Category: GATE Biotechnology 2010 Previous Year Papers with Solutions

Category: GATE Biotechnology 2010 Previous Year Papers with Solutions

GATE Biotechnology 2010 Previous Year Papers with Solutions

Q.65 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)? Options: (A) HSIG (B) SGHI (C) IGSH (D) IHSG

Sibling Age Order Logical

Q.64 Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4, how many distinct 4-digit numbers greater than 3000 can be formed? Options: (A) 50 (B) 51 (C) 52 (D) 54

How Many Distinct 4-Digit Numbers

Q.63 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? Options: (A) 20 days (B) 18 days (C) 16 days (D) 15 days

Time and Work Problem

Q.62 If 137 + 276 = 435, how much is 731 + 672? Options: (A) 534 (B) 1403 (C) 1623 (D) 1513

Number Reasoning Puzzle

Q.61 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: Options: (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in warfare would be undesirable. (D) People in military establishments like to use chemical agents in war.

Main Idea of Passage on Modern Warfare and Chemical Agents

Q.60 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 Options: (A) 2 (B) 17 (C) 13 (D) 3

Set Theory Problem

Q.59 Choose the most appropriate word from the options given below to complete the following sentence: His rather casual remarks on politics __________ his lack of seriousness about the subject. Options: (A) masked (B) belied (C) betrayed (D) suppressed

His Casual Remarks on Politics

Q.58 Choose the most appropriate word from the options given below to complete the following sentence: If we manage to __________ our natural resources, we would leave a better planet for our children. Options: (A) uphold (B) restrain (C) cherish (D) conserve

Conserve Natural Resources

Q.57 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 Options: (A) fallow : land (B) unaware : sleeper (C) wit : jester (D) renovated : house

Unemployed Worker Analogy

Q.56 Which of the following options is the closest in meaning to the word below: Circuitous Options: (A) cyclic (B) indirect (C) confusing (D) crooked

Circuitous Meaning

During bioconversion of sucrose to citric acid by Aspergillus niger, final samples of 6 batches of fermentation broth were analyzed for citric acid content. The results (in g/L) were found to be 47.3, 52.2, 49.2, 52.4, 49.1 and 46.3. Q.55 The standard deviation for the above results is Options: (A) 2.49 (B) 3.0 (C) 1.84 (D) 5.91

Standard Deviation Calculation

During bioconversion of sucrose to citric acid by Aspergillus niger, final samples of 6 batches of fermentation broth were analyzed for citric acid content. The results (in g/L) were found to be 47.3, 52.2, 49.2, 52.4, 49.1 and 46.3. Q.54 The mean value of acid concentration will be Options: (A) 49.4 (B) 51.0 (C) 48.2 (D) 50.8

Mean Citric Acid Concentration

The standard redox potential values for two half-reactions are given below. The value for Faraday’s constant is 96.48 kJ V-1 mol-1 and Gas constant R is 8.3 J K-1 mol-1. NAD+ + H+ + 2e- → NADH    E° = −0.315 V FAD + 2H+ + 2e- → FADH2    E° = −0.219 V Q.53 The value of ΔG°′ given Keq is 1.7 at 23 °C will be Options: (A) −17.19 kJ mol-1 (B) −19.8 kJ mol-1 (C) +52.82 kJ mol-1 (D) −117.07 kJ mol-1

Calculation of ΔG° from Equilibrium Constant

The standard redox potential values for two half-reactions are given below. The value for Faraday’s constant is 96.48 kJ V-1 mol-1 and Gas constant R is 8.3 J K-1 mol-1. NAD+ + H+ + 2e- → NADH    E° = −0.315 V FAD + 2H+ + 2e- → FADH2    E° = −0.219 V Q.52 The ΔG° for the oxidation of NADH by FAD is Options: (A) −9.25 kJ mol-1 (B) −103.04 kJ mol-1 (C) +51.52 kJ mol-1 (D) −18.5 kJ mol-1

ΔG° for Oxidation of NADH

The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right-handed α-helix. Q.51 The number of amino acid residues present in the protein is Options: (A) 15 (B) 18 (C) 17 (D) 20

Transmembrane α-Helix

The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right-handed α-helix. Q.50 The number of α-helical turns permeating the membrane is Options: (A) 5.6 turns (B) 3.5 turns (C) 6.5 turns (D) 5.0 turns

Transmembrane Alpha Helix Turns Calculation

A culture of Rhizobium is grown in a chemostat (100 m3 bioreactor). The feed contains 12 g/L sucrose. Ks for the organism is 0.2 g/L and μmax = 0.3 h-1. Q.49 If Yx/s = 0.4 g/g for the above culture and steady state cell concentration in the bioreactor is 4 g/L, the resulting substrate concentration will be Options: (A) 2 g/L (B) 8 g/L (C) 4 g/L (D) 6 g/L

Calculation of Substrate Concentration

A culture of Rhizobium is grown in a chemostat (100 m3 bioreactor). The feed contains 12 g/L sucrose. Ks for the organism is 0.2 g/L and μmax = 0.3 h-1. Q.48 The flow rate required to result in steady state concentration of sucrose as 1.5 g/L in the bioreactor will be Options: (A) 15 m3 h-1 (B) 26 m3 h-1 (C) 2.6 m3 h-1 (D) 150 m3 h-1

Chemostat Flow Rate Calculation

Q.47 Match Group I with Group II Group I Group II P. Real Time-PCR 1. Biochips Q. 2-D Electrophoresis 2. Syber Green R. Affinity chromatography 3. Antibody linked sepharose beads S. Microarray 4. Ampholytes Options: (A) P-1, Q-2, R-4, S-3 (B) P-2, Q-3, R-4, S-1 (C) P-2, Q-4, R-3, S-1 (D) P-3, Q-2, R-1, S-4

Match Molecular Biology Techniques

Q.46 True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila. The F1 progeny was observed to be with curved wings and tan body. The F1 progeny was again allowed to breed and produced flies of the following phenotype: 45 curved wings tan body, 15 short wings tan body, 16 curved wings dark body, and 6 short wings dark body. The mode of inheritance is Options: (A) Typical Mendelian with curved wings and tan body being dominant (B) Typical non-Mendelian with curved wings and tan body not following any pattern (C) Mendelian with suppression of phenotypes (D) Mendelian with single crossover

Inheritance Pattern of Wing Shape

Q.45 Thermal death of microorganisms in the liquid medium follows first-order kinetics. If the initial cell concentration in the fermentation medium is 103 cells/ml and the final acceptable contamination level is 101 cells/ml, for how long should the medium be treated at temperature of 120 °C (thermal deactivation rate constant = 0.23 min-1) to achieve acceptable load? Options: (A) 48 min (B) 11 min (C) 110 min (D) 20 min

Calculation of Thermal Death Time

Q.44 Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion (a): Cytoplasmic male sterility (CMS) is invariably due to defect(s) in mitochondrial function. Reason (r): CMS can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non-CMS plant. Options: (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) (a) is false but (r) is true (D) (a) is true but (r) is false

Cytoplasmic Male Sterility

Q.43 Match the products in Group I with the microbial cultures in Group II used for their industrial production Group I Group II P. Gluconic acid 1. Leuconostoc mesenteroides Q. L-Lysine 2. Aspergillus niger R. Dextran 3. Brevibacterium flavum S. Cellulase 4. Trichoderma reesei Options: (A) P-2, Q-1, R-3, S-4 (B) P-1, Q-3, R-4, S-2 (C) P-2, Q-3, R-1, S-4 (D) P-3, Q-2, R-4, S-1

Match Industrial Products with Microbial Cultures

Q.42 A cell has five molecules of a rare mRNA. Each cell contains 4 × 105 mRNA molecules. How many clones one will need to screen to have 99% probability of finding at least one recombinant cDNA of the rare mRNA, after making cDNA library from such cell? Options: (A) 4.50 × 105 (B) 3.50 × 105 (C) 4.20 × 105 (D) 4.05 × 105

Calculation of Number of Clones Required

Q.41 Receptor R is over expressed in CHO cells and analysed for expression. 6 × 107 cells were incubated with its radioactive ligand (specific activity 100 counts per picomole). If the total counts present in the cell pellet was 1000 cpm, the average number of receptors R per cell is (assume complete saturation of receptors with ligand and one ligand binds to one receptor) Options: (A) 104 (B) 103 (C) 105 (D) 107

Calculation of Receptor Number

Q.40 The dissociation constant Kd for ligand binding to the receptor is 10-7 M. The concentration of ligand required for occupying 10% of receptors is Options: (A) 10-6 M (B) 10-7 M (C) 10-8 M (D) 10-9 M

Ligand Concentration Required

Q.39 Dizygotic twins are connected to a single placenta during their embryonic development. These twins Options: (A) have identical MHC haplotypes (B) have identical TH cells (C) have identical T cells (D) can accept grafts from each other (both (A) and (B))

Dizygotic Twins Single Placenta Myth

Q.38 A mutant Gα protein with increased GTPase activity would Options: (A) not bind to GTP (B) not bind to GDP (C) show increased signaling (D) show decreased signaling

Gα Protein GTPase Mutant

Q.37 Match Group I with Group II Group I Group II P. Staphylococcus aureus 1. Biofilms Q. Candida albicans 2. Bacteriocins R. Mycobacterium tuberculosis 3. Methicillin resistance S. Lactobacillus lactis 4. Isoniazid Options: (A) P-1, Q-4, R-2, S-3 (B) P-2, Q-3, R-1, S-4 (C) P-3, Q-1, R-4, S-2 (D) P-1, Q-2, R-4, S-3

Microbial Pathogens Matching

Q.36 Match the items in Group I with Group II Group I (Vectors) Group II (Maximum DNA packaging) P. λ phage 1. 35–45 kb Q. Bacterial Artificial Chromosomes (BACs) 2. 100–300 kb R. P1 derived Artificial Chromosomes (PACs) 3. ≤ 300 kb S. λ cosmid 4. 5–25 kb Options: (A) P-3, Q-4, R-1, S-2 (B) P-1, Q-3, R-2, S-4 (C) P-4, Q-3, R-2, S-1 (D) P-1, Q-2, R-3, S-4

Cloning Vectors Packaging Capacities

Q.35 The turnover numbers for the enzymes E1 and E2 are 150 s-1 and 15 s-1 respectively. This means Options: (A) E1 binds to its substrate with higher affinity than E2 (B) The velocity of reactions catalyzed by E1 and E2 at their respective saturating substrate concentrations could be equal, if concentration of E2 used is 10 times that of E1 (C) The velocity of E1 catalyzed reaction is always greater than that of E2 (D) The velocity of E1 catalyzed reaction at a particular enzyme concentration and saturating substrate concentration is lower than that of E2 catalyzed reaction under the same conditions

Enzyme Turnover Numbers Explained

Q.34 Consider the following statements: I. T4 DNA ligase can catalyze blunt end ligation more efficiently than E. coli DNA ligase. II. The ligation efficiency of T4 DNA ligase can be increased with PEG and ficoll. Options: (A) Only I is true (B) Both I and II are true (C) Only II is true (D) I is true and II is false

T4 DNA Ligase vs E. coli DNA Ligase

Q.33 Match the promoters listed in Group I with the tissues listed in Group II Group I Group II P. α-Amylase 1. Endosperm Q. Glutenin 2. Tuber R. Phaseollin 3. Aleurone S. Patatin 4. Cotyledon Options: (A) P-3, Q-1, R-4, S-2 (B) P-3, Q-4, R-1, S-2 (C) P-4, Q-2, R-1, S-3 (D) P-1, Q-3, R-2, S-4

Tissue-Specific Expression in Plants

Q.32 Match Group I with Group II Group I Group II P. Fibronectin 1. Uptake of amino acids and glucose Q. Insulin 2. Trypsin inhibitor R. α-Macroglobulin 3. Binds iron S. Transferrin 4. Cell attachment to substratum Options: (A) P-2, Q-1, R-4, S-3 (B) P-3, Q-2, R-1, S-4 (C) P-4, Q-2, R-1, S-3 (D) P-4, Q-1, R-2, S-3

Fibronectin, Insulin, α-Macroglobulin and Transferrin Functions

Q.31 Match the chemicals in Group I with the possible type/class in Group II Group I Group II P. Picloram 1. Vitamin Q. Zeatin 2. Auxin R. Thiamine 3. Amino Acid S. Glutamine 4. Cytokinin Options: (A) P-2, Q-4, R-1, S-3 (B) P-4, Q-1, R-2, S-3 (C) P-3, Q-1, R-2, S-4 (D) P-4, Q-2, R-1, S-3

Picloram Zeatin Thiamine Glutamine Matching

Q.30 Match the following antibiotics in Group I with their mode of action in Group II Group I P. Chloramphenicol Q. Norfloxacin R. Puromycin S. Rifampicin Group II 1. Binds to DNA gyrase 2. Binds to RNA polymerase 3. Inhibits peptidyl transferase 4. Mimics aminoacyl-tRNA (A) P-1, Q-3, R-2, S-4 (B) P-3, Q-1, R-2, S-4 (C) P-3, Q-1, R-4, S-2 (D) P-4, Q-2, R-3, S-1

Antibiotics Matching

Q.29 Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion (a): MTT assay is used to determine cell viability based on the principle of colour formation by DNA fragmentation. Reason (r): MTT assay is used to determine cell viability based on the colour development by converting tetrazolium soluble salt to insoluble salt. (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) (a) is true but (r) is false (D) (a) is false but (r) is true

DNA Fragmentation vs Tetrazolium Reduction

Q.28 A roller bottle culture vessel perfectly cylindrical in shape having inner radius (r) = 10 cm and length (l) = 20 cm was fitted with a spiral film of length (L) = 30 cm and width (W) = 20 cm. If the film can support 104 anchorage dependent cells per cm2, the increase in the surface area after fitting the spiral film and the additional number of cells that can be grown respectively are (A) 1200 cm2 and 12 × 107 cells (B) 600 cm2 and 6 × 107 cells (C) 600 cm2 and 8300 cells (D) 1200 cm2 and 8300 cells

Increase in Surface Area and Cell Yield

Q.27 An immobilized enzyme being used in a continuous plug flow reactor exhibits an effectiveness factor (η) of 1.2. The value of η being greater than 1.0 could be apparently due to (A) Substrate inhibited kinetics with internal pore diffusion limitation (B) External pore diffusion limitation (C) Sigmoidal kinetics (D) Instability of the enzyme

Why Immobilized Enzyme Effectiveness Factor

Q.26 Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion (a): Somatic embryogenesis in plants is a two step process comprising of embryo initiation followed by embryo production. Reason (r): Embryo initiation is independent of the presence of 2,4-dichlorophenoxyacetic acid whereas embryo production requires a high concentration of 2,4-dichlorophenoxyacetic acid. (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) (a) is true but (r) is false (D) (a) is false but (r) is true

Somatic Embryogenesis

Q.25 The formation of peptide cross-links between adjacent glycan chains in cell wall synthesis is called (A) Transglycosylation (B) Autoglycosylation (C) Autopeptidation (D) Transpeptidation

Cell Wall Peptide Cross-Links

Q.24 Nuclease-hypersensitive sites in the chromosomes are sites that appear to be (A) H2 and H4 histone free (B) H1 and H2 histone free (C) H3 and H4 histone free (D) Nucleosome free

Nuclease-Hypersensitive Sites

Q.23 Oxidation reduction reactions with positive standard redox potential (ΔE°′) have (A) Positive ΔG°′ (B) Negative ΔG°′ (C) Positive ΔE°′ (D) Negative ΔE°′

Oxidation-Reduction Reactions Positive ΔE°′

Q.22 The degree of inhibition for an enzyme catalyzed reaction at a particular inhibitor concentration is independent of initial substrate concentration. The inhibition follows (A) Competitive inhibition (B) Mixed inhibition (C) Un-competitive inhibition (D) Non-competitive inhibition

Enzyme Inhibition

Q.21 Identify the enzyme that catalyzes the following reaction α-Ketoglutarate + NADH + NH4+ + H+  →  Glutamate + NAD+ + H2O (A) Glutamate synthetase (B) Glutamate oxoglutarate aminotransferase (C) Glutamate dehydrogenase (D) α-ketoglutarate deaminase

Glutamate Dehydrogenase

Q.20 During lactic acid fermentation, net yield of ATP and NADH per mole of glucose is (A) 2 ATP and 2 NADH (B) 2 ATP and 0 NADH (C) 4 ATP and 2 NADH (D) 4 ATP and 0 NADH

Lactic Acid Fermentation

Q.19 In a chemostat operating under steady state, a bacterial culture can be grown at dilution rate higher than maximum growth rate by (A) Partial cell recycling (B) Using sub-optimal temperature (C) pH cycling (D) Substrate feed rate cycling

Chemostat Steady State

Q.18 Expressed Sequence Tag is defined as (A) A partial sequence of a cDNA randomly selected from cDNA library (B) The characteristic gene expressed in the cell (C) The protein coding DNA sequence of a gene (D) Uncharacterized fragment of DNA present in the cell

Expressed Sequence Tags

Q.17 Accession number is a unique identification assigned to a (A) Single database entry for DNA/Protein (B) Single database entry for DNA only (C) Single database entry for Protein only (D) Multiple database entry for DNA/Protein

Accession Numbers

Q.16 Somatic cell gene transfer is used for P. Transgenic animal production Q. Transgenic diploid cell production R. In-vitro fertilization S. Classical breeding of farm animals (A) P, R and S (B) P, Q and R (C) P and R (D) P only

Somatic Cell Gene Transfer Applications

Q.15 Lymphocytes interact with foreign antigens in (A) Bone marrow (B) Peripheral blood (C) Thymus (D) Lymph nodes

Lymphocyte-Antigen Interactions

Q.14 A neonatally thymectomized mouse, immunized with protein antigen shows (A) Both primary and secondary responses to the antigen (B) Only primary response to the antigen (C) Delayed type hypersensitive reactions (D) No response to the antigen

Neonatal Thymectomy

Q.13 A culture of bacteria is infected with bacteriophage at a multiplicity of 0.3. The probability of a single cell infected with 3 phages is (A) 0.9 (B) 0.27 (C) 0.009 (D) 0.027

Bacteriophage MOI 0.3

Q.12 Interferon-β is produced by (A) Bacteria infected cells (B) Virus infected cells (C) Both virus and bacteria infected cells (D) Fungi infected cells

Interferon-β Production

Q.11 Which one of the following DOES NOT belong to the domain of Bacteria? (A) Cyanobacteria (B) Proteobacteria (C) Bacteroids (D) Methanobacterium

Bacteria Domain

Q.10 An example for template independent DNA polymerase is (A) DNA Polymerase I (B) RNA polymerase (C) Terminal deoxynucleotidyl transferase (D) DNA polymerase III

Template-Independent DNA Polymerase

Q.9 Under stress conditions bacteria accumulate (A) ppGpp (Guanosine tetraphosphate) (B) pppGpp (Guanosine pentaphosphate) (C) Both ppGpp and pppGpp (D) Either ppGpp or pppGpp

Bacterial Stress Response

Q.8 During transcription (A) DNA Gyrase introduces negative supercoils and DNA Topoisomerase I removes negative supercoils (B) DNA Topoisomerase I introduces negative supercoils and DNA Gyrase removes negative supercoils (C) Both DNA Gyrase and DNA Topoisomerase I introduce negative supercoils (D) Both DNA Gyrase and DNA Topoisomerase I remove negative supercoils

DNA Supercoiling Management During Transcription

Q.7 In transgenics, alterations in the sequence of nucleotide in genes are due to P. Substitution Q. Deletion R. Insertion S. Rearrangement (A) P and Q (B) P, Q and R (C) Q and R (D) R and S

Transgenic Gene Alterations

Q.6 Peptidyl transferase activity resides in (A) 16S rRNA (B) 23S rRNA (C) 5S rRNA (D) 28S rRNA

Peptidyl Transferase

Q.5 Program used for essentially local similarity search is (A) BLAST (B) RasMol (C) ExPASy (D) SWISS-PROT

Program for Local Similarity Search

Q.4 Antibiotic resistance marker that CANNOT be used in a cloning vector in Gram negative bacteria is (A) Streptomycin (B) Ampicillin (C) Vancomycin (D) Kanamycin

Antibiotic Resistance Markers

Q.3 The bacteria known to be naturally competent for transformation of DNA is (A) Escherichia coli (B) Bacillus subtilis (C) Mycobacterium tuberculosis (D) Yersinia pestis

Naturally Competent Bacteria

Q.2 Ames test is used to determine (A) The mutagenicity of a chemical (B) Carcinogenicity of a chemical (C) Both mutagenicity and carcinogenicity of a chemical (D) Toxicity of a chemical

Ames Test

Q.1 Hybridoma technology is used to produce (A) Monoclonal antibodies (B) Polyclonal antibodies (C) Both monoclonal and polyclonal antibodies (D) B cells

Hybridoma Technology

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