questions

CSIR-NET 2011-2017

Detailed Analysis of

BIOCHEMISTRY QUESTIONS Second Edition

Aditya Arya, PhD Anamika Gangwar, PhD

CSIR-NET EXAM

Detailed Analysis of Biochemistry Questions

CSIR-NET EXAM

Detailed Analysis of Biochemistry Questions 2011- 2017

Aditya Arya, PhD Anamika Gangwar, PhD

Drawing Pin Publishing New Delhi, India Detailed analysis fo Biochemistry Questions, 1st Edition ISBN 000-0-0000-0000-0 Copyright © 2017. Drawing Pin Publishing. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any other storage and retrieval system, without prior written permission of the publisher. All the copyright related queries may be directly sent to the publisher at [email protected] Disclaimer Although, utmost care has been taken to avoid any errors in the preparation of answers keys, and most of the answers has been matched with the key provided by CSIR, however in case of any discrepancies or loss of any kind due to incorrect answers or solutions, authors and/or publishers shall not be responsible. Cover Design: Dr. Aditya Arya Typesetting Editor: Dr.Amit Kumar Copy Editor: Mr. Subhojit Paul Production Manager: Shashank Arya Printed in India

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Preface

This booklet has been prepared as a complementary book to Concise Biochemistry, authored by Dr. Aditya Arya. Considering CSIR-NET as one of most comprehensive examination for research and teaching eligibility, preparation is incomplete without practicing previous year questions. As the exam is very diverse and consist of various domains of life sciences, ample choice is provided in the paper that enables students to choose the domain of their interest and expertise. A comprehensive preparation of Biochemistry can enable students to attempt nearly 20% of the questions in part C. Lack of detailed and authentic solutions to the questions leaves the preparation incomplete and most students either remember the choice from key provided by the examiner or several other publishers which does no enhance knowledge. Besides CSIRNET this booklet may also be used by students preparing for GATE or M.Sc. entrance examinations for improving their analytical skills in biochemistry. This book has been prepared in consultation with experts and suitable justifications are provided in case of ambiguity in answers. This booklet will be updated on yearly basis with new questions and solutions. We wish you a good luck in your CSIR-NET

Authors

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Table of Contents

01. Review of Trend 02. Questions and Solutions June- 2011 03. Questions and Solutions December- 2011 04. Questions and Solutions June- 2012 05. Questions and Solutions December - 2012 06. Questions and Solutions June- 2013 07. Questions and Solutions December - 2013 08. Questions and Solutions June - 2014 09. Questions and Solutions December - 2014 10. Questions and Solutions June- 2015 11. Questions and Solutions December - 2015 12. Questions and Solutions June - 2016 13. Questions and Solutions December- 2016 14. Questions and Solutions June - 2017 15. References

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Review of Trend

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Introduction CSIR NET JRF-LS is one of the most important exam for those who seek research and college teaching as their career. After qualifying NET or national eligibility test, one is eligible for applying to the post of assistant professor or above in colleges and by qualifying JRF (junior research fellowship), one get the research fellowship for the tenure of five years. The pattern of this exam is set to enable students from various streams of science to attempt with equal ease, therefore as per the prescribed syllabus which as 13 units, a diverse variety of topics and subjects has been included. Nevertheless, in exam ample choice is provided to facilitate a fair competition. In section A, 15 out of 20 questions, in section B, 35 out of 50 questions and in section C, 25 out of 75 questions have to be attempted. In total only 75 out of 145 questions are required to be attempted. Question paper has a very defined and set pattern of questions and weightage of each unit is well maintained in the paper. Part A has most of the aptitude questions involving basic Physics, Chemistry, Biology and Mathematics. Section B and C has questions from the defined syllabus. It has been observed in the past years the questions are often arranged in the order of units prescribed in the syllabus. As Biochemistry is the first unit of the syllabus, the question from biochemistry section are found in the beginning of section B and Section C. Questions in section B are often memory based and require less efforts while questions of C part may involve technical details and experimental observations and therefore need a better comprehension. Biochemistry per se, is equally important as other units and has defined set of questions that encompass, about 3-4 questions (generally, question no 21- 25) each year in part B and 4-5 question each year in part C (Generally Q 71- 76). However, some questions from the 13th Unit that involves techniques are often related to basic concepts of

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biochemistry and thereby increasing the questions from this subjects and causing variable trend across years.

Analysis of Biochemistry section If we, look at the overall trend of questions over a period of 6 years, we will find that number of biochemistry questions fluctuated between 8 -17, with lowest in June 2011 (8 questions) and highest in June 2014 (17 questions). However, the basic biochemistry questions Q 21- 24 in part B and Q 71- 74 was persistent across all question papers. Year wise distribution of Questions from part A and B in biochemistry is illustrated below.

Fig 1. Graphical presentation of Biochemistry questions in CSIR exams 2011- 2016. a. total vs Biochem questions, b. Biochemitry questions only

As mentioned, the number of biochemistry questions were not variable much and reached a highest of 17 in Jun 2014. The relative numbers between part B and part C also varied and did not show an uniform trend. However, this information just hints that just like all other subjects (not analysed here) Biochemistry was also persistent in all the papers throughout A deeper analysis into the sub topics of biochemistry provides a better insight to develop strategy for preparation of this exam. Prescribed syllabus of CSIR-NET contains biochemistry in unit one and has more focus on chemical principles such as atomic structure, Biomolecular interactions, biomolecules and enzymes. In this analysis, Biochemistry was divided into following main topics as per the abundance of the questions. Biomolecular interactions and atomic structure, Bioenergetics, pH and Buffers Amino acid and Proteins, Carbohydrates, Lipids, Nucleic Acids and Enzymology. Understanding the pattern of questions asked from these topics may help in the determining the focus areas. Following graph depicts the topic wise distribution of questions across the period of 7 years. 9

Most questions were asked from the amino acid and proteins part or carbohydrates, Year wise trend also suggest that there has been successive increase in the number of questions asked on enzymology each year and proteins part. However other topics did not show a regular trend.

Fig 2. Topic wise distribution of questiosn in Biochemistry 2011- 2016

Section wise trend In the first section of atomic structure and biomolecular interactions, most of the questions that were asked were either from hydrogen bonding or vander waals interactions, formula pertaining to the strength of van der waals interactions was also asked. A few questions on basic structure of atom like number of neutron protons and electrons and isotopes of hydrogen and questions related to mole concept and calculations of number of particles in given mass were also common. The second section bioenergetics remained a preferred choice by the examiners for several years with highest 4 questions in 2013. Most of the questions on this sub-topic were numerical problems associated with the Gibbs free energy relationship, with the equilibrium constant and also with standard reduction potential.

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The question on pH and Buffers were common even in those days when CSIR paper used to be descriptive (before 2011). Two important concepts that pH of buffers is calculated using Henderson haselbatch equation and pH of dilute acids may be calculated using pKa values were prominent. However, a few questions on pH were associated with the ionization properties of amino acids. Fourth section on amino acids and proteins was particularly very important for CSIR exam and questions were based on the ionization properties of amino acids. Concepts of pKa, pKb, pKr and pI were really important to answer those questions. Other than that, questions based on the rules of proteolytic cleavage were often seen in several papers. Some questions were however on basic structural details and properties of the side chain of amino acids. Only one question was found on the metabolism of amino acids (a biochemical pathway of tyrosine synthesis from shikimic acid)

Fifth section, carbohydrates was mainly focused at the glycolytic and citric acid cycles, infact some questions were also asked on the kind of linkage and units of disaccharides. Recently in 2016 a question pertaining to the reason for oxygen utilization after exercise was asked (See solutions for details). Also, there was one question on the structures of the TCA cycle intermediates. Sixth section, lipids might have been underestimated in this analysis due to the fact that several questions on lipids were releated to membrane lipids and they were linked to unit 2 (cell biology), however, questions on cholesterol and membrane lipids to undergo diffusion, melting were also asked. One question was repeated on the melting point of lipids. In that question abbreviated form of phosphatidyl choline based lipids such as DPPC, DOPC etc, was given without expansion when asked for the first time. Perhaps, due to concerns raised by student, next time when this question was asked expansion of all abbreviated forms was also given. The seventh section, nucleic acids was much focused on the association of mole concept of nucleic acid. Interesting questions on determining the number of moles of DNA in a cell were asked. Also, basic properties of 11

Nucleic acid such as type of DNA and their dimensions were also important. Most important part was the biosynthesis of purines. Although complete knowledge was of purine biosynthesis pathway was not needed but the sources of various atoms in the purines were asked. This question was asked 3 times in past 6 years. Enzymology, which is the most important part of any biochemistryinvolving exam, so was true with CSIR-NET. At least one question was asked each year from enzymology without an exception across these 6 years. Questions were not numerical problems but understanding based.

Difficulty level and Preparation guide For a novice, who would not have studied biochemistry at any level these question might have been challenging. But it is unlikely, that a student would not study biochemistry, at any level if completed a postgraduate course in life sciences. There were very few questions that were memory based (such as the type of linkage in amylose, number of base pairs in one turn of B DNA) but most of the questions were analytical and comprehension based, especially those of amino acids and proteins. A number of numerical problems were often seen in bioenergetics which might have been difficult for those who are not well versed with the formulae of Gibbs energy and its concept. Metabolism was not asked to great depth except for few instances of glycolysis, TCA cycle and purine precursors, perhaps, to minimise the questions based on rote learning. A new trend was set in last couple of years that many questions on technique section were also based on the principles of biochemistry. Techniques such as SDS-PAGE, chromatography, circular dichroism, and proteolytic cleavage rules were often asked. Keeping this analysis and trend in mind, preparation of Biochemistry o CSIR –NET can be very focused and targeted with a few topics to be stressed more, and understanding rather than rote learning may be really helpful. Understanding the concepts to the crystal clear level might enable

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the quick ability to solve questions. Books such as Lehninzer Biochemistry by Nelson and Cox, Biochemistry by Stryer can prove really very helpful. Enthusiastic student may also refer Voet and voet for detailed description on biomolecules structure and metabolic aspects. Harpers Biochemistry is a concise book that provides a rapid concise view of metabolic pathway with their regulation and clinical relevance. This analysis would be reviewed after 1 year and updated for a better preparation. Wishing good luck and successful preparation of Biochemistry for CSIR NET. Dr. Aditya Arya

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June 2011

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Total Questions: 145 (75), Biochemistry: 8 [Part B: 3, Part C: 5] Part B 1. The free energy ∆G of a dissolved solute 1. Increases with solute concentration 2. Decreases with solute concentration 3. is independent of solute concentration 4. Depends only on temperature. Solution: (Correct answer – 1). The free energy of a chemical reaction or a chemical process is governed by the relationship derived by the energy Gibbs’ as below: ∆G = ∆Go + RT ln [Dissolved Solute/Undissolved solute] Here in this case the increase in dissolved solute is directly related to the value of Gibbs free energy change, hence increases the concentration of dissolved solute the Gibbs free energy change will also increase. 2. Routinely used glucose biosensor estimates blood glucose level by sens ing the concentration of 1. Glucose. 2. Oxygen. 3. δ-glucolactone. 4. H2O2 Solution: (Correct answer – 2). Actually, all the components are involved in the biosensor, so in present day context and modern equipmentments or research techniues any of the four options may be correct. Glucose Biosensor is one of the first types of Biosensors developed. The basis mechanism of Glucose biosensor lies in oxidation of glucose to the electrochemical couples. On oxidation glucose often converts into d-glucolactone. The chemical reactions that forms the basis of glucose detection and the process of glucose detection is illustrated in figure. Glucose reacts with glucose oxidase to form gluconic acid (intermediate formed is glucolactone). Glucose mediator reacts with surrounding oxygen to form hydrogen peroxide and Glucose oxidase (GOD). The final sensing in based on oxygen. Hence, higher the glucose content, higher will be the oxygen consumption. 14 14

Part C 3. The area of allowed regions in the Ramachandran map will be least for 1. Gly. 2. L-Ala. 3. L-Prol. 4. a-methyl L-valine. Solution: (Correct answer – 4). Ramachandran plot depicts the position of allowed set of phi and psi angles in a peptide, the allowed set of torsion angles are often represented as shaded area in the ramachandran plot and empty area represents the disallowed or forbidden angles in the peptide. Although, we may imagine any angle between 0 to 360 degrees, but as the side chains of adjoining amino acids come close they cause Van der waals repulsion (especially when they come closer than 5 A distance), hence in that condition when repulsion occurs that set of torsion angle will not be allowed. Therefore higher the complexity of the side chain, lower will be the allowed torsion angles. In this case, alpha methyl substituted valine is the most complex side chain (Review the structure of amino acids from basic biochemistry books) and thus it will show least area for

allowed regions on the ramachandran plot.

4. A protein in 100 mM KCl solution was heated and the observed Tm (midpoint of unfolding) was 60°C. When the same protein solution in 500 mM KCI was heated, the observed Tm was 65°C. What is the most probable reason for this increase in Tm? 1. Hydrophobic interaction is increased and electrostatic repulsion is decreased. 2. Hydrophobic interaction is decreased and electrostatic repulsion is increased. 3. Hydrogen-bonding is increased. 4. Van Der Waals interaction is increased 15

Solution: (Correct answer - 4). With increasing salt concentrations the charged side chains are stabilized and thefore they are neutralized which increased the number of Van der waals interactions. 5. An amino acid contains no ionizable group in its side chain (R). It is titrated from pH 0 to 14. Which of the following ionizable state is not observed during the entire titration in the pH range 0 - 14?

Solution: (Correct answer 4). In this question option 4 cannot ever exist, as this represents a state where amino group has lost its proton but carboxyl group still has a proton with it. As per the fundamental rules of organic chemistry, amino groups is basic, while the carboxyl group is acidic, So carboxyl will always lose its proton before amino group does. Hence state 4 is incorrect. Additionally, option 1 will exist when pH of solution is equal to pI of amino acid. Option 2 will exist when amino acid is kept in highly acidic solution (both will be protonated) and option 3 will exist when amino acid is conditions are alkaline (both will lose proton). 6. Precursors of the atoms in the purine skeleton are 1. N1, Asp; C2 and C8, formate; N3 and N9 guanidine C4, C5 and N7, Gly; C6, CO2. 2. N1, Asp; C2 and C8, citrate; N3 and N9, amide nitrogen of Gln, C4, C5 and N7; Gly; C6, C02. 3. N1, Asp; C2 and C8, formate; N3 and N9 amide nitrogen of Gln, C4, C5 and N7, Gly; C6, C02. 4. N1, Glu; C2 and C8, acetate; N3 and N9, amide nitrogen of Asn; C4, C5 and N7, Gly; C6, CO2. Solution: (Correct answer - 3 ). This is one of the most frequently asked concept in CSIR, Here are following guiding illustrations to understand the precursors of the

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4. 20 cholesterol : 10 sphingolipid Solution: (Correct answer - 1). This is a simple mathematical problem, if the estimated size of raft is 35 nm2 = 3500 A2 Let us assume that number of cholesterol molecules is 2y and that of sphingolipid is y (as they are in ratio 2:1), then total area of both can be given by equation 2y x 40 + y x 60 = 3500; On solving for y, we get y = 3500/ 140 = 25 Hence number of sphingolipid molecules = 25 and cholesterol (2y) = 50. 14. Pyruvate dehydrogenase is subject to feedback inhibition by its products in glycolysis. Some of the chemical compounds which might be involved in the process, are listed below: A. NADH B. FAD C. Acetyl-coA D. Acetaldehyde Which one of the following combinations of above chemical compounds is involved in feedback inhibition of pyruvate dehydrogenase? 1. A and B 2. B and C 3. C and D 4. A and C Solution: (Correct answer – 4). Following figure depicts the regulation of pyruvate dehydrogenase: a key regulatory enzyme of the glucose metabolism pathway. NADH and acetyl coA are known to be involved in regulation (from the choice given the question), However, FAD and acetaldehyde do not play a direct role in regulation hence, A and C is correct choice.

15. In response to a drug, changes in protein levels were examined in a cell line. A pulse chase experiment was performed using 35S labelled methionine. In comparison to untreated samples, the following observations were made; few minutes after stimulation, protein X accumulates and this was 67

followed by reduction in protein Y and Z. The correct interpretation of these observations would be: 1. Protein X is a protease which degrades Y and Z. 2. Protein X is a transcriptional repressor that controls expression of Y and Z 3. Expression of protein X and loss of Y and Z are unrelated. 4. Information is not sufficient to distinguish between the three possibilities stated above. Solution: (Correct answer - 4). Solution (Correct answer: 4) A reduction in level of protein Y and Z concurrent with production of X can be due to either protein X is proteases for Y, Z or it is a transcriptional repressor or infact the protein Y and Z may have degraded to due to some other reasons. More experiments should be carried out to confirm any of the statements (1-3), such as testing the purified X for proteolytic activity using gel, or using a model expression system with Y or Z promoters to check the role of X. Hence, in thus question the information is insufficient determine the reason. 16. A protein undergoes post-translation modification. In an experiment to identify the nature of modification, following experimental results were obtained. A. Protein moved more slowly is an SDS- PAGE. B. Isoelectric focusing [IEF] showed that there was no change in the pI C. Mass spectrometric analysis showed that the modification was on serine The modification that the protein undergoes is likely to be 1. Phosphorylation 2. Glycosylation 3. Ubiquitination 4.ADP- ribosylation Solution (Correct answer: 2): Phosphorylation and ADP ribosylation would induce charge and therefore change the isoelectric point of the protein, Ubiquitination and does not take place at serine (instead occurs through Lysine, methionine). However, all the modifications will change the molecular weight of the protein, hence result in the slower movement on SDS-PAGE. Glycosylation is least likely to add charge, but it adds mass to the protein.

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December 2015

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Total Questions: 145 (75), Biochemistry: 9 [Part B: 5, Part C: 4] Part B 1. Enzymes accelerate a reaction by which one of the following strategies? 1. Decreasing energy required to from the transition state. 2. Increasing kinetic energy of the substrate. 3. Increasing the free energy difference between substrate and the product. 3. Increasing the turn over number of enzymes. Solution: (Correct answer - 1). Enzymes are known to accelerate the reaction by decreasing the activation energy, i.e. the amount of energy required to form the transition state. 2. The genome of a bacterium is composed of a single DNA molecule which is 109 bp long. How many moles of genomic DNA is present in the bacterium? (Consider Avogadro No = 6 x 1023) 2. 1/6 X 10-14 1. 1/6 X 10-23 3. 6 X 1014 4. 6 X 1023 Solution: (Correct answer: 1) In order to calculate the number of moles, one should know the number of molecules or molecular weight. Here we know that each chromosome is one molecule of DNA. As 6.022 x 1023 molecules of DNA will make 1 mole DNA, So 1 molecule (chromosome) will make 1/6 x 10 -23 mole. (The information that it has 109 is not needed to solve this question)

3. The ionic strength of a 0.2 M Na2HPO4 solution will be? 1. 0.2 M 2. 0.4 M 3. 0.6 M 4. 0.8 M Solution: (correct answer is - 3).The ionic strength of the solution can be calcu69

lated using, the following formula µ = 1/2 Σcz2 Where, µ is the ionic strength, c is the concentration of particular type of ion and z is the charge on that ion. Note that cz2 has to be calculated for alll the type of ions produced by salt and then all the values have to be added to obtain Σcz2 . In this question - 0.2 M Na2HPO4 will produce 0.2 x 2 = 0.4 M Na+ and 0.2 M HPO42Hence, cz2 for Na = 0.4 x 1 = 0.4 and cz2 for HPO42- = 0.2 x 22 = 0.8 Therefore using the formula ionic strength µ = 0.5 x (0.4 + 0.8) = 0.6 M. 4. A cell line deficient in salvage pathway for nucleotide biosynthesis was fed with medium containing 15N labelled amino acids. Purines were then extracted. Treatment with which one of the following amino acids is likely to produce 15N labelled purines? 1. Aspartic acid 2. Glycine 3. Glutamine 4. Aspartamine Solution: (Correct answer: 4). Refer to the guiding figure below, it indicates that Glutamine contributes two nitrogen atoms of the purine and aspartate contributes one Nitrogen and Glycine also contributes one nitrogen. Hence, by labelling any one of the aspartate, Glutamine or Glycine would produce labelled purine.

[Note: this question was wrong in CSIR- as it had more than one true answers, else they should have asked which of the following cannot label a purine, in that case answer would be aspartamine]

5. Coupling of the reaction centers of oxidative phosphorylation is achieved by which one of the following? 1. Making a complex of all four reaction centers 2. Locating all four complexes in the inner membrane 3. Ubiquinones and cytochrome c 4. Pumping of protons Solution: (Correct answer - 3). Ubiquinones and cytochrome c provide an efficient relay of electrons from one complex to another thereby enabling the coupling 70 70

10. The following are four statements on peptide/protein conformation: A. Glycine has the largest area of conformationally allowed space in the Ramachandran plot of φ and ψ. B. A 20-residue peptide that is acetylated at the N-terminus and amidated at the C-terminus has φ = -60ο for all the residues. It can be concluded that conformation of the peptide is helix-turn-strand. C. The allowed values of φ and ψ for amino acids in a protein are not valid for short peptides. D. A peptide acetyl-A1-A2-A3-A4-CONH2 (A1-A4 are amino acid) adopts a well-defined β-turn. The dihedral angles of A2 and A3 determine the type of β-turn Choose the combination of correct statements. 1. A and B 2. B and C 3. A and D 4. C and D Solution: (Correct answer - 3): Glycine will certainly have the largest area of conformationally allowed regions on ramachandran plot, as there will be no hindrance due to the fact that glycine does not have any side chains. In a beta turn, which is formed by set of four amino acids, the determination of conformation or type of beta turn is mainly on the basis of A2 and A3 amino acids (Hence statement A and D are correct). A helix-turn-strand conformation may not be directly deduced from the ramachandran plot. Also, the allowed values for amino acids in protein are valid for short peptides as well. Infact, Ramachandran gave these values for di-alanine, so statement B and C are incorrect. 11. A researcher was investigating the substrate specificity of two different enzymes, X and Y, on the same substrate. Both the enzymes were subjected to treatment with either heat or an inhibitor which inhibits the enzyme activity. Following are the results obtained where a=inhibitor treatment, b= heat treatment and c=control.

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1. Only protein X is specific for the substrate, S 2. Only protein Y is specific for the substrate, S 3. Both X and Y are specific for the substrate, S 4. Both X and Y are non-specific for the substrate, S Solution (Correct answer 3): As it is apperant from the graph that the inhibitor treatment and denaturation caused the decrease in the activity of both enzymes (X and Y) and therefore both the enzymes are specific for their substrate. 12. Engineering of metabolic pathways in plants can be achieved by introduction and over expression of appropriate candidate gene(s) using transgenic technology. The figure given below represents a biochemical pathway in plants where a precursor molecule ‘A’ is converted into products ‘T’ and ‘X’ through a series of enzymatic reactions. Enzymes 1-5 are involved in this pathway. Scientists attempted to increase the level of ‘X’ by introducing an additional copy of the gene for enzyme ‘5’ under transcriptional control of a strong constitutive promoter. However, the developed transgenic plants did not display a proportionate increase in the level of ‘X’.

The following statements were proposed for explaining the above results: A. Enzyme ‘4’ has greater affinity for D than enzyme ‘3’. B. Feedback inhibition of enzyme ‘5’ by compound X. C. Substrate limitation for enzyme ‘5’. Which of the above statements could represent probable reasons for NOT obtaining a proportionate increase in the amount of X in the transgenic plants? 1. Only C 2. Only A and B 3. Only A 4. A, B and C Solution: (Correct answer - 4). The formation of product by an enzyme is affected by various factors such as utilization of its substrate by another enzyme and therefore it will limit the substrate availability for another enzyme, or by allosteric regulation by modulators, or feedback inhibition by the substrate. In the given question experimental details and observations are not enough to distinguish and comment on one of the possibilities, hence all the three possibilities may be resulting in non-increase of enzyme levels despite of enhanced gene expression. 85

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Total Questions: 145 (75), Biochemistry: 12 [Part B: 5, Part C: 7] Part B 1. The energy- rich fuel molecules produced in the TCA cycle are 1. 2 GTP, 2NADH and 1 FADH2 2. 1 GTP, 2 NADH and 2FADH2 3. 1 GTP, 3 NADH and 1 FADH2 4. 2 GTP and 3 NADH Solution: (Correct answer – 3). Refer to the TCA cycle in standard Biochemistry books, each cycle generates 1 GTP, 3 NADH and 1 FADH2. 2. Denaturation of highly helical protein having disulfide bridges and two phenylalanine can be monitored as a function of temperature by which of the following techniques? 1. Recording the circular dichroism spectra at various temperatures 2. Monitoring the absorbance at 214 nm at various temperatures 3. Estimating the –SH content during heat denaturation 4. Monitoring the ratio of absorbance at 214 nm at various temperatures Solution: (Correct answer – 1). Circular dichroism is the best technique to evaluate the real time changes in the secondary conformation of proteins under the influence of various environmental factors. (Option 1- verified from the CSIR key) 3. Glycerol is added to protein solutions to stabilize the preparation by 1. increasing the viscosity of the solution 2. Stabilizing the pH 3. Preferential hydration of proteins 4. Interacting and neutralizing the surface charge on the proteins

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Solution: (Correct answer: 3). This question can have more than one answer correct based on the context, if we consider the context of protein sample preparation for gel electrophoresis (in lamellae buffer), the purpose of adding glycerol is to increase the viscosity and specific gravity that allows the settling of protein sample in wells. But if we consider the context of storage the correct answer will be preferential hydration of proteins, as glycerol is a good cryoprotectant and prevents the crystal formation. (There are number of scientific evidences to support this view are provided in the end). 4. Protein stability is represented as Folded — Unfolded Prior to development of sensitive calorimeters, thermodynamic parameters of the process were determined by following equation

Where ∆Ho and ∆So are standard enthalpy change and standard entropy chnage respectively. Which one of the following statement is correct for estimating ∆G, ∆H and ∆S. 1. Determining the ratio of folded and unfolded protein at 37oC. 2. Plotting Keq as a function of ∆H 3. Plotting Keq against ∆S 4. Plotting Keq against temperature Solution: (Correct answer – 4). Considering this as an equation of line y = mx + c. If we plot ln Keq on Y axis and keep the standard Gibbs free energy change and standard entropy change shall be a constant value, also R is a constant. Hence, if we plot inverse of temperature (1/T) on X axis, slope would give us the ratio of standard Gibbs free energy change and R from where the standard enthalpy change may be calculated. Similarly from the intercept on the Y axis the value of standard entropy change /R can be derived. However, the precise graph will be between ln Eq and 1/T. 5. Rotenone is an inhibitor of the electron transport chain. The addition of rotenone to cells result in which of the following? 1. Generation of mitochondrial reactive oxygen species and block the ATP generation 2. Block in ATP generation but no generation of reactive oxygen species. 3. Generation of reactive oxygen species but no block in ATP generation. 87

4. Permeabilization of inner membrane to compounds which are usually not able to transverse the membrane. Solution: (Correct answer - 1) Rotenone inhibits the electron transport from complex I, as complex I is the primary source of ATP for electron transport in mitochondria. ATP generation will be largely inhibited. However, in presence of complex 2 electrons a little ATP can still be generated which is negligible. In presence of excess electrons, cellular oxygen is reduced to the radical, creating a reactive oxygen species, which can damage DNA and other components of the mitochondria. Hence option 1 is most appropriate. 6. Metachromatic leukodystrophy (MLD) is casued by a deficiency of arylsulphatase A and affects the CNS. MLD is 1. A lysosomal storage disorder 2. A disease due to dysfunctional mitochondria 3. Caused by loss of myelin sheath 4. Caused by a defect in proteins of the nuclear envelope Solution: (Correct answer – 1). Metachromatic leukodystrophy is an inherited disorder characterized by the accumulation of fats called sulfatides in cells. This accumulation especially affects cells in the nervous system that produce myelin, the substance that insulates and protects nerves. It is a lysosomal storage disease which is commonly listed in the family of leukodystrophies as well as among the sphingolipidoses as it affects the metabolism of sphingolipids. Sulfatide accumulation in myelin-producing cells causes progressive destruction of white matter (leukodystrophy) throughout the nervous system, including in the brain and spinal cord (the central nervous system) and the nerves connecting the brain and spinal cord to muscles and sensory cells that detect sensations such as touch, pain, heat, and sound (the peripheral nervous system). 7. Which of the following is NOT true 1. Beta oxidation of the long chain fatty acids occur in mitochondria 2. Fatty acid biosynthesis occurs in peroxisomes. 3. Peroxisomes utilizes H2O2 to oxidise a variety of substrates 4. 4.Peroxisomes import their repertoire of proteins using sorting signals. Solution: (Correct answer – 2). It is believed that very long chain (greater than C-22) fatty acids, branched fatty acids does not occur in mitochondria. As it is unclear in the question that what is exactly meant by long chain (statement 1 may or not be correct). However, fatty acid biosynthesis never take place in peroxisomes, it occurs in cytoplasm, Peroxisomes are involved in isoprenoid and cholesterol synthesis in animals.

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References

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Following books and research articles were refered to prepare the solutions and in most cases the key provided by the CSIR-NET was also used. We acknowledge the annonymous persons who prepraed the CSIR- Question paper and the key which we have used in this booklet for academic purpose. Text Books 1. E. E. Conn and P. K. Stumpf, Out lines of Biochemistry, John Wiley & Sons, New York. 2. A. L. Lehninger, Principles of Biochemsitry, CBS Publishers and Distributors. 3. R. K. Murry, D. K. Granner, P.A. Mayes, V. W. Rodwell, Harper’s Biochemistry, Prentice Hall International Inc., Latest Edition. 4. M. Cohn, K. S. Roth, Biochemstry and Disease, William and Wilkins Co., Baltiomore, Latest Edition. 5. . U. Satyanarayana, Biochemistry, Books and Allied (P) Ltd., Calcutta, Latest Edition. 6.. G. F. Zubay, W. W. Parson, D. E. Vance, Principles of Biochemsitry, WBC Publishers, England, Latest Edition. 7. . S. K. Sawhney, Randhir Singh Eds, Introductory Practical Biochemistry, Narosa Publishing House, New Delhi. 8.. D. T. Plummer, An Introduction to Practical Biochemistry, Tata McGraw Hill, New Delhi. 9. Arya, A., Concise Biochemistry: FUndamental principles. Drawing Pin Publishing, New Delhi. India. 10. Berg, J. M., Tymoczko, J. L. and Stryer, L. Biochemistry. Freeman, 7th edition, 2011. 11. Alberts, B., Johnson, A., Lewis, J., Raff, M., Roberts, K. and Walter P. Molecular Biology of the Cell. Garland Science, 6th edition 2014 12. Principles of Biochemistry With a Human Focus by Reginald H. Garrett, Charles M. Grisham 12. Experiments in Biochemistry: A Hands-On Approach by Shawn O. Farrell, Ryan T. Ranallo,

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Notes

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About The Book This booklet has been prepared as a complementary book to Concise Biochemistry, authored by Dr. Aditya Arya. Considering CSIR-NET as one of most comprehensive examination for research and teaching eligibility, preparation is incomplete without practicing previous year questions. As the exam is very diverse and consist of various domains of life sciences, ample choice is provided in the paper that enables students to choose the domain of their interest and expertise. A comprehensive preparation of Biochemistry can enable students to attempt nearly 20% of the questions in part C. Lack of detailed and authentic solutions to the questions leaves the preparation incomplete and most students either remember the choice from key provided by the examiner or several other publishers which does not enhance knowledge. Besides CSIR-NET this booklet may also be used by students preparing for GATE or M.Sc. entrance examinations for improving their analytical skills in biochemistry. This book has been prepared in consultation with experts and suitable justifications are provided in case of ambiguity in answers. This booklet will be updated on yearly basis with new questions and solutions.

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Non Fiction