Category: GATE Life Science 2019 Previous Year Papers with Solutions

Category: GATE Life Science 2019 Previous Year Papers with Solutions

GATE Life Science 2019 Previous Year Papers with Solutions – Complete Preparation Guide

The GATE Life Science (XL) 2019 paper remains a highly valuable resource for aspirants preparing for GATE and other Life Science competitive examinations. Practicing GATE Life Science 2019 previous year papers with detailed solutions helps students understand the conceptual depth, analytical requirements, and question framing style of the exam.

At Let’s Talk Academy, GATE Life Science preparation is strongly centered on previous year question (PYQ)–based learning, ensuring aspirants focus on high-yield and exam-relevant topics only.

Why GATE Life Science 2019 Previous Year Papers Are Important

GATE Life Science is not a rote-learning exam. The 2019 paper tested:

  • Fundamental biological concepts
  • Application-based and analytical thinking
  • Numerical problem-solving ability
  • Interpretation of experimental and theoretical data

By solving GATE Life Science 2019 PYQs, aspirants can:

  • Understand the real difficulty level of the exam
  • Identify frequently asked and repeating concepts
  • Learn the structure of MCQs, MSQs, and NAT questions
  • Improve time management and accuracy
  • Develop the correct GATE exam mindset

Previous year papers act as the most authentic reference point for preparation.

Subjects Covered in GATE Life Science 2019 PYQs

The GATE Life Science 2019 question paper covered all major domains, including:

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

Solving PYQs subject-wise helps strengthen weak areas, while year-wise practice improves exam confidence.

Importance of Solved GATE Life Science 2019 Papers with Solutions

Attempting questions without explanations limits learning. Solved GATE Life Science 2019 previous year papers with detailed solutions help aspirants to:

  • Understand why an answer is correct or incorrect
  • Strengthen concepts directly tested in the exam
  • Learn exam-oriented numerical solving approaches
  • Avoid common conceptual mistakes
  • Retain concepts for long-term application

Each properly analyzed PYQ improves clarity across multiple topics.

Best Websites for GATE Life Science Previous Year Papers with Answer Keys

This is a very common query among aspirants.

Best websites for GATE Life Science previous year papers with answer keys?

For accurate, authentic, and exam-oriented resources, Let’s Talk Academy is the most reliable and recommended platform for GATE Life Science aspirants.

Why Let’s Talk Academy Is the Best Choice

Let’s Talk Academy provides authentic GATE Life Science previous year papers with answer keys and detailed solutions, including the 2019 paper, through:

  • Officially curated PYQ PDFs
  • Topic-wise solved previous year questions
  • Concept-based answer key explanations
  • Video solutions prepared by expert Life Science faculty
  • Structured practice sets aligned with the GATE syllabus

All PYQs are strictly sourced from official GATE papers, ensuring complete accuracy and exam relevance.

Unlike random online sources, Let’s Talk Academy focuses on conceptual explanation rather than just final answers, making PYQs a true learning tool.

How to Use GATE Life Science 2019 PYQs Effectively

To maximize the benefit of previous year papers:

  • Begin with topic-wise PYQs during concept building
  • Progress to full-length GATE 2019 paper practice
  • Analyze every incorrect or guessed question carefully
  • Maintain a PYQ mistake and concept notebook
  • Revise weak topics immediately after analysis

Quality analysis is far more important than solving a large number of questions.

Who Should Practice GATE Life Science 2019 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

PYQs build a strong foundation useful across multiple Life Science competitive exams.

Why Let’s Talk Academy Emphasizes PYQ-Based Preparation

Let’s Talk Academy follows a previous year paper–driven approach because it:

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

This strategy has consistently helped aspirants improve performance and rank.

Final Conclusion

For GATE Life Science 2019, previous year question papers with detailed solutions are indispensable for serious aspirants. When sourced from a trusted and expert-driven platform like Let’s Talk Academy, PYQs become the strongest weapon for GATE preparation.

Make GATE Life Science PYQs the backbone of your strategy, analyze them deeply, and let real exam questions guide your path to success.

Q.105 A genetic cross was made between homozygous wild-type males (a+b+c+) and triple mutant females (a b c) of Drosophila melanogaster. Then the F1 males (a+b+c+) were backcrossed to the triple-mutant females which resulted in the following F2 progenies: Genotype Number a+bc 16 ab+c 115 abc 311 a+b+c 64 abc+ 61 a+b+c+ 317 a+bc+ 99 ab+c+ 17 Total 1000 The order of genes as determined from the above data was found to be "b a c" (note that the order is equivalent to "c a b" and the order of outside markers are arbitrary). The map distance between b and c is ________ centiMorgan (round off to 1 decimal place).

Drosophila Three-Point Cross

104. A distinctly large population of randomly mating laboratory mice contains 𝟑𝟔% albino mice, which is caused by a double recessive genotype (aa). The black coloured mice in the population is due to dominant genotype (AA/Aa). Considering the fact that this population is in Hardy-Weinberg equilibrium, the frequency of heterozygous alleles (Aa) in this population is ____ (round off to 2 decimal places). 

Hardy-Weinberg Equilibrium

Q.103 An enzyme that follows Michaelis-Menten kinetics catalyzes the conversion of 𝟑𝟓𝝁𝐌 substrate into product with a reaction velocity of 𝟏𝟎𝝁𝐌𝐬−𝟏. The 𝑲𝐦 and 𝒌cat for the substrate are 𝟏𝟒𝝁𝐌 and 𝟓𝟎𝟎 𝐬−𝟏 respectively. The total amount of enzyme taken for the enzyme reaction is ____ nM .

Calculate Total Enzyme Amount

Q.102 The molecular weight of double stranded DNA fragment present in mitochondria is 7.9 × 106 g mol-1. A pair of nucleotide contributes 0.34 nm to the length of DNA and the average molecular weight of a nucleotide is 350 g mol-1. The number of protein molecules, consisting of 200 amino acids each coded by this mitochondrial DNA is _____ molecules.

Calculate Number of Proteins

Q101. Match the hormones in Column I to their respective physiological effects in Column II Column I Column II (a) Melatonin (b) Oxytocin (c) Prolactin (d) Somatostatin (1) Inhibition of growth hormone secretion (2) Synthesis of milk in mammary gland (3) Secretion of milk, uterine contraction (4) Regulation of rhythms from pacemaker acid cells Options: (A) 5 → 4; 3 → 1 (B) 3 → 5; 4 → 2; 1 (C) 5 → 1; 2 → 3; 4 (D) 5 → 3; 4 → 2; 1

Match Hormones to Physiological Effects

Q.100 Which one of the following factors does NOT contribute to “allopatric speciation”? (A) The isolated population is exposed to different selection pressure as compared to parent population (B) A population become geographically isolated from the parent population (C) There is a gene flow between the two separated populations (D) Genetic drift occurs between the two separated populations

Allopatric Speciation Factors

Q.99 Closure of neural tube during neurulation is the process in which neural tube eventually forms a closed cylinder that separates from surface ectoderm. Which one of the following terms is linked to the failure in the closure of entire neural tube i.e. brain and spinal cord, over the body axis? (A) Craniorachischisis (B) Spina bifida (C) Anencephaly           (D) Cleft palate

Neural Tube Closure Failure

Q.98 Match the respective behaviour given in Column I with the appropriate explanation in Column II Column I                                                           Column II a) Agonistic behaviour                     1) Individual behaviour where fitness of other increases at the expense of self b) Hierarchical behaviour               2) Individuals contribute towards enhancement of mutual fitness c) Altruistic behaviour                     3) Aggressive behaviour of individuals within a population d) Cooperative behaviour                4) Existence of different levels of dominance within a population (A) a-4; b-1; c-2; d-3                    (B) a-3; b-4; c-1; d-2 (C) a-4; b-2; c-1; d-3                     (D) a-3; b-4; c-2; d-1

Agonistic Behaviour Matching

Q.97 Match the following terms in Column I with the appropriate explanation in Column II Column I                               Column II a) Spliceosome                   1) Cell scavengers b) Peroxisome                    2) Metabolism of long chain fatty acids c) Lysosome                         3) Microtubules d) Centrosome                    4) Post transcriptional modifications (A) a-3; b-1; c-4; d-2            (B) a-4; b-2; c-1; d-3 (C) a-3; b-2; c-1; d-4            (D) a-4; b-3; c-1; d-2

Spliceosome Peroxisome Lysosome Centrosome Functions

Q.96. Match the given vitamins in Column I with the related proteins in Column II Column I                     Column II a) Vitamin C              1) Thrombin b) Vitamin K             2) Rhodopsin c) Vitamin A              3) Pyruvate dehydrogenase d) Vitamin B1            4) Prolyl hydroxylase (A) a-4; b-1; c-2; d-3      (B) a-4; b-1; c-3; d-2 (C) a-2; b-1; c-4; d-3      (D) a-3; b-4; c-2; d-1

Vitamins Proteins Matching

Q.95 In a 1500 base pair double stranded B-DNA, 70% of purines are adenine bases. The total number of hydrogen bonds (Watson–Crick base pairing) in the DNA will be _____

Total Hydrogen Bonds

Q.94 The total number of complementary determining regions (CDRs) in the functional form of an IgM antibody is _____

Total Number of CDRs in Functional IgM Antibody

Q.93 Consider the given peptide, Ala-Glu-Val-Asn-Ile-Asp-Pro-Asp-Gln-Gly-Asp. The net charge on the peptide at pH 1.0 will be____

Net Charge on Ala-Glu-Val-Asn-Ile-Asp-Pro-Asp-Gln-Gly-Asp Peptide

Q.92 Which one of the following animals is named as a “living fossil”, where the animal is persisting above 400 million years without further major morphological evolution? (A) King crabs (B) Porcelain crabs (C) Horseshoe crabs (D) Hermit crabs

Living Fossil Horseshoe Crab

Q.91 The finches on the Galapagos Islands have widely differing beaks adapted according to their diets. This is an example of (A) Adaptive radiation (B) Parallel evolution (C) Adaptive convergence (D) Co-adaptation

Galapagos Finches Adaptive Radiation

Q.90 The phylum that is characterized by the animals having a “water vascular system” is (A) Cnidaria (B) Annelida (C) Mollusca (D) Echinodermeta

Phylum with Water Vascular System

Q.89 A species adapted to a specific habitat is called (A) Biome (B) Ecotone (C) Ecotype (D) Niche

Species Adapted to Specific Habitat

Q.88 Which one of the following parasites does NOT cause lymphatic filariasis? (A) Brugia malayi (B) Brugia timori (C) Wuchereria bancrofti (D) Mansonella streptocerca

Which Parasite Does NOT Cause Lymphatic Filariasis?

Q.87 In eukaryotic chromatin organization, which one of the histones seals off the nucleosome at the location at which linker DNA enters and leaves the nucleosome? (A) H1 (B) H2A-H2B (C) H3 (D) H4

Histone Seals Off Nucleosome

Q.86 Which one of the following animals possesses two chambered heart? (A) Chameleon (B) Peacock (C) Gold fish (D) Blue whale

Which Animal Possesses Two Chambered Heart?

Q.85 When grouping bacteria using numerical taxonomy approach, many characteristics are determined for each strain. If all characteristics are same, then the % similarity is 100. When comparing two bacterial strains, if 72 characteristics are the same and 44 characteristics are different, the % similarity (round off to 2 decimal places) between the two strains is _____.

Bacterial Strains % Similarity in Numerical Taxonomy

Q.84 Consider a spherical bacterial cell with diameter 2 μm and a eukaryotic cell of spherical shape with a diameter of 20 μm. The surface area to volume (in μm-1) ratio of the bacterial cell (round off to 1 decimal place) is ______ times more than that of the eukaryotic cell.

Surface Area to Volume Ratio

Q.83 Match the diseases with the appropriate causative agents Disease                                                             Causative agent (i) Rheumatic heart disease                        (p) Candida albicans (ii) Syphilis                                                     (q) Group A Streptococcus (iii) Genital warts                                          (r) Human papilloma virus (iv) Oral thrush                                              (s) Treponema pallidum (A) (i)-(p), (ii)-(q), (iii) –(s), (iv)-(r)            (B) (i)- (q), (ii)- (s), (iii)-(r), (iv)-(p) (C) (i)-(r), (ii)-(s), (iii)-(p), (iv)-(q)               (D) (i)-(s), (ii)-(q), (iii)- (p), (iv) –(r)

Match Diseases with Causative Agents

Q.82 Match the class of antibodies to appropriate description or function Antibodies                              Description / function (i) IgA                                     (p) Mucosal immunity (ii) IgG                                    (q) Immunity to parasitic infections (iii) IgD                                   (r) Most abundant antibody in the blood (iv) IgE                                    (s) Found on the B cell membranes (A) (i)-(s), (ii)-(r), (iii)-(q), (iv)-(p)                    (B) (i)-(p), (ii)-(q), (iii)-(s), (iv)-(r) (C) (i)-(p), (ii)-(r), (iii)-(s), (iv)-(q)                   (D) (i)-(p), (ii)-(s), (iii)-(q), (iv)-(r)

Match Antibodies to Functions

Q.81 Which of the following processes are involved in horizontal gene transfer (HGT)? (i) Conjugation (ii) Transduction (iii) Binary fission (iv) Transformation (v) Translesion synthesis (vi) Transversion (A) (i), (iii), (iv) (B) (i), (iv), (vi) (C) (ii), (iv), (v) (D) (i), (ii), (iv)

Horizontal Gene Transfer Processes

Q.80 Among the following characteristics listed below, select those that are the most appropriate for archaebacteria. (i) Peptidoglycan containing muramic acid and D-amino acids are present in cell wall (ii) Long chain branched alcohols (phytanols) bound to glycerol by ether linkages are found in the lipids of cytoplasmic membrane (iii) First amino acid to initiate new polypeptide chain is N-Formylmethionine (iv) Translation process is sensitive to the action of diphtheria toxin (A) (i), (iii) (B) (ii), (iii), (iv) (C) (ii), (iv) (D) (i), (ii), (iv)

Archaebacteria Characteristics

Q.79 Match the culture media to their uses Culture Media                                  Uses (i) Complex media                     (p) identifying certain bacteria from mixed culture (ii) Selective media                    (q) chemically undefined media for growing bacteria (iii) Differential media              (r) cultivating fastidious bacteria (iv) Enriched media                   (s) distinguishing one bacteria from another (A) (i)-(s), (ii)-(r), (iii)-(q), (iv)-(p)                  (B) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(r) (C) (i)-(q), (ii)-(r), (iii)-(p), (iv)-(s)                  (D) (i)-(p), (ii)-(s), (iii)-(q), (iv)-(r)

Culture Media Types in Microbiology

Q.78 Granules are inclusion bodies found in bacteria. Which of the following statements are true about granules? (i) Granules can be storage vessels for carbohydrates/lipids (ii) Some granules supply substances for a variety of metabolic processes (iii) Granules are dissolved in cytoplasm (iv) Granules are membrane-enclosed structures (A) (ii) and (iii) (B) (iii) and (iv) (C) (i) and (ii) (D) (i) and (iii)

Bacterial Granules Inclusion Bodies

Q.77 Which of the following cellular structures are absent in eukaryotes and are present at least in some prokaryotic organisms? (i) Respiratory enzymes (ii) Ribosomes (iii) Nucleoid (iv) Internal membranes (v) Pili (A) (iii) and (v)         (B) (i) and (iv) (C) (i), (iii) and (v)   (D) (iii), (iv)

Cellular Structures

Q.76 Folic acid synthesis in bacteria is competitively inhibited by sulfonamides. Which of the following essential components is replaced by sulfonamides during the process of folic acid synthesis? (A) Pteridine                            (B) Glutamic acid (C) Para-amino benzoic acid (D) Sulfamethaxazole

Folic Acid Synthesis in Bacteria

Q.75 A bacterial culture contains 500 organisms / mL in the exponential growth phase at 8 am in the morning. If you consider a generation time of 20 minutes, the total number of organisms / mL (round off to the nearest integer) after three hours (i.e. at 11 am in the morning) will be _______.

Bacterial Exponential Growth Calculation

Q.74 What is the enzyme involved in the following reaction? Triglyceride + 3H2O -----------> Glycerol + fatty acids (A) Glycerol phosphate dehydrogenase (B) Glycerol kinase (C) Lipase (D) Zymase

Enzyme in Triglyceride Hydrolysis

Q.73 Which of the following is a symbiotic bacterial genus that converts nitrogen into nitrogen containing compounds that are utilized by legume plants to synthesize amino acids? (A) Spirillum (B) Azotobacter (C) Rhizobium (D) Clostridium

Symbiotic Nitrogen Fixing Bacteria

Q.72 Dick test is used to assess the susceptibility to______________ . (A) diphtheria (B) scarlet fever (C) typhoid fever (D) tuberculosis

Dick Test Susceptibility Scarlet Fever

Q.71 Secondary metabolites such as penicillin from mold are produced during the _______ . (A) lag phase (B) idiophase (C) log phase (D) decline phase

Secondary Metabolites Penicillin Production Idiophase

Q.70 The highly contagious viral disease measles is caused by a (A) double-stranded DNA virus (B) single-stranded DNA virus (C) single-stranded RNA virus (D) double-stranded RNA virus

Single-Stranded RNA Virus Causing Highly Contagious Disease

Q.69 Microbes make up about ___________ of earth’s living material by weight. (A) 1/4th (B) 2/3rd (C) 3/4th (D) 4/5th

Microbes Make Up About What Fraction of Earth’s Living Material

Q.68 Which part of the lipopolysaccharides has toxic properties that make some bacterial infections a potentially serious medical problem? (A) Polysaccharide side-chains    (B) Lipid A portion (C) Repeat carbohydrate units      (D) Core region of polysaccharides

Lipopolysaccharides Toxic Properties

Q.67 Which of the following radiation methods is used in preserving food from spoilage by microorganisms? (A) Radio waves (B) Microwaves (C) Non-ionizing (D) Ionizing

Ionizing Radiation in Food Preservation

Q.66 Chloramphenicol and tetracycline are broad spectrum antibiotics which inhibit bacterial growth by targeting _________. (A) cell wall synthesis      (B) protein synthesis (C) RNA synthesis           (D) DNA synthesis

Chloramphenicol and Tetracycline Broad Spectrum Antibiotics

Q.65 A cell in G1 of Interphase has 12 chromosomes. In Anaphase-I of meiosis, the number of DNA molecules per cell will be__________

Number of DNA Molecules in Anaphase I Meiosis

Q.64 Plant weight is determined by a pair of alleles at each of the two independently assorting loci (Aa and Bb) that are additive and equal in their effects. The recessive alleles do not contribute towards plant weight. Plants with genotype aa bb are 1 g in weight, whereas plants with genotype AA BB weigh 3.4 g. Plant with genotype aa bb is crossed with a plant of genotype AA BB. The weight (in g, round off to one decimal place) of an individual plant in F1 progeny of this cross would be_________

Plant Weight Genetics

Q.63 Which one of the following methods is INCORRECT with respect to haploid plant production? It can be produced (A) from an unfertilized egg cell (B) from nucellar tissue (C) from isolated pollen culture (D) by distant hybridization, followed by selective elimination of chromosomes of one of the parents

Haploid Plant Production Methods

Q.62 Which one of the following statements is CORRECT with respect to endosperm development? It originates (A) from the fusion product of three haploid nuclei - one from male gametophyte and two from the female gametophyte (B) from the fusion product of three haploid nuclei - two from male gametophyte and one from the female gametophyte (C) from the fusion product of two haploid nuclei - one from male gametophyte and one from the female gametophyte (D) by a phenomenon called apomixis

Endosperm Development Origin Fusion Nuclei

Q.61 Match the options in Group I with that of Group II with respect to steps in signal transduction mechanism in plants Group I                                                          Group II P. Phospholipase C                                 1. stimulates release of calcium from intracellular stores Q. Inositol triphosphate                        2. regulates ion channels/activates various enzymes R. Diacylglycerol                                     3. hydrolyzes posphatidylinositol bisphosphate (PIP2) S. Phosphatidic acid                              4. phosphorylated to phosphatidic acid (A) P-3, Q-4, R-2, S-1               (B) P-4, Q-3, R-2, S-1 (C) P-2, Q-3, R-1, S-4               (D) P-3, Q-1, R-4, S-2

 Signal Transduction in Plants

Q.60 Match the disease with causative organism and affected crop Disease Causative organism Crop P. Powdery mildew 1. Albugo candida i. Corn Q. White rust 2. Phytophthora infestans ii. Tomato R. Downy mildew 3. Erysiphe orontii iii. Potato S. Late blight 4. Peronoscleropora phillippinensis iv. Mustard (A) P-3-i, Q-1-iv, R-2-iii, S-4-ii (B) P-3-ii, Q-1-iv, R-4-i, S-2-iii (C) P-4-ii, Q-1-i, R-3-iv, S-2-iii (D) P-2-iii, Q-3-ii, R-1-iv, S-4-i

Powdery Mildew White Rust Downy Mildew Late Blight Crops

Q.59 Cybrids may arise through the (A) fusion of a normal protoplast with another nucleated protoplast of different origin (B) fusion between a normal protoplast and a protoplast containing viable nucleus (C) elimination of one of the nuclei from heterokaryon formed from two protoplasts of different origin (D) fusion of a normal cell with another nucleated cell

 Cybrids in Protoplast Fusion

Q.58 Match the plant species and their corresponding families with their economically important products: Plant species                                               Family                                          Plant product P. Cannabis sativa                                      1. Lamiaceae                               i. Oil from seeds Q. Corchorus olitorius                               2. Cannabaceae                          ii. Eugenol from leaves R. Ocimum sanctum                                   3. Euphorbiaceae                      iii. Fibre from stem S. Ricinus communis                                   4. Tiliaceae                                 iv. Marijuana from leaves (A) P-2-iv, Q-3-ii, R-4-iii, S-1-i                    (B) P-3-iv, Q-2-i, R-1-ii, S-4-iii (C) P-1-iii, Q-4-iii, R-2-i, S-3-ii                    (D) P-2-iv, Q-4-iii, R-1-ii, S-3-i

Plant Species Families Economically Important Products

Q.57 Select correct combination of molecules given in Group I with their functions in Group II Group I                                        Group II P. Ubiquitin                        1. Cell wall loosening enzymes Q. Phytotropins                 2. Repressors in gibberellin signaling R. Extensins                       3. Mediator of protein degradation S. DELLA proteins            4. Noncompetitive inhibitors of polar transport of auxin (A) P-3, Q-4, R-1, S-2      (B) P-3, Q-1, R-2, S-4 (C) P-4, Q-3, R-1, S-2       (D) P-1, Q-2, R-4, S-3

Ubiquitin DELLA Proteins Phytotropins Extensins Functions

Q.56 Rubisco catalyzes conversion of ribulose 1,5-bisphosphate to P. two molecules of stable 3-phosphoglycerate Q. one molecule of 3-phosphoglycerate and one molecule of phosphoglycolate R. one molecule of ribulose 5-phosphophate S. one molecule of ribose 5-phosphate Choose only the correct combination (A) P and S (B) Q and S (C) P and Q (D) R and S

Rubisco Catalyzes Conversion

Q.55 If a species has 2n = 16 chromosomes, the number of chromosomes per cell in a double monosomic individual would be_________

 Double Monosomic Chromosomes

Q.54 Glyphosate is a broad spectrum herbicide. Upon application to leaves, it is translocated to meristematic areas and underground rhizomes by (A) phloem (B) xylem (C) border pits (D) tracheids

Glyphosate Translocation in Plants

Q.53 The plant cells are considered totipotent. The phenomenon of a mature cell reverting to the meristematic state and forming undifferentiated callus tissue is called (A) redifferentiation (B) dedifferentiation (C) organogenesis (D) recalcitrancy

 Plant Cells Totipotent

Q.52 Which one of the following statements is INCORRECT with respect to cyclic photophosphorylation? (A) ATP is generated without concomitant formation of NADPH (B) Electron flows from Photosystem I to cytochrome bf complex (C) Photosystem II does not participate in cyclic photophosphorylation (D) Cyclic photophosphorylation occurs when NADP+/NADPH ratio is high

 Cyclic Photophosphorylation

Q.51 Which one of the following statements is INCORRECT? (A) Nitrogen fixation is aerobic process (B) Dinitrogenase catalyzes reduction of nitrogen to ammonia (C) Root nodules are found in Glycine max (D) Nitrogen fixation is anaerobic process

Nitrogen Fixation Is Anaerobic or Aerobic?

Q. 50 Which one of the following combinations of polysaccharides is present in plant cell wall? (A) Only cellulose, hemicellulose and fibroin (B) Only cellulose, hemicellulose and lignin (C) Only cellulose, hemicellulose and pectin (D) Only cellulose, pectin and lignin

Polysaccharides in Plant Cell Wall

Q. 49 Arbuscules are highly branched structures, formed by (A) ectomycorrhizae (B) endomycorrhizae (C) arbutoid mycorrhizae (D) monotropoid mycorrhizae

Arbuscules in Mycorrhizae

Q.48 Which one of the following families shows bicollateral vascular bundle in the transverse section of the stem? (A) Rutaceae (B) Asteraceae (C) Cucurbitaceae (D) Malvaceae

Bicollateral Vascular Bundles

Q.47 Raphides are deposits of (A) calcium oxalate (B) calcium carbonate (C) silica bodies (D) protein bodies 

Raphides Are Deposits of Calcium Oxalate

Q.46 Which one of the following ecosystems is represented by an inverted pyramid of numbers? (A) Grassland (B) Pond (C) Desert (D) Parasitic forest

Inverted Pyramid of Numbers

Q.45. An uncharged protein (P) has an Asp in position 23 with a molecular weight of 6501 Da , as determined by mass spectrometry. The uncharged mutant of this protein ( 𝐏′) contains a single amino acid substitution with Asn at position 23. The molecular weight of 𝐏′, as determined by mass spectrometry (rounded off to one decimal place) is ____ Da.

Asp to Asn Mutation

Q.44 X different sized DNA fragments can be observed upon incomplete EcoRI digestion of identical DNA molecules with two EcoRI sites as shown below. The maximum value of X is ____ . 

How Many Different Sized DNA Fragments?

Q.43 The C-terminal carboxyl group and the N-terminal amino group in amino acids have a dissociation constant ( 𝐩𝐊𝐚 ) of 2.2 and 9.2 , respectively. The 𝐩𝐊𝐚 of side chain carboxyl group in glutamic acid is 4.2 and side chain amino group in lysine is 10.2 . The difference in isoelectric point (pI) of lysine and glutamic acid (rounded off to two decimal places) is ____ .

Isoelectric Point Difference

Q.42. Absence of detectable protein expression upon blunt-ended mutation-free cloning of an E.coli gene with its own promoter in E. coli cells can be due to (A)Cloning occurred in reverse orientation (B) Cloning occurred out of frame (C) Codon bias (D) Rapid degradation of expressed protein

Blunt-Ended Cloning Failure

Q.41 The crystal structure of a peptide has an ordered structural repeat of amino acids with a distance of ∼ 𝟔. 𝟓Å between the alternate 𝐂𝜶 atoms. Which one of the following pair of dihedral angles ( 𝚽 and 𝚿 ) accurately represents the peptide structure? (A) Φ ≈ −60∘,Ψ ≈ −50∘ (B) Φ ≈ −120∘, Ψ ≈ −50∘ (C) Φ ≈ −60∘, Ψ ≈ +120∘ (D) Φ ≈ −120∘, Ψ ≈ +120∘

Unraveling Peptide Crystal Structures

Q. 40 Which one of the following is an incorrect biomolecule-modification pair? (A)Lipid - Palmitoylation (B) DNA and Protein - Methylation (C) Protein - Glycosylation (D) RNA - Polyadenylation

Incorrect Biomolecule-Modification Pair

Q. 39 DpnI is used to digest the PCR product during site directed mutagenesis because (A) DpnI digests irrespective of methylation status of DNA (B) DpnI digests only unmethylated DNA (C) DpnI digests only methylated DNA (D) DpnI digests GC-rich sequences

DpnI in Site-Directed Mutagenesis

Q.38 The enzyme NOT involved in oxidation of the molecule shown below is (A)Δ3, Δ2- Enoyl –CoA isomerase (B) Propionyl-CoA caboxylase (C) Acyl CoA dehydrogenase (D)Enoyl CoA hydratase

Enzyme NOT Involved in Oxidation

Q.37. The ascending order of half-life for the radioactive isotopes, 125I, 3H, 14C and 32P, is: (A) 14C < 125I < 3H < 32P (B) 32P < 3H < 125I < 14C (C) 14C < 3H < 32P < 125I (D) 32P < 125I < 3H < 14C

Ascending Order of Half-Life

Q.36 A mixture of the following purified proteins, IgG, IgM and Fab fragment of immunoglobulins, is separated using gel filtration chromatography. The order of elution of these proteins (first to last) is (A)Fab fragment, IgM and IgG (B) IgM, IgG and Fab fragment (C) Fab fragment, IgG and IgM (D) IgG, IgM and Fab fragment

Gel Filtration Chromatography

Q.35 You are given a 0.1M solution of Glucose (stock solution). The stock solution required to make 0.5 ml of 0.005M Glucose solution (rounded off to three decimal places) in ml is _____.

How Much 0.1M Glucose Stock Solution

Q.34 The electrostatic interaction energy between a positively charged atom A and negatively charged atom B separated by 3 Å in water is -6 kJ/mol. Considering the relative permittivity of water to be 80, the electrostatic interaction energy in kJ/mol (rounded off to one decimal place) between atoms A and B in vacuum is _____.

Electrostatic Interaction Energy

Q.33 Allergies are due to a hyper immune response. Drugs given to counter allergies target (A)Glycine (B) Histamine (C) Insulin (D)Cellulose

Allergies Hyper Immune Response

Q.32 Nitric oxide synthase is responsible for generation of Nitric oxide, an important signaling molecule. The substrate for this enzyme is (A)Glycine (B) Lysine (C) Histidine (D)Arginine

Nitric Oxide Synthase Substrate

Q.31 Slow intravenous infusion of ethanol is a therapy to treat methanol poisoning. The underlying chemical reaction is an example of (A)Competitive inhibition (B) Non-competitive inhibition (C) Mixed inhibition (D)Enzyme activation

Ethanol in Methanol Poisoning

Q.30 The light-harvesting pigment NOT used by Cyanobacteria for photosynthesis is (A)Rhodopsin (B) Phycobilin (C) Phycoerythrobilin (D)Phycocyanobilin

Light-Harvesting Pigment Not Used by Cyanobacteria

Q.29 The dipeptide with least rotational barrier in the peptide bond is

Dipeptide with Least Rotational Barrier

Q.28 The example of substrate level phosphorylation in glycolysis is (A)Conversion of Glucose to Glucose-6-phosphate (B) Conversion of Glyceraldehyde-3-phosphate to 1,3-Bisphosphoglycerate (C) Conversion of 1,3-Bisphosphoglycerase to 3-Phosphoglycerate (D)Conversion of Dihydroxyacetone phosphate to Glyceraldehyde-3-phosphate

Substrate Level Phosphorylation in Glycolysis

Q.27 RAG recombinase is responsible for the formation of specific immune receptors. This process occurs in (A)T cells & B cells (B) Natural killer cells (C) Macrophages (D)Neutrophils

RAG Recombinase in T Cells and B Cells

Q.26 Catalase is found exclusively in (A)Lysosomes (B) Golgi apparatus (C) Peroxisomes (D) Mitochondria

Catalase Enzyme Location

Q.25 Given the standard reduction potentials, ECu2+/Cu0 = 0.34 V and EAg+/Ag0 = 0.80 V, the standard free energy change (ΔG0) for the reaction Cu (s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag (s) in kJ mol-1 (rounded off to one decimal place; F = 96500 C mol-1), is -88.8.

Standard Free Energy Change

Q.24 Based on molecular orbital theory, the number of paramagnetic species in the following list N2, N2+, N22-, O2, O2+, O2- and O22- is ____ (assume that there is no change in energy of the orbitals upon addition/removal of electrons in a molecule)

Number of Paramagnetic Species

Q.23 The pH of a 1.0 L buffer solution containing 0.2 mol of acetic acid (CH3COOH) and 0.3 mol of sodium acetate (CH3COONa) is 5.0. The Ka of acetic acid is P × 10-5, where the numerical value of P (rounded off to one decimal place) is ____

Calculating Ka for Acetic Acid Buffer

Q.22 A protein in denatured state (D) is in equilibrium with native state (N). At 360 K, both N and D states are equally populated. If the standard entropy change for the reaction at this temperature ΔS0 = -139 J K-1 mol-1, then the corresponding standard enthalpy change ΔH0 for the reaction in kJ mol-1 (rounded off to one decimal place) is ____

Protein Denaturation Thermodynamics

Q.21 A coordination complex Y upon reaction with AgNO3 solution does NOT give any precipitation. Complex Y possesses two isomers, of which one has zero dipole moment. The crystal field stabilization energy of Y is either -0.8 Δ0  or -0.8 Δt. The magnetic moment for Y is found to be 3.9 Bohr Magneton. The coordination complex Y is (A) [Ti(NH3)4(Cl)2] (B) [Co(NH3)4(H2O)2]Cl2 (C) [Co(NH3)2(Cl)2] (D) [Co(NH3)4(Cl)2]

[Co(NH3)4(H2O)2]Cl2 Coordination Complex

Q. 20 The most stable coordination complexes 𝑷 and 𝑸 formed in the following reaction are 𝐅𝐞𝟑+ + 𝐇𝐠𝟐+ + 𝐒𝐂𝐍− ⟶ 𝐏 +Q

Most Stable Coordination Complexes Formed

Q. 19 The major product formed in the following reaction is

Complete E2 Elimination Explanation

Q. 18 The reactants P and Q in the following reaction are

Complete Mechanism Based Explanation

Q. 17 Match the type of reaction in Group-1 with the most appropriate description in Group-2 Group 1                                                               Group 2 P Hydroboration-oxidation                           1 Electrophilic aromatic substitution Q Nucleophilic aromatic substitution        2 Oxaphosphetane intermediate R Wittig reaction                                            3 Meisenheimer complex S Friedel-Crafts reaction                              4 Anti-Markownikoff's product (A) P-2, Q-4, R-1, S-3 (B) P-4, Q-3, R-1, S-2 (C) P-4, Q-3, R-2, S-1 (D) P-2, Q-1, R-4, S-3

Hydroboration Wittig Friedel-Crafts Reaction Matching Solved

Q.16 For the reaction mechanism, 2X ⇌ Y For this step, assume Keq = [Y] / [X]2 Y → P : rate constant for this step The rate law is: (A) d[P]/dt = Keq[Y] (B) d[P]/dt = k[X]2 (C) d[P]/dt = kKeq[Y] (D) d[P]/dt = kKeq[X]2

Rate Law Determination Using Pre-Equilibrium Approximation

Q. 15 The total number of stereoisomers possible for the following structure is ____

Total Number of Stereoisomers

Q. 14 The total number of tautomers for the following molecule (including the structure provided below) is ____ .

Total Number of Tautomers

Q. 13 According to VSEPR theory, the species that has the smallest F-X-F angle (where 𝐗 = central atom)is (A) BF3 (B) PF3 (C) BF4− (D) IF4−

Smallest F–X–F Bond Angle by VSEPR Theory

Q. 12 The boiling point of halogens from 𝐅𝟐 to 𝐈𝟐 increases due to (A) decrease in electron affinity (B) decrease in ionization potential (C) dipole-dipole interaction (D) induced dipole - induced dipole interaction

Why Boiling Point of Halogens Increases from F₂ to I₂

Q.11. The compound that provides a carboxylic acid, upon treatment with 𝐁𝐫𝟐/𝐍𝐚𝐎𝐇 followed by acidification, is.

Compound Giving Carboxylic Acid with Br₂/NaOH

Q. 10 X is an online media provider. By offering unlimited and exclusive online content at attractive prices for a loyalty membership, X is almost forcing its customers towards its loyalty membership. If its loyalty membership continues to grow at its current rate, within the next eight years more households will be watching X than cable television. Which one of the following statements can be inferred from the above paragraph? (A)Most households that subscribe to X's loyalty membership discontinue watching cable television (B) Non-members prefer to watch cable television (C) Cable television operators don't subscribe to X's loyalty membership (D) The X is cancelling accounts of non-members

X Online Media Provider Loyalty Inference Question

Q. 9 Mola is a digital platform for taxis in a city. It offers three types of rides - Pool, Mini and Prime. The Table below presents the number of rides for the past four months. The platform earns one US dollar per ride. What is the percentage share of revenue contributed by Prime to the total revenues of Mola, for the entire duration? Type            Month                      January February  March    April Pool                  170          320        215        190 Mini                 110           220         180        70 Prime               75             180        120         90 (A) 16.24 (B) 23.97 (C) 25.86 (D) 38.74

Mola Taxi Rides Revenue Share

Q. 8 While teaching a creative writing class in India, I was surprised at receiving stories from the students that were all set in distant places: in the American West with cowboys and in Manhattan penthouses with clinking ice cubes. This was, till an eminent Caribbean writer gave the writers in the once-colonised countries the confidence to see the shabby lives around them as worthy of being "told". The writer of this passage is surprised by the creative writing assignments of his students, because ____ . (A) Some of the students had written stories set in foreign places (B) None of the students had written stories set in India (C) None of the students had written about ice cubes and cowboys (D) Some of the students had written about ice cubes and cowboys

Creative Writing Class in India

Q. 7 Two pipes 𝑷 and 𝑸 can fill a tank in 6 hours and 9 hours respectively, while a third pipe 𝑹 can empty the tank in 12 hours. Initially, P and R are open for 4 hours. Then P is closed and Q is opened. After 6 more hours R is closed. The total time taken to fill the tank (in hours) is ____ . (A) 13.50 (B) 14.50 (C) 15.50

Solving Tank Filling Pipes Problem

Q. 6 Fiscal deficit was 𝟒% of the GDP in 2015 and that increased to 𝟓% in 2016. If the GDP increased by 𝟏𝟎% from 2015 to 2016, the percentage increase in the actual fiscal deficit is____ . (A) 37.50 (B) 35.70 (C) 25.00 (D) 10.00

Fiscal Deficit Increased

Q. 5 Are there enough seats here? There are ____ people here than I expected. (A) many (B) most (C) least (D) more

Solve the Grammar Puzzle

Q. 4 The product of three integers 𝐗, 𝐘 and Z is 𝟏𝟗𝟐. 𝐙 is equal to 4 and P is equal to the average of X and Y. What is the minimum possible value of P? (A) 6 (B) 7 (C) 8 (D) 9.5

Minimum Possible Value of P When XYZ=192

Q. 3 If 𝐈𝐌𝐇𝐎 = 𝐉𝐍𝐈𝐏;𝐈𝐃𝐊 = 𝐉𝐄𝐋; and 𝐒𝐎 = 𝐓𝐏, then 𝐈𝐃𝐂 =____ . (A) JDE (B) JED (C) JDC (D) JCD

If IMHO=JNIP, IDK=JEL, SO=TP Then IDC=?

Q. 2 A final examination is the ____ of a series of evaluations that a student has to go through. (A) culmination (B) consultation (C) desperation (D) insinuation

Final Examination is the Culmination of Evaluations

Q. 1 Once the team of analysts identify the problem, we ____ in a better position to comment on the issue. Which one of the following choices CANNOT fill the given blank? (A) will be (B) were to be (C) are going to be (D) might be

Which One Cannot Fill the Blank

Latest Courses

IIT JAM / CUET- PG

IIT JAM / CUET- PG

Are you aspiring to crack the IIT JAM Biotechnology exam with flying colors? Choosing the right coaching institute is crucial for your success. Let's Talk Academy (LTA) has established itself as the best coaching institute for IIT JAM Biotechnology, helping students achieve their dreams through expert guidance, structured study material, and personalized mentorship.

CSIR UGC NET Life Science Course

CSIR UGC NET Life Science Course

Let's Talk Academy is the number one CSIR NET Life Science coaching in Jaipur for Life Science. At Let's Talk Academy, you can be sure to get the right coaching strictly in adherence to the CSIR NET Life Science course.  CSIR NET Life Science is an exam that requires in-depth knowledge, consistent practice, and the right guidance to crack. If you are looking for a place where you can get the best CSIR NET Life Science coaching, you are at the right place.

ICMR JRF Life Science

ICMR JRF Life Science

If you are serious about cracking the ICMR Life Science JRF exam, Let’s Talk Academy is your ultimate destination. Join now and take the first step toward a successful research career!