Mutation and recombination

2011

2008

11. Apoptosis is characterized by

17. Which one of the following is the growth factor used for growth of tissues and organs in plant tissue culture?

(a) necrosis (b) programmed cell death (c) membrane leaky syndrome (d) cell cycle arrest process

(a) Cysteine

(b) Cytokinin

(c) Cytidylate

(d) Cyclic AMP

Cell Biology 4.3

18. Some living cells (e.g. plant cell) have the capacity to give rise to whole organism. The term used to describe this property is (a) morphogenesis (b) androgenesis (c) totipotency (d) organogenesis

2007 19. A bioremedial solution to reduce oxides of nitrogen and carbon in flue gases is to integrate flue gas emission to (a) micro-algal culture

2004 22. In the cell cycle of a typical eukaryote, the sequence of events operating at the time of cell division is (a) S phase  G2 phase  G1 phase  M phase (b) S phase  M phase  G1 phase  G2 phase (c) S phase  G2 phase  M phase  G1 phase (d) S phase  G1 phase  M phase  G2 phase

2003 23. In the course of cell cycle, the level of the protein cyclin abruptly falls during

(b) fish culture

(a) G1 phase

(c) mushroom culture

(b) S phase

(d) sericulture

(c) G2 phase

2005 20. The number of replicons in a typical mammalian cell is (a) 40-200 (b) 400

(d) M phase 24. An animal cell line was transfected with DNA extracted from a tumorous tissue. Which one of the following will be diagnostic of its tumorous transformation ?

(c) 1000-2000

(a) altered cell shape

(d) 50000-100000

(b) contact inhibition

21. Identify the following antibiotics with their modes of action. Antibiotic A. Ampicillin B. Tetracycline C. Nystatin D. Anthramycin Mode of action 1. inhibition of protein synthesis 2. inhibition of cell wall synthesis 3. damage to cytoplasmic membrane 4. damage to DNA structure

(c) anchorage-independent cell division (d) increased duration of cell cycle

2002 25. Plastome is (a) A type of plastid (b) An organellar genome (c) Plasmalemma protein (d) None of (a), (b), (c)

2001 26. When thymidine synthesis is inhibited, eukaryotic cells will be arrested at

(a) A  1, B  2, C  4, D  3

(a) End of S phase

(b) A  2, B  1, C  3, D  4

(b) End of M phase

(c) A  1, B  2, C  3, D  4

(c) Gl/S interphase

(d) A  3, B  4, C  2, D  1

(d) G2/M interphase

4.4 Cell Biology

ANSWERS 1. (d)

2. (b)

3. (c)

4. (c)

5. (b)

6. (b)

7. (a)

8. (a)

9. (b)

10. (a)

11. (b)

12. (b)

13. (c)

14. (d)

15. (a)

16. (d)

17. (b)

18. (c)

19. (a)

20. (d)

21. (b)

22. (c)

23. (d)

24. (c)

25. (b)

26. (a)

EXPLANATIONS 1. G2 phase, or Gap 2 phase, is the third and final subphase of Interphase in the cell cycle directly preceding mitosis. It follows the successful completion of S phase, during which the cell's DNA is replicated. G2 phase ends with the onset of prophase, the first phase of mitosis in which the cell's chromatin condenses into chromosomes. 2. During chemiosmosis, the free energy from the series of reactions that make up the electron transport chain is used to pump hydrogen ions across the membrane, establishing an electrochemical gradient. Hydrogen ions in the matrix space can only pass through the inner mitochondrial membrane through a membrane protein called ATP synthase. ATP synthesis in chloroplasts is very similar to that in mitochondria: Electron transport is coupled to the formation of a proton (H+) gradient across a membrane. The energy in this proton gradient is then used to power ATP synthesis. Two types of processes that contribute to the formation of the proton gradient are: (a) processes that release H+ from compounds that contain hydrogen, and (b) processes that transport H+ across the thylakoid membrane. 3. All G p ro te ins are no t trime ric. GPCR dimerization also was proposed to participate in arrestin-mediated desensitization based on the size of the cytoplasmic tip of rhodopsin and the presence of two cup-like domains in arresting both of which were implicated in receptor binding which one arrestin binds to two receptors in a dimer. 4. During apoptotic the mitochondrial membrane potential (mmp) decreases. 5. The cytokinetic organelle in plant cells is Phragmoplast.

6. Digestion of cell wall  protoplast viability  cybrid formation  osmotic stabilizer 7. The plot of swelling of protoplasts and bursting during endosmosis when protoplasts are kept or placed in distilled water is a sigmoidal curve where number of intact protoplasts decrease with time. So, the plot is represented by graph in (a). 8. In the G protein associated signal cascade system, binding of hormone to a cell surface receptor, like the 7TM (a 7 transmembrane alpha helices) which is a part of G protein. This will activate production of second messengers like cAMP by the enzyme adenylate cyclase. The cAMP in turn binds and activates down stream signal proteins line cAMP dependent protein kinase A (PKA). 9. MTT- binds to succinate dehydrogenase Annexin V-specifically binds to phosphatidyl serine Methotrexateallosterically dihydrofolate reductase

inhibits

Taxol-suppresses microtubule dynamics. 10. Plants can make glucose from fatty acids, but this is only because they are able to use the glyoxylate cycle instead of the Krebs cycle. 11. Programmed cell death (PCD) is a death of a cell in any form, mediated by an intracellular program. PCD is carried out in a regulated process which usually confers advantage during an organisms life cycle. PCD also serves an important function during both plant and metazoan (multicellular animals) tissue development. Apoptosis is a form of programmed cell d e a t h . H o w e v e r, n e c r o s i s i s a n o n physiological process that occurs as a result of infection or injury.

Cell Biology 4.5

12. Meiosis is a two part cell division process in organism that sexually reproduces which results in gametes with one half the number of chromosomes of the parent cell. Chromosomal replication does not occur between meiosis I and meiosis II; meiosis I proceeds directly to meiosis II without going through interphase. The second part of the meiosis i.e. meiosis II resembles mitosis more than meiosis I. Chromosomal number which have already been reduced to haploid (n) by the end of meiosis I, remains unchanged after this division. In meiosis II, the phases are again analogous to mitosis: prophase II, metaphase II, anaphase II and telophase II. Meiosis begins with two haploid (n=2) cell and end with 4 haploid (n=2) cells. 13. Mutagenicity and carcinogenicity are not mutually exclusive biological characteristics. The Ames test is an assay of chemicals to determine their mutagenic potential. A positive result implies the carcinogenic property of chemicals as well. The variable tested is the chemical’s potential to cause a reversion to growth and estimate frame-shift & point mutations. This allows detection of cancer-causing property based on mutagenic action. 14. Transpeptidation is a reaction involving the transfer of one or more amino acids from one peptide chain to another, as by transpeptidase action, or of a peptide chain itself, as in bacterial cell wall synthesis. The enzyme transpeptidase is also known as DD-transpeptidase, a bacterial enzyme that cross-links the peptidoglycan chains to form rigid cell walls. 15. Cytoplasmic male sterility (CMS) is a condition under which a plant is unable to produce functional pollen. CMS is a maternally inherited trait that is often associated with unusual open reading frames (ORFs) found in mitochondrial genomes. Male fertility can be restored by nuclear-encoded fertility restorer (Rf) gene(s). These fertility-restorer genes are thought to block or compensate for cytoplasmic dysfunctions that are phenotypically expressed during pollen development. On the one hand, sterility results from mitochondrial genes causing cytoplasmic dysfunction, and on the other, fertility restoration relies on nuclear genes that suppress cytoplasmic dysfunction.

16. Caspases, or cysteine-aspartic proteases or cysteine-dependent aspartate-directed proteases are a family of cysteine proteases that play essential roles in apoptosis (programmed cell death), necrosis, and inflammation. 17. Cytokinins (CK) are a class of plant growth substances (phytohormones) that promote cell division, or cytokinesis, in plant roots and shoots. They are involved primarily in cell growth and differentiation, but also affect apical dominance, axillary bud growth, and leaf senescence. Folke Skoog discovered their effects using coconut milk in the 1940s at the University of Wisconsin Madison. 18. Totipotency is the ability of a single cell to divide and produce all the differentiated cells in an organism, including extraembryonic tissues. Totipotent cells include spores and zygotes. 19. The advantages of using a microalgal-based system in integration with flue gas emission are that:



High purity CO2 gas is not required for algal culture. Flue gas containing varying amounts of CO2 can be fed directly to the microalgal culture. This will simplify CO2 separation from flue gas significantly.



Some combustion products such as NOx or SOx can be effectively used as nutrients for microalgae. This could simplify flue gas scrubbing for the combustion system.



Microalgae culturing may yield high value commercial products. Sale of these high value products can offset the capital and the operation costs of the process.



The envisioned process is a renewable cycle with minimal negative impacts on environment.

20. A replicon is a DNA molecule, or a region of DNA or RNA, that replicates from a single origin of replication. There are 50000–100000 replicons in a typical mammalian cell. 21. Ampicillin inhibits the third and final stage of bacterial cell wall synthesis in binary fission, which ultimately leads to cell lysis. Nystatin forms pores in the membrane that lead to K+ leakage and death of the fungus. Tetracycline antibiotics are protein synthesis inhibitors, inhibiting the binding of aminoacyltRNA to the mRNA-ribosome complex.

4.6 Cell Biology

Anthramycin exerts its biological activity through covalent binding to the C2-NH2 of guanine within the minor groove of DNA thus damaging the actual structure. 22. In eukaryotes, the cell cycle consists of four discrete phases: G 1, S, G 2, and M. The S or synthesis phase is when DNA replication occurs, and the M or mitosis phase is when the cell actually divides. The other two phases — G1 and G2, the so-called gap phases — are less dramatic but equally important. During G1, the cell conducts a series of checks before entering the S phase. Later, during G2, the cell similarly checks its readiness to proceed to mitosis.

23.

24. Tumor tissue generally show anchor dependent nature. So Anchor independent test should be performed. 25. The plastome is the genetic material that is found in plastids in plant cells (for example in the chloroplast). It composes part of the entire genome of photosynthetic organisms. 26. The arrest of thymidine kinase synthesis occurred after completion of the delayed mitosis.

„„

5

Microbiology

C HAPTER 2016 1. Bacteria with two or more flagella at one or both ends are called

6. Match the microorganisms in Group I with their fermentation products in Group II. Group I

Group II

(a) amphitrichous

(b) peritrichous

P. Leuconostoc mesenteroids 1. Cobalamin

(c) lophotrichous

(d) atrichous

Q. Rhizopus oryzae

2. Which family of viruses has single stranded DNA? (a) Herpesviridae

(b) Poxviridae

(c) Retroviridae

(d) Parvoviridae

3. Which of the following are TRUE for Treponema pallidum? P. It is the causative agent of syphilis Q. It is a spirochete R. It is a non-motile bacterium S. It is generally susceptible to penicillin Choose the correct combination. (a) P, Q and R only (b) P, Q and S only (c) P, R and S only (d) Q, R and S only 4. Which combination of the following statements is CORRECT for cyanobacteria? P. They can perform oxygenic photosynthesis Q. Usually filamentous forms are involved in nitrogen fixation R. Nitrogen fixation occurs in heterocysts S. They cannot grow in a mineral medium exposed to light and air (a) P, Q and R

2. Sorbose

R. Gluconobacter suboxydans 3. Dextran S. Streptomyces olivaceus

5. Butanol (a) P-5, Q-4, R-2, S-1

(b) P-5, Q-3, R-2, S-4

(c) P-3, Q-4, R-1, S-2

(d) P-3, Q-4, R-2, S-1

2014 7. Gram-positive bacteria are generally resistant to complement-mediated lysis because (a) thick peptidoglycan layer prevents insertion of membrane attack complex into the inner membrane (b) Gram-positive bacteria import the membrane attack complex and inactivate it (c) membrane attack complex is degraded by the proteases produced by the Gram-positive bacteria (d) Gram-positive bacteria cannot activate the complement pathway 8. A bacterium belonging to cocci group has a diameter of 2 m. The numerical value of the ratio of its surface area to volume (in m–1) is ______ 9. Which of the following essential element(s) is/are required as major supplement to enhance the bioremediation of oil spills by the resident bacteria?

(b) P, S and R

(a) Sulfur

(c) Q, R and S

(b) Nitrogen and phosphorus

(d) P, Q and S

(c) Iron

2015 5. Which one of the following organisms is used for the determination of phenol coefficient of a disinfectant ?

4. Lactic acid

(d) Carbon

2013 10. Phylum proteobacteria is subdivided into -,  -, -and -proteobacteria based on

(a) Salmonella typhi

(a) G + C content

(b) Escherichia coli

(b) 23S rRNA sequences

(c) Candida albicans

(c) tRNA sequences

(d) Bacillus psychrophilus

(d) 16S rRNA sequences

5.2 Microbiology

11. Match the commercial microbial sources in Group I with the products in Group II. Group I P. Corynebacteriumlilium Q. Klebsiellaoxytoca R. Aspergillusniger S. Alcaligeneseutrophus Group II 1. 2,3-Butane di-ol 2. Poly- -hydroxybutyric acid

2010 15. Under stress conditions bacteria accumulate (a) ppGpp (Guanosine tetraphosphate) (b) pppGpp (Guanosine pentaphosphate) (c) both ppGpp pppGpp (d) either ppGpp or pppGpp 16. Which one of the following DOES NOT belong to the domain of Bacteria? (a) Cyanobacteria (b) Proteobacteria

3. Glutamic acid

(c) Bacteroids

4. Citric acid

(d) Methanobacterium

(a) P-3,Q-1,R-2,S-4

17. Match Group I with Group II

(b) P-3,Q-1,R-4,S-2

Group I

(c) P-1,Q-3,R-2,S-4

P. Staphylococcus aureus

1. Biofilms

(d) P-1,Q-3,R-4,S-2

Q. Candida albicans

2. Bacteriocins

R. Mycobacterium tuberculosis

3. Methicillin resistance

12. Match the cell structures in Group I with the organismsin Group II.

Group II

Group I

S. Lactobacillus lactis

P. Endospores

(a) P-1, Q-4, R-2, S-3

(b) P-2, Q-3, R-1, S-4

Q. Bipolar flagella

(c) P-3, Q-1, R-4, S-2

(d) P-1, Q-2, R-4, S-3

R. Pseudomurine in cell wall S. Periplasmic flagella

4. Isoniazid

18. Match the products in Group I with the microbial cultures in Group II used for their industrial production

Group II

Group I

1. Methanobacterium

P. Gluconic acid

1. Leuconostoc mesenteroids

3. Spirillum

Q. L – Lysine

2. Aspergillus niger

4. Clostridium

R. Dextran

3. Brevibacterium flavum

(a) P-4, Q-3, R-1, S-2

S. Cellulase

4. Trichoderma reesei

(b) P-4, Q-3, R-2, S-1

(a) P-2, Q-1, R-3, S-4

(b) P-1, Q-3, R-4, S-2

(c) P-3, Q-4, R-1, S-2

(c) P-2, Q-3, R-1, S-4

(d) P-3, Q-2, R-4, S-1

2. Treponema

(d) P-4, Q-1, R-3, S-2

2011 13. The lipopolysaccharides present in bacterial cell wall has lipid A which is connected to

Group II

2009 19. Match the products in Group 1 with their producer organisms in Group 2: Group 1

Group 2

(a) O-polysaccharide

P. Ethanol from glucose 1. Aspergillus niger

(b) core polysaccharide

Q. Probiotics mesenteroides

2. Leuconostoc

R. Citric acid cerevisiae

3. Saccharomyces

S. Sauerkraut

4. Bifidobacterium

(c) both with O-polysaccharide and core polysaccharide (d) rhamnose-mannose disaccharide 14. Gas vacuoles are present in (a) Anabaena flos-aquae

(a) P-1, Q-3, R-2, S-4

(b) Bacillus subtilis

(b) P-3, Q-4, R-1, S-2

(c) Acanthurus nigrofuscus

(c) P-3, Q-4, R-2, S-1

(d) Mycobacterium tuberculosis

(d) P-1, Q-4, R-3, S-2

Microbiology 5.3

2008

2003

20. Match the products in group 1 with their producer organisms given in group 2

22. Microbes bring about biological transformation of xenobiotic compounds by

Group 1

(a) degradation

P. Ethanol

(b) conjugation

Q. L-Lysine

(c) detoxification

R. Biopesticide

(d) all of these

S. Vancomycin

2001

Group 2

23. Match the products in Column A with their corresponding organisms in Column B

1. Streptomyces orientalis 2. Saccharomyces cerevisiae

Column A

Column B

3. Corynebacterium glutamicum

(a) Aspergillus niger

4. Bacillus thuringiensis

(b) Saccharomyces cerevisiae 2. Citric acid

1. Lysine

P

Q

R

S

(c) Penicillium chrysogenum 3. Acetone/butanol

(a) 2

3

4

1

(d) Lactobacillus casei

4. Vitamin B12

(b) 3

4

1

2

5. Erythromycin

(c) 4

1

2

3

(e) Corynebacterium glutamicum

(d) 2

1

4

3

6. -Lactam 7. Diacetyl

2007 21. A bioremedial solution to reduce oxides of nitrogen and carbon in flue gases is to integrate flue gas emission to (a) micro-algal culture

8. Ethanol

2000 24. Which of the following eukaryotic organisms has been proven to be of great industrial importance ?

(b) fish culture

(a) Penicillium chrysogenum

(c) mushroom culture

(b) Saccharomyces cerevisiae

(d) sericulture

(c) Bacillus subtilis (d) Streptomyces griseus

ANSWERS 1. (c)

2. (d)

3. (b)

4. (a)

5. (a)

6. (d)

7. (a)

8. (3 to 3)

9. (b)

10. (d)

11. (b)

12. (a)

13. (b)

14. (a)

15. (c)

16. (a)

17. (c)

18. (c)

19. (b)

20. (a)

21. (a)

22. (d)

23. (*)

24. (b)

5.4 Microbiology

EXPLANATIONS 1. There are 4 types of flagellar distribution on bacteria-

1. Monotrichous - Single polar flagellumExample: Vibrio cholerae 2. Amphitrichous- Single flagellum on both sidesExample: Alkaligens faecalis 3. Lophotrichous- Tufts of flagella at one or both sides- Example: Spirillum 4. Peritrichous- Numerous falgella all over the bacterial body- Example: Salmonella Typhi 2. The virus family Parvoviridae contains 3 genera: the parvoviruses , the dependoviruses, and the densoviruses (which only infect insects). Parvoviridae is the only family of DNA viruses that infect humans that is single stranded. The DNA strand packaged in the virion can be either negative or positive sense, though in case of some strains, it is always negative. 3. Treponema pallidum is a spirochaete bacterium with subspecies that cause treponemal diseases such as syphilis, bejel, pinta, and yaws. It is a Gram-negative bacteria which is spiral in shape. It is generally susceptible to penicillin. 4. Cyanobacteria are photosynthetic, and so can manufacture their own food. Cyanobacteria get their name from the bluish pigment phycocyanin, which they use to capture light for photosynthesis. They also contain chlorophyll a, the same photosynthetic pigment that plants use. Nitrification cannot occur in the presence of oxygen, so nitrogen is fixed in specialized cells called heterocysts. These cells have an especially thickened wall that contains an anaerobic environment. e.g.,: Filamentous form of Nostoc. 5. Either Salmonella typhi or Staphylococus aureus. 6.

Group I

Group II

P. Leuconostoc mesenteroids

3. Dextran

Q. Rhizopus oryzae

4. Lactic acid

R. Gluconobacter suboxydans

2. Sorbose

S. Streptomyces olivaceus

1. Cobalamin

7. Gram positive bacteria have a thick layer of peptidoglycan as their cell wall which provides strength and support to the bacteria. Complement-mediated lysis and attack system is based on the formation of complement proteins which form membrane attack complex that insert into the plasma membrane of the cell and undergo cell lysis. But, due to the thick protection of gram positive bacteria by peptidoglycan on the outer cell membrane, complement proteins cannot enter the cell and cause lysis. 8. Gram-positive anaerobic cocci (GPAC) are a heterogeneous group of organisms defined by their morphological appearance which are spherical-shaped bacterium. Given, diameter of the cocci = 2m Therefore, radius = 1m Hence, surface area: volume of the cocci = 4r2/ (4/3r3) = 3/r = 3m 9. Oil spills in seas and oceans can be remedied by the use of certain special kinds of bacteria which degrade polyhydrocarbons. Such bacteria are utilized in the process of bioremediation to prevent serious impacts on marine environment caused by spillage. Such bacteria require nitrogen and phosphorus as major supplement for enhancing the rate of degradation of saturated and unsaturated hydrocarbons. 10. The phylum Proteobacteria is the largest and most diverse group of bacteria. They have many different shapes and ways of living. Some are freeliving, some are parasitic, and others form mutualistic relationships with other organisms, such as those found in the rumen. Many Proteobacteria are responsible for diseases in animals and humans. The largest and most diverse Bacterial phylum is the Proteobacteria. All are Gram-negative and show relationship based on 16s RNA comparison. 11. Mostly, Glutamic acid is produced by the species of Corynebacterium particularly Corynebacterium lilium. Klebsiella oxytoca produces butanol and in this case 2, 3-butane-diol. Citric acid is a common product of fermentation of the fungi Aspergillus niger. Alkaligenes eutrophus commonly produces butyric acids along with other derivatives through its fermentation. 12. Clostridium is a genus of Gram-positive bacteria, belonging to the Firmicutes. They are obligate anaerobes capable of producing endospores.

Microbiology 5.5

Spirillum is genus of spiral-shaped bacteria of the family Spirillaceae. Spirillum is microbiologically characterized as a gram-negative, motile helical cell with tufts of whiplike flagella at each end. Treponemes are helically coiled, corkscrewshaped cells, 6 to 15 m long and 0.1 to 0.2 m wide. They have an outer membrane which surrounds the periplasmic flagella, a peptidoglycan-cytoplasmic membrane complex, and a protoplasmic cylinder. Multiplication is by binary transverse fission. Methanobacteria are methane producing microorganisms (methanogens) and have pseudomurine in their cell wall. 13. The Gram negative cell envelope contains an additional outer membrane composed by phospholipids and lipopolysaccharides which face the external environment. The highly charged nature of lipopolysaccharide confers an overall negative charge to the Gram negative cell wall. Lipopolysaccharide is comprised of a hydrophilic polysaccharide and a hydrophobic component known as lipid A which is responsible for the major bioactivity of endotoxin. Lipopolysaccharide can be recognized by immune cells as a pathogen-associated molecule through Toll-like receptor 4. Lipid A usually gets attached to the core polysaccharide of Gram negative cell wall. 14. Gas-vacuoles consisting of collections of gascylinders, and continuity of the plasma membrane with lamellae, are observed at all stages of growth of Anabaena flos-aquae. It is known to produce endotoxins, the toxic chemicals released when cells die. Once released (lysed), and ingested, these toxins can damage liver and nerve tissues in mammals. 15. Stress is a stringent response that occurs in bacteria due to amino acid starvation or fatty acid limitation. Other reasons involve heat shock or iron limitation. This response is signalled by p/ ppGpp as an alarmone. It causes diversion of resources away from growth of cells towards the quenching of stress by promoting amino acid synthesis for enhancing survival. It continues until nutrients are restored. p/ppGpp causes inhibition of RNA synthesis in cases of AA shortage. It results in AA conservation by decreasing the translational machinery. Another mechanism of p/ppGpp action is the upregulation of genes for AA uptake and biosynthesis. It functions by influencing overall gene expression pattern for survival in stress conditions.

16. Besides cyanobacteria, all the others named were typically sharing the common feature belonging to bacteria domain – i.e. oxygen intolerance. However, cyanobacterium is a phylum which obtains energy through photosynthesis by performing oxygenic photosynthesis and reducing the atmosphere into an oxidising one. The ability to undergo oxygenic reactions distinctly distinguishes cyanobacteria from other bacterial domains. Thereby, cyanobacteria play a very important role in initiating biodiversity and are believed to evolve into plant chloroplast and eukaryotic algae by the process of endosymbiosis. Cyanobacterium due to its stand-out property of oxygenic photosynthesis in an oxygenic environment is thus distinct from other preferably an-oxygenic bacterial domains. 17. Methicillin-resistant Staphylococcus aureus (MRSA) are a type of staphylococcus or “staph” bacteria that are resistant to many antibiotics. Staph bacteria, like other kinds of bacteria, normally lives on skin and in nose, usually without causing problems. MRSA is different from other types of staph because it cannot be treated with certain antibiotics such as methicillin (as it is methicillin resistant). Biofilms formed by Candida albicans are resistant towards most of the available antifungal drugs. Therefore, infections associated with Candida biofilms are considered as a threat to immunocompromised patients. Also, the fungal pathogen Candida albicans is frequently associated with catheter-based infections because of its ability to form resilient biofilms. Isoniazid is an organic compound that is the firstline medication in prevention and treatment of tuberculosis. Isoniazid is a prodrug and must be activated by a bacterial catalase-peroxidase enzyme that in Mycobacterium tuberculosis is called KatG. Bacteriocins are the bioactive peptides produced by certain bacteria like lactobacillus lactis that inhibit the growth of other bacteria. It is a gram positive bacterium used in the production of milk and cheese. 18. Microbial species such as Aspergillus Niger have been utilized in many researches of gluconic acid fermentation. Strains of Brevibacterium flavum are used for the synthesis of L- lysine. Dextran is synthesized from sucrose by certain lactic-acid bacteria, the best-known being Leuconostoc mesenteroides and Streptococcus mutans.

5.6 Microbiology

Trichoderma reesei is an industrially important cellulolytic filamentous fungus. T. reesei has the capacity to secrete large amounts of cellulolytic enzymes (cellulases and hemicellulases).

 Some combustion products such as NOx or SOx can be effectively used as nutrients for microalgae. This could simplify flue gas scrubbing for the combustion system.

19. P>3: Saccharomyces cerevisiae is used extensively in batch fermentations to convert sugars to ethanol for the production of beverages and biofuels.

 Microalgae culturing may yield high value commercial products. Sale of these high value products can offset the capital and the operation costs of the process.

Q>4: Probiotics are live microorganisms thought to be beneficial to the host organism. Lactic acid bacteria (LAB) and bifidobacteria are the most common types of microbes used as probiotics.

 The envisioned process is a renewable cycle with minimal negative impacts on environment.

R>1: In production technique of citric acid, cultures of A. niger are fed on a sucrose or glucose containing medium to produce citric acid. S>2: Sauerkraut (sour cabbage) is finely shredded cabbage that has been fermented by various lactic acid bacteria, including Leuconostoc, Lactobacillus, Pediococcus etc. 20. P>2: Saccharomyces cerevisiae is used extensively in batch fermentations to convert sugars to ethanol for the production of beverages and biofuels. Q>3: Industrially, L-lysine is usually manufactured by a fermentation process using Corynebacterium glutamicum R>4: Bacillus thuringiensis (or Bt) is a Grampositive, soil-dwelling bacterium, commonly used as a biological pesticide. S>1: Streptomyces orientalis produces the antibiotic Vancomycin, which is used for the prevention and treatment of infections caused by Gram-positive bacteria, especially staphylococci. 21. The advantages of using a microalgal-based system in integration with flue gas emission are that:

 High purity CO2 gas is not required for algal culture. Flue gas containing varying amounts of CO2 can be fed directly to the microalgal culture. This will simplify CO2 separation from flue gas significantly.

22. microbes use replication procedure for xenobiotic compound transformation. A xenobiotic is a chemical which is found in an organism but which is not normally produced or expected to be present in it. It can also cover substances which are present in much higher concentrations than are usual. Specifically, drugs such as antibiotics are xenobiotics in humans because the human body does not produce them itself, nor are they part of a normal diet. However, the term xenobiotics is very often used in the context of pollutants such as dioxins and polychlorinated biphenyls and their effect on the biota, because xenobiotics are understood as substances foreign to an entire biological system, i.e. artificial substances, which did not exist in nature before their synthesis by humans. 24. Its complex genome and advanced splicing machinery makes it as a choice of industrial exploitation. Its been used for wine making, baking and brewing since ancient times. It is believed that it was originally isolated from the skin of grapes (one can see the yeast as a component of the thin white film on the skins of some dark-colored fruits such as plums; it exists among the waxes of the cuticle). It is one of the most intensively studied eukaryotic model organisms in molecular and cell biology, much like Escherichia coli as the model bacterium. It is the microorganism behind the most common type of fermentation.

„„

Immunology

6

6.1

Immunology

C HAPTER

(c) the joining of V, D and J segments (d) mutations in complementarity-determining regions 2. Find the INCORRECT combination. (a) Surface immunoglobulins – B cell antigen receptor (b) Affinity maturation – isotype switching (c) Fc region of antibodies – binding to complement proteins (d) Spleen, the secondary lymphoid organ – no connection with the lymphatic system 3. Which of the following statement(s) is/are CORRECT for antigen activated effector T cells? P. CD4+ cells make contact with macrophages and stimulate their microbicidal activity Q. CD4+ cells make contact with B cells and stimulate them to differentiate into plasma cells R. CD8+ cells make contact with B cells and stimulate them to differentiate into plasma cells S. CD8+ cells make contact with virus infected cells and kill them (a) Q only

(b) Q and S only

(c) P, Q and S only

(d) P, Q, R and S

2015 4. Anergy refers to (a) mitochondrial dysfunction (b) allergy to environmental antigens (c) unresponsiveness to antigens (d) a state of no energy 5. Based on the heavy chain, which one of the following antibodies has multiple subtypes ? (a) IgM

(b) IgD

(c) IgE

(d) IgG

(b) C3b

(c) C5a

(d) C5b

7. Production of monoclonal antibodies by hybridoma technology requires (a) splenocytes (b) osteocytes (c) hepatocytes (d) thymocytes 8. ABO blood group antigens in humans are differentiated from each other on the basis of (a) sialic acid (b) lipids (c) spectrin (d) glycoproteins 9. Three distinct antigens X, Y and Z were used to raise antibodies. Antigen Z was injected in a mouse on day zero followed by the administration of antigens X and Y on day 28. A second injection of antigen X was administered on day 70. The antibody titers were monitored in the serum every day and the results are shown below:

0 14

28

Antigen X

(b) the addition of N and P nucleotides

(a) C3a

Antigen Z + Antigen Y

(a) the addition of switch region nucleotides

Antigen Z

1. Junctional diversity of antibody molecules results from

6. Which one of the following complement proteins is the initiator of the membrane attack complex ?

Serumantibody titer

2016

42

56 70 Days

84 98

Which one of the following statements regarding the antibody titers in the serum is INCORRECT ? (a) Z-specific IgG will be high on day 14 (b) X-specific antibody titer will be high on day 84 (c) X-specific IgG will be high on day 42 (d) Y-specific IgG will be high on day 84

Immunology

6.2

10. Match the cells in Group I with their corresponding entries in Group II. Group I P. Mast cells

Group II 1. Activation of the complement pathway

Q. Natural killer 2. Expression of CD 56 cells R. Neutrophils

3. Containsazurophilic granules

S. Dendritic cells 4. Defense against helminthic infection 5. Production of antibodies specific to bacteria 6. Contains long membranous projections (a) P-4, Q-2, R-3, S-5 (b) P-4, Q-2, R-3, S-6 (c) P-3, Q-1, R-2, S-5 (d) P-3, Q-1, R-2, S-6 11. Match the antibiotics in Group I with their modes of action in Group II. Group I P. Chloramphenicol

Group II 1. Inhibits protein synth-esis by acting on 30S ribosomal subunit

Q. Rifampicin

2. Interferes with DNA replication by inhibiting DNA gyrase

R. Tetracycline

3. Inhibits protein synth-esis by acting on 50S ribosomal subunit

S. Quinolone

4. Interferes with RNA polymerase activity 5. Inhibits  -lactamase activity

(a) P-1, Q-2, R-3, S-5

(b) P-3, Q-4, R-1, S-3

(c) P-3, Q-2, R-1, S-4

(d) P-1, Q-4, R-3, S-2

2014

13. Reactions between antibodies and antigens that are detected by precipitate formation in an agar gel are referred as (a) immunoprecipitation assay (b) immunodiffusion assay (c) immunoaggregation assay (d) immunofixation assay 14. Cholera toxin increases cAMP levels by (a) modifying Gi protein (b) modifying Gs protein (c) binding to adenylate cyclase (d) activating cAMP phosphodiesterase 15. In a mouse genome, the numbers of functional V, J, V, D, J gene segments are 79, 38, 21, 2 and 11, respectively. The total number of possible combinations for  T cell receptors are ___×106. 16. Match the items in Group I with Group II Group–I Group–II P. Receptor tyrosine 1. Inactivation of kinase G-proteins Q. Cyclic GMP (cGMP) 2. Reception of insulin signal R. GTPase activating 3. Thyroid hormone protein (GAP) S. Nuclear receptor P (a) 1 (b) 2 (c) 3

Q 3 4 1

R 4 3 4

4. Receptor guanylyl cyclase S 2 1 2

(d) 2 4 1 3 17. Match the immunoglobulin class in Group I with its properties in Group II Group–I Group–II P. IgG 1. Major antibody in external secretions such as bronchial mucus Q. IgA 2. Protects against parasites R. IgE 3. Antibody that appears first in serum after exposure to an antigen

(c) innate immunity

P (a) 4 (b) 3 (c) 4

Q 1 2 3

4. Antibody present in highest concentration in serum R S 2 3 1 4 1 2

(d) antibody mediated response

(d) 1

4

3

12. Prior exposure of plants to pathogens is known to increase resistance to future pathogen attacks. This phenomenon is known as (a) systemic acquired resistance (b) hypersensitive response

S. IgM

2

Immunology

6.3

2013

2012

18. Protein A, which has strong affinity to Fc region of immunoglobulin, is extracted from

24. Idiotypic determinants of an antibody are associated with the

(a) Saccharomyces cerevisiae

(a) constant region of the heavy chains

(b) Staphylococcus aureus

(b) constant region of the light chains

(c) Streptococcus pyogenes

(c) variable region

(d) Streptococcus sanguis

(d) constant regions of light and heavy chains

19. The first humanized monoclonal antibody approved for the treatment of breast canceris (a) Rituximab (b) Cetuximab (c) Bevacizumab (d) Herceptin 20. The role of an adjuvant is to (a) prolong the persistence of antigen

25. Identification of blood groups involves (a) precipitation

(b) neutralization

(c) opsonization

(d) agglutination

26. B-lymphocytes originate from the bone marrow whereas T-lymphocytes originate from (a) thymus

(b) bone marrow

(c) spleen

(d) liver

27. A humanized antibody is one in which the

(b) cross link the antigen

(a) heavy and light chains are from human

(c) increase the size of antigen

(b) heavy chain is from human and light chain is from mouse

(d) avoid inflammation 21. Endogenous antigens are presented on to the cell surface along with (a) MHC-II (b) MHC-I (c) Fcreceptor (d) complement receptor 22. Which one of the following is an ABC transporter? (a) multidrug resistance protein (b) acetylcholine receptor (c) bacteriorhodopsin (d) ATP synthase 23. Match the antibioticsinGroup Iwith the targets in Group II. Group I P. Sulfonamide Q. Quinolones R. Erythromycin

(c) light chain is from human and heavy chain is from mouse (d) CDRs are from mouse and the rest is from human 28. Nude mice refers to (a) mice without skin (b) mice without thymus (c) knockout mice (d) transgenic mice 29. Heat inactivation of serum is done to inactivate (a) prions (b) mycoplasma (c) complement (d) pathogenic bacteria 30. Match the products in Group I with the applications in Group II. Group I

Group II

P. Digoxin

1. Muscle relaxant

Q. Stevioside

2. Anti-cancer agent

R. Atropine

3. C a r d i o v a s c u l a r disorder

2. Peptide chain elongation

S. Vinblastine

4. Sweetener

3. Folic acid biosynthesis

Codes :

S. Cephalosporin Group II 1. Peptidoglycan synthesis

4. Topoisomerase

P

Q

R

S

(a) P-3, Q-4, R-1, S-2

(a) 1

4

3

2

(b) P-2, Q-4, R-3, S-1

(b) 3

2

1

4

(c) P-4, Q-1, R-2, S-3

(c) 3

4

1

2

(d) P-3, Q-4, R-2, S-1

(d) 2

3

1

4

Immunology

6.4

2011

(c) (A) is true but (R) is false

31. Members of the antibody protein family that have common structural features are collectively known as

(d) (A) is false but (R) is true 36. Match items in Group I with Group II. Group I

(a) haptens

P. Alzheimer's disease

(b) allergens

Q. Mad cow disease

(c) antigens

R. Sickle cell anaemia

(d) immunoglobulins

S. Swine flu

32. In ABO blood group system, antigenic determinants are

Group II 1. H1N1

(a) nucleic acid

2. Hemoglobin

(b) carbohydrate

3. Prions

(c) lipid

4. Amyloid P

Q

R

S

(a) 4

3

2

1

(b) 3

4

1

2

(a) 1 and  2 domains only

(c) 2

1

4

3

(b)  1 and 2 domains only

(d) 1

2

3

4

(d) protein 33. The polymorphic domains for Class II MHC proteins are

(c) 1 and  1 domains only

2010

(d) 2 and  2 domains only 34. Match the viruses in Group I with their host cell receptors in Group II. Group I

Group II

P. Hepatitis A virus

1. Heparan sulphate

Q. Human immunode- 2. Acetylcholine ficiency virus

receptor

37. Hybridoma technology is used to produce (a) monoclonal antibodies (b) polyclonal antibodies (c) both monoclonal and polyclonal antibodies (d) B cells 38. Interferon- is produced by (a) bacteria infected cells

R. Rabies virus

3. CD4 protein

(b) virus infected cells

S. Herpes simplex virus type I

4. Alpha-2 macroglobulin

(c) both virus and bacteria infected cells

P

Q

R

S

(a) 1

3

2

4

(b) 3

4

1

2

(c) 4

3

2

1

(d) 2

3

1

4

35. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion : I g M is found in serum as a pentameric protein consisting of five I g M monomers. Reason : The pentameric form of IgM is due to cross-linking of I g M monomers via peptide bond.

(d) fungi infected cells 39. A neonatally thymectomized mouse, immunized with protein antigen shows (a) both primary and secondary responses to the antigen (b) only primary response to the antigen (c) delayed type hypersensitive reactions (d) no response to the antigen 40. Lymphocytes interact with foreign antigens in (a) Bone marrow

(b) Peripheral blood

(c) Thymus

(d) Lymph nodes

41. Dizygotic twins are connected to a single placenta during their embryonic development. These twins (a) have identical MHC haplotypes (b) have identical TH cells

(a) both (A) and (R) are true and (R) is the correct reason for (A)

(c) have identical T cells

(b) both (A) and (R) are true but (R) is not the correct reason for (A)

(d) can accept grafts from each other (both (a) and (b)

Immunology

2007 42. Match the recombinant products in group 1 with their therapeutic applications in group 2 Group 1 P. Human growth hormone Q. Platelet growth factor R. Factor VIII S. Erythropoietin Group 2 1. Pituitary dwarfism 2. Chemotherapy inducred thrombocy-topenia

(a) Both [a] and [r] are true and [r] is the correct reason for [a] (b) Both [a] and [r] are true but [r] is not the correct reason for [a] (c) Both [a] and [r] are false (d) [a] is true but [r] is false 47. Match the items on the left column with those on the right Left P. Programmed cell death at site of infection Q. Hormone upregulated during flooding stress R. Target for herbicide glyphosate

3. Haemophilia

S. Pathogen-derived resistance

4. Anaemia associated with chronic renal failure

Right

(a) P-l, Q-2, R-3, S-4

1. TMV coat protein

(b) P-2, Q-1, R-3, S-4

2. EPSP synthase

(c) P-l, Q-4, R-3, S-2

3. Hyper-sensitive response

(d) P-2, Q-4, R-3, S-l

4. Ethylene

43. Presence of CX2–4 CX X 8HX 3H sequence in a protein suggest that it is (a) a protein kinase (b) GTP binding protein (c) zinc finger protein (d) lipase 44. Oils rich in PUFA are NOT desirable for bio-diesel production because (a) they form epoxides in presence of oxygen (b) they do not form epoxides in presence of oxygen (c) they have high ignition temperature (d) they solidify at low temperature

2006 45. Which of these mice fail to develop a thymus ? (a) Motheaten mice (b) Beige mice

6.5

(a) P-l, Q-2, R-4, S-3 (b) P-3, Q-4, R-2, S-l (c) P-l, Q-4, R-2, S-3 (d) P-3, Q-2, R-4, S-l

2004 48. Reverse vaccinology indicates (a) From antigenic protein to vaccine development (b) From antigenic polysaccharide to vaccine development (c) From antibody to vaccine development (d) From genome sequence to vaccine development 49. In a heterogeneous population of cells containing T-cells, B-cells and macrophages, the cells are separated in the following scheme Heterogeneous cell suspension

Cell suspension left in a petriplate for 3 hrs

Supernatant cells passed through nylon wool column

(c) Knock out mice (d) Nude mice

2005 46. Determine the correctness or otherwise of the fo llowing Asse rtion[ a] an d Reason [ r] Assertion: An antigen recognized by one immunoglobulin subtype is not recognized by any other subtype. Reason: Immunoglobulin subtypes differ from each other body in the variable and in the constant regions.

Nylon wool adhered cell

Flow through

Identity the major population of cells present in petri-plate nylon wool adhered and nylon wool column flow through respectively (a) Macrophage, B-cell, T-cell (b) T-cell, B-cell, macrophage (c) Macrophage, T-cell, B-cell (d) B-cell, T-cell, macrophage

Immunology

6.6

2003 50. Match the activity spectrum of the following antibiotics P Actinomycin D

1. Antifungal

Q Daunorubicin

2. Antituberculosis

R Rifamycin

3. Antitumor

S Griseofulvin

4. Antiprotosoal

(a) A very high level expression is always obtained (b) Light promote more antibody production (c) Such antibodies are free from other antigen of animal origin (d) Functional antibody cannot be produced in plants 53. Viral replication within cells is inhibited by

(a)

(b)

(c)

(d)

P-3

P-3

P-3

P-2

Q-4

Q-1

Q-1

Q-1

R-2

R-4

R-2

R-4

2001

S-1

S-2

S-4

S-3

54. Antibody diversity is generated by

(a) IL – 4

(b) IL – 1

(c) IFN

(d) TNF

(a) Protein splicing

2002 51. Recombinant live attenuated vaccine against hepatitis B was prepared from

(b) Somatic mutation (c) Allelic exclusion (d) Inter-chromosomal recombination

(a) Plasma of chronically infected individual (b) Recombinant yeast expressing hepatitis B surface antigen (c) Recombinant vaccinia virus expressing hepatitis B surface antigen (d) Transgenic plants expressing hepatitis B surface antigen 52. Which of the following statements is most appropriate for recombinant antibody production in transgenic plants?

2000 55. Detection of which hormone is the commonly used test for pregnancy in humans? (a) LH (b) FSH (c) Charionic gonadotropin (d) Estrogen

ANSWERS 1. (b)

2. (b)

3. (c)

4. (c)

5. (d)

6. (d)

11. (b)

12. (a)

13. (b)

14. (b)

19. (d)

20. (a)

21. (b)

22. (a)

23. (d)

24. (c)

29. (c)

30. (c)

31. (d)

32. (b)

33. (c)

39.(b)

40.(d)

41.(d)

42. (a)

49. (a)

50. (a)

51. (c)

52. (c)

7. (a)

8. (d)

9. (d)

10. (b)

16. (d)

17. (a)

18. (b)

25. (d)

26. (b)

27. (d)

28. (b)

34. (c)

35. (c)

36. (a)

37.(a)

38.(c)

43. (c)

44. (a)

45. (d)

46. (c)

47. (b)

48. (d)

53. (c)

54. (c)

55. (c)

15. (1.38 to 1.40)

Immunology

6.7

EXPLANATIONS 1. Generation of junctional diversity starts as the proteins, recombination activating gene-1 and -2 (RAG1 and RAG2), along with DNA repair proteins, such as Artemis, are responsible for single-stranded cleavage of the hairpin loops and addition of a series of palindromic, ‘P' nucleotides. Subsequent to this, the enzyme, terminal deoxynucleotidyl transferase (TdT), adds further random ‘N' nucleotides. The newly synthesised strands anneal to one another, but mismatches are common. 2. Affinity maturation is the process by which THC cell-activated B cells produce antibodies with increased affinity for antigen during the course of an immune response. With repeated exposures to the same antigen, a host will produce antibodies of successively greater affinities once the memory has been generated. Whereas Isotope switching refers to the generation of IgA, IgD, IgM, IgE, IgG from IgM in response to a particular Ag. For e.g.: IgM will undergo isotope switching to IgE in response to Allergen reaction. 3. Activated effector T cells or CD4+ cells initially contact with the macrophages which possess its receptor and stimulate microbial activity. It also make contact with B cells and stimulate differentiation into plasma cells. On contrary CD8+ cells make contact with B cells and stimulate differentiation into plasma cells. CD4 is a co-receptor that assists the T cell receptor (TCR) in communicating with an antigenpresenting cell. Using its intracellular domain, CD4 amplifies the signal generated by the TCR by recruiting an enzyme, the tyrosine kinase Lck, which is essential for activating many molecular components of the signaling cascade of an activated T cell. 4. Inability of T cells to produce or responds to antigens. 5. Only IgG and IgA has subtypes . IgG has 4 subtypes: Gamma1, Gamma2, Gamma3, Gamma4 6. Main 4 complement proteins are the initiator of the membrane attack complex are (C5b, C6,C7 and C8). 7. Splenocytes obtained from Spleen. 8. Generaly oligosaccharide is present. 9. Y- specific IgG will be high on day 84

10.

Group I

Group II

P. Mast cells

4. Defense against helminthic infection

Q. Natural

2. Expression of CD 56

killer cells R. Neutrophils 3. Contains azurophilic granules S. Dendritic cells 11.

Group I

6. Contains long membranous projections Group II

P. Chloramphenicol 3. Inhibits protein synthesis by acting on 50S ribosomal subunit Q. Rifampicin

4. Interferes with RNA polymerase activity

R. Tetracycline

1. Inhibits protein synthesis by acting on 30S ribosomal subunit

S. Quinolone

2. Interferes with DNA replication by inhibiting DNA gyrase

12. Systemic acquired resistance (SAR) is a mechanism of induced defense that confers longlasting protection against a broad spectrum of microorganisms. SAR requires the signal molecule salicylic acid (SA) and is associated with accumulation of pathogenesis-related proteins, which are thought to contribute to resistance. Using the model plant Arabidopsis, it was discovered that the isochorismate pathway is the major source of SA during SAR. In response to SA, the positive regulator protein NPR1 moves to the nucleus where it interacts with TGA transcription factors to induce defense gene expression, thus activating SAR. 13. Immunodiffusion in gels encompasses a variety of techniques, which are useful for the analysis of antigens and antibodies. An antigen reacts with a specific antibody to form an antigenantibody complex, the composition of which depends on the nature, concentration and proportion of the initial reactants. Immunodiffusion in gels are classified as single diffusion and double diffusion. In ouchterlony double diffusion, both antigen and antibody are allowed to diffuse into the gel. This assay is frequently used for comparing different antigen preparation.

6.8

Immunology

14. Cholera toxin is produced by the CTXf bacteriophage residing within the bacteria. Cholera toxin consists of an A subunit coupled to a B subunit: the A subunit consists of an A1 domain containing the enzymatic active site, and an A2 domain that has an a-helical tail, while the B subunit contains five identical peptides that assemble into a pentameric ring surrounding a central pore. The A and B subunits of cholera toxin are produced in the cytosol by two genes that overlap by one base, and are then assembled into a toxin in the periplasm located between the inner membrane and the outer cell wall of the bacteria. When cholera toxin is released from the bacteria in the infected intestine, it binds to the intestinal cells known as enterocytes through the interaction of the pentameric B subunit of the toxin with the GM1 ganglioside receptor on the intestinal cell, triggering endocytosis of the toxin. Next, the A/B cholera toxin must undergo cleavage of the A1 domain from the A2 domain in order for A1 to become an active enzyme. Once inside the enterocyte, the enzymatic A1 fragment of the toxin A subunit enters the cytosol, where it activates the G protein GS? through an ADP-ribosylation reaction that acts to lock the G protein in its GTPbound form, thereby continually stimulating adenylate cyclase to produce cAMP. The high cAMP levels activate the cystic fibrosis transmembrane conductance regulator (CFTR), causing a dramatic efflux of ions and water from infected enterocytes, leading to watery diarrhea. 15. Given, Number of functional gene segments, V = 79 Number of functional gene segments, J = 38 Number of functional gene segments, V = 21 Number of functional gene segments, D = 2 Number of functional gene segments, J = 11 So, total number of possible combinations for  and  chains are: (79C  38C) + (21C  2C  11C) = 1.375  106 16. Tyrosine kinase is used to receive insulin hormone signals and regulates the level of insulin in the body. Cycli GMP is a part of 7 helical transmembrane domain protein comprising of Gcoupled receptors whose receptor protein is guanylyl cyclase. GTPase activating protein or GAP protein is used for the inactivation of Gproteins and stop the signaling. Nuclear receptor receives thyroid hormone as a signal for signal transduction.

17. The bulk of the immunoglobulins is found in the IgG fraction, which also contains most of the antibodies. The IgM molecules are apparently pentamers-aggregates of five of the IgG molecules. Electron microscopy shows their five subunits to be linked to each other by disulfide bonds in the form of a pentagon. The IgA molecules are found principally in milk and in secretions of the intestinal mucosa. Some of them contain, in addition to a dimer of IgG, a "secretory piece" that enables the passage of IgA molecules between tissue and fluid; the structure of the secretory piece is not yet known. The IgM and IgA immunoglobulins and antibodies contain 10 to 15 percent carbohydrate; the carbohydrate content of the IgG molecules is 2 to 3 percent. 18. Protein A is a cell wall component produced by several strains of Staphylococcus aureus that consists of a single polypeptide chain and contains little or no carbohydrate. The Protein A molecule contains four high-affinity binding sites capable of interacting with the Fc region from IgG of several species including human and rabbit. Optimal binding occurs at pH 8.2, although binding is also effective at neutral or physiological conditions (pH 7.0 to 7.6). The interaction between Protein A and IgG is not equivalent for all species. Even within a species, Protein A interacts with some subclasses of IgG and not others. For instance, human IgG1, IgG2 and IgG4 bind strongly, while IgG3 does not bind. There are also many instances in which monoclonal antibodies do not bind to Protein A, especially the majority of rat immunoglobulins and mouse IgG1. 19. Herceptin is approved for the treatment of earlystage breast cancer that is Human Epidermal growth factor Receptor 2-positive (HER2+) and has spread into the lymph nodes, or is HER2+ and has not spread into the lymph nodes. If it has not spread into the lymph nodes, the cancer needs to be estrogen receptor/progesterone receptor (ER/PR)-negative or have one high risk feature. Herceptin has two approved uses in metastatic breast cancer: i) Herceptin in combination with the chemotherapy drug Taxol (paclitaxel) is approved for the first line treatment of Human Epidermal growth factor Receptor 2-positive (HER2+) metastatic breast cancer and ii) Herceptin alone is approved for the treatment of HER2+ breast cancer in patients who have received one or more chemotherapy courses for metastatic disease.

Immunology

20. Adjuvants are compounds that enhance the specific immune response against co-inoculated antigens. Adjuvants can be used for various purposes: (i) to enhance the immunogenicity of highly purified or recombinant antigens; (ii) to reduce the amount of antigen or the number of immunizations needed for protective immunity; (iii) to improve the efficacy of vaccines in newborns, the elderly or immunocompromised persons; or (iv) as antigen delivery systems for the uptake of antigens by the mucosa. Some of the features involved in adjuvant selection are: the antigen, the species to be vaccinated, the route of administration and the likelihood of sideeffects. 21. T cell recognition of antigen requires that a complex form between peptides derived from the protein antigen and cell surface glycoproteins encoded by genes within the major histocompatibilty complex (MHC). MHC class II molecules present both extracellular (exogenous) and internally synthesized (endogenous) antigens to the CD4 T cells subset of lymphocytes. Endogenous antigens are internally generated molecules that become presented on the cell surface in the complex with class I histocompatibilty molecules (MHC I). Endogenous antigens may result from exogenous viral or bacterial infections that have altered the host cell. In autoimmune disorders, endogenous, selfmolecules induce autoimmune attack by CD8+ Tc/ CTLs that have escaped negative selection in the thymus. 22. The nine multidrug resistance proteins (MRPs) represent the major part of the 12 members of the MRP/CFTR subfamily belonging to the 48 human ATP-binding cassette (ABC) transporters. Cloning, functional characterization, and cellular localization of most MRP subfamily members have identified them as ATP-dependent efflux pumps with broad substrate specificity for the transport of endogenous and xenobiotic anionic substances localized in cellular plasma membranes. Multidrug resistance protein (MRP) is a broad specificity, primary active transporter of organic anion conjugates that confers a multidrug resistance phenotype when transfected into drugsensitive cells. The protein was the first example of a subgroup of the ATP-binding cassette superfamily whose members have three membrane-spanning domains (MSDs) and two nucleotide binding domains. 23. Sulfonamides were the first drugs acting selectively on bacteria which could be used systemically. Today they are infrequently used,

6.9

in part due to widespread resistance. The target of sulfonamides, and the basis for their selectivity, is the enzyme dihydropteroate synthase (DHPS) in the folic acid pathway. Quinolones are molecules structurally derived from the heterocyclic aromatic compound quinoline, the name of which originated from the oily substance obtained after the alkaline distillation of quinine. They have antimicrobial activity because of their action against the topoisomerase enzyme. Erythromycin is a macrolide antibiotic used to treat bacterial infections. By binding to the 50s subunit of the bacterial 70s rRNA complex, protein synthesis and subsequent structure and function processes critical for life or replication are inhibited. They prevent peptidoglycan synthesis. Cephalosporins are medicines that kill bacteria or prevent their growth. Cephalosporins disrupt the synthesis of the peptidoglycan layer of bacterial cell walls. The peptidoglycan layer is important for cell wall structural integrity. The final transpeptidation step in the synthesis of the peptidoglycan is facilitated by transpeptidases known as penicillin-binding proteins (PBPs). PBPs bind to the D-Ala-D-Ala at the end of muropeptides (peptidoglycan precursors) to crosslink the peptidoglycan. Beta-lactam antibiotics mimic the D-Ala-D-Ala site, thereby competitively inhibiting PBP crosslinking of peptidoglycan. 24. Immunoglobulin idiotypes are serologically defined determinants associated with the variable (V) region of antibody molecules 25. All the blood grouping reactions are agglutination reactions. In these the antibodies against A and B antigens when added, they bind to the cell surface and result in dumping of cells. These clumps or settling down of cells is referred to as agglutination. 26. All T cells originate from haematopoietic stem cells in the bone marrow. B cells undergo their complete maturation in the thymus while, T cells during their early stages of maturation migrate to the thymus and then undergo a process of maturation. They are released as competent and mature T cells from Thymus. 27. A humanized chain is a chain in which the complementarity determining regions (CDR) of the variable domains are foreign (originating from one species other than human, or synthetic) whereas the remaining chain is of human origin. 28. A nude mouse is a laboratory mouse from a strain with a genetic mutation that causes a deteriorated or absent thymus, resulting in an

6.10

Immunology

inhibited immune system due to a greatly reduced number of T cells. 29. Heat-inactivation (heating to 56  C for 30 minutes) of serum is done to inactivate complement, a group of proteins present in sera that are part of the immune response. 30. Digoxin- cardiovascular disorder Stevioside- sweetener Atropine- Muscle relaxant Vinblastine- anticancer agent 31. Immunoglobulins (Ig) are glycoproteins molecules that are produced by plasma cells in response to an immunogen and which functions as antibodies. The immunoglobulins derived their name from the findings that they migrate with globular proteins when antibody containing serum is placed in an electric field. Immunoglobulins generally assume as one of the two roles-It acts as plasma membrane bound antigen receptor on the surface of B cells. 32. The sequence of oligosaccharides determines whether the antigen is A, B, or A1. The ABO blood group antigens are attached to oligosaccharide chains that project above the RBC surface. These chains are attached to proteins and lipids that lie in the RBC membrane. The ABO locus has three main allelic forms: A, B, and O. The A allele encodes a glycosyltransferase that produces the A antigen (N-acetylgalactosamine is its immunodominant sugar), and the B allele encodes a glycosyltransferase that creates the B antigen (D-galactose is its immunodominant sugar). The O allele encodes an enzyme with no function, and therefore neither A or B antigen is produced, leaving the underlying precursor (the H antigen) unchanged. These antigens are incorporated into one of four types of oligosaccharide chain, type 2 being the most common in the antigen-carrying molecules in RBC membranes. Thus, in ABO system of blood grouping, the antigenic determinants are of carbohydrate origin. 33. Although similar to Class I, the MHC Class II molecule is composed of two membrane spanning proteins. Each chain is approximately 30 kDa in size, and made of two globular domains. The domains are named α-1, α-2, β-1 and β-2. The two regions farthest from the membrane are α-1 and β-1. The two chains associate without covalent bonds. The MHC molecules ability to present a wide range of antigenic peptides for T cell recognition requires a compromise between

broad specificity and high affinity. Class II molecules are dimers consisting of an α and β polypeptide chain. Each chain contains an immunoglobulin like region, next to the cell membrane. The antigen binding cleft, composed of two α -helices above a β -pleated sheet, specifically binds short peptides, about 15 to 24 residues long. The amino acid sequence around the binding site, which specifies the antigen binding properties, is the most variable site in the MHC molecule. 34. Hepatitis A virus (HAV), an atypical Picornaviridae that causes acute hepatitis in humans, usurps the HAV cellular receptor 1 (HAVCR1) to infect cells. HAVCR1 is a class 1 integral membrane glycoprotein that contains two extracellular domains: a virus-binding immunoglobulin-like (IgV) domain and a mucinlike domain that extends the IgV from the cell membrane. One of the early hallmarks of HIV infection is the impairment of a variety of CD4+ T-cell functions including T-cell colony formation, autologous mixed lymphocyte reactions, expression of interleukin-2 (IL-2) receptors, and IL-2 production. The nicotinic acetylcholine receptor (nAChR) is the first identified receptor for rabies virus. Rabies virus antigen was detected at sites coincident with nAChR. The nicotinic acetylcholine receptor (nAchR) is located at the postsynaptic muscle membrane. Heparan sulfate is a ubiquitously distributed polysulfated polysaccharide that is involved in the initial step of herpes simplex virus type 1 (HSV-1) infection. The virus interacts with cellsurface heparan sulfate to facilitate host-cell attachment and entry. 3-O-Sulfonated heparan sulfate has been found to function as an HSV-1 entry receptor. 35. Immunoglobulin M, or IgM for short, is a basic antibody that is produced by B cells. IgM is by far the physically largest antibody in the human circulatory system. It is the first antibody to appear in response to initial exposure to antigen. The spleen is the major site of specific IgM production.the same time because the large size of most antigens hinders binding to nearby sites. 36. Alzheimer’s is a form of dementia which happens due to the deposition of amyloid beta (Aß). This happen in neural juncture of brain. Mad cow is a fatal neurodegenative disease caused in cattles due to a misfolded protein called Prions. Sicklecell disease is a recessive genetic blood disorder. It causes abnormal, rigid, sickle shape of red blood cells due to mutation of the haemoglobin

Immunology 6.11

gene. Swine flu is a infection caused by various types of swine influenza virus. The known SIV strains include influenza C and the sub-types of influenza A known as H1N1, H1N2, H3N1, H3N2 and H2N3.

two separate and distinct placentas and membranes, both amnion and chorion. These twins can accept grafts from each other as they have identical MHC haplotypes and identical TH cells.

37. Hybridoma technology is used to combine the properties of site specific antigen activity of B cells with the immortality property of Myeloma M (cancerous) cells. Thus, the cells from a successful hybridoma experiment have single specificity of B cells with indefinite life span property. Anitbodies therefore produced are all of single specificity and hence are monoclonal in nature. Using monoclonal Ab’s, cells are categorised on the basis of expression of certain defined markers. Prostate specific Ag and other organ-associated Ag are used with the produced monoclonal Ab for determining nature of a primary tumour.

42. HGH is used in dwarfism, Erythropoetin used in Annemia treatment, Factor VIII is been used in Haemophilia treatment and Plalet derived growth factor used in chemotherapeutic drug induced thrombocytopenia treatment.

38. Interferon beta is produced mainly by fibroblasts and some epithelial cell types. The synthesis of IFN beta can be induced by common inducers of interferons, including viruses, double- stranded RNA, and bacteria. It is also induced by some cytokines such as TNF (Tumor Necrosis Factor) and IL1 (Inter Leukin1). 39. A neonatally thymectomized mouse, appeared healthy until about 2 to 4 months of age. Thymectomy after 3 weeks of age was not associated with any significant impairment of homograft immunity. 40. Lymphocytes constitute 20- 40% of the body’s white blood cells and 99% of the cells in the lymph. Lymphocytes continually circulate in the blood and lymph and are capable of migrating into the tissue spaces and lymphoid organs, thereby integrating the immune system to a high degree. Lymphocytes interact with foreign antigens in lymph nodes. Lymph nodes are the sites where immune responses are mounted to antigens in lymph. They are encapsulated bean shaped structures containing a reticular network packed with lymphocytes, macrophages, dendritic cells. Lymph nodes are the first organized lymphoid structure to encounter antigens that enter the tissue spaces. 41. A twin is one of two offspring produced in the same pregnancy and developed from one oocyte (monozygotic) or from two oocytes (dizygotic) fertilized at the same time. Dizygotic twins or fraternal twins are heterologous twins developed from two separate oocytes fertilized at the same time. They may be of the same or opposite sex, differ both physically and genetically, and have

43. A zinc finger protein is a DNA-binding protein domain consisting of zinc fingers ranging from two in the Drosophila regulator ADR1, the more common three in mammalian Sp1 up to nine in TF 111A. They occur in nature as the part of transcription factors conferring DNA sequence specificity as the DNA-binding domain. 44. PUFAs are long-chain fatty acids containing two or more double bonds. They provide structural & functional characteristics, and are involved in a wide range of biological components including membranes (in phospholipids). Equally important PUFAs serve as precursors for conversion into metabolites that regulate critical biological functions.Biodiesel is a non-petroleum and environmentally-friendly alternative to petroleum-based diesel fuel and not fatty acid.PUFA is fatty acid it is not used as it forms epoxide in presence of oxygen. 45. A nude mouse is a laboratory mouse from a strain with a genetic mutation that causes a deteriorated or absent thymus, resulting in an inhibited immune system due to a greatly reduced number of T cells. The phenotype, or main outward appearance of the mouse is a lack of body hair, which gives it the “nude” nickname. The nude mouse is valuable to research because it can receive many different types of tissue and tumor grafts, as it mounts no rejection response. 46. The immunoglobulins can be divided into five different classes, based on differences in the amino acid sequences in the constant region of the heavy chains. The valency of Immunoglobulin refers to the number of antigenic determinants that an individual immunoglobulin molecule can bind. The valency of all immunoglobulins is at least two and in some instances more than that. 47. Hypersensitive response is characterized by the rapid death of cells in the local region surrounding an infection carried out as a programmed mechanism for cell death.

6.12

Immunology

Environmental cues such as flooding, drought, chilling, wounding, and pathogen attack induce ethylene formation in plants. Glyphosate is a chemical herbicide which kills plants by inhbiting the shikimate pathway. It targets EPSP synthase, the enzyme that catalyzes the conversion of shikimate-3-phosphate and phosphoenolpyruvate into EPSP. 48. Reverse vaccinology is development of vaccine that employs bioinformatics. It indicates, from genome sequence to vaccine development and hence the vaccine is better and more specific. 49. B-cells selectively adhere to the fiber of nylon wool while T-cells do not and hence found in column flow resulting in successful recovery of both cell types. Whereas macrophage attaches to the petri plates and hence its major population will be there. 50. Griseofulvin is an antifungal drug that is administered orally. It is used both in animals and in humans,to treat fungal infections of the skin (known as ringworm) and nails. It is derived from the mold Penicillium griseofulvum. Daunorubicin is chemotherapeutic of the anthracycline family that is given as a treatment for some types of cancer. It was initially isolated from Streptomyces peucetius. The rifamycins are a group of antibiotics that are synthesized either naturally by the bacterium Amycolatopsis mediterranei or artificially. They are a subclass of the larger family Ansamycin. Rifamycins are particularly effective against mycobacteria, and are therefore used to treat tuberculosis, leprosy, and mycobacterium avium complex (MAC) infections. 51. A vaccine vector, or carrier, is weekend virus or bacterium into which harmless genetic material from another disease causing organism can be inserted. The vaccine virus, the virus that causes cowpox, is now used to make recombinant vector vaccines. Vaccinia is relatively large and has ample room to accept additional genetic fragments.

52. Its cross reactivity free in plant as such antibodies can be susceptible for cross reactivity in animal system. 53. Interferons (IFNs) are proteins made and released by host cells in response to the presence of pathogens such as viruses, bacteria, parasites ortumor cells. They allow for communication between cells to trigger the protective defenses of the immune system that eradicate pathogens or tumors. 54. Allelic exclusion has been observed most often in genes for cell surface receptors and has been extensively studied in immune cells such as B lymphocytes. In B lymphocytes, successful heavy chain gene rearrangement of the genetic material from one chromosome results in the shutting down of rearrangement of genetic material from the second chromosome. If no successful rearrangement occurs, rearrangement of genetic material on the second chromosome takes place. If no successful rearrangement occurs on either chromosome, the cell dies. As a result of allelic exclusion, all the antigen receptors on an individual lymphocyte will have the same amino acid sequence in the variable domain of the heavy chain protein. As the specificity of the antigen receptor is modulated by the variable domain of the light chain encoded by one of the immunoglobulin light chain loci, the specificities of B cells containing the same heavy chain recombination event can differ according to their light chain recombination event. 55. A pregnancy test attempts to determine whether a woman is pregnant. Markers that indicate pregnancy are found in urine and blood, and pregnancy tests require sampling one of these substances. The first of these markers to be discovered, human chorionic gonadotropin (hCG), was discovered in 1930 to be produced by the trophoblast cells of the fertilised ovum (blastocyst). While hCG is a reliable marker of pregnancy, it cannot be detected until after implantation this results in false negatives if the test is performed during the very early stages of pregnancy.

Genetics - I : Classical and Population Genetics

7

7.1

Genetics - I : Classical and Population Genetics

C HAPTER

(a) Non-random mating

2016 1. Match the type of chromosomal inheritance (Column-I) with the corresponding genetic disease or trait (Column-II). Column-I P. Autosomal recessive inheritance Q. Autosomal dominant inheritance R. X-linked inheritance S. Y-linked inheritance Column-II 1. Huntington disease 2. Hairy ears 3. Cystic fibrosis 4. Hemophilia Codes: P

Q

R

S

(a) 1

4

3

2

(b) 4

3

2

1

(c) 3

1

4

2

(d) 4

2

3

1

2. A crossing was performed between the genotypes DdEeFfgg and ddEeFfGg. Assuming that the allelic pairs of all genes assort independently, the proportion of progeny having the genotype ddeeffgg is expected to be ____________%.

2015 3. A heterozygous tall plant (Tt) was crossed with a homozygous dwarf plant (tt). The resultant seeds were collected. If five seeds are chosen at random, then the probability (in %) that exactly two of these seeds will yield dwarf plants is ______.

2014 4. In a relatively large but finite and closed population of sexually reproducing diploid organisms, the frequency of homozygous genotype PP changes from 0.40 to 0.50 and that of pp changes from 0.40 to 0.41 in a span of 10 generations. Which of the following is the most likely cause for the above change in frequency of the PP genotype?

(b) Random genetic drift (c) Selection (d) Combination of non-random mating and random genetic drift 5. Consider a population of 10,000 individuals, of which 2500 are homozygotes (PP) and 3000 are heterozygotes (Pp) genotype. The frequency of allele p in the population is ______. 6. In a genetic cross between the genotypes WWXX and wwxx, the following phenotypic distributions were observed among the F2 progeny: WX, 562; wx, 158; Wx, 38; and wX, 42. Likewise, a cross between XXYY and xxyy yielded the following results: XY, 675; xy, 175; Xy, 72; and xY, 78. Similarly, a cross between WWYY and wwyy yielded: WY, 292; wy, 88; Wy, 12; and wY, 8. In all the genotypes, capital letters denote the dominant allele. Assume that the F1 progeny were self-fertilized in all three crosses. Also, double cross-over does not occur in this species. Which of the following is correct? (a) Relative position: W-X-Y Distances: W-X = 5 map units, X-Y = 17 map units (b) Relative position: X-Y-W Distances: X-Y = 15 map units, Y-W = 11 map units (c) Relative position: Y-W-X Distances: Y-W = 5 map units, W-X = 11 map units (d) Relative position: X-W-Y Distances: X-W = 5 map units, W-Y = 10 map units 7. A dioecious plant has XX sexual genotype for female and XY for male. After double fertilization, what would be the genotype of the embryos and endosperms? (a) 100% ovules will have XXX endosperm and XX embryo (b) 100% ovules will have XXY endosperm and XY embryo (c) 50% ovules will have XYY endosperm and XY embryo, while other 50% will have XXY endosperm and YY embryo

Genetics - I : Classical and Population Genetics

7.2

2013 8. Of the two diploid species, species I has 36 chromosomes and species II has 28 chromosomes. How many chromosomes would be found in an allotriploid individual? (a) 42 or 54

(b) 46 or 50

(c) 74 or 86

(d) 84 or 108

9. Hypophosphatemia is manifested by an X-linked dominant allele.What proportion of the offsprings from a normal male and an affected heterozygous female will manifest the disease? (a) ½ sons and ½ daughters (b) all daughters and no sons

Common Data for Questions 13 and 14: Red-green colour blindness is inherited as a recessive X-linked trait. 13. What will be the probability of having the colourblind son to a woman with phenotypically normal parents and a colour-blind brother, and married to a normal man? (Assume that she has no previous children) (a) 100 % (b) 50 % (c) 25 %

(d) 12.5 %

14. What will be the probability of having the colourblind daughter to a phenotypically normal woman, who already had one colour-blind son, and is married to a colour-blind man?

(c) all sons and no daughters

(a) 75 %

(b) 50 %

(d) ¼ daughters and ¼ sons

(c) 25 %

(d) 15 %

Statement for Linked Answer Questions 15:

2012 10. Match the terms in Group I with the ploidy in Group II. Group I P. Disome Q. Monosome R. Nullisome S. Trisome Codes : (a) (b) (c) (d)

P 4 4 2 1

Q 2 3 3 4

1. 2. 3. 4. R 3 1 4 3

Group II 2n + 1 2n –1 n –1 n+1

S 1 2 1 2

11. A disease is inherited by a child with a probability 1 . In a family with two children, the 4 probability

of

that exactly one sibling is affected by this disease is 1 3 (a) (b) 4 8 9 7 (c) (d) 16 16

2011 12. Diploid Drosophila has eight chromosomes. Which one of the following terms should NOT be used to describe Drosophila with sixteen numbers of chromosomes? (a) Polyploid

The abdomen length (in millimeters) was measured in 15 male fruit flies, and the following data were obtained: 1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3 and 2.2. 15. The value of standard deviation (SD) will be (a) 0.061

(b) 0.25

(c) 0.61

(d) 0.85

2010 16. In transgenics, alterations in the sequence of uncleotide in genes are due to P. Substitution Q. Deletion R. Insertion S. Rearrangement (a) P and Q

(b) P, Q and R

(c) Q and R

(d) R and S

17. True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila. The F1 progeny was observed to be with curved wings and tan body. The F1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 short wings tan body, 16 curved wings dark body and, 6 short wings dark body. The mode of inheritance is (a) Typical Mendelian with curved wings and tan body being dominant

(b) Aneuploid

(b) Typical non-Mendelian with curved wings and tan body not following any pattern

(c) Euploid

(c) Mendelian with suppression of phenotypes

(d) Tetraploid

(d) Mendelian with single crossover

Genetics - I : Classical and Population Genetics

7.3

2004

2000

18. Some of the genes from viruses introduced into plants in fully functional form often exhibit Mendelian inheritance because

19. PKU is one of the best known hereditary disorders in amino acid metabolism. The defect is attributed to a lesion in one of the following enzymatic activities

(a) the genes are stably integrated in chromosomes (b) the genes are stably maintained in vectors

(a) Phenylalanine ammonia lyase

(c) the genes are co-expressed with chromosomal genes

(b) Phenylalanine hydroxylase

(d) the genes are not interrupted by introns

(d) Phenylalanine transaminase

(c) Tryosine hydroxylase

ANSWERS 1. (c)

2. (1.3 : 1.8)

10. (a)

11. (b)

4. (c) 5. (0.6 to 0.6) 6. (c)

3. (31%)

12. (b)

13. (b)

14. (b)

15. (b)

16. (b)

7. (d)

8. (b)

9. (a)

17. (a)

18. (a)

19. (b)

EXPLANATIONS 1. The common Chromosomal inheritances including Autosomal Recessive inheritance, Autosomal dominant inheritance, X linked inheritance, Y linked inheritance are associated with Cystic Fibrosis, Huntington Disease, Hemophilia, Hairy ears.

Therefore n = s + t. P =

=

2. Crossing of DdEeFfgg & ddEeFfGg. was performed hence.

P =

Dd  dd : Dd dd Dd dd i.e. dd = 0.5 Ee  Ee : EE Ee eE ee i.e. ee = 0.25 Ff  Ff : Ff fF fF ff i.e. ff = 0.25 gg  Gg : gG gG gg gg i.e. gg = 0.50



ddeeffgg = (0.5) (0.25) (0.25) (0.5) = 0.0156 = 0.0156  100% =1.56%

3.

T

t

t

Tt

tt

t

Tt

tt

The probability of dwarf seeds from resultant seeds = 0.5 = 50% Then, probability of selection of 2 seeds which yield dwarf plant from selected 5 seeds By applying Bionomial theorem

n! P = a s bt s!  t ! n = The total number of events s = The number of times outcome a occurs t = The number of times out come b occurs

3 2 5!  1   1      3!  2!  2   2  

5 4  1 1 1 1 1      2  2 2 2 2 2  20 64

20  100  31.25 64 4. The genotype of PP changes from 0.4 to 0.5 while in case of pp; it changes from 0.4 to 0.41 in a closed population of sexually reproducing diploids. So, it is clearly suggestive of the fact that the genotype PP is naturally selected and adapted by the environment and so its genotype has a higher frequency. %P =

5. We have, according to Hardy-Weinberg's law of population dynamics and equilibrium, P2 + 2PQ + Q2 = 1 Where, P is the allelic genotype or the frequency of 'P' and Q is the allelic frequency of 'p'. Now, Total population = 10000 Therefore, and Therefore,

P2 = 2500/10000 = 0.25 2PQ = 3000/10000 = 0.3 Q2 = 1 – P2 – 2PQ = 1 – 0.25 – 0.3 = 0.45

So, Q = 0.671 which is the frequency of the allele p 6. In the various genetic crosses, it can be seen that W and Y are more close to each other than X and

Genetics - I : Classical and Population Genetics

7.4

Y. At the same time, X and W are close to each other while Y is farthest from X. So, the relative positioning of the genes are Y-W-X and y-w-x. Now, Relative distances of genes with respect to each other are as follows: Distance of W gene from Y, Y - W = 5 units and W from X, W - X = 11 units 7. Dioecious plants have male (staminate) flowers on one plant, and female (pistillate) flowers on another plant. Some of these plants are polygamo-dioecious, with some male flowers on female plants and some female flowers on male plants. A dioecious plant having XX femal genotype and XY mal genotype undergoes double fertilization. So, on double fertilization, we have: 50% ovules : XXX endosperm and XX embryo And rest 50% ovules : XXY endosperm and XY embryo 8. Given, Species I has 36 (2n = 36) chromosomes and species II has 28 (2n = 28) chromosomes. Now, species I and II are crossed to obtain an allotriploid individual. That means it has a total chromosome of 3n. When the two diploid species are crossed, the chromosome set of one species would not undergo reduction division for the purpose of obtaining the allotriploid hybrid. That means the 3n would be contributed by 2n and n. So, the 2n can be from species I or II and so the n can be from species II or I respectively. Hence, the required chromosome number would be: Species I (2n) + Species II (n) = 3n



chromosomal pairs for a species (2n–2). A trisomy is a type of polysomy in which there are three instances of a particular chromosome, instead of the normal two (2n+1). 1 3 11. Given: P(E) = , P E = 4 4 3 1 3 P (x = 1) = 2 C1       = 4 4 8

 

12. Aneuploidy is an error in cell division that results in the daughter cells having the wrong number of chromosomes. It is the second major category of chromosomes mutation in which chromosome number is abnormal. An aneuploid is an individual organism whose chromosome number differs from the wild type by part of a chromosome set. Generally, the aneuploid chromosome set differs from wild type by one or small number of chromosomes. Aneuploids have chromosome number either greater or smaller than of wild type. Aneuploid nomenclature is based on the number of copies of the specific chromosome in the aneuploid state. So, for drosophila with 16 chromosomes, it is not aneuploid in condition. 13. The probability of having the colorblind son to woman with phenotypically normal parent and a colorblind brother, and married to a normal man than may be the woman is a carrier of blind trait or else not. Then we can say mother can be XRX a carrier or XX normal and have a cross with a normal man. X

36 + 14 = 50

Y

Or Species I (n) + Species II (2n) = 3n

XR XRX

XRY



X

XY

18 + 28 = 46

9. Given, Hypophosphatemia is caused due to Xlinked dominant allele. Now, when a normal male (XY) and a diseased female (X H X) are crossed together, the F1 generation will have offspring: XX, XHX, XY and XHY. So from this it can be seen that half of the male offspring and half of the female offspring bear the diseased allele. 10. Disome is a chromosome set having members paired (as in a normal somatic cell, n+1). Monosome is a chromosome having no homologue, especially an unpaired X chromosome (2n–1, because only one copy of some specific chromosome is present instead of the usual two found in its diploid progenitor. especially an unpaired X chromosome). Nullisome is a genetic condition involving the lack of one of the normal

XX

X OR

Y

X XX

XY

X XX

XY

Therefore, there is 1 chance of having a colorblind son out of 4 possible progenies. Hence, probability of one child to be colorblind = ¼ = 0.25 = 25% . Again 2 sons could be born, out of which 1 could be colorblind; Probability for colorblind son = ½ = 50% 14. Let say woman can be a carrier or normal and married to a colorblind man. So, woman can be either XRX or XX and man is XRY Therefore, XRX * XRY and XX * XRY XR R

X

R

X R

X X

Y XRY

R

X R

X

R

X X

X X X X RX

XY

Y XY

XY

Genetics - I : Classical and Population Genetics

Therefore, out of 4 children, there is only one chance of having a colorblind daughter. The probability for 1 out 4 children = ¼ = 0.25 = 25% ; Now there are 2 probable daughters. Probability for daughter to be colorblind = ½ = 50% 15. We know, Standard deviation, S.D. =  (Variance) = σ2 Now, Variance = 0.061 (from previous question) Hence, Standard deviation = (0.061) = 0.25 (approx.) 16. Transgenics is based on insertion of foreign genes from a different species, or deletion of existing nucleotides from the gene sequence in normal host cells. Also substitution of bases by transition( Pu by Pu/ Py by Py) or transversion( Pu replaced by Py of vice-versa). Usually, the process of random gene rearrangement is not included in transgenic experiments, because it is a natural process not yielding desired transgenic traits. Thus, in actual practice, insertion, deletion and/ or substitution is used for yielding transgenics.

7.5

17. Let us denote the dominant curved wings allele by C, short wings by c, tan body by D, dark body by d. Assumption is made that curved wings and tan body are recessive traits and that they follow Mendelian genetics. Now, F1 progeny CcDd (Since, F1 progeny was observed to be with curved wings and tan body) Cross breedÆ CcDd X CcDd F2 progeny  CCDD/CcDd/CCDd/CcDD ccDD/ccDd Ccdd/CCdd ccdd (45) 9

(15) :

3

:

(16)

(6)

3.2

: 1.2

which approximately becomes 9: 3: 3: 1, which is the phenotypic ratio of F2 generation according to Mendelian genetics 18. Mendelian inheritance is observed as the viral genes have been stably integrated and thus expresses itself. 19. Phenylketonuria (PKU) is an autosomal recessive metabolic genetic disorder characterized by a mutation in the gene for the hepatic enzyme phenylalanine hydroxylase (PAH), rendering it nonfunctional.

8

C HAPTER

Genetics - II : Molecular Genetics

8.1

Genetics - II : Molecular Genetics

2016

S. 5’-A T G G A C G T G C T T C C C A A t

1. What will be the binding status of regulatory proteins in lac operon when concentrations of both lactose and glucose are very low in the culture medium? (a) Only the repressor remains bound to the operator (b) Only the cyclic AMP-Catabolic Activator Protein (cAMP-CAP) complex remains bound to the CAP binding site (c) Neither the repressor nor cAMP-CAP complex remain bound to their respective binding sites (d) Both the repressor and cAMP-CAP complex remain bound to their respective binding sites 2. Based on their function, find the ODD one out.

G C A T C G G G C –3 T. 5’-A T G G A C G a G C T T C C C A A A G C A T C G G G C –3 [Point mutations are shown in the lower case or ‘–’ within the sequences] Which of the above mutant sequences DO NOT have frame-shift? (a) P, Q and S

(b) P, S and T

(c) Q, R and S

(d) Q, S and T

2015 5. How many 3-tuples are possible for the following amino acid sequence ? MADCMWDISEASE

(a) miRNA

(b) siRNA

(a) 4

(c) shRNA

(d) snRNA

(c) 11

3. Which set of the following events occurs during the elongation step of translation?

(b) 5 (d) 12

6. Match the drugs in Group I with their mechanism of action in Group II.

P. Attachment of mRNA with the smaller subunit of ribosome

Group I

Q. Loading of correct aminoacyl-tRNA into the A site

Q. Colchicine

R. Formation of a peptide bond between the aminoacyl-tRNA in the A site and the peptide chain that is attached to the peptidyl-tRNA in the P site

S. Methotrexate

S. Dissociation of the ribosomal subunits

2. Inhibits microtubule depolymerization

T. Translocation of peptidyl-tRNA from the A site to the P site of the ribosome

3. Inhibits DNA replication

(a) P, Q and R

(b) P, Q and T

5. Inhibits dihydrofolate reductase

(c) Q, R and T

(d) R, S and T

6. Inhibits microtubule polymerization

4. A DNA sequence, 5-A T G G A C G T G C T T C C C A A A G C A T C G G G C –3, is mutated to obtain P. 5-A T G G A C G T G C T T C a C A A A G C A T C G G G C –3 Q. 5-A T G G A C G T G C T T C C C g A A A G C A T C G G G C –3 R. 5-A T G G A C G T G C T T C C – A A A G C A T C G G G C –3

P. Paclitaxel R. Etoposide Group II 1. Inhibits protein translation

4. Alkylates DNA

(a) P-1, Q-6, R-3, S-4

(b) P-2, Q-6, R-3, S-5

(c) P-1, Q-3, R-6, S-5

(d) P-2, Q-3, R-6, S-4

7. An in vitor translation system can synthesize peptides in all three reading frames of the RNA template. When 5  -UCUCUCUC---(UC) n --UCUCUCUC - 3 was used as the template in this in vitro translation system, the synthesized peptides contained 50% each of serine and leucine. When 5-CCUCCUCCU---(CCU)n---CCUCCU-3 was used as the template, the synthesized peptides

Genetics - II : Molecular Genetics

8.2

contained 33.3% each of serine, leucine, and proline. Deduce the condon for proline.

PER TURN when the molecule is laid flat on the surface is _______ and__________, respectively.

(a) UCU

(b) CUC

(a) 187, 10.69

(b) 195, 10.25

(c) CCU

(d) UCC

(c) 200, 10.00

(d) 187, 10.50

Mutants

8. Cytoplasmic extract from the wild type strain of a bacterium has the ability to convert a colorless substrate (S) to a colored product (P) via three colorless intermediates X, Y and Z, in that order. Each step of the pathway involves a specific enzyme coded by a distinct gene. Four mutant strains (a–, b–, c–, d–) were isolated, whose extracts are incapable of producing the colored product in the presence of S. In a series of experiments, extracts from the individual mutants were incubated with X, Y or Z and secored for color development. The data are summarized in the table below. (Yes : color developed, No : no color developed) Compounds

11. The percentage SIMILARITIES and IDENTITIES, respectively, between the two peptide sequences given below will be ______% and _____%. Peptide I : Ala-Ala-Arg-Arg-Gln-Trp-Leu-ThrPhe-Thr-Lys-Ile-Met-Ser-Glu Peptide II: Ala-Ala-Arg-Glu-Gln-Tyr-Ile-SerPhe-Thr-Lys-Ile-Met-Arg-Asp (a) 80, 80

(b) 80, 60

(c) 60, 60

(d) 90, 60

12. For their efficient translation, eubacterial mRNAs possess a Shine-Dalgarno sequence for its recognition by an anti-Shine-Dalgarno sequence (ASD) in the ribosomes. The correct statement is (a) ASD is present in 5S rRNA

X

Y

Z

a–

No

No

No

b–

No

Yes

Yes

c–

Yes

Yes

Yes

(d) ASD is formed by the interaction of the 16S rRNA with the 23S rRNA upon docking of the 50S subunit on the 30S subunit of the ribosomes

d–

No

No

Yes

13. The active site in the alpha/beta barrel structures is usually located

Based on the data, which one of the following is the correct order of enzymes involved in the pathway ? d c b a (a) S   X   Y   Z  P a d b c (b) S   X   Y   Z  P b a c d (c) S   X   Y   Z  P c b d a (d) S   X   Y   Z  P

2014 9. The 5’ ends of the mature forms of the prokaryotic mRNAs and tRNAs are (a) a triphosphate group in mRNAs and a monophosphate group in tRNAs (b) triphosphate groups in both mRNAs and tRNAs (c) monophosphate groups in both mRNAs and tRNAs (d) a monophosphate group in mRNAs and a triphosphate group in tRNAs 10. Topological winding number of a 2.0 kb covalently closed circular DNA was found to be 191 with a writhing number of –4. Hence, its LINKING NUMBER and the NUMBER OF BASE PAIR

(b) ASD is present in 23S rRNA (c) ASD is present in 16S rRNA

(a) inside the barrel (b) at the amino side of the strands (c) at the carboxy side of the strands (d) at any arbitrary site

2013 14. Under alkaline conditions, DNA is more stable than RNA because (a) RNA forms secondary structures (b) RNA is a single stranded molecule (c) RNA has uracil in place of thymidine (d) RNA is susceptible to hydrolysis 15. Which one of the following modifications is common toboth protein and DNA? (a) SUMOylation

(b) Nitrosylation

(c) Methylation

(d) Ubiquitination

16. The nucleotide analogue used in DNA sequencing by chain termination method is (a) 1,3-dideoxy nucleoside triphosphate (b) 2,3-dideoxy nucleoside triphosphate (c) 2,4-dideoxy nucleoside triphosphate (d) 2,5-dideoxy nucleoside triphosphate

Genetics - II : Molecular Genetics

17. The RNA primer synthesized during the replication process in bacteria is removed by

8.3

24. The direction of shell coiling in the snail Limnaea peregra is a classic example of

(a) DNA gyrase

(a) chromosomal inheritance

(b) primase

(b) extra-chromosomal inheritance

(c) DNA polymerase I

(c) chromosomal translocation (d) homologous recombination

(d) DNA polymerase II 18. Match the entries in Group Iwith the entries in Group II. Group I

Group II

P. RNAse P

1. Polyadenylation

Q. RNase H

2. Splicing

R. snRNAs

3. Ribozymes

S. CstF

4. DNA-RNAhybrids

25. Which one of the following DNA sequences carries an invert repeat? (a) ATGAGCCCCGAGTA TACTCGGGGCTCAT (b) ATGAGCCGGCTCTA TACTCGGCCGAGAT (c) ATGAGCCGAGCCTA ACTCGGCTCGGAT

(a) P-3, Q-4, R-2, S-1

(d) ATGAGCCTATGGTA:

(b) P-4, Q-3, R-2, S-1

TACTCGGATACCAT

(c) P-3, Q-2, R-1, S-4 (d) P-2, Q-4, R-1, S-3 19. The total number of fragments generated by the complete and sequential cleavage of the polypeptide given below by Trypsin followed by CNBr is  Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-GlyArg-Cys-Ala-Gln 20. In a genetic study, 80 people were found to have alleles for polydactyly. Only 36 of them were polydactylous. What is the extent of penetrance percentage? 

26. In zinc finger proteins, the amino acid residues that coordinate zinc are (a) Cys and His (b) Asp and Glu (c) Arg and Lys (d) Asp and Arg

2011 27. Molecular chaperones are class of proteins that facilitate (a) the proper folding of newly synthesized proteins

2012

(b) unfolding of newly synthesized proteins

21. In mismatch correction repair, the parental DNA strand is distinguished from the daughter strand by 

(c) degradation of newly synthesized proteins

(a) acetylation

(b) phosphorylation

(c) methylation

(d) glycosylation

22. A protein is phosphorylated at a serine residue. A phosphomimic mutant of the protein can be generated by substituting that serine with (a) glycine

(b) alanine

(c) aspartate

(d) threonine

23. A truncated polypeptide is synthesized due to a nonsense mutation. Where would you introduce another mutation to obtain a full-length polypeptide? (a) Ribosomal protein gene

(d) targeting of newly synthesized proteins 28. Diphtheria toxin, tetracycline and streptomycin inhibit (a) DNA repair

(b) DNA replication

(c) transcription

(d) translation

29. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion : N-methyl-N'-nitro-N-nitrosoguanidine (NTG) is an effective chemical mutagen. Reason : Mutations induced by NTG mainly are the GC  AT transitions. (a) both (A) and (R) are true and (R) is the correct reason for (A)

(b) Transfer RNA gene

(b) both (A) and (R) are true but (R) is not the correct reason for (A)

(c) DNA repair gene

(c) (A) is true but (R) is false

(d) Ribosomal RNA gene

(d) (A) is false but (R) is true

8.4

Genetics - II : Molecular Genetics

30. Determine the correctness of the following statements I. Enhancer sequences are those DNA sequences that are involved in increasing the rate of DNA replication. II. Enhancer sequences work by binding with eukaryotic gene activator factors. (a) only I is true

R. 3 – 5 exonuclease activity S. 5 – 3 exonuclease activity (a) P and Q only (b) Q and R only (c) Q and S only (d) P and S only

2010

(b) only II is true (c) both I and II are true (d) both I and II are false 31. Match the terms in Group I with their associated functions in Group II. Group I

34. Peptidyl transferase activity resides in (a) 16S rRNA (b) 23 rRNA (c) 5S rRNA (d) 28 rRNA

P. Shine-Dalgarno sequences Q. Leucine zipper R. Aminoacyl tRNA synthetase S. RNA interference (RNAi) Group II 1. Aminoacylation of tRNA 2. Gene silencing

35. During transcription (a) DNA Gyrase introduces negative supercoils and DNA Topoisomerase I removes negative supercoils (b) DNA Topoisomerase I introduces negative supercoils and DNA Gyrase removes negative supercoils

3. Ribosome binding and facilitation of translation initiation

(c) both DNA Gyrase and DNA

4. Transcription factors

(d) both DNA Gyrase and DNA Topoisomearse I remove negative supercoils

Topoisomearse I introduce negative supercoils

P

Q

R

S

(a) 3

4

1

2

(b) 4

3

2

1

(c) 2

3

1

4

(a) H2 and H4 histone free

(d) 3

2

4

1

(b) H1 and H2 histone free

32. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion : Isopropylthiogalactoside (IPTG) is a gratuitous inducer of lactose operon. Reason : Gratuitous inducers are chemical analogs which behave like natural inducer but they do not serve as substrate for the enzymes that are subsequently synthesized. (a) both (A) and (R) are true and (R) is the correct reason for (A) (b) both (A) and (R) are true but (R) is not the correct reason for (A) (c) (A) is true but (R) is false (d) (A) is false but (R) is true 33. Which of the following characteristics with respect to bacterial DNA polymerase III are TRUE?

36. Nuclease-hypersensitive sites in chromosomes are sites that appear to be

the

(c) H3 and H4 histone free (d) Nucleosome free 37. Determine the correctness or otherwise of the following Asertion (a) and the Reason (r) Assertion : MTT assay is used to determine cell viability based on the principle of colour formation by DNA fragmentation. Reason : MTT assay is used to determine cell viability based on the colour development by converting tetrazolium soluble salt to insoluble salt. (a) both (a) and (r) are true and (r) is the correct reason for (a) (b) both (a) and (r) are true and (r) is not the correct reason for (a)

P. Initiation of chain synthesis

(c) (a) is true but (r) is false

Q. 5 – 3 polymerization

(d) (a) is false but (r) is true

Genetics - II : Molecular Genetics

38. Match the following antibiotics in Group I with their mode of action in Group II Group I Group II P. Chloramphenicol 1. Binds to DNA gyrase Q. Norfloxacin 2. Binds to RNA Polymerase R. Puromycin 3. Inhibits peptidyl transferase S. Rifampicin 4. Mimics aminoacyl tRNA (a) P-1, Q-3, R-2, S-4 (b) P-3, Q-1, R-2, S-4 (c) P-3, Q-1, R-4, S-2 (d) P-4, Q-2, R-3, S-1

2009 39. Which of the following are commonly used as reporter genes? P. NPTgene Q. Luciferase gene R. CFTR gene S. GFPgene (a) Q, S (b) R,S (c) P, R (d) P, Q 40. Determine the correctness or otherwise of the following Assertion (a) and Reason (r): Assertion (a): IPTG (Isopropylthiogalactoside) is a gratuitous inducer of lac operon. Reason (r) : IPTG is an efficient inducer, but not a substrate of lac operon (a) Both (a) and (r) are true and (r) is the correct reason for (a) (b) Both (a) and (r) are true but (r) is not the correct reason for (a) (c) (a) is true but (r) is false (d) (a) is false but (r) is true

2008 41. Zinc fingers are characteristics of (a) blood clotting proteins (b) RNA chaperones (c) DNA binding proteins (d) lysosomal hydrolases

2007 42. Si RNA(s) interfere at (a) transcriptional level (b) post-transcriptional level (c) DNA replication level (d) translational level 43. Presence of CX2–4 CX X 8HX 3H sequence in a protein suggest that it is (a) a protein kinase (b) GTP binding protein (c) zinc finger protein (d) lipase

8.5

44. Mobile genetic elements present in human genome are P. long interspersed elements (LINEs) Q. short interspersed elements (SINEs) R. P elements S. IS elements (a) Q, R (b) P, Q (c) P, R

(d) Q, S

2006 45. The enzymes that can be used in 5 end labeling of DNA are (P) alkaline phosphatase (Q) DNA ligase (R) terminal transferase (S) polynucleotide kinase (a) P, S (b) R, Q (c) P, R

(d) R, S

2005 46. Expression of which of the following reporter genes does not require addition of specific substrate for detection ? (a) Luciferase (b)  -Glucuronidase (c)  -Glucosidase (d) Green fluorescent protein Common Data for Questions 47 and 48: Normal primary hepatocytes can be artificially immortalized. Certain spontaneous mutants of immortalized hepatocytes are sensitive to ionizing radiation. 47. Which of the following genes are involved in Immortalization of primary hepatocytes? (a) Telomerase and Cyclin D (b) NFB and Thymidine kinase (c) Cyclin D and myc (d) Telomerase and Ras 48. What would happen to the mutant cells by ionizing radiation? (a) Apoptosis (b) Necrosis (c) Cell growth arrest (d) Cell proliferation

2004 49. Expression of antisense RNA of ACC synthase in transgenic tomato plants inhibited the synthesis of ethylene resulting in (a) Change in colour from green to red (b) Change in aroma (c) Change in colour from red to green (d) None of these

Genetics - II : Molecular Genetics

8.6

2003 50. Which of the following processes require energy? (a) ligation (b) transformation (c) restriction digestion

56. A heterologous protein for its expression in the milk of a transgenic animal should be under the control of the promoter of the gene coding for (a)  -Globin

(b)  -Lactoglobulin

(c) Preproinsulin

(d) lac Z

57. Reverse genetics means

(d) hybridization 51. The full-length coding sequence of an eukaryotic gene was expressed in bacteria and the protein was purified. However, in the functional assay, no activity was detected for the purified protein. The reason could be (a) the host bacteria produced an enzyme that inhibited the activity of the expressed eukaryotic protein (b) the purified protein was contaminated with bacteria (c) the host bacteria did not produce the essential co-factors (d) no post-translational modification on the protein expressed in bacteria

2002 52. RNA is very much susceptible to hydrolysis in alkali because (a) It contains uracil residues in its structure (b) Its 2 -OH group participate in intramolecular cleavage of phosphodiester backbone (c) Cleavage occurs in the glycosylic bonds of purine bases (d) Cleavage occurs in the glycosylic bonds of pyrimidine bases

2001

(a) Finding the function of a ORF (b) Finding the gene responsible for a trait (c) RNA dependent DNA synthesis (d) Converting somatic cell to a germ cell

2000 58. The G + C content of bacteriophage 13 double standed DNA is 68%. What would you expect the G + C content of its mRNA ? (a) About 68%

(b) About 34%

(c) About 32%

(d) About 86%

59. Rho-dependent and rho-independent transcription termination mechanisms operate in prokaryotes. Rho independent termination mechanism involves (a) Binding of the rho protein upstream of the termination element (b) No protein factors and only RNA secondary structure and run of ‘U' s (c) Presence of UGA or UAA stop codons (d) Binding of accessory factors at termination signal 60. In eukaryotes the ribosomal RNA genes are transcribed by (a) Reverse transcriptase (b) RNA dependent RNA polymerase (c) RNA polymerase I

53. Efficient expression of a heterologous protein product is influenced by (a) Transcriptional efficiency

(d) RNA polymerase III 61. The term protein splicing refers to

(b) Copy number of the plasmid

(a) Removal of intervening sequences between the genes

(c) Codon bias

(b) Splicing out of introns from RNA

(d) All of these

(c) Removal of intervening protein sequences from the translated protein

54. The fundamental feature of the genetic code which allows the expression of a protein in any host is its (a) Triplet nature

(b) Universality

(c) Degeneracy

(d) Redundancy

55. The major groove of DNA is lined by (a) N3 of purine and N1 of pyrimidine (b) N7 of purine and O2 of pyrimidine (c) N7 of purine and C4 of Pyrimidine (d) None of these

(d) Joining (splicing) of two different gene products to generate a novel protein 62. The double stranded DNA molecule of a virus was found by electron microscopy to have a length of 34 M. (a) How many nucleotide pairs are there in one molecule ? (1) (b) How many complete turns of the two polynucleotide chains are present in such a double helix ? (1)

Genetics - II : Molecular Genetics

8.7

ANSWERS 1. (d)

2. (d)

3. (c)

4. (b)

5. (c)

6. (b)

7. (c)

8. (d)

9. (a)

10. (a)

11. (b)

12. (c)

13. (c)

14. (d)

15. (c)

16. (b)

17. (c)

18. (a)

19. (5)

20. (45)

21. (c)

22. (c)

23. (b)

24. (b)

25. (b)

26. (a)

27. (a)

28. (d)

29. (a)

30. (b)

31. (a)

32. (a)

33. (b)

34. (b)

35. (a)

36. (d)

37. (d)

38. (c)

39. (a)

40. (b)

41. (c)

42. (b)

43. (c)

44. (b)

45. (a)

46. (a)

47. (d)

48. (c)

49. (d)

50. (a)

51. (a, d)

52. (b)

53. (d)

54. (c)

55. (c)

56. (b)

57. (a)

58. (b)

59. (b)

60. (c)

61. (c)

62. (*)

EXPLANATIONS 1. The binding status of regulatory proteins, in lac operon when concentrations of both lactose and glucose are very low in the culture medium will be both the repressor and cAMP-CAP complex remain bound to their respective binding sites. 2. miRNA is associated with RNA interference, nuclear export. siRNA and shRNA also functions in RNA interference. snRNA is involved in spliceosomal reaction. 3. On the ribosome, the anticodon of the incoming aminoacyl-tRNA is matched against the mRNA codon positioned in the A-site. The translational machinery select the aminoacyl-tRNA carrying the matching anticodon out of 64 different combinations. During this proof-reading, aminoacyl-tRNAs with non-cognate anticodons are thrown out of the ribosome and replaced by new amino acyl-tRNAs that are to be checked. When the right aminoacyl-tRNA enters the Asite the growing polypeptide in the P-site is almost immediately linked to the new amino acid in the A-site via a peptide bond. Formation of the peptide bond is catalysed by the ribosome itself. The reaction leaves an empty tRNA in the ribosomal P-site and the new peptidyl-tRNA in the A-site. 4. A frameshift mutation is a genetic mutation caused by insertions or deletions of a number of nucleotides in a DNA sequence that is not divisible by three. Due to the triplet nature of gene expression by codons, the insertion or deletion can change the reading frame resulting in a completely different translation from the original. Therefore P, S and T do not have frameshift. 5. Number of 3 tuples = L – k + 1 L  Total length of sequence = 13 k  tuples = 3 Then 3 tuples = 13 – 3 + 1 = 11

6.

Group I P. Paclitaxel Q. Colchicine R. Etoposide

Group II 2. Inhibits microtubule depolymerization 6. Inhibits microtubule polymerization 3. Inhibits DNA replication

S. Methotrexate 5. Inhibits dihydrofolate reductase 7. Only two codons are possible in 1st case : UCU CUC 3 codons are possible in 2nd case : CCU CUC UCC Due to wobble hypothesis the unique amino acid codon in 2nd case is only CCU hence a new amino acid is synthesized. c b d a 8. S   X   Y   Z  P

9. In the more matured forms of RNAs in case of prokaryotic organisms, the 5' end of mRNA consists of a triphosphate group while there is a single monophosphate group present at the 5' terminal of tRNA molecules. 10. Given topological winding no. of a closed circular DNA = 191 Writhing number = - 4 So, linking number = Winding No. + Writhing No. = 191 – 4 = 187 when the molecule is laid flat on the surface Number of base pair per turn = Bases/ winding no. = 2kb/ 191 = 2048/ 191 = 10.722 11. Peptide I has 15 sequences out of which 9 are identical to the 15 sequences of Peptide II. So, the percentage identity between the two sequences will be = 9/15  100 % = 60 % Now, arginine and glutamine, leucine and isoleucine and glutamate and aspartate are similar amino acids = 3 similar + 9 identical amino acids between the two peptides So percentage similarity = 12/15  100% = 80%

8.8

Genetics - II : Molecular Genetics

12. Shine-Dalgarno (SD) sequence is a ribonucleotide sequence for the formation of the preinitiation complex between a 30S ribosomal subunit and an mRNA molecule. It is a seven-nucleotide, pyrimidine-rich sequence near the 3? terminus of 16S (prokaryotic) or 18S eukaryote ribosomal RNA, found in many organisms, that forms base pairs with a complementary three- to seven-nucleotide, purine-rich sequence (the Shine-Dalgarno sequence) preceding the initiation codon in many mRNA molecules, thereby allowing the initiating ribosomal subunit to bind the mRNA so as to discriminate between AUG or GUG (i.e. initiation) triplets and to select the one at the beginning of a cistron. Anti Shine Dalgarno sequences are present in 16s rRNA molecules for the recognition of SD sequences in eubacterial mRNA. 13. The most frequent and most regular of the domain structures are the alpha / beta domains. These domains consist of a central parallel or mixed betasheet surrounded by alpha-helix. All the glycolytic enzymes are alpha / beta structure proteins. There are three maim classes of alpha / beta structure proteins, TIM barrel, Rossman fold, and horseshoes fold (leucine-rich motif). The alpha/beta barrel fold is adopted by most enzymes performing a variety of catalytic reactions, but with very low sequence similarity. The cluster centers are found to occur in the middle of the strands or in the C-terminal of the strands. In most cases, the residues forming the clusters are part of the active site or are located close to the active site. The active site is majorly present at the carboxy-terminal side of the  sheets or the  strands. 14. DNA is the repository of genetic information gathered over incalculable millions of years and it is stored in a stable form inside the cell. However, chemical molecules have some degree of instability and often breakdown with time. But DNA molecules have to remain stable. The presence of the –OH group on the second carbon chain of the ribose sugar makes a ribopolynucleotide less stable than a deoxyribose molecule. The presence of 2-OH group on the ribose sugar makes it susceptible to nucleophilic attack in the presence of OH (ions) on the 5phosphorous atom, thus causing breakage of the phosphodiester link and forming a 2 ,3 cyclic phosphate. The –OH ion facilitates the reaction because it can generate a 2–O ion from the 2OH group, which attacks the phosphorous atom and converts the phosphodiester group into a 2, 3  –cyclic nucleotide, thus breaking the polynucleotide chain. Hydrolysis of the cyclic

nucleotide produces a mixture of 2  and 3  nucleotides at the breakpoint. Moreover, the double stranded DNA has relatively small grooves as opposed to the larger grooves on RNA molecules. This provides ample docking space for damaging enzymes. 15. The addition of a methyl group (-CH 3 ) to a molecule is known as methylation. Extensive methylation of cytosine in DNA is correlated with reduced transcription. DNA methylation is one of several epigenetic mechanisms that cells use to control gene expression. Proteins can be methylated on the side-chain nitrogen of arginine and lysine residues or on carboxy-termini. Protein methylation is a way of subtly changing the primary sequence of a peptide so that it can encode more information. This common posttranslational modification is implicated in the regulation of a variety of processes including protein trafficking, transcription and protein-protein interactions. 16. DNA sequencing is the determination of the precise sequence of nucleotides in a sample of DNA. The most popular method for doing this is called the dideoxy method or Sanger method. DNA is synthesized from four deoxynucleotide triphosphates. The dideoxy method gets its name from the critical role played by synthetic nucleotides that lack the -OH at the 3 carbon atom. A dideoxynucleotide can be added to the growing DNA strand but when it is, chain elongation stops because there is no 3 -OH for the next nucleotide to be attached to. For this reason, the dideoxy method is also called the chain termination method. The DNA sample is divided into four separate sequencing reactions, containing the four standard deoxynucleotides (dATP, dGTP, dCTP and dTTP) and the DNA polymerase. To each reaction is added only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP, or ddTTP). These dideoxynucleotides are the chain-terminating nucleotides, lacking a 3 -OH group required for the formation of a phosphodiester bond between two nucleotides during DNA strand elongation. Incorporation of a dideoxynucleotide into the nascent (elongating) DNA strand therefore terminates DNA strand extension, resulting in various DNA fragments of varying length. The dideoxynucleotides are added at lower concentration than the standard deoxynucleotides to allow strand elongation sufficient for sequence analysis. 17. DNA Polymerase I is a template-dependent DNA polymerase which catalyzes 53 synthesis of DNA. DNA Polymerase I catalyzes the template-

Genetics - II : Molecular Genetics

directed polymerization of nucleotides into duplex DNA in a 5  3  direction. In addition to polymerase activity, this DNA polymerase exhibits 3 to 5 and 5 to 3 exonuclease activity. It is able to utilize nicked circular duplex DNA as a template and can unwind the parental DNA strand from its template. Because of its exonuclease activity in the 3 to 5 direction, the polymerase can remove or excise the RNA primers synthesized during replication. 18. Ribonuclease P (RNase P) is the endoribonuclease that generates the mature 5-ends of tRNA by removal of the 5-leader elements of precursortRNAs. Although it carries out a biochemically simple reaction, RNase P is a complex ribonucleoprotein particle composed of a single large RNA and at least one protein component. RNase H (Ribonuclease H) is an endoribonuclease that specifically hydrolyzes the phosphodiester bonds of RNA which is hybridized to DNA. The snRNAs, along with their associated proteins, form ribonucleoprotein complexes (snRNPs), which bind to specific sequences on the pre-mRNA substrate. This intricate process results in two sequential transesterification reactions. These reactions will produce a free lariat intron and ligate two exons to form a mature mRNA. This results in the splicing. CstF is recruited by cleavage and polyadenylation specificity factor (CPSF) and assembles into a protein complex on the 3' end to promote the synthesis of a functional polyadenine tail, which results in a mature mRNA molecule ready to be exported from the cell nucleus to the cytosol for translation. 19. Given sequence: Phe-Trp-Met-Gly-Ala-Lys-LeuPro-Met-Asp-Gly-Arg-Cys-Ala-Gln First it is digested with Trypsin and than with Cyanogen Bromide in a process of sequential cleavage. Cyanogen bromide hydrolyzes peptide bonds at the C-terminus of methionine residues. Trypsin cleaves peptide chains mainly at the carboxyl side of the amino acids lysine or arginine, except when either is followed by proline. After trypsin digestion, fragments generated are: Phe-Trp-Met-Gly-Ala-Lys; Cys-Ala-Gln; Leu-ProMet-Asp-Gly-Arg After CNBr digestion, the fragments generated are: Phe-Trp-Met; Gly-Ala-Lys; Cys-Ala-Gln; Leu-ProMet; Asp-Gly-Arg 20. Given, No. of polydactylous persons = 36 No. of people having the allele for polydactyly = 80

8.9

Therefore, the extent of penetrance in the population = 36/80 x 100 % = 45 % 21. Mismatch repair deals with correcting mismatch of the normal bases using BER or NER enzyme systems. The system assumes that parental strands are methylated and freshly synthesized daughter strands are non methylated. 22. To investigate the effect of serine 78 phosphorylation on p21 activity, replacement of serine 78 with aspartic acid is done, creating the phosphomimic p21s78D. 23. Non sense mutations causes premature stop to be introduced that would lead to truncated or incomplete protein synthesis. If the protein synthesis has to be continued, another mutation in t-RNA gene would continue be advised to get full length polypeptide 24. The direction of shell coiling in snail is a dassical example for extra chromosomal inheritance, as in this it is determined by maternal gene effects and not that of the offspring. 25. An inverted repeat (or IR) is a sequence of nucleotides followed downstream by its reverse complement. It is a sequence found in identical (but inverted) form, for example, at the opposite ends of a transposon. Here, the DNA sequence which carries an inverted repeat is: ATGAGCCCCGAGTA TACTCGGGGCTCAT 26. Zinc fingers coordinate zinc ions with a combination of cysteine and histidine residues. They can be classified by the type and order of these zinc coordinating residues (e.g., Cys2His2, Cys4, and Cys6). 27. The folding of many newly synthesized proteins in the cell depends on a set of conserved proteins known as molecular chaperones. These prevent the formation of misfolded protein structures, both under normal conditions and when cells are exposed to stresses such as high temperature. Significant progress has been made in the understanding of the ATP-dependent mechanisms used by the Hsp70 and chaperonin families of molecular chaperones, which can cooperate to assist in folding new polypeptide chains. 28. Diphtheria toxin consists of a single polypeptide. Proteolysis yields two fragments (A and B) which are held together by a disulfide bond. The toxin binds to EGF-like domain of Heparin-binding EGF-like growth factor (HB-EGF) through

8.10

Genetics - II : Molecular Genetics

fragment B and is internalized with HB-EGF by receptor-mediated endocytosis. Diphtheria toxin catalyzes the ADP-ribosylation of, and inactivates, the elongation factor eEF-2. In this way, it acts to inhibit translation during eukaryotic protein synthesis. Tetracyclines are antibiotics which inhibit the bacterial growth by stopping protein synthesis. Three different specific mechanisms of tetracycline resistance have been identified so far: tetracycline efflux, ribosome protection and tetracycline modification. Streptomycin is a water-soluble aminoglycoside derived from Streptomyces griseus. Streptomycin is an aminoglycoside antibiotic that is currently used largely in the therapy of active tuberculosis. Like other aminoglycosides, streptomycin is thought to act by binding to bacterial ribosomes and inhibiting protein synthesis. Nevertheless, streptomycin is considered bacteriocidal as well as bacteriostatic. 29. N-methyl-N’-nitro-N-nitrosoguanidine is a highly reactive chemical that introduces alkyl radicals into biologically active molecules and thereby prevents their proper functioning. It could be used as an antineoplastic agent, but it might be very toxic, with carcinogenic, mutagenic, teratogenic, and immunosuppressant actions. It could also be used as a component of poison gases. This substance is reasonably anticipated to be a human carcinogen. The mutations induced by this mutagen are mostly purine-pyrimidine transitions of GC Æ AT. 30. Enhancer sequences are regulatory DNA sequences that, when bound by specific proteins called transcription factors, enhance the transcription of an associated gene. Regulation of transcription is the most common form of gene control, and the activity of transcription factors allows genes to be specifically regulated during development and in different types of cells. Transcription factors can bind to enhancer sequences located upstream or downstream from an associated gene, resulting in stimulation or enhancement of transcription of the related gene. Enhancer sequences act upon genes on the same DNA molecule; however, enhancer sequences can be located thousands of base pairs away from the transcription start site of the gene being regulated. Because DNA is folded and coiled in the nucleus, the enhancer may actually be located near the transcription start site in the folded state. 31. The Shine-Dalgarno sequence is a ribosomal binding site in the mRNA. The leucine zipper is

a three dimensional structural motif in proteins. These motifs are usually found as part of a DNAbinding domain in various transcription factors, and are involved in regular gene expression. Aminoacyl tRNA synthetase is an enzyme that catalyzes the esterification of a specific amino acid or its precursor to one of all its compatible cognate tRNAs to form an aminoacyl tRNA. RNAi works for the gene silencing (Small interfering RNA is mainly used in gene silencing and Antisense technology). 32. Gratuitous inducer is an analogue of a natural inducer that is capable of inducing an operon while not serving as a substrate for the enzyme being induced. IPTG is a gratuitous inducer. It induces gene expression but itself is not metabolized by the enzyme. 34. Peptidyl transferase is an amino acyltransferase and forms peptide links between adjacent amino acids using tRNA during translation process. This peptidyl transferase activity is mediated by 23 ribosomal RNA. This ribozyme 23 rRNA is the transferase component in bacteria. Also in eukaryotic cells, the same 23 rRNA holds the subunit containing peptidyl transferase component acting as the ribozyme. 35. DNA gyrase catalyse ATP dependant –ve supercoiling as these are topoisomerases which influence the topological state of DNA. It is type II topoisomerase which induce –ve supercoils (or relax +ve supercoils) by looping the template and cutting one of the double helices and then 2 ends are twisted for –ve supercoils. Type I topoisomerase unwinds –ve supercoils to control protein synthesis and facilitate DNA replication, and ease out problems due to unwinding of double helical forms. 36. Region of eukaryotic chromosome DNA that is specifically vulnerable to nuclease attack perhaps because it is not wrapped in histone as nucleosomes are called the nuclease hypersensitive sites in the chromosomes. Thus they are H2 and H4 free, H1 and H2 free, H3 and H4 histone free. 37. The MTT assay is a colorimetric assay which is used for measuring the activity of cellular enzymes that reduce the tetrazolium dye, MTT, to its insoluble formazan dye, giving a purple color. These assays measure cellular metabolic activity via NAD(P)H-dependent cellular oxidoreductase enzymes and may, under defined conditions, reflect the number of viable cells present. MTT (3-(4,5-dimethylthiazol-2-yl)-2,5-

Genetics - II : Molecular Genetics

diphenyltetrazolium bromide, a yellow tetrazole), is reduced to purple formazan in living cells. A solubilization solution (usually either dimethyl sulfoxide, an acidified ethanol solution, or a solution of the detergent sodium dodecyl sulfate in diluted hydrochloric acid) is added to dissolve the insoluble purple formazan product into a colored solution. Reduction of MTT and other tetrazolium dyes increases with cellular metabolic activity due to elevated NAD(P)H flux. 38. Chloramphenicol binds to the peptidyl transferase enzyme to inhibit transfer of the growing polypeptide to the next amino acid occupying the “Acceptor” site. Norfloxacin is a nalidixic acid analogue and one of the most potent DNA gyrase inhibitors. Puromycin is a secondary metabolite of Streptomyces alboniger that blocks protein biosynthesis. Puromycin is a structural analogue of the 32 end of aminoacyl-transfer RNA. Rifampicin inhibits bacterial RNA polymerase by binding to the beta subunit of RNA polymerase. 39. Commonly used reporter genes that induce visually identifiable characteristics usually involve fluorescent and luminescent proteins. Examples include the gene that encodes jellyfish green fluorescent protein (GFP), which causes cells that express it to glow green under blue light, the enzyme luciferase, which catalyzes a reaction with luciferin to produce light, and the red fluorescent protein from the gene dsRed. 40. IPTG (I-8000, I-8050) is an artificial inducer of the lac operon. IPTG binds to the lac repressor and inactivates it, preventing its binding to the lac operator and thus inducing the expression of beta-galactosidase. IPTG cannot be hydrolyzed and broken down by the cell and is therefore called a "gratuitous" inducer. 41. Zinc fingers function as interaction modules that bind DNA, RNA, proteins, or other small, useful molecules. 42. RNA interference (RNAi) also called post transcriptional gene silencing (PTGS), is a biological process in which RNA molecules inhibit gene expression, typically by causing the destruction of specific mRNA molecules. Two types of small ribonucleic acid (RNA) molecules: microRNA (miRNA) and small interfering RNA (siRNA) - are central to RNA interference. 43. A zinc finger protein is a DNA-binding protein domain consisting of zinc fingers ranging from two in the Drosophila regulator ADR1, the more common three in mammalian Sp1 up to nine in TF 111A. They occur in nature as the part of

8.11

transcription factors conferring DNA sequence specificity as the DNA-binding domain. 44. Mobile genetic elements are the nucleotide sequences that can change their position in the genome. There are two categories of mobile genetic elements: DNA mobile genetic elements or DNA transposons and retroelements. There are 3 kinds of retroelements:

 LINE( Long Interspersed Nuclear Elements)  SINE( Short Interspersed Nuclear Elements)  LTR(retrovirus like elements with Long Terminal Repeats) 45. Alkaline phosphatase is a hydrolase enzyme responsible for removing phosphate groups from many types of molecules, including nucleotides, proteins, and alkaloids. The process of removing the phosphate group is called dephosphorylation. Polynucleotide kinase (or PNK) is a T7 bacteriophage (or T4 bacteriophage) enzyme that catalyzes the transfer of a gamma-phosphate from ATP to the free hydroxyl end of the 5' DNA or RNA.The resulting product could be used to endlabel DNA or RNA, or in a ligation reaction. 46. Luciferase bioluminescence color can vary between yellow-green ( λmax = 550 nm) to red (λmax = 620) There are currently several different mechanisms describing how the structure of luciferase affects the emission spectrum of the photon and effectively the color of light emitted. One mechanism proposes that the color of the emitted light depends on whether the product is in the keto or enol form. The mechanism suggests that red light is emitted from the keto form of oxyluciferin, while green light is emitted from the enol form of oxyluciferin. 47. Telomerase contains a catalytic unit with reverse transcriptase activity and an RNA part that provides template for telomere extension. Telomerase is normally expressed only in stem cells such as those found in hematopoietic tissues and germline cells but is nearly absent in most somatic cells. Induction of telomerase synthesis bypasses normal cellular senesce in cancer cells and endows them with unlimited replicative potential. 48. Ionizing radiation is generally harmful and potentially lethal to living things but can have health benefits in radiation therapy for the treatment of cancer and thyrotoxicosis. Its most common impact is the induction of cancer with a latent period of years or decades after exposure. High doses can cause visually dramatic radiation burns, and/or rapid fatality through acute

8.12

Genetics - II : Molecular Genetics

radiation syndrome. Controlled doses are used for medical imaging and radiotherapy. Some scientists suspect that low doses may have a mild hormetic effect that can improve health. 49. ACC synthase catalyzes the synthesis of ACC, a precursor for ethylene. Ethylene serves as a hormone in plants for stimulating or regulating the ripening of fruit, the opening of flowers, and the abscission (or shedding) of leaves and hence none of the given options is suitable. 50. ligation is the ATP dependent procedure. The mechanism of DNA ligase is to form two covalent phosphodiester bonds between 3 hydroxyl ends of one nucleotide, (“acceptor”) with the 5  phosphate end of another (“donor”). ATP is required for the ligase reaction. 51. Prokaryotes do not carry out the same kinds of post- translational modifications such as glycosylation, phosphorylation as eukaryotes do. This affects a protein’s activity or stability, or at least its response to antibodies. 52. While DNA contains deoxyribose, RNA contains ribose (in deoxyribose there is no hydroxyl group attached to the pentose ring in the 2 position). These hydroxyl groups make RNA less stable than DNA because it is more prone to hydrolysis. 53. Heterologous protein expression depends upon good codon biasing, transcriptional efficiency and optimum copy number production of the expression vector system. 54. Degeneracy produces the ultimate amino acid combination for protein expression. 55. Each groove in DNA is lined by potential hydrogen-bond donor and acceptor atoms that enable specific interactions with proteins (seeFigure 27.7). In the minor groove, N-3 of adenine or guanine and O-2 of thymine or cytosine can serve as hydrogen acceptors, and the amino group attached to C-2 of guanine can be a hydrogen donor. In the major groove, N-7 of guanine or adenine is a potential acceptor, as are O-4 of thymine and O-6 of guanine. The amino groups attached to C-6 of adenine and C-4 of cytosine can serve as hydrogen donors.

56. β-Lactoglobulin is the major whey protein of cow and sheep’s milk (~3 g/l), and is also present in many other mammalian species; a notable exception being humans. Its structure, properties and biological role have been reviewed many times. 57. Reverse genetics is an approach to discover the function of a gene by analyzing the phenotypic effects of specific gene sequences obtained by DNA sequencing. This investigative process proceeds in the opposite direction of socalled forward genetic screens ofclassical genetics. Simply put, while forward genetics seeks to find the genetic basis of a phenotype or trait, reverse genetics seeks to find what phenotypes arise as a result of particular genetic sequences. 58. In RNA GC content will be one half of the Double strand DNA. So G+C content of RNA is 34%. 59. Intrinsic termination (also called Rhoindependent termination) is a mechanism in prokaryotes that causes RNA transcription to be stopped.[1] In this mechanism, the mRNA contains a sequence that can base pair with itself to form a stem-loop structure 7-20 base pairs in length that is also rich incytosine-guanine base pairs. These bases form three hydrogen bonds between each other and are therefore particularly strong.[2] Following the stem-loop structure is a chain of uracil residues. 60. RNA polymerase I (also called Pol I) is, in higher eukaryotes, the enzyme that only transcribes ribosomal RNA (but not 5S rRNA, which is synthesized by RNA Polymerase III), a type of RNA that accounts for over 50% of the total RNA synthesized in a cell. 61. Protein splicing is an intramolecular reaction of a particular protein in which an internal protein segment (called an intein) is removed from a precursor protein with a ligation of C-terminal and N-terminal external proteins (called exteins) on both sides.

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9

9.1

Advanced Biotechnology

C HAPTER 2016

1. Which one of the following is NOT a therapeutic agent based on nucleic acid for the treatment of genetic disorders?

6. A variety of genetic elements are used in the transgenic plant research. Match the genetic elements (Column-I) with their corresponding source (Column-II).

(a) Antisense oligonucleotide

Column-I

(b) Ribozyme

P. Ubiquitin1 promoter

(c) Aptamer

Q. Nos transcriptional terminator

(d) Avidin

R. bar selection marker gene

2. Which one of the following is NOT used for the measurement of cell viability in animal cell culture? (a) Trypan blue dye exclusion (b) Tetrazolium (MTT) assay (c) LDH activity in the culture medium (d) Coulter counter 3. A biological process is involved in the ________ treatment of industrial effluent. (a) primary

(b) secondary

(c) tertiary

(d) quaternary

4. Select the CORRECT combination of genetic components that are essential for the transfer of TDNA segment from Agrobacterium tumefaciens to plant cells. (a) Border repeat sequences and oncogenes (b) Border repeat sequences and vir genes (c) Opine biosynthetic genes and vir genes (d) Opine biosynthetic genes and oncogenes 5. Match the secondary metabolites (Column-I) with the corresponding plant species (Column-II). Column-I

Column-II

P. Morphine

1. Datura stramonium

Q. Pyrethrins

2. Catharanthus roseus

R. Scopolamine

3. Papaver somniferum

S. Vincristine

4. Tagetes erecta

Codes:

S. gus reporter gene Column-II 1. Agrobacterium tumefaciens 2. Streptomyces hygroscopicus 3. Escherichia coli 4. Zea mays Codes: P

Q

R

S

(a) 2

1

3

4

(b) 2

3

4

1

(c) 3

4

1

2

(d) 4

1

2

3

7. In animal cell culture, a CO2 enriched atmosphere in the incubator chamber is used to maintain the culture pH between 6.9 and 7.4. Which one of the following statements is CORRECT? (a) Higher the bicarbonate concentration in the medium, higher should be the requirement of gaseous CO2 (b) Lower the bicarbonate concentration in the medium, higher should be the requirement of gaseous CO2 (c) Higher the bicarbonate concentration in the medium, lower should be the requirement of gaseous CO2 (d) CO 2 requirement is independent of bicarbonate concentration in the medium

2015

P

Q

R

S

(a) 4

3

1

2

(b) 3

4

1

2

(c) 2

3

4

1

(a) oriC

(b) Selection marker

(d) 4

1

2

3

(c) CMV promoter

(d) Ribosome binding site

8. Which one of the following features is NOT required in a prokaryotic expression vector ?

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(c) NO2–

(d) NH4+

2kb

11. Which one of the following is a second generation genetically engineered crop ?

(b) 2kb

(a) Bt brinjal (c) Golden rice

(c)

(d) Bt rice

2kb HaeIII

(b) Roundup soyabean

5 –GACCTGTGG------------------ATACGGGAT –3 3 –CTGGACACC-------------------TATGCCCTA –5 Primers P. 5 –GACCTGTGG–3 Q. 5 –CCACAGGTC–3 R. 5 –TAGGGCATA–3 S. 5 –ATCCCGTAT–3 (a) P and R

(b) P and S

(c) Q and R

(d) Q and S

Marker Sample

Marker Sample

13. A linear double stranded DNA of length 8 kbp has three restriction sites. Each of these can either be a BamHi or a HaeIII site. The DNA was digested completely with both enzymes. The products were purified and subjected to an endfilling reaction using the Klenow fragment and [ –32P]-dCTP. The products of the end-filling reaction were purified, resolved by electrophoresis, stained with ethidium bromide (EtBr) and then subjected to autoradiography. The corresponding images are shown below.

5kb

5kb

3kb 2kb

3kb 2kb

1kb

1kb

300 900

5-GGATCC-3 BamHI 3-CCTAGG-5  5-GGCC-3 3-CCGG-5 

Autoradiograph

HaeIII

2kb HaeIII

2kb

12. Choose the appropriate pair of primers to amplify the following DNA fragment by the polymerase chain reaction (PCR).

E tB r

2kb

(d) 2kb

HaeHI

(a)

3kb

2kb

1kb BamHI

(b) N2O

3kb

1kb BamHI

(a) N2

3kb

1kb HaeIII

10. Which one of the following is NOT a product of denitrification in Pseudomonas?

BamHI

(d) > 1

HaeHI

(c) 1

BamHI

(b) < 1

BamHI

(a) 0

The numbers below each band in the sample lane in the autoradiograph represent their mean signal intensity in arbitrary units. Which one of the following options is the correct restriction map of the DNA ? BamHI

9. In DNA sequencing reactions using the chain termination method, the ratio of ddNTPs to dNTPs should be

BamHI

9.2

3kb

1kb

14. Assuming random distribution of nucleotides, the average number of fragments generated upon digestion of a circular DNA of size 4.3  105 bp with AluI(5-AGCT-3) is ______  103. 15. Plasmid DNA (0.5 g) containing an ampicillin resistance marker was added to 200 l of competent cells. The transformed competent cells were diluted 10,000 times, out of which 50 l was plated on agar plates containing ampicillin. A total of 35 colonies were obtained. The transformation efficiency is _____ 106 cfu. g–1.

2014 16. The statistical frequency of the occurrence of a particular restriction enzyme cleavage site that is 6 bases long can be estimated to be (a) once every 24 bases (b) once every 256 bases (c) once every 1024 bases (d) once every 4096 bases 17. The growth medium for mammalian cells contains serum. One of the major functions of serum is to stimulate cell growth and attachment. However, it must be filter sterilized to (a) remove large proteins (b) remove collagen only (c) remove mycoplasma and microorganisms (d) remove foaming agents

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18. The plasmid DNA was subjected to restriction digestion using the enzyme EcoRI and analysed on an agarose gel. Assuming digestion has worked (the enzyme was active), match the identity of the DNA bands shown in the image in Group I with their identity in Group II.

(a) the coat of the intact seeds blocks the pungent volatiles from being released (b) the pungent chemicals are stored as inactive conjugates and compartmentalized from the enzymes that convert them into active chemicals (c) the pungent chemicals are formed only after the reaction with atmospheric oxygen

Uncut EcoRI A

(d) the pungent chemicals are formed only after the reaction with atmospheric carbondioxide

B D C

Group–I

Group–II

P. Bands labeled as A 1. Nicked Q. Band labeled as B

2. Supercoiled

R. Band labeled as C

3. Concatemers

S. Band labeled as D

4. Linear

21. The length of the minimum unique stretch of DNA sequence that can be found only once in a 3 billion base pairs long genome is (a) 14

(b) 15

(c) 16

(d) 18

2013 22. Human genome sequencing project involved the construction of genomic library in (a) bacterial artificial chromosome

P

Q

R

S

(b) pBR322

(a) 3

1

2

4

(c) bacteriophage

(b) 1

4

3

2

(d) pcDNA3.1

(c) 4

3

1

2

(d) 4

1

2

3

23. In nature, the horizontal gene transfer across bacteria is mediated by

19. Match the following plant sources with their secondary metabolites and medical uses. Source plant

Secondary metabolites

P. Belladona

1. Menthol

Q. Foxglove

2. Atropine

R. Pacific yew

3. Digitalin

S. Eucalyptus

4. Taxol

Medical use

(a) gene cloning followed by transformation (b) conjugation and transformation (c) conjugation only (d) transformation only 24. In nature, Agrobacterium tumefaciens mediated infection of plant cells leads to P. crown gall disease in plants Q. hairy root disease in plants R. transfer of T-DNA into the plant chromosome

A. Cancer treatment

S. transfer of Ri-plasmid into the plant cell

B. Heart disease C. Eye examination D. Cough P

9.3

(a) S only

(b) P and R only

(c) Q and S only

(d) Q only

25. Match the herbicides in Group Iwith the target enzymesin Group II.

Q

R

S

(a) 2-C

3-B

4-A

1-D

Group I

(b) 3-C

2-A

1-D

4-B

P. Glyphosate

1. Nitrilase

(c) 2-C

4-B

1-A

3-D

Q. Bromoxynil

2. Acetolactatesyn thetase

(d) 1-B

4-C

2-D

3-A

R. Sulphonylureas

3. Dehalogenase

S. Dalapon

4. 5-Enolpyruvylshikimate

20. The pungency of mustard seeds is primarily due to secondary metabolites such as isothiocyanate and nitrile. The pungency is usually felt only when the seeds are crushed. This is because of

Group II

3-phosphate synthase (a) P-4, Q-1, R-2, S-3

(b) P-2, Q-1, R-4, S-3

(c) P-4, Q-3, R-2, S-1

(d) P-3, Q-2, R-4, S-1

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9.4

26. A complete restriction digestion of a circular plasmid (5000bp) was carried out with HindIII, BamHI and EcoRIindividually. Restriction digestion yielded following fragments. Plasmid + HindIII’ 1200bp and3800bp Plasmid + BamHI’ 5000bp The number of sites for EcoRI, BamHIandHindIIIpresent on this plasmid are (a) EcoRI-2, BamHI-1, HindIII-2 (b) EcoRI-1, BamHI-1, HindIII-2 (c) EcoRI-3, BamHI-2, HindIII-1 (d) EcoRI-2, BamHI-2, HindIII-1 Statement for Linked Answer Q. 27 and 28: A DNA fragment of 5000bp needs to be isolated from E.coli (genome size 4  103kb) genomic library. 27. The minimum number of independent recombinant clones required to represent this fragment ingenomic library are (a) 16  10

(b) 12  10

(c) 8  10

(d) 1.25  102

2

(a) Inactivated Sendai virus (b) Ca2+ at alkaline pH (c) Polyethylene glycol (d) Colchicine

Plasmid + EcoRI’ 2500bp

2

32. Which one of the following is NOT a protoplast fusion inducing agent?

2

28. The number of clones to represent this fragment in genomic library with a probability of 95% are (a) 5.9  103

(b) 4.5  103

(c) 3.6  103

(d) 2.4  103

2012 29. The basis for blue-white screening with pUC vectors is (a) intraallelic complementation (b) intergenic complementation (c) intragenic suppression (d) extragenic suppression 30. Dimethyl sulfoxide (DMSO) is used as a cryopreservant for mammalian cell cultures because (a) it is an organic solvent (b) it easily penetrates cells (c) it protects cells by preventing crystallization of water (d) it is aiso utilized as a nutrient 31. Protein-DNA interactions in vivo can be studied by

33. A single base pair of DNA weighs 1.1  10–21 grains. How many picomoles of a plasmid vector of length 2750 bp are contained in 1 ug of purified DNA? (a) 0.30

(b) 0.55

(c) 0.25

(d) 0.91

34. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: The production of secondary metabolites in plant cell cultures is enhanced by the addition of elicitors. Reason: Elicitors induce the expression of enzymes responsibie for the biosynthesis of secondary metabolites. (a) Both (A) and (R) are true but (R) is not the correct reason for (A) (b) Both (A) and (R) are true and (R) is the correct reason for (A) (c) (A) is true but (R) is false (d) (A) is false but (R) is true 35. Determine the correctness or otherwise of the following Assertion (A) and Reason (R) Assertion: In direct somatic embry-ogenesis, embryos are developed without going through callus formation. Reason: This is possible due to the presence of pre-embryonically determined cells. (a) Both (A) and (R) are true but (R) is not the correct reason for (A) (b) (A) is false but (R) is true (c) (A) is true but (R) is false (d) Both (A) and (R) are true and (R) is the correct reason for (A)

2011 36. Embryonic stem cells are derived from

(a) gel shift assay

(a) fertilized embryo

(b) southern hybridization

(b) unfertilized embryo

(c) chromatin immunoprecipitation assay

(c) sperm

(d) fluorescence in situ hybridization assay

(d) kidney

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37. Yeast artificial chromosomes (YAC's) are used for cloning

43. The study of evolutionary relationships is known as

(a) large segments of DNA

(a) genomics

(b) mRNA

(b) proteomics

(c) bacterial DNA

(c) phylogenetics

(d) yeast DNA

(d) genetics

38. The product commercially produced by animal cell culture is (a) insulin (b) tissue plasminogen activator (c) interferon

9.5

44. Protein-protein interactions are studied by P. DNA foot printing Q. Yeast two hybrid system R. Ligase chain reaction S. Mass spectrometry

(d) hepatitis B vaccine 39. Hydrated synthetic seeds which are produced by ion exchange reaction involve mixing the somatic embryos in a solution of

(a) P and S only

(b) Q and S only

(c) P and R only

(d) Q and R only

45. Identify the CORRECT statements

(a) sodium alginate and dropping it in a solution of calcium nitrate

P. 5 and 3 ends of the transcripts can be mapped by utilizing polymerase chain reaction

(b) calcium alginate and dropping it in a solution of sodium nitrate

Q. S1 nuclease can cleave the DNA strand of a DNA-RNA hybrid

(c) calcium alginate and dropping it in a solution of ammonium nitrate

R. T4 polynucleotide kinase is used for labeling 3 end of DNA

(d) mannitol and dropping it in a solution of sodium nitrate

S. Baculovirus (Autographa californica) can be used as an insect expression vector

40. Shoot organogenesis by tissue culture results into

(a) P and Q only

(b) R and S only

(a) a bipolar structure that has no vascular connection with the explant

(c) P and S only

(d) Q and R only

(b) a monopolar structure that has a strong connection with the pre-existing vascular tissue of the explant (c) a monopolar structure that has no vascular connection with the explant (d) a bipolar structure that has a strong connection with the pre-existing vascular tissue of the explant 41. ‘Hairy roots' induced in vitro by the infection of Agrobacterium rhizogenes, are characterized by

46. HAT (hypoxanthine, aminopterin and thymidine) is used for selecting the hybridomas based on the following I. Only hybridoma will grow since it inherited the HGPRT genes from B-cells and can synthesize DNA from hypoxanthine. II. Myeloma cells will not grow in cultures since de novo synthesis is blocked by aminopterin and due to the lack of HGPRT enzyme. (a) only I is true

P. a high degree of lateral branching

(b) only II is true

Q. genetic instability of culture

(c) both I and II are true

R. an absence of geotropism

(d) I is true and II is false

S. poor biomass production (a) P and R only

(b) P and Q only

(c) Q and R only

(d) R and S only

42. Restriction endonucleases which recognize and cut same recognition sequences are known as (a) isoschizomers (b) isozymes (c) isoaccepting endonucleases (d) abzymes

Statement for Linked Answer Questions 47 and 48: A 200 l of polymerase chain reaction has 100 template DNA molecules and the reaction was performed for 10 cycles. 47. How many molecules of amplicons will be generated? (a) 1.024  104

(b) 1.024  105

(c) 2.048  104

(d) 2.048  105

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9.6

48. How many molecules of amplicons will be present in 0.1 l of reaction? (a) 102.4

(b) 1024

(c) 51.2

(d) 512

(a) both (a) and (r) are true and (r) is the correct reason for (a) (b) both (a) and (r) are true and (r) is not the correct reason for (a) (c) (a) is true but (r) is false

2010 49. The bacteria known to be naturally competent for transformation of DNA is (a) Escherichia coli

(d) (a) is false but (r) is true 55. Match Group I with Group II Group I P. Fibronectin

(b) Bacillus subtilis (c) Mycobacterium tuberculosis (d) Yersinia pestis 50. Antibiotic resistance marker that CANNOT be used in a cloning vector in Gram negative bacteria is

Group II 1. Uptake of amino acids and glucose

Q. Insulin

2. Trypsin inhibitor

R. -Macroglobulin

3. Binds iron

S. Transferrin

4. Cell attachment to substratum

(a) Streptomycin

(b) Ampicillin

(a) P-2, Q-1, R-4, S-3

(b) P-3, Q-2, R-1, S-4

(c) Vancomycin

(d) Kanamycin

(c) P-4, Q-2, R-1, S-3

(d) P-4, Q-1, R-2, S-3

51. An example for template independent DNA polymerase is

56. Match the promoters listed in Group I with the tissues listed in Group II Group I

(a) DNA Polymerase I

Group II

(b) RNA polymerase

P. –Amylase

1. Endosperm

(c) Terminal deoxynucleotidyl transferase

Q. Glutenin

2. Tuber

(d) DNA polymerase III

R. Phaseollin

3. Aleurone

S. Patatin

4. Cotyledon

52. Somatic cell gene transfer is used for P. transgenic animal production Q. transgenic diploid cell production

(a) P-3, Q-1, R-4, S-2

(b) P-3, Q-4, R-1, S-2

(c) P-4, Q-2, R-1, S-3

(d) P-1, Q-3, R-2, S-4

57. Consider the following statements,

R. in-vitro fertilization S. classical breeding of farm animals (a) P, R and S

(b) P, Q and R

(c) P and R

(d) P only

53. Expressed Sequence Tag is defined as (a) a partial sequence of a codon randomly selected from DNA library (b) the characteristic gene expressed in the cell (c) the protein coding DNA sequence of a gene (d) uncharacterized fragment for DNA presence in the cell 54. Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion : Somatic embryogenesis in plants is a two step process comprising of embryo initiation followed by embryo production. Reason : Embryo initiation is independent of the presence of 2, 4 dichlorop-henoxyacetic acid whereas embryo production requires a high concentration of 2, 4-dichloroph-enoxyacetic acid.

I. T4 DNA ligase can catalyze blunt end ligation more efficiently than E.coli DNA ligase II. The ligation efficiency of T4 DNA ligase can be increased with PEG and ficoll. (a) only I is true (b) both I and II are true (c) only II is true (d) I is true and II is false 58. Match the items in Group I with Group II Group I

Group II

(Vectors)

(Maximum DNA packaging)

P.  phage

1. 35–45 kb

Q. Bacterial Artificial Chromosomes (BACs)

2. 100–300 kb

R. P1 derived Artificial Chromosomes (PACs)

3.  300 kb

S.  cosmid

4. 5–25 kb

(a) P-3, Q-4, R-1, S-2

(b) P-1, Q-3, R-2, S-4

(c) P-4, Q-3, R-2, S-1

(d) P-1, Q-2, R-3, S-4

Advanced Biotechnology

59. A mutant G protein with increased GTPase activity would

9.7

66. Which of the following statements are true about glyphosate tolerant transgenic plants?

(a) not bind to GTP

P. Transgenic plants detoxify glyphosate.

(b) not bind to GDP

Q. Transgenic plants produce an altered enzyme that is not affected by glyphosate.

(c) show increased signaling (d) show decreased signaling 60. A cell has five molecules of a rare mRNA. Each cell contains 4  105 mRNA molecules. How many clones one will need to screen to have 99% probability of finding at least one recombinant cDNA of the rare mRNA, after making cDNA library from such cell? (a) 4.50  105

(b) 3.50  105

(c) 4.20  105

(d) 4.05  105

61. Match Group I with Group II Group I P. Real Time-PCR

Group II 1. Biochips

R. Transgenic plants sequester glyphosate in vacuoles. S. Transgenic plants overcome the inhibition of aromatic amino acid biosynthesis. (a) P, Q

(b) R,S

(c) Q,S

(d) P, R

67. Match the items in Group 1 with their functions in Group 2: Group 1 P. rol genes Q. Opines R. Virulence genes

Q. 2-D Electrophoresis 2. Syber Green

S. Aux and cyt genes

R. Affinity chromatography

3. Antibody linked sephrose beads

Group 2

S. Micoroarray

4. Ampholytes

2. Tumor formation

1. Food and energy source

(a) P-1, Q-2, R-4, S-3

3. Hairy root induction

(b) P-2, Q-3, R-4, S-1

4. T-DNA transfer and integration

(c) P-2, Q-4, R-3, S-1

(a) P-4, Q-3, R -2, S-1

(b) P-3, Q-2, R-4, S-1

(d) P-3, Q-2, R-1, S-4

(c) P-1, Q-3, R-4, S-2

(d) P-3, Q-1, R-4, S-2

2009 62. To produce plants that are homozygous for all traits, the best choice is (a) Protoplast culture (b) Cell suspension culture (c) Anther and pollen culture (d) Apical merisem culture 63. Restriction endonucleases from two different organisms that recognize the same DNA sequence for cleavage are called (a) Isoschizomers (c) Concatamers

(b) Isozymes (d) Palindromes

64. Baculovirus expression system is used to express heterologous genes in (a) Mammals

(b) Plants

(c) Insects

(d) Yeasts

65. Virus resistant transgenic plants can be developed by the expression of (a) Cowpea trypsin inhibitor (b) Crystalline toxin protein (c) Defective movement protein (d) Snowdrop lectin

68. Which of the following statements hold true for pluripotent stem cells (PSCs) under in vitro conditions? P. PSCs can be maintained in an undifferentiated state Q. PSCs exhibit abnormal and unstable karyotypes R. PSCs can differentiate into a wide variety of cell types S. PSCs cannot be passaged continuously (a) P, Q

(b) P, R

(c) Q, R

(d) Q, S

69. Match the items in Group 1 with correct options in Group 2 : Group 1

Group 2

P. DNA footprinting

1. Protein-protein interaction

Q. Yeast two-hybrid system

2. VNTR

R. DNA fingerprinting 3. DNA binding protein S. SAGE

4. Transcriptome analysis

(a) P-1, Q-2, R-4, S-3

(b) P-3, Q-1, R-2, S-4

(c) P-3, Q-4, R-1, S-2

(d) P-4, Q-2, R-1, S-3

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9.8

70. Match the products in Group 1 with their possible applications in Group 2: Group 1

Group 2

P. Erythropoietin

1. Blood clot

Q. Anti-fibrin 99

2. Binding and transport of iron

R. Collagenase

3. Anaemia

S. Transferrin

4. Animal cell separation

(a) P-3, Q-1, R-4, S-2 (b) P-3, Q-4, R-1, S-2

75. Determine the correctness or otherwise of following Assertion [a] and Reason [r] Assertion: The enzymatic degradation of cell wall to obtain single cell called protoplast has helped immensely in developing somatic cell genetics in plants Reason: In plants or animals, fusion to two cells must occur through the plasma membrane (a) Both [a] and [r] are true and [r] is the correct reason for [a]

(c) P-2, Q-3, R-1, S-4

(b) Both [a] and [r] are true but [r] is not the correct reason for [a]

(d) P-2, Q-1, R-4, S-3

(c) [a] is true but [r] is false

71. A RNA polymerization assay was performed using 3 H UTP as the labelled nucleotide with a specific activity of 500 Ci/ mol (1 Ci = 2.2  106 counts per min). After 10 min incubation, the trichloroacetic acid-insoluble radioactivity was found to be 692521 counts per min as determined in a liquid scintillation counter working at 60% efficiency for 3H. The amount of UTP incorporated into the RNA will be (a) 15 nmol

(b) 105 nmol

(c) 150 nmol

(d) 50 nmol

2008 72. Parthenogenetic embryos in plants are those which are formed by (a) unfertilized eggs

(b) fertilized eggs

(c) sporophytic cells

(d) male gametophyte

73. Multiplication of genetically identical copies of a cultivar by asexual reproduction is known as (a) aclonal propagation (b) vegetative propagation (c) polyclonal propagation (d) clonal propagation 74. Identify the correct statement for the ‘HAT medium’ (P) Includes drug aminopterin to block major pathway for synthesis of deoxyribonucleotides (Q) Hypoxanthine is precursor for thymidine (R) Includes drug aminopterin to block major pathway for synthesis of polypeptides (S) Cells can grow in presence of aminopterin only if they have enzymes thymidine kinase and hypoxanthine-guanine phosphoribosyl transferase

(d) [a] is false but [r] is true 76. Which of the following statements are true with respect to a special complex called 'dicer'? (P) It consists of deoxyribonuclease and DNA fragments (Q) It consists of ribonuclease and RNA fragments (R) It is involved in gene silencing (S) It triggers apoptosis (a) P, R

(b) Q, R

(c) P, S

(d) Q, S

77. Match the items in groupl with the terms given in group 2 Group 1

Group 2

P. Lactobacillus and

1. Prebiotics

Bifidobacteria Q. Polychlorobenzenes

2. Probiotics

(PCBs) R. Fructo-oligosaccharides 3. Antibiotics S.  -Lactams

4. Xenobiotics

P

Q

R

S

(a) 2

4

1

3

(b) 3

4

1

2

(c) 4

1

2

3

(d) 1

3

4

2

78. A polymerase chain reaction was performed beginning with 400 template DNA molecules in a 100 l reaction. After 20 cycles of polymerase chain reaction, how many molecules of the amplified product will be present in O.1 l of reaction ? (a) 2.19  104 (b) 4.19  104

(a) P, Q

(b) P, S

(c) 2.19  105

(c) R, S

(d) Q, S

(d) 4.19  105

Advanced Biotechnology

2007

9.9

85. Meristems escape virus invasion because

79. Which of the following reagents is used for harvesting anchorage-dependent animal cells from culture vessels ?

(a) vascular system is absent in the meristem (b) of low metabolic activity in the meristem

(a) Trypsin/Collagensase

(c) the ‘virus inactivating system’ has low activity in the meristem

(b) Trypsin/Collagen

(d) of low endogenous auxin level

(c) Collagen/Fibronectin

2006

(d) DMSO 80. Protein binding regions of DNA are identified by one of the following techniques

86. A thermostable DNA polymerase that can carry out both reverse transcription reaction and polymerizations has been isolated from

(a) finger printing

(a) Thermococcus litoralis

(b) foot printing

(b) Thermus aquaticus

(c) southern blotting

(c) Thermotoga maritima

(d) western blotting

(d) Thermus thermophilus

81. Plant secondary metabolites (a) help to increase the growth rate of plant

87. When present in tissue culture medium, gibberellin

(b) help in plant reproduction processes

(a) helps to break dormancy of buds and bulbs

(c) provide defense mechanisms against microbial attack

(b) promotes dormancy development in buds and bulbs

(d) make the plant susceptible to unfavorable conditions

(c) is regarded as plant growth inhibitor

82. Oils rich in PUFA are NOT desirable for bio-diesel production because (a) they form epoxides in presence of oxygen (b) they do not form epoxides in presence of oxygen (c) they have high ignition temperature (d) they solidify at low temperature 83. Gynogenesis is a process of development of haploid plants

(d) prevents normal recognition of auxin molecule 88. To promote attachment and spreading of anchorage-dependent animal cells, the surface of the culture vessel needs to be coated with (a) trypsin

(b) collagen

(c) pronase

(d) polyglycol

89. For amplificatio of GC rich sequences by polymerase chain reacstion, identify the reagent that binds and stabilizes AT sequences and destabilizes GC regions.

(a) from a fertilized cell of female gametophyte

(a) Tetramethyl ammonium chloride

(b) from an unfertilized cell of female gametophyte

(b) Betaine

(c) from isolated pollen grains

(c) 7-deaza-2-deoxyguanosine triphosphate

(d) by selective elimination of chromosomes following distant hybridization

(d) Sodium dodecyl sulphate

84. Match the following marker genes in group 1 with suitable selecting agent in group 2

90. Which of the following statements is INCORRECT about immobilized plant cell cultures?

Group 1

Group 2

P. npt II

1. Glyphosate

(b) Cells remain active for long periods

Q. aroA

2. Phosphinothricin

R. hpt

3. Kanamycin

(c) Cell products or inhibitors can be removed easily

S. bar

4. Hygromycin B

(d) It provides low shear resistance to cells

(a) P-1, Q-2, R-4, S-3 (b) P-3, Q-2, R-4, S-1

(a) It is possible to use high cell densities

91. All the cells that participate in immune responses originate from a population of

(c) P-2, Q-3, R-4, S-1

(a) neutrophils

(b) stem cells

(d) P-3, Q-1, R-4, S-2

(c) macrophages

(d) lymphocytes

9.10

Advanced Biotechnology

92. Identify the natural plant growth regulators from the following list.

96. Match items in group 1 with correct options from those given in group 2

(P) Zeatin

Group 1

(Q) Benzylaminopurine (BAP)

P. Amperometric biosensor

(R) Indole-3-acetic acid (IAA)

Q. Evanescent wave biosensor

(S) 2, 4-Dichlorophenoxyacetic acid

R. Calorimetric biosensor

(a) P, Q

(b) Q, S

S. Potentiometric biosensor

(c) P, R

(d) R, S

Group 2

93. What are the experimental steps needed for screening an expression library for a clone encoding a protein X that has been isolated and purified ?

1. Light beam 2. Flux of redox electrons 3. Field effect transistors 4. Exothermic reaction

(P) m-RNA isolation

(a) P-3, Q-4, R-2, S-1

(Q) Antibody preparation

(b) P-2, Q-l, R-4, S-3

(R) Cloning into an appropriate vector

(c) P-3, Q-2, R-4, S-1

(S) Western blotting on transferred plaques

(d) P-2, Q-4, R-3, S-1

(a) P, S

(b) Q, S

(c) Q, R

(d) R, S

94. Majority of the cereals are highly recalcitrant to Agrobacterium-mediated transformation, and so direct transfor-mation methods have been developed to transform such plants. Which of the following direct transformation methods is applicable to intact plant tissues? (a) Calcium chloride and PEG-mediated transformation

97. Which of the following would result in somaclonal variation in micropropagated plants? (P) Propagation by axillary branching in the absence of plant growth regulators (Q) Cell suspension maintained for five years before induction of somatic embryogenesis (R) Cullus induction using 20 M 2, 4-Dichlorophenoxyacetic acid followed by shoot organogenesis

(b) Liposome-mediated transformation

(S) Shoot organogenesis from an explant in the absence of an intermediate callus phase

(c) Electroporation

(a) P, Q

(b) Q, R

(d) Transformation using microprojectiles

(c) P, S

(d) Q, S

95. Match items in group 1 with correct options from those given in group 2 Group 1 P. Tissue plasminogen activator Q. Gamman interferon R. Podophyllotoxin S. Polyhydroxyalkanoate Group 2 1. Immunomodulator 2. Biodegradable plastic 3. Clot dissolution 4. Anti-tumor agent (a) P–4, Q–3, R–l, S–2 (b) P–l, Q–3, R–4, S–2 (c) P–3, Q–l, R–2, S–4 (d) P–3, Q–l, R–4, S–2

Common Data for Questions 98 and 99 : Lignocellulosic biomass was subjected to microbial composting. The microbial consortium produced an extracellular enzyme xylanase, which was a glycoprotein having a molecular weight of 68 kDa and a positive charge. An aqueous extract of the enzyme could be easily prepared from the compost. 98. What techniques would you recommend for confirming the molecular weight of the purified enzyme? (P) Isoelectric focusing (Q) SDS-PAGE (R) Native PAGE (S) Gel filtration (a) P, Q (b) Q, S (c) R, S (d) P, S

Advanced Biotechnology 9.11

99. If Con A sepharose column was used for the purification of enzyme, then separation would be based on

105. Which of the following fluorescent probes is used to monitor the progress of ampli fication in Real time PCR ?

(a) molecular exclusion

(a) SYBR green

(b) affinity binding

(b) Rhodamine

(c) ion exchange

(c) FITC

(d) hydrophobic interaction

(d) Cyan blue

Statement for Linked Answer Questions 100 and 101 : DNA content of Caenorhabditis elegans was analysed and found to contain 1.0  108 bp. 100. How many standard -phage vectors carrying 20kb DNA fragments or YACs carrying 250 kb DNA fragments are theoretically required to constitute a complete C. elegans genomic library? (a) 500 -phage vectors or 40 yeast clones

106. A linear DNA fragment is 100% labeled at one end and has 3 restriction sites for EcoRI. If it is partially digested by EcoRI so that all possible fragments are produced, how many of these fragments will be labeled and how many will not be labeled ? (a) 4 labeled; 6 unlabeled (b) 4 labeled; 4 unlabeled (c) 3 labeled; 5 unlabeled

(b) 400 -phage vectors or 5000 yeast clones

(d) 3 labeled; 3 unlabeled

(c) 5000 -phage vectors or 400 yeast clones (d) 5  104 -phage vectors or 4000 yeast clones 101. How many -phage vectors/yeast clones should be prepared in order to ensure that every sequence is included in the library?

107. Somatic embryo from cotyledon explant would develop in the following sequential stages, (a) cotyledonary  heart  globular  torpedo (b) globular  torpedo  heart  cotyledonary

(a) 25  10 -phage vectors/2000 yeast clones

(c) globular  heart  torpedo  cotyledonary

(b) 20  10 -phage vectors/1600 yeast clones

(d) cotyledonary  globular  heart torpedo

(c) 5  104 -phage vectors/4000 yeast clones

108. Though the right border (RB) and left border (LB) of T-DNA are identical, the DNA transfer is specific for the DNA left of the RB (the T-DNA), rather than for the DNA left of the LB because

3 3

(d) 10  104 -phage/10000 yeast clones

2005 102. Cells of meristemoid are best described as

(a) the sequence context at the RB defines the direction of transfer

(a) differentiated and non dividing (b) dedifferentiated and dividing

(b) the sequence context at the LB defines the direction of transfer

(c) differentiated and dividing (d) dedifferentiated and non dividing 103. The transplastomic plants bear no risk for gene transfer through pollens as (a) the pollens degenerate before fertilization (b) the transformed mitochondrial DNA is lost during pollen maturation (c) the transformed chloroplast DNA is lost during pollen maturation (d) the transformed genomic DNA are inherited maternally 104. The mobility of DNA in agarose electrophoresis is solely based on its (a) charge (b) conformation (c) size (d) none of these

gel

(c) the nuclear location sequence (NLS) of VirD2 protein drives the excised T-strand (d) the endonuclease activity of VirD2 protein allows nicking at RB 109. Identical sized RNA transcript is detected by Northern blot analysis of UDP glucuronosyl transferase obtained from human liver and kidney. Microarray analysis of the same samples shows equal spot intensity, ohereas Western blot detects a 55 kDa strong band in liver, but a very faint band in kidney of same size. The regulation of UDP glucuronosyl transferase is (a) transcriptionally controlled (b) post-transcriptionally controlled (c) translationally controlled (d) post-translationally controlled

Advanced Biotechnology

9.12

110. Expression in poor amount and in inactive form of cDNA of a euharyotic protein in Escherichia coli using its expression vector is due to

117. Nick translation of DNA is a method for making DNA probes. Identify from below what is NOT required for nick translation method

(P) the absence of capping mechanism of mRNA

(a) DNA polymerase

(Q) codon bias

(b) DNAase

(R) absence of polyadenylation

(c) Primers

(S) absence of proper glycosylation

(d) Deoxyribonucleotides

(a) P, Q

(b) Q, R

(c) Q, S

(d) P, S

Common Data for Questions 111, 112, 113: A recombinant SV40 virus delivers c-myc cDNA, which has a unique sal I site, into muscle cells. Southern analysis of Sal I digested total genomic DNA of the muscle cell using c-myc cDNA porbe generates a smear. 111. The DNA smear obtained on Southern blot is due to (a) head to head concatamer of viral DNA (b) head to tail concatamer of viral DNA (c) tail to tail concatamer of viral DNA (d) random integration of viral DNA 112. Western blot analysis of c-myc expression of such transformed cells last for (a) transiently

118. During the functioning of biosensor which of the following sequences of events occurs (a) Enzymatic/cellular reaction  detector  transducer (b) Enzymatic/cellular reaction  transducer  detector (c) Enzymatic/cellular reaction  pressure gauge  time (d) Enzymatic/cellular reaction  vibrator  mechanical signal 119. Somatic embryogenesis is a procedure in plant tissue culture methodology described best as (a) Formation of both shoot and root meristem (b) Formation of stable embryos (c) Formation of axillary buds (d) None of the above

(b) upto five generations (c) upto 10 generations (d) more than 100 generations 113. Which of the following types of cancer will be observed in such transformed cells ?

120. The culture fluids of 1000 to 5000 colonies of hybridoma are screened for monoclonal antibody by P. Western blot analysis Q. Antigen capture analysis

(a) Adenoma

(b) Melanoma

R. Northern blot analysis

(c) Sarcoma

(d) Hepatoma

S. Antibody capture analysis Choose the correct pair from the following:

2004 114. Expression of hundreds of different genes in DNA microarray technology is monitored by using (a) Radioactive probe (b) Visible chromogenic probe (c) UV absorbing probe (d) Fluorescent probe 115. Transfer of T - DNA from Ti plasmid into plant cell is mediated by (a) mob gene

(b) vir gene

(c) nif gene

(d) octopine gene

116. For the growth of T-cell, the growth factor needed would be (a) Epidermal growth factor (b) Interleukin-2 (c) Fibroblast growth factor (d) TNF- 

(a) P, Q

(b) Q, R

(c) R, S

(d) Q, S

121. Tobacco leaf discs are transfected with Agrobacterium tumefaciens strain containing binary vector (GUS as reporter gene) with selectable marker neo (kanamycin resistant gene) and then regenerated to plants. The plants are kanamycin resistant but leaf tissues are negative to GUS assay. The explanations are (a) The plants are transformed for both genes by GUS gene is turned off (b) The plants are transformed for only neo gene not the GUS gene (c) The plants are not transformed at all, but the development of kanamycin resistance is due to somaclonal variation (d) All of these

Advanced Biotechnology 9.13

122. The restriction endonuclease HaeIII recognizes the sequence GG  CC and the point of cleavage is given by the arrow. If you want to clone a piece of DNA in a plasmid digested by HaeIII, what will be restriction enzyme of choice ? (a) SmaI (CC  GGG)

(a) A  4, B  2, C  1, D  3 (b) A  2, B  4, C  1, D  3 (c) A  1, B  4, C  2, D  3 (d) A  4, B  2, C  3, D  1

2003

(b) NotI(GC  GGCCGC)

127. To be a cloning vector, a plasmid does NOT require

(c) SalmIII (GG  CC)

(a) an origin of replication

(d) PstI (CTGCA  G)

(b) an antibiotic resistance marker

123. For the sequence of ds DNA given below, identify the set of primers required to amplity this DNA by PCR 3 GACTCCA.............TACAACC 5 CTGAGGT.................ATGTTGG 3 (a) 5 GGTTGTA and 5 GACTCCA (b) 5 CTGAGGT and 5 CCCAACAT (c) 5 ACTCAGT and 5 ATGTTGG (d) None of these

(c) a restriction site (d) to have a high copy number 128. In animal cell cultures, the addition of serum to media is essential for providing (a) amino acids for protein synthesis (b) nucleotides for DNA synthesis (c) growth factors (d) all of these 129. For protoplast fusion to be successful in plant cells

124. Agrobacterium based transformation of protoplasts obtained from dicots is based on the fact that (a) These exhibit strong chromosomal structures (b) These have two cotyledons

(a) fusion agents other than polyethylene glycol should be used (b) cell wall of the two strains of cells should not be damaged

(c) These exhibit strong wound response

(c) DNA between the two cells should be compatible

(d) These have long tap root system

(d) osmolarity of the medium is not important

125. What would be the effect of addition of 2,4-D on the production of berberine by cell culture of Thalictrum minus (a) To stimulate growth and thereby urease secondary metabolite production

130. In order to identify the person who committed a crime, forensic experts will need to extract DNA from the tissue sample collected at the crime scene and conduct one of the following procedures for DNA finger-printing analysis

(b) Stimulate dedifferentiation and thereby decrease secondary metabolite production

(a) cut the DNA and hybridize with specific microsatellite probes

(c) Stimulate proliferation and reduce secondary metabolite production

(b) cut the DNA and subclone the fragments

(d) None of these

(d) (b) followed by (c)

126. Match the following genetic elements with their functions. Genetic

Functions

elements A. neoR

1. Facilitates inducible expression of genes in eukaryotes

B. SV40

2. Facilitates expression eukaryotes

of

constitutive genes in

(c) determine the sequence of the subclones Answer Q. 131 – 132 based on the data given below: Dr. Singh isolated a new 5-kb gene and wants to determine its sequence using a sequencer that can sequence upto 500 bases in a single reaction. Therefore, she decides to create subclones having suitable-size inserts for sequencing 131. Which one of the following will be the most appropriate restriction enzyme for this subcloning?

C. LTR

3. Allows amplification of gene

(a) 8-bp cutter

D. dhfr

4. Provides way of selecting eukaryotic cells, which have received foreign DNA

(b) 6-bp cutter (c) 3-bp cutter (d) 5-bp cutter

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Advanced Biotechnology

132. To generate the minimum number of subclones needed for sequencing, what should be the size of the insert in these subclones ? (a) 1000 bp

(b) 500 bp

(c) 250 bp

(d) 2000 bp

133. All of the following are true about DNA microarray technology except (a) an electron microscope is used to gather data from the arrays (b) the technology is used to assess transcription from multiple genes simultaneously (c) the technology works best for organisms whose genome is completely sequenced (d) the technology is derived from computer chip manufacture 134. You have cut the genome of a double-stranded viral genome with a restriction endonuclease and electrophoresed the products on an agarose gel. You observe only one band on the gel, equivalent to the size of the genome. This is because (a) there are no introns in the genome (b) the introns contain the recognition sites and have already been spliced out (c) all of restriction fragments are too small to detect (d) restriction endonucleases do not cut RNA, and this virus has an RNA genome 135. The restriction endonuclease Eco 52I recognizes the sequence C/GGCGG and cuts between the first C and the first G, indicated by the slash. DNA cut by which of the following enzymes (given with their recognition sequences and cut sites) could be cloned into a plasmid digested with Eco 52I ?

137. What is the primary purpose of neomycin in creating mice with knock-outs in gene X? (a) neomycin selects for the survival of embryonic stem cells (ES) that have incorporated the mutant gene X anywhere in the genome (b) neomycin selects for the survival of ES cells that have incorporated the mutant gene in the place of the wild-type gene (c) neomycin prevents Candida infection during ES cell culture that does not have gene X (d) neomycin makes the gene X knock - out mice resistant to Candida infection 138. All of the following produced by animal cells in culture and help the cells adhere to the culture dish except (a) glycoproteins

(b) collagen

(c) phospholipase A

(d) hyaluronic acid

139. The following are useful to introduce genes into crop plants except (a) Ti plasmid

(b) Particle gun

(c) breeding

(d) auxin

2002 140. Callus formation from mature tissue explant occurs through (a) Dedifferentiation (b) Redifferentiation (c) Both (a) and (b) of the above (d) None of (a) and (b) 141. Large scale clonal propagation practically means raising a population of plantlets from (a) A single cell (b) A single explant

(a) EcoRI (G/AATTC)

(c) Many explants from a single plant

(b) XmaIII (C/GGCGG)

(d) Many explants from a group of plants

(c) SmaI (CCC/GGG) (d) SacII (CCGC/GG) 136. If bacterial cells are transformed with a mixture of linear and circular molecules resulting from a ligation reaction designed to produce a recombinant molecule (a) no recombinant molecule will ever be detected (b) both linear and circular molecules will replicate equally well

142. T4 Polynucleotide kinase is used for (a) Labelling 3ends of DNA (b) Labelling 5 ends of DNA (c) Creating blunt ends of DNA (d) Dephosphorylation of DNA 143. Plant secondary metabolites production in suspension culture is mainly targeted for (a) Obtaining metabolites in aseptic condition

(c) none of the plasmids will express the antibiotic resistance gene located on the plasmid

(b) Enhanced in vitro production of desired metabolite

(d) the circular molecules will be amplified by the cells

(c) Enhanced production of all metabolites (d) Obtaining new metabolites

Advanced Biotechnology 9.15

144. In baculovirus expression vector foreign genes are expressed from the promoter of

Column - A

Column - B

(a) Diosgenin

1. Holoside

(a) Polyhedron gene

(b) Ajmalicine

2. Pyrrolizidine alkaloid

(b) Bacteriophage T7 gene

(c) Shikonin

3. Indole alkaloid

(c) E. coli lacZ gene

(d) Digoxin

4. Naphthaquinone nucleus

(d) Yeast phosphoglycerate kinase gene

(e) Scopolamine

5. Cardenolide

145. The length of each boarder sequence in Ti plasmid is about (a) 25 million base pairs

6. Saponin 7. Tropane alkaloid

2001

(b) 200 kilo base pairs

151. To PCR amply the sequence

(c) 25 kilo base pairs

ATCTTCTACG...............AAGCTTGCGG

(d) 25 base pairs 146. Enzyme used in ‘cycle’ sequencing of DNA is

TAGAAGATGC...............TTCGAACGCC

(a) T7 DNA polymerase

the required primers are

(b) T4 DNA polymerase

(a) ATCTTCTA and CGAACGCC

(c) Klenow DNA polymerase

(b) ATCTTCTA and CCGCAAGC

(d) Taq DNA polymerase

(c) TAGAAGAT and CGAACGCC

147. Cells deficient in hypoxanthine guanine phosphoribosyl transferase (HPRT) enzyme rely on

(d) TAGAAGAT and CCGCAAGC 152. Resistance to herbicide chlorsulfuron in plants is due to a change in

(a) Synthesis of purine deoxynucleotides by salvage pathway

(a) Glutamine synthetase

(b) Synthesis of purine deoxynucleotides by de novo pathway

(c) Acetolactate synthase

(c) Supply of hypoxanthine in the culture medium (d) Supply of thymidine in the culture medium 148. Enhanced axillary branching for multiple shoot productions is promoted by (a) 2, 4 – D (b) Abscisic acid

(d) DNA polymerase 153. To isolate a gene coding for glucagon, the cDNA library has to be constructed using mRNA isolated from (a) Intestine

(b) Pancreas

(c) Pituitary

(d) Brain

154. The essential component of Ti plasmid required for integration into plant genome is

(c) Gibberellic acid

(a) Origin of replication

(d) Benzyl adenine 149. Match the organisms in Column A with the product in Column B. Column - A

(b) Threonine deaminase

Column - B

(a) Thermus aquaticus 1. Beer

(b) Tumor inducing gene (c) Nopaline utilization gene (d) All of these 155. Positional cloning approach exploits information

(b) Acetobacter aceti

2. Bioinsecticides

(a) On the location of the gene in the genome

(c) Bacillus

3. Hind III

(b) On the status of its expression

thermogenesis (d) Saccharomyces

(c) About the position of promoter relative to MCS 4. Taq 1

carlbergensis (e) Hemophilus

5. Vinegar

influenza 150. Match the secondary metabolites in Column A with their most appropriate chemical characteristics in Column B.

(d) About the position of the restriction sites 156. Hormone pairs required for a callus to differentiate are (a) Auxin and cytokinin (b) Auxin and gibberellin (c) Ethylene and gibberellin (d) Cytokinin and gibberellin

Advanced Biotechnology

9.16

157. A gene can not be isolated from a human genomic DNA library by functional complementation in E. coli because of (a) Non-functional promoter (b) The absence of splicing machinery (c) Coupled transcription and translation (d) Codon bias 158. Embryo rescue is a useful technique to (a) Grow/generate hybrids between different plant species (b) Complete the growth of embryos susceptible to defects in seed development

(c) Break the dormancy of seeds (d) All of the above 159. (a) Write the reaction catalyzed by penicillin G acylase. (1) (b) Name any two techniques by which penicillin G acylase may be immobilized. (2) (c) Why are mammalian cells cultured in CO2 incubators? (1) (d) Mention an important post-translational modification absent in prokaryotes making them unsuitable hosts for expressing human genes. (1)

160. (a) The sequences at the cloning site of three vectors are given below. The Bam HI (GGATCC) and Hind III. (AAGCTT) sites are underlined. Only the sequences around the restriction sites are shown. The symbol “__________” indicates rest of the sequence. Vector 1: _______ Promoter ... ATGGGTCGCGGATCCGGCTGC .. AAGCTT ______ Vector 2 : _______ Promoter ... ATGGGTCGGGATCCGGCTGCT .. AAGCTT______ Vector 3 : _______ Promoter ... ATGGGTCGGATCCGGCTGCTA.. AAGCTT ______ Which one of the above three vectors is appropriate to clone the following ORF _______ ATGCCCAACACCCGGATCCCG.. TAAAAGCTT _______ for expression? Give the reason in one sentence. (b) Draw the restriction map of the plasmid given the following data (the gel pattern shown below is not to scale). The size of each DNA fragment (in kb) is indicated next to it (3) EcoRI

SalI

5.4 –

5.4 –

HindIII

EcoRI & HindIII

SalI & HindIII

EcoRI & SalI 3.6 –

2.1 – 1.9 1.4 –

2.1 – 1.4 – 1.3 –

1.9 – 1.4 – 1.2 –

0.6 –

0.9 –

2000 161. The substrate for restriction enzyme is (a) Single stranded RNA

1.8 –

163. In human populations, 4% of the individuals are homozygous recessive to a specific RFLP marker. What percentage of individuals do you expect to be heterozygous for this trait ?

(b) Partially double stranded RNA

(a) 4%

(b) 8%

(c) Cell wall proteins

(c) 16%

(d) 32%

(d) Double stranded DNA 162. DNA helicases catalyse the reaction (a) DNA supercoiling (b) DNA unwinding (c) Cleavage of DNA helix (d) Interconversion of DNA topoisomers

164. Many plasmids have Ampr marker. This implies (a) The plasmids contain genes for ampicillin biosynthesis (b) Ampicillin is required for bacterial growth after transformation (c) The plasmid contains the gene encoding  -lactamase (d) Ampicillin is essential for cell survival

Advanced Biotechnology 9.17

165. Which of the following has been produced commercially from mammalian cell cultures?

(a) Use relatively simple techniques

(a) Plasminogen activator

(b) Generally end up with hazardous waste material

(b) Antibacterial antibiotics

(c) Are relatively slow

(c) Insulin

(d) Are unobtrusive and non-disruptive

(d) Renin 166. Mung bean nuclease could be used for (a) DNA synthesis (b) nucleotide hydrolysis

172. cDNA made from the mRNA of an organism was used to make a cDNA library in a vector that allows the expression as a fusion with a reporter tag. What percentage of the cDNA clones is likely to give rise to correct gene products ?

(c) trimming single stranded regions in DNA

(a) 10%

(d) removal of phosphate groups from the ends of the DNA

(b) 30%

167. Phage T7 promoter containing plasmids are used for overexpression of cloned genes because (a) their convenient size (b) their single stranded nature (c) Exquisite specificity of T7 RNA polymerase to phage promoters (d) T7 infects E. coli and lysogenizes the cell 168. Yeast artificial chromosome (YAC) is used for (a) cloning large segments of DNA (b) cloning only, yeast genomic sequences (c) cloning of only cDNA sequences (d) all DNA except plant DNA sequences 169. Which one of the following is not a requirement for a PCR reaction ? (a) DNA template

(b) Taq polymerase

(c) NTPs

(d) MgCl2

170. Plant breeders have an advantage over animal breeders in reproducing a desired type offspring because the plant breeders can employ (a) Gene mutations (b) Hybridization (c) Clonal propagation (d) Selection 171. Which one of the following options related to the following statement is incorrect? In comparison to physical/chemical methods of clean up, bioremediation methods

(c) 50% (d) 100% 173. Commonly used reporter gene in plant expression vectors is (a) Ti gene of Agrobacterium tumifaciens (b) GUS gene (c)  -lactamase gene (d) - amylase gene 174. (a) Match the column A with those in column B (A) Chemical sequencing of DNA

(1) Southern

(B) DNA blotting

(2) Temin, Baltimore and Dulbecco

(C) Monoclonal antibodies (3) F. Sanger (D) Reverse transcription (4) Maxam and Gilbert (E) Protein sequencing

(5) Kohler and milstein

(F) Polymerase chain reaction

(6) KMullis

(b) If all the steps in a PCR reaction were to work at 100% efficiency, how many micrograms of 1 Kb product will be generated from 1 p mole of DNA template after 10 cycles (1 bp = 660 Da) 175. (a) What is somatic embryogenesis ? (b) What is the difference between direct and indirect somatic embryogenesis ? (c) State two methods for direct DNA transfer into plant cells.

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Advanced Biotechnology

ANSWERS 1. (d)

2. (d)

3. (b)

4. (b)

5. (b)

11. (c)

12. (b)

13. (a)

14. (1.68 to 1.7)

20. (b)

21. (c)

22. (a)

23. (b)

30. (c)

31. (c)

32. (d)

40. (b)

41. (a)

50. (c)

6. (d)

7. (a)

8. (c)

9. (b)

10. (d)

15. (2.8)

16. (d)

17. (c)

18. (a)

19. (a)

24. (b)

25. (a)

26. (a)

27. (c)

28. (d)

29. (a)

33. (b)

34. (d)

35. (d)

36. (a)

37. (a)

38. (b)

39. (a)

42. (a)

43. (c)

44. (b)

45. (c)

46. (c)

47. (b)

48. (c)

49. (a)

51. (c)

52. (b)

53. (a)

54. (c)

55. (d)

56. (a)

57. (b)

58. (c)

59. (d)

60. (b)

61. (c)

62. (c)

63. (a)

64. (c)

65. (c)

66. (c)

67. (d)

68. (b)

69. (b)

70. (b)

71. (*)

72. (a)

73. (d)

74. (b)

75. (b)

76. (b)

77. (a)

78. (d)

79. (a)

80. (b)

81. (c)

82. (a)

83. (b)

84. (d)

85. (a)

86. (d)

87. (a)

88. (b)

89. (b)

90. (b)

91. (d)

92. (c)

93. (b)

94. (b)

95. (d)

96. (b)

97. (b)

98. (b)

99. (b)

100. (c)

101. (c)

102. (c)

103. (d)

104. (c)

105. (a)

106. (c)

107. (c)

108. (a)

109. (b)

110. (d)

111. (b)

112. (c)

113. (c)

114. (d)

115. (b)

116. (b)

117. (c)

118. (b)

119. (d)

120. (a)

121. (c)

122. (c)

123. (a)

124. (c)

125. (d)

126. (a)

127. (b)

128. (d)

129. (c)

130. (a)

131. (c)

132. (b)

133. (a)

134. (d)

135. (b)

136. (d)

137. (a)

138. (c)

139. (d)

140. (c)

141. (c)

142. (b)

143. (b)

144. (a)

145. (d)

146. (d)

147. (b)

148. (a)

149. (*)

150. (*)

151. (b)

152. (c)

153. (b)

154. (b)

155. (a)

156. (a)

157. (b)

158. (d)

159. (*)

160. (*)

161. (d)

162. (b)

163. (d)

164. (c)

165. (a)

166. (c)

167. (c)

168. (a)

169. (c)

170. (c)

171. (b)

172. (c)

172. (b)

174. (*)

175. (*)

EXPLANATIONS 1. Gene therapy is the therapeutic delivery of nucleic acid polymers into a patient's cells as a drug to treat disease. Avidin is a protein derived from both avians and amphibians that shows considerable affinity for biotin, a co-factor that plays a role in multiple eukaryotic biological processes. 2. The culture viability is merely a ratio of viable cells in the culture to total cells in the culture. For the purpose, the hemocytometer will only be used to obtain a viability count, not a total cell count (cells/mL). The viability will be determined by the trypan blue assay , and it can be used with the total cell count (obtained from a Coulter counter) to determine a total count of viable cells, as : Number of viable cells = (% viability)*(Total number of cells) 3. Secondary treatment is the portion of a sewage treatment sequence removing dissolved and colloidal compounds measured as BOD. Secondary treatment is traditionally applied to

the liquid portion of sewage after primary treatment has removed settleable solids and floating material. Secondary treatment is typically performed by indigenous, aquatic microorganisms in a managed aerobic habitat. Bacteria and protozoa consume biodegradable soluble organic contaminants (e.g. sugars, fats, and organic short-chain carbon molecules from human waste, food waste, soaps and detergent) while reproducing to form cells of biological solids. 4. The T-DNA transfer process of Agrobacterium tumefaciens is activated by the induction of the expression of the Ti plasmid virulence loci by plant signal molecules such as acetosyringone. The overdrive proteins from acetosyringoneinduced Agrobacterium cells interact with T-DNA border and overdrive sequences. The vir gene products act in trans to mobilize the T-DNA element from the bacterial Ti plasmid. The T-DNA is bounded by 25-base pair direct repeat sequences, which are the only sequences on the

Advanced Biotechnology 9.19

element essential for transfer. Thus, specific reactions must occur at the border sites to generate a transferable T-DNA copy. 5. Secondary Metabolites including Morphine, Pyrethrins, Scopolamine, Vincristine are extracted from Papaver somniferum, Tagetes erecta, Datura stramonium, Catharanthus roses respectively. 6. Certain genetic elements used in transgenic research includes Ubiquitin I promoter, Nos Transcriptional terminator, bar Selection Marker Gene, gus Reporter gene. They are obtained from Zea mays, Agrobacterium tumifaciens, Streptomyces hygroscopicus, Escherichia coli respectively. 7. Bicarbonate is often included to act as a buffer system in a conduction with the CO2 environment (5-10%) in which the cell should be cultured. This allows the cultures to be maintained at the normally optimum pH range of 6.9 - 7.4. The CO2 is provided by a controlled atmosphere inside an incubator or alternatively in a sealed flask. 8. CMV promoter is plant expression vector.

ddNTPs 1 9. About dNTP = 300 s 10. NH 4 product is only in anaerobic processes like waste water treatment. 11. 2nd generation crop are the value added crops i.e; increased nutritional value. Instead of insect resistance (Bt) and herbicide tolerance (HT). 12. Respective primers will be anneal to 3 ends of the two template strands; keeping antiparallel nature of DNA duplex and complimentarity in the mind. 13. Hae III : GG|CC CC | GG BamHI :

G | GATC CCTAG | G

14. Average number of fragments generated

1 1 1 1 5 =       4.3  10 4 4 4 4 4.3  105 256 = 1.68  103

=

15.

DNA = 0.5 g Dilution =

50 = 2.5  10–5 200  10000

TE =

=

Number of colonies Dilution  Amount of DNA

35 2.5 105  0.5

= 2.8  106

16. The number of bases present in the restriction site of the restriction enzyme = 6 We know, there are four different types of nitrogenous bases in a DNA molecule. So, the statistical frequency of occurrence of the restriction site would be = 46 = 4096 17. Serum is commonly used as a supplement to cell culture media. It supports cell growth and product formation by providing a broad spectrum of macromolecules, nutrients and vitamins as well as certain growth factors and hormones. However, serum must be filter sterilized with 0.45 filter paper before using in the media to grow mammalian cells because it contains Mycoplasma and microorganisms as contaminants. 18. The bands which are labeled as A were uncut and separated at three different regions on the agarose gel after digestion with EcoRI. It shows th at th e plasm id A se parates is having concatemers. Plasmid B labeled as B is nicked while Plasmid C which is labeled as C has a very thick band which suggests that Plasmid C is supercoiled. Band D was the only plasmid which was the digested product of Eco RI are linear because EcoRI has already digested the plasmid. 19. Belladona plant produces a type of secondary metabolite called atropine which helps in the ocular examination procedure. Foxglove produces a secondary metabolite called digitalin that helps in the treatment of cardiac diseases. Pacific yew is another plant with medicinal use which produces taxol that is extensively used in cancer research and treatment purposes. Eucalyptus is used in the production of menthol that helps in the curing of cough and cold. 20. There are several species of mustard. Black mustard (Brassica nigra) produces small seeds that become black at maturity. They have an extremely rich and pungent flavor, stronger than yellow mustard. White mustard (Sinapis alba) produces large yellow-colored seeds with a bitter taste that is not as pungent as other varieties. Indian mustard or mustard greens (Brassica juncea) is a green vegetable whose very tasty leaves are used in the same way as spinach. The

9.20

Advanced Biotechnology

chemicals that are responsible for pungency are stored in inactive conjugate form in compartments within the seeds and are protected from enzymes. But when seeds are crushed, the pungent chemicals come in contact with the enzymes and are converted into active chemicals which give the pungent smell. 21. The length of the unique stretch of DNA sequence that can be found only once in 3 billion base pair long genome is a 16 bases long DNA sequence from which 16s rRNA is transcribed. 22. Several types of DNA library resources were sponsored by the DOE before and during the Human Genome Program (HGP). These included both prokaryotic and eukaryotic vector systems, and clone libraries representing single chromosomes. Bacterial Artificial Chromosomes (BACs) became the most broadly used resource for several reasons. The large size was a good match for capabilities of high throughput sequencing centres. A bacterial artificial chromosome (BAC) is an engineered DNA molecule used to clone DNA sequences in bacterial cells (for example, E. coli). BACs are often used in connection with DNA sequencing. Segments of an organism’s DNA, ranging from 100,000 to about 300,000 base pairs, can be inserted into BACs. The BACs, with their inserted DNA, are then taken up by bacterial cells. As the bacterial cells grow and divide, they amplify the BAC DNA, which can then be isolated and used in sequencing DNA. 23. Natural genetic transformation is believed to be the essential mechanism for the attainment of genetic plasticity in many species of bacteria. During bacterial evolution, the ability of Bacteria and Archaea to adapt to new environments most often results from the acquisition of new genes through horizontal transfer rather than by the alteration of gene functions through numerous point mutations. Horizontal gene transfer is defined to be the movement of genetic material between bacteria other than by descent in which information travels through the generations as the cell divides. It is most often thought of as a sexual process that requires a mechanism for the mobilization of chromosomal DNA among bacterial cells. However, because they are unable to reproduce sexually, bacterial species have acquired several mechanisms by which to exchange genetic materials. It can happen through transformation, conjugation and transduction. In bacteria, it mainly occurs by transformation and conjugation.

24. Agrobacterium tumefaciens is a gram-negative rod-shaped bacterium. It infects plants usually through an open wound. Once it enters its plant host, it injects a section of its DNA called the TDNA which is derived from its Ti (tumor inducing) plasmid into its host. The T-DNA is then integrated into the plant’s genome, and has two effects on the plant host. The T-DNA first directs the plant cells to make auxins and cytokinins, which causes the cells to become irregularly shaped and form a visible tumour called a gall. The T-DNA then directs the plant cell to start making opines (usually nopaline or agropine) which A. tumefaciens use as an energy source. 25. Glyphosate kills plants by interfering with the synthesis of the aromatic amino acids phenylalanine, tyrosine and tryptophan. It does this by inhibiting the enzyme 5-enolpyruvylshikimate-3-phosphate synthase (EPSPS), which catalyzes the reaction of shikimate-3-phosphate (S3P) and phosphoenolpyruvate to form 5-enolpyruvyl-shikimate-3-phosphate (ESP). Bromoxynil is a nitrile herbicide and one of its common degradation products (3, 5-dibromo-4hydroxybenzoic acid) have been shown to undergo metabolic reductive dehalogenation by the microorganism. Sulfonylureas is used as Chlorimuron (Classic) for soybeans. Their site of action is Acetolactate synthase (ALS) enzyme. Dalapon is a synthetic herbicide and plant growth regulator. It is responsible for inhibition of lipid synthesis by acting on dehalogenase enzyme. 26. The given plasmid sequence is 5000bp long. When it is digested with Hind III, it produces 2 fragments of 1200bp and 3800bp. This means that Hind III cuts the plasmid in two different sites. So Hind III has 2 sites in plasmid. When digested with Bam HI, it produces one fragment of 5000bp. This means Bam HI has only 1 recognition site on the plasmid sequence. Eco RI cuts the fragment in two equal halves of 2500bp each. Hence Eco RI also has 2 sites of recognition on the plasmid. 27. Given, size of the genome of E. coli = 4  103kb = 4  106 bp Size of the DNA fragment = 5000 bp Therefore, minimum number of independent recombinant clones required to represent the DNA fragment in a genomic library = (4  106)/5000 = 8  102 28. Given, Probability of finding the fragment in the library = 95% = 0.95

Advanced Biotechnology 9.21

29. Both pUC and the bacterial own genome produce a faulty gene product of Lac Z gene. The lac Z fragment, whose synthesis can be induced by IPTG, is capable of intra-allelic complementation with a defective form of  -galactosidase enzyme encoded by host chromosome (mutation lacZDMI5) 30. DMSO is used in cell freezing media to protect cells from ice crystal induced mechanical injury. 31. The strength of ChIP assays is their ability to capture a snapshot of specific protein: DNA interactions occurring in a system and to quantitate the interactions using quantitative polymerase chain reaction (qPCR). 32. Fusion of freely isolated protoplasts from different sources with the help of fusion inducing chemical agents is known as induced fusion. It includes use of high pH or Ca++ treatment, Use of PEG (Polyethylene glycol) which induces protoplast aggregation and subsequent fusion and the use of inactivated Sendai virus for fusion which occur in four temperature-dependent stages. Hence Colchicine is not a protoplast fusion inducing agent. Colchicine is used to prevent gout attacks in adults, and to relieve the pain of gout attacks when they occur. Colchicine is also used to treat familial Mediterranean fever (FMF; an inborn condition that causes episodes of fever, pain, and swelling of the stomach area, lungs, and joints) in adults and children 4 years of age and older. 33. Given, weight of single base pair of DNA = 1.1 * 10–21 grains Length of plasmid vector= 2750bp Purified DNA= 1ug Thus, 0.55 picomoles of a plasmid vector of length 2750bp are contained in 1ug of purified DNA. 34. Elicitors are compounds, which activate chemical defense in plants. Various biosynthetic pathways are activated in treated plants depending on the compound used. They are compounds that when introduced into a living organism signals the activation or synthesis of another compound. Thus (R) is true but (A) is false. 35. Somatic embryogenesis can be initiated either directly without going through the callus phase from predetermined embryonic cells or indirectly through callus proliferation and differentiation into embryonic cells within the callus tissue

36. Embryonic stem cells possess the capacity to divide for long periods and retain their ability to make all cell types within the organism. These are termed pluripotent stem cells. They are derived from a fertilized embryo. Cells are obtained from an embryo in the blastula phase, when they are still only a few days old. Because they have only begun to differentiate, these cells have the capability of developing into any cell in the human body which makes them potentially important in medicine. 37. The yeast artificial chromosome (YAC) is an artificially constructed system that can undergo replication. The design of a YAC allows extremely large segments of genetic materials to be inserted. Subsequent round of replication produce many copies of the inserted sequence, is a genetic procedure known as cloning. The reason the cloning vector is called the YAC has to do with the structure of vector.YACs are shuttle vector that can be amplified in bacteria and employed for the cloning and the manipulation of large DNA inserts (up to 3mb pairs) in the yeast-Saccharomyces cerevisiae. Large DNA (>100 Kb) is ligated between two arms. 38. Tissue plasminogen activator (t-PA) is a naturally occurring protein that catalyze the conversion of the inactive proenzyme plasminogen into active serine protease plasmin. It is one of the two major endogenous mammalian enzymes with this property and is found in blood and tissues from various organs. Although it exerts a similar enzymatic action, it is structurally different from the other major endogenous enzymes, urokinase. t-PA can be used to treat some people who are having a stroke caused by a blood clot. It plays an important role in cell migration and tissue remodeling. 39. Synthetic seeds or artificial seeds are seed-like structures derived from somatic embryos under in vitro condition after encapsulation by hydrogen. These are tissue culture derived somatic embryos encased in a protective coating. Hydrated synthetic seeds are produced in those plant species where the somatic embryos are recalcitrant and sensitive to desiccation. Hydrated synthetic seeds are produced by encapsulating the somatic embryos in hydrogel capsules. The somatic embryos are mixed with sodium alginate (2 %) and the suspension is dropped into the calcium salts solution (200mM). The principle involved is when sodium alginate dropped into the calcium salt solutions it form

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Advanced Biotechnology

round firm beads due to the ion exchange between Na+ in sodium alginate and Ca 2+ in calcium slat solutions and sodium alginate form calcium alginate in 20-30 minutes. The calcium salt used can be calcium nitrate. 40. Shoot organogenesis is one of the in vitro plant regeneration pathway. It has been widely employed in plant biotechnology for invitro micropropagation and genetic transformation, as well as in study of plant development. Organogenesis continues until the definitive characteristics of the organs or plant parts are achieved. 41. Hairy roots are produced by infecting sterile plant with a natural genetic engineer, Agrobacterim rhizogenes. Genes for auxin synthesis and sensitivity are engineered into plant cells leading to gravity- insensitive mass root production. It is very useful for products produced in roots. Aggregation and shear sensitivity are a major problem for scale up. 42. Restriction endonucleases that recognize the same sequence are isoschizomers. The first example discovered is called a prototype and all subsequent enzymes that recognize the same sequence are isoschizomers of the prototype. For example, Sph I (CGTAC/G) and Bbu I (CGTAC/ G) are isoschizomers of each other. Isoschizomers are pairs of restriction enzymes specific to the same recognition sequence. 43. Phylogenetics deals with evolutionary relatedness of the species. It calculates the distances amongst the species from its origin and can track down any ancestral relationship. 44. Yeast two hybrid screening is a molecular biology technique used to discover protein-protein interaction and protein-DNA interaction by testing for physical interactions. Mass spectrometry (MS) is an analytical technique that measures the mass-to-charge ratio of charged particles. 45. Rapid amplification of cDNA ends (RACE) polymerase chain reactions (PCR) are used for independent high-throughput verification of transcriptional starting sites (TSSs) determined by genome-wide assays. 52 -RACE PCR is a wellestablished and widely used method to specifically amplify the 52 end of a transcript, facilitating mapping of the TSS and the approximate location of promoter elements. Conventionally, this mapping is done by cloning the 52 -RACE PCR product into a bacterial vector and sequencing a few clones by classic

electrophoresis-based Sanger sequencing. Nowa-days, even 3’ end transcripts are being mapped using 3’-RACE PCR. Baculoviruses are a very diverse group of viruses with double-stranded, circular, supercoiled genomes, with sizes varying from about 80 to over 180 kb which encode between 90 and 180 genes. Baculoviruses have evolved to initiate infection in the insect midgut. Baculovirus-insect cell expression systems have the capacity to produce many recombinant proteins at high levels and they also provide significant eukaryotic protein processing capabilities. 46. Hybridoma technology is a technology of forming hybrid cell lines (called hybridomas) by fusing a specific antibody-producing B cell with a myeloma (B cell cancer) cell that is selected for its ability to grow in tissue culture and for an absence of antibody chain synthesis. The antibodies produced by the hybridoma are all of a single specificity and are therefore monoclonal antibodies Fused cells are incubated in HAT medium (hypoxanthine-aminopterin-thymidine medium) for roughly 10 to 14 days. Aminopterin blocks the pathway that allows for nucleotide synthesis. Hence, unfused myeloma cells die, as they cannot produce nucleotides by the de novo or salvage pathways because they lack HGPRT. Removal of the unfused myeloma cells is necessary because they have the potential to outgrow other cells, especially weakly established hybridomas. Unfused B cells die as they have a short life span. In this way, only the B cell-myeloma hybrids survive, since the HGPRT gene coming from the B cells is functional. 47. Let N0 be the initial number of molecules present in the aliquot and let N be the end product of PCR amplification. Let n be the number of cycles performed by the PCR. We, know N = N0  2n for n number of cycles since in each cycle the number of amplicons double the number of parent strands in an exponential manner. Here, N0 = 100 for a 200 μL solution (given), n = 10 Therefore,

N = 100  210 = 100  1024 = 1.024  105

Hence, the required amplicons generated after 10 cycles of PCR run = 1.024  105

Advanced Biotechnology 9.23

48. From previous question, we have Number of amplicons in 200 μL of the solution after PCR run = 1.024  105 Therefore, number of amplicons in 0.1 μL solution = 1.024  105 x 0.1/ 200 = 51.2 Hence, the required number of amplicons present in 0.1 μL of the solution = 51.2 49. E. coli is the bacteria which are known to be naturally competent for transformation of DNA. It offers many advantageous such as: Genetic simplicity: E. coli cells only have about 4,400 genes whereas the human genome project has determined that humans contain approximately 30,000 genes. E. coli, live their entire lifetime in a haploid state. Growth rate: E. coli grows rapidly at a rate of one generation per twenty minutes under typical growth conditions. Safety: E. coli is naturally found in the intestinal tracts of humans and animals where it helps provide nutrients (vitamins K and B12) to its host. Conjugation and the genome sequence: The E. coli genome was the first to be completely sequenced. Genetic mapping in E. coli was made possible by the discovery of conjugation. Ability to host foreign DNA: E. coli is readily transformed with plasmids and other vectors, easily undergoes transduction, and preparation of competent cells (cells that will take up foreign DNA) is not complicated. 50. Vancomycin is the antibiotic resistance marker that cannot be used in a cloning vector in gram negative bacteria. Vancomycin acts by inhibiting proper cell wall synthesis in Gram-positive bacteria. Vancomycin treats only gram positive infections (commonly a skin infection, an example of a Gram positive infection). It has little/no use against a gram negative bacteria because vancomycin acts by binding to parts of the cell wall that are present in a gram positive bacteria but not in a gram negative bacteria. 51. TDT do not require a template as it adds Nnucleotides to the variable exons during antibody gene recombination. It functions by catalysing the addition of nucleotides to the 3’ terminus using 3’ – overhang as the substrate. Also it can act independently on blunt or recessed 3’ ends. It is expressed in immature, pre- B , pre –T lymphoid cells and uses cobalt as cofactor, also Mg/Mn 2+ presence in-vitro. It is used in adding nucleotides labelled with radioactive isotopes in TUNEL assay (terminal deoxynucleotidyl transferase dUTP Nick End Labelling).

52. In somatic cell gene transfer, the therapeutic genes are transferred into the somatic cell or body, of a patient. It involves using a vector such as a virus to deliver therapeutic gene to the appropriate target cells. This technique is currently the basis for cloning animals (creating transgenic animals) and is used in vitro. 53. EST is a unique stretch of DNA within a coding region of a gene that is useful for identifying fulllength genes and serves as a landmark for mapping. An EST is a sequence tagged site (STS) derived from cDNA library. EST’s sequence can be submiited or accessed from 3 universal databases- GenBank, DDBJ, EMBL. NCBI also has a database exclusively for EST generation are individual from a cDNA library. 54. Somatic embryogenesis is a process where a plant or embryo is derived from a single somatic cell or group of somatic cells. Somatic embryos are formed from plant cells that are not normally involved in the development of embryos, i.e. ordinary plant tissue. In plants, somatic embryogenesis comprises of embryo initiation followed by embryo production. Embryo initiation occurs on a medium rich in auxin, which induces differentiation of localized meristematic cells. The auxin typically used is 2,4-Dichlorophenoxyacetic acid (2,4-D). thus embryo initiation is dependent on 2,4-D. While these are transferred to a medium containing low or no auxin, then embryo production does not require high concentration of 2, 4-D. 55. Fibronectin is a glycoprotein that can act as general cell adhesion molecule by anchoring cells to collagen or proteoglycan substrates. Insulin is a hormone, produced by beta cells of the pancreas, and is central to regulating carbohydrate and fat metabolism in the body Insulin is a hormone that has profound effects on metabolism. alpha-2-Macroglobulin,is a large plasma protein found in the blood. It inhibits the thrombin action at the time of couagulation by blocking trypsin enzyme. Transferrins is a iron binding glycoprotein. 56. Patatin class- I promoter is a tissue specific potato promoter which expresses the potato gene mainly in tuber. Phaseollin is a promoter for cotyledon tissue. Genes encoding High Molecular Weight (HMW) glutenin, a wheat seed storage protein, is expressed in developing endosperm. Alpha amylase is a promoter for aleurone tissue.

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Advanced Biotechnology

57. T 4DNA Ligase catalyzes the formation of a phosphodiester bond between juxtaposed 5'phosphate and 3'-hydroxyl termini in duplex DNA or RNA. The enzyme repairs single-strand nicks in duplex DNA, RNA, or DNA/RNA hybrids. It also joins DNA fragments with either cohesive or blunt termini, but has no activity on singlestranded nucleic acids. It can ligate blunt-ended DNA with much greater efficiency than E. coli DNA ligase. Unlike E. coli DNA ligase, T4DNA ligase cannot utilize NAD and it has an absolute requirement for ATP as a cofactor. The rates of blunt-end and cohesive-end ligation of DNA by T4DNA ligase are increased by orders of magnitude in the presence of high concentrations of a variety of nonspecific polymers such as polyethylene glycol, Ficoll, bovine plasma albumin, or glycogen. Blunt-end ligation of small self-complementary oligodeoxyribonucleotides is also stimulated. 58. The cloning sites of the vectors are following: Lambda phage has size of 5-24kb, BACs have size of 5300kb, PACS have size of 100-300kb and Cosmids have size of 35-45kb. 59. Alpha, Beta, and Gamma are the three subunits of the G-protein. The active form is the AlphaGTP complex. The intrinsic GTPase activity converts the bound GTP to GDP in the a - subunit, returning it to its original conformation. This results in the a - subunit diffusing from the adenylate cyclase and reassociating with the b & g subunits. Adenylate cyclase catalyses the conversion of ATP to cAMP which is then used to activate another, initially inactive, enzyme. The mutant G alpha protein with increased GTPase activity shows decreased signaling because of the group of proteins present which are specific to G alpha subunit called the regulator of G protein signaling. 60. Number of clones which have 99% probability = log10 (1– 0.99)/ log10 (1– 5/4 * 105) = log10 (0.01)/ log10 (0.9999875) = –4.605/–1.25 * 10–5 = 3.5 * 10–5 61. Real time quantitative PCR allows the sensitive, specific and reproducible quantification of nucleic acids. Real time PCR allows precise quantification of specific nucleic acids in a complex mixture even if the starting amount of the material is at a very low concentration. This is accompanied by monitoring the amplification of a target sequence in real time using fluorescent technology.

A common approach uses SYBR GREEN I. It binds to any double stranded DNA which could result in inaccurate data. 2-D electrophoresis is used to separate and display all gene products present. This technique separate proteins in two steps according to two independent properties: the first-dimension is isoelectric focussing (IEF), which separated proteins according to their more isoelectric points (pI); the second dimension is SDS-polyacrylamide gel electrophoresis (SDS-PAGE) which separates protein according to their molecular weights (MW). The procedure involves placing the sample in gel with a pH gradient and applying a potential difference across it. There are two alternative methods to create pH gradient Æ carries ampholytes and immobilized pH gradient (IPG) gels. Affinity chromatography is a powerful tool for the purification of specific biomolecules, including proteins. The basic principle is that a biospecific ligand is immobilized to a solid support or resin to which a solution containing the protein of interest is passed over. Ligands are often based on biological functional pairs, such as enzymes and substrates or antigens and antibodies. The specific ligand binds the protein of interest and all non-specific molecules are washed away. The protein is eluted in a specific buffer, either by pH or ionic strength shift or by competitively displacement elution. Examples of resin areÆCNBr-agarose, antibody linked sepharose beads, etc. Microarray is a multiplex lob on a chip. It is 2-D array on a solid substrate (usually a glass slide or silicon thin-film cell) that assays large amounts of biological material using high-throughput screening methods. Types of microarray include DNA microarrays (biochips are used), Protein microarrays, etc. 62. In anther culture, the anther swells and dehisces along its upper margin, lengthwise. This phenomenon helps to expose the pollen grain. Alternatively, huge amount of pollen grains can be isolated manually and can be cultured aseptically very easily. Therefore, pollen is more suitable material than egg cell for the production of haploid. The development and production of haploid plant in vitro is very important for the study of fundamental and applied aspects of genetics in the higher plants. Production of homozygous diploid by doubling the chromosome number of haploid in vitro makes a pure line in single step and such

Advanced Biotechnology 9.25

homozygous pure line is of great importance in plant breeding. Also it is easy to induce cell division in immature pollen cells in some species. By anther and pollen culture, haploids can be produced in large numbers very quickly. 63. Isoschizomers are pairs of restriction enzymes specific to the same recognition sequence. For example, Sph I (CGTAC/G) and Bbu I (CGTAC/ G) are isoschizomers of each other. The first enzyme to recognize and cut a given sequence is known as the prototype, all subsequent enzymes that recognize and cut that sequence are isoschizomers. Isoschizomers are isolated from different strains of bacteria and therefore may require different reaction conditions. 64. Baculo v irus e x p re ssio n in inse ct ce lls represents a robust method for producing recombinant glycoproteins. Baculovirusproduced proteins are currently under study as therapeutic cancer vaccines with several immunologic advantages over proteins derived from mammalian sources. 65. Cowpea Trypsin Inhibitor: The explanation for this is called co-suppression. The plant has ways of knowing that the viral coat protein should not be produced, and it has ways of eventually shutting down the protein’s expression. When the virus tries to infect the plant, the production of its essential coat protein is already blocked. 66. Statement Q: When glyphosphate expressed in chloroplasts of transgenic plants, the insensitive enzymes produced by confer tolerance to the herbicide glyphosphate. Statement S: Glyphosphate acts by inhibiting enolpyruvyl-shikimate-3-phosphate synthase (EPSPS), an enzyme in the pathway leading to biosynthesis of aromatic amino acids. Since transgenic plants exhibit resistance to glyphosphate, there is no inhibition of the synthesis of aromatic amino acids. 67. P>3: The rolA, rolB and rolC genes are plant oncogenes that are carried in plasmids of the plant pathogen Agrobacterium rhizogenes. Following agrobacterial infection, these genes are transferred into the plant genome and cause tumor formation and hairy root disease. Q>1: Agrobacterium rhizogenes causes hairy root disease in plants. The transformed roots start producing opines which serve as carbon or nitrogen source for the infecting bacteria.

R>4: Genes in the virulence region are grouped into the operons virABCDEFG, which code for the enzymes responsible for mediating transduction of T-DNA to plant cells. S>2: Agrobacterium rhizogenes transfers a piece of oncogenic DNA, i.e. T-DNA. The T-DNA contains onc genes, the expression of which is responsible for the tumorous character of the transformed plant cells. Three onc genes are of prime importance: cyt, aux1, aux2. 68. In cell biology, pluripotency (from the Latin plurimus, meaning very many, and potens, meaning having power) refers to a stem cell that has the potential to differentiate into any of the three germ layers: endoderm (interior stomach lining, gastrointestinal tract, the lungs), mesoderm (muscle, bone, blood, urogenital), or ectoderm (epidermal tissues and nervous system). Pluripotent stem cells can give rise to any fetal or adult cell type. However, alone they cannot develop into a fetal or adult organism because they lack the potential to contribute to extraembryonic tissue, such as the placenta. 69.  DNA footprinting is a method of investigating the sequence specificity of DNAbinding proteins in vitro. This technique can be used to study protein-DNA interactions both outside and within cells.

 Yeast two-hybrid is based on the reconstitution of a functional transcription factor (TF) when two proteins or polypeptides of interest interact.  VNTRs have become essential to forensic crime investigations, via DNA fingerprinting and the CODIS database. When removed from surrounding DNA by the PCR or RFLP methods, and their size determined by gel electrophoresis or Southern blotting, they produce a pattern of bands unique to each individual. When tested with a group of independent VNTR markers, the likelihood of two unrelated individuals having the same allelic pattern is extremely improbable.  Serial analysis of gene expression (SAGE) is a technique used by molecular biologists to produce a snapshot of the messenger RNA population in a sample of interest in the form of small tags that correspond to fragments of those transcripts. It has been successfully used to describe the transcriptome of other diseases and in a wide variety of organisms.

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Advanced Biotechnology

70.  Erythropoietin (EPO) is a hormone that prevents anemia (low blood count) by making red blood cells. Anemia causes fatigue and low energy levels. It occurs when there are not enough red blood cells to carry oxygen from the lungs to supply all the body’s needs. Erythropoietin has its primary effect on red blood cells progenitors and precursors by promoting their survival through protecting these cells from apoptosis.

 Anti-fibrin 99 is involved in animal cell separation.  Collagenase enzyme helps in wound healing by assisting in blood clot formation.  Transferrins are iron-binding blood plasma glycoproteins that control the level of free iron in biological fluids. 71. Since 60%  692551 cpm So, 100% = 1154201.67% cpm Number of moles of 3H UTP =

1154201.67 cpm   mol 500  2.2  106 cpm

= 1050  10–12 mole = 1.05 nmole There is some error in options. 72. Parthenogenesis is a form of asexual reproduction in which growth and development of embryos occur without fertilization. In plants, parthenogenesis means development of an embryo from an unfertilized egg cell, and is a component process of apomixis. 73. By clonal propogation or micropropagation thousands of plants can be produced from a single piece of plant tissue explant. They are genetically identical. 74. STATEMENT P: HAT Medium (hypoxanthineaminopterin-thymidine medium) is a selection medium for mammalian cell culture, which relies on the combination of aminopterin, a drug that acts as a powerful folate metabolism inhibitor by inhibiting dihydrofolate reductase, with hypoxanthine and thymidine which are intermediates in DNA synthesis. Aminopterin blocks DNA de novo synthesis. STATEMENT S: The thymidine present in the HAT medium that can be absorbed by the cells and phosphorylated by thymidine kinase (TK) into TMP. By additional phosphorylation reactions, TMP can be used to make thymidine triphosphate (TTP), one of the four nucleotide precursors that are used by DNA polymerases to create DNA.

Hypoxanthine-guanine phosphoribosyltransferase (HGPRT) reacts with hypoxanthine absorbed from the medium with PRPP, liberating pyrophosphate, to produce IMP(Inosine Mono Phosphate) by a salvage pathway. Therefore, the use of HAT medium for cell culture is a form of artificial selection for cells containing working TK and HGPRT. 75. Somatic fusion, also called protoplast fusion through plasma membrane, is a type of genetic modification in plants by which two distinct species of plants are fused together to form a new hybrid plant with the characteristics of both, a somatic hybrid. Hybrids have been produced either between the different varieties of the same species (e.g. between non-flowering potato plants and flowering potato plants) or between two different species (e.g. between wheat triticum and rye secale to produce Triticale). 76.  Dicer is an endoribonuclease in the RNase III family that cleaves double-stranded RNA (dsRNA) and pre-microRNA (miRNA) into short double-stranded RNA fragments about 20-25 base pairs long, with a two-base overhang on the 3' end. Dicer contains two RNase III domains and one PAZ domain; the distance between these two regions of the molecule is determined by the length and angle of the connector helix and influences the length of the siRNAs it produces.

 Dicer facilitates the formation of the RNAinduced silencing complex (RISC), whose catalytic component argonaute is an endonuclease capable of degrading messenger RNA (mRNA). 77.  Probiotics are live bacteria that may confer a health benefit on the host. Lactic acid bacteria (LAB) and bifidobacteria are the most common types of microbes used as probiotics.

 A xenobiotic is a chemical which is found in an organism but which is not normally produced or expected to be present in it. The term xenobiotics is very often used in the context of pollutants such as dioxins and polychlorinated biphenyls.  Fructo-oligosaccharide (FOS) and lactulose have been classified as prebiotics. Prebiotics are non-digestible food ingredients that stimulate the growth and/or activity of bacteria in the digestive system in ways claimed to be beneficial to health.

Advanced Biotechnology 9.27

 β-Lactam antibiotics (beta-lactam antibiotics) are a broad class of antibiotics, consisting of all antibiotic agents that contains a -lactam ring in their molecular structures. 78. In PCR, the template DNA after ‘n’ cycle = (original DNA)  2n = 400  220 But, this is amount in 100l. In 0.1 l we have, 400  220  0.1 100 = 4.194304  105

=

79. Cells are harvested when the cells have reached a population density which suppresses growth. For adherent cultures, we use trypsin or collagenase. Trypsin, collagenase, or pronase, usually in combination with EDTA, causes cells to detach from the growth surface, thereby aiding in harvesting of anchorage dependent animal cells from culture vessels. 80. DNA footprinting is a method of investigating the sequence specificity of DNA-binding proteins in vitro. This technique can be used to study protein-DNA interactions both outside and within cells. The regulation of transcription has been studied extensively, and yet there is still much that is not known. Transcription factors and associated proteins that bind promoters, enhancers, or silencers to drive or repress transcription are fundamental to understanding the unique regulation of individual genes within the genome. Techniques like DNA footprinting will help elucidate which proteins bind to these regions of DNA and unravel the complexities of transcriptional control. 81. Secondary metabolites often play an important role in plant defense against herbivory and other interspecies defenses. Humans use secondary metabolites as medicines, flavorings, and recreational drugs. 82. PUFAs are long-chain fatty acids containing two or more double bonds. They provide structural & functional characteristics, and are involved in a wide range of biological components including membranes (in phospholipids). Equally important PUFAs serve as precursors for conversion into metabolites that regulate critical biological functions.Biodiesel is a non-petroleum and environmentally-friendly alternative to petroleum-based diesel fuel and not fatty acid.PUFA is fatty acid it is not used as it forms epoxide in presence of oxygen.

83. A form of asexual reproduction related to parthenogenesis is gynogenesis. Here, offspring are produced by the same mechanism as in parthenogenesis, but with the requirement that the egg merely be stimulated by the presence of sperm in order to develop. However, the sperm cell does not contribute any genetic material to the offspring. Since gynogenetic species are all female, activation of their eggs requires mating with males of a closely related species for the needed stimulus. 84.

Marker Gene

Enzyme encoded

Selective agent(s)

hpt

Hygromycin Hygromycin B phosphotransferase

npt II

Neomycin Kanamycin phosphotransferase

aro A

EPSP synthase

Glyphosate (Roundup)

bar

Phosphinothricin acetyltransferase

Phosphinothricin (bialaphos)

85. The various reasons attributed to the escape of the meristems by virus invasion are: (a)Viruses move readily in a plant body through the vascular system which in meristems is absent, (b)A high metabolite activity in the actively dividing meristematic cells does not allow virus replication and (c)A high endogenous auxin level in shoot apices may inhibit virus multiplication. 86. TthPlus DNA polymerase is isolated from the Thermus thermophilus strain. TthPlus DNA polymerase is a single 92 kDa polypeptide showing a 52 -32 exonuclease activity but lacking 32 -52 exonuclease activity. It catalyzes the polymerization of nucleotides into double-stranded DNA in the presence of MgCl2. Its efficiency has been shown more particularly on large DNA fragments up to 12 kb (using lambda phage DNA as a template). TthPlus DNA polymerase is also capable of catalyzing the polymerization of DNA using a RNA template in the presence of Mn2+. The ability of TthPlus DNA polymerase to reverse transcribe at elevated temperatures (70 C) minimizes the problems encountered with strong secondary structures in RNA since they are unstable at higher reaction temperatures. Higher temperatures also result in increased specificity of primer hybridization and extension. In coupled RT/

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Advanced Biotechnology

PCR assays, TthPlus is about 50-100 times more efficient than regular Taq DNA polymerase.

the best known of the auxins, and has been the subject of extensive studies by plant physiologists.

87. Gibberellins (GAs) are plant hormones that regulate growth and influence various developmental processes, including stem elongation, germination, dormancy, flowering, sex expression, enzyme induction, and leaf and fruit senescence.

93. Required for the sandwich assay, the capture antibody binds to the antigen of interest and generates a detectable signal for quantification. This antibody is usually polyclonal in nature, and is conjugated with an enzyme (most commonly, horseradish peroxidase), so that, upon substrate addition, a colored product is formed. The density of the color can then be measured spectrophotometrically as a function of antigen concentration. The western blot (sometimes called the protein immunoblot) is a widely accepted analytical technique used to detect specific proteins in the given sample of tissue homogenate or extract. It uses gel electrophoresis to separate native proteins by 3-D structure or denatured proteins by the length of the polypeptide. The proteins are then transferred to a membrane (typically nitrocellulose or PVDF), where they are stained with antibodies specific to the target protein.

88. Type I collagen, a fibrous protein abundant in connective tissues including tendon, ligament, dermis and blood vessel, is the major component and the primary determinant of tensile strength of the extracelluar matrix (ECM). It is widely used as a thin layer on tissue-culture surfaces to enhance the attachment and proliferation of a variety of cells including endothelial cells, fibroblasts, hepatocytes, epithelial cells and etc. In addition, collagen I can self-assemble into a 3-D superamolecular gel in vitro, making it an ideal biological scaffold to promote more in vivo-like cellular morphology and function. 89. Betaine improves the PCR amplification of GC-rich DNA sequences. Betaine reduces the formation of secondary structure caused by GC-rich regions and, therefore can be used to assist the amplification of GC rich sequences by PCR. It facilitates an isostabilizing effect on the differential thermal stability of AT and GC base pairs, which may result in a larger destabilizing effect on GC pairs and stabilzes AT rich sequence by binding with them and causing conformational change. 90. The efficiency of the production process depends on the rate of release of products rather than actual rate of biosynthesis. The immobilization process reduces biosynthetic capacity. The process does not allow easy removal of cell products or inhibitors as release of products/ by-products from cell into medium is hindered. 91. A lymphocyte is a type of white blood cell in the vertebrate immune system.Under the microscope, lymphocytes can be divided into large lymphocytes and small lymphocytes. Large granular lymphocytes include natural killer cells (NK cells). Small lymphocytes consist of T cells and B cells. 92. Zeatin is a plant hormone derived from the purine adenine. It is a member of the plant growth hormone family known as cytokinins. Zeatin was first discovered in immature corn kernels from the genus Zea. It promotes growth of lateral buds and stimulates cell division to produce bushier plants if sprayed on meristems. Again the Indole-3-acetic acid (IAA) is the most common, naturallyoccurring, plant hormone of the auxin class. It is

94. Lipofection (or liposome transfection) is a technique used to inject genetic material into a cell by means of liposomes, which are vesicles that can easily merge with the cell membrane since they are both made of a phospholipid bilayer. Lipofection generally uses a positively charged (cationic) lipid to form an aggregate with the negatively charged (anionic) genetic material and hence is the best mechanism to transform intact plant tissues. 95. Tissue plasminogen activator is a protein involved in the breakdown of blood clots. The importance of IFNγ in the immune system stems in part from its ability to inhibit viral replication directly, and most importantly from its immunostimulatory and immunomodulatory effects. podophyllotoxin is the pharmacological precursor for the important anticancer drug etoposide. Polyhydroxyalkanoates or PHAs are linear polyesters produced in nature by bacterial fermentation of sugar or lipids, used in the production of biodegradable plastics. 96. Amperometric biosensors function by the production of a current generated by flux of redox electrons when a potential is applied between two electrodes. Evanescent wave biosensors are fiber-optic sensors which detect the refractive index of the light beams. Many enzyme catalysed reactions are exothermic, generating heat, this principle is used as a basis for measuring the rate of reaction and, hence, analyze concentration by calorimetric biosensors. Potentiometric biosensors make use of ion-selective electrodes in order to transduce the biological

Advanced Biotechnology 9.29

reaction into an electrical signal. A recent development from ion-selective electrodes is the production of ion-selective field effect transistors (ISFETs) and their biosensor use as enzyme-linked field effect transistors. 97. Somaclonal variation is not restricted to, but is particularly common in, plants regenerated from callus. The variations can be genotypic or phenotypic, which in the latter case can be either genetic or epigenetic in origin. Typical genetic alterations are: changes in chromosome numbers (polyploidy and aneuploidy), chromosome structure (translocations, deletions, insertions and duplications) and DNA sequence (base mutations). Typical epigenetic related events are: gene amplification and gene methylation. 98. Gel filtration is a separation based on size and same is the purpose of SDS-PAGE i.e. to separate proteins according to their size, and no other physical feature. 99. Con A has varying affinities for various polypeptides and hence affinity binding will be used. 100. Number of -phage vectors = Number of yeast clones =

1.0  108 = 5  103 20  103

108 = 400 250  103

101. -phage vectors =

in(1  p) = in(1  a / b)

=

in(0.0001) in(0.0001) = in(1  0.0002) in(0.9998)

in(0.0001)  20  103 bp  in  1  108 bp  

= Yeast clones =

9.210 = 5  104 2.0  10 4

in(0.0001)

103. A transplastomic plant is a genetically modified plant in which the new genes have not been inserted in the nuclear DNA but in the DNA of the chloroplasts. The major advantage of this technology is that in many plant species plastid DNA is not transmitted through pollen, which prevents gene flow from the genetically modified plant to other plants. 104. The separation of the fragments in agarose gel is accomplished by exploiting the mobilities with which different sized molecules are able to pass through the gel. Longer molecules migrate more slowly because they experience more resistance within the gel. Because the size of the molecule affects its mobility, smaller fragments end up nearer to the anode than longer ones in a given period. 105. Real-time quantitative reverse transcriptase– polymerase chain reaction (RT–PCR) is the method of choice for rapid and reproducible measurements of cytokine or growth factor expression in small samples. Fluorescence detection methods for monitoring real-time PCR include fluorogenic probes labelled with reporter and quencher dyes, such as Taqman probes or Molecular Beacons and the dsDNA-binding dye SYBR Green I. SYBR Green binds to the minor groove of the dsDNA only. In the solution the unbound dye exhibits little fluorescence. This fluorescence is enhanced when the dsDNA is bound with dye. 106. Let the DNA be the following line:

 250  103  in  1   108 

=

9.210 in(1  0.0025)

=

9.210 in(0.9975)

9.210 = 4000 2.5  10 3 102. The meristematic cells give rise to various organs of the plant, and keep the plant growing. The Shoot Apical Meristem (SAM) gives rise to organs like the leaves and flowers. The cells of the apical =

meristems - SAM and RAM (Root Apical Meristem) - divide rapidly and are considered to be indeterminate, in that they do not possess any defined end fate. In that sense, the meristematic cells are frequently compared to the stem cells in animals, which have an analogous behavior and function.

The cut portions are E.CoRI digestion site. So the partial digestion can generate 3 smaller fragments with label and 5 fragments without any level. 107. In the globular stage, the embryo develops radial patterning through a series of cell divisions, with the outer layer of cells differentiating into the ‘protoderm.’ The globular embryo can be thought of as two layers of inner cells with distinct developmental fates; the apical layer will go on to produce cotyledons and shoot meristem, while

9.30

Advanced Biotechnology

the lower layer produces the hypocotyl and root meristem. Bilateral symmetry is apparent from the heart stage; provascular cells will also differentiate at this stage. In the subsequent torpedo and cotyledonary stages of embryogenesis, the embryo completes its growth by elongating and enlarging. 108. virulence genes are grouped genes for synthesis of opines and plant hormones are contained on the T-DNA; T-DNA is bordered by 25 bp direct repeats; only the TR border is required for transfer although the Ti border increased the efficiency when it is present; no sequences other that the borders are required for transfer. 109. Regulation of UDP glucuronosyl transferase is the result of differential UGT gene expression and this is in response to those chemicals that are primarily eliminated by conjugation with glucuronic acid and hence more evident in liver and hence it can be stated that its regulation is post transcriptionally controlled. 110. The process of 52 capping is vital to creating mature messenger RNA, which is then able to undergo translation. Capping ensures the messenger RNA’s stability while it undergoes translation in the process of protein synthesis. Glycosylation confers stability to the proteins and is also required for the folding of proteins. Absence of either of these mechanisms result in poor expression and inactivity. 111. The replication intermediates of parvoviruses thus are concatemers of head-to-head or tail-totail structure. Surprisingly, in case of the novel human bocavirus, neither head-to-head nor tailto-tail DNA sequences were detected in clinical isolates; in contrast head-to-tail DNA sequences were identified by PCR and sequencing. Thereby, the head-to-tail sequences were linked by a novel sequence which can be detected by southern blot. 112. The Myc family of genes are proto-oncogenes implicated in cancer. Yamanaka et al. and Jaenisch et al. demonstrated that c-myc is a factor implicated in the generation of mouse iPS cells and Yamanaka et al. demonstrated it was a factor implicated in the generation of human iPS cells. However, Thomson et al., Yamanaka et al., and unpublished work by Johns Hopkins University reported that c-myc was unnecessary for generation of human iPS cells. Usage of the “myc” family of genes in induction of iPS cells is troubling for the eventuality of iPS cells as clinical therapies, as 25% of mice transplanted with c-myc-induced iPS cells developed lethal

teratomas. N-myc and L-myc have been identified to induce instead of c-myc with similar efficiency. C-Myc cells after transformation last for 10 generations. 113. Sarcoma is a cancer that arises from transformed cells of mesenchymal origin. Thus, malignant tumors made of cancerous bone, cartilage, fat, muscle, vascular, or hematopoietic tissues are, by definition, considered sarcomas. This is in contrast to a malignant tumor originating from epithelial cells, which are termed carcinoma. Sarcomas are quite rare. Common malignancies, such as breast, colon, and lung cancer, are almost always carcinoma. 114. A DNA microarray (also commonly known as DNA chip or biochip) is a collection of microscopic DNA spots attached to a solid surface. Scientists use DNA microarrays to measure the expression levels of large numbers of genes simultaneously or to genotype multiple regions of a genome. Each DNA spot contains picomoles (10"12 moles) of a specific DNA sequence, known as probes (or reporters or oligos). These can be a short section of a gene or other DNA element that are used to hybridize a cDNA or cRNA (also called anti-sense RNA) sample (called target) under highstringency conditions. Probe-target hybridization is usually detected and quantified by detection of fluorophore-, silver-, or chemiluminescencelabeled targets to determine relative abundance of nucleic acid sequences in the target. 115. Genes in the virulence region are grouped into the operons vir ABCDEFG, which code for the enzymes responsible for mediating transduction of T-DNA to plant cells 116. IL-2 is necessary for the growth, proliferation, and differentiation of T cells to become ‘effector’ T cells. IL-2 is normally produced by T cells during an immune response. Antigen binding to the T cell receptor (TCR) stimulates the secretion of IL-2, and the expression of IL-2 receptors IL2R. The IL-2/IL-2R interaction then stimulates the growth, differentiation and survival of antigen-specific CD4+ T cells and CD8+ T cells As such, IL-2 is necessary for the development of T cell immunologic memory, which depends upon the expansion of the number and function of antigen-selected T cell clones. 117. This process is called nick translation because the DNA to be processed is treated with DNase to produce single-stranded “nicks.” This is followed by replacement in nicked sites by DNA polymerase I, which elongates the 3' hydroxyl

Advanced Biotechnology 9.31

terminus, removing nucleotides by 5'-3' exonuclease activity, replacing them with dNTPs. Thus primers are not required. 118. the sensitive biological element (biological material (e.g. tissue, microorganisms, organelles, cell receptors, enzymes, antibodies, nucleic acids, etc.), a biologically derived material or biomimic component that interacts (binds or recognises) the analyte, the transducer or the detector element (works in a physicochemical way; optical, piezoelectric, electrochemical, etc.) that transforms the signal resulting from the interaction of the analyte with the biological element into another signal. Biosensor reader device with the associated electronics or signal processors that are primarily responsible for the display of the results in a user-friendly way. 119. Somatic embryogenesis is a process where a plant or embryo is derived from a single somatic cell or group of somatic cells. Somatic embryogenesis is thus not fully characterized by the given options. Hence none of the above is the right answer. 120. Monoclonal antibodies bind to a single epitope within a target antigen. Western blot procedures allow for flexibility in choosing gel electrophoresis and blotting conditions, it is possible to modify buffers to retain enough higher order protein structure for detection by some antibodies. Thus can be brought to usage in screening for monoclonal antibodies. 121. The plants are not transformed at all, but the development of kanamycin resistance is due to the somaclonal variation. 122. The plasmid contains a genetic sequence corresponding to the recognition site of a restriction endonuclease, such as HaeIII. After foreign DNA fragments, which have also been cut with the same or similar restriction endonuclease, have been inserted into host cell, the restriction endonuclease gene is expressed by applying heat, or by introducing a biomolecule, such as arabinose. 123. The polymerase starts replication at the 3' end of the primer which means 5' end of the template and hence A is the correct option. 124. Agrobacterium infection method has been extensively used to transfer foreign DNA into a number of dicotlyledonous species with the exception of soybean (Glycine max). However, this technique of gene transfer was not successful in monocotyledons for reasons yet unknown most probable being the lack of wound response of monocotyledonous cells.

125. Adding of 2,4-D (1 M), in the cultured cells grew rapidly, producing little berberine. On the other hand, the berberine-producing activity was remarkably enhanced by simultaneous administration of auxin and cytokinin, although cell growth was inferior.] 126. neoR stands for the eukaryotic gene expression marker, SV40 facilitate gene expression in eukaryotes, LTR induces new gene expression and dhfr is a gene amplification factor. 127. for a cloning vector, origin of replication, a restriction site, high copy number and a selectable marker are required. A selectable marker can be of any type not necessarily to be an antibiotic resistance site. 128.  Serum provides the basic nutrients (both in the solution as well as bound to the proteins) for cells.

 Serum provides several growth factors and hormones involved in growth promotion and specialized cell function.  It provides several binding proteins like albumin, transferrin, which can carry other molecules into the cell. For example: albumin carries lipids, vitamins, hormones, etc into cells.  It also supplies proteins, like fibronectin, which promote attachment of cells to the substrate. It also provides spreading factors that help the cells to spread out before they begin to divide.  It provides protease inhibitors which protect cells from preolysis.  It also provides minerals, like Na+, K+, Zn2+, Fe2+, etc.  It increases the viscosity of medium and thus, protects cells from mechanical damages during agitation of suspension cultures.  It also acts a buffer. 129. Protoplasts can be obtained routinely from many plant species, bacteria, yeasts and filamentous fungi. Protoplasts from different strains can sometimes be persuaded to fuse and so overcome the natural sexual mating barriers. However, the range of protoplast fusions is severely limited by the need for DNA compatibility between the strains concerned. Fusion of protoplasts can be enhanced by treatment with the chemical polyethylene glycol, which, under optimum conditions, can lead to extremely high frequencies of recombinant formation which can be increased still further by ultraviolet

9.32

Advanced Biotechnology

irradiation of the parental protoplast preparations. Protoplast fusion can also occur with human or animal cell types. 130. The vast majority of a human’s DNA will match exactly that of any other human, making distinguishing between two people rather difficult. DNA finger printing uses a specific type of DNA sequence, known as a microsatellite, to make identification much easier. Microsatellites are short pieces of DNA which repeat many times in a given person’s DNA. In a given area, microsatellites tend to be highly variable, making them ideal for DNA finger printing. By comparing a number of microsatellites in a given area, one can identify a person relatively easily. 131. Total gene size = 5000bp, Because sequencer can sequence upto 500 bases in a single reaction. Then the fragments are = 5000/500 = 10 Therefore 48 = 6536, 46 = 4096 , 45 = 1024, 43 = 64 Only 3 – bp cutter will generate more than 10 fragments ( 5000/64 = 78). Rest other cutter will generate less than 10 fragments. 132. insert size should be 500bp

5 ´-hydroxyl terminus of double- and singlestranded DNA and RNA, as well as nucleoside 3´-monophosphates. This modified version exhibits full kinase activity with no 3´ phosphatase activity. Highlights of T4 Polynucleotide Kinase: (1) 5´ phosphorylation of DNA/RNA for subsequent ligation (2) End-labeling of DNA or RNA (3) 5´ phosphorylation of 3´ phosphorylated mononucleotides to generate a substrate (pNp) that can be added to the 3´ end of DNA or RNA by ligase activity (4) 5´ end labeling of 3´ phosphorylated oligos 143. The secondary metabolism can be regulated to maximize the production of target compounds. 144. Baculoviruses are insect pathogenic and insect specific viruses belonging to the group of Nuclear polyhedrosis viruses. Cloning vectors exploit the extraordinary strength of the polyhedrin gene promoter.

134. RE doesn’t recognize RNA.

145. The border sequence of Ti plasmid is known to be around 25bp long which is sufficient for cleavage and overdrive sequences.

135. Check for cutting site specification. Cut by XmaIII can be cloned into a plasmid with Eco521.

146. Taq polymerase is necessary for DNA cycle sequencing.

136. Bacterial replication machinery is adapted for both linear and circular molecule in forms of genomic DNA and plasmid DNA but not equally.

147. The antibiotic aminopterin inhibits the de novo nucleotide synthesis pathway. However, normal cells survive in this medium as they are able to use the salvage pathway for nucleic acid synthesis. But if the cells are unable to produce the enzyme Hypoxanthine-Guanine Phospho Ribosyl Transferase (HGPRT), they are unable to utilize the salvage pathway and therefore die in aminopterin containing medium.

133. Array data is naked eye visible entity.

137. Common approach to inactivate a selected gene is to insert an antibiotic resistance gene into the cloned gene of interest. This both inactivates the gene of interest and allows identification of ES cells that have been successfully transfected. In the engineered insertion construct, the target gene is disrupted with the neoR gene which confers neomycin resistance. 138. A phospholipase is an enzyme that hydrolyzes phospholipids into fatty acids and other lipophilic substances. There are four major classes, termed A, B, C and D, distinguished by the type of reaction which they catalyse. 139. It’s a hormone (auxin). 140. Mature tissue explant first undergoes dedifferentiation and then redifferentiation in order to produce a viable callus in cell culture. 141. Clonal propagation is the asexual propagation of the plantlets. It needs less time and a single plant to generate a pure verity of the plant. 142. T4 Polynucleotide Kinase catalyzes the transfer and exchange of Pi from the γ position of ATP to the

148. 2,4-D, being a phenoxy auxin, promoted quick callus formation from the induced roots and also at the base of the shoots. 151. For sequencing we need complementary forward and reverse primer to bind with target sequence. 152. Chlorsulfuron is a selective systemic herbicide, absorbed by the leaves and roots and moves rapidly through the plant. Chlorsulfuron blocks acetolactate synthase and thus prevents the plant from producing a branched chain amino acids. This inhibits cell division in the root tips and shoots of sensitive plants. 153. The pancreas is a glandular organ in the digestive system and endocrine system of vertebrates. It is both an endocrine gland producing several important hormones,

Advanced Biotechnology 9.33

including insulin, glucagon, somatostatin, and pancreatic polypeptide, and a digestive organ, secreting pancreatic juice containing digestive enzymes that assist the absorption of nutrients and the digestion in the small intestine. These enzymes help to further break down the carbohydrates, proteins, and lipids in the chyme. 154. The Ti plasmid contains most of the genes required for tumor formation. Wounded plants exude phenolic compounds that stimulate the expression of the virulence genes(vir -genes), which are also located on the Ti plasmid. The vir genes encode a set of proteins responsible for the excision, transfer and integration of the T-DNA into the plant genome. The genes in the T-DNA region are responsible for the tumorigenic process. 155. Positional cloning is a method of gene identification in which a gene for a specific phenotype is identified only by its approximate chromosomal location (but not the function); this is known as the candidate region. Initially, the candidate region can be defined using techniques such as linkage analysis, and positional cloning is then used to narrow the candidate region until the gene and its mutations are found. Positional cloning typically involves the isolation of partially overlapping DNA segments from genomic libraries to progress along the chromosome toward a specific gene. 156. Auxin helps in generation of the root while cytokinin helps in shoot generation. 157. Ecoli doesn’t have any splicing system so that it is difficult to achieve the human protein purity in E.Coli cells. 158. Embryo Rescue is one of the earliest and successful forms of in-vitro culture techniques that is used to assist in the development of plant embryos that might not survive to become viable plants. Embryo rescue plays an important role in modern plant breeding, allowing the development of many interspecific and intergeneric food and ornamental plant crop hybrids. This technique nurtures the immature or weak embryo, thus allowing it the chance to survive. Plant embryos are multicellular structures that have the potential to develop into a new plant. The most widely used embryo rescue procedure is referred to as embryo culture, and involves excising plant embryos and placing them onto media culture. Embryo rescue is most often used to create interspecific and intergeneric crosses that would normally

produce seeds which are aborted. Interspecific incompatibility in plants can occur for many reasons, but most often embryo abortion occurs In plant breeding, wide hybridization crosses can result in small shrunken seeds which indicate that fertilization has occurred, however the seed fails to develop. Many times, remote hybridizations will fail to undergo normal sexual reproduction, thus embryo rescue can assist in circumventing this problem. 161. Restriction enzymes are commonly classified into three types, which differ in their structure and whether they cut their DNA substrate at their recognition site, or if the recognition and cleavage sites are separate from one another. To cut DNA, all restriction enzymes make two incisions, once through each sugar-phosphate backbone (i.e. each strand) of the DNA double helix. 162. Helicase proteins are motor proteins that move directionally along a nucleic acid phosphodiester backbone, separating two annealed nucleic acid strands (i.e., DNA, RNA, or RNA-DNA hybrid) using energy derived from ATP hydrolysis. 163. According to the Hardy–Weinberg’s principle, in the simplest case of a single locus with two alleles: the dominant allele is denoted A and the recessive a and their frequencies are denoted by p and q; freq(A) = p; freq(a) = q; p + q = 1. If the genotype frequencies are in Hardy–Weinberg proportions resulting from random mating, then we will have freq(AA) = p 2 for the AA homozygotes in the population, freq(aa) = q2 for the aa homozygotes, and freq(Aa) = 2pq for the heterozygotes. So, q2 = 0.04, q = 0.2, p = 1-0.2 = 0.8, p2+q2+2pq=1, 2pq = 1-(p2 + q2) = 0.32 = 32% 164. An ampicillin resistance gene (abbreviated ampR) is commonly used as a selectable marker in routine biotechnology. bacterial AmpR gene when complete, encodes b-lactamase conferring resistance to ampicillin. 165. Plasminogen activator - protease produced in the kidney that converts plasminogen to plasmin and so initiates fibrinolysis. The approval of Chinese hamster ovary (CHO)-derived tissue plasminogen activator (tPA, Activase) in 1986 revolutionized medicine and raised the possibility of using mammalian cell culture for the manufacturing of protein therapeutic products. 166. Mung Bean Nuclease acts as a preferential DNA and RNA endonuclease, hydrolyzing singlestranded DNA and RNA into 5'-phosphate and 3'-hydroxyl-containing products. If used at very

9.34

Advanced Biotechnology

high concentrations, Mung Bean Nuclease is also capable of degrading double-stranded DNA, RNA or DNA/RNA fragments from both ends. Due to its single-stranded DNA and RNA endonuclease activity, Mung Bean Nuclease is often used during transcript mapping, duplex or hybrid DNA and RNA blunt ending, and cDNA preparation. 167. A novel Eschericha coli expression system directed by bacteriophage T7 RNA Polymerase utilized for overexpression of the cloned gene. The recombinant cell contains the plasmid with a bacteriophage promoter, the T7 promoter, to regulate the expression of the target gene. This promoter is recognized only by T7 RNA polymerase, whose gene has been fused into the host chromosome and is under control of the lacUV5 promoter. Therefore, the target gene on the plasmid can be expressed only in the presence of T7 RNA polymerase, which is induced by isopropyl-beta-D-thiogalactopyranoside (IPTG). 168. These artificial vectors is amplified in yeast and hence is the name. YACs can hold huge inserts

ranging from 100kb to 3000kb, i.e. the entire chromosomes are being replicated in this vector. 169. dNTPs are required in place of NTPs in PCR. 170. Clonal propagation is pure genetic variation with respect to the parents. Thus its only possible in plants where the generation time is low as compare to animals. 171. Bioremediation employs the process of natural metabolic flux input for the metabolic degradation of the target. Thus it’s unobtrusive and non disruptive and not end up with hazardous waste material 172. For all cDNA clones 50% may take gene in reverse orientation. 173. The GUS reporter system (GUS: betaglucuronidase) is a reporter gene system, particularly useful in plant molecular biology. Several kinds of GUS reporter gene assay are actually available, depending on the substrate used. The term GUS staining refers to the most common of these, a histochemical technique.

10

C H A P TE R

Enzyme Science and Technology

7 2016 1. Which one of the following statements is NOT true? (a) In competitive inhibition, substrate and inhibitor compete for the same active site of an enzyme (b) Addition of a large amount of substrate to an enzyme cannot overcome uncompetitive inhibition

at this wavelength. In an assay, 25 L of a sample of enzyme (containing 5 g protein per mL) was added to a mixture of pyruvate and NADH to give a total volume of 3 mL in a cuvette of 1 cm pathlength. The rate of decrease in absorbance at 340 nm was 0.14 min–1. The specific activity of the enzyme will be _________ mol.min–1.mg–1. 5. Analysis of a hexapeptide using enzymatic cleavage reveals the following result:



Amino acid composition of the peptide is: 2R, A,V, S, Y



(d) In non-competitive inhibition, Km of an enzyme for its substrate remains constant as the concentration of the inhibitor increases.

Trypsin digestion yields two fragments and the compositions are: (R, A, V) and (R, S, Y)



2. Choose the CORRECT combination of True (T) and False (F) statements about microcarriers used in animal cell culture.

Chymotrypsin digestion yields two fragments and the compositions are: (A, R, V, Y) and (R, S)



Digestion with carboxypeptidase A yields no cleavage product.

(c) A transition state analogue in enzyme catalyzed reaction increases the rate of product formation

P. Higher cell densities can be achieved using microcarriers

Given:

Q. Microcarriers increase the surface area for cell growth

Chymotrypsin cleaves at carboxyl side of Y.

Trypsin cleaves at carboxyl side of R.

R. Microcarriers are used for both anchorageand nonanchorage-dependent cells

Carboxypeptidase A cleaves at amino side of the C-terminal amino acid (except R and K) of the peptide.

S. Absence of surface charge on microcarriers enhances attachment of cells

The correct amino acid sequence of the peptide is:

(a) P-T, Q-F, R-T and S-F (b) P-T, Q-T, R-F and S-F (c) P-F, Q-F, R-T and S-T (d) P-F, Q-T, R-F and S-T 3. In an assay of the type II dehydroquinase of molecular mass 18 kDa, it is found that the Vmax of the enzyme is 0.0134 mol.min–1 when 1.8 g enzyme is added to the assay mixture. If the Km for the substrate is 25 M, the kcat/Km ratio will be _____ 104 M–1.s–1. 4. The activity of lactate dehydrogenase can be measured by monitoring the following reaction: Pyruvate + NADH  Lactate + NAD+ The molar extinction coefficient of NADH at 340 nm is 6220 M–1.cm–1. NAD+ does not absorb

(a) RSYRVA (b) AVRYSR (c) SRYVAR (d) SVRRYA 6. An enzyme is immobilized on the surface of a nonporous spherical particle of 2 mm diameter. The immobilized enzyme is suspended in a solution having bulk substrate concentration of 10 mM. The enzyme follows first order kinetics with rate constant 10 s–1 and the external mass transfer coefficient is 1 cm.s –1. Assume steady state condition wherein rate of enzyme reaction (mmol.L–1.s–1) at the surface is equal to mass transfer rate (mmol.L –1 .s –1 ). The substrate concentration at the surface of the immobilized particle will be ________ mM.

10.2 Enzyme Science and Technology

2015 7. The interaction between an antigen (Ag) and a single-chain antibody (Ab) was studied using Scatchard analysis. The result is shown below.

Y [Ab-Ag] [Ag]

2014

X [Ab-Ag] The affinity of interaction and the total concentration of antibody, respectively, can be determined from (a) slope and Y-intercept (b) Y-intercept and slope (c) X-intercept and slope (d) slope and X-intercept 8. A single subunit enzyme converts 420 moles of substrate to product in one minute. The activity of the enzyme is  10–6 Katal. 9. Which one of the following graphs represents uncompetitive inhibition ? 0

(b)

in hi bi to r

1 [v0]

wi th

(a)

tor ibi nh i tor th ibi wi h n ti ou th i w 1 0 [s]

1 [v0]

t ou th wi

(c)

tor ibi h in

1 [s]

wi th

in hi bi to r

0

1 [v0]

tor ibi h in

t ou th wi

1 [s] nh ibi tor

0

t ou th wi

0

11. Which of the following is employed for the repeated use of enzymes in bioprocesses? (a) polymerization (b) immobilization (c) ligation (d) isomerization 12. The degree of reduction of ethanol is _______ 13. Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde3-phosphate (G-3-P) in a reversible reaction. At 298 K and pH 7.0, the equilibrium mixture contains 40 mM DHAP and 4 mM G-3-P. Assume that the reaction started with 44 mM DHAP and no G-3-P. The standard free-energy change in kJ/mol for the formation of G-3-P [R = 8.315 J/mol.K] is _________.

2013 14. The catalytic efficiency for an enzyme is definedas (a) k cat

vmax (b) k cat

kcat (c) k m

kcat (d) v max

15. The activity of an enzyme was measured by varying the concentration of the substrate (S) in the presence of three different concentrations of inhibitor (I) 0, 2 and 4 mM.The double reciprocal plot given belowsuggests that the inhibitor (I) exhibits [l] = 4 mM [l] = 2 mM

1/vo

[l] = 0 mM

0

1/[s]

(a) substrate inhibition

wi th i

(d)

1 [v0]

10. The Ki of novel competitive inhibitor designed against an enzyme is 2.5 M. The enzyme was assayed in the absence or presence of the inhibitor (5 M) under identical conditions. The Km in the presence of the inhibitor was found to be 30 M. The Km in the absence of the inhibitor is ____ M.

tor ibi h in

1 [s]

(b) uncompetitive inhibition (c) mixed inhibition (d) competitive inhibition

Enzyme Science and Technology 10.3

Common Data for Questions 16 and 17: The binding efficiency of three different receptorsR1, R2 and R3 were tested against a ligand using equilibrium dialysis, with a constant concentration of receptor and varying concentrations of ligand. The Scatchard plot of receptor titration with different concentration of ligand is given below (r is moles of bound ligand per moles of receptor and c is concentration of free ligand) 4

19. Identify the statement that is NOT applicable to an enzyme catalyzed reaction (a) Enzyme catalysis involves propinquity effects (b) The binding of substrate to the active site causes a strain in the substrate (c) Enzymes do not accelerate the rate of reverse reaction (d) Enzyme catalysis involves acid-base chemistry 20. An enzymatic reaction is described by the following rate expression.

R1

3 R2 r/c 2 R3

=

R4

1 0

m s km  s 

1

2

3

r

16. The number of ligand binding sites present on receptors R1 and R3, respectively are (a) 1 and 4 (b) 1 and 1 (c) 4 and 1 (d) 2 and 2 17. Which one of the receptors has thehighestaffinity for the ligand? (a) R1 (b) R2 (c) R3

s2 ks

Which one of the following curves represents this expression?

(a)

s

(b)

s 



(d) R4

2012 18. The activity of an enzyme is expressed in International Units (IU). However, the S.I. unit for enzyme activity is Katal. One Katal is (a) 1.66  104 IU (b) 60 IU (d) 106 IU (c) 6  107 IU

(c) s

(d) s





Directions (Q. 21 – 22) : The purification data for an enzyme is given below Step

Volume (ml) Total protein (mg) Total activity (Units) Specific activity (Units/mg)

P Cell-free extract

17

177

102

0.58

Q Q-Sepharose

14

18.8

72

3.83

R Phenyl Sepharose

26

9.2

45

4.89

S Sephacryl S-200

7

4.1

30

7.32

21. The fold purification for each step is (a) P – 0.1, Q – 0.66, R – 0.84, S – 1.26 (b) P – 1.0, Q – 0.52, R – 0.67, S – 0.8 (c) P – 1, Q – 6.6, R – 8.4, S – 12.6 (d) P – 100, Q – 66, R – 84, S – 12 22. The yield (%) for each step is (a) P – 10, Q – 7.2, R – 4.5, S – 2.0 (b) P – 34, Q – 24, R – 15, S – 1 (c) P – 3.4, Q – 2.4, R – 1.5, S – 0.1 (d) P – 100, Q – 71, R – 44, S – 29

10.4 Enzyme Science and Technology

Directions (Q. 23 – 24) : An E. coli cell of volume 10–12 cm3 contains 60 molecules of lac repressor. The repressor has a binding affinity (Kd) of 10–8 M and 10–9 M with and without lactose respectively, in the medium.

(b) The velocity of reactions catalyzed by E1 and E2 at their respective saturating substrate concentrations could be equal, if concentration of E2 used is 10 times that of E1

23. The molar concentration of the repressor in the cell is

(c) The velocity of E1 catalyzed reaction is always greater than that of E2

(a) 1.1 nM (b) 1 nM (c) 10 nM (d) 100 nM 24. Therefore the lac-operon is (a) repressed and can only be induced with lactose. (b) repressed and cannot be induced with lactose.

(d) The velocity of E1 catalyzed reaction at a particular enzyme concentration and saturating substrate concentration is lower than that of E2 catalyzed reaction under the same conditions 29. The dissociation constant Kd for ligand binding to the receptor is 10–7 M. The concentration of ligand required for occupying10% of receptors is

(c) not repressed.

(a) 10–6 M

(b) 10–7 M

(d) expressed only when glucose and lactose are present.

(c) 10–8 M

(d) 10–9 M

2010 25. Identify the enzyme that catalyzes the following reaction

-Ketoglutarate + NADH + NH4+ + H+ ?  Glutamate + NAD+ + H2O (a) Glutamate synthetase

30. Receptor R is over expressed in CHO cells and analysed for expression. 6  10 7 cells were incubated with its radioactive ligand (specific activity 100 counts per picomole). If the total counts present in cell pellet was 1000 cpm, the average number of receptors R per cell is (assume complete saturation of receptors with ligand and one ligand binds to one receptor)

(b) Glutamate oxoglutarate aminotransferase

(a) 104

(b) 105

(c) Glutamate dehydrogenase

(c) 106

(d) 107

(d) -ketoglutarate deaminase 26. The degree of inhibition for an enzyme catalyzed reaction at a particular inhibitor concentration is independent of initial substrate concentration. The inhibition follows (a) competitive inhibition (b) mixed inhibition (c) un-competitive inhibition (d) non-competitive inhibition 27. An immobilized enzyme being used in a continuous plug flow reactor exhibits an effectiveness factor () of 1.2. The value of  being greater than 1.0 could be apparently due to (a) substrate inhibited with internal pore diffusion limitation

2009 31. One unit of glucoamylase enzyme activity is defined as the amount of enzyme required to produce (imol of glucose per min in a 4% solution of Lintner starch at pH 4.5 and 60C. If in a reaction mixture with 1ml of the crude enzyme preparation containing 8 mg protein and 9 ml of 4.44% starch, 0.6 mc of glucose/ml-min is produced, what will be the specific activity of the crude enzyme preparation? (a) 1 unit/mg protein (b) 1.5 units/mg protein (c) 0.25 units/mg protein (d) 0.75 units/mg protein

(b) external pore diffusion limitation

2008

(c) sigmoidal kinetics

32. Which of the following techniques is best suited for immobilizing an affinity ligand?

(d) unstability of the enzyme 28. The turnover numbers for the enzymes, E1 and E2 are 150 s–1 and 15 s–1 respectively. This means (a) E1 binds to its substrate with higher affinity than E2

(a) Physical adsorption (b) Gel entrapment (c) Cross-linking with a polymer (d) Covalent linkage to a spacer arm

Enzyme Science and Technology 10.5

33. A DNA fragment of 4500 bp has to be tailed with dT residues by using dTTP and the enzyme ‘terminal transferase’. The stock solution of dTTP that is used as a substrate has a concentration of 150 M. Ten l of this stock solution is added to a total volume of 200 l reaction. What will be the concentration of dTTP in the reaction? (a) 7.5 M

(b) 75 M

(c) 0.75 M

(d) 0.075 M

34. Evaluate the Michaelis constant for the following lipase catalyzed trans-esterification reaction for the production of biodiesel k1

  Oil-lipase complex Vegetable oil +Lipase  k  2

2   Biodiesel +Glycerol

k

where, k1 = 3  108 M–1 s–1; k = 4  103 s–1 and k2 = 2  103 s–1 (a) 4.2  10 M

(b) 14.0  10 M

(c) 6.4  10 M

(d) 1.4  10–4 M

–3 –6

–4

35. In a chemostat, evaluate the dilution rate at the cell wash-out condition by applying Monod’s model with the given set of data: max = 1h–1; Yx/s = 0.5 g g–1; Ks = 0.2 g L–1; S0 = 10 g L–1 (a) 1.00 h–1

(b) 0.49 h–1

(c) 0.98 h–1

(d) 1.02 h–1

Common Data Questions Common Data for Questions 36 and 37: An enzyme (24000 Da) undergoes first-order deactivation kinetics while catalyzing a reaction according to Michaelis-Menten kinetics (Km= 10–4 M). The enzyme has a turnover number if 104 molecules substrate/min-(molecule enzyme) and a deactivation constant (kd) of 0.1 min–1 at the reaction conditions. The reaction mixture initially contains 0.6 mg/l of active enzyme and 0.02 M of the substrate. 36. The time required to convert 10% of the substrate will be rapproximately

38. After completion of Nick translation reaction, 10 l of reaction was spotted on a glass-fiber filter that upon counting resulted into 4.2  104 cpm in reaction. Another 10 l was processed for TCA precipitation to determine redioisotope incorporation. The TCA precipitated sample gave 2.94  104 cpm. What is the percent of [-32P]-dCTP incorporation into the DNA sample? (a) 40%

(b) 50%

(c) 60%

(d) 70%

39. If 2.94  104 cpm of TCA precipitable counts of the 10 l sample were taken from 1/10 dilutioa of the 100 l Nick Translation reaction containing 1 g of -Interferon DNA, what is the specific activity of the labeled product? (a) 1.47  106 cpm/ g

(b) 1.47  107 cpm/ g

(c) 2.94  10 cpm/ g

(d) 2.94  107 cpm/ g

6

Statement for Linked Answer Questions 40 and 41: A double reciprocal plot was created from the specific growth rate and limiting-substrate concentration data obtained from a chemostat experiment. A linear regression gave values of 1.25 hr and 100 mg-hr-l–1 for the intercept and slope, respectively. 40. The respective values of the Monod kinetic constants m (hr–1) and Ks (mg/l) are as follows: (a) 0.08, 8

(b) 0.8, 0.8

(c) 0.8, 80

(d) 8, 8

41. The same culture (with the m and Ks values as computed above) is cultivated in a 10-litre chemostat being operated with a 50 ml/min sterile feed containing 50 g/l of substrate. Assuming an overall yield coefficient of 0.3 g-DCW/g-substrate, the respective values of the outlet biomass and substrate concentration are (a) 15 g/l, 48 mg/l

(b) 15 g/l, 0.48 g/l

(c) 48 g/l, 15 g/l

(b) 4.8 g/l, 4.8 g/l

(a) 16 min

(b) 24 min

2007

(c) 32 min

(d) 8 min

42. Determine the correctness or otherwise of the following Assertion [a] and Reason [r]

37. The maximum possible conversion for the enzymatic reaction will be (a) 100% (b) 50% (c) 25% (d) 12.5% Statement for Linked Answer Questions 38 and 39: A Nick Translation reaction in a final volume of 100 l was carried out by using 25 Ci of labeled [-32P]– dCTP for labeling a 1.2 Kb -Interferon DNA fragment.

Assertion : Enzymatic method of tissue dispersion is milder than chemical and mechanical methods. Reason : Enzymes work at optimal temperature and pH (a) Both [a] and [r] are true and [r] is the correct reason for [a] (b) Both [a] and [r] are true but [r] is not the correct reason for [a] (c) [a] is true but [r] is false (d) [a] is false but [r] is true

10.6 Enzyme Science and Technology

Common Data for Questions 43 and 44 :

(a) A  4, B  1, C  2, D  3

Enzyme X converts substrates S1 and S2 (which are similar but not identical) to products P 1 and P2 respectively

(b) A  1, B  4, C  2, D  3

43. Km values of enzyme X for substrate S1 and S2 are 0.1 mM and 0.01 mM, respectively This suggest that P. enzyme X has more affinity towards S1

(c) A  2, B  3, C  1, D  4 (d) A  3, B  4, C  2, D  1 48. Using the Hill equation for an enzyme [S]0 = (v0 Km/ Vmax– v0)1/n and the plot of log 10 (v0/Vmax – v0) vs log 10 [S]0 one can find out

Q. enzyme X has low affinity towards S1

(P) Vmax form the intercept on the ordinate

R. enzyme X has more affinity towards S2

(Q) Km form the intercept on the ordinalte

S. enzyme X has low affinity towards S2

(R) ‘n’ from the slope

(a) P, Q

(b) R, S

(S) Km from the intercept on the abscissa

(c) Q, S

(d) Q, R

(a) P, Q

(b) Q, R

(c) R, S

(d) P, S

44. What would happen if enzyme X is incubated with a mixture of 0.1 mM of S1 and S2? (a) Products P1 and P2 are produced at equal concentrations (b) Only product P2 is produced (c) More P2 and less P1 are produced (d) More P1 and less P2 are produced

2006 45. Immobilization of enzymes using entrapment method requires (P) photosensitive polyethylene glycol dimethacrylate (Q) CNBr activation of sepharose (R) polyfunctional reagent like hexamethylene diisocyanate (S) radiation of polyvinyl alcohol (a) P, Q

(b) R, S

(c) P, S

(d) Q, S

46. Immobilization of enzymes (P) increases the specificity of the enzyme for its reactants (Q) facilitates reuse of the enzyme in batch reactions (R) makes it unsuitable for its use in a continuous reactor system (S) decreases the operational cost of the industrial process (a) Q, S

(b) Q, R

(c) R, S

(d) P, Q

2005 47. Match the following products with their starting substrates A. Sake

1. apple juice

B. cider

2. grape juice

C. wine

3. barley

D. lager

4. rice

Statement for Linked Answer Questions 49a and 49b An aliquot of competent E. coli cells were used for determination of cell density by plate count method and another aliquot was used for transformation by plasmid DNA 49a. E. coli cell culture (1 ml) was diluted 1:1000000 and 200l of this was used for plating. After 12h incubation of the plate, the number of colony forming units (CFU) was 150. What is the total CFUper ml in original culture ? (a) 7.5  108

(b) 1.5  108

(c) 1.5  106

(d) 3.0  106

49b. Isolated plasmid DNA (5  g) was used for transformation of 100 l competent E. coli cells to which 900 l of SOC medium was added. An aliquot of 50ul was plated on a selective plate. After overnight incubation, 300 colonies were observed. Calculate the efficiency of transformation and the percentage of transformed cells per ml of parent culture (a) 6.0  105 colonies per g of plasmid DNA, 0.01% (b) 1.2  105 colonies per g of plasmid DNA, 0.02% (c) 1.2  106 colonies per g of plasmid DNA, 0.008% (d) 6.0  106 colonies per g of plasmid DNA, 0.01% Statement for Linked Answer Questions 50a and 50b. HMGCoA reductase that binds HMGCoA, is the major rate limiting step in the cholesterol biosynthetic pathway. Several inhibitors of this enzyme are used as potential drugs. The assay of the enzyme is based on labeling the enzyme with radiolabeled HMGCoA and counting (cpm) the labeled enzyme substrate complex in the presence (test) and in the absence (control) of the inhibitor. A bland is set up that contains no enzyme.

Enzyme Science and Technology 10.7

50a. The percent inhibition for this enzyme is calculated from the equation

Fermentation products

Microbial cultures

(a) {[cpm (control) – cpm (test)]/[cpm (control) – cpm (blank)]}  100

(A) Ethanol

(P) Aspergillus niger

(B) Streptomycin

(Q) Zymomonas mobiles

(b) {[cpm (control) – cpm (test)]/[cpm(blank) – cpm (control)])}  100

(C) Citric acid

(R) Streptomyces griseus

(c) {[cpm (test) – cpm (control)]/[cpm (control) – cpm (blank)])}  100

(D) Cellulase

(S) Trichoderma reesei

(d) {[cpm (control) – cpm (blank)]/[cpm (test) – cpm (control)])}  100

(b) A-Q; B-R; C-P; D-S

50b. An inhibitor is considered active if it causes more than 65% inhibition. The. cpm Values respectively of control, test and blank samples for inhibitors W, X, Y and Z are given below. State which of the inhibitors is active.

(d) A-S; B-R; C-Q; D-P

(a) X – 8000, 4000 and 100 (b) W – 7000, 1400 and 135 (c) Y – 7500, 5000 and 90 (d) Z – 7200, 2800 and 200

2004 51. An immobilized enzyme being used in a continuous plug flow reactor exhibits an effectiveness factor () of 1.2. The value of  being greater than one could be apparently due to one of the following reasons. Identify the correct reason. (a) The enzyme follows substrate inhibited kinetics with internal pore diffusion limitation (b) The enzyme experiences external film diffusion limitation (c) The enzyme follows sigmoidal kinetics (d) The immobilized enzyme is operationally unstable 52. The degree of inhibition for non-competitive inhibition of an enzyme catalyzed reaction (a) Increases with increases in substrate concentration (b) Reaches a maxima with increase in substrate concentration and then decreases (c) Is independent of substrate concentration (d) Decreases with increase in substrate concentration 53. The two columns given below indicate some of the fermentation products and the microbial cultures used for their production. Identify the correct set of groups from the four options.

(a) A-P; B-Q; C-R; D-S (c) A-Q; B-R; C-S; D-P 54. An enzyme following Michaelis – Menten kinetics with Vm = 2.5 mmol m–3 s–1 and Km = 5.0 mM was used to carry out the reaction in a batch stirred reactor. Starting with an initial substrate concentration of 0.1 M. The time required for 50% conversion of the substrate will be about (a) 01 hr

(b) 06 hr

(c) 02 hr

(d) 12 hr

55. Inversion of sucrose by immobilized invertase follows a substrate inhibited kinetics. The reaction rate (v) in mol m–3 hr–1 can be expressed as V = 800 [S]/ {400 + 50 [S] + [S]2}, where [S] is the sucrose concentration. The immobilized invertase preparation is used in a CSTR with 100 mol m–3 sucrose concentration in the feed stream. If the reaction velocity passes through a maxima at [S] = 20 mol m–3, the feed flow rate for a reactor volume of 1m3 to get the maximum productivity from the reactor should be (a) 0.11 m3 hr–1

(b) 1.10 m3 hr–1

(c) 5.05 m3 hr–1

(d) None of these

56. Phytase an enzyme produced by Aspergillus niger can be adsorbed on microcrystalline cellulose powder (MCCP). The adsorption follows a Langmuir isotherm and the maximum concentration of the protein that can be obtained on the adsorbent is 70 mg cm–3. At a concentration of 50mgl –1 of protein in the solution the concentration of protein on the adsorbent reaches 35 mg cm–3. It is desired to recover 90% of the addition of MCCP to the solution. The concentration of the protein adsorbed on the solid at equilibrium will be (a) 21.4 mg cm–3 (b) 214 mg cm–3 (c) 2.14 mg cm–3 (d) None of these

10.8 Enzyme Science and Technology

57. Examine the data given below in the table on purification of a protein X Step

Volume

Crude cell– free extract Ammonium sulfate precipitation Ion-exchange Chromatography

(ml) 500

Protein Enzyme concentration activity (mg/ml) (units/ml) 12.0 5.0

125

3.0

11.0

2002 62. The typical coenzyme methanogenes is

present

in

the

(a) Coenzyme A (b) Coenzyme Q (c) Coenzyme M (d) None of these 63. Ethanol concentration is lowest in

10

9.0

75.0

(a) Wine (b) Beer

The yield percent and purification factor respectively at the end of the experiment will be approximately (a) 25 and 3

(b) 30 and 20

(c) 20 and 30

(d) 75 and 30

(c) Brandy (d) Rum 64. The following cross-linking agents may be used for the immobilization of enzymes (a) Glutaraldehyde

2003

(b) Cyanogen bromide

58. Enzyme papain is used with success to

(c) Thionyl chloride

(a) increase meat production

(d) All of these 65. The following culture systems are used for growing large amount of anchorage dependent animal cells except

(b) leaven bread (c) ripen papaya fruit (d) tenderize meat 59. Which one of the following reactions is used for the purpose of recycling enzymes inbioprocesses ?

(a) Roller bottle (b) Airlift fermenter (c) Hollow fibre reactor

(a) Isomerization

(d) Microcarrieres

(b) Immobilization (c) Phosphorylation

2001

(d) Polymerization

66. The factor (s) likely to increase the rate of reaction catalyzed by a surface immobilized enzyme is/are

60. Match the industrial application of the following enzymes P Penicillinase

1. Pharmaceutical

Q Pectinase

2. Leather

R Trypsin

3. Wine

S Rennin

2. Dairy

(a) Increased agitation of the bulk liquid containing the substrate (b) Continued replacement of the bulk liquid containing the substrate

(a)

(b)

(c)

(d)

(c) Increased concentration of the substrate in the bulk liquid

P-4

P-l

P-l

P-4

(d) All of these

Q-3

Q-3

Q-2

Q-2

R-l

R-2

R-3

R-3

S-2

S-4

S-4

S-l

61. Which of the following techniques is NOT ideal for immobilizing cell-free enzymes ? (a) physical entrapment by encapsulation (b) covalent chemical bonding to surface carriers (c) physical bonding by flocculation (d) covalent chemical bonding by cross-linking the precipitate

67. Which of the following cases are likely to lead to faster rates of catalysis by an enzyme immobilized on a negatively charged support ? (a) A positively charged substrate and a negatively charged product (b) A negatively charged substrate and a positively charged product (c) A positively charged substrate and a positively charged product (d) None of the above

Enzyme Science and Technology 10.9

68. (a) A chemostat is operating at steady state at a dilution rate of 0.1 hr–1 and a limiting nutrient concentration of 10 M. If max for the organism is 0.5 hr–1. Calculate the Monod constant for the nutrient. (b) Why asparaginase is used in anticancer

(d) Enzymes which hydrolyse anti- bodies 70. Which one of the following is not a protease ? (a) Proteosome (b) Trypsin (c) Chrymotrypsin (d) Peptidyl tRNA hydrolase

therapy?

71. The Pasteur Effect is

2000

(a) Inhibition of glucose utilization and lactate accumulation in glycolysis

69. Abzymes are (a) Enzymes that are highly specific like antibodies (b) Antibodies that have catalytic activities

(b) Sterilisation of milk (c) Vaccine production

(c) Also referred to as zymogens

(d) Heat treament of bacteria

ANSWERS 1. (c)

2. (b)

3. (8.6 : 9.4)

4. (525.0 : 555.0)

7. (d)

8. (7)

9. (a)

10. (90 or 10)

11. (b)

12. (6 to 6)

13. (2.46 to 2.50)

14. (c)

5. (b)

6. (7.5 : 7.5)

15. (d)

16. (d)

17. (a)

18. (c)

19. (c)

20. (a)

21. (c)

22. (d)

23. (d)

24. (b)

25. (c)

26. (d)

27. (*)

28. (b)

29. (c)

30. (b)

31. (*)

32. (d)

33. (a)

34. (d)

35. (c)

36. (*)

37. (*)

38. (d)

39. (d)

40. (c)

41. (a)

42. (b)

43. (d)

44. (c)

45. (c)

46. (a)

47. (a)

48. (d)

49a. (a)

49b. (b)

50a. (a)

50b. (b)

51. (a)

52. (c)

53. (b)

54. (b)

55. (d)

56. (*)

57. (b)

58. (d)

59. (b)

60. (b)

61. (d)

62. (c)

63. (b)

64. (d)

65. (b)

66. (b)

67. (a)

68. (*)

69. (b)

70. (b)

71. (d)

EXPLANATIONS 1. Enzyme inhibitors which resembled the transition state structure would bind more tightly to the enzyme than the actual substrate. Transition state analogs can be used as inhibitors in enzymecatalyzed reactions by blocking the active site of the enzyme. Eg: Saquinavir. 2. Higher cell densities can be achieved using microcarriers which increases the surface area for cell growth and also mediates the culture of different type of cells. They are used for anchorage dependant cells. Eg: Cytodex I. Absence of surface charge doesn't help in anchorage. 3.

Now,



= 2.233  10

–10

mol s , Km = 25  m. –1

1.8 = 0.1  mol = 0.1  10–6 mol. 18

Kcat =

Vmax 2.233  10 10   2.233  10 3 Et 0.1  10 6

K cat 2.233  10 3 = = 0.0893  103 Km 25  106 = 8.93  101 M–1s–1

4. Specific Activity of enzyme =

     Change in OD. Per min × 1000 × total reaction mix in ml   Extinction coefficient × Vol. of enzyme in sample in ml   × concentration of protein in g/ml   

=

0.14  1000  3 = 540 (6220  25  10 3  5  10 3 )

Vmax = 0.0134  mol min–1 = 02.233  10–4  mol s–1

Et =

10.10 Enzyme Science and Technology

5. Trypsin cleaves at carboxy side of R. So R is written on the C termini. On the other hand, Trypsin digested fragment includes RAV And RSY. Any of the fragment can be placed depending on the chymotrypsin digestion which is performed after trypsinisation. Chymotripsinisation produced ARVY and SR after cleaving C termini of another R. So the resultant amino acid sequence will be AVRYSR. 6.

V1 =

Vmax A[s] , V2  K L A([s o ]  [s]) K m  [s]

Reaction rate Mass transfer rate The given condition is V1 = V2 at steady state. Vmax= KLso = 1  10 = 10.

Further

10[s] = [10 – s]  s2 + 10s – 100 = 0 10  [s]



12. In ethanol, there are 6 hydrogen atoms which are present in it. So, the degree of reduction of ethanol is 6 as there are six hydrogen atoms gained by the ethanol molecule. Hence, degree of reduction for ethanol is 6.

Solving for s Using



 b  b 2  4ac 2a s = 6.180 or – 16.180

13. The standard Gibb's free energy change for the formation of G-3-P is given by:

s = 6.180

7. Scatchard analysis is plot of ratio of concentration of bound ligand to unbound ligand verses the bound ligand concentration. The slope = affinity for ligand binding and Y intercept = total concentration of antibody.

 Ab – Ag =  Ag 



 Ab – Ag .ka  Ag

+ nka

Then compare with equation y = mx + c 8. 1 Katal = The one mole of substrate converted by enzymes in 1 second. 420  moles = 7 10–6 katal 60 seconds 9. Graph A represent uncompetitive inhibition

1 Km 1 1 = + V V V [S] max max

G0 = - RT lnK 

G0 = - 8.315  298  ln 0.4



G0 = 2,270.449 J/mol = 2.27 kJ/mol

14. The ratio k(cat)/K(M)—often referred to as the “specificity constant”—is a useful index for comparing the relative rates of an enzyme acting on alternative, competing substrates. However, an alternative description, “catalytic efficiency”, is frequently used to compare the reactivity of two enzymes acting on the same substrate. K

1 K2  E + S  ES   E+ P K1

=

10.

11. Enzymes or 'biocatalysts' are remarkable discovery in the field of bioprocess technology. Biocatalysis has been widely accepted in diverse sectors owing to their ease of production, substrate specificity and green chemistry. However, for large extent commercialization of these bio-derived catalysts, their reusability factor becomes mandatory, failing which they would no longer be economic. Consequently, immobilized enzymes with functional efficiency and enhanced reproducibility are used as alternatives in spite of their expensiveness. Immobilized biocatalysts can either be enzymes or whole cells. Enzyme immobilization is confinement of enzyme to a phase (matrix/ support) different from the one for substrates and products. Inert polymers and inorganic materials are usually used as carrier matrices.

K2 here is also known as kcat /km, the catalytic efficiency of enzyme. ...(1)

K m1 1 = V V max

I  1 1  ...(2) 1  KI  [S]  V   min Then equation (1) – (2) and by simplification

5   Km = 30 1   = 90 2.5   Note : If we take Km in the absence of the inhibitor 30 M then Km in the presence of the inhibitor can be calculated as per above formulae 10 M.

kcat/km = k2 = VMax /[E]0 15. If a reversible inhibitor can bind to the enzyme active site in place of the substrate, it is described as a “competitive inhibitor.” In pure competitive inhibition, the inhibitor is assumed to bind to the free enzyme but not to the enzyme-substrate (ES) complex. At the macroscopic level, the effect of competitive inhibition is to increase the substrate concentration required to achieve a given reaction velocity; in other words, to raise the Km. Here, with increase in the concentration of the inhibitor

Enzyme Science and Technology 10.11

[I], the rate of the reaction changes and so does the slope of the plot. Hence it is a competitive inhibition. 16. From the graph we can see that the value of r is same for all the receptors. ‘r’ is the moles of bound ligand per moles of receptor. So, both R1 and R3 have the same number of sites for ligand binding. Now, in the graph, r = 2. Therefore the number of ligand binding sites is 2. 17. Analysing the graph it can be seen that r/c value is the maximum for Recpetor1. Here, r is the moles of bound ligand per moles of receptor and c is concentration of free ligand. So, more the value of r/c, more will be the efficiency of binding the ligand by a receptor. Since r/c for Recpetor1 is 4 which is the highest, so Receptor1 is the most efficient binder of the ligand. 7

18. One katak is 6  10 IU units 19. Enzymes accelerate the rate of the reverse reaction as well as the forward reaction, it would be helpful to ignore any back reaction by which E1P might form ES. The velocity of this back reaction would be given by v = k–2[E][P]. 25. Glutamate dehydrogenase (GLDH) is an enzyme, present in most microbes and the mitochondria of eukaryotes, as are some of the other enzymes required for urea synthesis, that converts glutamate to α-ketoglutarate, and vice versa. The α-ketoglutarate to glutamate reaction does not occur in mammals as glutamate dehydrogenase equilibrium favours the production of ammonia and α-ketoglutarate. NAD+ or NADP+ is a cofactor for the glutamate dehydrogenase reaction, producing α -ketoglutarate and ammonium as a byproduct. 26. A non- competitive inhibitor can combine with an enzyme molecule to produce a dead- end complex, regardless of whether a substrate molecule is bound or not. Hence, inhibitor binds at a different site from the substrate. Km (Michelis Menton) constant remains unchanged but the value of Vmax (maximum velocity) is altered. 27. The degree of effectiveness factor is determined by Hill equation for Non Michaelis-Menten kinetics. The Hill equation accounts for allosteric binding at sites other than the active site. n is the “Hill coefficient.” When n < 1, there is negative effectiveness factor; When n = 1, there is no effect; When n > 1, there is positive effectively.

28. Turnover number is expressed in molecules of product / active site / second. It can be defined as the number of molecules of a substrate acted upon in a period of 1 minute by a single enzyme molecule; with the enzyme working at a maximum rate.here the statement shows the dP/dt of the two enzymes at unit time. Again dP/dt is the velocity of the reaction i.e. the E1 has greater velocity than of E2. 29. Given, Dissociation constant, Kd= 10–7M The reaction is: L + R Æ LR Thus,

Kd for the reaction = [L][R]/[LR]

or

[LR]/[LR][R] = 10/100 [L] = 1/9 * 10–7M = 10–8M

Therefore,

30. Number of ligands = total radioactivity/ radioactivity in each ligand = 1000 cpm/ 100 cpm/pmole = 10 pmole = 10 * 10

–12

* 6.023 *10–12 * 6.023 * 1023 ligands

Thus, total ligands= 6.023 * 1012 Average number of receptors, R per cell = total receptor/number of cells = 6.023 * 1012/ 6 * 107 = 105 32. Affinity chromatography utilizes the specific interactions between two molecules for the purification of a target molecule. A ligand having affinity for a target molecule is covalently attached to an insoluble support and functions as bait for capturing the target from complex solutions. The affinity ligand can be any molecule that will bind the target without also binding other molecules in the solution. 33. Stock solution= 150 M Dilution factor= 10/200 l = 1/20 l Concentration of d TTP= (1/20)*150 M= 7.5 M 34. KM = =

K 1  K 2 K1 4  104 S 1  2  103 S 1 3  108 M 1 S 1

42000 3  108 = 1.4  10–4M =

10.12 Enzyme Science and Technology

35. We know, from Monod's equation

μ = μmax [S0/(KS + S0)] ⇒ μ = 1 * [10/(0.2 +10)] = 10/10.2 = 0.98 At wash out, Dilution rate D = μ Therefore, the required dilution rate = 0.98 hr–1 38. % incorporation is calculated by incorporation into precipitated form. (The polymerized form will precipitate). =

precipitate form cpm total cpm

=

2.94  104 cpm  100  70% 4.2  104 cpm

39. 2.94  105 cpm is from 0.1 dilution of 10 l sample. Thus, 10 l sample will have 2.94  106 cpm but, this 10 l sample is taken from 100 l which contain 1 g of -interferon. So, 100 l of 1 g interferon sample will have 2.94  107 cmp/g. 42. A : Enzymatic method is mild. Chemical and mechanical methods may tend to destroy the tissue. R :Enzymes show their catalytic activity at a particular optimum p H and temperature. 43. The Michaelis constant (Km) is a means of characterizing an enzyme's affinity for a substrate. The Km in an enzymatic reaction is the substrate concentration at which the reaction rate is half its maximum speed. Thus, a low Km value means that the enzyme has a high affinity for the substrate (as a "little" substrate is enough to run the reaction at half its max speed). 44. Because this enzyme has a high KM, less substrate is converted into product. 45. Photosensitive Polyethylene glycol dimethylacrylate is used in free radical copolymer crosslinking reactions which cross links enzyme with matrix and hence used in immobilzation of enzymes. Polyvinyl alcohol is used for formation of spherical structured beads in immobilisation process but after the process is over it needs to be degraded and for this purpose gamma radiation induced degradation of polyvinyl alcohol is carried out. 46. The immobilized enzyme is easily removed from the reaction making it easy to recycle the biocatalyst. Immobilized enzymes typically have

greater thermal and operational stability than the soluble form of the enzyme. This reduces the operational cost. 47. Sake or saké is an alcoholic beverage of that is made from fermented rice. Cider is a fermented alcoholic beverage made from fruit juice, most commonly and traditionally apple juice. Grape juice is obtained from crushing and blending grapes into a liquid. The juice is fermented and made into wine, brandy or vinegar. Barley is used as a source of fermentable material for beer called lager and certain distilled beverages. 48. V m can be plotted and calculated from the intercept on the ordinate while Km will be given by Hill equation from the intercept of the abscissa. 49a. Since, 1 ml = 1000 l New volume = 103  106 l = 109 l. Since, only 200 l is used we have a dilution factor of =

109 l = 0.5  107 200 l

Number of colony in original one ml culture would be = 0.5  107  150 = 7.5  108. 50a. The amount of radioactivity will be in order control >test > blank % inhibition =

control  test control  blank

decrease in enzyme complex =

formation due to inhibitor maximum enzyme- substrate complex formation in absence of inhibitor

50b. For each a, b, c, and d options respectively. The value of % inhibition is For,

a=

8000  4000 = 50.6% 8000  100

b=

7000  1400 = 81.5% 7000  135

c=

7500  5000 = 33.7% 7500  90

d=

7200  2800 = 62.86% 7200  200

Enzyme Science and Technology 10.13

51. The effect of intraparticle diffusion in artificial membranes on the kinetic behavior of immobilized enzymes has been considered theoretically in terms of an effectiveness factor. When the immobilized enxymes obey a Michaelis-Menten relationship, an approximate equation is proposed for the effectiveness factor. For a substrate inhibited enzymatic reaction, the effectiveness factor may exceed unity and display multiple steady-state behavior. Thus, the over-all rate of reaction in the artificial membrane is faster in this case than that when the substrate concentration in the interior is at the same as that at the exterior surface. In the case of a product inhibited enzymatic reaction, the effectiveness factor is always less than that which corresponds to a Michaelis-Menten relationship. 52. Non-competitive inhibitors are considered to be substances which when added to the enzyme alter the enzyme in a way that it cannot accept the substrate. 53. A. niger can ferment Citric Acid, Ethanol can be produced by Z. mobilis, Streptomycin is produced by S.griseus, Cellulose is produced by T. ressei. 54. By MM equation V= Vm .S/Km +S. 55. V =

800[ s] 400  50[ s]  [ s]2

S0 = 100 mol/m3  V0 = 5.195 ml/m3 hr. i.e. Vmax when S = 20 ml/m3,



 dp  F Vmax = 8.889 ml/m3.hr. V =   D = dt V

Now, Km =

Vmax = 4.444 2

Now, V0 =Vmax

S = 7.27 ml/m3. hr Km  S

58. Papain has utility in breaking down tough meat fibres and has been used for thousands of years in its native South America. It is sold as a component of powdered meat tenderiser available in most super markets. Papain, in the form of such a powdered meat tenderizer, made into a paste with water, is also a home remedy treatment for, bee, and yellow jacket (wasps) stings; mosquito bites; and possibly stingray wounds, breaking down the protein toxins in the venom.

59. immobilized enzymes are reusable in long term industrial utilization. 60. Beta-lactamases (penicillinase) are enzymes produced by some bacteria that provide resistance to beta-lactam antibiotics like penicillins, cephamycins and carbapenems. Beta-lactamase provides antibiotic resistance by breaking the antibiotic s structure(a four-atom ring known as a beta-lactam). Through hydrolysis , the lactamase enzyme breaks the β-lactam ring open, deactivating the molecule’s antibacterial properties. Pectinases have also been used in wine production since the 1960s. The function of pectinase in brewing is twofold, first it helps breakdown the plant (typically fruit) material and so helps the extraction of flavours from the mash. Secondly the presence of pectin in finished wine causes a haze or slight cloudiness, Pectinase is used to break this down and so clear the wine. One of the steps in leather industry is ‘bating’ which is done to make the leather soft and supple by using proteolytic bating enzymes of pancreatic or bacterial origin. Trypsin and alkaline proteases are commonly used. Rennin is an enzyme generated in the stomachs of young mammals to help them digest milk. It is also the active component in rennet, a substance which is used to make cheeses and other dairy products such as junket. 61. Covalent chemical bonding by cross-linking the precipitate is not ideal for immobilizing cell-free enzymes. 62. Coenzyme M is a coenzyme required for methyltransfer reactions in the metabolism of methanogens. The coenzyme is an anion with the formula HSCH 2CH 2 SO –3. It is named 2mercaptoethanesulfonate and abbreviated HS–CoM. The cation is unimportant, but the sodium salt is most available. Mercaptoethanesulfonate contains both a thiol, which is the main site of reactivity, and a sulfonate group, which confers solubility in aqueous media. 63. Amount of ethanol in brandy: 35 to 60%, rum: 40%, beer: 4 to 6%. 64. Commonly used methods for the covalent immobilisation of enzymes includes activation of Sepharose by cyanogen bromide. Conditions are chosen to minimise the formation of the inert carbamate. Glutaraldehyde is used to cross-link

10.14 Enzyme Science and Technology

enzymes or link them to supports. It usually consists of an equilibrium mixture of monomer and oligomers. The product of the condensation of enzyme and glutaraldehyde may be stabilised against dissociation by reduction with sodium borohydride. 65. Airlift fermenter is a fermenter in which circulation of the culture medium and aeration is achieved by injection of air into some lower part of the fermenter. Usually not suitable for animal cell production. 66. Continuous change in the substrate concentration in bulk liquid does not allow enzyme to attain the reverse feedback condition. 67. It creates the substrate based higher influx leading to a positively charged substrate and a negatively charged product. 68. Abzyme is a monoclonal antibody with catalytic activity. Abzymes are usually artificial constructs, but are also found in normal and in patients with autoimmune diseases such as systemic lupus erythematosus, where they can bind to and hydrolyze DNA.

69. Proteasomes are protein complexes inside all eukaryotes and archaea, and in some bacteria. In eukaryotes, they are located in the nucleus and the cytoplasm. The main function of the proteasome is to degrade unneeded or damaged proteins by proteolysis, a chemical reaction that breaks peptide bonds. 70. Pasteur effect : While the oxygen concentration is low, the product of glycolysis, (pyruvate), is turned into ethanol and carbon dioxide, and the energy production efficiency is low (2 moles of ATP per mole of glucose). If the oxygen concentration grows, pyruvate is converted to acetyl CoA that can be used in the citric acid cycle, which increases the efficiency to 36 moles of ATP per mole of glucose. Therefore, about 15 times as much glucose must be consumed anaerobically as aerobically to yield the same amount of ATP. Thus it is heat treatment for bacteria.

11

Biochemical Engineering and Bioprocess Technology

C H A P TE R 2016

1. Prandtl number is the ratio of (a) thermal diffusivity to momentum diffusivity (b) mass diffusivity to momentum diffusivity (c) momentum diffusivity to thermal diffusivity (d) thermal diffusivity to mass diffusivity 2. Fed batch cultivation is suitable for which of the following? P. Processes with substrate inhibition Q. Processes with product inhibition R. High cell density cultivation (a) P and Q only

(b) P and R only

(c) Q and R only

(d) P, Q and R

3. The power required for agitation of non-aerated medium in fermentation is ________ kW. Operating conditions are as follows: Fermentor diameter = 3 m Number of impellers = 1 Mixing speed = 300 rpm Diameter of the Rushton turbine = 1 m Viscosity of the broth = 0.001 Pa.s Density of the broth = 1000 kg.m–3 Power number = 5 4. Which one of the following is the most suitable type of impeller for mixing high viscosity (viscosity > 105 cP) fluids? (a) Propeller

(b) Helical ribbon

(c) Paddle

(d) Flat blade turbine

5. Which of the following events occur during the stationary phase of bacterial growth?

6. The empirical formula for biomass of an unknown organism is CH1.8 O0.5 N0.2. To grow this organism, ethanol (C2H5OH) and ammonia are used as carbon and nitrogen sources, respectively. Assume no product formation other than biomass. To produce 1 mole of biomass from 1 mole of ethanol, the number of moles of oxygen required will be ___________. 7. Saccharomyces cerevisiae is cultured in a chemostat (continuous fermentation) at a dilution rate of 0.5 h–1. The feed substrate concentration is 10 g.L–1. The biomass concentration in the chemostat at steady state will be _________g.L–1. Assumptions: Feed is sterile, maintenance is negligible and maximum biomass yield with respect to substrate is 0.4 (g biomass per g ethanol). Microbial growth kinetics is given by

 mS KS  S where  is specific growth rate (h–1), m = 0.7 h–1, Ks = 0.3 g.L–1 and s is substrate concentration (g.L–1). =

8. Decimal reduction time of bacterial spores is 23 min at 121 C and the death kinetics follow first order. One liter medium containing 10 5 spores per mL was sterilized for 10 min at 121C in a batch sterilizer. The number of spores in the medium after sterilization (assuming destruction of spores in heating and cooling period is negligible) will be ___________107. 9. A bioreactor is scaled up based on equal impeller tip speed. Consider the following parameters for small and large bioreactors: Parameters

P. Rise in cell number stops Q. Spore formation in some Gram-positive bacteria such as Bacillus subtilis R. Cell size increases in some Gram-negative bacteria such as Escherichia coli S. Growth rate of bacterial cells nearly equals their death rate T. Decrease in peptidoglycan crosslinking (a) P, Q and S only

(b) P, S and T only

(c) Q, R and S only

(d) P, R and T only

Small bioreactor

Large bioreactor

Impeller speed

N1

N2

Diameter of impeller

D1

D2

Power consumption

P1

P2

Assuming geometrical similarity and the bioreactors are operated in turbulent regime, what will be P2/P1? (a) (D1/D2)2

(b) (D2/D1)2

(c) (D1/D2)5

(d) (D2/D1)5

11.2

Biochemical Engineering and Bioprocess Technology

10. An enzyme converts substrate A to product B. At a given liquid feed stream of flow rate 25 L.min–1 and feed substrate concentration of 2 mol.L–1, the volume of continuous stirred tank reactor needed for 95% conversion will be _____________ L. Given the rate equation: – rA =

0.1CA 1  0.5CA

where – rA is the rate of reaction in mol.L–1.min–1 and CA is the substrate concentration in mol.L–1 Assumptions: Enzyme concentration is contant and does not undergo any deactivation during the reaction.

16. Samples of bacterial culture taken at 5 PM and then the next day at 5 AM were found to have 104 and 107 cells.mL–1, respectively. Assuming that both the samples were taken during the log phase of cell growth, the generation time of this bacterium will be _____ h. 17. Biomass is being produced in a continuous stirred tank bioreactor of 750 L capacity. The sterile feed containing 8 g.L–1 glucose as substrate was fed at a flow rate of 150 L.h–1. The microbial system follows Monod's model with m = 0.4 h–1, Ks = 1.5 g.L–1 and Yx/s = 0.5 g cell mass-g substrate –1. Determine the cell productivity (g.L –1.h–1) at steady state.

2015

(a) 0.85

11. Which one of the following relations holds true for the specific growth rate () of a microorganism in the death phase ?

(b) 0.65

(a)  = 0

(b) < 0

(c) = max

(d) 0 < < max

12. In a fed-batch culture, 200 g . L–1 glucose solution is added at a flow rate of 50 L . h–1. The initial culture volume (at quasi steady state) and the initial cell concentration are 600 L and 20 g . L–1, respectively. The yield coefficient (Yx/s) is 0.5 g cell mass . g substrate–1. The cell concentration (g . L–1) at quasi steady state at t = 8 h is (a) 40

(b) 52

(c) 60

(d) 68

13. A synchronous culture containing 1.8  10 5 monkey kidney cells was seeded into three identical flasks. The doubling time of these cells in 24 h. After 24 h, the cells from all the three flasks were pooled and dispensed equally into each well of three 6-well plates. The number of cells in each well will be ______  104. 14. The diameters of a large and a small vessel are 1.62 m and 16.2 cm, respectively. The vessels are geometrically similar and operated under similar volumetric agitated power input. The mixing time in the small vessel was found to be 15 s. Determine the mixing time (in seconds) in the large vessel. (a) 15

(b) 30

(c) 61

(d) 122

15. Oxygen transfer was measured in a stirred tank bioreactor using dynamic method. The dissolved oxygen tension was found to be 80% air saturation under steady state conditions. The measured oxygen tensions at 7s and 17s were 55% and 68% air saturation, respectively. The volumetric mass transfer coefficient KLa is _____s–1

(c) 0.45 (d) 0.25 18. Saccharomyces cerevisiae produces ethanol by fermentation. The theoretical yield of ethanol from 2.5 g of glucose is ______ g.

2014 19. The unit for specific substrate consumption rate in a growing culture is (a)

g L.h

(b)

g h

(c)

g g.h

(d)

gmoles L.h

20. In a batch culture of Penicillium chrysogenum, the maximum penicillin synthesis occurs during the (a) lag phase

(b) exponential phase

(c) stationary phase

(d) death phase

21. Since mammalian cells are sensitive to shear, scale-up of a mammalian cell process must consider, among other parameters, the following (given N = rotations/time, D = diameter of impeller) (a)  ND (b)  N2D (c)  ND2 (d) none of these 22. A T-flask is seeded with 105 anchorage-dependent cells. The available area of the T-flask is 25 cm2 and the volume of the medium is 25 ml. Assume that the cells are rectangles of size 5 μm × 2 μm. If the cells grow to monolayer confluence after 50 h, the growth rate in number of cells/(cm2.h) is ____ ×105.

Biochemical Engineering and Bioprocess Technology 11.3

23. Consider a continuous culture provided with a sterile feed containing 10 mM glucose. The steady state cell density and substrate concentration at three different dilution rates are given in the table below. Dilution rate (h–1)

0.248

Substrate concentration (mM) 0.067

0.208

1.667

0

10

Cell density (g L–1)

0.05 0.5 5

The maximum specific growth rate m (in h–1), will be _____. 24. Lysine is being produced in a lab-scale reactor by a threonine auxotroph. After 2 weeks of operation it was observed that the concentration of lysine in the reactor was gradually decreasing. Microbiological assays of reactor samples showed absence of contamination and recorded data showed no change in the operating conditions. The most probable reason for decrease in lysine concentration may be attributed to (a) accumulation of ethanol (b) growth of revertants (c) production of citric acid (d) unutilized phosphoenol pyruvate

2013 25. A callus of 5 g dry weight was inoculated on semi-solid medium for growth. The dry weight of the callus was found to increase by 1.5 fold after 10 days of inoculation. The growth index of the culture is  26. A chemostat is operated at a dilution rate of 0.6 h–1. At steady state, the biomass concentration in the exit stream was found to be 30 g l–1. The biomass productivity (g l–1h–1) after 3h of steady state operation will be  27. A batch bioreactor is to be scaled up from 10 to 10,000 liters. The diameter of the large bioreactor is 10 times that of the small bioreactor. The agitator speed in the small bioreactor is 450 rpm. Determine the agitator speed (rpm) of the large bioreactor with same impeller tip speed as that of the small bioreactor.  28. Calculate the percentage sequence identity for the pairwise alignment given below. HELLO– YELLOW 29. In a batch culture, the specific rate of substrate utilization is 0.25 g (g cell mass)–1 h–1 and specific rate of product formation is 0.215 g (g cell mass)–1 h–1. Calculate the yield of product from the substrate (Yp/s). 

30. The maximum cell concentration (g l–1) expected in a bioreactor with initial cell concentration of 1.75 g l–1 and an initial glucose concentration of 125 g l–1 is (Yx/s = 0.6 g cell/g substrate) 31. A fed batch culture was operated with intermittent addition of glucose solution at a flow rate of 200 ml h–1. The values of Ks, m and D are 0.3 g l–1, 0.4 h–1 and 0.1 h–1, respectively. Determine the concentration of growth limiting substrate (gl–1) in the reactor at quasi-steady state.  Statement for Linked Answer (Q. 32 - 33): During sterilization of a fermentation medium in a given bioreactor heating = 12.56, cooling = 7.48 and the total value of required for whole sterilization process is 52, where is the design criteria. 32. What is the value of holding? (a) 31.96 (b) 42.32 (c) 52.43 (d) 61.18 33. What is the holding period (min) at a k value of 3.36 min–1? (a) 10.6

(b) 9.5

(c) 8.4

(d) 7.2

2012 34. In an exponentially growing batch culture of Saccharomyces cerevisiae, the cell density is 20 gl–1 (DCW), the specific growth rate () is 0.4h–1 and substrate uptake rate () is 16 gl–1h–1. The cell yield coefficient Yx/s will be (a) 0.32

(b) 0.64

(c) 0.80

(d) 0.50

35. Match the downstream processes in Group I with the products in Group II. Group I

Group II

P. Solvent extraction

1. Lactic acid

Q. Protein-A linked affinity 2. Penicillin chromatography R. Extractive distillation 3. Monoclonal antibody S. Salting out

4. Lipase

Codes : P

Q

R

S

(a) 2

3

1

4

(b) 4

1

2

3

(c) 4

1

3

2

(d) 2

4

1

3

11.4

Biochemical Engineering and Bioprocess Technology

36. A bacterial culture (200 l containing 1.8  109 cells) was treated with an antibiotic Z (50 g per ml) for 4 h at 37C. After this treatment, the culture was divided into two equal aliquots. Set A: 100 l was plated on Luria agar. Set B: 100 l was centrifuged, the cell pellet washed and plated on Luria agar. After incubating these two plates for 24 h at 37C, Set A plate showed no colonies, whereas the Set B plate showed 0.9  109 cells. This experiment showed that the antibiotic Z is (a) bacteriostatic

(b) bacteriocidal

(c) bacteriolytic

(d) apoptotic

Directions (Q. 37 – 38) :  -Galactosidase bound to DEAE-cellulose is used to hydrolyze lactose to glucose and galac lose in a plug fiow bioreactor with a packed bed of volume 100 liters and a voidage () of 0.55. The Km and Vmax for the immobilized enzyme are 0.72 gl –1 and 18 gl –1 h–1, respectively. The lactose concentration in the field stream is 20 gl–1, and a fractional conversion of 0.90 is desired. Diffusional limitations may be ignored. 37. The residence time required for the steady state reactor operation will be (a) 0.1 h

(b) 0.4 h

(c) 1.0 h

(d) 1.1 h

38. The feed flow rate required for the above bioconversion will be (a) 50 lh–1

(b) 55 lh–1

(c) 137 lh–1

(d) 550 lh–1

(a) P, Q and S only (b) Q, R and S only (c) P, Q and R only (d) S only 41. Wash out (as defined by D = max) of a continuous stirred tank fermenter is characterized by (X = biomass, S = substrate concentration in bioreactor, S0 = substrate concentration in feed, P = product concentration in bioreactor) (a) X = 0, S = 0, P = 0 (b) X = 0, S = S0, P = 0 (c) X = 0, S < 0, P = 0 (d) X < 0, S < 0, P < 0 42. Match the microbial growth characteristics in Group I with the corresponding features in Group II. Group I P. Growth associated product formation Q. Non growth associated product formation R. Product inhibition S. Substrate inhibition Group II 1. Specific growth rate decreases with increasing product concentration 2. Specific product formation rate is constant 3. Specific product formation rate is proportional to specific growth rate 4. Specific growth rate decreases with increasing substrate concentration P

Q

R

S

(a) 1

2

4

3

P. all components of a cell grow at the same rate

(b) 3

2

1

4

Q. specific growth determined by cell number or cell mass would be the same

(c) 2

1

3

4

(d) 2

3

4

1

2011 39. In balanced growth phase of a cell

R. the growth rate is independent of substrate concentration S. the growth rate decreases with decreasing substrate concentration (a) P, Q and S only

(b) Q, R and S only

(c) P, Q and R only

(d) P only

40. Substrate consumption in lag phase of microbial growth is primarily used for P. turn over of the cell material Q. maintenance of intracellular pH R. motility S. increase in cell number

43. In a well aerated and agitated microbial culture, the ‘supply' of oxygen is equal to ‘demand' (uptake) of the growing culture. The KLa for such a system will be (KLa = volumetric mass transfer coefficient, C* = dissolved oxygen concentration in liquid in equilibrium with gaseous oxygen, C = instantaneous value of dissolved oxygen concentration, ‘r' = specific oxygen uptake rate per unit weight of cells, X = dry weight of the cells per unit volume) (a) (r X)/ (C* – C) (b) (r)/ X(C* – C) (c) (C* – C)/ (r X) (d) (X)/ r(C* – C)

Biochemical Engineering and Bioprocess Technology 11.5

44. Structured William's model P. can describe the changes in intracellular components of the cell during growth Q. cannot describe the death phase of the cells R. can describe the variation of size of cells in the different phases of growth S. cannot describe the lag period of growth Which one of the following is correct? (a) P, Q and S only

(b) P, Q and R only

(c) Q, R and S only

(d) P, R and S only

45. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion : In synchronous culture, majority of the cells move to next phase of the cell cycle simultaneously. Reason : Synchronous culture could be obtained by starving cells for essential nutrient components. (a) both (A) and (R) are true and (R) is the correct reason for (A) (b) both (A) and (R) are true but (R) is not the correct reason for (A) (c) (A) is true but (R) is false (d) (A) is false but (R) is true 46. Maximum specific growth rate ( max ) of a microorganism is calculated by taking the (ln = loge, X = biomass, t = time) (a) slope of ln X vs t of the growth cycle (b) slope of ln X vs t during the exponential growth phase (c) slope of X vs t (d) slope of X vs t during the exponential phase of growth Common Data for Questions 47 and 48: A microorganism grows in a continuous ‘chemostat' culture of 60 m 3 working volume with sucrose as the growth limiting nutrient at dilution rate, D = 0.55 h–1. The steady state biomass concentration is 4.5 Kg dry biomass m–3 and the residual sucrose concentration is 2.0 Kg m–3. The sucrose concentration in the incoming feed medium is 10.0 Kg m–3. 47. What would be the yield YX/S (Kg biomass/Kg substrate)? (a) 0.562

(b) 0.462

(c) 0.362

(d) 0.162

48. What would be the sucrose concentration in the input feed for the output to be 45 Kg biomass h–1? (a) 3.225 Kg m–3

(b) 4.425 Kg m–3

(c) 5.115 Kg m–3

(d) 6.525 Kg m–3

2010 49. A culture of bacteria is infected with bacteriophage at a multiplicity of 0.3. The probability of a single cell infected with 3 phages is (a) 0.9

(b) 0.27

(c) 0.009

(d) 0.027

50. In a chemostat operating under steady state, a bacterial culture can be grown at dilution rate higher than maximum growth rate by (a) partial cell recycling (b) using sub-optimal temperature (c) pH cycling (d) substrate feed rate cyling 51. A roller bottle culture vessel perfectly cylindrical in shape having inner radius (r) = 10 cm and length (l) = 20 cm was fitted with a spiral film of length (L) = 30 cm and width (W) = 20 cm. If the film can support 105 anchorage dependent cells per cm2, the increase in the surface area after fitting the spiral film and the additional number of cells that can be grown respectively are (a) 1200 cm2 and 12  107 cellls (b) 600 cm2 and 6  107 cells (c) 600 cm2 and 8300 cells (d) 1200 cm2 and 8300 cells 52. Thermal death of microorganisms in the liquid medium follows first order kinetics. If the initial cell concentration in the fermentation medium is 10 8 cells / ml and the final acceptable contamination level is 10–3 cells, for how long should 1m3 medium be treated at temperature of 120 (thermal deactivation rate constant = 0.23 / min) to achieve acceptable load? (a) 48 min

(b) 11 min

(c) 110 min

(d) 20 min

Common Data for Question (53 - 54): A culture of Rhizobium is grown in a chemostat (100 m3 bioreactor). The feed contains 12 g/L sucrose, KS for the organism is 0.2 g/L and m = 0.3 h–1. 53. The flow rate required to result in steady state concentration of sucrose as 1.5 g/L in the bioreactor will be (a) 15 m3 h–1

(b) 26 m3 h–1

(c) 2.6 m3 h–1

(d) 150 m3 h–1

54. If YX/S = 0.4 g / g for the above culture and steady state cell concentration in the bioreactor is 4 g/ L the resulting substrate concentration will be (a) 2 g/L

(b) 8 g/L

(c) 4 g/L

(d) 6 g/L

11.6

Biochemical Engineering and Bioprocess Technology

2009 55. A culture vessels in which physical, physicochemical and physiological conditions, as well as cell concentration, are kept constant is known as (a) Cell concentrator

(b) Biostat

(c) Batch bioreactor

(d) Incubator

56. Which of the following statements are true about bioreactors? P. Continuous bioreactors provide less degree of control and uniform product quality than batch bioreactors. Q. Batch bioreactors are ideally suited for reaction with substrate inhibition. R. Choice of a bioreactor is dictated by kinetic considerations.

P

Q

R

S

(a) 4

2

1

3

(b) 3

1

4

2

(c) 3

4

2

1

(d) 2

1

4

3

60. A bacterial culture with an approximate biomass composition of CH1.8O0.5N0.2 is grown aerobically on a defined medium containing glucose as the sole carbon source and ammonia being the nitrogen source. In this fermentation, biomass is formed with a yield coefficient of 0.35 gram dry cell weight per gram of glucose and acetate is produced with a yield coefficient of 0.1 gram acetate per gram of glucose. The respiratory coefficient for the above culture will be (a) 0.90

S. Fed batch bioreactors are also called semibatch bioreactors.

(b) 0.95

(a) P, Q

(b) Q, S

(c) 1.00

(c) R, S

(d) P, R

(d) 1.05

57. Determine the correctness or otherwise of the following Assertion (a) and Reason (r): Assertion (a): Bacterial growth is called synchronous when majority of the cells are in same stage of the bacterial cell cycle. Reason (r) : Synchronous culture can be obtained by growing bacteria in an enriched medium. (a) Both (a) and (r) are true and (r) is the correct reason for (a) (b) Both (a) and (r) are true but (r) is not the correct reason for (a) (c) (a) is true but (r) is false (d) (a) is false but (r) is true

2008 58. Diauxic pattern of biomass growth is associated with (P) multiple lag phases (Q) sequential utilization of multiple substrates (R) simultaneous utilization of multiple substrates (S) absence of lag phase (a) P, R

(b) P, Q

(c) R, S

(d) Q, S

59. Match the bioreactor components in group 1 with the most appropriate function given in group 2 Group 1 P. Marine type impeller

Group 2 1. Recirculation of medium

61. A bacterial culture having a specific oxygen uptake rate of 5 mmol O2 (g-DCW)–1 hr–1 is being grown aerobically in a fed-batch bioreactor. The maximum value of the volumetric oxygen transfer coefficient is 0.18s–1 for the stirred tank bioreactor and the critical dissolved oxygen concentration is 20% of the saturation concentration (8mg/ml). The maximum density to which the cells can be grown in the fed-batch process without the growth being limited by oxygen transfer, is approximately. (a) 14 g/l

(b) 26 g/l

(c) 32 g/l

(d) 65 g/l

2007 62. The specific growth rate () of a microorganism in death phase is (a) 0 (zero)

(b) max

(c) less than zero

(d) greater than zero

63. Main functions of baffles in a bioreactor are P. to prevent a vortex Q. to increase aeration R. to reduce interfacial area of oxygen transfer S. to reduce aeration rate (a) P, Q

(b) Q, R

(c) R, S

(d) P, S

64. How many kilograms of ethanol is produced from 1 kilogram of glucose in ethanol fermentation?

Q. Draft tube

2. Aeration of medium

R. Diaphragm valve

3. Animal cell cultivation

(a) 2.00

(b) 0.20

S. Sparger

4. Sterile operation

(c) 0.51

(d) 0.05

Biochemical Engineering and Bioprocess Technology 11.7

65. Downstream processing of an industrial process yielded a highly purified bioactive protein. This protein was subjected to cleavage by trypsin. Chromatographic separation of products resulted in 4 peptides (P, Q, R, S) with the following amino acid sequences P. phe-val-met-val-arg Q. ala-ala-try-gly-lys

69. The antigen is expressed as transmemb-rane protein with a single epitope on its extracellular domain. The cells that survived (assume 100% transfection and expression of protein) were incubated with a radio labeled Fab fragment (specific activity: 100 cpm/picomole) against this epitope. After washing, the cell pellet has 1000 cpm. The average number of epitopes present on a single recombinant alga are

R. val-phe-met-ala-gly-lys

(a) 6  109

(b) 1  109

S. phe-gly-try-ser-thr

(c) 6  103

(d) 1  106

Chemical cleavage of the same protein with cyanogenbromide and chromatographic separation resulted in three peptides (i, ii, iii) with the following sequences (i) ala-gly-lys-phe-gly-try-ser-thr (ii) ala-ala-pry-gly-lys-phe-val-met

2006 70. During cultivation of microorganisms in a fermenter, various parameters are controlled by appropriate sensor (probe). Match each probe in group 1 with the appropriate response mechanism in group 2.

(iii)val-arg-val-phe-met

Group 1 (Probe)

The order of the peptides that gives the primary structure of the original protein is

P. Thermistor

(a) P, Q, R, S

(b) Q, P, R, S

(c) Q, R, P, S

(d) R, Q, P, S

Statement for Linked Answer Questions 66 and 67 : In a Fed-batch culture glucose solution is added with a flow rate of 2 m3/day. The initial volume of the culture is 6m3. 66. The volume of culture at the end of second day (neglect loss due to vaporization) is (a) 6 m

Group 2 (Response) 1. Activation of acid alkali pump

Q. Oxygen electrode R. Metal rod

2. Activation of vegetable oil pump 3. Activation of hot cold water pump

S. pH electrode

4. Increase/decrease in stirrer motor speed

(a) P-2, Q-3, R-1, S-4

(b) P-l, Q-2, R-4, S-3

(c) P-3, Q-2, R-4, S-1

(d) P-3, Q-4, R-2, S-1

71. Which one of the following monolayer culture systems have the highest surface area: medium ratio?

3

(b) 8 m3 (c) 10 m3 (d) 12 m3 67. What would be the dilution rate of the system at the end of second day? (a) 2.00

(b) 0.20

(c) 0.02

(d) 0.01

Statement for Linked Answer Questions 68 & 69 : Absence of cellulosic cell wall, high -carotene content and GRAS status make Dunaliella salina a good model system for producing edible vaccines. 109 Cells of D. salina were electroporated with a high expression DNA vector containing an antigenic gene. 68. If 103 cells survived after electroporation, how many cells were killed during this process (round of to the nearest number) ? (a) 109

(b) 108

(c) 106

(d) 105

(a) Roux bottle

(b) Spiracell roller bottle

(c) Hollow fibres

(d) Plastic bag/film

Statement for Linked Answer Questions 72 & 73 : A bioreactor of working volume 50 m3 produces a metabolite (X) in batch culture under given operating conditions from a substrate (S). The final concentration of metabolite (X) at the end of each run was 1.1 kg m–3. The bioreactor was operated to complete 70 runs in each year. 72. What will be the annual output of the system (production of metabolite (X)) in kg per year ? (a) 55

(b) 3850

(c) 45.5

(d) 77

73. What will be the overall productivity of the system in kg year–1 m–3 ? (a) 19250

(b) 38.50

(c) 3850

(d) 77

11.8

Biochemical Engineering and Bioprocess Technology

2005 74. Aeration in a biorector is provided by (a) impeller

(b) baffles

(c) sparger

(d) all of these

75. In a bioreactor baffles are incorporated to

ions and the air was sparged in. After 10 minutes, the air flow was stopped and a 10 ml sample was taken and titrated. The concentration of sodium sulfite in the sample was found to be 0.20 M. The oxygen uptake rate for this aerated system will work out to be

(a) prevent vortex and to improve aeration efficiency

(a) 0.08 gl–1 s–1

(b) maintain uniform suspension of cells

(c) 0.8 gl–1 s–1

(c) minimize the size of air bubble for greater aeration

(d) None of these

(d) maintain uniform nutrient medium

2004 76. During the media preparation for cultivation of cells, insoluble precipitates of calcium phosphates are often formed. Identify which method can be adopted to avoid this problem?

(b) 0.008 gl–1 s–1

2003 80. The separation principle of dialysis used in the recovery of fermentation products is (a) diffusion

(b) charge

(c) turbulence

(d) solubility

81. Identify the statement that is NOT correct

(a) Hold the pH at 5.6

(a) Penicillin fermentation is an aerobic process

(b) Hold the pH at 7.5

(b) Penicillin biosynthesis affected by phosphate concentration

(c) Add calcium salt first and then phosphate source (d) None the above 77. Batch fermentation of glucose to ethanol yields a productivity of 4.5 g1 –1 hl –1. If the yeast cell concentration in the fermentation brought is 5% (v/v) and the intracellular NAD/NADH concentration in the yeast cells in 10 M the cycling rate of NAD+  NADH will be

(c) Lysine stimulates penicillin synthesis (d) Penicillin production shows a distinct catabolite repression by glucose 82. A batch culture fermentation was being conducted with streptomyces rimosus. Analysis of samples collected indicated doubling of cell number per unit time. Youor inference would be that the culture is in the

(a) 50,000 cycles hr–1

(a) lag phase

–1

(b) log phase

(b) 20,000 cycles hr –1

(c) 100 cycles hr

(c) stationary phase

(d) None of these

(d) death phase

78. The kinetics of the disintegration of baker’s yeast cells in a bead mill is described as dp/dt = K. (Pm – P), where P is the concentration of protein released and Pm is the maximum protein concentration achievable. K is the first order rate constant and is 0.5 hr–1 for the system studied. The time required for the release of 90% of the intracellular proteins will be (a) 10 hr

(b) 2 hr

(c) 4.6 hr

(d) None of these

79. Measurement of K1a in a bioreactor can be carried out by sodium sulfite oxidation method, that is based on the oxidation of sodium sulfite of sodium sulfate in the presence of a catalyst (Cu ++ or Co ++ ). In a typical experiment, a laboratory fermenter was filled with 5 liter of 0.5 M sodium sulfite solution containing 0.003 MCu++

83. To optimize the bioreactor system, which one of the following conditions is least important for anaerobic fermentation ? (a) culture agitation to maintain oxygen supply (b) restriction of the entry of contaminating organisms (c) control of parameters like pH and temperature (d) maintenance of constant culture volume 84. Power number, also called Newton's number, is defined as a dimensionless parameter relating to (a) turbulent flow (b) the relative velocity between the nutrient solution and individual cells (c) energy required by the stirred reactors (d) none of the above

Biochemical Engineering and Bioprocess Technology 11.9

85. The selection of the appropriate purification method in the product recovery after microbial fermentation depends on the (a) nature and the stability of the end products produced (b) type of the side products present (c) degree of purification required (d) all of these

2002 86. The precursor for penicillin-G biosynthesis during fermentation process is (a) Phenylacetic acid (b) Phenoxyacetic acid (c) Acetic acid (d) None of these 87. In large scale fermentation process, air is sterilized by (a) Jute fiber (b) Membrane (c) Cotton fiber (d) Glass wool fiber 88. (a) Name two metal ions which play important role in citric acid fermentation. (b) How is the agitator speed in a fermenter correlated with the power drawn by the agitator? (c) How does the sulfanilamide kill the bacteria ? (d) Human insulin gene cloned from a cDNA library into pUC19 could not be expressed. Justify the reason. (e) Mention the specific role of acetosyringone in Agrobacterium mediated plant transformation ? 89. (a) The volume of a chemostat system is 1000/. The feed flow rate to the reactor is 200 l/h and the glucose concentration in the feed is 5 g/l. Determine cell and glucose concentration in the effluent of the reactor under steady state conditions. Use the following constants for the cells : = 0.3 h–1, Ks = 0.1 g/l, max Yx/s = 0.4 (g dw cells/g glucose) (b) Find out the dilution rate which gives maximum biomass productivity.

2001 90. The limiting factor in the production of large quantities of ethanol as bio-fuel is (a) Lack of a balanced medium (b) Ethanol toxicity to cells (c) Expensive downstream processing steps (d) Only (b) and (c) of the above 91. If the fractional recovery at each step of unit operation is 0.8, the recovery after 4 steps will be (a) 0.24

(b) 3.23

(c) 0.41

(d) 0.82

92. Which of the following statements applies to the operation of a fed-batch process ? (a) The volume of the culture increases with time (b) It helps controlling repressive effects of the nutrient being fed (c) It eliminates the need for oxygen supply (d) Only (a) and (b) of the above 93. In large-scale fermentation, the preferred method of sterilization is (a) Chemicals (b) Radiation (c) Filtration (d) Heat 94. Which of the following is not true of aerobic digestion ? (a) It generates most sludge (b) It generally incurs higher running cost (c) It may generate a usable fuel (d) Requires a shorter residence time

2000 95. Choose the correct completion of the following statement. A fermenter sterilisation in situ is less efficient than continuous heat sterilisation because (a) it uses higher temperature for longer time (b) it uses longer heating time during which heat is lost (c) it uses larger volume and hence takes longer to cool the medium (d) it uses steam as the heating source

11.10 Biochemical Engineering and Bioprocess Technology

96. In secondary metabolism two distinct phasestrophophase and idiophase refer respectively to

(d) A pressure cycle is a type of air lift fermentor (e) Monoclonal antibodies are used extensively in diagnosis of haematopoietic cancers

(a) Growth and production phase (b) Early and late phase (c) Primary and secondary metabolism (d) Lag phase and log phase 97. Write whether the following statements are true or false: (a) Three important characteristics in performance of biosensors are selectivity, sensitivity and stability (b) Activated sludge process is one of the most common anaerobic sewage treatment method

98. A protein PZ is present in genetically engineered bacteria at 5% of the total protein (0.1 pico gram) per cell. 1 ml of log phase culture contains 2  10 8 cells while stationary phase culture contains 1  109 cells. The molecular weight of the protein is 30,000 Da and the Avogadro number is6.02  1023 molecules/mole. What is the number of molecules of PZ per cell? Calculate the amount of protein in milligrams in one litre each of log phase and stationary phase cultures.

(c) In a fermentor, impellors increase oxygen demand by producing high shear forces.

ANSWERS 1. (c)

2. (b)

3. (625.0 : 625.0)

4. (b)

5. (a)

6. (1.9 : 2.0)

7. (3.65 : 3.75)

8. (3.6 : 3.8)

9. (b)

10. (4986.0 : 4989.0)

11. (b)

12. (b)

13. (2)

14. (c)

15. (0.07339)

16. (1.2121)

17. (b)

18. (1.20)

19. (c)

20. (c)

21. (a)

22. (2 to 2)

23. (0.795 to 0.805)

24. (b)

25. (0.5)

26. (18)

27. (45)

28. (66.5 to 66.7)

29. (0.81 to 0.91)

30. (76.7 to 76.9)

31. (0.1)

32. (a)

33. (b)

34. (d)

35. (a)

36. (a)

37. (*)

38. (*)

39. (a)

40. (c)

41. (b)

42. (b)

43. (a)

44. (b)

45. (b)

46. (b)

47. (a)

48. (*)

49. (d)

50. (a)

51. (b)

52. (c)

53. (b)

54. (a)

55. (b)

56. (c)

57. (c)

58. (b)

59. (b)

60. (b)

61. (d)

62. (a)

63. (a)

64. (c)

65. (c)

66. (c)

67. (b)

68. (a)

69. (a)

70. (d)

71. (c)

72. (b)

73. (d)

74. (c)

75. (a)

76. (d)

77. (b)

78. (c)

79. (b)

80. (a)

81. (c)

82. (b)

83. (a)

84. (c)

85. (d)

86. (a)

87. (d)

88. (*)

89. (*)

90. (d)

91. (c)

92. (d)

93. (d)

94. (c)

95. (b)

96. (a)

97. (*)

98. (*)

Biochemical Engineering and Bioprocess Technology 11.11

EXPLANATIONS 

1. Prandlt No is the ratio of Momentum Diffusivity and Thermal Diffusivity.

3

300  5 = 5 1000     1 = 625000W  60  P = 625 kW

2. A fed-batch culture is a semi-batch operation in which the nutrients necessary for cell growth and product formation are fed either intermittently or continuously via one or more feed streams during the course of an otherwise batch operation. The types of bioprocesses for which fed-batch culture is effective can be summarized as follows:

Where NP = Power No. = 5 P = Power required S = Density 1000 kg/m3

1. Substrate inhibition,

n = Speed in Hz =

2. High cell density (High cell concentration) 3. Glucose effect (Crabtree effect) 5. Auxotrophic mutants 6. Expression control of a gene who has repressible promoter in recombinant cells 7. Extension of operation time, supplement of water lost by evaporation, and decreasing viscosity of culture broth. NP =

P Sn 3 D5

6. The equation for biomass production is

C2 H5OH  a NH3  bO2  CH1.8 O0.5 N0.2  dCO 2  eH 2O Given 1 : ? Now making elemental balance on both sides. LHS C= 2 N = a i.e. a = 0.2 RHS C = 1 + d = 2 i.e. d = 1 N = 0.2 Now, putting value of a & d in equation (i)

300  5 Hz 60

D = Diameter of turbine = 1 m 4. For mixing high viscosity fluids 10000cP helical ribbon impeller is generally used. 5. Stationary phase of Bacterial growth characterises of cell size increase of certain gram +ve bacteria, cell divion is halted, spore formation occurs in gram +ve bacteria and the growth rate becomes equal to the death rate owing to nutrient limitation, spore formation, starvation util and unless the culture is stirred. Peptidogylcan cross linking is also remarkably reduced.

4. Catabolite repression

3.

P = NP Sn3 D5

1

C2 H5OH  0.2NH3  bO 2  CH1.8 O0.5 N0.2  Co 2  eH2 O Now, H balance LHS H = 6 + 0.6 = 6.6 RHS H = 1.8 + 2e = 6.6  e = 2.4 and finally O2 balance LHS O = 1 + 2b RHS 0 = 0.5 + 2 + 2.4 (i.e. e = 2.4) = 4.9 hence 1 + 2b = 4.9  2b = 3.9  b = 1.95  Final equation C2 H5OH  0.2 NH3  1.95O 2  CH1.8 O0.5 N 0.2  CO 2  2.4 H2 O

11.12 Biochemical Engineering and Bioprocess Technology

XA = 0.95

 ksD  X = YX / S  SR   m  D  

7. (1)

–rA =

g biomass (2) HereYX/S = 0.4 g ethanol SR = kS = D= m =

(ii)

 0 CAO X A  0.1C AO (1  X A )     1  0.5CAO (1  X A )  Putting values & solve



X = 3.70 2.303 Dr = Kd

8.



Kd 

 Now





xt 11. ln x   t if xt = 5 105 , x0 = 6.5  105 0

2.303 , Dr  23min Dr

then  is –ve  < 0 12. For Quasi steady state

Kd = 0.100 min N0 = 105  1000 = 108 spores in 1 litre

x = x0 + FyX/S S0 t x0 = 20  600

N0 = kdt, t = 10 min N1

= 12000 F = 50 YX/S = 0.5

108 108 ln = 0.1  10  = e1 = 2.718 N1 N1

S0 = 200 t= 8

 N1 = 3.678  107 spores 9. Scaled up criteria is equal to impeller tip speed So N1D1 = N2D2 

N2 D1 = N1 D2

...(i)



3

P2 N  D  N D  2  2 P1 = N D  N1   D1  3 2 3 1

5 2 5 1

3

13. Total cells after 24 hours = 1.8  105  2 = 3.6  105 Total number of wells = 6  3 = 18

10. For CSTR

V C AO X A = 0 rA here 0 = 25 L/min CAO = 2 ml/L

 Number of cells in each well

5

=

3.6  105 = 2  104 18 1/6

5

D  D   N D  =  1   2   2  1   D2   D1   N1 D2  2 P2 D  =  2 P1  D1 

x = 52000 g/1000 lit = 52 g/lit.

Now, power equation is given by P2  N 23 D52

(i)

V=

V = 4987.5 L

–1

ln

0 CAO X A

 0.1CA     1  0.5CA  further CA = CAO (1 – XA)

10 gL–1 0.3 g L–1 0.5 h–1 0.7 h–1

(0.3)(0.5)   X = 0.4  10   (0.7  0.5)  

(3) 

V=

0.1CA 1  0.5CA

 D2  t2  =  t1  D1 

14.

Now,



 N1     N2 

2/3

P1 P2 = V1 V2 N13 D12 = N 32 D 22 2/3



N1  D2    N2 =  D1  = 4.6415

 t2 = 15(10)1 / 6 (4.6415)2/3  61

Biochemical Engineering and Bioprocess Technology 11.13

 80 – 55  ln    80 – 68  = 0.07339 15. KLa = 17 – 7 16. G =

t 3.3 log

N2 N1

N2 = 107 N1 = 104, t = 12hr G = generation time Then G = 1.2121 h

 KsD  17. Cell productivity = DYx / s  SR –  m – D   150 D = 750 = 0.2

SR = 8 KS = 1.5

m = 0.4 Yx/s = 0.5 Then after calculation cell productivity = 0.65 18. C6H12O6  2C2H5OH + 2CO2 180 246 = 92 So, Yield =

92 g ethanol/g glucose 180

92 So, for 2.5 gm = 2.5  180 = 1.27 gm

19. Specific substrate concentration is defined as the amount of substrate consumed in grams per unit gram of substrate provided to the growing culture per unit time or per hour. So, the unit of specific substrate concentration is gg-1h-1. 20. In batch cultures of microorganisms, especially bacteria, the exponential phase is concerned with the multiplication and high metabolic activities of the microbes along with growth. But as the microbe advances to the stationary phase, metabolic activities reach the maximum limit with stagnancy in multiplication and growth of the bacteria. During this phase, there occurs maximum production of secondary metabolites such as toxins. So, Penicillium chrysogenum synthesizes penicillin at maximum concentrations during the stationary phase of the batch culture.

21. Shear speed of an impeller along with the diameter of the impeller blade have profound impact on the shear stress generated in the bioreactor system. Mammalian cells are very shear sensitive. So, for scale up of a small scale bioreactor where mammalian cells are used as feed, the major parameters to be considered would be the shear speed and agitator diameter. This is because, while scaling up, impeller speed has to be considered for controlling the shear speed which is given by: TI = ND 22. Given, Seeding of T-flask with = 10 5 no. of anchorage dependent cells Area of flask available = 25cm2 and volume of flask = 25mL Size of cells = 5m  2m Area of the T-flask covered by a single cell = 10m2 = 10  10–8 cm2 Now, anchorage dependent cells multiply every hour per unit area So, the growth rate of the cells = 2  105 cells/cm2h 23. At dilution rate = 5h-1, cell density = 0. This means cell wash out occurs at this dilution rate The mass balance on the microorganisms in a CSTR of constant volume is: [Rate of accumulation of cells, g/s] = [Rate of cells entering, g/s] – [Rate of cells leaving, g/s] + [Net rate of generation of live cells, g/s] So, maximum specific growth rate max = 0.8 h-1 24. In a lab-scale bioreactor, lysine is produced from threonine auxotroph but it was found that concentration of lysine was gradually decreasing. The most probable reason was found to be revertants. Reversion is the process by which mutant DNA is changed so that the protein inactivated by the first mutation is reactivated by the second. There are three classes of reversion: Class I, class II and class III. Revertant is a mutant gene, individual, or strain that regains a former capability (as the production of a particular protein) by undergoing further mutation. 25. Given, Initial dry weight = 5 g After inoculation for 10 days, dry weight = 1.5  5 = 7.5 g Therefore Growth index of the callus =(Difference or change in weight)/ (Originalweight) = (7.5 – 5)/5 = 0.5

11.14 Biochemical Engineering and Bioprocess Technology

26. Given, Biomass concentration in the exit stream = 30 g/L

where, = rate of substrate consumed per unit time.

Dilution rate of the chemostat = 0.6 h-1

max

Therefore, Biomass productivity of the chemostat = 30  0.6 gL–1h–1 = 18 gL–1h–1 27. Let the agitator speed in the large bioreactor be VL Now, we have for scale up: VL  D L = V S  D S Where, DL=diameter of the large bioreactor, VS = agitator speed of the small bioreactor = 450 rpm, DS = diameter of the small bioreactor, n = scale up ratio = 10,000/10 = 1000 Now, DL = 10 DS Therefore, VL  10 DS = 450 DS



VL = 45

Hence the agitator speed of the large bioreactor is 45 rpm. 28. Given, two sequences HELLO- and YELLOW The % identity between the two sequences would be: No. of identical positions/ Total no. of positions

= maximum rate of substrate consumed.

KS = Monod’s constant. S = steady state substrate concentration. In steady state, mass balance equation is given by:Input – Output – Consumption = 0

 (F/V)*SR –

=0

 D(S) – {(

*S)/(KS + S)} = 0

max

Therefore, D(S) = {(

max

*S)/(KS + S)} = {(

= {(0.4S)/(0.3 + S)}

 0.1 = 0.4S/ (0.3 + S)  0.4S = 0.03 + 0.1S  0.3S = 0.03  S = 0.1 Therefore, the concentration of the limiting substrate = 0.1 g/L 32. Given,

 heating = 12.56,  cooling = 7.48

Now we have,

Therefore, percentage identity



29. Given, for a batch culture Specific rate of substrate utilization = 0.25 g/g/h Specific rate of product formation = 0.215 g/g/h Yield of product (YP/S) = 0.215/.25 = 0.86 30. Given, Initial cell concentration, X = 1.75 g/L Initial concentration of substrate glucose, S = 125 g/L Given, Yield of biomass, YX/S = 0.6 g/g Now, Concentration of biomass produced = YX/S x S = 125 x 0.6 g/L = 75 g/L Therefore, maximum concentration of cell expected in the bioreactor = 75 + 1.75 g/L = 76.75 g/L 31. Given, flow rate for the glucose solution, F = 200 mL/hr KS = 0.3 g/L; m = 0.4 h-1; D = 0.1 h-1 Therefore, Volume of the reactor, V = F/D = 200/0.1 mL = 2000 mL = 2L We have, from Monod’s equation of substrate consumption =(

max

* S)/ (KS + S)

*S)/(KS + S)}

 DS = 0.4S/ (0.3 + S)

No. of identical positions = 4 (ELLO); Total no. of positions = 6 = 4/6  100 = 66.667 %

max



 total =  heating   cooling   holding 52 = 12.56 + 7.48 +  holding  holding = 31.96

33. Given k value = 3.36 min–1 Holding value of design criteria,  holding = 31.96 Therefore, time of holding, t =  holding /k = 31.96/3.36 min = 9.512 min. 34. Yield coefficient = mass of new cells formed/ mass of substrate consumed New cells mass formed = cell density X specific growth rate = 20  0.4 = 8 Substrate consumed = 16 g/l/h 8 Yield coefficient = = 0.5. 16 35. Solvent extraction is a method for processing materials by using a solvent to separate the parts. It can be used for penicillin extraction. Affinity chromatography is the preferred method of bioselective adsorption and subsequent recovery of a compound from an immobilized ligand. Each is designed for highly specific and efficient purification of proteins and related compounds like monoclonal antibody.

Biochemical Engineering and Bioprocess Technology 11.15

Extractive distillation is defined as distillation in the presence of a miscible, high boiling, relatively non-volatile component, and the solvent that forms no azeotrope with the other components in the mixture. It can be used for lactic acid recovery. Salting out is a purification method that utilizes the reduced solubility of certain molecules in a solution of very high ionic strength. Crude lipase can be purified by using salting out method. 36. Bacteriostatic antibiotics limit the growth of bacteria by interfering with bacterial protein production, DNA replication, or other aspects of bacterial cellular metabolism. Since after incubation, plate A showed no colonies and plate B showed 0.9 * 109 colonies (half of the bacterial culture before inoculating), thus it shows that antibiotic Z is bacteriostatic which has stopped the bacteria from reproducing. 39. Balanced growth of the cell depends upon the concentration of the medium component concentration. The components are been uptaken in uniform rate which yields the specific growth rate constant. When it goes to decline phase, concentration of the component decreases. 40. In lag phase, environmental condition is adapted by bacteria themselves. In this phase each bacterium is matured enough but cannot divide. Synthesis of enzymes and other molecules occur in this phase. So, the substrate consumed by microbes in lag phase is involved in the motility of the micro-organisms, maintenance of intracellular pH and cell materials required for growth and physiology of the microbes in the upcoming phases. 41. Dilution rate is defined as rate of flow of medium into a continuous culture system. It is equal to the rate of cell division when the number of cell remains constant in vessels because the cells that are replaced by medium outflow by cell division equally. In continuous culture, rate of dilution regulates the rate of microbial growth through concentration of growth limiting nutrient in the medium. This is the advantage of this culture system. 42. At growth associated product formation specific product formation rate is proportional to specific growth rate when Non growth associated product

formation shows the constant sp. Growth rate. Product induced inhibition can be seen at increase in product concentration with decrease in sp growth rate, this phenomena is vice versa with substrate inhibition. 43. The “gas out-gas in” method can directly determine the volumetric mass transfer coefficient in a bioreactor system and provide estimates of the volumetric microbial oxygen uptake rate and the average oxygen saturation concentration at the gas-liquid interface. The rate of oxygen transfer from the bubble air to the liquid phase is described by the equation: dCL/ dt = KLa (C* – CL)

Ö

KLa = dCL/dt * 1/ (C* – CL)

Ö

KLa = r*X/(C* – CL)

Where r*X = dC L/dt, r is the specific oxygen uptake rate per unit weight of cells and X is the dry weight of cells per unit volume. For reliable measurements, deconvolution of the oxygen probe measurements must be made. 44. Williams proposed a simple structured model that divides the biomass into two parts: Part 1 is composed of intermediates, enzymes, and other entities involved in the formation of materials used for synthesis of structural and genetic material; Part 2 is the structural and genetic part of the biomass. It can be used to describe the change in intracellular components during cell growth. In the different phases of cell growth, this model enables analytical depiction and explanation of the variation in the size of the cells. But the model is not structured for reasoning with the death phase of the cells. 45. Synchronous growth of a bacterial population is that during which all bacterial cells of the population are physiologically identical and in the same stage of cell division cycle at a given time. Synchronous growth helps studying particular stages or the cell division cycle and their interrelations. A synchronous culture can be obtained either by manipulating environmental conditions such as by repeatedly changing the temperature or by adding fresh nutrients to cultures as soon as they enter the stationary phase, or by physical separation of cells by centrifugation or filtration. Synchronous cultures rapidly lose synchrony because not all cells in the population divide at exactly the same size, age or time.

11.16 Biochemical Engineering and Bioprocess Technology

Thus, the increase in the surface area after fitting of surface film = 30 * 20 = 600 cm2

46. By Monod model, ln X Vs t graph is linear and slope = μ, where μ is specific growth rate, μmax is maximum specific growth rate. It represents the log phase of the growth.

Also, the film can support 10 5 anchorage dependent cells per cm2

47. We know,

Thus, the film will add = 105 cells/cm2 * 600 cm2 = 6 * 107 cells.

Yield of biomass, YX/S = Biomass produced/ Substrate consumed Now,

52. Initial number of cells= 108 cells/ml * 1m3 But, we know, 1m3 = 106cm3

Substrate consumed,

Also,

S = Initial substrate concentration (Si) – Final substrate concentration (Sf )

103cm3 = 1000ml

Therefore, total initial number of cells = 108 cells * 106

= Si – Sf

= 1014 cells 3

Si = 10 kg/ m and Sf = 2 kg/ m

3

Final acceptance of contamination

Biomass produced, X = 4.5 kg dry biomass/ m3 Therefore,

= 10–3 cells/ml or 103 cells/m3 Thus, using the first order rate constant equation: t = 2.303/K log10 [1014]/103

YX/S = X/ (Si – Sf) = 4.5 (10 – 2)

= 2.303/ 0.23 per min log10 (1014/103)

= 0.562 Kg of biomass/ kg of the substrate.

= 10 * 11 log10 (10)

48. Data is insufficient to solve the question. 49. Probability of bacteria getting infected by a single bacteriophage= 0.3 i.e. P(A)= 0.3 Given, 3 phages, So, probability of bacteria getting infected by 3 phages = P(A) * P(B) * P(C)

t = 110 min. 53. By using the Monod Equation under steady state, We have, =

max

[S]/[S] + KS

= 0.3 * 0.3 * 0.3

= 0.3 hr–1 * 1.5g/l/ [1.5g/l] + [0.2g/l]

= 0.027

= 0.2648 hr–1

50. Chemostat is operated by supplying an essential growth limiting nutrient at a constant rate with the result that the culture density and growth rate adjust themselves to the supply. At steady state,

D = Also,

D = F/V

Therefore, equating the two, we get, F = V = 0.2648 * 10

dx/dt= 0 Therefore, μ= D This equation shows that by simply varying the medium supply rate the growth rate can be varied.This equation holds true if μ eD or μ–D = positive. Here, Dc = μ max, i.e. the critical dilution rate is defined as the lowest dilution rate at which washout occurs. Thus, for a bacterial culture to grow at dilution rate higher than maximum growth rate, the partial cell recycling is needed under steady state condition. 51. Given, Inner radius (r) = 10 cm Length (l) = 20 cm Length of spiral film (L)= 30 cm Width (W) = 20 cm

(by using the Monod model)

F = 26.5 m3hr–1 54. We know, Yield, YX/S = – ΔX/ΔS YX/S = 0.4 g/L

Therefore, ΔX = Xfinal – Xinitial = 4 g/L From the equation, YX/S = – ΔX/ΔS 0.4 = –[4/( Sfinal – 12)] 0.4 Sfinal – 4.8 = –4, 0.4 Sfinal = 0.8, Sfinal = 2 Final substrate concentration, Sfinal = 2 g/L

Biochemical Engineering and Bioprocess Technology 11.17

55. Biostat keeps all the given parameters constant. 56. Bioreactor choice should be dependent upon the kinetic properties if the reactor and its stability condition. Fed batch reactors are semi batch reactor as it holds the nutrient media for a specific retention time. 57. Assertion: A synchronous or synchronized culture is a microbiological culture or a cell culture that contains cells that are all in the same growth stage. Reason: Synchronous cultures can be obtained in several ways:

 External conditions can be changed, so as to arrest growth of all cells in the culture, and then changed again to resume growth.  Cell growth can also be arrested using chemical growth inhibitors.  Cells in different growth stages have different physical properties. Cells in a culture can thus be physically separated based on their density or size, for instance. This can be achieved using centrifugation (for density) or filtration (for size).  In the Helmstetter-Cummings technique, a bacterial culture is filtered through a membrane. Most bacteria pass through, but some remain bound to the membrane. Fresh medium is then applied to the membrane and the bound bacteria start to grow. Newborn bacteria that detach from the membrane are now all at the same stage of growth; they are collected in a flask that now harbors a synchronous culture. 58. When both the substrates are present in a batch bioreactor, microbial cells do not consume both substrates simultaneously or additively but instead grow on the sugars sequentially. The sequential consumption results in an interesting pattern of the cells first consuming one preferred substrate and then after an intermediate lag phase, consuming the remaining less preferred substrate. That is, the less preferred substrate is not utilized until the preferred substrate is completely consumed and is no longer available in the growth medium. This sequential utilization of two substrates in batch cultures has been observed in numerous experiments by Monod, who has termed this phenomenon as the "diauxic pattern of growth". It has multiple lag phases.

59. Sparger is being used for the aeration of the medium in bioreactor. Diaphram valve is being used for the air sterilization in large scale animal or microbial cell culture condition. Draft tube has been used for the feed circulation in semi fedbatch culture. Marine type turbine is being used to reduce shear stress on fluid medium in animal cell culture.

g acetate g dry cell g glucose g glucose The respiratory coefficient

60. Yx/s = 0.35 g/g, YP/S = 0.1

Respiratory quotient =

M oles of CO2 formed M oles of CO2 consumed

5 = 1.01 4.95 C6 H12 O6 + b NH3 + O2  CH1.8O0.5 N0.2 + eH2O + f CO2 =

C : 6a = 1 + f

...(1)

H : 12a + 3b = 1.8 + 2e

...(2)

O : 6a + 2d = 0.5 + e + 2F

...(3)

b = 0.2 N : b = 0.2



12a = 1.2 + 2e 6a = 1 + F

Let if

a = 1, i.e. F = 5 e = 5.4 d = 4.95 b = 0.2 a= 1

61. KLa = 0.185 S–1 = 5.139  10–5 hr–1 Critical dO2 is 20% of the saturate conc.(8 mg/ml) Cell density. CL = 1.6  103.m/l C = (8  103 mg/l) do2 = KLa (C – CL)

1/ OUR = 1 / o

2

= qo2X. qo2 = mmo2/gm sw. hr. = 5  10–3 gm dm hr  KLa (C – CL) = q02X.



X=

K La (C  CL ) q02



X=

5.139  10 5 (8  1.6)  10 3 5  10 3

= 65.779 g/lit.

11.18 Biochemical Engineering and Bioprocess Technology

62. Specific growth rate is defined as the increase in cell mass per unit time. In death phase, no cellular growth is observed. During this, the cell population continuously decreases with time and the population is said to have entered into death phase of the growth cycle. Therefore the value of specific growth constant for death phase is zero. 63. The main roles of a baffle in a bioreactor are

 Increase aeration rate  to prevent a vortex 64. C6H12O6 + Zymase → 2 C2H5OH + 2 CO2 use mass balance process to calculate the ethanol production. 180g of Glucose yields = 92 g of ethanol Therefore, 1000g of glucose yields = 92/180 * 1000 = 0.51 kg 65. Cyanogen bromide hydrolyzes peptide bonds at the C-terminus of methionine residues. This reaction is used to reduce the size of polypeptide segments for identification and sequencing. The aspartate residue (Asp 189) located in the catalytic pocket (S1) of trypsins is responsible for attracting and stabilizing positively charged lysine and/or arginine, and is, thus, responsible for the specificity of the enzyme. This means that trypsin predominantly cleaves proteins at the carboxyl side (or "C-terminal side") of the amino acids lysine and arginine except when either is bound to a N-terminal proline. Although largescale mass spectrometry data suggest cleavage occurs even with proline. Trypsins are considered endopeptidases, i.e., the cleavage occurs within the polypeptide chain rather than at the terminal amino acids located at the ends of polypeptides. 66. Final volume = initial volume + Flow rate  Number of day = 6m3 + 2m3 day–1 (2day) = 10m3 67. Dilution rate at end of 2nd day

Flow rate = Total volume at end of 2nd day 2 m3 = = 0.2 10 m3 68. Number of cell died = Total cell – Number of cell survived = 109 – 103 = 9.9  108 = 109

69. 103 cpm in pellet is due to Fab binding to epitope present on recombinant living cells/algae Total Fab or epitope =

Radioactivity in pellet Specific activity

=

103 cpm 10 2 cpm( picomole) 1

103  10 12 102 = 10  10–12 mole = 10 picomole These 10 picomole epitopes are present on 103 cells (assuming) 100% transfection and expression of protein). Average number of epitopes on single recombinant alga

=

10  10 12 mole 103 = 10–14 mole = 6.023  1023  10–14 = 6  109 70. Thermistors monitor temperature and as per requirements very precisely activate hot/cold water pump to heat or cool down the system. The oxygen electrode directly measures the concentration of dissolved oxygen in the liquid phase and so as to regulate the oxygen level it increases the stirrer motor speed for more aeration whereas decreases speed if oxygen concentration has to be decreased. Metal rod is used to switch on/off the vegetable oil pump as vegetable oil has defoming action but its higher concentration can as well has negative effects. Thus metal rod maintains appropriate oil concentration in the fermentor. pH electrode measures the pH of the system liquid and thus varies acid/alkali pump to increase and decrease pH respectively. 71. Hollow fibers also provide a tremendous amount of surface area in a small volume. Cells grow on and around the fibers at densities of greater than 1 X 108 per mL. Hollow fiber cell culture is the only means to culture cells at in vivo like cell densities.

=

72. In single run metabolite (X) concentration is =1.1 kg m–3. Since, total number of runs = 70 Total concentration of (X) = 70  1.1 kg m–3 run–1 = 70  1.1 kg m–3 = 77 kg m–3

Biochemical Engineering and Bioprocess Technology 11.19 3

Now, total volume of batch culture = 50 m . Total amount will be = 50 m3  77 kg m–3 = 3850 kg per year 73. Overall productivity = Total number of runs  final concentration at each end = 70 runs  1.1 kg m–3

81. lysine repress the penicillin production by feedback inhibition.

= 77 kg m–3 year–1 74. Air is provided through sparger in bioreactors. 75. The main roles of baffles are prevent vertex formation, improve aeration and help in mixing. 76. The equilibrium solubility of four calcium salts (calcium oxalate hydrate, calcium citrate tetrahydrate, calcium phosphate, calcium glycerophosphate) were determined at controlled pH values (7.5, 6.0, 4.5 and < or = 3.0) and in distilled water. 77. P = 4.5 gm/l.hr Ethanol production yeast cell conc.(X) = 5% v/v NAD /NADH ceme in yeast cell = 10 m (m = ml/lit) +

= (me/t) the cycling rate of NAD+ will be.

NADH

Am3

v = Am3 x = 5% x1 = 10m

Glucose EtoH + Total NAD /NADH conc = 5  10–5(v/v) 1  cycle  Now, cycle rate =   5  10 5  hr = 20,000 cycle/hr. 78.

dP = K(Pm – P) dt K =

2.303  100 P  log 10   10 P  t

K =

2.303 t

t =

2.303  4.606 hour 0.5

83. anerobic fermentation doesn’t need agitation to maintain oxygen supply. 84. The power number, N P is a commonly-used dimensionless number relating the resistance force to the intertia force. The power number has different specifications according to the field of application e.g., for stirrers, the power number is defined as : NP =

P with n 3 d 5

P = power n = rotational speed

50 ml No.of cy dest = 10–5 ml/lit = cycle rate. 100 lit unit time 5  10–5/lit

NADH

82. Log phase shows the doubling in cell concentration with in a stipulated time unit.

 = fluid density

5 ml in 1 liter of broth, X= 100 NAD+/NADH = 10–5 ml/lit.

NAD+

80. Diffusion helps in dialysis procedure to separate compounds in medium and product. Dialysis works on the principles of the diffusion of solutes and ultrafiltration of fluid across a semi-permeable membrane. Diffusion is a property of substances in water; substances in water tend to move from an area of high concentration to an area of low concentration.

d = diameter of stirrer 85. The degree of protein purity required depends on the intended end use of the protein. For some applications, a crude extract is sufficient. However, for other uses, such as in foods and pharmaceuticals, a high level of purity is required. In order to achieve this, several protein purification methods are typically used, in a series of purification steps. Protein purification step also differs for intra-cellular and extracellular products. 86. Penicillium chrysogenum utilizes phenylacetic acid as a side chain precursor in penicillin G biosynthesis. During industrial production of penicillin G, phenylacetic acid is fed in small amounts to the medium to avoid toxic side effects. Phenylacetic acid is taken up from the medium and intracellularly coupled to 6-aminopenicillanic acid. 87. To prevent contamination of either the fermentation by air-borne microorganisms or the environment by aerosols generated within the fermenter, both air input and air exhaust ports have air filters such as glass fibre filters, mineral fibers filters.

11.20 Biochemical Engineering and Bioprocess Technology

90. For ethanol commercial utilization we need high purity which leads to heavy downstream processing cost along with high percentage of ethanol cause dehydration to the microbial inoculums leading to toxicity to cell. 91. The recovery after four steps 4

= (0.8) = 0.4096 ~ 0.41 92. Fed Batch process is based on feeding of growth limiting nutrient substrate to a culture. This process is used to control repressive effects of the particular nutrient which at higher concentration inhibit the growth of the culture or product. 93. Heat sterilization likely the steam sterilization is helpful in large scale production.

94. Organic ingredients intended for composting can alternatively be used to generate biogas through anaerobic digestion. Anaerobic digestion is fast overtaking composting in some parts of the world including central Europe as a primary means of downcycling waste organic matter. 95. It uses longer heating time during which heat is lost. 96. In secondary metabolism two phases are apparent: trophophase and idiophase. Trophophase is the growth phase of the culture; idiophase is the time when the secondary metabolites are formed. The success of the idiophase is dependent on the trophophase.

12

Bioinformatics

C H A P TE R 2016 1. Which one of the following is NOT an algorithm for building phylogenetic trees? (a) Maximum parsimony (b) Neighbor joining

7. In an affine gap penalty model, if the gap opening penalty is -20, gap extension penalty is -4 and gap length is 8, the gap score is _________. 8. The amino acid substitution matrices in decreasing order of stringency for comparing protein sequences are

(c) Maximum likelihood

(a) PAM250, PAM120, PAM100

(d) Bootstrap

(b) PAM100, PAM120, PAM250

2015

(c) PAM250, PAM100, PAM120

2. Identify the file format given below: >Pl; JMFD

(d) PAM120, PAM250, PAM100

2013

Protein X – Homo sapiens MKALTARQQEVFDLIRDHISRTLRQQGDWL

9. The suitable substitution matrix to align closely related sequences is

(a) GDE

(b) FASTA

(a) PAM 250 or BLOSSUM 80

(c) NBRF

(d) GCG

(b) PAM 40 or BLOSSUM 80

3. Consider the following multiple sequence alignment of four DNA sequences. A

C

T

A

A

C

T

G

A

G

T

C

A

G

C

T

Shannon's entropy of the above alignment is ____ 4. An isolated population on an island has the following genotypic frequencies :

Genotype AA

Aa

aa

Frequency 0.3 0.4 0.3 Assuming that there are only two alleles (A and a) for the gene, the genotypic frequency of AA in the next generation will be _______. 5. How many rooted and unrooted phylogenetic trees, respectively, are possible with four different sequences ? (a) 3 and 15

(b) 15 and 3

(c) 15 and 12

(d) 12 and 3

2014 6. The algorithm for BLAST is based on (a) Dynamic Programming (b) Hidden Markov Model (c) k-tuple analysis (d) Neural Network

(c) PAM 120 or BLOSSUM 40 (d) PAM 250 or BLOSSUM 40 10. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: UPGMA method produces ultrametric tree. Reason:Sequence alignment is converted into evolutionary distances in UPGMA method. (a) Both (A) and (R) are true and (R) is the correct reason for (A) (b) Both (A) and (R) are true and (R) is not the correct reason for (A) (c) (A) is true but (R) is false (d) (A) is false but (R) is true 11. Match the entries in the Group Iwith the entries inGroup II. Group I

Group II

P. Threading

1. Gene duplication

Q. FASTA

2. Fold prediction

R. Profile

3. HMM

S. Paralogs

4. k-tuple

(a) P-2, Q-1, R-3, S-4 (b) P-2, Q-4, R-3, S-1 (c) P-3, Q-4, R-2, S-1 (d) P-1, Q-4, R-3, S-2

12.2

Bioinformatics

2012

(a) Multiple sequence alignment

12. An example of a derived protein structure database is

(b) Homology modeling

(a) Pfam

(b) SCOP

(c) GEO

(d) Prosite

13. An example of a program for constructing a phylogenetic tree is (a) phylip

(b) phrap

(c) prodom

(d) PHDsec

14. Synteny refers to

(c) Phylogeny (d) Docking 20. Match the item in Group I with an appropriate description in Group 2: Group 1

Group 2

P. UPGMA

1. Protein sequence database

Q. CLUSTAL

2. Phylogenetic analysis

R. SWISS-PROT

3. 3-D structure visualization

S. RasMOl

4. Multiple sequence alignment

(a) gene duplication from a common ancestor (b) a tree representation of related sequences (c) the extent of similarity between two sequences (d) local conservation of gene order 15. While searching a database for similar sequences, E value does NOT depend on the (a) sequence length (b) number of sequences in the database (c) scoring system (d) probability from a normal distribution

2011 16. The most widely used program for multiple sequence alignment is (a) BLAST (b) FASTA (c) CLUSTAL (d) Chime

(a) P-4, Q-1, R-2, S-3 (b) P-2, Q-4, R-1, S-3 (c) P-2, Q-3, R-1, S-4 (d) P-2, Q-1, R-4, S-3

2008 21. In bioinformatics, the term ‘BLAST’ refers to (a) database retrieval tool (b) computational tool for sequence homology searching and alignment (c) computational tool to view genomic sequences (d) computational tool to view protein structures 22. Match the terms in group 1 with their possible explanations in group 2

2010

Group 1

17. Program used for essentially local similarity search is

P. Orthologs Q. Paralogs

(a) BLAST

R. Proteome

(b) RasMol

S. Transgenic

(c) ExPASY

Group 2

(d) SWISS-PROT

1. A cell or an organsim having foreign gene

18. Accession number is a unique identification assigned to a (a) single database entry for DNA/Protein

2. The complement of a protein expressed by a genome

(b) single database entry for DNA only

3. Genes from different spe-cies related to each other

(c) single database entry for Protein only

4. Genes from same species related to each other

(d) multiple database entry for DNA/Protein

2009 19. The method used for prediction of three dimensional structure of a protein from known structure(s) of one or more related proteins is

P

Q

R

S

(a) 2

4

1

3

(b) 4

3

2

1

(c) 3

4

2

1

(d) 1

2

3

4

Bioinformatics

12.3

2007

2006

23. Match the methods available on world wide web in group 1 for performing the jobs listed in group 2

24. For prediction of three dimensional structure of protein

Group 1 P. Boxshade

Group 2 1. Searching family data base

(P) homology modeling tries many possible alignments

Q. BCM launcher 2. Finding alignments

(Q) threading first identifies homologues

R. Prosite

3. Displaying alignments

(R) threading evaluates many rough models

S. PSI-BLAST

4. Searching for multiple alignments

(S) homology modeling optimizes one model (a) Q, S

(a) P-l, Q-3, R-2, S-4

(b) P, Q

(b) P-2, Q-3, R-2, S-4

(c) R, S

(c) P-3, Q-4, R-l, S-4

(d) Q, R

(d) P-3, Q-2, R-l, S-4

ANSWERS 1. (d)

2. (c)

8. (b)

9. (b)

10. (b)

11. (b)

18. (a)

19. (b)

20. (b)

21. (b)

4. (0.25)

5. (b)

6. (c)

12. (b)

13. (a)

14. (d)

15. (d)

22. (c)

23. (d)

24. (c)

3. (3.80-3.82 bits)

7. (– 52 to – 52) 16. (c)

17. (a)

EXPLANATIONS and genotype frequency = p2 1. Bootstrapping is a resampling analysis that involves taking columns of characters out of your analysis, rebuilding the tree, and testing if the same nodes are recovered. 2. GDE  Start with # $ @ etc FASTA  Start with > Description NBRF Start with > P1; GCG  Start with Anotation lines 3.

= 0.25 5. Here n = 4 Number of rooted trees (NR) =

 2n – 3  ! 2n – 2 (n – 2)!

Number of unrooted trees (NU) =

(2n – 5)! 2 (n – 3)! n –3

 NR = 15 and NU = 3

H(p) = – Pi log2 Pi P1 = 0 P2 = 0.5 P3 = 0.25 P4 = 0.25  Pi = 1 H(p) = 0 + (– 0.5 log2 0.5)+ (– 0.25log2 0.25) + (– 0.25 log 2 0.25) = 3.80 bits

4.

= 0.5  0.5

f(AA) = 0.3, f(Aa) = 0.4, f (a,a) = 0.3 ( Allele frequency in next generation = p 1 = f(AA) + f (Aa) = 0.3 + 0.2 = 0.5) 2

6. The BLAST program sets a size k for k-tuple subwords. The program then looks for diagonals in the comparison matrix between query and search sequence along which many k-tuples match. This program also aims at identifying core similarities for later extension. The core similarity is defined by a window with a certain match density on DNA or with an amino acid similarity score above some threshold for proteins. Independent of the exact definition of the core similarity, BLAST rests on the precomputation of all strings which are in the given sense similar to any position in the query. n BLAST nomenclature this set of strings is called the neighborhood of the query. The code to generate this neighborhood is in fact exceedingly fast.

12.4

Bioinformatics

7. Gap score = 2  Gap opening penalty + Gap extension penalty – Gap length Here, Gap opening penalty = –20, Gap extension penalty = –4 and gap length = 8 Therefore, Gap score = 2  (–20) – 4 – 8 = – 52 8. Substitution matrices like PAM100, PAM120 and PAM250 are constructed by observing the frequencies of amino acid replacements in large samples of protein sequences. For a given replacement, the PAM value is proportional to the natural log of the frequency with which that replacement was observed to occur. One PAM unit is defined as the amount of sequence divergence corresponding to a 1% amino acid replacement rate. For closely-related sequences, it is appropriate to use a PAM100 matrix, in which PAM units have been extrapolated to 100% replacement. In other words, 100% of the positions show at least one replacement. For most database searches, a PAM250 matrix is preferred, since larger databases will tend to have sets of more distantly-related sequences. So, decreasing order of stringency of point accepted mutation matrices would be: PAM100 > PAM120 > PAM250. 9. Point accepted mutation (PAM) matrix is used for species having deletion or insertion at any particular point in the nucleotide sequence. Block substitution matrix (BLOSUM) is used for species whose sequences have undergone substitution. PAM 40 means that 40 % of the sequences are mutated and 60% are conserved while BLOSUM 80 means that 80% of the sequences are conserved and rest 20% are mutated. So, PAM 40 and BLOSUM 80 can be used for less divergent or closely related species for performing the alignment. 10. The unweighted pair-group method with arithmetic mean (UPGMA) is a popular distance analysis method. UPGMA is the simplest method for constructing trees. The great disadvantage of UPGMA is that it assumes the same evolutionary speed on all lineages, i.e. the rate of mutations is constant over time and for all lineages in the tree. This is called a ‘molecular clock hypothesis’. This would mean that all leaves (terminal nodes) have the same distance from the root. In reality the individual branches are very unlikely to have the same mutation rate. Sequence alignment is converted into evolutionary distances. UPGMA is “ultrametric”, meaning that all the terminal nodes (i.e. the sequences/taxa) are equally distance from

the root. In molecular terms, this means that UPGMA assumes a molecular clock, i.e. all lineages are evolving at a constant rate. 11. Threading is the method of three dimensional structure predictions of proteins by generating threads that help to determine folds in a protein. FASTA is a program for rapid alignment of pairs of protein and DNA sequences. Rather than comparing individual residues in the two sequences, FASTA instead looks for matching sequence patterns or words, called k-tuples, and then attempts to build a local alignment based upon these word matches. Profile HMMs turn a multiple sequence alignment into a positionspecific scoring system suitable for searching databases for remotely homologous sequences. Profile HMM analyses complement standard pairwise comparison methods for large-scale sequence analysis. Paralogs are genes related by duplication within a genome. Orthologs retain the same function in the course of evolution, whereas paralogs evolve new functions, even if these are related to the original one. 12. Protein structural database: Primary database: PDB Secondary database: SCOP, CATH 13. A program for constructing phylogenetic tree is PHYLIP. It is a Phylogeny Inference Package. It is a free computational phylogenetic package which has programs for inferring evolutionary tree constructions. 14. Synteny can be defined as the local conservation of gene order. It is the condition of two or more genes being located on the same chromosome whether or not there is demonstrable linkage between them. 15. The Expect value (E) is a parameter that describes the number of hits one can "expect" to see by chance when searching a database of a particular size. It decreases exponentially as the Score (S) of the match increases. Essentially, the E value describes the random background noise. Virtually identical short alignments have relatively high E values. This is because the calculation of the E value takes into account the length of the query sequence and also depends upon the number of sequences in the database. These high E values make sense because shorter sequences have a higher probability of occurring in the database purely by chance. Thus the E value does not depend on the probability from a normal distribution.

Bioinformatics

16. The Clustal series of programs are widely used in molecular biology for the multiple alignment of both nucleic acid and protein sequences and for preparing phylogenetic trees. The popularity of the programs depends on a number of factors, including not only the accuracy of the results, but also the robustness, portability and user-friendliness of the programs. New features include NEXUS and FASTA format output, printing range numbers and faster tree calculation. Although, Clustal was originally developed to run on a local computer, numerous Web servers have been set up, notably at the EBI (European Bioinformatics Institute). 17. BLAST (Basic Local Alignment Search Tool) is a program which is used for local similarity search. BLAST is an algorithm for comparing primary biological sequence information, such as the amino-acid sequences of different proteins or the nucleotides of DNA sequences. A BLAST search enables a researcher to compare a query sequence with a library or database of sequences, and identify library sequences that resemble the query sequence above a certain threshold. Different types of BLASTs are available according to the query sequences like nBLAST (nucleotide BLAST), pBLAST (protein BLAST) etc. 18. An accession number is a unique identifier assigned to a particular genome or protein sequence to uniquely identify it in a database. It is assigned to the entire sequence record (whether DNA or protein) when the record is submitted to GenBank. The NCBI Entrez System is searchable by accession number using the Accession [ACCN] search field. 19. Homology modeling, also known as comparative modeling of protein, refers to constructing an atomic-resolution model of the “target” protein from its amino acid sequence and an experimental three-dimensional structure of a related homologous protein (the “template”). Homology modeling relies on the identification of one or more known protein structures likely to resemble the structure of the query sequence, and on the production of an alignment that maps residues in the query sequence to residues in the template sequence. It has been shown that protein structures are more conserved than protein

12.5

sequences amongst homologues, but sequences falling below a 20% sequence identity can have very different structure. 20.  UPGMA (Unweighted Pair Group Method with Arithmetic Mean) is a simple agglomerative or hierarchical clustering method. Inbioinformatics, UPGMA is used for the creation of phenetic trees (phenograms). In a phylogenetic context, UPGMA assumes a constant rate of evolution (molecular clock hypothesis).  CLUSTAL is a general purpose multiple sequence alignment program for DNA or proteins.  SWISS-PROT is a curated protein sequence database which strives to provide a high level of annotation (such as the description of the function of a protein, its domain structure, posttranslational modifications, variants, etc), a minimal level of redundancy and a high level of integration with other databases.  RasMol is a computer program written for molecular graphics visualization intended and used primarily for the depiction and exploration of biological macromolecule structures 21. In bioinformatics, Basic Local Alignment Search Tool, or BLAST, is an algorithm for comparing primary biological sequence information, such as the amino-acid sequences of different proteins or the nucleotides of DNA sequences. A BLAST search enables a researcher to compare a query sequence with a library or database of sequences, and identify library sequences that resemble the query sequence above a certain threshold. 22. Orthologs, or orthologous genes, are genes in different species that originated by vertical descent from a single gene of the last common ancestor. Paralogous sequences provide useful insight into the way genomes evolve. The genes encoding myoglobin and hemoglobin are considered to be ancient paralogs. 23.  BOXSHADE is a program for pretty-printing multiple alignment output. The program itself doesn't do any alignment, you have to use a multiple alignment program like ClustalW or Pileup and use the output of these programs as input for BOXSHADE. Of course, you can also use manually aligned sequences (in a proper format).

12.6

Bioinformatics

 BCM Launcher, i.e. Baylor College of Medicine Search Launcher is an on-going project to organize molecular biology-related search and analysis services available on the WWW by function by providing a single point-of-entry for related searches (e.g., a single page for launching protein sequence searches using standard parameters). It helps in finding alignments.  PROSITE is a database of protein families and domains. It is based on the observation that, while there is a huge number of different proteins, most of them can be grouped, on the basis of similarities in their sequences, into a limited number of families. Proteins or protein domains belonging to a particular family generally share functional attributes and are derived from a common ancestor. PROSITE currently contains patterns and profiles specific for more than a thousand protein families or domains. Each of these signatures comes with documentation providing background information on the structure and function of these proteins. 24. Protein threading, also known as fold recognition, is a method of protein modeling (i.e.

computational protein structure prediction) which is used to model those proteins which have the same fold as proteins of known structures, but do not have homologous proteins with known structure. It differs from the homology modeling method of structure prediction as it (protein threading) is used for proteins which do not have their homologous protein structures deposited in the Protein Data Bank (PDB), whereas homology modeling is used for those proteins which do. Threading works by using statistical knowledge of the relationship between the structures deposited in the PDB and the sequence of the protein which one wishes to model. Homology modeling, also known as comparative modeling of protein, refers to constructing an atomicresolution model of the “target” protein from its amino acid sequence and an experimental threedimensional structure of a related homologous protein (the “template”). Homology modeling relies on the identification of one or more known protein structures likely to resemble the structure of the query sequence, and on the production of an alignment that maps residues in the query sequence to residues in the template sequence.

SOLVED PAPER - 2017 INSTRUCTIONS 1. Total of 65 questions carrying 100 marks, out of which 10 questions carrying a total of 15 marks are in General Aptitude (GA) 2. The Engineering Mathematics will carry around 15% of the total marks, the General Aptitude section will carry 15% of the total marks and the remaining 70% of the total marks. 3. Types of Questions (a) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all papers and sections. These questions are objective in nature, and each will have a choice of four options, out of which the candidate has to mark the correct answer(s). (b) Numerical Answer Questions of 1 or 2 marks each in all papers and sections. For these questions the answer is a real number, to be entered by the candidate using the virtual keypad. No choices will be shown for these type of questions. 4. For 1-mark multiple-choice questions, 1/3 marks will be deducted for a wrong answer. Likewise, for 2-marks multiple-choice questions, 2/3 marks will be deducted for a wrong answer. There is no negative marking for numerical answer type questions.

TECHNICAL SECTION (Q.1 – 25) : Carry One Mark Each 1. An enzyme catalyzes a reaction by (a) decreasing the energy of the substrate. (b) decreasing the activation energy of the reaction. (c) decreasing product stability. (d) increasing the activation barrier of the reaction. 2. Natural proteins are composed primarily of 20 -amino acids. Which one of the following statements is true for any of these amino acids in a solultion of pH 1.5? (a) Only the amino group is ionized. (b) Only the carboxylic acid group is ionized. (c) Both amino and carboxylic acid groups are ionized. (d) Both amino and carboxylic acid groups are neutral. 3. Which one of the following organisms is an indicator of fecal contamination? (a) Escherichia coli (b) Streptococcus lactis (c) Bacillus subtilis (d) Lactobacillus acidophilus 4. The surface area (in m2) of the largest sphere that can fit into a hollow cube with edges of length 1 meter is ________. Given data :  = 3.14

5. In a thin layer chromatography experiment using a silica gel plate, a compound showed migration of 12.5 cm and the solvent front showed migration of 18 cm. The R f value for the compound is________. 6. lim x 0

sin( x) is __________. x

7. Which one of the following mechanisms is used by human pathogens to evade host immune responses? (a) Somatic hypermutation (b) Antibody production (c) Antigenic variation (d) Complement activation 8. If the nucleotide composition (%) of a viral genome is A = 10, U = 20, C = 40, and G = 30, which one of the following is this genome? (a) Double stranded RNA (b) Single stranded RNA (c) Single stranded DNA (d) Double stranded DNA 9. Which one of the following techniques can be used to determine the structure of a 15kDa globular protein at atomic resolution? (a) Raman spectroscopy (b) IR spectroscopy (c) UV spectroscopy (d) NMR spectroscopy

2

Solved Paper - 2017

10. Assertion (a) : Gram negative bacteria show staining with saffranin. Reason (r) : Gram negative bacteria have an outer membrane with lipopolysaccharides. (a) Both (a) and (r) are true and (r) is the correct reason for (a) (b) Both (a) and (r) are true but (r) is not the correct reason for (a)

18. In eukaryotes, cytokinesis is inhibited by (a) cytochalasin D

(b) vinblastine

(c) nocodazole

(d) colchicine

19. A proto-oncogene is suspected to have undergone duplication in a certain type of cancer. Of the following techniques, which one would verify the gene duplication? (a) Northern blotting

(c) Both (a) and (r) are false.

(b) Southern blotting

(d) (a) is true but (r) is false.

(c) South western blotting

11. The plant hormone indole-3-acetic acid is derived from (a) histidine (b) tyrosine (c) tryptophan (d) proline 12. Enzyme-linked immunosorbent assay (ELISA) is used for

(d) Western blotting 20. A polymerase chain reaction (PCR) was set up with the following reagents : DNA template, Taq polymerase, buffer, dNTPs, and Mg2+. Which one of the following is missing in the reaction mixture? (a) Helicase (b) Single-stranded binding proteins

(a) quantifying antibody levels in blood

(c) Primers

(b) determining the molecular weight of an antigen

(d) Reverse transcriptase

(c) purifying proteins from biological fluids (d) determining the molecular weight of an antibody 13. Macrophages eliminate pathogenic bacteria upon activation by (a) NK cells

(a) both Km and Vmax will remain the same. (b) K m will remain the same but V max will increase. (c) Km will increase but Vmax will remain the same.

(b) basophils (c) CD4+ T cells

(d) both Km and Vmax will increase.

(d) plasma cells 14. If a protein contains four cysteine residues, the number of different ways they can simultaneously form two intra-molecular disulphide bonds is______. 15. Secretory proteins synthesized by ER-associated ribosomes traverse through (a) mitochondria (b) peroxisomes (c) the Golgi apparatus (d) the nucleus 16. For y = f(x), if

21. An enzymatic reaction exhibits Michaelis-Menten kinetics. For this reaction, on doubling the concentration of enzyme while maintaining [S] >> [E0].

dy d2 y  0 at x = 0, and y = 1  0, 2 dx dx

at x = 1, the value of y at x = 2 is __________. 17. During protein synthesis, tRNAs are NOT involved in (a) charging

(b) initiation

(c) elongation

(d) termination

22. A 5 liter chemostat is fed fresh medium at 0.2 liters/minute having a substrate concentration of 25 grams/liter. At steady state, the outgoing stream has substrate concentration of 2.5 grams/ liter. The rate of consumption (grams/liter/ minute) of the substrate in the reactor is ______. 23. The transcription factor X binds a 10 base pair DNA stretch. In the DNA of an organism, X was found to bind at 20 distinct sites. An analysis of these 20 binding sites showed the following distribution:

Base

Position in the binding site

A

1 11

2 0

3 0

4 0

5 16

6 2

7 8 4 0

9 4

10 3

T

3

0

19

0

1

3

4 20

2

4

G

4

20

0

0

2

4

6

0

12

2

C

2

0

1

20

1

11 6

0

2

11

Solved Paper - 2017

3

What is the consensus sequence for the binding site of X?

29. Match the proteins in Group I with cellular processes in Group II

(a) NGTCNNNTNN

Group I

(b) AGTCACNTGC

P. p53

1. DNA packaging

(c) CACCTANCTG

Q. Lysozyme

2. Apoptosis

(d) ANNNACGNGC

R. Tubulins

3. Hydrolysis of polysaccharides

S. Histones

4. Chromosome segregation

24. A bacterium has a genome of size 6 million base pairs. If the average rate of DNA synthesis is 1000 base pairs/second, the time taken (in minutes) for replication of the genome will be _______. 25. At the transcription strat site of a gene, any of the four nucleotides can occur with equal probability p. The Shannon Entropy S, given by S =  4i1 pt ln pi, for this start site is________.

Group II

(a) P-4, Q-2, R-3, S-1 (b) P-2, Q-3, R-1, S-4 (c) P-4, Q-3, R-1, S-2 (d) P-2, Q-3, R-4, S-1 30. The circular dichroism spectra of three proteins P, Q, and R are given below :

Given data : ln(2) = 0.69 (Q.26 – 55) : Carry Two Marks Each 26. Which one of the following graphs represents the kinetics of protein precipitation by addition of ammonium sulphate? On the Y-axis, [Protein] represents the concentration of free protein in solution.

(a)

(b)

(c)

(d)

27. The distribution of marks scored by a large class in an exam can be represented as a normal distribution with mean and standard deviation . In a follow-up exam in the same class, everyone scored 5 marks more than their respective score in the earlier exam. For this follow-up exam, the distribution of marks can be represented as a normal distribution with mean 2 and standard deviation 2. Which one of the following is correct? (a)

=

2;

 > 2

(b)


2

(c)

>

2;

 < 2

(d)


1) is equal to (up to two decimal places) ______.

27. Which of the following cytokines are endogenous pyrogens? P. Tumor necrosis factor-α Q. Interleukin-I R. Transforming growth factor-β S. Interleukin-10 (a) P and Q only

(b) P and R only

(c) R and S only

(d) Q and S only

28. Match the classes of RNA molecules in Group I with their functions in Group II. Group I

Group II

P. snoRNA

1. Protects germline from transposable elements

Q. piRNA

2. Blocks translation of selected mRNA

R. miRNA

3. Template elongation

S. snRNA

4. Modification and processing of rRNA

for

telomere

5. Splicing of RNA transcripts (a) P-3, Q-5, R-2, S-4 (b) P-1, Q-3, R-2, S-5 (c) P-1, Q-4, R-5, S-2 (d) P-4, Q-1, R-2, S-5

SOLVED PAPER 2018

5

Reason : Eukaryotic slice sites are difficult to predict. (a) Both [a] and [r] are false (b) [a] is true but [r] is false (c) Both [a] and [r] are true and [r] is the correct reason for [a] (d) Both [a] and [r] are true but [r] is not the correct reason for [a] 30. Which one of the following amino acids is catalyzed by activated macrophages to produce reactive nitrogen species? (a) Arginine

(b) Asparagine

(c) Cysteine

(d) Histidine

31. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r] Assertion : The association constant in water for the G-C base pair is three times lower than that for the A-T base pair. Reason : There are three hydrogen bonds in the G-C base pair and two in the A-T base pair. (a) Both [a] and [r] are true and [r] is the correct reason for [a] (b) [a] is false but [r] is true (c) Both [a] and [r] are false (d) Both [a] and [r] are true and [r] is not the correct reason for [a] 32. Which one of the combinations of the following statements is true about antibody structure? P. Limited proteolysis of rabbit IgG with the enzyme pepsin generates two antigen-binding regions (Fab) and an Fc fragment Q. Limited proteolysis of rabbit IgG with the enzyme papain generates a single bivalent antigen-binding region F(ab′)2 and peptide fragments R. The Fc fragment of IgG can self-associate and crystallize into a lattice S. The F(ab′)2 fragment of IgG is composed of both light and heavy chains (a) P and Q only (b) P and R only (c) R and S only (d) Q and S only

(d) Invariant chain in endoplasmic reticulum is involved in transport of peptides and loading of class I MHC 34. Which of the following are true about bacterial superoxide dismutase? P. Present in obligate aerobes Q. Present in facultative aerobes R. Present in aerotolerant anaerobes S. Absent in obligate aerobes (a) P and Q only (b) P, Q and R only (c) P and R only (d) Q and S only 35. Which of the following are true with regard to anaerobic respiration in bacteria? P. The final electron acceptor is an inorganic substance other than molecular oxygen Q. The number of ATP molecules produced per glucose molecule is more than that produced in aerobic respiration R. The number of ATP molecules produced per glucose molecule is less than that produced in aerobic respiration S. Only substrate level phosphorylation is used to generate ATP (a) P and S only (b) Q and S only (c) P and R only (d) P, Q and S only 36. Shear stress versus shear rate behaviour of four different types of fluids (I, II, III and IV) are shown in the figure below. II

–2

Assertion : Ab initio gene finding algorithms that predict protein coding genes in eukaryotic genomes are not completely accurate.

33. Which one of the following statements is true with regard to processing and presentation of protein antigens? (a) In the class II MHC pathway, protein antigens in the cytosol are processed by proteasomes (b) In the class I MHC pathway, extacellular protein antigens are endocytosed into vesicles and processed (c) In the class I MHC pathway, transporter associated antigen processing (TAP) protein is required for translocating processed peptides generated in the cytosol

Shear stress (Nm )

29. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]

IV I III

–1

Shear rate (S )

6

SOLVED PAPER 2018

Which one of the following options is correct? (a) I-Newtonian, II-Bingham plastic, III-Dilatant, IV-Pseudoplastic (b) I-Pseudoplastic, II-Dilatant, III-Newtonian, IVBingham plastic (c) I-Newtonian, II-Pseudoplastic, III-Bingham plastic, IV-Dilatant (d) I-Newtonian, II-Bingham plastic, IIIPseudoplastic, IV-Dilatant 37. An analysis of DNA-protein interactions was carried out using all DNA-protein complexes in the protein data bank (PDB). The frequency distribution of four amino acid residues, represented as P, Q, R and S, occurring in noncovalent interactions between the protein and DNA backbone is shown below.

Frequency (%)

60

Group I

Group II

P. Citric acid

1. Trichoderma viride

Q. Cellulase

2. Clostridium

R. Vitamin B12

3. Aspergillus niger

acetobutylicum S. Butanol

4. Propionibacterium freudenreichii

(a) P-4, Q-3, R-1, S-2 (b) P-3, Q-1, R-2, S-4 (c) P-2, Q-1, R-4, S-3 (d) P-3, Q-1, R-4, S-2 40. 5′ capping of mRNA transcripts in eukaryotes involves the following events: P. Addition of GMP on the 5′ end Q. Removal of γ-phosphate of the triphosphate on first base at the 5′ end

50

R. 5′-5′ linkage between GMP and the first base at 5′ end

40 30

S. Addition of methyl group to N7 position of guanine

20

Which one of the following is the correct sequence of events?

10 0

39. Match the industrial products mentioned in Group I with their producer organisms in Group II

P

S R Q Which one of the following is correct? (a) P-Lys, Q-Arg, R-Gln, S-Glu (b) P-Gln, Q-Glu R-Lys, S-Arg

(a) P, Q, R, S

(b) P, R, Q, S

(c) Q, P, R, S

(d) Q, P, S, R

41. Calculate the following integral (up to two decimal places)



(c) P-Asn, Q-Asp, R-Arg, S-Lys

1

0

(d) P-His, Q-Glu, R-Gln, S-Lys 38. A pedigree of an inheritable disease is shown below.

( x + 3) ( x + 1) dx = _______

42. The probability distribution for a discrete random variable X is given below.

X 1 2 3 4 P ( X ) 0.3 0.4 0.2 0.1 The expectation value of X is (up to one decimal place)______ 43. If 1 + r + r2 + r3 +....∞ = 1.5, then 1 + 2r + 3r2 + 4r3 + ....∞ = (up to two decimal places) _____

Unaffected Female

Carrier Female

Affected Female

Unaffected Male

Carrier Male

Affected Male

What type of inheritance does the disease follow? (a) Autosomal dominant (b) X-linked dominant (c) X-linked recessive (d) Autosomal recessive

44. Moist heat sterilization of spores at 121°C followes first order kinetics as per the expression:

dN = − kd N dt where, N is the number of viable spores, t is the dN is the rate time, kd is the rate constant and dt of change of viable spores.

SOLVED PAPER 2018

7

If kd value is 1.0 min–1, the time (in minutes) required to reduce the number of viable spores from an initial value of 1010 to a final of 1 is (up to two decimal places) ______ 45. An aqueous solution containing 6.8 mg/L of an antibiotic is extracted with amyl acetate. If the partition coefficient of the antibiotic is 170 and the ratio of water to solvent is 85, then the extraction factor is ______ 46. A microbial strain is cultured in a 100 L stirred fermenter for secondary metabolite production. If the specific rate of oxygen uptake is 0.4 h–1 and the oxygen solubility in the broth is 8 mg/L, then the volumetric mass transfer coefficient (KLa) (in s–1) of oxygen required to achieve a maximum cell concentration of 12 g/L is (up to two decimal places) ______ 47. In a chemostat, the feed flow rate and culture volume are 100 ml/h and 1.0 L, respectively. With glucose as substrate, the values of μmax and Ks are 0.2h–1 and 1 g/L, respectively. For a glucose concentration of to g/L in the feed, the effluent substrate concentration (in g/L) is ______ 48. Mammalian cells in active growth phase were seeded at a density of 1 × 105 cell/ml. After 72 hours, 1 × 10 6 cells/ml were obtained. The population doubling time of the cells in hours is (up to two decimal places) _______ 49. Yeast converts glucose to ethanol and carbon dioxide by glycolysis as per the following reaction: C6H12O6 → 2C2H5OH + 2CO2 Assuming complete conversion, the amount of ethanol produced (in g) from 200 g of glucose is (up to two decimal places)______

50. At the end of a batch culture, glucose solution is added at a flow rate of 200 ml/h. If the culture volume after 2 h of glucose addition is 1000 ml, the initial culture volume (in ml) is ______ 51. Consider the following alignment of two DNA sequences: AGTAAC AA--AC

Assuming an affine gap scoring scheme of an identity matrix for substitution, a gap initiation penalty of 1 and a gap extension penalty of 0.1, the score of the alignment is (up to one decimal place) ______ 52. First order deactivation rate constants for soluble and immobilized amyloglucosidase enzyme are 0.03 min–1 and 0.005 min–1, respectively. The ratio of half-life of the immobilized enzyme to that of the soluble enzyme is (rounded off to the nearest integer)______ 53. Consider a simple uni-substrate enzyme that follows Michaelis-Menten kinetics. When the enzyme catalyzed reaction was carried out in the presence of 19 nM concentration of an inhibitor, there was no change in the maximal velocity. However, the slope of the Lineweaver-Burk plot increased 3-fold. The dissociation constant for the enzyme-inhibitor complex (in nM) is ______ 54. The product of complete digestion of the plasmid shown below with EcoRI and HaeIII was purified and used as a template in a reaction containing Klenow fragment of DNA polymerase, dNTPs and [α–32P]-dATP in a suitable reaction buffer. The product thus obtained was purified and subjected to gel electrophoresis followed by autoradiography. 3 kb HaeIII

EcoRI

EcoRI

1 kb HaeIII HaeIII 5 kb

5′ -GAATTC-3′ 3′ -CTTAAG-5 ′ 5-GGCC-3 ′ ′ 3-CCGG-5 ′ ′

The number of bands that will appear on the X-ray film is______ 55. A rod shaped bacterium has a length of 2 μm, diameter of 1 μm and density the same as that of water. If proteins constitute 15% of the cell mass and the average protein has a mass of 50 kDa, the number of proteins in the cell is ______ (1 Da = 1.6 × 10–24g)

8

SOLVED PAPER 2018

ANSWERS General Aptitude (GA) 1. (a)

2. (c)

3. (b)

4. (c)

5. (d)

6. (a)

7. (c)

8. (d)

9. (c)

10. (a)

Technical Section 1. (c)

2. (c)

3. (b)

4. (d)

5. (c)

6. (d)

7. (c)

8. (b)

9. (b)

10. (d)

11. (d)

12. (c)

13. (b)

14. (c)

15. (a)

16. (d)

17. (b)

18. (b)

19. (c)

20. (b)

21. (– 10)

22. (0.32)

23. (50)

24. (1)

25. (4096)

26. (d)

27. (a)

28. (d)

29. (c)

30. (a)

31. (b)

32. (c)

33. (c)

34. (b)

35. (c)

36. (d)

37. (a)

38. (c)

39. (d)

40. (c)

41. (5.33)

42. (2.1)

43. (2.25)

44. (23.04)

45. (2)

46. (0.167)

47. (1)

48. (21.67)

49. (102.22)

50. (600)

51. (1.8)

52. (6)

53. (5)

54. (2)

55. (2945000)

EXPLANATIONS GENERAL APTITUDE 1. Stares - To look at someone for a long time Stairs - A construction designed to bridge a large vertical distance by dividing it into smaller vertical distances, called steps

Now perimeter of equilateral triangle = 3a = 3 × 2 = 6 5. (i) Volume of cuboid = (8 × 10 × 6) cm3 = 480 cm3 (ii) Volume of cube = (8 × 8 × 8) cm3 = 512 cm3 (iii) Volume of cylinder = πr 2 h

2. Errant means misbehaving, exhibiting inappropriate behaviour / offending conduct. 3. cos x

0

π π 2

3π 2

From the curve it is clear that sin x and cos x ⎛π ⎞ both are decreasing in the interval ⎜⎝ , π⎟⎠ 2



3 =

∴ ∴

a2 = 4 a=2

3 2 a 4

(iv) Volume of Sphere =

3 2 a 4

4 3 ⎛4 2⎞ πr = ⎜ π × ( 7 ) ⎟ cm3 3 ⎝3 ⎠ = 1436.75 cm3

Hence, the descending order of their volumes is

x sin x

4. Area of equilateral triangle =

⎛ 22 ⎞ ⎜ 7 × 7 × 7 × 7 ⎟ cm3 = 1078 cm3 ⎝ ⎠

(iv), (iii), (ii), (i). 6.

Average speed =

2(S1 × S2 ) S1 + S2

=

2(60 × 90) km/h (60 + 90)

=

2 × 60 × 90 = 72 km / h 150

7. Number of parallelogram = 4C2 × 5C2 =

4! 5! × 2! × 2! 3! × 2!

= 6×

5×4 = 60 2

SOLVED PAPER 2018

8.

9 100,000

B (2,50,000)

A (3,30,000)

150000 + y 100000 – y

60000 + y

y 80000 – y

90000 – y

80,000

90,000 90000 + y

C (2,60,000)

From above diagram,

∴ 6,30,000 = 2y + 1,50,000 + 100000 + 80000 + 60000 + 90000 + 90000

∴ 63000 = 2y + 57000 ∴ 6,30,000 – 5,70,000 = 2y ∴ 60000 = 2y ∴

y = 30000

Students who failed to clear the test = 150000 + 60000 + 90000 + 4y = 300000 + 4 × 30000

( y = 30000)

= 420000 Students who failed to clear the test = 420000 2

x – 1 = –x

9.

x–

x2 + Now,

x4 +

1 =–1 x

1

x2

= 0.2 2

1

⎛ 1 ⎞ = ⎜ x2 + ⎟ −2 4 x x2 ⎠ ⎝

= (3)2 – 2

=9–2 =7

BIOTECHNOLOGY 1. Probability of getting head (p) = 0.6

⇒ Probability of getting tail (q) = 11 – p = 0.4 Desired outcomes = {HT, TH, HH} = 0.6 × 0.4 + 0.4 × 0.6 × 0.6 × 0.6 = 0.24 + 0.24 + 0.36 = 0.84

∴ Option (c) is correct.

2. During G1, cells synthesize all of the molecules needed for a new round of cell division. The serum in growth media contains everything needed for this, including growth factors, which are important signals for cell division. Therefore, starving cells of serum can prevent them from dividing. Human embryonic kidney cells 293, also often referred to as HEK 293 or less precisely as HEK cells, are a specific cell line originally derived from human embryonic kidney cells grown in tissue culture. Cell Cycle Arresting Compounds rely on differing mechanisms of action to prevent the normal progression of the cell cycle. Some of these compounds interfere with CDK/cyclin complexes leaving cells stuck at the G2/M phase border, while others affect CaMKII phosphorylation, inducing arrest at the G1 phase. Other mechanisms of action include interference with RNA function and inhibition of protein synthesis. Many of these compounds ultimately induce apoptosis as a result of their interruption of the cell cycle. The cell cycle checkpoints play an important role in the control system by sensing defects that occur during essential processes such as DNA replication or chromosome segregation, and inducing a cell cycle arrest in response until the defects are repaired. 3. Telomeres are sections of DNA found at the ends of each of our chromosomes. They consist of the same sequence of bases repeated over and over. In humans the telomere sequence is TTAGGG. This sequence is usually repeated about 3,000 times and can reach up to 15,000 base pairs in length. 4. Wherever a gene exists on a DNA molecule, one strand is the coding strand (or sense strand or non-template strand), and the other is the noncoding strand (also called the antisense strand, anticoding strand, template strand, or transcribed strand). Sequence of the transcribed strand is complimentary to the sequence of sense strand. Hence the sequence of mRNA will also be same as that of sense strand (Ts will be replaced with Us). 5. Botulinum toxin (BTX) is a neurotoxic protein produced by the bacterium Clostridiumb botulinum and related species. It prevents the release of the neurotransmitter acetylcholine from axon endings at the neuromuscular junction and thus causes flaccid paralysis. Infection with the bacterium causes the disease botulism.

10

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Cholera toxin (also known as choleragen) is protein complex secreted by the bacterium Vibrio cholerae. CTX is responsible for the massive, watery diarrhea characteristic of cholera infection. Streptolysin O is a toxin secreted by Streptococcus pyogenes and is a prototype member of poreforming bacterial cytolysins. It is used permeabilize cell membranes to permit cellular uptake of large or charged molecules. It is used to study macromolecule delivery. It is a potential anticancer agent and is used to study suicide cancer gene therapy. Diphtheria toxin is an exotoxin (extracellular protein) secreted by Corynebacterium diphtheriae, the pathogenic bacterium that causes diphtheria, inhibits protein synthesis and kills susceptible cells. The gene that encodes DT (tox) is present in some corynephages, and DT is only produced by C. diphtheriae isolates that harbor tox+ phages. 6. Molecular mechanics describes the potential energy of a molecule in terms of a set of potential energy equations that are reminiscent of classical mechanics in physics. The potential energy equations to calculate the energy, the “atom types” that define different atoms in a molecule, and the parameters/constants used in the equations are known as a force-field. The total potential energy of a molecular system is the sum of individual potential components. Simple molecular mechanics force fields include bond stretching, angle bending, torsion, and van der Waals interactions in their make-up. The sums extend over all bonds, bond angles, torsion angles, and non-bonded interactions in the molecule. Modern force fields add other terms for greater accuracy. 7. BLAST is actually a family of programs (all included in the blastall executable). These include: Nucleotide-nucleotide BLAST (blastn) This program, given a DNA query, returns the most similar DNA sequences from the DNA database that the user specifies. Protein-protein BLAST (blastp) This program, given a protein query, returns the most similar protein sequences from the protein database that the user specifies. Position-Specific Iterative BLAST (PSIBLAST) (blastpgp)

This program is used to find distant relatives of a protein. Nucleotide 6-frame translation-protein (blastx) This program compares the six-frame conceptual translation products of a nucleotide query sequence (both strands) against a protein sequence database. Nucleotide 6-frame translation-nucleotide 6frame translation (tblastx) This program is the slowest of the BLAST family. It translates the query nucleotide sequence in all six possible frames and compares it against the six-frame translations of a nucleotide sequence database. The purpose of tblastx is to find very distant relationships between nucleotide sequences. Protein-nucleotide 6-frame translation (tblastn) This program compares a protein query against the all six reading frames of a nucleotide sequence database. Large numbers of query sequences (megablast) When comparing large numbers of input sequences via the command-line BLAST, “megablast” is much faster than running BLAST multiple times. It concatenates many input sequences together to form a large sequence before searching the BLAST database, then postanalyzes the search results to glean individual alignments and statistical values. 8. Anion exchange resins exchange the negative ions readily, so when various peptides with different charges pass through it, then they will be eluted in order from strong negative charge to positive charge. Therefore the order of elution will be R (Strongly negative), followed by Q (Weakly negative) and the last P (Positively charged) peptides. 9. When the van der Waals attraction between two atoms exactly balances the repulsion between their two electron clouds, the atoms are said to be in van der Waals contact. Each type of atom has a van der Waals radius at which it is in van der Waals contact with other atoms. The van der Waals radius of an H atom is 0.1 nm, and the radii of O, N, C, and S atoms are between 0.14 and 0.18 nm. Two covalently bonded atoms are closer together than two atoms that are merely in van der Waals contact. For a van der Waals

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interaction, the internuclear distance is approximately the sum of the corresponding radii for the two participating atoms. Thus the distance between a C atom and an H atom in van der Waals contact is 0.27 nm, and between two C atoms is 0.34 nm. In general, the van der Waals radius of an atom is about twice as long as its covalent radius. For example, a C—H covalent bond is about 0.107 nm long and a C—C covalent bond is about 0.154 nm long. 10. Glycolysis occurs in most organisms in the cytosol of the cell. The most common type of glycolysis is the Embden–Meyerhof–Parnas (EMP pathway). In mammalian cells, both peroxisomes and mitochondria contain a beta-oxidative pathway. Beta-oxidation is a key pathway for the breakthrough of fatty acids. In yeasts and plants, fatty acid oxidation occurs uniquely in peroxisomes. Conversion of pyruvate to acetyl CoA, the citric acid cycle takes place in the matrix of the mitochondria. Oxidative phosphorylation is a process occurring in the intermembrane space of the mitochondria that results in the formation of ATP from the flow of electrons to oxygen. 11. The main components of the innate immune system are (1) physical epithelial barriers, (2) phagocytic leukocytes, (3) dendritic cells, (4) a special type of lymphocyte called a natural killer (NK) cell, and (5) circulating plasma proteins. Memory B-cells is NOT a principal component of innate immunity 12. Electrospray ionization mass spectrometry (ESIMS) was employed to monitor the heme release and the conformational changes of myoglobin (Mb) under different solvent conditions, and to observe ligand bindings of Mb. ESI-MS, complemented by circular dichroism and fluorescence spectroscopy, was used to study the mechanism of acid- and organic solvent-induced denaturation by probing the changes in the secondary and the tertiary structure of Mb. The light microscope can be used to provide information about the activity of cells and to look at very small structures such as nanostructures. 13. A trickling filter, is a fixed-bed, biological reactor that operates under (mostly) aerobic conditions. Pre-settled wastewater is continuously ‘trickled’ or sprayed over the filter. As the water migrates through the pores of the filter, organics are aerobically degraded by the biofilm covering the filter material.

11

14. Paracrine signaling is a form of cell-to-cell communication in which a cell produces a signal to induce changes in nearby cells, altering the behavior of those cells. Autocrine : mode of hormone action to which hormons bind to receptors on to the cell and affects the cell that produces it. Paracrine : describes hormone action where hormones are released from cells and bind to receptor on nearby cells and affects their function. Hence the answer is Paracrine. 15. In the presence of physiological concentrations of monovalent and divalent cations, actin monomers polymerize into filaments. The actin filament is structurally polarized and the two ends are not identical. 16. The standard error (SE) of a parameter is the standard deviation of its sampling distribution or an estimate of the standard deviation ]. If the parameter or the statistic is the mean, it is called the standard error of the mean (SEM). The standard error of the sample mean depends on both the standard deviation and the sample size, by the simple relation SE = SD/√(sample size). The standard error falls as the sample size increases, as the extent of chance variation is reduced—this idea underlies the sample size calculation for a controlled trial, for example. By contrast the standard deviation will not tend to change as we increase the size of our sample. 17. In situ hybridization (ISH) is a powerful technique for localizing specific nucleic acid targets within fixed tissues and cells, allowing you to obtain temporal and spatial information about gene expression and genetic loci. 18. Construction of the phylogenetic tree : (1) Distance methods (2) Character methods: (a) Maximum parsimony and (b) Maximum likelihood. Distance based methods (eg. Neighbour Joining (NJ)): find a tree such that branch lengths of paths between sequences (species) fit a matrix of pairwise distances between sequences. Maximum Parsimony : find a phylogenetic tree that explains the data, with as few evolutionary changes as possible. Maximum likelihood : find a tree that maximizes the probability of the genetic data given the tree.

12

SOLVED PAPER 2018 2

to 360° curls, shepherd’s crooks, which are formed by the host’s root hairs. Purified bacterial signal molecules, the nodulation factors (NFs), which are lipochitooligosaccharides, induce root hair deformation in the appropriate host legume and have been proposed to be a key player in eliciting root hair curling.

19. cos x + 2cos x + 1 = 0 Let

2

cos x = y

∴ y + 2y + 1 = 0 ∴ ∴ ∴

2

(y + 1) = 0 y = −1

cos x = – 1

–1



x = cos (–1)



x = π – cos (1)

–1

=π–0



x=π



x = 180°

∴ Option (c) is correct. 20. The pyrethrins are a class of organic compounds (mixture of six chemicals) normally derived from Chrysanthemum cinerariifolium flowers that have potent (toxic) insecticidal activity by targeting the nervous systems of insects. Pyrethrins are commonly used to control mosquitoes, fleas, flies, moths, ants, and many other pests. 4 −6 ⎤ 21. The given matrix is A = ⎡ ⎢ −3 2 ⎥ ⎣ ⎦



A = 4(2) − (−3)(−6) = 8 – 18 = – 10

23. A = εcl, l = 2cm, A = 0.56, ε = 5600 m–1cm–1 ∴ concentration of tryptophan = A/(εl) 0.56 = 5 × 10–5 M 2 × 5600 = 50 × 10–6 M = 50 μm

=

24. When stem cells undergoes the asymmetric cell divisions then these gives rise to one cell of the same type while another differentiated into a specialized cell types such as neurons etc. Thus when one stem cells undergoes 10 asymmetric cell divisions only one cell will have same property of stem cell while the rest cells will have any specified cell type like neurons. 25. DNA have 4 bases = C, G, A, T Now, restriction enzyme recognizes 6 bp sequence. So, Avg. length (bp) of fragment generated = 46 = 4096 (bp) 26. A critical step in establishing a successful nitrogenfixing symbiosis between rhizobia and legume plants is the entrapment of the bacteria between root hair cell walls, usually in characteristic 180°

27. In essence, all endogenous pyrogens are cytokines, molecules that are a part of the immune system. They are produced by activated immune cells and cause the increase in the thermoregulatory set point in the hypothalamus. Major endogenous pyrogens are interleukin 1 ( α and β ) and interleukin 6 (IL-6). Minor endogenous pyrogens include interleukin-8, tumor necrosis factor-β, macrophage inflammatory protein-α and macrophage inflammatory proteinβ as well as interferon- α, interferon- β , and interferon-γ. Tumor necrosis factor-α also acts as a pyrogen. It is mediated by interleukin 1 (IL-1) release. These cytokine factors are released into general circulation, where they migrate to the circumventricular organs of the brain due to easier absorption caused by the blood–brain barrier’s reduced filtration action there. The cytokine factors then bind with endothelial receptors on vessel walls, or interact with local microglial cells. When these cytokine factors bind, the arachidonic acid pathway is then activated. 28. Small nucleolar RNAs (snoRNAs) are a class of small RNA molecules that primarily guide chemical modifications of other RNAs, mainly ribosomal RNAs, transfer RNAs and small nuclear RNAs. In animals, small RNA molecules termed PIWIinteracting RNAs (piRNAs) silence transposable elements (TEs), protecting the germline from genomic instability and mutation. piRNAs have been detected in the soma in a few animals, but these are believed to be specific adaptations of individual species. MicroRNAs (miRNA) regulate the expression of protein-coding genes in animals and plants. They function by binding to mRNA transcripts with complementary sequences and inhibit their expression. The level of sequence complementarity between the microRNA and mRNA transcript varies between animal and plant systems. Owing to this subtle difference, it was initially believed that animal and plant microRNAs act in different ways. Recent developments revealed that, although differences still remain in the two kingdoms, the differences are smaller than fi rst

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thought. It is now clear that both animal and plant microRNAs mediate both translational repression of intact mRNAs and also cause mRNA degradation. A small nuclear RNA (snRNA) is one of many small RNA species confined to the nucleus; several of the snRNAs are involved in splicing or other RNA processing reactions. Many snRNAs of eukaryotic cells are involved in the maturation of primary transcripts of messenger RNAs (mRNAs). Certain snRNAs are involved in the formation and function of the spliceosome, a large ribonucleoprotein structure in which pre-mRNA splicing takes place. 29. Ab initio gene finding in eukaryotes, especially complex organisms like humans, is considerably more challenging for several reasons. First, the promoter and other regulatory signals in these genomes are more complex and less well-understood than in prokaryotes, making them more difficult to reliably recognize. Two classic examples of signals identified by eukaryotic gene finders are CpG islands and binding sites for a poly(A) tail. Second, splicing mechanisms employed by eukaryotic cells mean that a particular proteincoding sequence in the genome is divided into several parts (exons), separated by non-coding sequences (introns). (Splice sites are themselves another signal that eukaryotic gene finders are often designed to identify.) A typical proteincoding gene in humans might be divided into a dozen exons, each less than two hundred base pairs in length, and some as short as twenty to thirty. It is therefore much more difficult to detect periodicities and other known content properties of protein-coding DNA in eukaryotes. 30. L-arginine-dependent production of reactive nitrogen intermediates (RNls: nitric oxide, nitrite, and nitrate) by mammalian macrophages has been proposed to occur via an L-arginine oxidative deimination pathway and is known to be responsible for certain antineoplastic and antimicrobial effect or functions. 31. Assertion : The association constant in water for the G-C base pair is three times lower than that for the A-T base pair. Reason : There are three hydrogen bonds in the G-C base pair and two in the A-T base pair. Assertion is false. Reason is correct. The ‘‘reason statement’’ is itself an explanation why the assertion is false. 32. Porter fragmented rabbit IgG with the proteolytic enzyme papain in the presence of the reducing agent cysteine. Papain hydrolyzed peptide bonds

13

in IgG and produced three fragments (I, II, and III). The three fragments had similar molecular weights (50 kD) but different charges. Two of the three fragments were identical and retained the ability to bind antigen. These two fragments were called the Fab fragments (for fragments of antigen-binding). Because the intact IgG is bivalent and the two Fab fragments each could bind antigen, Porter concluded that the Fab fragments must be univalent. The third fragment produced by papain digestion did not bind with antigen and crystallized during cold storage. Porter called this piece the Fc fragment (for fragment crystallizable). Thus the ratio of Fab to Fc is 2:1. Edelman confirmed Porter’s results by cleaving and electrophoresing human IgG into two antigenically different fractions equivalent to the two fragments from rabbit IgG (Figure 4-2). In similar studies, Alfred Nisonoff used pepsin, which hydrolyzes different sites on the IgG molecule than does papain. IgG treated with pepsin yielded one large fragment with a molecular weight (100 kD) double that of one Fab fragment, and many small fragments. Nisonoff called the large fragment F(ab9)2. This fragment also could bind antigen, but unlike the Fab fragment, it led to a visible serologic reaction. It had both of the antigen-binding sites of IgG (the chains remained linked) and could be treated further with reducing agents to yield two Fab-like fragments called Fab9. Collectively, the two enzymes cleave at about the same region of the IgG molecule. Papain splits the molecule on one side, and pepsin on the other side, of the bond that holds Fab fragments together.

14

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33. Endogenous antigens are processed by cytosolic pathway. In the cytoplasm antigen is first processed by proteasomes and then smaller peptides are transported inside endoplasmic reticulum by TAP protein. Protein antigens in the cytosol are processed by proteasomes in the class I MHC pathway. In the class II MHC pathway, extracellular protein antigens are endocytosed into vesicles and processed. Invariant chain in endoplasmic reticulum is involved in transport of peptides and loading of class II MHC pathway. 34. Distribution of superoxide dismutase, catalase and peroxidase in prokaryotes with different O2 tolerances. Group Obligate aerobes and most facultative anaerobes (e.g. Enterics) Most aerotolerant anaerobes (e.g. Streptococci) Obligate anaerobes (e.g. Clostridia, Methanogens, Bacteroides)

Superoxide Cata- Perodismutase lase xidase

The following are true with regard to anaerobic respiration in bacteria: (P) The final electron acceptor is an inorganic substance other than molecular oxygen (R) The number of ATP molecules produced per glucose molecule is less than that produced in aerobic respiration 36. Curve II is bingham plastic follows the τ = τ0 + k

du dy

Now, for rest type i.e. I, III, IV they follows the n

⎛ du ⎞ τ= k⎜ ⎟ ⎝ dy ⎠ n = flow behaviour index Now if n = 1 then fluid is newtonian which is curve I. If n < 1 then fluid is pseudoplastic which is curve III.

Yes

If n > 1 then fluid is Dilatant which is curve IV.

Yes No

Now, curve for newtonian (I) will be standard line passing through origin. Curve for pseudoeplastic (III) will be concave downward passing through origin.

Yes

No

No

Yes

No No

35. For aerobic, the cells involved include those in most organisms and body cells; however, anaerobic may occur in muscle cells and red blood cells, and is some types of bacteria and yeast. Lactic acid is produced during Anaerobic, and none is produced during aerobic. A high amount of energy is produced in aerobic respiration with 38 glucose molecules, and low energy with only 2 glucose molecules during anaerobic. In addition, the reactants for aerobic respiration is both oxygen and glucose, yet for anaerobic the reactant is just glucose. The reaction site in the cell for aerobic is in the cytoplasm or mitochondria, and just in the cytoplasm for anaerobic respiration. Both aerobic and anaerobic respiration involves a first stage called glycolysis. The additional stage in anaerobic is fermentation, but in aerobic there is the Krebs cycle and electron transfer chain. Finally, combustion is complete in aerobic respiration, and incomplete during anaerobic respiration.

Curve for Dilatant (IV) will be concave upward through passing through origin. 37. DNA has net negative charge due to the presence of phosphate groups. Lys, Arg and Gln are basic amino acids whereas Glu is acidic amino acid. Arg (Q) attracts more phosphate and forms extensive hydrogen bonding with its five H-bond donors present in its structure that stabilizes the interactions in DNA-protein complexes reflecting the highest percentage frequency in the graph. Followed by Arg the next higher frequency is shown for Lys (P) due to its basic nature and hydrogen bonding capacity which is followed by the frequency of Gln (R ). Glu (s) due to negative charge has least frequency. 38. It can’t be autosomal dominant as carriers are not possible in such cases. It can’t be X-linked dominant. If that was the case, all female carriers actually would have been affected. It can’t be autosomal recessive as affected children in such cases are born when both the parents are carriers. Answer is X-Linked Recessive as most of the following traits of X-linked recessive inheritance are fulfilled by the given pedigree chart: 1. More males than females are affected 2. Affected sons are usually born to unaffected mothers, thus the trait skips generations

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15

3. Approximately 1/2 of carrier mothers’ sons are affected

42.

Producing Organism

Citric acid

Aspergillus niger

Cellulase

Trichoderma viride

Vitamin B12 Propionibacterium freudenreichii Butanol

Clostridium acetobutylicum

4

= 1 × 0.3 + 2 × 0.4 + 3 × 0.2 + 4 × 0.1 = 0.3 + 0.8 + 0.6 + 0.4 = 2.1 43. The given series 1, r, r2, r3, ... ∞ is infinite G.P. series with common ratio p(r) Now

40. The starting point for capping with 7methylguanylate is the unaltered 5′ end of an RNA molecule, which terminates at a triphosphate group. This features a final nucleotide followed by three phosphate groups attached to the 5 ′ carbon. The capping process is initiated before the completion of transcription, as the nascent pre-mRNA is being synthesized. 1. One of the terminal phosphate groups is removed by RNA triphosphatase, leaving a bisphosphate group (i.e. 5′(ppN)[pN]n); 2. GTP is added to the terminal bisphosphate by mRNA guanylyltransferase, losing a pyrophosphate from the GTP substrate in the process. This results in the 5′–5′ triphosphate linkage, producing 5′(Gp)(ppN)[pN]n;

41.

3

E(x) = 1 × p (x = 1) + 2 × p (x = 2) + 3 × p (x = 3) + 4 × p (x = 4)

5. All daughters of affected fathers are carriers Product

2

p( x) 0.3 0.4 0.2 0.1

4. It is never passed from father to son 39.

1

x

∴ ∴

S = n

a 1−r

1 1−r 1.5 – 1.5 r = 1 0.5 = 1.5 r 1.5 =



r=

1 3

1 1 + + ...∞ = 1.5 3 9 Now sum of 1 + 2r + 3r2 + .... + ∞



1+

...(i)

= 1 + 2 ⎛⎜ 1 ⎞⎟ + 3 ⎛⎜ 1 ⎞⎟ + ...∞ ⎝3⎠ ⎝9⎠ = 1.5 × 1.5

( from eq. (i) )

3. The 7-nitrogen of guanine is methylated by mRNA (guanine-N7-)-methyltransferase, with S-adenosyl-L-methionine being demethylated to produce S-adenosyl-L-homocysteine, resulting in 5′(m7Gp) (ppN)[pN]n (cap-0); 4. Cap-adjacent modifications can occur, normally to the first and second nucleotides, producing up to 5′(m7Gp) (ppN*) (pN*)[pN]n (cap-1 and cap-2) 5. If the nearest cap-adjacent nucleotide is 2′-Oribose methyl-adenosine (i.e. 5 ′ (m7Gp) (ppAm)[pN]n), it can be further methylated at the N6 methyl position to form N 6 methyladenosine, resulting in 5 ′ (m7Gp) (ppm6Am)[pN]n.

45. (Separation factor) Extraction factor is one distribution ratio divide by another, Here, partition coefficient of antibiotic is 170, and distribution ratio of water to solvent is 85.

∫ ( x + 3)( x + 1)dx

46.

= 2.25 Here Given, N0 = 1010; Nt = 1; Kd = 1 min–1

Therefore

=

1

∫ ⎡⎣ x 0

2

=

Partition coefficient of antibiotic Distribution coefficient of water to solvent

=

170 =2 85

qo2 = 0.4 h–1 = 0.4/3600 = 1.111 × 10–4 sec–1

+ 4 x + 3 ⎤⎦ dx

⎡ x3 4 x 2 ⎤ + 3x⎥ = ⎢ + 2 ⎣⎢ 3 ⎦⎥ 0

⎡1 ⎤ = ⎢ + 2 + 3⎥ ⎣3 ⎦ = 5.33

t = ln(1010/1) = 23.03 min.

∴ Extraction factor

1 0

ln (N0/Nt) = Kdt

44.

x = 12 g/L

∴ ∴ ∴ ⇒

OUR = qo2 X = 1.333 × 10–3

ΔC = 8 mg/L = 8 × 10–3 gm/L OUR = kLaΔC KLa =

1.333 × 10 −3 = 0.167 8 × 10 −3

16

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S=

47.

k SD μ−D

D = Dilution rate =

∴ 48.

S=

1 × 0.1 =1 0.2 − 0.1

but

F 100 ml = = 0.1 V 1000 ml

⎛x ⎞ ln ⎜ t ⎟ = n ln 2, ⎝ xo ⎠ We know that number of cycles n = t/td t = time given = 72 hrs t d = doubling time xt = 106 cells/ml xo = 105 cells/ml So,

td =

Because final concentration is half of the initial concentration for both types of enzymes. Therefore



ln c1 c0 K t1 2 = 1 11 2 ln c2 c0 K 2 t2

t11 2

∴ ∴

=

K1 K2

t11 2

=

0.03 0.005

t11 2

=6

t21 2 t21 2 t21 2

Slope = Km/Vm, as Vm = constant

K m2 = 3 K m1 Now,

49.

Data for immobilized enzyme K2 = 0.005, half life = t21 2



53.

t ln 2 = 72 × 0.6931 = 21.67 hrs 2.303 ⎛x ⎞ ln ⎜ t ⎟ ⎝ xo ⎠

C6H12O6 → 2C2H5OH + 2CO2 1 mol of glucose produces → 2 mole of ethanol MW of glucose = 180, MW of ethanol = 46, As per equation 180 gm glucose produces = 2 × 46 i.e., 92 gm of ethanol Hence, 1 gm of glucose will produce 92/180 gm of ethanol Therefore 200 gm glucose will produce = 200 × (92/180) gm ethanol = 102.22 gm ethanol 50. Let x = initial culture volume in ml. glucose addition = 200 ml/h, so in 2 hr total 400 ml glucose added. Final culture volume = 1000 = x + 400 ⇒ x = 600 ml 51. For, affine gap scoring scheme, penalty is calculated using following formula = n (A + B) where n = number of gaps A = gap initiation penalty = 1 B = gap extension penalty = 0.1 Total gap penalty = 2(1 + 0.1) = 2.2 Now in the given structure the total number of matches = 4. ∴ Score of Alignment = Match score – Total gap penalty = (4 × 1) – 2.2 = 4 – 2.2 = 1.8 52. Data for soluble enzyme : K1 = 0.03, half life = t11 2

ln c1 c0 ln c2 c0 = 1

⇒ ⇒

K m2 = K ⎡1 + [ I ] ⎤ = 3 K m1 ⎥ m1 ⎢ ⎣ KI ⎦ 2KI = [I] = 10 nM KI = 5 nM

54. From the diagram, we observe that one of the enzymes (EcoRI) used is a sticky end enzyme and the other (HaeII) is a blunt end cutter. For incorporation of the radioactive ATP, Klenow fragment needs a free 3’-OH end and a template to read. This will only be provided at the sticky ends generated by EcoRI enzyme. EcoRI will generate two such ends on two different DNA fragments. Hence we will finally see two bands on the X-ray film. S = 1 kg/l

55.

l = 2 × 10–6 mt D = 1 × 10–6 mt

π 2 Dl 4 = 1.5708 × 10–18 m3

Volume of bacterium =

= 1.5708 × 10–15 kg [  1 m3 = 1000 kg] = 1.5708 × 10–12 gm 15% of this is proteins = 2.356 × 10–13 gm. Now, average weight of one protein = 50 kDa = 50 × 1000 × 1.6 × 10–24 = 8 × 10–20 gm.



No. of proteins =

2.356 × 10 −13 8 × 10 −20

= 2.945 × 106 = 2945000

2

Solved Paper - 2019

Solved Paper - 2019

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For X 1. Until Iran came along, India had never been defeated in kabaddi 2. The fishermen, to whom the flood victims owed their lives, were rewarded by the government. 3. Initial Stage

Time taken = t Distance x = velocity × time 80t Ÿ t

x

x 80

...(1)

For y time taken = t Distance y = 100t

h

1.10h

h

t

y 100 10

...(2)

x + y = 540 km

...(3)

From (1) and (2)

4.

5.

6.

7.

r 1.10r After increasing 10% for radius & height cone, 1 2 Vi r h 3 1 1.331 2 2 r h Vf 1.1r 1.1h 1.1 3 3 Percentage increase 1.331 2 Sr h Vf  Vi 3 u 100 u 100 1 2 Vi Sr h 3 1.331 1 100 33.1% 33 1 Number are 2, 4, 5, 7, and 10. The correct order of arrangement 10, 5, 4, 7, and 2 Thus an arrangement follows given three condition. 'Some students were not involved in the strike'. If this statement is true, then the logically necessary conclusion is : (4) Some who were not involved in the strike were students. 10. Based on the given paragraph, the prestige of a head-hunter depended upon: the number of head he could gather. Second number from right = 7 N Train A 80km/h

x

t

x 80

y 100

x

0.8y 0.8

x + y = 540 Ÿ x = 0.8y + y = 540 1.8y = 540 Ÿ y = 300km Time taken y

y 100

300 100

3hrs 3h

Time at which these trains = 7.00 AM + 3hrs = 10.00 AM 8. Total population = 1400 million Number of people whose having own mobile phones = 70% of 1400 = 0.7 × 1400 = 980 million Number of people whose accesses the internet = 294 million Number of people who buy goods from e-commercial portals = Half of internet users 294 2

147 million

Percentage buyers

147 million 1400 million illi

100

10. 10.5%

7 AM y

Train B 100km/h S

9. Based on the given para, NOT correct pairings is : (C) baaj, institution 10. From the given passage, the statements can be inferred as: 'A reduction in interest rates on small saving schemes follow only after a reduction in repo rate by the Reserve Bank of India'.

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9

1. Oxygen's atomic weight is 16.00 amu.

7. In mathematics, a geometric series is a series with a constant ratio between successive terms.

Oxygen's molecular weight is 32.00 amu.

In General we write a Geometric Sequence like this:

1 mole of oxygen is 6.02 u 1023 atoms of oxygen

{a, ar, ar2, ar3, ... }

1 amu = 1.661 u 10

Where : a is the first term, and r is the factor between the terms (called the "common ratio")

–24

g;

Then, Mass of 1 kmol of oxygen molecules is = 32.00 amu u 1000 u 1.661 u 10–24 u 6.02 u 1023 = 319.97504 u 100 = 31997.504 g or 31998.00 g 2. PROSITE is a protein database. It consists of entries describing the protein families, domains and functional sites as well as amino acid patterns and profiles in them. PROSITE offers tools for protein sequence analysis and motif detection. The database ProRule builds on the domain descriptions of PROSITE. 3. Given data is 12, 10, 16, 8, 90, 50, 30, 24. The proper ordered data set [increasing order] is 8, 10, 12, 16, 24, 30, 50, 90 ? Median value = Average of middle most observations 16

24 2

20

4. The successful replication of human immunodeficiency virus type 1 (HIV-1) requires viral genome reverse transcription to generate proviral DNA and integration into the target cell genome. These events are carried out by the viral encoded enzymes, reverse transcriptase (RT) and integrase (IN), respectively. The RT is a heterodimer of p66 and p51 subunits, which can copy both RNA and DNA templates thus producing double stranded proviral DNA. 5. Match of the human diseases in Group I with the causative agents in Group II is given bellow: Group-I P. Amoebiasis

Group-II 3. Entamoeba histolytica

Q. African sleeping 4. Trypanosoma gambiense sickness R. Kala azar

1. Leishmania donovani

S. Chagas' disease

2. Trypanosoma cruzi

6. In biochemical reactions total mass, number of atoms of each element and total energy remain conserved however Total moles need NOT be conserved in a biochemical reaction.

8. The eukaryotic cell cycle consists of four distinct phases: G1 phase, S phase (synthesis), G2 phase (collectively known as interphase) and M phase (mitosis and cytokinesis). DNA synthesis in eukaryotes (chromosome replication) occurs during S phase of the mitotic cell cycle. 9. First, calculate the number of electrons each atom of the molecule can donate to reach a full valence shell. For example, Hydrogen = 1, Carbon = 4, Oxygen = – 2 (since it receives electrons instead of donating), Nitrogen = – 3. This value is the same as the charges of the elements. After finding these values, calculate the sum of these values for all atoms of the molecule. For example, for ethanol (C2H4O2) = (2 × 4) + (4 × 1) + (2 × – 2) = 8. Secondly, divide this value by the number of Carbon atoms in the molecule. For acetic acid (C2H4O2 i.e. 2 Carbon atoms), Degree of reduction = 8 / 2 = 4. 10. Most of the commercially available culture media include phenol red as a pH indicator, which allows constant monitoring of pH. During the cell growth, the medium changes color as pH is changed due to the metabolites released by the cells. At low pH levels, phenol red turns the medium yellow, while at higher pH levels it turns the medium purple. Medium is bright red for pH 7.4, the optimum pH value for cell culture. 11. Antibody is not a part of the human nonspecific defense system however interferon, mucous and saliva are parts of the human nonspecific defense system. In animals, there are two types of defenses against foreign invaders: specific and nonspecific. Specific immune responses can distinguish among different invaders. The response is different for each invader. With nonspecific defenses, the protection is always the same, no matter what the invader may be. Whereas only vertebrates have specific immune responses, all animals have some type of nonspecific defense. Examples of nonspecific defenses include physical barriers, protein defenses, cellular defenses, inflammation, and fever.

10

Solved Paper - 2019

x 8 x 6 x 2 64 l lim 12. lim x 8 x 8 x 8 x 8 lim im x 8 8 8 16 x

Sample value = 10 PL Activity of 10 PL Sample = 5 Pmol min–1 Total No. of Sample of 10uL each in 1 ml 1000 100 10 Total Activity of 1ml Sample = 5 × 100

8

13. The general definition of the heat transfer coefficient is: h = q /'T ; Where: q : heat flux, W/m2; i.e., thermal power per unit area, q = dQ/dA h : heat transfer coefficient, W/(m2 · K) 'T : difference in temperature between the solid surface and surrounding fluid area, K It is used in calculating the heat transfer, typically by convection or phase transition between a fluid and a solid. The heat transfer coefficient has SI units in watts per square meter kelvin: W/(m2K). 14. Nonsense mutation : changes an amino acid to a STOP codon, resulting in premature termination of translation. Missense mutation : changes an amino acid to another amino acid. This may or may not affect protein function, depending on whether the change is "conservative" or "nonconservative," and what the amino acid actually does. "Silent" mutation : does not change an amino acid, but in some cases can still have a phenotypic effect, e.g., by speeding up or slowing down protein synthesis, or by affecting splicing. Frameshift mutation : Deletion or insertion of a number of bases that is not a multiple of 3. Usually introduces premature STOP codons in addition to lots of amino acid changes. A synonymous mutation is a change in the DNA sequence that codes for amino acids in a protein sequence, but does not change the encoded amino acid. 15. Enzymes do not change the equilibrium state of a biochemical reaction. 'G and Keq remain the same. Instead, the enzyme reduces the activation energy needed for the reaction to proceed, and thus increase the rate of reaction. Say for example, Keq = 1, and you start with reactants only. Under normal conditions it may take 100 years, but eventually the concentration of reactants will equal the concentration of products. Now start over, but add an enzyme. In this case we reach equilibrium in 100 seconds, but it still results in a 1:1 ratio of reactants to products. 16. Given data : Protein Conc. of crude enzyme

10

mg ml

? Specific Activity

17.

18.

19. 20.

= 500 Pmol min–1. Total Activity of1ml sample Pr otein concontration

500 50 mg 1 10 Internal reserves are catabolized during endogenous metabolism in a batch bacterial cultivation. Many bacteria are able to accumulate polymers that are considered to function as energy reserves. These compounds may accumulate during growth or at the end of the growth phase and provide either energy or a source of a component no longer available from the environment. Bacillus thuringiensis (Bt) crops are plants genetically engineered (modified) to contain the endospore (or crystal) toxins of the bacterium, Bt to be resistant to certain insect pests. Currently, the most common Bt crops are corn and cotton. The crystal, referred to as Cry toxins, is proteins formed during sporulation of some Bt strains and aggregate to form crystals. Such Cry toxins are toxic to specific species of insects belongs to orders: Lepidoptera, Coleoptera, Hymenoptera, Diptera, and Nematoda. Intracellular carbon source can NOT be a limiting substrate if Monod's growth kinetics is applicable. The wave equation is an important second-order linear partial differential equation for the description of waves — as they occur in classical physics - such as mechanical waves (e.g. water waves, sound waves and seismic waves) or light waves. It arises in fields like acoustics, electromagnetics, and fluid dynamics. The wave equation in one space dimension can be written as follows:

This equation is typically described as having only one space dimension x, because the only other independent variable is the time t. Nevertheless, the dependent variable u may represent a second space dimension, if, for example, the displacement u takes place in y-direction, as in the case of a string that is located in the x-y plane.

Solved Paper - 2019

11

21. Paul Andersen describes mechanisms that increase the genetic variation within a population. He begins by discussing how horizontal transfer can move genetic material between bacteria. Transformation, transduction, and conjugation in bacteria are all included. He also explains how crossing over, random assortment, and random fertilization can maintain genetic variation in eukaryotes. 22. Method - I : AT = – A

A is skew – symmetric 0

Given A

P 0

A A

?

6

0

6

P

0

6 0

0 P

0

6

6

P

0

0

24. The number of possible rooted trees in a phylogeny of three species is 3. A rooted phylogenetic tree is a directed tree with a unique node - the root corresponding to the (usually imputed) most recent common ancestor of all the entities at the leaves of the tree. The root node does not have a parent node, but serves as the parent of all other nodes in the tree. The root is therefore a node of degree 2 while other internal nodes have a minimum degree of 3. 25. A DNA microarray (also commonly known as DNA chip or biochip) is a collection of microscopic DNA spots attached to a solid surface. Scientists use DNA microarrays to measure the expression levels of large numbers of genes simultaneously or to genotype multiple regions of a genome. 26. Given D.E is

x y

Ÿ y.dy = x.dx o variable - seprate D.E.

dx x2 2

y.dy y y2 2

27. For plasmid manipulations other site-directed mutagenesis techniques have been supplanted, An example of these techniques is the Quik change method, wherein a pair of complementary mutagenic primers are used to amplify the entire plasmid in a thermocycling reaction using a highfidelity non-strand-displacing DNA polymerase such as pfu polymerase. The reaction generates a nicked, circular DNA. The template DNA must be eliminated by enzymatic digestion with a restriction enzyme such as DpnI, which is specific for methylated DNA.

= (–100)[0.25] + (0)(0.5) + (100)(0.2) + (500)(0.05) ŸP = –6

23. Tetracycline antibiotics inhibit bacterial protein synthesis by preventing the association of aminoacyl-tRNA with the bacterial ribosome.

x

y2 1  2 2

? Mean of x = 6x.p(x)

A = –A

dy dx

1 1 x2 c c Ÿ 2 2 2 2 2 2 Ÿ x = y – 1 Ÿ y = x2 + 1 0

28. Given table is clearly discrete probability distribution; since 6P (x) = 1; where x is discrete R.V.

0 P

T

Given x = 0, y = 1

c

...(1)

= – 25 + 20 + 25 Ÿ E(x) = 20 ? E(x2) = 6x2.p(x) = (–100)2[0.25] + (0)2[0.5] + (100)2[0.2] + (500)2(0.05) 1000 2 5 1000 250000 4 10 100 = 2500 + 2000 + 12500 = 17000 ? Variance of x = E(x2) – [E(x)]2 = 17000 – (200)2 = 16,600 ? S.D of x

var iance

16600

129

29. Given data: min. delectable Absorbance = 0.02 Molar extinction co.effi. of protein = 10000 Lmol–1 cm–1, length of cuvette path length = 1 cm. As per the bur Lambert law. A = ecl. C = minimum of Protein in Sample that can be measured reliably. A 0.02 C 2 10 6 mol m / lit ell 10000 1 C = 2 μM 30. Molecular mass of Protein = 22 kDa = 22000 Da Avg. molecular col of Amino Acid = 110 Da ? No. of Amino Acid in given Protein

22000 200 20 110 So, No. of nucleotide = 200 × 3 = 600. 600 0.6 kb So, the size of the cDNA = 1000

12

Solved Paper - 2019

31. Given 1 C6H12O6 + 0.48NH3 + 3O2 o 0.48C6H10O3N + 3.12 CO2 + 4.32 H2O Moles: MW:

1

0.48

3

180

0.48

3.12

4.32

144 Group-I

Working volume = 100000 L.

Group-II

Amount of yeast biomass. = 100000 L × 50 g/L

(Spectroscopic methods) (Biomolecular applications)

= 50,00,000 gm

P. Infrared

1. Identification of functional groups Q. Circular Dichroism 2. Determination of secondary structure R. Nuclear Magnetic 4. determination of 3-D

= 5000 kg Total moles of yeast biomass 5000 34.722 kmol. 34 144 Now, 1 kmol of yeast biomass required 1 2.083 kmol of glucose 0.48 ? 34.722 kmol of b/yeast biomass required

Resonance

structure

36. ? Given data : Pressure drop 'p = 2 atm. Inlet feed Pressure pi = 3 atm. Filtrate Pressure pf = 1 atm.

= 34.722 × 2.083

retandate Pressure pr = pi – 'p = 3 – 2 = 1 atm.

= 72.33 kmol glucose.

Now average transmembrane pressure drop

? Total amount of glucose required = 72.33 × 180

Pi  Pr  Pf 2 = 2 – 1 = 1 atm.

32. Correct Match of instruments in Group I with their corresponding measurements Group I (Measurement)

Group II (Instrument)

P. Manometer

2. Pressure difference

Q. Rotameter

4. Air flow rate

R. Tachometer

1. Agitator Speed

S. Haemocytometer

3. Cell Number

33. Histo-blood group ABO system transferase is an enzyme with glycosyltransferase activity, which is encoded by the ABO gene in humans. It is ubiquitously expressed in many tissues and cell types. ABO determines the ABO blood group of an individual by modifying the oligosaccharides on cell surface glycoproteins. Variations in the sequence of the protein between individuals determine the type of modification and the blood group. 34. In eukaryotes, the 5c cap (cap-0), found on the 5c end of an mRNA molecule, and consists of a guanine nucleotide connected to mRNA via an unusual 5c to 5c triphosphate linkage. This guanosine is methylated on the 7 position directly after capping in vivo by a methyltransferase. It is referred to as a 7-methylguanylate cap, abbreviated m 7G. 35. Correct Match of methods in Group I with the applications in Group II.

31 1 2

'Pm

= 13019.4 kg,

37. Pro amino acid residues will destabilize an D-helix when inserted in the middle of the helix because Proline does not have the D—NH group that acts as a stabilizing H-bond donor in the middle of the helix and Proline is not able to adopt the ideal I and \ angles for an D = helix. 38. f(t) = t2 + 2t + 1 Ÿ L[f(t)] = L[t2 + 2t + 1] = L[t2] = 2L[t] + L[1]; Using linearity property

2!

2

3

2

s

s

L t t 39.

1 ;sin ceL t n s 2

2

3

2

s

Purple Pp 3 Purple

s

n n! s

n 1

1 . s

X

Purple

?

Pp

:

1 White

40. The statements: (Q) E-value < 10–10 indicates sequence homology & (R) A higher BLAST score indicates higher sequence similarity, are CORRECT when a protein sequence database is searched using the BLAST algorithm.

Solved Paper - 2019

13

41. The CORRECT statements about the function of fetal bovine serum in animal cell culture are (P) It stimulates cell growth; (Q) It enhances cell attachment; (R) It provides hormones and minerals.

47. Given data: Difference in conc. of solute Between two compartment = 1.6 fold C2 C1

42. Using trapezoid rule, we have b

d f x dx a

h 2

y0

yn

2 y1

y2

... y n 1

Given y0 = 0.37, y1 = 0.51, y2 = 0.71, y3 = 1.0, y4 = 1.40, 1 y5 = 1.95, y6 = 2.71 & h = 3 1

? From 1 l;

f x dx 1

1 3 2

43. The hexapeptide P has an isoelectric point (pI) of 6.9. Hexapeptide Q is a variant of P that contains valine instead of glutamate at position 3. The two peptides are analyzed by polyacrylamide gel electrophoresis at pH 8.0. The correct statement is P will migrate faster than Q towards the anode. 44. Proportional Control action is the simplest and most commonly encountered of all continuous control modes. In this type of action, the controller produces an output signal which is proportional to the error. Hence, the greater the magnitude of the error, the larger is the corrective action applied. 45. The statement 'Hydrogen bonds are weaker than van der Waals interactions' is incorrect. Van der Walls intermolecular forces form weak bonds compared to hydrogen bonds. 46. Given data : Phenol Conc. in 10000 ppm,

= 10000 mgL –1 = 10 gL–1 = SR Phenol Conc. out = 10 ppm, = 10 mg L–1 = 0.01 gL = S –1

Input feed rate of wastewater = 80 Lh–1. = F μm = 1 h–1, Ks = 100 mg L–1, kd = 0.01 h–1 = 0.1 gL–1

0.091

S 0.01 1 Ks S 0.1 0 1 0.01 00 0.0 01 F F 80 ŸV V 0.091 0.091

E

ln

C2 C1

R RT

= ln 1.6 × 1.987 × 310

Now, Scale-up factor

³1 f x dx | 2.37

D

E = Energy required for active transport in cal mol–1

48. Given data: V1 = 1 L, N1 = 600 rpm, V2 = 8000 L

1

μm

C2 E , ln C RT 1 R = 1.987 cal mol–1 k–1 T = 37 + 273 = 310 K

= 289.50 cal mol–1

1 3.08 2 5 3 5.57 6

Now, D

Now, Active transport eq. is given by.

0.37 32.271

+ 2[0.51 + 0.71 + 1.0 + 1.40 + 1.95]]

1.6

0.091 0.0 879.1lit

D2 D1

?

Impeller diameter of large fermentor Impeller diameter of small fermentor

D2 8000 3 D1 1

20 Ÿ

D2 D1

3

8000 . 1

? D2 = 20D1 Now, the Scale-up criteria is constant impeler tip speed NiDi, So, N1D1 = N2 D2

N2

N1

D1 D2

600

1 20

30 rpm.

? N2 = 30 rpm. 49. All three factors i.e. (P) Mg2+ concentration, (Q) pH and (R) Annealing temperature, affect the fidelity of DNA polymerase in polymerase chain reaction 50. Correct Match of organelles in Group I with their functions in Group II. Group-I (organelles) Group-II (functions) P. Lysosome

1. Digestion of foreign substances

Q. Smooth ER R. Golgi apparatus

3. Lipid synthesis 2. Protein targeting

S. Nucleolus

5. rRNA synthesis

51. The statements, (P) Substrate concentration inside the chemostat is equal to that in the exit stream; (Q) Optimal dilution rate is lower than critical dilution rate; (S) Cell recirculation facilitates operation beyond critical dilution rate; are ALWAYS CORRECT about an ideal chemostat.

14

Solved Paper - 2019

52. Assertion [a] : It is possible to regenerate a whole plant from a single plant cell. Reason [r] : It is easier to introduce transgenes in to plants than animals. Both [a] and [r] are true but [r] is not the correct reason for [a]. 53. Given data: Kd = 4 min–1 Nt = Desired Probability of containation = 10–3 No = total No. of spores = 10000 L × 1000 ml × 106 =1013 ? ln

No Nt

k d t nd , where t nd No Nt kd

ln ? The holding time t nd

? tnd = 9.21

holding time 13

10

3 ln 10 4

54. Tumor cells that are able to replicate endlessly are fused with mammalian cells that produce a specific antibody which result in fusion called hybridoma that continuously produce antibodies. Those antibodies are named monoclonal because they come from only 1 type of cell, which is the hybridoma cell. 55. Many factors influence the duration of lag time, including inoculum size, the physiological history of the cells, and the precise physiochemical environment of both the original and the new growth medium. Therefore, all three factors i.e. (P) Inoculum size; (Q) Inoculum age; (R) Medium composition can influence the lag phase of a microbial culture in a batch fermenter.

S OLVED PAPER - 2020 INSTRUCTIONS 1. Total of 65 questions carrying 100 marks, out of which 10 questions carrying a total of 15 marks are in General Aptitude (GA) 2. The Engineering Mathematics will carry around 15% of the total marks, the General Aptitude section will carry 15% of the total marks and the remaining 70% of the total marks. 3. Types of Questions ( a ) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all papers and sections. These questions are objective in nature, and each will have a choice of four options, out of which the candidate has to mark the correct answer(s). ( b ) Numerical Answer Questions of 1 or 2 marks each in all papers and sections. For these questions the answer is a real number, to be entered by the candidate using the virtual keypad. No choices will be shown for these type of questions. 4. For 1-mark multiple-choice questions, 1/3 marks will be deducted for a wrong answer. Likewise, for 2-marks multiple-choice questions, 2/3 marks will be deducted for a wrong answer. There is no negative marking for numerical answer type questions.

GENERAL APTITUDE Q. 1 – Q. 5 carry one mark each 1. Rajiv Gandhi Khel Ratana Award was conferred ______ Mary Kom, a six time world champion in boxing, recently in ceremony ____ the Rashtrapati Bhawan (the President’s official residence) in New Delhi. (a) with, at

(b) on, in

(c) to, at

(d) on, at

2. Despite a string of poor performances, the chances of K.L Rahul’s selection in the team are _____ (a) Slim

(b) Bright

(c) Obvious

(d) Uncertain

3. Select the word that fits the analogy: Cover: Uncover:: Associate: _____ (a) Unassociate (b) Inassociate (c) Misassociate (d) Dissociate 4. Hit by floods, the kharif (summer sown) crops in various parts of the country have been affected. Officials believe that the loss in production of the kharif crops can be recovered in the output of the rabi (winter sown) crops so that the country can achieve its food-grain production target of 291 million tons in the crop year 2019-20 (July-June). They are hopeful that good rains in July- August will help the soil retain moisture for a longer

period, helping winter sown crops such as wheat and pulses during the November-February period. Which of the following statements can be inferred from the given passage? (a) Officials declared that the food grain production target will be met due to good rains (b)

Officials want the food grain production target to be met by the November-February period.

(c) Officials feel that the food grain production target cannot be met due to floods (d) Officials hope that the food grain production target will be met due to a good rabi produce 5. The difference between the sum of the first 2n natural numbers and the sum of the first n odd numbers is _____ (a) n2 – n

(b) n2 + n

(c) 2n – n

(d) 2n2 + n

2

Q. 6 – Q. 10 carry two marks each 6. Repo rate is the rate at which Reserve Bank of India (RBI) lends commercial banks, and reverse repo rate at which RBI borrows money from commercial banks. Which of the following statements can be inferred from the above passage? (a) Decrease in repo-rate will increase cost of borrowing and decrease lending by commercial banks

2

Solved Paper- 2020

(c) Increase in repo-rate will decrease cost of borrowing and decrease lending by commercial banks (d) Increase in repo-rate will decrease cost of borrowing and increase lending by commercial banks 7. P,Q,R,S,T,U,V and W are seated around a circular table I. S is seated opposite to W II. U is seated at the second place to the right of R III. T is seated at the third place to the right of R

70 Profit percentage

(b) Decrease in repo-rate will decrease cost of borrowing and increase lending by commercial banks

60

Company P Company Q

50 40 30 20 10 0

2013 2014 2015 2016 2017 2018 Year (a) 15 : 17 (b) 16 : 17

(c) 17 : 15

IV. V is a neighbour of S Which of the following must be true? (a) P is a neighbour of R (b) Q is a neighbour of R (c) P is not seated opposite to Q (d) R is the left neighbour of S 8. The distance between Delhi and Agra is 233 km. A car P started travelling from Delhi to Agra and another car Q started from Agra to Delhi along the same road 1 hour after the car P started. The two cars crossed each other 75 minutes after the car Q started. Both cars were travelling at constant speed. The speed of car P was 10km/hr more than the speed of car Q. How many kilometres the car Q had travelled when the cars crossed each other? (b) 75.2

(d) 17 : 16

BIOTECHNOLOGY Q. 1 – Q. 25 carry one mark each 1. Protein P becomes functional upon phosphorylation of a serine residue. Replacing this serine with _____ will result in a phosphomimic mutant of P. (a) alanine (c) phenylalanine

(b) aspartic acid (d) lysine

2. Ras protein is a (a) trimeric GTPase involved in relaying signal from cell surface to nucleus. (b) monomeric GTPase involved in relaying signal from cell surface to nucleus. (c) trimeric GTPase involved in regulation of cytoskeleton. (d) monomeric GTPase involved in regulation of cytoskeleton. 3. Which of the following statements are CORRECT?

(a) 66.6 (d) 116.5 (c) 88.2 9. For a matrix M = [mij]: i, j = 1, 2, 3, 4: the diagonal elements are all zero and m ij = –m ji. The minimum number of elements required to fully specify the matrix is _______. (a) 0

(b) 6

(c) 12

(d) 16

10. The profit shares of two companies P and Q are shown in the figure. If the two companies have invested a fixed and equal amount every year, then the ratio of the total revenue of company P to the total revenue of company Q, during 2013-2018 is ______

[P] Viruses can play a role in causing human cancer [Q] A tumor suppressor gene can be turned off without any change-in its DNA sequence [R] Alteration in miRNA expression levels contributes to the development of cancer (a) P and Q only

(b) Q and R only

(c) P and R only

(d) P, Q and R

4. Which class of antibody is first made by developing B cells inside bone marrow? (a) IgG (c) IgA

(b) IgE (d) IgM

Solved Paper- 2020

3

5. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r] regarding mammalian cells. Assertion [a]: Cells use Ca2+, and not Na+, for cellto-cell signaling Reason [r]: In the cytosol, concentration of Na+ is lower than that of Ca2+ (a) Both [a] and [r] are true and [r] is the correct reason for [a]. (b) Both [a] and [r] are true but [r] is not the correct reason for [a]. (c) Both [a] and [r] are false. (d) [a] is true but [r] is false. 6. Vincristine and vinblastine, two commercially important secondary metabolites from Catharanthus roseus, are examples of (a) alkaloids. (c) terpenoids.

(b) flavonoids. (d) steroids.

7. DNA synthesized from an RNA template is called (a) recombinant DNA (b) transcript (c) T-DNA (d) complementary DNA 8. Two monomeric His-tagged proteins of identical molecular weight are present in a solution, pIs of these two proteins are 5.6 and 6.8. Which one of the following techniques can be used to separate them? (a) Denaturing polyacrylamide gel electrophoresis (b) Size-exclusion chromatography (c) Ion-exchange chromatography (d) Nickel affinity chromatography 9. During a positive-negative selection process, transformed animal cells expressing _______ are killed in presence of ganciclovir in the medium. (a) pyruvate kinase (b) viral thymidine kinase (c) viral serine/threonine kinase (d) viral tyrosine kinase 10. A vector derived from which one of the following viruses is used for high-frequency genomic integration of a transgene in animal cells? (a) (b) (c) (d)

Adenovirus Adeno-associated virus Lentivirus Herpes simplex virus

11. Which one of the following statements about Agrobacterium Ti plasmid is CORRECT? (a) Vir genes are located within the T-DNA segment (b) Phytohormone biosynthesis genes are located outside the T-DNA segment

(c) Opine catabolism genes are located within the T-DNA segment (d) Opine biosynthesis genes are located within the T-DNA segment 12. Which of the following types of molecules act as biological catalysis? [P] Protein

[Q] RNA

[R] Phospholipid (a) P and Q only

(b) P and R only

(c) Q and R only

(d) P, Q and R

13. Which one of the following media components is used to maintain pH in, mammalian cell culture? (a) CaCl2

(b) MgSO4

(c) NaCl

(d) NaHCO3

14. Which of the following are energy transducing membranes? [P] Plasma membrane of bacteria [Q] Inner membrane of chloroplasts [R] Inner membrane of mitochondria (a) P and Q only (b) P and R only (c) Q and R only (d) P, Q and R 15. Amino acid sequences of cytochrome c and ribulose 5-phosphate epimerase from 40 organisms were chosen and phylogenetic trees were obtained for each of these two protein families. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r] Assertion [a]: The two trees will not be identical Reason [r]: The nature and frequency of mutations in the two families are different (a) Both [a] and [r] are true and [r] is the correct reason for [a]. (b) Both [a] and [r] are true but [r] is not the correct reason for [a]. (c) Both [a] and [r] are false. (d) [a] is false but [r] is true. 16. A microorganism isolated from a salt-rich (salt concentration ~2 M) lake was found to possess diglycerol tetraethers, with polyisoprenoid alcohol side chains, as the major lipid component of its cell membrane. The isolated organism is (a) a planctomycete (b) a cyanobacteria (c) a unicellular amoeba (d) an archaea

4

Solved Paper- 2020

17. A function f is as follows:

19. The elemental composition of dry biomass of a yeast species is CH 1.6O0.4N0.2S0.0024P0.017. The contribution of carbon to the dry biomass is _____ % (round off to 2 decimal places).

(a)

Time

[Given: atomic weights of H, C, N, O, P and S are 1, 12, 14, 16, 31 and 32, respectively] 20. Solvents A and B are completely immiscible. Solute S is soluble in both these solvents. 100 g of S was added to a container which has 2 kg each of A and B. The solute is 1.5 times more soluble in solvent A than in solvent B. The mixture was agitated thoroughly and allowed to reach equilibrium. Assuming that the solute has completely dissolved, the amount of solute in solvent A phase is _________ g.

(b)

Time

(c)

21. The number of molecules of a nucleotide of molecular weight 300 g/mol present in 10 picomoles is ________ × 1012 (round off to 2 decimal places). 22. To facilitate mass transfer from a gas to a liquid phase, a gas bubble of radius r is introduced into the liquid. The gas bubble then breaks into 8 bubbles of equal radius. Upon this change, the ratio of the interfacial surface area to the gas phase volume for the system changes from 3/r to 3n/r. The value of n is _______. 4 23. The largest eigenvalue of the matrix  2 

1 1 

is _______ . 24. A normal random variable has mean equal to 0, and standard deviation equal to 3, The probability that on a random draw the value of this random variable is greater than 0 is __________ (round off to 2 decimal places). 25. A variable Y is a function of t. Given that Y (t = 0)= 1 dY in the interval t = [0, 1] can dt be approximated as _________ .

and Y(t = 1) = 2,

Temperature

Z for X = 1 X and Y = 0 is _______ (answer is an integer),

18. Given that Z = X2 + Y2, the value of

Temperature

The function f is a continuous function when c is equal to _______ (answer is an integer).

26. A block of ice at 0°C is supplied heat at a constant rate to convert ice to superheated steam. Which one of the following trajectories correctly represents the trend of the temperature of the system with time? Assume that the specific heat of H2O is not a function of temperature.

Temperature

if x < 1 if x  1

Time

(d)

Temperature

15 f (x)    cx

Q. 26 – Q. 55 carry two marks each

Time

27. The DNA sequence shown below is to be amplified by PCR: 5'GCTAAGATCTGAATTTTCC...TTGGGCA ATA ATGTAGCGC3' 3'CGATTCTAGACTTAAAAGG ... AACCCGTTAT TACATCGCG5' Which one of the following pair of primers can be used for this amplification? (a) 5GGAAATTCAGATCTTAGT3 and 5TGGGCAATAATGTAGCGC3 (b) 5GCTAAGATCTGAATTTTC3 and 5GCGCTACATTATTGCCCA3 (c) 5CGGAAATTCAGATCTTAG3 and 5GCGCTACATTATTGCCCA3 (d) 5GCTAAGATCTGAATTTTC3 and 5TGGGCAATAATGTAGCGC3

Solved Paper- 2020

5

28. Which of the following statements about immune response are CORRECT? [P] T cells are activated by antigen-presenting cells [Q] Foreign peptides are not presented to helper T cells by Class II MHC proteins [R] Dendritic cells are referred to as professional antigen-presenting cells (a) P and R only

(b) P and Q only

(c) Q and R only

(d) P, Q and R

29. Which of the following statements are CORRECT about eukaryotic cell cycle? [P] CDKs can phosphorylate proteins in the absence of cyclins [Q] CDKs can be inactivated by phosphorylation [R] Degradation of cyclins is required for cell cycle progression [S] CDKs are not involved in chromosome condensation (a) P and R only (b) P and S only (c) P, Q and R only (d) Q and R only 30. W, X and Y are the intermediates in a biochemical pathway as shown below: S  W  X  Y  Z Mutants auxotrophic for Z are found in four different complementation groups, namely Z1, Z2, Z3 and Z4. The growth of these mutants on media supplemented, with W, X, Y or Z is shown below (Yes: growth observed; No: growth not observed): Mutants Zl Z2 Z3 Z4

Media supplemented with W X Y Z No No Yes Yes No Yes Yes Yes No No No Yes Yes Yes Yes Yes

What is the order of the four complementation groups in terms of the step they block? Z1 Z2 Z3 Z4 (a) S ......  W  X  Y  Z Z4 Z2 Z1 Z3 (b) S ......  W  X  Y  Z Z3 Z1 Z2 Z4 (c) S ......  W  X  Y  Z Z4 Z1 Z2 Z4 (d) S ......  W  X  Y  Z

31. In tomato plant, red (R) is dominant over yellow (r) for fruit color and purple (P) is dominant over green (p) for stem color. Fruit color and stem color assort independently. The number of progeny plants of different fruit/stem colors obtained. from a mating are as follows: Red fruit, purple stem – 145 Red fruit, green stem – 184 Yellow fruit, purple stem – 66 Yellow fruit, green stem – 47 What are the genotypes of the parent plants in this mating? (a) RrPp × Rrpp (b) RrPp × RrPp (c) RRPP × rrpp (d) RrPP x Rrpp 32. Some of the cytokinins used in plant tissue culture media are given below: [P] BAP [Q] Zeatin [R] Kinetin [S] 2iP Which of these are synthetic analogs? (a) P and Q only (b) Q and S only (c) R and S only (d) P and R only 33. Carl Woese used the gene sequence of which one of the following for phylogenetic taxonomy of prokarvotes’? (a) A ribosomal RNA of large ribosomal subunit (b) A ribosomal RNA of small ribosomal subunit (c) A ribosomal protein of large ribosomal subunit (d) A ribosomal protein of small ribosomal subunit 34. A list of pathogens (Group I) and a list of antimicrobial agents (Group II) used to treat their infections are given below. Match the pathogens with the corresponding anti-microbial agents. Group I Group II P. Influenza A virus 1. Isoniazid Q. Fungus 2. Amantadine R. Plasmodium 3. Fluconazole S. Mycobacterium 4. Artemisinin 5. Iodoquinol (a) P-4, Q-3, R-2, S-5 (b) P-5, Q-2, R-4, S-1 (c) P-2, Q-3, R-4, S-1 (d) P-2, Q-3, R-1, S-5 35. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: Dam methylase protects E. coli DNA from phage endonucleases Reason [r]: E. coli Dam methylase methylates the adenosine residue in the sequence “GATC” (a) Both [a] and [r] are true and [r] is the correct reason for [a] (b) Both [a] and [r] are true but [r] is not the correct reason for [a] (c) Both [a] and [r] are false (d) [a] is false but [r] is true

6

Solved Paper- 2020

36. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: Embryonic stem cells are suitable for developing knockout mice Reason [r]: Homologous recombination is more frequent in embryonic stem cells than that in somatic cells (a) Both [a] and [r] are false (b) Both [a] and [r] are true, and [r] is the correct reason for [a] (c) Both [a] and [r] are true, but [r] is not the correct reason for [a] (d) [a] is true, but [r] is false 37. The schematic of a plasmid with a gap in one of the strands is shown below:

ATCGCGATG TAGC AC 3OH 5OH

39. Which of the following strategies are used by cells for metabolic regulation? [P] Phosphorylation - dephosphorylation [Q] Allostery [R] Feedback inhibition (a) P and Q only (b) P and R only (c) Q and R only (d) P, Q and R 40. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: A zygote and its immediate descendant cells are unspecialized and are called totipotent Reason [r]: Totipotent cells retain the capacity to differentiate into only a few cell types (a) Both [a] and [r] are false (b) Both [a] and [r] are true but [r] is not the correct reason for [a] (c) Both [a] and [r] are true and [r] is the correct reason for [a] (d) [a] is true but [r] is false 41. Which of the following statements about gene therapy are CORRECT? [P] Affected individuals, but not their progeny, can be cured through germline gene therapy [Q] Affected individuals, as well as their progeny, can be cured through germline gene therapy [R] Affected individuals, but not their progeny, can be cured through somatic gene therapy

Which of the following enzyme (s) is/are required to fill the gap and generate a covalently closed circular plasmid? [P] DNA ligase [Q] Alkaline phosphatase [R] DNA polymerase [S] Polynucleotide kinase (a) P only (b) P, R and S only (c) P and R only (d) P, Q and R only 38. Match sub-cellular organelles listed in Group I with their features listed in Group II; Group I Group II [P] Mitochondrion 1. Single-membrane enclosed [Q] Chloroplast 2. Double-membrane enclosed [R] Nucleus 3. Maternal inheritance [S] Endoplasmic 4. Endosymbiotic origin reticulum (a) P-1, Q-4, R-2, S-3 (b) P-2, Q-3, R-4, S-1 (c) P-3, Q-4, R-2, S-1 (d) P-3, Q-1, R-4, S-2

[S] Affected individuals, as well as their progeny, can be cured through somatic gene therapy (a) P and R only (b) P and S only (c) Q and R only (d) Q and S only 42. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: A genetically engineered rice that produces beta-carotene in the rice grain is called Golden rice Reason [r]: Enabling biosynthesis of provitamin A in the rice endosperm gives a characteristic yellow/orange color (a) Both [a] and [r] are false (b) Both [a] and [r] are true but [r] is not the correct reason for [a] (c) Both [a] and [r] are true and [r] is the correct reason for [a] (d) [a] is true but [r] is false

Solved Paper- 2020

7

43. The sequence of a 1 Mb long DNA is random. This DNA has all four bases occurring in equal proportion. The number of nucleotides, on average, between two successive EcoRI recognition site GAATTC is ________ . 44. E. coli was grown In 15N medium for several generations. Cells were then transferred to 14 N medium, allowed to grow for 4 generations and DNA was isolated immediately. The proportion of total DNA with intermediate density is ______ (round off to 2 decimal places). 45. A batch reactor is inoculated with 1 g/L biomass, Under these conditions, cells exhibit a lag phase of 30 min. If the specific growth, rate in the log phase is 0.00417 min–1, the time taken for the biomass to increase to 8 g/L is ___________ min (round off to 2 decimal places). 46. The system of linear equations cx + y = 5 3x + 3y = 6 has no solution when c is equal to __________ . 47. The amino acid sequence of a peptide is Phe-LeuIle-Met-Ser-Leu. The number of codons that encode the amino acids present in this peptide is given below: Phe: 2 codons Leu: 6 codons Ile: 3 codons Met: 1 codon Ser: 4 codons The number of unique DNA sequences that can encode this peptide is ______. 48. Assume that a cell culture was started with five human fibroblast cells. Two cells did not divide even once whereas the other three cells completed three rounds of cell division. At this stage, the total number of kinetochores in all the cells put together is __________ . 49. Growth of an organism on glucose in a chemostat is characterized by Monod model with specific growth rate = 0.45 h–1 and Ks = 0.5 g/L. Biomass from the substrate is generated as YXS = 0.4 g/g. The chemostat volume is 0.9 L and media is fed at 1 L/h and contains 20 g/L of glucose. At steady state, the concentration of biomass in the chemostat is __________ g/L.

50. A function f is given as: f (X) = 4X – X2 The function f is maximized when X is equal to ________ . 51. An infinite series S is given as: S = 1 + 2/3 + 3/9 + 4/27 + 5/81 + .... (to infinity) The value of S is _________ (round off to 2 decimal places). 52. Protein A and protein B form a covalent complex. Gel filtration chromatography of this complex showed a peak corresponding to 200 kDa. SDSPAGE analysis of this complex, with and without beta-mercaptoethanol, showed, a single band corresponding to molecular weight 50 and 25 kDa, respectively. Given that the molecular weight of protein A is 25 kDa, the molecular weight of protein B is _______ kDa. 53. The concentrations of ATP, ADP and inorganic phosphate in a cell are 2.59, 0.73 and 2.72 mM. respectively. Under these conditions, free energy change for the synthesis of ATP at 37°C is _________ kJ/mol (round off to 2 decimal places), Given: free energy change for ATP hydrolysis under standard conditions is –30.5 kJ/mol and R = 8.315 kJ/mol.K 54. An algorithm was designed to find globins in protein sequence databases. A database which has 78 globin sequences was searched using this algorithm. The algorithm retrieved 72 sequences of which only 65 were globins. The sensitivity of this algorithm is __________ % (round off to 2 decimal places). 55. The mitochondrial electron transfer chain oxidizes NADH with oxygen being the terminal electron acceptor. The redox potentials for the two halfreactions are given below: NAD  H  2e   NADH, E0  0.32V 1 O2  2H  2e   H2O, E0  0.816V 2 The free energy change associated with the transfer of electrons from NADH to O2 is _______ kJ/mol (round off to 2 decimal places).

Given: F = 96500 C/mol.

8

Solved Paper- 2020

ANSWERS General Aptitude 1. (c)

2. (b)

3. (d)

4. (d)

5. (b)

6. (d)

7. (c)

8. (b)

9. (b)

10. (b)

Biotechnology 1. (b)

2. (b or d)

3. (d)

4. (d)

5. (d)

6. (a)

7. (d)

8. (c)

9. (b)

10. (c)

11. (d)

12. (a)

13. (d)

14. (b)

15. (a)

16. (d)

17. (15 to 15)

18. (2 to 2)

19. (51.01 to 52.99)

20. (60 to 60)

21. (5.99 to 6.05) 22. (2 to 2)

23. (3 to 3)

24. (0.49 to 0.51) 25. (1 to 1)

26. (a)

27. (b)

28. (a)

29. (d)

30. (b)

31. (a)

32. (d)

33. (b)

34. (c)

35. (d)

36. (b)

37. (b)

38. (c)

39. (d)

40. (d)

41. (c)

42. (c)

43. (4096 to 4096)44. (0.11 to 0.14) 45. (526.01 to 529.99)

47. (864 to 864)

48. (1196 to 1196)49. (0 to 0)

53. (48.50 to 49.50 or 18521.20 to 18521.30)

50. (2 to 2)

46. (1 to 1)

51. (2.20 to 2.30) 52. (25 to 25)

54. (83.30 to 83.40)

55. (-219.30 to -219.20)

EXPLANATIONS GENERAL APTITUDE 1. Rajiv Gandhi Khel Ratana Award was conferred on Mary Kom, a six time world champion in boxing, recently in ceremony at the Rashtrapati Bhawan (the President's official residence) in New Delhi.

6. Decrease in repo-rate will decrease cost of borrowing and increase lending by commercial banks is the correct inference. 7.

V S V

2. Despite means without being affected by something.

R

3. Dissociate is the opposite of Associate.

T U V can take any of the two positions.

4. Officials hope that the food grain production target will be met due to a good rabi produce. 5. Sum of first n natural numbers

In above diagram, P can not be seated opposite to Q. y x 8. Delhi Agra Q X P

 n  n  1  =  2 Sum of first 2n natural numbers  2n  2n  1 =  2 Sum of first n odd numbers = n2

Difference of (1) and (2) D = 2n2 + n – n2 = n2 + n

W

....(1) ...(2)

233 km

VP = VQ + 10 x = VP × 2.25 y = VQ × 1.25 x + y = 233

...(i)

Solved Paper- 2020

VP × 2.25 + VQ × 1.25 = 233 VP =

491 km/hr 7

VQ =

421 km/hr 7

9

...(ii)

Distance travelled by Q  y = VQ × 1.25 

421  1.25 7

y = 75.17 km 9. Total no. of elements in in 4 × 4 matrix is 16; of which four diagonal elements are zero. Remaining 12 elements can be divided in to two equal groups of 6 elements each. Both groups are opposite signed elements to meet the given condition. 10. Let the fixed investment annually be ‘A’. Total revenue of company P (2013 to 2018) = 1.1A + 1.2A + 1.4A + 1.4A + 1.5A + 1.4A = 8A Total revenue of Company Q (2013 to 2018) = 1.2A + 1.3A + 1.3A + 1.5A + 1.6A + 1.6A = 8.5 A

BIOTECHNOLOGY 1. Protein P becomes functional upon phosphorylation of a serine residue. Replacing this serine with aspartic acid will result in a phosphomimic mutant of P. 2. These proteins are monomeric proteins of approximately 20–30 kDa with conserved structures and functions. These proteins bind guanine nucleotides and function as a molecular switch by shuttling between GTP-bound and GDP-bound forms. Ras is a family of related proteins which is expressed in all animal cell lineages and organs. All Ras protein family members belong to a class of protein called small GTPase, and are involved in transmitting signals within cells (cellular signal transduction). Ras is the prototypical member of the Ras superfamily of proteins, which are all related in 3D structure and regulate diverse cell behaviours. When Ras is ‘switched on’ by incoming signals, it subsequently switches on other proteins, which ultimately turn on genes involved in cell growth, differentiation and survival. Mutations in ras genes can lead to the production of permanently activated Ras proteins. As a result, this can cause

unintended and overactive signaling inside the cell, even in the absence of incoming signals. Because these signals result in cell growth and division, overactive Ras signaling can ultimately lead to cancer. 3. All three sentences are correct i.e. [P] Viruses can play a role in causing human cancer [Q] A tumor suppressor gene can be turned off without any change-in its DNA sequence [R] Alteration in miRNA expression levels contributes to the development of cancer 4. IgM is not only the first class of antibody to appear on the surface of a developing B cell. It is also the major class secreted into the blood in the early stages of a primary antibody response, on first exposure to an antigen. (Unlike IgM, IgD molecules are secreted in only small amounts and seem to function mainly as cell-surface receptors for antigen.) In its secreted form, IgM is a pentamer composed of five four-chain units, giving it a total of 10 antigen-binding sites. Each pentamer contains one copy of another polypeptide chain, called a J (joining) chain. The J chain is produced by IgMsecreting cells and is covalently inserted between two adjacent tail regions. 5. Assertion [a]: Cells use Ca2+, and not Na+ , tor cell-to-cell signalling is true but the Reason [r]: In the cytosol, concentration of Na+ is lower than that of Ca2+ is false. 6. Vinblastine and vincristine are alkaloids found in the Madagascar periwinkle, Catharanthus roseus (formerly known as Vinca rosea). The natives of Madagascar traditionally used the vinca rosea to treat diabetes. In fact it has been used for centuries throughout the world to treat all kinds of ailments from wasp stings, in India, to eye infections in the Caribbean. They work by preventing mitosis in metaphase. These alkaloids bind to tubulin, thus preventing the cell from making the spindles it needs to be able to divide. This is different from the action of taxol which interferes with cell division by keeping the spindles from being broken down. Vinblastine is mainly useful for treating Hodgkin’s disease, advanced testicular cancer and advanced breast cancer. Vincristine is mainly used to treat acute leukemia and other lymphomas. 7. Complementary DNA (cDNA) is the DNA produced on an RNA template by the action of reverse transcriptase (RNA-dependent DNApolymerase). The sequence of the cDNA becomes complementary to the RNA sequence.

10

Solved Paper- 2020

8. Given two proteins have similar molecular weight. The ISO electric points are 5.6 & 6.8. Now, we know that ISO electric focusing can resolve protein that differ in PI values as little as 0.01.  Isoelectric focusing is the 1st step in two-dimensional gate electrophoresis.  So, given proteins can be separated by using Denaturing Poly a chefamide. 9. During a positive-negative selection process, transformed animal cells expressing viral thymidine kinase are killed in presence of ganciclovir in the medium. 10. A vector derived from Lentivirus is used for highfrequency genomic integration of a transgene in animal cells. The kinetics of lentivirus integration into the target genome defines the degree of genotypic mosaicism found in the resulting animal. Integration events occurring subsequent to the first cell division will, indeed, limit the presence of individual proviruses to only a subset of cells and, hence, condition their rate of transmission to the first generation (G1) progeny. 11. Opines are low molecular weight compounds found in plant crown gall tumours or hairy root tumours produced by pathogenic bacteria of the genus Agrobacterium and Rhizobium.[1] Opine biosynthesis is catalyzed by specific enzymes encoded by genes contained in a small segment of DNA (known as the T-DNA, for ‘transfer DNA’), which is part of the Ti plasmid (in Agrobacterium) or Ri plasmid (in Rhizobium), inserted by the bacterium into the plant genome. The opines are used by the bacterium as an important energy, carbon and nitrogen source.

14. [P] Plasma membrane of bacteria and [R] inner membrane of mitochondria are energy transducing membranes. 15. As given per paragraph; Both [a] and [r] are true and [r] is the correct reason for [a] i.e. Assertion [a]: The two trees will not be identical and Reason [r]: The nature and frequency of mutations in the two families are different are true and [r] is the correct reason for [a]. 16. Halophiles are organisms that thrive in high salt concentrations. They are a type of extremophile organism. The name comes from the Greek word for “salt-loving”. Halophiles can be found anywhere with a concentration of salt five times greater than the salt concentration of the ocean, such as the Great Salt Lake in Utah, Owens Lake in California, the Dead Sea, and in evaporation ponds.

lim f ( x) = lim f ( x) x 1

17.

x 1

15 = c

c  15 18. z = x + y 2

2

z = 2x x  z    =21=2  x  at x1& y0 19. The molecular weight of yiest can found using composition CH1.6O0.4N0.2S0.0024P0.017 = 12 + (1.6 × 1) + (0.4 × 16) + (0.2 × 14) + (0.0024 × 32) + (0.017 × 31)

12. Protein and RNA, types of molecules act as biological catalysis.

= 23.404

13. The most commonly used buffer in tissue culture media is the bicarbonate + carbon dioxide buffer but phosphate or tris buffers have also been used. Each of these buffers has particular disadvantages for use with in vitro biological systems. The bicarbonate+ CO2 buffer, pKa of 6"3 at 37 °, has the obvious limitation that one component is in the gaseous phase and tissue cultures must be maintained in a closed system. Although the phosphate buffer has a more suitable pKa, 6"9 at 37 °, phosphates form insoluble complexes with essential divalent cations when used at concentrations necessary for a suitable buffering capacity.

Now, amount of carbon present =

12 100 23.404 = 51.27%

20. Solvent A = 2 kg Solvent B = 2 kg Now, 100 gm solute S is added. Solubility of solute S in solvent A is 1.5 times than in solvent B. Let assume x amount of solute S dissolved in solvent B  1.5 x amount of solute S dissolve in solvent A. Now, 1.5x + x = 100 gm  x = 40 gm

Solved Paper- 2020

11

|A – I| = 0

So, amount of solute in solvant A = 1.5x = 1.5 × 40 = 60 gm 21. Molecular weight of nucleotide = 300 gm 1 mole contains 6.023 × 1023 molecules 1 Pico mole = 10–12 moles 

Pico mole =

10–11

moles

Interfacial surface area of large bubble

4 3 r 3 Interfacial surface area  Ratio of of large bubble Gas phase volume 4r 2 3  = 4 r r 3 3 Now, large bubble breaks into 8 small bubble of equal radius.

r 8 Now, interfacial surface area of small bubble So, the radius of small bubble =

4 r   3 8

3

Interfacial surface area of small bubbles Gas phase volume

=

3 3 8  r r   8

As per the problem statement the ratio is changed from

So,

3n 3  8  r r n=8

 4 1    2 1

23. Let A = 

2 – 3– 2 + 6 = 0 (– 3) (– 2) = 0  = 3, 2 Largest eigen value = 3 24. Mean ± SD 0±3

3

0

–3

Probability of random variable more than 0 (shaded area) = 0.5 or 50% Y2 – Y 1 = 2 – 1 = 1 25. Y2 = 2 Y1 = 1 t1 = 0

 Y=

2

So,

=0

t2 = 1

2

Gas phase volume of small bubble =

3 n to 3 r r

1

2 – 5 + 6 = 0

= 4r2

Volume of large bubble =

r 4   8 = 3 4 r   3 8

2

(4 – ) (1 –) + 2 = 0

= 6.023 × 1012 molecules 22. Radius of original gas bubble = r

 Ratio of

1

4 –  + 2 – 4 + 2 = 0

 10–11 moles contain = 6.023 × 1023 × 10–11

r = 4   8

4

2 1 =1 10

26. When ice block which is at 0°C supplied heat at constant rate than it posses a heat of fusion at 0°C. Until entire block of ice over come the heat of fusion than temperature of the melted water is constant upon supplying heat. Then the temperature of water will rise linearly with the supplied heat. Again near vaporization point, the water does not get starting vaporize until the latent heat of vaporization will not be overcome by the supplied heat so during this temperature is constant for some time and after that temperature is increasing again. So option (a) shows this behaviour. 27. The 3’ end of a primer has to be towards the inside of the target sequence to be amplified. In other words, a primer should bind to the 3’ end of any template strand for its amplification. Considering this rule plus the anti-parallel nature of complimentary strands of DNA the answer is B i.e. 5GCTAAGATCTGAATTTTC3 and 5GCGCTACATTATTGCCCA3 pair of primers can be used for the given amplification.

12

28. Statements about immune response i.e. [P] T cells are activated by antigen-presenting cells and [R] Dendritic cells are referred to as professional antigen-presenting cells are CORRECT. 29. Statements i.e. [Q] CDKs can be inactivated by phosphorylation [R] Degradation of cycling is required for cell cycle progression are CORRECT about eukaryotic cell cycle. 30. Auxotrophic mutants are strains that require a growth supplement that the organism isolated from nature (the wild-type strain) does not require i.e. mutant strain of microorganism that will proliferate only when the medium is supplemented with some specific substance not required by wild-type organisms. 32. Synthetic analogs are defined as the artificial compounds differing from the natural compounds in their structure, resemblance and function. Synthetic refers to any material made from nonnatural sources. Cytokinins used in plant tissue culture media i.e. [P] BAP and [R] Kinetin are synthetic analogs. 33. Carl Woese used the gene sequence of a ribosomal RNA of small ribosomal subunit for phylogenetic taxonomy of prokaryotes. n the early 1970s, Woese realized that the sequence of 5S ribosomal RNA contained too few nucleotides (120) to provide a way to classify thousands of organisms. This led him to take on the daunting task of analysing 16S ribosomal RNA, which contains more than 1,500 nucleotides 34. Correct match of the pathogens with corresponding microbial agent is: Group I Group II (Pathogens) (microbial agent) (P) Influenza A virus 2. Amantadine (Q) Fungus 3. Fluconazole (R) Plasmodium 4. Artemisinin (S) Mycobacterium 1. Isoniazid 35. Assertion [a]: Dam methylase protects E.coli DNA from phage endonucleases is FALSE but Reason [r]: E.coli Dam methylase methylates the adenosine residue in the sequence “GATC” is true. DAM methylase, an abbreviation for deoxyadenosine methylase, is an enzyme that adds a methyl group to the adenine of the sequence 5'GATC-3' in newly synthesized DNA. Immediately after DNA synthesis, the daughter strand remains unmethylated for a short time. It is an orphan methyltransferase that is not part of a restrictionmodification system and regulates gene expression. Dam methylase is unique to prokaryotes and bacteriophages

Solved Paper- 2020

Most laboratory strains of E. coli contain three site-specific DNA methylases. The methylase encoded by the dam gene (Dam methylase) transfers a methyl group from Sadenosylmethionine (SAM) to the N6 position of the adenine residues in the sequence GATC (1,2). 36. Both i.e. Assertion [a]: Embryonic stem cells are suitable for developing knockout mice and Reason [r]: Homologous recombination is more frequent in embryonic stem cells than that in somatic cells are true and [r] is the correct reason for [a]. 37. The DNA polymerase will add new nucleotides at the 3’OH end, Polynucleotide will transfer phosphate group to the 5’OH end and Ligase will create the phosphodiester bond to create the final covalently linked double stranded plasmid. 38. Correct match of sub-cellular organelles with their features is: Group I (sub-cellular organelles)

Group II (features)

(P) Mitochondrion

3. Maternal inheritance

(Q) Chloroplast

4. Endosymbiotic origin

(R) Nucleus

2. Double-membrane enclosed

(S) Endoplasmic reticulum

1. Single-membrane enclosed

39. (P) Phosphorylation and dephosphorylation are important posttranslational modifications of native proteins, occurring site specifically on a protein surface. These biological processes play important roles in intracellular signal transduction cascades and switching the enzymatic activity. (Q) Allostery is the process by which biological macromolecules (mostly proteins) transmit the effect of binding at one site to another, often distal, functional site, allowing for regulation of activity. (R) Feedback inhibition is defined as the process in which the end product of a reaction inhibits or controls the action of the enzyme that helped produce it. In other words, it refers to a situation wherein the end products formed at the end of a sequence of reactions participate in suppressing the activity of the enzymes that helped synthesis the end product(s). Therefore (P) Phosphorylation and dephosphorylation, (Q) Allostery and (R) Feedback inhibition, strategies are used by cells for metabolic regulation.

Solved Paper- 2020

13

40. Assertion [a]: A zygote and its immediate descendant cells are unspecialized and are called totipotent is TRUE but Reason [r]: Totipotent cells retain the capacity to differentiate into only a few cell types is FALSE 41. Statements, given bellow about gene therapy are CORRECT: [Q] Affected individuals, as well as their progeny, can be cured through germline gene therapy [R] Affected individuals, but not their progeny, can be cured through somatic gene therapy 42. Both i.e. Assertion [a]: A genetically engineered rice that produces beta-carotene in the rice grain is called Golden rice and Reason [r]: Enabling biosynthesis of provitamin A in the rice endosperm gives a characteristic yellow/orange color, are true and [r] is the correct reason for [a]. 43 Sequence of DNA = 1 Mb = 1024 base. DNA has all four bases. Occuring in equal proportion. So, the number of nucleotides between two successive EcoRI recognition site GAATTC is equal to the total nucleotides on 1 Mb sequence of DNA. This equal to 1024 × 4 = 4096 number of nucleo tides. 44. Let assume E.coli was grown in 15N medium for n generation. So, the total DNA forms cells after growing in 15N medium. Nt1 = N02n

Now, these Nt1 cells transferred to 14N medium and allows to grow for 4 generations. So, the total DNA from cells after growing in 14N medium. Nt2 = Nt1 24 = N02n24 = N02(n + 4) So, the proportion of total DNA with intermediate density. 1  Nt1  1  N 0 2 n  1  1  =       0.5  Nt 2  0.5  N 0 2 n 4  0.5  2 4 

= 0.125 Because after transfering cells from 15N medium to

14N

medium the DNA is

Parent 15N) and 14N medium).

1 heavy (from the 2

1 light (Newly grown i.e. from 2

45. Initial conc. of biomass = 1 g/L = X0 Lag phase of cells = 30 min Specific growth rate in log phase,  = 0.00417 min–1 Now, final conc. of biomass = Xt = 8 g/L We know that,  ln

dx = X dt

Nt = t N0



t=

ln N t N 0 ln  8 1   0.00417

= 498.67 min So, total time to reach cell conc 8 g/L = lag phase time + log phase time = 30 min + 498.67 min = 528.67 min 46. cx + y – 5 = 0 3x + 3y – 6 = 0 No solution condition is as follows

a1 b1 c1   a2 b2 c2

c 1 5   3 3 6 c 1 48. 5 human fibroblast cells from these 2 are not divide while 3 undergone 3 rounds of cell division. So, total number of cells after 3 rounds of cell division. = 2 + 23 + 23 + 23 = 26 cells. So, total number of Kinetochores in all cells = 26 × 46 = 1196 49. Given,  = 0.45 h–1, KS = 0.5 g/L, Yxis = 0.4 g/g Feed rate F = 1 L/hr, Chemostate volume V = 0.9 L Glucose concentration in feed = 20 g/L Dilution rate D =

F  1.11 V

As dilution rate is higher than the volume of chemostate itself so, during the process the biomass as well as substrate will carried away or it will be displaced from the chemostate. So, no biomess will be grown. (washout will occurs) So, concentration of biomass in chemostate = O g/L.

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Solved Paper- 2020

50. f(x) = 4x – x

2

=

f(x) = 0 4 – 2x = 0

x2 51. S = 1 

2 3 4 5     ............ 3 9 27 81

2 3 4 S = 1  21  31  4 1  ............ 3 3 3 S = 1

2 3 n   ............ n1 32 1 331 3

S = 1 + 0.6666 + 0.3333 + 0.1481 + 0.0617 + 0.0247 + 0.0096 + 0.0037 + 0.0014 + ...... S = 2.2 (approximately) 53. ATP synthesis equal is given by ADP + Pi  ATP The concentration of each species is given as, [ADP] = 0.73 mm, [Pi] = 2.72 mm, [ATP] = 2.59 mm Temp T = 273 + 37 = 310 K R = 8.315 KJ/(ml K) G° = + 30.5 KJ/ml G = G° + RT ln K Now, K per the ATP synthesis reaction is given by K =

 ATP  1M  ADP  1M   Pi  1M 

(2.59 mm)  1000

 0.73 mm  2.72 mm 

K = 1304.4  G° = G° + RT ln K = + 30.5 + (8.315 × 310 × ln 1304.4) = 18521.26 KJ/mol. 54. Database has 78 globin sequences. The algorithm retrived 72 sequences from 78 globin sequences. Further, out of 72 retrived sequences 65 were globins.  72  65   Sensitivity of algorithm =     100  78  72  = 83.33%

55. Two half-cell reactions are given by NAO+ + H+ 2e–  NAOH 1/2 U2 +

2H+

+

2e–

 H2O

E0 = –0.32 V E0 = 0.816 V

NAO+ + 3H+ + 1/2 O2 + 4e–  NADH + H2O E0 = 0.496 Number of electron transfered n = 4, F = 96500 C/mgl  96500   f ° = – nFE0 = 4   (0.496) kJ/ml  1000  = – 191.46 KJ/ml

SOLVED PAPER - 2021 1. 2. 3.

4.

INSTRUCTIONS Total of 65 questions carrying 100 marks, out of which 10 questions carrying a total of 15 marks are in General Aptitude (GA) The Engineering Mathematics will carry around 15% of the total marks, the General Aptitude section will carry 15% of the total marks and the remaining 70% of the total marks. Types of Questions (a) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all papers and sections. These questions are objective in nature, and each will have a choice of four options, out of which the candidate has to mark the correct answer(s). (b) Numerical Answer Questions of 1 or 2 marks each in all papers and sections. For these questions the answer is a real number, to be entered by the candidate using the virtual keypad. No choices will be shown for these type of questions. (c) Multiple Select Questions (MSQ) carrying 1 or 2 marks each in all the papers and sections. These questions are objective in nature, and each will have choice of four answers, out of which ONE or MORE than ONE choice(s) is / are correct. For 1-mark multiple-choice questions, 1/3 marks will be deducted for a wrong answer. Likewise, for 2-marks multiple-choice questions, 2/3 marks will be deducted for a wrong answer. There is no negative marking for numerical answer type questions & multiple selection questions.

Which of the following is the CORRECT observation about the above two sentences?

GENERAL APTITUDE Q.1 to Q.5 : Multiple Choice Question (MCQ), carry ONE mark each (for each wrong answer: - 1/3).

(a) (i) is grammatically correct and (ii) is unambiguous (b) (i) is grammatically incorrect and (ii) is unambiguous

1. The ratio of boys to girls in a class is 7 to 3. Among the options below, an acceptable value for the total number of students in the class is: (a) 21

(b) 37

(c) 50

(d) 73

2. A polygon is convex if, for every pair of points, P and Q belonging to the polygon, the line segment PQ lies completely inside or on the polygon.

(c) (i) is grammatically correct and (ii) is ambiguous, (d)(i) is grammatically incorrect and (ii) is ambiguous 4.

Which one of the following is NOT a convex polygon?

(a)

(b)

(c)

(d)

A circular sheet of paper is folded along the lines in the directions shown. The paper, after being punched in the final folded state as shown and unfolded in the reverse order of folding, will look like_______. (a)

(b)

(c)

(d)

3. Consider the following sentences: (i) Everybody in the class is prepared for the exam. (ii) Babu invited Danish to his home because he enjoys playing chess.

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SOLVED PAPER - 2021

5. _____ is to surgery as writer is to _____ Which one of the following options maintains a similar logical relation in the above sentence?

9. Given below are two statements 1 and 2, and two conclusions I and II. Statement 1: All bacteria are microorganisms.

(a) Plan, outline

Statement 2: All pathogens are microorganisms.

(b) Hospital, library

Conclusion I: Some pathogens are bacteria.

(c) Doctor, book

Conclusion II: All pathogens are not bacteria.

(d) Medicine, grammar

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

Q.6 to Q.10 : Multiple Choice Question (MCQ), carry TWO marks each (for each wrong answer: - 2/3). 6. We have 2 rectangular sheets of paper, M and N, of dimensions 6 cm  1 cm each. Sheet M is rolled to form an open cylinder by bringing the short edges of the sheet together. Sheet N is cut into equal square patches and assembled to form the largest possible closed cube. Assuming the ends of the cylinder are closed, the ratio of the volume of the cylinder to that of the cube is _____ (a)

 2

(c)

9 

(b)

3 

(d) 3

7. Items Cost (`) Profit % Marked Price (`) P

5,400

---

5,860

Q

---

25

10,000

Details of prices of two items P and Q are presented in the above table. The ratio of cost of item P to cost of item Q is 3:4. Discount is calculated as the difference between the marked price and the selling price. The profit percentage is calculated as the ratio of the difference between selling price and cost, to the cost Selling price  Cost    100 .  Pr ofit%   Cost

The discount on item Q, as a percentage of its marked price, is ____ (a) 25 (b) 12.5

(a) Only conclusion I is correct (b) Only conclusion II is correct (c) Either conclusion I or II is correct. (d) Neither conclusion I nor II is correct. 10. Some people suggest anti-obesity measures (AOM) such as displaying calorie information in restaurant menus. Such measures sidestep addressing the core problems that cause obesity: poverty and income inequality. Which one of the following statements summarizes the passage? (a) The proposed AOM addresses the core problems that cause obesity. (b) If obesity reduces, poverty will naturally reduce, since obesity causes poverty. (c) AOM are addressing the core problems and are likely to succeed. (d) AOM are addressing the problem superficially.

BIOTECHNOLOGY Q.1 to Q.17 : Multiple Choice Question (MCQ), carry ONE mark each (for each wrong answer – 1/3). 1. Coronavirus genome consists of (a) double-stranded DNA (b) double-stranded RNA (c) negative-sense single-stranded RNA (d) positive-sense single-stranded RNA 2. The enzyme that transcribes the eukaryotic genes encoding precursor ribosomal RNAs (pre-rRNAs) of 28S, 18S and 5.8S rRNAs is

(c) 10

(a) RNA polymerase I

(d) 5

(b) RNA polymerase II

8. There are five bags each containing identical sets of ten distinct chocolates. One chocolate is picked from each bag. The probability that at least two chocolates are identical is _____ (a) 0.3024

(b) 0.4235

(c) 0.6976

(d) 0.8125

(c) RNA polymerase III (d) RNA polymerase IV 3. Number of unrooted trees in a phylogeny of five sequences is (a) 3

(b) 15

(c) 105

(d) 945

SOLVED PAPER - 2021

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4. Which one of the following methods is used to test the significance of a predicted phylogeny? (a) Bootstrap (b) Maximum likelihood (c) Maximum parsimony (d) Minimum evolution 5. The Cartesian coordinates (x, y) of a point A with



(c)

2



(b) 2,2 3

2, 3



(d) 2 2, 2 2

(b) adaptive immunity in prokaryotes





(c) innate immunity in eukaryotes



6. The order of genes present in a chromosome is as follows.

L M

(b) H2 (d) O2

(a) adaptive immunity in eukaryotes

3, 2 2



(a) CO2 (c) N2

11. CRISPR-Cas system is associated with

  polar coordinates  4,  is  4

(a)

10. Under standard temperature (T) and pressure (P) conditions, 128 g of an ideal gas molecule A occupies a volume of 1L. The gas molecule A obeys the relationship RT = 0.25PV. R and V are universal gas constant and ideal gas volume, respectively. The molecule A is

N O P Q

Which one of the following rearrangements represents a paracentric inversion? (a)

L O N

M P Q

(b)

L M

N P O Q

(c)

L M M

N N O P Q

(d)

L M N

O P Q

(d) innate immunity in prokaryotes 12. The process by which intracellular macromolecules are supplied for lysosomal degradation during nutrient starvation is (a) apoptosis

(b) autophagy

(c) phagocytosis

(d) pinocytosis

13. The process and instrumentation diagram for a feedback control strategy to maintain the level (h) of a liquid by regulating a valve (V) in a tank is shown below. F1 is inlet liquid flow rate, F2 is outlet liquid flow rate, LT is the liquid level transmitter, LC is the liquid level controller, hsp is the setpoint value of the liquid level, hm is the measured value of the liquid level and PV is the valve pressure.

7. Which one of the following statements is INCORRECT about hybridoma production? (a) Hybridoma cells can use hypoxanthine and thymidine (b) DNA synthesis in myeloma cells is blocked by aminopterin (c) Hybridoma cells are made to produce polyclonal antibodies (d) Polyethylene glycol is used to fuse myeloma cells to B-cells d 8.  ln(2 x) is equal to dx 1 1 (b) (a) 2x x 1 (c) (d) x 2 9. Which one of the following techniques/tools is NOT used for inserting a foreign gene into a cell? (a) DNA microarray (b) Electroporation (c) Gene gun (d) Microinjection

The manipulating variable(s) is/are (a) F1 only

(b) F2 only

(c) hm and PV only

(d) hsp and PV only

14. A protein without its prosthetic group is known as (a) apoprotein

(b) hemoprotein

(c) holoprotein

(d) lipoprotein

15. The enzyme which adds phosphate group to the free 5' terminus of a DNA sequence is (a) adenosine kinase (b) alkaline phosphatase (c) polynucleotide kinase (d) terminal deoxynucleotidyl transferase

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SOLVED PAPER - 2021

(d) Trypticase soy agar is a complex medium 17. The cellular process which utilizes RNA-induced silencing complex to block gene expression is (a) RNA editing (b) RNA interference (c) RNA polyadenylation (d) RNA splicing Q.18 & Q.19 : Multiple Select Question (MSQ), carry ONE mark each (no negative marks). 18. Which of the following layer(s) is/are formed from the inner cell mass of the blastocyst? (a) Ectoderm

25. The enzyme  -amylase used in starch hydrolysis has an affinity constant (Km) value of 0.005 M. To achieve one-fourth of the maximum rate of hydrolysis, the required starch concentration in mM (rounded off to two decimal places) is ____. Q.26 to Q.34 : Multiple Choice Question (MCQ), carry TWO marks each (for each wrong answer – 2/3). 26. Which one of the following represents nongrowth associated product formation kinetics in a bioprocess system? X and P denote viable cell and product concentrations, respectively. (a)

(b) Endoderm (c) Mesoderm (d) Trophectoderm 19. Which of the following cell organelle(s) is/are surrounded by a single phospholipid membrane? (a) Golgi apparatus

(b) Lysosome

(c) Mitochondria

(d) Nucleus

Q.20 to Q.25 : Numerical Answer Type (NAT), carry ONE mark each (no negative marks). 20. The sum of the infinite geometric series

1 1 1    ... (rounded off to one decimal 3 32 33 place) is _____. 1

(c)

X

(b)

P

Concentration

(c) Sabouraud dextrose agar is a differential medium

Time

X

(d)

P Time

Concentration

(b) Nutrient broth is a defined medium

Concentration

(a) Luria-Bertani broth is a synthetic medium

valve is closed and pressure P in R1 and R2 are 9 and 6 atm, respectively. Later, when the valve is opened, the system reaches equilibrium. If the temperature T of both the reactors is maintained constant, the final equilibrium pressure in atm of the system is _____.

Concentration

16. Which one of the following is CORRECT about microbial growth medium?

X P Time

X P Time

27. Match enzymes in Group I with their corresponding industrial application in Group II. Group I

Group II

21. Three balls, colored in blue, green and red, are successively transferred from box A to box B in the order BLUE-GREEN-RED. The probability of a reverse transfer of the balls to the box A in the same order (rounded off to two decimal places) is _____.

P. Amylase

1. Laundry detergent

Q. Invertase

2. Fruit juice clarification

R. Pectinase

3. Liquefaction of sucrose

S. Xylanase

4. Pulp and paper processing

22. Decimal reduction time of a bacterial strain is 20 min. Specific death rate constant in min–1 (rounded off to two decimal places) is _____.

(c) P-1, Q-2, R-3, S-4 (d) P-1, Q-4, R-2, S-3

 x  sin 2 x  (rounded off to two 23. The value of lim  x0  x  sin 5 x  decimal places) is _____. 24. A system consists of two reactors, connected by a valve. The first reactor (R1) contains an ideal gas A of volume 5 L and the second reactor (R2) has an ideal gas B of volume 10 L. Initially, the

(a) P-2, Q-3, R-4, S-1 (b) P-1, Q-3, R-2, S-4 28. Match separation methods in Group I with associated properties in Group II. Group I

Group II

P. Centrifugation

1. Density

Q. Dialysis

2. Diffusivity

R. Solvent extraction

3. Size

S. Ultrafiltration

4. Solubility

(a) P-4, Q-2, R-1, S-3

(b) P-3, Q-1, R-2, S-4

(c) P-1, Q-3, R-2, S-4

(d) P-1, Q-2, R-4, S-3

SOLVED PAPER - 2021

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29. Match the autoimmune diseases in Group I with the corresponding primarily affected organ in Group II. Group I

Group II

(3) poky   × wild-type

 All progeny are poky

(4) poky   × poky  All progeny are poky Which one of the following explains the inheritance mode of poky?

P. Hashimoto's disease

1. Brain

Q. Juvenile diabetes

2. Pancreas

R. Multiple sclerosis

3. Skeletal muscle

(a) Episomal inheritance (b) Mendelian inheritance

S. Myasthenia gravis

4. Thyroid

(c) Mitochondrial inheritance

(a) P-1, Q-2, R-3, S-4

(b) P-3, Q-1, R-2, S-4

(c) P-4, Q-2, R-1, S-3

(d) P-1, Q-2, R-4, S-3

30. Match hypersensitivity types in Group I with their corresponding condition in Group II. Group I P. Type I

Group II 1. Erythroblastosis fetalis

Q. Type II

2. Host reaction to bee venom

R. Type III

3. Systemic lupus erythematosus

S. Type IV

4. Tuberculin reaction

(a) P-2, Q-3, R-1, S-4

(b) P-3, Q-1, R-4, S-2

(c) P-2, Q-3, R-4, S-1

(d) P-2, Q-1, R-3, S-4

31. Which of the following combinations of plant hormones and their associated function are CORRECT? Hormone

Function

P. Abscisic acid

Breaks seed dormancy

Q. Auxin

Induces cell division

R. Ethylene

Stimulates ripening of fruits

S. Gibberellin

Promotes seed dormancy

(a) P and R only

(b) P and S only

(c) Q and R only

(d) Q and S only

32. Which one of the following tools is used to compare all the possible six-open reading frames of a given nucleotide query sequence with all the available six-open reading frames of the nucleotide sequence database? (a) BLASTN (b) BLASTX

34. Tertiary structure of a protein consisting of  -helices and  -strands can be determined by

(a) circular dichroism spectroscopy (b) mass spectrometry (c) nuclear magnetic resonance spectroscopy (d) UV spectroscopy Q.35 to Q.38 : Multiple Select Question (MSQ), carry TWO mark each (no negative marks). 35. Which of the following statement(s) is/are CORRECT about Agrobacterium tumefaciens? (a) It contains tumor inducing plasmid (b) It causes crown gall disease in dicotyledonous plants (c) It is a Gram-positive soil bacterium (d) It is used in generating transgenic plants 36. Which of the following antimicrobial agent(s) is/are growth factor analog(s)? (a) 5-Fluorouracil

(d) TBLASTX 33. In Neurospora crassa, a mutation in the poky gene results in a slow growth phenotype (poky). The results of four crosses are given below. (1) wild-type   × wild-type wild-type

 All progeny are

 All progeny are wild-

(b) Isoniazid

(c) Sulfanilamide (d) Tetracycline 37. Which of the following chemical messenger(s) is/are derivative(s) of tryptophan? (a)  -amino butyric acid (b) Indole acetic acid (c) Melatonin (d) Serotonin 38. Which of the following nucleus/nuclei is/are NMR active? (a) 1H (c) 16O

(c) TBLASTN

(2) wild-type   × poky type

(d) X-linked inheritance

(b) (d)

13C 32S

Q.39 to Q.55 : Numerical Answer Type (NAT), carry TWO marks each (no negative marks). 39. In a Mendel's dihybrid experiment, a homozygous pea plant with round yellow seeds was crossed with a homozygous plant with wrinkled green seeds. F1 intercross produced 560 F2 progeny. The number of F2 progeny having both dominant traits (round and yellow) is _____.

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SOLVED PAPER - 2021

40. A 0.1 mL aliquot of a bacteriophage stock having a concentration of 4 × 109 phages mL–1 is added to 0.5 mL of E. coli culture having a concentration of 2 × 108 cells mL–1. The multiplicity of infection is _____. 41. If the area of a triangle with the vertices (k, 0), (2, 0) and (0, –2) is 2 square units, the value of k is _____. 42. In a chemostat with a dilution rate of 0.8 h–1, the steady state biomass concentration and the specific product formation rate are 8 mol m–3 and 0.2 (mol product)(mol biomass) –1 h –1 , respectively. The steady state product concentration in mol m–3 is _____.

49. Calculate the following integral 2 / 4



0

sin x dx  _______.

50. A feed stream (F1) containing components A and B is processed in a system comprising of separation unit and a mixer as shown below in the schematic diagram. The mole fractions of the components A and B are xA and xB, respectively. If F1 + F2 = 100 kgh–1, the degrees of freedom of the system is _____.

43. If the values of two random variables (X, Y) are (121, 360), (242, 364) and (363, 362), the value of correlation coefficient between X and Y (rounded off to one decimal place) is _____. 1 1  1 1 A 44. The determinant of matrix  1 1  1 1

1 1 1 1  1 1 1 3

F4 XA4 XB4 F1

F3

XA1

XA3

XB1

XB3

Separation Unit

46. The specific growth rate of a mold during exponential phase of its growth in a batch cultivation is 0.15 h–1. If the cell concentration at 30 h is 33 g L–1, the cell concentration in gL–1 (rounded off to the nearest integer) at 24 h is _____. 47. A sedimentation tank of height 100 cm is used in a conventional activated sludge process to separate a suspension of spherical shaped granular sludge biomass of 0.5 mm diameter. The viscosity of the liquid is 1 cP. The difference in density between the suspended biomass and the liquid is 0.1g cm–3. If the biomass reach their terminal velocity instantaneously, the biomass settling time in min (rounded off to two decimal places) is ____. 48. In a random mating population, Y and y are dominant and recessive alleles, respectively. If the frequency of Y allele in both sperm and egg is 0.70, then the frequency of Y/y heterozygotes (rounded off to two decimal places) is_____.

XA5 XB5

Mixer

F6 XA6 XB6

F2

is _____. 45. It is desired to scale-up a fermentation from 1L to 1000 L vessel by maintaining a constant powerto-volume ratio. The small fermenter is operated at an agitator speed of 300 rotations per minute (rpm). If the value of scale up factor is 10, agitator speed in rpm (rounded off to the nearest integer) for the large fermenter is____ .

F5

XA2 XB2

51. A batch cultivation of E. coli follows zeroth order Monod's growth kinetics. The cell growth is terminated when the residual dissolved oxygen concentration attains 10% of its saturation value and oxygen mass transfer coefficient (kLa) reaches its maximum value (80 h–1). The saturation value of dissolved oxygen concentration is 0.007 kg m– 3. If the maximum specific growth rate and yield





coefficient YX / O2 are 0.2 h–1 and 1.5 (kg cells) (kg O 2) –1 , respectively, then the final cell concentration in kg m–3 (rounded off to two decimal places) at the end of the batch cultivation is _____. 52. Milk flowing through a stainless steel inner tube (40 mm inner diameter) of double tube-type heater is to be heated from 10 °C to 85 °C by saturated steam condensing at 120 °C on the outer surface of the inner tube. Total heat transferred (Q) is 146200 kcal h–1 and the overall heat transfer coefficient is 750 kcal h–1 m–2° C–1. The total length of the heating tube in m (rounded off to one decimal place) is _____.

SOLVED PAPER - 2021

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53. A DNA solution of 50 g mL–1 concentration gives an absorbance of 1.0 at 260 nm. An aliquot of 20 L from a 50 L purified plasmid solution is diluted with distilled water to a total volume of 1000 L. The diluted plasmid solution gives an absorbance of 0.550 at 260 nm. The concentration

55. A bacterium produces acetic acid from ethanol as per the following reaction 2CH3CH2OH + 2O2  2CH3COOH + 2H2O The thermodynamic maximum yield of acetic acid from ethanol in gg–1 (rounded off to two decimal places) is _____.

of the purified plasmid in g L1 (rounded off to two decimal places) is _____. 54. The possible number of SalI restriction sites in a 9 kb double-stranded DNA, with all four bases occurring in equal proportion (rounded off to the nearest integer) is ____ .

ANSWERS General aptitude (GA) 1. (c)

2. (a)

3. (c)

4. (a)

5. (c)

6. (c)

7. (c)

8. (c)

9. (d)

10. (d)

1. (d)

2. (a)

3. (b)

4. (a)

5. (d)

6. (b)

7. (c)

8. (b)

9. (a)

10. (d)

11. (b)

12. (b)

13. (a)

14. (a)

15. (c)

16. (d)

17. (b)

18. (a; b; c)

19. (a; b)

20. (1.5 to 1.5)

21. (0.16 to 0.18)

22. (0.10 to 0.13)

23. (0.25 to 0.25)

24. (7 to 7)

25. (1.60 to 1.80)

26. (c)

27. (b)

28. (d)

29. (c)

30. (d)

31. (c)

32. (d)

33. (c)

34. (c)

35. (a; b; d)

36. (a; b; c)

37. (b; c; d)

38. (a; b)

39. (315 to 315)

40. (4 to 4)

41. (0 or 4)

42. (2 to 2)

43. (0.5 to 0.5)

44. (8 to 8)

45. (64 to 65)

46. (13 to 14)

47. (1.20 to 1.30) 48.(0.41 to 0.43)

49. (2 to 2)

50. (6 to 6)

51. (3.70 to 3.80)

52. (23.0 to 24.0)

53. (1.37 to 1.38)

Biotechnology

55. (1.90 to 2.00)

54. (2 to 2)

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EXPLANATIONS GENERAL APTITUDE 1. Given: Ratio of boys to girls if 7 to 3  Multiples of 10 are : 10, 20, 30, 40, 50, 60, 70, ..........  An acceptable value for the total number of students is 50.

Where h = 1, r is radious of base of cylender But, perimeter of base is 2r = 6 cm  r=

6 3 = 2  2

9  3 3  V =    (1)  V = cm  

2.

1 cm 6 cm Sheet N

Here line segment does not lie inside of on the polygon. Hence the figure in option (B) is not a convex polygon 3. (i) is grammatically correct and (ii) is ambiguous. Statement 2 is ambiguous because we do not know who enjoys playing chess, Babu or Danish!! Statement 1 is grammatically correct. 4. A circular sheet of paper is folded along the lines in the directions shown. The paper, after being punched in the final folded state as shown and unfolded in the reverse order of folding, will look like figure (a). 5. Doctor is to surgery as writer is to book i.e. option (a) maintains a similar logical relation in the sentence 6. Given,

1 cm 6 cm

1

1 cm

Volume of unit cube = 1 cm3 Required ratio =

9 9 =   (1)

7. Given: Ratio of cost of item P to cost of item Q=3:4

Sheet M

Cost of item P = 5400 Cost of item Q = 7200 Profit % on item Q = 25

1

 Selling price of item Q = 7200 

125 = 9000 100

 Discount of item Q = Marked price – Selling price

6 cm Volume of cylinder V = r2h

= 10,000 – 9000 = 1000  Discount % = 1000  100 = 10 10000

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3. Formula for unrooted tree in phylogeny (2n  5)! where n is number of taxas 2 (n  3)! For 5 sequences,

8.

=

Total number of cases in sample space = 10  10  10  10  10 =105 Event A  At least two chocolates are identical Probability of A, ie., P (atleast two are identical) = 1– P (all different) P(A) = 1 



10  9  8  7  6 105

= 1 – 0.3024 = 0.6976 9. Using Venn diagrams, the different possibilities are M

M

M B

B

P

(ii)

(iii)

y = r sin  = 4 



(iv)

From figure (i), conclusion I is incorrect Hence, neither conclusion I nor II is correct

Superficial: shallow, cursory mean lacking in depth or solidity. superficial implies a concern only with surface aspects or obvious features.

2. To eukaryotes RNA pol I catalyze the formation of r-RNA such as 28S, 18S and 5.8S rRNA exception SS rRNA is catalyzed by RNA pol III. RNA pol II catalyzes mRNA formation and other regulatory RNA formation. RNA pol III catalyzes transfer RNA formation and 5S rRNA formation. r-RNA is responsible for ribosome formation, which is involved in protein synthesis.

NOPQ

L M

NPOQ

PEG is used to fuse myeloma cells to B-cell. It acts as Fusogen. Aminopterin block de novo synthesis of purine formation, so therefore DNA replication is stop.

1. Coronaviruses are enveloped virus with positivesense SS RNA genome and a nucleocapsid of helical symmetry.

Coronavirus can cause pneumonia viral or bacterial and bronchitis than may be lethal to humans.

L M

7. Hybridoma technology used to form monoclonal antibodies not poly clonal antibodies, which binds to single epitope of an antigen.

BIOTECHNOLOGY

Genome size –26 to 32 kbp. They cause respiratory tract infections. SARS, MERS and COVID -19 are lethal.



Because paracentric inversion do not involve centromere. It occurs within a single arm of chromosome.

From figure (ii), conclusion II is incorrect 10. As AOM are not addressing the core problems, they are superficial.

1 = 2 2 2

 Required point A is 2 2, 2 2 , option (c) is correct. 6.

P

(i)

(2  5  5)! 5! 5  4  3  2! = 2 = = 15 53 2 (5  3)! 2 2! 4  2! 4. Boot’s strapping is on method which provides confidence of the phylogenetic tree created showing the number of times a branch will be observed while constructing the tree.  5. r = 4,  = 4 1  x = r cos  = 4  = 2 2 2

M P B

B P

n3

Hybridoma cells use hypoxanthine and Thymidine due to presence of enzyme hypoxanthine guanine phosphoribosyl transfer rase (HGPRT) Thymidine kinases (TKs). d 1 d 1 1 2 =  l n(2 x) =  (2 x) = dx 2 x dx 2x x 9. A microarray is a tool used to detect the expression of thousand of genes or it is used to check the gene expression of thousand genes at a time. It is a Nucleic acid hybridization based method to detect expression of genes.

8.

Electroporation with the help of electric current temporary pores are formed DNA insert into the cells. Gene Gun: DNA coated on Gold or Tungsten particle and transfer into plant cells. Microinjection: It deliver foreign DNA into living cells through glass micropipette.

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10. Given, RT = 0.25 PV We know that, RT = nPV Where n = number of moles m (m = given mass in g and M is molar M mass in g)

And n =

 0.25 =

128 M

128  M= = 512 g 0.25 512 = 16 32 11. This system was discovered from prokaryotic organism showing resistance to phages and viruses whose guide sequence is already present in the palindromic repeat sequence.

So, A is O2 as

height setpoint) and sends a pressure signal (Pv) to the valve. This valve top pressure moves the valve stem up and down, changing the flow rate through the valve (F1). If the controller is designed properly, the flow rate changes to bring the tank height close to the desired setpoint. In this process and instrumentation diagram dashed lines are used to indicate signals between different parts of instrumentation. 14. Apoprotein means protein part of an enzyme. It lacks prosthetic group which may be metal ions or vitamin B complex such as NAD+, FAD, Biotin, TPP etc. Lipoprotein: It is a lipid protein complex and involve in lipid transport such an HDL, LDL, ULDL etc. Hemoprotein: It contains Heme such as Hemoglobin and its transport oxygen to body.

Hence is called an adaptive immunity and not innate.

Halo protein: It is a combination of protein and Non protein part of enzyme.

12. During starvation autophagy occurs it remove damage protein and organelles to prevent cell damage and intracellular molecules are digested to provide the nutrients of cell needs.

15. Polynucleotide kinase add 5´ terminal phosphate to DNA. The kinases enzyme responsible for the Transfer of phosphate.

Apoptosis is a programmed cell death process catalyzed by Fas, FasL, caspases lead to cell death. Phagocytosis is the entrapment and engulfment of antigen; it is done by phagocytic cell known as macrophage. Pinocytosis: It is a process by which liquid droplets are injected by living cells. 13. The measured variable for a feedback control strategy is the tank height. Which input variable is manipulated depends on what is happening in the 2 processes (process 1 and process 2). Let us consider two different processes- process 1 and process 2.Assume that the process 2 regulates the flow-rate F2 and process 1 controls flow rate F1. This could happen, only, if process 2 is a steam generation system and process 1 is a deionization process. Process 2varies the flow rate of water (F2) depending on the steam demand. As far as the tank process is concerned, F2 is a “wild” (disturbance) stream because the regulation of F2 is determined by another system. In this case, the F1 would be used as the manipulated variable; that is, F1 is adjusted to maintain a desired tank height. Notice that the level transmitter (LT) sends the measured height liquid in the tank (hm) to the level controller (LC). The LC compares the measured level with the desired level (hsp, the

Holoenzyme: Apoprotein + Prosthetic group.

Alkaline phosphates remove phosphate from 5’ terminal phosphate of DNA molecule. Tdt enzyme odd nucleotides to 3’ end of the DNA molecule. It makes DNA sticky. Adenosine kinase add phosphate group to Adenosine nitrogenous base. 16. Trypticase soy agar is a complex media to grow many types of bacteria, it grows certain pathogenic bacteria which have high nutritional requirements. Sabouraud Dextrose Agar (SDA) is a selective media for the isolation of Dermatophytes such as fungi and yeast. LB and N.B is a complex media because nutrient not defined, it contains beef extract and peptone where exact chemical composition is unknown. 17. RNA interference is a biological process where RNA molecule inhibits gene expression or translation, mi-RNA, si-RNA play a important role, double stranded RNA molecule form cleaved by DICER enzyme, that leads to RNA induced silencing complex (RISC) formation. RNA poly-adenylation: Where 50-250 adenine residues bind to 3 end of mRNA catalyze by PdyAdenylate Polymerase (PAP). RNA Splicing: It removes introns and join the exons which is coding region.

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RNA editing: RNA molecules are edit by Guide RNA, by addition and deletion in RNA leads to truncated protein formation.

After removal of valve – the environmental condition is same then Vtotal = V1 + V2 = 5 + 10 = 15

18. Gastrulation is the formation of the three layers of the embryo: ectoderm, endoderm, and mesoderm. The endoderm gives rise to the lining of the digestive system and respiratory system. The ectoderm gives rise to the nervous system and epidermis. The mesoderm gives rise to the muscle and skeletal systems.

ntotal = n1 + n2 For ideal gas PV = nRT P = pressure, V = volume, R = gas constant, T = temperature

19. Vacuole, lysosome, Golgiapparatus Endoplasmic reticulum are single membrane bound organelles present only in eukaryotic cell.

n = number of moles For n1 and n2 , P1 = 9 atm, P2 = 6 atm V1 = 5L, V2 = 10L

Nucleus, mitochondria and chloroplast are double membrane bound organelles present only in eukaryotic cells.

n1 =

So therefore, lysosome and Golgi apparatus are correct.

n1 = 1.81, n2 = 2.41 n = (n1 + n2 ) = 1.81 + 2.41 = 4.22 So for the two gas system P * V* = n, RT 4.22  0.08205  303 p* =  7 atm 15L  25. Given km = 0.005 M and  = m 4

So,

1 1 1 1   = = 0.1666  0.17 3 2 1 6

 Xt  22. l n  X  = –kdt  o

kd = 0.115 min–1 = 0.12 min–1 sin 2 x   sin 2 x       x  1  x     1  x   lim  lim    23. x  0   sin 5 x   = x  0   sin 5 x    x  1  5     1  5  

sin ax   1 2  a = 0.25  lim x0  x 15 24. Given a system with two reactor =

R1 having ideal gas A – 5L at 9 atm R2 having ideal gas B –10L at 6 atm

(using MM kinetics)

m ms = 4 0.005  s  0.005 + s = 4s

i.e., no change in both directions.

 0.1  ln  1  = –kd(20)

m s m = k s 4 m

0.005 =s 3  1.67 mM



26.

X P Time Mixed growth Associated

X P

Time Growth Associated

Concentration

3 = = 1.5 1 2 1 3 21. The probability that three balls (red, green, blue, respectively)successively, transferred from box A to box B is 1

and value of R = 0.08205 L.atm.mol–1.k–1

Concentration

a  Required sum is = 1– r

Let temperature T = 303K (constant),

Concentration

1 1 1 20. 1   2  3  ....... is infinite geometric series 3 3 3 1 with common ratio r = (i.e., –1 < r < 1) and 3 first term, a = 1

P1 V1 PV , n2 = 2 2 RT RT

X P Time Non-growth Associated

27. Amylases: are the enzyme used in detergent formulation to improve the detergency. It is used to digest the starch grains. Xylanases: It is used in paper and pulp industry, it act as bio bleaching agent, increase the pulp fibrillation and water retention. It reduce the beating time in virgin pulps.

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Pectinases: It is involved in fruit juice clarification. This enzyme used to breakdown pectin polymer into monomeric form galacturonic acid.

Gibberellins, it is responsible for seed germination, stem elongation, leaf expansion, pollen maturation and induction of flowering.

Invertase: It catalyze the hydrolysis of sucrose into glucose and fructose, it is responsible for liquefaction of sucrose.

Abscisic acid is a growth inhibitor hormone in plants. It acts as antagonist to gibberellic acid. It is responsible for stress tolerance in plants.

28. Centrifugation is a technique used for the separation of particles from a solution according to their size, shape, density, viscosity of the medium and rotor speed. The particles are suspended in a liquid medium and placed in a centrifugal tube. The tube is then placed in a rotor and spun at a defined speed. Ultra-Filtration (UF) is a pressure-driven barrier to suspended solids, bacteria, viruses, entotoxins and other pathogens to produce water with very high purity and low silt density. Ultra Filtration (UF) is a variety of membrane filtration in which hydrostatic pressure forces a liquid against a semi permeable membrane. 29. Myasthenia gravis is a rare autoimmune disease at the neuromuscular junction of skeletal muscle. Auto-anti bodies block the acetylcholine receptors. Diabetes it affects the pancreatic beta cells which forms insulin, and the insulin formation defective result in hyperglycemia. Multiple sclerosis, it affects brain which matter, Th1, Tc and autoantibodies damages the brain white matter. Hashimoto’s diseases: Thyroid protein and cells are damaged by TDTH and autoantibodies.

32. Variants of BLAST • BLASTN: Compares a DNA query to a DNA database. Searches both strands automatically. It is optimized for speed, rather than sensitivity. • BLASTP: Compares a protein query to protein database. • BLASTX: Compares a DNA query to a protein database, by translating the query sequence in the 6 possible frames and comparing each against the database (3 reading frames from each strand of the DNA) searching. • TBLASTN: Compares a protein query to a DNA database, in the 6 possible frames of the database. • TBLASTX: Compares the protein encoded in DNA database, in the 6*6 possible frames of both query and database sequences (Note that all the combinations of frames may have different scores). • BLAST 2: Also called advanced BLAST. It can perform gapped alignments. 33.   poky  progency poky  wild type  progency wild type 

Type III: Immune complex mediated, in includes systemic lupus erythema matosus, Arthritis and glomerulonephritis.

Mitochondrial inheritance 34. Tertiary structure of a protein consisting of ahelices and ß-strands can be determined by NMR. Several methods are currently used to determine the structure of a protein, including X-ray crystallography, NMR spectroscopy, and electron microscopy. Each method has advantages and disadvantages. In each of these methods, the scientist uses many pieces of information to create the final atomic model.

Type IV: Cell mediated hypersensitivity TDTH cell mediated, it includes tubercular lesion contact dermatitis and graft rejection.

35. Agrobacterium tumefaciens is a • Gram Negative Bacteria • Rod shaped bacterial

30. Type I: 19E mediated hypersensitivity 19E Ab binds to mast cells release Histamine that leads to systemic and localized anaphylaxis. Type II: 19G mediated hypersensitivity, it includes erythroblastosis fetalis, Autoimmune hemolytic anemia.

31. Auxin promote cell elongation, inhibits growth of lateral buds, it induces cell division, cell expansion and cell differentiation. Ethylene act as plant growth regulator it stimulates senescence, ripening and abscission. It plays an important role in ripening of fruits.

• Present in soil • It infects dicots plants. • It has Ti-plasmid. • It is responsible for crown gall Tumor in bacteria.

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• The T-DNA (Transfer DNA) is entering to plant cause Tumor in plants. • The T-DNA contains Auxin, cytokinin and opines genes. 36. Option A B D is correct, because they are growth factor analogue, they inhibit PABA and stop purine synthesis. But tetracycline is an antibiotic which inhibits protein synthesis by blocking A site of Ribosome therefore tetracycline drug is protein synthesis inhibitor not growth factor analogue. 37. Melatonin is a hormone formed by PINEAL GLAND, It is produced from tryptophan amino acid. It is involved in circadian rhythms or diurnal variations; it performs and Neurotransmitters also. Serotonin: It is Neurotransmitter derive from tryptophan. It is formed in intestinal cells. It is a vaso­constrictor, and it controls behavior patterns sleep, blood pressure and body temperature. Indole Acetic acid: It is a phytohormone. Derived from tryptophan. It regulates growth in plants. It is responsible for cell division and elongation process.  - amino butyric Acid: It is deriving from glutamate amino acid not from truptophan amino acid. 38. Except 1H all other elements needs to be present in stable isotope form in order to be NMR active the element which changes its spin position when placed in a magnetic field. Example: 12C and 16O are not NMR active. RRYY

rryy

round yellow

wrinkled green

RY

Ry

rY

ry

RY

RRYY

RRYy

RrYY

RrYy

Ry

RRYy

RRyy

RrYy

Rryy

rY

RrYY

RrYY

rrYY

rrYy

ry

RrYy

Rryy

rrYy

rryy

9 Both dominant trait = 16

Number of pea plant =

41. Area of triangle with the vertices A (x1, y1), B (x2, y2) and C (x3, y3) is 1 x1 1 1 x2 2 1 x3

y1 2

y2 y3

1 k 0



1 1 2 0 2 =2 2 1 0 2



1  2(k  2) = 2 2

 |2 – k|= 2  2–k=2  k=0 42. Given, D =  = 0.8h–1 for chemostat, D = dilution,  = Specific growth rate of biomass production 1 dP =0.2 h –1 , Q P = Specific product X dt formation rate constant

Qp =

1 dp Qp dp dt = 0.2 = 0.25, X Yp/X = = =  1 dX dX 0.8 X dt Yp/x = Product yield (g of prod/ g of biomass)

Product Concentration P = Yp/X  X = 0.25  8 = 2 mol m3

43. n  3

_______ F1

RrYy

4  109 cells m l 1  0.1m l = 2  108 cells m l 1  0.5ml = 4 (Ans)

x = 8 mol/m3

It is inhibitory neurotransmitter.

39.

40. Multiplicity of infection

9  560 = 315 16

X 1  121 Y1  360 X 2  242 Y2  364 X 3  363 Y3  362

1 X = 3 [121  242  363] = 242

1 Y = 3 [360  364  362] = 362

x =

1  xi  X n



=



2

=

1 (121)2  (0)2  (121)2 3

2  121 = 98.796 3

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1 y =  yi  Y n



=



2

1 (2)2  22  02 = 3

8  1.632 3

33 = X0e015  30 X0 = 0.367 g/L Xt at 24h, Xt = 0.367  e015  24 = 13.43 g/L 47. Given,

1 Cov(x,y) =  xi yi  XY n

Biomass (cell) Diameter

1 = [121 ]–242  3

= 0.5 mm =

P = Particle density, = solvent density

121 [360 + 728 + 1086] – 242  362 3 = 87684.6667 – 87604 = 80.667



cov( x, y) 80.6667 =  x y (98.796)  (1.632) = 0.5

 r=

1

44.

1

(P – ) = 0.1 gcm–3  = viscosity = 1Cp = 0.01 g/cms g = 9.8 m/s = 980 cm/s So we know,Velocity of sedimentation exposed by Stokes equation

1 1

Vs =

1 1 1 1 1 1 1 1 1

1

1 3

0 2 2 2 = 0 0 2 2

=1  2  2  2 = 8

D1 1 N1 = 300 rpm, D = 10 2 N2 = ?

P    D2i V

dx = 2tdt

And

x=0t=0

 2  t = 2 4  /2

 Given integral becomes

 sin t  2tdt

= 2[(t)(–cost)–(1)(–sint)] 0 / 2 ( Integration by parts)

2

= 2 [(0 – 0) + (1 – 0)] = 2

6

50. Consider F1 as independent variable and F2 as dependent

= N2 = 64.63

Answer is 64 to 65 46. Given, exponential growth rate = 0.15h–1 Cell concentration at 30hr is 33gL

–1

Cell concentration at 24h is _______ . We know, Xt = X0eut



0

2 2

2

27  10 102

x = t2

x=

D  (300)3  1  = N32 D 

3

Heterozygotes = 2pq 49. Put

Scale up factor –10

So,  D =  D

Y = 0.70 and y = (1 – 0.70) = 0.30 = 2  0.70  0.30 = 0.14  0.3 = 0.42

45. Given, 1L to 1000 L

 

100 cm 100 1 min = 1.22 min  = Vs 1.36 60

y = recessive

is determinant of upper triangular matrix.

r 1

980  0.1  (0.05)2 = 1.36 cm/s 18  0.01

48. Y = dominant

0 0 0 2

For power to volume ratio

18 

Now time,

1 1 1 1

 

g( p  )D2P

=

(Applying R2 + R1; R3 + R1; R4 – R1)

0.5 = 0.05 cm 10

F1 + F2 = 100 kgh–1 F = 100 – F1  (F2  dependent) F 1 – F 2 = F3 F3 = F1 – (100 – F1) F3 = 2F1 –100 (F3  dependent)

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F4 XA4 XB4 F1

F3

XA1

XA3

XB1

XB3

F5

Separation Unit

XA5

Mixer

XB5

F6 XA6 XB6

F2

In mixer, mole fraction again changes xA6 + xB6 = 1 xB6  dependent xA6  independent Number of independent variables = Degree of freedom = 6 [F1, F4, xA1, xA4, xA5, xA6] 51. Given the cell growth will terminate at 10% saturated oxygen concentration kLA = 80h–1,

XA2

 = 0.2h–1,

XB2

YX / o2 = 1.5 kg.cells/kg of O2

Between F4 and F5, consider F4 as independent variable Also, F3 – F4 = F5 F5 = 2F1 – 100 – F4 (F5  dependent) F6  F5 + F2 F6 = 2F1 – 100 – F4 + 100 – F1 (F6  dependent)

C* = 0.007 kg.m–3, C = C*  10% We know for mass transfer equation, XqO2 = K La (C* – C)  constant This formula will follow at equilibrium or at no growth

Hence, F1 and F4 are independent variables

So, in the case at 10% saturation

Given, Mole fraction of A  xA

We know, qO2 = oxygen up take rate constant.

Mole fraction of B  xB Now,

x A1 + x B1 = 1

1 dCO2 1 dX dCO2   = X dt X dt dX

=

or, x B1 = 1–x A1 Hence XA1 is independent and XB1 is dependent

 0.2 = Y = = 0.133 X / O2 1.5

F1

F3

XA1 XB1

XA3 XB3 F2 XA2 XB2

Division in flow, does not result in change in concentration. Hence, xA1 = xA3 and xB1 = xB3 Similarly, xA2 = xAl and xB2 = xB1 = 1 – xA1 Hence, xA3, xB3, xA2, xB2 are dependent variables. In separation unit, flow divides concentration changes Hence xA4 + xB4 = 1 Considering xA4 independent xB4 = 1 – xA4 xA4  independent, xB4  dependent Similarly, xA5  independent xB5  dependent

Hence, X(0.133) = 80(0.007 – 0.0007) X=

80(0.007  0.0007) = 3.789 0.133

52. Given,

40 mm 10ºC

120ºC steam

85ºC

We know, Overall heat transfer across heat exchanger Q = U  Aoverall TLMTD U = overall heat transfer coefficient A = Area LMTD = Log mean temperature difference Aoverall = 2r.L = 2   20  10-3.L = 0.1256L T1 = (120 – 10) = 110ºC T2 = (120 – 85) = 35ºC TLMTD =

110  35 T1  T2 = = 65.5ºC  110   T1  l n   ln 35   T2 

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Given, U = 750 kcal h .m .ºC –1

–2

Q = 146200 kWh–1 Q = U  Aoverall  TLMTD 146200 = 750  0.1256.L  65.5 146200 = L = 23.69 m 750  0.1256  65.5

First condition OD = 1 and C1 = 50g/ml We know from Beer-Lambert Law, OD = .C.l Where, C = Concentration  = constant l = path length Then 1 = l  50 1 = l 50

But as given in second condition OD = 0.55 As we know OD = .C.l 1 C 50 2

Hence C2 = 27.5g/ml Now concentration of purified plasmid =

27.5  Dilution factor 1000

=

The possible number of restriction size in 9kb double-stranded DNA is 9000 9  10 3 = = 2.19  2 4096 46

55. 2CH3CH2OH + 2O2  2CH3COOH + 2H2O To find maximum yield of acetic acid = YP/S max

53. Given,

0.55 =

54. Sal I restriction size = 6 base power frequency of sal I restriction size = 46 = 4096

27.5  50 = 1.37 g/l 1000

Ethanol  C2H6O  46 Acetic acid  C2H4O2  60 rs =

4262 = 6, 2

42422 =4 2 Maximum number of moles

rp =

 rs 26 We know, fmax = jr = = 1.5 mole p 24 Maximum yield yPS = =

1.5  MW of product MW of substrate

1.5  60 = 1.95 g/g 46