GATE Solved Question Papers for Life Science [XL] by AglaSem.Com

SOLVED PAPERS

GATE LIFE SCIENCES (XL)

A comprehensive study guide for GATE

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.com

CONTENTS

• • • • • •

Introduction Question Paper Pattern Design of Questions Marking Scheme Syllabus Previous Year Solved Papers o Solved Question Paper 2014 o Answer Key 2014 o Solved Question Paper 2013 o Answer Key 2013 o Solved Question Paper 2012 o Answer Key 2012

Introduction The Graduate Aptitude Test in Engineering (GATE) is the all India level examination conducted by the Indian Institute of Science and seven Indian Institutes of Technology (IITs). A crucial part of GATE preparation is to solve and practice using previous year GATE papers. Solving previous year GATE papers help the candidate in understanding the exam pattern, knowing the level of difficulty of questions, and analyzing preparation. While attempting to solve any previous year GATE paper, it is advisable that it is done in a simulated test environment. This means, that the candidate sets a timer to countdown to test time, makes sure there is no other distraction, and then sits down to take the test as if he / she is in the exam hall. After attempting the paper, check how many questions you could get right in the first attempt. Analyse the strong and weak areas of preparation, and accordingly devise a study schedule or revision pattern. After going through those areas where in the first attempt could not do well, try the next paper. Almost all of the engineering colleges in India take admission in M.Tech courses on the basis of GATE scores. Apart from that, PSUs also recruit students directly on this basis. To score high in this elite examination is tough, but quite achievable.

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Question Paper Pattern In all the papers, there will be a total of 65 questions carrying 100 marks, out of which 10 questions carrying a total of 15 marks are in General Aptitude (GA). In the papers bearing the codes AE, AG, BT, CE, CH, CS, EC, EE, IN, ME, MN, MT, PI, TF and XE, the Engineering Mathematics will carry around 13% of the total marks, the General Aptitude section will carry 15% of the total marks and the remaining percentage of the total marks is devoted to the subject of the paper. In the papers bearing the codes AR, CY, EY, GG, MA, PH and XL, the General Aptitude section will carry 15% of the total marks and the remaining 85% of the total marks is devoted to the subject of the paper. GATE would contain questions of two different types in various papers: (i) Multiple Choice Questions (MCQ) carrying 1 or 2 marks each in all papers and sections. These questions are objective in nature, and each will have a choice of four answers, out of which the candidate has to mark the correct answer(s). (ii) Numerical Answer Questions of 1 or 2 marks each in all papers and sections. For these questions the answer is a real number, to be entered by the candidate using the virtual keypad. No choices will be shown for this type of questions.

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Design of Questions The questions in a paper may be designed to test the following abilities: (i) Recall: These are based on facts, principles, formulae or laws of the discipline of the paper. The candidate is expected to be able to obtain the answer either from his/her memory of the subject or at most from a one-line computation. (ii) Comprehension: These questions will test the candidate's understanding of the basics of his/her field, by requiring him/her to draw simple conclusions from fundamental ideas. (iii) Application: In these questions, the candidate is expected to apply his/her knowledge either through computation or by logical reasoning. (iv) Analysis and Synthesis: In these questions, the candidate is presented with data, diagrams, images etc. that require analysis before a question can be answered. A Synthesis question might require the candidate to compare two or more pieces of information. Questions in this category could, for example, involve candidates in recognising unstated assumptions, or separating useful information from irrelevant information.

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Marking Scheme For 1-mark multiple-choice questions, 1/3 marks will be deducted for a wrong answer. Likewise, for2-marks multiple-choice questions, 2/3 marks will be deducted for a wrong answer. There is no negative marking for numerical answer type questions. General Aptitude (GA) Questions In all papers, GA questions carry a total of 15 marks. The GA section includes 5 questions carrying 1 mark each (sub-total 5 marks) and 5 questions carrying 2 marks each (sub-total 10 marks). Question Papers other than GG, XE and XL These papers would contain 25 questions carrying 1 mark each (sub-total 25 marks) and 30 questions carrying 2 marks each (sub-total 60 marks). The question paper will consist of questions of multiple choice and numerical answer type. For numerical answer questions, choices will not be given. Candidates have to enter the answer (which will be a real number, signed or unsigned, e.g. 25.06, -25.06, 25, -25 etc.) using a virtual keypad. An appropriate range will be considered while evaluating the numerical answer type questions so that the candidate is not penalized due to the usual round-off errors. GG (Geology and Geophysics) Paper Apart from the General Aptitude (GA) section, the GG question paper consists of two parts: Part A and Part B. Part A is common for all candidates. Part B contains two sections: Section 1 (Geology) and Section 2 (Geo-physics). Candidates will have to attempt questions in Part A and either Section 1 or Section 2 in Part B. Part A consists of 25 multiple-choice questions carrying 1-mark each (sub-total 25 marks and some of these may be numerical answer type questions). Each section in Part B (Section 1 and Section 2) consists of 30 multiple choice questions carrying 2 marks each (sub-total 60 marks and some of these may be numerical answer type questions). XE Paper (Engineering Sciences)

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In XE paper, Engineering Mathematics section (Section A) is compulsory. This section contains 11 questions carrying a total of 15 marks: 7 questions carrying 1 mark each (subtotal 7 marks), and 4 questions carrying 2 marks each (sub-total 8 marks). Some questions may be of numerical answer type questions. Each of the other sections of the XE paper (Sections B through G) contains 22 questions carrying a total of 35 marks: 9 questions carrying 1 mark each (sub-total 9 marks) and 13 questions carrying 2 marks each (sub-total 26 marks). Some questions may be of numerical answer type. XL Paper (Life Sciences) In XL paper, Chemistry section (Section H) is compulsory. This section contains 15 questions carrying a total of 25 marks: 5 questions carrying 1 mark each (sub-total 5 marks) and 10 questions carrying 2-marks each (sub-total 20 marks). Some questions may be of numerical answer type. Each of the other sections of the XL paper (Sections I through M) contains 20 questions carrying a total of 30 marks: 10 questions carrying 1 mark each (sub-total 10 marks) and 10 questions carrying 2 marks each (sub-total 20 marks). Some questions may be of numerical answer type. Note on Negative Marking for Wrong Answers For a wrong answer chosen for the multiple choice questions, there would be negative marking. For1-mark multiple choice questions, 1/3 mark will be deducted for a wrong answer. Likewise, for 2-mark multiple choice questions, 2/3 mark will be deducted for a wrong answer. However, there is no negative marking for a wrong answer in numerical answer type questions.

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Syllabus for General Aptitude (GA) Verbal Ability: English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction. Numerical Ability: Numerical computation, numerical estimation, numerical reasoning and data interpretation.

Syllabus for Life Sciences (XL) Section H: Chemistry (Compulsory) Atomic structure and periodicity: Planck’s quantum theory, wave particle duality, uncertainty principle, quantum mechanical model of hydrogen atom; electronic configuration of atoms; periodic table and periodic properties; ionization energy, election affinity, electronegativity, atomic size. Structure and bonding: Ionic and covalent bonding, M.O. and V.B. approaches for diatomic molecules, VSEPR theory and shape of molecules, hybridisation, resonance, dipole moment, structure parameters such as bond length, bond angle and bond energy, hydrogen bonding, van der Waals interactions. Ionic solids, ionic radii, lattice energy (Born-Haber Cycle). s.p. and d Block Elements: Oxides, halides and hydrides of alkali and alkaline earth metals, B, Al, Si, N, P, and S, general characteristics of 3d elements, coordination complexes: valence bond and crystal field theory, color, geometry and magnetic properties. Chemical Equilibria: Colligative properties of solutions, ionic equilibria in solution, solubility product, common ion effect, hydrolysis of salts, pH, buffer and their applications in chemical analysis, equilibrium constants (Kc, Kp and Kx) for homogeneous reactions, Electrochemistry: Conductance, Kohlrausch law, Half Cell potentials, emf, Nernst equation, galvanic cells, thermodynamic aspects and their applications. Reaction Kinetics: Rate constant, order of reaction, molecularity, activation energy, zero, first and second order kinetics, catalysis and elementary enzyme reactions. Thermodynamics: First law, reversible and irreversible processes, internal energy, enthalpy, Kirchoff’s equation, heat of reaction, Hess law, heat of formation, Second law,

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entropy, free energy, and work function. Gibbs-Helmholtz equation, ClausiusClapeyron equation, free energy change and equilibrium constant, Troutons rule, Third law of thermodynamics. Basis of Organic Reactions Mechanism: Elementary treatment of SN1, SN2, E1 and E2 reactions, Hoffmann and Saytzeff rules, Addition reactions, Markonikoff rule and Kharash effect, Diels-Alder reaction, aromatic electrophilic substitution, orientation effect as exemplified by various functional groups. Identification of functional groups by chemical tests Structure-Reactivity Correlations: Acids and bases, electronic and steric effects, optical and geometrical isomerism, tautomerism, conformers, concept of aromaticity Section I: Biochemistry Organization of life.Importance of water. Cell structure and organelles. Structure and function of biomolecules: Amino acids, Carbohydrates, Lipids, Proteins and Nucleic acids. Biochemical separation techniques and characterization: ion exchange, size exclusion and affinity chromatography, electrophoresis, UV-visible, fluorescence and Mass spectrometry. Protein structure, folding and function: Myoglobin, Hemoglobin, Lysozyme, Ribonuclease A, Carboxypeptidase and Chymotrypsin. Enzyme kinetics including its regulation and inhibition, Vitamins and Coenzymes. Metabolism and bioenergetics. Generation and utilization of ATP. Metabolic pathways and their regulation: glycolysis, TCA cycle, pentose phosphate pathway, oxidative phosphorylation, gluconeogenesis, glycogen and fatty acid metabolism. Metabolism of Nitrogen containing compounds: nitrogen fixation, amino acids and nucleotides. Photosynthesis: the Calvin cycle. Biological membranes. Transport across membranes. Signal transduction; hormones and neurotransmitters. DNA replication, transcription and translation. Biochemical regulation of gene expression. Recombinant DNA technology and applications: PCR, site directed mutagenesis and DNA-microarray.

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Immune system. Active and passive immunity. Complement system. Antibody structure, function and diversity. Cells of the immune system: T, B and macrophages. T and B cell activation. Major histocompatibilty complex. T cell receptor. Immunological techniques: Immunodiffusion, immunoelectrophoresis, RIA and ELISA. Section J: Botany Plant Systematics: Systems of classification (non-phylogenetic vs. phylogenetic – outline), plant groups, molecular systematics. Plant Anatomy: Plant cell structure, organization, organelles, cytoskeleton, cell wall and membranes; anatomy of root, stem and leaves, meristems, vascular system, their ontogeny, structure and functions, secondary growth in plants and stellar organization. Morphogenesis & Development: Cell cycle, cell division, life cycle of an angiosperm, pollination, fertilization, embryogenesis, seed formation, seed storage proteins, seed dormancy and germination. Concept of cellular totipotency, clonal propagation; organogenesis and somatic embryogenesis, artificial seed, somaclonal variation, secondary metabolism in plant cell culture, embryo culture, in vitro fertilization. Physiology and Biochemistry: Plant water relations, transport of minerals and solutes, stress

physiology,

photosynthesis,

stomatal

physiology,

photorespiration;

signal

respiration,

transduction,

Flowering:

N2 metabolism,

photoperiodism

and

vernalization, biochemical mechanisms involved in flowering; molecular mechanism of senencensce and aging, biosynthesis, mechanism of action and physiological effects of plant growth regulators, structure and function of biomolecules, (proteins, carbohydrates, lipids, nucleic acid), enzyme kinetics. Genetics: Principles of Mendelian inheritance, linkage, recombination, genetic mapping; extrachromosomal inheritance; prokaryotic and eukaryotic genome organization, regulation of gene expression, gene mutation and repair, chromosomal aberrations (numerical and structural), transposons. Plant Breeding and Genetic Modification: Principles, methods – selection, hybridization, heterosis; male sterility, genetic maps and molecular markers, sporophytic and gametophytic self incompability, haploidy, triploidy, somatic cell hybridization, marker-assisted selection, gene transfer methods viz. direct and vector-

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mediated, plastid transformation, transgenic plants and their application in agriculture, molecular pharming, plantibodies. Economic Botany: A general account of economically and medicinally important plants- cereals, pulses, plants yielding fibers, timber, sugar, beverages, oils, rubber, pigments, dyes, gums, drugs and narcotics. Economic importance of algae, fungi, lichen and bacteria. Plant Pathology: Nature and classification of plant diseases, diseases of important crops caused by fungi, bacteria and viruses, and their control measures, mechanism(s) of pathogenesis and resistance, molecular detection of pathogens; plant-microbe beneficial interactions. Ecology and Environment: Ecosystems – types, dynamics, degradation, ecological succession; food chains and energy flow; vegetation types of the world, pollution and global warming, speciation and extinction, conservation strategies, cryopreservation, phytoremediation. Section K: Microbiology Historical Perspective: Discovery of microbial world; Landmark discoveries relevant to the field of microbiology; Controversy over spontaneous generation; Role of microorganisms in transformation of organic matter and in the causation of diseases. Methods in Microbiology: Pure culture techniques; Theory and practice of sterilization; Principles of microbial nutrition; Enrichment culture techniques for isolation of microorganisms; Light-, phase contrast- and electron-microscopy. Microbial Taxonomy and Diversity: Bacteria, Archea and their broad classification; Eukaryotic microbes: Yeasts, molds and protozoa; Viruses and their classification; Molecular approaches to microbial taxonomy. Prokaryotic and Eukaryotic Cells: Structure and Function: Prokaryotic Cells: cell walls, cell membranes, mechanisms of solute transport across membranes, Flagella and Pili, Capsules, Cell inclusions like endospores and gas vesicles; Eukaryotic cell organelles: Endoplasmic reticulum, Golgi apparatus, mitochondria and chloroplasts. Microbial Growth: Definition of growth; Growth curve; Mathematical expression of exponential growth phase; Measurement of growth and growth yields; Synchronous growth; Continuous culture; Effect of environmental factors on growth.

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Control of Micro-organisms: Effect of physical and chemical agents; Evaluation of effectiveness of antimicrobial agents. Microbial Metabolism: Energetics: redox reactions and electron carriers; An overview of metabolism; Glycolysis; Pentose-phosphate pathway; Entner-Doudoroff pathway; Glyoxalate pathway; The citric acid cycle; Fermentation; Aerobic and anaerobic respiration; Chemolithotrophy; Photosynthesis; Calvin cycle; Biosynthetic pathway for fatty acids synthesis; Common regulatory mechanisms in synthesis of amino acids; Regulation of major metabolic pathways. Microbial Diseases and Host Pathogen Interaction: Normal microbiota; Classification of infectious diseases; Reservoirs of infection; Nosocomial infection; Emerging infectious diseases; Mechanism of microbial pathogenicity; Nonspecific defense of host; Antigens and antibodies; Humoral and cell mediated immunity; Vaccines; Immune deficiency; Human diseases caused by viruses, bacteria, and pathogenic fungi. Chemotherapy/Antibiotics: General characteristics of antimicrobial drugs; Antibiotics: Classification, mode of action and resistance; Antifungal and antiviral drugs. Microbial Genetics: Types of mutation; UV and chemical mutagens; Selection of mutants; Ames test for mutagenesis; Bacterial genetic system: transformation, conjugation, transduction, recombination, plasmids, transposons; DNA repair; Regulation of gene expression: repression and induction; Operon model; Bacterial genome with special reference to E.coli; Phage λ and its life cycle; RNA phages; RNA viruses; Retroviruses; Basic concept of microbial genomics. Microbial Ecology: Microbial interactions; Carbon, sulphur and nitrogen cycles; Soil microorganisms

associated

with

vascular

plants.

Section L: Zoology Animal world:Animal diversity, distribution, systematics and classification of animals, phylogenetic relationships. Evolution: Origin and history of life on earth, theories of evolution, natural selection, adaptation, speciation. Genetics: Principles of inheritance, molecular basis of heredity, mutations, cytoplasmic inheritance, linkage and mapping of genes. Biochemistry

and

Molecular

Biology: Nucleic

acids,

proteins,

lipids

and

carbohydrates; replication, transcription and translation; regulation of gene expression,

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organization of genome, Kreb’s cycle, glycolysis, enzyme catalysis, hormones and their actions, vitamins. Cell Biology: Structure of cell, cellular organelles and their structure and function, cell cycle, cell division, chromosomes and chromatin structure. Eukaryotic gene organization and expression (Basic principles of signal transduction). Animal Anatomy and Physiology: Comparative physiology, the respiratory system, circulatory system, digestive system, the nervous system, the excretory system, the endocrine system, the reproductive system, the skeletal system, osmoregulation. Parasitology and Immunology: Nature of parasite, host-parasite relation, protozoan and helminthic parasites, the immune response, cellular and humoral immune response, evolution of the immune system. Development

Biology: Embryonic

development,

cellular

differentiation,

organogenesis, metamorphosis, genetic basis of development, stem cells. Ecology: The ecosystem, habitats, the food chain, population dynamics, species diversity, zoogerography, biogeochemical cycles, conservation biology. Animal Behaviour: Types of behaviours, courtship, mating and territoriality, instinct, learning and memory, social behaviour across the animal taxa, communication, pheromones, evolution of animal behaviour. Section M: Food Technology Food Chemistry and Nutrition: Carbohydrates: Structure and functional properties of mono- oligo-polysaccharides including starch, cellulose, pectic substances and dietary fibre; Proteins: Classification and structure of proteins in food; Lipids: Classification and structure of lipids, Rancidity of fats, Polymerization and polymorphism; Pigments: Carotenoids, chlorophylls, anthocyanins, tannins and myoglobin; Food flavours: Terpenes, esters, ketones and quinones; Enzymes: Specificity, Kinetics and inhibition, Coenzymes, Enzymatic and non-enzymatic browning; Nutrition: Balanced diet, Essential amino acids and fatty acids, PER, Water soluble and fat soluble vitamins, Role of minerals in nutrition, Antinutrients, Nutrition deficiency diseases. Food Microbiology: Characteristics of microorganisms: Morphology, structure and detection of bacteria, yeast and mold in food, Spores and vegetative cells; Microbial growth in food: Intrinsic and extrinsic factors, Growth and death kinetics, serial dilution method for quantification; Food spoilage: Contributing factors, Spoilage bacteria,

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Microbial spoilage of milk and milk products, meat and meat products; Foodborne disease: Toxins produced by Staphylococcus, Clostridium and Aspergillus; Bacterial pathogens: Salmonella, Bacillus, Listeria, Escherichia coli, Shigella, Campylobacter; Fermented food: Buttermilk, yoghurt, cheese, sausage, alcoholic beverage, vinegar, sauerkraut and soya sauce. Food Products Technology: Processing principles: Canning, chilling, freezing, dehydration, control of water activity, CA and MA storage, fermentation, hurdle technology, addition of preservatives and food additives, Food packaging, cleaning in place and food laws.; Grain products processing: Milling of rice, wheat, and maize, parboiling of paddy, production of bread, biscuits, extruded products and breakfast cereals, Solvent extraction, refining and hydrogenation of oil; Fruits, vegetables and plantation products processing: Extraction, clarification concentration and packaging of fruit juice, Production of jam, jelly, marmalade, squash, candies, and pickles, pectin from fruit waste, tea, coffee, chocolate and essential oils from spices; Milk and milk products processing: Pasteurized and sterilized milk, cream, butter, ghee, ice-cream, cheese and milk powder; Animal products processing: Drying and canning of fish, post mortem changes, tenderization and freezing of meat, egg powder. Food Engineering: Mass and energy balance; Momentum transfer: Flow rate and pressure drop relationships for Newtonian fluids flowing through pipe, Characteristics of non-Newtonian fluids – generalized viscosity coefficient and Reynolds number, Flow of compressible fluid, Flow measurement, Pumps and compressors; Heat transfer: Heat transfer by conduction, convection, radiation, boiling and condensation, Unsteady state heat transfer in simple geometry, NTU- effectiveness relationship of co-current and counter current double pipe heat exchanger; Mass transfer: Molecular diffusion and Fick’s Law, Steady state mass transfer, Convective mass transfer, Permeability of films and laminates; Mechanical operations: Energy requirement and rate of operations involved in size reduction of solids, high pressure homogenization, filtration, centrifugation, settling, sieving, flow through porous bed, agitation of liquid, solid-solid mixing, and single screw extrusion; Thermal operations: Energy requirement and rate of operations involved in process time evaluation in batch and continuous sterilization, evaporation of liquid foods, hot air drying of solids, spray and freeze-drying, freezing

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and crystallization; Mass transfer operations: Properties of air-water vapor mixture; Humidification and dehumidification operations.

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GATE Previous Year Solved Papers Life Sciences – XL

2012 - 14

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GATE 2014 Solved Paper Life Science - XL Duration: 180 minutes

Maximum Marks: 100

Read the following instructions carefully. 1. To login, enter your Registration Number and password provided to you. Kindly go through the various symbols used in the test and understand their meaning before you start the examination. 2. Once you login and after the start of the examination, you can view all the questions in the question paper, by clicking on the View All Questions button in the screen. 3. This paper consists of 7sections: GA (General Aptitude), H (Chemistry), I (Biochemistry), J (Botany), K (Microbiology), L (Zoology) and M (Food Technology). Section GA (General Aptitude) and Section H (Chemistry) are compulsory. Attempt any 2 sections out of the 5 optional Sections I through M. There are 10 questions carrying 15 marks in the compulsory General Aptitude (GA) section. Questions 1 to 5 of this section carry 1 mark each, and questions 6 to 10 carry 2 marks each. There are 15 questions carrying 25 marks in Section H (Chemistry), which is compulsory. Questions 1 to 5 of this section carry 1 mark each and questions 6 to 15 carry 2 marks each. Each of the optional sections (Sections I through M) contains 20 questions carrying 30 marks. In these sections, questions 1 to 10 are of 1 mark, while questions 11 to 20 are of 2 marks. 4. Depending upon the GATE paper, there may be useful common data that may be required for answering the questions. If the paper has such useful data, the same can be viewed by clicking on the Useful Common Data button that appears at the top, right hand side of the screen. 5. The computer allotted to you at the examination center runs specialized software that permits only one answer to be selected for multiple-choice questions using a mouse and to enter a suitable number for the numerical answer type questions using the virtual keyboard and mouse. 6. Your answers shall be updated and saved on a server periodically and also at the end of the examination. The examination will stop automatically at the end of 180 minutes. 7. In each paper a candidate can answer a total of 65 questions carrying 100 marks. 8. The question paper may consist of questions of multiple choice type (MCQ) and numerical answer type. 9. Multiple choice type questions will have four choices against A, B, C, D, out of which only ONE is the correct answer. The candidate has to choose the correct answer by clicking on the bubble (⃝) placed before the choice. 10. For numerical answer type questions, each question will have a numerical answer and there will not be any choices. For these questions, the answer should be enteredby using the virtual keyboard that appears on the monitor and the mouse. 11. All questions that are not attempted will result in zero marks. However, wrong answers for multiple choice type questions (MCQ) will result in NEGATIVE marks. For all MCQ questions a wrong answer will result in deduction of⅓ marks for a 1-mark question and ⅔ marks for a 2-mark question. 12. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE. 13. Non-programmable type Calculator is allowed. Charts, graph sheets, and mathematical tables are NOT allowed in the Examination Hall. You must use the Scribble pad provided to you at the examination centre for all your rough work. The Scribble Pad has to be returned at the end of the examination. Declaration by the candidate: “I have read and understood all the above instructions. I have also read and understood clearly the instructions given on the admit card and shall follow the same. I also understand that in case I am found to violate any of these instructions, my candidature is liable to be cancelled. I also find at the start of the examination that all the computer hardware allotted to me is in proper working condition”.

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GATE 2014

General Aptitude - GA

Q. 1 – Q. 5 carry one mark each. Q.1

Choose the most appropriate word from the options given below to complete the following sentence. A person suffering from Alzheimer’s disease (A) experienced (C) is experiencing

Q.2

short-term memory loss. (B) has experienced (D) experiences

Choose the most appropriate word from the options given below to complete the following sentence. ____________ is the key to their happiness; they are satisfied with what they have. (A) Contentment

Q.3

(B) Ambition

(C) Perseverance

(D) Hunger

Which of the following options is the closest in meaning to the sentence below? “As a woman, I have no country.” (A) Women have no country. (B) Women are not citizens of any country. (C) Women’s solidarity knows no national boundaries. (D) Women of all countries have equal legal rights.

Q.4

In any given year, the probability of an earthquake greater than Magnitude 6 occurring in the Garhwal Himalayas is 0.04. The average time between successive occurrences of such earthquakes is ____ years.

Q.5

The population of a new city is 5 million and is growing at 20% annually. How many years would it take to double at this growth rate? (A) 3-4 years

(B) 4-5 years

(C) 5-6 years

(D) 6-7 years

Q. 6 – Q. 10 carry two marks each.

Q.6

In a group of four children, Som is younger to Riaz. Shiv is elder to Ansu. Ansu is youngest in the group. Which of the following statements is/are required to find the eldest child in the group? Statements 1. Shiv is younger to Riaz. 2. Shiv is elder to Som. (A) Statement 1by itself determines the eldest child. (B) Statement 2 by itself determines the eldest child. (C) Statements 1 and 2 are both required to determine the eldest child. (D) Statements 1 and 2 are not sufficient to determine the eldest child.

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GATE 2014 Q.7

General Aptitude - GA

Moving into a world of big data will require us to change our thinking about the merits of exactitude. To apply the conventional mindset of measurement to the digital, connected world of the twenty-first century is to miss a crucial point. As mentioned earlier, the obsession with exactness is an artefact of the information-deprived analog era. When data was sparse, every data point was critical, and thus great care was taken to avoid letting any point bias the analysis. From “BIG DATA” Viktor Mayer-Schonberger and Kenneth Cukier The main point of the paragraph is: (A) The twenty-first century is a digital world (B) Big data is obsessed with exactness (C) Exactitude is not critical in dealing with big data (D) Sparse data leads to a bias in the analysis

Q.8

The total exports and revenues from the exports of a country are given in the two pie charts below. The pie chart for exports shows the quantity of each item as a percentage of the total quantity of exports. The pie chart for the revenues shows the percentage of the total revenue generated through export of each item. The total quantity of exports of all the items is 5 lakh tonnes and the total revenues are 250 crore rupees. What is the ratio of the revenue generated through export of Item 1 per kilogram to the revenue generated through export of Item 4 per kilogram?

Exports Item 6 16% Item 5 12%

Item 4 22%

(A) 1:2

Q.9

Item 1 11%

Item 6 19% Item 2 20%

Item 2 20%

Item 5 20%

Item 3 19%

(B) 2:1

Item 1 12%

Item 4 6%

(C) 1:4

Item 3 23%

(D) 4:1

X is 1 km northeast of Y. Y is 1 km southeast of Z. W is 1 km west of Z. P is 1 km south of W. Q is 1 km east of P. What is the distance between X and Q in km? (A) 1

Q.10

Revenues

(B) √2

(C) √3

(D) 2

10% of the population in a town is HIV+. A new diagnostic kit for HIV detection is available; this kit correctly identifies HIV+ individuals 95% of the time, and HIV− individuals 89% of the time. A particular patient is tested using this kit and is found to be positive. The probability that the individual is actually positive is _______

END OF THE QUESTION PAPER

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GATE 2014

Life Sciences - XL

H : CHEMISTRY (COMPULSORY) Q. 1 – Q. 5 carry one mark each. Q.1

Q.2

− + Hybridizations of nitrogen in NO+ 2 , NO3 , NH4 respectively are

(A) sp, sp2 and sp3

(B) sp, sp3 and sp2

(C) sp2, sp and sp3

(D) sp3, sp2 and sp

Potassium metal crystallizes in the body-centered cubic structure. The number of atoms per unit cell is equal to (A) one

Q.3

(B) two

(C) three

(D) four

Assuming ideal condition, the solution that has the highest freezing point is (A) 0.002 M aqueous solution of copper nitrate (B) 0.001 M aqueous solution of potassium dichromate (C) 0.001 M aqueous solution of sodium chloride (D) 0.002 M aqueous solution of magnesium chloride

Q.4

The major product formed in the following reaction is OH Br2 CS2, < 5 oC

(A)

(B)

(C)

OH

(D)

OH

OH Br

Br

Br

Br

Br

Q.5

OH

Br

Br

The acid that undergoes decarboxylation most readily upon heating is

(A)

(B)

(C) O

O Ph

(D)

COOH

Ph

OH

O OH

Ph

CH2COOH

Ph

GATE Previous Year Solved Papers by

CH2COOH

GATE 2014

Life Sciences - XL

Q. 6 – Q. 15 carry two marks each. Q.6

A ball of mass 330 g is moving with a constant speed, and its associated de Broglie wavelength is 1 × 10−33 m. The speed of the ball is _________ m s–1. (h = 6.6 × 10−34 J s)

Q.7

Diphosphonic acid (H4P2O5) has no P ̶ P bond. This acid is (A) tetrabasic

Q.8

(B) tribasic

(C) dibasic

(D) monobasic

The magnetic moment of an octahedral Co(II) complex is approximately 4.0 µ B (atomic number of Co is 27). The CFSE for this complex, in Δo units, is _______ .

Q.9

3+

The complex ion [Cr(H2O)6]

(atomic number of Cr is 24) exhibits

(A) slightly distorted octahedral geometry (B) tetragonally elongated octahedral geometry (C) tetragonally compressed octahedral geometry (D) perfect octahedral geometry

Q.10

Assuming ideal behavior, the density of fluorine gas at 20 °C and 0.3 atm is ____ g L–1. (Molecular weight of F2 = 38 g mol–1, R = 0.082 L atm mol–1 K–1)

Q.11

For a first order reaction, the time required for 50% completion is 20 minutes. The time required for 99.9% completion of the reaction is ______ minutes.

Q.12

At 298 K, the bond dissociation energies of C–H, C–C and C=C are 415, 344 and 615 kJ mol–1, respectively. The enthalpy of atomization of carbon is 717 kJ mol–1 and that of hydrogen is 218 kJ mol–1. The heat of formation of naphthalene at 298 K is ______ kJ mol–1.

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GATE 2014 GATE 2014 Q.13

Sciences– XL-H - XL CHEMISTRYLife (COMPULSORY)

The Fisher projection that represents (2R,3S)-2,3-dihydroxybutanoic acid is (A)

(B)

(C)

CH3

CH3

HO

H

H

HO

H

HO

COOH

Q.14

(D) CH3

CH3 OH

H

OH

HO

H

H

H

OH

H

OH

COOH

COOH

COOH

A hydrocarbon that undergoes ozonolysis (reaction with ozone followed by reduction with Me2S) to form formaldehyde and glyoxal is

(A)

Q.15

(B)

(C)

(D)

The order of acidity of the following acids is OMe

Me

NO2

COOH

COOH

COOH

(2)

COOH

(1)

(3)

(4)

(A) 3 > 2 > 1 > 4

(B)

1>4>3>2

H

(C)

4>3>2>1

(D)

3>4>2>1

END OF THE QUESTION PAPER

GATE Previous Year Solved Papers by

GATE 2014

Life Sciences - XL

I : BIOCHEMISTRY Q. 1 – Q. 10 carry one mark each. Q.1

During an enzyme catalyzed reaction, the equilibrium constant (A) increases (B) decreases (C) remains unchanged (D) can increase or decrease, depending on the enzyme

Q.2

A mixture of Arginine, Phenylalanine and Histidine was fractionated using cation exchange chromatography at neutral pH. The amino acids were eluted with an increasing salt gradient. Identify the correct order of elution. (A) Arginine, Histidine, Phenylalanine (B) Phenylalanine, Histidine, Arginine (C) Histidine, Phenylalanine, Arginine (D) Arginine, Phenylalanine, Histidine

Q.3

Which one of the following proteases does NOT cleave on the carboxyl side of any Arginine residue in a protein? (A) Trypsin (C) Thrombin

Q.4

(B) Proteinase K (D) Chymotrypsin

The receptor for epinephrine is a (A) Tyrosine kinase receptor (C) G-protein-coupled receptor

Q.5

(B) Serine-threonine kinase receptor (D) Ligand activated transcription factor

Choose the option with two reducing sugars. (A) Lactose and Maltose (C) Maltose and Trehalose

Q.6

(B) Trehalose and Sucrose (D) Lactose and Sucrose

The affinity of an antibody can be determined quantitatively by (A) MALDI-TOF MS (C) SDS-PAGE

Q.7

(B) isoelectric focusing (D) equilibrium dialysis

Which one of the following molecules is an allosteric activator of phosphofructokinase-1? (A) Fructose 1,6-bisphosphate (C) Glucose 6-phosphate

(B) Fructose 2,6-bisphosphate (D) Citrate

Q.8

For a single substrate enzyme, a reaction is carried out at a substrate concentration four times the value of Km. The observed initial velocity will be _________ % of Vmax.

Q.9

Consider the following biochemical reaction: Fructose 6-phosphate + ATP

Fructose 1,6-bisphosphate + ADP

The equilibrium constant under biochemical standard conditions (K'eq) for the above reaction is 254. The standard free energy change (∆G'°) for the conversion of fructose 6-phosphate is _______ kJ/mol.

GATE Previous Year Solved Papers by

GATE 2014 Q.10

Life Sciences - XL

Given below is the hydropathy plot of a monomeric transmembrane protein. 180

Hydropathy index (kJ/mol)

120 60

20

40

60

100 120

80

140

0 -60

-120 -180

First amino acid in the window

How many transmembrane α-helices are present in the protein? (A) 1

(B) 2

(C) 4

(D) 5

Q. 11 – Q. 20 carry two marks each. Q.11

An aqueous solution contains two compounds X and Y. This solution gave absorbance values of 1.0 and 0.4 at 220 and 280 nm, respectively, in a 1 cm path length cell. Molar absorption coefficients (ε) of the compounds X and Y are as shown in the table below.

Compound X Compound Y

ε220 (M-1cm-1)

ε280 (M-1cm-1)

1000 800

200 400

The concentration of Y in the solution is___________ mM. Q.12

A purified oligomeric protein was analyzed by SDS-PAGE under reducing and non-reducing conditions. A one litre solution of 1 mg/ml concentration has 4.01 × 1018 molecules of the oligomeric protein. Based on the data shown below, deduce the total number of polypeptide chains that constitute this protein. 1

2

75 kDa

50 kDa

1 : Reducing condition 2 : Non-reducing condition

25 kDa

(A) 2

(B) 4

(C) 6

(D) 12

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GATE 2014 Q.13

Life Sciences - XL

The concentration of Mg2+ ions outside a cell is twice the concentration inside. If the transmembrane potential of the cell is –60 mV (inside negative), the free energy change of transporting Mg2+ ions across the membrane against the concentration gradient at 37 °C is __________ kJ/mol. Faraday constant: 96.5 kJ/V mol

Q.14

Match the entries in Group I with those in Group II Group I P) J chain Q) Serpin R) β2-microglobulin S) Artemis

Group II 1) VDJ recombinase complex 2) Component of MHC class I 3) B cell co-receptor complex 4) C1 complement inhibitor 5) Component of MHC class II 6) Multimerization of IgA and IgM

(A) P-3, Q-4, R-5, S-1 (C) P-6, Q-4, R-2, S-1

Q.15

(B) P-6, Q-5, R-2, S-3 (D) P-3, Q-4, R-1, S-6

The kinetic data for a single substrate enzyme is shown below. The concentration of inhibitor [I] used in the reaction was equal to the Ki of the inhibitor. The Km value of an uninhibited reaction is 2 × 10-5 M. In the presence of the inhibitor, the observed Km value is _____× 10-5 M.

[I] = Ki

1

― (M-1 s) V

Control

1

― (M-1) [S]

Q.16

One litre of phosphate buffer was prepared by adding 208 grams of Na2HPO4 (Mol. wt. 142) and 71 grams of NaH2PO4 (Mol. wt. 120) in water. If the pKa for the dissociation of H2PO4– into HPO42– and H+ is 6.86, the pH of the buffer will be _____________.

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GATE 2014 Q.17

Life Sciences - XL

Shown below is an electrospray ionization mass spectrum of a protein:

Intensity

2417.67 2072.42

2901

3626

1813.50

m/z The numbers written on top of the peaks are the m/z values. The mass of the protein deduced from the given data is ___________ kDa. Q.18

A human gene has only three exons (I, II and III in the given order). Total RNA was isolated from cultured human kidney cells and reverse transcribed. The resultant cDNA was used as a template in a PCR reaction containing a forward primer specific to Exon I and a reverse primer specific to Exon III. When the PCR product was analyzed by gel electrophoresis, two bands were observed of sizes 2.5 kb and 1 kb. However, when Northern blotting was performed with the same total RNA using a radiolabeled probe specific to Exon II, only one band was observed. Based on these observations, which one of the following statements is FALSE ? (A) Northern blotting with a probe specific to Exon III will show two bands. (B) The gene codes for two mRNA splice variants. (C) If the forward primer were specific to Exon II, two bands will be observed. (D) The Exon II is 1.5 kb in size.

Q.19

Using the Sanger’s dideoxy chain termination method, a particular exonic region of a protein coding gene was sequenced for two individuals referred to as Subject 1 and Subject 2. The figure below shows a segment of the autoradiogram corresponding to a small window of the DNA sequence.

Subject 1

Subject 2

A T G C A T G C

Which one of the following interpretations is correct for the sequenced DNA fragments? (A) Subject 2 has two allelic variants. (B) Subject 1 has the sequence 5’- TAGTCGGA -3’. (C) Subject 2 has the sequence 5’- AGGCTAGAT -3’. (D) Subject 1 has a single nucleotide deletion in the gene.

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A 7 kb DNA molecule of a specific sequence has two EcoRI and one PvuII restriction endonuclease sites. The restriction sites are shown below. The DNA was completely digested with both EcoRI and PvuII. The digestion product was purified and added to an appropriately buffered reaction mixture at 37 °C, which contained the Klenow fragment of DNA polymerase I and α-32P dNTPs. After one hour, the DNA in the reaction product was purified and analyzed by electrophoresis. The bands were visualized by both ethidium bromide (EtBr) staining and autoradiography. The result is shown below.

ph

ng

4.0 kb 3.0 kb

DNA Marker

EcoRI :

GAATT C C TTAAG

PvuII :

CA GC TG GTCGAC

2.0 kb 1.5 kb

1.0 kb 0.5 kb 0.3 kb

Which one of the following restriction maps is in agreement with the above result? (A)

(B)

3.0 kb

7.0 kb plasmid

(C)

2.0 kb

PvuII

EcoRI

EcoRI

7.0 kb linear DNA

0.5 kb

1.5 kb

(D)

7.0 kb plasmid

3.0 kb

PvuII 2.0 kb

END OF SECTION I GATE Previous Year Solved Papers by

EcoRI

7.0 kb linear DNA

EcoRI

Q.20

Life Sciences - XL

0.5 kb

1.5 kb

GATE 2014

Life Sciences - XL

J : BOTANY Q. 1 – Q. 10 carry one mark each. Q.1

Plant which grows attached to another plant species but is not a parasitic is known as (A) Endophyte

Q.2

(B) Halophyte

(C) Epiphyte

(D) Lithophyte

An ideal cybrid should have (A) both nuclear genome and cytoplasmic genome equally from both the parents (B) nuclear genome from one of the parents and cytoplasmic genome from other parent (C) nuclear genome predominantly/exclusively from one of the parents and cytoplasmic genome equally from both the parents (D) nuclear genome equally from both the parents and cytoplasmic genome predominantly/ exclusively from one of the parents

Q.3

Transmission Electron Micrograph of fungal cell can usually be distinguished from plant cell due to lack of P and having less abundant Q. Find the correct combination of P and Q. (A) P- Plastid; Q-Vacuoles (B) P- Plastid; Q-Mitochondria (C) P- Plastid; Q-Endoplasmic reticulum (D) P- Mitochondria; Q-Plastid

Q.4

Identify the CORRECT answer RNA interference (RNAi) P. is an event of post transcriptional gene silencing Q. works through RNA induced silencing complex (A) P only

Q.5

(C) Both P and Q

(D) neither P nor Q

(B) Sepal

(C) Petiole

(D) Tepal

Find the odd one out (A) Petal

Q.6

(B) Q only

Plantibody is the (A) Antibody expressed in transgenic plant (B) Transgenic plant that expresses antibody (C) Antibody against plant based antigen (D) Transgenic plant that expresses antigen

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GATE 2014 Q.7

In a typical oil-seed crop, the matured seeds are enriched with (A) Phospholipid

Q.8

Life Sciences - XL

(B) Galactolipid

(A) P-4 Q-2 R-1 S-3

Column II 1. Garcinia sp. 2. Rocella tinctoria 3. Crocus sativus 4. Curcuma sp. (B) P-3 Q-4 R-1 S-2

(C) P-2 Q-3 R-2 S-1

(D) P-3 Q-1 R-2 S-4

The semi-dwarf trait of corn, wheat and rice plants used in breeding program during 1960s resulted in green revolution. Later this ‘green-revolution gene’ has been identified to be involved in either signal transduction pathway or biosynthesis of (A) Auxin

Q.10

(D) Sphingolipid

Match the following products (Column I) with the corresponding plant species (Column II )

Column I P. Saffron Q. Gamboge R. Litmus S. Turmeric

Q.9

(C) Neutral lipid

(B) Gibberellin

(C) Cytokinin

(D) Ethylene

In classical model to explain the plant-pathogen interaction, the host will not develop the disease upon the pathogen attack when (A) The resistance gene (R) is non-functional

(B) The avirulence gene (Avr) is non-functional

(C) Both R and Avr are non-functional

(D) Both R and Avr are functional

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GATE 2014

Life Sciences - XL

Q. 11 – Q. 20 carry two marks each. Q.11

Select the CORRECT combination from the promoter (Column I), transcription machinery used (Column II) and target tissue type (Column III) to express a foreign gene in a plant system. Column I

Column II

P. Ubiquitin Q. Napin R. RbcL S. RbcS

Q.12

1. Chloroplast 2. Nucleus 3. Mitochondria

Column III i. Leaf ii. Seed

(A) P-1-i, Q-3-ii, R-2-i, S-3-ii

(B) P-3-i, Q-1-i, R-2-ii, S-1-ii

(C) P-2-i, Q-2-ii, R-1-i, S-2-i

(D) P-1-ii, Q-3-i, R-2-ii, S-3-ii

In a plant species, flower colour purple is dominant over white. One such purple-flowered plant upon selfing produced 35 viable plants, of which 9 were white-flowered and the rest were purpleflowered. What fraction of these purple-flowered progeny is expected to be pure purple-flowered line? (A) 1/2

(B) 1/3

(C) 1/4

(D) 2/3

GATE Previous Year Solved Papers by

GATE 2014 Q.13

Life Sciences - XL

Following diagram represents the sequence of genes in a normal chromosome of a plant species: GH IJ

KLMN

Match the CORRECT combination for chromosomal mutation using Column - I and Column - II. Column-I P.

GH IKL

Q.

G J

R.

GH IJ

Column-II

JM N

1. Tandem duplication

KLHIM N

2.

Deletion

3.

Pericentric inversion

KLKLM N 4. Non-reciprocal translocation

S.

Q.14

GH J

KLMN

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

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

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

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

Match the nuclei status of mutant plant (Column-I) with the typical chromosome number (ColumnII), when the wild type plant species is having 2N = 46 chromosomes. Column-I P. Q. R. S.

Q.15

Column-II

Trisomic Triploid Monosomic Monoploid

1. 2. 3. 4.

23 45 47 69

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

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

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

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

Match the following reporter genes used in plant transformation experiments with the source of gene and detection/assay system Reporter gene

Source of gene

P. β-glucuronidase Q. Green fluorescence protein R. Luciferase S. Chloramphenicol acetyl transferase

1. Aequorea victoria 2. Photinus pyralis 3. E. coli

Detection/assay i. Radioactive assay ii. Fluorimetric iii. Fluorescence iv. Luminescence

(A) P-3-i, Q-1-ii, R-2-iii, S-3-iv

(B) P-3-ii, Q-1-iii, R-2-iv, S-3-i

(C) P-2-ii, Q-1-iii, R-3-iv, S-1-i

(D) P-1-ii, Q-2-iii, R-3-i, S-3-iv

GATE Previous Year Solved Papers by

GATE 2014 Q.16

Life Sciences - XL

Find the CORRECT statements in the context of Global warming effect on plant photosynthesis. P. Decreasing aqueous solubility of dissolved CO2 compared to dissolved O2 Q. Decreasing oxygenase activity of Rubisco relative to carboxylation R. Enhancing the ratio of CO2 to O2 in air equilibrated solution S. Increasing photorespiration relative to photosynthesis

(A) P & Q

Q.17

(B) R & S

(C) P & R

(D) P & S

Statements given below are either TRUE (T) or FALSE (F). Find the correct combination. P. Regulation of cell cycle progression depends on cyclin dependent kinase (CDK) and protease activity. Q. In photosynthesis, oxidation of water produces O2 and releases electrons required by photosystem I (PSI). R. Photorespiratory reaction occurring in oxidative photosynthetic carbon (C2) pathway involves a cooperative interaction among three organelles: chloroplast, peroxisome and mitochondria. S. Ethylene acts as a promoter of senescence and cytokinins act as a senescence antagonist. (A) P-T, Q-F, R-T, S-F

(B) P-T, Q-T, R-T, S-F

(C) P-T, Q-F, R-F, S-T

(D) P-T, Q-F, R-T, S-T

GATE Previous Year Solved Papers by

GATE 2014 Q.18

Life Sciences - XL

Match the following diagrams P, Q, R, and S with the inflorescence type (Column I) and the corresponding plant species (Column II).

v

P

Q

Column I 1. Umbel 2. Raceme 3. Compound determinate 4. Spike (A) P-2-i Q-3-iv R-4-ii S-1-iii

Q.19

R

S

Column II i. Pedicularis sp ii. Smilacina sp. iii. Epilobium sp. iv. Pelargonium sp.

(B)

(C)

(D)

P-3-ii Q-2-iii R-4-i S- 1-iv

P-1-iii Q-3-ii R-4-iv S-2-i

P-1-iv Q-4-i R-2-iii S-3-ii

Find the right combination for P, Q, R and S with respect to gametophyte development in flowering plants.

Vegetative Cell

P Microsporocyte

4 Microspores

Q

Mitosis

R (A)

P-Meiosis, Q -Generative cell , R- Pollen Tube, S- 2 Sperm Cells

(B)

P- Meiosis, Q- Pollen Tube, R- Generative Cell, S- 2 Sperm Cells

(C)

P-Mitosis, Q- Generative Cell, R- Pollen Tube, S- 2 sperm Cells

(D)

Growth

Mitosis

P- Growth, Q- 2 Sperm Cells, R- Pollen Tube, S- Generative Cell

GATE Previous Year Solved Papers by

S

GATE 2014 Q.20

Life Sciences - XL

Match the definition (Column I) with the type of plant community (Column II) Column I P. The process of occupation of a particular area by different plant communities from their birth to maturity Q. A major ecological unit of vegetation R. A smaller unit of plant association S. A subdivision of plant association which is related to minor differences in temperature and moisture relations

(A) P-1 Q-3 R-4 S-2

(B) P-3 Q-2 R-1 S-4

(C) P-4 Q-1 R-2 S-3

Column II 1. Formation 2. Consociation 3. Faciation 4. Plant succession

(D) P-2 Q-4 R-3 S-1

END OF THE QUESTION PAPER

GATE Previous Year Solved Papers by

GATE 2014

Life Sciences - XL

K: MICROBIOLOGY Q. 1 – Q. 10 carry one mark each. Q.1

Most viral capsids have (A) 08 faces

Q.2

(C) 16 faces

(D) 20 faces

(C) tRNA

(D) cDNA

Intergenic suppression involves mutation in (A) rRNA

Q.3

(B) 12 faces

(B) mRNA

Which one of the following proteins does NOT bind to a gaseous ligand? (A) Leghemoglobin (C) Nitrogenase

(B) Carbonic anhydrase (D) NADPH oxidase

Q.4

A bacterial culture (5 ×10 8 cells/ml) is maintained in a chemostat of working volume 10 L. If the doubling time of the bacteria is 50 min, the required rate of flow of nutrients (in ml/min) is _______________.

Q.5

Rheumatic fever is an example of (A) autoimmune disease (C) immunodeficiency disease

Q.6

(B) type IV hypersensitive reaction (D) neurodegenerative disorder

Oxygenases that catalyse the initial step in the degradation of polycyclic aromatic hydrocarbons by using molecular oxygen belong to which enzyme class? (A) Hydrolase (C) Lyase

Q.7

(B) Transferase (D) Oxido-reductase

Which one of the following is NOT involved in horizontal gene transfer? (A) Conjugation (C) Transduction

Q.8

(B) Transformation (D) Mutation

The principle of immunization was first explained by (A) Edward Jenner (C) Louis Pasteur

Q.9

Lysozyme catalyzes the breakdown of (A) NAG-NAM

Q.10

(B) lipopolysaccharide

(C) teichoic acid

(D) lipoprotein A

Which one of the following microscopic techniques can be used to study the contour of proteins? (A) SEM

XL-K

(B) Elie Metchnikoff (D) Robert Koch

(B) TEM

(C) AFM

(D) Confocal microscopy

GATE Previous Year Solved Papers by

1/3

GATE 2014

Life Sciences - XL

Q. 11 – Q. 20 carry two marks each. Q.11

Match compounds in Group I with inhibitory activities in Group II. Group I (P) Vancomycin (Q) Rifampin (R) Puromycin (S) Ciprofloxacin

Group II (i) Folate metabolism (ii) DNA synthesis (iii) Protein synthesis (iv) RNA synthesis (v) Cell wall synthesis

(A) P-v, Q-iv, R-iii, S-ii (C) P-iv, Q-i, R-iii, S-ii Q.12

(B) P-iv, Q-iii, R-i, S-ii (D) P-v, Q-iii, R-ii, S-iv

Match the organisms with the appropriate growth curves. (P) Bacteria (Q) Extracellular virus (R) Intracellular virus

Cell/virus  Cell/virus  number Number

(i)

(ii)

(iii)

Time (Time) (A) P-iii, Q-i, R-ii (C) P-ii, Q-iii, R-i Q.13

The length of a coding region in an mRNA is 897 bases. How many amino acids will be there in the polypeptide synthesized using this mRNA? (A) 297

Q.14

(B) P-ii, Q-i, R-iii (D) P-i, Q-ii, R-iii

(B) 298

(C) 299

(D) 897

Match the media in Group I for screening microbial isolates in Group II.

(P) (Q) (R) (S)

Group I Blood agar media Minimal media Skimmed milk agar media Bile salt media

(A) P-iii, Q-v, R-ii, S-i (C) P-i, Q-iii, R-ii, S-iv

(i) (ii) (iii) (iv) (v)

Group II Coliforms Protease producers Hemolytic microbes Lipase producers Autotrophs (B) P-iii, Q-ii, R-i, S-iv (D) P-ii, Q-i, R-iv, S-v

Q.15

During a bacterial growth experiment, the total viable cell count at 2 h and 6 h was 1×104 cells/ml and 1×109 cells/ml, respectively. The specific growth rate (in h-1) of the culture is _____________.

Q.16

The concentration of sodium chloride in the cytoplasm of a Halobacterium sp. was found to be 250 ng/nl. The molarity (in M) of sodium chloride is __________.

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GATE 2014 Q.17

Life Sciences - XL Match organisms in Group I with shapes in Group II and flagellar arrangements in Group III.

(P) (Q) (R) (S)

Group I Salmonella typhi Saccharomyces cerevisiae Aquaspirillum serpens Vibrio cholerae

Group II (i) helical (ii) rod (iii) curved rod (iv) ovoid

Group III non-motile amphitrichous peritrichous polar

(1) (2) (3) (4)

(A) P-ii-3, Q-iv-1, R-i-2, S-iii-4 (B) P-iii-1, Q-iv-2, R-ii-4, S-i-3 (C) P-i-2, Q-ii-4, R-iii-2, S-iv-3 (D) P-ii-2, Q-iii-1, R-i-3, S-iv-4 Q.18

Lethal dose curves of different microorganisms (1, 2, 3 and 4) are shown below. Which of these microorganisms are the most lethal?

Host death (%)

100 1

2

75 50

3

4

25

10

20 30

40

50 60

70

Number of microorganisms Number of microorganism

(A) 1 & 3 only Q.19

(B) 1 & 2 only

(D) 2 & 3 only

Match items in Group I with sterilization methods in Group II.

(P) (Q) (R) (S)

Group I Ampicillin 1% glucose in phosphate buffer Plastic syringe Luria broth

(A) P-iv, Q-iii, R-v, S-ii (C) P-i, Q-ii, R-v, S-iii Q.20

(C) 3 & 4 only

(i) (ii) (iii) (iv) (v)

Group II 70 % alcohol treatment Autoclaving at 15 psi for 15 min Autoclaving at 10 psi for 20 min Membrane filtration γ-ray irradiation (B) P-iii, Q-iv, R-ii, S-v (D) P-v, Q-ii, R-iii, S-i

Which of the following statements are TRUE regarding recA mutants of E. coli? (P) Exhibit much reduced recombination (Q) Do not survive UV irradiation (R) Show no effect on doubling time (S) Exhibit pleiotropy (A) P, Q & R only (C) P, R & S only

(B) P, Q & S only (D) Q, R & S only

END OF SECTION-K

GATE Previous Year Solved Papers by

GATE 2014

Life Sciences - XL

L : ZOOLOGY

Q. 1 – Q. 10 carry one mark each. Q.1

Small geographic areas with high concentrations of endemic species and a large number of endangered and threatened species are known as (A) endemic sinks (C) biodiversity hot spots

Q.2

Which ONE of the following animals has “Osculum” as an excretory structure? (A) Hydra

Q.3

(B) Sponge

(C) Jelly Fish

(D) Sea pen

During development of which ONE of the following organisms, bilateral meroblastic cleavage is found? (A) Mollusc

Q.4

(B) critical communities (D) endemic metapopulations

(B) Fish

(C) Bird

(D) Amphibian

The mitochondrion is NOT considered a part of the endomembrane system on account of which ONE of the following reasons? (A) It does not undergo structural changes (B) It is not derived from the ER or Golgi (C) It does not synthesize proteins (D) It is not attached to the outer nuclear envelope

Q.5

The end products of glycolysis include ATP, (A) CO2 and H2O (C) NADH and pyruvate

Q.6

(B) H2O and pyruvate (D) CO2 and NADH

The TATA box is found in the vicinity of the transcription start site. The role of this box is to (A) serve as a ribosome recruitment site (B) serve as RNA polymerase binding site (C) provide 3-D structural integrity to a DNA molecule (D) act as a terminator sequence

Q.7

Which ONE of the following processes does NOT occur in prokaryotic gene expression, but occurs in eukaryotic gene expression? (A) Transcription of mRNA, tRNA, and rRNA (B) Binding of RNA polymerase to the promoter (C) Addition of a poly-A tail to the 3' end and the 5' capping of an mRNA (D) Translation begins as soon as transcription is initiated

Q.8

In Graves’ disease, the presence of auto antibodies against which ONE of the following molecules is the direct cause of hyperthyroidism? (A) Thyroperoxidase (C) Thyroid stimulating hormone

Q.9

(B) Thyroxine (D) Thyroid stimulating hormone receptor

In mammals, the two important organs associated with the production and elimination of urea are (A) gastrointestinal tract and lungs (C) kidneys and lungs

(B) gastrointestinal tract and liver (D) liver and kidneys

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Life Sciences - XL

Q.10 Some endocrine glands produce hormones that stimulate functions of other endocrine glands. Which ONE of the following hormones specifically acts to increase secretion of other hormones? (A) Thyroxine

(B) Prolactin

(C) ACTH

(D) ADH

Q. 11 – Q. 20 carry two marks each. Q.11 If the recombination frequency between X - Y loci is 12, X - Z loci is 4, and Y - Z loci is 8, then the order of the loci on the chromosome is (A) X-Y-Z

(B) Y-X-Z

(C) X-Z-Y

(D) Z-Y-X

Q.12 A cross is made between a white eyed-miniature winged female with a red eyed-normal winged male of Drosophila melanogaster. Further crossing of F1 female offspring from this cross with a white eyed-miniature winged male fly gave 95 white eyed-normal winged, 102 red eyed-miniature winged, 226 red eyed-normal winged and 202 white eyed-miniature winged offspring in F2 generation. What is the percent frequency of recombination between the two genes? (A) 20.11

(B) 31.52

(C) 49.10

(D) 34.12

Q.13 A green fluorescent protein (GFP) encoding gene is fused to a gene encoding specific protein for expression in cells. What is the advantage of using GFP over staining cells with fluorescently labeled antibodies that bind to the target protein? (A) It bleaches less compared to fluorescent probes. (B) It allows imaging at higher resolution than fluorescent probes. (C) It provides more precise location of the protein than fluorescent probes. (D) Its fusion allows tracking the location of the protein in living cells, while staining usually requires fixation of cells. Q.14 A newborn was accidentally given a drug that destroyed the thymus. Which ONE of the following would be the most likely outcome? (A) Lack of class I MHC molecules (B) Inability to rearrange antigen receptors (C) Inability to differentiate to mature T cells (D) Reduction in T-independent number of B cells Q.15 One individual has a parasitic worm infection and another is responding to an allergen such as pollen. Which ONE of the following features is common to both of them? (A) Increase in cytotoxic T cell population (B) Risk of developing an autoimmune disease (C) Reduced innate immune response (D) Increased levels of IgE

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GATE 2014

Life Sciences - XL

Q.16 Five dialysis bags (DB1-DB5), impermeable to sucrose, were filled with various concentrations of sucrose. The bags were placed in separate beakers containing 0.6 M sucrose solution. Every 10 minutes, the bags were weighed and the percent change in mass of each bag was plotted as a function of time.

DB1 DB2 DB3 DB4 DB5 10 20 30 40 50 60 Time (minutes)

Which plot in the graph (X-axis representing time in minutes and Y-axis representing mass change in percentage) represent(s) bags that contain a solution that is hypertonic at 50 minutes? (A) DB2

(B) DB4

(C) DB3

(D) DB4 and DB5

Q.17 Which ONE of the following combinations of products will result, when 3 molecules of acetyl CoA is fed into TCA cycle? (A) 1 ATP, 2 CO2, 3 NADH, and 1 FADH2 (B) 3 ATP, 6 CO2, 9 NADH, and 3 FADH2 (C) 3 ATP, 3 CO2, 3 NADH, and 3 FADH2 (D) 38 ATP, 6 CO2, 3 NADH, and 12 FADH2 Q.18 A DNA fragment shown below has restriction sites I and II, which create fragments X, Y, and Z. Which ONE of the following agarose gel electrophoresis patterns represents the separation of these fragments?

X (A)

(B)

X Y Z

Z

Y (C)

(D)

Z

X

X

Z

Y

Y

Z Y X

Q.19 Theoretically, it is possible to resurrect the extinct woolly mammoth by which ONE of the following methods? (A) Transferring cell nuclei from the frozen tissue into enucleated unfertilized eggs of a suitable mammal (B) Introducing sequenced mammoth genome into donor eggs of a suitable mammal (C) Transferring mammoth nuclear material into stem cells (D) Collection of oocytes from ovaries of the frozen mammoth for in vitro fertilization and transfer of fertilized eggs into animals such as elephants

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GATE 2014

Life Sciences - XL

Q.20 Regions of higher abundance of cholesterol molecules on the plasma membrane will (A) be more fluid (B) result in clogged arteries as it can detach from the plasma membrane (C) be more rigid than the surrounding membrane (D) have higher rates of lateral movement of proteins into and out of plasma membrane

END OF THE QUESTION PAPER

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GATE 2014

Life Sciences - XL

M : FOOD TECHNOLOGY

Q. 1 – Q. 10 carry one mark each. Q.1

Which one of the following is NOT a source of caffeine? (A) Coffee

Q.2

(B) Cocoa beans

(C) Corn syrup

(D) Tea leaves

Yoghurt is prepared using a pair of microorganisms. Choose the correct pair from the following: (A) Lactobacillus bulgaricus, Streptococcus thermophilus (B) Lactobacillus lactis, Streptococcus thermophilus (C) Lactobacillus bulgaricus, Streptococcus lactis (D) Lactobacillus lactis, Streptococcus lactis

Q.3

Choose the target organism for milk pasteurization from the following: (A) Mycobacterium tuberculosis (C) Clostridium botulinum

Q.4

(B) Coxiella burnetii (D) Bacillus cereus

Hypobaric storage is also known as _______ (A) Modified atmospheric storage (B) Controlled atmospheric storage (C) Low pressure storage (D) Modified aseptic package

Q.5

In a solution of vegetable oil (molecular mass = 292 kg kmol-1) and ethanol (molecular mass = 46 kg kmol-1), the concentration of vegetable oil in the solution is measured to be 60% (total mass basis). Therefore, mole fraction of ethanol in the solution is _______.

Q.6

An experiment started with 4 numbers of bacterial cells. After nth generation, number of cells becomes 128. Therefore, value of ‘n’ is _______.

Q.7

One ton of refrigeration means one of the following options: (A) Cooling provided by one kg of ice in one hour (B) Cooling provided by one ton of ice in one hour (C) Energy extracted to freeze one ton of water in one day (D) Coefficient of performance is unity

Q.8

Fruit juice is flowing in a circular pipe (inner diameter 2 cm) at a mass flow rate of 2 kg s-1 and at a temperature of 25°C. The density and viscosity of the juice at 25°C are 1045 kg m-3 and 0.5 Pa s, respectively. Take π = 22/7. The Reynolds number for this flow will be _______.

Q.9

Shear stress (τ, Pa) and shear rate (γ, s-1) relationship of a pseudoplastic fluid follows the Power law equation given by, τ = kγn = 2.6 γ0.45, where ‘n’ and ‘k’ are flow behavior index and consistency index respectively. The apparent viscosity (µ a) of the fluid at a shear rate of 5 s-1 is _______ Pa s.

Q.10

In a sterilization process, D121.1 value of the target organism is 0.22 minute. Time required for 99.999% inactivation of the target organism at 121.1°C will be _____ minutes.

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Life Sciences - XL

Q. 11 – Q. 20 carry two marks each. Q.11

A centrifuge having diameter of 10 cm is rotating at 10000 rpm. Take π = 22/7 and g = 9.81 m s-2. The ratio of centrifugal force to gravitational force will be _________.

Q.12

Match the items under Group I with items under Group II Group I

Group II

P. Threonine Q. Pyridoxine phosphate R. Xylose S. Oleic acid

1. Fatty acid 2. Sugar 3. Amino acid 4. Co-enzyme

(A) P-4, Q-3, R-1, S-2 (C) P-1, Q-2, R-3, S-4 Q.13

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

Match the items under Group I with items under Group II Group I

Group II

P. Iron Q. Calcium R. Zinc S. Iodine

1. Osteoporosis 2. Anemia 3. Goiter 4. Dwarfism

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

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

Q.14

In a counter-current double pipe heat-exchanger, milk is cooled from 110 to 40°C using chilled water as coolant. Water enters at 5°C and leaves at 60°C. Heat flux for the system with overall heat transfer coefficient of 950 W m-2 K-1 will be ______ W m-2.

Q.15

Saturated steam at 100°C is injected at 0.2 kg s-1 into air stream flowing at 3 kg s-1 and 25°C. Air contains 0.012 kg moisture per kg dry air. If the atmospheric pressure is 101.1 kPa, absolute humidity of air will be _____kg kg-1.

Q.16

In an evaporator, milk is concentrated from 9.8% TSS to 52% TSS. Assume the solutes in the milk are non-volatile. The amount of vapour produced for 100 kg feed will be _______ kg.

Q.17

Water enters a cylindrical tank at a steady uniform rate of 0.1 m3 s-1; simultaneously water is discharged from the tank through an orifice (area 0.05 m²) located at the bottom of the tank. Initial level of water in the tank from the bottom is 5 m. If the acceleration due to gravity = 9.81 m s-2 and coefficient of discharge = 0.30, the final value of the steady-state height of water level from the bottom of tank is _______ m.

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GATE 2014 Q.18

Life Sciences - XL

Match the following between Group I and Group II in relation to pretreatments. Group I

Group II

P. Ascorbic acid dip Q. Heat blanching R. Deaeration S. Rendering

1. Sogginess in fruits 2. Minimizes fruit oxidation 3. Melting of fat in meat 4. Removal of odours 5. Minimizes destruction of vitamin C

(A) P-1, Q-2, R-3, S-4 (C) P-1, Q-3, R-4, S-5

(B) P-2, Q-1, R-5, S-3 (D) P-3, Q-4, R-5, S-2

Q.19

A chocolate mix at 100°C is flowing through a 2 cm diameter and 4 m long stainless steel tube at 13.2 kg per minute. The density of the mix is 1750 kg m-3 and its viscosity at 100°C is 2 Pa s. Take π = 22/7. The pressure drop for this flow will be ________ Pa.

Q.20

In a tray dryer, 100 kg of a vegetable material in a suitably reduced form is dried to yield a final product of 75 kg. The dried sample of 5 g, when kept in an oven at 105°C for 24 hours results in 3.56 g of dry matter. The moisture content of the vegetable, before drying, in dry basis is_______%.

END OF THE QUESTION PAPER

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GATE 2014 - Answer Keys General Aptitude– GA Q. No. 1 2 3 4

Key / Range D A C 25 to 25

Q. No. 5 6 7 8

Key / Range A A C D

Q. No. 9 10

Key / Range C 0.48 to 0.49

Key / Range 1.9 to 2.1 C ‐0.8 to ‐0.8 D 0.40 to 0.55

Q. No. 11 12 13 14 15

Key / Range 190 to 240 440 to 470 B A D

Key / Range 80 to 80 13.6 to 13.8 A 0.82 to 0.84 B 13.3 to 13.4 C

Q. No. 15 16 17 18 19 20

Key / Range 4 to 4 7.2 to 7.3 14.4 to 14.6 C A B

Key / Range D B D C B D C

Q. No. 15 16 17 18 19 20

Key / Range B D D B B C

Life Sciences – XL (Section – H) Q. No. 1 2 3 4 5

Key / Range A B C B C

Q. No. 6 7 8 9 10

Life Sciences – XL (Section – I) Q. No. 1 2 3 4 5 6 7

Key / Range C B D C A D B

Q. No. 8 9 10 11 12 13 14

Life Sciences – XL (Section – J) Q. No. 1 2 3 4 5 6 7

Key / Range C C A C C A C

Q. No. 8 9 10 11 12 13 14

__________________________________________________________ GATE Previous Year Solved Papers by

Life Sciences – XL (Section – K) Q. No. 1 2 3 4 5 6 7

Key / Range D C D 200 to 200 A D D

Q. No. 8 9 10 11 12 13 14

Key / Range C A C A C B A

Q. No. 15 16 17 18 19 20

Key / Range 2.8 to 2.9 4.2 to 4.4 A B A B

Key / Range D D C C B D C

Q. No. 15 16 17 18 19 20

Key / Range D A B B A C

Key / Range 0.078 to 0.080 81 to 82 0.55 to 0.58 B 256000 to 256000 87.0 to 87.5

Life Sciences – XL (Section – L) Q. No. 1 2 3 4 5 6 7

Key / Range C B A B C B C

Q. No. 8 9 10 11 12 13 14

Life Sciences – XL (Section – M) Q. No.

Key / Range

Q. No.

Key / Range

Q. No.

1 2 3 4

C A B C

8 9 10 11

254 to 255 1.05 to 1.08 1.0 to 1.2 5590 to 5600

15 16 17 18

5 6

0.80 to 0.82 5 to 5

12 13

19 20

7

C

14

B A 39900 to 40000

___________________________________________________________ GATE Previous Year Solved Papers by

GATE 2013 Solved Paper Life Sciences - XL Duration: Three Hours

Maximum Marks:100

Paper specific instructions:

1. There are a total of 65 questions carrying 100 marks. The question paper consists of questions of multiple choice type and numerical answer type. Multiple choice type questions will have four choices for the answer with only one correct choice. For numerical answer type questions, the answer is a number and no choices will be given. A number as the answer should be entered using the virtual keyboard on the monitor. 2. There are Seven sections: GA (General Aptitude), H (Chemistry), I (Biochemistry), J (Botany), K (Microbiology), L (Zoology) and M (Food Technology). 3. Section GA (General Aptitude) and Section H (Chemistry) are compulsory. Attempt any two optional Sections I through M. 4. There are 10 questions carrying 15 marks in General Aptitude (GA) section, which is compulsory. Questions Q.1 – Q.5 carry 1 mark each, and questions Q.6 – Q.10 carry 2 marks each. 5. There are 15 questions carrying 25 marks in Section H (Chemistry), which is compulsory. Questions Q.1 - Q.5 carry 1 mark each and questions Q.6- Q.15 carry 2 marks each. The 2 marks questions include one pair of common data questions and one pair of linked answer questions. The answer to the second question of the linked answer questions depends on the answer to the first question of the pair. If the first question in the linked pair is wrongly answered or is not attempted, then the answer to the second question in the pair will not be evaluated. 6. Each of the other sections (Sections I through M) contains 20 questions carrying 30 marks. Questions Q.1 - Q.10 carry 1 mark each and questions Q.11 - Q.20 carry 2 marks each. 7. Questions not attempted will result in zero mark. Wrong answers for multiple choice type questions will result in NEGATIVE marks. For all 1 mark questions, ⅓ mark will be deducted for each wrong answer. For all 2 marks questions, ⅔ mark will be deducted for each wrong answer. However, in the case of the linked answer question pair, there will be negative marks only for wrong answer to the first question and no negative marks for wrong answer to the second question. There is no negative marking for questions of numerical answer type. 8. Calculator is allowed. Charts, graph sheets or tables are NOT allowed in the examination hall. 9. Do the rough work in the Scribble Pad provided.

GATE Previous Year Solved Papers by

GATE 2013

Life Sciences - XL

General Aptitude (GA) Questions Q. 1 – Q. 5 carry one mark each. Q.1

If 3 ≤ ≤ 5 and 8 ≤ ≤ 11 then which of the following options is TRUE?

(A) (B) (C)

(D) Q.2

≤ ≤

≤ ≤ ≤ ≤

≤ ≤

The Headmaster ___________ to speak to you. Which of the following options is incorrect to complete the above sentence? (A) is wanting (B) wants (C) want (D) was wanting

Q.3

Mahatama Gandhi was known for his humility as (A) he played an important role in humiliating exit of British from India. (B) he worked for humanitarian causes. (C) he displayed modesty in his interactions. (D) he was a fine human being.

Q.4

All engineering students should learn mechanics, mathematics and how to do computation. I II III IV Which of the above underlined parts of the sentence is not appropriate? (A) I

Q.5

(B) II

(C) III

(D) IV

Select the pair that best expresses a relationship similar to that expressed in the pair: water: pipe:: (A) cart: road (C) sea: beach

(B) electricity: wire (D) music: instrument

Q. 6 to Q. 10 carry two marks each. Q.6

Velocity of an object fired directly in upward direction is given by = 80 − 32 , where is in seconds. When will the velocity be between 32 m/sec and 64 m/sec?

(A) (1, 3/2)

(B) (1/2, 1)

(C) (1/2, 3/2)

(D) (1, 3)

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(time)

GATE 2013 Q.7

Life Sciences - XL

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

Q.8

(B) 0.45

(C) 0.5

(D) 0.4

Following table gives data on tourists from different countries visiting India in the year 2011. Country USA England Germany Italy Japan Australia France

Number of Tourists 2000 3500 1200 1100 2400 2300 1000

Which two countries contributed to the one third of the total number of tourists who visited India in 2011?

(A) USA and Japan (B) USA and Australia (C) England and France (D) Japan and Australia

Q.9

If | −2 + 9| = 3 then the possible value of | − | −

(A) 30

Q.10

(B) -30

(C) -42

would be: (D) 42

All professors are researchers Some scientists are professors Which of the given conclusions is logically valid and is inferred from the above arguments: (A) All scientists are researchers (B) All professors are scientists (C) Some researchers are scientists (D) No conclusion follows

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GATE 2013

Life Sciences - XL

H:CHEMISTRY (Compulsory) Q. 1 – Q. 5 carry one mark each. Q.1

N(CH3)3 and N(SiH3)3are congeners, but around N-atom the former has pyramidal geometry whereas the latter is nearly planar. The bonding responsible for planarity of N(SiH3)3is (A)pπpπ

Q.2

(B) pπdπ

(C) dπdπ

(D) δ

The type of electronic transition responsible for the yellow colour of K2CrO4 is (A)metal to ligand charge transfer (B)ligand to metal charge transfer (C)intra-ligand charge transfer (D)d-d transition

Q.3

The given equation

(∆ )

= ∆

are the enthalpy, temperature and heat capacity at constant pressure, where , and respectively, is called (A) Clausius-Clapeyron equation (C) Kirchhoff’s equation

(B) Hess’s law (D) Trouton’s rule

Q. 4 - Q. 5 are questions with numerical answer. Q.4

The number of 2-center2-electron bonds in anhydrous AlCl3 is ___________

Q.5

When dissolved in water, the number of H+ ions released from a molecule of H3BO3is _________

Q. 6 - Q. 15 carry two marks each. Q.6

In NaCl crystal, the arrangement and coordination number of the ions are (A)fcc and 6

Q.7

(B) fcc and 4

(C) hcp and 6

The solubility product (Ksp) of Ca3(PO4)2 is 1.3 ×10 solubility of Ca3(PO4)2 (in units of M) is (A) 6.5 ×10

(B)1.6 ×10

(D) hcp and 4

. In a 0.02 M solution of Ca(NO3)2, the

(C) 8.0 ×10

(D) 4.0 ×10

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GATE 2013 Q.8

Q.9

Life Sciences - XL

Identify the CORRECT product in the following reaction:

(A)

(B)

(C)

(D)

The major product obtained in the following reaction is

(A)

(B)

(C)

(D)

Q. 10- Q. 11 are questions with numerical answer. Q.10

Iodine forms an anionic species Q in aqueous solution of iodide(I). The number of lone pair(s) of electrons on the central atom of Q is __________

Q.11

The rate of a chemical reaction is tripled when the temperature of the reaction is increased from 298 K to 308 K. The activation energy (in kcal mol1 K1, up to one decimal place) for the reaction is(Given R = 1.987cal mol1 K1) _____________

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GATE 2013

Life Sciences - XL

Common Data Questions Common Data for Questions 12 and 13: Consider the following SN2 reaction of optically pure 1-chloro-3-ethylcyclopentane (X).

Q.12

Q.13

The structure of Y in the above reaction is (A)

(B)

(C)

(D)

The absolute configuration of1-chloro-3-ethylcyclopentane (X) shown above is (A) (1S,3R)

(B) (1S,3S)

(C) (1R,3R)

(D)(1R,3S)

Linked Answer Questions Statement for Linked Answer Questions 14 and 15: The molar conductance at infinite dilution of sodium acetate, sodium sulfate and sulfuric acid solutions are 91.0 ×10 , 259.8×10 and 859.3×10 S m2 mol1, respectively.

Q.14

The molar conductance at infinite dilution (in S m2 mol1)of acetic acid is (A)1028 ×10

Q.15

(B) 820.4 ×10

(C)690.5 ×10

If the molar conductance of an acetic acid solution is15.2 ×10 dissociation of acetic acid in the solution will be (A) 3.89

(B) 2.20

(C) 1.85

(D)390.8×10 Sm2 mol1,then the percentage (%) (D) 1.48

END OF SECTION - H

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GATE 2013

Life Sciences - XL

I:BIOCHEMISTRY Q. 1 – Q. 10 carry one mark each. Q.1

Which one of the following statements is TRUE when a cell is kept in a hypotonic solution? (A) Water moves out of the cell (B) Size of the cell remains same (C) No movement of water takes place (D) Size of the cell increases

Q.2

Which one of the following amino acids has a higher propensity for cis peptide bond formation? (A) Histidine (B) Cysteine (C) Glycine (D) Proline

Q.3

The length of an -helix composed of 36 amino acid residues is (A) 10 Å (B) 54 Å (C) 27 Å (D) 360 Å

Q.4

The order n for a given substrate concentration in an enzyme catalyzed reaction following Michaelis-Menten kinetics, is (A) n = 1

Q.5

(B) n = 0

(C) n is not defined

(D) 0 ≤ n ≤ 1

Which one of the following amino acid residues is specifically recognised by chymotrypsin during peptide bond cleavage? (A) Phe (B) Leu (C) Val (D) Asp

Q.6

The terminal electron acceptor during mitochondrial respiration is (A) O2

(B) FAD+

(C) NAD+

(D) ATP

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GATE 2013

Q.7

Life Sciences - XL

During the biosynthesis of urea in the urea cycle, the two nitrogen atoms are derived from (A) Two free ammonium groups (B) Free ammonium group and aspartate (C) Both nitrogen atoms are derived from arginine (D) One nitrogen atom is derived from citrulline and one from glutamate.

Q.8

An enzyme has two binding sites for an inhibitor molecule. When the inhibitor binds to the first site, the dissociation constant of the inhibitor for the second site increases, leading to negative co-operativity. The Hill coefficient for such an inhibitor is (A) equal to one

Q.9

(B) greater than one

(C) less than one

(D) less than zero

An oligonucleotide was sequenced by the dideoxy method of Sanger and the following autoradiogram was obtained

The sequence of the oligonucleotide is (A) (B) (C) (D) Q.10

3’-GTCCTGTACA-5’ 5’-GTCCTGTACA-3’ 5’-ACATGTCCTG-3’ 3’-AATTTCCCGG-5’

In different types of tissue transplantations, the rate of graft rejection in decreasing order is (A) Isograft > Xenograft > Allograft (B) Allograft > Isograft > Xenograft (C) Xenograft > Autograft > Allograft (D) Xenograft > Allograft > Isograft

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GATE 2013

Life Sciences - XL

Q. 11 - Q. 20 carry two marks each. Q.11

You have prepared 1.0 liter of 0.5 M acetate buffer (pH=5.0). The dissociation constant of acetic acid is 1.7×10-5 M. What would be the acetate ion concentration in the buffer? (A) 0.1M (B) 0.25 M (C) 0.315 M (D) 0.415 M

Q.12

The following figures show the plot of reaction rate versus substrate concentration (mM) for an enzyme catalyzed reaction in the presence and absence of an inhibitor. Match the possible reaction types with the plots. (P) Competitive inhibition (R) Michaelis-Menten

(Q) Substrate inhibition (S) Non-competitive inhibition

(A) P – i, Q – iii, R – ii, S – iv (B) P – iii, Q – ii, R – i, S – iv (C) P – iii, Q – iv, R – i, S – ii (D) P – iv, Q – ii, R – i, S – iii

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GATE 2013 Q.13

Life Sciences - XL

Arrange the following in the decreasing order of their permeability coefficients across a lipid bilayer membrane. (i) Urea (ii) Glucose (iii) H2O (iv) Na+ (v) Tryptophan (A) (i), (iii), (v), (ii), (iv) (C) (iii), (i), (v), (ii), (iv)

Q.14

(B) (iii), (v), (ii), (iv), (i) (D) (i), (iii), (iv), (v), (ii)

Arrange the following in the increasing order of amount of ATP generated by metabolism of one molecule of the following compounds. (i) Anaerobic catabolism of starch with 300 glucose units (ii) Aerobic catabolism of glucose (iii) Aerobic catabolism of acetate (iv) Aerobic catabolism of palmitate (A) (ii), (iv), (iii), (i) (C) (iv), (ii), (i), (iii)

Q.15

(B) (iii), (ii), (i), (iv) (D) (iii), (ii), (iv), (i)

Match the following enzymes with their regulatory mechanism (a) (b) (c) (d)

Phosphofructokinase Glycogen synthase β-galactosidase Lactate dehydrogenase

1. Product inhibition 2. Control of enzyme synthesis 3. Allosteric interaction 4. Covalent modification

(A) (a)-3, (b)-2, (c)-1, (d)-4 (B) (a)-3, (b)-4, (c)-2, (d)-1 (C) (a)-4, (b)-3, (c)-1, (d)-4 (D) (a)-4, (b)-1, (c)-2, (d)-3 Q.16

A researcher wants to clone 3 DNA fragments of sizes 1.1 Mb, 0.097 Mb and 0.045 Mb. The choice of the vectors for cloning each of the fragments are (A) Cosmid, bacteriophage  bacteriophage P1 (B) Yeast artificial chromosome, bacteriophage P1cosmid (C) Bacterial artificial chromosome, bacteriophage , yeast artificial chromosome (D) Only plasmids

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GATE 2013 Q.17

Life Sciences - XL

Which of the four restriction enzymes given below cut the following DNA sequence? 5’-CCGATATCTCGAGGGC-3’ (P) (Q) (R) (S) (A) P & Q

Q.18

BamH1 (3’-CCTAG^G-5’) XhoI (3’-GAGCT^C-5’) EcoRI (3’-CTTAA^G-5’) EcoRV (3’-CTA^TAG-5’) (B) P, R & S

(C) Q & S

(D) P & R

You have expressed the following protein that has an isoelectric point of 6.0. The best order of protein purification methodologies to obtain a pure protein is?

(A) Gel filtration chromatography, Anion exchange chromatography at pH=4.0, Ammonium sulphate precipitation (B) Cation exchange chromatography at pH=9.0, Ni-affinity chromatography, Gel filtration chromatography (C) Anion exchange chromatography at pH=8.0, Ni-affinity chromatography, Gel filtration chromatography (D) Ammonium sulphate precipitation, Anion exchange chromatography at pH=4.0, Niaffinity chromatography

Q.19

An enzyme of 40 kDa is added to a substrate solution in a molar ratio of 1:3. The concentration of the enzyme in the mixture is 12 mg/ml. What would be the corresponding substrate concentration? (A) 0.4 mM

Q.20

(B) 0.12 mM

(C) 0.9 mM

(D) 0.3 mM

A patient suffering from pneumonia and tuberculosis was found to have very low CD4+ T cells. In all probability the PRIMARY causative infectious agent belongs to (A) Klebsiella family (B) Mycobacterium family (C) Retrovirus family (D) Streptococcus family

END OF SECTION - I

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GATE 2013

Life Sciences - XL

J:BOTANY Q. 1 – Q. 10 carry one mark each. Q.1

Bast fibres are present in (A) Xylem

Q.2

(B) Phloem

(C) Collenchyma

(D) Parenchyma

During cellular respiratory process, pyruvate must be oxidized to acetyl CoA and CO2 before it enters the citric acid cycle. The corresponding simplified equation is Pyruvate + NAD+ + CoASH

Acetyl-S-CoA + NADH + CO2

This oxidation reaction occurs in mitochondria and is carried out in presence of the enzyme (A) Pyruvate kinase (C) Pyruvate decarboxylase Q.3

(B) Pyruvate dehydrogenase (D) Pyruvate carboxylase

Which one of the following statements having the terms - gene, chromosome and genome is CORRECT? (A) The rice gene contains about 50,000 genomes located on 12 different chromosomes (B) The rice genome contains about 50,000 genes located on 12 different chromosomes (C) The rice chromosome contains about 50,000 genes located on 12 different genomes (D) The rice genome contains about 50,000 chromosomes located on 12 different genes

Q.4

The aflatoxin found in post-harvested grains is injurious to health due to (A) Aspergillus

Q.5

(B) Alternaria

(B) Spindle formation (D) Chromosome movement to pole

In the symbiotic nitrogen fixation process, Leghemoglobin present in the nodule helps in fixing nitrogen in presence of the enzyme (A) Nitrate synthetase (C) Glutathione synthetase

Q.7

(D) Phytopthora

Identify the event that exclusively occurs in meiotic cell division (A) Chromatid formation (C) Synapsis

Q.6

(C) Fusarium

(B) Nitrate synthase (D) Nitrogenase

Considering environment and ecosystem, identify the INCORRECT statement (A) In detrital food chain, it is very difficult to measure the energy flow although the use of radioactive tracers give some idea about the energy flow (B) The change in species composition, community structure and function over time is ecological succession (C) The slower rise of environmental temperature may be attributed to the increase in pollution in environment which reflects enough solar energy back to other spacer to slow down global warming (D) Photoperiodism has no relation with the environment and ecosystem, rather it is a biological event taking place in every plant

Q.8

The two enzymes commonly used for isolation of protoplasts from plants are (A) Cellulase and Lipase (C) Pectinase and Cellulase

(B) Cellulase and Amylase (D) Pectinase and Lipase

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GATE 2013

Q.9

Life Sciences - XL

For successful transfer of a foreign gene from the engineered Ti-plasmid to the plant genome, few cis-acting DNA elements and trans-acting protein factors are very much essential. Select the CORRECT combination from the following (A) Opine catabolism genes, Left border sequence, Right border sequence (B) Opine catabolism genes, Left border sequence, Virulence genes (C) Hormone biosynthetic genes, Right border sequence, Virulence genes (D) Left border sequence, Right border sequence, Virulence genes

Q.10

In naturally occurring cytoplasmic male sterility, the molecular determinant is located in (A) Chloroplast (C) Golgi complex

(B) Endoplasmic reticulum (D) Mitochondria

Q. 11 - Q. 20 carry two marks each. Q.11

Identify the floral formula with the family and the corresponding plant species

P.

General Floral Formula

Family

Plant

P0 or 2-3, A3 or 3+3, G1 or (3)

1. Liliaceae

i. Cocos nucifera

P3+3, A3+3 or 3 /

2. Cannaceae

ii. Musa paradisiaca

+

Q. R. S.

Q.12

+

P3+3, G(3)

+

P3+3, A3+3, G(3)

3. Graminae (Poaceae)

iii. Maranta bicolor

+

P3+3, A6 or 5, G(3)

4. Palmae (Arecaceae)

iv. Canna indica

5. Musaceae

v. Allium cepa

6. Marantaceae

vi. Oryza sativa

(A)

(B)

(C)

(D)

P-1-iv Q-2-iii R-6-i S-4-ii

P-2-v Q-4-vi R-3-iii S-5-iv

P-3-v Q-6-iii R-5-i S-2-ii

P-3-vi Q-4-i R-1-v S-5-ii

A plant of genotype GGHH is crossed with another plant of the genotype gghh. If the F1 is test crossed, what percentage (%) of the test cross progeny will have the genotype gghh when the two genes are - (P) unlinked, (Q) completely linked with no crossing over, (R) 10 m.u. (genetic map unit) apart, (S) 24 m.u. apart? (A)

(B)

(C)

(D)

P-25 Q-25 R-25 S-25

P-25 Q-50 R-45 S-38

P-50 Q-50 R-90 S-76

P-25 Q-50 R-10 S-24

GATE Previous Year Solved Papers by

GATE 2013 Q.13

Life Sciences - XL

Which two of the following are the CORRECT statements ? P. Q. R. S.

Nondisjunction in the parental meiosis is not essential to produce diploid organisms Nondisjunction in the parental meiosis is essential to produce aneuploid organisms Nondisjunction in the parental meiosis is essential to produce hexaploid organisms Nondisjunction in the parental meiosis is essential to produce tetraploid organisms

(A) P,Q Q.14

(B) Q,R

(C) R,S

Identify the correct matching by taking one item from each column Column -I P. Morphine Q. Nicotine R. Atropine S. Vinblastine

(A) P-1-i Q-2-ii R-3-iii S-4-iv Q.15

(D) P,S

Column -II 1. 2. 3. 4. 5. 6.

Column -III

Cinchona officinalis Hyoscyamus niger Papaver somniferum Nicotiana tabacum Coffea arabica Catharanthus roseus

i. ii. iii. iv. v. vi.

Antineoplastic Respiratory paralysis Antibacterial Narcotic analgesic Anticholinergic Antifungal

(B)

(C)

(D)

P-2-iii Q-3-i R-1-iv S-5-v

P-3-iv Q-4-ii R-2-v S-6-i

P-4-iii Q-1-v R-5-i S-2-vi

Which two of the following are the INCORRECT statements ? P. All plants fix CO2 by the action of ribulose biphosphate carboxylase. The reaction occurs in the bundle sheath of C4 plants Q. All plants fix CO2 by the action of ribulose biphosphate carboxylase. The reaction occurs in the mesophyll cells of C4 plants R. Phosphoenol pyruvate + CO2 Oxaloacetate + Pi, catalyzed by the enzyme phosphoenol pyruvate carbxylase occurs in the mesophyll cells of C4 plants S. Phosphoenol pyruvate + CO2 Oxaloacetate + Pi, catalyzed by the enzyme phosphoenol pyruvate dehydrogenase occurs in the mesophyll cells of C4 plants (A) P,Q

Q.16

(C) Q,S

(D) P,R

Select the CORRECT set comprising only the synthetic analogues of auxin and cytokinin (A) IAA and Kinetin

Q.17

(B) Q,R

(B) 2, 4-D and Zeatin

(C) IAA and Zeatin

(D) 2, 4-D and Kinetin

Which two of the following are the INCORRECT statements ? P. In monocotyledonous stems with closed vascular bundle, secondary growth takes place without any cambium Q. In dicotyledonous stems hypodermis is collenchymatous R. Mobilization of storage reserves takes place during post-germination of embryo S. Osborne’s classification of seed storage proteins is based upon solubility in n-hexane (A) P,R

(B) Q,S

(C) Q,R

(D) P,S

GATE Previous Year Solved Papers by

GATE 2013 Q.18

Life Sciences - XL

Identify two CORRECT characteristic features of (P) Autogamy and (Q) Allogamy from the following statements 1. Plant species usually does not depend on external agents 2. Plant species usually depends on external agents 3. Plant species normally produces progeny that are healthier and better adapted in nature 4. Plant species normally produces weaker progeny in several generations

Q.19

(A)

(B)

(C)

(D)

P-1,3 Q-2,4

P-2,4 Q-1,3

P-1,4 Q-2,3

P-2,3 Q-1,4

Identify the correct matching by taking one item from each column Column-I P. Q. R. S.

Q.20

Coenzyme Holoenzyme Prosthetic group Apoenzyme

Column-II 1. 2. 3. 4. 5. 6.

Holoenzyme + Apoenzyme Non-protein part in an active enzyme Only the protein part in an active enzyme Cofactor in an enzymatic reaction Apoenzyme + Prosthetic group Holoenzyme + inorganic phosphate

(A)

(B)

(C)

(D)

P-4 Q-5 R-2 S-3

P-4 Q-3 R-1 S-2

P-4 Q-3 R-5 S-6

P-4 Q-1 R-3 S-5

Identify two CORRECT statements from the following which are related to the ion transportation in the root system of plant P. The proton pumps and H+-pyrophosphatase appear to work in anti-parallel with the vacuolar ATPase to create a proton gradient across the tonoplast Q. Calcium is one of the important ion whose concentration is strongly regulated by the concentration of apoplastic spaces in the millimolar level R. Solute move through both apoplast and symplast, and xylem parenchyma cells participate in xylem loading S. The limitation of Nerst equation for relating the membrane potential to the distribution of ion at equilibrium is that it cannot distinguish between active and passive transport (A) P,S

(B) Q,R

(C) R,S

(D) Q,S

END OF SECTION - J

GATE Previous Year Solved Papers by

GATE 2013

Life Sciences - XL

K:MICROBIOLOGY Q. 1 – Q. 10 carry one mark each. Q.1

In 1976, Tonegawa’s experiment gave clue about gene rearrangement during differentiation of Bcells. The two different types of cells used in this experiment were (A) HeLa cells and fibrosarcoma cells (B) embryonic cells and fibroblasts (C) adult myeloma cells and HeLa cells (D) embryonic cells and adult myeloma cells

Q.2

To which one of the following groups, the antibiotics kanamycin, streptomycin and gentamicin belong (A) cephalosporins

Q.3

Which one of the following transport mechanism is NOT employed by prokaryotes? (B) Group translocation

(D) Active transport

(B) Bacillus subtilis (C) Escherichia coli

(D) Clostridium tetani

The theoretical maximum number of ATP molecules produced from aerobic oxidation of glucose by eukaryotic cells is (A) 38

Q.7

(C) Endocytosis

The most common indicator organism of faecal pollution in water is (A) Clostridium botulinum

Q.6

(D) quinolones

end of rRNA end of rRNA end of tRNA end of tRNA

(A) Passive diffusion Q.5

(C) aminoglycosides

Shine Dalgarno’s sequence present in mRNA binds to (A) 3 (B) 5 (C) 5 (D) 3

Q.4

(B) macrolides

(B) 24

(C)12

(D) 8

Which one of the following DO NOT use water as an electron source during photosynthesis? (A) Sulfate reducing bacteria (B) Methanogenic bacteria (C) Green and purple bacteria (D) Nitrifying bacteria

Q.8

If the radius of a spherical coccus is 0.8 µm, the value of (A) 7.45

Q.9

(C) 3.75

(D) 0.85

The enzyme that catalyzes the reduction of nitrogen to ammonia is (A) nitrogenase

Q.10

(B) 4.05

surface area in µm-1 will be volume

(B) nitrate reductase

(C) nitrite reductase

(D)deaminase

Chemostat is a continuous culture system in which sterile medium is fed into the culture vessel at the same rate as the spent medium is removed. If in a chemostat culture, the flow rate is 30 ml h-1 and volume of the medium inside the vessel is 100 ml, the dilution rate in h-1 is (A) 3.33

(B) 1.50

(C) 0.75

(D) 0.30

GATE Previous Year Solved Papers by

GATE 2013

Life Sciences - XL

Q. 11 - Q. 20 carry two marks each. Q.11

In an experiment the structural genes lacZYAof lac operon were found to be constitutively expressed. The following explanations were given for the constitutive expression (P) absence of a functional repressor due to mutation in the repressor gene lacI (Q) mutation in the operator that can no longer bind the repressor (R) mutation in the lacAgene Which of the following is CORRECT (A) Only P

Q.12

(B) Only Q

(C) Both P & Q

(D) Only R

Which one of the following is TRUE about siderophores in bacteria? (A) Siderophores are secreted only when soluble iron is available in the medium (B) Siderophores form complex with ferrous ions in the medium (C) Siderophores are the only route of iron uptake (D) Siderophores form complex with ferric ions in the medium

Q.13

In a population containing fast and slow growing bacteria, the slow growing bacteria can be enriched by supplementing the medium with (A) chloramphenicol

Q.14

(B) penicillin

(C) penicillin & chloramphenicol

(D) rifampin

When the supply of tryptophan is plentiful, the tryptophan operon is repressed because the (A) repressor protein-corepressor complex is bound at the operator (B) repressor protein is synthesized in large quantity (C) repressor-corepressor complex is not formed (D) repressor becomes inactive and therefore has reduced specificity for the operator

Q.15

Match the scientists in Group I with their contributions in microbiology in Group II Group I P. Robert Hooke Q. Paul Ehrlich R. Antony van Leeuwenhoek S. Sergei Winogradsky

(A) P-I, Q-II, R-III, S-V (C) P-II, Q-I, R-III, S-IV

Group II I. II. III. IV. V.

Proved that microbes can cause diseases First to observe cells First to observe bacteria Used the first synthetic chemotherapeutic agent Linked specific bacteria to biogeochemical transformations (B) P-II, Q-IV, R-III, S-V (D) P-V, Q-III, R-IV, S-II

GATE Previous Year Solved Papers by

GATE 2013 Q.16

Life Sciences - XL

Match the infectious agents in Group I with the associated diseases in Group II Group I

Group II

P. Bordetellapertusis Q. Mycobacterium leprae R. Haemophilusinfluenzae S. Rubella

I. II. III. IV. V.

Mumps Meningitis Tuberculosis Whooping cough Hansen’s disease

(A) P-IV, Q-V, R-II, S-I (C) P-I, Q-III, R-V, S-IV Q.17

(B) P-V, Q-III, R-I, S-IV (D) P-III, Q-II, R-V, S-I

Match the microscopes in Group I with their working principles in Group II Group I

Group II

P. Phase contrast Q. Dark field R. Bright field S. Electron microscopy

I. II. III. IV. V.

(A ) P-V, Q-II, R-III, S-V (C) P-III, Q-II, R-V, S-I

Q.18

(B) 5.0

(C) 1.0

(D) 0.2

How many electrons are accepted when sulfate acts as the terminal electron acceptor in bacteria such as Desulfovibrio? (A) 8

Q.20

(B)P-III, Q-I, R-IV, S-V (D) P-II, Q-I, R-III, S-V

In a phenol coefficient test for determining the efficacy of a disinfectant X, the maximum effective dilution for X and phenol was found to be 1/450 and 1/90, respectively. Calculate the phenol coefficient for X. (A) 10.0

Q.19

Light reaches the specimen only from the sides Uses fluorescent dyes Difference in the refractive index of cells from their surrounding medium Difference in contrast between specimen and its surrounding medium Uses electrons instead of photons as energy source

(B) 6

(C) 4

(D) 2

A bacterial population increases from 103 cells to 109 cells in 10 h. Calculate the number of generations per hour. (A) 20

(B)10

(C)4

(D)2

END OF SECTION - K

GATE Previous Year Solved Papers by

GATE 2013

Life Sciences - XL

L:Zoology Q. 1 – Q. 10 carry one mark each. Q.1

Which one of the following provides the strongest support for the theory of “descent with modification”? (A) Early embryonic forms of diverse organisms (examples: fishes, birds and mammals) appear similar (B) Ability of fishes and whales to swim (C) Variation in flower colour in a given species (D) Skin colour variation among individuals in a human population

Q.2

Which one of the following is an example of sympatric speciation? (A) Origin of new species among wasps that pollinate figs (B) Emergence of a new species among finches that migrated to an island and thus isolated from their ancestors (C) Evolution of birds’ and bats’ wings (D) Speciation of squirrels separated by a wide river

Q.3

The primary difference between glycogen and cellulose is in the (A)types of constituent monosaccharides (B)number of monomers per molecule (C)configuration of the monomers (D)susceptibility to acid hydrolysis

Q.4

Control mechanisms operate at any of the several steps involved in gene expression. Which one of the following is the key mode of regulation during the cell cycle? (A) Transcription (B) mRNA processing (C) Activation of protein function resulting from protein-protein interaction (D) mRNA export

Q.5

Testicular feminization syndrome is a genetic condition wherein an individual with a XY genotype will have an external female-like phenotype. This is caused by (A) Functional loss of androgen receptor (B) Increased production of estrogen and its receptor (C) Functional loss of Mullerian inhibiting hormone (D) Functional loss of androgen receptor and Mullerian inhibiting hormone

Q.6

Which one of the following defects do you expect to see if you were able to specifically block apoptosis in the developing limb bud of a frog embryo? (A) The digits will remain connected through a web-like extension (B) The bones will not form, and the limb would look like a paddle (C) The limb would look normal but would be larger in size (D) The anterior-posterior polarity of the limb will be lost

Q.7

The formation of antigen-antibody complex helps in disposing antigen through the following pathways EXCEPT: (A) Neutralizing the antigen by blocking its activity (B) By directly hydrolyzing the antigen (C) By promoting the precipitation of antigen (D) By activating cell lysis pathway

GATE Previous Year Solved Papers by

GATE 2013 Q.8

Life Sciences - XL

The term “ecological succession” refers to: (A) A process wherein newer species populate a region that was devoid of flora and fauna (B) A transition phase wherein one particular set of species is replaced by another set of species (C) Changes in the community due to a disturbance in the habitat (D) All the above

Q.9

Which one of the following options provide example for the term “habituation” in behavioral ecology? (A) A fish transferred to a fish tank startles initially for a hand clap, but not later (B) Migratory birds from the temperate zone migrating towards the tropical part during the winters (C) Adult kingfisher birds are more successful in catching fishes than their younger siblings (D) Female lizard getting used to a new male lizard during the courtship period

Q.10

Among the following cell structure-function pairs, identify the correctly paired one (A) Microvilli – engulfment of foreign bodies (C) Peroxisomes – cellular respiration

(B)Cytoskeleton – cell migration (D) Nucleolus – mRNA transcription

Q. 11 - Q. 20 carry two marks each. Q.11

Which of the following most accurately states the goal of systematics? (A) Classification scheme should reflect phylogenetic relationship (B) All animals should be classified based on the relatedness at the early embryonic stage (C) All animals should be grouped based on DNA sequence data (D) Classification of animals should be based on morphological characters

Q.12

Among the following options, choose the one that is probably a cause of rapid diversification of animal groups during the Cambrian explosion. (A) Adaptation of organisms to live in the salty environment of ocean (B) Emergence of coelom (C) The movement of animals to land (D) The accumulation of sufficient atmospheric oxygen to support the metabolism of actively moving animals

Q.13

A newly discovered, recessively-inherited disease-susceptibility trait (DS) is observed only in cotton plants with white flowers, although the flower colour (R) and DS are independently inherited. In a breeding programme, one variety that is homozygous for the absence of DS, but heterozygous for R was mated to another having white flowers but heterozygous for DS. What is the probability that a given plant among the cross progeny will be susceptible to the disease? (A) 25 %

Q.14

(B) 12.5 %

(C) 75 %

(D) 0 %

In a new species of moth, the genes for body colour (black, B, is dominant over grey, b), wing size (normal wing, W, is dominant over vestigial, w) and eye colour (red, R, is dominant over white, r) are linked. In this species, only one cross-over event has been observed between any two homologous chromosomes during meiosis. In a cross between BB; ww and bb; WW, 5 % of the progeny were black with normal wings. In a separate cross between RR; WW and rr; ww, 15 % of the progeny were red-eyed with vestigial wings. In a third cross between BB; rr and bb; RR, 10 % of the progeny were black-coloured with red eyes. Which among the following is the correct order of these three genetic loci? (A) Body colour – Eye colour – Wing size (C) Wing size – Body colour – Eye colour

(B) Eye colour – wing size – Body colour (D) Eye colour – Body colour – Wing size

GATE Previous Year Solved Papers by

GATE 2013 Q.15

In vertebrates, the variations in the structure and function of nephrons are directly linked to the osmoregulatory requirements of the organisms depending on the habitats they live in. From the options given below, identify the correct combination that truly represents the adaptation seen in desert mammals: i. Long loops of Henle ii. Short loop of Henle iii. Hyperosmotic urine iv. Large volume of urine v. Removal of nitrogen as uric acid (A) Options i and iii

Q.16

Life Sciences - XL

(B) Options ii and iii

(C) Options i and iv

(D) Options i and v

In Drosophila, mutations in homeotic genes result in which one of the following developmental defects? (A) The anterior portion of the embryo will not develop (B) Several segments in the anterior region of the embryo will be lost (C) Segmentation will be lost, and the embryo will have only one segment (D) Segment-specific identities will be lost

Q.17

Retroviruses, like the influenza virus, escape the detection of pre-existing antibodies in the host by generating surface antigen variants. They do so: (A) By editing the surface antigen post-translationally (B) Because RNA polymerase of these viruses display high mutation rate during RNA synthesis (C) Because the surface antigens are attacked by the proteases present in the host cell (D) Because DNA polymerase of the host mutates the viral genome in the infected cell

Q.18

The relative number of individuals in each age group is an important demographic factor for the study of future growth trends, and this is normally depicted in the form of an age-structure pyramid. Based on your understanding on population dynamics, match the following distribution properties of age structure (Group 1) with the individual forecast given (Group 2): Group 1 I). Uniform age distribution in the pyramid II). Distribution skewed towards younger age groups III). Distribution skewed towards older age groups IV). Reduction in the number of males in the middle age group Group 2 i). Emigration in the recent past and the possible increase in the older age groups in the near future. ii). Likely to be a stable population iii). Possible unemployment in the near future iv). Increased government expenditure on medical needs and social security related issues in the near future (A) I – iii; II – ii; III– i; IV – iv (C) I – ii; II – i; III – iii; IV – iv

Q.19

(B) I – ii; II – iii; III – iv; IV – i (D) I – i; II – iii; III – ii; IV – iv

Like any other trait, animal behavior also evolves by natural selection. Which one of the following examples is NOT true with regard to the evolution of behavior by natural selection? (A) The behavioral trait is determined only by genes (B) The behavioral trait varies within the population of that species (C) The reproductive success partly dependent upon the behavioral trait (D) The behavioral trait is influenced by the genotype

GATE Previous Year Solved Papers by

GATE 2013 Q.20

Life Sciences - XL

Among the following molecular process-biological effect pairs, identify the mismatched pair. (A) Histone deacetylation – activation of gene expression (B) Protein phosphorylation – signal transduction (C) DNA methylation – sex-specific control of gene expression (D)Proteolytic cleavage – activation of signaling by peptide hormones

END OF SECTION - L

GATE Previous Year Solved Papers by

GATE 2013

Life Sciences - XL

M: FOOD TECHNOLOGY Q. 1 – Q. 10 carry one mark each. Q.1

Kawashiorkor disease is caused due to the deficiency of (A) lysine (C) vitamin K

Q.2

(B) unsaturated fatty acids (D) protein

Which of the following statements is TRUE in case of oxidative rancidity of vegetable oils and fats? (A) It is caused by the reaction of saturated fatty acids and oxygen (B) It involves polymerization of fatty acids (C) It is caused by the reaction of unsaturated fatty acids with oxygen (D) It is caused by oxidative enzymes

Q.3

The food borne disease, Q fever is caused by the organism (A) Clostridium perfringens (C) Bacillus cereus

Q.4

(B) Coxiella burnetti (D) Staphylococcus aureus

The primary bacterial spoilage of poultry meat at low temperature, with characteristic sliminess at outer surface, is caused by (A) Pseudomonas spp. (C) Bacillus spp.

Q.5

(B) Aspergillus spp. (D) Candida spp.

The weight gain (in gram) per gram protein consumed is called (A) Net Protein Ratio (NPR) (C) Protein Efficiency Ratio (PER)

Q.6

(B) Biological Value (BV) (D) Chemical Score (CS)

Which of the following carbohydrates is NOT classified as dietary fibre? (A) Agar (C) Sodium alginate

Q.7

(B) Pectin (D) Tapioca starch

In the extruder barrel, the compression is achieved by back pressure created by the die and by (A) increasing pitch and decreasing diameter of the screw (B) using the tapered barrel with constant pitch (C) increase in the clearance between barrel surface and screw (D) opening of the die

Q.8

The brown colour of bread crust during baking is due to Maillard reaction between (A) aldehyde groups of sugars and amino groups of proteins (B) aldehyde groups of sugars and vitamins (C) aldehyde groups of sugars and salt (D) starch and yeast

Q.9

Blanching influences vegetable tissues in terms of (A) enzymes production (B) alteration of cytoplasmic membrane (C) stabilization of cytoplasmic proteins (D) stabilization of nuclear proteins

GATE Previous Year Solved Papers by

GATE 2013 Q.10

Life Sciences - XL

When garlic is cut or processed, the crushed garlic odour is due to the formation of (A) diacetyl (C) ethyl butyrate

(B) diallyl disulfide (D) benzaldehyde

Q. 11 - Q. 20 carry two marks each. Q.11

Match the toxicants of plant foods in Group I with their main plant source given in Group II. Group I Group II P) Gossypol 1) Khesari Dahl (Lathyrus sativus) Q) Vicine 2) Cotton seeds R) Glucosinolates 3) Fava beans S) BOAA (beta-N- Oxalyl Amino L-Alanine) 4) Rapeseeds (A) P-2, Q-3, R-4, S-1 (C) P-3, Q-1, R-2, S-4

Q.12

Match the products in Group I with the enzymes used for their preparation given in Group II. Group I Group II P) Aspartame 1) Lipase Q) Cocoa butter substitute 2) Glucose isomerase R) High fructose corn syrup 3) Thermolysin S) Lactose free milk 4) Invertase 5) Beta galactosidase (A) P-2, Q-1, R-4, S-3 (C) P-1, Q-3, R-2, S-4

Q.13

(B) P-3, Q-1, R-2, S-5 (D) P-1, Q-2, R-4, S-5

Match the food items in Group I with the type of colloidal dispersion given in Group II. Group I Group II P) Mayonnaise 1) Sol Q) Tomato ketchup 2) Emulsion R) Cake 3) Gel S) Curd 4) Solid foam (A) P-4, Q-1, R-2, S-3 (C) P-2, Q-3, R-4, S-5

Q.14

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

(B) P-3, Q-1, R-2, S-5 (D) P-2, Q-1, R-4, S-3

[a] Assertion: In the presence of sucrose, the temperature and time for gelatinization of starch increases. [r] Reason: Sucrose, due to its hygroscopic nature, competes with starch for water needed for gelatinization. (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

Q.15

Thermal death of viable spores of Bacillus subtilis in a food sample follows a first order kinetics with a specific death rate constant of 0.23 min-1 at 100°C. The time (in minutes) required to kill 99% of spores in the food sample at 100°C will be (A) 10

Q.16

(B) 20

(C) 23

(D) 60

How much skim milk (in kg) containing 0.1% fat should be added to 500 kg of cream containing 50% fat to produce standardized cream containing 36% fat? (A) 140

(B) 165

(C) 195

(D) 210

GATE Previous Year Solved Papers by

GATE 2013 Q.17

Life Sciences - XL

Which of the following statements is NOT CORRECT in relation to muscle proteins ? (A) Actin and myosin interact to form actomyosin which is responsible for muscle contraction (B) Collagen contributes to the toughness of muscles due to its abundant presence (C) Elastin, a constituent of ligaments, is tougher than collagen (D) Actomyosin is not the main state of actin and myosin in post-mortem muscles

Q.18

A cold storage plant is used for storing 50 tonnes of apples in perforated plastic crates. During the storage, apples are cooled down from 28°C to storage temperature of 2°C. (Specific heat of the apple = 0.874 kCal kg-1 °C-1). If the required cooling is attained in 16 hours, the refrigeration plant capacity (in Tons) is (A) 19

Q.19

(C) 29

(D) 32

An actively growing culture of Acetobacter aceti is added to the vigorously aerated fermented fruit juice medium containing 10 g l-1 ethanol to produce vinegar. After some time, the ethanol concentration in the medium is 0.8 g l-1 and acetic acid produced is 8.4 g l-1. What is the conversion efficiency of the process with respect to theoretical yield? (A) 30

Q.20

(B) 24

(B) 50

(C) 70

(D) 90

An enzyme catalyzed reaction (following Michaelis-Menten kinetics) exhibits maximum reaction velocity (Vm) of 75 nmol l-1 min-1. The enzyme at a substrate concentration of 1.0x10-4 M shows the initial reaction velocity of 60 nmol l-1 min-1. The Km value of the enzyme in molar concentration (M) is (A) 2.5 x 10-5 (C) 2.5 x 10-4

(B) 5.0 x 10-5 (D) 5.0 x 10-4

END OF THE QUESTION PAPER

GATE Previous Year Solved Papers by

GATE 2013 - Answer Keys General Aptitude– GA Q. No. 1 2 3 4

Key / Range B C C D

Q. No. 5 6 7 8

Key / Range B C C C

Q. No. 9 10

Key / Range B C

Key / Range A D C C 3

Q. No. 11 12 13 14 15

Key / Range 19.8 to 20.2 A D D A

Key / Range C B D C B C D

Q. No. 15 16 17 18 19 20

Key / Range B B C C C C

Key / Range C D D D B A C

Q. No. 15 16 17 18 19 20

Key / Range C D D C A B

Life Sciences – XL (Section – H) Q. No. 1 2 3 4 5

Key / Range B B C 8 1

Q. No. 6 7 8 9 10

Life Sciences – XL (Section – I) Q. No. 1 2 3 4 5 6 7

Key / Range D D B D A A B

Q. No. 8 9 10 11 12 13 14

Life Sciences – XL (Section – J) Q. No. 1 2 3 4 5 6 7

Key / Range B B B A C D D

Q. No. 8 9 10 11 12 13 14

__________________________________________________________ GATE Previous Year Solved Papers by

Life Sciences – XL (Section – K) Q. No. 1 2 3 4 5 6 7

Key / Range D C A C C A C

Q. No. 8 9 10 11 12 13 14

Key / Range C A D C D B A

Q. No. 15 16 17 18 19 20

Key / Range B Marks to All B B A D

Key / Range D A B A D D Marks to All

Q. No. 15 16 17 18 19 20

Key / Range A D B B A A

Key / Range A B B A B Marks to All A

Q. No. 15 16 17 18 19 20

Key / Range B C D B C A

Life Sciences – XL (Section – L) Q. No. 1 2 3 4 5 6 7

Key / Range A Marks to All C C A A B

Q. No. 8 9 10 11 12 13 14

Life Sciences – XL (Section – M) Q. No. 1 2 3 4 5 6 7

Key / Range D Marks to All B A C D B

Q. No. 8 9 10 11 12 13 14

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GATE 2012 Solved Paper Life Sciences - XL

Duration: Three Hours

Maximum Marks: 100

Read the following instructions carefully.

1. Do not open the seal of the Question Booklet until you are asked to do so by the invigilator. 2. Take out the Optical Response Sheet (ORS) from this Question Booklet without breaking the seal and read the instructions printed on the ORS carefully. 3. On the right half of the ORS, using ONLY a black ink ball point pen, (i) darken the bubble corresponding to your test paper code and the appropriate bubble under each digit of your registration number and (ii) write your registration number, your name and name of the examination centre and put your signature at the specified location. 4. This Question Booklet contains 28 pages including blank pages for rough work. After you are permitted to open the seal, please check all pages and report discrepancies, if any, to the invigilator. 5. There are a total of 65 questions carrying 100 marks. All these questions are of objective type. Each question has only one correct answer. Questions must be answered on the left hand side of the ORS by darkening the appropriate bubble (marked A, B, C, D) using ONLY a black ink ball point pen against the question number. For each question darken the bubble of the correct answer. More than one answer bubbled against a question will be treated as an incorrect response. 6. Since bubbles darkened by the black ink ball point pen cannot be erased, candidates should darken the bubbles in the ORS very carefully. 7. This Question Booklet contains Seven sections: GA (General Aptitude), H (Chemistry), I (Biochemistry), J (Botany), K (Microbiology), L (Zoology) and M (Food Technology). 8. Section GA (General Aptitude) and Section H (Chemistry) are compulsory. Attempt any two optional sections I through M. Using a black ink ball point pen, mark the sections you have chosen by darkening the appropriate bubbles provided on the left hand side of the ORS. Also, write the codes of the optional sections in the boxes provided. In case the candidate does not bubble section codes corresponding to Optional Section-1 or Optional Section-2 or both, the corresponding sections will NOT be evaluated. 9. Questions Q.1 – Q.10 belong to Section GA (General Aptitude) and carry a total of 15 marks. Questions Q.1 – Q.5 carry 1 mark each, and questions Q.6 – Q.10 carry 2 marks each. 10. There are 15 questions carrying 25 marks in Section H (Chemistry), which is compulsory. Questions Q.1–Q.5 carry 1 mark each and questions Q.6–Q.15 carry 2 marks each. The 2 marks questions include one pair of common data questions and one pair of linked answer questions. The answer to the second question of the linked answer questions depends on the answer to the first question of the pair. If the first question in the linked pair is wrongly answered or is unattempted, then the answer to the second question in the pair will not be evaluated. 11. Each of the other sections (Sections I through M) contains 20 questions carrying 30 marks. Questions Q.1–Q.10 carry 1 mark each and questions Q.11–Q.20 carry 2 marks each. 12. Unattempted questions will result in zero mark and wrong answers will result in NEGATIVE marks. For all 1 mark questions, ̃ mark will be deducted for each wrong answer. For all 2 marks questions, ̄ mark will be deducted for each wrong answer. However, in the case of the linked answer question pair, there will be negative marks only for wrong answer to the first question and no negative marks for wrong answer to the second question. 13. Calculator is allowed whereas charts, graph sheets or tables are NOT allowed in the examination hall. 14. Before the start of the examination, write your name and registration number in the space provided below using a black ink ball point pen. Name Registration Number

XL

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GATE 2012

Life Sciences - XL

General Aptitude (GA) Questions (Compulsory) Q. 1 – Q. 5 carry one mark each. Q.1

If (1.001)1259 = 3.52 and (1.001)2062 = 7.85, then (1.001)3321 = (A) 2.23

Q.2

(B) 4.33

(C) 11.37

(D) 27.64

One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT? I requested that he should be given the driving test today instead of tomorrow. (A) (B) (C) (D)

Q.3

requested that should be given the driving test instead of tomorrow

Which one of the following options is the closest in meaning to the word given below? Latitude (A) Eligibility

Q.4

(B) Freedom

(C) Coercion

(D) Meticulousness

Choose the most appropriate word from the options given below to complete the following sentence: Given the seriousness of the situation that he had to face, his ___ was impressive. (A) beggary

Q.5

(B) nomenclature

(C) jealousy

(D) nonchalance

Choose the most appropriate alternative from the options given below to complete the following sentence: If the tired soldier wanted to lie down, he ___ the mattress out on the balcony. (A) should take (B) shall take (C) should have taken (D) will have taken

Q. 6 - Q. 10 carry two marks each. Q.6

One of the legacies of the Roman legions was discipline. In the legions, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them. Which one of the following statements best sums up the meaning of the above passage? (A) Thorough regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances. (B) The legions were treated inhumanly as if the men were animals. (C) Discipline was the armies’ inheritance from their seniors. (D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them.

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Q.7

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

Q.8

Life Sciences - XL

(B) 1/16

(C) 7/16

(D) 9/16

The data given in the following table summarizes the monthly budget of an average household. Category Food Clothing Rent Savings Other expenses

Amount (Rs.) 4000 1200 2000 1500 1800

The approximate percentage of the monthly budget NOT spent on savings is (A) 10% Q.9

(C) 81%

(D) 86%

There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is (A) 2

Q.10

(B) 14%

(B) 3

(C) 4

(D) 8

Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money value of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is (A) 5

(B) 6

(C) 9

(D) 10

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Life Sciences - XL

H : CHEMISTRY (Compulsory) Q. 1 – Q. 5 carry one mark each. Q.1

Q.2

Q.3

Q.4

Among the following, the most reactive diene in the Diels-Alder reaction is (A)

(B)

(C)

(D)

The molecule that does NOT absorb the microwave radiation is (A) CO2

(B) H2O

(C) CO

(D) NO

The hybridization of atomic orbitals of sulphur in SF4 is (A) dsp2

(B) sp3d2

(C) sp3d

(D) sp3 +

3+

(A) Al

3+

(C) Al

Q.5

-

2+

The ionic size of Na , F , Mg 2+

> Mg

2+

> Mg

+

and Al

-

> Na > F -

> F > Na

+

3+

varies as -

+

2+

(B) F > Na > Mg +

-

2+

(D) Na > F > Mg

3+

> Al

3+

> Al

The non-aromatic compound/ion is (A)

(B)

(C)

(D)

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Life Sciences - XL

Q. 6 - Q. 15 carry two marks each. Q.6

+

As predicted by MO theory, the bond order and magnetic nature of NO are (A) three and paramagnetic (C) two and paramagnetic

Q.7

(B) two and diamagnetic (D) three and diamagnetic –14

The value of ionic product of water changes with temperature. It is 1  10 at 60 C. The CORRECT statement with respect to H and S is (A) H is negative and S is negative (C) H is positive and S is negative

Q.8

–13

(B) H is positive and S is zero (D) H is negative and S is positive

10 micrograms of the enzyme carbonic anhydrase (molecular weight = 30,000 g/mole) removes 300 milligrams of carbon dioxide per minute from the cells. The turnover number of the enzyme is (A) 20 min–1 (C) 2  1010 min–1

Q.9

at 25 oC and 1  10

(B) 2  107 min–1 (D) 7.2  1010 min–1

The iodide which reacts most slowly with cyanide ion as a nucleophile in a SN2 reaction is (A)

(B)

CH2I

CH3CH2CH2CH2I

(C)

(D) (CH3)2CH-I

I

Q.10

The value of Ka of acetic acid is 1.7  10–5 mol/dm3. The pH of a buffer solution prepared by mixing 100 ml of 0.1 M acetic acid with a solution of 100 ml of 0.2 M sodium acetate is (A) 4.1 (C) 5.1

Q.11

(B) 4.5 (D) 5.5

The value of standard half cell potential of Cu+2, Cu couple (E0Cu(+2), Cu) is 0.34 V. A wire of pure copper is immersed into a solution of copper nitrate. If the measured cell potential against standard hydrogen electrode at 298 K is 0.24 V, the molar concentration of copper nitrate is {Assume activity of Cu+2 = [Cu+2]}. (A) 4.1  10–4 M (C) 3.4  10–2 M

(B) 2.0  10–2 M (D) 1.8  10–1 M

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Life Sciences - XL

Common Data Questions Common Data for Questions 12 and 13: [FeCl4]2– (I), [CoCl4]2– (II) and [NiCl4]2– (III) are paramagnetic tetrahedral complexes. Q.12

The order of values of crystal field stabilization energy is (A) I > III > II (B) III > I > II (C) I > II > III

Q.13

(D) II > III > I

The order of values of spin only magnetic moment is (A) III > II > I

(B) III > I > II

(C) II > I > III

(D) I > II > III

Linked Answer Questions Statement for Linked Answer Questions 14 and 15: Bromine water is decolourised upon reaction with (E)-3-hexene. Q.14

Q.15

It is due to (A) electrophilic addition of bromine to C=C (C) electrophilic allylic bromination

(B) nucleophilic addition of bromine to C=C (D) nucleophilic allylic bromination

The structure of the product obtained is (A)

(B) Br

Br

Br

Br

(C)

(D) Br

Br

Br

Br

END OF SECTION - H

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Life Sciences - XL

I : BIOCHEMISTRY Q. 1 – Q. 10 carry one mark each. Q.1

Four proteins (P1, P2, P3 and P4) have 17, 10, 21 and 14 percent hydrophobic amino-acids respectively. The order of precipitation of these proteins using ammonium sulphate will be (A) P3, P1, P4, P2

Q.2

(B) Tryptophan

(B) Rotenone

(C) Oligomycin

(B) Tyr-Ser

(D) Histidine

(D) Antimycin A

(C) Asn-Ser

(D) Ala-Gly

Match the hormones in Group I with their metabolic precursor in Group II Group I

Group II

P. Q. R. S.

1. 2. 3. 4.

17- estradiol Thromboxane A2 Epinephrine Retinoic acid

Arachidonic acid Tyrosine -carotene Cholesterol (B) P-1, Q-3, R-2, S-4 (D) P-1, Q-2, R-4, S-3

Upon stimulation of a eukaryotic cell, the intracellular calcium (Ca2+) is released from (A) Endoplasmic reticulum (C) Peroxisome

Q.9

(C) Phenylalanine

The pair of amino-acids which does NOT undergo post-translational modification is

(A) P-4, Q-2, R-3, S-1 (C) P-4, Q-1, R-2, S-3 Q.8

(D) Ala-Arg

Which one of the following compounds does NOT block electron transport?

(A) Asn-His Q.7

(C) Gly-Pro

Which one of the following amino-acids has highest fluorescence quantum yield () in aqueous solution?

(A) Cyanide Q.6

(B) Pro-His

State with high degree of secondary structure and loss of tertiary structure State with complete loss of secondary structure Completely unfolded state Loss of quaternary structure

(A) Tyrosine Q.5

(D) P2, P4, P1, P3

Which one of the following closely defines ‘Molten Globule’ state of a protein? (A) (B) (C) (D)

Q.4

(C) P2, P4, P3, P1

Which one of the following pairs of amino-acids in the protein has high propensity to take up the α-helix conformation? (A) Gly-Asp

Q.3

(B) P3, P1, P2, P4

(B) Nucleus (D) Mitochondria

Leguminous plants maintain a very low concentration of free oxygen in their root nodules because (A) (B) (C) (D)

the nitrogen fixing bacteria living in the root nodules are anaerobic of binding of oxygen to leghemoglobin reductase enzyme of the nitrogenase complex helps in removal of O2 nitrogenase enzyme of the nitrogenase complex helps in removal of O2

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Q.10

Life Sciences - XL

The membrane of mature B cells have (A) both IgG and IgM (C) both IgM and IgE

(B) both IgG and IgD (D) both IgM and IgD

Q. 11 - Q. 20 carry two marks each. Q.11

An amino-acid has one proton donating group in the side chain (R). The pK COOH, pKNH2 and pKR values for this amino-acid are 2.19, 9.67 and 4.25, respectively. Which one of the following statements about this amino-acid is CORRECT? (A) (B) (C) (D)

Q.12

Majority of the molecules will have a net charge of -1 at pH of 7.0 Majority of the molecules will have a net charge of 0 at pH of 4.25 All the molecules will have a deprotonated R group at pH of 3.22 During titration with a strong base, deprotonation will start with the R group

Which one of the following bacterial toxins does NOT have ADP-ribosyl transferase activity? (A) Pertussis toxin (C) Pseudomonas Exotoxin A

Q.13

The CORRECT pair of amino-acid sequence and the corresponding target organelle is (A) KDEL - Golgi (C) SKL - Peroxisome

Q.14

(8 Acetyl CoA) and (8 Acetyl CoA + CO2), respectively (5 Propionyl CoA + 1 CO2) and (5 Propionyl CoA +1 Acetyl CoA), respectively (5 Propionyl CoA + 1 CO2) and (5 Propionyl CoA + 2 CO2), respectively (8 Acetyl CoA) and (7 Acetyl CoA + 1 Propionyl CoA), respectively

Pick the correctly matched pairs. P. Q. R. S.

Immature B cells Activated B cells Pre B cells Mature B cells

(A) P and R (C) Q and S Q.16

(B) K-K/R-X-K/R - Lysosome (D) NPVY - Endoplasmic reticulum

-oxidation of a 16 carbon fatty acid and a 17 carbon fatty acid leads to formation of (A) (B) (C) (D)

Q.15

(B) Diphtheria toxin (D) S. aureus α-toxin

— — — —

Terminal deoxynucleotidyl transferase Class switching Surrogate light chain Recombination activating gene 1 (B) Q and R (D) Q and P

The actual free energy change of a given biochemical reaction carried out under standard conditions with 1 M initial concentration of each of the reactants and products will be (A) (B) (C) (D)

equal to zero equal to standard free energy change for the reaction less than the standard free energy change for the reaction greater than the standard free energy change for the reaction

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Q.17

Life Sciences - XL

Match the enzymes in Group I with their corresponding activity in Group II Group II

Group I P. Flippase

1. Catalyzes the movement of any phospholipid across the lipid bilayer down its concentration gradient

Q. Floppase

2. Catalyzes the translocation of amino-phospholipids from the extracellular to the inner leaflet

R. Lipase

3. Catalyzes the translocation of membrane phospholipids from cytosolic to the extracellular leaflet

S. Scramblase

4. Degradation of phospholipids from the lipid bilayer including the inner and outer leaflets

(A) P-2, Q-3, R-4, S-1 (C) P-4, Q-1, R-2, S-3 Q.18

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

Match the antibiotics in Group I with their mechanism of action in Group II Group I

Group II

P. Tetracyclines

1. Inhibits bacterial protein synthesis by blocking peptidyl transfer

Q. Chloramphenicol

2. Inhibits bacterial protein synthesis by blocking the A-site on the ribosome 3. Misreads the genetic code and inhibits initiation of protein synthesis 4. Inhibits protein synthesis by blocking peptidyl transferase on 80S ribosome

R. Cycloheximide S. Streptomycin (A) P-2, Q-1, R-3, S-4 (C) P-4, Q-3, R-1, S-2

The kinetics of an enzyme in the presence (+I) or absence (-I) of a reversible inhibitor is described in the following graph. (M min)

+I

–1

Q.19

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

-I

1 V0

-4 -3 -2 -1

1

2

3

4

1 (  103 M 1 ) [S] If concentration of the reversible inhibitor in +I experiment was equal to 3.0 × 10–3 M, then the dissociation constant for the enzyme-inhibitor complex is (A) 1 × 10–3 M

(B) 2 × 10–3 M

(C) 3 × 10–3 M

(D) 4 × 10–3 M

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Q.20

Life Sciences - XL

The figure below is a schematic of a linear double stranded DNA containing the indicated restriction sites. The DNA was completely digested with PvuII and EcoRI. The products were purified and added to an appropriately buffered reaction mixture containing dNTP mix, α-32P dATP, and Klenow fragment of E. coli DNA polymerase I. The Klenow reaction products were analyzed by gel electrophoresis and autoradiography. Which of the following products depicts the expected result?

3 kb

2 kb

Pvull CAGCTG

EcoRI

Pvull

GTCGAC

1 kb GAATTC EcoRI CTTAAG

5 kb 4 kb

(A)

3 kb

5 kb 4 kb

(B)

(C)

5 kb 4 kb

5 kb 4 kb

(D)

3 kb

3 kb

3 kb

2 kb

2 kb

2 kb

2 kb

1 kb

1 kb

1 kb

1 kb

END OF SECTION - I

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Life Sciences - XL

J : BOTANY Q. 1 – Q. 10 carry one mark each. Q.1

The swollen base of a petiole is known as (A) Ligule

Q.2

(B) Companion cells

(B) 4

(C) 6

(D) 8

(B) Cynodon dactylon (D) Hordeum vulgare

(B) Clostridium

(C) Rhodospirillum

(D) Rhizobium

Carbon dioxide and other ‘greenhouse gases’ act by destroying ozone in the stratosphere trapping heat in the earth’s atmosphere allowing more visible light to reach the earth’s surface reducing the amount of radiant energy which reaches the surface of the earth

Which of the following best represents the flow of energy through an ecosystem? (A) (B) (C) (D)

Q.10

(D) Subsidiary cells

Which one of the following is a free-living photosynthetic nitrogen fixer?

(A) (B) (C) (D) Q.9

(C) Bulliform cells

Which one of the following performs normal C3 photosynthesis when water is available, but switches to crassulacean acid metabolism (CAM) during salt or drought stress?

(A) Frankia Q.8

(D) Dendrogram

The number of nucleosomes present in a 30 nm solenoid structure of a chromatin is

(A) Mesembryanthemum crystallinum (C) Eleucine coracana Q.7

(C) Phenogram

-Ketothiolase Acetoacetyl CoA carboxylase Acetoacetyl CoA reductase PHB synthase

(A) 2 Q.6

(B) Idiogram

Cytoplasmic male sterility via the chloroplast genome can be induced by the expression of Pha A gene encoding (A) (B) (C) (D)

Q.5

(D) Stipule

Parenchyma cells associated with sieve tube members are called (A) Albuminous cells

Q.4

(C) Pulvinus

An estimate of phylogenetic relationships among the taxa is commonly represented in the form of a (A) Cladogram

Q.3

(B) Hastule

Producers  Primary consumers  Secondary consumers Sun  Producers  Secondary consumers  Primary consumers Sun  Producers  Primary consumers  Secondary consumers Secondary consumers  Primary consumers  Producers  Sun

Which one of the following drugs is obtained from the capsule of Papaver somniferum? (A) Papain

(B) Codeine

(C) Digoxin

(D) Bromelain

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Life Sciences - XL

Q. 11 - Q. 20 carry two marks each. Q.11

Which of the following statements are TRUE for plant growth regulators? P. The release of cellulase and polygalacturonase into the cell wall is promoted by abscissic acid Q. The early biosynthetic steps of gibberellic acid, up to the formation of ent-kaurene take place in the plastid R. The naturally occurring zeatin belongs to the aromatic group of cytokinins S. Induction of protease inhibitors as a result of wounding and pathogen attack is activated by jasmonic acid (A) P, R

Q.12

(B) Q, S

(C) Q, R

(D) P, Q

Which of the following statements are TRUE for the transposable elements? P. Barbara McClintock discovered the autonomous and non-autonomous transposable elements in Maize Q. Variations in flower pigmentation in Antirhinum are due to the presence of transposable elements Ac and Ds R. The Ac transposable element is 4563 bp long and has an 11 bp inverted repeats S. Ds produces the transposase and mobilize the Ac elements (A) Q, S

Q.13

(B) P, Q

(D) R, S

Match the recombinant proteins produced through molecular farming with their applications. Recombinant proteins P. Hirudine Q. Trichosanthin R. Somatotrophin S. -Interferon

Q.14

(C) P, R

Applications 1. 2. 3. 4. 5. 6.

HIV therapy Anticoagulant Growth hormone Hypertension Cystic fibrosis Treatment for hepatitis-B

(A)

(B)

(C)

(D)

P-2 Q-3 R-4 S-1

P-4 Q-1 R-5 S-2

P-1 Q-5 R-3 S-2

P-2 Q-1 R-3 S-6

Which of the following statements are CORRECT for somatic cell hybridization? P. For fusion of protoplast, dimethylsulfoxide (DMSO) is used as a fusogen Q. The enzyme ‘Cellulase Onozuka’ used for protoplast isolation is sourced from Trichoderma viride R. The first report of somatic hybrid plants resulted from the fusion of protoplasts of Nicotiana glauca and N. tabacum S. Viability of isolated protoplasts can be determined by Evan’s blue staining (A) P, Q

(B) Q, S

(C) R, S

(D) Q, R

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GATE 2012 Life Sciences - XL ________________________________________________________________

Q.15

Which of the following statements are TRUE on transgene approach? P. T-DNA integration occurs mainly through non-homologous recombination Q. The Gateway cloning depends on recombination technology as opposed to standard uses of restriction enzymes and DNA ligase R. The localization of -glucuronidase (GUS) activity as a result of expression of GUS reporter gene can be visualized in a histochemical assay using the X-gal S. The green fluorescent protein gene (GFP) is isolated from the bacterium Photinus pyralis (A) P, Q

Q.16

(B) Q, R

(C) P, S

(D) R, S

Identify the free radicals (marked as ‘?’) in sequence from the inter-conversion of reactive oxygen species as shown below. O2  ?  H2O2  ?  H2O P. O2  Q. OH  R. HO2 S. 1 O 2 (A) P, Q

Q.17

(B) R, Q

(C) P, R

(D) Q, S

With respect to adhesion and cohesion of stamens, identify the INCORRECT statements. P. Adnation of stamens to petals is described as epiphyllous stamens Q. In Calotropis, stamens and carpels are united to form gynostegium R. In syngenesious stamens, filaments are united to form a bundle while the anthers are free S. Synandrous stamens found in Cucurbita represent the union of filaments as well as anthers (A) P, Q

Q.18

(B) P, R

(C) Q, S

(D) Q, R

Identify the CORRECT statements in plant secondary metabolism. P. Tropane alkaloids in Atropa belladonna are synthesized from tyrosine Q. Antioxidative food ingredient rosmarinic acid is obtained from cell suspension cultures of Coleus blumei R. Thiophenes are produced from hairy root cultures of Tagetes patula S. Cyanidin, the principal anthocyanin responsible for red color in Rosa hybrida is produced from cinnamaldehyde (A) P, S

Q.19

(B) R, S

(C) P, Q

(D) Q, R

Which of the following statements are TRUE for respiration? P. The conversion of one molecule of pyruvate to three molecules of CO2 generates four molecules of NADH Q. Fructose 6-phospate is the principal substrate for glycolysis R. The oxidation of glucose 6-phosphate to 6-phosphogluconate is the first step in the oxidative pentose phosphate pathway S. The mitochondrial ‘alternative oxidase’ provides an alternative pathway for transfer of electrons from ubiquinone to oxygen utilizing proton pumping complex of the respiratory chain (A) P, R

(B) P, S

(C) Q, R

(D) Q, S

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Q.20

Life Sciences - XL

Match the name of the disease with the causal organism. Disease

Causal organism

P. Black rot of sugarcane Q. Stem rot of jute R. Tikka disease of groundnut S. Crown gall of grapes

1. 2. 3. 4. 5. 6.

Cercospora personata Macrophomina phaseolina Ceratocystis adiposa Synchytrium endobioticum Agrobacterium tumefaciens Colletotrichum corchorum

(A)

(B)

(C)

(D)

P-1 Q-3 R-6 S-5

P-2 Q-3 R-1 S-5

P-3 Q-2 R-1 S-5

P-2 Q-6 R-3 S-4

END OF SECTION - J

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Life Sciences - XL

K : MICROBIOLOGY Q. 1 – Q. 10 carry one mark each. Q.1

Which ONE of the following components is NOT an electron acceptor during anaerobic respiration? (A) Lactate

Q.2

An auxotrophic mutant arises spontaneously in a wild type E.coli culture growing in a rich medium. Which ONE of the following techniques will ensure the isolation of the auxotrophic mutant? (B) Streaking for single colonies (D) Direct microscopic observation

Which ONE of the following mutants is used to carry out genetic analysis to determine the function of an essential gene? (A) Knock out mutant (C) Insertion mutant

Q.5

(C) recessive

(D) cis-dominant

(B) 2  1020

(C) 1  1050

(D) 1  2100

Which ONE of the following statements about E.coli is NOT true ? (A) (B) (C) (D)

Q.8

(B) trans-dominant

A rich medium is inoculated with a bacterium that divides every 30 minutes. The number of bacteria at the end of 50 hours is (A) 2  1010

Q.7

(B) Deletion mutant (D) Temperature sensitive mutant

An E.coli mutant constitutive for the lac operon was mated with a wild type strain. The merodiploid thus obtained was inducible by lactose. This observation indicates that the original mutation is (A) dominant

Q.6

(D) Sulphate

nutritional requirement phylogenetic relationships pathogenic properties morphology

(A) Replica plating (C) Pour plating method Q.4

(C) Nitrate

Bergey’s Manual of Systematic Bacteriology groups bacteria into species according to their (A) (B) (C) (D)

Q.3

(B) Carbonate

E.coli was the first disease- causing bacterium identified by Robert Koch E.coli is part of the normal microbiota of humans Certain E.coli strains can cause bloody diarrohea E.coli is beneficial to human

Antibody coated pathogens are recognized by effector cells through (A) CD4 receptor

(B) FC receptor

(C) CD8 receptor

(D) IFN gamma receptor

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Q.9

Life Sciences - XL

Match the disease in Group I with their corresponding organism in Group II Group I P. African sleeping sickness Q. Rocky mountain spotted fever R. Mumps S. Filariasis (A) (B) (C) (D)

Q.10

Group II I. Rubulavirus II. Trypanosoma brucei III. Wuchereria bancrofti IV. Rickettsia rickettsii V. Leishmania donovani

P-III, Q-V, R-II, S- I P-II, Q-I, R-III, S- IV P-II, Q-IV, R-I, S- III P-I, Q-V, R-II, S- IV

Select the technique most appropriate to demonstrate that lactose induces the synthesis of -galactosidase enzyme. (A) (B) (C) (D)

Northern Blot Western Blot Quantitative PCR Southern Blot

Q. 11 - Q. 20 carry two marks each. Q.11

Frederick Griffith used smooth (S) and rough (R) strains of Streptococcus pneumoniae in his classical experiment that showed DNA might be the genetic element. Which ONE of the following observations gave the clue for this discovery? (A) (B) (C) (D)

Q.12

Entry of  phage lysogen to lytic phase is triggered by (A) (B) (C) (D)

Q.13

R strain became S strain when mixed with heat killed S strain R strain remained R strain when mixed with heat killed S strain S strain became R strain when mixed with heat killed R strain R strain became S strain when mixed with live S strain

mutation in the genome loss of co-operativity in binding of  repressor increase in the  repressor concentration decrease in recA function

Match the Phylum in Group I with their characteristic motility appendage listed in Group II Group I P. Q. R. S.

Archaezoa Amoebozoa Ciliophora Apicomplexa

Group II I. Flagella II. Fimbriae III. Pseudopods IV. Cilia V. Pili

(A) P-V, Q-II, R-IV, S-IV (B) P-II, Q-I, R-IV, S-III (C) P-I, Q-III, R-IV, S-I (D) P-III, Q-II, R-IV, S-V

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Q.14

Life Sciences - XL

Ten bacteria were inoculated into a rich medium. If at the end of ten hours the total number of cells is 104, then the number of elapsed generations and the generation time respectively is (A) 10, 120 minutes

Q.15

An E.coli mutant defective for an enzyme is unable to grow on acetate but grows on glycerol as the sole carbon source. Which ONE of the following enzymes is likely to be defective in this mutant? (B) Glyceraldehyde 3-phospahte dehydrogenase (D) Isocitrate lyase

Which one of the following pairs of bacterial species fixes atmospheric Nitrogen? (A) (B) (C) (D)

Q.18

Clostridia and Rhizobia Clostridia and Lactobacillus Rhizobia and Enterococcus Actinomycetes and Mycoplasma

Nalidixic acid inhibits gyrase activity. Resistance to this antibiotic arises mainly due to (A) nonsense mutation in the gyrase gene (C) missense mutation in the gyrase gene

Q.19

(B) deletion mutation in the gyrase gene (D) degradation of the gyrase gene product

Transformation of normal cyanobacterial cells into heterocysts involves (A) (B) (C) (D)

Q.20

(D) 40, 15 minutes

complementary strand of RNA double stranded RNA complementary strand of DNA double stranded DNA

(A) Isocitrate dehydrogenase (C) Pyruvate dehydrogenase

Q.17

(C) 20, 30 minutes

The first step in the replication of a virus with the reverse transcriptase deals with the synthesis of (A) (B) (C) (D)

Q.16

(B) 10, 60 minutes

synthesis of nitrogenase and retention of photosystem I synthesis of nitrogenase and loss of photosystem I loss of nitrogenase but retention of photosystem I loss of both nitrogenase and photosystem I

Methane belched (eructation) out by cattle arises from the carbon dioxide produced (A) (B) (C) (D)

during normal respiration oxidation of food stuff occurring in mitochondria lactic acid fermentation occurring in muscles bacterial fermentation occurring in the gut

END OF SECTION - K

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Life Sciences - XL

L : ZOOLOGY Q. 1 – Q. 10 carry one mark each. Q.1

Trees in the equatorial region of earth supply oxygen into the atmosphere that sustains species living in distant polar regions. This relationship is called as (A) mutualism (C) commensalism

Q.2

(B) symbiosis (D) parasitism

A swimmer is preparing to swim ‘non-stop’ across the English channel (a distance of 34 kilometers). Consumption of which of the following category of food/s should the swimmer increase to accomplish this feat? (A) Proteins (C) Proteins and Carbohydrates

Q.3

(B) Fats (D) Carbohydrates

Some species of beetles and fishes can survive at sub freezing temperature. They accomplish this by maintaining cellular integrity using one of the following mechanisms. (A) Expressing anti-freeze proteins (B) Accumulating fats (C) Increasing accumulation of complex polyols (D) Reducing the availability of total water in the body

Q.4

Zygotic genes required for the formation of a group of adjacent segment in the developing Drosophila embryo is called (A) Maternal gene

Q.5

(B) Pair rule gene

(D) Gap gene

A typical receptor senses extracellular stimuli by virtue of its localization on plasma membrane. The receptor to which of the following ligands is an exception to this rule? (A) -amino butyric acid (C) Estrogen

Q.6

(C) Homeotic gene

(B) Acetylcholine (D) Luteinizing hormone

The relationship between genes and enzymes was first suggested by the discovery of (A) in-born errors of metabolism in human (B) sexual phenotype in insects (C) metabolic pathways in fungi (D) gene regulation in bacteria

Q.7

The blind spot in the retina is blind because of which of the following reasons? (A) It is the region where the optical nerve leaves the retina. (B) The opsin is not expressed in this region. (C) It lies in the shadow of pupil. (D) It is the junction between rods and cones.

Q.8

Lamprey, a jawless fish, belongs to which one of the following ‘Classes’? (A) Myxini (C) Conodonta

Q.9

(B) Cephalaspilomorphi (D) Anaspida

Chorionic gonadotropin plays an important role in the establishment and maintenance of pregnancy and is synthesized in the placenta of (A) Cattle

(B) Pigs

(C) Mice

(D) Human

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Q.10

Life Sciences - XL

Glucose and hexanoic acid, each having six carbon atoms can undergo complete biological oxidation. In terms of net ATP generation, which of the following statements is CORRECT? (A) (B) (C) (D)

Glucose produces more ATP than hexanoic acid Only glucose can generate ATP Both glucose and hexanoic acid produce same amount of ATP Hexanoic acid produces more ATP than glucose

Q. 11 - Q. 20 carry two marks each. Q.11

Match the following evolutionary biologists with their respective theory I) August Weisman i) Neutral theory of molecular evolution II) Jean-Baptiste Lamark ii) Handicap principle III) Amotz Zahavi iii) Germ plasm theory IV) Motoo Kimura iv) Inheritance of acquired characteristics (A) I-ii, II-iv, III-i, IV-iii (C) I-iii, II-iv, III-i, IV-ii

Q.12

(B) I-iii, II-iv, III-ii, IV-i (D) I-iii, II-i, III-iv, IV-ii

A female cat with a mutant phenotype was bred with a wild-type male cat. All progeny (4 males and 4 females) show the mutant phenotype. On the other hand, all progeny (4 males and 4 females) from the reciprocal cross between a mutant male and a wild-type female show the wild-type phenotype. Which of the following explain the inheritance pattern of the mutation? (A) Recessive (C) Mitochondrial inheritance

Q.13

If all the nucleotides have equal probability of occurrence in a 4 Mbp long DNA sequence, then how many times will the site of EcoRI, restriction endonuclease occur? (A) 976

Q.14

(B) Linked inheritance (D) Autosomal inheritance

(B) 46

(C) 64

Seasonally breeding animals and birds measure the day length, i.e. photoperiod and use these measurements as predictive information to prepare themselves for breeding. Besides melatonin, which of the following hormones is involved in this biological process? (A) Gonadotropin releasing hormone (C) Thyroxine

Q.15

(B) Growth hormone (D) Adrenocorticotropic hormone

Red-Green color blindness is an X-linked recessive disorder. In a population which is in the HardyWeinberg equilibrium, the incidence of occurrence of this in males is 1:1000. What will be the expected incidence of affected homozygous females? (A) 1 in 1002000

Q.16

(D) 1000

(B) 1 in 2000000

(C) 1 in 1001000

(D) 1 in 1000000

Golgi apparatus is also termed as cellular post office, since it packages and transports cellular proteins across various organelles and outside the cell. In general, the Golgi is perinuclear in location and is closely associated with the endoplasmic reticulum. A chemical compound, Monensin inhibits all trafficking from Golgi. If Golgi is visualized by immunofluorescence microscopy after treatment with this compound, the Golgi will be (A) absent

(B) normal

(C) swollen

(D) fragmented

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Q.17

Life Sciences - XL

In an individual, three distinct proteins bind oxygen depending on the location and development stage. While hemoglobin is the major oxygen binding protein in adults, myoglobin is present in skeletal muscles and fetal hemoglobin is present in fetal stage only. The following graph shows the oxygen binding capacity of these proteins. The A, B and C plots represent oxygen binding capacity of A

Saturation

1.0

B 0.5

C

10

20

30

40

50

O2 pressure (pO2) (A) (B) (C) (D) Q.18

A patient comes with symptoms of autonomic hemolysis. The diagnostic tests reveal that he has auto-antibodies to red blood cells (RBCs). Which one of the following mechanisms is the cause of this condition? (A) (B) (C) (D)

Q.19

hemoglobin, fetal hemoglobin and myoglobin, respectively fetal hemoglobin, hemoglobin and myoglobin, respectively hemoglobin, myoglobin and fetal hemoglobin, respectively myoglobin, fetal hemoglobin and hemoglobin, respectively

Neutrophils release granzymes which lyse RBCs Complement is activated and membrane attack complex lyse RBCs Cytotoxic T-cells lyse RBCs Interleukin-2 binds to the receptor on RBCs

The nerve impulse at the neuromuscular junction results in discharge of acetylcholine (Ach) from its vesicles into the synaptic cleft. Ach gets degraded by acetylcholine esterase and is present in which one of the following locations? (A) Post synaptic membrane (C) Presynaptic membrane

Q.20

(B) Both pre and post – synaptic clefts (D) Synaptic cleft

Increasing estradiol (E2) hormone from ovarian follicles prior to ovulation has been hypothesized to play a critical role for induction of pheromones. These pheromones render females sexually receptive to males to facilitate mating. An investigator performs experiments in sheep in which females are gonadectomized, then treated with E2 or vehicle alone and allowed to breed. Which one of the findings listed below will validate the hypothesis that pheromones are induced by E2? (A) (B) (C) (D)

Sexual receptivity is regained only in vehicle treated females. Sexual receptivity is regained only in E2 treated females Sexual receptivity was regained irrespective of E2 treatment Sexual receptivity is not regained by any treatment

END OF SECTION - L

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Life Sciences - XL

M : FOOD TECHNOLOGY Q. 1 – Q. 10 carry one mark each. Q.1

Among the following fatty acids, which group is known as essential fatty acids? (A) (B) (C) (D)

Q.2

Cellulose, the structural polysaccharide of plant, is a polymer of (A) (B) (C) (D)

Q.3

9,11-Octadecadienoic and 9,11,13-Octadecatrienoic 9,12-Octadecadienoic and 9,12,15-Octadecatrienoic 9-Octadecenoic and 9,11-Octadecadienoic 9,11-Octadecadienoic and 9-Eicosenoic -D-Glucose -D-Glucose -D-Galactose -D-Galcturonic acid

The important role of carotenoids in the human diet is their ability to serve as precursors of (A) Vitamin C

Q.4

(B) Vitamin D

(B) Saccharomyces cerevisiae (D) Aspergilus niger

Which one of the microorganisms given below is NOT RESPONSIBLE for ropy or stringy fermentation of milk? (A) (B) (C) (D)

Q.6

Alcaligenes viscolactis Enterobacter aerogenes Streptococcus cremoris Streptococcus lactis

A mild heat treatment of foods that destroys pathogens and extends its shelf life is called (A) Baking (C) Sterilization

Q.7

(B) Blanching (D) Pasteurization

The most common and least expensive plastic film used for packaging of solid food materials is (A) Polyethylene (C) Polypropylene

Q.8

(B) Polystyrene (D) Polyvinylchloride

Reassociation of amylose and formation of crystalline structure upon cooling of cooked starch solution is termed as (A) Synersis (C) Retrogradation

Q.9

(B) Gelatinization (D) Denaturation

Thermal destruction of microorganisms follows a kinetics of (A) Zero order

Q.10

(D) Vitamin K

Which one of the following microorganisms is used in the preparation of bread? (A) Candida utilis (C) Saccharomyces cevarum

Q.5

(C) Vitamin A

(B) First order

(C) Second order

(D) Fractional order

100 kg tomato juice containing 5% Total Solids (w/w) is concentrated to 25% Total Solids (w/w). The total amount of water removed from tomato juice in kg is (A) 65

(B) 70

(C) 75

(D) 80

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Life Sciences - XL

Q. 11 - Q. 20 carry two marks each. Q.11

Which one of the following is NOT A CORRECT statement? (A) Meatiness is the taste produced by compounds such as glutamate in products like cheese, soy sauce. (B) Astringency is a dry mouth feel in the oral cavity that is most associated with phenolic compounds. (C) Saltiness is a taste that is mainly produced by chloride ions. (D) Sourness is related to acidity and is sensed by hydrogen ion channels in the human tongue. The following plot represents the Lineweaver-Burk equation of an enzymatic reaction both in the presence and the absence of inhibitor. Here, V is the velocity of reaction and S is the substrate concentration.

V–1

Q.12

Inhibition No inhibition

S–1 The nature of inhibition shown in the plot is (A) (B) (C) (D) Q.13

Non-competitive Anti-competitive Competitive Mixed type

Make the correct match of the food constituents in Group I with their nature given in Group II. Group I P) Ascorbic Acid Q) Phenyl alanine R) Dextrose S) Haemoglobin (A) P-4, Q-3, R-1, S-2 (C) P-3, Q-4, R-2, S-1

Q.14

Group II 1) Sugar 2) Chelate 3) Amino Acid 4) Antioxidant (B) P-4, Q-1, R-3, S-2 (D) P-4, Q-2, R-1, S-3

Make the correct match of the fermented food products in Group I with the microorganisms in Group II. Group I P) Yoghurt Q) Cheese R) Sauerkraut S) Kefir (A) P-1, Q-4, R-2, S-3 (C) P-3, Q-4, R-2, S-1

Group II 1) Lactobacillus acidophilus and Lactobacillus delbrueckii 2) Leuconostoc mesenteroides and Lactobacillus plantarum 3) Lactobacillus delbrueckii and Streptococcus thermophillus 4) Lactobacillus casei and Streptococcus thermophillus (B) P-4, Q-3, R-1, S-2 (D) P-3, Q-2, R-4, S-1

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Q.15

Life Sciences - XL

Match the following between organelle or cellular components of a bacterium cell in Group I with the constituents and functionalities in Group II. Group I P) Cytoplasmic membrane Q) Flagellum R) Cell wall S) Ribosome (A) (B) (C) (D)

Q.16

P-3, Q-2, R-4, S-1 P-4, Q-2, R-1, S-3 P-3, Q-4, R-2, S-1 P-2, Q-3, R-4, S-1

Thermal death time (TDT) of Clostridium botulinum at 121C is 2.78 min with a z-value of 10C. The TDT of the microorganism at 116C (in min) is (A) 5.270

Q.17

Group II 1) Protein synthesis 2) Peptidoglycan 3) Phospholipid bilayer 4) Motility of cell

(B) 8.791

(C) 1.390

Make the correct match between specific food processing operations in Group I with their mechanism of action in Group II. Group I P) Ball Mill Q) Roller Mill R) Flash Peeling S) Abrasive Peeling

Group II 1) Compression and shear 2) Pressure bursting 3) Friction and shear 4) Impact and shear

(A) P-4, Q-2, R-1, S-3 (C) P-4, Q-3, R-2, S-1 Q.18

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

650 g of a wet food containing 405 g water is dried in a tray dryer to a final moisture content of 6.8 % (dry basis). It is observed that the drying process occurs under constant rate period and it takes 8 h. The rate of drying (in kg/h) is (A) 128.79

Q.19

(D) 0.712

(B) 126.35

(C) 77.81

(D) 0.0485

Air at 1 atmospheric pressure (101.325 kPa) and 30C with absolute humidity of 0.0218 kg/kg of dry air is flowing in a drying chamber. The saturated vapor pressure of water ( pw0 , in kPa) is related to temperature (T, in C) as given below

ln pw0  18.6556 

5217.635 T  273

The relative humidity of air (in percentage) is (A) 62.82 Q.20

(B) 68.22

(C) 86.62

(D) 81.80

The total solids content in a milk sample is 18 %. It is desired to produce 1000 kg of sweetened condensed milk (SCM) having 40 % sugar, 25 % moisture and rest milk solids. What is the ‘Sugar Ratio’ (in percentage) in the SCM in terms of sugar and water content in the final product? (A) 48.19

(B) 61.54

(C) 54.16

(D) 56.14

END OF THE QUESTION PAPER

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GATE 2012 - Answer Keys General Aptitude– GA Q. No. 1 2 3 4

Key / Range D B B D

Q. No. 5 6 7 8

Key / Range A A C D

Q. No. 9 10

Key / Range A A

Key / Range D Marks to All B C C

Q. No. 11 12 13 14 15

Key / Range A Marks to All D A C

Key / Range A B D A D C D

Q. No. 15 16 17 18 19 20

Key / Range B B A B C A

Key / Range B C B B C D Marks to All

Q. No. 15 16 17 18 19 20

Key / Range A A B D A C

Life Sciences – XL (Section – H) Q. No. 1 2 3 4 5

Key / Range D A C B A

Q. No. 6 7 8 9 10

Life Sciences – XL (Section – I) Q. No. 1 2 3 4 5 6 7

Key / Range A D A B C D C

Q. No. 8 9 10 11 12 13 14

Life Sciences – XL (Section – J) Q. No. 1 2 3 4 5 6 7

Key / Range C A B A C A C

Q. No. 8 9 10 11 12 13 14

_________________________________________________________ GATE Previous Year Solved Papers by

Life Sciences – XL (Section – K) Q. No. 1 2 3 4 5 6 7

Key / Range A B A D C D A

Q. No. 8 9 10 11 12 13 14

Key / Range B C B A B C B

Q. No. 15 16 17 18 19 20

Key / Range C D A C A D

Key / Range B D D B C A C

Q. No. 15 16 17 18 19 20

Key / Range D C D B A B

Key / Range C B D C C A C

Q. No. 15 16 17 18 19 20

Key / Range C B B D D B

Life Sciences – XL (Section – L) Q. No. 1 2 3 4 5 6 7

Key / Range C B A D C A A

Q. No. 8 9 10 11 12 13 14

Life Sciences – XL (Section – M) Q. No. 1 2 3 4 5 6 7

Key / Range B A C B D D A

Q. No. 8 9 10 11 12 13 14

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