GATE Biotechnology 2013 Previous Year Papers with Solutions Archives - Page 4 of 7 - CSIR NET LIFE SCIENCE COACHING | NTA NET LIFE SCIENCE | CSIR LIFE SCIENCE

An Institute for CSIR NET Life Science- DBT JRF/ M Sc Entrance
letstalkacademy@gmail.com

Q.65 All professors are researchers Some scientists are professors Which of the given conclusions is logically valid and is inferred from the above arguments: (A) All scientists are researchers (B) All professors are scientists (C) Some researchers are scientists (D) No conclusion follows

All Professors Are Researchers Some Scientists Are Professors

Q.64 If |−2X + 9| = 3, then the possible value of |−X| − X2 would be: (A) 30 (B) −30 (C) −42 (D) 42

Find Possible Values of |−X| − X²

Q.63 The following table gives data on tourists from different countries visiting India in the year 2011. Country Number of Tourists USA 2000 England 3500 Germany 1200 Italy 1100 Japan 2400 Australia 2300 France 1000 Which two countries contributed to one third of the total number of tourists who visited India in 2011? (A) USA and Japan (B) USA and Australia (C) England and France (D) Japan and Australia

Tourist Data Interpretation

Q.62 In a factory, two machines M1 and M2 manufacture 60% and 40% of the autocomponents respectively. Out of the total production, 2% of M1 and 3% of M2 are found to be defective. If a randomly drawn autocomponent from the combined lot is found defective, what is the probability that it was manufactured by M2? (A) 0.35 (B) 0.45 (C) 0.5 (D) 0.4

Bayes Theorem Factory

Q.61 Velocity of an object fired directly in upward direction is given by V = 80 − 32t, where t (time) is in seconds. When will the velocity be between 32 m/sec and 64 m/sec? (A) (1, 3/2) (B) (1/2, 1) (C) (1/2, 3/2) (D) (1, 3)

When is Velocity 32-64 in V=80-32t?

Q.60 Select the pair that best expresses a relationship similar to that expressed in the pair: water: pipe:: (A) cart: road (B) electricity: wire (C) sea: beach (D) music: instrument

Water Pipe Analogy Question

Q.59 All engineering studentsI should learnII mechanics, mathematicsIII and how to do computation.IV Which of the above underlined parts of the sentence is not appropriate? (A) I (B) II (C) III (D) IV

All Engineering Students Should Learn

Q.58 Mahatama Gandhi was known for his humility as (A) he played an important role in humiliating exit of British from India. (B) he worked for humanitarian causes. (C) he displayed modesty in his interactions. (D) he was a fine human being.

Mahatma Gandhi Humility

Q.57 The Headmaster ___________ to speak to you. Which of the following options is incorrect to complete the above sentence? (A) is wanting (B) wants (C) want (D) was wanting

Headmaster to Speak

Q.56 If 3 ≤ X ≤ 5 and 8 ≤ Y ≤ 11, then which of the following options is TRUE? (A)3⁄5 ≤ X⁄Y ≤ 8⁄5 (B)3⁄11 ≤ X⁄Y ≤ 5⁄8 (C) 3⁄11 ≤ X⁄Y ≤ 8⁄5 (D) 3⁄5 ≤ X⁄Y ≤ 8⁄11

Inequality of Ratios

During sterilization of a fermentation medium in a given bioreactor, Vheating = 12.56, Vcooling = 7.48 and the total value of V required for the whole sterilization process is 52, where V is the design criteria. Q.55 What is the holding period (min) at a k value of 3.36 min-1? (A) 10.6 (B) 9.5 (C) 8.4 (D) 7.2

Fermentation Medium Sterilization

During sterilization of a fermentation medium in a given bioreactor, Vheating = 12.56, Vcooling = 7.48 and the total value of V required for the whole sterilization process is 52, where V is the design criteria. Q.54 What is the value of Vholding? (A) 31.96 (B) 42.32 (C) 52.43 (D) 61.18

Bioreactor Sterilization

A DNA fragment of 5000 bp needs to be isolated from E. coli (genome size 4 × 103 kb) genomic library. Q.53 The number of clones required to represent this fragment in a genomic library with a probability of 95% are: (A) 5.9 × 103 (B) 4.5 × 103 (C) 3.6 × 103 (D) 2.4 × 103

E. coli Genomic Library

A DNA fragment of 5000 bp needs to be isolated from E. coli (genome size 4 × 103 kb) genomic library. Q.52 The minimum number of independent recombinant clones required to represent this fragment in a genomic library are: (A) 16 × 102 (B) 12 × 102 (C) 8 × 102 (D) 1.25 × 102

Minimum Recombinant Clones

The binding efficiency of three different receptorsR1, R2 and R3 were tested against a ligand using equilibrium dialysis, with a constant concentration of receptor and varying concentrations of ligand. The Scatchard plot of receptor titration with different concentration of ligand is given below (ris moles of bound ligand per moles of receptor and c is concentration of free ligand) Q.51 Which one of the receptors has thehighestaffinity for the ligand? (A) R1 (B) R2 (C) R3 (D) R4

Identifying the Highest Ligand Affinity Receptor

The binding efficiency of three different receptorsR1, R2 and R3 were tested against a ligand using equilibrium dialysis, with a constant concentration of receptor and varying concentrations of ligand. The Scatchard plot of receptor titration with different concentration of ligand is given below (ris moles of bound ligand per moles of receptor and c is concentration of free ligand) Q.50 The number of ligand binding sites present on receptors R1 and R3, respectively are (A)1 and 4 (B)1 and 1 (C)4 and 1 (D)2 and 2

Determining Ligand Binding Sites

A solution was prepared by dissolving 100 mg of protein X in 100 ml of water. Molecular weight of protein X is 15,000 Da; Avogadro’s number = 6.022 × 1023. Q.49 The number of molecules present in this solution is: (A) 40.15 × 1019 (B) 6.023 × 1019 (C) 4.015 × 1019 (D) 0.08 × 1019

Number of Molecules

A solution was prepared by dissolving 100 mg of protein X in 100 ml of water. Molecular weight of protein X is 15,000 Da; Avogadro’s number = 6.022 × 1023. Q.48 Calculate the molarity (μM) of the resulting solution. (A) 66.6 (B) 6.6 (C) 0.67 (D) 0.067

Calculate Protein X Solution Molarity

Q.47 A fed batch culture was operated with intermittent addition of glucose solution at a flow rate of 200 ml h−1. The values of Ks, μm and D are 0.3 g l−1, 0.4 h−1 and 0.1 h−1, respectively. Determine the concentration of growth limiting substrate (g l−1) in the reactor at quasi-steady state. __________

Growth-Limiting Substrate Concentration

Q.46 The maximum cell concentration (g l−1) expected in a bioreactor with initial cell concentration of 1.75 g l−1 and an initial glucose concentration of 125 g l−1 (Yx/s = 0.6 g cell/g substrate) __________.

Maximum Cell Concentration

Q.45 One percent of the cars manufactured by a company are defective. What is the probability (upto four decimals) that more than two cars are defective, if 100 cars are produced? __________

Probability More Than 2 Defective Cars in 100

Q.44 In a genetic study, 80 people were found to have alleles for polydactyly. Only 36 of them were polydactylous. What is the extent of penetrance percentage? ________

Polydactyly Penetrance Calculation

Q.43 The total number of fragments generated by the complete and sequential cleavage of the polypeptide given below by Trypsin followed by CNBr is _________ Phe-Trp-Met-Gly-Ala-Lys-Leu-Pro-Met-Asp-Gly-Arg-Cys-Ala-Gln

Total Fragments from Trypsin-CNBr Cleavage of Polypeptide

Q.42 A complete restriction digestion of a circular plasmid (5000bp) was carried out with HindIII,BamHIandEcoRIindividually. Restriction digestion yielded following fragments. Plasmid + HindIII→ 1200bp and3800bp Plasmid + BamHI→5000bp Plasmid + EcoRI→ 2500bp The number of sites for EcoRI,BamHIandHindIIIpresent on this plasmid are (A)EcoRI-2, BamHI-1, HindIII-2 (B)EcoRI-1, BamHI-1, HindIII-2 (C)EcoRI-3, BamHI-2, HindIII-1 (D)EcoRI-2, BamHI-2, HindIII-1

Decoding Plasmid Restriction Digestion

Q.41 The solution to dy/dx + y cot x = cosec x is y = (c + x) cot x y = (c + x) cosec x y = (c + x) cosec x cot x y = (c + x) cosec x / cot x

Solution of Differential Equation

Q.40 The solution of the following set of equations is x + 2y + 3z = 20 7x + 3y + z = 13 x + 6y + 2z = 0 x = −2, y = 2, z = 8 x = 2, y = −3, z = 8 x = 2, y = 3, z = −8 x = 8, y = 2, z = −3

Solving a System of Linear Equations

Q.39 The Laplace transform of f(t) = 2t + 6 is 1 / s + 2 / s2 3 / s − 6 / s2 6 / s + 2 / s2 −6 / s + 2 / s2

Laplace Transform of f(t) = 2t + 6

Q.38 Evaluate limx→∞ x tan 1/x ∞ 1 0 −1

limx→∞ x tan(1/x)

Bioinformatics Matching

Q.36 Determine the correctness or otherwise of the following Assertion (a) and Reason (r). Assertion:UPGMA method produces ultrametric tree. Reason:Sequence alignment is converted into evolutionary distances in UPGMA method. (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

UPGMA Ultrametric Trees

Q.35 Match the entries in Group Iwith the entries in Group II. Group I Group II P. RNAse P 1. Polyadenylation Q. RNase H 2. Splicing R. snRNAs 3. Ribozymes S. CstF 4. DNA-RNAhybrids (A) P-3, Q-4, R-2, S-1 (B) P-4, Q-3, R-2, S-1 (C) P-3, Q-2, R-1, S-4 (D) P-2, Q-4, R-1, S-3

RNA Processing Enzyme Matching

Q.34 The activity of an enzyme was measured by varying the concentration of the substrate (S) in the presence of three different concentrations of inhibitor (I) 0, 2 and 4 mM.The double reciprocal plot given belowsuggests that the inhibitor (I) exhibits (A) substrate inhibition (B) uncompetitive inhibition (C) mixed inhibition (D) competitive inhibition

Identification of Enzyme Inhibition

Q.33 Match the herbicides in Group Iwith the target enzymesin Group II. Group I Group II P. Glyphosate 1. Nitrilase Q. Bromoxynil 2. Acetolactatesynthetase R. Sulphonylureas 3. Dehalogenase S. Dalapon 4. 5-Enolpyruvyl shikimate3-phosphate synthase (A) P-4, Q-1, R-2, S-3 (B) P-2, Q-1, R-4, S-3 (C) P-4, Q-3, R-2, S-1 (D) P-3, Q-2, R-4, S-1

Herbicide Enzyme Targets Matching

Q.32 Match the entries inGroup I with the enzymes inGroup II. Group I Group II P. NAD+ 1. Glutathione peroxidase Q. Selenium 2. Nitrogenase R. Pyridoxal phosphate 3. Lactate dehydrogenase S. Molybdenum 4. Glycogen phosphorylase (A) P-3, Q-2, R-4, S-1 (B) P-4, Q-1, R-3, S-2 (C) P-3, Q-1, R-4, S-2 (D) P-3, Q-4, R-2, S-1

Enzyme Cofactor Matching

Q.31 In nature, Agrobacterium tumefaciens mediated infection of plant cells leads to P. crown gall disease in plants Q. hairy root disease in plants R. transfer of T-DNA into the plant chromosome S. transfer of Ri-plasmid into the plant cell (A) S only (B) P and R only (C) Q and S only (D) Q only

Agrobacterium tumefaciens Crown Gall Disease

Q.30 Match the antibioticsinGroup Iwith the targets inGroup II. Group I Group II P. Sulfonamide 1. Peptidoglycan synthesis Q. Quinolones 2. Peptide chain elongation R. Erythromycin 3. Folic acid biosynthesis S. Cephalosporin 4. Topoisomerase (A) P-3, Q-4, R-1, S-2 (B) P-2, Q-4, R-3, S-1 (C) P-4, Q-1, R-2, S-3 (D) P-3, Q-4, R-2, S-1

Antibiotic Targets Matching

Q.29 Match the cell structures in Group I with the organismsin Group II. Group I Group II P. Endospores 1. Methanobacterium Q. Bipolar flagella 2. Treponema R. Pseudomurine in cell wall 3. Spirillum S. Periplasmic flagella 4. Clostridium (A) P-4, Q-3, R-1, S-2 (B) P-4, Q-3, R-2, S-1 (C) P-3, Q-4, R-1, S-2 (D) P-4, Q-1, R-3, S-2

Bacterial Cell Structures Matching

Q.28 Determine the correctness or otherwise of the following Assertion (a) and Reason (r). Assertion: Immobilization of plant cells can enhance secondary metabolite production during bioreactor cultivation. Reason: Immobilization protects the plant cells from shear forces in the bioreactor. (A) Both (a) and (r) are true and (r) is the correct reason for (a). (B) Both (a) and (r) are true but (r) is not the correct reason for (a). (C) (a) is true but (r) is false. (D) (a) is false but (r) is true.

Plant Cell Immobilization Enhances Secondary Metabolites

Q.27 Match the entries in the Group I with the elution conditions in Group II. Group I Group II P. Ion-exchange chromatography 1. Isocratic solvent Q. Hydrophobic column chromatography 2. Ampholytes R. Gel filtration chromatography 3. Increasing gradient of salt S. Chromatofocusing 4. Decreasing gradient of polarity (A) P-4,Q-1,R-2,S-3 (B) P-4,Q-3,R-1,S-2 (C) P-3,Q-4,R-1,S-2 (D) P-3,Q-4,R-2,S-1

Chromatography Elution Matching

Q.26 Match the commercial microbial sources in Group I with the products in Group II. Group I Group II P. Corynebacteriumlilium 1. 2,3-Butane di-ol Q. Klebsiellaoxytoca 2. Poly-β-hydroxybutyric acid R. Aspergillusniger 3. Glutamic acid S. Alcaligeneseutrophus 4. Citric acid (A) P-3,Q-1,R-2,S-4 (B) P-3,Q-1,R-4,S-2 (C) P-1,Q-3,R-2,S-4 (D) P-1,Q-3,R-4,S-2

Corynebacterium lilium Glutamic Acid, Klebsiella 2,3-Butanediol Match Explained

Q.25 In a batch culture, the specific rate of substrate utilization is 0.25 g (g cell mass)−1 h−1 and specific rate of product formation is 0.215 g (g cell mass)−1 h−1. Calculate the yield of product from the substrate (Yp/s). __________

Batch Culture Yp/s Calculation

Q.24 Calculate the percentage sequence identity for the pairwise alignment given below. __________ H E L L O - Y E L L O W

HELLO- vs YELLOW Pairwise Alignment

Q.23 A batch bioreactor is to be scaled up from 10 to 10,000 liters. The diameter of the large bioreactor is 10 times that of the small bioreactor. The agitator speed in the small bioreactor is 450 rpm. Determine the agitator speed (rpm) of the large bioreactor with same impeller tip speed as that of the small bioreactor. __________

Bioreactor Scale-Up

Q.22 A chemostat is operated at a dilution rate of 0.6 h−1. At steady state, the biomass concentration in the exit stream was found to be 30 g l−1. The biomass productivity (g l−1 h−1) after 3 h of steady state operation will be __________.

Chemostat Biomass Productivity

Q.21 A callus of 5 g dry weight was inoculated on semi-solid medium for growth. The dry weight of the callus was found to increase by 1.5 fold after 10 days of inoculation. The growth index of the culture is _________

Callus Growth Index

Q.20 One of the eigenvalues of P = [ 10 −4 18 −12 ] is 2 4 6 8

Finding the Eigenvalues of a 2×2 Matrix

Q.19 Hypophosphatemia is manifested by an X-linked dominant allele. What proportion of the offsprings from a normal male and an affected heterozygous female will manifest the disease? ½ sons and ½ daughters all daughters and no sons all sons and no daughters ¼ daughters and ¼ sons

Hypophosphatemia X-Linked Dominant

Q.18 If u = log (ex + ey), then ∂u/∂x + ∂u/∂y = ex + ey ex − ey 1 / (ex + ey) 1

Finding ∂u/∂x + ∂u/∂y

Q.17 If P = [ 1 1 2 2 ], Q = [ 2 1 2 2 ] and R = [ 3 0 1 3 ], which one of the following statements is TRUE? PQ = PR QR = RP QP = RP PQ = QR

Matrix Multiplication Properties

Q.16 The suitable substitution matrix to align closely related sequences is (A) PAM 250 or BLOSSUM 80 (B) PAM 40 or BLOSSUM 80 (C) PAM 120 or BLOSSUM 40 (D) PAM 250 or BLOSSUM 40

PAM 40 or BLOSUM 80

Q.15 The RNA primer synthesized during the replication process in bacteria is removed by (A) DNA gyrase (B)primase (C) DNA polymeraseI (D) DNA polymerase II

RNA Primer Removal in Bacteria

Q.14 Of the two diploid species, species I has 36 chromosomes and species II has 28 chromosomes. How many chromosomes would be found in an allotriploid individual? (A) 42 or 54 (B)46 or 50 (C) 74 or 86 (D) 84 or 108

Allotriploid Chromosomes

Q.13 The catalytic efficiency for an enzyme is defined as kcat Vmax / kcat kcat / Km kcat / Vmax

Understanding Catalytic Efficiency in Enzymes

Q.12 Which one of the following is an ABC transporter? (A)multidrug resistance protein (B) acetylcholine receptor (C)bacteriorhodopsin (D) ATP synthase

Which One of the Following is an ABC Transporter?

Q.11 Phylum proteobacteria is subdivided into α-, β-, γ-, δ- and ε-proteobacteria based on (A) G + C content (B) 23S rRNA sequences (C) tRNA sequences (D) 16S rRNA sequences

Proteobacteria Classification

Q.10 In nature, the horizontal gene transfer across bacteria is mediated by (A) gene cloning followed by transformation (B)conjugation and transformation (C) conjugation only (D) transformation only

Horizontal Gene Transfer in Bacteria

Q.10 In nature, the horizontal gene transfer across bacteria is mediated by (A) gene cloning followed by transformation (B)conjugation and transformation (C) conjugation only (D) transformation only

Natural Bacterial HGT

Q.9 The nucleotide analogue used in DNA sequencing by chain termination method is (A) 1',3'-dideoxy nucleoside triphosphate (B) 2',3'-dideoxy nucleoside triphosphate (C) 2',4'-dideoxy nucleoside triphosphate (D) 2',5'-dideoxy nucleoside triphosphate

Sanger Sequencing Chain Terminator

Q.7 Endogenous antigens are presented on to the cell surface along with (A) MHC-II (B) MHC-I (C) Fcγreceptor (D) complement receptor

Endogenous Antigen Presentation

Q.6 The role of an adjuvant isto (A) prolong the persistence of antigen (B) cross link the antigen (C) increase the size of antigen (D) avoid inflammation

Vaccine Adjuvant Role

Q.5 Which one of the following aminoacids in proteins does NOT undergo phosphorylation? (A) Ser (B) Thr (C) Pro (D) Tyr

Protein Phosphorylation Sites

Q.4 The first humanized monoclonal antibody approved for the treatment of breast canceris (A) Rituximab (B) Cetuximab (C) Bevacizumab (D) Herceptin

Pioneer Humanized Antibody

Q.3 Protein A, which has strong affinity to Fc region of immunoglobulin, is extracted from (A)Saccharomyces cerevisiae (B)Staphylococcus aureus (C)Streptococcus pyogenes (D)Streptococcus sanguis

Protein A Extraction Source

Q.2 Which one of the following modifications is common toboth protein and DNA? (A) SUMOylation (B) Nitrosylation (C)Methylation (D) Ubiquitination

Protein-DNA Shared Modifications

Q.1 Under alkaline conditions, DNA is more stable than RNA because (A) RNA forms secondary structures (B) RNA is a single stranded molecule (C)RNA has uracil in place of thymidine (D)RNA is susceptible to hydrolysis

DNA vs RNA Stability in Alkali

Latest Courses

IIT JAM / CUET- PG

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

CSIR UGC NET Life Science Course

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

ICMR JRF Life Science

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