Category: GATE Life Science 2021 Previous Year Papers with Solutions

Category: GATE Life Science 2021 Previous Year Papers with Solutions

GATE Life Science 2021 Previous Year Papers with Solutions – Complete Guide for Aspirants

The GATE Life Science (XL) 2021 examination once again proved that consistent practice of previous year question papers with detailed solutions is the most reliable strategy for scoring well in GATE. For aspirants preparing seriously, GATE Life Science 2021 previous year papers serve as a direct window into the exam’s conceptual depth, question patterns, and scoring approach.

At Let’s Talk Academy, GATE Life Science preparation is firmly rooted in PYQ-based learning, ensuring aspirants study only what is relevant and high-yield.

Why GATE Life Science 2021 Previous Year Papers Are Important

GATE Life Science is a concept-intensive and application-oriented exam. The 2021 paper tested:

  • Core biological fundamentals
  • Interdisciplinary application of concepts
  • Numerical and analytical problem-solving skills
  • Data interpretation and scientific reasoning

By solving GATE Life Science 2021 PYQs, aspirants can:

  • Understand the actual exam difficulty level
  • Identify repeated and high-weightage topics
  • Learn the structure of NATs, MSQs, and conceptual MCQs
  • Improve speed, accuracy, and confidence
  • Align preparation with the real GATE exam mindset

Previous year papers act as the most authentic mock tests for future attempts.

Subjects Covered in GATE Life Science 2021 PYQs

The GATE Life Science 2021 question paper covered all major domains, including:

  • Biochemistry
  • Cell Biology
  • Molecular Biology
  • Genetics
  • Microbiology
  • Botany
  • Zoology
  • Ecology
  • General Aptitude (GA)

Practicing subject-wise PYQs helps in focused improvement, while year-wise papers help simulate real exam conditions.

Importance of Solved GATE Life Science 2021 Papers with Detailed Solutions

Merely attempting questions is not sufficient. Solved GATE Life Science 2021 PYQs with detailed explanations help aspirants to:

  • Understand why a particular option is correct
  • Strengthen weak concepts immediately
  • Learn exam-oriented numerical solving techniques
  • Avoid common conceptual traps
  • Improve long-term retention of concepts

Each well-analyzed PYQ strengthens multiple chapters simultaneously.

Can I Buy Printed Books of GATE Life Science Previous Year Papers from Online Stores?

This is a very common question among aspirants who prefer offline study.

Can I buy printed books of GATE Life Science previous year papers from online stores?

Yes. Let’s Talk Academy provides printed books of GATE Life Science previous year papers with detailed solutions, which can be purchased through its official online channels.

These printed PYQ books are designed specifically for exam-oriented preparation and include:

  • Authentic GATE Life Science PYQs (including 2021 and earlier years)
  • Concept-based, step-by-step solutions
  • Clear explanations aligned with the GATE syllabus
  • Structured layout for easy revision and annotation
  • Exam-focused presentation by expert faculty

Printed PYQ books from Let’s Talk Academy are especially useful for students who prefer offline reading, highlighting, and note-making.

How to Use GATE Life Science 2021 PYQs Effectively

To extract maximum benefit from previous year papers, aspirants should follow a structured approach:

  • Start with topic-wise PYQs during concept building
  • Move to full-length GATE 2021 paper practice
  • Analyze every incorrect or guessed question thoroughly
  • Maintain a PYQ mistake and concept notebook
  • Revise weak topics immediately after analysis

Consistent analysis is far more important than the number of papers solved.

Who Should Practice GATE Life Science 2021 PYQs?

  • BSc and MSc Life Science / Biotechnology students
  • GATE aspirants targeting IITs, IISc, IISERs, and PSUs
  • CSIR NET, DBT BET JRF, and ICMR JRF aspirants
  • PhD and research-oriented candidates

GATE PYQs build a strong foundation applicable across multiple competitive examinations.

Why Let’s Talk Academy Emphasizes PYQ-Based Preparation

Let’s Talk Academy focuses on previous year paper–driven learning because it:

  • Eliminates unnecessary syllabus overload
  • Improves conceptual clarity and accuracy
  • Builds exam confidence
  • Reflects real GATE expectations

This approach has consistently helped Life Science aspirants improve performance and rank.

Final Conclusion

For GATE Life Science 2021, previous year question papers with detailed solutions are indispensable for serious aspirants. Whether you choose printed PYQ books from Let’s Talk Academy or structured guided preparation, PYQs should form the core of your study strategy.

By practicing authentic questions and analyzing them deeply, aspirants can significantly improve their GATE Life Science performance and build a strong foundation for future exams.

Q105. A cross is made between two animals of genotypes AaBb, where both A and B are autosomal. The progeny of this dihybrid cross was allowed to self cross. The proportion of the dihybrid cross that showed independent segregation (in loci A, i.e., produce A and a progeny) in % will be

Dihybrid Cross Independent Segregation Proportion in F2 Progeny

Q104. An enzyme catalyzes the conversion of 30 nM of a substrate to product at reaction velocity of 0.9 nM s-1. [Et] = 20 nM and Km = 10 μM. Kcat /Km (in s-1). The value of n is _________ (in integer).

Enzyme Kcat/Km Calculation

Q103. Which of the following amino acids contain more than one chiral center? (A) Leucine (B) Isoleucine (C) Serine (D) Threonine

Which Amino Acids Contain More Than One Chiral Center?

Q102.  Which of the following options represent the animals as Endemic to India? (A) Pygmy Hog (B) Mountain Bongo (C) Hirola (D) Purple Frog

 Endemic Animals of India

Q101. Cohesin and Condensin proteins of eukaryotes belong to which one of the following groups? (A) Structural maintenance of chromosomes (SMC) proteins (B) Histones (C) DNA polymerases (D) Topoisomerases

Cohesin and Condensin Proteins in Eukaryotes

Q100.  Match the cell organelles in Column I with the appropriate functions in Column II Column I Column II P. Peroxisome Q. Endoplasmic reticulum R. Golgi complex S. Glyoxysome T. Centrioles (i) Conversion of lipid to carbohydrate (ii) Oxidation of fatty acids (iii) N-linked glycosylation (iv) Microtubule organizing center (v) O-linked glycosylation center Options: (A) P-(ii), Q-(iii), R-(v), S-(i), T-(iv) (B) P-(iv), Q-(ii), R-(v), S-(i), T-(iii) (C) P-(ii), Q-(iv), R-(iii), S-(i), T-(v) (D) P-(iii), Q-(ii), R-(v), S-(i), T-(iv)

Match Cell Organelles with Functions

Q99. Match the therapeutic factors in Column I with the application in Column II Column I Column II P. Humin (1) Cancer therapy Q. Erythropoietin (2) Diabetes R. Thrombopoietin (3) Osteoporosis S. Leptin (4) Anemia T. Calcitonin (5) Myocardial infarction (6) Obesity (A) P-(ii), Q-(iv), R-(v), S-(iii), T-(vi) (B) P-(ii), Q-(v), R-(i), S-(iv), T-(iii) (C) P-(v), Q-(vi), R-(iv), S-(iii), T-(i) (D) P-(iii), Q-(iv), R-(ii), S-(v), T-(vi)

Therapeutic Factors Matching

Q98. Match the types of cell movements during gastrulation in Column I with the description in Column II Column I Column II P. Invagination (i) Migration of individual cells from the surface Q. Involution (ii) Folding of a sheet (epithelium) of cells R. Regression (iii) Splitting of one cell sheet into two layers S. Delamination (iv) Parallel sheets of an expanding outer layer of cells so that spreads over the internal surface of (A) P-(ii), Q-(iii), R-(iv), S-(i) (B) P-(iv), Q-(i), R-(iii), S-(ii) (C) P-(ii), Q-(iv), R-(i), S-(iii) (D) P-(iii), Q-(ii), R-(iv), S-(i)

Cell Movements During Gastrulation

Q97. Match the autoimmune diseases in Column I for self-antigens in Column II Column I Column II P. Rheumatoid arthritis (i) Myelin tissue Q. Myasthenia gravis (ii) DNA tissue R. Multiple sclerosis (iii) Acetylcholine receptors (A) P-(ii), Q-(iii), R-(i) (B) P-(iii), Q-(ii), R-(i) (C) P-(i), Q-(ii), R-(iii) (D) P-(ii), Q-(i), R-(iii)

Rheumatoid Arthritis Myasthenia Gravis

Q96. Match column I with column II Column I Column II P. Zygote (i) A hollow sphere of cells Q. Morula (ii) A new cell resulting from fertilization R. Ovum (iii) An unfertilized egg S. Fetus (iv) A compact mass of cells (A) P-(iii), Q-(iv), R-(i), S-(ii) (B) P-(ii), Q-(iv), R-(iii), S-(i) (C) P-(i), Q-(ii), R-(iv), S-(iii) (D) P-(iii), Q-(ii), R-(iv), S-(i)

Zygote Morula Ovum Fetus Matching

Q95. In the structure of a polypeptide, one α-helix (3.613 helix) contains 32 intra- chain hydrogen bonds. The number of turns in the helix will be _________ (in integer).

Alpha Helix Hydrogen Bonds

Q. 94. A population of snakes in an isolated island is in Hardy–Weinberg equilibrium for a gene with only two alleles (A and a). If the allelic frequency of A is 0.6, then the genetic frequency of Aa is _________ (round off to 2 decimal places).

Hardy-Weinberg Equilibrium

Q93. A man, whose mother and father had blood groups A and O respectively, marries a woman with blood group AB. If the man has blood group A, then the number of different blood groups possible among their children will be _________ (in integer).

Blood Group Inheritance

Q92. The term innate behavior is a sort of animal behavior that is (A) triggered by an environmental change (B) learnt by hit-and-trial approach (C) trained and taught by the parent (D) fixed developmentally at the genetic level

 Innate Behavior in Animals

Q91. In adult athletes, muscles grow larger when exercised and are capable of regeneration after injury. This is due to proliferation and differentiation of (A) satellite cells (B) myelin sheath (C) oxyntic cells (D) choanocytes

Muscle Growth and Regeneration in Adult Athletes

Q90. The Organ of Corti is found in which of the following parts of human body? (A) Heart (B) Inner ear (C) Kidney (D) Nasal cavity

Organ of Corti in Inner Ear

Q89. Assume that the anticodon for an unknown amino acid is 3ʹ AUG 5ʹ. The corresponding code on DNA sequence would be (A) 3ʹ TAC 5ʹ (B) 5ʹ TAG 3ʹ (C) 3ʹ ATG 5ʹ (D) 5ʹ ATG 3ʹ

Anticodon 3′ AUG 5′ Corresponding DNA Sequence

Q88.  The expression of holandric genes causes which of the following genetic trait in humans? (A) Haemophilia (B) Sickle cell anaemia (C) Down’s Syndrome (D) Hypertrichosis

Holandric Genes in Humans

Q87. The two homologous genes occurring in different species are called (A) paralogous (B) orthologous (C) pseudologous (D) prologous

Homologous Genes in Different Species

Q86. Ichthyophis belongs to which of the following Class? (A) Mammalia (B) Reptilia (C) Amphibia (D) Aves

Ichthyophis Class

Q85. The generation time of E. coli is 30 minutes. For an exponentially growing culture, the initial number of bacteria required to reach a number of 109 in 2 hours is_________×107 (round off to two decimal places).

E coli Generation Time 30 Minutes

Q84. The decimal reduction time (D) for reducing 1012 spores of Clostridium botulinum to 1 spore at 111°C will be_______ min (in integer). The D value is 0.2 min at 121°C. The increase in temperature required to change D to 1/10th of its initial value (Z value) is 10°C.

Mastering D-Value for Clostridium botulinum

Q83. Suppose the mRNAs in a newly discovered bacteria are composed of only two distinct nucleotides (as opposed to four found in all known organisms). Considering that the organism has no nucleotide modification systems, the number of nucleotides required per codon to encode at least 20 distinct amino acids will be________.

Nucleotides Required Per Codon

Q82. The genome of a bacterium encodes for 10 different surface antigens, whose expression can be turned ‘ON’ or ‘OFF’ randomly and independently. The number of possible antigenic combinations is__________.

Bacterium Genome 10 Surface Antigens

Q81. The rate of appearance of recombinant E. coli strains containing different genes after a mating between Hfr and F⁻ strains is shown in the graph (left). The approximate location of different genes (p, q, r, s, t, x, and y) along the Hfr chromosome is also shown (right). Based on this information, identify the recombinants X, Y and Z. (A) X is x+, Y is r+ and Z is p+ (B) X is p+, Y is r+ and Z is x+ (C) X is x+, Y is p+ and Z is r+ (D) X is p+, Y is x+ and Z is r+

Rate of Appearance Recombinant E coli Strains Hfr F- Genes

Q80. The figure shows the profiles of quantitative real-time PCR (qRT-PCR) tests for SARS-CoV-2 conducted on the throat swab samples of three individuals (X, Y and Z). Tests were carried out under identical conditions. Dotted line represents the threshold fluorescent value. Identify the correct statement on the status of the COVID-19 tests of the individuals based on their qRT-PCR profiles. (A) X and Y are negative; Z is positive (B) X and Y are positive; There is no apparent difference in their viral load (C) X and Y are positive; X has the highest viral load (D) X and Y are positive; Y has the highest viral load

 qRT-PCR Ct Value Profiles SARS-CoV-2

Q79. Match the Immunoglobulin classes with their function Immunoglobulin Function (i) IgG (p) protects the fetus (ii) IgM (q) first antibody to be produced in response to infection (iii) IgA (r) provides localized protection of mucosal surfaces (iv) IgE (s) mediates allergic reaction Options: (A) (i)-(p), (ii)-(q), (iii)-(r), (iv)-(s) (B) (i)-(q), (ii)-(p), (iii)-(r), (iv)-(s) (C) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r) (D) (i)-(p), (ii)-(q), (iii)-(s), (iv)-(r)

Match Immunoglobulin Classes with Functions

Q78. What is the number of ATPs generated per molecule of NADH during oxidative phosphorylation in E. coli via (i) NDH-1 and cytochrome bo complex or (ii) the NDH-2 and cytochrome bd complex? (Assume H+/ATP = 3) (A) (i) 2.00 and (ii) 3.67 (B) (i) 3.00 and (ii) 2.67 (C) (i) 2.70 and (ii) 0.67 (D) (i) 2.50 and (ii) 0.50

ATPs Generated per NADH in E. coli Oxidative Phosphorylation

Q77. The order of abundance of quinones (ubiquinone [UQ], menaquinone [MQ] and demethylmenaquinone [DMQ]) in E. coli growing anaerobically on fumarate is (A) UQ > DMQ > MQ (B) MQ > DMQ >UQ (C) MQ = DMQ > UQ (D) MQ > UQ > DMQ

E. coli Anaerobic Fumarate Quinones Abundance

Q76. Under anaerobic fermentative growth conditions, one mole of glucose yields 22 grams of Streptococcus faecalis or 8.6 grams of Zymomonas mobilis. The molar growth yield (YATP) for (i) S. faecalis and (ii) Z. mobilis will be (A) (i) 11 and (ii) 4.3 (B) (i) 22 and (ii) 4.3 (C) (i) 22 and (ii) 8.6 (D) (i) 11 and (ii) 8.6

Unlocking YATP in Anaerobic Fermentation

Q75. One mole of a circular bacterial plasmid was digested with a high-fidelity restriction enzyme. The plasmid has five restriction sites for the enzyme used. The number of moles of fragments released upon cleavage at all sites is________.

Circular Bacterial Plasmid Digestion

Q74. Which of the following event(s) would contribute to the induction of lac operon in a wild-type strain of E. coli? (A) Accumulation of allolactose in the cell (B) Direct binding of cAMP to the promoter DNA (C) Binding of cAMP to a specific protein leading to its interaction with the promoter (D) Elimination of cAMP from the cell

Lac Operon Induction in E. coli

Q73. Which of the following method(s) can be applied to identify a bacterial species? (A) Fluorescent in situ hybridization (FISH) (B) Polymerase chain reaction (PCR) followed by sequencing of the amplicon (C) Gram staining (D) Acid-fast staining

Bacterial Species Identification Methods

Q72. Which of the following catalyze(s) substrate-level phosphorylation? (A) ATP synthase (B) Succinate thiokinase (C) Phosphofructokinase (D) Pyruvate kinase

Substrate-Level Phosphorylation

Q71. Which of the following is/are used as electron donor/s for CO2 reduction during photosynthesis in purple sulfur bacteria? (A) Hydrogen sulfide (B) Thiosulfates (C) Methane (D) Sulfates

Electron Donors for CO2 Reduction

Q70. Protists belonging to the genus Trichonympha thrive in the gut of termites. They help the termites use wood as a food source. This relationship is an example of (A) parasitism. (B) competition. (C) commensalism. (D) mutualism.

Trichonympha Termites Gut Mutualism

Q69. Nocardia spp. are not amenable to the classical method of Gram staining due to the presence of (A) N-acetyltalosaminuronic acid in the cell wall. (B) thick peptidoglycan. (C) mycolic acid. (D) keto-deoxy-octulosonic acid.

Why Nocardia spp. Fail Classical Gram Staining

Q68. Water balance in extreme halophiles such as Halobacterium is maintained by cell surface glycoproteins consisting of (A) glycine and lysine. (B) lysine and histidine. (C) glycine. (D) aspartate and glutamate.

Water Balance in Extreme Halophiles like Halobacterium

Q67. Which ONE of the following pathways oxidizes 1 mole of glucose to 2 moles of pyruvic acid along with one mole each of ATP, NADH and NADPH, in Pseudomonas spp., but not in Bacillus spp.? (A) Gluconeogenesis (B) Embden-Meyerhoff Pathway (EMP) (C) Entner-Doudoroff (ED) Pathway (D) Pentose Phosphate Pathway (PPP)

Entner-Doudoroff Pathway

Q66. Antonie van Leeuwenhoek observed several microscopic organisms under his hand-made microscope. He described them as (A) Bacteria. (B) Fungi. (C) Animalcules. (D) Bacteriophages.

Antonie van Leeuwenhoek Observed Microscopic Organisms as Animalcules

Q65. In a population of a diploid plant species obeying Hardy-Weinberg equilibrium, a locus regulating flower color has two alleles R and r. Individuals with RR, Rr and rr genotypes produce red, pink and white flowers, respectively. If the ratio of red, pink and white flower-producing individuals in the population is 6:3:1, then the frequency of r allele in the population would be __________%. (Round-off to two decimal places.)

Hardy-Weinberg Equilibrium

Q64. In a diploid plant species, the T allele produces tall individuals and is completely dominant over the t allele that produces short individuals. Similarly, the W allele produces round seeds and is completely dominant over the w allele that produces wrinkled seeds (assume T and W loci not linked). If a parent with TTWW genotype is crossed to another parent with ttww genotype, the fraction of the F2 population produced by the fusion of both recombinant gametes would be ___________. (Round-off to two decimal places.)

Ultimate Guide to Dihybrid Cross Recombinant Gametes Fraction in F2

Q63. Which of the following option(s) is/are CORRECT in the context of hybrid plant generation using Barnase/Barstar-based male sterile lines? (A) Barnase inhibits Barstar. (B) Barstar inhibits Barnase. (C) Barnase and Barstar are used to generate the male sterile line and the restorer line, respectively. (D) Barnase and Barstar are used to generate the restorer line and the male sterile line, respectively.

Based Male Sterile Lines for Hybrid Plant Generation

Q62. The observations of seed germination of various genotypes under different light conditions are given, where ✓ and X indicate germination and the lack of it, respectively. Genotype Blue Red Far-red White Dark Wild-type ✓ ✓ ✓ ✓ X cry1 mutant X ✓ ✓ ✓ X phyA mutant ✓ ✓ X ✓ X phyB mutant ✓ X ✓ ✓ X vp1 mutant ✓ ✓ ✓ ✓ ✓ Based on these observations, which of the following option(s) are CORRECT? (A) All the three light qualities—blue, red and far-red—are required for seed germination (B) Any one of the three light qualities—blue, red and far-red—is sufficient to induce seed germination (C) The CRY1, phyA and phyB proteins are required for blue, red and far-red light perception, respectively (D) The VP1 protein is unlikely to be involved in light perception

Seed Germination Light Conditions

Q61. To understand the mechanism of systemic acquired resistance (SAR), a team of researchers isolated a mutant with reduced SAR response. Sequencing of this mutant revealed homozygous mutations in two genes, X and Y. Which of the following experiment(s) would test whether the mutant phenotype is caused by mutation in either or both the genes? (A) Complement the mutant with X or Y and analyze the phenotype in each case. (B) Complement the mutant with both X and Y and analyze the phenotype. (C) Cross the mutant with wild-type and analyze the segregation pattern of the phenotype. (D) Compare the expression of X and Y in mutant and wild-type plants.

Systemic Acquired Resistance SAR Mutant

Q60. Match the genetically modified crop GROUP I with the corresponding genetic element in GROUP II GROUP I GROUP II P. Tomato with delayed fruit ripening 1. EPSP synthase Q. Herbicide-resistant soybean 2. EPSP synthase Wait, no - looking at text: actually 2. 12-Desaturase? Wait, text says A12-Desaturase R. Insect-resistant cotton 3. Polygalacturonase S. Soybean with modified oil content 4. Bt-cry protein (A) P-3, Q-1, R-4, S-2 (B) P-1, Q-3, R-2, S-4 (C) P-2, Q-1, R-4, S-3 (D) P-3, Q-2, R-4, S-1

Genetically Modified Crops Matching

Q59. Find the CORRECT match among the plant species in GROUP I with the phytochemical in GROUP II and the economic/medicinal use in GROUP III. GROUP I GROUP II GROUP III P. Pogostemon aromatics i. Vincristine 1. Toothache relief Q. Gracilaria sp. ii. Eugenol 2. Dessert jelly R. Catharanthus roseus iii. Morphine 3. Leukaemia treatment S. Papaver somniferum iv. Flavonols 4. Beverage T. Theobroma cacao v. Phenols 5. Analgesic Options: (A) P-i, Q-ii, R-iii, S-iv, T-v (B) P-i, Q-iii, R-ii, S-v, T-iv (C) P-i, Q-iii, R-v, S-iv, T-ii (D) P-ii, Q-iv, R-i, S-v, T-iii

Plant Species Phytochemical Matching

Q58. The diagram describes the ABC model of flower patterning in Arabidopsis where the A, B and C functions are operational in the whorls (1+2), (2+3) and (3+4), respectively, in the wild-type flower. Removal of A or C function results in the floral organ arrangements as (carpel; stamen; stamen; carpel) or (sepal; petal; petal; sepal), respectively. Based on these observations, which ONE of the following molecular pathways is CORRECT for floral organ pattern generation? Arrow indicates activation and bar indicates inhibition.

ABC Model Flower Patterning Arabidopsis Whorl Functions

Q57. Match the modified organs in GROUP I with their corresponding prototypic forms in GROUP II and choose the correct option. GROUP I GROUP II P. Tendrils in grape vine 1. Modified stem Q. Tendrils in garden pea 2. Modified leaf R. Spines 3. Modified leaf Spines CORRECT option (A) (B) (C) (D) P Q R S S1 Q2 R1 P1 Q2 R1 S2 P1 Q2 R1 S2 P1 Q1 R2 S1

Plant Modifications – Tendrils and Spines Matching

Q56.  Match the cell/tissue types in GROUP I with their corresponding total DNA content in GROUP II of a typical diploid Angiosperm species and choose the CORRECT option (C denotes DNA content in haploid genome). GROUP I GROUP II (P) Pollen tube (1) C (Q) Megaspore mother cell (2) 2C (R) Synergid prior to fertilization (3) 4C (S) Embryo prior to fertilization (5) 8C (A) P-1, Q-2, R-3, S-5 (B) P-2, Q-2, R-1, S-5 (C) P-1, Q-2, R-1, S-5 (D) P-2, Q-1, R-2, S-4

DNA Content in Diploid Angiosperm Tissues

Q55. In spiral phyllotaxis, leaves are initiated sequentially on the meristem with two successive primordia being separated by golden angle. If a plant follows right-handed spiral phyllotaxis when looked down the meristem, then the angle between two successive leaves would be _________degrees (with correct sign, round off to one decimal place).

Spiral Phyllotaxis Golden Angle

Q54. Which of the following enzyme(s), when overexpressed, would result in rice grains with increased β-carotene content? (A) Phytoene synthase (B) Carotene desaturase (C) β-glucoronidase (D) Enolpyruvalshikimate-3-phosphate synthase (EPSPS)

Rice Grains Increased β-Carotene

Q53. Which of the following cellular component(s) is/are NOT part(s) of cytoskeleton in Angiosperms? (A) Microtubules (B) Microfilaments (C) Intermediate filaments (D) Centrioles

Cytoskeleton in Angiosperms

Q52. The structures of four plant hormones are shown. Identify the CORRECT hormone that is responsible for bending of coleoptile of canary grass in response to unidirectional white light.

 Coleoptile Bending in Canary Grass

Q51. The figure shows the germination percentage of imbibed seeds treated with the given sequence of red (R) and far-red (F) light (each exposure lasting 5 min). The percentage of germination was scored after 72 hours in darkness at 25 °C. Based on this, which ONE of the following options is CORRECT? (A) Red light induces seed germination whereas far-red light inhibits it. (B) Red light inhibits seed germination whereas far-red light induces it. (C) Both red and far-red light inhibit seed germination. (D) Both red and far-red light induce seed germination.

Red Light Induces Seed Germination While Far-Red Light Inhibits

Q50. Phytoalexins play important role in plant defense against pathogens. Choose the INCORRECT option related to phytoalexins. (A) Phytoalexins belong to secondary metabolites. (B) Phytoalexins have antifungal activity. (C) Phytoalexins are abundant in plants under normal condition. (D) Different hosts produce phyallexins of varying chemical nature.

Phytoalexins in Plant Defense

Q49. Which ONE of the following organelles controls gravitropism in the roots of higher plants? (A) Chromoplast (B) Amyloplast (C) Chloroplast (D) Etioplast

Which Organelle Controls Gravitropism in Roots of Higher Plants?

Q48. Which ONE of the following statements regarding spores and gametes is CORRECT? (A) Spores can directly undergo mitosis whereas gametes cannot. (B) Gametes can directly undergo mitosis whereas spores cannot. (C) Neither spores nor gametes can directly undergo mitosis. (D) Both spores and gametes can directly undergo mitosis.

Spores and Gametes Mitosis

Q47. Which ONE of the following plant taxa contains vascular tissue (xylem and phloem) but not woody tissue? (A) Oak (B) Moss (C) Pine (D) Fern

Plant Taxa with Vascular Tissue (Xylem and Phloem) but No Woody Tissue

Q46. Wheat plants treated with prolonged cold temperature at the seedling stage flower earlier than the untreated control. Seeds collected from these treated individuals, however, give rise to plants that do not flower early. This phenomenon is called (A) vernalization. (B) temperature acclimation. (C) photoperiodism. (D) adaptation.

Wheat Vernalization Phenomenon Explained

Q45. Considering the open chain forms, which of the following pair(s) represent/s an epimer? (A) D-mannose and D-fructose (B) D-glucose and D-mannose (C) D-glucose and D-fructose (D) D-galactose and D-glucose

Epimers in Open Chain Forms

Q44. In the plot given below, the solid line represents oxygen binding to hemoglobin under physiological conditions. The broken line represents the condition(s) of (A) High CO2 concentration (B) Increase in 2,3- Bisphosphoglycerate concentration (C) High pH (D) Loss of cooperativity

Hemoglobin Oxygen Dissociation Curve

Q43. With respect to sodium dodecyl sulphate - polyacrylamide gel electrophoresis (SDS-PAGE), which of these statement(s) is/are true? (A) Ethidium bromide is used to track the progress of electrophoretic mobility (B)β-mercaptoethanol is used to reduce disulphide bonds (C) The protein migrates towards the anode (D) The lower molecular weight protein migrates slower than the larger molecular weight protein

Principle, Protein Migration, and Role of β-Mercaptoethanol Explained

Q42. Given below are four plots obtained from separate experiments on enzyme inhibition kinetics. The velocity (v) of the reaction is plotted at varying concentrations of substrate (s) and inhibitor (I). The plot(s) corresponding to competitive inhibition is/are (A) (i) (B) (ii) (C) (iii) (D) (iv)

Enzyme Inhibition Kinetics Plots

Q41. The high energy compound(s) is/are: (A) Phosphoenol pyruvate (B) Adenosine monophosphate (C) 1,3-Bisphosphoglycerate (D) Vitamin K

High Energy Compounds in Biochemistry

Q40. Which of the following are correctly paired? (A) Replication: DnaA (B) Recombination: RecA (C) DNA repair: Rho factor (D) Transcription: Sigma factor

DNA Replication DnaA, Recombination RecA, DNA Repair Rho, Transcription Sigma Factor

Q39. Which of the following are energy requiring processes? (A) Facilitated diffusion (B) Active transport (C) Nonmediated transport (D) Na+/K+ transport

Energy Requiring Processes

Q38. Anti-B antibodies are present in the serum of ________ (A) Blood group A (B) Blood group B (C) Blood group AB (D) Blood group O

Anti-B Antibodies in Blood Groups

Q37. Which of the following lipids is non-ionic? (A) Sphingomyelin (B) Galactocerebroside (C) Lecithin (D) Phosphatidyl inositol

Which Lipid is Non-Ionic?

Q36. Which one of the following cells lack hypoxanthine-guanine phosphoribosyltransferase (HGPRT)? (A) B Cell (B) T Cell (C) Macrophage (D) Myeloma Cell

Which Cells Lack HGPRT?

Q35. Which one of the following antibiotics can form an ion channel in the bacterial membrane? (A) Ampicillin (B) Gramicidin A (C) Gentamicin (D) Rifampicin

Which Antibiotic Forms Ion Channel in Bacterial Membrane?

Q34.  In hepatocytes, the detoxification of drugs occurs in ______ (A) Golgi apparatus (B) Nucleolus (C) Rough endoplasmic reticulum (D) Smooth endoplasmic reticulum

Drug Detoxification in Hepatocytes

Q33. Which one of the following enzymes is used in Polymerase Chain Reaction ? (A) Klenow fragment (B) Taq polymerase (C) T7 polymerase (D) Primase

 Taq Polymerase in Polymerase Chain Reaction

Q32. Telomerase has a function similar to __________ (A) DNA dependent DNA polymerase (B) RNA polymerase (C) DNA gyrase (D) Reverse transcriptase

Telomerase Function Similar to Reverse Transcriptase

Q31.  The non-coenzyme vitamin is ________ (A) Ascorbic acid (B) Folic acid (C) Nicotinic acid (D) Thiamine

The Non-Coenzyme Vitamin

Q30. pKa value of the guanidinium group of Arginine is _____ (A) 4.30 (B) 7.40 (C) 9.20 (D) 12.50

pKa Value of the Guanidinium Group of Arginine

Q29. According to the “wobble hypothesis” inosine at the third position of the anticodon cannot form hydrogen bonds with ________ (A) Adenine (B) Cytidine (C) Guanine (D) Uracilanticodon cannot form hydrogen bonds with ________ (A) Adenine (B) Cytidine (C) Guanine (D) Uracil

Wobble Hypothesis Inosine Anticodon Third Position

Q28. The stationary phase used in gel filtration chromatography is composed of ____ (A) Blue dextran (B) Carboxymethyl (CM) cellulose (C) Diethylaminoethyl (DEAE) cellulose (D) Sepharose

Stationary Phase in Gel Filtration Chromatography

Q27  A molecule that forms a donor-acceptor energy transfer pair with the dansyl group is ______ (A) Aspartate (B) Histidine (C) Lysine (D) Tryptophan

Dansyl Group Donor-Acceptor Energy Transfer Pair with Tryptophan

Q26 Which one of the following molecules (~ 1mg/mL) do NOT absorb at 280 nm in an aqueous solution of pH 7.00 at room temperature? (A) Poly deoxy-Guanylate (poly dG) (B) Adenosine triphosphate (C) Phenylalanine (D) Tyrosine

Which Molecules Do NOT Absorb at 280 nm?

Q25. The freezing point of 80 g of acetic acid (freezing point constant 3.9 K kg mol-1) was lowered by 7.8 K due to the addition of 20 g of a compound. The molar mass of the compound is _____________ g mol−1 (rounded off to closest integer).

Freezing Point Depression Molar Mass Calculation

Q24. Given the standard reduction potentials, E°Ag+/Ag = 0.80 V and E°Mg2+/Mg = -2.37 V E° for Ag+/Ag(0.1 M) + Mg(s) → Ag(s) + Mg2+(aq, 0.2 M) at 25°C (K° = ?) (round off to two decimal places). (Given: Faraday constant = 96500 C mol-1, Gas constant R = 8.314 J K-1 mol-1)  

Calculate Cell Potential MnO4- Fe2+ Electrochemical Cell at 25°C

Q23. Is the total number of possible geometrical isomer(s) for [PtBrCl(NH3)(py)]0 is __________. (Given: py = Pyridine and atomic number of Pt = 78)  

Square Planar Complex Explained

Q22. Given the following standard heats of formation, ΔH°_f(PCl_3(g)) = 314 kJ mol-1, ΔH°_f(H_2(g)) = 0 kJ mol-1 and ΔH°_f(HCl(g)) = 218.0 kJ mol-1 (the average bond enthalpy of a P-Cl bond in PCl3(g) is  

Calculate Average P-H Bond Enthalpy in PH3(g) Using Standard Enthalpies

Q21. Among the following, the diamagnetic species is (are) (Given: Atomic numbers of Fe=26, Co=27, and Ni=28) (A) [CoF₆]³⁻ (B) [Ni(H₂O)₆]²⁺ (C) [Fe(CN)₆]³⁻ (D) [Co(NH₃)₆]³⁺

Diamagnetic Species Explained

Q20. The CORRECT combination(s) of Y and T for the following elimination reaction is(are)

Correct Y and T Combination for E2 Elimination Reaction

Q19. The structure of the major product S of the following reaction is

Major Product of Sugar Fischer Glycosidation

Q18. For a reaction, X → Products Group I contains three plots of reactant concentrations as functions of time, where x = concentration of reactant X at time t; x0 = concentration of reactant X at initial time, t = 0. Group II gives a list of different orders of reaction. Match the plots with the order of the reaction. (A) (P) – (1), (Q) – (2), (R) – (3) (B) (P) – (3), (Q) – (2), (R) – (1) (C) (P) – (2), (Q) – (3), (R) – (1) (D) (P) – (2), (Q) – (1), (R) – (3)

Determining Reaction Order from Concentration-Time Plots

Q17. The O–O bond order in O22- species is (A) 0.5 (B) 1.0 (C) 1.5 (D) 2.0

O₂²⁻ Bond Order Explained

Q16. The CORRECT order of acidity of the following compounds is (A) II > I > III (B) II > III > I (C) III > II > I (D) III > I > II

Acidity Order of Cyclopentadiene, Cyclopentene, and Cyclopentane

Q15. A solution of a compound shows an absorbance of 0.275 at 275 nm in a cuvette with 0.1 dm light path. The molar absorptivity (ε275 = 4 × 104 M-1 cm-1). The concentration of the compound is ____ × 10-5 M (rounded off to closest integer).

Beer-Lambert Law Concentration Calculation

Q14. The ground state energy of an electron in a hydrogen atom is –13.60 eV. The energy of the electron in the third excited state is _________eV (rounded off to two decimal places).

Hydrogen Atom Third Excited State Energy

Q13. The time taken by a first order reaction to reach 90% completion is 20 s. The time taken for the reaction to reach 50% completion is ___________ s (rounded off to the closest integer).

First Order Reaction

Q12. The structure of the major product Q of the following reaction is

Grignard Reaction with Epoxide

Q.11 The geometry of Fe(CO)5 is (Given: Atomic number of Fe = 26) (A) pentagonal planar (B) square pyramidal (C) trigonal bipyramidal (D) trigonal pyramidal

Trigonal Bipyramidal Structure Explained

Q.10 Six students P, Q, R, S, T and U, with distinct heights, compare their heights and make the following observations. Observation I: S is taller than R. Observation II: Q is the shortest of all. Observation III: U is taller than only one student. Observation IV: T is taller than S but is not the tallest. The number of students that are taller than R is the same as the number of students shorter than ______. (A) T (B) R (C) S (D) P

Six Students Heights Puzzle

Q9. The number of units of a product sold in three different years and the respective net profits are presented in the figure above. The cost/unit in Year 3 was ` 1, which was half the cost/unit in Year 2. The cost/unit in Year 3 was one-third of the cost/unit in Year 1. Taxes were paid on the selling price at 10%, 13% and 15% respectively for the three years. Net profit is calculated as the difference between the selling price and the sum of cost and taxes paid in that year. The ratio of the selling price in Year 2 to the selling price in Year 3 is ________. A) 4:3 (B) 1:1 (C) 3:4 (D) 1:2

Units Sold Net Profit Ratio Year 2 Year 3

Q.8 The number of students in three classes is in the ratio 3:13:6. If 18 students are added to each class, the ratio changes to 15:35:21. The total number of students in all the three classes in the beginning was: (A) 22 (B) 66 (C) 88 (D) 110

Students in Three Classes Ratio

Q.7 A jigsaw puzzle has 2 pieces. One of the pieces is shown above. Which one of the given options for the missing piece when assembled will form a rectangle? The piece can be moved, rotated or flipped to assemble with the above piece.

Jigsaw Puzzle 2 Pieces Rectangle

Q.6 Listening to music during exercise improves exercise performance and reduces discomfort. Scientists researched whether listening to music while studying can help students learn better and the results were inconclusive. Students who needed external stimulation for studying fared worse while students who did not need any external stimulation benefited from music. Which one of the following statements is the CORRECT inference of the above passage? (A) Listening to music has no effect on learning and a positive effect on physical exercise. (B) Listening to music has a clear positive effect both on physical exercise and on learning. (C) Listening to music has a clear positive effect on physical exercise. Music has a positive effect on learning only in some students. (D) Listening to music has a clear positive effect on learning in all students. Music has a positive effect only in some students who exercise.

Listening to Music During Exercise Improves Performance

Q.5 Pen : Write :: Knife : _________ Which one of the following options maintains a similar logical relation in the above? (A) Vegetables (B) Sharp (C) Cut (D) Blunt

Pen : Write :: Knife Analogy Solution

Q4 If (x-1/2)2 - (x-3/2)2 = x+2 , then the value of x: (A) 2 (B) 4 (C) 6 (D) 8

Solve (x-1)^2 – (x-2)^2 = 2

Q3 If θ is the angle, in degrees, between the longest diagonal of the cube and one of the edges of the cube, then cos θ = (A) 1/2 (B) 1/√2 (C) √2 / 3 (D) 1 / √3

Cosine of Angle Between Cube Space Diagonal and Edge

Q.2 A transparent square sheet shown above is folded along the dotted line. The folded sheet will look like ________.

Transparent Square Sheet Folded Along Dotted Line

Q.1 Gauri said that she can play the keyboard __________ her sister. (A) as well as (B) as better as (C) as nicest as (D) as worse as

Gauri Keyboard Comparison

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