Category: GATE Life Science 2023 Previous Year Papers with Solutions

GATE Life Science 2023 Previous Year Papers with Solutions – Complete Guide for Serious Aspirants

The GATE Life Science (XL) 2023 paper clearly demonstrated that success in GATE depends on conceptual clarity, interdisciplinary understanding, and rigorous practice of previous year questions. For aspirants aiming to strengthen their preparation, GATE Life Science 2023 previous year papers with detailed solutions are among the most reliable and exam-relevant resources available.

At Let’s Talk Academy, PYQ-based learning forms the backbone of GATE Life Science preparation, ensuring students focus only on what truly matters for the exam.

Why GATE Life Science 2023 Previous Year Papers Are Important

GATE Life Science questions are designed to test:

Core biological concepts
Application-based problem-solving
Numerical and analytical ability
Data interpretation and scientific reasoning

By solving GATE Life Science 2023 PYQs, aspirants can:

Understand the actual exam difficulty level
Identify high-frequency and repeated concepts
Learn the pattern of MSQs, NATs, and numerical questions
Improve time management and accuracy
Build confidence by practicing real GATE-level questions

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

Subjects Covered in GATE Life Science 2023 PYQs

The 2023 paper included questions from all major Life Science domains, including:

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

Solving subject-wise PYQs helps in targeted preparation, while full-length papers simulate the real exam environment.

Importance of Solved GATE Life Science PYQs with Detailed Explanations

Merely attempting questions is not enough. Detailed, concept-oriented solutions are essential to extract full learning value.

Solved PYQs with explanations help aspirants:

Understand why an option is correct or incorrect
Revise linked concepts instantly
Learn exam-oriented numerical solving techniques
Avoid common conceptual traps
Improve long-term retention of concepts

One well-analyzed PYQ can reinforce multiple chapters simultaneously.

Where to Get Solved GATE Life Science Previous Year Papers with Detailed Explanations?

This is one of the most frequently asked questions by GATE aspirants.

Where to get solved GATE Life Science previous year papers with detailed explanations?

Below are the most trusted and effective sources:

1. Let’s Talk Academy Solved PYQ Resources

Let’s Talk Academy provides expert-curated GATE Life Science PYQs with detailed explanations, including:

Topic-wise solved PYQ PDFs
Concept-based video solutions
Step-by-step numerical problem explanations
Exam-oriented discussion of tricky questions

All solutions are prepared strictly from official GATE question papers, ensuring authenticity and accuracy.

2. Official GATE Question Papers (Base Resource)

The official GATE website releases original question papers every year. While these papers are 100% authentic, they usually do not include detailed solutions, making them ideal only when supplemented with expert explanations.

3. Reputed Educational Platforms

Some established academic platforms provide solved GATE Life Science PYQs in PDF or online formats. However, aspirants should always verify the quality and correctness of explanations before relying on them.

4. Self-Study with Standard Textbooks

For disciplined self-study aspirants:

Download official GATE papers
Solve them under exam conditions
Verify answers using standard reference books
Note repeated concepts and revise weak areas

This method works best when combined with structured guidance.

How to Use GATE Life Science 2023 PYQs Effectively

To maximize benefit from previous year papers:

Begin with topic-wise PYQs during concept building
Progress to full-length 2023 paper practice
Analyze every incorrect or guessed question
Maintain a PYQ error and concept notebook
Revisit difficult questions after revision

Effective analysis matters more than the number of questions solved.

Who Should Practice GATE Life Science 2023 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-focused candidates

GATE PYQs strengthen fundamentals applicable across multiple national-level exams.

Why Let’s Talk Academy Recommends PYQ-Centric Preparation

Based on years of mentoring Life Science aspirants, Let’s Talk Academy strongly advocates PYQ-first learning because it:

Filters out irrelevant syllabus content
Improves conceptual clarity
Enhances exam confidence
Aligns preparation with real GATE expectations

This approach consistently delivers measurable improvement in accuracy and performance.

Final Words

For GATE Life Science 2023, previous year question papers with detailed explanations are not optional—they are essential. Knowing where to get authentic solved PYQs and how to analyze them correctly can significantly improve your rank.

Whether you are preparing through coaching or self-study, make GATE Life Science PYQs the foundation of your preparation strategy and let real exam questions guide your success.

Q103 The length of a double helical DNA molecule is 13.6 km. If the DNA double helix weighs 1 × 10-18 g per 1000 nucleotide pairs and 3.4 Å per base pair, then weight of the double helical DNA molecule (in nanogram) will be ____ (in integer).

Length of Double Helical DNA 13.6 km

Drosophila Three Point Cross

Q.101 Which of the following statements is/are TRUE for Colchicine? (A) It binds to tubulin molecule and disrupts the assembly/polymerization of microtubule. (B) It inhibits crossover of chromosomes during meiosis. (C) It inhibits chromosome condensation during Prophase. (D) It blocks mitotic cells in Metaphase.

Colchicine Mechanism – Microtubule Disruption and Mitotic Arrest

Q.100 Among the following statements, which is/are TRUE regarding the replication of DNA? (A) Replication is bidirectional and conservative in nature. (B) Replication in eukaryotes takes place at multiple Ori sites simultaneously. (C) Both the strands replicate in discontinuous manner. (D) One strand replicates in continuous while the other replicates in discontinuous manner.

DNA Replication in Eukaryotes

$Q.99 Which of the following techniques is/are used for determining the three-dimensional structure of proteins? (A) Cryo-electron Microscopy (B) Circular Dichroism (C) Nuclear Magnetic Resonance Spectroscopy (D) X-ray Diffraction$

Techniques for Determining the Three-Dimensional Structure of Proteins

Q.98 The presence of excess glucose has been known to prevent the induction of lac operon as well as other operon controlling enzymes involved in carbohydrate metabolism in E. coli. Which of the following processes define(s) the phenomenon? (A) Catabolite repression (B) Attenuation (C) Glucose effect (D) Feedback inhibition

Excess Glucose Prevents Lac Operon Induction in E. coli

Q97. Match the standard stated cofactors in Column I with their respective enzymes in Column II Column I Column II P. Cu2+ (i) Nitratenase Q. Se (ii) Cytochrome oxidase R. Ni2+ (iii) Pyruvate kinase S. K+ (iv) Glutathione peroxidase T. Mo (v) Urease (A) P-(i); Q-(ii); R-(iii); S-(iv); T-(v) (B) P-(ii); Q-(iv); R-(i); S-(iii); T-(v) (C) P-(i); Q-(ii); R-(iii); S-(iv); T-(v) (D) P-(i); Q-(iii); R-(ii); S-(v); T-(iv)

Matching Cofactors to Enzymes

Q.96 Match the immunological statements in Column I with the appropriate descriptions from Column II Column I Column II P. Active acquired immunity (i) Complement proteins and interferons Q. First line of defense (ii) Direct contact with pathogens that enter body R. Passive natural immunity (iii) Surface barriers S. Second line of defense (iv) Antibodies pass through placenta (A) P-(ii); Q-(iii); R-(iv); S-(i) (B) P-(i); Q-(ii); R-(iii); S-(iv) (C) P-(iv); Q-(ii); R-(i); S-(iii) (D) P-(i); Q-(iv); R-(ii); S-(iii)

Match Immunological Statements Column I Column II

Q95 Match the syndromes listed in Column I with the cause/symptoms listed in Column II Column I Column II P. Prader-Willi syndrome (i) a collection of signs and symptoms due to prolonged exposure to corticosteroids like Q. Down syndrome (ii) a syndrome of inadequate reabsorption in the proximal renal tubule of the kidney R. Cushing syndrome (iii) a deletion of a part of chromosome 15 S. Turner syndrome (iv) a genetic disorder caused by the presence of all or part of a third copy of chromosome 21 T. Fanconi syndrome (v) a genetic condition in which a female has partially or completely missing X chromosome Codes: (A) P-(ii); Q-(iv); R-(v); S-(iii); T-(i) (B) P-(i); Q-(iii); R-(i); S-(v); T-(iv) (C) P-(iii); Q-(iv); R-(i); S-(v); T-(ii) (D) P-(iv); Q-(v); R-(iii); S-(i); T-(ii)

Matching Genetic Syndromes Causes

Q94 Match the hormones/precursors listed in Column I with their chemical type in Column II and tissue of origin listed in Column III Column I Column II Column III P. Glucagon (i) Tryptophan derivative a. Anterior pituitary Q. Pregnenolone (ii) Peptide b. Pineal R. FSH (iii) Steroid c. Adrenal S. Melatonin (iv) Glycoprotein d. Pancreas Options: (A) P-(ii)-(d); Q-(iii)-(a); R-(iv)-(b); S-(i)-(c) (B) P-(i)-(d); Q-(iv)-(a); R-(i)-(b); S-(iii)-(c) (C) P-(ii)-(c); Q-(i)-(b); R-(iv)-(d); S-(iii)-(a) (D) P-(iv)-(a); Q-(ii)-(d); R-(i)-(b); S-(iii)-(c)

Hormones Chemical Classification and Tissue Origin Matching

Q.93 In an experiment, excess amount of bicod mRNA (more than wild-type expression level) was injected into the posterior pole of a wild-type Drosophila embryo at pre-blastodermal stage. Out of the following options, which one represents the best expected phenotype in the resulted developing embryo? (A) Normal embryo with head structure at anterior and tail structure at posterior pole (B) Head structure only at posterior pole of the embryo (C) Tail structure at anterior and head structure at posterior poles of the embryo (D) Head structure at both anterior and posterior poles of the embryo

Bicoid mRNA Injection Posterior Pole Drosophila Embryo Phenotype

Q.92 A mature rat sperm cell has 2.5 μg of genomic DNA that is equivalent of a haploid genome. Compared to this sperm cell, the amount of genomic DNA (in μg) in a somatic cell, which is in the G2 phase of cell cycle, will be _____ (in integer).

Rat Sperm Cell DNA Content and G2 Phase Somatic Cell Genomic DNA Amount

Q.91 In a population of 1000 wild dogs in a grassland, 360 and 480 dogs had black body colour with genotypes BB and Bb, respectively. In the same population, remaining dogs were white in colour with a genotype of bb. Based on this data, the frequency of allele “b” in the population is ______ (round off to one decimal place).

Frequency of Allele b in Wild Dogs Population

Causative Agents of Filariasis

Organisms That Obtain Energy from Inorganic Compounds

Q.88 Which of the following animals show “Bottle cells” during the gastrulation stage of development? (A) Snails (B) Amphibians (C) Birds (D) Mammals

Bottle Cells During Gastrulation in Amphibians

True Ecological Population

Innate Behavior of an Animal

Q.85 Which one of the following animals has “Book Lungs” as a respiratory organ? (A) Earthworm (B) Scorpion (C) Octopus (D) Starfish

Which Animal Has Book Lungs as Respiratory Organ?

Q.84 During the exponential growth, it took 6 hours for the population of bacterial cells to increase from 2.5 × 106 to 5 × 108. The generation time of the bacterium, rounded off to the nearest integer, is ________ minutes.

Bacterial Generation Time Calculation

Q.83 The spontaneous, and induced mutations in bacteria can be distinguished by (A) fluctuation test (B) replica plating (C) disc diffusion test (D) use-dilution test

Distinguishing Spontaneous and Induced Mutations in Bacteria

Q.82 Which of the following statements about the primary and secondary adaptive immune responses to an antigen is/are correct? (A) IgM antibodies appear first in response to the initial exposure of the antigen. (B) Majority of the antibodies produced in response to the second exposure of the same antigen are IgM isotype. (C) Second exposure of the same antigen stimulates production of memory cells. (D) Primary antibody response has shorter lag phase than secondary antibody response.

Primary and Secondary Adaptive Immune Responses to Antigens

Q.81 Which of the following antibiotics is/are isolated from Streptomyces spp.? (A) Gentamicin (B) Nystatin (C) Polymyxins (D) Tetracyclines

Antibiotics Isolated from Streptomyces spp

Q.80 Which of the following is/are non-membrane bound inclusion bodies? (A) Carboxysomes (B) Cyanophycin granules (C) Poly-β-hydroxybutyrate granules (D) Polyphosphate granules

Non-Membrane Bound Inclusion Bodies in Prokaryotes

Q.79 Which of the following genus is/are a spirochete(s)? (A) Borrelia (B) Leptospira (C) Spirulina (D) Treponema

Which Genus is Spirochete

Q78 Which one of the following conjugations will result in formation of merodiploid? (A) F- donor × F- recipient (B) Hfr donor × F- recipient (C) F+ donor × F- recipient (D) F+ donor × Hfr recipient

Bacterial Conjugation Merozygote Formation

Q.77 Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]. Assertion [a]: Endospore can survive heat that would rapidly kill vegetative cells of the same species. Reason [r]: In endospore, the protoplasm is reduced to minimum volume as a result of the accumulation of calcium-dipicolinic acid complexes and small acid-soluble spore proteins, which forms a cytoplasmic gel and a thick cortex. (A) Both [a] and [r] are true and [r] is the correct reason for [a] (B) Both [a] and [r] are true and [r] is not the correct reason for [a] (C) Both [a] and [r] are false (D) [a] is true but [r] is false

Endospore Heat Resistance

Q.76 A suspension of photosynthetic green algae was illuminated in the presence of 14CO2 for few seconds. The first metabolite in the Calvin cycle to be radiolabeled will be (A) glyceraldehyde (B) 1,3-bisphosphoglycerate (C) 3-phosphoglycerate (D) ribulose 1,5-bisphosphate

First Labeled Metabolite Calvin Cycle

Antibiotic Resistance Mechanism in Microbes

Q74 | Match the test in Group I with its application in Group II Group I Group II P. Ouchterlony-Fulthorpe test Q. Limulus amoebocyte lysate test R. Weil-Felix reaction test S. Complement-fixation test 1. IgM detection 2. Determining antigen-antibody specificity 3. Endotoxin detection 4. Rickettsial infection diagnosis (A) P-2, Q-3, R-4, S-1 (B) P-2, Q-1, R-4, S-3 (C) Q-1, R-2, S-4 (D) P-4, Q-3, R-2, S-1

Mastering Immunological Test Matching

Batch Culture of Microbes

Q.72 Phagocytosis was first described by (A) Elie Metchnikoff (B) Robert Hooke (C) Robert Koch (D) Paul Ehrlich

Who First Described Phagocytosis?

Catabolite Repression in Bacteria

Q.70 The correct sequence of metabolic intermediates in Krebs cycle is (A) α-ketoglutarate → fumarate → succinate → malate (B) fumarate → malate → succinate → α-ketoglutarate (C) α-ketoglutarate → succinate → fumarate → malate (D) succinate → α-ketoglutarate → malate → fumarate

Krebs Cycle Sequence of Metabolic Intermediates

Q.69 Microbial plastics are made from (A) polyhydroxyalkanoates (B) polystyrene (C) polyurethane (D) polyvinyl chloride

Microbial Plastics Made from Polyhydroxyalkanoates

Q.68 Which one of the statements about bacterial flagella is correct? (A) Flagella varies in length ranging from 0.5 to 2 μm. (B) Flagella are adjacent fibrils with regular patterns. (C) Flagella helps in conjugation. (D) Flagella originates from basal body.

Bacterial Flagella Structure

Q.67 Which one of the following converts sulfate to hydrogen sulfide? (A) Beggiatoa (B) Desulfovibrio (C) Thiobacillus (D) Thiothrix

Q.66 Monkey pox is caused by a (A) double-stranded DNA virus (B) single-stranded DNA virus (C) double-stranded RNA virus (D) single-stranded RNA virus

Monkeypox Double-Stranded DNA Virus

Q.65 The frequencies for autosomal alleles A and a are p = 0.5 and q = 0.5, respectively, where A is dominant over a. Under the assumption of random mating, the mating frequency among dominant parents is _________. (Rounded off to two decimal places)

Autosomal Alleles A and a Frequencies p=0.5 q=0.5

Q.64 A cytoplasmic male-sterile female plant with the restorer (nuclear) genotype rr is crossed to a male-fertile male plant with the genotype RR. Both RR and Rr can restore the fertility, whereas rr cannot. When an F1 female plant with Rr genotype was test-crossed to a male-fertile male plant with the rr genotype, the percentage of the population that is male fertile would be _______ %. (Answer in integer)

Ultimate Guide to Cytoplasmic Male Sterility Test Cross

Q.63 In a diploid angiosperm species, flower colour is regulated by the R gene. RR and Rr genotypes produce red flowers, whereas the rr genotype produces white flowers. If two individual plants are randomly selected from a large segregating population of a genetic cross between RR and rr parents, the probability of both the plants producing red flowers is _______. (Rounded off to two decimal places)

Probability Both Plants Red Flowers Diploid Angiosperm R Gene

Q.62 Blue light can directly induce opening of stomata. Blue light also triggers photosynthesis in the guard cells, which indirectly induces stomatal opening. Which one or more of the following experimental approaches would test the direct effect of blue light on stomatal opening? (A) Application of low photon fluxes of red light followed by high fluence rate of blue light. (B) Application of high fluence rates of red light followed by low photon fluxes of blue light. (C) Application of high fluence rates of blue light followed by high photon fluxes of red light. (D) Inhibition of photosynthetic electron transport by dichlorophenyldimethylurea (DCMU).

Blue Light Stomatal Opening

Q.61 A drought tolerant rice genotype was found to be associated with a missense mutation in the gene A. Which one or more of the following experiments is/are appropriate to validate whether the mutation in A is the causal factor for drought tolerance? (A) Introduce the same mutation in a drought sensitive rice genotype and test if it becomes drought tolerant. (B) Delete the wild-type A in drought sensitive plant and test if it becomes drought tolerant. (C) Determine the stability of the protein encoded by the wild-type and the mutant forms of A. (D) Repair the mutation in the drought tolerant rice genotype and test if it becomes drought sensitive.

Validating Missense Mutation in Drought Tolerant Rice Genotype

Q60. Which one of the options given correctly matches the alkaloids in Group I with their source plants in Group II? Group I Group II P. Cocaine Q. Caffeine R. Morphine S. Atropine 1. COCA 2. Nightshade 3. COCA 4. Poppy (A) P-3; Q-1; R-4; S-2 (B) P-1; Q-3; R-4; S-2 (C) Q-1; R-3; S-4 (D) P-4; Q-2; R-1; S-3

Which Option Matches Alkaloids in Group I with Plants in Group II?

Q.59 Consider the following four experimental observations (i, ii, iii, iv) on the effect of the FT gene on flowering transition in the shoot apical meristem (SAM) of Arabidopsis thaliana. i) The FT promoter is active in leaves alone. ii) The ft null mutation causes delayed flowering transition of the SAM. iii) Expressing a recombinant FT protein fused to nuclear localization signal sequence under the endogenous promoter does not rescue the delayed-flowering phenotype of the ft null mutant. iv) Downregulation of FT transcript in the SAM by RNA interference in the wild-type background does not alter flowering transition. Which one of the following conclusions best explains the above observations? (A) FT protein resident in leaves causes flowering transition of the SAM. (B) FT transcript moves from leaves to the meristem and promotes flowering. (C) FT protein moves from leaves to the SAM and promotes flowering. (D) Both FT transcript and FT protein are required in the SAM to promote flowering.

FT Protein Movement from Leaves to SAM

Q.58 Central vascular cylinder or stele consists of the primary vascular system (xylem and phloem) and the associate fundamental tissue. Match the schematics of stele in Group I (xylem shown in green, and phloem shown as ) with their respective types in Group II. (A) P-2; Q-4; R-1; S-3 (B) P-5; Q-1; R-4; S-2 (C) P-5; Q-3; R-1; S-2 (D) P-3; Q-4; R-2; S-5

Primary Vascular Stele Types

Q57 Match the following rice diseases in Group I with their causal agents in Group II Group I Group II (P) Blast smut (1) Sarocladium oryzae (Q) Sheath blight (2) Rhizoctonia solani (R) Sheath smut (3) Ustilaginoidea virens (S) Downy mildew (4) Ustilago tragia (A) P-5, Q-3, R-2, S-1 (B) P-4, Q-2, R-5, S-3 (C) Q-5, R-3, S-1 (D) P-5, Q-4, R-1, S-2

Rice Diseases and Causal Agents

Q56 Periderm is a protective tissue found in stems and roots of gymnosperm and woody dicotyledons. It contributes to the increased thickness by secondary growth. Match the periderm components given in Group I with the cell tissue in Group II. Group I Group II (P) Phelloids (1) Tissue resembling cortical parenchyma (Q) Phellogen (2) Cork cambium (R) Phelloderm (3) Cork-like cells (S) Periderm (4) Cork cells (A) P-4; Q-3; R-1; S-2 (B) P-3; Q-2; R-4; S-1 (C) P-2; Q-1; R-3; S-4 (D) Q-1; R-3; S-2

Periderm Structure in Gymnosperms

Q55 Match the tasks given in Group I with the associated techniques conventionally used as listed in Group II Group I Group II Ploidy analysis Profiling DNA methylation Identifying non-coding RNAs Identifying SNPs Satellite DNA isolation RNA sequencing Exome sequencing Fluorescence in situ hybridization Bisulfite sequencing Density-gradient centrifugation (A) P-2, Q-1, R-3, S-4, T-5 (B) Q-4, R-1, S-2, T-5 (C) P-5, Q-4, R-1, S-2, T-3 (D) P-3, Q-5, R-1, S-2, T-4

Best Technique Match

Q.54 The ovule of a diploid species with 2n = 8 undergoes double fertilization. If the pollen is contributed by an individual with meiotic nondisjunction, the chromosome number of the zygote will be _________.

Ovule Diploid 2n=8 Double Fertilization

Q.53 Which one or more processes listed below DOES NOT/DO NOT produce carbon dioxide during fermentation? (A) Brewing wine using yeast. (B) Baking bread using yeast. (C) Making yogurt using lactobacillus. (D) Making cheese using fungus.

Which Fermentation Processes Do Not Produce Carbon Dioxide?

Q.52 Regulation of phosphoenolpyruvate carboxylase (PEPCase) governs CO2 fixation in both C4 and CAM (crassulacean acid metabolism) plants. Which one or more of the following statements with respect to PEPCase activity is/are CORRECT? (A) PEPCase in C4 plants is inactivated by dephosphorylation during the day. (B) PEPCase in CAM plants is inactivated by dephosphorylation during the day. (C) PEPCase in C4 plants is inactivated by dephosphorylation at night. (D) PEPCase in CAM plants is inactivated by dephosphorylation at night.

PEPCase Regulation in C4 and CAM Plants

Q.51 Which one or more of the following statements is/are NOT CORRECT with respect to pollen development in angiosperm? (A) Tapetal cell wall in all angiosperms breaks down to release the cytoplasmic content. (B) Tapetal cell wall in all angiosperms remains intact. (C) Tapetal cell wall breaks down in some angiosperm species, whereas it remains intact in others. (D) Within an angiosperm species, the tapetal cell wall breaks down in some individuals and not in others.

Pollen Development in Angiosperms

Q.50 The schematic depicts an unexpanded plant cell within a hypocotyl with the arrangement of cellulose microfibrils marked on its cell wall. Which one of the following shapes would most likely result from the expansion of this cell if the pattern of the cellulose fibrils does not change?

How Cellulose Microfibril Orientation Controls Plant Cell Expansion

Q.49 Although Pseudomonas syringae infection in plants is actively inhibited by the endogenous salicylic acid (SA) of host origin, a successful infection is still established because the bacterium secretes coronatine, an effector molecule. Which one of the following best describes the mode of action of coronatine? (A) Coronatine inhibits SA biosynthesis. (B) Coronatine promotes the biosynthesis of jasmonic acid (JA), and JA signaling in turn inhibits SA response. (C) Coronatine is a structural analogue of SA, which binds to the SA receptor and inhibits its function. (D) Coronatine is a structural analogue of jasmonic acid (JA), which activates JA signaling to inhibit SA response.

Coronatine Mode of Action

Q.48 Rotenone is a chemical often used to kill insect pests on crop plants and fishes in lakes. Rotenone acts by inhibiting electron transport from the NADH dehydrogenase enzyme in Complex I to ubiquinone in the mitochondrial electron transport chain. Which one of the following explains why plants can tolerate rotenone application? (A) The Complex I in plants is resistant to rotenone. (B) Plants inactivate rotenone by enzymatic degradation. (C) Plants have specific channels that efflux rotenone out of the cell. (D) Plants have additional NAD(P)H dehydrogenases that are resistant to rotenone.

Why Plants Tolerate Rotenone Application

Q.47 Which one of the following statements on Casparian strips is correct? (A) Casparian strips are specific to vascular plants found in epidermal cells. (B) Casparian strips are modifications mostly found in shoot tissue. (C) Casparian strips act as a cellular barrier to allow selective nutrient uptake and exclusion of pathogens. (D) Casparian strips are common in root endodermal cells of non-vascular plants.

Casparian Strips Function in Root Endodermis

Q.46 If the extracellular concentration of sodium ion (Na+) is ten times more than its intracellular concentration, then the sodium equilibrium potential at 20 °C in mV is ____________ (rounded off to two decimal places). Assume that the membrane is permeable only to Na+ ions. [Use R = 1.987 cal deg−1 mol−1 and F = 23062 cal mol−1 V−1]

Sodium Equilibrium Potential Calculation

Q.45 An organism uses only the glycerophosphate shunt pathway to transport cytosolic NADH to mitochondria. For every two electrons transported, complex I, complex III, and complex IV of the electron transport chain in this organism transport 2.5, 1.5, and 2.0 protons (H+), respectively. The H+ to ATP ratio of FOF1-ATPase of this organism is 4.0. Terminal electron acceptor is oxygen. The number of ATP molecules synthesized by oxidizing NADH from glycolysis is __________ (rounded off to two decimal places).

Glycerophosphate Shuttle ATP Yield from Glycolytic NADH

Q.44 A 100 ml solution of pH 10 was well-mixed with a 100 ml solution of pH 4. The pH of the resultant 200 ml solution is ___________ (rounded off to two decimal places).

Calculate pH of 100 ml pH 10 + 100 ml pH 4 Mixture

Q.43 B cells produce two forms of an immunoglobulin: (i) membrane-bound form, known as B cell receptor (BCR) and (ii) soluble form, known as antibody. Which of the following statements is/are CORRECT about BCR and antibody produced by the same B cell? (A) BCR and antibody have identical antigen binding site (B) BCR and antibody recognize different epitopes (C) BCR and antibody are encoded by the same gene (D) BCR and antibody are formed by differential splicing

B Cell Receptor vs Antibody

Q.42 Which of the following technique(s) can be used to separate proteins according to their molecular weights from a mixture of proteins? (A) Ion exchange chromatography (B) Size exclusion chromatography (C) Sodium dodecylsulfate – polyacrylamide gel electrophoresis (SDS-PAGE) (D) Sucrose density gradient centrifugation

Protein Separation by Molecular Weight

Q.41 Four statements about lipids are given below as options. Choose the statement(s) which is/are CORRECT. (A) Cholesterol is amphipathic (B) Self-assembly of phospholipids in water is due to hydrophobic effect (C) The temperature at which the gel phase changes to liquid crystalline phase increases with an increase in the degree of unsaturation of fatty acyl tails (D) The choline head group of lipids is positively charged

Cholesterol Amphipathic, Phospholipid Self-Assembly, Unsaturation Transition Temperature, Choline Charge Explained

Q.40 Match the cell types listed in Group I with associated processes listed in Group II. Group I Group II (p) NK cells (q) B cells (r) Mast cells (s) Neutrophils (i) Antibody production (ii) First cells to be recruited at the site of infection (iii) Antibody-dependent cell-mediated cytotoxicity (iv) Histamine production Option Matching (A) (p) – (ii); (q) – (i); (r) – (iii); (s) – (iv) (B) (p) – (ii); (q) – (i); (r) – (iv); (s) – (iii) (C) (p) – (iii); (q) – (i); (r) – (iv); (s) – (ii) (D) (p) – (iii); (q) – (ii); (r) – (i); (s) – (iv)

Matching immune cell types with their functions

Q.39 A linear DNA fragment of 5 kilobase (kb) when completely digested with EcoRI produces 2.5 kb, 1.5 kb, and 1 kb fragments. Complete digestion of the same 5 kb fragment with XbaI produces 3.5 kb and 1.5 kb fragments. Which one of the following sets of fragments will be obtained if the 5 kb fragment is fully digested with EcoRI and XbaI simultaneously? (A) 3 kb and 2 kb (B) 2 kb and 1 kb (C) 2 kb, 1.5 kb, 1 kb, and 0.5 kb (D) 2.5 kb, 1.5 kb, 0.75 kb, and 0.25 kb

5 kb Linear DNA EcoRI XbaI Double Digest

Q.38 Enzyme activity profiles as a function of time in the absence or presence of different types of feedback mechanisms are shown in the figure below. Match the following feedback mechanisms with the corresponding profiles in the figure. (p) No feedback mechanism (q) Negative feedback mechanism with short delay (r) Negative feedback mechanism with long delay (s) Positive feedback mechanism (A) (p) – (i); (q) – (ii); (r) – (iii); (s) – (iv) (B) (p) – (iv); (q) – (iii); (r) – (ii); (s) – (i) (C) (p) – (iv); (q) – (ii); (r) – (iii); (s) – (i) (D) (p) – (i); (q) – (iii); (r) – (ii); (s) – (iv)

Enzyme Activity Profiles and Feedback Mechanisms

Q.37 Which one of the following amino acids has more than two acid-base groups? (A) Alanine (B) Leucine (C) Phenylalanine (D) Tyrosine

Which Amino Acid Has More Than Two Acid-Base Groups?

Q.36 Intracellular concentrations of ATP, ADP, and inorganic phosphate in four cell types are given below. Which one of these cell types has the most negative ΔG for ATP hydrolysis? Cell type ATP (mM) ADP (mM) Inorganic phosphate (mM) L 3.0 1.8 5.0 K 3.9 1.3 3.0 B 2.7 0.7 2.7 M 7.2 0.9 8.0 (A) L (B) K (C) B (D) M

Most Negative ΔG for ATP Hydrolysis

Q.35 DNA in a 1 cm long chromatin contains 5 × 109 base pairs. The fold compaction of this DNA within the chromatin is _______ (in integer).

DNA Fold Compaction in Chromatin

Q.34 A lyophilized sample of 20 nanomoles of an oligonucleotide is dissolved in water and the volume of the solution is made up to 200 µL. The concentration (in µM) of the oligonucleotide in this solution is ______ (in integer).

Accurate Molarity Calculation of an Oligonucleotide

Q.33 A protein has seven cysteine residues. The maximum number of disulfide bonds of different combinations that can possibly be formed by these seven cysteine residues is _______ (in integer).

Maximum Disulfide Bonds from Seven Cysteine Residues

Q.32 Which of the following molecules is/are second messenger(s) produced by the phosphoinositide signaling cascade? (A) Phosphatidylinositol 4,5-bisphosphate (B) Inositol 1,4,5-triphosphate (C) Inositol 1,3,5-triphosphate (D) Diacylglycerol

Phosphoinositide Signaling Cascade Second Messengers

Q.31 Given below are four reactions of the glycolytic pathway catalyzed by the enzymes E1, E2, E3, and E4, as indicated. Which of these enzymes is/are NOT part of the gluconeogenesis pathway? Fructose 6-phosphate —(E1)—> Fructose 1,6-bisphosphate Fructose 1,6-bisphosphate —(E2)—> Dihydroxyacetone phosphate + Glyceraldehyde 3-phosphate 3-Phosphoglycerate —(E3)—> 2-Phosphoglycerate Phosphoenolpyruvate —(E4)—> Pyruvate Options: E1 E2 E3 E4

Glycolysis vs Gluconeogenesis

Q.30 In a population, the probability of a susceptible individual getting infected with SARS-CoV-2 is low when a majority of individuals in the population becomes immune to this virus. This phenomenon is known as _______. (A) innate immunity (B) adaptive immunity (C) active immunity (D) herd immunity

Herd Immunity Explained

$Q.29 When cell components are fractionated by sedimentation, the correct order (from lower to higher gravitational force, g) in which the components get separated is _______. (A) nuclei, mitochondria, microsomes, and ribosomes (B) microsomes, mitochondria, ribosomes, and nuclei (C) nuclei, ribosomes, mitochondria, and microsomes (D) ribosomes, microsomes, mitochondria, and nuclei$