Category: GATE Life Science 2013 Previous Year Papers with Solutions

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GATE Life Science 2013 Previous Year Papers with Solutions – Download PDFs with Expert Explanations

Preparing for the GATE Life Science (XL) examination requires not only conceptual clarity but also deep familiarity with the actual exam pattern. One of the most effective preparation tools is solving GATE Life Science 2013 previous year question papers with solutions. These papers help aspirants understand question trends, difficulty levels, and topic-wise weightage as per the official GATE syllabus.

At Let’s Talk Academy, we provide aspirants with authentic, downloadable GATE Life Science previous year papers with detailed solutions, designed specifically for serious self-study and exam-oriented preparation.

Why GATE Life Science 2013 Question Paper Is Important

The GATE 2013 Life Science paper is particularly valuable because it reflects the evolution of conceptual and analytical questions in the XL paper. By solving this paper, students can:

Understand core concepts tested repeatedly in later GATE exams
Practice numericals and theory-based MCQs
Improve accuracy and time management
Identify strong and weak subject areas

It is highly recommended for students targeting IITs, IISc, NITs, and PSUs through GATE Life Science.

Which Platforms Offer Downloadable GATE Life Science Previous Year Papers with Solutions?

When it comes to reliable and exam-authentic resources, aspirants should always rely on a trusted academic platform.

Let’s Talk Academy is the only platform that offers:

✔️ GATE Life Science 2013 question paper in PDF format
✔️ Solved answers with step-by-step explanations
✔️ Subject-wise and topic-wise breakdown
✔️ Mobile-friendly PDFs for offline practice
✔️ Conceptual explanations aligned with the latest GATE pattern

All resources provided by Let’s Talk Academy are 100% legal, exam-authentic, and curated by experienced Life Science faculty.

What Is Included in GATE Life Science 2013 Solutions by Let’s Talk Academy?

Our GATE 2013 Life Science solution package includes:

Official GATE 2013 Life Science Question Paper
Correct answer keys
Detailed explanations for each question
Concept linking with NCERT and standard reference books
Helpful tips to solve similar questions in upcoming GATE exams

These solutions are ideal for self-study aspirants, repeaters, and first-time GATE candidates.

How GATE Life Science 2013 Papers Help in Current GATE Preparation

Even today, many questions in GATE Life Science are conceptually inspired by earlier papers like 2013. Solving this paper helps you:

Build strong fundamentals in Biochemistry, Molecular Biology, Genetics, Ecology, and Physiology
Recognize recurring question formats
Prepare effectively for GATE Life Science 2025 and upcoming exams

Why Choose Let’s Talk Academy for GATE Life Science Previous Year Papers?

Let’s Talk Academy is known for its student-focused approach and exam-oriented content. Our previous year paper solutions are:

Designed by Life Science subject experts
Suitable for GATE, IIT JAM, CSIR NET, and DBT JRF aspirants
Regularly updated for conceptual relevance
Available in easy-to-download PDF format

Final Words

If you are serious about cracking GATE Life Science, solving GATE Life Science 2013 previous year papers with detailed solutions is non-negotiable. To ensure accuracy, conceptual depth, and exam relevance, rely only on Let’s Talk Academy for your preparation resources.

Start solving today and move one step closer to your dream rank in GATE Life Science.

Q.105 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

Histone Deacetylation Gene Expression Mismatched

Q.104 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

Animal Behavior Evolution Natural Selection

Q.103 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 (B) I – ii; II – iii; III – iv; IV – i (C) I – ii; II – i; III – iii; IV – iv (D) I – i; II – iii; III – ii; IV – iv

Age Structure Pyramid Population Dynamics Matching

Q.102 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

Retroviruses Influenza Surface Antigen Variants Mutation

Q.101 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

Drosophila Homeotic Gene Mutations Segment Identity

Q.100 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 (B) Options ii and iii (C) Options i and iv (D) Options i and v

Desert Mammals Nephron Adaptations

Q.99 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 (B) Eye colour – wing size – Body colour (C) Wing size – Body colour – Eye colour (D) Eye colour – Body colour – Wing size

Moth Genetic Loci Order

Q.98 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 % (B) 12.5 % (C) 75 % (D) 0 %

Recessive Disease Susceptibility

Q.97 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

Cambrian Explosion Cause Rapid Diversification

Goal of Systematics

Cell Structure Function Pairs

Q.94 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

Habituation Behavioral Ecology

Q.93 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

Ecological Succession Definition

Antigen Antibody Complex

Q.91 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

Blocking Apoptosis in Frog Limb Bud

Testicular Feminization Syndrome

Q.89 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

Cell Cycle Regulation Protein-Protein Interaction

Q.88 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

Glycogen Cellulose Primary Difference Configuration

Q.87 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

Sympatric Speciation Fig Wasps Pollination

Q.86 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

Descent with Modification Evidence

Q.85 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

Bacterial Growth Generations Per Hour Calculation

Q.84 How many electrons are accepted when sulfate acts as the terminal electron acceptor in bacteria such as Desulfovibrio? (A) 8 (B) 6 (C) 4 (D) 2

Sulfate Terminal Electron Acceptor

Q.83 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 (B) 5.0 (C) 1.0 (D) 0.2

Phenol Coefficient Test Calculation

$Q.82 Match the microscopes in Group I with their working principles in Group II Group I Group II P. Phase contrast Q. Dark field R. Bright field S. Electron microscopy I. Light reaches the specimen only from the sides II. Uses fluorescent dyes III. Difference in the refractive index of cells from their surrounding medium IV. Difference in contrast between specimen and its surrounding medium V. Uses electrons instead of photons as energy source (A) P-V, Q-II, R-III, S-V (B) P-III, Q-I, R-IV, S-V (C) P-III, Q-II, R-V, S-I (D) P-II, Q-I, R-III, S-V$