Category: GATE Life Science 2010 Previous Year Papers with Solutions

Category: GATE Life Science 2010 Previous Year Papers with Solutions

GATE Life Science 2010 Previous Year Papers with Solutions

Q.100. During immune response, helper T-cell memory against the antigen appears earlier than the B memory cells. Which one of the following is the primary reason for this phenomenon? (A) Affinity of antibody molecules produced by B cells is weaker than those of T cells {B) B memory cells proliferate at a rate slower than that of T cells (C) B-cell activation requires helper T cells (D) Thymic selection more rapidly enhances the T cell population than B cell population

Helper T Cell Memory Forms Before B Cell Memory:

Q.99. In the field of community ecology, the lerm "competitive exclusion" refers to two species that cannot co-exist (A) in a community if the niches are identical. (B) in two different communities if the niches are identical. (C') if the ecosystem is imbalanced. (D) in the event of a volcanic eruption.

Competitive Exclusion Principle

Q.98. Which one of the following options is NOT a viable strategy for developing a female contraceptive? The administration of (A) a combination of synthetic progesterone and estrogen (B) synthetic progesterone alone (C) ormeloxifene - a selective estrogen recepior modulator (D) a synthetic oxytocin

Which Option Is NOT a Viable Female Contraceptive Strategy?

Q.97. An alien species has been discovered with very similar genetie makeup as that of the existing species on planet earth with certain differences. The genetic material of this new species is referred to as DNA *. The building blocks of the genetic material is known as Nucleotide *. The proteins of the new species (Protcin*) is made up of Amino Acids *. It has also been discovered that the new species hus 5 distinet Nucleotide* as opposed to the four for species on planet earth. The new species has 40 different Amino Acids* as opposed lo the 20 for species on planet earth. What should be the codon length for this new species (the same for species of planet earth is 3)? It may be assumed that the average codon degeneracy of the new species is very similar to that of species of planet earth. (A) 2 (B) 3 (C) 4 (D) 5

Alien Genetic Code: Codon Length Calculation

Q.96. Neurogen (Ngn) a newly discovered protein in chicken. is produced by the notochord and the floor plate (FP). Ngn induces cells of the neural tube (NT) to become neurons. It is known that from ventral to dorsal direction cells at different levels give rise to distinct types of neuronal cells. Which of the following observations will cast a doubt in the claim that Ngn is a morphogen? (A) Ngn is a cytosolic protein (B) Artificial mis-expression of Ngn at identical level through out NT does not affect the neuronal cell types formed in the NT (C) Ngn is an integral membrane protein (D) All of the above

Neurogen Protein Morphogen Role

Q.95. In allopatric mode of specialion, a new species forms due to (A) Geographic isolation (B) Genetic drift (C) Formation of a few fertile individuals thar can not mare with other members of the same species living in the same geographical area (D) The formation of allopolyploid condition

Allopatric Speciation Defined

Q.94 As compared lo peplide hormones, steroid hormones take more lime to activate a celluiar response because (A) steroid hormones show non-specific binding with diverse sets of receptors. (B) steroid hormone acts through a receptor which is a transcription factor. (C) cells that respond to steroid hormones are dormant in nature. (D) peptide hormones are not transported through plasma while steroid hormones are.

Steroid Hormones Take Longer

Q.93 In the context of cell differentiation, lateral inhibition is referred 1o as the (A) formation of two distinct cell types within a uniform field, (B) inhibilion of formation of a distinct cell type next to an existing cell type. (C) inhibition of stem cells towards self-renewal. (D) inhibition of erythopoesis in the lateral plate mesoderm.

Lateral Inhibition in Cell Differentiation

Q.92.Which one of the following results from Mendel's monohybrid cross is the strongest evidence against the blending theory? (A) 3:1 ratio of phenolypes in the FI generation (B) All progeny of the FI generation exhibited the dominant phenotype (C) The recessive phenolype showed up in the F2 progeny (D) The observation of incomplete dominance

Mendel Monohybrid Cross

Q91. Enzymes catalyze biochemical reactions by Sequestering the product(s) Decreasing the ΔG of the reaction Increasing the ΔG of the reaction Stabilizing the transition state of the reaction

How Enzymes Catalyze Reactions

Q90. In case of parasites that require multiple hosts to complete their life cycle, what does definitive host mean? It is the host that harbors the sexual stages of the parasite. It is the host in which the parasite reproduces asexually. It is the host in which the parasite feeds. It is the host in which the parasite remains in a dormant stage.

Definitive Host in Parasite Life Cycles

Smallest Biological Unit of Evolution

Q88. Identify the phylum that is characterized by the animals that have segmented appendages. Cnidaria Porifera Arthropoda Mollusca

Phylum with Segmented Appendages

Q87. Acoelomates are characterized by The absence of cavity surrounding the internal organs The presence of huge body cavity, as in case of terrestrial animals The presence of air sacs, as in case of birds The absence of brain in a group of extinct species

Acoelomates Characteristics

Q86. From the perspective of developmental origin, which of the following structures is homologous to a tortoise shell? Exoskeleton of a lobster Bones of a fish Skull of humans Feathers of birds

Tortoise Shell Homology

Q85. Match the resistance mechanism in Group I with the antibiotic in Group II Group I Group II P. β-Lactamases Q. Enhanced folate metabolism R. Drug efflux S. Phosphorylation of the drug T. Mutant RNA polymerase 1. Aminoglycosides 2. Penicillins 3. Sulfa drugs 4. Tetracyclins 5. Nalidixic acid 6. Rifamycin P-2, Q-3, R-4, S-5, T-6 P-3, Q-4, R-1, S-6, T-5 P-2, Q-3, R-4, S-1, T-6 P-1, Q-2, R-3, S-4, T-6

Antibiotic Resistance Mechanisms Matching

Q84. A mutant of E. coli grows normally on glucose or on glycerol but not on acetate. The most likely metabolic pathway that is defective in this mutant is Glyoxylate cycle Hexose monophosphate shunt Krebs cycle Entner–Doudoroff pathway

E. coli Mutant Can’t Grow on Acetate

Q83. Match the microorganism in Group I with the application in Group II Group I Group II P. Aspergillus oryzae Q. Brevibacterium flavum R. Thiobacillus ferrooxidans S. Saccharomyces cerevisiae T. Rhizobium meliloti 1. Metal ore leaching 2. Glucoamylase producer 3. Bread making 4. Glutamic acid producer 5. Penicillin producer 6. Symbiotic nitrogen fixer P-1, Q-6, R-4, S-5, T-2 P-2, Q-4, R-1, S-3, T-6 P-4, Q-1, R-6, S-3, T-5 P-6, Q-2, R-3, S-5, T-1

Microorganisms Matching

Q82. A bacterial culture was diluted 1000 fold and 0.1 ml of this diluted sample was spread per plate on nutrient agar. In a triplicate run, the number of colonies formed is 121, 93 and 86. The number of colony forming units/ml in the original bacterial culture is 106 105 103 102

CFU/ml Calculation

Q81. Match the pathogen in Group I with the corresponding disease in Group II Group I Group II P. Bacteria Q. Virus R. Fungi S. Protozoa 1. Measles 2. Candidiasis 3. Malaria 4. Bovine spongiform encephalitis 5. Tuberculosis P-1, Q-2, R-4, S-5 P-1, Q-4, R-2, S-3 P-5, Q-1, R-4, S-2 P-5, Q-1, R-2, S-3

Pathogen Disease Matching

Q80. Which one of the following events occurs in prokaryotes but NOT in eukaryotes? Protein phosphorylation RNA polymerase and promoter interaction Control of transcription by attenuation Formation of Okazaki fragments

Which Event Occurs in Prokaryotes But Not Eukaryotes?

Q79. In a lacOc lacZ− / lacO+ lacZ+ partial diploid, of the two lacZ enzymes, only the mutant enzyme (lacZ−) is synthesized constitutively. This observation shows that lacOc mutation is trans-dominant trans-recessive cis-dominant cis-recessive

lac Oc Mutation is Cis-Dominant

Q78. MHC-I groove can be loaded with peptides of only 8–10 amino acids because MHC-I groove is closed on both ends Fragments of only 8–10 amino acids are generated in MHC-I bearing cells β₂-microglobulin of MHC-I prevents the binding of large peptides to MHC-I β polypeptides of MHC-I prevent binding of 8–10 amino acid long peptides to MHC-I

Why MHC-I Binds Only 8-10 Amino Acid Peptides

Q77. Superantigens elicit a very strong T cell response because they Bind to the specific antigen binding site on the T cell receptors (TCR) Bind to the site on T cell receptor (TCR) that is outside the antigen-specific binding site Directly activate the T cell without the help of antigen presenting cells Directly induce cytokine secretion by macrophages

Superantigens Elicit a Very Strong T Cell

Q76. Match the scientist from Group I with the corresponding contribution listed in Group II Group I Group II P. Robert Koch Q. Walter Hesse R. Louis Pasteur S. Ferdinand Cohn 1. Discovery of endospores 2. Disproved spontaneous generation 3. Discovery of causative agent of tuberculosis 4. Use of agar as solid media 5. Invention of microscope P-5, Q-3, R-4, S-2 P-3, Q-4, R-2, S-5 P-3, Q-4, R-1, S-5 P-3, Q-4, R-2, S-1

Microbiology Pioneers Matching

Q75. The quickest way to determine bacterial growth in terms of viable cells is through Most probable number (MPN) technique Spread plate method Pour plate method Slide culture technique

Quickest Way to Determine Bacterial Growth

Q74. A growing bacterial culture with a doubling time of 20 min reaches cell density of 2 × 108 cells/ml in 3 hours. How much time would it take to reach the cell density of 1 × 106 cells/ml? 200 min 180 min 160 min 90 min

Bacterial Growth Doubling Time

Q.73.Pasteurization of milk is carried oul by (A) boiling for 5 min (B) heating at 63C for 15 min (C) heating at 72°C for 30 min (D) heating al 63C: for 30 min

Correct Temperature and Time for Pasteurization

Q.72. The resistance of a lambda lysogenic E.coli to re-infection by lambda is mediated by (A) blocking entry of the incoming lambda DNA (B) degrading the incoming lambda DNA (C) blocking transcription of the incoming lambda DNA (D) triggering mutation of the lambda receptor of the host

Superinfection Immunity in Lambda Lysogens

Q71. Wild type E. coli forms purple colored colonies on EMB-lactose plate. This is due to Increase in pH of the medium Decrease in pH of the medium Secretion of purple colored pigment Secretion of β-galactosidase

Why Wild-Type E. coli Forms Purple Colonies

Q70. Ames test is performed to detect Mutagen pH Nutrient stress Salinity

Ames Test

Q69. A mutant phenotype due to a nonsense mutation can be rescued by a mutation in tRNA gene. This rescue is an example of Induced mutation Suppressor mutation Spontaneous mutation Deletion mutation

Nonsense Mutation Rescue by tRNA

Q68. The recombination frequencies between three genes x, y and z are as follows: x-y: 2.6%, y-z: 1.4% and x-z: 1.2%. Then the gene order is x-z-y x-y-z y-x-z z-x-y

Determining Gene Order from Recombination Frequencies

Q67. Bacterial cell lysis by lysozyme is due to the Hydrolysis of α-1,4-glycosidic bonds between the N-acetylglucosamine and N-acetylmuramic acid Inhibition of cell wall synthesis Hydrolysis of pentapeptide bridges Hydrolysis of β-1,4-glycosidic bonds between the N-acetylglucosamine and N-acetylmuramic acid

Bacterial Cell Lysis by Lysozyme

Q66. An electron microscope has higher resolution as compared to the light microscope. This is because (A) the wavelength of an electron is longer than the wavelength of light (B) the wavelength of an electron is shorter than the wavelength of light (C) the electrons can penetrate the sample better (D) they use different stains

Why Electron Microscope Has Higher Resolution Than Light Microscope

Q.65 In the context of plant development, which of the following statements are CORRECT? P. Cell migration is absent Q. Apoptosis plays a major role R. Pattern formation continues throughout life S. Homeotic changes are caused by mutations in non-homeodomain proteins Options: (A) P, Q, R (B) Q, R, S (C) P, Q, S (D) P, R, S

Plant Development Key Features

Q.64 Study the following diagram depicting the plant cell cycle and match the following Stages of Cell Cycle Type of Cyclin P. Late G1 phase 1. Cyclin B Q. Beginning of S-phase 2. Cyclin E R. Prior to mitotic phase 3. S-Cyclin S. Early G1 phase 4. Cyclin D Options: (A) P-4, Q-3, R-1, S-2 (B) P-2, Q-3, R-1, S-4 (C) P-1, Q-4, R-3, S-2 (D) P-3, Q-1, R-2, S-4

Plant Cell Cycle and Cyclins

Q63. Which of the following statements in plant respiration are CORRECT? P. The oxidative Pentose Phosphate Pathway can accomplish the oxidation of glucose in the stroma of mitochondria Q. ATP is produced in the reaction step of TCA cycle catalyzed by succinyl-CoA synthetase R. In addition to Cytochrome c oxidase, an alternative oxidase enzyme resistant to cyanide reduces oxygen molecule in the electron transport system S. In Glyoxylate cycle acetyl-CoA reacts with citrate to form α-ketoglutarate (A) P, R (B) P, Q (C) Q, R (D) Q, S

Plant Respiration Correct Statements

Q62. Identify the CORRECT statements In Arabidopsis, vernalization is associated with P. Chromatin modification at the FLC (FLOWERING LOCUS C) locus Q. Degradation of the FLC protein R. Inactivating the FLC protein by post-translational modification S. Down-regulation of FLC transcript (A) Q, S (B) P, S (C) P, R (D) Q, R

Vernalization in Arabidopsis

Q61. Name the structures given below in the order of their appearance and identify corresponding glycosidic linkages (A) Amylose, Cellulose; (α1→4), (β1→6) (B) Cellulose, Dextran; (β2→4), (α3→6) (C) Starch, Cellulose; (α1→6), (α1→4) (D) Amylopectin, Amylose; (α1→6), (α1→4)

Amylopectin and Cellulose Glycosidic Linkages

Q60. Regarding the relationships between two organisms in an ecosystem, match the following Group I (Relationship) Group II (Definition) P. Commensalism Q. Mutualism R. Parasitism S. Amensalism 1. Both organisms are benefited 2. One impeding the success of the other 3. One organism benefits but the other is unaffected 4. One benefited, other is harmed (A) P-1, Q-2, R-3, S-4 (B) P-2, Q-3, R-4, S-1 (C) P-3, Q-1, R-4, S-2 (D) P-1, Q-4, R-3, S-2

Relationships Between Two Organisms in an Ecosystem

Q.59 From the structures given below, identify the compounds Group I (Structure) Group II (Compound) 1. Ethylene 2. Indole butyric acid 3. Nicotine 4. Indole acetic acid 5. Gibberellic acid 6. Menthol Options (A) P-6, Q-3, R-4, S-1 (B) P-5, Q-2, R-3, S-1 (C) P-4, Q-3, R-2, S-6 (D) P-1, Q-2, R-5, S-6

Identify Compounds from Organic Structures

Q.58 Identify the Correct Match Group II (Type of Fixation) 1. Basifixed 2. Longitudinal 3. Dorsifixed 4. Adnate 5. Porous 6. Versatile Options (A) P-1, Q-4, R-6, S-3 (B) P-2, Q-3, R-5, S-6 (C) P-1, Q-2, R-6, S-5 (D) P-4, Q-3, R-5, S-6

Types of Anther Fixation

Q57. Match the following Group I Group II Group III P. Sorghum Q. Castor R. Mushroom S. Cotton 1. Gossypol 2. Strychnine 3. Dhurrin 4. Bungarotoxin 5. Ricin 6. α-Amanitin i. Protein ii. Glycosidic conjugate iii. Alkaloid iv. Polyphenol v. Lipid vi. Cyclic peptide (A) P-3-ii, Q-5-i, R-6-vi, S-1-iv (B) P-2-iii, Q-4-iv, R-1-ii, S-6-v (C) P-2-vi, Q-5-v, R-1-iv, S-6-ii (D) P-2-i, Q-3-iii, R-4-iv, S-1-v

Sorghum Dhurrin, Castor Ricin, Mushroom α-Amanitin, Cotton Gossypol Matching

Q56. Identify the CORRECT statements from the following P. Lenticels are the small pores present on the surface of the stem or branches of woody plants Q. Glyoxysomes contain chlorophyll molecules in their thylakoid membranes R. The enzyme ribulose 1,5 bisphosphate carboxylase is otherwise known as carboxydehydrase S. 18 ATP and 12 NADPH molecules are utilized for fixing 6 molecules of CO2 in the dark reaction of photosynthesis (A) P, Q (B) P, R (C) Q, R (D) P, S

 Lenticels Glyoxysomes Rubisco Calvin Cycle ATP NADPH

Q55. Free radical scavenging activity of a medicinally important plant extract can be quantified by (A) ABTS (2,2′-azino-bis-(3-ethyl benzothiazoline-6-sulphonic acid)) method (B) Bradford method (C) Walkley and Black method (D) Kjeldahl method

Free Radical Scavenging Activity

Q54. The amount of chemical energy available to consumers in an ecosystem is best represented by (A) Gross primary production (B) Net primary production (C) Respiration (D) Photosynthesis

Chemical Energy Available to Consumers in Ecosystem

Q53. The possible genotypes of endosperms borne on a heterozygous (Rr) plant will be (A) RRR, RRr, Rrr, rrr (B) RRR, Rrr (C) RR, Rr, rr (D) Rr

Possible Genotypes of Endosperms

Q52. In the arrangement of floral parts in a bud, identify the INCORRECT statement (A) Valvate: where the petals or sepals do not overlap but simply touch one another by their margins (B) Scabrous: petals rough and harsh to touch (C) Epicalyx: an extra calyx found in some flowers outside the calyx (D) Imbricate: where sepals and petals overlap each other at the margin

Floral Aestivation Types in Buds

Q51. A sign is hammered into a tree trunk 2 meters above the tree’s base. If the tree is 10 meters tall and elongates 1 meter each year, how high will the sign be after 10 years? (A) 12 meters (B) 7 meters (C) 4 meters (D) 2 meters

Tree Sign Height After Growth

Q50. Which of the following statements is CORRECT? (A) Plants adapted to cold environment have higher ratio of "unsaturated to saturated" fatty acids in their membrane compared to those adapted to hot environment (B) Plants adapted to cold environment have lower ratio of "unsaturated to saturated" fatty acids in their membrane compared to those adapted to hot environment (C) Plants adapted to cold environment have same ratio of "unsaturated to saturated" fatty acids in their membrane compared to those adapted to hot environment (D) Plants do not have any unsaturated fatty acids in the membrane

Unsaturated to Saturated Fatty Acids

Q49. Arabidopsis and rice have diploid chromosome numbers of 10 and 24, respectively. Assuming no crossing over taking place, genetic variation among F2 individuals in a genetic cross is likely to be (A) Same in both species but not zero (B) More in Arabidopsis (C) More in rice (D) Zero in both the species

Genetic Variation in F2 Arabidopsis

Q48. Which one of the following is NOT a plant hormone? (A) Abscisic acid (B) Brassinosteroid (C) Ethylene (D) Cytokine

Which One of the Following is NOT a Plant Hormone?

Q47. A population growing exponentially can be described by the differential equation dN/dt = rN, where dN/dt represents the rate of population growth, N is the size of the population, r is the intrinsic rate of increase, and t is time. According to this equation, the per capita rate of growth is (A) Highest at large N (B) Constant (C) Lowest at large N (D) Highest at small N

Per Capita Rate of Growth

Q46. When changes in the phenotype or gene expression occur without changes in the underlying DNA sequence, the phenomenon is called (A) Mutation (B) Eugenics (C) Epigenetics (D) Epistasis

Epigenetics: Changes in Phenotype

Q45. Match the metabolic pathways in Group 1 with the corresponding enzymes in Group 2 Group 1 Group 2 P. β-Oxidation Q. Glycolysis R. Gluconeogenesis S. Calvin cycle 1. Ribulose bisphosphate carboxylase 2. Phosphofructokinase 1 3. Phosphoenol pyruvate carboxykinase 4. Thiolase 5. Phosphofructokinase 2 (A) P-4, Q-2, R-3, S-5 (B) P-3, Q-2, R-4, S-1 (C) P-3, Q-1, R-5, S-2 (D) P-4, Q-2, R-3, S-1

β-Oxidation Glycolysis Gluconeogenesis

Q44. The metabolic disorders, Alkaptonuria and Phenylketonuria are caused by defects in the enzymes P. Glucose-6-phosphatase Q. Phenylalanine hydroxylase R. Homogentisate 1,2-dioxygenase S. Tyrosinase (A) Q, R (B) P, R (C) P, Q (D) Q, S

Alkaptonuria Phenylketonuria Enzyme Defects

Q43. Match the proteins in Group 1 with their correct functions in Group 2 Group 1 Group 2 P. Shaker protein Q. Bacteriorhodopsin R. Porin S. ABC transporter 1. Inner membrane receptor 2. Active transport 3. Voltage gated K+ channel 4. Light driven H+ pump 5. Membrane fusion 6. β-barrel simple diffusion channel (A) P-4, Q-2, R-3, S-5 (B) P-5, Q-3, R-4, S-6 (C) P-6, Q-1, R-5, S-4 (D) P-3, Q-4, R-6, S-2

Shaker Protein Bacteriorhodopsin Porin

Q42. The rise per residue of α-helix is about 1.5 Å. A protein spans 4 nm bilayer 7 times through its transmembrane α-helical domain. Approximately, how many amino acid residues constitute the transmembrane domain of the protein? (A) 105 (B) 450 (C) 30 (D) 190

Alpha Helix Rise Per Residue

Q41. Match the parameters in Group 1 with the correct options in Group 2 Group 1 Group 2 P. Km Q. kcat/Km R. pKa S. Ki 1. Catalytic efficiency of the enzyme 2. Affinity of enzyme to the inhibitor 3. Affinity of enzyme to the substrate 4. Maximum buffering capacity (A) P-3, Q-1, R-2, S-4 (B) P-3, Q-1, R-4, S-2 (C) P-1, Q-2, R-4, S-3 (D) P-1, Q-4, R-2, S-3

Km kcat/Km pKa Ki Matching

Q40. In the two half reactions: Acetaldehyde + 2 H+ + 2 e− → Ethanol     ΔE° = −0.16 V NADH + H+ → NAD+ + 2 H+ + 2 e−     ΔE° = −0.32 V (F = 23,063 cal/V) The ΔG° for the coupled reaction will be (A) +7,400 cal (B) −7,400 cal (C) −22,200 cal (D) +22,200 cal

Calculation of ΔG° for Coupled Redox Reaction

Q39. Match the items in Group 1 with the most appropriate separation techniques in Group 2 Group 1 Group 2 P. Mixture of glycine and albumin Q. Mixture of 20 and 60 kDa proteins R. Histones from nuclear extract S. Lectins 1. Gas chromatography 2. Dialysis 3. Affinity chromatography 4. Size exclusion chromatography 5. Thin layer chromatography 6. Cation exchange chromatography (A) P-1, Q-4, R-3, S-5 (B) P-5, Q-3, R-6, S-1 (C) P-2, Q-4, R-6, S-3 (D) P-6, Q-5, R-2, S-4

Protein Separation Techniques

Q38. Multiple RNA polymerase transcribes a DNA template, unwinding about 1.5 turns of DNA template per transcription bubble. From the structural information of classical B-DNA, how many transcription bubbles are possible for a 180 base pair DNA molecule? (A) 12 (B) 27 (C) 6 (D) 270

How Many Transcription Bubbles

Q37. Match the items in Group 1 with those in Group 2 Group 1 Group 2 P. Isotype switching Q. Clonal anergy R. Class II MHC S. Self tolerance 1. VH domain 2. Non-responsive to self antigen 3. Non-responsive TH cells 4. β2-microglobulin (A) P-1, Q-4, R-3, S-2 (B) P-2, Q-4, R-1, S-3 (C) P-1, Q-3, R-4, S-2 (D) P-2, Q-1, R-3, S-4

Isotype Switching, Clonal Anergy

Q36. Chymotrypsin and lysozyme are involved respectively in P. Removal of successive carboxyl terminal residues Q. Hydrolytic cleavage of peptide bond R. Cleavage of glycosidic C–O bond S. Oxygen transport in blood (A) P, Q (B) Q, R (C) Q, S (D) R, S

Chymotrypsin and Lysozyme Functions

Q35. Specific precursor for all prostaglandins is (A) Oleic acid (B) Arachidonic acid (C) Palmitic acid (D) α-Linolenic acid

 Specific Precursor for All Prostaglandins

Q34. A mixture of proteins (W, X, Y, Z) elute from Sephadex G-200 column in the order W, X, Y, Z. The protein with maximum electrophoretic mobility on SDS-PAGE will be (A) W (B) X (C) Y (D) Z

Sephadex G-200 Elution Order

Q33. Lineweaver–Burk plot is a plot of (A) 1/v0 vs 1/[S] (B) v0 vs [S] (C) v0 vs 1/[S] (D) 1/v0 vs [S]

Lineweaver–Burk Plot

Q32. Approximate molecular weight (kDa) of the product after translation of a 390 bases mRNA will be (A) 48 (B) 26 (C) 39 (D) 14

Approximate Molecular Weight

Q31. One of the following compounds is NOT a neurotransmitter (A) Dopamine (B) Glutamic acid (C) Histidine (D) Glycine

Which Compound is NOT a Neurotransmitter

Q30. Amino acids responsible for N-linked and O-linked glycosylation of proteins are (A) Asparagine and Aspartic acid (B) Glutamine and Serine (C) Glutamic acid and Serine (D) Asparagine and Threonine

O-Linked Glycosylation of Proteins

Q29. A DNA-binding motif is (A) Helix-loop-helix (B) Helix-turn-helix (C) Helical wheel (D) Loop-helix-loop

DNA-Binding Motif Explained

Q28. Positive selection of T cells ensures (A) MHC restriction (B) Self tolerance (C) TCR engagements (D) Activation by co-stimulatory signal

Positive Selection of T Cells

Q27. In tryptophan operon, tryptophan acts as (A) Repressor (B) Activator (C) Co-repressor (D) Co-activator

Tryptophan Operon

Q.26 Nucleolus is involved in the synthesis of (A) rRNA (B) tRNA (C) DNA (D) mRNA

Nucleolus rRNA Synthesis

Reaction Sequence Problem Q.25 The structure of Y in the above reaction sequence is: (A) Structure A (B) Structure B (C) Structure C (D) Structure D

Structure of Y in Reaction Sequence

Reaction Sequence Problem Q.24 The structure of X in the above reaction sequence is: (A) Structure A (B) Structure B (C) Structure C (D) Structure D

Structure of X in Friedel–Crafts Acylation

The ionization constants of phosphorous acid (H3PO3) are: Ka1 = 3 × 10−2, Ka2 = 1.7 × 10−7 Q.23 The structure of phosphorous acid is: O || HO — P — OH | H HO — P — OOH | H HO — P — O | O | H H | H — P = O | OOH

Structure and Ionization Constants

The ionization constants of phosphorous acid (H3PO3) are: Ka1 = 3 × 10−2, Ka2 = 1.7 × 10−7 Q.22 For a dilute solution of phosphorous acid in a pH 5 buffer, the predominant species is: H3PO3 H2PO3− HPO32− PO33−

Phosphorous Acid (H3PO3) at pH 5

Q.21 Reaction of Na2SO3 with 2 equivalents of HCl produces a gas X. Solution of X in water is acidic in nature. X is: O2 Cl2 SO2 H2S

Na2SO3 with HCl Reaction

Q.20 The ionization energy follows the order: O2+ > O2 > O2− > O22− O2 > O2+ > O22− > O2− O2 > O2− > O2+ > O2 O22− > O2 > O2− > O2+

Ionization Energy Order

Q.19 Structure of the compound Y in the following reaction sequence is: Cyclohexene →(Br2 / AcOH)→ X →(KOtBu)→ Y Bromocyclohexene Cyclohexene 1,3-Cyclohexadiene Benzene

Bromination and E2 Elimination

Q.18 Among heptan-1-ol, heptan-2-ol, heptan-3-ol and heptan-4-ol, compounds those exhibit optical activity are: heptan-2-ol and heptan-3-ol heptan-2-ol and heptan-4-ol heptan-3-ol and heptan-4-ol heptan-1-ol and heptan-4-ol

Optical Activity in Heptanol Isomers

Q.17 For a particular reaction, the use of a catalyst reduces the activation energy (Ea) to one-third its original value. The ratio of rate constants (kcatalyzed / kuncatalyzed) is: 1 1/3 exp (2Ea / 3RT) exp (Ea / 3RT)

Catalyst Reduces Activation Energy

Q.16 Consider the following equilibrium reaction: CO (g) + Cl2 (g) ⇌ COCl2 (g) 0.60 atm of CO and 1.10 atm of Cl2 were mixed in a constant volume reaction vessel at a particular temperature. After the equilibrium was established, 0.10 atm of COCl2 was observed. The equilibrium constant for the reaction is: 0.02 0.15 0.2 6.6

Equilibrium Constant (Kp) Calculation

Q.15 The compound that is aromatic, among the choices, is:

Which Compound is Aromatic?

Nickel Oxidation State

Q.13 Molecule that has no lone pair of electrons on the central atom (among the choices) is: XeF4 PF3 ClF3 BF3

BF3 No Lone Pairs

Q.12 A battery delivers a steady current of 1.25 A for 90 minutes. The total charge ‘Q’ (in Coulomb units) is: 6750 1012.5 112.5 12.5

Battery Charge Calculation

Q.11 For a spontaneous process, the total entropy change (ΔSsystem + ΔSsurroundings) is: equal to zero greater than zero less than zero for endothermic process less than zero for exothermic process

Second Law of Thermodynamics

Q.10 Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause. Which of the following statements best sums up the meaning of the above passage: Modern warfare has resulted in civil strife. Chemical agents are useful in modern warfare. Use of chemical agents in warfare would be undesirable. People in military establishments like to use chemical agents in war.

Modern Warfare Chemical Agents Passage

Q.9 Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e., brothers and sisters). All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts: Hari’s age < Gita’s age > Irfan’s age + Saira’s age. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. There are no twins. In what order were they born (oldest first)? HSIG SGHI IGSH IHSG

Age Puzzle Solution

Q.8 If 137 + 276 = 435, how much is 731 + 672? 534 1403 1623 1513

Number Pattern Question

Q.7 Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4, how many distinct 4-digit numbers greater than 3000 can be formed? 50 51 52 54

How Many Distinct 4-Digit Numbers Greater Than 3000?

Q.6 5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall? 20 days 18 days 10 days 15 days

Skilled Workers Wall

Q.5 Choose the most appropriate word from the options given below to complete the sentence: If we manage to ______ our natural resources, we would leave a better planet for our children. uphold restrain cherish conserve

Conserve Natural Resources

Q.4 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is: 2 17 13 3

Number of Persons Playing

Q.3 Which of the following options is the closest in meaning to the word below: Circuitous cyclic indirect confusing crooked

Circuitous Meaning and Closest Synonym

Q.2 Choose the most appropriate word from the options given below to complete the sentence: His rather casual remarks on politics ______ his lack of seriousness about the subject. masked belied betrayed suppressed

Casual Remarks Politics

Q.1 The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed : Worker fallow : land unaware : sleeper wit : jester renovated : house

Unemployed Worker Analogy

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