GATE Biotechnology Chapter - Wise Solved Papers (2000 - 2020) [First ed.] 9788126595440, 9788126577231
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Table of contents :
Cover
Title
Copyright
Contents
Note to the Aspirants
Contents
Chapter 1: Engineering Mathematics
Questions
Answers with Explanation
Chapter 2: Biochemistry
Questions
Answers with Explanation
Chapter 3: Microbiology
Questions
Answers with Explanation
Chapter 4: Cell Biology
Questions
Answers with Explanation
Chapter 5: Molecular Biology and Genetics
Questions
Answers with Explanation
Chapter 6: Analytical Techniques
Questions
Answers with Explanation
Chapter 7: Immunology
Questions
Answers with Explanation
Chapter 8: Bioinformatics
Questions
Answers with Explanation
Chapter 9: Recombinant DNA Technology
Questions
Answers with Explanation
Chapter 10: Plant Biotechnology
Questions
Answers with Explanation
Chapter 11: Animal Biotechnology
Questions
Answers with Explanation
Chapter 12: Bioprocess Engineering and Process Biotechnology
Questions
Answers with Explanation
Appendix: GATE BIOTECHNOLOGY 2020
Questions
ANSWER KEY
Answers with Explanations
Back Cover
Citation preview
GATE
About the Book
The book comprises previous years' (2000 to 2020)* GATE questions in Biotechnology covered chapter-wise with detailed solutions. Each chapter begins with a detailed year-wise analysis of topics on which questions are based and is followed by explanation and concepts for that chapter. The book is designed to make the students well-versed with the pattern of examination, level of questions asked and the concept distribution of questions and thus bring greater focus to their preparation. It aims to be a must-have resource for an essential step in their preparation, which is, solving and practicing previous years' papers.
Salient Features l
Chapter-wise coverage of GATE Biotechnology previous years' questions (2000 to 2019).*
l
Solved GATE 2020 Questions included chapter-wise as appendix at the end of the book.
l
Detailed topic-wise explanation of questions for each chapter.
l
All questions marked for level of difficulty (i.e. 1 Mark or 2 Marks).
l
Detailed solutions for all questions, tagged topic-wise.
l
Packaged with unique scratch code for free 7-day subscription for online GATE tests.
* Biotechnology was introduced as a separate GATE paper from year 2010 onwards. Earlier it was an optional section in GATE Life Sciences paper, so questions for years 2000-2009 are included from GATE Life Sciences papers.
BIOTECHNOLOGY CHAPTER-WISE SOLVED PAPERS (2000-2020)
The Graduate Aptitude Test in Engineering (GATE) is an All-India level examination for engineering graduates aspiring to pursue Master's or Ph.D. programmes in India. The Public Sector Undertakings (PSU's) also use GATE as a recruiting examination. The examination is highly competitive, and the GATE score plays an important role in fulfilling the academic and professional aspirations of the students. This book is aimed at supporting the efforts of GATE aspirants in achieving a high GATE score.
GATE
BIOTECHNOLOGY
CHAPTER-WISE SOLVED PAPERS *
(2000-2020)
Inside
l
Wiley India Pvt. Ltd. Customer Care +91 120 6291100 [email protected] www.wileyindia.com www.wiley.com
l
l
ISBN 978-81-265-7723-1
l
9 788126 577231
* Biotechnology introduced as a separate GATE paper from year 2010 onwards.
Detailed Exam Analysis Chapter-wise and Topic-wise. * Solved Questions from previous years’ (2000 – 2019) papers. Solved GATE 2020 Questions Included Chapter-wise as Appendix Unique scratch code in the book that provides access to w Free online test with analytics. w 7-Day free subscription for topic-wise GATE tests. w Instant correction report with remedial action.
GATE
BIOTECHNOLOGY
CHAPTER-WISE SOLVED PAPERS
(2000-2020)
GATE BIOTECHNOLOGY CHAPTER-WISE SOLVED PAPERS
(2000-2020)
Copyright © 2020 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 First Edition: 2019 Second Edition: 2020 ISBN: 978-81-265-7723-1 ISBN: 978-81-265-9544-0 (ebk) www.wileyindia.com Printed at:
Note to the Aspirants About the Examination Graduate Aptitude Test in Engineering (GATE) is an examination conducted jointly by the Indian Institute of Science (IISc), Bangalore and the seven Indian Institutes of Technology (at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee) on behalf of the National Coordination Board (NCB)GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD), and Government of India. Qualifying in GATE is a mandatory requirement for seeking admission and/or financial assistance to:
• M aster’s programs and direct Doctoral programs in Engineering/Technology/Architecture. • Doctoral programs in relevant branches of Science, in the institutions supported by the MHRD and other Government agencies. • Many Public Sector Undertakings (PSUs) have been using the GATE score in their recruitment process.
Graduate Aptitude Test in Engineering (GATE) is basically an examination on the comprehensive understanding of the candidates in various undergraduate subjects in Engineering/ Technology/Architecture and post-graduate level subjects in Science. GATE is conducted for 24 subjects (also referred to as “papers”) and is generally held at the end of January or early February of the year. The examination is purely a Computer Based Test (CBT). The GATE score reflects the relative performance level of the candidate in a particular subject, which is quantified based on the several years of examination data.
Eligibility for GATE Biotechnology
Bachelor’s degree in Engineering/Technology or Master’s degree (M.Sc) in any relevant science subject or must be in the final year of the programme.
About the Book This book GATE Chapter-wise Biotechnology Solved Papers is designed as a must have resource for the students preparing for the M. E./M.Tech./M.S./Ph.D. in Bio-Sciences or BioEngineering. It offers Chapter-wise solved previous years’ GATE Biotechnology questions for the years 2000–2020*. This book will help students become well-versed with the pattern of examination, level of questions asked and concept distribution in questions. Key Features of the Book
• C omplete solutions provided for every question, tagged for the topic on which the question is based. • Chapter-wise analysis of GATE questions provided at the beginning of the book to make students familiar with chapter-wise marks distribution and weightage of each. • Topic-wise analysis of GATE Questions provided at the beginning of each chapter to make students familiar with important topics on which questions are commonly asked. • Unique scratch code in the book that provides access to Free online test with analytics. 7-Day free subscription for topic-wise GATE tests. Instant correction report with remedial action. These features will help students develop problem-solving skills and focus in their preparation on important chapters and topics. ■
■
■
Validity
The GATE score is valid for THREE YEARS from the date of announcement of the results.
*Biotechnology was introduced as a separate GATE paper from year 2010 onwards. Earlier it was an optional section in GATE Life Sciences paper so questions for years 2000–2009 are included from GATE Life Sciences papers.
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GATE 2012
GATE 2013
GATE 2014
GATE 2015
GATE 2016
GATE 2017
GATE 2018
GATE 2019
GATE 2020
4 3 1
2 3 1 2
3
Microbiology
Cell Biology
Molecular Biology and Genetics
Analytical Techniques
Immunology
Bioinformatics
Recombinant DNA Technology
Plant Biotechnolgy
Animal Biotechnology
Bioprocess Engineering and Process Biotechnology
3
4
5
6
7
8
9
10
11
12
3
1
2
Biochemistry
2
0
Engineering Mathematics
6
1
0
3
0
1
2
7
2
3
2
3
0
1
1
2
3
6
2
5
1
0
4
0
8
0
2
1
0
0
0
5
2
1
9
2
4
0
0
2
2
4
0
7
0
1
2
3
6
0
1
5
2
0
1
2
0
2
6
5
6
1
0
1
1
3
1
2
0
2
3
5
2
1
2
1
4
2
0
7
2
0
5
4
1
1
1
3
1
5
0
2
1
2
3
3
7
1
0
4
2
2
0
4
0
4
3
4
6
1
0
1
1
1
2
2
1
3
3
4
7
2
1
3
0
2
3
4
1
1
3
2
1
0
1
2
0
3
2
5
2
2
4
3
3
0
1
4
0
2
4
2
2
4
3
5
1
1
1
2
2
1
3
2
3
1
3
5
7
1
0
1
2
4
0
5
0
4
3
2
4
1
1
1
2
0
0
2
1
3
4
4
9
2
1
3
1
1
3
3
1
0
3
4
3
1
1
4
1
1
1
2
3
2
1
5
3
2
3
6
2
1
1
3
2
1
3
3
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks Mark Marks
GATE 2011
1
S.No.
Chapter Name
The table given below depicts the chapter-wise question and marks distribution of previous years’ (2011–2020) GATE Biotechnology papers. This will help students understand the relative weightage of each chapter in terms of number of questions asked and thus bring focus in their preparation.
GATE Biotechnology: Chapter-Wise Question Distribution Analysis 2011–2020
Contents
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Note to the Aspirants
iii
Chapter 1: Engineering Mathematics Questions Answers with Explanation
1 2 6
Chapter 2: Biochemistry Questions Answers with Explanation
19 19 28
Chapter 3: M icrobiology Questions Answers with Explanation
43 43 47
Chapter 4: Cell Biology Questions Answers with Explanation
55 55 58
Chapter 5: Molecular Biology and Genetics Questions Answers with Explanation
65 65 75
Chapter 6: Analytical Techniques Questions Answers with Explanation
91 91 94
Chapter 7: Immunology Questions Answers with Explanation
101 101 107
Chapter 8: Bioinformatics Questions Answers with Explanation
119 119 122
Chapter 9: Recombinant DNA Technology Questions Answers with Explanation
129 129 141
Chapter 10: Plant Biotechnology Questions Answers with Explanation
163 163 168
Chapter 11: Animal Biotechnology Questions Answers with Explanation
179 179 183
Chapter 12: Bioprocess Engineering and Process Biotechnology Questions Answers with Explanation
191 192 212
Appendix: GATE 2020 Biotechnology Paper
249
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CHAPTER1
Engineering Mathematics
Syllabus Linear Algebra— Matrices and Determinants, Systems of Linear Equations, Eigen Values and Eigen Vectors. Calculus: Limit, Continuity and Differentiability, Partial Derivatives, Maxima and Minima, Sequences and Series, Test for Convergence, Fourier Series. Differential Equations— Linear and Nonlinear First Order ODEs, Higher Order ODEs with Constant Coefficients, Cauchy’s and Euler’s Equations, Laplace Transforms, PDE-Laplace, Heat and Wave Equations. 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 of Trapezoidal and Simpson’s Rule, Single and Multistep Methods for Differential Equations.
Chapter Analysis Topic Linear Algebra Matrices and Determinants Systems of Linear Equations Eigen Values and Eigen Vectors Calculus Limit, Continuity and Differentiability Partial Derivatives Maxima and Minima Sequences and Series Test for Convergence Fourier Series Differential Equations Linear and Nonlinear First Order ODEs, Higher Order ODEs with Constant Coefficients Cauchy’s and Euler’s Equations Laplace Transforms, PDE-Laplace, Heat and Wave Equations Probability and Statistics Probability 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 of Trapezoidal and Simpson’s Rule Single and Multistep Methods for Differential Equations Others Mensuration Vectors
GATE GATE GATE GATE GATE GATE GATE GATE GATE 2011 2012 2013 2014 2015 2016 2017 2018 2019 1
2 1 1
2 1
1 1
2 1 1
2 1 1
1
1
1
1
1
1
1
1
1
1
1 1
1
1
1 1
1 1 1
1
1
2
1 2
1
1
1
1
1
1
1
1
1
1
2
1
1 1
1 1 1
2 GATE Biotechnology Chapter-wise Solved Papers
Questions 1. Value of the determinant mentioned below is 1 4 1 2 (A) 24 (C) -24
0 −1 7 0 1 −1 0 2
0 2 1 1 (B) -30 (D) -10 (GATE 2011; 2 Marks)
Statement for Linked Answer Questions 2 and 3: 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. 2. Variance (Vx) for this population of fruit flies as calculated from the above data shall be (A) 0.85 (B) 0.25 (C) 0.061 (D) 0.08 (GATE 2011; 2 Marks)
7. If u = log (ex + ey), then (A) ex + ey (C)
⎡1 1 ⎤ ⎡2 1⎤ ⎡3 0 ⎤ 6. If P = ⎢ ⎥ , Q = ⎢ 2 2⎥ and R = ⎢1 3⎥ which one 2 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ of the following statements is TRUE? (A) PQ = PR (B) QR= RP (C) QP = RP (D) PQ = QR (GATE 2013; 1 Mark)
(D) 1 (GATE 2013; 1 Mark)
(B) 4 (D) 8 (GATE 2013; 1 Mark)
9. Evaluate lim x tan x →∞
(A) ∞ (C) 0
1 x (B) 1 (D) -1 (GATE 2013; 2 Marks)
10. The Laplace transform of f (t) = 2t + 6 is (A)
1 2 + s s2
(C)
6 2 + s s2
(B)
3 6 − s s2
6 2 + s s2 (GATE 2013; 2 Marks) (D) −
11. The solution of the following set of equations is x + 2y + 3z = 20 7x + 3y + z = 13
(B) 1 (D) 3 (GATE 2012; 2 Marks)
5. 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 mean is unchanged, variance changes. (D) the mean changes, the variance is unchanged. (GATE 2012; 2 Marks)
1 ex + e y
(A) 2 (C) 6
4. What is the rank of the following matrix?
(A) 0 (C) 2
(B) ex - ey
⎡10 −4 ⎤ 8. One of the eigen values of P = ⎢ ⎥ is ⎣18 −12⎦
3. The value of standard deviation (SD) will be (A) 0.061 (B) 0.25 (C) 0.61 (D) 0.85 (GATE 2011; 2 Marks) 5 3 −1 6 2 −4 14 10 0
∂u ∂u + = ∂x ∂y
x + 6y + 2z = 0 (A) x = -2, y = 2, z = 8 (C) x = 2, y = 3, z = -8
(B) x = 2, y = -3, z = 8 (D) x = 8, y = 2, z = -3 (GATE 2013; 2 Marks)
dy + y cot x = cosec x is dx (A) y = (c + x) cot x (B) y = (c + x) cosec x
12. The solution to
(C) y = (c + x) cosec x cot x (D) y = (c + x)
cos ec x cot x
(GATE 2013; 2 Marks)
13. One percent of the cars manufactured by a company are defective. What is the probability (upto four decimals) that more than two cars are defective, if 100 cars are produced? (GATE 2013; 2 Marks)
Chapter 1 • Engineering Mathematics 3
1 −4 ⎤ 14. The eigen values of A = ⎢⎡ ⎥ are ⎣ 2 −3⎦ (A) 2 ± I (C) -1 ± 2i
(B) -1, -2 (D) non-existent (GATE 2014; 1 Mark)
15. If an unbiased coin is tossed 10 times, the probability that all outcomes are same will be __________ × 10–5. (GATE 2014; 1 Mark) 16. The solution for the following set of equations is, 5x + 4y + 10z = 13
21. Which of the following statements is true for the series given below? 1 1 1 1 sn = 1 + + + + ... + n 2 3 4 (A) sn converges to log( n ) (B) sn converges to
n
(C) sn converges to exp( n ) (D) sn diverges (GATE 2014; 2 Marks)
x 22. The graph of the function F ( x ) = for 2 k1 x + k2 x + 1 0 σ2
(C) μ > μ2 ; σ < σ2
(D) μ < μ2 ; σ = σ2 (GATE 2017; 2 Marks)
Chapter 1 • Engineering Mathematics 5
41. The angle (in degrees) between the vectors x = iˆ − ˆj + 2kˆ and y = 2iˆ − ˆj − 1.5kˆ is ____________.
(GATE 2017; 2 Marks)
42. Consider an infinite number of cylinders. The first cylinder has a radius of 1 meter and height of 1 meter. The second one has a radius of 0.5 meter and height of 0.5 meter. Every subsequent cylinder has half the radius and half the height of the preceding cylinder. The sum of the volumes (in cubic meters) of these infinite numbers of cylinders is ________. Given data: π = 3.14. (GATE 2017; 2 Marks) 43. The value of c for which the following system of linear equations has an infinite number of solutions is ________. ⎡1 2⎤ ⎡ x ⎤ ⎡ c ⎤ ⎢1 2⎥ ⎢ y ⎥ = ⎢ 4 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ (GATE 2017; 2 Marks) 44. For the probability density P ( x ) = 0.5e −0.5 x , the integral
∫
∞
0
P ( x )dx = ________. (GATE 2017; 2 Marks)
45. Consider an unfair coin. The probability of getting heads is 0.6. If you toss this coin twice, what is the probability that the first or the second toss is heads? (A) 0.56 (B) 0.64 (C) 0.84 (D) 0.96 (GATE 2018; 1 Mark) 46. Standard error is (A) the probability of a type I error in a statistical test. (B) the error in estimating a sample standard deviation. (C) the standard deviation of a variable that follows standard normal distribution. (D) the standard deviation of distribution of sample means. (GATE 2018; 1 Mark) 47. Which one of the following is the solution for cos 2 x + 2 cos x + 1 = 0 , for values of x in the range of 0° < x < 360°? (A) 45° (B) 90° (C) 180° (D) 270° (GATE 2018; 1 Mark) ⎡ 4 −6 ⎤ 48. The determinant of the matrix ⎢ ⎥ is _________. ⎣ −3 2 ⎦ (GATE 2018; 1 Mark)
49. The variable z has a standard normal distribution. If P(0 ≤ z ≤ 1) = 0.34, then P (z2 > 1) is equal to (upto two decimal places) _______ . (GATE 2018; 1 Mark) 50. Calculate the following integral (up to two decimal places)
∫ (x 1
0
+ 3) ( x + 1) dx = ____________. (GATE 2018; 2 Marks)
51. 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) ________. (GATE 2018; 2 Marks) 52. If 1 + r + r2 + r3 + … ∞ = 1.5, then, 1 + 2r + 3r2 + 4r3 + … ∞ = (up to two decimal places ___________. (GATE 2018; 2 Marks) 53. The median value for the dataset (12, 10, 16, 8, 90, 50, 30, 24) is __________ . (GATE 2019; 1 Mark) 54. Which of the following are geometric series? P. 1, 6, 11, 16, 21, 26, .... Q. 9, 6, 3, 0, –3, –6, ... R. 1, 3, 9, 27, 81, .... S. 4, –8, 16, –32, 64, .... (A) P and Q only (C) Q and S only
(B) R and S only (D) P, Q and R only (GATE 2019; 1 Mark)
⎛ x 2 − 64 ⎞ is ______________. 55. The solution of lim ⎜ x →8 ⎝ x − 8 ⎟ ⎠ (GATE 2019; 1 Mark) 56. Which one of the following equations represents a one-dimensional wave equation? (A)
∂u ∂2u = C2 2 ∂t ∂x
(B)
∂2u ∂2u = C2 2 2 ∂t ∂x
(C)
∂2u ∂u = C2 2 ∂x ∂t
(D)
∂2u ∂2u + =0 ∂t 2 ∂x 2
(GATE 2019; 1 Mark)
6 GATE Biotechnology Chapter-wise Solved Papers ⎡ 0 6⎤ 57. Matrix A = ⎢ ⎥ will be skew-symmetric when, ⎣ p 0⎦ p = __________. (GATE 2019; 1 Mark) dy x 58. What is the solution of the differential equation = , dx y with the initial condition, at x = 0, y = 1? (A) x 2 = y 2 + 1 (B) y2 = x2 + 1 x −y =0 (C) y = 2 x + 1 (D) 2
2
2
2
The standard deviation (s) for the casino payout is Rs. ________ (rounded off to the nearest integer). (GATE 2019; 2 Marks) 60. The Laplace transform of the function f (t ) = t 2 + 2t + 1 is (A)
1 3 1 4 4 1 + 2 + (B) + + 3 s s s s3 s 2 s
(C)
1 2 1 2 2 1 + 2 + (D) + + 3 s s s s3 s 2 s
(GATE 2019; 2 Marks)
(GATE 2019; 2 Marks)
59. A new game is being introduced in a casino. A player can lose Rs. 100, break even, win Rs. 100 or win Rs. 500. The probabilities (P(X)) of each of these outcomes (X) are given in the following table:
61. Calculated using trapezoidal rule for the values given in the table is ____________ (rounded off to 2 decimal places).
X (in Rs.)
–100
0
100
500
P (X)
0.25
0.5
0.2
0.05
X
–1
–2/3
–1/3
0
1/3
2/3
1
f (x)
0.37
0.51
0.71
1.0
1.40
1.95
2.71
(GATE 2019; 2 Marks)
ANSWER KEY 1. (C)
2. (C)
3. (B)
4. (C)
7. (D)
8. (C)
9. (B)
10. (C)
11. (B)
12. (B)
13. (0.0794) 14. (C)
15. (195.3 × 10-5) 16. (B)
17. (0)
18. (D)
19. (B)
20. (C)
21. (D)
22. (A)
23. (-60)
24. (1.43)
25. (2.5)
26. (A)
27. (1.1)
28. (D)
29. (A) 30. (24.9) 31. (3)
32. (C)
33. (1.705)
34. (14.645) 35. (160) 36. (+3)
41. (90°)
42. (3.588 m3)
43. (4)
44. (1)
45. (C)
53. (20)
54. (B)
40. (D)
5. (D)
6. (A)
37. (3.14 m2) 38. (1)
39. (1)
46. (D)
47. (C)
48. (-10) 49. (0.32)
50. (5.33)
51. (2.1)
52. (2.25)
55. (16)
56. (B)
57. (−6)
59. (129)
60. (D)
61. (2.37)
58. (B)
Answers with Explanations 1. Topic: Matrices and Determinants 1 0 −1 0 4 7 0 2 1 1 −1 1 2 0 2 1 7 0 2 4 0 2 4 7 2 4 7 0 = 1 1 −1 1 − 0 1 −1 1 − 1 1 1 1 − 0 1 1 −1 0 2 1 2 2 1 2 0 1 2 0 2
= 1[7( −1 − 2) − 0(1 − 0) + 2( 2 − 0)] + 0 − 1[4(1 − 0) − 7(1 − 2) + 2(0 − 2)] + 0 = [7( −3) + 2( 2)] − [4 − 7( −1) + 2( −2)] = −21 + 4 − [4 + 7 − 4] = −17 − 7 = −24 Answer (C)
Chapter 1 • Engineering Mathematics 7
2. Topic: Mean, Median, Mode and Standard Deviation From the given data, we can form the following table: The mean can be calculated by step-deviation method taking 2.0 as assumed mean. The number of
Determinant can be calculated as A = (5 × 2) − (6 × 3) = 10 − 18 = −8 ≠ 0 Thus, rank of the given matrix is 2.
xi
fi
fi xi
( xi − x )
( xi − x )
f i ( xi − x )
1.9
1
1.9
–0.17
0.0289
0.0289
2.4
2
4.8
0.33
0.1089
0.2178
2.1
2
4.2
0.03
0.0009
0.0018
2.0
3
6.0
–0.07
0.0049
0.0147
2.2
2
4.4
0.13
0.0169
0.0338
1.7
1
1.7
–0.37
0.1369
0.1369
1.8
1
1.8
–0.27
0.0729
0.0729
2.3
2
4.6
0.23
0.0529
0.1058
1.6
1
1.6
–0.47
0.2209
0.2209
15
31
2
Answer (C)
2
5. Topic: Statistics Let x be the mean of previous data set. Then, Mean, x = x=
1 n ∑ xi n i =1 1 (14 + 18 + 14 + 14 + 10 + 29 + 33 + 31 + 25) 9
= 20.88 (i) As given, on adding 20 to each value, new mean ( x2 ) would be:
0.8335
∑f
= N = 15
i
9
∑fx
i =1
i i
=
= 31
Variance (Vx), σ 2 =
∑fx
i i
i =1
31 = = 2.07 15
N 1 N
9
∑ f (x i =1
i
i
− x) = 2
Answer (C) 3. Topic: Mean, Median, Mode and Standard Deviation
σ2 =
Answer (B)
A = 5(0 + 40) − 3(0 + 56) − 1(60 − 28)
5 3 6 2
1 n ∑ ( xi − x )2 (iii) n i =1
For new data set, variation can be calculated by:
σ2 2 =
4. Topic: Matrices and Determinants Determinant A for the matrix can be calculated as:
Now, for sub-matrix
[From (i)]
Variance of previous data can be calculated as:
Standard deviation (SD), σ = Variance
= 5( 40) − 3(56) − 1(32) = 200 − 168 − 32 = 200 − 200 = 0
n ⎤ 1⎡ n xi + ∑ a ⎥ ∑ ⎢ n ⎣ i =1 i =1 ⎦
x2 = 20.88 + 20 = 40.88 (ii)
1 × 0.8335 = 0.06 15
= 0.06 = 0.25
1 n ∑ ( xi + a) n i =1
na 1 n = ∑ xi + = x +a n n i =1
9
Mean, x =
x2 =
1 n 2 ( xi + 20) − ( x2 )] [ ∑ n i =1
=
1 n 2 ∑ [( xi + 20) − ( x + 20)] n i =1
=
1 n 2 xi + 20 − x − 20)] [ ∑ n i =1
=
1 n 2 ∑ [ xi − x )] = σ2 n i =1
[From (ii)]
[From (iii)]
σ2 2 = σ2 Answer (D)
8 GATE Biotechnology Chapter-wise Solved Papers 6. Topic: Matrices and Determinants Given, ⎡1 1 ⎤ ⎡2 1⎤ ⎡3 0 ⎤ P=⎢ , Q=⎢ and R = ⎢ ⎥ ⎥ ⎥ ⎣ 2 2⎦ ⎣ 2 2⎦ ⎣1 3⎦ ⎡1 P×Q = ⎢ ⎣2 ⎡4 =⎢ ⎣8
1⎤ ⎡2 1⎤ × 2⎥⎦ ⎢⎣ 2 2⎥⎦ 3⎤ 6 ⎥⎦
⎡1 P×R = ⎢ ⎣2 ⎡4 =⎢ ⎣8
1 ⎤ ⎡3 0 ⎤ × 2⎥⎦ ⎢⎣1 3⎥⎦ 3⎤ 6 ⎥⎦
= 1 / 2 ⎡⎣( −2) ± 196 ⎤⎦ = 1 / 2 [( −2) ± 14] = 1 / 2 [12] or 1/2 [ −16] = 6 or − 8 Therefore, λ = 6. Answer (C) 9. Topic: Limits, Differentiability and Continuity lim x tan
1 x
Let us assume that, x =
1 n
x →∞
Thus, if x → ∞, n → 0 Now,
Thus, PQ = PR
sin n lim 1 n→ 0 n lim tan n = n →0 n lim cos n
Answer (A) 7. Topic: Partial Derivatives
n→ 0
Given, u = log(e + e ) Now, on differentiating it with respect to x and y and summing the solution we get: x
y
{
} {
}
x y x y ∂u ∂u ∂ log(e + e ) ∂ log(e + e ) + = + ∂x ∂y ∂x ∂y x y ∂( e x + e y ) ∂( e + e ) ∂y ∂x = + x (e x + e y ) (e + e y )
=
ex ey + (e x + e y ) (e x + e y )
=
ex + e y =1 (e x + e y )
We know, that, lim n→ 0
sin n = 1 and lim cos n = 1 [∵ cos 0 = 1] n→ 0 n
Therefore, sin n n =1 lim cos n lim n→ 0 n→ 0
Answer (B) 10. Topic: Laplace Transform Given, f (t ) = 2t + 6 On applying the Laplace transform, we have L { f (t )} = L {2t } + L {6} Answer (D)
8. Topic: Eigen Values and Eigen Vectors
=
2 6 + s2 s Answer (C)
Given, ⎡10 −4 ⎤ P=⎢ ⎥ ⎣18 −12⎦ For a 2 × 2 matrix, the Eigen values can be given by:
λ = 1 / 2 ⎡( a11 + a22 ) ± 4 a12 a21 + ( a11 − a22 ) 2 ⎤ ⎣ ⎦ Therefore, λ for the given matrix
λ = 1 / 2 ⎡(10 − 12) ± 4 × ( −4) × (18) + (10 + 12) 2 ⎤ ⎣ ⎦ = 1 / 2 ⎡⎣( −2) ± −288 + 484 ⎤⎦
11. Topic: System of Linear Equations Given, x + 2y + 3z = 20 7x + 3y + z = 13 x + 6y + 2z = 0
(i) (ii) (iii)
(A) On using x = -2, y = 2 and z = 8 in equations (i), (ii) and (iii), we get (–2) + 2(2) + 3(8) = 26 ≠ 20 7(–2) + 3(2) + 8 = 0 ≠ 13 (–2) + 6(2) + 2(8) = 26 ≠ 0
Chapter 1 • Engineering Mathematics 9
Hence, none of the equations are satisfied with these values. (B) On using x = 2, y = –3 and z = 8 in equations (i), (ii) and (iii), we get 2 + 2(–3) + 3(8) = 20 7(2) + 3(–3) + 8 = 13 2 + 6(–3) + 2(8) = 0 Hence, all the equations are satisfied with these values. (C) On using x = 2, y = 3 and z = –8 in equations (i), (ii) and (iii), we get 2 + 2(3) + 3(–8) = −16 ≠ 20 7(2) + 3(3) + (–8) = 15 ≠ 13 2 + 6(3) + 2(–8) = 4 ≠ 0 Hence, none of the equations are satisfied with these values. (D) On using x = 8, y = 2 and z = -3 in equations (i), (ii) and (iii), we get 8 + 2(2) + 3(–3) = 3 ≠ 20 7(8) + 3(2) + (–3) = 59 ≠ 13 8 + 6(2) + 2(–3) = 14 ≠ 0 Hence, none of the equations are satisfied with these values. Thus, the correct option is (B). Answer (B) 12. Topic: Linear First Order ODEs, Higher Order ODEs with Constant Coefficients dy + y cot x = cosec x dx To find the solution of the given differential equation, it can be rewritten as dy = cosec x − y cot x dx dy = cosec x.dx − y cot x.dx
We are given a sample size of 100 cars and told that 1% of the cars are defective and asked what the probability of more than 2 cars is being defective. Let’s look at a success as failure of a car, then p = 0.01 and q = 0.99 P (k > 2) = 1 – P (k = 0) – P (k = 1) – P (k = 2) (i) Now, P (k = 0) = 100C0 × (0.01)0 × (0.99)(100–0) = 0.3660323412732289 (ii) P (k = 1) = 100C1 × (0.01)1 × (0.99)(100–1) = 0.36972963764972616
(iii)
P (k = 2) = C2 × (0.01) × (0.99) = 0.18486481882486308
(iv)
100
⇒ y = log (cosec x − cot x ) − y log sin x + c cos x ⎞ ⎛ 1 ⇒ y = log ⎜ − − y log sin x + c ⎝ sin x sin x ⎟⎠ ⇒ y = log (1 − cos x ) − log sin x − y log sin x + c ⇒ y = ( x + c) cosec x Answer (B) 13. Topic: Poisson Distribution This problem uses the binomial probability formula, P (k success in n trials) = nCk ⋅ pk ⋅ q(n–k)
(100–2)
On summing (ii), (iii) and (iv), we get P (k = 0) + P (k = 1) + P (k = 2) = 0.920626798 (v) From Eqs. (i) and (v), we get P(k > 2 ) = 1 − 0.920626798 = 0.079373202 = approx. 0.0794 Answer (≅ 0.0794) 14. Topic: Eigen Values For eigen values, A−I =0 1 −4 λ 0 − =0 2 −3 0 λ 1 − λ −4 − 0 =0 2 − 0 −3 − λ ∴ (1 − λ)( −3 − λ) − ( 2)( −4) = 0 ⇒ ( −3 − λ + 3λ + λ 2 ) + 8 = 0 ⇒ λ2 + 2 λ + 5 = 0 ⇒ λ=
−2 ± 4 − 4(1)(5) 2
⇒ λ=
−2 ± −16 −2 ± 16i 2 = 2 2
⇒ λ=
−2 ± 4i = −1 ± 2i 2
On integrating the equation, we get,
∫ dy = ∫ (cosec x − y cot x ) dx
2
Answer (C) 15. Topic: Probability An unbiased coin has two sides, H and T and probability of coming either of them is 1/2. On tossing the unbiased coin 10 times, the probability of getting same outcome each time would be: P (all outcomes all 10 times) = P (head all 10 times) + P (tail all 10 times)
10 GATE Biotechnology Chapter-wise Solved Papers 1 1 + 210 210 1 1 2 = + = = 0.001953 = 195.3 × 10 −5 1024 1024 1024 Answer (195.3 × 10−5)
P=
16. Topic: System of Linear Equations For equations: 5x + 4y + 10z = 13 x + 3y + z = 7 4x – 2y + z = 0
18. Topic: System of Linear Equations For equations: 2x + 3y = 4 4x + 6y = 0 a1 2 1 = = a2 4 2 b1 3 1 = = b2 6 2
(i) (ii) (iii)
From option (A), x = 2, y = 1, z = 1, putting these values of x, y and z in Eqs (i), (ii) and (iii) we get: 5(2) + 4(1) + 10 (1) = 24 2 + 3(1) +1 = 5 4(2) – 2(1) + 1 = 7 None of the equations satisfies with these values, hence this option is wrong. From option (B), x = 1, y = 2, z = 0, putting these values of x, y and z in Eqs (i), (ii) and (iii) we get: 5(1) + 4(2) + 10(0) = 13 1 + 3(2) + 0 = 7 4(1) – 2(2) + 0 = 0 All the equations satisfy with these values, hence this option is correct. From option (C), x = 1, y = 0, z = 2, putting these values of x, y and z in Eqs (i), (ii) and (iii) we get:
c1 4 = =∞ c2 0 As,
a1 b1 c1 = ≠ , thus, these equations have no solution. a2 b2 c2 Answer (D)
19. Topic: Partial Derivatives Given, concertation profile of a chemical, c ( x, t ) =
∂c = ∂x = = = =−
t →∞
∂e
c0
∂x
2 π Dt c0 2 π Dt c0 2 π Dt c0 2 π Dt
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
e
e
e
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
c0 Dt 2 π Dt
e
⎛ x2 ⎞ ∂⎜− ⎝ 2 Dt ⎟⎠ ∂x ⎛ −2 x ⎞ ⎜⎝ ⎟ 2 Dt ⎠ ⎛ −2 x ⎞ ⎜⎝ ⎟ 2 Dt ⎠
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
sin t ⎡ ⎤ = 0⎥ ⎢∵ lim t →∞ e 2 t ⎣ ⎦ Answer (0)
⋅x
⎡ x2 ⎤ ⎡ ⎡⎢ − x 2 ⎤⎥ ⎤ ⎥ ⎢− c0 2x ⎞ ⎥ ∂2c 2 Dt ⎥⎦ ⎢⎣ 2 Dt ⎥⎦ ⎛ ⎢ ⎣ ⎢e = − xe + − ⎜⎝ ⎟ 2 Dt ⎠ ⎥ ∂x 2 Dt 2 π Dt ⎢ ⎣ ⎦
=−
c0 Dt 2 π Dt
e
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
⎡ x2 ⎤ ⎢1 − ⎥ ⎣ Dt ⎦
⎤ c( x , t ) ⎡ x 2 ⎢ − 1⎥ (i) Dt ⎣ Dt ⎦ Now, change in concertation with respect to t is given by =
lim e −2t sin t = lim
e
On double differentiating with respect to x we get:
17. Topic: Limit, Continuity and Differentiability sin t t →∞ e 2 t 2t ⎤ sin t 1 ⎡ = lim × ⎢lim 2t ⎥ t →∞ t 2 ⎣ t →∞ e ⎦ =0
2 π Dt
⎡ x2 ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
Now, change in concertation with respect to x is given by
5(1) + 4(0) + 10(2) = 25 1 + 3(0) + 2 = 3 4(1) – 2(0) + 2 = 6 None of the equations satisfies with these values, hence this option is wrong. From option (D), x = 0, y = 1, z = 2, putting these values of x, y and z in Eqs (i), (ii) and (iii) we get: 5(0) + 4(1) + 10(2) = 24 0 + 3(1) + 2 = 5 4(0) – 2(1) + 2 = 0 Only one equation satisfies with these values, hence this option is wrong. Answer (B)
c0
∂c = ∂t
⎡ x2 ⎤ ⎡ ⎤ ∂ ⎢ −1/ 2 ⎢⎢⎣ − 2 Dt ⎥⎥⎦ ⎥ t ⋅e ⎥ 2 π D ∂t ⎢ ⎣ ⎦
c0
Chapter 1 • Engineering Mathematics 11 ⎡ x2 ⎤ ⎡ x2 ⎤ ⎡ ⎤ 2 ⎥ ⎢− ⎥ ⎢− 1 −3/ 2 ⎞ ⎥ −1/ 2 ⎢⎣ 2 Dt ⎥⎦ ⎛ ⎢⎣ 2 Dt ⎥⎦ ∂ ⎛ − x ⎞ ⎢ = + − t t ⋅e e ⎜ ⎟ ⎝ 2 ⎠⎥ ∂t ⎜⎝ 2 Dt ⎟⎠ 2π D ⎢ ⎣ ⎦
c0
⎡ x ⎤ ⎥ ⎢− ⎢⎣ 2 Dt ⎥⎦
On its integration we get: ∞
∫ 1
2
=
Thus, this series is divergent.
⎡ ∂ ⎛ − x2 ⎞ 1 ⎤ ⎢ ⎜ ⎟− ⎥ 2 π Dt ⎣ ∂t ⎝ 2 Dt ⎠ 2t ⎦
c0 e
Answer (D) 22. Topic: Maxima and Minima x F ( x ) = k1 x 2 + k2 x + 1
⎡ − x 2 ( −1) 1 ⎤ = c( x , t ) ⎢ − ⎥ 2 2t ⎦ ⎣ 2 Dt
=
⎤ c( x , t ) ⎡ x 2 ⎢ − 1⎥ (ii) 2t ⎣ Dt ⎦
On taking k1 common in the denominator. F (x) =
From Eqs. (i) and (ii) ∂c D ∂2c = ∂t 2 ∂x 2 Answer (B) 20. Topic: Limit, Continuity and Differentiability y = xx (i) Taking log on both sides log y = log xx ⇒ log y = x log x On differentiating with respect to x, we get: 1 dy d log x dx =x + log x y dx dx dx ⇒
∞
∞ ⎡ n1/ 2 ⎤ dn ⇒ ∫ n −1/ 2 dn = ⎢ ⎥ =∞ n ⎣1 / 2 ⎦1 1
1
x / k1 ⎛ k 1⎞ k1 ⎜ x 2 + 2 x + ⎟ k k ⎝ 1 1⎠
It’s clear from the equation above that natural frequency of the function (ωn) is 1/k1. Now, at x = 0 0 / k1 F (0 ) = =0 ⎛ 2 k2 1⎞ k1 ⎜ 0 + 0 + ⎟ k1 k1 ⎠ ⎝ Thus, graph is starting from the origin. Now, at x = ∞ x / k1 F (∞ ) = lim x →∞ ⎛ k 1⎞ k1 ⎜ x 2 + 2 x + ⎟ k k ⎝ 1 1⎠ Taking x2 common in the denominator.
1 dy 1 = x ⋅ + log x y dx x
F (∞ ) = lim
x →∞
⇒
dy = y (1 + log x ) dx
⇒
dy = x x (1 + log x ) dx
= lim
[From (i)]
x →∞
Answer (C)
21. Topic: Test for Convergence A sequence of partial sums is said to be a convergent sequence if its limit exists and is finite. In this case, lim sn = s then, n →∞
∞
∫
ai = s
i =1
Likewise, the sequence of partial sums is said to be divergent sequence if its limit doesn’t exist or is plus or minus infinity. Now, the series give is:
sn = 1 +
Here, an =
1 n
1 2
+
1 3
+
1 4
+ +
1 n
x / k12 ⎛ k 1 1 1⎞ x 2 ⎜1 + 2 + k1 x k1 x 2 ⎟⎠ ⎝ 1 / k12 ⎛ k 1 1 1⎞ x ⎜1 + 2 + k1 x k1 x 2 ⎟⎠ ⎝
As x → ∞, F (∞ ) = 0 Thus, at x = ∞, value of function is 0 again. Therefore, only graph of option (A) is satisfying these conditions. Answer (A) 23. Topic: Matrices and Determinants 0⎤ ⎡3 0 ⎢ A = 2 5 0 ⎥ ⎢ ⎥ ⎢⎣6 −8 −4 ⎥⎦ = 3 (–20–0) – 0 (–8–0) + 0 (–16 –30) = –60 Answer (–60)
12 GATE Biotechnology Chapter-wise Solved Papers 24. Topic: Sequence and Series It’s a geometric progression of common ratio r. Sum of infinite geometric progression can be calculated as: a S∞ = 1− r Here, a=1 r = 0.3 Putting in the formula, S∞ =
1 = 1.43 1 − 0.3
28. Topic: Limits, Continuity and Differentiability n
⎛ x⎞ ⎜1+ ⎟ as n→∞ ⎝ n⎠ Binomial expansion of the above expression: ⎛ ⎜⎝1 +
n
x⎞ n n( n − 1) x 2 + ⎟⎠ = 1 + x + n n n2 2 !
⎛ n2 ⎞ (1 − 1 / n) x 2 ⎛ n⎞ = 1+ ⎜ ⎟ x + ⎜ 2 ⎟ + ⎝ n⎠ 1 2! ⎝n ⎠ As n→∞ coefficient in n = 1
Answer (1.43) 25. Topic: System of Linear Equations For a pair of linear equation to have infinite solutions, a1 b1 c1 = = a2 b2 c2 2 1 3 = = 5 b 7.5
1+ x +
Answer (D) 29. Topic: Matrices and Determinants Given ⎡ 4 2⎤ A= ⎢ ⎥ ⎣ 1 3⎦ ⎡ 4 2⎤ ⎡ 4 2⎤ A2 = ⎢ ⎥ ⎥⎢ ⎣ 1 3⎦ ⎣ 1 3⎦
Therefore, b = 2.5. Answer (2.5)
⎡ 4 × 4 + 2 × 1 2 × 4 + 2 × 3⎤ = ⎢ ⎥ ⎣1× 4 + 3 ×1 1× 2 + 3 × 3 ⎦
26. Topic: Eigen Values ⎡ 1 1⎤ A = ⎢ ⎥ ⎣ −2 4 ⎦
⎡18 14 ⎤ = ⎢ ⎥ ⎣ 7 11⎦
For Eigen values, A − I = 0
⎡ 4 × 3 2 × 3⎤ 3A = ⎢ ⎥ ⎣ 1 × 3 3 × 3⎦
1 ⎤ ⎡ 1 1 ⎤ ⎡ λ 0 ⎤ ⎡1 − λ =⎢ −⎢ ⇒ A−I = ⎢ ⎥ ⎥ ⎥ ⎣ −2 4 ⎦ ⎣ 0 λ⎦ ⎣ −2 4 − λ⎦
⎡12 6 ⎤ = ⎢ ⎥ ⎣ 3 9⎦
∴ (1 − λ)( 4 − λ) − ( −2)(1) = 0 ⇒ λ 2 − 5λ + 6 = 0
⎡18 14 ⎤ ⎡12 6 ⎤ A2 + 3A = ⎢ ⎥ ⎥+⎢ ⎣ 7 11⎦ ⎣ 3 9⎦
⇒ λ 2 − 3λ − 2λ + 6 = 0 ⇒ λ ( λ − 3) − 2( λ − 3) = 0
⎡18 + 12 14 + 6 ⎤ = ⎢ ⎥ ⎣ 7 + 3 11 + 9 ⎦
⇒ ( λ − 3)( λ − 2) = 0 ⇒ λ = 3; λ = 2
⎡30 20 ⎤ = ⎢ ⎥ ⎣10 20 ⎦
Answer (A) 27. Topic: Random Variables From the table, we know that, F (0) = 0.5, thus P (0) = 0.5 F (1) = 0.6, thus P (1) = 0.1 F (2) = 0.8, thus P (2) = 0.2 F (3) = 1.0, thus P (3) = 0.2 Mean value of X is
Answer (A) 30. Topic: Standard Deviation Mean, x =
1 N
N
∑x
i
i =1
1 300 x = (55 + 75 + 67 + 88 + 15) = = 60 5 5
3
∑ XP (X) = 0(0.5) + 1(0.1) + 2(0.2) + 3(0.2) x=0
x 2 x3 + + = ex 2 ! 3!
= 0 + 0.1 + 0.4 + 0.6 = 1.1 Answer (1.1)
Standard deviation,
σ=
1 N
N
∑ (x i =1
i
− x )2
Chapter 1 • Engineering Mathematics 13
1 (55 − 60)2 + (75 − 60)2 + (67 − 60)2 σ= 5 2 2 + (88 − 60 ) + (15 − 60 ) =
1 ( −5)2 + (15)2 + (7)2 + (28)2 + ( −45)2 5
The integral converges for s > 0. The integral can be computed by doing integration by parts twice or by looking in an integration table. s F ( s) = 2 s + a2 Answer (C) 33. Topic: Limit, Continuity and Differentiability Solving by integration by parts:
1 = (25 + 225 + 49 + 784 + 2025) 5
∴
Answer (24.9)
B
∫ (1 − x ) (2 − x) = ∫ (1 − x) dx + (2 − x)dx (i)
= 24.9
1
(1 − x ) ( 2 − x)
=
A B + (ii) (1 − x ) ( 2 − x )
To find A, multiply equation (ii) with (1 – x).
(1 − x ) = A (1 − x ) + B (1 − x ) (1 − x ) ( 2 − x) (1 − x) (2 − x) B (1 − x ) 1 = A+
31. Topic: Eigen Values ⎡2 1 ⎤ A = ⎢ ⎥;I= ⎣ 5 −2⎦
A
dx
1 = (3108) = 621.6 5
⎡1 0 ⎤ ⎢0 1⎥ ⎣ ⎦
(2 − x)
⎡ λ 0⎤ λ I = ⎢ ⎥ ⎣ 0 λ⎦
(2 − x)
1 = A( 2 − x ) + B (1 − x )
⎡2 − λ A− λI = ⎢ ⎣ 5
1 ⎤ −2 − λ ⎥⎦
Put (1 – x) = 0, therefore, x = 1.
1 = A( 2 − 1) + B (1 − 1)
By using, characteristics equation of matrix A, A − λ I = 0 1 ⎤ ⎡2 − λ =0 ⎢ 5 −2 − λ⎥⎦ ⎣
∴ A = 1(iii) To find B, multiply equation (i) with (2 – x).
(2 − x ) = A (2 − x ) + B (2 − x ) (1 − x ) ( 2 − x) (1 − x) (2 − x)
= − ( 2 − λ) ( 2 + λ) − 5 = 0
(
A (2 − x ) 1 +B = (1 − x ) (1 − x )
)
= − 4 − λ2 − 5 = 0
λ2 − 9 = 0
1 = A( 2 − x ) + B (1 − x )
∴ λ2 = 9 λ = ±3
Put (2 – x) = 0, therefore, x = 2.
1 = A( 2 − 2) + B (1 − 2)
Thus, positive Eigen value = 3 Answer (3) 32. Topic: Laplace Transform The Laplace Transform of a function f (t) is defined by ∞
F ( s) = L [ f (t )] ( s) = ∫ e − st f (t )dt 0
if the integral exists. The notation L[f(t)](s) means take the Laplace transform of f(t). The functions f(t) and F(s) are partner functions. Here, f(t) = cos (at) ∞
F ( s) = ∫ e − st cos ( at ) dt 0
∴ B = −1 (iv) On putting values of A and B from (iii) and (iv) in equation (i), we get 0.9
0.9
dx
−1
1
∫ (1 − x ) (2 − x) = ∫ (1 − x) dx + (2 − x)dx 0
0
0.9
=
∫ 0
1 dx − (1 − x )
0.9
1
∫ (2 − x)dx 0
0.9
0.9
⎡ −1 ⎤ ⎡1 ⎤ = ⎢ log 1 − x ⎥ − ⎢ log 2 − x ⎥ − 1 − 1 ⎣ ⎦0 ⎣ ⎦0
1 1 ⎧ ⎫ = log ax + b + c ⎬ ⎨∵ ∫ ax + b a ⎩ ⎭
14 GATE Biotechnology Chapter-wise Solved Papers = [ − log 0.1 + log 1] − [ − log 1.1 + log 2] = [ 2.303 + 0] − [ −0.09303 + 0.693] = 1.705
Answer (1.705)
34. Topic: Single and Multistep Methods for Differential Equation d2 y For differential equation, − y = 0 , general solution dx 2 is of type y = c1e + c2 e Auxiliary equation, m1 x
m2 x
36. Topic: Maxima and Minima aS V= S2 b+S+ c Given: a = 4; b = 1 and c = 9 Putting these values in equation we get: S 9 For maximum value of V
((
On differentiation of Eq. (i) dy = c1e x − c2 e − x (ii) dx Given, y(0) = 1 Putting this condition in (i) 1 = c1e 0 + c2 e 0 c1 + c2 = 1 (iii)
dy = 3 at x = 0 dx Putting this condition in (ii) 3 = c1e 0 − c2 e 0 Also, given
))
2 − 36 S 2S + 9 9 +99+S9+S S+2 S36 36 − 36(S ( 2S +) 9) dVdV = 0= =0 = 2 2 dSdS 9 +99+S9+S S+2 S 2
−x
((
))
324 + 324 S +S36 S S− 72 S S− 3−2342S4=S 0= 0 324 + 324 + 36 − 72 2 2 3636 S S= 324 = 324 2
2
2
2
S 2 S=2 9= 9 S = ±3 (ii) On double differentiation of (i)
(
)
9 + 9S + S 2 ( −72S ) − ⎡⎣( −36 S 2 + 324)( 2)( 2S + 9) ⎤⎦ d 2V = 4 dS 2 9 + 9S + S 2
(
)
From Eq. (ii) for S = +3
c1 − c2 = 3 (iv) On solving Eqs. (iii) and (iv) c1 = 2 and c2 = −1
⎡ ⎤ (9 + 9(3) + 3 ) (−72 × 3) − ⎢((2−(363)(+3)9)+ 324)(2)⎥ 2
2 2
d 2V = dS 2
⎣
⎦
(9 + 9(3) + 3 )
2 4
(9 + 27 + 9) ( −216) − [( −36(9) + 324)( 2)(6 + 9)] (9 + 27 + 9)4 ( 45) 2 ( −216) − [0] 216 = =− = −0.1067 < 0 2
=
On putting these values in Eq. (i) y = 2e x − 1e − x Using x = 2
454
y = 2e 2 − e −2
2025
From Eq. (ii) for S = –3
= 2(7.39) − 0.135 = 14.78 − 0.135 = 14.645
36 S (i) 9 + 9S + S 2
On differentiation of Eq. (1)
y = c1e + c2 e (i)
=
dV d 2V = 0 and 0
( −9) 4
81
Thus, V is maximum at S = +3. Answer (160)
Answer (+3)
Chapter 1 • Engineering Mathematics 15
37. Topic: Mensuration For maximum surface area, radius of sphere will be half of the edge of the cube that is 0.5 m.
μ2 = μ + a (ii) ∴ μ < μ2 Standard deviation of previous exam is given by:
Thus, Surface area of sphere = 4π r2 = 4 × 3.14 × (0.5)2 = 3.14 m2 Answer (3.14 m2) 38. Topic: Limits, Continuity and Differentiability sin( x ) = x −
1 n ∑ ( xi − μ)2 n i =1
For new exam, standard deviation would be given by:
x3 x5 x7 + − + 3! 5! 7 !
σ2 =
⎤ ⎡ x 2 x 4 x6 sin( x ) = x ⎢1 − + − + ⎥ ! ! ! 3 5 7 ⎣ ⎦
1 n 2 ( xi + 5) − ( μ2 )] [ ∑ n i =1
=
1 n 2 ∑ [( xi + 5) − ( μ + 5)] n i =1
lim
⎤ x ⎡ x 2 x 4 x6 sin( x ) = lim ⎢1 − + − + ⎥ x→0 x→0 x x 3 ! 5 ! 7 ! ⎣ ⎦
=
1 n 2 xi + 5 − μ − 5)] [ ∑ n i =1
⎤ ⎡ x 2 x 4 x6 sin( x ) = lim ⎢1 − + − + ⎥ x→0 x → 0 x ⎣ 3! 5! 7 ! ⎦
=
1 n 2 ∑ [ xi − μ)] = σ n i =1
lim
On applying limit, we get,
Answer (D) Answer (1)
39. Topic: Single and Multistep Methods for Differential Equations Given, at x = 0, d2 y dy = 0 and =0 2 dx dx Thus, it shows that given point x is inflection point as well as stationary point with constant function value (y) of 1 for all values of x. Answer (1) 40. Topic: Standard Deviation Let µ be the mean of x1, x2, …., xn marks scored in the previous exam by n number of students. Then, 1 n Mean, µ = ∑ xi (i) n i =1 Given, everyone scored 5 marks more than their respective score in the earlier exam, therefore, new mean (µ2) of the class would be: µ2 =
1 n ∑ ( xi + a) n i =1
=
⎤ 1⎡ a⎥ ∑ xi + ∑ n ⎢⎣ i =1 i =1 ⎦
=
1 n na xi + = μ+ a ∑ n i =1 n
n
[From (ii)
∴ σ2 = σ
sin( x ) lim =1 x→0 x
σ=
41. Topic: Vector Algebra The two vectors given are: x = iˆ − ˆj + 2kˆ y = 2iˆ − ˆj − 1.5kˆ Dot product of vectors x and y would be: x ⋅ y = iˆ − ˆj + 2kˆ ⋅ 2iˆ − ˆj − 1.5kˆ
(
)(
)
= (1 × 2) + {( −1) × ( −1)} + {2 × ( −1.5} = 2 +1− 3 = 0 Now, we know that, x⋅ y cos θ = = 0 x y Thus, θ = cos −1 0 = 90° Answer (90°) 42. Topic: Sequence and Series We know that, volume of cylinder = π r 2 h Sum of the volumes of these infinite number of cylinders, ∞
Vtotal = ∑ Vi i =1
= V1 + V2 + V3 +
n
= ⎣⎡π (1) 2 1⎤⎦ + ⎣⎡π (0.5) 2 0.5⎤⎦ + ⎣⎡π (0.25) 2 0.25⎤⎦ + [From (i)]
= 3.14 [1 + 0.125 + 0.015625 + ] (i) It’s a geometric progression (GP) with first term, a = 1
16 GATE Biotechnology Chapter-wise Solved Papers And, common ratio, r = 0.125 Sum of infinite terms a 1 1 = = = 1.143 (1 − r ) (1 − 0.125) 0.875
=
Putting this in (i) we get, Vtotal = 3.14 × 1.143 = 3.588 m3 Answer (3.588 m3) 43. Topic: Matrices and Determinants
46. Topic: Statistics Square root of variance of an estimator is known as standard error of the estimator and indicates the amount of deviation or error that has been committed while estimating the parameter of distribution of sample means. Usually standard error is inversely proportional to N, the sample size. Answer (D) 47. Topic: Maxima and Minima Given, cos 2 x + 2 cos x + 1 = 0 Let us take cos x = z (i) Putting this in equation, we get,
⎡1 2⎤ ⎡ x ⎤ ⎡ c ⎤ ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣1 2⎦ ⎣ y ⎦ ⎣ 4 ⎦
⇒ z2 + 2z + 1 = 0
⎡ x × 1 + 2 × y⎤ ⎡c ⎤ ⎢1 × x + 2 × y ⎥ = ⎢ 4 ⎥ ⎣ ⎦ ⎣ ⎦
⇒ ( z + 1) 2 = 0 From (i)
⎡ x + 2 y⎤ ⎡c⎤ ⎢ x + 2 y ⎥ = ⎢4⎥ ⎣ ⎦ ⎣ ⎦
⇒ z = cos x = −1 ⇒ x = cos −1 ( −1)
∴ x + 2 y = c and, x + 2 y = 4
⇒ x = π − cos −1 (1) ⇒ x = π −0
For infinite number of solutions both equations will be equal. Therefore, c = 4. Answer (4) 44. Topic: Poisson, Normal and Binomial Distribution Given, P ( x ) = 0.5e −0.5 x
∫
∞
0
∞
P ( x )dx = ∫ 0.5e −0.5 x dx
⇒ x = π = 180 Answer (C) 48. Topic: Matrices and Determinants Given matrix: ⎡ 4 −6 ⎤ A= ⎢ ⎥ ⎣ −3 2 ⎦
0
A = 4(2) – (–3) (–6)
∞
= 8 – 18 = –10
⎡1 ⎤ As, ∫ e ax dx = ⎢ e ax ⎥ 0 ⎣ a ⎦0 ∞
∴
∫
∞
0
Answer (–10) ∞
1 −0.5 x ⎤ ⎡ 0.5 e −0.5 x dx = ⎢ −(0.5) × e ⎥ 0.5 ⎣ ⎦0 ∞
= ⎡⎣ −e −0.5 x ⎤⎦ 0
= ⎡⎣ −e −∞ + e 0 ⎤⎦
∞ 0
49. Topic: Poisson, Normal and Binomial Distribution P (0 ≤ z ≤ 1) = 0.34 P (z > 1) = 0.5 – 0.34 = 0.16 P (z2 > 1) = 2 × 0.16 = 0.32 Answer (0.32)
= 0 +1 = 1 Answer (1) 45. Topic: Probability Probability of getting head (p) = 0.6 Probability of getting tail (q) = 1 – 0.6 = 0.4 Favorable outcomes = [HT, TH, HH] = 0.6 × 0.4 + 0.4 × 0.6 + 0.6 × 0.6 = 0.24 + 0.24 + 0.36 = 0.84 Answer (C)
50. Topic: Limit, Continuity and Differentiability
∫ ( x + 3) ( x + 1) dx = ∫ ( x 1
1
0
0
2
)
+ 4x + 3 dx 1
⎤ ⎡ x3 4x 2 =⎢ + + 3x ⎥ 2 ⎦0 ⎣3 ⎡1 ⎤ = ⎢ + 2 + 3⎥ = 5.33 3 ⎣ ⎦ Answer (5.33)
Chapter 1 • Engineering Mathematics 17
51. Topic: Poisson, Normal and Binomial Distribution X
1
2
3
4
P(X)
0.3
0.4
0.2
0.1
On increasing the data set in increasing order, we get 8, 10, 12, 16, 24, 30, 50, 90 Now, we know, Median value = Average of middle most observations
Expected value of X = E(X) = µ
∴ Median =
μ = E ( X ) = ∑ i =1 XP ( X ) n
Answer (20)
E(X) = 1 × P(X = 1) + 2 × P(X = 2) + 3 × P(X = 3) + 4 × P(X = 4) = 1 × 0.3 + 2 × 0.4 + 3 × 0.2 + 4 × 0.1 = 0.3 + 0.8 + 0.6 + 0.4 = 2.1 Answer (2.1) 52. Topic: Sequences and Series The given series 1, r, r2, r3, … ∞ is infinite geometric progression series with a common ration of p(r). We know that, a Sn = 1− r 1 ⇒ 1.5 = 1− r
1 + r + r + r + … ∞ = 1.5 2
3
Answer (16) (i)
3
3
From Eqs. (ii) and (iii), we get (1 − r )Sn = 1 + r + r 2 + r 3 + ⋅⋅⋅⋅ ∞ Sn = =
=
⎛ x 2 − 64 ⎞ ⎛ x 2 − 82 ⎞ lim ⎜ ⇒ lim ⎜ ⎟ x →8 ⎝ x − 8 ⎠ x →8 ⎝ x − 8 ⎟ ⎠ x →8
rSn = r + 2r + 3r + ⋅⋅⋅⋅ ∞ (iii) 2
55. Topic: Limit, Continuity and Differentiability
⇒ lim
Let, Sn = 1 + 2r + 3r + 4 r + ⋅⋅⋅⋅ ∞ (ii) 2
54. Topic: Sequence and Series P: 1, 6, 11, 16, 21, 26, ... is an arithmetic progression with a common difference of 5. Q: 9, 6, 3, 0, –3, –6, … is an arithmetic progression with a common difference of –3. R: 1, 3, 9, 27, 81, … is a geometric progression with a common ratio of 3. S: 4, –8, 16, –32, 64, ... is a geometric progression with a common ratio of –2. Answer (B)
( x − 8)( x + 8) ⇒ lim ( x + 8) x →8 ( x − 8) ⇒ 8 + 8 = 16
⇒ 1.5 − 1.5r = 1 ⇒ 1.5 − 1 = 1.5r 0.5 1 ⇒r= = 1.5 3 Given,
16 + 24 = 20 2
1 + r + r 2 + r 3 + ⋅⋅⋅⋅ ∞ (1 − r ) 1.5 ⎛ 1⎞ ⎜⎝1 − ⎟⎠ 3
56. Topic: Laplace Transform The mathematical description of the one-dimensional waves (both traveling and standing) can be expressed as: ∂2u ∂2u = C2 2 2 ∂t ∂x with u is the amplitude of the wave at position x and time t, and C is the velocity of the wave. This equation is called the classical wave equation in one dimension and is a linear partial differential equation. It tells us how the displacement u can change as a function of position and time and the function. Answer (B) 57. Topic: Matrices and Determinants For A to be a skew-symmetric matrix, it must satisfy the following:
1.5 × 3 = 2.25 2
A T = −A = A Answer (2.25)
53. Topic: Mean, Median, Mode and Standard Deviation Given data set is 12, 10, 16,8, 90, 50,30, 24
⎡ 0 6⎤ Given, A=⎢ ⎥ ⎣ p 0⎦ ⎡ 0 −6 ⎤ ⇒ −A = ⎢ ⎥ ⎣− p 0 ⎦
18 GATE Biotechnology Chapter-wise Solved Papers ⎡0 p ⎤ AT = ⎢ ⎥ ⎣6 0 ⎦ For skew-symmetric matrix, –A = AT, therefore, ⎡0 ⎢6 ⎣
p⎤ ⎡ 0 6⎤ = 0 ⎥⎦ ⎢⎣ − p 0 ⎥⎦
1000 2 5 + 10000 × + 250000 × 4 10 100 = 2500 + 2000 + 12500 = 17000
=
(ii)
( )
Therefore, Variance of X, σ 2 = X 2 − X
2
From (i) and (ii)
Thus, p = –6.
σ 2 = 17000 − ( 20 ) = 16600 2
Answer (–6)
Standard deviation of X, σ = σ = 16600 129
58. Topic: Differential Equations Given differential equation: dy x = dx y
Answer (129) 60. Topic: Laplace Transforms f (t ) = t 2 + 2t + 1
⇒ y ⋅ dy = x ⋅ dx
L[ f (t )] = L ⎡⎣t 2 + 2t + 1⎤⎦
On integrating on both sides we get,
= L ⎡⎣t 2 ⎤⎦ + 2 L [t ] + L [1]
⇒ ∫ x ⋅ dx = ∫ y ⋅ dy x2 y2 = + c (i) 2 2 Given, x = 0, y = 1. On putting these values in (i) we get: ⇒
0 1 = +c 2 2 1 ⇒c=− 2
Using a linearity property, = Since, L ⎡⎣t n ⎤⎦ =
⇒
⇒ L [t ( t ) ] =
2! 2 1 + + s3 s 2 s n! s n +1 2 2 1 + + s3 s 2 s
On putting this value of c in (i) we get: ⇒
2
Answer (D)
2
1 x y = − 2 2 2
61. Topic: Integration of Trapezoidal and Simpson’s Rule Using trapezoid rule, we have
⇒ x2 = y2 −1
b
h
∫ f ( x)dx ≈ 2 [( y
⇒ y2 = x2 + 1
0
Answer (B) 59. Topic: Probability and Statistics The given table is showing discrete probability distributions. We know, ∑ P( X ) = 1 , where X can be any discrete random variable. Thus, mean of X, X = ∑ X ⋅ P( X ) = (–100) [0.25] + (0) (0.5) + (100) (0.2) + (500) (0.05) = –25 + 20 +25 = 20 (i) Now, mean of X2 = X 2 = ∑ X 2 ⋅ P( X ) = (–100)2 [0.25] + (0)2 (0.5) + (100)2 (0.2) + (500)2 (0.05)
+ yn ) + 2( y1 + y2 + ... + yn −1 )]
a
Given, y0 = 0.37, y1 = 0.51, y2 = 0.71, y3 = 1.0, y4 = 1.40, y5 = 1.95, y6 = 2.71 and h =
1 3
On putting these values in equation above, we get: ⎛ 1⎞ 1 ⎜⎝ ⎟⎠ ⎡(0.37 + 2.71) + 2(0.51 + 0.71 + 1.0 + ⎤ 3 f ( x ) dx ≈ ∫−1 1.40 + 1.95) ⎥⎦ 2 ⎢⎣ ≈
1 [3.08 + 2(5.57)] ≈ 2.37 6 Answer (2.37)
CHAPTER2
Biochemistry
Syllabus Biomolecules — Structure and Functions; Biological Membranes — Structure, Action Potential and Transport Processes; Enzymes — Classification, Kinetics and Mechanism of Action; Basic Concepts and Designs of Metabolism (Carbohydrates, Lipids, Amino Acids and Nucleic Acids) Photosynthesis, Respiration and Electron Transport Chain; Bioenergetics.
Chapter Analysis Topic Biomolecules: Structure and Functions Biological Membranes: Structure, Action Potential and Transport Processes Enzymes: Classification, Kinetics and Mechanism of Action Basic Concepts and Designs of Metabolism: (Carbohydrates, Lipids, Amino Acids and Nucleic Acids) Photosynthesis, Respiration and Electron Transport Chain Bioenergetics
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Questions 1. During lactic acid fermentation, net yield of ATP and NADH per mole of glucose is (A) 2 ATP and 2 NADH. (B) 2 ATP and 0 NADH. (C) 4 ATP and 2 NADH. (D) 4 ATP and 0 NADH. (GATE 2010; 1 Mark) 2. Identify the enzyme that catalyzes the following reaction.
3. The degree of inhibition for an enzyme catalyzed reaction at a particular inhibitor concentration is independent of initial substrate concentration. The inhibition follows (A) (B) (C) (D)
competitive inhibition. mixed inhibition. uncompetitive inhibition. non-competitive inhibition. (GATE 2010; 1 Mark)
α-Ketoglutarate + NADH + NH +4 + H + → Glutamate + NAD + H2O 4. +Oxidation reduction reactions with positive standard + + + redox potential (ΔE°) have α-Ketoglutarate + NADH + NH 4 + H → Glutamate + NAD + H 2 O (A) positive ΔG°. (A) Glutamate synthetase (B) negative ΔG°. (B) Glutamate oxoglutarate aminotransferase (C) positive ΔE+. (C) Glutamate dehydrogenase (D) negative ΔE+. (D) α-Ketoglutarate deaminase (GATE 2010; 1 Mark) (GATE 2010; 1 Mark)
20 GATE Biotechnology Chapter-wise Solved Papers 5. Match the chemicals in Group class in Group II. Group I P. Picloram Q. Zeatin R. Thiamine S. Glutamine (A) P-2; Q-4; R-1; S-3 (B) P–4; Q-l; R-2; S-3 (C) P-3; Q-l; R-2; S-4 (D) P-4; Q-2; R-l; S-3
I with the possible type/ Group II 1. Vitamin 2. Auxin 3. Amino Acid 4. Cytokinin
Statement for Linked Answer Questions 10 and 11: The standard redox potential values for two half-reactions are given below. The value for Faraday’s constant is 96.48 kJ V−1 mol−1 and Gas constant R is 8.31 J K−1, mol−1. NAD+ + H+ + 2e− ↔ NADH –0.315 V FAD + 2H+ + 2e− ↔ FADH2
–0.219 V
10. The ΔG° for the oxidation of NADH by FAD is (A) –9.25 kJ mol−1 (B) – 103.04 kJ mol−1 (C) + 51.52 kJ mol−1 (D) –18.5 kJ mol−1 (GATE 2010; 2 Marks)
(GATE 2010; 2 Marks) 6. 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. (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. (C) The velocity of E1 catalyzed reaction is always greater than that of E2. (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. (GATE 2010; 2 Marks) 7. The dissociation constant Kd for ligand binding to the receptor is 10−7 M. The concentration of ligand required for occupying 10% of receptors is (A) 10−6 M (B) 10−7 M (C) 10−8 M (D) 10−9 M (GATE 2010; 2 Marks) Common Data for Questions 8 and 9: The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right handed α-helix. 8. The number of α-helical turns permeating the membrane is (A) 5.6 turns (B) 3.5 turns (C) 6.5 turns (D) 5.0 turns (GATE 2010; 2 Marks) 9. The number of amino acid residues present in the protein is (A) 15 (B) 18 (C) 17 (D) 20 (GATE 2010; 2 Marks)
11. The value of ΔG°, if given Keq is 1.7, at 23°C will be (A) –17.19 kJ mol−1 (B) –19.8 kJ mol−1 (C) +52.82 kJ mol−1 (D) –117.07 kJ mol−1 (GATE 2010; 2 Marks) 12. An alternative to glycolysis pathway is (A) glyoxylate pathway. (B) pentose phosphate pathway. (C) citric acid cycle. (D) gluconeogenesis. (GATE 2011; 1 Mark) 13. In N-linked glycosylation, the oligosaccharide chain is attached to protein by (A) asparagine. (B) arginine. (C) serine. (D) threonine. (GATE 2011; 1 Mark) 14. Match items in Group I with Group II. Group I Group II P. Glycolytic pathway 1. Chloroplast Q. Eukaryotic oxidative 2. Glyoxysomes metabolism R. Glyoxylate cycle 3. Mitochondria S. Calvin cycle 4. Cytosol (A) (B) (C) (D)
P–1; Q–2; R–3; S–4 P–2; Q–3; R–4; S–l P–4; Q–3; R–2; S–l P–3; Q–4; R–l; S–2 (GATE 2011; 2 Marks)
15. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion: A very low amount of inhibitor can act as an activator for allosteric enzymes. Reason: Allosteric enzymes follow Michaelis–Menten kinetics.
Chapter 2 • Biochemistry 21
(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. (GATE 2011; 2 Marks)
(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. (GATE 2012; 2 Marks)
16. During photorespiration under low CO2 and high O2 levels, O2 reacts with ribulose 1,5- bisphosphate to yield (A) one molecule each of 3-phosphoglycerate and 2-phosphoglycolate. (B) two molecules of 3-phosphoglycerate. (C) two molecules of 2-phosphoglycolate. (D) one molecule each of 3-phosphoglycerate and glyoxylate. (GATE 2012; 1 Mark)
22. Match the high energy compounds in Group I with the biosynthetic pathways for the molecules in Group II.
17. The activity of an enzyme is expressed in International Units (IU). However, the SI unit for enzyme activity is Katal. One Katal is (A) 1.66 × 104 IU (B) 60 IU (C) 6 × 107 IU (D) 106 IU (GATE 2012; 1 Mark) 18. 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. (GATE 2012; 1 Mark) 19. The molarity of water in a water: ethanol mixture (15 : 85, v/v) is approximately (A) 0.85 (B) 5.55 (C) 8.5 (D) 55.5 (GATE 2012; 1 Mark) 20. 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 (GATE 2012; 2 Marks) 21. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: Plants convert fatty acids into glucose. Reason: Plants have peroxisomes.
Group I
Group II
P. GTP
1. Fatty acid
Q. UTP
2. Phospholipid
R. CTP
3. Protein
S. Acyl coenzyme A
4. Peptidoglycan
(A) (B) (C) (D)
P–3; Q–2; R–4; S–1 P–2; Q–4; R–3; S–1 P–4; Q–3; R–1; S–2 P–3; Q–4; R–2; S–1 (GATE 2012; 2 Marks)
23. 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
(A) (B) (C) (D)
P–5; Q–2; R–4; S–1 P–4; Q–1; R–3; S–2 P–4; Q–2; R–1; S–3 P–2; Q–1; R–3; S–5 (GATE 2012; 2 Marks)
24. An enzymatic reaction is described by the following rate expression. Vm S v= K M + S + S 2 / KS Which one of the following curves represents this expression? (B) (A) S
S
V
V
22 GATE Biotechnology Chapter-wise Solved Papers (C)
(C) P–3.4, Q–2.4, R–1.5, S–0.1 (D) P–100, Q–71, R–44, S–29 (GATE 2012; 2 Marks)
(D) S
S
V
V
(GATE 2012; 2 Marks) Common Data for Questions 25 and 26: In a muscle, the extracellular and intracellular concentrations 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 25o C and that the membrane potential is –60 mV, with the interior more negatively charged than the exterior. (R = 8.314 J mol−1 K−1; F = 96.45 kJ mol−1 V−1) 25. The free energy change for the transport of three Na+ out of the cell is (A) +1.5 kJ mol−1 (B) +17.4 kJ mol−1 (C) +18.9 kJ mol−1 (D) +36.3 kJ mol−1 (GATE 2012; 2 Marks) 26. The free energy change for the transport of two K into the cell is (A) +8.0 kJ mol−1 (B) +11.6 kJ mol−1 (C) +19.6 kJ mol−1 (D) +31.2 kJ mol−1 (GATE 2012; 2 Marks) +
Common Data for Questions 27 and 28: The purification data for an enzyme is given below: Step
P
Volume Total (mL) protein (mg) 177
Total Specific activity activity (Units) (Units/ mg)
Cell-free extract
17
102
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
29. Which one of the following is an ABC transporter? (A) Multidrug resistance protein (B) Acetylcholine receptor (C) Bacteriorhodopsin (D) ATP synthase (GATE 2013; 1 Mark) 30. The catalytic efficiency for an enzyme is defined as Vmax (A) kcat (B) kcat kcat (C) kcat (D) KM vmax (GATE 2013; 1 Mark) 31. 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 phosphate
3. Lactate dehydrogenase
S. Molybdenum
4. Glycogen phosphorylase
+
(A) (B) (C) (D)
P–3; Q–2; R–4; S–1 P–4; Q–1; R–3; S–2 P–3; Q–1; R–4; S–2 P–3; Q–4; R–2; S–1 (GATE 2013; 2 Marks)
0.58
32. 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 below suggests that the inhibitor (I ) exhibits
[I ] = 4 mM
27. The fold purification for each step is (A) (B) (C) (D)
P–0.1, Q– 0.66, R–0.84, S–1.26 P–1.0, Q–0.52, R–0.67, S–0.8 P–1, Q–6.6, R–8.4, S–12.6 P–100, Q–66, R–84, S–12 (GATE 2012; 2 Marks)
28. 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
[I ] = 2 mM
1/Vo
[I ] = 0 mM
0
1/[S ]
(A) substrate inhibition. (B) uncompetitive inhibition.
Chapter 2 • Biochemistry 23
(C) α-D-Glucose only. (D) β-D-Glucose only.
(C) mixed inhibition. (D) competitive inhibition. (GATE 2013; 2 Marks)
(GATE 2014; 1 Mark)
Common Data for Questions 33 and 34: A solution was prepared by dissolving 100 mg of protein X in 100 mL of water. Molecular weight of protein X is 15,000 Da; Avogadro’s number = 6.022 × 1023.
38. 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 (GATE 2014; 1 Mark)
33. Calculate the molarity (µM) of the resulting solution. (A) 66.6 (B) 6.6 (C) 0.67 (D) 0.067 (GATE 2013; 2 Marks) 34. The number of molecules present in this solution is (A) 40.15 × 1019 (B) 6.023 × 1019 (C) 4.015 × 1019 (D) 0.08 × 1019 (GATE 2013; 2 Marks) Common Data for Questions 35 and 36: The binding efficiency of three different receptors R1, 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
R1
3
R2
r/c 2
R3
1
R4
0
1
r
2
3
35. 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 (GATE 2013; 2 Marks) 36. Which one of the receptors has the highest affinity for the ligand? (A) R1 (B) R2 (C) R3 (D) R4 (GATE 2013; 2 Marks) 37. The product(s) resulting from the hydrolysis of maltose is/are (A) a mixture of α-D-Glucose and β-D-Glucose. (B) a mixture of D-Glucose and L-Glucose.
39. The reactions leading to the formation of amino acids from the TCA cycle intermediates are (A) carboxylation. (B) isomerization (C) transamination. (D) decarboxylation. (GATE 2014; 1 Mark) 40. Triose phosphate isomerase converts dihydroxy acetone phosphate (DHAP) to glyceraldehyde-3-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 freeenergy change in kJ mol−1 for the formation of G-3-P [R = 8.315 J mol−1 K−1] is __________. (GATE 2014; 2 Marks) 41. Match the following photoreceptors with their prosthetic groups and spectral specificity Photoreceptor
Moiety that absorbs light
Absorption (nm)
P. Phototropin
1. Chromobilin
a. 400–500
Q. Cryptochrome
2. FAD
b. 600–800
R. Phytochrome
3. FMN
c. 500–600
(A) (B) (C) (D)
P–3 –a; Q–2–a; R–1–b P–1–b; Q–1 –a; R–3–b P–3–a; Q–1–a; R–1–c P–2–c; Q–1–c; R–1–a (GATE 2014; 2 Marks)
42. If a plant is shifted to cold temperature, which of the following changes would take place in its membrane? (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. (GATE 2014; 2 Marks)
24 GATE Biotechnology Chapter-wise Solved Papers 43. 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)
46. A single subunit enzyme converts 420 µ moles of substrate to product in one minute. The activity of the enzyme is _________× 10−6 Katal. (GATE 2015; 1 Mark)
Number of intact protoplast
47. 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 (GATE 2015; 1 Mark) 48. Which one of the following graphs represents uncompetitive inhibition? (A)
Time
Number of intact protoplast
(B)
tor
ibi
nh hi
t
1 [V0]
Wi
ou
ith
W 0
tor
ibi
h t in
1 [s]
Time
(B) inh t
ou
ith
W 0
ib inh
1 [s]
ibi
to
r
(C)
Time
r
ito
ith
1 [V0]
W
Number of intact protoplast
ibi
tor
(C)
W
Number of intact protoplast
1 [V0]
ith
inh
(D)
tor
ibi
h t in
ou
ith
W 0
1 [s] ibit
or
(D)
44. The active site in the alpha/beta barrel structures is usually located (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. (GATE 2014; 2 Marks) 45. Levinthal’s paradox is related to (A) protein secretion. (B) protein degradation. (C) protein folding. (D) protein trafficking. (GATE 2015; 1 Mark)
th
1 [V0]
Wi
(GATE 2014; 2 Marks)
inh
Time
ou
ith
W 0
tor
ibi
h t in
1 [s]
(GATE 2015; 2 Marks) 49. The Ki of a novel competitive inhibitor designed against an enzyme is 2.5 µM. The enzyme was assayed in the absence or presence of the inhibitor (5 µM) under identical conditions. The KM in the presence of the inhibitor was found to be 30 µM. The KM in the absence of the inhibitor is __________ µM. (GATE 2015; 2 Marks)
Chapter 2 • Biochemistry 25
50. 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 mol−1. (GATE 2015; 2 Marks) 51. 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. (D) Within the matrix of mitochondria and within the stroma of chloroplast. (GATE 2016; 1 Mark) 52. Disaccharide molecules that contain β (1 →4) glycosidic linkage are (A) sucrose and maltose. (B) sucrose and isomaltose. (C) maltose and isomaltose. (D) lactose and cellobiose. (GATE 2016; 1 Mark) 53. 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 i nhibition. (C) A transition state analogue in enzyme catalyzed reaction increases the rate of product formation. (D) In non-competitive inhibition, KM of an enzyme for its substrate remains constant as the concentration of the inhibitor increases. (GATE 2016; 1 Mark) 54. The equilibrium potential of a biological membrane for Na+ is 55 mV at 37°C. Concentration of Na+ inside the cell is 20mM. 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) (GATE 2016; 2 Marks) 55. 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 µmol min−1 when 1.8 µg enzyme is added to the
assay mixture. If the KM for the substrate is 25 µM, the kcat/KM ratio will be ___________ × 104 M−1 s−1. (GATE 2016; 2 Marks) 56. Analysis of a hexapeptide using enzymatic cleavage reveals the following result: • Amino acid composition of the peptide is: 2R, A, V, S, Y • Trypsin digestion yields two fragments and the compositions are: (R, A, V) and (R, S, Y) • Chymotrypsin digestion yields two fragments and the compositions are: (A, R, V, Y) and (R, S) • Digestion with carboxypeptidase A yields no cleavage product. Given: Trypsin cleaves at carboxyl side of R. Chymotrypsin cleaves at carboxyl side of Y. Carboxypeptidase A cleaves at amino side of the Cterminal amino acid (except R and K) of the peptide. The correct amino acid sequence of the peptide is: (A) RSYRVA (B) AVRYSR (C) SRYVAR (D) SVRRYA (GATE 2016; 2 Marks) 57. 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. (GATE 2017; 1 Mark) 58. Natural proteins are composed primarily of 20 α-amino acids. Which one of the following statements is true for any of these amino acids in a solution 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 i onized. (D) Both amino and carboxylic add groups are neutral. (GATE 2017; 1 Mark) 59. If a protein contains four cysteine residues, the number of different ways they can simultaneously form two intramolecular disulfide bonds is ___________. (GATE 2017; 1 Mark) 60. An enzymatic reaction exhibits Michaelis–Menten kinetics. For this reaction, on doubling the concentration of enzyme while maintaining [S]>> [E0], (A) both KM and Vmax will remain the same. (B) KM will remain the same but Vmax will increase. (C) KM will increase but Vmax will remain the same. (D) both KM and Vmax will increase. (GATE 2017; 1 Mark)
26 GATE Biotechnology Chapter-wise Solved Papers 61. Which one of the following amino acids has three ionizable groups? (A) Glycine (B) Leucine (C) Valine (D) Lysine (GATE 2017; 2 Marks)
P. Bond stretching R. Torsional bond rotation
Q. Bond angle bending S. N on-bonded interactions (A) P and Q only (B) P, Q and R only (C) P, Q and S only (D) P, Q, R and S (GATE 2018; 1 Mark)
62. If the chemical composition of proteins in an organism is CH1.5 O0.3 N 0.3S0.004 . the mass percentage of carbon in the proteins is _______. Given data: Atomic weights (Da) of C = 12, H = 1, O = 16, N = 14, and S = 32. (GATE 2017; 2 Marks) 63. The interaction energy E between two spherical particles is plotted as a function of the distance (r) between them. When r < a, where a is a constant, the net force between the spherical particles is repulsive. When r ≥ a. they attract via van der Waals attraction. Which one of the following plots correctly represents the interaction energy between the above two particles? (A) E(r)
0
(B)
a
r
65. Which one of the following is INCORRECT about protein structures? (A) A protein fold is stabilized by favorable noncovalent interactions. (B) All parts of a fold can be classified as helices, strands or turns. (C) Two non-covalent atoms cannot be closer than the sum of their van der Waals radii. (D) The peptide bond is nearly planar. (GATE 2018; 1 Mark) 66. Which one of the following metabolic processes in mammalian cells does NOT occur in the mitochondria? (A) Citric acid cycle (B) Oxidative phosphorylation (C) Fatty acid β-oxidation (D) Glycolysis (GATE 2018; 1 Mark) 67. Yeast converts glucose to ethanol and carbon dioxide by glycolysis as per the following reaction:
E(r)
C6H12O6 → 2C2H5OH + 2CO2
0
(C)
r
E(r)
0
(D)
a
a
r
a
r
68. Consider a simple uni-substrate enzyme that follows Michaelis–Menten kinetics. When the enzyme catalyzed reaction was carried out in the presence of 10 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 __________. (GATE 2018; 2 Marks)
E(r)
0
Assuming complete conversion, the amount of ethanol produced (in g) from 200 g of glucose is (up to two decimal places)_________. (GATE 2018; 2 Marks)
(GATE 2017; 2 Marks) 64. Which of the following components constitute a molecular mechanics force field?
69. 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) (GATE 2018; 2 Marks)
Chapter 2 • Biochemistry 27
70. The mass of 1 kmol of oxygen molecules is ___________ g (rounded off to the nearest integer). (GATE 2019; 1 Mark) 71. Which one of the following need NOT be conserved in a biochemical reaction? (A) Total mass (B) Total moles (C) Number of atoms of each element (D) Total energy (GATE 2019; 1 Mark) 72. Which one of the following statements is CORRECT for enzyme catalyzed reactions? (DG is Gibbs free energy change, Keq is equilibrium constant) (A) Enzymes affect DG, but not Keq (B) Enzymes affect Keq, but not DG (C) Enzymes affect DG and Keq (D) Enzymes do not affect DG or Keq (GATE 2019; 1 Mark) 73. Protein concentration of a crude enzyme preparation was 10 mg mL−1. 10 mL of this sample gave an activity of 5 mmol min−1 under standard assay conditions. The specific activity of this crude enzyme preparation is ______________ units mg−1.
74. Which one of the following amino acid residues will destabilize an a-helix when inserted in the middle of the helix? (A) Pro (B) Val (C) Ile (D) Leu (GATE 2019; 2 Marks) 75. In general, which one of the following statements is NOT CORRECT? (A) Hydrogen bonds result from electrostatic interactions. (B) Hydrogen bonds contribute to the folding energy of proteins. (C) Hydrogen bonds are weaker than van der Waals interactions. (D) Hydrogen bonds are directional. (GATE 2019; 2 Marks) 76. The difference in concentrations of an uncharged solute between two compartments is 1.6-fold. The energy required for active transport of the solute across the membrane separating the two compartments is_____cal mol−1 (rounded off to the nearest integer). (R = 1.987 cal mol−1 K−1, T = 37°C). (GATE 2019; 2 Marks)
(GATE 2019; 1 Mark)
ANSWER KEY 1. (B)
2. (C)
3. (D)
4. (B)
5. (A)
6. (C)
7. (C)
8. (A)
9. (D)
10. (D)
11. (A)
12. (B)
13. (A)
14. (C)
15. (C)
16. (A)
17. (C)
18. (C)
19. (C)
20. (A)
21. (A)
22. (D)
23. (B)
24. (A)
25. (D)
26. (D)
27. (C)
28. (D)
29. (A)
30. (C)
31. (C)
32. (D)
33. (66.6 µM)
34. (4.01 × 10 )
35. (D)
36. (A)
-1
19
39. (C)
40. (5.7 kJ mol )
41. (A)
42. (A)
45. (C)
46. (7 × 10 Katal)
47. (C)
48. (A)
50. (– 47.7 kJ mol ) 51. (B)
52. (D)
53. (C)
54. (158.49 mM)
55. (8.93 × 10 M s )
56. (B)
57. (B)
58. (A)
59. (3)
60. (B)
61. (D)
62. (53.03%)
63. (D)
64. (D)
65. (B)
66. (D)
67. (102.22 g)
68. (5 nM)
69. (2.94 × 10 ) 70. (32000 g)
71. (B)
72. (A)
74. (A)
75. (C)
37. (A)
38. (C)
43. (A)
44. (C)
-6
-1
49. (10 µM) 4
-1
-1
73. (50 µ mol min mg ) −1
−1
6
76. (289.51 cal mol ) −1
28 GATE Biotechnology Chapter-wise Solved Papers
Answers with Explanation 1. Topic: Metabolism—Fermentation Lactic acid fermentation is observed in some bacteria. It is also observed in the muscles of animals, wherein, during vigorous activity, when the demand for ATP is high and oxygen is in short supply, ATP is mainly synthesized via anaerobic glycolysis. In this process, ATP is generated rapidly as compared to the slower process of oxidative respiration. Under these conditions, the enzyme lactate dehydrogenase (LDH) catalyzes the oxidation of NADH by pyruvate to yield NAD+ and lactic acid.
spontaneous redox reaction, Gibbs free energy, ΔG° must be negative, in accordance with the following equation: ΔG° = −nFE° Answer (B)
5. Topic: Biomolecules—Structure and Function • Picloram is a synthetic auxin or also called as auxin mimic. It is used as herbicide to kill susceptible plants by imitating the plant growth hormone auxin (indole acetic acid). On its administration in effective doses, it can cause uncontrolled and disorganized growth of 2+ plant that leads to its death. dehydrogenase , Mg , FMN Pyruvic acid + NADH + H + ⎯Lactic ⎯⎯⎯⎯⎯⎯⎯⎯ → Lactic acid + NAD + 2+ • Zeatin is the natural cytokinin found in corndehydrogenase , Mg , FMN Pyruvic acid + NADH + H + ⎯Lactic ⎯⎯⎯⎯⎯⎯⎯⎯ → Lactic acid + NAD + kernels and coconut milk. Zeatin was first isolated from young corn seed endosperm in 1963. It is the Under anaerobic conditions only 2 molecules of ATP are most common cytokinin in plants. produced (as a result of glycolysis). • Thiamine is Vitamin B1. It acts as coenzyme for Answer (B) many different enzymes that break carbon-to 2. Topic: Metabolism—Amino Acids carbon bonds and are involved in carbohydrate Glutamate dehydrogenase (GDH) is located in the mitometabolism of pyruvic acid to CO2 and H2O. It is chondria. GDH catalyzes the reversible NADP+-linked essential for synthesis of neurotransmitter acetylchooxidative deamination of L-glutamate into alpha ketogluline. Its deficiency in body causes Beri-Beri, which tarate and ammonia in two steps. is partial paralysis of smooth muscle of alimentary • The first step involves a Schiff base intermediate canal, causing digestive disturbances and skeletal being formed between ammonia and alpha ketoglutamuscle paralysis. rate. This Schiff base intermediate is crucial because • Glutamine is a standard neutral amino acid. It is it establishes the alpha carbon atom in glutamate’s found abundantly in muscles and gets converted into stereochemistry. glutamic acid essential for cerebral function. It helps • The second step involves the Schiff base intermedito treat arthritis, autoimmune diseases, fibrosis, etc. ate being protonated, which is done by the transfer of Answer (A) a hydride ion from NADPH resulting in L-glutamate. Answer (C) 6. Topic: Enzymes—Kinetics The turnover number of an enzyme is the number of reac 3. Topic: Enzymes—Kinetics tion processes (turnovers) that each active site catalyzes In non-competitive inhibition, inhibitors do not compete per unit time. It is also defined as the catalytic constant, with the substrate for the same binding site of the enzyme; kcat of an enzyme as: rather, the inhibitor binds to a site other than the enzyme’s V active site. This changes the conformation of the active kcat = max site in such a manner that the substrate can no longer bind [ E ]T to the active site. This type of inhibition depends only Here, kcat = Turnover number, on the inhibitor concentration and is independent of the concentration of the substrate. This is because in the presVmax = Maximum rate of reaction when all the enzyme ence of a non-competitive inhibitor, a certain fraction of catalytic sites are saturated with substrate and the enzyme molecules is necessarily inactive at any given ET = Total enzyme concentration or concentration of total instant. The inactivation of cytochrome oxidase by cyaenzyme catalytic sites. nide is an example of non-competitive inhibition. Thus, with this relation, we know that for the same Answer (D) concentration of both the enzyme, velocity of E1 will 4. Topic: Bioenergetics be greater than that of E2, as turnover number of E1 is higher. A redox reaction is spontaneous, if the standard electrode potential for the redox reaction is positive. And, for a Answer (C)
Chapter 2 • Biochemistry 29
11. Topic: Electron Transport Chain
7. Topic: Enzymes—Kinetics
Given, Keq = 1.7
Ka
Ligand + Receptor [LR ] K (L)
d
(R)
[LR ] Ka = or [ ][R ] L Association constant
[L][R ] Kd = [LR ] Dissociation constant
Given, 10 % receptor is occupied,
Answer (A)
K × [LR ] [ L] = d [R ]
12. Topic: Metabolism—Glycolysis
0.1 × 10 −7 = ≈ 10 −8 0.9 Answer (C) 8. Topic: Biological Membrane—Structure We know, In a right handed α-helix, in a pitch of 5.1Å, number of turn = 1 Given, Width of the lipid bilayer membrane = 30 Å For 30 Å, number of turns = 30/5.1 = 5.6 turns Answer (A) 9. Topic: Biomolecules—Structure of Proteins We know, In a right handed α-helix, number of residues in 1 turn = 3.6 Number of residues in 5.6 turns = 5.6 × 3.6 = 20 Answer (D) 10. Topic: Electron Transport Chain Given, F = is 96.48 kJ V−1 mol−1 R = 8.31 J K−1 mol−1
FA D + + 2H + + 2e− ←⎯ → FA D H 2
−0.315 V −0.219 V
Oxidation of NADH by FAD ΔEcell = Ered − Eox = −0.219 − ( −0.315) = 0.096 V ΔG = − nF ΔEcel l
= −2 × 96.48 × 0.096 = −18.5kJm ol−1
ΔG = ΔG + RT ln K eq = −17.2 kJm ol−1
[L] + 0.9[R ] 0.1[LR ]
N A D + H + + 2e− ←⎯ → NADH
ΔG = ?
= −18.5 + 8.31× 296 × ln1.7
Kd = 10−7 M
+
Temperature, T = 23°C = 23 + 273 = 296 K
Answer (D)
The pentose phosphate pathway or phosphogluconate pathway or hexose monophosphate shunt is a metabolic pathway alternate to glycolysis. It produces NADPH and pentoses (5-carbon sugars) as well as ribose 5-phosphate, the last one is a precursor for the synthesis of nucleotides. Answer (B) 13. Topic: Biomolecules—Structure N-linked glycosylation or N glycosylation refers to the attachment of glycan (oligosaccharide) to the amide (N-group) of asparagine of the protein. Answer (A) 14. Topic: Metabolism—Carbohydrates • Glycolysis or glycolytic pathway converts glucose to pyruvate through a series of intermediate metabolites which occurs in cytoplasm. • Eukaryotic oxidative metabolism occurs at the internal folded mitochondrial membranes called cristae. • The glyoxylate cycle occurs in plant’s glyoxysomes (special peroxisomes). This cycle lets seeds to use lipids as a source of energy to form the shoot system during germination. • The Calvin cycle (C3 cycle) is the set of chemical reactions that occurs in chloroplasts during photosynthesis. The reaction is light-independent because it takes place after the energy has been trapped from sunlight. Answer (C) 15. Topic: Enzymes—Kinetics An allosteric enzyme is an enzyme that contains a region to which small, regulatory molecules (called effectors) may bind in addition to and separate from the substrate binding site and thereby affect the catalytic activity. Allosteric enzymes are an exception to the Michaelis– Menten model. Because they have more than two subunits and active sites, they do not obey the Michaelis–Menten kinetics, but instead have sigmoidal kinetics. Allosteric
30 GATE Biotechnology Chapter-wise Solved Papers enzymes have two states: a low affinity state dubbed the T state and the high affinity R state. Inhibitors work by preferentially binding to the T state of an allosteric enzyme, causing the enzyme to maintain this low affinity state. Allosteric Inhibition
Allosteric Activation
Enzyme Enzyme Active site
Allosteric Altered active site site Activator
Inhibitor Substrate
Substrate
19. Topic: Biomolecules Molarity is defined as number of moles of solute per liter of solution Molarity = (weight of solute/mol. mass of solute) × (1000/vol. of solution in mL) Weight of water = 15mL = 15g (taking density of water to be 1) Molecular mass of water (H2O) = 2 + 16 = 18 Volume of solution = 100 mL Hence, molarity = (15/18) × (1000/100) = 0.833 × 10 = 8.33 M Answer (C)
Altered active site
Active site
Answer (C) 16. Topic: Photorespiration Photorespiration is process in which RuBisCO oxygenates RuBP rather than addition of carbon dioxide. It decreases the efficiency of photosynthesis and reduced the photosynthetic output by 25%. During photorespiration, oxygen is added to ribulose-1, 5-bisphosphate to produce one molecule of 3-phosphoglycerate and one molecule of 2-phosphoglycolate. Answer (A) 17. Topic: Enzymes—Mechanism of Action Enzyme activity of any enzyme is expressed in terms of international unit. One international unit of enzyme activity can be defined as amount of enzyme consuming or producing 1 micro-mole of substrate or product per minute under standard conditions. In contrast, Katal is the unit and which gives the amount of enzyme required to convert 1 mol of substrate per second. Enzyme unit (IU) =
μmol 10 −6 mol 1 = = Katal 60 sec min 6 × 10 7
∴ 1 Katal = 6 × 10 7 IU Answer (C) 18. Topic: Enzymes—Mechanism of Action Enzymes are biological catalyst which speed up the rate of reaction by lowering the activation energy of the system. They show proximity and propinquity effects and being protein in nature they often show acid-base chemistry. They do not change the equilibrium constant and free energy of the system. Thus, they catalyze both forward as well as backward reactions in order to achieve equilibrium state. Answer (C)
20. Topic: Biomolecules—Protein Structure and Function Zinc finger proteins are among the most abundant proteins in eukaryotic genomes. Their functions are extraordinarily diverse and include DNA recognition, RNA packaging, transcriptional activation, regulation of apoptosis, protein folding and assembly and lipid binding. The zinc ion is coordinated by two histidine residues and two cysteine residues. Answer (A) 21. Topic: Metabolism—Lipids Fatty acid degradation in most organisms occurs through the beta-oxidation cycle. In mammals, betaoxidation occurs in both mitochondria and peroxisomes, whereas in plants and most fungi the beta-oxidation cycle occurs only in the peroxisomes. In plant cells, peroxisomes play a variety of roles including converting fatty acids to sugar and assisting chloroplasts in photorespiration. In animal cells, peroxisomes protect the cell from its own production of toxic hydrogen peroxide. Answer (A) 22. Topic: Metabolism • GTP-binding protein regulators are proteins that regulate the function of small G proteins. • In the step of peptidoglycan synthesis, the N-acetyl-glucosamine-1-phosphate attacks UTP (Uridine triphosphate), a pyrimidine nucleotide, has the ability to act as an energy source. • Two minor phospholipids are phosphatidylglycerol (PG) and phosphatidylinositol (PI). They are derived from phosphatidic acid instead of 1,2-diglyceride. Phosphatidic acid is activated by activation using CTP to form the high energy compound CDP-diglyceride. • Acyl coenzyme A is a group of coenzymes involved in the metabolism of fatty acids. Answer (D) 23. Topic: Metabolism • Pantothenate (vitamin B5) is the key precursor for the biosynthesis of coenzyme A (CoA), an essential
Chapter 2 • Biochemistry 31
•
•
•
cofactor involved in a many metabolic reactions, including the synthesis of phospholipids, the synthesis and degradation of fatty acids, and the operation of the tricarboxylic acid cycle (TCA or Krebs’ cycle). FAD and FMN are called as flavins, since they are derived from riboflavin (Vitamin B2). Flavoproteins of electron transport chain (ETC), including FMN in Complex I and FAD in Complex II. Vitamin B6 (pyridoxine) is a water-soluble vitamin that is naturally present in many foods. It helps in carboxylation reactions. Tetrahydrafolate (THF) is derived from the vitamin folic acid (folate). We eat folate and use the enzyme dihydrofolate reductase to convert it into tetrahydrofolate, which is the active form that carries 1-carbon groups in a variety of reactions. Answer (B)
24. Topic: Enzymes—Kinetics Given, Vm S v= K M + S + S 2 / Ks
[ Na + ]out + Z ⋅ F ⋅ Δψ [ Na + ]in
= 8.314 × 298 × ln
150 mM + 1 × 96.45 × 103 × (60 × 10 −3 ) 12 mM
= 2477.572 × ln 12.5 − 5787 = 2477.572 × 2.526 − 5787 = 6257.674 + 5787 = 12044.674 J mol −1 = 12.04 kJ mol −1 For 3 Na+ ions, it will be three times, thus
ΔG3 Na + = 12.04 × 3 Answer (D) 26. Topic: Biological Membrane—Action Potential Free energy change for transport of one K+ ions from outside to inside that is against the concentration gradient, must be positive. It can be calculated as:
Vm S K M + S (1 + S / K s )
⎞ ⎛ This is factor 1+ s of the substrate inhibition, Si. In ⎜⎝ K ⎟⎠ s substrate inhibition, the velocity of enzymatic reaction increases with increase in substrate concentration, till velocity reaches vm. After achieving maximum velocity, any further increase in substrate concentration, starts decreasing the velocity. Thus, only graph of option (A) is satisfying the condition. Answer (A)
25. Topic: Biological Membrane—Action Potential Given, Temperature, T = 25o C = 25 + 273 = 298 K Membrane potential, Δψ = - 60 mV = - 60 × 10−3 V (As interior is more negatively charged than exterior, we can take this value as + 60 × 10−3 V for calculation) R = 8.314 J mol−1 K−1 F = 96.45 kJ mol−1 V−1 = 96.45 × 103 J mol−1 V−1 Na+ = 150 mM K+ = 2.7 mM 12 mM = Na+ 140 mM = K+ 2K+
Cell
ΔGNa + = RT ln
= 36.12 kJ mol −1
It can also be written as: v=
Free energy change for transport of one Na+ ions from inside to outside that is against the concentration gradient, must be positive. It can be calculated as:
Na+ – K+ pump
3Na+
ΔGK + = RT ln
[K + ]in + Z ⋅ F ⋅ Δψ [K + ]out
= 8.314× 298× ln
140m M + 1× 96.45× 103 × (60× 10−3 ) 2.7 m M
= 2477.575 × ln 51.85 + 5787 = 2477.575 × 3.95 + 5787 = 9779.94 + 5787 = 15566.94 J mol −1 = 15.57 kJ mol −1 For 2 K+ ions, it will be two times, thus ΔG2K + = 15.57 × 2 = 31.14 kJ mol −1 Answer (D) 27. Topic: Enzymes—Kinetics Fold purification can be calculated at each step by dividing specific activity of each step by initial specific activity. Therefore, At P, Fold purification =
0.58 =1 0.58
At Q, Fold purification =
3.83 = 6.6 0.58
At R, Fold purification =
4.89 = 8.43 0.58
32 GATE Biotechnology Chapter-wise Solved Papers
At S, Fold purification =
7.32 = 12.62 0.58 Answer (C)
•
28. Topic: Enzymes—Kinetics For percentage yield calculation at each step, divide total enzyme activity at each step by starting total activity multiply by100. At P, Percentage yield =
102 × 100 = 100% 102
At Q, Percentage yield =
72 × 100 = 70.59% 102
At R, Percentage yield =
45 × 100 = 44.12% 102
30 At S, Percentage yield = × 100 = 29.41% 102 Answer (D) 29. Topic: Biological Membrane—Transport Process ABC (ATP-binding cassette) transporters are the members of membrane transport system which use ATP to transport material across the plasma membrane. They play crucial role in developing multidrug resistance in a cell. They do so by actively pumping out the drug of the cell. Answer (A) 30. Topic: Enzyme—Kinetics The catalytic efficiency or specificity constant refers to the measure of efficiency of an enzyme, i.e., how efficiently an enzyme can convert a substrate into the product. It can be calculated as kcat/KM. where, kcat is a turnover number of an enzyme, i.e. no. of times each enzyme site converts a substrate into its product in unit time, and KM is a Michaelis constant which represents the substrate concentration at which rate of reaction is half of maximum velocity (1/2 Vmax). Answer (C) 31. Topic: Bioenergetics • Lactate dehydrogenase (LDH) is an enzyme required during the process of turning sugar into energy for cells. LDH or LD is found in nearly all living cells including animals, plants, and prokaryotes. It catalyzes the conversion of lactate to pyruvic acid and reverse, as it converts NAD+ to NADH and reverse too. • Glutathione peroxidase or GPx is a seleniumcontaining antioxidant enzyme that catalyzes the
•
reduction of hydrogen peroxide (H2O2) to water and oxygen and in turn oxidizes glutathione to glutathione disulfide. Glycogen phosphorylase is an enzyme that catalyzes the phosphorolysis of glycogen to produce glucose 1-phosphate, having a cofactor called pyridoxal-5′– phosphate, that is located in the active site and bound to a K681 residue with a Schiff base linkage. It shuttles the phosphate group onto the substrate. Nitrogenases are enzymes that are produced by certain bacteria. These enzymes are responsible for the reduction of nitrogen (N2) to ammonia (NH3). It is composed of mainly two component proteins called the Fe protein and the Mo-Fe protein. Answer (C)
32. Topic: Enzymes—Kinetics A substance that competes directly with a normal substrate for an enzyme’s substrate-binding site is known as a competitive inhibitor. Such an inhibitor usually resembles the substrate so that it specifically binds to the active site but differs from the substrate so that it cannot react as the substrate does. Its equation will be: 1 ⎛ α KM ⎞ 1 1 + =⎜ ⎟ Vo ⎝ Vmax ⎠ [S ] Vmax A plot of this equation is linear and has a slope of α KM/Vmax, a 1/[S] intercept of −1/α KM, and a 1/Vo intercept of 1/Vmax. The double-reciprocal plots for a competitive inhibitor at various concentrations of I intersect at 1/Vmax on the 1/Vo axis, a property that is diagnostic of competitive inhibition. 1/Vo Increasing [I ]
a=4 a=2 a = 1 (no inhibitor)
1/Vmax
Slope = aKM /Vmax a=1+
–1/aKM
•
0
[I ] K1
1/[S ]
The double-reciprocal plot consists of a family of parallel lines with slope KM/Vmax, 1/Vo intercepts of α ′/Vmax, and 1/[S] intercepts of −α ′/KM is a Lineweaver –Burk plot of Michaelis–Menten enzyme in the presence of uncompetitive inhibitor.
Chapter 2 • Biochemistry 33 1/Vo
35. Topic: Enzymes—Kinetics A receptor has a limited number of binding sites and is therefore saturated at high ligand concentration. As
Increasing [I ]
a′ = 2
a′/Vmax
4 3
a′ = 1 (no inhibitor) Slope = KM /Vmax a′ = 1 + 0
–a′/KM
•
[I ] K′1
1/Vo a = 2.0 a′ = 1.5
a = 1.5 a′ = 1.25
Increasing [I ]
a = a ′ = 1 (no inhibitor) Slope = aKM /Vmax [I ] a =1+ K1 [I ] a′ = 1 + K ′1
1 – a′ , a – a′ (a – 1)KM (a – 1)Vmax
–
0
a′ aKM
1/[S ]
Answer (D) 33. Topic: Biomolecules Given, Weight of protein = 100 mg = 0.1 g Molecular weight of protein = 15000 Da = 15 × 103 g Volume = 100 mL = 0.1 L NA = 6.022 × 1023 Molarity(μM) = =
R2
r/c 2
R3
1
R4
1/[S ]
I n a Lineweaver –Burk plot of Michaelis –Menten enzyme in the presence of mixed inhibitor, lines intersect to the left of the 1/Vo axis
a′/Vmax
R1
a′ = 1.5
0
1
34. Topic: Biomolecules Number of molecules = NA × Molarity 23
= 6.022 × 10 × 66.6 × 10
3
shown in the plot between ratio of moles of bound ligand per moles to the concentration of free ligand (r/c) and moles of bound ligand per moles (r), is same starting from the same point for all the receptors. Thus, all the receptors (R1, R2 and R3) must have same number of two binding sites for ligand. Answer (D) 36. Topic: Enzymes–Kinetics More is the value of r/c, more is the efficiency of binding the ligand by a receptor. On analyzing the plot, it is clear that value of r/c is highest for receptor R1. Thus, receptor R1 has got the highest affinity for the ligand. Answer (A) 37. Topic: Biomolecules—Structure and Function Maltose (C12H22O11) is a disaccharide composed of two D-glucose residues joined together by α (1→4) glycosidic bond. It is found in sprouting grains. It is found less commonly in nature than sucrose and lactose. On hydrolysis, maltose splits into two molecules of glucose. CH2OH
CH2OH O
H
H
O OH
OH HO
+ H2O
OH
O
H
OH
Weight of protein Molecular weight of protein 0.1 = 66.6μM 15 × 103 Answer (66.6 µM)
2
r
OH Maltose
CH2OH
CH2OH O
H
O OH +
OH HO
OH
OH HO
H
−6
= 4.01 × 1019 molecules Answer (4.01 × 1019 molecules)
OH a-D-Glucose
OH b -D-Glucose
Answer (A)
34 GATE Biotechnology Chapter-wise Solved Papers 38. Topic: Biomolecules—Structure and Function The amino acid side chains in globular proteins are spatially distributed according to their polarities: •
•
•
he non-polar residues Val, Leu, Ile, Met, and Phe T occur mostly in the interior of a protein, out of contact with the aqueous solvent. The hydrophobic effects that promote this distribution are largely responsible for the three-dimensional structure of native proteins. The charged polar residues Arg, His, Lys, Asp, and Glu are usually located on the surface of a protein in contact with the aqueous solvent. The uncharged polar groups Ser, Thr, Asn, Gln, and Tyr are usually on the protein surface but also occur in the interior of the molecule. When buried in the protein, these residues are almost always hydrogen bonded to other groups; in a sense, the formation of a hydrogen bond neutralizes their polarity. Answer (C)
39. Topic: Metabolism—Carbohydrate During exercise, some of the pyruvate generated by increased glycolytic flux is directed toward oxaloacetate synthesis as catalyzed by pyruvate carboxylase. Pyruvate can accept an amino group from glutamate (a transamination reaction) to generate alanine (the amino acid counterpart of pyruvate) and the citric acid cycle intermediate α -ketoglutarate (the ketone counterpart of glutamate). Both of these mechanisms help the citric acid cycle efficiently catabolize the acetyl groups derived—also from pyruvate—by the reactions of the pyruvate dehydrogenase complex. The end result is increased production of ATP to power muscle contraction. Answer (C) 40. Topic: Bioenergetics At, Temperature, T = 298 K and pH = 7 Triose phosphate
DHAP G-3-P Isomerase 44 mM
0 mM
R = 8.315 J mol−1 K−1 At equilibrium, Concentration of [DHAP] = 40 mM Concentration of [G-3-P] = 4 mM Standard free energy change ( ΔG °) = − RT ln =
[Products] [Reactants ]
−8.315 J [4 mM ] × 298K × ln K ⋅ mol [40 mM ]
=−
[4 mM ] 8.315 J × 298K × ln K ⋅ mol [40 mM ]
= −2477.5 × ( −2.3) = 5698.4 J mol −1 = 5.7 kJ mol −1 Answer (5.7 kJ mol–1) 41. Topic: Photosynthesis • Phototropins are blue-light receptors controlling a range of responses that serve to optimize the photosynthetic efficiency of plants. These include phototropism, light-induced stomatal opening, and chloroplast movements in response to changes in light intensity. It contains two light, oxygen, and voltage domains in the N-terminus, which bind a chromophore flavin mononucleotide (FMN) and a serine/ threonine kinase domain in the C-terminus, which is required for autophosphorylation of the protein. The action spectrum of plant phototropin is in the UV-A and blue light range (360-500 nm). • The blue light receptors cryptochromes mediate various light responses in plants. The photoexcited cryptochrome molecules undergo a number of biophysical and biochemical changes, including electron transfer, phosphorylation, and ubiquitination, resulting in conformational changes to propagate light signals. The structure of cryptochrome involves a fold very similar to that of photolyase, with a single molecule of FAD non-covalently bound to the protein. Its activity is fairly uniform throughout its range of maximal response of UV-A and blue light (390-480 nm). • The phytochrome molecule is the photoreceptor for red light responses. It exists in two forms, Pr and Pfr. The Pr form absorbs at a peak of 666 nm. The Pfr form absorbs at a peak of 730 nm. Phytochrome is a soluble chromoprotein with a molecular mass of 250 kDa it occurs as a dimer made up of two subunits, each of 125 kDa. Each subunit consists of two components — light-absorbing pigment molecule, chromophore and a polypeptide chain, Apo protein. The chromophore is a linear tetrapyrrole similar to phycocyanin termed phytochromobilin and it is a ring attached to the protein through thioether-linkage to a cysteine residue. Answer (A) 42. Topic: Biological Membrane—Structure In nature and in the laboratory, a decrease in temperature may cause liquids to solidify and an increase in temperature may cause solids to liquefy. In all instances the lipids are fluid at temperatures at which the organism
Chapter 2 • Biochemistry 35
normally exists. In general, the higher this temperature. the higher the lipid melting point; the lower the temperature, the lower the lipid melting point. It is well known that straight chain saturated fatty acids have higher melting point that same length unsaturated fatty acids. At low temperatures, the degree of unsaturation of fatty acids is increased through complex biosynthesis pathways. Cold acclimation increases the ratio of unsaturated to saturated fatty acids. Answer (A) 43. Topic: Biological Membrane—Action Potential
A protoplast when placed in distilled water which acts as hypotonic solution for it, will absorb water by endosmosis, so that the increased volume of water in the cell will increase pressure, making the protoplasm push against the cell wall, a condition known as turgor. They swell quickly and, ultimately cell membrane burst. Thus, the number of intact cells will decrease with time. Thus, graph in option (B) and (C) are wrong. The protoplasts do not start to bust immediately. For first few minutes, it just absorbs water and swell and when the intake of water exceeds its limit then the cells burst. Thus, graph (A) is correct. Answer (A)
44. Topic: Biomolecules—Protein Structure and Function Alpha-Beta domains are the most commonly observed domain structure. In general, they consist of parallel or mixed beta sheets connected and surrounded by alpha helices. Alpha-beta barrels must have at least 8 parallel beta strands (can be more) that form a closed circle and enclose a hydrophobic core. Cross connections between the parallel beta strands are alpha helices as in the betaalpha-beta motif. All alpha helices are on one side of the beta sheet. The active site of all alpha-beta barrels is found in a pocket formed when carboxy ends of parallel strands loop to connect to amino ends of adjacent alpha helices. Answer (C) 45. Topic: Biomolecules—Protein Structure and Function Levinthal’s paradox is an apparent contradiction between the number of possible conformations for a protein chain and the fact that proteins can fold to their native conformation quickly. The paradox is resolved by thermodynamic and kinetic principles which create a folding pathway for the unfolded protein. The paradox is reopened by intrinsically disordered regions of proteins which are inherently random. Multiple techniques like NMR, circular dichroism, X-ray crystallography and cryo-EM are used together to confirm the protein structure. Answer (C)
46. Topic: Enzymes—Kinetics Katal (kat) is defined as the amount which will catalyze the transformation of one mole of substance per second. Given, Concentration of substrate = 450 µ moles = 450 ×10−6 moles Time = 1 min = 60 s We know, moles Enzymatic activity, Ea = time =
420 × 10 −6 = 7 × 10 −6 Katal 60 Answer (7 × 10−6 Katal)
47. Topic: Biomolecules—Protein Structure and Function The charged polar residues Arg, His, Lys, Asp and Gl are usually located on the surface of a protein in contact with the aqueous solvent. In contrast, amino acids with non-polar side groups avoid water and aggregate to form the water insoluble core of proteins. Answer (C) 48. Topic: Enzymes—Kinetics In uncompetitive inhibition, the inhibitor binds directly to the enzyme–substrate complex but not to the free enzyme. An uncompetitive enzyme inhibitor affects catalytic activity such that both that apparent KM and the apparent Vmax decrease. The Michaelis–Menten equation for uncompetitive inhibition and the equation for its double-reciprocal plot are: Vo =
(Vmax / α ′ )[S ] Vmax [S ] = K M + α ′[S ] ( K M / α ′ ) + [S ] K 1 1 α′ = M + vo Vmax [S ] Vmax
The double-reciprocal plot consists of a family of parallel lines with slope KM/Vmax, 1/Vo intercepts of α′/Vmax, and 1/[S] intercepts of −α′/KM. Thus graph (A) is a Lineweaver–Burk plot of Michaelis–Menten enzyme in the presence of uncompetitive inhibitior. Graph (B) is a Lineweaver–Burk plot of Michaelis– Menten enzyme in the presence of competitive inhibitor. Lines intersect on the 1/Vo axis at 1/Vmax. Graph (C) is a Lineweaver–Burk plot of Michaelis– Menten enzyme in the presence of non-competitive inhibitor. Graph (D) is a Lineweaver–Burk plot of Michaelis– Menten enzyme in the presence of mixed inhibitor. Lines intersect to the left of the 1/Vo axis. Answer (A)
36 GATE Biotechnology Chapter-wise Solved Papers 49. Topic: Enzymes—Kinetics A competitive inhibitor therefore reduces the concentration of free enzyme available for substrate binding. Given, Ki = 2.5 µM Concentration of inhibitor, [I] = 5 µM K Mapp in presence of inhibitor = 30 µM We know, ⎡ I ⎤ K Mapp = K M ⎢1 + ⎥ K i ⎦ ⎣ KM is measured in the absence of inhibitor.
KM =
K Mapp
⎡ I ⎤ ⎢1 + K ⎥ i ⎦ ⎣ Putting the values in the formula we get: 30 30 KM = = = 10μM 5 ⎡ ⎤ 3 + 1 ⎢⎣ 2.5⎥⎦ Answer (10 µM) 50. Topic: Bioenergetics Given, Standard free energy change (ΔG′°) for ATP hydrolysis = –30 kJ mol−1 In vivo concentrations of [ATP] = 7.09 mM [ADP] = 1.04 mM [Pi] = 7.9 mM Temperature = 37°C = 37 + 273 = 310 K Reaction: A TP A D P + Pi
K =
[A D P][Pi] [1.04][7.9] = [A TP] [7.09]
of H+ to the thylakoid lumen leads to the formation of proton gradient. Synthesis of ATP in mitochondria takes place in its inner membrane where an ionic gradient is established. Oxidative phosphorylation is the process by which ATP formation is driven by energy released from electron removed during substrate oxidation. Answer (B) 52. Topic: Biomolecules—Carbohydrate Structure and Function A disaccharide is a molecule formed from the combination of two monosaccharides by dehydration s ynthesis. Their molecules yield two molecules of the same or of different monosaccharides when hydrolyzed. The systematic name for lactose is o-β -D-galactopyranosyl(1→ 4)-D-glucopyranose. The glycosidic bond links C1 of the β-anomer of galactose to O4 of glucose. HOCH2 HO
H OH
= −30 + 0.008314 × 310ln[1.04 × 10−3 ] = −30 + 2.57 × −6.9 = −30 − 17.7 = −47.7 kJm ol−1
Answer (–47.7 kJ mol–1)
51. Topic: Electron Transport Chain ATP synthesis occurring due to oxidation-reduction in the proton gradient formed across the mitochondrial membrane. Higher concentration of protons [H+] is present in the thylakoid lumen (interior space of thylakoid). The free energy of the proton gradient is tapped by chloroplast ATP synthase. The light reactions of photosynthesis establish a proton gradient across the thylakoid membrane that leads to ATP formation. Removal of H+ from the stroma and addition
O
H
H OH
O OH H H
H H
OH
H
Galactose
OH
Glucose Lactose
Cellobiose is a disaccharide consists of two b-glucose residues that are linked by β (1→ 4) glycosidic bonds. HOCH2 H
H OH H
HOCH2
O H
HO
H H
O
H OH
O OH H H
OH
Glucose
= 1.04m M =1.04 × 10 M
ΔG = ΔG° + RT ln K
H
H
−3
Now, we know,
HOCH2 O
H Cellulose
OH
Glucose
Answer (D) 53. Topic: Enzymes—Kinetics A transition state is observed when the reactants are at the crest of the hump. At this state, they are ready to be converted to products. As the transition state is converted to products, the affinity of the enzyme for the bound molecule(s) decreases as the products are unable to bind to the buttressing group and get expelled. Competitive inhibitor
Enzyme Non-competitive inhibitor Enzyme
Substrate is not able to bind to the active site
Chapter 2 • Biochemistry 37
(a) Competitive inhibition (b) Non-competitive inhibition • In competitive inhibition, the inhibitor binds directly to enzyme substrate complex but not to the free enzyme. • Hence, even if a large amount of substrate added to an enzyme, it cannot overcome uncompetitive inhibition. • In non-competitive inhibition (also called mixed inhibition), both enzyme and enzyme-substrate complex bind inhibitor. The inactivator (irreversible inhibitor) reduces the concentration of functional enzyme at all substrates concentration. Hence, Vmax decreases and KM remains unchanged. Answer (C) 54. Topic: Biological Membrane—Action Potential + ΔG o = RT ln [ Na ]in + Z ⋅ F ⋅ ψ [ Na + ]out
At equilibrium, ΔG = 0 [Na + ]in RT ln = −Z ⋅ F ⋅ψ [Na + ]out Given, Equilibrium potential,ψ = 55m V = 55 × 10−3 V Gas constant, R = 1.98 cal.mol .K Temperature, T = 37 + 273 = 310 K Faraday constant F = 23062 cal V−1 mol−1 −1
−1
+ Inside concentration of Na+, [N a ]in = 20m M
Z for Na+ = 1 Putting all these in equation above we get: 20 1 × 23062 × 55 × 10 −3 ln = − 1.98 × 310 [ Na + ]out ln
+ 20 or ln [ Na ]o = 2.07 = − 2 . 07 20 [ Na + ]out
[N a+ ]o = e2.07 20
1.8 µg of enzyme =
1.8 × 10 −6 = 10−10 mol 18 × 103 ∴ [ E0 ] = 10 −10 mol
Vmax of the enzyme = 0.0134 µmol min−1 =
0.0134 × 10−6 mol s −1 60
We know, Vmax = kcat [ E0 ] kcat =
Vmax 0.0134 × 10 −6 = = 2.23 s −1 [ E0 ] 60 × 10 −10
Now, to calculate turnover number, KM for the substrate = 25 µM = 25 × 10−6 M
kcat 2.23 = = 8.93 × 104 M −1 s −1 K M 25 × 10 −6 Answer (8.93 × 104 M–1s–1) 56. Topic: Biomolecules—Structure and Function For any peptide amino side lies at left side from where sequence started and carboxyl group at right side where sequence ends. • Option A: Let us assume the amino acid sequence to be RSYRVA. On its trypsin digestion that cleaves at carboxyl side of R as shown: R ↓ SYR ↓ VA Three fragments (R), (S, Y, R) and (V, A) will be generated which are clearly not satisfying the condition given in the question. Thus, this option is wrong. • Option B: Let us assume the amino acid sequence to be AVRYSR. (a) On its digestion with trypsin that cleaves at carboxyl side of R as shown: AVR ↓ YSR ↓
[ Na + ]o = 7.942 20
[N a+ ]o = 158.49m M
Amount of enzyme added = 1.8 µg To find concentration of enzyme [E0], 18 × 103 g of enzyme = 1 mol
Answer (158.49 mM)
55. Topic: Enzymes—Kinetics Molecular mass of type II dehydroquinase = 18kDa = 18 × 103 Da = 18 × 103 g
Two fragments (A,V, R) and (Y, S, R) will be generated which is satisfying the condition given in question. (b) On its digestion with chymotrypsin that cleaves at carboxyl side of Y as shown: AVRY ↓ SR Two fragments (A,V, R,Y) and (S, R) will be generated which is satisfying the condition given in question.
38 GATE Biotechnology Chapter-wise Solved Papers (c) On its digestion with carboxypeptidase that cleaves at amino side of C (except R and K), no cleavage product will be formed, as its C-terminal has R which is an exception.
•
AVRYSR Thus, this the correct option. Option C: Let us assume the amino acid sequence to be SRYVAR. On its trypsin digestion that cleaves at carboxyl side of R as shown: SR ↓ YVAR ↓
•
Two fragments (S, R) and (Y, V, A, R) will be generated which are clearly not satisfying the condition given in the question. Thus, this option is wrong. Option D: Let us assume the amino acid sequence to be SVRRYA. On its trypsin digestion that cleaves at carboxyl side of R as shown: SVR ↓ R ↓ YA Three fragments (S, V, R), (R) and (Y, A) will be generated which are clearly not satisfying the condition given in the question. Thus, this option is wrong. Answer (B)
57. Topic: Enzymes—Mechanism of Action Activation energy is the minimal kinetic energy needed for a reactant to undergo a chemical reaction. It is the difference in energy between the substrate and the transition state. Apart from bringing the substrates in proper orientation, enzymes also lower the activation energy of a chemical reaction by decreasing the “randomness” of the collisions between molecules. The enzyme action can thus be summarized as: E + S → ES → EP → E + P An enzyme is responsible for providing a lower energy pathway from substrate to product. It, however, does not affect the overall free energy change for the reaction. Answer (B) 58. Biomolecules—Protein Structure and Function All proteins are composed of 20 standard amino acids. The common amino acids are known as α-amino acids as they have a primary amino group (—NH2) as a substituent of the α-carbon atom, the carbon next to the carboxylic acid group (—COOH). The 20 amino acids differ in the structures of their side chains (R groups). The amino and carboxylic groups of amino acids can readily ionize such molecules that carry charged groups of opposite polarity are called dipolar ions or zwitterions. R H2N
Ca H
COOH
The α -amino acids generally have two, of for those ionizable side chains, three acid-base groups. When the pH is very low, these groups are fully protonated. When the pH is very high, these are unprotonated. H H H
N+ C
H H
H
H
H
C O
N+ C
H
H
H
H
C O
Low pH
H
O
N C
H H
C O–
Zwitterion neutral pH
O
O–
High pH
Answer (A) 59. Topic: Biomolecules—Structure and Function Two cysteine residues can be linked by a disulfide bond to form cystine. Thus, four cysteine residues can form two disulfide bonds. Let us assume that in N number of different ways four cysteine residues can form two disulphide bonds: n! N= ( n − 2 p)! p ! 2 p where, n = number of cysteine residue = 4 p = number of disulphide bond to be formed = 2 4! N= ( 4 − 2 × 2)! 2! 22 4 ×3× 2 = =3 (0)! × 2 × 4 Answer (3) 60. Topic: Enzymes—Kinetics In 1913, Leonor Michaelis and Maud Menten derived a mathematical relationship between substrate concentration and the velocity of enzyme reactions. They measured the amount of product formed (or substrate consumed) in a given time period and gave the following equation, also known as Michaelis–Menten equation. [S ] V = Vm ax [S ] + K M where, V = initial velocity of the reaction; Vmax = maximum rate of reaction; KM = Michaelis-Menten constant and [S ] = substrate concentration. According to the equation, when the substrate concentration [S] is set at a value equivalent to KM, then the velocity of the reaction (V) becomes equal to Vmax/2, or one-half the maximal velocity. Thus, KM = [S], when V = Vmax/2. Michaelis constant (KM) is the substrate concentration when the reaction velocity is one-half of Vmax. It is constant for a given enzyme and is independent of substrate or enzyme concentration. Answer (B)
Chapter 2 • Biochemistry 39
61. Topic: Biomolecules—Structure and Function The five complex amino acids include Glutamic acid (Glu), Aspartic acid (Asp), Lysine (Lys), Arginine (Arg) A
H
H3N+
C
COOH
and Histidine (His). Each of these has 3 ionizable groups. The three ionization points for Lys is shown below:
B
H
H3N+
C
CH2
COO–
CH2 pK = 2.17
CH2
C
H
H2N
C
COO–
CH2 pK = 9.04
CH2
D
H
H2N
C
COO–
CH2 pK = 12.48
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
CH2
NH3+
NH3+
NH3+
NH2
Net charge: 2+
Net charge: 1+
Net charge: 0
Net charge: 1–
Answer (D) 62. Topic: Biomolecules—Protein Structure and Function Chemical composition of protein = CH1.5 O0.3 N 0.3S0.004 Mass of protein = (1 × 12) + (1.5 × 1) + (0.3 × 16) + (0.3 × 14) + (0.004 × 32) = 12 + 1.5 + 4.8 + 4.2 + 0.128 = 22.628 g 1 × 12 × 100 22.628 = 53.03%
Mass % of carbon =
Answer (53.03%) 63. Topic: Biomolecules—Structure and Function Lennard-Jones potential is the most known realistic model of a two-particle interaction. Strong repulsive forces Seperation at energy minimum
Intermolecular potential energy, E(r)
rP1; JMFD Protein X - Homo sapiens MKALT ARQQEVFDLIRDHISRTLRQQGDWL (A) GDE (B) FASTA (C) NBRF (D) GCG (GATE 2015; 1 Mark) 23. 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 (GATE 2015; 2 Marks) 24. Consider the following multiple sequence alignment of four DNA sequences. ACTA ACTG AGTC AGCT
Shannon’s entropy of the above alignment is ________. (GATE 2015; 2 Marks) 25. Which one of the following is not an algorithm for building phylogenetic trees? (A) Maximum parsimony (B) Neighbor joining (C) Maximum likelihood (D) Bootstrap (GATE 2016; 1 Mark) 26. Which one of the following BLAST search programs is used to identify homologs of a genomic DNA query in a protein sequence database? (A) blastp (B) blastn (C) blastx (D) tblastn (GATE 2018; 1 Mark) 27. Identify the character-based method(s) used for the construction of a phylogenetic tree. P. Maximum parsimony Q. Neighbor joining R. Maximum likelihood S. Bootstrapping (A) Q only (B) P and R only (C) Q and S only (D) S only (GATE 2018; 1 Mark) 28. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion: Ab initio gene finding algorithms that predict protein coding genes in eukaryotic genomes are not completely accurate. Reason: Eukaryotic splice 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). (GATE 2018; 2 Marks) 29. 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) __________. (GATE 2018; 2 Marks)
122 GATE Biotechnology Chapter-wise Solved Papers 30. Which one of the following is a database of protein sequence motifs? (A) PROSITE (B) TrEMBL (C) SWISSPROT (D) PDB (GATE 2019; 1 Mark)
P. A larger E-value indicates higher sequence similarity. Q. E-value < 10 −10 indicates sequence homology. R. A higher BLAST score indicates higher sequence similarity. S. E-value > 10 −10 indicates sequence homology.
31. The number of possible rooted trees in a phylogeny of three species is _________. (GATE 2019; 1 Mark)
(A) P, Q and R only (C) P, R and S only
(B) Q and R only (D) P and S only (GATE 2019; 2 Marks)
32. Which of the following statements are CORRECT when a protein sequence database is searched using the BLAST algorithm?
Answer Key 1. (C)
2. (C)
3. (B)
4. (B)
5. (B)
6. (A)
7. (A)
8. (C)
9. (C)
10. (C)
11. (A)
12. (D)
13. (66.67%)
14. (B)
15. (B)
16. (B)
17. (C)
18. (B)
19. (–52) 20. (C)
21. (B)
22. (C)
23. (B)
24. (3.80 bits)
25. (D)
26. (C)
27. (B)
28. (C)
29. (1.8)
31. (3)
32. (B)
30. (A)
Answers with Explanations 1. Topic: Data Mining and Analytical Tools for Genomic and Proteomics Many computational methodologies and algorithms are used for 3-D Protein Structure Prediction (3-D-PSP). These methods can be divided in four main classes: (a) first principle methods without database information; (b) first principle methods with database information; (c) fold recognition and threading methods; and (d) comparative modeling methods and sequence alignment strategies. Deterministic computational techniques, optimization techniques, data mining and machine learning approaches are typically used in the construction of computational solutions for the prediction of protein structure. Answer (C) 2. Topic: Sequence Analysis • Boxshade is a program for creating visually pleasing images of multiple alignments of protein or DNA sequences. The program itself doesn’t does any alignment. Output of multiple alignment programs like ClustalW or Pileup is used as input for Boxshade. It requires a file containing the aligned sequences in msf and aln formats. • The BCM (Baylor College of Medicine) Search Launcher is an integrated set of World Wide Web (WWW) pages that organize molecular biology-related search and analysis services available on the WWW by function and provide a single point of entry for related searches.
•
•
he PROSITE database consists of a large collecT tion of biologically meaningful signatures that are described as patterns or profiles. Each signature is linked to a documentation that provides useful biological information on the protein family, domain or functional site identified by the signature. PSI-BLAST allows users to construct and perform a NCBI BLAST search with a custom, position-specific, scoring matrix which can help find distant evolutionary relationships. Users can specify pattern files to restrict search results using the PSI-BLAST functionality under ‘more options’. Answer (C)
3. Topic: Sequence Analysis BLAST (Basic Local Alignment Search Tool) finds regions of local similarity between sequences. The program compares nucleotide or protein sequences to sequence databases and calculates the statistical significance of matches. BLAST can be used to infer functional and evolutionary relationships between sequences as well as help identify members of gene families. Answer (B) 4. Topic: Data Mining and Analytical Tools for Genomic and Proteomics The term “homology modeling”, also called comparative modeling or sometimes template-based modeling (TBM), refers to modeling a protein 3D structure using
Chapter 8 • Bioinformatics 123
a known experimental structure of a homologous protein (the template). It is a procedure that generates a previously unknown protein structure by “fitting” its sequence (target) into a known structure (template), given a certain level of sequence homology (at least 30%) between target and template. Answer (B) 5. Topic: Sequence Analysis • UPGMA (unweighted pair group method with arithmetic mean) is a straightforward approach to construct a phylogenetic tree from a distance matrix. UPGMA subtly assumes a constant substitution rate, over time and phylogenetic lineages (known as the molecular clock hypothesis). • 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 multiple alignment is built up progressively by a series of pairwise alignments, following the branching order in a guide tree. • 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 domains structure, post-translational modifications, variants, etc.), a minimal level of redundancy and high level of integration with other databases. • RasMol is a computer program written for molecular graphics visualization intended and used mainly to depict and explore biological macromolecule structures, such as those found in the Protein Data Bank. Answer (B) 6. Topic: Sequence and Structure Databases • The Basic Local Alignment Search Tool (BLAST) finds regions of local similarity between sequences. The program compares nucleotide or protein sequences to sequence databases and calculates the statistical significance of matches. BLAST can be used to infer functional and evolutionary relationships between sequences as well as help identify members of gene families. • RasMol is a program developed for molecular graphics visualization. It is largely envisioned to illustrate and explore structure of biological macromolecule found in the Protein Data Bank. • ExPASy is the SIB (Swiss Institute for Bioinformatics) Bioinformatics Resource Portal that provides access to scientific databases and software tools in different areas of life sciences including proteomics, genomics, phylogeny, systems biology, population genetics, transcriptomics etc. • 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, post-translational modifications, variants, etc.), a minimal level of redundancy and a high level of integration with other databases. Answer (A) 7. Topic: Data Mining and Analytical Tools for Genomics and Proteomics Studies In bioinformatics, an accession number is a unique identifier given to a DNA or protein sequence record to allow for tracking of different versions of that sequence record and the associated sequence over time in a single data repository. Because of its relative stability, accession numbers can be utilized as foreign keys for referring to a sequence object, but not necessarily to a unique sequence. Answer (A) 8. Topic: Data Mining and Analytical Tools for Genomics and Proteomics Studies Phylogenetic is the branch of science that deals in the study of evolutionary relationship among different organisms at species, taxa or individual level. These studies are based on physical as well as genetic similarities among the individuals. Answer (C) 9. Topic: Sequence and Structure of Databases • Clustal is series of computer programmes widely used in bioinformatics for multiple sequence alignment. • BLAST (Basic Local Alignment Search Tool) is a bioinformatics tool used for comparing primary biological sequence such as amino-acid sequences of proteins, nucleotide sequences of DNA, RNA etc. • FASTA is a specific text-based format of DNA and protein sequences designed for software package. • Chime is a free software people used for viewing the three-dimensional structure of a molecule. Answer (C) 10. Topic: Sequence and Structure of Databases Derived or secondary database is collection of data related to biomolecular structure of protein, DNA, etc., which are derived from primary database, for example, e.g., SCOP (Structural Classification of Protein) database. Answer (B) 11. Topic: Sequence and Structure Databases • PHYLIP (PHYLogeny Inference Package) is a group of programs used for studying and inferring the phylogenetic trees. It handles data that are nucleotide sequences, protein sequences, gene frequencies, restriction sites, restriction fragments, distances, discrete characters, and continuous characters.
124 GATE Biotechnology Chapter-wise Solved Papers •
•
•
Phrap is a bioinformatics program used to study the assembly, alignment and comparison of DNA sequences. It allows use of the entire read and not just the trimmed high-quality part. It uses a combination of user-supplied and internally computed data quality information to improve assembly accuracy in the presence of repeats. ProDom is a protein database consists of domain family entries. It is generated from the global comparison of all available protein sequences. PHDsec is a software used to predict secondary structure of biomolecules. It is based on sequence family alignments. Answer (A)
12. Topic: Data Mining and Analytical Tool for Genomic and Proteomic Studies The E-value of expected value is the parameter that describes expected number of score or hits one can get by chance while searching a database of particular size. It depends on the length of sequence, number of sequences present in the database and on the scoring system, but not on probability from a normal distribution. Answer (D) 13. Topic: Sequence Alignment Given sequences are HELLO– YELLOW The percentage identity between two sequences can be calculated by dividing number of identical positions by the total number of positions. Here, number of total positions = 6 Number of identical positions (E L L O) = 4 Therefore,
Percentageidentity =
4 × 100 = 66.67% 6
Answer (66.67%)
14. Topic: Scoring Matrices Empirical matrices are based on surveys of actual amino acid substitutions among related proteins. Most widely used matrices are PAM and BLOSUM. BLOSUM looks directly at mutations in motifs of related sequences, while PAM’s extrapolate evolutionary information based on closely related sequences. • Margaret Dayhoff developed this scoring matrix, based on an evolutionary model, where a closely related group of sequences sharing 85% identity was aligned and the frequency of substitution derived. These frequencies were normalized to those expected if 1% of the sequence were to change, giving the PAM1 (percentage acceptable mutations) matrix.
•
The lower PAM will tend to find short alignments of highly similar sequences. PAM 40 matric represents 40 mutations per 100 amino acids. It is used for scoring amino acids alignment. The BLOSUM (BLOcks SUbstitution Matrix) matrix is a substitution matrix used for sequence alignment of proteins. BLOSUM matrices are used to score alignments between evolutionarily divergent protein sequences. BLOSUM 80 is used for less divergent alignments, and BLOSUM 45 is used for more divergent alignments. Answer (B)
15. Topic: Sequence Alignment UPGMA (Unweighted Pair Group Method with Arithmetic mean) produce rooted trees and require a constant-rate assumption, that is, it assumes an ultrametric tree in which the distances from the root to every branch tip are equal. It is the only method of phylogenetic reconstruction dealt with which the resulting trees are rooted. Answer (B) 16. Topic: Sequence and Structure Databases • Threading (protein sequence) also known as fold recognition, is a method of protein modeling 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. • FASTA is a DNA and protein sequence alignment software package. In number theory, a prime k-tuple is a finite collection of values representing a repeatable pattern of differences between prime numbers. The FASTA 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-tuple match. • Hidden Markov Models (HMM) is used to recognize and pinpoint the location in target sequences of local descriptors of protein structure, LDPS. When threading is done through the Hidden Markov models together with the secondary structure, it is able to assign 58.5% of the descriptors to their proper locations. • Orthologous and paralogous genes are two types of homologous genes, that is, genes that arise from a common DNA ancestral sequence. Orthologous genes diverged after a speciation event, while paralogous genes diverge from one another within a species. Answer (B) 17. Topic: Sequence and Structure Databases BLAST is an acronym for ‘Basic Local Alignment Search Tool’. Members of the BLAST family of database
Chapter 8 • Bioinformatics 125
search programs take as input a query deoxyribonucleic acid (DNA) or protein sequence and search a DNA or protein sequence database for similarities that may indicate homology. The programs implement variations of the BLAST algorithm, which is a heuristic method for rapidly finding local alignments with scores sufficiently high to be statistically significant. The BLAST algorithm first seeks near-perfect matches to words within a query sequence, and then extends these matches to determine whether they are contained within longer, high-scoring local alignments. BLAST approximates the rigorous Smith–Waterman local alignment algorithm; it permits a chance of missing weak sequence similarities in exchange for greatly increased speed. An algorithm is described for generation of the long sequence written in a four-letter alphabet from the constituent k-tuple words in the minimal number of separate, randomly defined fragments of the starting sequence. Answer (C) 18. Topic: Scoring Matrices Sequence identity is the amount of characters which match exactly between two different sequences. Hereby, gaps are not counted and the measurement. In the given peptide sequences, Peptide I: Ala-Ala-Arg-Arg-Gln-Trp-Leu-Thr-Phe-ThrLys-Ile-Met-Ser-Glu Peptide II: Ala-Ala-Arg-Glu-Gln-Tyr-Ile-Ser-Phe-ThrLys-Ile-Met-Arg-Asp 9 out of 15 amino acid are identical.
∴
% Identity =
9 × 100 = 60% 15
Now, among peptide I and II, Arg and Gln, Leu and Ile and Glu and Asp are similar amino acids. Thus, % similarity between two peptides will be
=
9+ 3 × 100 = 80% 15 Answer (B)
19. Topic: Scoring Matrices Two types of gap penalties are used: a gap insertion (Gi) penalty (which is normally high) and a gap-extension (Ge) penalty (smaller). Gap scores follow an affine scoring scheme with the equation: G = Gi + ( K + Ge ) where, G is the gap score to be calculated Gi the insertion penalty = –20 Ge the gap extension penalty = –4
And, K is the number of residues which form the extension = 8
G = −20 + (8 × −4) = −52 Answer (–52) 20. Topic: Data Mining and Analytical Tools for Genomic and Proteomic Studies Length of genome = 3 × 109 Let the minimum unique stretch of DNA sequence that can be found only once in the genome = x The DNA sequence is evenly made up of G, A, T, and C nucleotides (i.e., 25% of each), thus it is likely to occur after every 4x sequence. Now, it should occur once in 3 billion base pairs. ∴ 4 x = 3 × 10 9 Taking log on both sides. ⇒ ⇒ ⇒ ⇒
log 4 x = log(3 × 109 ) x log 44 = log 33 + log 10 10 99 x log = log + log 9 00.602 x = 0.477 + 9 .602 x = 0.477 + 9 9.477 x = 9.477 = 15 15.7 ≈ 16 x = 0.602 = .7 ≈ 16 0.602 Answer (C)
21. Topic: Scoring Matrices Margaret Dayhoff developed PAM (Point Accepted Mutations), a scoring matrix for protein sequences. It is based on an evolutionary model, where a closely related group of sequences sharing 85% identity was aligned and the frequency of substitution derived. These frequencies were normalized to those expected if 1% of the sequence were to change, giving the PAM1 (percentage acceptable mutations) matrix. This matrix is multiplied by itself to extrapolate the substitution frequencies over longer periods. The PAM100 matrix corresponds to 100 accepted mutations per 100 residues, but since the same residue might change more than once, two sequences with this level of mutations will have about 50% identities. The lower PAM matrices will tend to find short alignments of highly similar sequences, while higher PAM matrices will find longer, weaker local alignments. As the PAM250 matrix corresponds to a level of about 20% identical residues. Thus, amino acid substitution matrices in decreasing order of stringency for comparing protein sequences would be PAM100, PAM120, PAM250. Answer (B) 22. Topic: Biomolecular Sequence File Format • It’s a NBRF format, also known as PIR format consists of One line starting with a “>” (greater-than) sign, followed by a two-letter code describing the
126 GATE Biotechnology Chapter-wise Solved Papers sequence type (P1, F1, DL, DC, RL, RC, or XX), followed by a semicolon, followed by the sequence identification code (the database ID-code). • The GDE format (.gde) is a multiple sequence format of the Genetic Data Environment program. The name of each “sequence” in a GDE flat file indicates the type of data: # – DNA/RNA % – protein @ – mask sequence “ – text • FASTA format is a text-based format for representing either nucleotide sequences or peptide sequences, in which base pairs or amino acids are represented using single-letter codes. A sequence record in a FASTA format consists of a single-line description (sequence name), followed by line(s) of sequence data. The first character of the description line is a greater-than (“>”) symbol, for example, >AB000263 |acc = AB000263|descr = Homo sapiens mRNA for prepro cortistatin like peptide, complete cds.|len = 368 ACAAGATGCCATTGTCCCCCGGCCTCCTGCTGCTGC • A sequence file in GCG format contains exactly one sequence, begins with annotation lines and the start of the sequence is marked by a line ending with two dot (“..”) characters. This line also contains the sequence identifier, the sequence length and a checksum. This format should only be used if the file was created with the GCG package. An example sequence in GCG format is: ID AB000263 standard; RNA; PRI; 368 BP. XX AC AB000263; XX DE Homo sapiens mRNA for prepro cortistatin like peptide, complete cds. XX SQ Sequence 368 BP; AB000263 Length: 368 Check: 4514 .. Answer (C) 23. Topic: Phylogeny Number of different sequences, n = 4
(2n − 3)! ( 2 × 4 − 3)! Number of rooted trees = n − 2 = 4−2 2 (n − 2)! 2 ( 4 − 2)!
=
(5)! = 15 22 (2)!
Number of unrooted trees =
(2 × 4 − 5)! ( 2n − 5)! = 2n −3 ( n − 3)! 24 −3 (4 − 3)!
(3)! =3 21 (1)!
=
Answer (B) 24. Topic: Sequence Alignment The Shannon entropy equation provides a way to estimate the average minimum number of bits needed to encode a string of symbols, based on the frequency of the symbols. It can be calculated by: n
H ( A) = −∑ pi log 2 pi i =0
p1 = 0 p2 = 0.5 p3 = 0.25 p4 = 0.25 H ( A) = 0 + ( −0.5 log 2 0.5) + ( −0.25 log 2 0.25) + ( −0.25 log 2 0.25) = 3.80 bits Answer (3.80 bits) 25. Topic: Phylogeny Based on the information obtained from fossils, the scientists can construct phylogenetic trees or evolutionary trees. These trees help in determining the history of an organism over time. • Maximum parsimony predicts the evolutionary tree or trees that minimize the number of steps required to generate the observed variation in the sequences from common ancestral sequences. • Neighbor joining is a bottom-up (agglomerative) clustering method for the creation of phylogenetic trees. • Maximum likelihood estimation is a method of estimating the parameters of statistical model, given observations. • Bootstrap is a free and open-source front-end framework for designing websites and web applications. Answer (D) 26. Topic: Sequence and Structure Databases Biological databases contain millions of sequences and to generate optimal alignments between two sequences, the sheer number of alignments to be made requires a faster method. To meet this requirement, BLAST was developed.
Chapter 8 • Bioinformatics 127
• •
•
•
blastp is used for protein from protein database specified by the user. blastn is used for DNA query where it returns the most similar DNA sequences from the DNA database specified by the user. blastx is used to compare products of translation in a nucleotide query sequence (translated into all six frames — three forward and three on the reverse strand) and protein database-query set. tblastn is used to compare protein query against nucleotide sequence (translated into all six frames) database. Answer (C)
27. Topic: Phylogeny The data used in molecular phylogeny fall into two categories. The first category character data is which provides information about individual operational taxonomic units (OTU). For character-based methods, the position in the sequence is the character and the residue at that position is the character state. The other category is where a measure called distance, a quantitative description of the difference between two individual OTUs, is used. The most common algorithms used in phylogeny to generate the tree topology are maximum parsimony and maximum likelihood methods, which work on character data, and distance methods. Answer (B) 28. Topic: Molecular Dynamics and Simulations Ab initio gene locaters use statistical models to predict genes and their exon-intron structures from the genome sequence alone. They depend on two types of sequence information: signal sensors and content sensors. Signal sensors denote short sequence motifs, like splice sites, branch points, polypyrimidine tracts, start codons and stop codons. Exon detection rest on the content sensors, which refer to the patterns of codon usage that are unique to a species and allow coding sequences to be distinguished from the surrounding non-coding sequences by statistical detection algorithms. To repeat, a desired goal from ab initio gene finders is to use the output to discover novel proteins. Incorrect predicted exons manifest in incorrect translations to amino acid residues. Answer (C) 29. Topic: Sequence Alignment Affine gap penalties consist of an “open gap” penalty to begin an indel, and an “extend gap” penalty for each subsequent gap character inserted into the alignment. Affine gap penalties are calculated by the formula:
Affine gap penalty = n (A + B) n = number of gaps = 2 A = open gap penalty = 1 B = extend gap penalty = 0.1 Putting these in formula above: Affine gap penalty = 2 (1 + 0.1) = 2.2 Total number of matches = 4 Score of alignment = Match score – Gap penalty = 4 – 2.2 = 1.8 Answer (1.8) 30. Topic: Sequence Analysis • 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. • TrEMBL is a computer-annotated protein sequence database. It contains the translations of all coding sequences (CDS) present in the EMBL/GenBank/ DDBJ Nucleotide Sequence Databases and also protein sequences extracted from the literature or submitted to UniProtKB/Swiss-Prot. The database is enriched with automated classification and annotation. • UniProtKB/Swiss-Prot is the manually annotated and reviewed section of the UniProt Knowledgebase (UniProtKB). It is a high quality annotated and non-redundant protein sequence database, which brings together experimental results, computed features and scientific conclusions. • The Protein Data Bank (PDB) is the single worldwide database for the three-dimensional structural data of large biological molecules, such as proteins and nucleic acids. Answer (A) 31. Topic: Phylogeny Number of species, thus different sequences, n = 3 Number of rooted trees =
( 2n − 3)! 2n − 2 ( n − 2)!
( 2 × 3 − 3)! 23− 2 (3 − 2)! 3! = =3 2
=
Answer (3)
128 GATE Biotechnology Chapter-wise Solved Papers 32. Topic: Sequence Analysis BLAST (Basic Local Alignment Search Tool) is a program that finds similar protein or nucleotide sequences to the target sequence. The score (S) is a measure of the similarity of the query to the sequence shown. The Expect value (E) is a measure of the reliability of the S score. The E-value 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. The lower the E-value, or the closer it is to zero, the more “significant” the match is. The typical threshold for a good E−value from a BLAST search is 10−5 or lower. Answer (B)
CHAPTER9
Recombinant DNA Technology
Syllabus Restriction and Modification Enzymes; Vectors—Plasmid, Bacteriophage and other Viral Vectors, Cosmids, Ti Plasmid, Yeast Artificial Chromosome; Mammalian and Plant Expression Vectors; cDNA and Genomic DNA Library; Gene Isolation, Cloning and Expression; Transposons and Gene Targeting; DNA Labeling; DNA Sequencing; Polymerase Chain Reactions; DNA Fingerprinting; Southern and Northern Blotting; In situ Hybridization; RAPD, RFLP; Site-Directed Mutagenesis; Gene Transfer Technologies; Gene Therapy.
Chapter Analysis Topic Restriction and Modification Enzymes Vectors (Mammalian and Plant Expression Vectors): Plasmid, Bacteriophage and other Viral Vectors, Cosmids, Ti Plasmid, Yeast Artificial Chromosome cDNA and Genomic DNA Library Gene Isolation, Cloning and Expression Transposons and Gene Targeting
DNA Labeling DNA Sequencing Polymerase Chain Reactions DNA Fingerprinting Southern and Northern Blotting In situ Hybridization RAPD, RFLP Site-Directed Mutagenesis Gene Transfer Technologies Gene Therapy
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Questions 1. The substrate for restriction enzyme is (A) single stranded RNA. (B) partially double stranded RNA. (C) cell wall proteins. (D) double stranded DNA. (GATE 2000; 1 Mark)
2. In human populations 4% of the individuals are homozygous recessive to a specific RFLP marker. What percentage of individuals does you expect to be heterozygous for this trait? (A) 4% (B) 8% (C) 16% (D) 32% (GATE 2000; 1 Mark)
130 GATE Biotechnology Chapter-wise Solved Papers 3. 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 β -lactamase. (D) ampicillin is essential for cell survival. (GATE 2000; 1 Mark) 4. Mung bean nuclease could be used for (A) DNA synthesis. (B) nucleotide hydrolysis. (C) trimming single stranded regions in DNA. (D) removal of phosphate groups from the ends of the DNA. (GATE 2000; 1 Mark) 5. Phage T7 promoter containing plasmids are used for over-expression 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. (GATE 2000; 1 Mark) 6. 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. (GATE 2000; 1 Mark) 7. Which one of the following is not a requirement for a PCR reaction? (A) DNA template (B) Taq polymerase (C) NTPs (D) MgCl2 (GATE 2000; 1 Mark) 8. 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? (A) 10% (B) 30% (C) 50% (D) 100% (GATE 2000; 1 Mark) 9. Commonly used reporter gene in plant expression vectors is (A) Ti gene of Agrobacterium tumifaciens. (B) Gus gene.
(C) β-lactamase gene. (D) α-amylase gene. (GATE 2000; 1 Mark) 10. Which one of the following is not a protease? (A) Proteosome (B) Trypsin (C) Chrymotrypsin (D) Peptidyl tRNA hydrolase (GATE 2000; 1 Mark) 11. Match the Columns. Column I Column II P. Chemical sequencing 1. Southern of DNA Q. DNA blotting 2. Temin, Baltimore and Dulbecco R. Monoclonal antibodies 3. F. Sanger S. Reverse transcription 4. Maxam and Gilbert T. Protein sequencing 5. Kohler and Milstein U. Polymerase chain reaction 6. K. Mullis (GATE 2000; 2 Marks) 12. 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 pmole of DNA template after 10 cycles (1 bp = 660 Da)? (GATE 2000; 2 Marks) 13. For PCR to amplify the sequence ATCTTCTACG……….AAGCPTGCGG TAGAAGATGC………TTCGAACGCC The required primers are (A) ATCTTCTA and CGAACGCC (B) ATCTTCTA and CCGCAAGC (C) TAGAAGAT and CGAACGCC (D) TAGAAGAT and CCGCAAGC (GATE 2001; 1 Mark) 14. The heterozygosity of any locus can be ascertained by (A) RFLP analysis. (B) SNPs. (C) FISH analysis. (D) either RFLP analysis or SNPs. (GATE 2001; 1 Mark) 15. BAC, which can be used to clone large DNA fragments, is derived from (A) ColE plasmid. (B) F plasmid. (C) 2μ. (D) Mu phage. (GATE 2001; 1 Mark)
Chapter 9 • Recombinant DNA Technology 131
16. 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. (GATE 2001; 1 Mark) 17. The essential component of Ti plasmid required for integration into plant genome is (A) origin of replication. (B) tumor inducing gene. (C) nopaline utilization gene. (D) all of the above. (GATE 2001; 1 Mark) 18. A gene cannot 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. (GATE 2001; 1 Mark)
19. The sequences at the cloning site of three vectors are given below. The BamHI (GGATCC) and HindIII (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 (GATE 2001; 2 Marks)
20. 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. EcoRI
SalI
5.4 –
5.4 –
HindIII
EcoRI & HindIII
SalI & HindIII
EcoRI & SalI
3.6 – 2.1 –
2.1 –
1.9 –
1.9 – 1.8 –
1.4 –
1.4 –
1.4 –
1.3 –
1.2 –
0.6 –
0.9 – (GATE 2001; 2 Marks)
21. T4 polynucleotide kinase is used for (A) labeling 3′ ends of DNA. (B) labeling 5′ ends of DNA. (C) creating blunt ends of DNA. (D) dephosphorylation of DNA. (GATE 2002; 1 Mark)
23. In baculovirus expression vector, foreign genes are expressed from the promoter of (A) polyhedrin gene. (B) bacteriophage T7 gene. (C) E.coli lacZ gene. (D) Yeast phosphoglycerate kinase gene.
22. A plasmid cloning vector should contain all of the following sequences except (A) origin of replication. (B) inducible promoter. (C) selectable marker gene. (D) multiple cloning sites. (GATE 2002; 1 Mark)
(GATE 2002; 1 Mark) 24. The length of each boarder sequence in Ti plasmid is about (A) 25 million base pairs. (B) 200 kilo base pairs. (C) 25 kilo base pairs. (D) 25 base pairs. (GATE 2002; 1 Mark)
132 GATE Biotechnology Chapter-wise Solved Papers 25. Enzyme used in ‘cycle’ sequencing of DNA is (A) T7 DNA polymerase. (B) T4 DNA polymerase. (C) Klenow DNA polymerase. (D) Taq DNA polymerase. (GATE 2002; 1 Mark) 26. Human insulin gene cloned from a cDNA library into pUC19 could not be expressed. Justify the reason. (GATE 2002; 1 Mark) 27. Which of the following processes require energy? (A) Ligation (B) Transformation (C) Restriction digestion (D) Hybridization (GATE 2003; 1 Mark) 28. To be a cloning vector, a plasmid does not require (A) an origin of replication. (B) an antibiotic resistance marker. (C) a restriction site. (D) to have a high copy number. (GATE 2003; 1 Mark) 29. 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 fingerprinting analysis (A) cut the DNA and hybridize with specific micro-satellite probes. (B) cut the DNA and subclone the fragments. (C) determine the sequence of the subclones. (D) (B) followed by (C). (GATE 2003; 2 Marks) Common Data for Questions 30 to 31: 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 inserted for sequencing 30. Which one of the following will be the most appropriate restriction enzyme for this subcloning? (A) 8 bp cutter (B) 6 bp cutter (C) 3 bp cutter (D) 5 bp cutter (GATE 2003; 2 Marks) 31. 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 (GATE 2003; 2 Marks)
32. 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. (GATE 2003; 2 Marks) 33. 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. (GATE 2003; 2 Marks) 34. The restriction endonuclease Eco52I 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 Eco52I? (A) EcoRI (G/AATTC) (B) XmaIII (C/GGCGG) (C) SmaII (CCCC/GGG) (D) SacII (CCGC/GG) (GATE 2003; 2 Marks) 35. 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. (C) none of the plasmids will express the antibiotic resistance gene located on the plasmid. (D) the circular molecules will be amplified by the cells. (GATE 2003; 2 Marks) 36. Expression of hundreds of different genes in DNA microarray technology is monitored by using (A) radioactive probe. (B) visible chromogenic probe.
Chapter 9 • Recombinant DNA Technology 133
(C) UV absorbing probe. (D) fluorescent probe. (GATE 2004; 1 Mark) 37. Transfer of T-DNA from Ti plasmid into plant cell is mediated by (A) mob gene. (B) vir gene. (C) nif gene. (D) octoptine gene. (GATE 2004; 1 Mark) 38. Nick translation of DNA is a method for making DNA probes. Identify from below what is not required for nick translation method. (A) DNA polymerase (B) DNAase (C) Primers (D) Deoxyribonucleotides (GATE 2004; 1 Mark) 39. 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) SmaII (CC↓GGG) (B) NotI (GC↓GGCCGC) (C) SalmIII (GG↓CC) (D) PstI (CTGCA↓G) (GATE 2004; 2 Marks) 40. For the sequence of dsDNA given below, identify the set of primers required to amplify this DNA by PCR. 3′ GACTCCA ……… TACAACC 5′ 5′ CTGAGGT ……… ATGTTGG 3′ (A) 5′ GGTTGTA and 5′ GACTCCA (B) 5′ CTGAGGT and 5′ CCAACAT (C) 5′ ACTCAGT and 5′ ATGTTGG (D) None of the above (GATE 2004; 2 Marks) 41. Match the following genetic elements with their functions. Genetic elements Functions P. neoR 1. F acilitates inducible expression of genes in eukaryotes. Q. SV40 2. F acilitates constitutive expression of genes in eukaryotes. R. LTR 3. Allows amplification of gene. S. dhfr. 4. P rovides way of selecting eukaryotic cells, which have received foreign DNA.
P Q R S (A) 4 2 1 3 (B) 2 4 1 3 (C) 1 4 2 3 (D) 4 2 3 1 (GATE 2004; 2 Marks) 42. The mobility of DNA in agarose gel electrophoresis is solely based on its (A) charge. (B) conformation. (C) size. (D) none of the above. (GATE 2005; 1 Mark) 43. Which of the following fluorescent probes is used to monitor the progress of amplification in Real time PCR? (A) SYBR green (B) Rhodamine (C) FTTC (D) Cyan blue (GATE 2005; 1 Mark) 44. Expression of which of the following reporter genes do not require addition of specific substrate for detection? (A) Luciferase (B) β-Glucuronidase (C) β-Glucosidase (D) Green fluorescent protein (GATE 2005; 1 Mark) 45. 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 (D) 3 labeled; 3 unlabeled (GATE 2005; 2 Marks) 46. 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 (A) the sequence context at the RB defines the direction of transfer. (B) the sequence context at the LB defines the direction of transfer. (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. (GATE 2005; 2 Marks)
134 GATE Biotechnology Chapter-wise Solved Papers 47. 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, whereas Western blot detects a 55 kDa strong band in liver, but a very faint 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. (GATE 2005; 2 Marks) 48. Expression in poor amount and in inactive form of cDNA of a eukaryotic protein in Escherichia coli using its expression vector is due to P. the absence of capping mechanism of mRNA. Q. codon bias. R. absence of polyadenylation. S. absence of proper glycosylation. (A) P, Q (B) Q, R (C) Q, S (D) P, S (GATE 2005; 2 Marks) Common Data for Questions 49, 50 and 51: A recombinant SV40 virus delivers c-myc cDNA, which has a unique SalI site, into muscle cells. Southern analysis of SalI digested total genomic DNA of the muscle cells using c-myc cDNA probe generates a smear. 49. The DNA smear obtained on Southern blot is due to (A) head to head concatemer of viral DNA. (B) head to tail concatemer of viral DNA. (C) tail to tail concatemer of viral DNA. (D) random integration of viral DNA. (GATE 2005; 2 Marks) 50. Western blot analysis of c-myc expression of such transformed cells last for (A) transiently. (B) upto five generations. (C) upto 10 generations. (D) more than 100 generations. (GATE 2005; 2 Marks) 51. Which of the following types of cancer will be observed in such transformed cells? (A) Adenoma (B) Melanoma (C) Sarcoma (D) Hepatoma (GATE 2005; 2 Marks)
Statement for Linked Answer Questions 52 and 53: 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. 52. E.coli cell culture (1 mL) was diluted 1 : 1000000 and 200 μL of this was used for plating. After 12 h incubation of the plate, the number of colony forming units (CFU) was 150. What is the total CFU per ml in the original culture? (B) 1.5 × 108 (A) 7.5 × 108 (C) 1.5 × 106 (D) 3.0 × 106 (GATE 2005; 2 Marks) 53. Isolated plasmid DNA (5ng) was used for transformation of 100 μL competent E.coli cells to which 900 μL of SOC medium was added. An aliquot of 50 μL 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 μg of plasmid DNA, 0.01% (B) 1.2 × 105 colonies per μg of plasmid DNA, 0.02% (C) 1.2 × 106 colonies per μg of plasmid DNA, 0.008% (D) 6.0 × 106 colonies per μg of plasmid DNA, 0.1% (GATE 2005; 2 Marks) 54. A thermostable DNA polymerase that can carry out both reverse transcription reaction and polymerization has been isolated form (A) Thermococcus litoralis. (B) Thermus aquaticus. (C) Thermotoga maritime. (D) Thermus thermophilus. (GATE 2006; 1 Mark) 55. For amplification of GC rich sequences by polymerase chain reaction, identify the reagent that hinds and stabilizes AT sequences and destabilizes GC regions. (A) Tetramethyl ammonium chloride (B) Betaine (C) 7-deaza-2’-deoxyguanosine triphosphate (D) Sodium dodecyl sulphate (GATE 2006; 1 Mark) 56. Match items in Group I with correct options from those given in Group II. Group I Group II P. VNTR sequence 1. Gene regulation on the same chromosome Q. Leader sequence 2. Ribosome binding site R. SD sequence 3. DNA fingerprinting S. cis-acting sequence 4. Functions in attenuation
Chapter 9 • Recombinant DNA Technology 135
(A) Finger printing (B) Foot printing (C) Southern blotting (D) Western blotting (GATE 2007; 1 Mark)
P Q R S (A) 3 1 4 2 (B) 2 3 1 4 (C) 3 4 2 1 (D) 3 1 2 4 (GATE 2006; 2 Marks) 57. What are the experimental steps needed for screening an expression library for clone encoding a protein X that has been isolated and purified? P. mRNA isolation Q. Antibody preparation R. Cloning into an appropriate vector S. Western blotting on transferred plaques (A) P, S (B) Q, S (C) Q, R (D) R, S (GATE 2006; 2 Marks) 58. The enzymes that can be used in 5′ end labeling of DNA are (A) alkaline phosphatase. (B) DNA ligase. (C) terminal transferase. (D) polynucleotide kinase. (GATE 2006; 2 Marks) Statement for Linked Answer Questions 59–60: DNA content of Caenorhabditis elegans was analyzed and found to contain 1.0 × 108 bp. 59. How many standards λ -phage vectors carrying 20 kb DNA fragments or YACs carrying 250 kb DNA fragments are theoretically required to constitute a complete C. Elegans genomic library? (A) 500 λ-phage vectors or 40 yeast clones (B) 400 λ-phage vectors or 5000 yeast clones (C) 5000 λ-phage vectors or 400 yeast clones (D) 5 × 104 λ-phage vectors or 4000 yeast clones (GATE 2006; 2 Marks) 60. How many λ-phage vectors/yeast clones should be prepared in order to ensure that every sequence is included in the library? (A) 25 × 103 λ-phage vectors/2000 yeast clones (B) 20 × 103 λ-phage vectors/1600 yeast clones (C) 5 × 104 λ-phage vectors/4000 yeast clones (D) 10 × 104 λ-phage vectors/10000 yeast clones (GATE 2006; 2 Marks) 61. Protein binding regions of DNA are identified by one of the following techniques.
62. 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 concentration of 150 μM. Ten μL of this stock solution is added to a total volume of 200 μL reaction. What will be the concentration of dTTP in the reaction? (A) 7.5 μM (B) 75 μM (C) 0.75 μM (D) 0.075 μM (GATE 2008; 2 Marks) 63. A polymerase chain reaction was performed beginning with 400 template DNA molecules in a 100 μL reaction. After 20 cycles of polymerase chain reaction, how many molecules of the amplified product will be present in 0.1 μL of reaction? (B) 4.19 × 104 (A) 2.19 × 104 (C) 2.19 × 105 (D) 4.19 × 105 (GATE 2008; 2 Marks) 64. Restriction endonucleases from two different organisms that recognize the same DNA sequence for cleavage are called (A) isoschizomers. (B) isozymes. (C) concatemers. (D) palindromes. (GATE 2009; 1 Mark) 65. Baculovirus expression system is used to express heterologous genes in (A) mammals. (B) plants. (C) insects. (D) yeasts. (GATE 2009; 1 Mark) 66. Which of the following are commonly used as reporter genes? P. NPTgene Q. Luciferase gene R. CFTR gene S. GFP gene (A) Q, S (B) R, S (C) P, R (D) P, Q (GATE 2009; 2 Marks) 67. Match the items in Group I with their functions in Group II. Group I Group II P. rol genes 1. Food and energy source Q. Opines 2. Tumor formation R. Virulence genes 3. Hairy root induction S. aux and cyt genes 4. T-DNA transfer and integration
136 GATE Biotechnology Chapter-wise Solved Papers (A) (B) (C) (D)
(A) (B) (C) (D)
P–4; Q–3; R–2; S–1 P–3; Q–2; R–4; S–1 P–1; Q–3; R–4; S–2 P–3; Q–1; R–4; S–2
Only I is true. Both I and II are true. Only II is true. I is true and II is false
(GATE 2009; 2 Marks)
(GATE 2010; 2 Marks)
68. Match the items in Group I with correct options in Group II. Group I Group II 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 (GATE 2009; 2 Marks)
73. Match the items in Group I with Group II. Group I Group II (Vectors) (Maximum DNA packaging) P. λ phage 1. 35–45 kb Q. B acterial Artificial 2. 100–300 kb Chromosomes (BACs) R. P1 derived Artificial 3. ≤ 300 kh Chromosomes (PACs) S. λ 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 (GATE 2010; 2 Marks)
69. The bacteria known to be naturally competent for transformation of DNA is (A) Escherichia coli. (B) Bacillus subtilis. (C) Mycobacterium tuberculosis. (D) Yersinia pestis. (GATE 2010; 1 Mark) 70. Antibiotic resistance marker that cannot be used in a cloning vector in Gram negative bacteria is (A) streptomycin. (B) ampicillin. (C) vancomycin. (D) kanamycin. (GATE 2010; 1 Mark) 71. Expressed Sequence Tag is defined as (A) a partial sequence of a codon randomly selected from cDNA library. (B) the characteristic gene expressed in the cell. (C) the protein coding DNA sequence of a gene. (D) uncharacterized fragment of DNA presence in the cell. (GATE 2010; 2 Marks) 72. Consider the following statements. 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.
74. 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 (GATE 2010; 2 Marks) 75. Yeast artificial chromosomes (YAC’s) are used for cloning (A) large segments of DNA. (B) mRNA. (C) bacterial DNA. (D) yeast DNA. (GATE 2011; 1 Mark) 76. ‘Hairy roots’ induced in vitro by the infection of Agrobacterium rhizogenes, are characterized by P. a high degree of lateral branching. Q. genetic instability of culture. R. an absence of geotropism. S. poor biomass production. (A) P and R only (B) P and Q only (C) Q and R only (D) R and S only (GATE 2011; 1 Mark) 77. Restriction endonucleases which recognize and cut same recognition sequences are known as (A) isoschizomers. (B) isozymes. (C) isoaccepting endonucleases. (D) abzymes. (GATE 2011; 1 Mark)
Chapter 9 • Recombinant DNA Technology 137
78. Identify the correct statements. P. 5′ and 3′ ends of the transcripts can be mapped by utilizing polymerase chain reaction. Q. S1 nuclease can cleave the DNA strand of a DNARNA hybrid. R. T4 polynucleotide kinase is used for labeling 3’ end of DNA. S. Baculovirus (Autographa californica) can be used as an insect expression vector. (A) P and Q only (B) R and S only (C) P and S only (D) Q and R only (GATE 2011; 2 Marks) Statement for Linked Answer Questions 79–80: A 200 µL of polymerase chain reaction has 100 template DNA molecules and the reaction was performed for 10 cycles. 79. How many molecules of amplicons will be generated? (B) 1.024 × 105 (A) 1.024 × 104 4 (C) 2.048 × 10 (D) 2.048 × 105 (GATE 2011; 2 Marks) 80. How many molecules of amplicons will be present in 0.1 µL of reaction? (A) 102.4 (B) 1024 (C) 51.2 (D) 512 (GATE 2011; 2 Marks) 81. The basis for blue-white screening with pUC vectors is (A) intraallelic complementation. (B) intergenic complementation. (C) intragenic suppression. (D) extragenic suppression. (GATE 2012; 1 Mark) 82. Protein-DNA interactions in vivo can be studied by (A) gel shift assay. (B) Southern hybridization. (C) chromatin immunoprecipitation assay. (D) fluorescence in situ hybridization assay. (GATE 2012; 1 Mark) 83. A single base pair of DNA weighs 1.1 × 10–21 grams. How many picomoles of a plasmid vector of length 2750 bp are contained in 1 µg of purified DNA? (A) 0.30 (B) 0.55 (C) 0.25 (D) 0.91 (GATE 2012; 2 Marks) 84. Human genome sequencing project involved the construction of genomic library in
(A) bacterial artificial chromosome. (B) pBR322. (C) bacteriophage. (D) pcDNA. (GATE 2013; 1 Mark) 85. 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. (GATE 2013; 1 Mark) 86. 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. S. transfer of Ri-plasmid into the plant cell. (A) S only (B) P and R only (C) Q and S only (D) Q only (GATE 2013; 2 Marks) 87. A complete restriction digestion of a circular plasmid (5000bp) was carried out with HindIII, BamHI and EcoRI individually. Restriction digestion yielded following fragments. Plasmid + HindIII → 1200bp and 3800bp Plasmid + BamHI → 5000bp Plasmid + EcoRI → 2500bp The number of sites for EcoRI, BamHI and HindIII present 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 (GATE 2013; 2 Marks) 88. 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-Gly-ArgCys-Ala-Gln (GATE 2013; 2 Marks) Statement for Linked Answer Questions 89–90: A DNA fragment of 5000bp needs to be isolated from E.coli (genome size 4 × 103 kb) genomic library. 89. The minimum number of independent recombinant clones required to represent this fragment in genomic library are
138 GATE Biotechnology Chapter-wise Solved Papers (A) 16 × 102 (C) 8 × 102
(B) 12 × 102 (D) 1.25 × 102 (GATE 2013; 2 Marks)
90. 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 3 (C) 3.6 × 10 (D) 2.4 × 103 (GATE 2013; 2 Marks) 91. 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. (GATE 2014; 1 Marks) 92. The plasmid DNA was subjected to restriction digestion using the enzyme EcoRI and analyzed 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. Uncut
EcoRI
A
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 (A) P–3; Q–1; R–2; S–4 (B) P–1; Q–4; R–3; S–2 (C) P–4; Q–3; R–1; S–2 (D) P–4; Q–1; R–2; S–3 (GATE 2014; 2 Marks) 93. Which one of the following features is not required in a prokaryotic expression vector?
(A) oriC (B) Selection marker (C) CMV promoter (D) Ribosome binding site
(GATE 2015; 1 Mark)
94. In DNA sequencing reactions using the chain termination method, the ratio of ddNTPs to dNTPs should be (A) 0 (B) < 1 (C) 1 (D) > 1 (GATE 2015; 1 Mark) 95. Choose the appropriate pair of primers to amplify the following DNA fragment by the polymerase chain reaction (PCR). 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 (GATE 2015; 1 Mark) 96. Plasmid DNA (0.5 µg) containing an ampicillin resistance marker was added to 200 µL of competent cells. The transformed competent cells were diluted 10,000 times, out of which, 50 µL was plated on agar plates containing ampicillin. A total of 35 colonies were obtained. The transformation efficiency is _________ × 106 cfu⋅µg–1. (GATE 2015; 2 Marks) 97. Match the reagents in Group I with their preferred cleavage sites in Group II. Group I Group II P. Cyanogen bromide 1. Carboxyl side of methionine Q. o-Iodosobenzoate 2. Amino side of methionine R. Hydroxylamine 3. Carboxyl side of tryptophan S. 2-Nitro-5- 4. Amino side of cysteine thiocyanobenzoate 5. Asparagine-glycine bonds (A) P–1; Q–3; R–5; S–4 (B) P–2; Q–3; R–1; S–4 (C) P–1; Q–2; R–5; S–4 (D) P–4; Q–2; R–5; S–3 (GATE 2015; 2 Marks) 98. 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↓CT3′) is __________ × 103. (GATE 2015; 2 Marks)
Chapter 9 • Recombinant DNA Technology 139
5 kb
5 kb
3 kb
3 kb
2 kb
2 kb
1 kb
1 kb
Sample
Marker
Sample
Marker
99. 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 end-filling reaction using the Klenow fragment and [α32P]-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.
5′-GGATCC-3′ 3′-CCTAGG-5′
BamHI
300 5′-GGCC-3′ 3′-CCGG-5′
900
HaeIII
Autoradiograph
EtBr
2 kb HaeIII
(D)
2 kb
2 kb
102. A variety of genetic elements are used in the transgenic plant research. Match the genetic elements (Group I) with their corresponding source (Group II). Group I Group II P. Ubiquitin1 promoter 1. Agrobacterium tumefaciens Q. Nos transcriptional terminator 2. Streptomyces hygroscopicus R. bar selection marker gene 3. Escherichia coli S. gus reporter gene 4. Zea mays
HaeIII 3 kb
3 kb
(A) P–2; Q–1; R–3; S–4 (B) P–2; Q–3; R–4; S–1 (C) P–3; Q–4; R–1; S–2 (D) P–4; Q–1; R–2; S–3 (GATE 2016; 2 Marks)
1 kb
3 kb BamHI
2 kb
101. Select the correct combination of genetic components that are essential for the transfer of T-DNA 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. (GATE 2016; 2 Marks)
1 kb
BamHI
HaeIII
(C)
2 kb BamHI
2 kb
3 kb HaeIII
BamHI
(B)
2 kb
1 kb HaeIII
2 kb
BamHI
BamHI
(A)
BamHI
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?
(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. (GATE 2016; 1 Mark)
1 kb
103. A 1.2 kb DNA fragment was cloned into BamHI and EcoRI sites located on a 2.8 kb cloning vector. The BamHI and EcoRI sites are adjacent to each other on the vector backbone. The vector contains an XhoI site located 300 bp upstream of the BamHI site. An internal XhoI site is present in the gene sequence as shown in the figure. The resultant recombinant plasmid is digested with EcoRI and XhoI and analyzed through 1% agarose gel electrophoresis. Assuming complete digestion with EcoRI and XhoI, the DNA fragments (in base pairs) visible on the agarose gel will correspond to:
(GATE 2015; 2 Marks) 100. 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.
BamHI EcoRI Xhol 300 bp Vector backbone 2.8 kb
BamHI 500 bp
Xhol
Insert 1.2 kb
EcoRI
140 GATE Biotechnology Chapter-wise Solved Papers (A) 2800, 700 and 500 (C) 2500, 700 and 800
(B) 2800, 700 and 800 (D) 2500, 1200 and 300 (GATE 2016; 2 Marks)
104. 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 (B) Southern blotting (C) South western blotting (D) Western blotting (GATE 2017; 1 Mark)
107. Shown below is a plasmid vector (P) and an insert (Q). The insert was cloned into the BamHI site of the vector. The recombinant plasmid was isolated and digested with BamHI or Xhol. The results from the digestion experiments are shown in (R).
ol
mH I
Xh
BamHI
Ba
DN
Al
ad
de
r
105. 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 (C) Primers (D) Reverse transcriptase (GATE 2017; 1 Mark)
106. A recombinant protein is to be expressed under the control of the lac promoter and operator in a strain of E. coli having the genotype lacI+ crp+. Even in the absence of inducer IPTG, low levels of expression of the recombinant protein are seen (leaky expression). Which one of the following should be done to minimize such leaky expression? (A) Addition of lactose to the medium (B) Removal of all glucose from the medium (C) Addition of excess glucose to the medium (D) Addition of allo-lactose to the medium (GATE 2017; 2 Marks)
6 kb 5 kb 5 kb plasmid vector
BamHI Xhol 0.3 kb
4 kb
BamHI
3 kb 2 kb
1 kb insert
1 kb
EcoRI P
Which one of the following explains the digestion results shown in (R)? (A) The insert did not ligate to the vector. (B) One copy of the insert ligated to the vector. (C) The insert ligated to the vector as two tandem copies. (D) The insert ligated to the vector as two copies but not in tandem. (GATE 2017; 2 Marks) 108. Which one of the following cannot be a recognition sequence for a type II restriction enzyme? (A) GAATTC (B) AGCT (C) GCGGCCGC (D) ATGCCT (GATE 2017; 2 Marks)
Q
R
109. EcoRI restriction sites on a 10kb DNA fragment are shown below.
0 kb 2 kb 4 kb
7 kb
10 kb
Upon partial digestion, what are the lengths (in kb) of all the possible DNA fragments obtained? (A) 2, 3, 4, 5, 6, 7, 8, and 10 (B) 2, 3, 4, 5, 6, and 7 (C) 2, 3, 4, and 7 (D) 2 and 3 (GATE 2017; 2 Marks)
Chapter 9 • Recombinant DNA Technology 141
110. Which one of the following techniques is used to monitor RNA transcripts, both temporally and spatially? (A) Northern blotting (B) In situ hybridization (C) Southern blotting (D) Western blotting (GATE 2018; 1 Mark)
113. Which one of the following techniques can be used to compare the expression of a large number of genes in two biological samples in a single experiment? (A) Polymerase chain reaction (B) DNA microarray (C) Northern hybridization (D) Southern hybridization
111. Genomic DNA isolated from a bacterium was digested with a restriction enzyme that recognizes a 6-base pair (bp) sequence. Assuming random distribution of bases, the average length (in bp) of the fragments generated is _________ . (GATE 2018; 1 Mark)
(GATE 2019; 1 Mark)
112. 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
5′-GAATTC-3′ 3′-CTTAAG-5′
HaeIII
5′-GGCC-3′ 3′-CCGG-5′
1 kb HaeIII
5 kb
114. For site-directed mutagenesis, which one of the following restriction enzymes is used to digest methylated DNA? (A) KpnI (B) DpnI (C) XhoI (D) MluI (GATE 2019; 2 Marks) 115. The molecular mass of a protein is 22 kDa. The size of the cDNA (excluding the untranslated regions) that codes for this protein is _________ kb (rounded off to 1 decimal place). (GATE 2019; 2 Marks) 116. Which of the following factors affect the fidelity of DNA polymerase in polymerase chain reaction? P. Mg2+ concentration Q. pH R. Annealing temperature (A) P and Q only (B) P and R only (C) Q and R only (D) P, Q and R
The number of bands that will appear on the X-ray film is _________ . (GATE 2018; 2 Marks)
(GATE 2019; 2 Marks)
Answer Key 1. (D)
2. (D)
3. (C)
4. (C)
5. (C)
6. (A)
7. (C)
8. (C)
9. (B)
10. (D)
11. (P–4; Q–1; R–5; S–2; T–3; U–6)
12. (675.84 µg)
13. (B)
14. (B)
15. (B)
16. (B)
17. (B)
18. (B)
19. (Vector 3)
20. (5.4 kb) 21. (B)
22. (B)
23. (A)
24. (D)
25. (D)
26. (*)
27. (A)
28. (B)
29. (A)
30. (C)
31. (B)
32. (A)
33. (D)
34. (B)
35. (D)
36. (D)
37. (B)
38. (C)
39. (C)
40. (B)
41. (A)
42. (C)
43. (A)
44. (D)
45. (A)
46. (A)
47. (B)
48. (C)
49. (B)
50. (C)
51. (C)
52. (A)
53. (C)
54. (D)
55. (B)
56. (D)
57. (C)
58. (A)
59. (C)
60. (C)
61. (B)
62. (A)
63. (D)
64. (A)
65. (C)
66. (A)
67. (D)
68. (B)
69. (B)
70. (C)
71. (A)
72. (B)
73. (C)
74. (D)
75. (A)
76. (A)
77. (A)
78. (C)
79. (B)
80. (C)
81. (A)
82. (C)
83. (B)
84. (A)
85. (B)
86. (B)
87. (A)
88. (5)
89. (8 × 10 )
91. (D)
94. (B)
95. (B)
96. (2.8 × 106 cfu ⋅µg–1)
101. (B)
102. (D)
103. (C)
104. (B)
112. (2)
113. (B)
111. (4096 bp)
90. (2.4 × 10 )
92. (A)
93. (C)
97. (A)
98. (1.6 × 103)
99. (A)
100. (D)
105. (C)
106. (C)
107. (C)
109. (A)
110. (B)
114. (B)
115. (0.6 kb) 116. (D)
2
3
108. (D)
142 GATE Biotechnology Chapter-wise Solved Papers
Answers with Explanations 1. Topic: Restriction and Modification Enzymes Restriction enzymes are DNA-cutting enzymes found in mostly bacteria and some other prokaryotes. Each enzyme recognizes one or a few target sequences and cuts double stranded DNA at or near those sequences, called restriction sites. When it finds its target sequence, a restriction enzyme will make a double-stranded cut in the DNA molecule and produce either staggered or blunt ends. Answer (D) 2. Topic: RFLP Marker Hardy-Weinberg equation explains the basic principle of population genetics. It is an expression of Hardy-Weinberg equilibrium, which states that the amount of genetic variation in a population will remain constant from one generation to the next if the disturbing factors are absent. The binomial expansion of (p + q)2 is p2 + 2pq + q2 = 1 where, p = frequency of the dominant A allele in the population; q = frequency of the recessive a allele in the population; p2 = frequency of the homozygous genotype AA; q2 = frequency of the homozygous genotype aa and 2pq = frequency of the heterozygous genotype Aa. Given, Frequency of the homozygous recessive genotype, q2 = 0.04 Thus, q = 0.2 We know, p + q = 1 p = 1 – 0.2 = 0.8 Now, by Hardy-Weinberg equation, p2 + 2pq + q2 = 1 2pq = 1– (p2 + q2) 2pq = 1– (0.64 + 0.04) 2pq = 1– (0.68) 2pq = 0.32 Hence, percentage of individuals heterozygous for this trait would be 32%. Answer (D) 3. Topic: Bacterial Plasmid The ampr gene on the plasmid codes for an enzyme β -lactamase that is secreted into the periplasmic space of the bacterium, where it catalyzes hydrolysis of the β -lactam ring of ampicillin, with concomitant detoxification of the drug. Answer (C)
4. Topic: Restriction and Modification Enzymes Mung bean nuclease is a nuclease derived from sprouts of the mung bean Vigna radiata that removes nucleotides in a step-wise manner from single-stranded DNA molecules (ssDNA) and is used in biotechnological applications to remove such ssDNA from a mixture also containing double-stranded DNA (dsDNA). This enzyme is useful for transcript mapping, removal of single-stranded regions in DNA hybrids or single-stranded overhangs produced by restriction enzymes, etc. Answer (C) 5. Topic: Bacterial Plasmid 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). Answer (C) 6. Topic: Yeast Artificial Chromosome Yeast artificial chromosomes (YACs) are the products of a recombinant DNA cloning methodology to isolate and propagate very large segments of DNA in a yeast host. They provide the largest insert capacity of any cloning system. This system, developed by Burke and Olson in 1987, supports the propagation of exogenous DNA segments hundreds of kilobases in length. YACs representing contiguous stretches of genomic DNA (YAC contigs) have provided a physical map framework for the human, mouse, and even Arabidopsis genomes. Answer (A) 7. Topic: Polymerase Chain Reactions For PCR there are five chemical components needed, including a DNA template, DNA polymerase enzyme, primers, nucleotides and reaction buffer. These are described here in detail. • The DNA template is that particular DNA sequence that is to be copied. • There are two requirements for a suitable DNA polymerase enzyme for PCR. First, one is needed that has a good activity rate around 75°C. Second, it should be able to withstand temperatures of 95–100°C so that more enzyme does not have to be added at the beginning of each new cycle. The first to be discovered with these characteristics was Taq polymerase, isolated from Thermus aquaticus, a thermophilic eubacterium.
Chapter 9 • Recombinant DNA Technology 143
•
•
•
rimers are short oligonucleotides of DNA, usually P around 8–60 base pairs in length. They can be random sequences if the project’s goal is for general genomic studies. However, if the purpose is to amplify a certain section of DNA in the genome, such as a known gene, then primers of specific sequences must be used. The four different deoxyribonucleotide triphosphates (dNTPs and not NTPs), adenine (A), guanine (G), cytosine (C), and thymine (T) are needed to provide the building blocks for DNA replication. DNA polymerase will add each complementary base to the new growing DNA strand according to the original strand’s sequence following normal A–T and C–G pairings. Finally, a reaction buffer is used to provide a stable pH. It may also contain magnesium chloride, if not, then MgCl2 must be added separately. Answer (C)
8. Topic: cDNA and Genomic DNA Library There are 50% chances of cDNA clone to express correct gene product. A cDNA library is generally better if the size of the inserts (that is the amount of continuous cDNA in each clone) is large, ideally full-length. Ideally, no member of the library should include cDNAs derived from different mRNAs (this could be confusing). The library should be sufficiently large that it contains the cDNA of interest (or, more precisely, it should have enough independently derived clones that it contains the cDNA of interest). Answer (C) 9. Topic: Plant Expression Vectors 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. The GUS gene has been commonly used in plants but luciferase and GFP are becoming more common. Answer (B) 10. Topic: Restriction and Modification Enzymes Proteolytic enzyme, also called protease, proteinase, or peptidase, any of a group of enzymes that break the long chainlike molecules of proteins into shorter fragments (peptides) and eventually into their components, amino acids. Examples of proteases include: fungal protease, pepsin, trypsin, chymotrypsin, papain, bromelain, and
subtilisin. Proteasomes are protein complexes which degrade unneeded or damaged proteins by proteolysis, a chemical reaction that breaks peptide bonds. While, function of peptidyl-tRNA hydrolase is completely different. It releases tRNA from peptidyl-tRNA by cleaving the ester bond between the peptide and the tRNA. Answer (D) 11. Topic: Polymerase Chain Reactions • Maxam–Gilbert sequencing is a method of DNA sequencing developed by Allan Maxam and Walter Gilbert in 1976–1977. This method is based on nucleobase-specific partial chemical modification of DNA and subsequent cleavage of the DNA backbone at sites adjacent to the modified nucleotides. • The Southern blot is a technique that uses gel electrophoresis combined with labeled probes for a DNA sequence of interest that allows for the detection of repeat expansions within specific genes. It was introduced by Edwin Southern in 1975 as a method to detect specific sequences of DNA in DNA samples. • In 1975, Georges Köhler and César Milstein succeeded in making fusions of myeloma cell lines with B cells to create hybridomas that could produce antibodies, specific to known antigens and that were immortalized. • In 1970, the independent and simultaneous discovery of reverse transcriptase in retroviruses (then RNA tumor viruses) by David Baltimore and Howard Temin revolutionized molecular biology and laid the foundations for retrovirology and cancer biology. For their achievements, they shared the 1975 Nobel Prize in Physiology or Medicine (with Renato Dulbecco). • In the course of identifying the amino groups, Sanger figured out ways to order the amino acids. He was the first person to obtain a protein sequence. By doing so, Sanger proved that proteins were ordered molecules and by analogy, the genes and DNA that make these proteins should have an order or sequence as well. • Polymerase chain reaction is a revolutionary method developed by Kary Mullis in the 1980s. PCR is based on using the ability of DNA polymerase to synthesize new strand of DNA complementary to the offered template strand. Answer (P–4; Q–1; R–5; S–2; T–3; U–6) 12. Topic: Polymerase Chain Reactions After 10 cycles, number of templates will be 210 = 1024 Molecular weight of 1 DNA template = 1 pmole Molecular weight of 1024 DNA template = 1024 pmole
144 GATE Biotechnology Chapter-wise Solved Papers We know, 1 pmole of 1000 bp DNA = 0.66 µg 1024 pmole of 1000 bp DNA = 0.66 × 1024 µg = 675.84 µg Answer (675.84 µg) 13. Topic: Polymerase Chain Reactions PCR primers are short fragments of single stranded DNA (15–30 nucleotides in length) that are complementary to DNA sequences that flank the target region of interest. The purpose of PCR primers is to provide a “free” 3′-OH group to which the DNA polymerase can add dNTPs. Thus, for given sequence primer would be: 5′ATCTTCTACG ………. AAGCTTGCGG3′ 3′CGAACGCC5′ 5′ATCTTCTACG3′ 3′TAGAAGATGC ……… TTCGAACGCC5′ Answer (B) 14. Topic: RFLP In population genetics, the concept of heterozygosity is commonly extended to refer to the population as a whole, i.e., the fraction of individuals in a population that are heterozygous for a particular locus. It can also refer to the fraction of loci within an individual that are heterozygous. The large single nucleotide polymorphism (SNP) typing projects have provided an invaluable data resource for human population geneticists. Almost all of the available SNP loci, could influence the allelic distributions in the sampled loci. Answer (B) 15. Topic: Vectors Bacterial artificial chromosomes (BACs) are designed for the cloning of large DNA insert (typically 100 to 300 kb) in E. coli host. BAC vectors contain a single copy F plasmid origin of replication (ori). The F (fertility) plasmid is relatively large and vectors derived from it have a higher capacity than normal plasmid vectors. F plasmid has F (fertility) factor which controls the replication and maintain low copy number. Also, conjugation can take place between F+ bacteria (male) and F− bacteria (female) to transfer F-plasmid via pilus. Answer (B) 16. Topic: cDNA and Genomic Library Glucagon is a pancreatic hormone of 29 amino acids that regulates carbohydrate metabolism and glicentin is an intestinal peptide of 69 amino acids that contains the sequence of glucagon flanked by peptide extensions at the amino and carboxy termini. The glucagon gene encodes a precursor containing glucagon and two additional,
structurally related, glucagon-like peptides separated by an intervening peptide. These peptides are encoded in separate exons. Thus, it is more fruitful to use mRNA of pancreas. Answer (B) 17. Topic: Ti plasmid The Ti plasmid has three important regions: • T-DNA region: This region has the genes for the biosynthesis of auxin (aux), cytokinin (cyt) and opine (ocs), and is flanked by left and right borders. A set of 24 kb sequences present on either side (right and left) of T-DNA are also transferred to the plant cells. It is clearly established that the right border is more critical for T-DNA transfer. • Virulence region: The genes responsible for the transfer of T-DNA into host plant are located outside T-DNA and the region is referred to as vir or virulence region. • Opine catabolism region: This region codes for proteins involved in the uptake and metabolisms of opines. Besides the above three there is ori region that responsible for origin of DNA replication which permit the Ti plasmid to be stably maintain in A. tumefaciens. Answer (B) 18. Topic: cDNA and Genomic Library Functional complementation is the process of compensating a missing function in a mutant cell by a particular DNA sequence for restoring the wild-type phenotype. If the mutant cells are non-viable, the cells carrying the clone of interest can be positively selected and isolated. Several methods of screening are based on detection of biological activity of the protein encoded by the target cDNA. Prokaryotic translation machinery lack the splicing mechanism, thus, E.coli would not express cDNA of human genome that efficiently. Answer (B) 19. Topic: Restriction and Modification Enzymes Vector 3 is best suited for cloning the given ORF. It is because, in the given ORF, the restriction site of BamHI is cutting at 2nd position (G/GATCC), same is in the case of vector 3 and also, the restriction site of HindIII is cutting at 2nd position (A/AFCTT), same as in the case of vector 3. Vector 3: _________Promoter.. ATGGGTCGGATCCGGCTGCTA .. AAGCTT ORF: __________ ATGCCCAACACCCGGATCCCG.. TAAAAGCTT __________ Answer (Vector 3)
Chapter 9 • Recombinant DNA Technology 145
20. Topic: Restriction and Modification Enzymes The gel pattern of the plasmid indicates following points: • EcoRI has single restriction site. On cutting the plasmid with EcoRI only, a single fragment of 5.4 kb is produced which is the total length of the plasmid.
EcoRI
0.6 kb
1.3 kb
EcoRI
1.4 kb
2.1 kb
5.4 kb
Hind III
• •
Hind III
Hind III
S alI also has single restriction site. On cutting the plasmid with SalI only, a single fragment of 5.4 kb is produced which is the total length of the plasmid.
n cutting the plasmid with both SalI and HindIII, O four fragments of 1.9 kb, 1.4 kb, 1.2 kb and 0.9 kb are produced. Fragments of length 1.9 kb and 1.4 kb are intact and fragment of 2.1 kb has been further broken down into 1.2 kb and 0.9 kb. It indicates that SalI site is present in between this fragment. 1.9 kb Hind III
Hind III 5.4 kb 1.4 kb 1.2 kb SalI
•
indIII has three restriction sites on this plasmid. H On cutting the plasmid with HindIII only, three fragments of 2.1 kb, 1.9 kb and 1.4 kb are produced.
Hind III
0.9 kb SalI
•
Thus, the complete restriction map of plasmid would be: EcoRI
1.9 kb Hind III
Hind III
0.6 kb
1.3 kb 1.4 kb
Hind III
Hind III
5.4 kb
1.4 kb
1.2 kb
2.1 kb Hind III
•
n cutting the plasmid with both EcoRI and O HindIII, four fragments of 2.1 kb, 1.4 kb, 1.3 kb and 0.6 kb are produced. Fragments of length 2.1 kb and 1.4 kb are intact and fragment of 1.9 kb has been further broken down into 1.3 kb and 0.6 kb. It indicates that EcoRI site is present in between this fragment.
0.9 kb
Hind III
SalI
Answer (5.4 kb) 21. Topic: Restriction and Modification Enzymes T4 polynucleotide kinase catalyzes the transfer of the γ -phosphate from ATP to the 5′-OH terminus of single
146 GATE Biotechnology Chapter-wise Solved Papers and double stranded DNAs and RNAs oligonucleotides or nucleoside bearing a 3′-phosphate group. T4 PNK is widely used to end-label 5′ termini of short oligonucleotide probes, DNA and RNA molecules. Under certain conditions, the reaction with polynucleotides can be made reversible, permitting exchange of the γ -phosphate of ATP with the 5′ terminal phosphate of a polynucleotide, thus circumventing the need to dephosphorylate the substrate with alkaline phosphatase. Answer (B) 22. Topic: Plasmid Some important characteristics of a cloning vector include: • It must be small in size. • It must be self-replicating inside host cell. • It must possess restriction site for restriction endonuclease enzymes. • Introduction of donor DNA fragment must not interfere with replication property of the vector. • It must possess some marker gene such that it can be used for later identification of recombinant cell. • It must possess multiple cloning site. Expression vectors contain the promoter, operator, repressor-binding, ribosome-binding and terminator sequences necessary for gene expression and protein production from the inserted DNA segment of interest. Answer (B) 23. Topic: Expression Vectors The baculovirus expression vector system (BEVS) has been widely used to produce a large number of recombinant proteins, and is becoming one of the most powerful, robust, and cost-effective systems for the production of eukaryotic proteins. It uses the promoter for polyhedron gene to improve recombination efficiencies. Answer (A) 24. Topic: Ti Plasmid T-DNA left and right border repeat sequences define and delimit T-DNA. T-DNA border repeat sequences (T-DNA borders) contain 25 bp that are highly conserved in all Ti- and Ri-plasmids. Within Agrobacterium, nucleotides 4 to 25 remain within the T-DNA at the left border (LB), whereas at the right border (RB) nucleotides 1 to 3 remain intact. Answer (D) 25. Topic: Polymerase Chain Reactions The enzyme most commonly used in cycle sequencing is Taq, isolated from Thermus aquaticus, a bacterium that lives in the hot springs of Yellowstone National Park. Because the high temperature used for the denaturation step in each cycle would destroy most DNA polymerase
enzymes, a special heat-stable enzyme must be used in the annealing stage, one that remains active even after repeated exposure to very high temperatures. Answer (D) 26. Topic: cDNA and Genomic Libraries pUC19 is a general-purpose cloning vector that accepts DNA inserts of upto 10 kb. Insulin is transcribed as proinsulin mRNA. After removal of the precursor signal peptide, proinsulin is post-translationally cleaved into three peptides: the B chain and A chain peptides, which are covalently linked via two disulfide bonds to form insulin, and C-peptide. But, being prokaryotic plasmid pUC19, it lacks translation machinery required for splicing, thus it cannot express human insulin. Answer (*) 27. Topic: Restriction and Modification Enzymes DNA ligase in E. coli, as well as most prokaryotes, uses energy gained by cleaving nicotinamide adenine dinucleotide (NAD) to create the phosphodiester bond. It does not ligate blunt-ended DNA except under conditions of molecular crowding with polyethylene glycol, and cannot join RNA to DNA efficiently. Answer (A) 28. Topic: Plasmid Some important characteristics of a cloning vector include: • It must be small in size. • It must be self-replicating inside host cell. • It must possess restriction site for restriction endonuclease enzymes. • Introduction of donor DNA fragment must not interfere with replication property of the vector. • It must possess multiple cloning site. • It must have high copy number. • It must possess some marker gene such that it can be used for later identification of recombinant cell. It needs not necessarily to be an antibiotic resistance marker. Answer (B) 29. Topic: DNA Fingerprinting DNA fingerprinting is a method used to identify an individual from a sample of DNA by looking at unique patterns in their DNA. • The first step of DNA fingerprinting is to extract DNA from a sample of human material, from tissues collected at crime scene. • Then, restriction enzymes are used to cut the DNA. This resulted in thousands of pieces of DNA with a variety of different lengths.
Chapter 9 • Recombinant DNA Technology 147
•
•
•
•
These pieces of DNA are then separated according to size by a process called gel electrophoresis. The smallest DNA molecules will go farthest from where the original sample. Once the DNA has been sorted, the pieces of DNA are transferred or ‘blotted’ out of the fragile gel on to a robust piece of nylon membrane and then ‘unzipped’ to produce single strands of DNA. Next the nylon membrane was incubated with radioactive probes. Probes are small fragments of minisatellite DNA tagged with radioactive phosphorous. The probes only attach to the pieces of DNA that they are complementary to the minisatellites in the genome. The minisatellites that the probes have attached to are then visualized by exposing the nylon membrane to X-ray film. Answer (A)
30. Topic: Gene Isolation, Cloning and Expression Total gene size = 5 kb = 5000 bp Given that, sequencer can sequence only upto 500 bases in a single reaction. Thus, number of subclones that has be generated = 5000 / 500 = 10 Now, we can choose a restriction enzyme that can produce more than 10 subclones. • Restriction enzyme with 8 bp cutter has probability of finding the restriction site at every 48 = 6536 Thus, number of fragments expected to be produced = 5000/6536 = 0.765 • Restriction enzyme with 6 bp cutter has probability of finding the restriction site at every 46 = 4096 Thus, number of fragments expected to be produced = 5000/4096 = 1.2 • Restriction enzyme with 3 bp cutter has probability of finding the restriction site at every 43 = 64 bp Thus, number of fragments expected to be produced = 5000/64 = 78.12 • Restriction enzyme with 5 bp cutter has probability of finding the restriction site at every 45 = 1024 bp Thus, number of fragments expected to be produced = 5000/1024 = 4.8 Hence, only 3 bp cutter can generate more than 10 fragments. Answer (C) 31. Topic: Gene Isolation, Cloning and Expression Given that, sequencer can sequence only upto 500 bases in a single reaction.
Thus, number of subclones that has been generated = 5000 / 500 = 10 Thus, to generate a minimum number of 10 subclones, the size of insert should be 500 bp. Answer (B) 32. Topic: Microarray In microarray, thousands of spotted samples known as probes (with known identity) are immobilized on a solid support (a microscope glass slides or silicon chips or nylon membrane). The spots can be DNA, cDNA, or oligonucleotides. These are used to determine complementary binding of the unknown sequences thus allowing parallel analysis for gene expression and gene discovery. An experiment with a single DNA chip can provide information on thousands of genes simultaneously. An orderly arrangement of the probes on the support is important as the location of each spot on the array is used for the identification of a gene. The probe sequences are spotted onto polycarbonate slides with a mini-microarray printer, and after the hybridization, the results are visible with the naked eye. Answer (A) 33. Topic: Restriction and Modification Enzymes Restriction enzymes are DNA-cutting enzymes. Each enzyme recognizes one or a few target sequences and cuts DNA at or near those sequences. As on cutting the viral genome with a restriction endonuclease giving only a single band on gel on electrophoresis. There can be two possibilities; Either the viral genome has double stranded DNA, but it does not have the required restriction site or the viral genome is double stranded RNA that is not recognized by the restriction enzyme. Answer (D) 34. Topic: Restriction and Modification Enzymes Of the given restriction endonuclease, the restriction site of XmaIII (C/GGCGG) is exactly same as that of Eco52I (C/GGCGG). Therefore, it would be apt to clone a DNA cut by XmaIII into a plasmid digested with Eco52I. Answer (B) 35. Topic: Plasmid In E. coli, DNA passes intact through a weakened cell envelope; closed circular plasmid reaches the cytoplasm in a state ready to begin replication, which is why it transforms better. Linearized DNA would need to be religated or recombine in order to establish itself before the intracellular nucleases degrade it. Thus, circular recombinant molecules are likely to be amplified by the cells. Answer (D)
148 GATE Biotechnology Chapter-wise Solved Papers 36. Topic: Microarray In microarray, the unknown DNA molecules are cut into fragments by restriction endonucleases; fluorescent markers are attached to these DNA fragments. These are then allowed to react with probes of the DNA chip. Then the target DNA fragments along with complementary sequences bind to the DNA probes. The remaining DNA fragments are washed away. The target DNA pieces can be identified by their fluorescence emission by passing a laser beam. A computer is used to record the pattern of fluorescence emission and DNA identification. Answer (D)
40. Topic: Polymerase Chain Reactions PCR primers are short fragments of single stranded DNA (15–30 nucleotides in length) that are complementary to DNA sequences that flank the target region of interest. The purpose of PCR primers is to provide a “free” 3′-OH group to which the DNA polymerase can add dNTPs. Thus, for given sequence primer would be: 5′CTGAGGT3′ 3′GACTCCA ……… TACAACC5′ 5′CTGAGGT ………. ATGTTGG3′ 3′TACAACC5′ Answer (B)
37. Topic: Ti Plasmid The processing and transfer of T-DNA are mediated by products encoded by the vir (virulence) region, which is also resident on the Ti plasmid. Those vir genes, whose products are directly involved in T-DNA processing and transfer, are tightly regulated so that expression occurs only in the presence of wounded plant cells, the targets of infection. Control of gene expression is mediated by the VirA and VirG proteins, a two component regulatory system. VirA detects the small phenolic compounds released by wounded plants resulting in autophosphorylation. VirA phosphorylation of VirG then leads to activation of vir gene transcription. Answer (B)
41. Topic: Transposons and Gene Targeting • A neomycin resistance gene (neoR) is commonly included in DNA plasmids used by molecular biologists to establish stable mammalian cell lines expressing cloned proteins in culture; many commercially available protein expression plasmids contain neo as a selectable marker. Non-transfected cells will eventually die off when the culture is treated with neomycin or similar antibiotic. • Constitutive promoters are used routinely to drive ectopic gene expression. Some of the commonly used constitutive promoters include simian virus 40 (SV40), cytomegalovirus immediate-early promoter (CMV), human Ubiquitin C promoter (UBC), human elongation factor 1α promoter (EF1A), mouse phosphoglycerate kinase 1 promoter (PGK), and chicken β -actin promoter coupled with CMV early enhancer (CAGG) for mammalian systems, and copia transposon promoter (COPIA) and actin 5C promoter (ACT5C) for Drosophila systems. • LTR retrotransposons are class I transposable element characterized by the presence of Long Terminal Repeats (LTRs) directly flanking an internal coding region. As retrotransposons, they mobilize through reverse-transcription of their mRNA and integration of the newly created cDNA into another location. The long terminal repeat (LTR) regions have been shown to be important determinants for pathogenicity and tissue specificity, by virtue of their ability to interact with various transcription factors and induce the expression of genes. • Dihydrofolate reductase (dhfr) is an enzyme that converts dihydrofolate to tetrahydrofolate and is involved in purines and thymidylate synthesis. Amplification of the dihydrofolate reductase gene has been shown to be a common mechanism of resistance in cultured cells, when grown in the presence of stepwise increasing concentrations of methotrexate (MTX). The drug resistant cells with amplified dhfr genes generally exhibit a proportional increase in dhfr transcripts as well as in dhfr protein. Answer (A)
38. Topic: Microarray Nick translation is a technique for radioactively labeling double-stranded DNA, making it suitable as a hybridization probe for detecting specific genomic sequences. The endonuclease DNase I is used to create nicks in a DNA probe fragment. Following DNase I treatment, DNA polymerase I is used to add nucleotide residues to the free 3′-hydroxyl ends created during the DNase I nicking process. As the DNA polymerase I extends the 3′ ends, the 5′ to 3′ exonuclease activity of the enzyme removes bases from the 5′-phosphoryl terminus of the nick. The sequential addition of bases onto the 3′ end with the simultaneous removal of bases from the 5′ end of the downstream annealed strand results in translation of the nick along the DNA molecule. When performed in the presence of a radioactive deoxynucleoside triphosphate (such as [α-32P]dCTP), the newly synthesized strand becomes labeled. Thus, primers are not required in this process. Answer (C) 39. Topic: Restriction and Modification Enzymes Of the given restriction endonuclease, the restriction site of HaeIII (GG↓CC) is exactly same as that of SalmIII (GG↓CC). Therefore, it would be apt to clone a DNA cut by HaeIII into a plasmid digested with SalmIII. Answer (C)
Chapter 9 • Recombinant DNA Technology 149
42. Topic: Southern and Northern Blotting Agarose gel electrophoresis is the most effective way of separating DNA fragments of varying sizes ranging from 100 bp to 25 kb. To separate DNA using agarose gel electrophoresis, the DNA is loaded into pre-cast wells in the gel and a current applied. The phosphate backbone of the DNA (and RNA) molecule is negatively charged, therefore when placed in an electric field, DNA fragments will migrate to the positively charged anode. Because DNA has a uniform mass/charge ratio, DNA molecules are separated by size within an agarose gel in a pattern such that the distance traveled is inversely proportional to the log of its molecular weight. Answer (C) 43. Topic: Polymerase Chain Reactions Real-time polymerase chain reaction (RT PCR) is a useful technique to direct online monitoring and helps in the determination of the generated PCR product. There are two detection methods of RT-PCR, the first is based on sequence-specific probe such as TaqMan probe, molecular beacon; the second is based on generic non-sequencespecific double-stranded DNA-binding dye such as SYBR green. Answer (A) 44. Topic: DNA Labeling Green fluorescent protein (GFP) has been widely used to tag proteins and to study the dynamic changes of cellular processes in living cells. It emits green light following excitation of an internal fluorophore composed of a SerTyr-Gly sequence positioned near the protein’s amino terminus. Excitation of GFP-expressing cells can be performed by exposure to long-wave UV light, making detection of GFP activity simple and obviating the need for specific substrates. Answer (D) 45. Topic: Restriction and Modification Enzymes Given, its partially digested means incomplete digestion of DNA, thus not all sites would have been targeted by restriction enzyme. Thus, there is possibility of generation of all probable fragments. EcoRI
EcoRI
EcoRI
Thus, number of labeled fragments that can be generated =4 Number of unlabeled fragments that can be generated = 6 Answer (A)
46. Topic: Ti Plasmid T-DNA on the Ti plasmid is flanked by 25-base-pair (bp) imperfect direct repeat sequences. The ends of the T-DNAs are precisely defined with respect to these 25-bp sequences. In tumors induced by nopaline-type Ti plasmids, whose T-DNA borders have been studied most extensively, the right ends of T-DNAs from three independent tumor lines are within 3 bp of the right 25-bp sequence and the left ends are all within 100 bp of the left 25-bp sequence. Mutational studies have provided direct evidence of the importance of the right border region in T-DNA transformation. In nopaline-type Ti plasmids, deletions of the T-DNA right border region resulted in attenuation or abolition of transformation activity. Conversely, in insertion experiments, a fragment containing the nopaline T-DNA right border sequences was shown to be capable of promoting the transfer and integration of adjacent sequences from several locations around the Ti plasmid into the plant genome. Answer (A) 47. Topic: Southern and Northern Blotting The UDP glucuronosyltransferase (UGT) content of cells and tissues is a major determinant of our response to those chemicals that are primarily eliminated by conjugation with glucuronic acid. There are marked interindividual differences in the content of UGTs in the liver and other organs. The mechanisms that lead to these differences are unknown but are most likely the result of differential UGT gene expression. Several transcription factors involved in the regulation of UGT genes have been identified. These include factors such as Hepatocyte Nuclear Factor 1, CAAT-Enhancer Binding Protein, Octamer transcription Factor 1 and Pbx2, which appear to control the constitutive levels of UGTs in tissues and organs. In addition, UGT gene expression is also modulated by hormones, drugs and other foreign chemicals through the action of proteins that bind and/or sense the presence of these chemicals. Answer (B) 48. Topic: cDNA and Genomic DNA Library Bacteria have been a common expression system for recombinant proteins, but the flaw of the system is that insoluble and inactive proteins are co-produced due to codon bias, protein folding, phosphorylation, glycosylation, mRNA stability and promoter strength. They have limited eukaryotic post-translational machinery function, which is considered as a key disadvantage for producing the eukaryotic phosphoproteins, such as, serine kinase, threonine kinase, tyrosine kinases. • Glycosylation is a major post-translational modification. It is responsible for the formation of cellular
150 GATE Biotechnology Chapter-wise Solved Papers
•
glycans which are often attached to proteins and lipids. Glycosyltransferase and glycosidases are enzymes responsible for glycosylation of many proteins. Glycoproteins, which are commonly distributed in eukaryotic cells, are rarely presented in prokaryotic organisms because cellular organelles essential for glycosylation are missing in these organisms. Also, transfer RNA (tRNA) of cells reflects the codon bias of its mRNA. Any heterologous gene with abundant codons, which are rarely used in bacteria, may not be properly expressed in bacteria, and may lead to translation errors. Answer (C)
49. Topic: Southern and Northern Blotting When multiple copies of c-myc cDNA are introduced into muscle cells by SV40, they are efficiently integrated into one or a very few random site(s) within the host genome as a concatemer. Sequence analysis of the concatemer shows that the multiple copies of the DNA sequences are not randomly oriented within the concatemer, but rather are all oriented in the same direction (a head-to-tail concatemer). These highly ordered concatemers arose through homologous recombination. Concatemeric DNA must be resolved and unit length genome excised in order to assemble mature virus. Answer (B) 50. Topic: Southern and Northern Blotting c-myc promotes cell proliferation and transformation by activating growth promoting genes, including the ODC and cdc25A genes. Western blot analysis of the fusion proteins was performed by using a mAb to human c-myc. It will last for upto 10 generations. Answer (C) 51. Topic: Gene Isolation, Cloning and Expression 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. Answer (C) 52. Topic: Gene Isolation, Cloning and Expression We know, Volume of cell culture = 1 mL = 1000 µL Given, it was diluted 1 : 1000000, thus, new volume = 103 × 106 µL = 109 µL Only 200 µL of this dilution was taken, Thus, dilution factor is given by =
10 9 μL = 0.5 × 10 7 200 μL
Number of colony forming units after 12 h incubation of diluted culture = 150 Therefore, total number of colony forming units in original culture = 150 × 0.5 × 107 = 7.5 × 108 CFU Answer (A) 53. Topic: Gene Isolation, Cloning and Expression Given, to 100 μL competent E.coli, 900 μL of SOC medium was added, thus culture got diluted by 100/1000, that is, 1/10. Only 50 µL of this dilution was taken from 1000 µL, Thus, dilution factor is given by =
103 μL = 20 50 μL
Number of colonies after overnight incubation of diluted culture = 300 Therefore, total number of colonies in original culture = 300 × 20 = 6000 Transformation efficiency can be calculated by = (no. of transformed colonies on plate/ng of DNA plated) × 1000 ng/µg = (6000/5) × 1000 ng/µg = 1.2 × 106 colonies per μg Answer (C) 54. Topic: Polymerase Chain Reactions Tth polymerase, isolated from Thermus thermophilus HB-8, is 94 kDa thermostable polymerase lacking 3′→5′ exonuclease (proofreading) activity. Tth polymerase catalyzes the polymerization of nucleotides into duplex DNA from a DNA template (DNA polymerase) in the presence of Mg2+, and from an RNA template (reverse transcriptase) in the presence of Mn2+. Answer (D) 55. Topic: Polymerase Chain Reactions As GC base pair has 3 hydrogen bonds while AT pair has only 2 bonds, so GC pair exhibits stronger base-base interactions, leading to stable self-complementary superstructures such as hairpin-loop and dimer, producing regions with higher melting temperatures, and making the DNA strand harder to be opened and amplified. To tackle such problems, additives such as dimethylsulfoxide (DMSO), formamide, glycerol and betaine are often used in such amplifications due to their ability of reducing the effect of hydrogen bonding, destabilizing the secondary structure, and enhancing the specificity of the amplification reaction. The concentration of additives like betaine, lower the stability of dsDNA and thus make the GC-rich region less stable, easier to be opened and more ready for the primers to bind and extend. Betaine has more advantages over other additives: it is not as viscid as glycerol,
Chapter 9 • Recombinant DNA Technology 151
which is hard to be transferred; it is not as inhibitive to Taq polymerase as formamide, which will require more enzymes when a higher concentration of formamide is applied. Furthermore, betaine is compatible with various downstream procedures such as product-purification, sequencing, etc. Answer (B) 56. Topic: DNA Fingerprinting • Variable number tandem repeats (VNTR), also called mini-satellites, are among the families of repetitive DNA dispersed in the genome. Each repeating unit comprises a sequence of 16–64 base pairs. The repeating units in each group have conserved sequences and probes can be prepared to recognize them. Mini satellites from an individual, analyzed by Southern blotting, give a characteristic banding pattern. Except for identical twins, it is highly unlikely that two individuals with the same pattern will be found because it is an individual characteristic, much like fingerprints. The method, called DNA fingerprinting, is used to identify a particular person in forensic cases, or to establish parenthood. • The 5′ untranslated region (5′ UTR) (also known as a leader sequence or leader RNA) is the region of an mRNA that is directly upstream from the initiation codon. This region is important for the regulation of translation of a transcript by differing mechanisms in viruses, prokaryotes and eukaryotes. • The Shine-Dalgarno (SD) Sequence is a ribosomal binding site in bacterial and archaeal messenger RNA, generally located around 8 bases upstream of the start codon AUG. The RNA sequence helps recruit the ribosome to the messenger RNA (mRNA) to initiate protein synthesis by aligning the ribosome with the start codon. • Attenuation is a regulatory feature found throughout Archaea and Bacteria causing premature termination of transcription. Attenuators are 5’-cis acting regulatory regions which fold into one of two alternative RNA structures which determine the success of transcription. The folding is modulated by a sensing mechanism producing either a Rho-independent terminator, resulting in interrupted transcription and a non-functional RNA product; or an anti-terminator structure, resulting in a functional RNA transcript. Answer (D) 57. Topic: Gene Isolation, Cloning and Expression Most methods for screening libraries of cDNA clones rely on the selective binding of either nucleic acid probes to cDNA molecules or antibodies to the polypeptide gene product encoded by the cDNA. When trying to identify a clone within a cDNA library, which may contain a desired protein (X) gene, one useful technique includes
immunological screening, using antibodies raised against isolated protein. Antibody screening can be carried out on a cDNA cloned into a wide range of vectors, including plasmids and phage-based vectors. Answer (C) 58. Topic: Restriction and Modification Enzymes There are two ways to label a DNA molecule by the ends. End labeling can be performed at the 3′- or 5′-end. • Labeling at the 3′ end is performed by filling 3′-end recessed ends with a mixture or labeled and unlabeled dNTPs using Klenow or T4 DNA polymerases. Both reactions are template dependent. Terminal deoxynucleotide transferase incorporates dNTPs at the 3′ end of any kind of DNA molecule or RNA. Labels incorporated at the 3′-end of the DNA molecule prevent any further extension or ligation to any other molecule, but this can be overcome by labeling the 5′-end of the desired DNA molecule. • 5′-end labeling is performed by enzymatic methods (T4 polynucleotide kinase exchange and forward reactions), by chemical modification of sensitized oligonucleotides with phosphoramidite, or by combined methods. Answer (A) 59. Topic: cDNA and Genomic DNA Library Given, DNA content of Caenorhabditis elegans = 1.0 × 108 bp Length of λ -phage vectors = 20 kb = 20 × 103 bp Number of λ -phage vectors required =
1.0 × 108 = 5000 20 × 103
Length of yeast vectors = 250 kb = 250 × 103 bp Number of yeast clones required =
1.0 × 108 = 400 250 × 103 Answer (C)
60. Topic: cDNA and Genomic DNA Library Number of λ -phage vectors required to ensure that every sequence is included in the library =
ln(1 − p) ln(1 − 0.9999) = ln(1 − a / b) ⎛ 20 × 103 ⎞ ln ⎜1 − 108 ⎟⎠ ⎝
=
ln(0.0001) ln(0.0001) = ln (1 − 0.0002 ) ln (0.9998)
=
−9.210 = 5 × 10 4 −2 × 10 −4
152 GATE Biotechnology Chapter-wise Solved Papers Number of yeast vectors required to ensure that every sequence is included in the library =
ln(1 − p) = ln(1 − a / b)
=
ln(0.0001) ln(0.0001) = ln (1 − 0.0025 ) ln (0.9975)
=
−9.210 = 4000 −2.5 × 10 −3
ln(1 − 0.9999) ⎛ 250 × 103 ⎞ ln ⎜1 − 108 ⎟⎠ ⎝
Answer (C) 61. Topic: DNA Fingerprinting DNA footprinting is a molecular technique used to identify the specific DNA sequence that binds to a protein. It is based on the observation that when a protein binds to DNA, the DNA is protected from chemicals that would otherwise cleave it. It is mainly used to identify the transcription factors which binds to promoter, enhancer or silencer region of gene to regulate its expression. Answer (B) 62. Topic: Restriction and Modification Enzymes Concentration of stock solution = 150 μM Dilution factor = 10/200 = 1/20 Concentration of dTTP = (1/20) × 150 = 7.5 μM Answer (A) 63. Topic: Polymerase Chain Reactions Number of DNA template = 400 In polymerase chain reactions, the number of DNA template after n cycles = (No. of initial DNA template) 2n Thus, after 20 cycle, the number of DNA templates = (400) 220 It is in 100 µL of total volume. But, on taking 0.1 µL volume of it, 400 × 220 × 0.1 100 = 4.19 × 105 =
Answer (D)
64. Topic: Restriction and Modification Enzymes • Isoschizomers (same cut) are distinct restriction endonucleases that recognize the same DNA sequence and make the same cut. • Isozymes are enzymes that differ in amino acid sequence but catalyze the same chemical reaction. • Concatemer is a molecule made up of multiple copies of the same genome strung together in tandem.
•
palindromic sequence is a sequence made up of A nucleic acids within double helix of DNA and/or RNA that is the same when read from 5’ to 3’ on one strand and 5’ to 3’ on the other, complementary strand. Answer (A)
65. Topic: Expression Vectors The Baculovirus Expression Vector System (BEVS) is one of the most powerful and versatile eukaryotic expression systems available. The BEVS is a helper-independent viral system which has been used to express heterologous genes from many different sources, including fungi, plants, bacteria and viruses, in insect cells. Answer (C) 66. Topic: Gene Isolation, Cloning and Expression A reporter gene is a nonendogenous gene encoding an enzyme or fluorescent protein whose expression is controlled by a promoter for a separate gene of interest. Examples of commonly used reporter systems include GUS, GFP, luciferase and anthocyanin, and depending on the purpose of the experiment, these may be used as a transcription tag, a transcriptional fusion or a translational fusion. Reporter genes have been successfully used to characterize native and heterologous gene expression as well as protein trafficking. Answer (A) 67. Topic: Ti Plasmid • The Ri TL-DNA carries approximately 18 potential genes, 11 of which four genes, rol A, B, C and D, have a fundamental role in the induction of hairy root syndrome in tobacco, 10, 12–16 although nothing is known about the functions of the proteins encoded by these genes. • 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, inserted by the bacterium into the plant genome. The opines are used by the bacterium as an important source of nitrogen and energy. • T-DNA carries genes responsible for conferring unlimited growth and ability to synthesise opines upon the transformed plant tissue. The genes responsible for T-DNA transfer are located in a separate part of the Ti plasmid called vir (virulence) segment. Two of these genes (virA and virG) are expressed constitutively at a low level of control. The vir gene expresses a protein that forms a conjugative plasmid through which T-DNA is transferred to the nucleus. • Aux gene in Ri plasmid are necessary for auxinautotrophic cell division. The Ri TR-DNA carries the
Chapter 9 • Recombinant DNA Technology 153
genes responsible for opine synthesis and the genes representing two steps of auxin biosynthesis, referred to as aux1 and aux2 (also referred to as tms1 and tms2, respectively). Similarly, cyt gene is required for cytokine synthesis. Answer (D) 68. Topic: DNA Fingerprinting • DNA footprinting is the technique used to locate the protein binding site on the DNA. Regulatory proteins (such as repressors, promoters etc.) binds to particular sequence on the DNA which may affect the gene expression. • DNA fingerprinting is a method used to identify an individual from a sample of DNA by looking at unique patterns in their DNA. Answer (B)
71. Topic: DNA Sequencing An expressed sequence tag (EST) is a short stretch of unique DNA sequence that is used for the identification of an expressed gene. Though EST sequences are generally only 200 to 500 nucleotides in length, this is largely sufficient to identify the full-length complementary DNA (cDNA). ESTs are derived by sequencing of a single segment of random clones from a cDNA library. A single sequencing reaction and automation of DNA isolation, sequencing, and analysis have allowed the rapid determination of many ESTs. Answer (A)
Answer (B)
72. Topic: Restriction and Modification Enzymes DNA ligase catalyzes the joining of DNA strands together by catalyzing the formation of a phosphodiester bond. E. coli and phage T4 encode an enzyme, DNA ligase, which seals single-stranded nicks between adjacent nucleotides in a duplex DNA chain. Although the reactions catalyzed by the enzymes of E. coli and T4 infected E. coli are very similar, they differ in their cofactor requirements. The T4 enzyme requires ATP, while the E. coli enzyme requires NAD+. T4 DNA ligase can ligate either cohesive or blunt ends of DNA, oligonucleotides, as well as RNA and RNA-DNA hybrids, but not single-stranded nucleic acids. It can also ligate blunt-ended DNA with much greater efficiency than E. coli DNA ligase. Also, the rates of blunt-end and cohesive-end ligation of DNA by T4 DNA ligase can be increased by orders of magnitude in the presence of high concentrations of a variety of non-specific polymers such as polyethylene glycol, Ficoll, bovine plasma albumin or glycogen. Answer (B)
70. Topic: Gene Isolation, Cloning and Expression • Vancomycin is a large, complex molecule produced by the soil actinomycete Streptomyces orientalis. It is too large a molecule to pass through pores in the outer membrane of Gram-negative cell walls and is therefore not effective against most Gram-negative bacteria. It can be used to treat infections caused by methicillin-resistant staphylococci and enterococci. It is also the drug of choice against antibiotic-induced pseudomembranous colitis (enteritis with the formation of false membranes in stool). Because it is poorly absorbed through the gastrointestinal tract, it must be administered intravenously. • Streptomycin and Kanamycin are aminoglycosides that attacks target organism by inhibiting their protein synthesis. • Ampicillin is a penicillin that inhibits cell wall synthesis of target organism. Answer (C)
73. Topic: Vectors • The λ DNA molecule is 49 kb in size. The derivatives of DNA of bacteriophage lambda serve as cloning vectors for use in E. coli. The DNA is packaged inside the phage head so there is an upper and lower limit to the amount of DNA that can be packaged. Effective packaging takes place between 5–35 kb. The two types of lambda vectors used are insertion and substitution type. • Bacterial artificial chromosomes (BAC) vectors are based on fertility plasmid of bacteria. These are used for mapping of large eukaryotic genome. They can accommodate large DNA fragments of size 300–350 kb. • A P1-derived artificial chromosome (PAC) is a DNA construct that is derived from the DNA of P1 bacteriophage. It can carry large amounts of other sequences of about 100–300 kb for a variety of bioengineering purposes.
69. Topic: Gene Transfer Uptake of DNA occurs only at a certain stage in a cell’s growth cycle, in response to high cell density and depletion of nutrients. In this stage, a protein called competence factor is released into the medium and apparently facilitates the entry of DNA. When competence factor from one culture is used to treat a culture that lacks it, cells in the treated culture become competent to receive DNA—they can now take up DNA fragments. However, not all bacteria can become competent, thus not all bacteria can be transformed. Natural transformation was observed in organisms from a wide variety of genera, including Acinetobacter, Bacillus, Hemophilus, Neisseria, and Staphylococcus.
154 GATE Biotechnology Chapter-wise Solved Papers •
osmid vectors are hybrid vectors constructed by C inserting “cos” site of phage Lambda into plasmid DNA. Cosmid is inserted into lambda protein heads. They infect bacteria and replicate like plasmids. They can clone DNA fragments up to 45 kb of length. Answer (C)
74. Topic: cDNA and Genomic DNA Library Number of rare mRNA molecules in a cell = 5 Total number of m mRNA molecules in a cell = 4 × 105 Number of clones need to screen to have 99% probability of finding at least one recombinant cDNA of the rare mRNA can be calculated as:
=
log e (1 − 0.99) log e (1 − 5 / 4 × 105 )
log e (0.01) log e (0.9999875)
=
−4.605 = 3.7 × 10−5 −1.25 × 10−5
79. Topic: Polymerase Chain Reactions In PCR amplification,
N = N 0 × 2n Template DNA, N0 = 100 No. of cycles, n = 10 Amplicons, N 10 N = 100 × 2
log e (0.01) = log e (1 − 0.0000125) =
or DNA-RNA hybrids. TdT (Terminal deoxynucleotidyl transferase) is a DNA polymerase enzyme which is effective for ssDNA and on 3’–hydroxyl ends of DNA strands whereas T4 PNK (T4 Polynucleotide Kinase enzyme is coded in the genome of the T4 bacteriophage which is effective for wide variety of DNA and RNA, and it labeling site is 5′. Answer (C)
log N = log 100 + log 210 = 2 + 10(0.301) = 5.01 antilog(log N ) = antilog5.01 N = 1.024 × 105
Answer (D)
75. Topic: Yeast Artificial Chromosome Yeast artificial chromosomes (YAC’s) are the hybrid chromosomes genetically engineered in the lab. DNA of yeast Saccharomyces cerevisiae is cut and ligated into a bacterial plasmid. YAC have very high insert value of 100–1000 kb and thus they can be used to clone and assemble the entire genomes of an organism. In addition, they are useful in expressing the eukaryotic proteins which required post-translational changes. Answer (A) 76. Topic: Gene Isolation, Cloning and Expression Rhizobium rhizogenes (Agrobacterium rhizogenes) is a gram-negative soil bacterium that initiate hairy root disease in dicot plants. Distinctive hairy root characters, high growth rate, branching and, most noticeably, absence of geotropism, were shown by roots. Answer (A) 77. Topic: Restriction and Modification Enzymes Restriction endonucleases that recognize the same sequence are called isoschizomers, e.g., SphI (CGTAC/G) & BbuI (CGTAC/G) and SmaI (CCC/GGG) & XmaI (C/ CCGGG), etc. Answer (A) 78. Topic: DNA Labeling S1 nuclease is an endonuclease, which is used extensively to remove single-stranded regions from DNA or DNARNA hybrids and does not degrade double-stranded DNA
Answer (B)
80. Topic: Polymerase Chain Reactions In 200 µL volume no. of amplicons generated = 1.024 × 105 In 0.1 µL volume no. of amplicons generated =
1.024 × 105 × 0.1 = 51.2 amplicons 200 Answer (C)
81. Topic: Gene Isolation, Cloning and Expression The basis for blue-white screening with pUC vectors is intraallelic complementation. LacZ gene codes for β galactosidase which breaks X-gal into X and galactose and produce blue colour. LacZ gene is composed of two regions; ω -peptide and α -peptide region. ω -peptide is carried by E.coli, whereas α -peptide is carried by pUC vectors. When both the peptide chains are expressed together, they complement each other to form a functional β galactosidase enzyme. Answer (A) 82. Topic: DNA Labeling Chromatin immunoprecipitation assay is one type of immune-precipitation experimental technique used to study the interaction of DNA and protein in in vivo conditions. It mainly aims for detecting the association of protein with DNA such as presence of transcription factor, inducer, histones etc. Secondly, it aims for detecting the specific locations of such association in the genome. Answer (C)
Chapter 9 • Recombinant DNA Technology 155
83. Topic: Gene Isolation, Cloning and Expression Given, Molecular weight of single base pair = 1.1 × 10−21 g Molecular weight of a mole of base pair
= 1.1 × 10−21 × 6.022 × 1023 = 662 g mole−1 Molecular weight of 2750 base pairs = 662 g mole −1 × 2750 = 1.821×106 g mole −1 Picomoles of a plasmid vector in 1 µg of purified DNA
= =
87. Topic: Restriction and Modification Enzymes The circular plasmid is 5000 bp long. It has restriction sites for HindIII, BamHI and EcoRI. On complete digestion with these enzymes individually results obtained are: • On complete digestion of plasmid with HindIII, fragments of size 1200 bp and 3800 bp are obtained. These two fragments sum up to give the complete plasmid. Thus, the plasmid must be having two restriction sites for HindIII as shown:
Weight × 1012 Molecular weight
1200 bp
1× 10 −6 g ×1012 = 0.55p mole 1.821× 106 g mole −1 Answer (B)
84. Topic: cDNA and Genomic DNA Library Human genome project was an international research project launched with the motive of sequencing the entire genome sequence of human beings. It used bacterial artificial chromosome (BAC) for constructing genome library due to its large insert size of 150-350 kbp. Answer (A)
HindIII 5000 bp
3800 bp
•
85. Topic: DNA Sequencing Chain termination method, given by Sanger is used to determine the sequence of DNA. It uses 2′, 3′–dideoxyribosenucleotides to terminate the chain elongation during in vitro DNA replication. P
On complete digestion of plasmid with BamHI, fragments of size 5000 bp are obtained. Thus, the plasmid must be having one restriction site for BamHI as shown: BamHI
P P
5000 bp
P P
HindIII
P CH2
O
Base
5′
CH2
O
Base
5′ 3′ OH
Normal dNTP (extends DNA strand)
•
3′ H
No OH ddNTP (Terminates synthesis)
n complete digestion of plasmid with EcoRI, O fragments of size 2500 bp are obtained. Thus, the plasmid must be having two restriction sites for EcoRI at equidistant as shown: 2500 bp
Answer (B) 86. Topic: Ti Plasmid Agrobacterium tumefaciens naturally form tumors in dicot plants only. It is a bacterium that has the potential to transfer genes in host plant cells. It causes wound infections in dicot plants and leads to the formation of root-knot (tumors). Its infection in plant cells leads to crown gall (white masses of callus tissue or small swellings) disease. During infection the T-DNA, a mobile segment of Ti or Ri plasmid, is transferred. Answer (B)
EcoRI
5000 bp
EcoRI
2500 bp
Answer (A)
156 GATE Biotechnology Chapter-wise Solved Papers 88. Topic: DNA Sequencing The digestive enzyme trypsin has the greatest specificity and is therefore the most valuable member of the arsenal of endopeptidases used to fragment polypeptides. It cleaves peptide bonds on the C side (toward the carboxyl terminus) of the positively charged residues Arg and Lys if the next residue is not Pro. Thus, on digestion of the given peptide with trypsin we obtain: Phe–Trp–Met–Gly–Ala Lys–Leu–Pro–Met–Asp–Gly Arg–Cys–Ala–Gln Several chemical reagents promote peptide bond cleavage at specific residues. The most useful of these, cyanogen bromide (CNBr), cleaves on the carboxyl side of Met residues. Thus, after trypsin digestion, treatment with CNBr, would give: Phe–Trp Met–Gly–Ala Lys–Leu–Pro Met–Asp–Gly Arg–Cys–Ala–Gln Thus, a total of 5 fragments will be formed. Answer (5) 89. Topic: cDNA and Genomic DNA Library Given, Size of DNA fragment = 5000 bp = 5kb Genome size of E. coli = 4 × 103 kb Minimum order of recombinant clones
Genome size Insert size 4 × 103 kb = = 8 × 102 5 kb
Recombinant clone =
Answer (8 × 102) 90. Topic: cDNA and Genomic DNA Library Number of clones to represent this fragment with a probability of 95% Number of clones =
ln(1 − P ) ⎛ a⎞ ln ⎜1 − ⎟ ⎝ b⎠
Where, Probability, P = 1 Average size of fragment, a = 5kb Genomic size, b = 4 × 103 kb Putting these values in the formula above, we get,
Number of clones =
ln(1 − P) ⎛ a⎞ ln ⎜1 − ⎟ ⎝ b⎠
ln(1 − 0.95) ln 0.05 = 5 ⎛ ⎞ ln 0.99875 ln ⎜1 − 3 ⎟ ⎝ 4 × 10 ⎠ −2.99 = = 24 × 102 = 2.4 × 103 −0.00125 Answer (2.4 × 103) =
91. Topic: Restriction and Modification Enzymes If a DNA sequence is evenly made up of G, A, T, and C nucleotides (i.e. 25% of each), the probability of getting a six base sequence would be:
1 1 1 1 1 1 1 × × × × × = 4 4 4 4 4 4 4096 That means the six base nucleotide is likely to occur once after every 4096 nucleotides on an average. Answer (D) 92. Topic: Restriction and Modification Enzymes • Multimers are concatemers/fused products of several plasmids recombined together. These plasmids are several times as large as the individual plasmid and therefore run very slowly in agarose and form ladder like structure on agarose gel. Thus, band A is representing concatemers. • A small, compact supercoiled knot of covalently closed circular DNA is the native confirmation that occurs when extra twists are introduced in the strand. It sustains less friction against the agarose matrix than does a large, floppy open circle of open circular DNA. Therefore, for the same over-all size, supercoiled DNA runs faster than open-circular DNA due to its confirmation. Thus, band C is representing supercoiled DNA. • If the plasmid is cut once with a restriction enzyme, the supercoiled and open-circular conformations are all reduced to a linear conformation. Linear DNA runs through a gel end first and thus sustains less friction than open-circular DNA, but more than supercoiled. Nicked plasmid is a large floppy circle is the slowest migrating form in an agarose gel. Thus, band B is representing nicked open-circular DNA and band D on right is representing linear conformation. Answer (A) 93. Topic: Vectors • DNA sequence sometimes referred to Ori or simply as origin that signals for the origin of replication. In E.coli, ori is some 250 nucleotides in length for the chromosomal origin (oriC). • Selection markers are generally antibiotic resistance genes found in bacteria that confers a trait suitable for artificial selection.
Chapter 9 • Recombinant DNA Technology 157
•
•
he human cytomegalovirus (CMV) promoter is T commonly used for transient mammalian expression vectors to drive gene expression. However, its use in the formation of stable cell lines is less common. A ribosome binding site is a sequence of nucleotides upstream of the start codon of an mRNA transcript which engages a ribosome during protein translation’s initiation. In prokaryotes, the ribosome binding site (RBS), which promotes efficient and accurate translation of mRNA, is called the Shine-Dalgarno sequence after the scientists that first described it. Answer (C)
94. Topic: DNA Sequencing Chain-terminator method is a technique for determining the nucleotide sequence of a DNA using dideoxy nucleotides so as to yield a collection of daughter strands of all different lengths. It is also called as the dideoxy method. The DNA to be sequenced is incubated with DNA polymerase I, a suitable primer, and the four dNTP substrates (reactants in enzymatic reactions) for the polymerization reaction. The key component of the reaction mixture is a small amount of a 2′,3′-dideoxynucleoside triphosphate (ddNTP), which lacks the 3′-OH group of deoxynucleotides. When the dideoxy analog is incorporated into the growing polynucleotide in place of the corresponding normal nucleotide, chain growth is terminated because addition of the next nucleotide requires a free 3′-OH group. By using only a small amount of the ddNTP (approx. ddNTP:dNTP 1:100 ), a series of truncated chains is generated, each of which ends with the dideoxy analog at one of the positions occupied by the corresponding base. Answer (B) 95. Topic: Polymerase Chain Reactions Primer is a short polynucleotide sequence that is complementary to the 3′ end of the template DNA, and thus becomes the 5′ end of the new strand. Thus, (P) 5′- GACCTGTGG - 3′ which is complimentary to the 3′ end of lower strand and (S) 5′- ATCCCGTAT - 3′ which is complimentary to the 3′ end of upper stand will act as primer. They will anneal to 3′ ends of the two template strands following the antiparallel nature of DNA duplex. Answer (B) 96. Topic: Gene Isolation, Cloning and Expression Transformation efficiency can be calculated as follows:
Transmission efficiency No.of colonies Total volume × Dilution factor = × Wt.of DNA Sample Volume Given, Weight of plasmid DNA = 0.5 µg Volume of competent cell, i.e., total volume = 200 µL
Dilution factor = 104 Sample volume = 50 µL Number of colonies obtained = 35 Putting these values in the formula above we get: Transformation efficiency per µg =
35 200μL × × 10 4 0.5μg 50μL
= 2.8 × 106 cfu ⋅µg–1 Answer (2.8 × 106 cfu ⋅µg–1) 97. Topic: DNA Sequencing • Cyanogen bromide (CNBr) reacts specifically with Met residues in a polypeptide, resulting in cleavage of the peptide bond on their C-terminal side. The newly formed C-terminal residue forms a cyclic structure known as a peptidyl homoserine lactone. • o-Iodosobenzoate in a polypeptide attacks on the carboxyl side of the tryptophan. • Hydroxylamine in a polypeptide cleaves the peptide bond between asparagine and glycine. • 2-Nitro-5-thiocyanobenzoate attacks on the amino side of cysteine in a polypeptide. Answer (A) 98. Restriction and Modification Enzymes If a DNA sequence is evenly made up of G, A, T, and C nucleotides (i.e. 25% of each), the probability of getting the sequence “AGCT” by chance would be: 1 1 1 1 1 × × × = 4 4 4 4 256 That means about every 256 nucleotides on an average. On digestion of circular DNA of 4.3 × 105 by AluI, chances of getting restriction site is 1/256. Thus, average number of fragments generated would be
1 × 4.3 × 105 = 1.6 × 103 256 Answer (1.6 × 103) 99. Topic: Southern and Northern Blotting The DNA was digested completely with both enzymes, thus every single restriction site will be cut. BamHI and HaeIII, BamHI produces sticky cuts at cleavage site (G↓GATCC) and HaeI produces blunt ends at cleavage site (GG↓CC). In end-filling reaction, Klenow fragment will fill-in 5′ overhangs with [α32P]-dCTP of BamHI cleaved sites to form blunt ends and remove 3′ overhangs to form blunt ends. Thus, only BamHI cleavage restriction fragments will be labelled. As in the autoradiograph, two bands each at 3kb and 2kb are seen. Thus, the restriction map of the DNA in accordance to the result would be:
2 kb
dicotyledonous plants. The crown gall disease is caused due to the presence of Ti (tumor inducing) plasmid within the bacterial cell. It carries genes that help in infection. After the attachment of the bacteria to the host plant, the vir genes are induced and the T-DNA is transferred in to the cells of the host. This process is believed to be similar to bacterial conjugation where the T-DNA is transferred as single-stranded molecule and gets integrated in the plant chromosomal DNA. It has also been seen that the transfer process always has the 5′ end of the right border sequence of the T-DNA as the leading end followed by other components of the T-DNA. Answer (B)
HaeIII
(A)
BamHI
BamHI
158 GATE Biotechnology Chapter-wise Solved Papers
2 kb
3 kb
1 kb
Answer (A) 100. Topic: Gene Isolation, Cloning and Expression The lac operon is an example of an inducible operon which induces transcription of structural genes in the presence of lactose or galactose and hence it is called an inducer. In other words, the presence of substrate induces enzyme synthesis. The repressor proteins bind to the DNA only in the absence of lactose. Glucose and galactose cannot induce the lac operon hence lac operon will be expressed as long as lactose is present. At very low level, lac operon operates all the time. Otherwise there would be no permease to help lactose enter the cell. Repressors exert their influence by negative control. However, a positive control also exists which was discovered during investigation of glucose effect. The concentration of cAMP is dependent on the concentration of glucose in the medium. If glucose is low, cAMP is high. cAMP acts by binding to a protein and this cAMP-CAP complex recognize and binds to lac control region. The DNA conformation is changed and RNA polymerase is able to transcribe the lac operon. The presence of bound cAMP-CAP complex is necessary for transcription of operon even when lactose is present, and repressor is inactivated. Hence even when lactose is very low, lac operon is still operating and since glucose is also very low, cAMP is present in high concentration and cAMP-CAP complex binds to its specific site and exerts positive control. Once inside the cell, lactose acts as an inducer and induces the operon Answer (D)
102. Topic: Plant Expression Vector •
•
•
•
101. Topic: Plant Expression Vector Agrobacterium tumefaciens is a soil bacterium that causes crown gall disease in many species of
biquitin is a protein found in eukaryotic cells and U helps in processes such as protein turnover, chromatin structure, cell cycle control, DNA repair, and response to heat shock and other stresses. In 1992, Christensen et al. identified two out of the 8 to 10 loci encoding ubiquitin in Zea mays. The function of NOS terminator is to signal the termination of the herbicide tolerant gene expression. Termination sequence of the nopaline synthase gene is isolated from Agrobacterium tumefaciens. The bar gene was originally cloned from Streptomyces hygroscopicus, an organism which produces the tripeptide bialaphos as a secondary metabolite. The β -glucuronidase (GUS) gene is the most frequently used reporter gene in plants. GUS activity is found in many bacterial species, is common in all tissues of vertebrates and is also present in organisms of various invertebrate taxa. The transgenic GUS originates from the enterobacterial species Escherichia coli that are widespread in the vertebrate intestine, and in soil and water ecosystems. Answer (D)
103. Topic: Restriction and Modification Enzymes Xhol 700 bp 500 bp Xhol 300 bp Vector backbone 2.8 kb 2500 bp
EcoRI
BamHI
BamHI EcoRI BamHI
Xhol
EcoRI
500 bp 700 bp Insert 1.2 kb
300 bp Xhol
Recombinant protein 4 kb (4000 bp)
2500 bp
Chapter 9 • Recombinant DNA Technology 159
On insertion of 1.2kb DNA fragment in vector of 2.8kb, a recombinant vector of 4kb will be formed as shown in figure. Now, it has two XhoI restriction sites, one EcoRI site and one BamHI site. On complete digestion with XhoI and EcoRI, three fragments could be seen on analysis of 1% agarose gel electrophoresis. • One 2500bp fragment, with XhoI cut on one side and EcoRI cut on the other side. • One 800bp fragment, with XhoI cut on both sides. • One 700bp fragment, with XhoI cut on one side and EcoRI cut on the other side. Answer (C) 104. Topic: Southern and Northern Blotting Southern blot hybridization was developed by M. Southern in 1975 in which the locations of genes and other DNA sequences can be identified on restriction fragments separated by gel electrophoresis. Answer (B) 105. Topic: Polymerase Chain Reactions The polymerase chain reaction (PCR) is a technique in which a short region of a DNA molecule is copied many times in vitro in presence of DNA polymerase enzyme. For any region of DNA to be amplified, the sequences at the borders of the region should be known. This is required because in order to carry out a PCR, two short oligonucleotides (these act as primers and delimit the region to be amplified) should hybridize to a DNA molecule (that serves as a template), one to each strand of the double helix. Primer is a small segment of DNA, which binds to a complementary strand of DNA. Primers are necessary to start the amplification of DNA using DNA polymerase
enzyme and therefore are necessary in polymerase chain reaction (PCR). Answer (C) 106. Topic: Gene Isolation, Cloning and Expression Lac operon is an inducible prokaryote operon wherein, the presence of lactose induces the transcription of structural genes. The repressor proteins bind to the DNA only in the absence of lactose. Repressors exert negative control while ‘glucose effect’ exerts positive control on lac operon. When glucose is supplied with other substances such as lactose and galactose, the bacterial cells use glucose, ignoring other compounds. Glucose suppresses the production of catabolic enzymes such as β -galactosidase. Glucose tends to lower cAMP concentration in the cell. As long as glucose is present in abundance, cAMP levels remain lower than required to promote transcription of operon. The lac promoter of E. coli has been extensively used for the induced expression not only of the homologous structural genes in the lac operon but also of a variety of foreign genes cloned in vectors (see below). Expression of the genes cloned downstream from the lac promoter can be conveniently induced by adding IPTG. However, in the absence of an inducer, there is a basal level of expression from this promoter. This inherent “leakiness,” which is particularly bothersome when the product of the cloned gene is toxic to the cells, can be virtually eliminated 100% by growing the cells (e.g., transformed with pUC plasmids) in the presence of 0.4% glucose. Under this concentration of glucose, even the induction by IPTG is strongly repressed due to the phenomenon of glucose repression and inducer exclusion. Answer (C)
107. Topic: Restriction and Modification Enzymes BamHI
0.3 kb
BamHI XhoI 0.3 kb 0.7 kb 1 kb insert BamHI XhoI 0.3 kb 0.7 kb 1 kb insert
BamHI BamHI BamHI
0.7 kb
0.7 kb
XhoI 5 kb plasmid vector
XhoI
BamHI
0.3 kb
7 kb recombinant vector
EcoRI
5 kb EcoRI
BamHI
160 GATE Biotechnology Chapter-wise Solved Papers On insertion of two tandem copies of the insert, a total of 7kb recombinant vector will be formed as shown in figure. Now, it has two XhoI restriction sites and three BamHI sites. • On digestion with XhoI, two fragments one with 6kb and one with 1kb will be formed. • On digestion with BamHI, three fragments one with 5kb and two with 1kb each will be formed. Answer (C) 108. Topic: Restriction and Modification Enzymes Type II restriction enzyme recognize a single sequence of bases in dsDNA that are usually palindromic. A palindromic sequence is a sequence made up of nucleic acids within double helix of DNA or RNA that is the same when read from 5′ to 3′ on one strand and 5′ to 3′ on the other, complementary strand. Of the given options, only option (D) is non-palindromic, thus it cannot be recognition sequence for type II restriction enzyme. Answer (D) 109. Topic: Restriction and Modification Enzymes Given, its partial means incomplete digestion of DNA, thus not all sites would have been targeted by restriction enzyme. Thus, there is possibility of generation of all probable fragments. 10 kb 8 kb 2 kb
2 kb 0 kb
5 kb
2 kb 4 kb
7 kb
4 kb
10 kb
3 kb 6 kb 7 kb
Answer (A)
110. Topic: In situ Hybridization • In situ hybridization (ISH) can be used to detect and locate a specific sequence of DNA on the chromosome. The ability of DNA helix to renature is the basis for molecular hybridization. In this process, the DNA in the chromosome is denatured and is allowed to hybridize with a labeled (radioactive isotope or a fluorescent dye) complementary strand of DNA/ RNA (known as probe). After hybridization has occurred, the labeled strand or probe in the chromosome is located by autoradiography if the probe is radioactively labelled or by immunocytochemistry for fluorescently labeled probe. The location of the sequence of interest can thus be detected in the chromosome as the labeled probe reveals the actual location of the sequence in the cells.
•
•
•
orthern blotting is a technique by which RNA N molecule can be separated by electrophoresis and identified with a labeled DNA probe after being blotted onto a membrane. Southern blotting is a technique by which DNA molecules are identified with a labeled single-stranded DNA (or RNA) probe. Western blotting is a technique to identify specific proteins from a complex mixture. Answer (B)
111. Topic: Restriction and Modification Enzymes Restriction enzyme can recognize 6 bp sequence, e.g. Eco RI can recognize a 6 bp sequence of GAATTC. DNA contains 4 nucleotide bases viz, A, G, C, T. Thus, the probability of getting a particular 6 base pair sequence is: 1 1 1 1 1 1 1 × × × × × = . 4 4 4 4 4 4 4096 Thus, the same sequence can be found after 4096 bp. Thus, average length of the fragment generated would 4096 bp. Answer (4096 bp) 112. Topic: Restriction and Modification Enzymes EcoRI is a sticky end enzyme and makes cut between guanine and adenine and produce flanking site of 5 base pairs (AATTC), while HaeIII is a blunt end enzyme that makes cut between guanine and cytosine and produce blunt ends. As shown in diagram, these pieces have been generated: (a) 5kb (Both blunt end cut) GG……………………CC CC……………………GG (b) 3kb (One blunt end cut and one sticky end cut) GG…………….G CC……………..CTTAA (c) 1kb (One blunt end cut and one sticky end cut) AATTC…..CC G….GG Klenow fragment requires a free 3′–OH end and a template to read. On incorporation of Klenow fragment, Taq DNA polymerase, radioactive labeled ATP and rest normal NTP to the restricted pieces, radioactive adenine will base pair with thymine at the flanking site of only 3kb and 1kb pieces as 5kb piece has blunt end which cannot be read. GG…………….GAATT CC……………..CTTAA AATTC…..CC TTAAG….GG
Chapter 9 • Recombinant DNA Technology 161
Thus, on autobiography, only two bands will be seen on the X-ray film. Answer (2) 113. Topic: Polymerase Chain Reactions; DNA Fingerprinting; Southern and Northern Blotting A DNA array can be considered as a large parallel Southern or Northern blot analysis (instead of a gel, the probes are attached to an inert surface, which will become the microarray). mRNA is extracted from tissues or cells, reversed-transcribed and labeled with a dye (usually fluorescent), and hybridized on the array. Hybridization and washes are performed under high stringency conditions to minimize the likelihood of cross-hybridization between similar genes. The next step is to generate an image using laser-induced fluorescent imaging. The principle behind the quantification of expression levels is that the amount of fluorescence measured at each sequence-specific location is directly proportional to the amount of mRNA with complementary sequence present in the sample analyzed. These experiments do not provide data on the absolute level of expression of a particular gene (true concentrations of mRNA) but are useful to compare the expression level among conditions and genes. DNA microarrays can simultaneously measure the expression level of thousands of genes within a particular mRNA sample. Such high-throughput expression profiling can be used to compare the level of gene transcription. Answer (B) 114. Topic: Site-Directed Mutagenesis Enzymes such as DpnI require methylation at their recognition sites in order to efficiently cleave DNA. DpnI is often used for site directed mutagenesis. During this process, incorporation of a desired mutation into the plasmid of interest by PCR generates mutated plasmids
with no methylation (there are no methyltransferases in the PCR reaction). The template plasmid, on the other hand, should be derived from a dam+ E. coli strain and will therefore have methylated adenines in any GATC sequences found in the plasmid. When the PCR products are digested with DpnI, only the non-mutated and methylated template is destroyed leaving behind a pool of mutated plasmids which can later be verified by Sanger sequencing. Answer (B) 115. Topic: cDNA and Genomic DNA Library We know that, 100 kDa of protein = 2.7 kb of DNA Thus, 22 kDa of protein =
2.7 × 22 = 0.594 kb 0.6 kb 100 Answer (0.6 kb)
116. Topic: Polymerase Chain Reactions The proofreading capability of a DNA polymerase defines fidelity, which increases the accuracy of DNA sequence replication. High-fidelity DNA polymerases are enzymes with strong proofreading activity. PCR fidelity is a result of a complex process which is affected by many factors, including the enzyme, buffer conditions, thermal cycling parameters, and number of cycles. High dNTP concentrations increase error rate by driving the reaction in the direction of DNA synthesis and by decreasing error discrimination at the extension step. A reduction in dNTP and Mg2+ concentration leads to improvements in fidelity. In addition, post-replication DNA damage-induced errors at elevated temperatures can contribute to infidelity. Answer (D)
C H A P T E R 10
Plant Biotechnology
Syllabus 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 by Plant Suspension Cultures; Hairy Root Culture; Transgenic Plants; Plant Products of Industrial Importance.
Chapter Analysis Topic
GATE GATE GATE GATE GATE GATE GATE GATE GATE GATE 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019
Totipotency Regeneration of Plants
1
Plant Growth Regulators and Elicitors Tissue Culture and Cell Suspension Culture System: Methodology, Kinetics of Growth and Nutrient Optimization Production of Secondary Metabolites by Plant Suspension Cultures Hairy Root Culture
1 1
1 1
2
2 1
2
Transgenic Plants
1
1
1
Plant Products of Industrial Importance
1
Questions 1. 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. (GATE 2000; 1 Mark)
5. Resistance to herbicide chlorsulfuron in plants is due to change in (A) (B) (C) (D)
glutamine synthetase. acetolactate synthase theronine deaminase. DNA polymerase.
2. What is somatic embryogenesis? (GATE 2000; 1 Mark) 3. What is the difference between direct and indirect somatic embryogenesis? (GATE 2000; 2 Marks) 4. State two methods for direct DNA transfer into plant cells. (GATE 2000; 2 Marks)
(GATE 2001; 1 Mark) 6. Hormone pairs required for a callus to differentiate are (A) (B) (C) (D)
auxin and cytokinin. auxin and gibberellin. ethylene and gibberellin cytokinin and gibberellin. (GATE 2001; 1 Mark)
164 GATE Biotechnology Chapter-wise Solved Papers 7. 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. (GATE 2001; 1 Mark) 8. Callus formation from mature tissue explant occurs through (A) dedifferentiation. (B) redifferentiation. (C) both (A) and (B) of the above. (D) Neither (A) nor (B). (GATE 2002; 1 Mark) 9. Large scale clonal propagation practically means raising a population of plantlets from (A) a single cell. (B) a single explant. (C) many explants from a single plant. (D) many explants from a group of plants. (GATE 2002; 1 Mark) 10. Plant secondary metabolites production in suspension culture is mainly targeted for (A) obtaining metabolites in aseptic condition. (B) enhanced in vitro production of desired metabolite. (C) enhanced production of all metabolites. (D) obtaining new metabolites. (GATE 2002; 1 Mark) 11. Stable transformation of plants is reliably obtained by (A) Agrobacterium plasmid integration. (B) electroporation. (C) microinjection. (D) silicon carbide whisker. (GATE 2002; 1 Mark) 12. Which of the following statements is most appropriate for recombinant antibody production in transgenic plants? (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. (GATE 2002; 1 Mark)
13. Enhanced axillary branching for multiple shoot production is promoted by (A) 2,4-D (B) abscisic acid. (C) gibberellic acid. (D) benzyl adenine. (GATE 2002; 1 Mark) 14. Mention the specific role of acetosyringone in Agrobacterium mediated plant transformation. (GATE 2002; 1 Mark) 15. For protoplast fusion to be successful in plant cells (A) fusion agents other than polyethylene glycol should be used. (B) cell wall of the two strains of cells should not be damaged. (C) DNA between the two cells should be compatible. (D) osmolarity of the medium is not important. (GATE 2003; 1 Mark) 16. The following are useful to introduce genes into crop plants except (A) Ti plasmid. (B) particle gun. (C) breeding. (D) auxin. (GATE 2003; 2 Marks) 17. 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 (GATE 2004; 1 Mark) 18. Tobacco leaf discs are transferred with Agrobacterium tumefaciens strain containing binary vector (GUS as reporter gene) with selectable marker neo (kanamycin resistance gene) and then regenerated to plants. The plants are kanamycin resistant, but leaf tissues are negative to GUS assay. The explanation is (A) the plants are transformed for both genes but 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 the above. (GATE 2004; 2 Marks) 19. Expression of antisense RNA of ACC synthase in transgenic tomato plants inhibited the synthesis of ethylene resulting in
Chapter 10 • Plant Biotechnology 165
(A) (B) (C) (D)
change in color from green to red. change in aroma. charge in color from red to green. none of the above. (GATE 2004; 2 Marks)
20. What would he 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. (B) Stimulate dedifferentiation and thereby decrease secondary metabolite production. (C) Stimulate proliferation and reduce secondary metabolite production. (D) None of the above. (GATE 2004; 2 Marks) 21. 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. (C) these exhibits strong wound response. (D) these have long tap root system. (GATE 2004; 2 Marks) 22. Cells of meristemoid are best described as (A) differentiated and non-dividing. (B) dedifferentiated and dividing. (C) differentiated and dividing. (D) dedifferentiated and non-dividing. (GATE 2005; 1 Mark) 23. 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. (GATE 2005; 1 Mark) 24. Somatic embryo from cotyledon explant would develop in the following sequential stages. (A) Cotyledonary → Heart → Globular → Torpedo (B) Globular → Torpedo → Heart → Cotyledonary (C) Globular → Heart → Torpedo → Cotyledonary (D) Cotyledonary → Globular → Heart → Torpedo (GATE 2005; 2 Marks)
25. When present in tissue culture medium, gibberellin (A) helps to break dormancy of buds and bulbs. (B) promotes dormancy development in buds and bulbs. (C) is regarded as plant growth inhibitor. (D) prevents normal recognition of auxin molecule. (GATE 2006; 1 Mark) 26. Identify the natural plant growth regulators from the following list. P. Zeatin Q. Benzylaminopurine (BAP) R. Indole-3-acetic acid (IAA) S. 2, 4-Dichlorophenoxyacetic acid (A) P, Q (B) Q, S (C) P, R (D) R, S (GATE 2006; 1 Mark) 27. Majority of the cereals are highly recalcitrant to Agrobacterium mediated transformation, and so direct transformation 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. (B) Liposome-mediated transformation. (C) Electroporation. (D) Transformation using micro projectiles. (GATE 2006; 2 Marks) 28. 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. Callus induction using 20 μM 2,4-dichlorophenoxyacetic acid followed by shoot organogenesis. S. Shoot organogenesis from an explant in the absence of an intermediate callus phase. (A) P, Q (B) Q, R (C) P, S (D) Q, S (GATE 2006; 2 Marks) 29. Gynogenesis is a process of development of haploid plants (A) from a fertilized cell of female gametophyte. (B) from an unfertilized cell of female gametophyte. (C) from isolated pollen grains. (D) by selective elimination of chromosomes following distant hybridization. (GATE 2007; 2 Marks)
166 GATE Biotechnology Chapter-wise Solved Papers 30. Meristems escape virus invasion because (A) vascular system is absent in the meristem. (B) of low metabolic activity in the meristem. (C) the ‘virus inactivating system’ has low activity in the meristem. (D) of low endogenous auxin level. (GATE 2007; 2 Marks) 31. Parthenogenetic embryos in plants are those which are formed by (A) unfertilized eggs. (B) fertilized eggs. (C) sporophytic cells. (D) male gametophyte. (GATE 2008; 1 Mark) 32. Which one of the following is the growth factor used for growth of tissues and organs in plant tissue culture? (A) Cysteine (C) Cytidylate
(B) Cytokinin (D) cyclic AMP (GATE 2008; 1 Mark)
33. 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. (GATE 2008; 1 Mark) 34. Determine the correctness or otherwise of the following Assertion (A) and the 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 of two cells must occur through the plasma membrane. (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. (GATE 2008; 2 Marks) 35. 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. (GATE 2008; 2 Marks) 36. 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 meristem culture. (GATE 2009; 1 Mark) 37. 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. (GATE 2009; 1 Mark) 38. Which of the following statements are true about glyphosate tolerant transgenic plants? P. Transgenic plants detoxify glyphosate. Q. Transgenic plants produce an altered enzyme that is not affected by glyphosate. 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 (GATE 2009; 2 Marks) 39. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion: Somatic embryogenesis in plants is a twostep process comprising of embryo initiation followed by embryo production. Reason: Embryo initiation is independent of the presence of 2,4-dichlorophenoxyacetic acid, whereas embryo production requires a high concentration of 2,4-dichlorophenoxyacetic acid. (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. (GATE 2010; 2 Marks) 40. Match the promoters Listed in Group I with the tissues listed in Group II Group I Group II 1. Endosperm P. α-Amylase Q. Glutenin 2. Tuber R. Phaseollin 3. Aleurone S. Patatin 4. Cotyledon
Chapter 10 • Plant Biotechnology 167
(A) (B) (C) (D)
45. Match the herbicides in Group I with the target enzymes in Group II.
P–3; Q–1; R–4; S–2 P–3; Q–4; R–1; S–2 P–4; Q–2; R–1; S–3 P–1; Q–3; R–2; S–4
Group I Group II (GATE 2010; 2 Marks)
41. Shoot organogenesis by tissue culture results into (A) a bipolar structure that has no vascular connection with the explant. (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. (GATE 2011; 1 Mark) 42. Which one of the following is not a protoplast fusion inducing agent? (A) Inactivated Sendai virus (B) Ca2+ at alkaline pH (C) Polyethylene glycol (D) Colchicine (GATE 2012; 1 Mark) 43. 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 responsible 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. (GATE 2012; 2 Marks) 44. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: In direct somatic embryogenesis, 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). (GATE 2012; 2 Marks)
P. Glyphosate Q. Bromoxynil R. Sulphonylureas S. Dalapon (A) (B) (C) (D)
1. Nitrilase 2. Acetolactatesynthetase 3. Dehalogenase 4. 5-Enolpyruvyl shikimate 3-phosphate synthase
P–4; Q–1; R–2; S–3 P–2; Q–1; R–4; S–3 P–4; Q–3; R–2; S–1 P–3; Q–2; R–4; S–1 (GATE 2013; 2 Marks)
46. Match the following plant sources with their secondary metabolites and medical uses. Source plant Secondary metabolites Medical use P. Belladonna 1. Menthol a. Cancer treatment Q. Foxglove
2. Atropine
b. Heart disease
R. Pacific yew
3. Digitalin
c. Eye examination
S. Eucalyptus
4. Taxol
d. Cough
(A) (B) (C) (D)
P–2–c; Q–3–b; R–4–a; S–1–d P–3–c; Q–2–a; R–1–d; S–4–b P–2–c; Q–4–b; R–1–a; S–3–d P–1–b; Q–4–c; R–2–d; S–3–a (GATE 2014; 2 Marks)
47. 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 (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. (D) the pungent chemicals are formed only after the reaction with atmospheric carbon dioxide. (GATE 2014; 2 Marks) 48. Which one of the following is a second generation genetically engineered crop? (A) Bt brinjal (B) Roundup soyabean (C) Golden rice (D) Bt rice (GATE 2015; 1 Mark)
168 GATE Biotechnology Chapter-wise Solved Papers 49. Match the secondary metabolites (Group I) with the corresponding plant species (Group II). Group I Group II P. Morphine 1. Datura stramonium Q. Pyrethrins 2. Catharanthus roseus R. Scopolamine 3. Papaver somniferum S. Vincristine 4. Tagetes erecta (A) (B) (C) (D)
P–4; Q–3; R–1; S–2 P–3; Q–4; R–1; S–2 P–2; Q–3; R–4; S–1 P–4; Q–1; R–2; S–3
(C) P–3; Q–l; R–2; S–4 (D) P–2; Q–l; R–4; S–3 (GATE 2017; 2 Marks) 52. Which one of the following plant secondary metabolites is a natural insecticide? (A) Digitoxin (B) Pyrethrin (C) Salicylic acid (D) Avenacin A-1 (GATE 2018; 1 Mark)
(GATE 2016; 2 Marks) 50. The plant hormone indole-3-acetic acid is derived from (A) histidine. (B) tyrosine. (C) tryptophan. (D) proline. (GATE 2017; 1 Mark) 51. Match the plant hormones in Group I with functions in Group II. Group I Group II P. Gibberellic acid 1. Seed and bud dormancy Q. Zeatin 2. Fruit ripening R. Ethylene 3. Delaying leaf senescence S. Abscisic acid 4. Regulation of plant height (A) P–4; Q–3; R–2; S–l (B) P–4; Q–2; R–3; S–l
53. The Bt toxin gene from Bacillus thuringiensis used to generate genetically modified crops is (A) cry (B) cro (C) cdc (D) cre (GATE 2019; 1 Mark) 54. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion: It is possible to regenerate a whole plant from a single plant cell. Reason: It is easier to introduce transgenes into plants than animals. (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. (GATE 2019; 2 Marks)
ANSWER KEY 1. (C)
2. (*)
3. (*)
4. (*)
5. (B)
6. (A)
7. (D)
8. (C)
9. (C)
10. (B)
11. (A)
12. (C)
13. (A)
14. (*)
15. (C)
16. (D)
17. (B)
18. (D)
19. (D)
20. (D)
21. (C)
22. (C)
23. (D)
24. (C)
25. (A)
26. (C)
27. (B)
28. (B)
29. (B)
30. (A)
31. (A)
32. (B)
33. (D)
34. (B)
35. (C)
36. (C)
37. (C)
38. (C)
39. (C)
40. (A)
41. (B)
42. (D)
43. (B)
44. (D)
45. (A)
46. (A)
47. (B)
48. (C)
49. (B)
50. (C)
51. (A)
52. (B)
53. (A)
54. (B)
Answers with Explanation 1. Topic: Regeneration of Plants Plant breeding is the purposeful manipulation of plant species in order to create desired genotypes and phenotypes for specific purposes. Classical plant breeding uses deliberate interbreeding (crossing) of closely
or distantly related individuals to produce new crop varieties or lines with desirable properties. Plants are crossbred to introduce traits/genes from one variety or line into a new genetic background. Clonal propagation refers to the process of asexual reproduction by
Chapter 10 • Plant Biotechnology 169
multiplication of genetically identical copies of individual plants. The term clone is used to represent a plant population derived from a single individual by asexual reproduction. It is not possible in case of animal breeding. Answer (C) 2. Topic: Tissue Culture and Cell Suspension Culture System Somatic embryogenesis is an artificial developmental process that restructures somatic cells toward the embryogenic pathway. Thus, an embryo or plant is derived form a single somatic cell. It forms the basis of cellular totipotency in higher plants. Answer (*) 3. Topic: Tissue Culture and Cell Suspension Culture System Direct regeneration (embryogenesis) means that embryos origin directly from explant tissue creating an identical clone. In indirect embryogenesis, regenerants develop through the stage of callus — unorganized and very unstable tissue. One of the consequences of indirect regeneration is the high chance of mutation and often significant genetic diversity observed among the regenerated organisms. Indirect somatic embryogenesis
Direct somatic embryogenesis
Leaf explant
Leaf explant
3 weeks Callus induction 3 weeks
6 weeks
Embryo formation 3 weeks Shoot regeneration
Shoot elongation
Shoot regeneration
2 weeks
Rooting
Shoot elongation
Rooting 4 weeks
Answer (*) 4. Topic: Transgenic Plants Several techniques for direct DNA delivery are available, ranging from uptake of DNA into isolated protoplasts mediated by chemical procedures or electroporation, to
injection and the use of high-velocity particles to introduce DNA into intact tissues. • Physical methods: Electroporation, microprojectile, microinjection, liposome fusion and silicon carbide fibers. • Chemical methods: Polyethylene glycol (PEG) mediated and Diethylaminoethyl (DEAE) dextran mediated. Answer (*) 5. Topic: Plant Growth Regulators and Elicitors Chlorsulfuron- or sulfometuron-resistant plants have been obtained by selection in cultures of haploid tobacco (Nicotiana tabacum) protoplasts and in mutagenized Arabidopsis thaliana seedlings. Resistance is due to an altered acetolactate synthase (ALS) that is much less sensitive to chlorsulfuron than the ALS from the susceptible (S) biotype. ALS is the first enzyme in branched-chain amino acid synthesis and the target of sulfonylurea and imidazolinone herbicides. Herbicide tolerance was accompanied also by a reduced sensitivity to the regulative feedback inhibition by leucine and valine, with concentrations causing 50% inhibition about 10 and 3-fold higher, respectively, than those found for the wild-type enzyme. Answer (B) 6. Topic: Plant Growth Regulators and Elicitors Plants develop unorganized cell masses like callus in response to various biotic and abiotic stimuli. The combination of two growth-promoting hormones, auxin and cytokinin, induces callus from plant explants in vitro. The balance between two plant hormones, auxin and cytokinin, determines the state of differentiation and dedifferentiation. A high ratio of auxin-to-cytokinin or cytokinin-to-auxin induces root and shoot regeneration, respectively. Answer (A) 7. Topic: Tissue Culture and Cell Suspension Culture System The term embryo rescue refers to a number of in vitro techniques whose purpose is to promote the development of an immature or weak embryo into a viable plant. Embryo rescue and culture is a technique that has found wide acceptance and utilization. Embryo culture techniques also are used to rescue embryos from early maturing fruit varieties, to hasten maturation in some species and to overcome dormancy requirements in others. Many interspecific hybrid plants can be produced by embryo culture. Utilizing this technique wider range of hybridization can be done with success. From haploid plants arising from crosses between distantly related plants viable plants can be produced by embryo culture. Answer (D)
170 GATE Biotechnology Chapter-wise Solved Papers 8. Topic: Tissue Culture and Cell Suspension Culture System Formation of the callus tissue is the outcome of cell expansion and cell division of the cells of the explants. During the formation of callus tissue, the explants lose its original characteristic. So, under the influence of exogenously supplied hormone, the explants are triggered off a growth sequence in which cell enlargement and cell division predominate to form an unorganized mass of cells. In plant tissue culture, inducing organogenesis is an important way to regenerate plants from the culture. Organogenesis in plant tissue culture involves two distinct phases: dedifferentiation and redifferentiation. • Dedifferentiation begins shortly after the isolation of the explant tissues with an acceleration of cell division and a consequent formation of a mass of undifferentiated cells (called callus). • Redifferentiation, also called budding in plant tissue culture, may begin any time after the first callus cell forms. In this, process of tissue called organ primordia is differentiated from a single or a group of callus cells. The organ primordia give rise to small meristems with cells densely filled with protoplasm and strikingly large nuclei. The development (or growth) of an organ is monopolar. Polarity of the longitudinal axis of the organizing growing points of the organs can be seen some time after the formation of meristem tissues. Different types of specialized cells again differentiate. The vascular system is formed connecting the new organs to their parent explants or callus mass. Answer (C) 9. Topic: Tissue Culture and Cell Suspension Culture System Clonal propagation refers to the process of asexual reproduction by multiplication of genetically identical copies of individual plants. The different techniques of single cell and protoplast culture from a single plant enable thousand plants to be derived within a short space and time. The products of these rapid vegetative propagation can also be regarded as clones when it is established that the cells they comprise are genetically identical. Answer (C) 10. Topic: Production of Secondary Metabolites by Plant Suspension Cultures Plant cell culture systems represent a potential source of valuable secondary metabolites which can be used as food additives (flavors, fragrances, and colorants), nutraceuticals, and pharmaceutical. In vitro production of secondary metabolite in plant cell suspension cultures has been reported from various medicinal plants, and bioreactors are the key step for their commercial production.
The evolving importance of the secondary metabolites has resulted in a high level of interest in the possibility of altering their production through improving cultivation technology. Answer (B) 11. Topic: Transgenic Plants Plant genetic engineering has become one of the most important molecular tools in the modern molecular breeding of crops. Over the last decade, significant progress has been made in the development of new and efficient transformation methods in plants. Despite a variety of available DNA delivery methods, Agrobacterium and biolistic-mediated transformation remain the two predominantly employed approaches. In particular, progress in Agrobacterium-mediated transformation of cereals and other recalcitrant dicot species has been quite remarkable. Natural vector Agrobacterium tumefaciens is preferred because they can cope with whole plant tissues such as roots and leaves, which are easier to handle, more stable and require less of the lengthy steps that are required for plant regeneration. Transgenes are integrated at random positions in the plant genome, but procedures for targeted integration become more and more efficient. Answer (A) 12. Topic: Production of Secondary Metabolites by Plant Suspension Cultures The productions of antibodies in plants are popularly referred as plantibodies and it is special challenge because the molecules must fold and assemble correctly to recognize the antigens. Transgenic plants have a number of distinct advantages over culture-based systems for antibody production. The main advantage is the anticipated cost savings, reflecting the large amount of biomass that can be produced in a short time with no need for specialized equipment or expensive media. Also, there are minimal risks of contamination with human pathogens. Sowing, growing, and harvesting can be carried out using traditional agricultural practices and unskilled labor. Moreover, scale-up can be achieved rapidly and inexpensively, simply by cultivating more land. Heterologous proteins accumulate to high levels in plant cells, and plant-derived antibodies are virtually indistinguishable from those produced by hybridoma cells. Protein synthesis, secretion, and folding, as well as post-translational modifications such as signal peptide cleavage, disulfidebond formation and the initial stages of glycosylation, are very similar in plant and animal cells. Only minor differences in glycan structure have been identified, such as the absence of terminal sialic acid residues and the presence of the plant-specific linkages α (1,3)-fucose, and β (1–2) xylose. Answer (C)
Chapter 10 • Plant Biotechnology 171
13. Topic: Plant Growth Regulators and Elicitors • 2,4-Dichlorophenoxyacetic acid (2,4-D) is used in plant cell cultures as a dedifferentiation (callus induction) hormone. It is classified as an auxin plant hormone derivative. It is involved in shoot branching, and it has been established that it controls the shoot tip apical dominance. • Abscisic acid (ABA) functions in many plant developmental processes, including seed and bud dormancy, the control of organ size and stomatal closure. • Gibberellic acid is a simple gibberellin, a pentacyclic diterpene acid promoting growth and elongation of cells. It affects decomposition of plants and helps plants grow if used in small amounts, but eventually plants develop tolerance to it. • 6-Benzyladenine, also called 6-benzylaminopurine, is a synthetic cytokinin that stimulates cell division in plants. Among other actions, it spurs plant growth, sets blossoms, and improves fruit quality. Answer (A) 14. Topic: Transgenic Plants Acetosyringone is a phenolic compound secreted by wounded plant tissue and is known to be a potent inducer of Agrobacterium vir genes. In the presence of low concentration of acetosyringone, sugars such as glucose and galactose act synergistically to induce the vir genes. Induction of the vir gene by acetosyringone is enhanced by opines and antagonized by a growth inhibitor from corn. Several reports suggest that acetosyringone pre-induction of Agrobacterium or inclusion of acetosyringone in the co-cultivation medium can enhance significantly Agrobacterium mediated transformation Answer (*) 15. Topic: Transgenic Plants The potential for generating novel genetic combinations remains high, however, with useful targets including the introgression of disease resistance from compatible wild species into cultivated lines. Protoplasts of different species, genera and even families were compatible when fused. A number of protoplast combinations (soybean + corn, soybean + pea, soybean + tobacco, carrot + barley, etc.) provided fusion products which underwent cell division and callus formation. Answer (C) 16. Topic: Transgenic Plants A variety of genetic engineering techniques are described below: • Agrobacterium is a naturally occurring genetic engineering agent and is responsible for the majority of GE plants in commercial production. It contains tumor inducing (Ti) plasmid, that can be modified with the gene of interest.
•
aked DNA could be delivered to plant cells by N “shooting” them with microscopic pellets or biolistic gene gun to which DNA had been adhered. These are crude, but effective physical method of DNA delivery. • Mutation breeding involves exposing plants or seeds to mutagenic agents (e.g., ionizing radiation) or chemical mutagens (e.g., ethyl methanesulfonate) to induce random changes in the DNA sequence. The progeny are assessed for phenotypic expression of potentially valuable new traits. Auxin is a plant growth hormone. It does not bring any change in the plant genome. Answer (D)
17. Topic: Tissue Culture and Cell Suspension Culture In plant tissue culture, the developmental pathway of numerous well-organized, small embryoids resembling the zygotic embryos from the embryo genic potential somatic plant cell of the callus tissue or cells of suspension culture is known as somatic embryogenesis. Somatic embryos usually do not mature properly. Instead, due to environmental factors such as constant contact with inducing medium, somatic embryos often deviate from the normal developmental pattern by bypassing embryo maturation producing callus, undergoing direct secondary embryogenesis and/or germinating precociously. Answer (B) 18. Topic: Transgenic Plants All the three conditions can be possible for the plants to show kanamycin resistant but being negative to GUS assay. • It can be possible that the plants are transformed for both the genes, but GUS gene has been turned off. • This can also be a possibility that the plants have been transformed for only neo gene and not the GUS gene. • It is also possible that the plants are not transformed at all, but the development of kanamycin resistance is due to somaclonal variation. Somaclonal variation may alter the gene expression to increase the synthesis of secondary metabolites. Answer (D) 19. Topic: Tissue Culture and Cell Suspension Culture Polygalacturonase (PG) is the major enzyme responsible for pectin disassembly in ripening fruit. Transgenic tomato (Lycopersicon esculentum) fruits in which endogenous ethylene production is suppressed by the expression of an antisense 1-aminocyclopropane-1-carboxylic acid (ACC) synthase gene are used to re-examine the role of ethylene in regulating the accumulation of PG mRNA, enzyme activity, and protein during fruit ripening. The color and aroma change accompanying ripening of transgenic antisense ACC synthase fruit is thus strongly inhibited. Answer (D)
172 GATE Biotechnology Chapter-wise Solved Papers 20. Topic: Production of Secondary Metabolites by Plant Suspension Cultures Effects of auxin and cytokinin on cell growth and alkaloid production in cell suspension cultures of Thalictrum minus were examined in an attempt to increase the productivity of a medicinal compound, berberine. In Linsmaier and Skoog medium containing auxin such as 2,4-D, 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. In particular, for the combination of NAA and 6-benzylaminopurine, the yield of berberine was as high as after 2 weeks of culture. Furthermore, most of the berberine produced by the cells was released into the liquid medium, in which an excess of berberine crystallized. The results of the present experiments are suggestive of an advantage in adopting a two-stage culture method for the production of berberine in fermenter systems. Answer (D) 21. Topic: Transgenic Plants The species capable of inducing crown galls on a large variety of dicotyledonous plants were called Agrobacterium tumefaciens, those inducing hairy root disease A. rhizogenes, and the non-pathogenic strains A. radiobacter. Wounding promotes T-DNA transfer, presumably due to induction of phenolic compounds produced during cell repair or throughout the formation of new cells. In reaction, a signal received by virA activates a cascade of other vir protein machinery gene. Answer (C)
Meristematic cell
Initiation of somatic embryo development
Early proembryo
22. Topic: Totipotency Meristemoids developed in the calli and each of them at maturity had peripheral layers of meristematic cells enclosing a centrifugally differentiating mass of tracheid-like cells. Cells of meristemoids were smaller, more densely cytoplasmic, and more compactly arranged compared with callus. As the meristemoids developed, further cell divisions became more organized so that separate regions of meristematic activity could be distinguished within each meristemoid. Answer (C) 23. Topic: Transgenic Plants 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. Plastid transformation is becoming more popular and an alternative to nuclear gene transformation because of various advantages like high protein levels, the feasibility of expressing multiple proteins from polycistronic mRNAs, and gene containment through the lack of pollen transmission. In most species, plastids are usually strictly maternally inherited in most (80%) angiosperm plant species. It is also not influenced by polyploidy, gene duplication and recombination that are widespread features of the nuclear genomes of plants Answer (D) 24. Topic: Tissue Culture and Cell Suspension Culture Somatic embryos generally originate from single cells which divide to form a group of meristematic cells. The cells of meristematic mass continue to divide to give rise to globular, heart shaped, torpedo and cotyeledonary stages.
Globular somatic embryo
Cuticle
Callus/explant
Cotyledonary stage somatic embryo
Heart-shaped somatic embryo
Callus/explant
Torpedo-shaped somatic embryo
Answer (C)
Chapter 10 • Plant Biotechnology 173
25. Topic: Plant Growth Regulators and Elicitors Gibberellins (GAs) are plant hormones that regulate various developmental processes, including stem elongation, germination, dormancy, flowering, flower development and leaf and fruit senescence. They stimulate cell elongation and cause plants to grow taller. Answer (A) 26. Topic: Plant Growth Regulators and Elicitors • Zeatin is a natural cytokinin derived from adenine, which occurs in the form of a cis- and a trans-isomer and conjugates. It promotes callus initiation when combined with auxin. It causes auxiliary stems to grow and flower. • 6-Benzylaminopurine, benzyl adenine or BAP is a first-generation synthetic cytokinin that elicits plant growth and development responses, setting blossoms and stimulating fruit richness by stimulating cell division. • Indole-3-acetic acid (A, 3-IAA) is the most common, naturally occurring, plant hormone of the auxin class. It induces cell elongation and cell division with all subsequent results for plant growth and development. • 2,4-D is an herbicide that kills plants by changing the way certain cells grow. It selectively kills most broadleaf weeds by causing uncontrolled growth in them. Answer (C) 27. Topic: Transgenic Plants Liposome-mediated transformation technique is used for delivering foreign DNA in the intact plant tissue genome. Cationic liposomes are positively charged lipids and are increasingly used for DNA uptake due to their favorable interactions with negatively charged DNA and cell membranes. In this approach foreign DNA must be encapsulated in a spherical lipid bilayer termed a liposome to prepare lipoplexes. After endocytosis, the DNA is free to recombine and integrate into the host genome. As with other transformation systems, a variety of vectors including viral vectors can be applied with this system. This method is relatively non-toxic, is simple to perform with readily available chemical reagents, is highly reproducible and efficient and requires no sophisticated equipment. Answer (B) 28. Topic: Tissue Culture and Cell Suspension Culture A variety of factors may contribute to the phenomenon of somaclonal variation. The system by which the regeneration is induced, type of tissue, explant source, media components and the duration of the culture cycle are some of the factors that are involved in inducing variation during in vitro culture.
Regeneration systems can be ranked in order from high to low in terms of genetic stability, as follows: micropropagation by preformed structures, such as shoot tips or nodal explants; adventitiously derived shoots; somatic embryogenesis; and organogenesis from callus, cell and protoplast cultures. Cellular organization is a critical factor for plant growth, whereas in vitro the loss of cellular control, which gives rise to disorganized growth, is a characteristic of somaclonal variation. Answer (B) 29. Topic: Tissue Culture and Cell Suspension Culture Gynogenesis is a special form of sexual reproduction in which insemination is necessary but the head of the sperm penetrating into the ovum does not transform into male pronucleus; and the gynogenetic embryo develops at the expense of the ovum nucleus only. Haploid plants can be developed from ovary or ovule cultures. It is possible to trigger female gametophytes (megaspores) of angiosperms to develop into a sporophyte. The plants so produced are referred to as gynogenic haploids. Answer (B) 30. Topic: Tissue Culture and Cell Suspension Culture In infected plants, the apical meristems are generally either free or carry a very low concentration of the viruses. In older tissues the virus titer increases with increasing distance from the meristem-tips. The reasons proposed for the escape of meristem from virus invasion are: • Viruses readily move in a plant body through the vascular system which is absent in the meristem. The alternative method of cell-to-cell movement of the virus through plasmodesmata is rather too slow to keep pace with the actively growing tip. • High metabolic activity in the actively dividing meristem cells does not allow virus replication. • The ‘virus inactivating systems’ in the plant body, if any, has higher activity in the meristem than in any other region. Thus, the meristem is protected from infection. • A high endogenous auxin level in shoot apices may inhibit virus multiplication. Answer (A) 31. Topic: Tissue Culture and Cell Suspension Culture Parthenogenesis can be defined as the production of an embryo from a female gamete without any genetic contribution from a male gamete, with or without the eventual development into an adult. Thus, parthenogenic embryo is derived from unfertilized female reproductive cell. Answer (A) 32. Topic: Plant Growth Regulators and Elicitors Plant tissue culture is defined as culturing plant seeds, organs, explants, tissues, cells, or protoplasts on a chemically defined synthetic nutrient media under sterile and
174 GATE Biotechnology Chapter-wise Solved Papers controlled conditions of light, temperature, and humidity. The composition of the medium, particularly the plant hormones and the nitrogen source (nitrate versus ammonium salts or amino acids) have profound effects on the morphology of the tissues that grow from the initial explant. For example, an excess of auxin will often result in a proliferation of roots, while an excess of cytokinin may yield shoots. Answer (B) 33. Topic: Tissue Culture and Cell Suspension Culture Clonal propagation refers to the process of asexual reproduction by multiplication of genetically identical copies of individual plants. In nature, clonal propagation occurs by apomixis (seed development without meiosis and fertilization) and/or vegetative reproduction (regeneration of new plants from vegetative parts). Answer (D) 34. Topic: Tissue Culture and Cell Suspension Culture Protoplasts are naked plant cells without the cell wall, but they possess plasma membrane and all other cellular components. They represent the functional plant cells but for the lack of the barrier, cell wall. Protoplasts of different species can be fused to generate a hybrid and this process is referred to as somatic hybridization (or protoplast fusion). Cybridization is the phenomenon of fusion of a normal protoplast with an enucleated (without nucleus) protoplast that results in the formation of a cybrid or cytoplast (cytoplasmic hybrids). The recombinant protoplast can then be induced to reform a cell wall, proliferate, form callus, and regenerate. This technique can be used to fuse protoplasts of the same species or of different species. Answer (B) 35. Topic: Totipotency • Totipotency in this strict sense is demonstrated by the ability of an isolated cell to produce a fertile, adult individual. Consequently, a cell that is totipotent is also a one-cell embryo; that is, a cell that is capable of generating a globally coordinated developmental sequence. • Morphogenesis is the biological process that causes an organism to develop its shape. • Androgenesis is the process of development of an embryo containing only paternal chromosomes due to failure of the egg to participate in fertilization. • Organogenesis is the process by which the three germ tissue layers of the embryo, which are the ectoderm, endoderm, and mesoderm, develop into the internal organs of the organism. Answer (C) 36. Topic: Tissue Culture and Cell Suspension Culture By doubling the chromosome of the haploid plants produced by pollen or anther culture completely
homozygous diploid fertile plants are produced. Chromosome number of haploids can be doubled by various methods: • In chemically induced doubling of the haploid plantlets, the plantlets are treated with colchicine. In Nicotiana tabacum plantlets are immersed in 0.4% solution of colchicine upto 96 hours. • Plantlets are transferred to culture medium after treatment. In mature plants 0.4% colchicine in lanoline paste is used on the upper axillary buds. Terminal bud is removed. This encourages the growth of the lateral buds. • In regeneration by tissue culture, diploids may be produced due to endomitosis in callus. Answer (C) 37. Topic: Tissue Culture and Cell Suspension Culture Cell-to-cell or long-distance transport of viruses in plants is brought about by movement proteins (MPs) or coat protein or helper component-proteinase (HC-Pro) protein encoded by viruses. Interfering with cell-to-cell or long-distance transport would thus be an ideal strategy for developing virus resistance. In order to achieve this goal, defective movement proteins have been expressed in transgenic plants. In contrast to other strategies, this approach offers attractive possibility to confer broad-spectrum resistance to related and unrelated viruses. For example, a defective TMV MP expressed in transgenic tobacco plants was shown to confer varying levels of resistance to a number of viruses that are not members of the tobamovirus group. Answer (C) 38. Topic: Transgenic Plants Glyphosate herbicide kills plants by blocking the EPSPS enzyme, an enzyme involved in the biosynthesis of aromatic amino acids, vitamins and many secondary plant metabolites. There are several ways by which crops can be modified to be glyphosate-tolerant. One strategy is to incorporate a soil bacterium gene that produces a glyphosate tolerant form of EPSPS. Another way is to incorporate a different soil bacterium gene that produces a glyphosate degrading enzyme. Answer (C) 39. Topic: Tissue Culture and Cell Suspension Culture The process of formation of embryos from somatic cells is known as somatic embryogenesis and the embryos so formed are known as somatic embryos. Somatic embryo formation takes place in two phases, induction phase and expression phase. In first phase, they acquire embryonic competence and proliferate as embryonic cell. In second phase, they differentiate into embryonic cells. Generally, these are induced from the explants grown on a high concentration of an auxin 2,4-D. Complete plants can be formed from mature somatic embryos. In plants
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which do not produce seeds or produce very little seeds, somatic embryos are encapsulated in gel matrix along with growth substances and antibiotics. These are known as artificial seeds. Answer (C) 40. Topic: Plant Growth Regulators and Elicitors • During germination, the embryo triggers starch degradation by releasing gibberellins, which stimulate cells of the aleurone layer to secrete degradative enzymes into the endosperm. A number of enzymes vital to starch degradation have been identified from this system. One of the most important enzymes is α -amylase, an endohydrolase that is able to rapidly degrade the starch into soluble substrates for other enzymes to attack. • Gluten is a highly molecular complex of proteins – gladins and glutenins – together with other components of the endosperm cells Gladins and glutenins are typical storage proteins of wheat. • Phaseolin is a vascular glycoprotein that accumulates as the major storage protein in cotyledons of the bean seeds. Its genes are induced by unidentified factors at the onset of seed maturation in embryos. It is synthesized as a precursor with a short C-terminal propeptide. • Patatin is a rich glycoprotein found in the tubers of the potato plants that constitute lipid acyl hydrolase activity. On fusion of the promoter of the patatin genes that are expressed in tubers with the reporter gene encoding beta-glucuronidase (GUS), it has been found that patatin transcription shows a high degree of tuber specificity. Answer (A) 41. Topic: Tissue Culture and Cell Suspension Culture System Whole plants are regenerated from culture via two different processes: • Somatic embryogenesis in which cells and tissues develop into a bipolar structure containing both root and shoot axes with a closed vascular system (the type of embryogenesis that occurs in a seed). • Organogenesis in which cells and tissues develop into a unipolar structure, namely a shoot or a root with the vascular system of this structure often connected to parent tissues. Answer (B) 42. Topic: Tissue Culture and Cell Suspension Culture System Induced protoplast fusion is carried by three procedures namely; mechanical, chemical, and electrical. Chemo-fusion can be done by using various chemicals such as calcium at high pH, sodium nitrate, polyethylene glycol (PEG), polyvinyl alcohol etc. In addition, recently it has
been found that Sendai virus has an ability to induce syncytia formation in in vitro eukaryotic colonies. Hence, correct answer is (D), i.e., colchicine. Answer (D) 43. Topic: Production of Secondary Metabolites by Plant Suspension Cultures Elicitors are compounds, which activate chemical defense in plants. As a major response of plants to biotic and abiotic stress, the accumulation of secondary metabolites in plant tissue cultures can be stimulated by the elicitors. Answer (B) 44. Topic: Tissue Culture and Cell Suspension Culture System Direct somatic embryogenesis is the development of somatic embryos directly from the explant deprived of the development of an intermediate callus phase. The development of a callus usually is not required, if embryogenesis is the goal of the research. Two known exceptions to direct embryogenesis are carrot and alfalfa. Answer (D) 45. Topic: Plant Growth Regulators and Elicitors • Glyphosate is a crop desiccant and herbicide. Glyphosate (N-(phosphonomethyl) glycine), targets the shikimic acid pathway by inhibiting EPSPS (5-Enolpyruvyl shikimate 3-phosphate synthase). • Bromoxynil is an organic compound. It is classified as a nitrile herbicide. It is a white solid. It works by inhibiting the process of photosynthesis. The nitrilase also known as the CN‐hydrolases, is comprised of enzymes that catalyse the hydrolysis of non‐peptide carbon–nitrogen bonds. In enzymology, a bromoxynil nitrilase is an enzyme that catalyzes the chemical reaction: 3,5-dibromo-4-hydroxybenzonitrile + 2 H2O → 3,5-dibromo-4-hydroxy-benzoate + NH3 • Acetohydroxyacid synthase (AHAS), also called acetolactate synthase (ALS), is the first enzyme catalyzing the synthesis of branched amino acids valine, leucine, and isoleucine, and sulfonylurea herbicides inhibit acetolactate synthase, the first enzyme specific to the branched chain amino acid biosynthetic pathway. Sulphonylureas is also used as medicines that control blood sugar levels in patients suffering from type 2 diabetes by stimulating the production of insulin in the pancreas. • Dalapon (2,2-dichloropropionate) is a selective herbicide is widely used against monocotyledonous plants. It is readily removed from the soil by a variety of micro-organisms including species of Bacillus. Bacteria able to utilize 2,2-dichloropropionate are known to release chloride ions from this herbicide and form pyruvate by their dehydrogenase activity. Answer (A)
176 GATE Biotechnology Chapter-wise Solved Papers 46. Topic: Production of Secondary Metabolites by Plant Suspension Cultures • The chemicals atropine and scopolamine, which are derived from belladonna, have important medicinal properties. Atropine and scopolamine have almost the same uses, but atropine is more effective at relaxing muscle spasms and regulating heart rate. It’s also used to dilate the pupils during an eye exam. Atropine can also be an antidote for insecticides and chemical warfare agents. • The botanical name for foxglove is Digitalis purpurea. An extract from this poisonous plant when formulated into a medication with a controlled dosage, digitalis is valuable in treating heart failure. It helps a weakened heart to pump harder. • National Cancer Institute (NCI)-funded researchers unlocked the Pacific yew tree’s potential to treat cancer and developed the lifesaving compound paclitaxel (Taxol). Taxol, an antimitotic agent used to treat cancer, blocks cancer cell growth by stopping cell division, resulting in cell death. • Camphor, eucalyptus and menthol are three essential oils that are used in complementary and alternative medicine to treat cough. Camphor oil is extracted from the leaves, roots or stems of the Cinnamomum camphora tree, while eucalyptus oil comes from the leaves of the Eucalyptus globulus plant. Menthol is extracted from the leaves of various mint plants, such as peppermint. Answer (A) 47. Topic: Production of Secondary Metabolites by Plant Suspension Cultures Plants of the Brassica genus contain special sulfur compounds called glucosinolates (e.g., sinigrin and sinalbin), which are very important secondary metabolites. These glucosinolates are inactive until they react with water or any moisture. At this point, they are hydrolyzed to isothiocyanates (resulting in a pungent, irritating odor and characteristic flavor), glucose, and potassium bisulfate. Mustard seed oil comprises mainly erucic, arachidic, α -linolenic, oleic, and palmitic acids. Answer (B) 48. Topic: Transgenic Plants Second generation of genetically modified organisms (GMOs), referred to as value-enhanced crops (VEC), includes those plant varieties that have one or more output characteristic modified adding end-user value to the commodity. It should have improved nutritional features and processing characteristics. Golden rice is a transgenic variety of rice (Oryza sativa) that is so named because of its yellow color grains due to the presence of large quantities of b-carotene (provitamin A or inactive state of vitamin A). Beta carotene gets
converted to vitamin A (retinol) which is vital for good vision and strong immune system. The rice thus has added nutritional benefits that are particularly valuable for the poor people of the world who depend on rice as a primary food. This strain of genetically modified rice was developed by Ingo Potrykus and Peter Beyer to overcome the deficiency of vitamin A which can lead to night blindness. Answer (C) 49. Topic: Production of Secondary Metabolites by Plant Suspension Cultures • Morphine is an opiate alkaloid isolated from the plant Papaver somniferum. It has widespread effects in the central nervous system and on smooth muscle. • The Pyrethrin naturally occurs in Tagetes erecta and Chrysanthemum cinerariifolium flowers and belongs to a class of organic compounds. It has insecticidal activity and affects the nervous systems of insects. • Scopolamine is an alkaloid obtained from Solanaceae family, especially Datura and Scopolia. Scopolamine and its quaternary derivatives act as antimuscarinics like atropine but may have more central nervous system effects. • Vincristine is a vinca alkaloid that can be obtained from the Catharanthus roseus. It is a chemotherapy medication used to treat a number of types of cancer. Answer (B) 50. Topic: Plant Growth Regulators and Elicitors The most common naturally occurring auxin is indole-3acetic acid (IAA). The IAA auxin is mainly synthesized from tryptophan; however, there are some plants which do not require tryptophan for IAA synthesis (e.g., Zea mays). Answer (C) 51. Topic: Plant Growth Regulators and Elicitors • The treatment with Gibberellic acid is used for the regulation of plant height. • Zeatin is the natural cytokinin found in corn-kernels and coconut milk and is used in delaying leaf senescence. • Ethylene is a simple gaseous plant growth hormone regulator that is synthesized mainly in response to stress and in tissues undergoing senescence or ripening (climacteric ripening of fruits). • Abscisic acid is a plant hormone that acts as a stress hormone, increases dormancy, promotes abscission, inhibits development of lateral or secondary roots, etc. Answer (A) 52. Topic: Plant Products of Industrial Importance • Digitoxin is a phytosteroid and is a liquid soluble cardiac glycoside eliminated via the liver.
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Pyrethrins are a mixture of six chemicals that are toxic to insects and are found naturally in chrysanthemum flowers. • Salicylic acid is a colorless crystalline organic acid commonly used in organic synthesis. It also functions as a plant hormone. • Avenacin A-1, a type of saponin, is found in oat plants. It protects the plant against fungal pathogens. Answer (B) 53. Topic: Transgenic Plants The cry gene has been isolated from the bacteria Bacillus thuringiensis and introduced into cotton and corn plants where it expresses itself to produce the toxin. Whenever any insect visits the plant, it gets killed. This helps in protecting the crop from these insects. Bt cotton is thus a genetically modified crop in which proteins encoded by the genes cryIAc and cryIIAb control bollworms. cryIAb and cryIAc are known to control corn borer in corn plants and lepidopteran insects in brinjals, respectively. Answer (A)
54. Topic: Regeneration of Plants There are several methods for introducing genes into plants, including infecting plant cells with plasmids as vectors carrying the desired gene or shooting microscopic pellets containing the gene directly into the cell. In contrast to animals, there is no real distinction between somatic cells and germline cells. Somatic tissues of plants, e.g., root cells can be grown in culture, can be transformed in the laboratory with the desired gene. If all goes well, the transgene will be incorporated into the pollen and eggs and passed on to the next generation. In this respect, it is easier to produce transgenic plants than transgenic animals. Plant tissue culture refers to the in vitro culture of plant parts on nutrient medium under aseptic conditions. These plant parts are grown under controlled aseptic conditions in vitro on a nutrient medium are known as explants. The ability to generate whole plants from cells or explants is called totipotency. Answer (B)
C H A P T E R 11
Animal Biotechnology
Syllabus Animal Cell Culture— Media Composition and Growth Conditions; Animal Cell and Tissue Preservation; Anchorage and Non-Anchorage Dependent Cell Culture; Kinetics of Cell Growth; Micro and Macro-Carrier Culture; Hybridoma Technology; Stem Cell Technology; Animal Cloning; Transgenic Animals.
Chapter Analysis Topic Animal Cell Culture: Media Composition and Growth Conditions Animal Cell and Tissue Preservation Anchorage and Non-Anchorage Dependent Cell Culture Kinetics of Cell Growth Micro and Macro-Carrier Culture Hybridoma Technology Stem Cell Technology Animal Cloning
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Questions 1. Which of the following has been produced commercially from mammalian cell cultures? (A) Plasminogen activator (B) Antibacterial antibiotics (C) Insulin (D) Renin (GATE 2000; 1 Mark) 2. Detection of which hormone is the commonly used test for pregnancy in humans? (A) LH (B) FSH (C) Chorionic gonadotropin (D) Estrogen (GATE 2000; 1 Mark)
3. What are bispecific monoclonal antibodies? (GATE 2000; 1 Mark) 4. Explain in one sentence why you cannot have monoclonal antibodies which can react with mast cells. (GATE 2000; 2 Marks) 5. How bispecific monoclonal antibodies can be generated? Give two methods (GATE 2000; 2 Marks) 6. 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) β -globin. (B) β -lactoglobulin. (C) preproinsulin. (D) lacZ. (GATE 2001; 1 Mark)
180 GATE Biotechnology Chapter-wise Solved Papers 7. Positional cloning approach exploits information (A) on the location of the gene in the genome. (B) on the status of its expression. (C) about the position of promoter relative to MCS. (D) about the position of the restriction sites. (GATE 2001; 1 Mark) 8. Why are mammalian cells cultured in CO2 incubators? (GATE 2001; 1 Mark) 9. Cells deficient in hypoxanthine guanine phosphoribosyltransferase (HPRT) enzyme rely on (A) synthesis of purine deoxynucleotides by salvage pathway. (B) synthesis of purine deoxynucleotides by de novo pathway. (C) supply of hypoxanthine in the culture medium. (D) supply of thymidine in the culture medium. (GATE 2002; 1 Mark)
16. What is the primary purpose of neomycin in creating mice with knockouts 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 cell 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. (GATE 2003; 2 Marks) 17. All of the following are 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. (GATE 2003; 2 Marks)
10. The following culture systems are used for growing large amount of anchorage dependent animal cells except (A) roller bottle. (B) airlift fermenter. (C) hollow fiber reactor. (D) microcarriers. (GATE 2002; 1 Mark)
18. 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-α. (GATE 2004; 1 Mark)
11. What is transgene?
19. 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? (A) Hold the pH at 5.6 (B) Hold the pH at 7.5 (C) Add calcium salt first and then phosphate source (D) None of the above (GATE 2004; 1 Mark)
(GATE 2002; 1 Mark) 12. What are the characteristics of ‘normal’ primary animal cells? (GATE 2002; 2 Marks) 13. Name different methods for the separation of different cell types from a mixed population of animal cells. (GATE 2002; 2 Marks) 14. 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 the above. (GATE 2003; 1 Mark) 15. An animal cell line was transected with DNA extracted from a tumorous tissue. Which one of the following will be diagnostic for its tumorous transformation? (A) Altered cell shape (B) Contact inhibition (C) Anchorage-independent cell division (D) Increased duration of cell cycle (GATE 2003; 2 Marks)
20. The culture fluids of 1000 to 5000 colonies of hybridoma are screened for monoclonal antibody by P. western blot analysis. Q. antigen capture analysis. R. northern blot analysis. S. antibody capture analysis. Choose the correct pair from the following (A) P, Q (B) Q, R (C) R, S (D) Q, S (GATE 2004; 2 Marks) 21. The number of replicons in a typical mammalian cell is (A) 40–200 (B) 400 (C) 1000–2000 (D) 50,000–100,000 (GATE 2005; 1 Mark)
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Common Data for Questions 22 and 23: Normal primary hepatocytes can be artificially immortalized. Certain spontaneous mutants of immortalized hepatocytes are sensitive to ionizing radiation. 22. Which of the following genes are involved in immortalization of primary hepatocytes? (A) Telomerase and Cyclin D (B) NFκ B and Thymidine kinase (C) Cyclin D and myc (D) Telomerase and Ras (GATE 2005; 2 Marks) 23. What would happen to the mutant cells by ionizing radiation? (A) Apoptosis (B) Necrosis (C) Cell growth arrest (D) Cell proliferation (GATE 2005; 2 Marks) 24. To promote attachment and spreading of anchoragedependent animal cells the surface of the culture vessel needs to be coated with (A) trypsin. (B) collagen. (C) pronase. (D) polyglycol. (GATE 2006; 1 Mark) 25. A hybrid derived from the fusion of a myeloma cell (HPRT–) with an antibody secreting B-lymphocyte (HPRT+) can be selected to produce monoclonal antibody by growing in a medium containing (A) thiamine, hypoxanthine, aminopterin. (B) thymidine, histidine, aminopterin. (C) uridine, hypoxanthine, aminopterin. (D) thymidine, hypoxanthine, aminopterin. (GATE 2006; 2 Marks) 26. When electroporation is used for introducing DNA into mammalian cell 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 cell becomes approximately zero percent. S. the first step involves absorption of DNA on the cell membrane. (A) P, Q (B) Q, R (C) P, S (D) Q, S (GATE 2006; 2 Marks) 27. Which of the following reagents is used for harvesting anchorage-dependent animal cells from culture vessels?
(A) Trypsin/Collagenase (C) Collagen/Fibronectin
(B) Trypsin/Collagen (D) DMSO (GATE 2007; 1 Mark)
28. Identify the correct statements 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 hypoxanthineguanine phosphoribosyl transferase. (A) P, Q (B) P, S (C) R, S (D) Q, S (GATE 2008; 2 Marks) 29. Hybridoma technology is used to produce (A) monoclonal antibodies. (B) polyclonal antibodies. (C) both monoclonal and polyclonal antibodies. (D) B cells. (GATE 2010; 1 Mark) 30. Somatic cell gene transfer is used for P. transgenic animal production. Q. transgenic diploid cell production. 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 (GATE 2010; 1 Mark) 31. Receptor R is over expressed in CHO cells and analyzed for expression. 6 × 107 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) (A) 104 (B) 105 (C) 106 (D) 107 (GATE 2010; 2 Marks) 32. Embryonic stem cells are derived from (A) fertilized embryo. (B) unfertilized embryo. (C) sperm. (D) kidney. (GATE 2011; 1 Mark)
182 GATE Biotechnology Chapter-wise Solved Papers 33. The product commercially produced by animal cell culture is (A) insulin. (B) tissue plasminogen activator. (C) interferon. (D) hepatitis B vaccine. (GATE 2011; 1 Mark) 34. 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. (B) Only II is true. (C) Both I and II are true. (D) I is true and II is false. (GATE 2011; 2 Marks) 35. 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 also utilized as a nutrient. (GATE 2012; 1 Mark) 36. 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. (GATE 2014; 1 Mark) 37. 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. (GATE 2014; 2 Marks) 38. Production of monoclonal antibodies by hybridoma technology requires
(A) splenocytes. (C) hepatocytes.
(B) osteocytes. (D) thymocytes. (GATE 2015; 1 Mark)
39. 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 (GATE 2015; 2 Marks) 40. 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 (GATE 2016; 1 Mark) 41. 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) CO2 requirement is independent of bicarbonate concentration in the medium. (GATE 2016; 2 Marks) 42. Choose the correct combination of True (T) and False (F) statements about microcarriers used in animal cell culture. P. Higher cell densities can be achieved using microcarriers. Q. Microcarriers increase the surface area for cell growth. R. Microcarriers are used for both anchorage- and non-anchorage-dependent cells. S. Absence of surface charge on microcarriers enhances attachment of cells.
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P-T, Q-F, R-T and S-F P-T, Q-T, R-F and S-F P-F, Q-F, R-T and S-T P-F, Q-T, R-F and S-T (GATE 2016; 2 Marks)
43. A single stem cell undergoes 10 asymmetric cell divisions. The number of stem cells at the end is _______. (GATE 2018; 1 Mark) 44. Mammalian cells in active growth phase were seeded at a density of 1 × 105 cells/mL. After 72 hours, 1 × 106 cells/ mL were obtained. The population doubling time of the cells in hours is (up to two decimal places ____________. (GATE 2018; 2 Marks) 45. Which one of the following is used as a pH indicator in animal cell culture medium? (A) Acridine orange (B) Phenol red (C) Bromophenol blue (D) Coomassie blue (GATE 2019; 1 Mark)
46. Which of the following statements are CORRECT about the function of fetal bovine serum in animal cell culture? P. It stimulates cell growth. Q. It enhances cell attachment. R. It provides hormones and minerals. S. It maintains pH at 7.4. (A) P and Q only (B) P and S only (C) P, Q and R only (D) P, Q, R and S (GATE 2019; 2 Marks) 47. Antibody-producing hybridoma cells are generated by the fusion of a (A) T cell with a myeloma cell. (B) B cell with a myeloma cell. (C) Macrophage with a myeloma cell. (D) T cell and a B cell. (GATE 2019; 2 Marks)
Answer Key 1. (A)
2. (C)
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16. (A)
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20. (A)
21. (D)
22. (D)
23. (C)
24. (B)
25. (D)
26. (A)
27. (B)
28. (B)
29. (A)
30. (A)
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32. (A)
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37. (2 × 10 )
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44. (21.64 hrs)
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Answers with Explanations 1. Topic: Animal Cell Culture: Media Composition and Growth Composition Recombinant DNA technology is used in producing cDNA molecule complementary to the mature mRNA of gene encoding tissue plasminogen activator. This cDNA is then cloned into a synthetic plasmid and introduced into mammalian cells. Transformed mammalian cells are cultured in a suitable animal cell culture media. Tissue plasminogen activator producing cells are then transferred into fermenters to produce them in large quantity. Tissue plasminogen activator secreted into the culture media is isolated and purified. Tissue plasminogen activator is the first pharmaceutical product produced using mammalian cell culture. Answer (A)
2. Topic: Animal Cell Culture: Media Composition and Growth Composition A pregnancy test is used to determine if a woman is pregnant. This is done by analyzing a sample of a woman’s urine or blood for the presence of human chorionic gonadotropin (hCG). hCG is a hormone produced by the placenta and is usually only present in a woman’s body when she is pregnant. Answer (C) 3. Topic: Hybridoma Technology Bispecific antibodies are artificial proteins that can simultaneously bind to two different types of antigen. They are engineered to combine the variable domains from two distinct parental antibodies, resulting in an antibody
184 GATE Biotechnology Chapter-wise Solved Papers binding to two distinct antigens. A number of strategies have been developed to generate bispecific antibodies. They have been explored for cancer immunotherapy and drug delivery. Answer (*) 4. Topic: Hybridoma Technology Monoclonal antibodies can have monovalent affinity, in that they bind to the same epitope (the part of an antigen that is recognized by the antibody). Mast cells, a type of blood cell, play an important role in the body’s immune system. They reside in all body tissues and form part of the body’s initial defense system. Mast cells react to foreign bodies and injury by releasing a variety of potent chemical mediators, such as histamine, when activated. Monoclonal antibodies against human mast cell tryptase demonstrate shared antigenic sites on subunits of tryptase and selective localization of the enzyme to mast cells. Answer (*) 5. Topic: Hybridoma Technology Bispecific monoclonal antibodies are primarily produced by three method: • Quadroma technology based on the somatic fusion of two different hybridoma cell lines. • Chemical conjugation, which involves chemical cross-linkers. • Genetic approaches utilizing recombinant DNA technology. Answer (*) 6. Topic: Transgenic Animals Ovine β -lactoglobulin, a milk gene promoter is attached to the therapeutic protein coding region to direct expression of the gene specifically to the mammary gland of the animal. In this way, the transgenic protein is produced in the milk of the animal and can be harvested and purified with no adverse effects on the animal. An added advantage is that the protein can be purified using regular dairy techniques, as well as high technology chromatographic procedures. Answer (B) 7. Topic: Animal Cloning Positional cloning is the approach 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. During the course of positional cloning, one needs to determine whether the DNA segment currently under consideration is part of the gene. Answer (A) 8. Topic: Animal Cell Culture: Media Composition and Growth Conditions CO2 incubator is used to culture cells to provide it with the optimum temperature, moisture (sterile environment) and to maintain optimum pH. When the media contains carbonate buffer, the CO2 gas from the cylinder is let into the incubator in such a way that the pH remains constant. Typical temperature settings range from 4°C to 50°C, and CO2 concentrations run from 0.3 to 19.9 %. Non-corrosive stainless-steel interiors are standard, but some newer models feature antimicrobial copper surfaces to prevent contamination. Auto decontamination using heat or UV light is another new and attractive feature available in CO2 incubators. Answer (*) 9. Topic: Animal Cell Culture: Media Composition and Growth Conditions Hypoxanthine-guanine phosphoribosyltransferase (HPRT) is the enzyme which catalyzes salvage of the purine bases guanine and hypoxanthine into their respective monophosphate nucleoside, i.e., guanylic monophosphate (GMP) and inosine monophosphate (IMP). Thus, HPRT mutants are unable to utilize the purine salvage pathway, thus they depend on the synthesis of purine deoxynucleotides by de novo pathway. Answer (B) 10. Topic: Anchorage and Non-anchorage Dependent Cell Culture • Roller bottles are cylindrical vessels that revolve slowly (between 5 and 60 revolutions per hour) which bathes the cells that are attached to the inner surface with medium. They provide a very economical means for cultivating large quantities of anchoragedependent cells using essentially the same culture techniques as with traditional cell culture flasks but with considerably less labor. • Air lift/Gas lift bioreactors are an alternative to mechanically agitated systems that reduce shear stress and heat generation by eliminating the mechanical agitator. They contain an inner draft tube that creates the drafting force required for liquid circulation and possible improvement of bulk mixing. • A hollow fiber bioreactor is a 3D cell culturing system based on hollow fibers, which are small, semi-permeable capillary membranes arranged in parallel array with a typical molecular weight cut-off
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(MWCO) range of 10–30 kDa. These hollow fiber membranes are often bundled and housed within tubular polycarbonate shells to create hollow fiber bioreactor cartridges. The convention of this technique is the growth of anchorage dependent cells on the 3-dimensional surface of tissue culture plastic. To produce and maintain a very large population of anchorage-dependent cells, a microcarrier-based stirred tank bioreactor is commonly used. In this approach, the cells are exposed to harmful hydrodynamic shear stress in the bioreactor and the mass transfer rates of nutrients and gases in the bioreactor are often kept below an optimal level to prevent cellular damages from the shear stress. Answer (B)
11. Topic: Transgenic Animal A transgene is a gene or genetic material that has been transferred naturally, or by any of a number of genetic engineering techniques from one organism to another. The introduction of a transgene (called “transgenesis”) has the potential to change the phenotype of an organism. Answer (*) 12. Topic: Animal Cell Culture: Media Composition and Growth Conditions Primary culture refers to the stage of the culture after the cells are isolated from the tissue and proliferated under the appropriate conditions until they occupy all of the available substrate (i.e., reach confluence). At this stage, the cells have to be sub-cultured (i.e., passaged) by transferring them to a new vessel with fresh growth medium to provide more room for continued growth. Answer (*) 13. Topic: Animal Cell Culture: Media Composition and Growth Conditions Some important methods for the separation of different cell types from a mixed population of animal cells include: Tissue Disaggregation; Adherence and Filtration; Conventional and Zonal Centrifugation; Centrifugal Elutriation (Counter-Streaming Centrifugation); Unit Gravity Separation; Countercurrent Distribution; Electrophoresis and Fluorescence-Activated Cell Sorting. Answer (*) 14. Topic: Animal Cell Culture: Media Composition and Growth Conditions Serum is commonly used as a supplement to basal growth medium in cell culture. The most common type of serum is fetal bovine serum (FBS) because of its high content of embryonic growth promoting factors. In cell culture, serum provides a wide variety of macromolecular proteins, low molecular weight nutrients, carrier proteins for
water-insoluble components, and other compounds necessary for in vitro growth of cells, such as hormones and attachment factors. Serum also adds buffering capacity to the medium and binds or neutralizes toxic components. The selection of a serum supplement for cell culture applications is primarily dependent on the chemical definition of the basal medium, the type of cell to be grown, and the culture system being employed. Answer (D) 15. Topic: Anchorage and Non-anchorage Dependent Cell Culture Cell lines can often be most easily generated from cancer cells, but these cells differ from those prepared from normal cells in several ways. Cancer cell lines often grow without attaching to a surface, for example, and they can proliferate to a very much higher density in a culture dish. Similar properties can be induced experimentally in normal cells by transforming them with a tumor-inducing virus or chemical. The resulting transformed cell lines, in reciprocal fashion, can often cause tumors if injected into a susceptible animal. Answer (C) 16. Topic: Transgenic Animals A new sequence is given a marker gene, a gene that normal mice don’t have and that confers resistance to a certain toxic agent (e.g., neomycin) or that produces an observable change (e.g. color or fluorescence). Cells that were transformed with a vector containing the neomycin resistance gene are grown in a solution containing neomycin in order to select for the transformations that occurred via homologous recombination. Answer (A) 17. Topic: Anchorage and Non-anchorage Dependent Cell Culture • Glycoproteins help cells recognize each other. The unique patterns of oligosaccharides on the cell surface can be read by corresponding glycoproteins on another cell. It’s more like fitting a key into a lock; glycoproteins will often bind together if they’re a match. This binding process communicates to the cell that it has found its match. • Collagen promoted cell proliferation, cell survival under stress and promoted high cell adhesion to the cell culture surface. Coevolution of collagens and cell adhesion mechanisms has given rise to receptors that bind to specific motifs in collagens. These receptors may also recognize the different members of the large collagen family in a selective manner. • Phospholipases play central roles in regulating signaling during numerous cellular events including proliferation, G protein-coupled receptor signaling
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and neuronal activation by catalyzing the lysis of phosphorylated lipids. In turn, second messengers are released, which control downstream events. Four distinct families of phospholipases have been described and are named A, B, C and D and range in size from 13 to 250 kDa. Hyaluronic acid helps to make up the network structure of the extracellular matrix through its interactions with proteoglycans or collagen. When it is very concentrated, it can interact with itself creating meshes or frameworks that give the tissue specific viscoelastic properties. Answer (C)
18. Topic: Animal Cell Culture: Media Composition and Growth Conditions T-cell growth factor (TCGF), also known as interleukin-2, is a small glycoprotein which provides a necessary signal for the proliferation of activated T cells. The factor is released by T cells in response to antigen or lectin stimulation and requires the action of a second, macrophage-derived lymphokine, termed interleukin-1, for its production. Although the bioactivity of TCGF is antigen-nonspecific and often crosses species lines, specificity of the immune response is maintained by the requirement that a responding T cell be specifically activated before it expresses receptors for the growth factor. Answer (B) 19. Topic: Animal Cell Culture: Media Composition and Growth Conditions The approximate solubilities of calcium phosphates in cold water are monobasic phosphate, Ca(H2PO4)2, 71 mM; calcium dibasic phosphate, CaHPO4, 1.7 mM and calcium tribasic phosphate, Ca3(PO4)2, 0.06 mM. At pH 7.0, approximately 61% of the phosphate in the dibasic form and at pH 7.6, this increases to about 87%. The solubility of calcium phosphate in this form is approximately 1.7 mM and the highest concentration typically used in media formulations is 1.8 mM. This suggests that the precipitation of calcium byphosphate ions should only be a problem when the pH is made basic during activities such as base titration. The resulting formation of calcium tribasic phosphate would result in precipitation and possible loss of the calcium during filtration. EDTA has a log affinity for calcium of approximately 10.6 and it is often used to remove calcium from cell cultures media, especially when cells are being detached from substrates or when cell clumping is a problem. Citrate and albumin bind calcium with log affinities of approximately, 3.6 and 2, respectively. Answer (D)
20. Topic: Hybridoma Technology Some of methods that can be used for the screening of monoclonal antibodies are: • ELISA (Enzyme-linked immunosorbent assay), a sensitive immunoassay that uses an enzyme linked to an antibody or antigen as a marker for the detection of a specific protein, especially an antigen or antibody. • Sandwich ELISA, a modification of the standard ELISA in which an antibody bound to capture an antigen or antibody from a solution and a second, enzyme-linked antibody is used to detection. • Western blot, an assay that detects specific proteins within a protein mixture by a multistep process consisting of electrophoresis on a slab gel, transfer of the proteins on the gel to a membrane followed by identification of the specific proteins by antibody staining. • Flow cytometry, a cytometric technique in which cells suspended in a laminar fluid flow one at a time through a focus of exciting light, which is scattered in patterns characteristic to the cells and their components; cells are frequently labeled with fluorescent antibodies so that light is first absorbed and then emitted at altered frequencies. A sensor detecting the scattered or emitted light measures the size and molecular characteristics of individual cells; tens of thousands of cells can be examined per minute and the data gathered is processed by computer. Answer (A) 21. Topic: Kinetics of Cell Growth A replicon is a DNA molecule or RNA molecule, or a region of DNA or RNA, that replicates from a single origin of replication. The number of replicons in an average mammalian cell is about 10,000 –100,000 and the number of replicons active at any one time during S phase appears to be the main factor affecting changes in the DNA synthesis rate. Answer (D) 22. Topic: Animal Cell Culture: Media Composition and Growth Conditions Immortalized hepatocytes are defined as a population of indefinitely dividing parenchymal cells that retain critical liver functions. Since mature hepatocytes normally possess only limited growth potential when stimulated in vitro, immortalization strategies have been developed based mainly on the transduction or transfection of hepatocytes with well-known immortalizing genes. The most frequently used immortalization methods are (i) overexpression of viral oncogenes such as Ras (ii) forced
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expression of hTERT (human telomerase reverse transcriptase), or (iii) a combination of both. Answer (D) 23. Topic: Animal Cell Culture: Media Composition and Growth Conditions The in vitro premature growth arrest, observed in primary hepatocyte cultures could be related to a telomere-independent senescence mechanism, which remains to be fully elucidated, but is suggested to involve tumor suppressor proteins and cdks. Indeed, several studies support the contribution of cdki p21 and/or p16 to the inhibition of DNA synthesis in primary hepatocyte cultures. In this respect, it was demonstrated that the second cell cycle G1 block caused by chronic MAPK pathway activation in mitogen stimulated primary hepatocyte cultures is partly related to p21 induction. Of note, transient MAPK pathway inhibition allows the establishment of multiple replication rounds in these hepatocyte cultures. Answer (C) 24. Topic: Anchorage and Non-anchorage Dependent Collagens are a family of highly characteristic fibrous protein found in all multicellular animals and are critical in cell adhesion. Collagen Type I is found in several tissues including skin, connective tissue cartilage and bone. Collagen Type I is an attachment factor that facilitates the attachment and cytoplasmic spreading of all types of anchorage-dependent cells, when used as a thin layer on a tissue culture surface. As a gel, Collagen I enhances expression of cell-specific morphology and function. Answer (B) 25. Topic: Hybridoma Technology Hybridoma technology is a method for producing large numbers of identical antibodies (also called monoclonal antibodies). This process starts by injecting a mouse (or other mammal) with an antigen that provokes an immune response. A type of white blood cell, the B cell, produces antibodies that bind to the injected antigen. These newly produced antibodies are then harvested from the mouse. These isolated B cells are in turn fused with immortal cancer cells, a myeloma, to produce a hybrid cell line called a hybridoma, which has both the antibody-producing ability of the B-cell and the exaggerated longevity and reproductivity of the myeloma. The hybridomas can be grown in culture of HAT medium. HAT Medium (hypoxanthine-aminopterin-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
(a purine derivative) and thymidine (a deoxynucleoside) which are intermediates in DNA synthesis. The trick is that aminopterin blocks DNA de novo synthesis, which is absolutely required for cell division to proceed, but hypoxanthine and thymidine provide cells with the raw material to evade the blockage (the “salvage pathway”), provided that they have the right enzymes, which means having functioning copies of the genes that encode them. Answer (D) 26. Topic: Animal Cloning Electroporation can be used for both transient and stable transfection of mammalian cells. Cells are placed in suspension in an appropriate electroporation buffer and put into an electroporation cuvette. DNA is added, the cuvette is connected to a power supply, and the cells are subjected to a high-voltage electrical pulse of defined magnitude and length. The cells are then allowed to recover briefly before they are placed in normal (non-selecting) cell growth medium. Selection for permanently transfected cells and for cells carrying targeted gene insertions by homologous recombination can be accomplished by modified media. Answer (A) 27. Topic: Anchorage and Non-anchorage Dependent Cell Culture For the primary cells that are anchorage-dependent, adherent cells (such as solid tissues) require a surface to grow properly in vitro. These cells are mostly cultured in a flat uncoated plastic vessel, but sometimes a microcarrier, which may be coated with extracellular matrix (such as collagen and laminin) components to increase adhesion properties and provide other signals needed for growth and differentiation. The cell culture media is composed of a basal medium supplemented with appropriate growth factors and cytokines. Cells are grown and maintained at an appropriate temperature and gas mixture (typically, 37°C, 5% CO2 for mammalian cells) in a cell incubator. Growth media can vary in pH, glucose concentration, growth factors, and the presence of other nutrients depending on the cell types. Anchorage-dependent cells grow in monolayers and need to be sub-cultured at regular intervals with appropriate culture medium to maintain exponential growth. Sub-cultivation of monolayers involves the breakage of both inter- and intracellular cell-to-surface bonds. Most adherent primary cells require the digestion of their protein attachment bonds or separation from the monolayer or relevant tissue with a low concentration of a proteolytic enzyme such as trypsin/EDTA. After the cell’s dissociation and dispersion into a single-cell suspension, they are
188 GATE Biotechnology Chapter-wise Solved Papers counted and diluted to the appropriate concentration and transferred to fresh culture vessels (the composition of the media varies depending up on the cell types) where they will reattach and divide. Answer (B) 28. Topic: Animal Cell Culture: Media Composition and Growth Conditions HAT Medium (hypoxanthine-aminopterin-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 (a purine derivative) and thymidine (a deoxynucleoside) which are intermediates in DNA synthesis. The trick is that aminopterin blocks DNA de novo synthesis, which is absolutely required for cell division to proceed, but hypoxanthine and thymidine provide cells with the raw material to evade the blockage (the “salvage pathway”), provided that they have the right enzymes, which means having functioning copies of the genes that encode them. Answer (B) 29. Topic: Hybridoma Technology Hybridoma technology is used for fusing B cells with tumor plasma cells (myeloma) that grow easily and proliferate endlessly. The resulting hybrid cell is called a hybridoma. Hybridomas are long-term sources of large quantities of pure, identical antibodies, called monoclonal antibodies (MAbs) because they come from a single clone of identical cells. This method of production of MAbs was described by Jerne, Kohler and Milstein. Answer (A) 30. Topic: Animal Cloning Somatic gene transfer involves adding genes to cells other than egg or sperm cells. It can be used in embryonic stem cell research, or in regenerative medicine. Somatic cell nuclear transfer (SCNT) is a technique for cloning. The nucleus is removed from a healthy egg. This egg becomes the host for a nucleus that is transplanted from another cell, such as a skin cell. The resulting embryo can be used to generate embryonic stem cells with a genetic match to the nucleus donor (therapeutic cloning) or can be implanted into a surrogate mother to create a cloned individual, such as Dolly the sheep (reproductive cloning). Dolly (sheep) was the first animal to be cloned successfully from the somatic cell of an adult in 1997 at the Roslin Institute in Edinburgh, Scotland. Transgenic sheep were developed to produce animals with better growth rate and hence better meat production. Human genes for blood clotting (anti-hemophilic) factor IX have been introduced into the sheep and expressed in the mammary tissues, such factor IX is secreted in the milk.
It can then be separated from the milk and given to hemophilic patients. Answer (A) 31. Topic: Animal Cell Culture: Media Composition and Growth Conditions Given: Total no. of cells = 6 × 107 cells Specific activity = 100 counts/pmole (cpm/pmole) Total counts = 1000 cpm On complete saturation, [L] = [R] Avg. no. of receptor per cell = ? Specific acitivity = No.of ligands =
Total counts No.of ligands
1000 cpm 100 cpm/pmole
= 10 pmole = 10 × 10 −12 mole ⇒ 10 × 10 −12 × N A ⇒ 10 −11 × 6.023 × 10 23 ⇒ 6.023 × 1012 No. of ligands = No. of receptors = 6.022 × 1012 Avg. no. of receptor = =
Total receptor No.of cells 6.022 × 1012 = 105 6 × 10 7
Answer (B)
32. Topic: Stem Cell Technology Embryonic stem cells are derived from the inner cell mass of fertilized embryo, particularly from blastocyst stage. Blastocyst is the post-implantation stage in which developing embryo consists of 100 –150 cells. These cells form the inner cell mass and an outer layer termed as trophoblast. Cells of inner cell mass are pluripotent stem cells which have an ability to differentiate into any kind of cell of the body for example, cardiac cell, muscle cell, etc. Answer (A) 33. Topic: Animal Cell Culture: Media Composition and Growth Conditions Tissue plasminogen activator is the first pharmaceutical product produced using mammalian cell culture. These are used in acute myocardial infarction, acute stroke, pulmonary embolism, deep vein thrombosis, etc. Answer (B) 34. Topic: Hybridoma Technology HAT consists of hypoxanthine, aminopterin and thymidine in its culture medium which is used to select the
Chapter 11 • Animal Biotechnology 189
hybridomas. In this process first all spleen cells are fused with the myeloma cells using PEG (polyethylene glycol), i.e., hybrid cells. Spleen cells (B cells) produce antibodies and myeloma cells have high replication power. The unfused spleen cell will die after some time and added in HAT is culture to kill the unfused myeloma cells. During normal condition cells replicate using de novo pathway and if it is blocked they used salvage pathway to replicate but only in the presence of hypoxanthine and thymidine. Myeloma cells lack HPRT so, deno vo pathway is blocked and they use salvage pathway to replicate but due to presence of aminopterin in culture it is blocked too, and they get die. Hybridomas cell grow by using salvage pathway as they have HPRT and they replicate. Answer (C) 35. Topic: Animal Cell and Tissue Preservation Dimethyl sulfoxide (DMSO) is used as a cryopreservant for mammalian cell cultures because it reduces the formation of ice during freezing process and thus protects the death of the cell. In addition, it is also less toxic as compared to other alternatives available such as ethanol. Answer (C) 36. Topic: Animal Cell Culture: Media Composition and Growth Conditions Basal media are often supplemented with sera to provide the essential growth factors, hormones and carbohydrates necessary for cell proliferation and high protein production. Serum, generally fetal calf or newborn calf is added to basal media at concentrations of 1–10% Cell culture media must be sterilized before it can be delivered to an aseptic bioreactor or fermenter to prevent the growth of adventitious bacteria or other organisms. This sterilization is generally accomplished by normal flow or dead-end filtration using a validated, sterilizing-grade microporous membrane, typically carrying a nominal pore size rating of either 0.2 or 0.1 micron. Careful preparation of cell culture media is important to ensure sterility of the cell culture batch and subsequent purification processing. Mycoplasma has become widely recognized as the smallest bacterium to thrive in cell culture that cannot be retained by 0.2 µm sterilizing-grade filters. As cell culture media spans many media types among several cell lines, there exists a variety of constituents at a wide range of concentrations. Prefilters should be sized appropriately to handle batch-to-batch media variability and ensure that the sterile media fill in the bioreactor. Final sterilizing-grade filtration should remove bacteria (and mycoplasma) to preserve the aseptic barrier around the bioreactor. Answer (C) 37. Topic: Anchorage and Non-anchorage Dependent Cell Given, Initial number of cells, N0 = 105 Time for monolayer to confluence, t = 50 h
Area available for cells to grow, A = 25 cm2 Volume of the medium, V = 25 mL Cells are rectangular in shape with dimensions = 5 µm × 2 µm Area of cell = 5 µm × 2 µm = 10 µm2 = 10 × 10–8 cm2 [∵1 μm = 10 −4 cm ] = 10–7 cm2 Number of cells that could be present in the area available N= Growth rate =
25 cm 2 = 25 × 10 7 cells 10 −7 cm 2 No.of cells 25 × 10 7 = Area × Confluence time 25 × 50
= 2 × 105 cells per cm 2 per hour Answer (2 × 105 cells per cm2 per hour) 38. Topic: Hybridoma Technology Splenocytes are a mixture of spleen cells such as lymphocytes, macrophages, and dendritic cells. They are used in the production of monoclonal antibodies by hybridoma technology. Answer (A) 39. Topic: Animal Cloning Osmotic stabilizer → Digestion of cell wall → Protoplast viability → Cybrid formation. Cybrid is hybrid cell resulting from fusion of a cytoplast with a whole cell, thereby creating cytoplasmic hybrid. If the two species are distant and their nuclear and cytoplasmic genomes are incompatible, then cybrids are created. In such cases, the resulting cell will have functional nuclear genome of one species while it will be a hybrid for cytoplasmic traits. Answer (B) 40. Topic: Animal Cell Culture: Media Composition and Growth Conditions • Trypan blue dye exclusion is used as vital stain to selectively colour dead tissues or cells blue. • Tetrazolium (MTT) assay is used to determine both cell viability and number. MTT is a water-soluble yellow dye which when taken up by viable cells is reduced by mitochondrial dehydrogenases resulting in water-insoluble blue colored product called formazan that is dissolved further and measured calorimetrically. • The enzyme lactate dehydrogenase (LDH) catalyzes the oxidation of NADH by pyruvate to yield NAD+ and lactic acid. When cells undergo necrosis or apoptosis the cell membrane is damaged which results in the release of cytoplasmic enzyme LDH into the medium. Hence, the amount of this enzyme is used to assess cell viability.
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Larger organisms such as yeast can be directly counted using an electronic counter called Coulter counter. Coulter counter does not determine cell viability. It is generally used for counting bacteria, viruses, etc. Answer (D)
41. Topic: Animal Cell Culture: Media Composition and Growth Conditions Since bicarbonate is included as a buffer system with the CO2 environment in which cell should be cultured. A CO2 enriched atmosphere in the incubator chamber is used to maintain the culture pH between 6.9 and 7.4. The CO2 concentration is controlled by an incubator in a sealed flask. It is important to maintain the pH of the medium as in general, mammalian cell lines grow well at pH 7.4. The pH is controlled by the growth medium. It buffers the medium against changes in pH by rising CO2bicarbonate based buffers. The balance of dissolved CO2 and bicarbonate is responsible for maintaining pH of the medium as any change in atmospheric CO2 will alter the pH of the medium. Therefore, exogenous CO2 becomes necessary when a medium is buffered using CO2bicarbonate based buffer. Hence, if bicarbonate concentration is higher, CO2 required will also be higher. Answer (A) 42. Topic: Animal Cell Culture: Media Composition and Growth Conditions Higher cell densities can be achieved using microcarriers that increase the surface area for cell growth. They also assist in anchorage of dependent cells and mediate the culture of different type of cells. Microcarriers are ionic and their surface charge density has a critical value above which cell growth is inhibited and below which adhesion is poor. Answer (B) 43. Topic: Stem Cell Technology Asymmetric cell division in contrast to symmetric cell division gives rise to two daughter cells, but with different cellular providences. One cell is exact copy of the original mother cell and other programmed for differentiation into specialized cell type. Thus, after undergoing 10 asymmetric divisions, only one stem cell will be there at the end. Answer (*)
44. Topic: Animal Cell Culture: Media Composition and Growth Conditions Given, Initial cell concentration (N0) = 1 × 105 cells/mL Final cell concentration (Nt) = 1 × 106 cells/mL Time (t) = 72 hrs We know that, time (t ) log 2 Doubling time (td ) = ⎛N ⎞ log ⎜ t ⎟ ⎝ N0 ⎠ =
72 × 0.301 = 21.67 hrs ⎛ 106 ⎞ log ⎜ 5 ⎟ ⎝ 10 ⎠ Answer (21.67 hrs)
45. Topic: Animal Cell Culture: Media Composition and Growth Conditions Phenol red is used as a pH indicator. A solution of phenol red will have a yellow color at a pH of 6.4 or below and a red color at a pH of 8.2 and above. It is typically used in culture media to identify changes from neutral to acidic pH values. Phenol red in tissue culture media can act as a weak estrogen, especially with human breast cancer cells. Answer (B) 46. Topic: Animal Cell Culture; Media Composition and Growth Conditions Fetal bovine serum is the most common growth supplement in animal cell culture medium. It is rich in natural growth-factors required to stimulate cell growth, cell proliferation, differentiation and also for regulation of cellular activity. It provides a wide variety of macromolecular proteins, low molecular weight nutrients, carrier proteins or chelators for water-insoluble components, and other compounds necessary for in vitro growth of cells, such as hormones and attachment factors. Answer (C) 47. Topic: Hybridoma Technology Hybridoma cells result from the fusion of B lymphocytes and myeloma cells produce specific monoclonal antibodies. A hybridoma is a cell that can be grown in culture and that produces immunoglobulins that all have the same sequence of amino acids and consequently the same affinity for only one epitope on an antigen Answer (B)
C H A P T E R 12
Bioprocess Engineering and Process Biotechnology
Syllabus Chemical Engineering Principles Applied to Biological System – Principle of Reactor Design, Ideal and Non-Ideal Multiphase Bioreactors, Mass and Heat Transfer; Rheology of Fermentation Fluids; Aeration and Agitation; Media Formulation and Optimization; Kinetics of Microbial Growth – Substrate Utilization and Product Formation; Sterilization of Air and Media; Batch, Fed-batch and Continuous Processes; Various Types of Microbial and Enzyme Reactors; Instrumentation Control and Optimization; Unit Operations in Solid-Liquid Separation and Liquid-Liquid Extraction; Process Scale-up, Economics and Feasibility Analysis. Engineering Principle of Bioprocessing – Upstream Production and Downstream; Bioprocess Design and Development from Lab to Industrial Scale; Microbial, Animal and Plant Cell Culture Platforms; Production of Biomass and Primary/ Secondary Metabolites – Biofuels, Bioplastics, Industrial Enzymes, Antibiotics; Large Scale Production and Purification of Recombinant Proteins; Industrial Application of Chromatographic and Membrane Based Bioseparation Methods; Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes; Bioremediation – Aerobic and Anaerobic Processes for Stabilization of Solid/Liquid Wastes.
Chapter Analysis Topic Chemical Engineering Principles Applied to Biological System: Principle of Reactor Design, Ideal and Non-Ideal Multiphase Bioreactors, Mass and Heat Transfer Rheology of Fermentation Fluids
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Batch, Fed-batch and Continuous Processes Various Types of Microbial and Enzyme Reactors Instrumentation Control and Optimization Unit Operations in Solid-Liquid Separation and Liquid-Liquid Extraction Process Scale-up, Economics and Feasibility Analysis Engineering Principle of Bioprocessing: Upstream Production and Downstream Bioprocess Design and Development from Lab to Industrial Scale Microbial, Animal and Plant Cell Culture Platforms
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192 GATE Biotechnology Chapter-wise Solved Papers Production of Biomass and Primary/ Secondary Metabolites: Biofuels, Bioplastics, Industrial Enzymes, Antibiotics Large Scale Production and Purification of Recombinant Proteins Industrial Application of Chromatographic and Membrane Based Bioseparation Methods Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Bioremediation: Aerobic and Anaerobic Processes for Stabilization of Solid/Liquid Wastes
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Questions 1. Choose the correct completion of the following statement. A fermenter sterilization in situ is less efficient than continuous heat sterilization 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. (GATE 2000; 1 Mark) 2. In secondary metabolism, two distinct phases— trophophase and idiophase refer, respectively, to (A) growth and production phase. (B) early and late phase. (C) primary and secondary metabolism. (D) lag phase and log phase. (GATE 2000; 1 Mark) 3. Which of the following eukaryotic organisms has been proven to be of great industrial importance? (A) Penicillium chrysogenum (B) Saccharomyces cerevisiae (C) Bacillus subtilis (D) Streptomyces griseus (GATE 2000; 1 Mark) 4. Which one of the following options related to the following statements is incorrect? In comparison to physical/chemical methods of clean up, bioremediation methods (A) use relatively simple techniques. (B) generally end up with hazardous waste material.
(C) are relatively slow. (D) are unobtrusive and non-disruptive. (GATE 2000; 1 Mark) 5. The Pasteur Effect is (A) inhibition of glucose utilization and lactate accumulation in glycolysis. (B) sterilization of milk. (C) vaccine production. (D) heat treatment of bacteria. (GATE 2000; 1 Mark) 6. Write whether the following statements are true or false: P. Three important characteristics in performance of biosensors are selectivity, sensitivity and stability. Q. Activated sludge process is one of the most common anaerobic sewage treatment method. R. In a fermenter, impellers increase oxygen demand by producing high shear forces. S. A pressure cycle is a type of air lift fermenter. T. Monoclonal antibodies are used extensively in diagnosis of hematopoietic cancers. (GATE 2000; 5 Marks) 7. A protein PZ is present in genetically engineered bacteria at 5% of the total protein (0.1 picogram) in a cell. 1 mL cells of log phase culture contain 2 × 108 cells, while stationary phase culture contains 1 × 109 cells. The molecular weight of the protein is 30,000 Da and the Avogadro number is 6.02 × 1023 molecules/mole. What is the number of molecules of PZ per cell? Calculate the amount of protein in milligrams in one liter each of log phase and stationary phase cultures. (GATE 2000; 3 Marks)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 193
8. The limiting factor in the production of large quantities of ethanol as biofuel 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. (GATE 2001; 1 Mark) 9. 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 (GATE 2001; 1 Mark) 10. The factor(s) likely to increase the rate of reaction catalyzed by a surface immobilized enzyme is/are (A) increased agitation of the bulk liquid containing the substrate. (B) continued replacement of the bulk liquid containing the substrate. (C) increased concentration of the substrate in the bulk liquid. (D) all of the above. (GATE 2001; 1 Mark) 11. 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. (GATE 2001; 1 Mark) 12. 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 (GATE 2001; 1 Mark) 13. In large-scale fermentation, the preferred method of sterilization is (A) chemicals. (B) radiation. (C) filtration. (D) heat. (GATE 2001; 1 Mark)
14. 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) It requires a shorter residence time. (GATE 2001; 1 Mark) 15. A chemostat is operating at steady state at a dilution rate of 0.1 h−1 and a limiting nutrient concentration of 10 μM. If the μm for the organism is 0.5 h−1, calculate the Monod constant for the nutrient. (GATE 2001; 1 Mark) 16. Why asparaginase is used in anticancer therapy? (GATE 2001; 1 Mark) 17. Write the reaction catalyzed by penicillin G acylase. (GATE 2001; 1 Mark) 18. Name any two techniques by which penicillin G acylase may be immobilized. (GATE 2001; 2 Marks) 19. Match the products in Group I with their corresponding organisms in Group II. Group I Group II P. Aspergillus niger 1. Lysine Q. Saccharomvces cerevisiae 2. Citric acid R. Penicillium chrysogenum 3. Acetone/butanol S. Lactobacillus casei 4. Vitamin B12 T. Corynebacterium glutamicum 5. Erythromycin 6. b-Lactam 7. Diacetyl 8. Ethanol (GATE 2001; 5 Marks) 20. Match the unit operations in Group I with the most appropriate recovery stage in Group II. Group I Group II P. Drying 1. Pretreatment Q. Sedimentation 2. Purification R. Membranes 3. Formulation S. Cell disruption 4. Solid/liquid separation T. Chromatography 5. Solid/solid separation 6. Concentration (GATE 2001; 5 Marks) 21. The precursor for penicillin-G biosynthesis during fermentation process is (A) phenylacetic acid. (B) phenoxyacetic acid.
194 GATE Biotechnology Chapter-wise Solved Papers (C) acetic acid. (D) none of the above. (GATE 2002; 1 Mark) 22. Energy capture efficiency of the aerobic cells using glucose as a substrate is (A) 50% (B) 40% (C) 30% (D) 20% (GATE 2002; 1 Mark) 23. For scaling up for a bioreactor, the following parameter is assumed to be constant (A) airflow rate. (B) diameter of the impeller. (C) agitator speed. (D) volumetric mass transfer coefficient. (GATE 2002; 1 Mark) 24. Mechanism of separation of contaminants present in air by fibrous media are (A) interception. (B) inertial impaction. (C) diffusion. (D) all of the above. (GATE 2002; 1 Mark) 25. Ethanol concentration is lowest in (A) wine. (B) beer. (C) brandy. (D) rum. (GATE 2002; 1 Mark) 26. The following crosslinking agents may be used for the immobilization of enzymes. (A) Glutaraldehyde (B) Cyanogen bromide (C) Thionyl chloride (D) All of the above (GATE 2002; 1 Mark) 27. In large scale fermentation process, air is sterilized by (A) jute fiber. (B) membrane. (C) cotton fiber. (D) glass wool fiber. (GATE 2002; 1 Mark) 28. Name two metal ions which play important role in citric acid fermentation. (GATE 2002; 1 Mark) 29. How is the agitator speed in a fermenter correlated with the power drawn by the agitator? (GATE 2002; 1 Mark) 30. How does the sulfanilamide kill the bacteria? (GATE 2002; 1 Mark) 31. (a) The volume of a chemostat system is 1000 L. The feed flow rate to the reactor is 200 L h−1 and the
glucose concentration in the feed is 5 g L−1. Determine cell and glucose concentration in the effluent of the reactor under steady state conditions. Use the following constants for the cells: μmax = 0.3 h−1, KS = 0.1 g L−1, YX/S = 0.4 (g dw cells/g glucose) (GATE 2002; 3 Marks) (b) Find out the dilution rate which gives maximum biomass productivity. (GATE 2002; 2 Marks) 32. Match the organisms in Group I with the product in Group II. Group I Group II P. Thermus aquaticus 1. Beer Q. Acetobacter aceti 2. Bioinsecticides R. Bacillus thuringiensis 3. HindIII S. Saccharomyces carlsbergensis 4. TaqI T. Hemophilus influenza 5. Vinegar (GATE 2002; 5 Marks) 33. Match the secondary metabolites in Group I with their most appropriate chemical characteristics in Group II. Group I Group II P. Diosgenin 1. Holoside Q. Ajmalicine 2. Pyrrolizidine alkaloid R. Shikonin 3. Indole alkaloid S. Digoxin 4. Napthoquinone nucleus T. Scopolamine 5. Cardenolide 6. Saponin 7. Tropane alkaloid (GATE 2002; 5 Marks) 34. Enzyme papain is used with success to (A) increase meat production. (B) leaven bread. (C) ripen papaya fruit. (D) tenderize meat. (GATE 2003; 1 Mark) 35. Microbes bring about biological transformation of xenobiotic compounds by (A) degradation. (B) conjugation. (C) detoxification. (D) all of the above. (GATE 2003; 1 Mark) 36. Which one of the following reactions is used for the purpose of recycling enzymes in bioprocesses? (A) Isomerization (B) Immobilization (C) Phosphorylation (D) Polymerization (GATE 2003; 1 Mark)
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37. The separation principle of dialysis used in the recovery of fermentation products is (A) diffusion. (B) charge. (C) turbulence. (D) solubility. (GATE 2003; 1 Mark) 38. Identify the statement that is not correct. (A) Penicillin fermentation is an aerobic process. (B) Penicillin biosynthesis is affected by phosphate concentration. (C) Lysine stimulates penicillin synthesis. (D) Penicillin production shows a distinct catabolite repression by glucose. (GATE 2003; 1 Mark) 39. A batch culture fermentation was being conducted with Streptomyces rimosus, Analysis of samples collected indicated doubling of cell number per unit time. Your interference would be that the culture is in the (A) lag phase. (B) log phase. (C) stationary phase. (D) death phase. (GATE 2003; 2 Marks) 40. Match the industrial application of the following enzymes. Group I Group II P. Penicillinase 1. Pharmaceutical Q. Pectinase 2. Leather R. Trypsin 3. Wine S. Rennin 4. Dairy P Q R S (A) 4 3 1 2 (B) 1 3 2 4 (C) 1 2 3 4 (D) 4 2 3 1 (GATE 2003; 2 Marks) 41. 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. (GATE 2003; 2 Marks) 42. Match the activity spectrum of the following antibiotics. Group I Group II P. Actinomycin D 1. Antifungal Q. Daunorubicin 2. Antituberculosis R. Rifamycin 3. Antitumor S. Griseofulvin 4. Antiprotozoal
(A) (B) (C) (D)
P Q R S 3 4 2 1 3 1 4 2 3 1 2 4 2 1 4 3 (GATE 2003; 2 Marks)
43. Autoclaves are routinely used in laboratories for sterilization. They act by (A) disrupting cell membranes. (B) denaturing proteins. (C) changing physically membrane lipids. (D) all of the above. (GATE 2003; 2 Marks) 44. 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. (GATE 2003; 2 Marks) 45. 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 the above. (GATE 2003; 2 Marks) 46. 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. (GATE 2003; 2 Marks) 47. During the functioning of biosensor which of the following sequences of event 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 (GATE 2004; 1 Mark)
196 GATE Biotechnology Chapter-wise Solved Papers 48. An immobilized enzyme being used in a continuous plug flow reactor exhibits an effectiveness factor (h) of 1.2. The value of h 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 enzymes experiences external film diffusion limitation. (C) The enzyme follows sigmoidal kinetics. (D) The immobilized enzyme is operationally unstable. (GATE 2004; 1 Mark) 49. 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. Fermentation products Microbial cultures P. Ethanol 1. Aspergillus niger Q. Streptomycin 2. Zymomonas mobilis R. Citric acid 3. Streptomyces griseus S. Cellulase 4. Trichoderma reesei P Q R S (A) 1 2 3 4 (B) 2 3 1 4 (C) 2 3 4 1 (D) 4 3 2 1 (GATE 2004; 1 Mark) 50. Batch fermentation of glucose to ethanol yields a productivity of 4.5 g−1L h−1. If the yeast cell concentration in the fermentation broth is 5% (v/v) and the intracellular NAD+/NADH concentration in the yeast cells is 10 μM, the cycling rate of NAD+ NADH will be (A) 50,000 cycles h−1 (C) 100 cycles h−1
(B) 20,000 cycles h−1 (D) None of the above (GATE 2004; 2 Marks)
51. 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, m is the maximum protein concentration achievable, K is the first order rate constant and is 0.5 h−1 for the system studied. The time required for the release of 90% of the intracellular proteins will be: (A) 10 h (C) 4.6 h
(B) 0.2 h (D) None of the above (GATE 2004; 2 Marks)
52. Inversion of sucrose by immobilized invertase follows a substrate inhibited kinetics. The reaction rate (v) in mol m−3 h−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 maximum at [S] = 20 mol m−3 the feed flow rate for a reactor volume of 1 m3 to get the maximum productivity from the reactor should be (A) 0.11 mol m−3 h−1 (B) 1.10 m3 h−1 (C) 5.05 m3 h−1 (D) None of the above (GATE 2004; 2 Marks) 53. 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−1. At a concentration of 50 mg L−1 of protein in the solution, the concentration of protein on the adsorbent reaches 35 mg cm−1. It is desired to recover 90% of the protein from 1.5 liter of the cell free culture filtrate containing 220 mg L−1 protein by 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 the above (GATE 2004; 2 Marks) 54. Measurement of KLa in a bioreactor can be carried out by sodium sulfite oxidation method, that is, based on the oxidation of sodium sulfite to 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 M Cu++ ions and the air was sparged in. After 10 min, 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) 0.08 g L−1 s−1 (B) 0.008 g L−1 s−1 (C) 0.8 g L−1 s−1 (D) None of the above (GATE 2004; 2 Marks)
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55. Examine the data given below in the table on purification of a protein X. Step
Volume Protein Con(mL) centration (mg mL−1) 500 12.0
Enzyme Activity (unit mL−1) 5.0
Crude cellfree extract Ammonium 125 3.0 11.0 sulfate precipitation Ion-exchange 10 9.0 75.0 chromatography 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 (GATE 2004; 2 Marks) 56. Ultrafiltration process cannot be used for (A) fractionation of proteins. (B) desalting. (C) harvesting of cells. (D) selective removal of solvent. (GATE 2005; 1 Mark) 57. Aeration in a bioreactor is provided by (A) impeller. (B) baffles. (C) sparger. (D) all of the above. (GATE 2005; 1 Mark) 58. Match the following products with their starting substrates. Group I Group II P. Sake 1. Apple juice Q. Cider 2. Grape juice R. Wine 3. Barley S. Lager 4. Rice P Q R S (A) 4 1 2 3 (B) 1 4 2 3 (C) 2 3 1 4 (D) 3 4 2 1 (GATE 2005; 2 Marks) 59. Identify the following antibiotics with their modes of action. Group I (Antibiotic) Group II (Mode of action) P. Ampicillin 1. Inhibition of protein synthesis Q. Tetracycline 2. Inhibition of cell wall synthesis R. Nystatin 3. Damage to cytoplasmic membrane S. Anthramycin 4. Damage to DNA structure
(A) (B) (C) (D)
P Q R S 1 2 4 3 2 1 3 4 1 2 3 4 3 4 2 1 (GATE 2005; 2 Marks)
60. In a bioreactor baffles are incorporated to (A) prevent vortex and to improve aeration efficiency. (B) maintain uniform suspension of cells. (C) minimize the size of air bubble for greater aeration. (D) maintain uniform nutrient medium. (GATE 2005; 2 Marks) Statement for Linked Answer Questions 61 and 62: 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 blank is set up that contains no enzyme. 61. The percent inhibition for this enzyme is calculated from the equation (A) {[cpm (control) – cpm (test) / [cpm (control) – cpm (blank)]} × 100 (B) {[cpm (control) – cpm (test) / [cpm (blank) – cpm (control)]} × 100 (C) {[com (test) – cpm (control) / [cpm (control) – cpm (blank)]} × 100 (D) {[cpm (control) – cpm (blank) / [cpm (test) – cpm (control)]} × 100 (GATE 2005; 2 Marks) 62. 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. (A) X – 8000, 4000 and 100 (B) W – 7000, 1400 and 135 (C) Y – 7500, 5000 and 90 (D) Z – 7200, 2800 and 200 (GATE 2005; 2 Marks) 63. Which of the following statements is incorrect about immobilized plant cell cultures? (A) It is possible to use high cell densities. (B) Cells remain active for long periods. (C) Cell products or inhibitors can be removed easily. (D) It provides low shear resistance to cells. (GATE 2006; 1 Mark)
198 GATE Biotechnology Chapter-wise Solved Papers 64. During cultivation of microorganisms in a fermenter, various parameters are controlled by appropriate sensor (probe). Match each probe in Group I with the appropriate response mechanism in Group II. Group I (Probe) Group II (Response) P. Thermistor 1. Activation of acid/alkali pump Q. Oxygen electrode 2. Activation of vegetable oil pump R. Metal rod 3. Activation of hot/cold water pump S. pH electrode 4. Increase/decrease in stirrer motor speed P Q R S (A) 2 3 1 4 (B) 1 2 4 3 (C) 3 2 4 1 (D) 3 4 2 1 (GATE 2006; 2 Marks) 65. 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 (GATE 2006; 2 Marks) 66. Which one of the following monolayer culture systems have the highest surface area to medium ratio? (A) Roux bottle (B) Spiracell roller bottle (C) Hollow fibers (D) Plastic hag/film (GATE 2006; 2 Marks) 67. Match items in Group I with correct options from those given in Group II. Group I Group II P. Amperometric biosensor 1. Light beam Q. Evanescent wave biosensor 2. Flux of redox electrons R. Calorimetric biosensor 3. Field effect transistors S. Potentiometric biosensor 4. Exothermic reaction P Q R S (A) 3 4 2 1 (B) 2 1 4 3 (C) 3 2 4 1 (D) 2 4 3 1 (GATE 2006; 2 Marks)
68. 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 (GATE 2006; 2 Marks) Common Data Questions for 69 and 70: Lignocelluloses biomass was subjected to microbial compositing. 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. 69. 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 (GATE 2006; 2 Marks) 70. If Con A sepharose column was used for the purification of enzyme, the separation would be based on (A) molecular exclusion. (B) affinity binding. (C) ion exchange. (D) hydrophobic interaction. (GATE 2006; 2 Marks) Statement for Linked Answer Questions 71 and 72: A bioreactor of working volume 50 m3 produces a metabolite (X) in batch culture under given operating conditions from a substrate (S). Then 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. 71. 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 (GATE 2006; 2 Marks) 72. What will be the overall productivity of the system in kg year−1 m−3? (A) 19250 (B) 38.50 (C) 3850 (D) 77 (GATE 2006; 2 Marks)
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73. The specific growth rate ( m) of a microorganism in death phase is (A) 0 (zero). (B) mmax. (C) less than zero. (D) greater than zero. (GATE 2007; 1 Mark) 74. Plant secondary metabolites (A) help to increase the growth rate of plant. (B) help in plant reproduction processes. (C) provide defense mechanisms against microbial attack. (D) make the plant susceptible to unfavorable conditions. (GATE 2007; 1 Mark) 75. Oils rich in PUFA are not desirable for biodiesel 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. (GATE 2007; 2 Marks) 76. Match items in Group I with correct examples from those in Group II. Group I Group II P. Catabolic product 1. Griseofulvin Q. Bioconversion 2. Baker’s yeast R. Biosynthetic product 3. 6-Aminopenicillanic acid S. Cell mass 4. Ethanol (A) P–4; Q–3; R–2; S–1 (B) P–3; Q–4; R–1; S–2 (C) P–4; Q–3; R–1; S–2 (D) P–1; Q–4; R–3; S–2 (GATE 2007; 2 Marks) 77. A bioremedial solution to reduce oxides of nitrogen and carbon in flue gases is to integrate flue gas emission to (A) microalgal culture. (B) fish culture. (C) mushroom culture. (D) sericulture. (GATE 2007; 2 Marks) 78. Match the recombinant products in Group I with their therapeutic applications in Group II. Group I Group II P. Human growth hormone 1. Pituitary dwarfism Q. Platelet growth factor 2. Chemotherapy induced thrombocytopenia R. Factor VIII 3. Hemophilia S. Erythropoietin 4. Anemia associated with chronic renal failure
(A) P–1; Q–2; R–3; S–4 (B) P–2; Q–1; R–3; S–4 (C) P–1; Q–4; R–3; S–2 (D) P–2; Q–4; R–3; S–1 (GATE 2007; 2 Marks) 79. Match the following marker genes in Group I with suitable selecting agent in Group II. Group I Group II P. nptII 1. Glyphosate Q. aroA 2. Phosphinothricin R. hpt 3. Kanamycin S. bar 4. Hygromycin B (A) P–1; Q–2; R–4; S–3 (B) P–3; Q–2; R–4; S–1 (C) P–2; Q–3; R–4; S–1 (D) P–3; Q–1; R–4; S–2 (GATE 2007; 2 Marks) 80. Match each parameter in Group I with the appropriate measuring device in Group II. Group I Group II P. Pressure 1. Photometer Q. Foam 2. Rotameter R. Turbidity 3. Diaphragm gauge S. Flow rate 4. Rubber sheathed electrode (A) P–3; Q–4; R–1; S–2 (B) P–1; Q–3; R–2; S–4 (C) P–4; Q–1; R–2; S–3 (D) P–1; Q–2; R–3; S–4 (GATE 2007; 2 Marks) 81. 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 (GATE 2007; 2 Marks) 82. How many kilograms of ethanol is produced from 1 kilogram of glucose in ethanol fermentation? (A) 2.00 (B) 0.20 (C) 0.51 (D) 0.05 (GATE 2007; 2 Marks)
200 GATE Biotechnology Chapter-wise Solved Papers 83. 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 R. Val-Phe-Met-Ala-Gly-Lys S. Phe-Gly-Try-Ser-Thr Chemical cleavage of the same protein with cyanogen bromide 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-Try-Gly-Lys-Phe-Val-Met (iii) Val-Arg-Val-Phe-Met The order of the peptides that gives the primary structure of the original protein is (A) P, Q, R, S (B) Q, P, R, S (C) Q, R, P, S (D) R, Q, P, S (GATE 2007; 2 Marks) Statement for Linked Answer Questions 84 and 85: In a fed-batch culture glucose solution is added with a flow rate of 2 m3 per day. The initial volume of the culture is 6 m3. 84. The volume of culture at the end of second day (neglect loss due to vaporization) is (A) 6 m3 (B) 8 m3 (C) 10 m3 (D) 12 m3 (GATE 2007; 2 Marks) 85. 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 (GATE 2007; 2 Marks) 86. 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 (GATE 2008; 1 Mark)
88. Match the items in Group I with the terms given in Group II. Group I Group II P. Lactobacillus and Bifidobacteria 1. Prebiotics Q. Polychlorobenzenes (PCBs) 2. Probiotics R. Fructo-oligosaccharides 3. Antibiotics S. β-Lactams 4. Xenohiotics (A) P–2; Q–4; R–1; S–3 (B) P–3; Q–4; R–1; S–2 (C) P–4; Q–1; R–2; S–3 (D) P–1; Q–3; R–4; S–2 (GATE 2008; 2 Marks) 89. Match the coefficients in Group I with their corresponding downstream processing Steps given in Group II. Group I Group II P. Sedimentation coefficient 1. Aqueous two-phase extraction Q. Partition coefficient 2. Ultrafiltration R. Rejection coefficient 3. Dialysis S. Activity coefficient 4. Centrifugation (A) P–3; Q–1; R–4; S–2 (B) P–2; Q–1; R–4; S–3 (C) P–4; Q–3; R–1; S–2 (D) P–4; Q–1; R–2; S–3 (GATE 2008; 2 Marks) 90. Match the bioreactor components in Group I with the most appropriate function given in Group II. Group I Group II P. Marine type impeller 1. Recirculation of medium Q. Draft tube 2. Aeration of medium R. Diaphragm valve 3. Animal cell cultivation S. Sparger 4. Sterile operation (A) P–4; Q–2; R–1; S–3 (B) P–3; Q–1; R–4; S–2 (C) P–3; Q–4; R–2; S–1 (D) P–2; Q–1; R–4; S–3 (GATE 2008; 2 Marks) 91. Evaluate the Michaelis constant for the following lipase catalyzed transesterification reaction for the production of biodiesel. k−1
2 Vegetable oil + Lipase Oil-lipase complex ⎯k⎯ → Biodiesel + Glyc
k1 k−1 87. Which of the following techniques is best suited for 2 Vegetable oil + Lipase Oil-lipase complex ⎯k⎯ → Biodiesel + Glycerol immobilizing an affinity ligand? k1 8 −1 −1 where k1 = 3 × 10 M s ; k–1 = 4 × 104 s−1 and k2 = (A) Physical adsorption 2 × 103 s−1 (B) Gel entrapment (A) 4.2 × 10−3 M (B) 14.0 × 10−4 M (C) Crosslinking with a polymer (C) 6.4 × 10−6 M (D) 1.4 × 10−4 M (D) Covalent linkage to a spacer arm (GATE 2008; 2 Marks) (GATE 2008; 1 Mark)
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92. In a chemostat, evaluate the dilution rate at the cell washout condition by applying Monod’s model with the given set of data: μmax = 1 h−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 −1 (C) 0.98 h (D) 1.02 h−1 (GATE 2008; 2 Marks) 93. Match the products in Group I with their producer organisms given in Group II. Group I Group II P. Ethanol 1. Streptomyces orientalis Q. L-Lysine 2. Saccharomyces cerevisiae R. Biopesticide 3. Corynebacterium glutanicum S. Vancomycin 4. Bacillus thuringiensis (A) (B) (C) (D)
P–2; Q–3; R–4; S–1 P–3; Q–4; R–1; S–2 P–4; Q–1; R–2; S–3 P–2; Q–1; R–4; S–3 (GATE 2008; 2 Marks)
94. A bacterial culture with an appropriate 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 (B) 0.95 (C) 1.00 (D) 1.05 (GATE 2008; 2 Marks) 95. 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 0.18 s−1 for the stirred tank bioreactor and the critical dissolved oxygen concentration is 20% of the saturation concentration (8 mg mL−1). 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−1 (C) 32 g L−1
(B) 26 g L−1 (D) 65 g L−1 (GATE 2008; 2 Marks)
Statement for Linked Answer Questions 96 and 97: 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 h and 100 mg h L−1 for the intercept and slope, respectively.
96. The respective values of the Monod kinetic constants μmax (h−1) and KS (mg L−1) are as follows: (A) 0.08, 8 (B) 0.8, 0.8 (C) 0.8, 80 (D) 8, 8 (GATE 2008; 2 Marks) 97. The same culture (with the μmax and KS values as computed above) is cultivated in a 10-litre chemostat being operated with a 50 mL min−1 sterile feed containing 50 g L−1 of substrate. Assuming an overall yield coefficient of 0.3 g DCW per g substrate, the respective values of the outlet biomass and substrate concentration are (A) 15 g L−1, 48 mg L−1 (B) 15 g L−1, 0.48 g L−1 (C) 48 g L−1, 15 g L−1 (D) 4.8 g L−1, 4.8 g L−1 (GATE 2008; 2 Marks) 98. A culture vessel 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. (GATE 2009; 1 Mark) 99. Match the properties in Group I with the downstream operations in Group II. Group I Group II P. Size 1. Extraction Q. Density 2. Distillation R. Volatility 3. Filtration S. Solubility 4. Sedimentation (A) P–3; Q–4; R–2; S–1 (B) P–4; Q–1; R–2; S–3 (C) P–4; Q–3; R–1; S–2 (D) P–3; Q–2; R–4; S–1 (GATE 2009; 2 Marks) 100. 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. S. Fed batch bioreactors are also called semi-batch bioreactors. (A) P, Q (B) Q, S (C) R, S (D) P, R (GATE 2009; 2 Marks)
202 GATE Biotechnology Chapter-wise Solved Papers 101. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: Bacterial growth is called synchronous when majority of the cells are in same stage of the bacterial cell cycle. Reason: 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. (GATE 2009; 2 Marks) 102. Match the products in Group I with their possible applications in Group II. Group I Group II 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 (C) P–2; Q–3; R–1; S–4 (D) P–2; Q–1; R–4; S–3 (GATE 2009; 2 Marks) 103. Match the products in Group I with their producer organisms in Group II. Group I Group II P. Ethanol from glucose 1. Aspergillus niger Q. Probiotics 2. Leuconostoc mesenteroides R. Citric acid 3. Saccharomyces cerevisiae S. Sauerkraut 4. Bifidobacterium (A) P–1; Q–3; R–2; S–4 (B) P–3; Q–4; R–1; S–2 (C) P–3; Q–4; R–2; S–1 (D) P–1; Q–4; R–3; S–2 (GATE 2009; 2 Marks) 104. In 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 cycling. (GATE 2010; 1 Mark) 105. An immobilized enzyme being used in a continuous plug flow reactor exhibits an effectiveness factor (h) of 1.2. The value of η being greater than 1.0 could be apparently due to
(A) substrate inhibited kinetics with internal pore diffusion limitation. (B) external pore diffusion limitation. (C) sigmoidal kinetics. (D) unstability of the enzyme. (GATE 2010; 2 Marks) 106. 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 cells. (B) 600 cm2 and 6 × 107 cells (C) 600 cm2 and 8300 cells. (D) 1200 cm2 and 8300 cells. (GATE 2010; 2 Marks) 107. Match the products in Group I with the microbial cultures in Group II used with their industrial production. Group I Group II P. Gluconic acid 1. Leuconostoc mesenteroides Q. L-Lysine 2. Aspergillus niger R. Dextran 3. Brevibacterium flavum S. Cellulase 4. Trichoderma reesei (A) P–2; Q–1; R–3; S–4 (B) P–1; Q–3; R–4; S–2 (C) P–2; Q–3; R–1; S–4 (D) P–3; Q–2; R–4; S–1 (GATE 2010; 2 Marks) 108. Thermal death of microorganisms in the liquid medium follows first order kinetics. If the initial cell concentration in the fermentation medium is 108 cells mL−1 and the final acceptable contamination level is 10−3 cells, for how long should 1 m3 medium be treated at temperature of 120°C (thermal deactivation rate constant = 0.23 min−1) to achieve acceptable load? (A) 48 min (B) 11 min (C) 110 min (D) 20 min (GATE 2010; 2 Marks) Common Data for Questions 109 and 110: A culture of Rhizobium is grown in a chemostat (100 in3 bioreactor), The feed contains 12 g L–1 sucrose, KS for the organism is 0.2 g L−1 and µm is 0.3 h−1. 109. The flow rate required to result in steady state concentration of sucrose as 1.5 g L−1 in the bioreactor will be (A) 15 m3 h−1 (B) 26 m3 h−1 3 −1 (C) 2.6 m h (D) 150 m3 h−1 (GATE 2010; 2 Marks)
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110. If Y = 0.4 g g−1 for the above culture and steady state cell concentration in the bioreactor is 4 g L−1 the resulting substrate concentration will be (A) 2 g L−1 (B) 8 g L−1 −1 (C) 4 g L (D) 6 g L−1 (GATE 2010; 2 Marks)
116. Match the microbial growth characteristics in Group I with the corresponding features in Group II. Group I Group II P. Growth associated product formation
Statement for Linked Answer Questions 111 and 112: 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−1) were found to be 47.3, 52.2, 49.2, 52.4, 49.1 and 46.3.
Q. Non-growth associated product formation R. Product inhibition
111. The mean value of acid concentration will be (A) 49.4 (B) 51.0 (C) 48.2 (D) 50.8 (GATE 2010; 2 Marks)
S. Substrate inhibition
(A) (B) (C) (D)
112. The standard deviation for the above results is (A) 2.49 (B) 3.0 (C) 1.84 (D) 5.91 (GATE 2010; 2 Marks) 113. In balanced growth phase of a cell, P. all components of a cell grow at the same rate. Q. specific growth determined by cell number or cell mass would be the same. 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 (GATE 2011; 1 Mark) 114. Substrate consumption in lag phase of microbial growth is primarily used for P. turnover of the cell material. Q. maintenance of intracellular pH. R. motility. S. increase in cell number. (A) P, Q and S only (B) Q, R and S only (C) P, Q and R only (D) S only (GATE 2011; 1 Mark) 115. Wash out (as defined by D = µmax) 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 (GATE 2011; 1 Mark)
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–1; Q–2; R–4; S–3 P–3; Q–2; R–l; S–4 P–2; Q–l; R–3; S–4 P–2; Q–3; R–4; S–l (GATE 2011; 2 Marks)
117. In a well aerated and agitated microbial culture, the ‘supply’ of oxygen is equal to ‘demand’ (uptake) of the growing culture. The K L a for such a system will be ( K L a = 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) ( rX ) / (C * − C )
(B) ( r ) / X (C * − C )
(C) (C * − C ) / ( rX )
(D) ( X ) / r (C * − C ) (GATE 2011; 2 Marks)
118. 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. (GATE 2011; 2 Marks)
204 GATE Biotechnology Chapter-wise Solved Papers 119. Maximum specific growth rate (µmax) 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. (GATE 2011; 2 Marks) Common Data for Questions 120 and 121: A microorganism grows in a continuous ‘chemostat’ culture of 60 m3 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. 120. What would be the yield YX/S (kg biomass/kg substrate)? (A) 0.562 (B) 0.462 (C) 0.362 (D) 0.162 (GATE 2011; 2 Marks) 121. 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 (GATE 2011; 2 Marks) 122. In an exponentially growing batch culture of Saccharomyces cerevisiae, the cell density is 20 g L−1 (DCW), the specific growth rate (µ) is 0.4 h−1 and substrate uptake rate (v) is 16 g L−1 h−1. The cell yield coefficient YX/S will be (A) 0.32 (B) 0.64 (C) 0.80 (D) 0.50 (GATE 2012; 2 Marks) 123. Match the entries in Group I with the process parameters in Group II. Group I Group II P. Clark electrode 1. Liquid level Q. Redox probe 2. Dissolved oxygen concentration R. Load cell 3. Vessel pressure S. Diaphragm gauge 4. pH (anaerobic process) (A) P–2; Q–1; R–3; S–4 (B) P–4; Q–2; R–3; S–1 (C) P–2; Q–4; R–1; S–3 (D) P–2; Q–1; R–4; S–3 (GATE 2012; 2 Marks) 124. Match the downstream processes in Group I with the products in Group II.
Group I P. Solvent extraction Q. Protein-A linked affinity chromatography R. Extractive distillation S. Salting out (A) P–2; Q–3; R–1; S–4 (B) P–4; Q–1; R–2; S–3 (C) P–4; Q–1; R–3; S–2 (D) P–2; Q–4; R–1; S–3
Group II 1. Lactic acid 2. Penicillin 3. Monoclonal antibody 4. Lipase
(GATE 2012; 2 Marks) 125. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: Cell mass yield of a methylotrophic yeast is more on methanol compared to glucose. Reason: Methanol has a greater degree of reductance compared to glucose. (A) Both (A) and (R) are correct and (R) is the correct reason for (A). (B) (A) is correct, (R) is false. (C) (A) is false, (R) is correct. (D) Both (A) and (R) are correct but (R) is not the correct reason for (A). (GATE 2012; 2 Marks) 126. Match the entries in Group I with the methods of sterilization in Group II. Group I Group II P. Serum 1. Autoclave Q. Luria broth 2. Membrane filtration R. Polypropylene tubes 3. UV irradiation S. Biological safety cabinets 4. Gamma irradiation 5. Dry heat (A) P–5; Q–3; R–1; S–4 (B) P–1; Q–4; R–5; S–3 (C) P–2; Q–1; R–4; S–3 (D) P–4; Q–1; R–3; S–5 (GATE 2012; 2 Marks) 127. A bacterial culture (200 µL containing 1.8 × 109 cells) was treated with an antibiotic Z (50 µg per mL) for 4 h at 37°C. After this treatment, the culture was divided into two equal aliquots. Set A: 100 mL was plated on Luria agar. Set B: 100 mL 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
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(A) bacteriostatic. (C) bacteriolytic.
(B) bacteriocidal. (D) apoptotic. (GATE 2012; 2 Marks)
Statement for Linked Answer Questions 128 and 129: β-Galactosidase bound to DEAE-cellulose is used to hydrolyze lactose to glucose and galactose in a plug flow bioreactor with a packed bed of volume 100 liters and a voidage (e) of 0.55. The K′M and V ′max for the immobilized enzyme are 0.72 g L-1 and 18 g L-1 h-1, respectively. The lactose concentration in the field stream is 20 g L-1, and a fractional conversion of 0.90 is desired. Diffusional limitations may be ignored. 128. 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 (GATE 2012; 2 Marks) 129. The feed flow rate required for the above bioconversion will be (A) 50 L h−1 (B) 55 L h−1 (C) 137 L h−1 (D) 550 L h−1 (GATE 2012; 2 Marks) 130. 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 ______________. (GATE 2013; 1 Mark) 131. 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−1 h−1) after 3h of steady state operation will be _____________. (GATE 2013; 1 Mark) 132. 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. (GATE 2013; 1 Mark) 133. 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). (GATE 2013; 1 Mark)
134. Match the commercial microbial sources in Group I with the products in Group II. Group I Group II P. Corynebacterium lilium 1. 2,3-Butane di-ol Q. Klebsiella oxytoca 2. Poly-β-hydroxybutyric acid R. Aspergillus niger 3. Glutamic acid S. Alcaligenes eutrophus 4. Citric acid (A) P–3; Q–1; R–2; S–4 (B) P–3; Q–1; R–4; S–2 (C) P–1; Q–3; R–2; S–4 (D) P–1; Q–3; R–4; S–2 (GATE 2013; 2 Marks) 135. Determine the correctness or otherwise of the following Assertion (A) and Reason (R). Assertion: Immobilization of plant cells can enhance secondary metabolite production during bioreactor cultivation. Reason: Immobilization protects the plant cells from shear forces in the bioreactor. (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. (GATE 2013; 2 Marks) 136. 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) __________. (GATE 2013; 2 Marks) 137. 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, mm 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 (g L−1) in the reactor at quasi-steady state. (GATE 2013; 2 Marks) Statement for Linked Answer Questions 138 and 139: During sterilization of a fermentation medium in a given bioreactor Δheating = 12.56, Δcooling = 7.48 and the total value of Δ required for whole sterilization process is 52, where Δ is the design criteria. 138. What is the value of Δholding? (A) 31.96 (C) 52.43
(B) 42.32 (D) 61.18 (GATE 2013; 2 Marks)
206 GATE Biotechnology Chapter-wise Solved Papers 139. What is the holding period (min) at a k value of 3.36 min−1? (A) 10.6 (C) 8.4
(B) 9.5 (D) 7.2 (GATE 2013; 2 Marks)
140. The unit for specific substrate consumption rate in a growing culture is (A)
g L⋅ h
(B)
(C)
g g⋅ h
(D)
g h
g moles L⋅h (GATE 2014; 1 Mark)
141. 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. (GATE 2014; 1 Mark) 142. Which of the following is employed for the repeated use of enzymes in bioprocesses? (A) Polymerization (B) Immobilization (C) Ligation (D) Isomerization (GATE 2014; 1 Mark) 143. 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) nND (B) nN2D 2 (C) nND (D) none of these (GATE 2014; 1 Mark) 144. The degree of reduction of ethanol is __________. (GATE 2014; 1 Mark) 145. Which of the following essential element(s) is/are required as major supplement to enhance the bioremediation of oil spills by the resident bacteria? (A) Sulfur (B) Nitrogen and phosphorus (C) Iron (D) Carbon (GATE 2014; 1 Mark) 146. 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) 0.05 0.5 5
Cell density (g L−1) 0.248 0.208 0
Substrate concentration (mM) 0.067 1.667 10
The maximum specific growth rate mm (in h−1), will be ________. (GATE 2014; 2 Marks) 147. 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. (GATE 2014; 2 Marks) 148. Which one of the following relations holds true for the specific growth rate (µ) of a microorganism in the death phase? (A) µ = 0 (B) µ < 0 (C) µ = µmax (D) 0 < µ < µmax (GATE 2015; 1 Mark) 149. 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 7 s and 17 s were 55% and 68% air saturation, respectively. The volumetric mass transfer coefficient K L a is ________ s−1. (GATE 2015; 2 Marks) 150. Saccharomyces cerevisiae produces ethanol by fermentation. The theoretical yield of ethanol from 2.5 g of glucose is __________ g. (GATE 2015; 2 Marks) 151. 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 (GATE 2015; 2 Marks) 152. A synchronous culture containing 1.8 × 105 monkey kidney cells was seeded into three identical flasks. The doubling time of these cells is 24 h. After 24 h, the cells from
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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. (GATE 2015; 2 Marks) 153. 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 (GATE 2015; 2 Marks) 154. 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. (GATE 2015; 2 Marks) 155. 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 µm = 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. (A) 0.85 (B) 0.65 (C) 0.45 (D) 0.25 (GATE 2015; 2 Marks) 156. 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. (GATE 2016; 1 Mark) 157. 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 (GATE 2016; 1 Mark) 158. A biological process is involved in the ___________ treatment of industrial effluent. (A) primary (B) secondary (C) tertiary (D) quaternary (GATE 2016; 1 Mark)
159. In dead-end filtration, rate of filtration is (A) directly proportional to the square root of pressure drop across the filter medium. (B) inversely proportional to the pressure drop across the filter medium. (C) inversely proportional to the viscosity of the solution. (D) inversely proportional to the square of viscosity of the solution. (GATE 2016; 1 Mark) 160. The power required for agitation of non-aerated medium in fermentation is __________kW.
Operating conditions are as follows: Fermenter 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 (GATE 2016; 1 Mark)
161. 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 (GATE 2016; 1 Mark) 162. Which of the following events occur during the stationary phase of bacterial growth? 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 cross-linking. (A) P, Q and S only (C) Q, R and S only
(B) P, S and T only (D) P, R and T only (GATE 2016; 2 Marks)
163. The empirical formula for biomass of an unknown organism is CH1.8O0.5N0.2. 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 ______________. (GATE 2016; 2 Marks)
208 GATE Biotechnology Chapter-wise Solved Papers 164. 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).
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. (GATE 2016; 2 Marks)
μmax S KS + S where, m is specific growth rate (h−1), µmax = 0.7 h−1, KS = 0.3 g L−1 and S is substrate concentration (g L−1). (GATE 2016; 2 Marks)
169. 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 _______. (GATE 2017; 1 Mark)
167. 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. 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 constant and does not undergo any deactivation during the reaction. (GATE 2016; 2 Marks)
X(t) 0
Time (t)
0
Time (t)
0
Time (t)
0
Time (t)
(B)
0
(C)
Given the rate equation: − rA =
168. An enzyme is immobilized on the surface of a non-porous spherical particle of 2 mm diameter. The immobilized enzyme is suspended in a solution having bulk
0
X(t)
166. A bioreactor is scaled up based on equal impeller tip speed. Consider the following parameters for small and large bioreactors: Parameters 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 (GATE 2016; 2 Marks)
170. Growth of a microbe in a test tube is modeled as dX X⎞ ⎛ = rX ⎜1 − ⎟ , where, X is the biomass, r is the ⎝ dt K⎠ growth rate, and K is the carrying capacity of the environment (r ≠ 0; K ≠ 0). If the value of starting biomass is K/100, which one of the following graphs qualitatively represents the growth dynamics? (A)
X(t)
165. Decimal reduction time of bacterial spores is 23 min at 121°C and the death kinetics follow first order. Oneliter medium containing 105 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. (GATE 2016; 2 Marks)
0
(D) X(t)
Microbial growth kinetics is given by μ =
0
(GATE 2017; 2 Marks)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 209
171. A zero-order liquid phase reaction A ⎯k⎯ → B is being carried out in a batch reactor with k = 10−3 mol min−1. If the starting concentration of A is 0.1 moles liter−1, the time (in minutes) taken by the system before A is exhausted in a 100 liters reactor is _________. (GATE 2017; 2 Marks) 172. During anaerobic growth, an organism converts glucose (P) into biomass (Q), ethanol (R), acetaldehyde (S), and glycerol (T). Every mole of carbon present in glucose gets distributed among the products as follows: 1 (C-mole P) → 0.14 (C-mole Q) + 0.25 (C-mole R) + 0.3 (C-mole S) + 0.31 (C-mole T) From 1800 grams of glucose fed to the organism, the ethanol produced (in grams) is _______. Given data: Atomic weights (Da) of C = 12, H = 1, O = 16, and N = 14 (GATE 2017; 2 Marks) 173. Which one of the following bioreactor configurations is the basis for a trickling biological filter? (A) Stirred tank (B) Packed bed (C) Air lift (D) Fluidized bed (GATE 2018; 1 Mark) 174. Shear stress versus shear rate behavior of four different types of fluids (I, II, III and IV) are shown in the figure below.
Shear stress (N.m–2)
II
IV I III
Shear rate (s–1)
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, IV-Bingham plastic (C) I-Newtonian, II-Pseudoplastic, III-Bingham plastic, IV-Dilatant (D) I-Newtonian, II-Bingham plastic, III-Pseudoplastic, IV-Dilatant (GATE 2018; 2 Marks) 175. Moist heat sterilization of spores at 121°C follows first-order kinetics as per the expression:
where N is the number of viable spores, t is the time, kd is the rate constant and dN /dt is the rate of change of viable spores. 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 value of 1 is (up to two decimal places) ________. (GATE 2018; 2 Marks) 176. An aqueous solution containing 6.8 mg L−1 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 ________. (GATE 2018; 2 Marks) 177. 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−1, then the volumetric mass transfer coefficient (KLa) (in s−1) of oxygen required to achieve a maximum cell concentration of 12 g L−1 is (up to two decimal places) ________. (GATE 2018; 2 Marks) 178. In a chemostat, the feed flow rate and culture volume are 100 mL h−1 and 1.0 L, respectively. With glucose as substrate, the values of µmax and KS are 0.2 h−1 and 1 g L−1, respectively. For a glucose concentration of 10 g L−1 in the feed, the effluent substrate concentration (in g L−1) is __________. (GATE 2018; 2 Marks) 179. At the end of a batch culture, glucose solution is added at a flow rate of 200 mL h−1. If the culture volume after 2 h of glucose addition is 1000 mL, the initial culture volume (in mL) is __________. (GATE 2018; 2 Marks) 180. 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) ________. (GATE 2018; 2 Marks) 181. The degree of reduction for acetic acid (C2H4O2) is ____________. (GATE 2019; 1 Mark) 182. Which one of the following is the unit of heat transfer coefficient? (A) W m2 K−1 (B) W m−2 K (C) W m−2 K−1 (D) W m2 K (GATE 2019; 1 Mark)
210 GATE Biotechnology Chapter-wise Solved Papers 183. Which one of the following is catabolized during endogenous metabolism in a batch bacterial cultivation? (A) Internal reserves (B) Extracellular substrates (C) Extracellular products (D) Toxic substrates (GATE 2019; 1 Mark) 184. Which one of the following can NOT be a limiting substrate if Monod’s growth kinetics is applicable? (A) Extracellular carbon source (B) Extracellular nitrogen source (C) Dissolved oxygen (D) Intracellular carbon source (GATE 2019; 1 Mark) 185. Yeast biomass (C6H10O3N) grown on glucose is described by the stoichiometric equation given below:
(C) They have no offset. (D) There is no corrective action if the error is a constant. (GATE 2019; 2 Marks) 189. Phenolic wastewater discharged from an industry was treated with Pseudomonas sp, in an aerobic bioreactor. The influent and effluent concentration of phenol were 10,000 and 10 ppm, respectively. The inlet feed rate of wastewater was 80 L h−1. The kinetic properties of the organism are as follows: Maximum specific growth rate (mm) = 1 h−1 Saturation constant (KS) = 100 mg L−1 Cell death rate (kd) = 0.01 h−1 Assuming that the bioreactor operates under ‘chemostat’ mode, the working volume required for this process is ____________ L (rounded off to the nearest integer). (GATE 2019; 2 Marks)
190. The dimensions and operating condition of a lab-scale are as follows: C6 H12 O6 + 0.48NH 3 + 3O 2 → 0.48C6 H10 O3 N + 3.12CO 2 + 4.32Hfermenter 2O Volume = 1 L C6 H12 O6 + 0.48NH 3 + 3O 2 → 0.48C6 H10 O3 N + 3.12CO 2 + 4.32H 2 O Diameter = 20 cm The amount of glucose needed for the production of Agitator speed = 600 rpm −1 50 g L of yeast biomass in a batch reactor with a workRatio of impeller diameter to fermenter diameter = 0.3 ing volume of 100,000 L is _____________ kg (rounded off to the nearest integer). This fermenter needs to be scaled up to 8,000 L for a largescale industrial application. If the scale-up is based on (GATE 2019; 2 Marks) constant impeller tip speed, the speed of the agitator in the 186. Match the instruments in Group I with their correspondlarger reactor is __________rpm. Assume that the scale-up ing measurements in Group II. factor is the cube root of the ratio of fermenter volumes. Group I Group II (GATE 2019; 2 Marks) P. Manometer 1. Agitator speed 191. Which of the following statements is ALWAYS Q. Rotameter 2. Pressure difference CORRECT about an ideal chemostat? R. Tachometer 3. Cell number P. Substrate concentration inside the chemostat is equal S. Hemocytometer 4. Air flow rate to that in the exit stream. (A) P–4; Q–1; R–2; S–3 Q. Optimal dilution rate is lower than critical dilution (B) P–3; Q–4; R–1; S–2 rate. (C) P–2; Q–4; R–1; S–3 R. Biomass concentration increases with increase in (D) P–2; Q–1; R–4; S–3 dilution rate. (GATE 2019; 2 Marks) S. Cell recirculation facilitates operation beyond critical dilution rate. 187. In a cross-flow filtration process, the pressure drop (DP) (A) P and Q only (B) P, R and S only driving the fluid flow is 2 atm, inlet feed pressure (Pi) is (C) P and S only (D) P, Q and S only 3 atm and filtrate pressure (Pf) is equal to atmospheric (GATE 2019; 2 Marks) pressure. The average transmembrane pressure drop (DPm) is ___________ atm. 192. An industrial fermenter containing 10,000 L of medium (GATE 2019; 2 Marks) needs to be sterilized. The initial spore concentration in 188. Which one of the following statements is CORRECT about proportional controllers? (A) The initial change in control output signal is relatively slow. (B) The initial corrective action is greater for larger error.
the medium is 106 spores mL−1. The desired probability of contamination after sterilization is 10−3. The death rate of spores at 121°C is 4 min−1.
Assume that there is no cell death during heating and cooling phases. The holding time of the sterilization
Chapter 12 • Bioprocess Engineering and Process Biotechnology 211
process is ___________min (rounded off to the nearest integer). (GATE 2019; 2 Marks) 193. Which of the following factors can influence the lag phase of a microbial culture in a batch fermenter?
P. Inoculum size Q. Inoculum age R. Medium composition (A) P and Q only (C) P and R only
(B) Q and R only (D) P, Q and R (GATE 2019; 2 Marks)
ANSWER KEY 1. (B)
2. (A)
7. (6.26 × 1028; 1mg; 5mg)) 13. (D)
3. (B)
4. (B)
5. (A)
6. (P. True; Q. False; R. False; S. True; T. True)
8. (D)
9. (C)
10. (B)
11. (A) 12. (D)
15. (40 mM)
14. (C)
16. (*) 17. (*) 18. (*)
19. (P–2; Q–8; R–6; S–7; T–1)
20. (P–3; Q–4; R–6; S–5; T–2)
21. (A)
22. (B)
23. (D)
24. (D)
26. (D)
28. (*)
29. (*)
30. (*)
25. (B)
27. (D)
31. (a: 1.92 g L ; 0.384 g L h ; b: 0.2 h )
32. (P–4; Q–5; R–2; S–1; T–3) 33. (P–6; Q–3; R–4; S–5; T–7)
34. (D)
35. (D)
36. (B)
37. (A) 38. (C)
39. (B)
40. (B)
41. (A)
42. (A)
43. (D)
44. (C) 45. (D)
46. (C)
47. (B)
48. (A)
49. (B)
50. (B)
51. (C) 52. (D)
53. (D)
54. (B)
55. (B)
56. (C)
57. (C)
58. (A) 59. (B)
60. (A)
61. (A)
62. (B)
63. (D)
64. (D)
65. (C) 66. (C)
67. (B)
68. (A)
69. (B)
70. (B)
71. (B)
72. (D) 73. (A)
74. (C)
75. (A)
76. (C)
77. (A)
78. (A)
79. (D) 80. (A)
81. (A)
82. (C)
83. (B)
84. (C)
85. (B)
86. (B) 87. (D)
88. (A)
89. (D)
90. (B)
91. (D)
92. (C)
93. (A) 94. (C)
95. (D)
96. (C)
97. (A)
98. (B)
99. (A)
100. (C) 101. (C)
102. (A)
103. (B)
104. (D)
105. (A)
106. (B)
107. (C)
108. (C)
109. (B)
110. (A)
111. (A)
112. (A)
113. (B)
114. (C)
115. (B)
116. (B)
117. (A)
118. (A)
119. (B)
120. (A)
121. (A)
122. (D)
123. (C)
124. (A)
125. (A)
126. (C)
127. (A)
128. (D)
129. (A)
130. (0.5)
131. (18 g L−1 h−1)
132. (45 rpm)
133. (0.86)
134. (B)
135. (A)
136. (76.75 g L−1) 137. (0.1 g L−1) 138. (A)
139. (B)
140. (C)
141. (C)
142. (B)
147. (B)
148. (B)
152. (2 × 10 cells) 153. (B) 159. (C)
−1
146. (0.8 h ) −1
−1
−1
4
160. (625 kW)
165. (3.7 × 10 spores)
143. (A)
144. (6)
145. (B)
149. (0.0732 s )
150. (1.27 g)
151. (C)
154. (1.21 h)
155. (B)
156. (C)
157. (B)
158. (B)
161. (B)
162. (A)
163. (1.95)
164. (3.7 g L )
−1
−1
166. (A) 167. (4987.5 L) 168. (6.18 mM) 169. (0.9 g L–1 min–1)
7
170. (A)
−1
171. (104 min)
172. (345 g)
173. (B)
174. (D)
175. (23.03 min) 176. (2)
177. (0.167 s )
178. (1 g L )
179. (600 mL) 180. (6)
181. (4)
182. (C)
184. (D)
185. (13020.83 kg) 186. (C)
188. (B)
189. (80 L) 190. (30 rpm)
191. (D)
192. (5.2)
–1
–1
193. (D)
187. (2 atm)
183. (A)
212 GATE Biotechnology Chapter-wise Solved Papers
Answers with Explanations 1. Topic: Sterilization of Air and Media Continuous heat sterilization is more efficient than sterilization in situ as: • Less heat is used in the sterilizing of medium. • The fermenter does not have to be heated by steam. Answer (B) 2. Topic: Production of Biomass and Primary/Secondary Metabolites The growth phase is called the trophophase, whereas the production phase is termed the idiophase. As the exponential growth of the microorganisms cease (i.e. as the trophophase ends), they enter idiophase. Idiophase is characterized by secondary metabolism wherein the formation of certain metabolites, referred to as secondary metabolites (idiolites) occurs. Answer (A) 3. Topic: Production of Biomass and Primary/Secondary Metabolites • Saccharomyces cerevisiae is a single-celled eukaryote that is frequently used in scientific research. It is used in various industrial processes from bread making to beer brewing. • Penicillium chrysogenum is a eukaryote. It is a major industrial producer of the β-lactam antibiotic penicillin. • Bacillus subtilis is an attractive bacterial organism for industrial use mainly because of its straightforward genetic manipulation, favorable growth characteristics in industrial ‐scale fermentations. • Streptomyces griseus is a species of bacteria and is a well-known producer of antibiotics and other such commercially significant secondary metabolites. Answer (B)
concluded that this was due to the presence of a life-form that could function only in the absence of oxygen. This led to his introduction of the terms aerobic and anaerobic to designate organisms that live in the presence or absence of oxygen, respectively. Answer (A) 6. Topic: Multiple Topics P. True. There are certain static and dynamic attributes that every biosensor possesses. The optimization of these properties is reflected on the performance of the biosensor. They include: Selectivity, Reproducibility, Stability, Sensitivity and Linearity. Q. False. Activated-sludge method, sewage-treatment process in which sludge, the accumulated, bacteriarich deposits of settling tanks and basins, is seeded into incoming waste water and the mixture agitated for several hours in the presence of an ample air supply. R. False. In a fermenter, impellers increase aeration capacity by producing high shear forces. S. True. Pressure-cycle fermenter is of two types: Airlift type and Deep-shaft type. In air-lift type, the air is introduced centrally at the bottom making the medium less dense. In deep-shaft, air is introduced from the top. T. True. Monoclonal antibody-based treatment of cancer has been established as one of the most successful therapeutic strategies for both hematologic malignancies and solid tumors. The modulation of immune system interplay with tumor cells through targeting of T-cell receptors has emerged as a powerful new therapeutic strategy for tumor therapy and to enhance cancer vaccine efficacy. Answer (P. True; Q. False; R. False; S. True; T. True)
7. Topic: Kinetics of Microbial Growth 4. Topic: Bioremediation Total protein in a cell = 0.1 pg 5 Environmentally friendly and cost saving features are Quantity of Protein PZ per cell = 0.1 × = 0.005 pg = 5 × 10 −15 g 100 amongst the major advantages of bioremediation compared 5 … (i) .1 × = 0.005 pg = 5 × 10 −15 g to both chemical and physical methods of 0remediation. 100 Bioremediation is a means of cleaning up contaminated We know, 1 Da = 1.6 × 10 −24 g environments by exploiting the diverse metabolic abilities 5 × 10 −15 of microorganisms to convert contaminants to harmless Therefore, 5 × 10 −15 g = = 3.125 × 10 9 Da products by mineralization, generation of carbon (IV) 1.6 × 10 −24 oxide and water, or by conversion into microbial biomass. Molecular weight of protein PZ = 30000 Da It is a relatively simpler and slower technique. 3.125 × 10 9 9 3 . 125 × 10 Da = = 1.04 × 105 mol Answer (B) 30000 Given that Avogadro number = 6.02 × 1023 molecules per 5. Topic: Sterilization of Air and Media mole. So, Pasteur discovered that the fermentation process could be 1.04 × 105 mol = 6.02 × 10 23 × 1.04 × 105 arrested by passing air (i.e., oxygen) through the fermenting fluid, a process known today as the Pasteur effect. He
= 6.26 × 10 28 molecules
Chapter 12 • Bioprocess Engineering and Process Biotechnology 213
charged support. On catalysis of substrate to a negatively charged product by the enzyme, the product will repeal From (i) from the surface and would easily get detached. Thus, it Quantity of Protein PZ per mL = 2 × 108 × 5 × 10 −15 g = 10 −6 g would = 1 μg faster the rate of catalysis and ease the process of separation of product. 2 × 108 × 5 × 10 −15 g = 10 −6 g = 1 μg Answer (A) During stationary phase, number of cells per mL = 1 × 10 9 During log phase, number of cells per mL = 2 × 108
From (i) 12. Topic: Batch, Feb-Batch and Continuous Processes Quantity of Protein PZ per mL = 1 × 10 9 × 5 × 10 −15 g = 5 × 10 −6 gA=fed-batch 5 μg culture is a semi-batch operation in which × 10 9 × 5 × 10 −15 g = 5 × 10 −6 g = 5 μg the nutrients necessary for cell growth and product forAnswer (6.26 × 1028; 1mg; 5mg) 8. Topic: Production of Biomass and Primary/Secondary Metabolites The formation of high concentrations of organic chemicals, solvents such as ethanol, butanol or isobutanol all have detrimental effects on both growth and product accumulation. The accumulation results in multiple cellular changes, such as slower growth rates, formation of undesired by-products and low productivities. The main target in the cell of the biofuel molecules is the membrane, sensitive to the accumulating products. Unfortunately, the energy intensive downstream process in lignocellulosic bioethanol production still limits the ability to compete with conventional bioethanol or petroleum. A process setup that provides possibilities for heat integration would consequently result in a more efficient overall process and increase its competitiveness. Answer (D) 9. Topic: Engineering Principle of Bioprocessing The amount distillate recovered is equal to the fractional recovery. After each step of unit operation, fractional recovery is 0.8. Thus, value of fractional recovery after four steps will be 0.8 × 0.8 × 0.8 × 0.8 = 0.4096 ≈ 0.41 Answer (C) 10. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes An immobilized enzyme is the one which has been attached to or enclosed by an insoluble support medium (termed as carrier) or one where the enzyme molecules have been cross-linked to each other, without loss of catalytic activity. An important factor to enhance the rate of reaction is continuous replacement of the liquid containing the substrate and product. Separation of final desired product would help in overcoming feedback inhibition and help in saving cost of downstream processing. Answer (B) 11. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes A positively charged substrate will be more attracted towards the enzyme immobilized on the negatively
mation are fed either intermittently or continuously via one or more feed streams during the course of an otherwise batch operation. The culture broth is harvested usually only at the end of the operational period, either fully or partially (the remainder serving as the inoculum for the next repeated run). This process may be repeated (repeated fed-batch) a number of times if the cells are fully viable and productive. Thus, there are one or more feed streams but no effluent during the course of operation. Therefore, the culture volume increases during the course of operation until the volume is full. Thereafter, a batch mode of operation is used to attain the final results. Thus, the fed-batch culture is a dynamic operation. Answer (D) 13. Topic: Sterilization of Air and Media The most common technique for sterilization of largescale process equipment is moist-heat sterilization with steam. The steam-sterilization process typically involves the following three steps: • Displacement of air with steam and heating to sterilization temperature, usually 121°C (250°F). • Holding at sterilization temperature for a minimum of 15 min (although 30 min to an hour is more common). • Cool down, which includes introduction of sterile air to prevent a vacuum condition resulting from steam collapse as the equipment cools off. Answer (D) 14. Topic: Bioremediation Aerobic sludge digestion is one process that may be used to reduce both the organic content and the volume of the sludge. Under aerobic conditions, a large portion of the organic matter in sludge may be oxidized biologically by microorganisms to carbon dioxide and water. The process results in approximately 50% reduction in solids content. Aerobic sludge digestion facilities may be designed for batch or continuous flow operations. The amount of bacteria, and the volume of sludge are reduced through this process of endogenous respiration. Solids at concentrations of 2–45 ppm are removed, and the clarified liquid supernatant is decanted and recycled to the wastewater treatment plant. In a continuous flow system, an aeration tank is utilized, followed by a settling tank.
2
N
Penicillin G
CH3
O
COOH
214 GATE Biotechnology Chapter-wise Solved Papers The advantages of using aerobic digestion, as compared to the use of anaerobic digestion include: (1) simplicity of operation and maintenance; (2) lower capital costs; (3) lower levels of biochemical oxygen demand (BOD); (4) greater sludge fertilizer value; and (5) shorter retention periods. Disadvantages include: (1) higher operating costs, especially energy costs; (2) no useful byproduct such as methane gas that is produced in anaerobic digestion; (3) variability in the ability to dewater to reduce sludge volume; (4) lower reduction in volatile solids; and (5) unfavorable economics for larger wastewater treatment plants. Answer (C) 15. Topic: Kinetics of Microbial Growth For Monod growth model: Maximum specific growth rate, µm = 0.5 h−1 Dilution rate, D = 0.1 h−1 Limiting nutrient concentration, S = 10 mM For chemostat, KS =
S ( μ m − D ) 10(0.5 − 0.1) = = 40 μM D 0.1 Answer (40 mM)
16. Topic: Production of Biomass and Primary/Secondary Metabolites The amino acid, L-asparagine, is a nutritional requirement of both normal and cancer cells. Unlike normal cells, however, certain leukemic cells cannot synthesize asparagine and must consequently rely on an external supply in the plasma and tissues. The administration of the enzyme, L-asparaginase, destroys this free source of asparagine, starving and killing certain cancer cells. Answer (*) 17. Topic: Production of Biomass and Primary/Secondary Metabolites Penicillin G acylase (PGA) is one of very important industrial enzymes used in the production of polysynthetic b-lactam antibiotics. This enzyme catalyzes the hydrolysis of the amidic bond of penicillin G with the development of 6-aminopenicillanic acid which serves as the initial substance for the production of semisynthetic penicillins. O H N
C H2
H
H S N
Penicillin G
CH3
+
H2O
CH3
O
COOH
Penicillin G Acylase
H
H2N
H S N
O
O
CH3 CH3
+
C H2
H
Penicillin G Acylase
H
H2N
H S N
O
O
CH3 CH3
+
C H2
H
COOH 6-aminopenicillanic acid (6-APA)
Phenylacetic acid (PAA)
Answer (*) 18. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Three important methods of enzyme penicillin G acylase immobilization are: • Support binding, usually by covalent attachment, to a prefabricated carrier. The carrier can be a synthetic resin, a biopolymer or an inorganic material. • Entrapment via inclusion of an enzyme in a polymer network (gel lattice) or a membrane device such as a hollow fiber or a microcapsule. The physical restraints generally are too weak, however, to prevent enzyme leakage entirely additional covalent attachment is generally required. • Crosslinking of enzyme aggregates or crystal using a bifunctional reactant, to prepare carrier-less macroparticles. Answer (*) 19. Topic: Production of Biomass and Primary/Secondary Metabolites • Aspergillus niger is used for the production of citric acid by submerged fermentation from different sources of carbohydrates, such as molasses and starchbased media. • Saccharomvces cerevisiae is the most employed yeast used for ethanol production. Wide range of substrates has been used for ethanol production such as lignocellulose, molasses, sweat sorghum cane extract, starchbased substrate and other wastes. • Penicillium chrysogenum can be considered as an excellent cell factory for the production of β-lactam antibiotics as well as other secondary metabolites. It is because of the high capacity of its protein secretion machinery. • Lactobacillus casei strains naturally produced large amounts of diacetyl. It is an important flavoring compound that determines specific characteristics of products such as fermented milks and, at very low concentrations, is also responsible for the characteristic “buttery” aroma of milk products. • Corynebacterium glutamicum is used in industry to produce L-lysine through an aerobic process. It is because it is able to metabolize a variety of carbohydrates, alcohols and organic acids as carbon and
Chapter 12 • Bioprocess Engineering and Process Biotechnology 215
energy sources for growth and for amino acid production. It is also able to metabolize inorganic nitrogen, such as ammonia. • Closteridium acetobutylicum acting on starch produces both acetone and butanol. • Large scale industrial production of vitamin B12 occurs via microbial fermentation, predominantly utilizing Pseudomonas denitrificans, Propionibacterium shermanii, or Sinorhizobium meliloti. • Erythromycin is produced from the strain of actinomycete Saccharopolyspora erythraea formerly known as Streptomyces erythraeus. Answer (P–2; Q–8; R–6; S–7; T–1) 20. Topic: Unit Operations in Solid-Liquid Separation and Liquid-Liquid Extraction • Drying is an essential component of product formulation. It basically involves the transfer of heat to a wet product for removal of moisture. Most of the biological products of fermentation are sensitive to heat, and therefore require gentle drying methods. Based on the method of heat transfer, drying devices may be categorized as contact, convection, radiation dryers. • Sedimentation is widely used for yeast separation during alcohol production and in waste-water treatment. The rate of particle sedimentation is a function of both size and density. Larger the size, greater its density, and faster the rate of sedimentation. Therefore, for quick separation, the difference in density between the particle and the medium must be large with low viscosity. • Membrane filtration has become a common separation technique in industrial biotechnology. It can be conveniently used for the separation of biomolecules and particles, and for the concentration of fluids. The membrane filtration technique basically involves the use of a semi permeable membrane that selectively retains the particles/molecules that are bigger than the pore size while the smaller molecules pass through the membrane pores. • When the product is intracellular, it is necessary to disrupt the cells/cell membrane so as to release the product. Cell disruption can be achieved by mechanical and non-mechanical methods. Some of the methods used are ultra-sonication and application of lysozyme. • The biological products of fermentation (proteins, pharmaceuticals, diagnostic compounds and research materials) are very effectively purified by chromatography. It is basically an analytical technique dealing with the separation of closely related compounds from a mixture. Chromatography usually consists of a stationary phase and mobile phase.
Desired product in culture broth
Intracellular product
Extracellular product
Cell disruption, (Physical, chemical enzymatic methods) Broth with solids and liquid
Solid-liquid separation (flotation, flocculation, filtration, centrifugation, sedimentation)
Concentration (evaporation, liquidliquid extraction, membrane filtration, precipitation, adsorption)
Purification by chromatography (gel-filtration, ion-exchange, affinity, hydrophobic interaction)
Formulation (drying, freeze-drying, crystallization)
Final product
Answer (P–3; Q–4; R–6; S–5; T–2) 21. Topic: Production of Biomass and Primary/Secondary Metabolites 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. Answer (A)
216 GATE Biotechnology Chapter-wise Solved Papers 22. Topic: Mass and Heat Transfer 25. Topic: Engineering Principle of Bioprocessing In aerobic respiration, the oxidation of organic comAlcohol content in various standard drinks is as follows: pounds such as glucose takes place through the action • Vodka: 40–95% of oxygen into carbon dioxide and water. The energy • Gin: 36–50% released is sufficient to synthesize approximately 38 mol• Rum: 36–50% ecules of ATP. • Whiskey: 36–50% C6 H12 O6 + 38ADP + 38Pi + 6O 2 ⎯⎯ → 6CO 2 + 38H 2 O + 38ATP • Tequila: 50–51% • Brandy: 35–60% C6 H12 O6 + 38ADP + 38Pi + 6O 2 ⎯⎯ → 6CO 2 + 38H 2 O + 38ATP • Fortified wine: 16–24% Now, if 30.6 kJ of energy is stored in 1 molecule of ATP. • Unfortified wine: 14–16% Then the energy obtained from 38 molecules of ATP is • Beer: 4–8% 38 × 30.6 = 1162.8 kJ. From thermodynamic data, we have that the energy released from complete oxidation of • Malt beverage: 15% glucose = 2805 kJ per mol. Answer (B) Hence, the efficiency of aerobic respiration is 26. Topic: Immobilization of Biocatalysts (Enzymes and 1162.8 Cells) for Bioconversion Processes Efficiency = × 100 = 41.45% 2805 Crosslinking is performed by formation of intermolecAnswer (B) ular cross-linkages between the enzyme molecules by means of bi- or multifunctional reagents. The most com 23. Topic: Process Scale-Up, Economics and Feasibility monly used crosslinking reagent is glutaraldehyde as it Analysis is economical and easily obtainable in large quantities. The process characteristics which have been suggested to It combines amino groups through a Schiff base linkage. be maintained constant during scale-up include: The enzymes can be covalently linked to amine carrying • Reactor geometry supports activated by thionyl chloride and cyanogen bro• Volumetric mass transfer coefficient, (KLa) mide. • Maximum shear Answer (D) • Power input per unit volume of liquid, (Pg/V) 27. Topic: Sterilization of Air and Media • Volumetric gas flow rate per unit volume of liquid, Q/V Many ways have been suggested for sterilizing air. These or VVM include destruction of microorganisms by: dry heat-gas • Superficial gas velocity, vs fired or electrical; adiabatic compression; and irradiation • Mixing time or removal of microorganism. The removal of microor• Impeller Reynolds number, Rei = QNDi2 /μ ganisms involve: scrubbing, electrostatic precipitation, • Momentum factor sieving, and filtration fibrous or granular beds. Of these, only adiabatic compression, filtration through beds of Answer (D) fibrous materials, and filtration through beds of granular 24. Topic: Sterilization of Air and Media materials have found widespread usage on an industrial The random arrangement of fibers in the three-dimenscale. Filtration through beds of fibrous materials such sional space of the fibrous filters makes the structure as glass wool is by far the most common method of the of the pores present inside more complex, resulting in three. higher probability of capturing of particles. By using Answer (D) appropriate fibers and suitable manufacturing process, the structure of the filters can be designed such a way 28. Topic: Production of Biomass and Primary/Secondary Metabolites that the dust is loaded in the interior portion of the filters, rather than the dust is deposited only on the surface of the Citric acid is produced at industrial scale using strains of filters. This provides a low pressure drop across the filAspergillus niger. A. niger requires certain trace metals ter, resulting in low running cost. The fibrous filter media for growth. However, a limitation by other trace elements are known to capture the airborne particles of different is necessary for citric acid production, especially in the sizes due to Brownian diffusion, direct interception, submerged fermentation. The metals that should be in inertial impaction, gravity settling, and particle sieving limiting concentrations are Zn, Mn, Fe, Cu and heavy mechanisms. metals. Answer (D) Answer (*)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 217
32. Topic: Production of Biomass and Primary/Secondary Metabolites • The TaqI DNA polymerase found in Thermus aquaticus remains stable even at very high temperatures. Because of this stability it can be used in the process known as5the polymerase chain reaction, or PCR. PCR Power = Power Number × Density × ( Agitator speed ) 3 × (Impeller disiameter ) a technique used to amplify a piece of DNA by in 3 5 vitro enzymatic replication. ower = Power Number × Density × ( Agitator speed ) × (Impeller diameter ) • Acetobacter aceti is economically important because P = N p ρN i3 D 5 it is used in the production of vinegar by converting the ethanol in wine into acetic acid. Answer (*) • Bacillus thuringiensis (commonly known as Bt) is an 30. Topic: Sterilization of Air and Media insecticidal bacterium, marketed worldwide for conSulfonamides are bacteriostatic drugs. They do not kill bactrol of many important plant pests - mainly caterpillars teria but inhibit growth and multiplication by interfering of the Lepidoptera (butterflies and moths). with their enzyme systems. Sulfonamide is an organic com• The first established lager yeast strain is known as the pound consisting of an aniline derivatized with a sulfonamide bottom fermenting Saccharomyces carlsbergensis, group. By acting as a substrate analogue, it competitively which was originally termed Unterhefe No. 1 by Emil inhibits enzymatic reactions involving para-aminobenzoic Chr. Hansen and has been used in production in since acid (PABA). PABA is needed in enzymatic reactions that 1883. produce folic acid, which acts as a coenzyme in the synthe• HindIII is a type II site-specific deoxyribonuclease sis of purines and pyrimidines. Mammals do not synthesize restriction enzyme isolated from Hemophilus influenzae their own folic acid so are unaffected by PABA inhibitors, that cleaves the DNA palindromic sequence AAGCTT which selectively kill bacteria. in the presence of the cofactor Mg2+ via hydrolysis. The Answer (conceptual) cleavage of this sequence between the AA’s results in 5’ overhangs on the DNA called sticky ends: 31. (a) Topic: Engineering Principle of Bioprocessing 29. Topic: Aeration and Agitation For a standard cylindrical vessel geometry, the power consumed by the impeller(s) at a specified rotational speed, for a liquid with known density, can be determined from the Power Number correlation.
Given, volume of tank, V = 1000 L Initial concentration, S0 = 5 g L−1 Flow rate, F = 200 L h−1 For Monod growth model: Maximum specific growth rate, µm = 0.3 h−1 Yield coefficient, YX/S = 0.4 g g−1 Half-velocity constant, KS = 0.1 g L−1 Dilution rate, D =
F 200 L h −1 = = 0.2 h −1 V 1000 L
For steady state, Concentration of biomass, X = YX/S ( S0 − S ) Final substrate concentration, S can be calculated as: K ⋅D 0.1 × 0.2 0.02 = = = 0.2 g L−1 S= S μ m − D 0.3 − 0.2 0.1 Now, X = 0.4(5 − 0.2) = 0.4 × 4.8 = 1.92 g L−1 Now, we know, cell productivity = D × X = 0.2 h−1 × 1.92 g L−1 = 0.384 g L−1 h−1 Answer (1.92 g L−1; 0.384 g L−1 h−1)
(b) Topic: Engineering Principle of Bioprocessing From (a), it is clear that dilution rate which gives maximum biomass productivity = 0.2 h−1 Answer (0.2 h−1)
5 ′ − A ↓ A G C T T − 3′ 3′ − T T C G A ↑ A − 5 ′ Answer (P–4; Q–5; R–2; S–1; T–3)
33. Topic: Production of Biomass and Primary/Secondary Metabolites • Diosgenin, a phytosteroid sapogenin, is the product of hydrolysis by acids, strong bases, or enzymes of saponins, extracted from the tubers of Dioscorea wild yam, such as the Kokoro. • The Rauwolfia indole alkaloid, ajmalicine, has been found to have broad application in the treatment of circulatory diseases, especially in the relief of obstruction of normal cerebral blood flow. • Shikonin is a major component of zicao (purple gromwell, the dried root of Lithospermum erythrorhizon), a Chinese herbal medicine with various biological activities, including inhibition of human immunodeficiency virus (HIV) type 1 (HIV-1). Its chemical structure is a naphthoquinone derivative. • Digoxin is a cardiotonic glycoside obtained mainly from Digitalis lanata. It increases the force of contraction of the heart by inhibiting sodium-potassium adenosine triphosphatase (ATPase). • Scopolamine is a tropane alkaloid from Solanaceae, especially Datura and Scopola carniolica. It is a
218 GATE Biotechnology Chapter-wise Solved Papers medication used to treat motion sickness and postoperative nausea and vomiting. • Holoside is a compound containing one or more identical, glycosidically linked carbohydrates that yields only glycoses on hydrolysis. • Pyrrolizidine alkaloids (PAs), sometimes referred to as necine bases, are a group of naturally occurring alkaloids based on the structure of pyrrolizidine. Pyrrolizidine alkaloids are produced by plants as a defense mechanism against insect herbivores. Answer (P–6; Q–3; R–4; S–5; T–7) 34. Topic: Various Types of Microbial and Enzyme Reactors Papain is a proteolytic enzyme extracted from the raw fruit of the papaya plant. Proteolytic enzymes help break down proteins into smaller protein fragments called peptides and amino acids. This is why papain is a popular ingredient in meat tenderizer. It is also used as a stabilizing agent in the brewing industry, Answer (D) 35. Topic: Various Types of Microbial and Enzyme Reactors Transformation of xenobiotics include biodegradation, mineralization or transformation. • Mineralization is the complete decomposition of an organic compound into inorganic elements. • Biodegradation is a process of decomposition of organic compounds into inorganic elements taking place with the participation of living organisms with the simultaneous accretion of biomass. • Biotransformation is the process leading to the change of the structure of the original chemical compound to such degree that its original characteristic properties change as well. It modifies not only the physicochemical properties of compounds, such as solubility or (bio)availability, but also the toxicity level of the given xenobiotic. Answer (D) 36. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes An immobilized enzyme is an enzyme attached to an inert, insoluble material—such as calcium alginate (produced by reacting a mixture of sodium alginate solution and enzyme solution with calcium chloride). This can provide increased resistance to changes in conditions such as pH or temperature. It also lets enzymes be held in place throughout the reaction, following which they are easily separated from the products and may be used again. Answer (B)
37. Topic: Membrane Based Bioseparation Method Dialysis works on the principles of the diffusion of solutes and ultrafiltration of fluid across a semi-permeable membrane. Diffusion is a property by virtue of which substances in any liquid tend to move from an area of high concentration to an area of low concentration. Answer (A) 38. Topic: Production of Biomass and Primary/Secondary Metabolites (A) Correct. Penicillin was the first important commercial product produced by aerobic submerged fermentation process. Thus, sterile air is provided in the reactor continuously. (B) Correct. Extraction of penicillin is carried out by employing counter current extraction method. The pH of the liquid after separation of the mycelium is adjusted to 2.0 to 2.5 by adding phosphoric or sulphuric acid. This treatment converts penicillin into anionic form. (C) Incorrect. Lysin inhibits penicillin biosynthesis. It is because both lysine and penicillin biosynthesis involve L-aminoadipate as an intermediate. Lysine is known to inhibit homocitrate synthase, an early enzyme in its biosynthetic path. (D) Correct. Glucose repressed the formation of penicillin-synthesizing enzymes, but had no effect on the activity of these enzymes. This suggests that glucose represses but does not inhibit penicillin biosynthesis. Answer (C) 39. Topic: Kinetics of Microbial Growth • Lag phase is the initial phase is characterized by cellular activity but not growth. These cells increase in size, but no cell division occurs in the phase. • After the lag phase, bacterial cells enter the exponential or log phase. This is the time when the cells are dividing by binary fission and doubling in numbers after each generation time. • Eventually, the population growth experienced in the log phase begins to decline as the available nutrients become depleted and waste products start to accumulate. Bacterial cell growth reaches a plateau, or stationary phase, where the number of dividing cells equals the number of dying cells. • In the death phase, the number of living cells decreases exponentially, and population growth experiences a sharp decline. Answer (B) 40. Topic: Production of Biomass and Primary/Secondary Metabolites • Penicillinase (b-lactamase) is produced by most strains of Staphylococcus and disrupts certain types of
Chapter 12 • Bioprocess Engineering and Process Biotechnology 219
penicillins by hydrolyzing the b-lactam ring that is central to the antimicrobial activity of these drugs. It is used mainly on pharmaceutical industry to make drugs. • Pectinase is an enzyme that breaks down pectin, a polysaccharide found in plant cell walls. It is used extensively in winemaking, primarily to separate the juice from the mash. It also helps to clear wine, which pectin can make cloudy. • Leather processing involves several steps such as soaking, dehairing, bating and tanning. Bating process is traditionally an enzymatic process. The enzyme used comprises trypsin or crude enzyme pancreatin. • Rennin, also called chymosin, is a protein-digesting enzyme that curdles milk by transforming caseinogen into insoluble casein; it is found only in the fourth stomach of cud-chewing animals, such as cows. A commercial form of rennin, rennet, is used in manufacturing cheese and preparing junket. Answer (B) 41. Topic: Production of Biomass and Primary/Secondary Metabolites Anaerobic fermentation is a method cells use to extract energy from carbohydrates when oxygen or other electron acceptors are not available in the surrounding environment. Thus, agitation of culture to maintain oxygen supply is certainly not required for anaerobic fermentation process. Answer (A) 42. Topic: Production of Biomass and Primary/Secondary Metabolites • Actinomycin D is a well-known antibiotic of the actinomycin group that exhibits high antibacterial and antitumor activity. There are several mechanisms of its action that are responsible for its cytotoxic and antitumor action, these being associated with DNA functionality, leading to RNA and, consequently, protein synthesis inhibition. • Daunorubicin is an antitumour antibiotic which is highly active against the sleeping sickness protozoal parasite Trypanosoma rhodesiense, in vitro, but which lacks in vivo activity. • Rifamycin is a natural antibiotic produced by Streptomyces mediterranei. It is a commonly used antimycobacterial drug that inhibits prokaryotic DNA-dependent RNA synthesis and protein synthesis. It blocks RNA-polymerase transcription initiation. It has an activity spectrum against Gram-positive and Gram-negative bacteria, but is mainly used against Mycobacterium sp. (especially M. tuberculosis) in association with other agents to overcome resistance. • Griseofulvin is an effective antifungal medicine used to treat infections of the skin, scalp, nails, feet, groin
and other parts of the body. It is mostly used to treat infections caused by Tinea strains of fungi. Answer (A) 43. Topic: Sterilization of Air and Media The autoclave uses steam heat to kill any microbial life that may be present on a contaminated load by putting it under high pressure and temperature. Heat generated through application of high temperatures acts by disrupting membranes and denaturing proteins and nucleic acids. In order to kill a cell through heat, its temperature must be raised to the point where the proteins in the cell wall breakdown and coagulate. Steam is a very efficient medium for transferring heat, therefore it is an excellent way to destroy microbes. Answer (D) 44. Topic: Rheology of Fermentation Fluids The power number Np (also known as Newton number) is a commonly used dimensionless number relating the resistance force to the inertia force. The power-number has different specifications according to the field of application, e.g., for stirrers the power number is defined as: P ρN i 3 D 5 where P is power; r is fluid density; Ni is rotational speed and D is diameter of stirrer. Answer (C) Np =
45. Topic: Large Scale Production and Purification of Recombinant Proteins The choice of recovery process is based on the following criteria: • The intracellular or extracellular location of the product. • The concentration of the product in the fermentation broth. • The physical and chemical properties of the desired product (as an aid to select separation procedures). • The intended use of the product. • The minimal acceptable standard of purity. • The magnitude of biohazard of the product or broth. • The impurities in the fermenter broth. • The marketable price for the product. Answer (D) 46. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Flocculation is the process in which the destabilized particles are bound together by hydrogen bonding or Van der Waal’s forces to form larger particle flocs. During flocculation, further particulate removal takes place by entrapment. It is generally used for the immobilization of cells.
220 GATE Biotechnology Chapter-wise Solved Papers Enzymes or cells Bonding
Physical bonding
Flocculation
Covalent chemical bonding
Adsorption to surface carriers
Carrier-free method (cells only)
Physical entrapment
Surface carriers
Gelation (mostly cells)
Encapsulation (enzymes)
Cross-linking precipitate
Carrier-free method (mostly for enzymes)
Answer (C) 47. Topic: Bioremediation A Biosensor is an analytical device. The sensor which integrates the biological elements with the physiochemical transducer to produce an electronic signal is proportional to a single analyte and which is fetched into a detector. Analyte
Bioreceptor
Transducer Measurable signal
Electric signal
Detector
Answer (B) 48. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes The effectiveness factor is the ratio of the actual reaction rate to the rate at surface conditions in the presence of pore diffusion resistance. This effectiveness factor has a number of characteristics; one is that in most cases the effectiveness factor will be less than or equal to one because the present support diffusional limitations will always reduce the substrate concentration in the particle surface. Two specific cases demonstrate where the effectiveness factor can be more than one. • One is in the case of enzyme reactions which are controlled by substrate inhibition kinetics. That means at higher substrate concentration the reaction velocity is inhibited. Therefore, as the substrate concentration gradient is developed in the particle, the concentration drops and the reaction rate can be higher than that at the surface. Therefore, the effectiveness factor can be more than one. • The second case is due to the partitioning effect. If suppose the matrix is charged and the substrate is also
charged and if the two species have opposite charges, then the concentration of the substrate inside the enzyme particle will be more than that in the bulk. Therefore, the enzyme reaction rate within the matrix may be more than what is obtained at the matrix surface. Answer (A) 49. Topic: Production of Biomass and Primary/Secondary Metabolites • Zymomonas mobilis is a natural ethanologen with many desirable industrial biocatalyst characteristics, such as high specific productivity, high alcohol tolerance, a broad pH range for production (pH 3.5–7.5), and the generally regarded as safe status. • Streptomycin is an aminoglycoside antibiotic produced by the soil actinomycete Streptomyces griseus. It acts by binding to the 30S ribosomal subunit of susceptible organisms and disrupting the initiation and elongation steps in protein synthesis. It is bactericidal due to effects that are not fully understood. • Citric acid is produced mainly by submerged fermentation using Aspergillus niger or Candida sp. from different sources of carbohydrates, such as molasses and starch-based media. However, other fermentation techniques, e.g., solid state fermentation and surface fermentation, and alternative sources of carbon such as agro-industrial residues have been intensively studied showing great perspective to its production. • The filamentous fungi Trichoderma is an important fungus used to produce enzymes by fermentation process. This genus secretes large amounts of cellulase and hemicellulase enzymes capable of degrading carbohydrate polymer. Solid state fermentation is a popular technique that is often employed for the enzymes due to some practical and economic advantages. Answer (B)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 221
50. Topic: Production of Biomass and Primary/Secondary Metabolites For the reaction: NAD + → NADH
Given, [S] = 20 mol m−3 v=
800[20] 16000 16000 = = = 8.89 mL m −3 h 2 400 + 1000 + 400 1800 400 + 50[20] + [20]
Glucose ] Ethanol 16000 800 16000 NAD[+20 ← NADH v= = = = 8.89 mL m −3 h 2 Productivity of 400 conversion ethanol, 400 +to1000 + 400 1800 [20]glucose + 50[20] +of P = 4.5 g−1 h−1 ⎛ dp ⎞ Concentration of yeast cells in the fermentation broth, Now, v = ⎜ ⎟ ⎝ dt ⎠ X = 5% (v/v) We know, 5 = mL per litre of broth 100 V 8.89 K M = max = = 4.44 −5 2 2 5 10 = × L per litre of broth Concentration of NAD+/NADH in the yeast cells = [S] 20 V0 = Vmax = 8.89 = 7.27 mL m −3 h −1 10 mM = 10−5 g L−1 K M + [S] 4.44 + 20 1 No.of cycles Answer (D) Cycle rate = = = 20, 000 cycles h −1 Unit time 5 × 10 −5 Answer (B) 53. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes 51. Topic: Engineering Principles of Bioprocessing According to Langmuir adsorption isotherm, According to the kinetics of the disintegration of baker’s Ce 1 Ce yeast cells, = + qe qm qm dP = K ( Pm − P ) where Ce is the concentration at equilibrium, qe is the dt adsorption at equilibrium, and qm is the maximum On differentiation of the equation we get: adsorption. Substituting qe = 35 mg cm−1 and qm = ⎛ Pm ⎞ 2.303 70 mg cm−1, we get Ce = 1 mg L−1. Now, substituting log10 ⎜ K= ⎟ Ce = (0.9 × 220) mg L−1 in the following equation, we get t ⎝ Pm − P ⎠ Let us assume that maximum protein that can be released, Pm = 100P The concentration of protein released, P = 90P 2.303 ⎛ 100 P ⎞ log10 ⎜ ⎝ 100 P − 90 P ⎟⎠ t
Therefore,
K=
or
2.303 ⎛ 100 P ⎞ 2.303 log10 ⎜ K= = ⎝ 10 P ⎟⎠ t t
2.303 2.303 t= = = 4.606 h K 0.5 Answer (C)
qe1 =
(C0 − Ce )V (50 − 35) × 1.0 ⇒ 35 = ⇒ m = 0.428 mg m m
In the second case, qe2 =
(C0 − Ce )V ( 220 − (0.9 × 220)) × 1.5 = = 77.1 mg L−1 m 0.428 Answer (D)
54. Topic: Heat and Mass Transfer The amount of sodium sulfite reacted for 10 min = 0.5 – 0.2 = 0.3 mol/L
52. Topic: Immobilization of Biocatalysts (Enzymes and ++ According to the stoichiometric relation, Na 2SO3 + 12 O 2 ⎯Cu ⎯⎯ → Na 2S Cells) for Bioconversion Processes ++ Na 2SO3 + 12 O 2 ⎯Cu ⎯⎯ → Na 2SO 4 , the amount of oxygen required to react 800[S ] Reaction rate, v = 0.3 mol L−1 is 400 + 50[S ] + [S ]2 Given, [S0] = 100 mol m−3 (0.3) / 2 = 0.15 mol L−1 800[100] 80000 80000 Therefore, the oxygen uptake is = = = 5.195 mL m −3 h 2 400 + 5000 + 10000 15400 400 + 50[100] + [100] ⎛ 32 g O 2 /mol ⎞ OUR = (0.5 mol O 2 /L) × ⎜ = 0.008 g L−1 s −1 80000 80000 −3 ⎝ 10 × 60 s ⎟⎠ = = 5.195 mL m h 400 + 5000 + 10000 Answer (B) 154000 v=
222 GATE Biotechnology Chapter-wise Solved Papers 55. Topic: Large Scale Production and Purification of Recombinant Proteins Step
Protein Concentration (mg)
Enzyme Activity (unit)
Specific Activity = Total activity/ Total protein
Crude cell-free 6000 2500 extract Ammonium sul375 1375 fate precipitation Ion-exchange 90 750 chromatography Total 6465 4125 So, for the final step, % yield = 30 and fold purification = 20.
Purification = Specific activity × (initial specific activity)
0.42
Yield = (Total enzyme activity/ initial total enzyme activity) × 100 100
3.67
55
8.7
8.33
30
19.8
12.42
—
—
1
Answer (B) 56. Topic: Membrane Based Bioseparation Method Ultrafiltration (UF) is a variety of membrane filtration in which forces like pressure or concentration gradients lead to a separation through a semipermeable membrane. Suspended solids and solutes of high molecular weight are retained in the so-called retentate, while water and low molecular weight solutes pass through the membrane in the permeate (filtrate). This separation process is used in industry and research for purifying and concentrating macromolecular (103–106 Da) solutions, especially protein solutions. Other than this, it is used for filtration of effluent, in enzyme recovery, desalting and solvent exchange of protein and dialysis. Answer (C) 57. Topic: Aeration and Agitation • Sparger is a device for introducing air into fermenter. Aeration provides sufficient oxygen for organism in the fermenter. Fine bubble aerators must be used. Large bubbles will have less surface area than smaller bubbles which will facilitate oxygen transfer to a greater extent. There are three types of sparger viz. porous sparger, orifice sparger and nozzle sparger. • Baffles are metal strips that prevent vortex formation around the walls of the vessel. • An impeller is a rotating component of a centrifugal pump which transfers energy from the motor that drives the pump to the fluid being pumped by accelerating the fluid outwards from the center of rotation. Answer (C) 58. Topic: Production of Biomass and Primary/Secondary Metabolites • Sake, also spelled saki, Japanese alcoholic beverage made from fermented rice. Sake is light in color, is
non-carbonated, has a sweet flavor, and contains up to 18% alcohol. • Cider is an alcoholic beverage made from the fermented juice of apples. Cider alcohol content varies from 1.2% to 8.5%. • Wine is an alcoholic drink made from fermented grapes. Yeast consumes the sugar in the grapes and converts it to ethanol, carbon dioxide, and heat. • Lager is a type of beer conditioned at low temperatures. They are fermented from grain barley. Answer (A) 59. Topic: Production of Biomass and Primary/Secondary Metabolites • Ampicillin acts as an irreversible inhibitor of the enzyme transpeptidase, which is needed by bacteria to make the cell wall. It inhibits the third and final stage of bacterial cell wall synthesis in binary fission, which ultimately leads to cell lysis; therefore, ampicillin is usually bacteriolytic. • Tetracycline antibiotics are protein synthesis inhibitors. They inhibit the initiation of translation in variety of ways by binding to the 30S ribosomal subunit, which is made up of 16S rRNA and 21 proteins. They inhibit the binding of aminoacyl-tRNA to the mRNA translation complex. • Nystatin is an ionophore. It binds to ergosterol, a major component of the fungal cell membrane. When present in sufficient concentrations, it forms pores in the membrane that lead to K+ leakage, acidification, and death of the fungus. • Anthramycin works by inhibiting the synthesis of RNA and DNA of carcinoma cells. It is a competitive inhibitor of cell-free RNA and DNA
Chapter 12 • Bioprocess Engineering and Process Biotechnology 223
synthesis, and blocks the action of DNase I.Anthramycinmethyl-ether (AME) forms a complex with DNA which blocks off synthesis by prohibiting DNA binding with proper enzymes. Answer (B)
60. Topic: Aeration and Agitation The baffles are normally incorporated into agitated vessel of all sizes to prevent a vortex and to improve aeration efficiency. They are metal strips roughly one-tenth of the vessel diameter and attached radially to the walls. Answer (A)
61. Topic: Engineering Principle of Bioprocessing The amount of radioactivity should be in the order Control > Test > Blank ⎪⎧ [cpm (control) − cpm (test) ] ⎪⎫ Thus, percent inhibition can be calculated as ⎨ ⎬ ×100 ⎩⎪ [cpm (control) − cpm (blank )] ⎭⎪ Decrease in enzyme complex =
Formation due to inhibitor Maximum enzyme-substrate complex formaation in absence of inhibitor Answer (A)
62. Topic: Engineering Principle of Bioprocessing For X inhibitor, the percent inhibition will be: ⎧⎪ [8000 − 4000] ⎫⎪ ⎬ × 100 = 50.6% ⎨ ⎩⎪ [8000 − 100] ⎭⎪ For W inhibitor, the percent inhibition will be: ⎪⎧ [7000 − 1400] ⎪⎫ ⎬ × 100 = 81.6% ⎨ ⎩⎪ [7000 − 135] ⎭⎪ For Y inhibitor, the percent inhibition will be: ⎧⎪ [7500 − 5000] ⎫⎪ ⎬ × 100 = 33.7% ⎨ ⎩⎪ [7500 − 90] ⎭⎪ For Z inhibitor, the percent inhibition will be: ⎪⎧ [7200 − 2800] ⎪⎫ ⎬ × 100 = 62.8% ⎨ ⎩⎪ [7200 − 200] ⎭⎪ Maximum percent inhibition is of W that is 81.6%. Answer (B) 63. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Immobilized cells has many advantages viz: • The density of cells in bioreactor can be increased 2–4 times than in suspension culture. • The same biomass can be used for a long time. • The entrapped cells are protected against shear stress forces; therefore, a simple bioreactor may be used. • The plant cell cultures notably allow ease of continuous harvesting in the large scale, provided that metabolites are excreted into the medium. Answer (D)
64. Topic: Instrumentation Control and Optimization • Thermistors are used as temperature sensors. This can be placed in a pocket (tube) welded into the vessel wall. The probe is linked to a control unit which activates pump to circulate hot or cold water around the fermenter jacket or coil as required. • Oxygen tension (concentration of dissolved oxygen) in the medium is measured using oxygen electrode, which must be in direct contact with the medium and so is sealed through the fermenter wall. The electrode is linked to a control unit. • Metal rod, a foam probe is used as a feedback control system to control foam. It is fitted in the headspace at a certain level. When foam touches the probe, the control unit linked to it activates a pump to pump in antifoam mineral oil until the foam subsides and no longer makes contact with the probe. • The pH of the medium is measured using a pH electrode. This must be in direct contact with the medium and so is sealed through the fermenter wall using an air-tight seal. The electrode is linked to a control unit which activates pumps to pump in acid or alkali as required to correct pH and provide a source of nutrient. Answer (D) 65. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes • Reagents such as glutaraldehyde, diazobenzidine, hexamethylene diisocyanate, and toluene diisothiocyanate are required for cross-linking. • Cyanogen bromide (CNBr)-agarose and CNBractivated-Sepharose containing carbohydrate moiety
224 GATE Biotechnology Chapter-wise Solved Papers and glutaraldehyde as a spacer arm imparts thermal stability to covalently bound enzymes. • In entrapment method, enzyme can be physically entrapped into a polymer matrix, gel, or capsule. There are number of methods by which enzymes can be entrapped. Inclusion in gels: In this method, enzyme is immobilized by the polymerization techniques, e.g., radical formation (polyacrylamide gel), by radiation (polyvinyl alcohol and polyvinyl pyrrolidone gel), by photosensitive resin (polyethylene glycol dimethacrylate). Inclusion of fibers: A mixture of polymer and enzyme is extruded to have the enzyme trapped in fiber format. The material used is usually polyacetate or cellulose. Inclusion in microcapsules: The enzyme is entrapped by polymerize techniques at the air water interface, forming a microcapsule with enzyme molecules trapped within it. Monomer mixture used in microcapsule formation are polyamine and polybasic acid chloride. Answer (C) 66. Topic: Bioprocess Design and Development from Lab to Industrial Scale Hollow fiber membranes are generally solvent spun and as such can be considered a sub-group of solvent spinning. The small diameter of the fibers (in the order of 200 microns) generates an extremely high surfacearea-to-cartridge volume ratio in the range of 100–200 cm2 mL−1. Coupled with the high gross filtration rate of the more optimized polysulfone fibers, the rate of exchange of primary and secondary metabolites appears high enough to support any practical purpose. Answer (C) 67. Topic: Bioremediation • Amperometric biosensor are based on the electron’s movement, i.e., electronic current determination as a reaction of enzyme-catalyzed redox reaction. Generally, a normal contact voltage passes through the electrodes to analyze. In the enzymatic reaction which produces the substrate or product can transfer the electrons with the surface of electrodes to be reduced. • Optical biosensors, based on evanescent wave technology, are analytical devices that measure the interactions between biomolecules in real time, without the need for any labels. Specific ligands are immobilized to a sensor surface, and a solution of receptor or antibody is injected over the top. Binding is measured by recording changes in the refractive index, caused by the molecules interacting near the sensor surface within the evanescent field.
• Many enzyme catalyzed reactions are exothermic, generating heat which may be used as a basis for measuring the rate of reaction and, hence, the analyte concentration. This represents the most generally applicable type of calorimetric biosensor. • Potentiometric sensors measure the potential difference between two electrodes under the conditions of no current flow. The measured potential may then be used to determine the analytical quantity of interest, generally the concentration of some component of the solution. The signal of a general potentiometric sensor is based on the Nernst equation. Answer (B) 68. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Compared to free enzymes in solution, immobilized enzymes are more robust and more resistant to environmental changes. More importantly, the heterogeneity of the immobilized enzyme systems allow an easy recovery of both enzymes and products, multiple reuse of enzymes and continuous operation of enzymatic processes. Answer (A) 69. Topic: Large Scale Production and Purification of Recombinant Proteins • The technique of SDS-PAGE separates molecules based on molecular weight. Molecules with lower molecular weight travel faster in the gel than the molecules with higher molecular weight. • In native PAGE, the mobility depends on both the protein’s charge and its hydrodynamic size. • Gel filtration chromatography is an established method for determining the size and molecular mass of proteins. Fractionation is based on the diffusion of molecules into the pores of the resin. Larger proteins do not enter the pores of the resin as readily but pass through the fluid volume of the column faster than smaller proteins. These protein molecules elute from the column in the order of decreasing molecular mass. • Isoelectric focusing (IEF), also known as electrofocusing, is a technique for separating different molecules by differences in their isoelectric point (pI). Answer (B) 70. Topic: Large Scale Production and Purification of Recombinant Proteins Con A Sepharose 4B with immobilized Concanavalin A is an affinity chromatography resin for capture of glycosylated biomolecules including glycoproteins and glycolipids. The principle of affinity chromatography is that the stationary phase consists of a support medium (e.g., cellulose beads) on which the substrate (or sometimes a coenzyme) has been bound covalently, in such a way that
Chapter 12 • Bioprocess Engineering and Process Biotechnology 225
the reactive groups that are essential for enzyme binding are exposed. As the mixture of proteins is passed through the chromatography column, those proteins that have a binding site for the immobilized substrate will bind to the stationary phase, while all otter proteins will be eluted in the void volume of the column. Answer (B) 71. Topic: Batch, Fed-Batch and Continuous Processes In single run metabolite (X) concentration = 1.1 kg m−3 Given, total number of runs = 70 Thus, total metabolite concentration = Total number of runs × Final concentration at each end = 70 year−1 × 1.1 kg m−3 = 77 kg m−3 year−1 Annual output = Total concentration × Working volume = 77 kg m−3 year−1 × 50 m3 = 3850 kg per year Answer (B) 72. Topic: Batch, Fed-Batch and Continuous Processes Overall productivity can be calculated as = Metabolite concentration in one run × Total number of runs = 70 year−1 × 1.1 kg m−3 = 77 kg year−1 m−3 Answer (D) 73. Topic: Kinetics of Microbial Growth The death phase is characterized by a net loss of culturable cells. Even in the death phase there may be individual cells that are metabolizing and dividing, but more viable cells are lost than are gained so there is a net loss of viable cells. Thus, specific growth rate is zero in this phase. This could be caused by lack of nutrients, environmental temperature above or below the tolerance band for the species, or other injurious conditions. Answer (A) 74. Topic: Production of Biomass and Primary/Secondary Metabolites A plant cell produces two types of metabolites: primary metabolites involved directly in growth and metabolism (carbohydrates, lipids and proteins), and secondary metabolites considered as end products of primary metabolism and not involved in metabolic activity (alkaloids, phenolics, sterols, steroids, essential oils, lignin and tannins, etc.). They act as defense chemicals. Their absence does not cause bad effects in the plants. Answer (C) 75. Topic: Production of Biomass and Primary/Secondary Metabolites An ideal biodiesel composition should have fewer polyunsaturated and saturated fatty acids (SFAs). High levels
of polyunsaturated fatty acids (PUFAs) would negatively impact the oxidative stability and increase nitrogen oxide exhaust emissions, which do not suit diesel engines. Polyunsaturated fatty acids in cell membranes react with oxygen to produce peroxyl radicals, the primary free radical intermediate of lipid peroxidation. On the contrary, biodiesel derived from SFAs would have good oxidative stability, but poor fuel properties at low temperatures, which is a disadvantage in winter operation. Monounsaturated fatty acids (MUFAs) are the best components for biodiesel when considering the low temperature fluidity and oxidative stability. Answer (A) 76. Topic: Production of Biomass and Primary/Secondary Metabolites • Ethanol, an alcohol found in nature and in alcoholic drinks, is metabolized through a complex catabolic metabolic pathway. The first three steps of the reaction pathways lead from ethanol to acetaldehyde to acetic acid to acetyl-CoA. Once acetyl-CoA is formed, it is free to enter directly into the citric acid cycle. • Biotransformation of penicillin V to 6-aminopenicillanic acid using immobilized whole cells of E. coli express a highly active penicillin V acylase. • Griseofulvin is a tubulin inhibiting agent. It is a spirocyclic fungal natural product used in treatment of fungal dermatophytes. The physiologic effect of griseofulvin is by means of decreased mitosis, and microtubule inhibition. It is a fungistatic agent used to treat superficial fungal skin infections such as Tinea capitis and Tinea pedis. • Baker’s yeast is of the species Saccharomyces cerevisiae, which is the same species (but a different strain) commonly used in alcoholic fermentation, which is called brewer’s yeast. Baker’s yeast is also a single-cell microorganism found on and around the human body. Answer (C) 77. Topic: Bioremediation The microalgae cultivation method mimics the natural processes that underlie how fossil carbon and in particular oil were once formed. In the industrial version, flue gas is fed from a chimney into pools of water and microalgae that quickly produce biomass, predominantly from the carbon dioxide that is contained in the flue gas. In a dry state, the algae function as an energy carrier and can be used for the smelting processes at smelting plants. When the algae biomass is burned, metals that were previously in the flue gas are returned to the molten metal. It is possible to extract energy, clean water and flue gas, and produce metal ingots all at the same time. Answer (A)
226 GATE Biotechnology Chapter-wise Solved Papers 78. Topic: Large Scale Production and Purification of Recombinant Proteins • Pituitary dwarfism, or growth hormone deficiency, is a condition in which the pituitary gland does not make enough growth hormone. This results in a child’s slow growth pattern and an unusually small stature. For these individuals, treatment with injections of a synthetic version of the hormone may increase final height. • Platelet-derived growth factor like thrombopoietin, significantly promote the recovery of platelets and the formation of bone marrow colony-forming unit-megakaryocyte in chemotherapy induced thrombocytopenia. • Hemophilic people have low factor VIII level; thus they are at risk for bleeding longer after an injury/ surgery and for bleeding inside the body (especially into the joints and muscles). Administration of human factor VIII, also called antihemophilic factor in them is used to temporarily replace the missing factor VIII, so that the blood can clot, and the bleeding can stop. • Chronic renal failure (CRF) is a progressive loss in kidney function over a period of time. It is identified by higher amount of creatinine and lower glomerular filtration rate. Red cell production due to the erythropoietin deficiency is too low in CRF and causes development of anemia in this situation. Thus, the treatment can include injections of a genetically engineered form of EPO. Answer (A) 79. Topic: Microbial, animal and Plant Cell Culture Platforms • The most used selection agent is the neomycin phosphotransferase II (nptII) gene, which confers resistance to aminoglycoside antibiotics kanamycin, neomycin and G-418. • A high level of the aroA gene constitutive expression in transgenic plants gives resistance to glyphosate. The overproduction of this enzyme results in high enzymatic activities that enable the plant cell to survive even in the presence of the herbicide. • Genes that are frequently used to select transformed plant tissues include hpt, bar, and gox, that confer resistance to hygromycin, phosphinothricin, and glyphosate, respectively. Answer (D)
the pressure of a canister of gas, measuring atmospheric pressure, or recording the strength of the vacuum in a vacuum pump. • Foam control is the antifoam sensor that is used to control the foam in the fermenter. Anti-foam system is foam contact in the rubber sheathed electrode and electric circuit actuated a solenoid valve to allow the passage of sterile antifoam agent into the reactor. • Turbidity is measured using instruments called turbidimeters. Traditional turbidimeters shine light through a section of water and detect how much light is scattered from particulates in the water at a 90-degree angle from the incoming light. This type of scattered light measurement is called nephelometric. Turbidity can also be approximated in an instrument such as a colorimeter or spectrophotometer by measuring the decrease in transmitted light due to blockage by particles. • The rotameter is an industrial flow meter used to measure the flowrate of liquids and gases. The rotameter consists of a tube and float. The float response to flowrate changes is linear, and a 10–1 flow range or turndown is standard. Answer (A) 81. Topic: Aeration and Agitation Four baffles are normally incorporated into agitated vessels of all sizes to prevent a vortex and to improve aeration efficiency. Baffles are metal strips roughly one-tenth of the vessel diameter and attached radially to the wall. It is recommended that baffles should be so installed that a gap existed between them and the vessel wall, so that there was a scouring action around and behind the baffles thus minimizing microbial growth on the baffles and the fermenter walls. Answer (A) 82. Topic: Kinetics of Microbial Growth Molecular mass of glucose = 6 × 12 + 12 × 1 + 6 × 16 = 180 g Molecular mass of ethanol = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 g = 46 g Conversion of glucose to ethanol takes place by following reaction: C6 H12 O6 ⎯Zymase ⎯⎯→ 2C 2 H 5 OH + 2CO 2 If, 180 g of glucose yields 92 g of ethanol Then, 1000 g of ethanol will yield
80. Topic: Instrumentation Control and Optimization 92 × 1000 = 511.1 g = 0.51 kg • A diaphragm pressure gauge is a device that uses 180a diaphragm with a known pressure to measure pressure in a fluid. It has many different uses, such as monitoring
92 × 1000 = 511.1 g = 0.51 kg 180 Answer (C)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 227
83. Topic: Engineering Principle of Bioprocessing Trypsin is a pancreatic serine protease with substrate specificity based upon positively charged lysine and arginine side chains. It cleaves peptides on the C-terminal side of lysine and arginine amino acid residues. If a proline residue is on the carboxyl side of the cleavage site, the cleavage will not occur. Cyanogen bromide (CNBr) cleaves proteins on the C-terminal side of unoxidized methionine (Met) residues. (A) If the order of the peptides that gives the primary structure of the original protein is: P-Q-R-S. Phe-Val-Met-Val-Arg-Ala-Ala-Try-Gly-Lys-ValPhe-Met-Ala-Gly-Lys-Phe-Gly-Try-Ser-Thr On trypsin cleavage it will give: Phe-Val-Met-Val-Arg, Ala-Ala-Try-Gly-Lys, Val-Phe-Met-Ala-Gly-Lys, Phe-Gly-Try-Ser-Thr On CNBr cleavage it will give: Phe-Val-Met, Val-Arg-Ala-Ala-Try-Gly-Lys-Val-Phe-Met, Ala-Gly-Lys-Phe-Gly-Try-Ser-Thr which is not the case, thus option (A) is incorrect. (B) If the order of the peptides that gives the primary structure of the original protein is: Q-P-R-S. Ala-Ala-Try-Gly-Lys-Phe-Val-Met-Val-Arg-ValPhe-Met-Ala-Gly-Lys-Phe-Gly-Try-Ser-Thr On trypsin cleavage it will give: Ala-Ala-Try-Gly-Lys, Phe-Val-Met-Val-Arg, Val-Phe-Met-Ala-Gly-Lys, Phe-Gly-Try-Ser-Thr On CNBr cleavage it will give: Ala-Ala-Try-Gly-Lys-Phe-Val-Met, Val-Arg-Val-Phe-Met, Ala-Gly-Lys-Phe-Gly-Try-Ser-Thr which is the case, thus option (B) is correct. (C) If the order of the peptides that gives the primary structure of the original protein is: Q-P-R-S. Ala-Ala-Try-Gly-Lys-Val-Phe-Met-Ala-Gly-LysPhe-Val-Met-Val-Arg-Phe-Gly-Try-Ser-Thr On trypsin cleavage it will give: Ala-Ala-Try-Gly-Lys, Val-Phe-Met-Ala-Gly-Lys, Phe-Val-Met-Val-Arg, Phe-Gly-Try-Ser-Thr On CNBr cleavage it will give: Ala-Ala-Try-Gly-Lys-Val-Phe-Met, Ala-Gly-Lys-Phe-Val-Met,
Val-Arg-Phe-Gly-Try-Ser-Thr which is not the case, thus option (C) is incorrect. (D) If the order of the peptides that gives the primary structure of the original protein is: R-Q-P-S. Val-Phe-Met-Ala-Gly-Lys-Ala-Ala-Try-Gly-LysPhe-Val-Met-Val-Arg-Phe-Gly-Try-Ser-Thr On trypsin cleavage it will give: Val-Phe-Met-Ala-Gly-Lys, Ala-Ala-Try-Gly-Lys, Phe-Val-Met-Val-Arg, Phe-Gly-Try-Ser-Thr On CNBr cleavage it will give: Val-Phe-Met, Ala-Gly-Lys-Ala-Ala-Try-Gly-Lys-Phe-Val-Met, Val-Arg-Phe-Gly-Try-Ser-Thr which is not the case, thus option (D) is incorrect. Answer (B) 84. Topic: Batch, Fed-Batch and Continuous Processes Final volume = Initial volume + Flow rate × Number of days = 6 m3 + 2 m3 day−1 × 2 days = 6 + 4 = 10 m3 Answer (C) 85. Topic: Batch, Fed-Batch and Continuous Processes Dilution rate at the ⎫ Flow rate ⎬= end of 2 nd day ⎭ Total volume at the end of 2 nd day =
2 m3 = 0.2 10 m3 Answer (B)
86. Topic: Kinetics of Microbial Growth Diauxic growth is the phenomenon whereby a population of microbes, when presented with two carbon sources, exhibits bi-phasic exponential growth intermitted by a lag-phase of minimal growth. The preferred sugar is consumed first, which leads to rapid growth, followed by a lag phase. During the lag phase the cellular machinery used to metabolize the second sugar is activated and subsequently the second sugar is metabolized. Answer (B) 87. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes 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. Answer (D)
228 GATE Biotechnology Chapter-wise Solved Papers 88. Topic: Production of Biomass and Primary/Secondary Metabolites • Probiotics are live bacteria and yeasts that are beneficial in preventing several health conditions. They are usually consumed as supplements or yoghurts and are also referred to as “good bacteria.” They include Lactobacillus and Bifidobacteria. • A xenobiotic is a chemical substance found within an organism that is not naturally produced or expected to be present within the organism. It can also cover substances that are present in much higher concentrations than are usual. The term xenobiotics, however, is very often used in the context of pollutants such as dioxins and polychlorinated biphenyls and their effect on the biota. • Prebiotics are compounds in food that induce the growth or activity of beneficial microorganisms such as bacteria and fungi. The most common example is in the gastrointestinal tract, where prebiotics can alter the composition of organisms in the gut microbiome. Specifically, fructans and galactans are two oligosaccharide sources which have been found to stimulate the activity and growth of beneficial bacterial colonies in the gut. • An antibiotic is a type of antimicrobial substance active against bacteria and is the most important type of antibacterial agent for fighting bacterial infections. β-lactam antibiotics (beta-lactam antibiotics) are a class of broad-spectrum antibiotics. These include penicillin derivatives. Answer (A) 89. Topic: Engineering Principle of Bioprocessing • The sedimentation coefficient of a particle is used to characterize its behavior in sedimentation processes, notably centrifugation. It is defined as the ratio of a particle’s sedimentation velocity to the acceleration that is applied to it (causing the sedimentation). • A partition coefficient is the ratio of the concentration of a substance in one medium or phase (C1) to the concentration in a second phase (C2) when the two concentrations are at equilibrium. The media can be gases such as air, liquids such as water or olive oil, or complex mixtures such as blood or other tissues. • The rejection coefficient for the ultrafiltration of monodisperse solutes (polystyrene in cyclohexane at theta temperature) is calculated from average pore size and pore size distribution of mineral membranes, assuming a sieve mechanism for membrane crossflow and negligible adsorption. • The semi-equilibrium dialysis method has been used to infer solubilization equilibrium constants or, alternatively, activity coefficients of solutes solubilized into micelles of aqueous surfactant solution. Answer (D)
90. Topic: Various Types of Microbial and Enzyme Reactors • The leading face of the blades on a marine-blade impeller can be flat or concave, whereas their back sides are convex. This produces an axial flow. Marine blade impellers are used for applications that require gentle mixing without causing cell damage such as in animal cell cultures. • The use of a draft tube is recommended when the ratio of height to diameter of the recipient is greater than 1.3. This device is then able to circulate the suspension of solid from top to bottom and from bottom to top and, therefore, neutralizes the tendency for decantation caused by gravity. • Originally, the diaphragm valve was developed for use in industrial applications and/or pipe-organs. Later on, the design was adapted for use in the bio-pharmaceutical industry by using compliant materials that can withstand sanitizing and sterilizing methods. • Sparger is a device for introducing air into fermenter. Aeration provides sufficient oxygen for organism in the fermenter. Answer (B) 91. Topic: Production of Biomass and Primary/Secondary Metabolites Given, k1 = 3 × 108 M−1s−1; k–1 = 4 × 104 s−1 and k2 = 2 × 103 s−1 Now, we know, KM =
k −1 + k2 k1
KM =
4 × 10 4 s −1 + 2 × 103 s −1 3 × 108 M −1 s −1
=
42000 s −1 = 1.4 × 10 −4 M 3 × 108 M −1 s −1
Answer (D)
92. Topic: Kinetics of Microbial Growth From Monod’s equation, ⎡ S0 ⎤ μ = μmax ⎢ ⎥ ⎣ K S + S0 ⎦ Given, μmax = 1 h−1; YX/S = 0.5 g g−1; KS = 0.2 g L−1; S0 = 10 g L−1 Putting these values in Monod equation, we get, ⎡ ⎤ 10 −1 10 g L−1 μ = 1 h −1 ⎢ = h = 0.98 h −1 −1 −1 ⎥ ⎣ 0.2 g L + 10 g L ⎦ 10.2 At wash out, dilution rate, D = m. Therefore, the required dilution rate, D = 0.98 h−1 Answer (C)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 229
Moles of CO 2 formed 93. Topic: Production of Biomass and Primary/Secondary Respiratory quotient = Metabolites Moles of O 2 consumed • Saccharomyces cerevisiae is the most employed yeast for 5 ethanol production at industrial level though ethanol is = = 1.01 4.95 produced by an array of other yeasts, bacteria, and fungi. Answer (C) • The organism Corynebacterium glutamicum is used in industry to produce L-lysine through an aerobic 95. Topic: Batch, Fed-Batch and Continuous Processes process. The biosynthetic pathway is followed leading Given, from aspartate (which is itself derived from oxaloaceVolumetric mass-transfer coefficient, K L a = 0.18 s −1 = 5.139 × 10 −5 h −1 tate, a Krebs’ cycle intermediate) through a series of −1 K L a = 0.18 s = 5.139 × 10 −5 h −1 intermediates, all the way to L-lysine. • Bacillus thuringiensis is a naturally occurring soil Specific oxygen uptake ( qo ) = 5 mmol h−1 = 5 × 10−6 g bacterium that causes disease on insect pests. It is g−1 h−1 accepted in organic farming and is considered ideal Saturation concentration (CL) = 8 mg mL−1 = 8 g L−1 for pest management due to its low cost, ease of appliCritical dissolved O2 (dO2) is 20% of the saturate conc. cation, high virulence and narrow host specificity. (8 mg mL−1) • Vancomycin is produced by Streptomyces orientalis, is a glycopeptide antibiotic that complexes with the Thus, C * = 20% of CL d-Ala-d-Ala terminus of peptidoglycan precursors, 20 which inhibits the peptidoglycan coupling and transC= × 8 = 1.6 g L−1 peptidation reaction. It has characteristic antibacterial 100 spectrum against Gram-positive bacteria. We know, Answer (A) Oxygen uptake rate (OUR) = qo x 94. Topic: Kinetics of Microbial Growth OUR KL a = g dry cell g acetate YX/S = 0.35 and YP/S = 0.1 and, (C * −CL ) g glucose g glucose ⇒ OUR = K L a(C * −CL ) aC6 H12 O6 + bNH 3 + dO 2 ⎯⎯ → CH1.8 O0.5 N 0.2 + eH 2 O + fCO 2 Therefore, q0 x = K L a(C * −CL ) aC6 H12 O6 + bNH 3 + dO 2 ⎯⎯ → CH1.8 O0.5 N 0.2 + eH 2 O + fCO 2 Thus, on equating equations on both sides we will have, 5 × 10 −3 x = 5.139 × 10 −5 h −1 (8g L−1 − 1.6 g L−1 ) For C, 6a = 1 + f …(i) 5.139 × 10 −5 h −1 (8g L−1 − 1.6 g L−1 ) x= For H, 12a + 3b = 1.8 + 2e …(ii) 5 × 10 −3 g h −1 For O, 6 a + 2d = 0.5 + e + 2 f …(iii) 5.139 × 10 −5 h −1 (6.4 g L−1 ) For N, b = 0.2 . Putting this value of b in (ii) we get, x= = 65.779 g L−1 −6 −1 −1 5 10 g g h × 12a + 3 × 0.2 = 1.8 + 2e Answer (D) or 12a = 1.2 + 2e …(iv) Let us take a = 1 Thus, From (i) f = 5 From (iv), e = 5.4 …(v) From (ii) and (v), 12 + 3b = 1.8 + 2(5.4) 3b = 12.6 − 12 3b = 0.6 b = 0.2 From (iii), (i) and (iv)
6 + 2d 2d 2d d
= 0.5 + 5.4 + 2(5) = 15.9 − 6 = 9.9 = 4.95
96. Topic: Kinetics of Microbial Growth In double reciprocal plot or Lineweaver Burk plot, from Monod’s equation, K 1 1 1 = S + μ μ max S μ max y = mx + c Thus, the slope of the line (m) is KS/mmax, the intercept (c) is 1/mmax. Intercept (c) =
μmax =
1 = 1.25 h μmax 1 = 0.8 h −1 1.25 h
230 GATE Biotechnology Chapter-wise Solved Papers
Slope (m) =
KS = 100 mg h L−1 μ max
KS = 100 mg h L−1 0.8 h −1 KS = 100 mg h L−1 × 0.8 h −1 = 80 mg L−1 Answer (C)
• The separation of a component from a liquid mixture by treatment with a solvent in which the desired component is preferentially soluble is known as liquid-liquid extraction. Prior to starting a large-scale extraction, it is important to find out on a small scale the solubility characteristics of the product using a wide range of solvents Answer (A)
97. Topic: Kinetics of Microbial Growth From solution to previous question, μmax = 0.8 h−1, KS = 80 mg L−1 Given that total volume, V = 10 L = 10 × 103 mL Flow rate, F = 50 mL min−1 = 50 × 60 mL h−1 = 3000 mL h−1 Substrate concentration, S0 = 50 g L−1 Yield, Y = 0.3 g DCW g−1 substrate
0 mg L−1 × 0.3 h −1 0.8 h −1 − 0.3 h −1
100. Topic: Batch, Feb-Batch and Continuous Processes • False. Continuous bioreactors provide high degree of control and uniform product quality than batch bioreactors. Whereas the disadvantages include controlling or minimizing the production of non-growth-related products and loss of original product strain over time. • False. Continuous stirred tank bioreactor is ideally suited for reaction with substrate inhibition, while F 3000 mL h −1 −1 batch and fed-batch bioreactors are used for product We know, D = = = 0.3 h V 10 × 103 mL inhibition. From Monod’s equation, • True. Kinetic considerations dictate the reactor choice. Mode of operation is immaterial for a zero-order reacKS D S= tion, whereas batch or tubular reactor is desirable for μ max − D first-order or Michaelis-Menten type reaction kinetics. KS D 80 mg L−1 × 0.3 h −1 24 −1 −1 1 In the batch bioreactor, feed is introduced = mg L = 48 mg• S= = L True. = 0.048 g L−fed 0.5 μ max − D 0.8 h −1 − 0.3 h −1 either continuously or intermittently while there is no 24 continuous product removal. This leads to an increase mg L−1 = 48 mg L−1 = 0.048 g L−1 = 0.5 in the reactor volume with time. The products are dis We know, charged at the end of the cycle. These reactors are also called semi-batch reactors. X = Y[S0 − S ] = 0.3 [50 − 0.048] = 0.3 × 49.952 = 15 g L−1 Answer (C) Answer (A)
98. Topic: Various Types of Microbial and Enzyme Reactors Biostat is a culture vessel in which physical, physicochemical and physiological conditions, as well as cell concentration, are kept constant by perfusion, monitoring and feedback. Answer (B) 99. Topic: Engineering Principle of Bioprocessing • Filtration depends on the nature of the solid particles, particularly their size and shape, the size distribution and packing characteristics. • Factors influencing the rate of sedimentation over which one has little or no control are the difference in density between the cells and the liquid (increased temperature would lower media density but is of little practical use with fermentation broths), the diameter of the cells (could be increased by coagulation/flocculation), and the viscosity of the liquid. • Distillation is done for vapor-liquid separation in a column, to separate the lower boiling more volatile component from other less volatile components.
101. Topic: Kinetics of Microbial Growth A synchronous culture is one in which all the cells of a particular population are in the same stage of development and they divide simultaneously. Synchronization of cell division in a bacterial population can be achieved in several ways. • One of them is repeated alternation of the incubation temperature. A culture growing at the optimal temperature is exposed to a lower temperature, so that the growth is slowed down considerably. After some time, the culture is again brought to its optimal temperature. The time interval is adjusted according to the growth rate (generation time) of the organism. By lowering the temperature, cell division is delayed and all cells divide simultaneously when the temperature is raised to the optimal level. The treatment has to be continued through several cycles to obtain satisfactory results. • Another method of getting a synchronously dividing population of cells is by filtration through a membrane filter. Enrichment culture is the use of certain growth media to favor the growth of a particular microorganism over
Chapter 12 • Bioprocess Engineering and Process Biotechnology 231
others, enriching a sample for the microorganism of interest. This is generally done by introducing nutrients or environmental conditions that only allow the growth of an organism of interest. Answer (C) 102. Topic: Large Scale Production and Purification of Recombinant Proteins • The role of erythropoietin is to control red blood cell production by regulating the differentiation and proliferation of erythroid progenitor cells in the bone marrow. The most common use is in people with anemia (low blood count) related to kidney dysfunction. When the kidneys are not properly functioning, they produce less than normal amounts of erythropoietin, which can lead to low red blood cell production, or anemia. • Antifibrinolytics are a class of medication that are inhibitors of fibrinolysis. These lysine-like drugs interfere with the formation of the fibrinolytic enzyme plasmin from its precursor plasminogen by plasminogen activators (primarily t-PA and u-PA) which takes place mainly in lysine rich areas on the surface of fibrin. These drugs block the binding sites of the enzymes or plasminogen respectively and thus stop plasmin formation. • Collagenase is an enzyme that breaks down collagen in damaged tissues within the skin and helps the body generate new healthy tissue. • Transferrin is a plasma protein that transports iron through the blood to the liver, spleen and bone marrow. It transports ferric ions from the iron stores of intracellular or mucosal ferritin to bone marrow where erythrocyte precursors and other cells have transferrin surface receptors. Answer (A) 103. Topic: Production of Biomass and Primary/Secondary Metabolites • Alcoholic beverages are produced following the fermentation of sugars by yeasts, mainly (but not exclusively) strains of the species, Saccharomyces cerevisiae. In the presence of sugars, together with
other essential nutrients such as amino acids, minerals and vitamins, S. cerevisiae will conduct fermentative metabolism to ethanol and carbon dioxide (as the primary fermentation metabolites) as the cells strive to make energy and regenerate the coenzyme NAD+ under anaerobic conditions. • Bifidobacteria belong to a group of bacteria called lactic acid bacteria. Lactic acid bacteria are found in fermented foods like yogurt and cheese. Bifidobacteria are used in treatment as so-called “probiotics,” the opposite of antibiotics. They are considered “friendly” bacteria and are taken to grow and multiply in areas of the body where they normally would occur. • Citric acid is an organic acid of diverse economic use. It is produced almost entirely by fermentation of molasses sugar with selected strains of Aspergillus niger. • Sauerkraut is a traditional vegetable product usually produced by spontaneous fermentation that relies on lactic acid bacteria (LAB) naturally present in white cabbage. However, current trends on sauerkraut production propose the application of starter cultures of LAB such as Leuconostoc mesenteroides in order to ensure a uniform quality of the product. Answer (B) 104. Topic: Instrumentation Control and Optimization A chemostat consists of a culture into which: • Fresh medium is continuously introduced at a constant rate. • The culture volume is kept constant by continuous removal of culture at the same rate. • The supply of a single nutrient controls growth rate. The fermenter is called a chemostat as the growth rate is controlled by the availability of a single component of the medium (the limiting substrate). An important feature is that, it is the continuous introduction of fresh medium that feeds the limiting substrate to the culture and the dilution rate (that is, the rate of addition of fresh medium) determines the specific growth rate of the culture. Answer (D)
105. Topic: Immobilization of Biocatalysts for Bioconversion Processes Effectiveness factor ( η) =
Reaction rate with inter-particle diffusion limitation Reaction rate without diffusion limitation
Case
η = 1 , Conversion is limited by reaction rate
η < 1 , Conversion is diffusion limited (external pore)
η > 1 , Conversion is dependent upon internal-pore diffusion limited Answer (A)
232 GATE Biotechnology Chapter-wise Solved Papers 106. Topic: Various Types of Microbial and Enzyme Reactors Given, Dimensions of cylinder: Height, l = 20 cm Radius, r = 10 cm Dimensions of spiral firm: Width, W = 20 cm Length, L = 30 cm Area = W × L = 20 × 30 = 600 cm2 Support = 105 cells per cm2 Surface area is increased by spiral film, therefore, Increase in area = Area of spiral film = 600 cm2 Additional number of cells: In 1 cm2 number of cells it can support = 105 cells In 600 cm2 number of cells it can support = 600 × 105 cells = 6 × 107 cells Answer (B) 107. Topic: Production of Biomass and Primary/Secondary Metabolites • Gluconic acid is found in fruit juices and other food substances, such as honey, dairy products, rice, etc. It is produced by fungi Aspergillus niger, Penicillium spp. as well as bacteria, such as Acetobacter methanolicus, A. diazeotrophicus, Pseudomonas ovalis. • Lysine as an essential amino acid supplemented to fulfill the dietary requirement of poultry feed. Strain of Corynebacterium glutamicum and Brevibacterium flavum produce L-lysin. • Dextran for human application is usually produced by Leuconostoc mesenteroides. It is used mainly therapeutically as plasma volume expanders and anticoagulants. • Cellulase is produced by Aspergillus niger and Trichoderma reesei. It is used in detergents as it can digest small fibers. It is used in fruit juices as it can digest cellulose. Answer (C)
109. Topic: Chemical Engineering Principles Applied to Biological System Given: V = 100 m3 = 105 L, S = 12 g L−1, KS = 0.2 g L−1, mm = 0.3 h−1 Dilution rate, D = μ = F=
1 [N ] 1 [N ] kd = ln 0 ⇒ t = ln 0 t [N t ] kd [ N t ]
1 [108 ] 1 ( 25.33) = 110 min ln −3 = So, t = 0.23 [10 ] 0.23 Answer (C)
μm ⋅ S 0.3 × 12 ⋅V = × 100 = 26.5 m3 h −1 KS + S 0.2 + 12 Answer (B)
110. Topic: Chemical Engineering Principles Applied to Biological System Given: YX/S = 0.4 g g-1, X = 4 g L-1 YX/S = −
ΔX ( X f − X i ) ( X f − 0) = = ( Si − Sf ) ( Si − Sf ) ΔS
Sf = Si −
Xf X 4 = Si − f = 12 − = 2 g L−1 YX/S YX/S 0.4 Answer (A)
111. Topic: Chemical Engineering Principles Applied to Biological System 1 6 Mean( x ) = ∑ xi 6 i =1 47.3 + 52.2 + 49.2 + 52.4 + 49.1 + 46.3 = 6 296.5 = = 49.4 6 Answer (A) 112. Topic: Chemical Engineering Principles Applied to Biological System Standard deviation, σ =
108. Topic: Sterilization of Air and Media Given, initial no. of cells (N0) = 108 cells mL−1 Final no. of cells (Nt) = 10−3 Temperature = 120°C Volume of culture = 1 m3 = 1000 L Death constant (kd) = 0.23 min−1 Time (t) =? We know,
μ ⋅S F = m (Using Monod V KS + S equation)
1 n ∑ ( xi − x )2 n i =1
xi
( xi − x )
( xi − x ) 2
47.3 52.2 49.2 52.4 49.1 46.3
(–2.1) (2.8) (–0.2) (3) (–0.3) (–3.1)
4.41 7.84 0.04 9 0.09 9.61 30.99
1 (30.99) = 5.165 = 2.27 6 which is closer to option (A).
σ=
Answer (A)
Chapter 12 • Bioprocess Engineering and Process Biotechnology 233
113. Topic: Kinetics of Microbial Growth Different components of a cell grow at different rate and at different time of cell growth. It is only a log phase (exponential phase) in which different components of a cell grows at same rate. Cell growth curve is the exponential growth curve.
Non-growth associated product
[m]
log Growth associated product
Answer (B) 114. Topic: Kinetics of Microbial Growth Lag phase is a time period just after the inoculation and before the cell division start (log phase). During lag phase, cells grow in size but not in number. Cells are biochemically active in generating proteins, enzymes and other machinery in order to accommodate in the new medium and exploit the same for nutrition and cell division.
Time
• Product inhibition is a type of enzyme inhibition where the product of an enzyme reaction binds to the enzyme and inhibits its activity. • Substrate inhibition is a type of enzyme inhibition where there is a progressive decrease in its activity at high substrate concentrations. Answer (B)
Answer (C) 115. Topic: Kinetics of Microbial Growth When X equals zero means there are no cells coming out in the outlet stream which is called a washout situation. When the washout occurs X→ 0 and there should be no change in substrate concentration (S = S0) from the inlet because it is not being consumed for the production of cells (P = 0) maybe little bit is taken up for maintenance, but that may not be very apparent. At wash out condition, D = µmax and is known as Dcritical. At Dcritical, X → 0, S → S0, P → 0 (if X → 0)
X
S0
[X ]
[S]
S0
Dilution D rate
0ops
0 Dcrit = mm
Answer (B) 116. Topic: Kinetics of Microbial Growth • Growth associated products are formed in the log phase of growth along with the growth of the microbial cells and product concentration is almost directly proportional to microbial growth rate. • Non-growth associated products are formed in the stationary phase of growth when no growth of the microbial cells is taking place. Thus, product formation rate remains constant.
117. Topic: Aeration and Agitation Rate of oxygen uptake, rO 2 = K L a (C * − C ) ...(i) Also, we know, rO2 = r × X…(ii) From (i) and (ii),
r ⋅ X = K L a(C * − C ) KL a =
r⋅ X (C * − C ) Answer (A)
118. Topic: Batch, Fed-Batch and Continuous Processes A synchronous or synchronized culture is one in which the bulk of cells proceed through each cell cycle phase, i.e., G1, S, G2 and M. In the culture, all the cells grow at the same rate and stage. Synchronous cultures can be obtained in the following ways: • By changing the external conditions, so as to stop growth of all cells in the culture, and then again change it to resume the growth of cells. The freshly growing cells are now started to grow at the same growth stage, and in this way they are synchronized. • Alternate method is to remove an essential nutrient from the growth medium and afterwards add it again. It is done through using chemical growth inhibitors. After growth has completely arrested for all cells, the growth inhibitor is removed from the culture medium and the cells then started to grow synchronously Answer (A) 119. Topic: Kinetics of Microbial Growth Specific growth rate of bacteria means their growth rate with respect to time. During exponential growth phase
234 GATE Biotechnology Chapter-wise Solved Papers growth rate is independent of nutrient and substrate concentration, only cell number and mass conc. increases exponentially. In mathematical terms:
YX/S =
dX /dt = μ max X
X = X 0 at t = 0
ln
122. Topic: Chemical Engineering Principles Applied to Biological System The cell yield coefficient YX/S is given by
X = μmax t X0
where X = cell concentration, t = time of incubation, X0 = cell concentration at time zero or inoculum size, dX/dt = change in cell number X during time t, m = specific growth rate. pe
Exponential phase
ln[X ] mm
=
slo
lnX0 Time
Answer (B) 120. Topic: Kinetics of Microbial Growth Given, Volume, V = 60 m3 Sucrose concentration in the incoming feed medium, S0 = 10 kg m−3 Residual sucrose concentration, S = 2.0 kg m−3 Initial biomass concentration, X0 = 0 (if nothing is given) Steady state biomass concentration, X = 4.5 kg m−3
=
Biomass production rate Substrate uptake rate
μ ⋅ X 0.4 h −1 × 20 g L−1 = = 0.50 ν 16 g L−1 h −1 Answer (D)
123. Topic: Instrumentation Control and Optimization • Clark electrode is an electrode that measures oxygen on catalytic platinum. • Oxidation reduction potential (ORP) is measured using an electrochemical sensor called an ORP or REDOX sensor. Similar to pH sensors, the most common type of ORP sensor is a combination sensor with a measuring electrode and a reference electrode. The measuring cell, typically a noble metal like platinum or gold, detects changes in REDOX potential, while the reference provides a stable comparison signal. • A load cell is a transducer that is used to create an electrical signal whose magnitude is directly proportional to the force being measured. The various load cell types include hydraulic, pneumatic, and strain gauge. To measure level of an open tank, load cells can be used. • A diaphragm pressure gauge is a device that uses a diaphragm with a known pressure to measure pressure in a fluid. Common uses for diaphragm seals are to protect pressure sensors from the fluid whose pressure is being measured. Answer (C) 124. Topic: Engineering Principle of Bioprocessing
• Solvent extraction (liquid-liquid extraction) is a method to separate compounds based on their relative solubilities in two different immiscible liquids, usually 4.5 − 0 −1 water and an organic solvent. = = 0.562 kg kg 10 − 2 • In affinity chromatography separation of biomolecules Answer (A) can be achieved through their specific interactions. It is a technique of separating biochemical mixture based 121. Topic: Kinetics of Microbial Growth on a highly specific interaction between antibody and Given, rate of biomass production, rB = 45 kg h−1 antigen, enzyme and substrate, receptor and ligand, or Rate of biomass production can also be calculated by, nucleic acid and protein. The degree of purification rB = V × D × X can be thousands of times, and the achieved yield can also be usually very good. where X is biomass concentration; V is the volume = 60 m3; • Extractive distillation is one of the leading distillation D is the dilution rate = 0.55 h−1 processes for the separation of minimum- or maximumTherefore, boiling azeotropes and low-relative-volatility mixtures. rB h 45 kg 1 X = = × × = 1.362 kg m3 3 • S alting out (antisolvent crystallization/precipitation h VD 0.55 60 m crystallization/drowning out) is an effect based on the X 1.363 electrolyte-non-electrolyte interaction, in which the Sucrose concentration in the input feed, S0 = S + = 2.0 + = 4.425 kg m −3 Y 0 . 562 X/S X 1.363 non-electrolyte could be less soluble at high salt conS0 = S + = 2.0 + = 4.425 kg m −3 centrations. It is used as method of separating proteins. YX/S 0.562 Answer (A) Answer (A) We know,
YX/S =
X − X0 Biomass produced = S0 − S Substrate consumed
Chapter 12 • Bioprocess Engineering and Process Biotechnology 235
125. Topic: Production of Biomass and Primary/Secondary 128. Topic: Immobilization of Biocatalysts (Enzymes and Metabolites Cells) for Bioconversion Processes Methylotrophic yeasts are capable to metabolize comGiven: Volume, V = 100 L, Voidage, ε = 0.55, pounds like methanol and formaldehyde. They are very K′M = 0.72 g L−1, V ′max = 18 g L−1 h−1, S0 = 20 g L−1 prevalent in nature, in particular, in molds, fruit and other For fractional conversion of 0.90, S f = S0 − 0.9S0 = 20 − 18 = 2 −1 vegetable products, exudates of trees and theirSbarks. f = S0 − 0.9S0 = 20 − 18 = 2 g L They all use a common methanol-utilizing pathway. For plug flow reactor (PFR), Answer (A) Si − S f S K′ Residence time, τ = M ln i + 126. Sterilization of Air and Media Vmax Vmax Sf ′ ′
• Lysogeny broth (LB), a nutritionally rich medium, is primarily used for the growth of bacteria. culture media will require 121°C for 20 minutes to finally sterilize it in an autoclave. • Polypropylene tube is gamma radiation sterilized and non-pyrogenic. • Germicidal UV lamps have been used in Biological SafetyCabinetstokeepinteriorcleanwhennotinuse. • Membrane filtration is used in chemical and biotechnology processes, which is a means of filtering and cleaning substances. Answer (C)
127. Topic: Kinetics of Microbial Growth In absence of drug, bacteria show exponential first order kinetics growth. A bactericidal drug is one that kills the target organism by interfering with its metabolism. A bacteriolytic drug kills target organisms by lysing its plasma membrane. An apoptotic drug kills target organism by programming cell death. As shown in the graph, on adding these bacteriocidal, bacteriolytic and apoptotic drugs, there is exponential decrease in number of bacteria. On removing these drugs, bacterial growth becomes static. A bacteriostatic drug stops microbial growth without killing them. Thus, on adding bacteriostatic drug, number of bacteria becomes constant and does not multiply. On its removal, as happened in Set B, an exponential increase in bacterial number occurs as previously inhibited static bacteria starts multiplying and resuming growth. Thus, the given antibiotic is bacteriostatic.
Number of live bacteria
Add drug
Remove drug Inhibition
Bacteriostatic drug
Remove Kill Bactericidal or Bacteriolytic drug Time
Answer (A)
0.72 20 20 − 2 ln + 18 2 18 = 0.04 ln 10 + 1 = 0.04 × 2.302 + 1 = 0.092 + 1 = 1.1 h =
Answer (D) 129. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Working Volume = Total volume × Voidage = 100 L × 0.55 = 55 L We know, Residence time, τ = Therefore, D =
Volume (Working) 55 = D Flow rate
55 = 50 L h −1 1.1
Answer (A)
130. Topic: Kinetics of Microbial Growth Change in the callus Growth index = Initial amount Initial weight of inoculum = 5 g Dry weight increased by 1.5-fold of initial inoculum, Thus, final weight of callus = 5 × 1.5 = 7.5 g Putting in the formula above, we get: 7.5 − 5.0 2.5 = = 0.5 Growth index = 5 5 Answer (0.5) 131. Topic: Kinetics of Microbial Growth Given, dilution rate, D = m = 0.6 h−1; product biomass, X = 30 g L−1. So, Biomass productivity (BPx) = m × X = 0.6 × 30 = 18 g L−1 h−1 Answer (18 g L−1 h−1) 132. Topic: Aeration and Agitation We know,
1 Diameter of impeller ( D ) Let us take D1 = Diameter of small bioreactor, D2 = Diameter of large bioreactor, N1 = Agitator speed of small bioreactor, N2 = Agitator speed of large bioreactor Impeller speed ( N ) ∝
236 GATE Biotechnology Chapter-wise Solved Papers
Thus,
N1 D2 = N 2 D1 ⇒ N2 =
D1 × N1 D2
D1 1 = , N1 = 450 rpm. So, putting it in the D2 10 equation above, we get Given
1 N2 = × 450 = 45 rpm 10 Answer (45 rpm) 133. Topic: Kinetics of Microbial Growth Given that Specific rate of substrate consumption, rS = 0.25 g g −1 h −1 Specific rate of product formation, rP = 0.215 g g −1 h −1 Yield of product from the substrate can be calculated by, r 0.215 YP/S = P = = 0.86 rS 0.25 Answer (0.86) 134. Topic: Production of Biomass and Primary/Secondary Metabolites • Corynebacterium lilium, and Brevibacterium lactofermenturn are members of the same species. It is anticipated that all of these strains should be categorized in the species Corynebacterium glutamicum which are glutamic acid producing strains. • Klebsiella oxytoca is a Gram-negative, rod-shaped bacterium. A high production of 2,3-Butanediol occurs from the glycerol by the use of Klebsiella oxytoca. • Aspergillus niger (fungus) is mostly used in the commercial production of citric acid and in enzymes such as amylases, pectinases, and proteases which are of great use in various industries. • Alcaligenes eutrophus is a Gram-negative bacillus. It synthesizes polyhydroxybutyrate (PHB) naturally, which is a specific type of polyhydroxyalkanoates (PHA) that is helpful in the formation of biodegradable plastic on an industrial scale. Answer (B) 135. Topic: Immobilization of Biocatalyst (Enzymes and Cells) for Bioconversion Processes Cell cultures encapsulated in agarose and calcium alginate gels or entrapped in membranes are called immobilized plant cell cultures. Immobilization of plant cells allows better cell-to-cell contact and the cells are also protected from high shear stresses. These immobilized systems can effectively increase the productivity of secondary metabolites in a number of species. Elicitors can also be added to these systems to stimulate secondary metabolism.
Immobilization with suitable bioreactor system provides several advantages, such as continuous process operation. Production of metabolite can be more reliable, simpler, and more predictable. Isolation of the plant products can be quick and efficient, when matched with extraction from the whole plants. It protects the plant cells from shear forces in the bioreactor. Answer (A) 136. Topic: Kinetics of Microbial Growth Given that Initial concentration of substrate, X0 = 1.75 g L−1 Initial concentration of glucose, S0 = 125 g L−1 Yield of product (YX/S) = 0.6 g g−1 Concentration of biomass product = YX/S × S = 125 × 0.6 = 75 g L−1 Maximum final cell concentration, Xf = X0 + YX/S × S = 1.75 + 75 = 76.75 g L−1 Answer (76.75 g L−1) 137. Topic: Batch, Fed-Batch and Continuous Processes Given that Flow rate, F = 200 mL h−1 = 0.2 L h−1 Dilution rate, D = 0.1 h−1 Maximum specific growth rate, mm = 0.4 h−1 Half-velocity constant, KS = 0.3 g h−1 Substrate concentration at quasi-steady state can be calculated by D ⋅ KS 0.1 × 0.3 = = 0.1 g L−1 S= μm − D 0.4 − 0.1 Answer (0.1 g L−1) 138. Topic: Sterilization of Air and Media Given that Δ heating = 12.56 , Δ cooling = 7.48 and Δ total = 52 We know,
Δ total = Δ heating + Δ holding + Δ cooling Δ holding = Δ total − Δ heating − Δ cooling Δ holding = 52 − 12.56 − 7.48 = 31.96 Answer (A)
139. Topic: Sterilization of Air and Media We know Δ holding = kt Given, k = 3.36 min−1, so 31.96 = 3.36t 31.96 t= = 9.5 min 3.36 Answer (B) 140. Topic: Kinetics of Microbial Growth Specific substrate consumption rate in a growing culture is defined as amount of substrate consumed per unit biomass per unit time.
Chapter 12 • Bioprocess Engineering and Process Biotechnology 237
Specific substrate consumption rate Amount of substrate consumed (g) = Biomass (g) × Time (h) =
Degree of reduction =
g g⋅h Answer (C)
141. Topic: Kinetics of Microbial Growth Penicillium notatum produces the antibiotic penicillin which is a secondary metabolite. For primary metabolites like amino acids, proteins which are utilized in the growth of the microorganisms are synthesized in the log or exponential phase. While, secondary metabolites are synthesized only after the active stage of microbial growth, when the microorganism enters in the stationary phase. In this case, the environment may be altered in such a way that entry into stationary phase is encouraged. Penicillin producing molds, for example, produce more of the penicillin when provided with lactose instead of glucose. Answer (C) 142. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes Enzymes or biocatalysts have got remarkable significance in the field of bioprocess technology. They increase the rate of biochemical reaction within the cell without themselves undergoing any change at the end. However, for large extent commercialization of these bio-derived catalysts, their reusability factor becomes mandatory. Consequently, immobilized enzymes with functional efficiency and enhanced reproducibility are used as alternatives inspite 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. It refers to the linkage of an enzyme to an insoluble matrix by either adsorption or by chemical reaction such that the new product is insoluble in water but retains a certain degree of catalytic activity. Answer (B) 143. Topic: Process Scale-Up, Economics and Feasibility Analysis Tip speed of the impeller (vi) relates the fluid shear stress to the rotation speed of impeller. It can be calculated by using blade diameter and number of revolutions per minute (rpm). The relation is vi = π ⋅ N ⋅ D where N = rotations/time and D = diameter of impeller. Answer (A) 144. Topic: Production of Biomass and Primary/Secondary Metabolites For ethanol C2H5OH, number of carbon atom = 2 Number of electrons = 2 × 4 + 6 × 1 + 1 × (–2) = 12
Total no.of available e − 12 = =6 No.of carbon atom 2 Answer (6)
145. Topic: Bioremediation The term biostimulation is often used to describe the addition of electron acceptors, electron donors, or nutrients to stimulate naturally occurring microbial populations. Comprehensively, biostimulation could be perceived as including the introduction of adequate amounts of water, nutrients, and oxygen into the soil, in order to enhance the activity of indigenous microbial degraders. Biostimulation is usually paired under the enhanced bioremediation techniques along with bioaugmentation which includes the introduction of specific microorganisms (indigenous or nonindigenous) aimed at enhancing the biodegradation of target compound or serving as donors of the catabolic genes. Even though the diversity of natural microbial populations apparently displays the potential for contaminant remediation at polluted sites, factors such as lack of electron acceptors or donors, low nitrogen or phosphorus availability, or a lack of stimulation of the metabolic pathways responsible for degradation can inhibit or delay the remediation. Answer (B) 146. Topic: Batch, Fed-Batch and Continuous Processes We know that, specific growth rate (m) can be calculated by:
μ=
μm ⋅ S KS + S
μ ( KS + S )
μKS +μ S S where mm = maximum growth rate, S = substrate concentration, KS = constant. The rate of nutrient exchange is expressed as the dilution rate D. At steady state, the specific growth rate μ of the microorganism is equal to the dilution rate D. Therefore, DKS μm = +μ S (a) When, D = 0.05 and S = 0.067 0.05 × KS μm = + 0.05 = 0.75 KS + 0.05 ….(i) 0.067 (b) When, D = 0.5 and S = 1.667 0.5 × KS μm = + 0.5 = 0.3KS + 0.5 ….(ii) 1.667 Solving equation (i) and (ii) for mm we get: μm = 0.75 KS + 0.05 2.5 μm = 0.75 KS + 1.25 −1.5 μm = −1.20 1.20 μm = = 0.8 h −1 1 . 5 Answer (0.8 h−1) μm =
=
238 GATE Biotechnology Chapter-wise Solved Papers 147. Topic: Bioprocess Design and Development from Lab to Industrial Scale The most probable reason for decrease in lysine concentration may be attributed to the growth of revertant to non-auxotrophic state. The sequence change associated with a mutation can be reversed by a second mutation at the same genetic site where the first mutation had taken place, and the phenotype as well as the genotype of the organism revert back to wild type, is called a reverse or back mutation. The resulting derivative is called a revertant. Answer (B)
Oxygen tension at saturation C2 = 68% Putting values in the formula, we get ⎛ 80 − 55 ⎞ ⎛ 25 ⎞ ln ⎜ ln ⎜ ⎟ ⎝ 80 − 68 ⎟⎠ ⎝ 12 ⎠ 0.732 = = 0.0732 s −1 KL a = = 17 − 7 10 10 Answer (0.0732 s-1) 150. Topic: Media Formulation and Optimization CH O
6 12 6 Glucose (M.W.180 g)
2.5 g of glucose yields =
5 × 105 = ln 0.769 = −0.263 6.5 × 105
Therefore, m < 0. Answer (B) 149. Topic: Aeration and Agitation In dynamics method, Formula for volumetric mass transfer coefficient, ⎛ C * − C2 ⎞ ln ⎜ ⎝ C * − C1 ⎟⎠ KL a = t 2 − t1
Ethanol ( M.W.46g)
By unitary method, 180 g of glucose yields 92 g of ethanol
148. Topic: Kinetics of Microbial Growth x We know that μt = ln t x0 5 Given, xt = 5 ×10 , x0 = 6.5 ×105. Thus,
μt = ln
→ 2C 2 H 5 OH + CO 2 + H 2 O
92 × 2.5 = 1.27g 180 Answer (1.27 g)
151. Topic: Aeration and Agitation Given, Diameter of large vessel, DL = 1.62 m = 162 cm Diameter of small vessel, DS = 16.2 cm Mixing time in small vessel, tS = 15 s Mixing time in large vessel, tL = ? We know that, under similar volumetric agitated input, mixing time is proportional to the vessel diameter. t L ⎛ DL ⎞ = tS ⎜⎝ DS ⎟⎠
11/18
t L = 15 × (10 )
11/18
where, Oxygen tension at saturation C* = 80% Oxygen tension at saturation C1 = 55%
= 61.1s Answer (C)
152. Topic: Kinetics of Microbial Growth Doubling time 0.6 × 105 1.2 × 105 24 h cells cells 6-well plate
Single well
1.2 × 105 6
⇒ 2 × 104 cells
Doubling time 1.8 × 105 cells
0.6 × 105 1.2 × 105 24 h cells cells 6-well plate
Single well
1.2 × 105 6
⇒ 2 × 104 cells
Doubling time 0.6 × 105 1.2 × 105 24 h cells cells 6-well plate
Single well
1.2 × 105 6
⇒ 2 × 104 cells
Chapter 12 • Bioprocess Engineering and Process Biotechnology 239
After 24 h, cells will double in number to 1.2 ×105 cells in each flask. On dispensed these cells equally into 6-well 1.2 × 105 = 2 × 10 4 cells . plates, each well would have 6 Answer (2 × 104 cells) 153. Topic: Batch, Fed-batch and Continuous Processes Given, Initial concentration, S0 = 200 g L−1 Flow rate, F = 50 L h−1 Initial culture volume, V0 = 600 L Initial cell concentration = 20 g L−1 For 600 L, X0 = 20 × 600 = 12000 g Yield coefficient, YX/S = 0.5 g g−1 Time, t = 8 h At quasi steady state,
X = X 0 + F ⋅ YX/S ⋅ S0 ⋅ t = 12000 g + 50 L h −1 × 0.5 g g −1 × 200g L−1 × 8 h
= 12000 g + 40000 g = 52000 g Final volume, V = V0 + Ft = 600 + 50 × 8 = 1000 L Cell concentration =
X 52000 g = = 52 g L−1 V 1000 L Answer (B)
154. Topic: Kinetics of Microbial Growth Incubation time (From 5 PM to 5AM), t = 12 h Initial cell concentration, ci = 104 cells mL−1 Final cell concentration, cf = 107 cells mL−1 The rate of exponential growth of a bacterial culture is expressed as generation time, also the doubling time of the bacterial population. It can be calculated as: t ln 2 12 × 0.693 = c 10 7 ln f ln 4 ci 10 8.316 8.316 = = = 1.21 h 6.91 ln 103 Answer (1.21 h)
Generation time, G =
155. Topic: Kinetics of Microbial Growth Given, Volume of tank, V = 750 L Initial concentration, X0 = 8 g L−1 Flow rate, F = 150 L h−1 For Monod growth model: Maximum specific growth rate, µm = 0.4 h−1 Yield coefficient, YX/S = 0.5 g g−1 Half-velocity constant, KS = 1.5 g L−1
Dilution rate, D =
F 150 L h −1 = = 0.2 h −1 V 750 L
For steady state, Concentration of biomass, X = YX/S ( S0 − S ) Final substrate concentration, S can be calculated as: S=
KS ⋅ D 1.5 × 0.2 0.3 = = = 1.5 μm − D 0.4 − 0.2 0.2
Now, X = 0.5(8 − 1.5) = 0.5 × 6.5 = 3.25 g L−1 Now, we know, Cell productivity = D × X = 0.2 h−1 × 3.25 g L−1 = 0.65 g L−1 h−1 Answer (B) 156. Topic: Rheology of Fermentation Fluid The Prandtl number (Pr) is a dimensionless number, defined as the ratio of momentum diffusivity to thermal diffusivity. That is, the Prandtl number is given as: Pr = v/α where Pr = Prandtl’s number, v = momentum diffusivity (m2 s−1) and α = thermal diffusivity (m2 s−1). Answer (C) 157. Topic: Batch, Fed-Batch and Continuous Processes Fed batch process is a combination of batch and continuous processes and is used to solve the problem of substrate inhibition by optimizing the concentration of the nutrient medium in the tank. Answer (B) 158. Topic: Bioremediation Secondary biological process is involved in the treatment of industrial effluent. It is the biological treatment of water using a wide variety of microorganisms, primarily bacteria. These microorganisms convert biodegradable organic matter contained in wastewater into simple substances and additional biomass. Answer (B) 159. Topic: Large Scale Production and Purification of Recombinant Protein In dead-end filtration the flow of water is perpendicular to the membrane surface and is inversely proportional to the viscosity of the solution. All the water that is introduced in the dead-end cell passes through the membrane. Factors affecting rate of filtration can be spotted by Darcy’s law equation which states that there is a linear relationship between flow velocity (v) and hydraulic gradient (i) for any given saturated soil under steady laminar flow conditions. Q = KA
ΔP μL
240 GATE Biotechnology Chapter-wise Solved Papers where Q = Rate of water flow (Volume of filtrate per unit time; K = hydraulic conductivity, constant for the filter medium; A = cross-sectional area of filter medium; ΔP = Pressure drop across the filter medium; m = viscosity of the filtrate and L = thickness of the cake.
So,
Helical ribbons
Helical screws
Gate anchors
Paddles
Flat-blade turbines
102
5
P = 5 × 1000 × (5)3 × (1)5
103
10
P = NP × ρ × N × D where P is power required is to be calculated; NP , power number = 5 P, density of broth = 1000 kg m−3 N, number of revolutions per second = 300 rpm/60 = 5 rps D, Diameter of turbine = 1m Putting all these values in the formula above we get, 3
104
Propellers
P NP = ρ × N 3 × D5
105
Anchors
160. Topic: Aeration and Agitation The dimensionless power number is determined by the following formula:
106
Viscosity (centipoise)
Answer (C)
107
= 625 × 103 kg ⋅ m 2 ⋅ s −3 P = 625 × 103 Watt = 625 kW Answer (625 kW)
161. Topic: Aeration and Agitation Impellers are responsible for mixing action and imparting shear stress to the fluid in a process. According to figure shown (Mixing time for four different types of impellers (Zhao et al. 2011)), every impeller type can be used in a certain range of viscosity. A high Reynolds number (turbulence) can only be achieved by disrupting the uniform flow of horizontal layers of the viscous liquids in laminar flow. Using the right impeller is imperative to achieve this. An impeller used to mix a low viscosity fluid (turbulent impeller) will fail to agitate the fluid sufficiently to turn laminar flow into turbulent flow. It may lead to heat a build-up, which can cause more trouble. On the other hand, a laminar impeller has a much larger surface area, which can disrupt the uniformity of flow, turning a laminar flow into a turbulent flow with a sufficiently high Reynolds number.
1 Impeller type
Answer (B) 162. Topic: Kinetics of Microbial Growth Stationary phase of bacterial growth results from a condition where growth rate and death rate are equal. The rate of cell growth matches the rate of cell death because the number of new cells created is limited by the growth factor. Answer (A) 163. Topic: Media Formulation and Optimization → CH1.8 O0.5 N 0.2 + cCO 2 + dH 2 O C2 H 5 OH + a NH 3 + bO 2 ⎯⎯ C-Source
N-Source
Biomass
→ CH1.8 O0.5 N 0.2 + cCO 2 + dH 2 O C2 H 5 OH + a NH 3 + bO 2 ⎯⎯ C-Source N-Source Molecular mass of biomass = B1iomass × 12 + 1.8 × 1 + 0.5 × 16 + 0.2 × 14 = 1 × 12 + 1 . 8 × 1 + 0.5 × 16 + 0.2 × 14 = 12 + 1.8 + 8 + 2.8 = 24.6 g From 1 mole of ethanol, 1 mole of biomass is produced. On performing elemental balance on both sides of reaction: On LHS, No. of C atoms = 2 2=1+c c=1 No. of N atoms = a a = 0.2 No. of H atoms = 6 + 3a = 6 + 3 (0.2) = 6.6 1.8 + 2d = 6.6 2d = 6.6 – 1.8 = 4.8 d = 4.8/2 = 2.4
On RHS, No. of C atoms = 1 + c No. of N atoms = 0.2 No. of H atoms = 1.8 + 2d
Chapter 12 • Bioprocess Engineering and Process Biotechnology 241
No. of O atoms = 1 + 2b 1 + 2b = 0.5 + 2c + d 1 + 2b = 0.5 + 2(1) + 2.4 1 + 2b = 4.9 2b = 3.9 b = 1.95
No. of O atoms = 0.5 + 2c + d
N0 = e kd ⋅t Nt
N0 108 108 = = e kd ⋅t e 0.1×10 2.71 = 3.7 × 10 7 spores
Nt =
Answer (3.7 × 107 spores)
Final equation will be: 166. Topic: Chemical Engineering Principles Applied to Biological System → CH1.8 O0.55 N 0.2 + CO 2 + 2.4 H 2 O C2 H 5 OH + 0.2 NH 3 + 1.95O 2 ⎯⎯ We know that, C-Source
N-Source
Biomass
In laminar transition aeration P ⎧ N i2 ….(i) ∝⎨ In turbulent aeratiion V ⎩ N i3 Di2 Scaled up criteria is equal to the impeller speed, thus,
CH1.8 O0.55 N 0.2 + CO 2 + 2.4 H 2 O Biomass
Answer (1.95)
N1 D1 = N 2 D2
164. Topic: Batch, Fed-Batch and Continuous Processes Given, Dilution rate, D = 0.5 h−1 Feed substrate concentration, S0 = 10 g L−1 Growth rate, mm = 0.7 h−1 Growth rate constant, KS = 0.3 g L−1 Maximum biomass yield with respect to substrate, YX/S = 0.4 g g−1
Using Eq. (i) for turbulent aeration,
⎛ KS D ⎞ X = YX/S ⎜ S0 − μ m − D ⎟⎠ ⎝ 0.3 × 0.5 ⎞ ⎛ = 0.4 ⎜10 − ⎟ ⎝ 0.7 − 0.5 ⎠
= 0.4 (10 − 0.75) = 0.4 × 9.25 = 3.7 g L−1
Answer (3.7 g L−1)
165. Topic: Sterilization of Air and Media Decimal reduction time = 23 min at 121°C Death kinetics
N0 1 N 1 kd = ln 0 = ln t N t 23 N 0 / 10 =
1 10 1 ln = × 2.302 = 0.1 min −1 23 1 23
P2 ⎛ N 23 D22 ⎞ = P1 ⎜⎝ N13 D12 ⎟⎠
X = YX/S ( S0 − S ) From Monod’s kinetics,
N 2 D1 = …(ii) N1 D2
Therefore,
Using Eq. (ii), we have
P2 ⎛ D13 ⎞ ⎛ D22 ⎞ = P1 ⎜⎝ D23 ⎟⎠ ⎜⎝ D12 ⎟⎠
Thus,
P2 ⎛ D1 ⎞ = P1 ⎜⎝ D2 ⎟⎠
2
Answer (A) 167. Topic: Various Types of Microbial and Enzyme Reactors Given, Flow rate, F = 25 L min−1 Feed substrate concentration, S0 = 2 mol L−1 Conversion efficiency, h = 95% = 95/100 = 0.95 Final concentration, CA = (S0 – h S0) = (2 – 0.95 × 2) = 0.1 ml L−1 F We know, Rate of reaction, rA = μ ⋅ S ⋅ η = ⋅ S ⋅ η V F V = ⋅ S ⋅ η …(i) rA Given, the rate equation:
Initial spore concentration = 105 spores per mL In one-liter of medium, no. of spores =105 × 103 = 108 spores
− rA =
0.1CA 1 + 0.5CA
Putting this in Eq. (i) above, we get:
105 × 103 = 108 spores Time of sterilization = 10 min at 121°C
V=
F ⋅ S0 ⋅ η 0.1CA 1 + 0.5CA
242 GATE Biotechnology Chapter-wise Solved Papers
25 × 2 × 0.95 V= ⎛ 0.1 × 0.1 ⎞ ⎜⎝ ⎟ 1 + 0.5 × 0.1⎠ 25 × 1.9 = 0.45 = 4987.5 L = 0 ⎛ .01⎞ ⎜⎝ ⎟ 1.05 ⎠
Answer (4987.5 L) 168. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes We know, the reaction rate for immobilized enzyme, V, is given by V=
Vmax A[S ] ...(i) K M + [S ]
where KM, rate constant = 10 s−1 A, Total surface area Vmax, maximal rate of reaction S, substrate concentration at the surface Assuming that all of the surface is equally accessible, the rate of flow of substrate to the surface is found to be proportional both to the surface area and the difference in substrate concentration between the bulk of the solution and the microenvironment next to the surface. It is given by the relationship
−10 + 22.3 −10 − 22.3 or 2 2 = 6.18 or − 16.65
S=
Substrate concentration cannot be in negative. Therefore, substrate concentration at the surface of the immobilized particle will be 6.18 mM. Answer (6.18 mM) 169. Topic: Chemical Engineering Principles Applied to Biological System Volume of chemostat (V) = 5 L Flow rate of fresh medium (F) = 0.2 L min−1 F 0.2 Dilution rate ( μ) = = = 0.04 min −1 V 5 Inlet substrate concentration (Si) = 25 g L−1 Outlet substrate concentration (So) = 2.5 g L−1 Rate of substrate consumption = μΔS = 0.04 (25 – 2.5) = 0.9 g L−1 min−1 Answer (0.9 g L−1 min−1) 170. Topic: Kinetics of Microbial Growth Given, dX X⎞ ⎛ = rX ⎜1 − ⎟ ⎝ dt K⎠ dX rX = (K − X ) dt K
V = K L A[S0 − S ] …(ii)
KL, mass transfer coefficient = 1 cm s−1 S0, bulk substrate concentration On equating Eqs. (i) and (ii), we get
Vmax A[S ] = K L A[S0 − S ] K M + [S ]
Vmax [S ] = K L [S0 − S ] …(iii) K M + [S ]
Further, Vmax = K L S0 = 1 × 10 = 10 mM cm s −1 Putting values in (iii) we get:
10 [S ] = 1[10 − S ] 10 + [S ]
1 dX r = (K − X ) X dt K
dX r = dt ...(i) X (K − X ) K Now,
dX = K [log X − log( K − X )] X (K − X ) X = K log (K − X ) Putting this value in (i) we get, K log
10 [S ] = [10 − S ][10 + S ] 10 S = 10 − S 2
2
log
S + 10 S − 100 = 0 It’s a quadratic equation, on solving for S we get,
X rt = 2 (K − X ) K rt
2
−10 ± 10 − 4 × ( −100) −10 ± 100 + 400 = 2 2 −10 ± 500 −10 ± 22.3 = = 2 2
X r r = dt = t (K − X ) K K
X 2 = eK (K − X )
2
rt
S=
rt
rt
X = e K ( K − X ) = Ke K − Xe K 2
rt
2
rt
X + Xe K = Ke K 2
2
rt rt ⎞ ⎛ 2 2 X ⎜1 + e K ⎟ = Ke K ⎠ ⎝ rt K2
2
rt
rt
rt
X = e K ( K − X ) = Ke K − Xe K 2
rt
2
Chapter 12 • Bioprocess Engineering and Process Biotechnology 243
rt
X + Xe K = Ke K 2
2
2
rt rt ⎞ ⎛ 2 2 X ⎜1 + e K ⎟ = Ke K ⎠ ⎝ rt
X =
Ke K
2
rt ⎞ ⎛ K2 ⎜1 + e ⎟ ⎠ ⎝
This equation is depicted by exponential graph (A). Answer (A) 171. Topic: Batch, Fed-Batch and Continuous Processes Given that for zero-order reaction, k = 10−3 mol min−1 Initial concentration of A = 0.1 mol L−1, volume of reactor = 100 L Initial concentration of A in reactor [A0] = 0.1 × 100 = 10 moles We know, for zero order kinetics A B
d[A ] d[A ] ∝ [A ]0 ⇒ − =k dt dt −( A f − A 0 ) = kt
As A f → 0,
−
A0 − Af = k ⋅ t t=
A0 − Af k
t=
10 − 0 = 10 4 min 10 −3 Answer (104 min)
172. Topic: Various Types of Microbial and Enzyme Reactors Anaerobic growth of organism Glucose fed = 1800 g 180 g of glucose contains = 1 mole of carbon 1800 g of glucose contains = 10 moles of carbon Every mole of carbon present in glucose distributes among the products as follows: 1 (C-mole of glucose) → 0.14 (C-mole of biomass) + 0.25 (C-mole of ethanol) + 0.3 (C-mole acetaldehyde) + 0.31 (C-mole glycerol) 1 C mole of glucose (C6H12O6) is converting into 0.25 carbon mole of ethanol (C2H5OH) To balance out the number of carbon atoms on each side, we need to multiply ethanol side by 3. Thus, 1 C mole of glucose = 3 × 0.25 moles of ethanol 1 (C mole of glucose = 0.75 × 10 moles of ethanol = 7.5 moles of ethanol 1 mole of ethanol = 46 g 7.5 moles of ethanol = 7.5 × 46 g = 345 g Answer (345 g)
173. Topic: Bioremediation Trickling biological filters is a method of secondary or biological treatment that use a packed bed of rocks covered with a slimy film of aerobic microorganisms such as Sphaerotilus and Beggiatoa. Wastewater is fed on the top. As it descends, the organic compounds diffuse through the microbial film and are decomposed. Answer (B) 174. Topic: Rheology of Fermentation Fluids A basic general equation for Newtonian fluid is given below: dv x dy where η is the apparent velocity. τ yx is shear stress and
τ yx = − η
dv x = γ is shear rate. η is a function of shear rate. dy So, η = f (γ ) . • For curve I, the relationship between the shear stress (τ) and the shear rate (γ) is a straight line, and the viscosity of the fluid remains constant as the shear rate is varied. Thus, it is representing Newtonian fluid. They are true fluids. Other fluids are non-Newtonian fluids. • A non-Newtonian fluid is a fluid for which the relationship τ/γ is not a constant. The variation of shear strength causes to change the viscosity of such fluids. • Curve III, a downward curve passing through the origin is characterized by an apparent viscosity that decreases with an increasing shear rate, thus, it is representing a pseudo-plastic fluid. • Curve IV, an upward curve passing through the origin, is characterized by an apparent viscosity that increases with an increasing shear rate, thus, it is representing a dilatant fluid. • Curve II represents a Bingham plastic fluid, which are fluids that require a minimum stress to be applied before they flow. These are strictly non-Newtonian, but once the flow starts they behave essentially as Newtonian fluids. Answer (D) 175. Topic: Sterilization of Air and Media Given, moist heat sterilization follows first order kinetics, thus, ⎛N ⎞ ln ⎜ 0 ⎟ = kd t ⎝ Nt ⎠ where N0 = number of initial viable spores = 1010 Nt = number of viable spores after time t = 1 kd = rate constant = 1.0 min−1 t = time required in min
244 GATE Biotechnology Chapter-wise Solved Papers Putting values in the formula above:
Thus, in 2 hours duration, amount of glucose added [C] = 200 × 2 = 400 mL Given, final culture volume [F] = 1000 mL
⎛ 1010 ⎞ =1t ln ⎜ ⎝ 1 ⎟⎠
⎛ 1010 ⎞ t = ln ⎜ = 23.03 min ⎝ 1 ⎟⎠
Answer (23.03 min) 176. Topic: Unit Operations in Liquid-Liquid Extraction
E= =
[ F ] = [ I ] + [C ] ⇒ 1000 = x + 400 ⇒ x = 1000 − 400 = 600 mL Answer (600 mL)
180. Topic: Immobilization of Biocatalysts (Enzymes and Cells) for Bioconversion Processes For soluble enzyme, K1 = 0.03, half-life = t(1/ 2 )se
Extraction factor,
Partition coefficient ( K D ) Distribution ratio of water to solvent ( D ) 170 =2 85
Answer (2)
177. Topic: Mass and Heat Transfer The volumetric mass-transfer coefficient (KLa) is used as a measure of the aeration capacity of a fermenter. Given, Specific oxygen uptake ( qo ) = 0.4 h−1 = 0.4 /3600 = 1.11 × 10−4 s−1 Concentration of biomass (X) = 12g/L Oxygen uptake rate (OUR) = qo X = 1.11 × 10−4 s−1 × 12g L−1 = 1.332 × 10−3 g L−1 s−1 KLa = OUR/(C* – CL) = OUR / DC where DC is solubility of oxygen in the medium = 8 mg L−1 = 8 × 10−3 g L−1 KLa = 1.332 × 10−3 / 8 × 10−3 = 0.167 s−1 Answer (0.167 s−1) 178. Topic: Engineering Principle of Bioprocessing F V where F = Feed flow rate = 100 mL h−1 V = Culture volume = 1 L = 1000 mL Dilution rate (D ) =
Therefore, D =
100 = 0.1 h −1 1000
Effluent substrate concentration [S ] = [S ] =
Ks D μmax − D
1 × 0.1 = 1 gL−1 0.2 − 0.1 Answer (1 g L−1)
179. Topic: Batch, Fed-Batch and Continuous Processes Let us assume the initial culture volume [I] to be x mL Flow rate of glucose = 200 mL per hour
For immobilized enzyme K 2 = 0.005, half-life = t(1/ 2 ) ie t(1/ 2 )se ln c1 / c0 K = 1 = ln c2 / c0 K 2 t(1/ 2 ) ie As final concentration is half of the initial concentration for both soluble and immobilized enzymes. t(1/ 2 )se K1 t(1/ 2 )se 0.03 = ⇒ = =6 K 2 t(1/ 2 ) ie t(1/ 2 ) ie 0.005 Answer (6) 181. Topic: Production of Biomass and Primary/Secondary Metabolites The degree of reduction, g, for organic compounds can be defined as the number of equivalents of available electrons per gram atom C. The degree of reduction for some key elements are C = 4, H = 1, O = –2 and N = –3. Thus, degree of reduction for acetic acid can be calculated as follows: Acetic acid (C2H4O2): 2 (4) + 4 (1) + 2 (–2) = 8 + 4 – 4 =8 γ = 8/ 2 = 4 (as acetic acid has 2 carbon atoms) Answer (4) 182. Topic: Mass and Heat Transfer Heat transfer coefficient 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 m−2 K−1). Answer (C) 183. Topic: Kinetics of Microbial Growth, Substrate Utilization and Product Formation When a carbon source is used up it does not necessarily mean that all growth stops. This is because dying cells can lyse and provide a source of nutrients. Growth on dead cells is called endogenous metabolism. Endoge-
Chapter 12 • Bioprocess Engineering and Process Biotechnology 245
nous metabolism occurs throughout the growth cycle, but it can be best observed during stationary phase when growth is measured in terms of oxygen uptake or evolution of carbon dioxide. Answer (A) 184. Topic: Kinetics of Microbial Growth, Substrate Utilization and Product Formation The kinds of organisms found in a given environment and the rates at which they grow can be influenced by a variety of factors, both physical and biochemical. Physical factors include pH, temperature, oxygen concentration, moisture, hydrostatic pressure, osmotic pressure, and radiation. Nutritional (biochemical) factors include availability of extracellular carbon, nitrogen, sulfur, phosphorus, trace elements, and, in some cases, vitamins. Monod equation is a kinetic model which describes microbial growth as a functional relationship between the specific growth rate and an essential substrate concentration.
μ = μ max ×
S ( KS + S )
where μ = Specific growth rate of bacteria
μmax = Specific growth rate of bacteria S = Concentration of limiting substrate or nutrient provided externally KS = Saturation constant, equal to the concentration of substrate giving growth rate of µmax Answer (D) 185. Topic: Media Formulation and Optimization Molecular weight of glucose (C6H12O6) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g Molecular weight of biomass (C6H10O3N) = (6 × 12) + (10 × 1) + (3 × 16) + (1 × 14) = 144 g From the reaction, 1 mole of glucose produces 0.48 mole of biomass. Thus, 180 g of glucose = 0.48 × 144 = 69.12 g of biomass 180 Now, to produce 1 g of biomass = = 2.604 g of 69.12 glucose is required Thus, to produce 50 g of biomass in 100,000 L of reactor = 2.604 × 50 × 105 g glucose is required = 13020.83 kg of glucose Answer (13020.83 kg)
186. Topic: Instrumentation Control and Optimization • A Manometer is an instrument used to measure pressures. A common simple manometer consists of a U-shaped tube of glass filled with some liquid. This device indicates the difference between two pressures (differential pressure), or between a single pressure and atmosphere (gage pressure), when one side is open to atmosphere. Typically, the liquid is mercury because of its high density. • The rotameter is an industrial variable area flowmeter used to measure the flowrate of liquids and gases. It works on the principle of upthrust force exerted by fluid and force of gravity. The rotameter consists of a tube and float. The buoyant force exerted on an immersed object is equal to the weight of liquid displaced by the object. The float response to flowrate changes is linear. • A tachometer is a sensor device used to measure the rotation speed of a shaft or a disk. This device indicates the revolutions per minute (RPM) performed by the object. It is usually restricted to mechanical or electrical instruments. • Hemocytometer, also known as Petroff-Hausser counting chamber can be used to measure microbial growth by direct microscopic counts. In this method, a known volume of medium is introduced into a specially calibrated, etched glass slide. A microbial suspension is introduced onto the chamber with a calibrated pipette. After the cells settle and the liquid currents have slowed, the microorganisms are counted in specific calibrated areas. Their number per unit volume of the original suspension is calculated by using an appropriate formula. Answer (C) 187. Topic: Unit Operations in Solid-Liquid Separation and Liquid-Liquid Extraction During cross flow filtration, transmembrane pressure (TMP) is the average applied pressure from the feed to the filtrate side of the membrane. It is given by: TMP =
Pi + Po − Pp 2
Given, pressure drop (∆P) that is the difference in pressure along the feed channel of the membrane from the inlet to the outlet is 2 atm. ΔP = PF − PR Inlet feed pressure, Pi = 3 atm Filtrate pressure, PF = atmospheric pressure = 1 atm
246 GATE Biotechnology Chapter-wise Solved Papers Thus, there is no permeate side pressure, (Pp). Therefore,
TMP =
3 +1 = 2 atm 2 Answer (2 atm)
188. Topic: Instrumentation Control and Optimization Proportional controller is called as gain controller, where the output position is related to the deviation of the set point and measured control process value. It produces an output, which is proportional to error signal. It is used to change the transient response as per the requirement. Advantages of proportional controller includes: • It helps in reducing the steady state error, thus makes the system more stable. • Slow response of the over damped system can be made faster with the help of these controllers. Disadvantages of proportional controller includes: • Due to presence of these controllers we get some offsets in the system. • It also increases the maximum overshoot of the system. Answer (B) 189. Topic: Bioremediation Given, Initial concentration = 10,000 ppm = 10,000 mg L−1 Final concentration = 10 ppm = 10 mg L−1 Inlet flow rate = 80 L h−1 Maximum specific growth rate, mm = 1 h−1 Saturation constant, KS = 100 mg L−1 Cell death rate, kd = 0.01 h−1 We know, Dilution rate, D = Also,
D = μ=
Flow rate Volume
μm ⋅ S KS + S
Substrate concentration, S = 10,000 mg L−1 – 10 mg L−1 = 9,990 mg L−1 Purring the values in the formula, we get, D= Volume required =
1 × 9990 = 0.99 h −1 100 + 9990 80 = 80 L 0.99 Answer (80 L)
190. Topic: Process Scale-Up, Economics and Feasibility Analysis Given, Volume of fermenter, V = 1 L …(i) Diameter of impeller, Di = 20 cm …(ii) Agitator speed, N1 = 600 rpm …(iii) Ratio of impeller diameter to fermenter diameter, Di/Dr = 0.3 …(iv) Volume of scaled-up fermenter, V = 8000 L …(v) Scale-up factor = 3 Volume of the reactor = 3 8000 = 20 For fermenter, from Eqs. (i) and (iv), Diameter of reactor, Dr = 20/0.3 = 66.67 cm For scaled-up reactor, on multiplying with scale up factor, Diameter of scaled-up reactor, Dr′ = 66.67 × 20 cm ⎛D ⎞ N 2 = N1 ⎜ r ⎟ ⎝ Dr′ ⎠
We know, for scale-up fermenter, Putting the values in the formula, ⎛ 66.67 ⎞ N 2 = 600 ⎜ = 30 rpm ⎝ 66.67 × 20 ⎟⎠
Answer (30 rpm) 191. Topic: Batch, Fed-Batch and Continuous Processes A chemostat, or bioreactor, is a continuous stirred-tank reactor (CSTR) used for continuous production of microbial biomass. It consists of a fresh water and nutrient reservoir connected to a growth chamber (or reactor), with microorganism. It is a growth vessel into which fresh medium is delivered at a constant rate and cells and spent medium overflow at the same rate. Thus, the culture is forced to divide to keep up with the dilution, and the system exists in a steady state where inputs match outputs. The chemostat is attractive due to the enormous amount of control that is possible: growth rate, cell density, and selection pressure are all independently set. In a chemostat setup, the rate at which the growth-limiting substrate is supplied to the culture vessel is called the dilution rate (D) and thus equals the specific growth rate. If the dilution rate exceeds the critical dilution rate at which maximum specific growth rate of the microorganism is observed, the culture will become unstable and washout will occur. Answer (D) 192. Topic: Sterilization of Air and Media During the sterilization process, Volume of culture = 104 L
Chapter 12 • Bioprocess Engineering and Process Biotechnology 247
Initial no. of cells, No = 106 spores mL−1 Final no. of cells, Nt = 10−3 Death constant, kd = 4 min−1 Time, t = ? 1 [N ] We know, kd = ln 0 t [N t ]
Therefore, t=
t=
1 [N 0 ] ln kd [ N t ]
1 [106 ] 1 ln = ( 20.72) = 5.2 min 4 [10 −3 ] 4 Answer (5.2 min)
193. Topic: Kinetics of Microbial Growth, Substrate Utilization and Product Formation In the lag phase, the organisms do not increase significantly in number, but they are metabolically active— growing in size, synthesizing enzymes, and incorporating
various molecules from the medium. During this phase the individual organisms increase in size, and they produce large quantities of energy in the form of ATP. The length of the lag phase is determined in part by characteristics of the bacterial species and in part by conditions in the media—both the medium from which the organisms are taken and the one to which they are transferred. Some species adapt to the new medium in an hour or two; others take several days. Organisms from old cultures, adapted to limited nutrients and large accumulations of wastes, take longer to adjust to a new medium than do those transferred from a relatively fresh, nutrient-rich medium. It depends to some extent on the size of inoculum also. High cell densities assist adaptation of resistant cells to the unfavorable growth conditions by some unspecified medium conditioning effect. Answer (D)
appendix
GATE BIOTECHNOLOGY 2020 Questions
Chapter 1: Engineering Mathematics 1. A function is as follows: ⎧15 if x < 1 f ( x) = ⎨ ⎩cx if x ≥ 1 The function f is a continuous function when c is equal to __________ (answer is an integer). (GATE 2020; 1 Mark) ∂Z for X = 1 and ∂X Y = 0 is __________ (answer is an integer). (GATE 2020; 1 Mark)
2. Given that Z = X 2 + Y 2, the value of
⎡ 4 1⎤ 3. The largest eigenvalue of the matrix ⎢ ⎥ is ⎣ −2 1⎦ __________. (GATE 2020; 1 Mark) 4. 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). (GATE 2020; 1 Mark) 5. A variable Y is a function of t. Given that Y (t = 0) = 1 dY and Y (t = 1) = 2, in the interval t = [0, 1] can be dt approximated as __________. (GATE 2020; 1 Mark) 6. The system of linear equations cx + y = 5 3x + 3y = 6 has no solution when c is equal to __________. (GATE 2020; 2 Marks) 7. A function f is given as: f (X ) = 4X − X 2 The function f is maximized when X is equal to __________. (GATE 2020; 2 Marks)
Ch wise GATE_Biotechnology_2020 paper.indd 249
8. 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). (GATE 2020; 2 Marks)
Chapter 2: Biochemistry 9. Which of the following types of molecules act as biological catalysts? 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 (GATE 2020; 1 Mark) 10. 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 (GATE 2020; 2 Marks) 11. 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−1 (round off to 2 decimal places). Given: free energy change for ATP hydrolysis under standard conditions is −30.5 kJ mol−1 and R = 8.315 kJ mol−1 K−1 (GATE 2020; 2 Marks) 12. The mitochondrial electron transfer chain oxidizes NADH with oxygen being the terminal electron acceptor. The redox potentials for the two half-reactions are given below: NAD + + H + + 2e − → NADH, E 0′ = −0.32 V 1 O 2 + 2H + + 2e − → H 2 O, E 0′ = 0.816 V 2
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250 GATE Biotechnology Chapter-wise Solved Papers The free energy change associated with the transfer of electrons from NADH to O2 is __________ kJ mol−1 (round off to 2 decimal places). Given: F = 96500 C/mol. (GATE 2020; 2 Marks)
Chapter 3: Microbiology 13. Which of the following statements are correct? 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 (GATE 2020; 1 Mark) 14. 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. (GATE 2020; 1 Mark) 15. A list of pathogens (Group I) and a list of anti-microbial 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) (B) (C) (D)
P−4; Q−3; R−2; S−5 P−5; Q−2; R−4; S−l P−2; Q−3; R−4; S−l P−2; Q−3; R−l; S−5
17. 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 cell-to-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. (GATE 2020; 1 Mark) 18. 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 (C) Q and R only
(B) P and R only (D) P, Q and R (GATE 2020; 1 Mark)
19. 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 (C) P, Q and R only
(B) P and S only (D) Q and R only (GATE 2020; 2 Marks)
20. Match sub-cellular organelles listed in Group I with their features listed in Group II. (GATE 2020; 2 Marks)
Chapter 4: Cell Biology 16. 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.
Ch wise GATE_Biotechnology_2020 paper.indd 250
(C) trimeric GTPase involved in regulation of cytoskeleton. (D) monomeric GTPase involved in regulation of cytoskeleton. (GATE 2020; 1 Mark)
Group I
Group II
P. Mitochondrion
1. Single membrane enclosed
Q. Chloroplast
2. Double-membrane enclosed
R. Nucleus
3. Maternal inheritance
S. Endoplasmic reticulum
4. Endosymbiotic origin
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APPENDIX: GATE BIOTECHNOLOGY 2020 251
Chapter 6: Analytical Techniques
(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 (GATE 2020; 2 Marks)
Chapter 5: Molecular Biology and Genetics 21. Protein P becomes functional upon phosphorylation of a serine residue, replacing this serine with __________ will result in a phosphomimic mutant of P. (A) alanine (B) aspartic acid (C) phenylalanine (D) lysine (GATE 2020; 1 Mark) 22. The number of molecules of a nucleotide of molecular weight 300 g mol−1 present in 10 picomoles is __________ × 1012 (round off to 2 decimal places). (GATE 2020; 1 Mark) 23. 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 × Rrpp (GATE 2020; 2 Marks) 24. E. coli was grown in 15N medium for several generations. Cells were then transferred to 14N 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). (GATE 2020; 2 Marks) 25. The amino acid sequence of a peptide is Phe−Leu−Ile− 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 __________. (GATE 2020; 2 Marks)
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26. 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 (GATE 2020; 1 Mark) 27. Protein A and protein B form a covalent complex. Gel filtration chromatography of this complex showed a peak corresponding to 200 kDa. SDS-PAGE 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. (GATE 2020; 2 Marks)
Chapter 7: Immunology 28. Which class of antibody is first made by developing B cells inside bone marrow? (A) IgG (B) IgE (C) IgA (D) IgM (GATE 2020; 1 Mark) 29. 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 (C) Q and R only
(B) P and Q only (D) P, Q and R (GATE 2020; 2 Marks)
Chapter 8: Bioinformatics 30. 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.
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252 GATE Biotechnology Chapter-wise Solved Papers (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. (GATE 2020; 1 Mark) 31. Carl Woese used the gene sequence of which one of the following for phylogenetic taxonomy of prokaryotes? (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 (GATE 2020; 2 Marks) 32. 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). (GATE 2020; 2 Marks)
Chapter 9: Recombinant DNA Technology 33. DNA synthesized from an RNA template is called (A) recombinant DNA. (B) transcript. (C) T-DNA. (D) complementary DNA. (GATE 2020; 1 Mark) 34. 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 (GATE 2020; 1 Mark) 35. A vector derived from which one of the following viruses is used for high-frequency genomic integration of a transgene in animal cells? (A) Adenovirus (B) Adeno-associated virus (C) Lentivirus (D) Herpes simplex virus (GATE 2020; 1 Mark)
Ch wise GATE_Biotechnology_2020 paper.indd 252
36. 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. (GATE 2020; 1 Mark) 37. The DNA sequence shown below is to be amplified by PCR: 5′ GCTAAGATCTGAATTTTCC……. TTGGGCAATAATGTAGCGC 3′ 3′ CGATTCTAGACTTAAAAGG……. AACCCGTTATTACATCGCG 5′ Which one of the following pair of primers can be used for this amplification? (A) 5′ GGAAATTCAGATCTTAGT 3′ and 5′ TGGGCAATAATGTAGCGC 3′ (B) 5′ GCTAAGATCTGAATTTTC 3′ and 5′ GCGCTACATTATTGCCCA 3′ (C) 5′ CGGAAATTCAGATCTTAG 3′ and 5′ GCGCTACATTATTGCCCA 3′ (D) 5′ GCTAAGATCTGAATTTTC 3′ and 5′ TGGGCAATAATGTAGCGC 3′ (GATE 2020; 2 Marks) 38. 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
Media supplemented with Mutants W
X
Y
Z
Z1
No
No
Yes
Yes
Z2
No
Yes
Yes
Yes
Z3
No
No
No
Yes
Z4
Yes
Yes
Yes
Yes
What is the order of the four complementation groups in terms of the step they block? 1 2 3 4 (A) S ........ ⎯Z⎯ →W ⎯Z⎯ → X ⎯Z⎯ → Y ⎯Z⎯ →Z 4 2 1 3 (B) S ........ ⎯Z⎯ →W ⎯Z⎯ → X ⎯Z⎯ → Y ⎯Z⎯ →Z
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APPENDIX: GATE BIOTECHNOLOGY 2020 253 3 1 2 4 (C) S ........ ⎯Z⎯ →W ⎯Z⎯ → X ⎯Z⎯ → Y ⎯Z⎯ →Z 4 1 2 3 (D) S ........ ⎯Z⎯ →W ⎯Z⎯ → X ⎯Z⎯ → Y ⎯Z⎯ →Z (GATE 2020; 2 Marks)
39. 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. (GATE 2020; 2 Marks) 40. The schematic of a plasmid with a gap in one of the strands is shown below: ATCGCGATG TAGC AC 3′ OH 5′ OH
(A) P and R only (C) Q and R only
(B) P and S only (D) Q and S only (GATE 2020; 2 Marks)
42. 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 __________. (GATE 2020; 2 Marks)
Chapter 10: Plant Biotechnology 43. Vincristine and vinblastine, two commercially important secondary metabolites from Catharanthus roseus, are examples of (A) alkaloids. (B) flavonoids. (C) terpenoids. (D) steroids. (GATE 2020; 1 Mark) 44. 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 (GATE 2020; 2 Marks)
(B) P, R and S only (D) P, Q and R only (GATE 2020; 2 Marks)
45. 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. (GATE 2020; 2 Marks)
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 S. Affected individuals, as well as their progeny, can be cured through somatic gene therapy
46. 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].
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 (C) P and R only
Ch wise GATE_Biotechnology_2020 paper.indd 253
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254 GATE Biotechnology Chapter-wise Solved Papers
49. 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 __________. (GATE 2020; 2 Marks)
(B)
(A)
Temperature
48. 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. (GATE 2020; 2 Marks)
53. 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.
Time
(C)
(D)
Time
Chapter 12: Bioprocess Engineering and Process Biotechnology 50. The elemental composition of dry biomass of a yeast species is CH1.6O0.4N0.2S0.0024P0.017. The contribution of carbon to the dry biomass is __________ % (round off to 2 decimal places). [Given: atomic weights of H, C, N, O, P and S are 1, 12, 14, 16, 31 and 32, respectively] (GATE 2020; 1 Mark) 51. 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
Ch wise GATE_Biotechnology_2020 paper.indd 254
Time
Temperature
47. Which one of the following media components is used to maintain pH in mammalian cell culture? (A) CaCh (B) MgSO (C) NaCl (D) NaHCO3 (GATE 2020; 1 Mark)
52. 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 __________ . (GATE 2020; 1 Mark)
Temperature
Chapter 11: Animal Biotechnology
dissolved, the amount of solute in solvent A phase is __________ g. (GATE 2020; 1 Mark)
Temperature
(C) Both [A] and [R] are true and [R] is the correct reason for [A]. (D) [A] is true but [R] is false. (GATE 2020; 2 Marks)
Time
(GATE 2020; 2 Marks) 54. A batch reactor is inoculated with 1 g L−1 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−1 is __________ min (round off to 2 decimal places). (GATE 2020; 2 Marks) 55. 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−1. Biomass from the substrate is generated as Yx/s = 0.4 g g−1. The chemostat volume is 0.9 L and media is fed at 1 L h−1 and contains 20 g L−1 of glucose. At steady state, the concentration of biomass in the chemostat is __________ g L−1. (GATE 2020; 2 Marks)
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APPENDIX: GATE BIOTECHNOLOGY 2020 255
ANSWER KEY 1. (15)
2. (2)
3. (3)
4. (0.5)
5. (1)
6. (1)
7. (2)
8. (0.75) 9. (A)
10. (D)
11. (18521.25) 12. (219.3) 13. (D)
14. (D)
15. (C)
16. (B)
17. (D)
19. (D)
20. (C)
21. (B)
22. (6.02) 23. (A)
24. (0.12)
25. (864) 26. (C)
27. (25kDa)
28. (D)
29. (A)
30. (A)
31. (B)
32. (83.3 %)
33. (D)
34. (B)
35. (C)
36. (D)
37. (B)
38. (B)
39. (D)
40. (B)
41. (C)
42. (4096) 43. (A)
44. (D)
45. (D)
46. (C)
47. (D)
48. (B)
49. (1196) 50. (51.24%) 51. (60)
53. (A)
54. (528.67)
52. (2)
18. (B)
55. (8.34)
Answers with Explanations 1. Topic: Limit, Continuity and Differentiability A function f (x) is continuous at x = 1 when Left Hand Limit of f (x) = Right Hand Limit of f (x) = f (1) Now,
4. Topic: Standard Deviation Since normal curves are symmetric about their mean, the area under the curve to the right of the mean is 0.5 and the area under the curve to the left of the mean is 0.5.
lim f ( x ) = lim− 15 = 15, lim+ f ( x ) = lim+ cx = c
x →1−
x →1
x →1
x →1
and f (1) = c(1) = c Therefore, f (x) is a continuous function when c is equal to 15. Answer (15)
2. Topic: Partial Derivatives Given, Z = X 2 + Y 2 Differentiating the given equation with respect to X, we get
area = 0.5
0 –3
∂Z ∂Z = 2X + 0 ⇒ = 2X ∂X ∂X For X = 1 and Y = 0, we get
Answer (2) 3. Topic: Eigenvalues and Eigenvectors The characteristic equation of a matrix A is given by 1
−2
1− λ
=0
⇒ ( 4 − λ )(1 − λ ) + 2 = 0
Now,
z
⇒ λ 2 − 3λ − 2λ + 6 = 0 ⇒ λ ( λ − 3) − 2( λ − 3) = 0 ⇒ ( λ − 3)( λ − 2) = 0 ⇒ λ = 2, 3 Hence, the largest eigenvalue of the matrix is 3. Answer (3)
dY dY (t ) = = 1+ 0 = 1 dt dt
Answer (1)
6. Topic: System of Linear Equations The system of linear equations a1 x + b1 y = c1 and a2 x + b2 y = c2 has no solution, if a1 b1 c1 = ≠ . a2 b2 c2
⇒ λ 2 − 5λ + 6 = 0
Ch wise GATE_Biotechnology_2020 paper.indd 255
+3
5. Topic: Limit, Continuity and Differentiability Given, Y is a function of t, Y (t = 0) = 1 and Y (t = 1) = 2. Let Y (t) = t + 1 Y ( t = 0) = 0 + 1 = 1 and Y (t =1) = 1 + 1 = 2
| A − λI |= 0 4−λ
0
Answer (0.5)
∂Z = 2(1) = 2 ∂X
⇒
area = 0.5
Given, cx + y = 5 3x + 3y = 6 On comparing with the standard equations, we get a1 = c; a2 = 3; b1 = 1; b2 = 3; c1 = 5 and c2 = 6
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256 GATE Biotechnology Chapter-wise Solved Papers On substituting n − 1 = k, that is, n = 1 + k
Substituting these values in above condition, we get c 1 5 = ≠ ⇒c=1 3 3 6
∞
Answer (1)
⎛ ∑ (k + 1 − 1) ⋅ ⎜⎝
k +1= 0
1⎞ ⎟ 3⎠
Given, f ( X ) = 4 X − X 2
⎛ 1⎞ −1 ⎜ ⎟ ⎝ 3⎠
−1
f ′′ ( X ) = −2 < 0 Therefore, x = 2 is a point of maximum.
1 2 3 4 5 + + + + + ... + ∞ 30 31 32 33 34 n
∞
a a ⋅ rn = ∑ 1− r n= 0 For a = 1,
∞
∑r
=
n
n= 0
1 1− r
Differentiating with respect to r ∞
∑ n⋅ r
n −1
=−
n= 0
1 1 × ( −1) = 2 (1 − r ) (1 − r ) 2
1 Substituting r = , we get 3 ∞
⎛ 1⎞
∑ n ⋅ ⎜⎝ 3⎟⎠
n −1
=−
n= 0
1 ⎛ ⎛ 1⎞ ⎞ ⎜⎝1 − ⎜⎝ 3 ⎟⎠ ⎟⎠
2
=
9 4
Substituting n = n − 1 + 1, we get ∞
⎛ 1⎞
n −1
⎛ 1⎞ +⎜ ⎟ ⎝ 3⎠
n −1
⎛ 1⎞ +⎜ ⎟ ⎝ 3⎠
n −1
∑ [(n − 1) + 1] ⋅ ⎜⎝ 3⎟⎠
=
9 4
=
9 4
=
9 4
n= 0
∞
n −1
∞
n −1
⎛ 1⎞ ( n − 1) ⋅ ⎜ ⎟ ∑ ⎝ 3⎠ n= 0 ⎛ 1⎞ ( n − 1) ⋅ ⎜ ⎟ ∑ ⎝ 3⎠ n= 0
Ch wise GATE_Biotechnology_2020 paper.indd 256
n
−1
∞
n
n
∞
⎛ 1⎞
∑ n ⎜⎝ 3⎟⎠ n= 0
5 2 3 4 + + + + ... + ∞ 3 9 27 81
=
k
9 4
k
n
∞ 9 ⎛ 1⎞ + ∑⎜ ⎟ = ⎝ ⎠ 3 4 n= 0
∞ a ⎛ 1⎞ Now, = ∑ ⎜⎝ ⎟⎠ = 1− r n= 0 3
Therefore,
8. Topic: Sequences and Series
k +1−1
n
∞ 9 ⎛ 1⎞ ⎛ 1⎞ n⎜ ⎟ + ∑⎜ ⎟ = ∑ ⎝ ⎠ ⎝ ⎠ 3 3 4 n= 0 n= 0
Answer (2)
∞ ⎛ 1⎞ ⇒ S = ∑ n⎜ ⎟ n= 0 ⎝ 3⎠
⎛ 1⎞ ∑ ⎜ ⎟ k +1= 0 ⎝ 3 ⎠
∞ ⎛ 1⎞ ⎛ 1⎞ + ∑ n⎜ ⎟ + ⎜ ⎟ ⎝ 3⎠ n= 0 ⎝ 3⎠
On differentiating Eq. (i), we get f ′( X ) = 4 − 2 X ⇒ 4 − 2X = 0 ⇒ X = 2 On double differentiating f (X), we get
∞
∞ 9 ⎛ 1⎞ ⎛ 1⎞ k⎜ ⎟ +∑ ⎜ ⎟ = ∑ ⎝ ⎠ ⎝ ⎠ 3 3 4 k = −1 k = −1
(i)
We know that for maximum and minimum value of f , f ′( X ) = 0
S=
+
∞
7. Topic: Maxima and Minima
S = 1+
k +1−1
∞
n
+
1 1−
1 3
=
3 2
3 9 = 2 4 n
9 3 ⎛ 1⎞ n⎜ ⎟ = − = S ∑ ⎝ ⎠ 3 4 2 n= 0 ⇒S=
9 3 3 − = = 0.75 4 2 4 Answer (0.75)
9. Topic: Enzymes Enzymes are the biggest class of biological catalysts which are basically proteins with catalytic activity. Ribozymes (ribonucleic acid enzymes) are RNA molecules that have the ability to catalyze specific biochemical reactions, including RNA splicing in gene expression, similar to the action of protein enzymes. Answer (A) 10. Topic: Basic Concepts and Designs of Metabolism All three strategies are used for metabolic regulation: phosphorylation/dephosphorylation, allostery and feedback inhibition: • Protein tyrosine phosphorylation and dephosphorylation are fundamental cellular signaling mechanisms that control cell growth and differentiation, mitogenesis, cell cycle, metabolism, gene transcription, cytoskeletal integrity, neuronal development, and the immune response. • Allosteric control is an extremely important mechanism for cellular regulation. Allosteric enzymes play a pivotal role in cells because they have two functions – they not only catalyze reactions in metabolic pathways, but also control the rates of these pathways. • Cells have evolved to use feedback inhibition to regulate enzyme activity in metabolism, by using the
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APPENDIX: GATE BIOTECHNOLOGY 2020 257
products of the enzymatic reactions to inhibit further enzyme activity. Answer (D) 11. Topic: Bioenergetics Given, Concentration of ATP = 2.59 mM = 2.59 × 10 −3 M Concentration of ADP = 0.73 mM = 0.73 × 10 −3 M Concentration of Pi = 2.72 mM = 2.72 × 10 −3 M T = 37oC = 37 + 273 = 310 K For ATP hydrolysis, ΔG 0 = −30.5 kJ mol −1 Thus, for ATP biosynthesis ( ADP + P → ATP ), ΔG 0 = 30.5 kJ mol −1 Putting the values in the formula below, ⎡ ( ATP) ⎤ ΔG = ΔG o + RT ln ⎢ ⎥ ⎢⎣ ( ADP) ( Pi ) ⎥⎦ ⎡ ⎤ 2.59 × 10 −3 = 30.5 + 8.315 × 310 ln ⎢ −3 −3 ⎥ ⎣ 0.73 × 10 × 2.72 × 10 ⎦ = 30.5 + 8.315 × 310 ln [1304.39] = 30.5 + 18490.75 ΔG = 18521.25 kJ mol−1 Answer (18521.25) 12. Topic: Bioenergetics 0 ENAD = − 0.32 V + / NADH
EH0 + / H O = 816 V 2
ΔG = − nFE = −( 2 moles)(96500 C mol −1 )( −0.32) + z ( −2 moles)(96500 C mol −1 )(0.816) = ( −61, 760) + ( −157, 488) = 219, 248 J mol −1 = 219.3 kJ mol −1 Answer (219.3) 13. Topic: Viruses: Structure and Classification Viruses can be oncogenic too causing cancer as one of the symptoms of their infection for example Human papilloma virus causing Cervical Cancer. A tumor suppressor gene can be turned off due to epigenetic alterations too like hypermethylation of promoter sequence. miRNAs can have both oncogenic and tumor suppressive functions. Answer (D) 14. Topic: Microbial Classification and Diversity The archaeal membrane lipid composition is one of the most remarkable distinguishing features of Archaea
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where the hydrocarbon chain consists of isoprenoid moieties which are ether linked to the enantiomeric glycerol backbone, glycerol-1-phosphate (G1P). Answer (D) 15. Topic: Microbial Diseases • Amantadine inhibits the replication of influenza A viruses by interfering with the uncoating of the virus inside the cell. Like rimantadine, it is an M2 inhibitor which blocks the ion channel formed by the M2 protein that spans the viral membrane. • Fluconazole is an antifungal medication used for a number of fungal infections. This includes candidiasis, blastomycosis, coccidiodomycosis, cryptococcosis, histoplasmosis, dermatophytosis, and pityriasis versicolor. • Artemisinin and its derivatives (collectively referred to as ARTs) rapidly reduce the parasite burden in Plasmodium falciparum infections, and antimalarial control is highly dependent on ART combination therapies (ACTs). • Isoniazid, also known as isonicotinyl hydrazide (INH), is an antibiotic used for the treatment of tuberculosis against Mycobacterium. Answer (C) 16. Topic: Cell Signaling Ras is a guanosine-nucleotide-binding protein. It’s a monomeric GTPase, which is related in structure to the Gα subunit of heterotrimeric G proteins. Ras signaling regulates many important physiologic processes within a cell, such as cell cycle progression, survival, apoptosis, etc. When a cell surface receptor receives a signal to grow, it transmits the signal to Ras. Activated Ras binds GTP and sends the growth signal to the nucleus. After the signal has been sent, GTP is hydrolyzed to GDP, and Ras becomes inactive once again. Answer (B) 17. Topic: Cell Signaling Cells use Ca2+ and not Na+ for cell to cell signaling because: • Ca2+ exerts allosteric regulatory effects on many enzymes and proteins. • It activates ion channels etc. But, in the cytosol of mammalian cell, the concentration of Na+ is 12 mM, whereas Ca2+ concentration is less than 0.0002 mM. Thus, assertion is true, but the reason is false. Answer (D)
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258 GATE Biotechnology Chapter-wise Solved Papers 18. Topic: Signal Transduction The energy transducing membranes are the plasma membrane of prokaryotic bacteria and blue-green algae, the inner membrane of mitochondria, and the thylakoid membrane of chloroplasts. The ATP synthase, called the proton-translocating ATPase, is present in closely similar forms in all energy-transducing membranes. When the complex is assembled in the membrane it catalyzes a hydrolysis or synthesis of ATP that is coupled to the obligatory translocation of protons across the membrane. Answer (B) 19. Topic: Cell Cycle P. Incorrect: Cyclins are named because they undergo a cycle of synthesis and degradation in each cell cycle. A cyclin forms a complex with CDK, which begins to activate but the complete activation requires phosphorylation, as well. Q. Correct: CDKs phosphorylate their substrates by transferring phosphate groups from ATP to specific stretches of amino acids in the substrates. CDKs must also be in a particular phosphorylation state — with some sites phosphorylated and others dephosphorylated — in order for activation to occur. R. Correct: Cell cycle progression is driven forward by the action of CDKs, which are activated by binding to cyclins. To drive the cell cycle in one direction, cyclins must be cyclically destroyed, and this is brought about via their ubiquitin mediated proteolysis. S. Incorrect: Chromosome condensation process is carried out with the help of protein complexes called condensins. Condensins are inactive until they are phosphorylated by MPF (mitosis promoting factor). MPF is a complex of CDK1 and cyclin B. Once they are phosphorylated, they begin the process of condensation. Answer (D) 20. Topic: Eukaryotic Cell Structure • Cytoplasmic inheritance in mammals is derived mostly, if not exclusively, from the maternal line. Mitochondria, and their DNA molecules (mtDNA), are the genetic units of this method of inheritance. Mitochondrial genes, in the contrast to genes in the nucleus, have an exclusively maternal mode of inheritance in mammals • Endosymbiotic origin is an evolutionary theory of the origin of eukaryotic cells from prokaryotic organisms,
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it holds that the organelles distinguishing eukaryote cells evolved through symbiosis of individual single-celled prokaryotes. The theory holds that mitochondria, plastids such as chloroplasts, and possibly other organelles of eukaryotic cells are descended from formerly free-living prokaryotes taken one inside the other in endosymbiosis. • The nuclear membrane, also called the nuclear envelope, is a double membrane layer that separates the contents of the nucleus from the rest of the cell. It is found in both animal and plant cells. • Organelles such as vacuole, lysosome, Golgi apparatus, endoplasmic reticulum, etc., are bounded by a single membrane. Answer (C) 21. Topic: Mutation and Mutagenesis A phosphomimic mutant is an amino acid substitution that mimics a phosphorylated amino acid leading to activation or deactivation. The amino acids that are phosphorylated are serine, tyrosine and threonine. Non-phosphorylated amino acids that appear chemically similar to phosphorylated serine is aspartic acid. Answer (B) 22. Topic: Molecular Structure of Genes and Chromosomes Molecular weight = 300 g mol−1 No. of mol = 10 picomoles = 10 × 10 −12 moles We know, 1 mol = 6.023 × 10 23 molecules 10 × 10 −12 moles = 6.023 × 10 23 × 10 × 10 −12 molecules = 6.023 × 1012 molecules Answer (6.02) 23. Topic: Mendelian Inheritance For red fruit, purple stem, the genotype can be RRPP, RrPp, RRPp, RrPPP – 145 For red fruit, green stem, the genotype can be RRpp, Rrpp – 184 For yellow fruit, purple stem, the genotype can be rrPP, rrPp – 66 For yellow fruit, green stem, the genotype can be rrpp – 47 The two genotypes with maximum number of offspring are the genotypes of the two parents. Thus, the answer is option (A). Answer (A)
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APPENDIX: GATE BIOTECHNOLOGY 2020 259
24. Topic: Microbial Genetics All DNA is initially 15N-labeled E.coli
Add 15N labeled cells to medium and grow for four generations 14N
1
2
3
4
Generation
75%
88%
100%
50%
25%
12%
20
40
100% 0
60
80
N14 N15
Time (min.)
• Generation 1: DNA isolated after one round of DNA replication produced a single band of intermediate density between the heavy 15N and 14N DNA. • Generation 2: DNA isolated after second round of DNA replication produced two bands. One was in the intermediate position, while the second was higher bring of lighter 14N. • Generation 3: Each hybrid DNA molecule from the second generation would give rise to a hybrid molecule and a light molecule in the third generation, while each light DNA molecule would only yield more light molecules. Thus, one-third (25%) DNA will form band at intermediate density and two-third (75%) DNA will form band at lighter density. • Generation 4: Similarly, in this generation, one-eighth (12.5%) DNA will form band at intermediate density and seven-eighth (87.5%) DNA will form band at lighter density. Thus, the proportion of total DNA with intermediate density = 0.12% Answer (0.12) 25. Topic: Translation Number of DNA sequences that encode this peptide can be calculated by: Phe × Leu × Ile × Met × Ser × Leu = 2 × 6 × 3 × 1 × 4 × 6 = 864 Answer (864)
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Gen 3
Density
ultracentrifuge
50%
Gen 4
0 –
Gen 2
N14
CsCl
+
Gen 1
Take a sample prior to adding to 14N medium (Generation 0)
Gen 0
N15
26. Topic: Principles of Chromatography: Ion Exchange Ion-exchange chromatography separates ions and polar molecules depending on their affinity to the ion exchanger which is based on the differences in net charge. Negatively charged molecules bind to positive supports and positively charged molecules bind to negative supports. Proteins are solubilized in buffers at pH above or lower than their iso-electric points to give them an overall negative or positive charge, respectively. The protein solution will then be subjected to ion exchange where in this overall charge will play a part in binding to the particular support and separating. If these proteins are dissolved in pH = 6, the protein pI = 5.6 will have negative charge and protein with pI = 6.8 will have a positive charge. Answer (C) 27. Topic: Gel Filtration Chromatography Gel filtration chromatography separates proteins, peptides, and oligonucleotides on the basis of size. Molecules move through a bed of porous beads, diffusing into the beads to greater or lesser degrees. Smaller molecules diffuse further into the pores of the beads and therefore move through the bed more slowly, while larger molecules enter less or not at all and thus move through the bed more quickly. Both molecular weight and threedimensional shape contribute to the degree of retention. SDS-PAGE (sodium dodecyl sulfate–polyacrylamide gel electrophoresis) is a variant of polyacrylamide gel
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260 GATE Biotechnology Chapter-wise Solved Papers
electrophoresis, an analytical method in biochemistry for the separation of charged molecules in mixtures by their molecular masses in an electric field. It uses sodium dodecyl sulfate (SDS) molecules to help identify and isolate protein molecules. Given, that both proteins A and B show peak of 200 kDa collectively. Since SDS-PAGE with beta-mercaptoethanol show single band of 50 kDa and without beta mercaptoethanol show single band of 25 kDa. The beta mercaptoethanol used in SDS PAGE is a reducing agent that cleaves disulphide bonds in proteins. This means A is a dimer having two subunits of 25 kDa each making it 50 kDa. Also, B is a trimer with 3 subunits of 50 kDa each. Answer (25 kDa)
28. Topic: Antibody: Structure and Classification B cells develop in the bone marrow from a common progenitor shared with T, NK, and some DC subsets. Progenitor B cells progress through the early stages of maturation, rearranging heavy- and light-chain genes at the pro- to pre-B cell stage until they express rearranged IgM receptors on the cell surface as immature B cells, at which point they exit the bone marrow to continue maturation in the peripheral immune system. Answer (D) 29. Topic: Cell Mediated Immunity • Correct: The T cell recognizes and interacts with the antigen-class II MHC molecule complex on the membrane of the antigen-presenting cell. • Incorrect: B cells recognize intact antigen; T cells recognize fragments of protein antigens that have been partly degraded inside the antigen-presenting cell. The peptide fragments are then carried to the surface of the presenting cell on special molecules called MHC proteins, which present the fragments to T cells. • Correct: Dendritic cells, macrophages, and B cells are referred to as “professional” antigen presenting cells due to their ability to present exogenous antigens using MHC II receptors. Answer (A) 30. Topic: Sequence Analysis: Phylogeny A phylogenetic tree is based on alignment of protein amino acid sequences. Since both these proteins cytochrome c and ribulose 5-phosphate belong to two different protein families, thus their two phylogenetic trees will not be identical. The two proteins are from different families and thus the frequency of the mutation will not coincide. Answer (A) 31. Topic: Sequence Analysis: Phylogeny Carl Woese proposed a three-domain system based on phylogenetic taxonomy of 16 S ribosomal RNA, a technique that revolutionized microbiology. 16 S ribosomal
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RNA is the component of the 30 S small subunit of a prokaryotic ribosome. Answer (B) 32. Topic: Sequence and Structure Database Given, Total number of globin sequence = 78 Number of globin in algorithm = 65 Putting these values in the formula below: Number of globins retrieved from algorithm ×100 Sensitivity = Total number of globin sequencies in database =
65 × 100 = 83.3% 78 Answer (83.3%)
33. Topic: cDNA and Genomic DNA Library In a process called reverse transcription, the enzyme reverse transcriptase synthesizes cDNA or complementary DNA from an RNA template. mRNA Reverse transcriptase
Single Stranded DNA DNA polymerase
Complementary DNA (double stranded)
Answer (D) 34. Topic: Gene Isolation, Cloning and Expression Ganciclovir is a synthetic analogue of 2′-deoxy-guanosine. It has to be phosphorylated by the viral kinases to be active. The initial monophosphorylation by thymidine kinase is crucial for accumulation ganciclovir tri-phosphate accumulation in the virus infected cells. Thus, animal cells expressing viral thymidine kinase will be killed and the ones without it, won’t because of the absence of initial mono phosphorylation of ganciclovir. Answer (B) 35. Topic: Vectors Lentiviral vectors are a type of retrovirus that can infect both dividing and nondividing cells because their preintegration complex (virus “shell”) can get through the intact membrane of the nucleus of the target cell. They can be used to provide highly effective gene therapy as lentiviruses can change the expression of their target cell’s gene for up to six months. They can be used for nondividing or terminally differentiated cells such as neurons, macrophages, hematopoietic stem cells, retinal photoreceptors, and muscle and liver cells, cell types for which previous gene therapy methods could not be used. Answer (C)
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APPENDIX: GATE BIOTECHNOLOGY 2020 261
36. Topic: Ti Plasmid The Ti plasmid contains several genes including the vir genes which control the process of infection of the plant and transfer of the T-DNA to the chromosome. The T-DNA contains the tumor-inducing (auxin) genes and a gene that expresses specific compounds, opines, which are used by the bacterium as a carbon source. T-DNA region Cytokinin Opine
Auxin
Right border
Left border
Opine catabolism vir genes
ori
Answer (D) 37. Topic: Polymerase Chain Reactions Two primers are used in each PCR reaction, and they are designed so that they flank the target region. That is, they are given sequences that will make them bind to opposite strands of the template DNA, just at the edges of the region to be copied. The primers bind to the template by complementary base pairing. Thus, primer for given DNA sequence will be: 5′ GCTAAGATCTGAATTTTCC…….TTGGGCAATA ATGTAGCGC 3′ 3′ ACCCGTTATTACATCGCG 5′ 3′ CGATTCTAGACTTAAAAGG…….AACCCGTTAT TACATCGCG 5′ 5′ GCTAAGATCTGAATTTTC 3′ Answer (B) 38. Topic: Site-Directed Mutagenesis • Z4 mutants can grow when on any W, X Y and Z if provided as media supplements. Thus, Z4 can be the starting intermediate. • Z2 mutants can grow if X, Y and Z are provided as media supplements, but they won’t show any growth if W is provided. Thus, Z2 can be the second intermediate. • Z1 mutants can grow only if Y and Z are provided as media supplements, but they won’t show any growth if W and X are provided. Thus, Z1 can be the third intermediate.
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• Z3 mutants can grow only if Z is provided as media supplements, but they won’t show any growth if W, X and Y are provided. Thus, Z3 can be the last intermediate. This, option (B) is correct. Answer (B) 39. Topic: Restriction and Modification Enzymes • Dam methylase is an orphan methyltransferase that is not part of a restriction-modification system but operates independently to regulate gene expression, mismatch repair, and bacterial replication amongst many other functions. Thus, the assertion is wrong. • Dam methylase adds a methyl group to the adenosine residue of the sequence 5′− GATC− 3′ in newly synthesized DNA. Thus, the reason is true. Answer (D) 40. Topic: Plasmids • Polynucleotide kinase 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. This is used since both the ends have hydroxyl group. This enzyme adds a phosphate group on the 5′ end. • DNA polymerase is used for the addition of the nucleotide that will pair with the template strand, • DNA ligase is required for ligating the end of the new DNA strand. Answer (B) 41. Topic: Gene Therapy • Germ line therapy is when the gene therapy is targeted in germ lines like sperms and eggs which can be used to produce progeny which would not carry the genetic defect post the successful gene therapy. • Somatic gene therapy is when the gene therapy is targeted at somatic cells which do not participate in reproduction and thus a somatic gene therapy doesn’t lead to cured progeny. Answer (C) 42. Topic: Restriction and Modification Enzymes There are four different possible nucleotides that may be inserted at any one position (G, A, T, or C). EcoRI has a six-base recognition site, So, they will cut once per every 46, or 4096 bases. Answer (4096) 43. Topic: Production of Secondary Metabolites by Plant Suspension Cultures Vincamine, vinblastine and vincristine are very important clinic alkaloids. They are produced naturally by plants: vincamine by Vinca minor, and vinblascine and vincristine by Madagascar periwinkle (Catharanthus roseus). Answer (A)
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262 GATE Biotechnology Chapter-wise Solved Papers 44. Topic: Tissue Culture and Cell Suspension Culture System • 6-Benzylaminopurine, benzyl adenine, BAP or BA is a first-generation synthetic cytokinin that elicits plant growth and development responses, setting blossoms and stimulating fruit richness by stimulating cell division. • Kinetin was the first artificial cytokinin to be discovered. It was isolated from herring sperm and named after its ability to promote cell division. Answer (D) 45. Topic: Totipotency Totipotent cells are the cells that have the ability to divide and produce all of the differentiated cells in an organism. Spores and zygotes are examples of totipotent cells. These cells have the capacity to divide and self-renew and can give rise to establish each type of cells in the body. Answer (D) 46. Topic: Transgenic Plants Golden rice is a variety of rice produced through genetic engineering to biosynthesize beta-carotene; a precursor of vitamin A. It was created by transforming rice with two beta-carotene biosynthesis genes: • psy (phytoene synthase) from daffodil Narcissus pseudonarcissus. • crtI (phytoene desaturase) from the soil bacterium Erwinia uredovora. The Golden Rice Solution β -Carotene Pathway Genes Added IPP Geranylgeranyl diphosphate Daffodil gene
Phytoene synthase
Phytoene Vitamin A Pathway is complete and functional
Single bacterial gene: performs both functions
β - carotene desaturase Lycopene Daffodil gene
Golden rice
Phytoene desaturase
Lycopene-beta-cyclase
β - carotene (vitamin A precursor)
Answer (C) 47. Topic: Animal Cell Culture: Media Composition Sodium bicarbonate or NaHCO3 dissociates into sodium and bicarbonate ions in water to act as a buffer because bicarbonate ions increase the pH of the solution making it more alkaline. For media containing 1.5 to 2.2 g L−1 sodium bicarbonate, use 5% CO2. The addition of bicarbonate ions to the media increases the pH to about 7.4 and changes the color to red. Answer (D)
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48. Topic: Stem Cell Technology Homologous recombination applied to mouse embryonic stem (ES) cells has revolutionized the study of gene function in mammals. Although most often used to generate knockout mice, homologous recombination has also been applied in mouse ES cells allowed to differentiate in vitro. It was first applied for the development of knockout mice through targeted gene inactivation Answer (B) 49. Topic: Animal Cell Culture During the cell cycle, two kinetochores are attached to spindle fibers on the opposite poles. In humans, there are 23 pairs of chromosomes. Given, N0 = 3, n = 3 (since 3 cells completed 3 rounds of cell division) Thus, Number kinetochores = N 0 × 2n = 3 × 23 = 24 Given, 2 cells did not divide even once, Thus, total number of kinetochores = 24 + 2 = 26 Since two kinetochores are attached to each spindle fibers. Thus, number of kinetochores = 26 × 2 = 52. For 23 pairs of chromosomes, total number of kinetochores = 52 × 23 = 1196 Answer (1196) 50. Topic: Production of Biomass and Primary/ Secondary Metabolites Given, No. of carbon atoms = 1 Atomic weight of carbon atom = 12 No. of hydrogen atoms = 1.6 Atomic weight of hydrogen atom = 1 No. of oxygen atoms = 0.4 Atomic weight of oxygen atom = 16 No. of nitrogen atoms = 0.2 Atomic weight of nitrogen atom = 14 No. of sulphur atoms = 0.0024 Atomic weight of sulphur atoms = 32 No. of phosphorous atoms = 0.017 Atomic weight of phosphorous atoms = 31 Total weight of biomass = 1(12) +1.6(1) + 0.4(16) + 0.2(14) + 0.0024(32) + 0.017(31) = 1.6 + 6.4 + 2.8 + 0.0768 + 0.527 + 12 = 23.4 Contribution of carbon to the dry biomass =
12 × 100 = 51.24 % 23.4 Answer (51.24%)
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APPENDIX: GATE BIOTECHNOLOGY 2020 263
51. Topic: Rheology of Fermentation Fluids Mass of solvent A in container = 2 kg Mass of solvent B in container = 2 kg Let us assume that, amount of solute S in B = x Then, amount of solute S in A = 100 − x ….(i) Or, as given solute S is 1.5 times more soluble in A than in B, Thus, amount of solute S in A can be 1.5 × B = 1.5 × x ….(ii) 100 g
A
2 kg
1.5
53. Topic: Mass and Heat Transfer The amount of energy required to melt ice and vaporize water are respectively called the latent heats of fusion and vaporization. • Saturation temperature describes the temperature at the boiling point; at this temperature steam can be saturated with (carry) an unspecified amount of water of identical temperature. • Superheated steam describes the condition at which steam has absorbed sensible heat beyond the latent heat, warming the steam beyond saturation temperature. The region of the constant temperature with time is the transitions point from ice to water and then to water to stream where the temperature doesn’t increase linearly with time.
B 2 kg
SENSIBLE HEAT
1
LATENT HEAT OF FUSION
SENSIBLE HEAT
LATENT HEAT OF VAPORIZATION
LIQUID WATER
BOILING WATER OR SATURATED STEAM
SENSIBLE HEAT
From (i) and (ii)
100 − x = 1.5 x ∴ x = 40 g So, the amount of solute in A becomes 1.5 × 40 = 60 g Answer (60)
52. Topic: Mass and Heat Transfer We know that, Radius of big sphere = R Surface area of big sphere = 4π R 2 Volume of big sphere = 4 / 3π R3 Now, Ratio of surface area to volume of big sphere =
4π R 2 3 = 4 / 3π R3 R
T E M P E R A T U R E
ICE
METLING OR FREEZING
….(i)
HEAT ADDED
Similarly, for radius of small sphere = r 4π r 2 3 Ratio of surface area to volume = = ….(ii) 4 / 3π r 3 r As the big gas bubble has broken down into 8 small bubbles of equal radius. Thus, 4 4 8 × π r 3 = π R3 3 3 R3 = 8r3 R = 2r
….(iii)
Answer (A) 54. Topic: Batch, Fed-Batch and Continuous Processes Given, Initial concentration, X0 = 1g L−1 Time of lag phase, tlag= 30 min Specific growth rate, μm = 0.00417 min−1 Final concentration, X = 8 g L−1 We know, ln 2 Doubling time, μ max
From (i), (ii) and (iii), n=2
ln Answer (2)
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SUPERHEATED STEAM
X = μ m tlog X0
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264 GATE Biotechnology Chapter-wise Solved Papers tlog =
1 X ln μm X 0
tlog =
8 1 ln 0.00417 1
tlog = 498.67 min Total duration = tlog + tlag = 498.67 + 30 = 528.67 min Answer (528.67) 55. Topic: Kinetics of Microbial Growth Given, Specific growth rate, μm = 0.45 h−1 Ks = 0.5 g L−1 Biomass, Yx/s = 0.4 g g−1
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Volume, V = 0.9 L, Flow rate, F = 1 L h−1 Concentration, S0 = 20 g L−1 Now, we know, F = DV D = F/V = 1/0.9 = 1.111 h−1 On using the following formula and putting the given values, X = ( S0 − SF ) Yx/s ⎧⎪ ⎡ K D ⎤ ⎫⎪ = ⎨20 − ⎢ s ⎥ ⎬ Yx/s ⎣ μ m − D ⎦ ⎭⎪ ⎩⎪ ⎧ ⎡ 0.5 × 1.11 ⎤ ⎫ = ⎨20 − ⎢ ⎥ ⎬ 0.4 ⎣ 0.45 − 1.11⎦ ⎭ ⎩ = 8.34 g L−1 Answer (8.34)
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