Category: GATE Life Science 2018 Previous Year Papers with Solutions

Category: GATE Life Science 2018 Previous Year Papers with Solutions

GATE Life Science 2018 Previous Year Papers with Solutions – Complete Preparation Guide

The GATE Life Science (XL) 2018 question paper remains one of the most valuable resources for aspirants preparing for GATE and other Life Science competitive examinations. Solving GATE Life Science 2018 previous year papers with detailed solutions helps candidates understand the depth of concepts, analytical level of questions, and the exact expectations of the GATE exam.

At Let’s Talk Academy, GATE Life Science preparation is built on a PYQ-focused strategy, ensuring aspirants concentrate on high-yield, exam-relevant content only.

Why GATE Life Science 2018 Previous Year Papers Are Important

GATE Life Science is a concept-driven examination that tests understanding rather than memorization. The 2018 paper assessed:

  • Fundamental Life Science concepts
  • Application-based and analytical thinking
  • Numerical problem-solving skills
  • Interpretation of biological data and experiments

By practicing GATE Life Science 2018 PYQs, aspirants can:

  • Understand the true difficulty level of the exam
  • Identify repeated and high-weightage topics
  • Learn the structure of MCQs, MSQs, and NAT questions
  • Improve speed, accuracy, and confidence
  • Develop the correct GATE exam approach

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

Subjects Covered in GATE Life Science 2018 PYQs

The GATE Life Science 2018 paper included questions from all major domains, such as:

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

Solving PYQs subject-wise helps strengthen weak areas, while full-length practice builds exam readiness.

Importance of Solved GATE Life Science 2018 Papers with Detailed Solutions

Attempting questions without explanations limits learning. Solved GATE Life Science 2018 previous year papers with detailed explanations help aspirants to:

  • Understand why a particular option is correct or incorrect
  • Strengthen concepts directly tested in the exam
  • Learn exam-oriented numerical-solving techniques
  • Avoid common conceptual mistakes
  • Improve long-term retention of important topics

Each properly analyzed PYQ improves clarity across multiple chapters.

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

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

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

For accurate, authentic, and concept-rich solutions, Let’s Talk Academy is the most reliable and recommended source.

Why Let’s Talk Academy Is the Best Choice

Let’s Talk Academy provides expert-curated GATE Life Science previous year papers with detailed explanations, including the 2018 paper, through:

  • Topic-wise solved PYQ PDFs
  • Step-by-step, concept-based solutions
  • Exam-oriented numerical explanations
  • Faculty-reviewed answer keys aligned with the GATE syllabus
  • Structured learning resources for self-study and guided preparation

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

Unlike random online sources, Let’s Talk Academy focuses on concept clarity and exam logic, not just final answers.

How to Use GATE Life Science 2018 PYQs Effectively

To extract maximum benefit from previous year papers:

  • Start with topic-wise PYQs during concept building
  • Move to full-length GATE 2018 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 solving a large number of questions.

Who Should Practice GATE Life Science 2018 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 conceptual foundation useful across multiple competitive examinations.

Why Let’s Talk Academy Emphasizes PYQ-Based Preparation

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

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

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

Final Conclusion

For GATE Life Science 2018, 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 tool for mastering concepts and improving exam performance.

Make GATE Life Science PYQs the foundation of your preparation strategy and let real exam questions guide your success.

Q.105 In a population which is in Hardy-Weinberg equilibrium, the frequency of occurrence of a disorder caused by recessive allele (q) is 1 in 1100. The frequency of heterozygotes in the population will be ______________. (Give the answer to three decimal places).

Hardy-Weinberg Equilibrium

Q.104 A population of rabbits was determined to have a birth rate of 200 and mortality rate of 50 per year. If the initial population size is 4000 individuals, after 2 years of non-interfered breeding the final population size will be _______________.

Rabbit Population Growth Calculation

Q.103 Which ONE of the following statements is NOT part of the classical Darwinian theory of evolution by natural selection? (A) A trait which is constantly used will get inherited by next generation. (B) Phenotypic variations exist among the individuals of a population of a species (C) Individuals that best fit into a given environment are more likely to survive (D) Each population can randomly acquire a distinct and separate suite of variations.

Darwinian Theory of Natural Selection

Q102. Match the organisms listed in Column I with the features listed in Column II Column I Column II (I) Tapeworm (i) Bioluminescence (II) Jellyfish (ii) Viviparous (III) Earthworm (iii) Lateral heart (IV) Trichinella (iv) Microvilli on body surface (A) I−iii; II−iv; III−i; IV−ii  (B) I−ii; II−iv; III−iii; IV−i  (C) I−iv; II−iii; III−ii; IV−i  (D) I−i; II−ii; III−iii; IV−iv

Tapeworm Jellyfish Earthworm Trichinella Matching

Q.101 In an experiment, nucleus from a Drosophila oocyte was transplanted into the anterior part of another oocyte, at a region opposite to the existing nucleus. Which ONE of the following phenotypes will the developing egg show? (A) A ventralized egg with no dorsal appendages (B) A dorsalized egg with two dorsal appendages (C) A ventralized egg with two dorsal appendages (D) A dorsalized egg with four dorsal appendages

Drosophila Oocyte Nucleus Transplantation

Q.100 Match the proteins/molecules listed in column I with cellular location mentioned in the column II. Column I Column II (I) Galactosyl transferase (i) Vesicles (II) Galactosyl oxidase (ii) Cytosol (III) Cytochrome oxidase (iii) Golgi complex (IV) Tubulin (iv) Mitochondria (A) I-ii; II-iii; III-iv; IV-i (B) I-i; II-iv; III-iii; IV-ii (C) I-iii; II-ii; III-i; IV-iv (D) I-iv; II-iii; III-ii; IV-i

Matching Proteins to Cellular Locations

Q.99 Which ONE of the following is TRUE in case of respiratory acidosis? (A) Increased rate of ventilation is a cause of respiratory acidosis (B) Blood pH more than 7 (C) Increased levels of carbon dioxide in blood (D) Acidosis can be compensated through reduction of bicarbonate levels in plasma

 Respiratory Acidosis

Q.98 In Drosophila melanogaster, a mutation in Ultrabithorax which defines the third segment of the thorax or T3 leads to development of four winged flies, as the halteres develop into a second pair of wings. Which ONE of the following phenotypes in fly will result from overexpression of Ultrabithorax in the second thoracic segment? (A) Four winged flies. (B) Two wings and two halteres flies. (C) Flies with four halteres. (D) Flies with two halteres.

Ultrabithorax Overexpression

Q.97  Glucose monomers are joined together by glycosidic linkages to form a cellulose polymer. During this process, changes in the free energy, total energy, and entropy respectively are represented correctly by which ONE of the following options? (A) +ΔG, +ΔH, +ΔS. (B) +ΔG, -ΔH, -ΔS. (C) -ΔG, +ΔH, +ΔS. (D) +ΔG, +ΔH, -ΔS.

Glucose Monomers Glycosidic Linkages

Q.96 Match the disorders/diseases listed in Column I to their respective causative agents listed in Column II. Column I                                                  Column II I) African tick bite fever                         i) Trypanosoma gambiense II) Yellow fever                                        ii) Zika virus III) Microcephaly                                    iii) Rickettsia sp. IV) Sleeping sickness                              iv) Flavivirus (A) I-iv, II-iii, III-ii, IV-i                                                                                                                                                                                                                                                                                                                                                                                (B) I-iii, II-iv, III-ii, IV-i (C) I-iii, II-iv, III-i, IV-ii (D) I-iii, II-i, III-iv, IV-ii

Disease-Causative Agent Matches

Q.95 Which ONE of the following is the primary function of lung surfactants? (A) Remove dust particles from bronchi (B) Provide immunity to respiratory tract (C) Prevent alveoli from collapsing by decreasing surface tension (D) Aid in carbon dioxide exchange

 Primary Function of Lung Surfactants

Q.94 Bowman’s capsules are present in which ONE of the following organs/ tissues? (A) Renal cortex (B) Urinary bladder (C) Renal medulla (D) Ureter

Bowman’s Capsule in Kidney

Q.93 An animal’s ability to escape from a predator by using the explored knowledge of home area is an example of (A) Latent learning (B) Insight learning (C) Mimicry (D) Imprinting

Latent Learning in Animals

Q.92 Cardiac and cerebral tissues are derived from the following germ layers respectively (A) Ectoderm and mesoderm (B) Mesoderm and ectoderm (C) Mesoderm and endoderm (D) Endoderm and ectoderm

Cardiac and Cerebral Tissues Germ Layers

Q.91 Which ONE of the following is NOT a characteristic of a cancer cell? (A) Increase in cell motility (B) Loss of contact inhibition (C) Decrease in apoptosis (D) Uncontrolled meiosis 

Which One is NOT a Characteristic of a Cancer Cell?

Q.90 Increase in the existent population of grey peppered moth, Biston betularia, during industrial revolution in Britain is an example of which ONE of the following evolutionary processes? (A) Neutral selection (B) Disruptive selection (C) Directional selection (D) Stabilizing selection

Peppered Moth Industrial Revolution

Q.89 Tay-Sachs disease is a human genetic disorder that is associated with defects in which ONE of the following cellular organelles? (A) Endoplasmic reticulum (B) Mitochondria (C) Golgi apparatus (D) Lysosome

Tay-Sachs Disease Lysosome Defect

Q.88 In a chemical reaction where the substrate and product are in equilibrium in solution, what will occur if an enzyme is added? (A) The equilibrium of the reaction will not change. (B) There will be a decrease in product formed. (C) Additional substrate will be formed. (D) The free energy of the system will change.

Effect of Enzymes on Chemical Equilibrium

Q.87 A zoologist recovered some tissue from preserved skin of a woolly mammoth. Further genetic analysis requires DNA isolation and increasing its amount. Which ONE of the following techniques would be most useful for increasing the amount of DNA? (A) RFLP analysis (B) Polymerase chain reaction (PCR) (C) Electroporation (D) Chromatography

Best Technique to Amplify Woolly Mammoth DNA

Q.86 Animals belonging to phylum Echinodermata are closer to chordates than other invertebrate phyla. Which ONE of the following reasons can account for this relatedness? (A) Highly evolved nervous system (B) Radially symmetric body plan (C) Deuterostomic development (D) Well-developed muscles

Why Echinodermata Are Closer to Chordates

Q.85 The following stoichiometric equation represents the conversion of glucose to lactic acid in a cell. Glucose + 2Pi + 2ADP → 2Lactic acid + 2ATP + 2H2O (-7000 cal) The free energy efficiency of energy transfer is ____% (to lactic acid/ATP = -7000 cal, the ΔG for ATP hydrolysis is -7 kcal/mol) (up to one decimal point).

Glucose to Lactic Acid Free Energy Efficiency Calculation

Q.84 The initial concentration of cells (N0) growing unrestricted in culture is 1.0×106 cells/ml. The time required for the concentration to become 10.0×106 cells/ml is ____ h (up to two decimal points).

Calculating Time for Cell Concentration

Q.83 A continuous cell culture being carried out in a stirred tank reactor is described in terms of S. The cell mass concentration of the X substrate in the sterile feed stream is SF = 10 g/L and yield coefficient YX/S = 0.5. (F - flow rate (ml/min) and feed stream and specific growth rate (h-1) = 0.35 in the steady state. The concentration of S is g/L (up to 1 decimal point).

Chemostat Steady State

Q.82  Active transport involves the movement of a biomolecule against a concentration gradient across the cell membrane using metabolic energy. If the extracellular concentration of a biomolecule is 0.005M and its intracellular concentration is 0.5M, the least amount of energy that the cell would need to spend to transport this biomolecule from the outside to the inside of the cell is ________ kcal/mol (up to 2 decimal points). (Temperature T = 298K and universal gas constant R= 1.98 cal/mol.K)

Active Transport Energy Calculation

Q.81 Which of the following combinations would improve the resolution of a microscope? (i) Increasing the half aperture angle of the objective lens (ii) Decreasing the wavelength of the illumination source (iii) Decreasing the numerical aperture of the objective lens (iv) Decreasing the refractive index of immersion medium (A) (i) and (ii) (B) (ii) and (iii) (C) (ii) and (iv) (D) (i) and (iii)

Improve Microscope Resolution

Q.80 Match the scientists to their area of major contribution Scientists Area of major contribution (i) Antonie van Leeuwenhoek (p) Taxonomy (ii) Carl Linnaeus (q) Antibiotic agents (iii) Louis Pasteur (r) Vaccination (iv) Sir Alexander Fleming (s) Microscopy (A) (i)-(p), (ii)-(q), (iii)-(r), (iv)-(s) (B) (i)-(s), (ii)-(p), (iii)-(q), (iv)-(r) (C) (i)-(q), (ii)-(s), (iii)-(r), (iv)-(p) (D) (i)-(p), (ii)-(q), (iii)-(s), (iv)-(r)

 Scientists and Their Major Contributions

Q.79 Match the precursors/intermediates with the corresponding metabolic pathways. Precursors/Intermediates Metabolic Pathway (i) Inosine monophosphate (p) L-Methionine biosynthesis (ii) Orotate (q) L-Tryptophan biosynthesis (iii) Ornithine (r) L-Arginine biosynthesis (iv) Homocysteine (s) Purine biosynthesis (A) (i)-(q), (ii)-(r), (iii)-(s), (iv)-(p) (B) (i)-(p), (ii)-(q), (iii)-(r), (iv)-(s) (C) (i)-(r), (ii)-(p), (iii)-(s), (iv)-(q) (D) (i)-(s), (ii)-(r), (iii)-(p), (iv)-(q)

Precursors Intermediates Metabolic Pathways Matching

Q.78 Match the microorganisms to their predominant modes of transmission. Microorganism Mode of Transmission (i) Bordetella pertussis (p) Vector-borne (ii) Dengue virus (q) Blood-borne (iii) Entamoeba histolytica (r) Droplet infection (iv) Hepatitis B virus (s) Contaminated food Options: (A) (i)-(q), (ii)-(p), (iii)-(s), (iv)-(q) (B) (i)-(s), (ii)-(q), (iii)-(p), (iv)-(r) (C) (i)-(r), (ii)-(q), (iii)-(p), (iv)-(s) (D) (i)-(s), (ii)-(r), (iii)-(p), (iv)-(q)

Microorganisms Predominant Modes of Transmission

Q.77 Match the antimicrobial agents in group I with their category/mode of action in group II Group I Group II (i) Fluoroquinolones (p) β-lactam antibiotic (ii) Tetracycline (q) Inhibition of nucleic acid synthesis (iii) Amphotericin B (r) Inhibition of protein synthesis (iv) Amoxicillin1 (s) Antifungal agent Options: (A) (i)-(p), (ii)-(q), (iii)-(r), (iv)-(s) (B) (i)-(q), (ii)-(r), (iii)-(s), (iv)-(p) (C) (i)-(r), (ii)-(s), (iii)-(p), (iv)-(q) (D) (i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)

Matching Antimicrobial Agents with Mode of Action

Q.76 Accumulating evidence suggest that Domain Archaea is more closely related to Domain Eukarya than to Domain Bacteria. Which of the following properties are shared between eukaryotes and archaea ? (i) Protein biogenesis (ii) Presence of sterol containing membranes (iii) Ribosomal subunit structures (iv) Adaptation to extreme environmental conditions (v) Fatty acids with ester linkages in the cell membrane (A) (ii), (iii) and (v) (B) (i), (ii), (iv), and (v) (C) (i) and (iii) (D) (iii) and (iv)

Domain Archaea Closer to Eukarya than Bacteria

Q.75 You have a 50 mg/mL stock solution of arginine. To prepare 1 liter of growth medium for an arginine auxotroph that requires 70 μg/mL of arginine, the volume of this stock solution that should be added is _____________ mL (up to 1 decimal point) .

Preparing Arginine Growth Medium for Auxotrophs

Q.74 In humans, the key stages in the life cycle of malarial parasites occur in (A) red blood cells and the liver (B) red blood cells and platelets (C) red blood cells and the pancreas (D) red blood cells and the gut

Life Cycle of Malarial Parasites in Humans

Q.73 The primary product of nitrogen fixation is (A) N2 (B) NH4+ (C) NO2- (D) NO3

Primary Product of Nitrogen Fixation

Q.72 Which of the following is NOT an accepted method for sterilization? (A) Autoclaving (B) X-rays (C) Gamma rays (D) UV rays

Which is NOT an Accepted Sterilization Method

Q.71 Binomial nomenclature has NOT yet been adopted for (A) bacteria (B) fungi (C) viruses (D) protozoa

Binomial Nomenclature Not Adopted for Viruses

Q.70 Analysis of DNA sequences suggest that eukaryotic mitochondrial genomes primarily originated from (A) fungi (B) protozoa (C) algae (D) bacteria

Eukaryotic Mitochondrial Genomes Origin from Bacteria

Q.69 Which of the following eukaryotic cellular components carries out intracellular degradation during autophagy? (A) Nucleus (B) Golgi bodies (C) Ribosomes (D) Lysosomes

Eukaryotic Cellular Components in Autophagy Degradation

Q.68 The oral polio vaccine (OPV) consists of (A) live attenuated virus (B) killed virus (C) viral toxin (D) viral capsid subunit

Oral Polio Vaccine

Q.67 Which of the following immune system components can function as an opsonin? (A) Antibodies (B) T-cell receptors (C) Histamines (D) Interferons

Which Immune System Component Functions as an Opsonin?

Q.66 David Baltimore’s classification of viruses is based on differences in (A) host cell receptors used by viruses (B) the pathways required to synthesize virus mRNA (C) the modes of transmission of viruses (D) the envelope proteins on the surface of viruses

David Baltimore’s Virus Classification

Q.65 Two true-breeding snapdragon (Antirrhinum majus) plants, one with red flowers and another with white flowers were crossed. The F1 plants were all with pink flowers. When the F1 plants were selfed, they produced three kinds of F2 plants with red, pink and white flowers in a 1:2:1 ratio. The probability that out of the five plants picked up randomly, two would be with pink flowers, two with white flowers and one with red flowers is _____%.

Snapdragon F2 Flower Color Probability

Q.64 In garden pea, dwarf plants with terminal flowers are recessive to tall plants with axial flowers. A true-breeding tall plant with axial flowers was crossed with a true-breeding dwarf plant with terminal flowers. The resulting F1 plants were testcrossed, and the following progeny were obtained: Tall plants with axial flowers = 320 Dwarf plants with terminal flowers = 318 Tall plants with terminal flowers = 79 Dwarf plants with axial flowers = 83 The map distance between the genes for plant height and flower position is______cM.

Garden Pea Map Distance

Q.63 Identify the CORRECT combination of statements with respect to chemical defense in plants. P. Pisatin, a phytoalexin produced by Ricinus communis is a constitutive defense compound Q. Phaseolus vulgaris produces Phaseolus agglutinin I, which is toxic to the cowpea weevil R. A single step non-enzymatic hydrolysis of cyanogenic glycoside releases the toxic hydrocyanic acid (HCN) to protect plant against herbivores and pathogens S. Avenacin, a triterpenoid saponin from oat prevents infection by Gaeumannomyces graminis, a major pathogen of cereal roots (A) P, Q (B) Q, S (C) R, S (D) P, S

Chemical Defense in Plants

Q.62 Match the plant alkaloids with their uses and source species. Alkaloid Use Source species P. Codeine 1. Analgesic i. Catharanthus roseus Q. Scopolamine 2. Stimulant ii. Coca nitida [*Coca*?] R. Vinblastine 3. Antineoplastic iii. Papaver somniferum S. Vinblastine 4. Anticholinergic iv. Senecio jacobaea Options: (A) P-2, Q-iv, R-1, S-3-ii (B) P-iii, Q-2, R-1, S-3-i (C) P-2, Q-1, R-3, S-4-i (D) P-3, Q-4, R-1, S-2

Plant Alkaloids Matching Uses and Sources

Q.61 Select the CORRECT combination by matching Group-I with Group-II Group-I Group-II Process Code Precursors Code P. Photrespiration 1. 1. Glutamate → Malonyl-CoA A. R. Amino acid degradation 2. 2. Acetyl-CoA → Malonyl-CoA B. S. Fatty acid synthesis 3. 3. Glyoxylate → Succinyl-CoA C. 4. Oxaloacetate → Succinyl-CoA D. Options: (A) P-1, Q-2, R-3 (B) P-2, Q-1, R-4 (C) P-3, Q-4, R-5 (D) P-4, Q-3, R-1

Photrespiration, Amino Acid Degradation & Fatty Acid Synthesis Precursors Matching

Q.60 Select the CORRECT combination by matching the disease, affected plant and causal organism. Disease Affected plant Causal organism P. Black rot 1. Corn i. Ustilago zeae *syn.* Ustilago maydis Q. Loose smut 2. Banana ii. Botryodiplodia theobromae *syn.* Lasiodiplodia theobromae R. Panama wilt 3. Watermelon iii. Ustilago maydis S. Banana fruit blotch 4. Apple iv. Colletotrichum musae v. Venturia inaequalis Options: (A) P-2-v, Q-1-iv, R-3-iii, S-4-i (B) P-3-ii, Q-4-iii, R-1-iv, S-2-iii (C) P-4-iii, Q-1-v, R-2-iii, S-3-ii (D) P-4-i, Q-3-iii, R-2-v, S-1-ii

Black Rot Watermelon Loose

Q.59 Match the fruit characters with their families and representative plant species. Fruit character Family Plant species P. Syconus 1. Moraceae i. Canavalia ensiformis Q. Capsule, opening by apical pores or valves 2. Fabaceae ii. Artabotrys odoratissimus R. Legume 3. Papaveraceae iii. Ficus religiosa S. An etaerio of drupe 4. Annonaceae iv. Papaver somniferum v. Pistacia vera vi. Citrus aurantium Options: P-2-iv, Q-3-ii, R-1-vi, S-4-v P-1-iii, Q-3-iv, R-2-i, S-4-ii P-3-i, Q-2-iii, R-4-ii, S-1-vi P-4-v, Q-1-ii, R-2-v, S-3-i

Syconus Capsule Legume Etaerio of Drupe

Q. 58 Select the CORRECT combination of the following statements. P. The cyclic electron transport chain involving PSI results in net production of both ATP and NADPH Q. The cyclic electron transport chain involving PSI results in net production of ATP R. Rubisco enzyme usually converts RuBP and CO2 into 2-phosphoglycolate and 3-phosphoglycerate S. Rubisco enzyme usually converts RuBP and O2 into 2-phosphoglycolate and 3-phosphoglycerate (A) P, Q (B) R, S (C) Q, S (D) P, R

Cyclic Electron Transport and Rubisco

Q.57 You are asked to design a genetic construct for high-level expression of a gene encoding the therapeutic protein 18 (TP18) via plastid transformation. Select the CORRECT set of genetic elements for this construct. (A) Actin1 promoter → TP18 coding sequence → Actin1 transcription terminator (B) Ubiquitin1 promoter → TP18 coding sequence → Ubiquitin1 transcription terminator (C) rbcS promoter → TP18 coding sequence → rbcS transcription terminator (D) rbcL promoter → TP18 coding sequence → rbcL transcription terminator

Plastid Transformation

Q. 56 Statements given below are either TRUE (T) or FALSE (F). Select the CORRECT combination. P. Mitosis occurs exclusively in diploid mother cell Q. Mitosis occurs both in diploid and haploid mother cells R. Meiosis occurs exclusively in diploid mother cell S. Meiosis occurs both in diploid and haploid mother cells (A) P-T, Q-F, R-T, S-F (B) P-F, Q-T, R-F, S-T (C) P-T, Q-F, R-F, S-T (D) P-F, Q-T, R-T, S-F

Mitosis vs Meiosis

Q. 55 The coding sequence of a gene 𝑿𝑳𝑹𝟏𝟖 has the single ORF of 783 bp . The approximate molecular weight of the XLR18 protein in kDa is ____ .

Calculate Protein Molecular Weight

Q. 54 Which of the following stains is used to visualize callose under the microscope? (A) Alcian blue (B) Aniline blue (C) Toluidine blue (D) Thymol blue 

Best Stain to Visualize Callose Under the Microscope

Q. 53 Identify the INCORRECT statement in connection with polar transport of auxin. (A) The putative influx carrier AUX1 is a cytosolic protein (B) Polar auxin transport in root tends to be both acropetal and basipetal in direction (C) Naphthylphthalamic acid (NPA) is an inhibitor of polar auxin transport (D) AUX1 and PIN1 proteins are located in the opposite ends of a cell for polar transport

Polar Auxin Transport

Q. 52 Which of the following plants produces Ylang-ylang oil? (A) Cananga odorata (B) Carcum copticum (C) Pandanus odoratissimus (D) Pimenta racemosa

Which Plant Produces Ylang-Ylang Oil?

Q. 51 Acid rain with a pH of 4.0 is more acidic than the rain with a pH of 6.0 by (A) 2 times (B) 10 times (C) 100 times (D) 1000 times

Acid Rain pH 4.0 vs 6.0

Q. 50 Which of the following statements is TRUE for an ecotone? (A) An ecotone is the synonym of an ecosystem (B) An ecotone is an interface zone of two or more ecosystems (C) An ecotone is a special feature of land biomes (D) An ecotone is exclusively characterized by decreased biodiversity

Understanding Ecotones

Q. 49 Identify the CORRECT statement. (A) Receptor-like kinases play role in gametophytic self-incompatibility in Brassicaceae (B) Receptor-like kinases play role in sporophytic self-incompatibility in Solanaceae (C) Ribonucleases play role in sporophytic self-incompatibility in Brassicaceae (D) Ribonucleases play role in gametophytic self-incompatibility in Solanaceae

Self Incompatibility in Plants

Q. 48 Which of the statements is TRUE for transposable elements 𝑨𝒄 and 𝑫𝒔 ? (A) Both 𝐴𝑐 and 𝐷𝑠 are autonomous because they encode their own transposase (B) Both 𝐴𝑐 and 𝐷𝑠 are non-autonomous because they do not encode their own transposase (C) Only 𝐴𝑐 is autonomous because it encodes its own transposase (D) Only 𝐷𝑠 is autonomous because it encodes its own transposase

Ac and Ds Elements

Q. 47 The genes for microRNA (miRNA) in plants are usually transcribed by (A) RNA polymerase I (B) RNA polymerase II (C) RNA polymerase III (D) RNA polymerase IV

Plant miRNA Genes

Q.46 Which of the following genera produces dimorphic seeds that help to broaden the time of germination in a variable habitat? (A) Xanthium (B) Pisum (C) Mangifera (D) Linum

Xanthium Dimorphic Seeds

Q. 45 The standard free energy (ΔG′) values of reactions catalyzed by citrate lyase and citrate synthetase are −670 cal/mol and −8192 cal/mol, respectively. Citrate lyase reaction: Citrate ⇌ Acetate + Oxaloacetate ΔG1′ = −670 cal/mol Citrate synthetase reaction: Acetyl-CoA + Oxaloacetate + H2O ⇌ Citrate + CoA ΔG2′ = −8192 cal/mol The standard free energy (in cal/mol) of acetyl-CoA hydrolysis is _____ (correct to integer number). Benzoic acid's mass spectrum shows a prominent base peak at m/z 105 due to the stable benzoyl cation. ​​ Correct Answer The base peak (100% relative abundance) in benzoic acid's electron ionization mass spectrum occurs at m/z 105. This fragment forms from the molecular ion (m/z 122) losing a hydroxyl radical (- OH, mass 17), yielding the resonance-stabilized benzoyl cation [C₆H₅CO]⁺. For Q.44, enter the integer 105.​​ Fragmentation Pathways Benzoic acid (C₆H₅COOH, MW 122) undergoes characteristic cleavages in EI-MS. Molecular ion peak appears at m/z 122 (weak, ~10-20% abundance).​ Base peak at m/z 105: [M - OH]⁺, highly stable due to resonance between phenyl and carbonyl.​​ m/z 77: C₆H₅⁺ (phenyl cation) from further loss of CO (mass 28) from m/z 105.​ Minor peaks: m/z 51 (C₄H₃⁺, tropylium-like), m/z 123 (M+1 from ¹³C isotope).​ Common Exam Options Explained Typical multiple-choice distractors test fragmentation knowledge: 105 (Correct): Most abundant due to benzoyl stability; 100% base peak.​​ 122: Molecular ion [M]⁺- ; present but weak as aromatics fragment easily.​ 77: Significant (phenyl ion) but secondary (~50-70% abundance).​​ 51 or 45: Smaller fragments (e.g., [C₆H₅COOH - COOH]⁺ or COOH⁺); low intensity.​ Why m/z 105 Dominates Resonance in [C₆H₅CO]⁺ delocalizes charge across oxygen and ring, enhancing survival to detector. This pattern repeats in aromatic carboxylic acids. Spectra from NIST or textbooks confirm this for benzoic acid identification.​​

Benzoic Acid Mass Spectrum

Q. 44 The mass spectrum of benzoic acid will generate the fragment as a base peak ( 𝟏𝟎𝟎% relative abundance) of 𝒎/𝒛 (mass to charge ratio) at ____ (correct to integer number).

Benzoic Acid Mass Spectrum

Q. 43 An enzyme follows Michaelis-Menten kinetics with substrate S. The fraction of the maximum velocity ( 𝑽max ) will be observed with the substrate concentration [ S ] = 𝟒𝑲𝐦 is ____ (correct to one decimal place). ( 𝑲𝐦 is Michaelis-Menten constant) 

Michaelis-Menten Kinetics

Q. 42 The secondary structure topology diagram of 400 amino acid long "Protein-X" is depicted in the figure. The start and end amino acid residue numbers of each 𝜶-helix are marked. The percentage (correct to integer number) of residues forming 𝜶-helix is ____ .

Percentage of α-Helix Residues

Q. 41 A polymerase reaction is carried out for 10 cycles in a volume of 1 ml with 5 molecules of template DNA. Assuming that the efficiency of the reaction is 𝟏𝟎𝟎%, the number of molecules of DNA present in 𝟏𝟎𝟎𝝁𝐥 at the end of the reaction is ____ (correct to integer number). 

PCR Amplification

Q. 40 Which one the following reaction mechanisms drives the conversion of low energy 3phosphoglyceraldehyde to high energy 1,3-bisphosphoglycerate? (A) Oxidation without anhydride bond formation (B) Oxidation coupled with anhydride bond formation (C) Substrate level phosphorylation (D) Formation of carboxylate

3-Phosphoglyceraldehyde to 1,3-Bisphosphoglycerate

Q. 39 The type-II hypersensitivity reaction is mainly mediated by ____ . (A) IgE (B) IgM (C) IgA (D) T cells

Type II Hypersensitivity Reaction

Q. 38 Which one of the following is NOT a neurotransmitter? (A) Adrenaline (B) Glutamate (C) Histamine (D) Histidine 

Which One of the Following Is NOT a Neurotransmitter?

Q. 37 Match the protein elution condition given in Group I with the appropriate chromatography matrices from Group II. Group I Group II P Increasing concentration of sodium chloride i Phenyl-Sepharose Q Increasing concentration of histidine ii Chromatofocusing R Decreasing concentration of ammonium sulphate iii DEAE-Sephacryl S Decreasing concentration of H + iv Ni-NTA (A) P-iii; Q-iv; R-i; S-ii (B) P-ii; Q-iv; R-i; S-iii (C) P-i; Q-ii; R-iii; S-iv (D) P-iv; Q-ii; R-iii; S-i

Protein Chromatography Matching

Q. 36 Among the reagents given below which one of the combination of reagents will NOT break the disulphide bonds in the immunoglobulin molecules? (P) Reduced glutathione (Q) Dithiothritol (R) Sodium dodecyl sulphate (S) Methionine (A) R&S (B) 𝑃&𝑅 (C) P& S (D) Q&R 

Reagents That Do Not Break Disulfide Bonds

Q. 35 Measurement of the absorbance of a solution containing NADH in a path length of 1 cm cuvette at 340 nm shows the value of 0.31. The molar extinction coefficient of NADH is 𝟔𝟐𝟎𝟎𝐌−𝟏𝐜𝐦−𝟏. The concentration of NADH in the solution is ____ 𝝁𝐌 (correct to integer number). 

Calculate NADH Concentration

Q. 34 The number of NADP +molecules required to completely oxidize one molecule of glucose to 𝐂𝐎𝟐 through pentose phosphate pathway is ____ (correct to integer number). 

Pentose Phosphate Pathway

Q. 33 The movement of protons through the 𝐅𝟎 𝐅𝟏-ATPase during mitochondrial respiration is required for ____ (A) the increase in pH of mitochondrial matrix. (B) changing the conformation of FoF1 -ATPase to expel the ATP. (C) importing Pi from inter membrane space. (D) decreasing the affinity of ADP to FoF1 -ATPase. 

FoF1-ATPase Protons in Mitochondria

Q. 32 Which one of the following properties of the myeloma cells is used in the hybridoma technology to generate monoclonal antibody? (A) lack of thymidylate synthase (B) over-expression of hypoxanthine-guanine phosphoribosyl transferase (C) over-expression of inosine 5'-monophosphate cyclohydrolase (D) lack of hypoxanthine-guanine phosphoribosyl transferase 

Hybridoma Technology

Q. 31 The glycosylation of the proteins occurs in ____ . (A) glyoxysomes (B) lysosomes (C) Golgi apparatus (D) plasma membrane

Protein Glycosylation

Q. 30 Which one of the following amino acids is responsible for the intrinsic fluorescence of proteins? (A) Pro (B) Met (C) His (D) Trp

Intrinsic Fluorescence of Proteins

Q. 29 Which one of the following profiles represent the phenomenon of cooperativity?

Which Velocity–Substrate Profile Represents Cooperativity?

Q. 28 Which one of the following conformations of glucose is most stable? (A) Boat (B) Half Chair (C) Chair (D) Planar

Most Stable Glucose Conformation

Q. 2 The substrate saturation profile of an enzyme that follows Michaelis-Menten kinetics is depicted in the figure. What is the order of the reaction in the concentration range between 0.8 to 1.4 M? (A) Zero (B) Fraction (C) First (D) Second

Order of Reaction in Michaelis

Q. 26 To which one of the following classes of enzymes does chymotrypsin belong? (A) Oxidoreductase (B) Hydrolase (C) Transferase (D) Isomerase

Chymotrypsin Enzyme Class

Q. 25 One mole of helium gas in an isolated system undergoes a reversible isothermal expansion at 25°C from an initial volume of 2.0 liters to a final volume of 10.0 liters. The change in entropy (ΔS) of the surroundings is ____ J K−1 (up to two decimal places). (Given: Gas constant R = 8.314 J K−1 mol−1)

Entropy Change of Surroundings

Q. 24 The activation energy (Ea) values for two reactions carried out at 25°C differ by 5.0 kJ mol−1. If the pre-exponential factors (A1 and A2) for these two reactions are of the same magnitude, the ratio of rate constants (k1/k2) is ____ (up to two decimal places). (Given: Gas constant R = 8.314 J K−1 mol−1)

Activation Energy and Rate Constants

Q. 23 The solubility product (Ksp) of Mg(OH)2 at 25°C is 5.6 × 10−11. Its solubility in water is S × 10−2 g/L, where the value of S is ____ (up to two decimal places). (Given: Molecular weight of Mg(OH)2 = 58.3 g mol−1)

Mg(OH)2 Solubility from Ksp

Q. 22 The standard reduction potential (E°) for the conversion of Cr2O72− to Cr3+ at 25°C in an aqueous solution of pH 3.0 is 1.33 V. The concentrations of Cr2O72− and Cr3+ are 1.0 × 10−4M and 1.0 × 10−3M, respectively. Then the potential of this half-cell reaction is (Given: Faraday constant = 96500 C mol−1, Gas constant R = 8.314 J K−1 mol−1) (A) 1.04 V (B) 0.94 V (C) 0.84 V (D) 0.74 V

Cr₂O₇²⁻ to Cr³⁺ Half-Cell Potential Calculation

Q. 21 The compound, which upon mono-nitration using a mixture of 𝐇𝐍𝐎𝟑 and 𝐇𝟐𝐒𝐎𝟒, does NOT give the meta-isomer as the major product, is

Aromatic Nitration

Q. 20 The two major products of the reaction

Hofmann Elimination Reaction

Q. 19 The major product 𝐌 in the reaction is

Ozonolysis Reaction

Q. 18 In [Mn(H2O)6]2+ (atomic number of Mn = 25), the d-d transitions according to crystal field theory (CFT) are Laporte forbidden and spin forbidden Laporte allowed and spin allowed Laporte forbidden and spin allowed Laporte allowed and spin forbidden

[Mn(H2O)6]2+ d-d Transitions

Q. 17 Among the following statements, (i) [NiCl4]2− (atomic number of Ni = 28) is diamagnetic (ii) Ethylamine is a weaker Lewis base compared to pyridine (iii) [NiCl2{P(C6H5)3}2] has two geometrical isomers (iv) Bond angle in H2O is greater than that in H2S, the CORRECT one is (i) (ii) (iii) (iv)

Why Bond Angle in H2O > H2S?

Q. 16 Among B−H, C−H, N−H and Si−H bonds in BH₃, CH₄, NH₃ and SiH₄, respectively, the polarity of the bond which is shown INCORRECTLY is Bδ+−Hδ− Cδ−−Hδ+ Nδ−−Hδ+ Siδ−−Hδ+

Incorrect Bond Polarity

Q. 15 The total number of possible stereoisomers for the compound with the structural formula CH₃CH(OH)CH=CHCH₂CH₃ is ____.

Total Stereoisomers

Q. 14 Two equivalents of 𝐏 react with one equivalent of 𝐐 to produce a major product 𝐑. The number of double bonds present in the major product 𝐑 is ____ .

Number of Double Bonds Explained

Q. 13 Hybridization of xenon in XeF2 is (A) sp (B) sp2 (C) sp3 (D) sp3d

Hybridization of Xenon in XeF2

Q. 12 For a 4s orbital of hydrogen atom, the magnetic quantum number (ml) is (A) 4 (B) 3 (C) 1 (D) 0

Magnetic Quantum Number ml for 4s Orbital Hydrogen

Q. 11 For the complete combustion of graphite and diamond in oxygen individually, the standard enthalpy change (ΔH°298) values are −393.5 kJ mol−1 and −395.4 kJ mol−1, respectively. Then, the ΔH°298 for the conversion of graphite into diamond is (A) +1.9 kJ mol−1 (B) −1.9 kJ mol−1 (C) +3.8 kJ mol−1 (D) −3.8 kJ mol−1

Enthalpy Change Calculation Explained

Q. 10 Consider the following three statements: (i) Some roses are red. (ii) All red flowers fade quickly. (iii) Some roses fade quickly. Which of the following statements can be logically inferred from the above statements? (A) If (i) is true and (ii) is false, then (iii) is false. (B) If (i) is true and (ii) is false, then (iii) is true. (C) If (i) and (ii) are true, then (iii) is true. (D) If (i) and (ii) are false, then (iii) is false.

Logical Inference in Syllogisms

Q. 9 Which of the following functions describe the graph shown in the below figure? (A) y = ||x| + 1| - 2 (B) y = ||x| - 1| - 1 (C) y = ||x| + 1| - 1 (D) y = ||x - 1| - 1|

Absolute Value Graph Explained

Q. 8 From the time the front of a train enters a platform, it takes 25 seconds for the back of the train to leave the platform, while travelling at a constant speed of 54 km/h. At the same speed, it takes 14 seconds to pass a man running at 9 km/h in the same direction as the train. What is the length of the train and that of the platform in meters, respectively? (A) 210 and 140 (B) 162.5 and 187.5 (C) 245 and 130 (D) 175 and 200

Solving Train Length and Platform Length

Q. 7 Given that a and b are integers and a + a²b³ is odd, which one of the following statements is correct? (A) a and b are both odd (B) a and b are both even (C) a is even and b is odd (D) a is odd and b is even

Solving Parity When a + a²b³ is Odd for Integers a and b

Q. 6 For integers a, b and c, what would be the minimum and maximum values respectively of a + b + c if log |a| + log |b| + log |c| = 0? (A) -3 and 3 (B) -1 and 1 (C) -1 and 3 (D) 1 and 3

Solving Logarithmic Constraint

Q. 5 A number consists of two digits. The sum of the digits is 9. If 45 is subtracted from the number, its digits are interchanged. What is the number? (A) 63 (B) 72 (C) 81 (D) 90

Solving the Two-Digit Number Puzzle

Q. 4 A rectangle becomes a square when its length and breadth are reduced by 10 m and 5 m, respectively. During this process, the rectangle loses 650 m² of area. What is the area of the original rectangle in square meters? (A) 1125 (B) 2250 (C) 2924 (D) 4500

Original Area Calculation

Q. 3 Seven machines take 7 minutes to make 7 identical toys. At the same rate, how many minutes would it take for 100 machines to make 100 toys? (A) 1 (B) 7 (C) 100 (D) 700

Seven Machines Seven Minutes

Q. 2 "Her ____ should not be confused with miserliness; she is ever willing to assist those in need." The word that best fills the blank in the above sentence is (A) cleanliness (B) punctuality (C) frugality (D) greatness

Her Frugality Should Not Be Confused with Miserliness

Q. 1 "Going by the ____ that many hands make light work, the school ____ involved all the students in the task." The words that best fill the blanks in the above sentence are (A) principle, principal (B) principal, principle (C) principle, principle (D) principal, principal

Principle vs Principal

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