GATE BT 2013 solutions (1-30)

GATE 2013 solutions (1-30)

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

SOL: If one recalls RNA primary structure, easy guess can be made, as it is having a free OH group in 2C and 3C it is susceptible to hydrolysis. SO the answer is D.

2) Which one of the following modifications is common to both protein and DNA?

(A) SUMOylation (B) Nitrosylation (C)Methylation (D) Ubiquitination

SOL:  Both DNA & Protein undergo methylation. DNA is methylated in DNA regulation, Protein methylation typically takes place on arginine or lysine amino acid( positive charged)  residues in the protein sequence.

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

SOL: studies involving mutation of genes coding for protein A resulted in a lowered virulence of S. aureus as measured by survival in blood, which has led to speculation that protein A-contributed virulence requires binding of antibody Fc regions.

4) The first humanized monoclonal antibody approved for the treatment of breast canceris

(A) Rituximab (B) Cetuximab (C) Bevacizumab (D) Herceptin

SOL: D

5) Which one of the following aminoacids in proteins does NOT undergo phosphorylation?

(A) Ser (B) Thr (C) Pro (D) Tyr

SOL:  ser,thr,tyr has OH group,where as pro doesn’t, so doesn’t undergo phosphorylation.

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

SOL: prolong the persistence of antigen , Nonreplicating, purified subunit or synthetic viral vaccines of the future are likely to be weak immunogens that will require immunopotentiation if they are to be effective. These marginal vaccines could be improved by combination with potent and safe immunologic adjuvants. The use of adjuvants should also reduce the amount of purified antigen required for successful immunization, thus making vaccine production more economical and more feasible.

7)  Endogenous antigens are presented on to the cell surface along with

(A) MHC-II (B) MHC-I (C)  Fcγreceptor (D) complement receptor

SOL:  T helper cells has MHC-II and Endogenous antigens are presented on to the cell surface.

8) Human genome sequencing project involved the construction of genomic library in

(A)bacterial artificial chromosome (B)pBR322 (C) bacteriophage (D) pcDNA3.1

SOL: BAC used for cloning large fragments, so it is used in HGS,

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

SOL: 2′,3′-dideoxy nucleoside triphosphate, Dna nucleotides add in 5’–à3’. So if no 3’ reaction terminated.

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

SOL:  Horizontal gene transfer (HGT) is defined as the transfer of genetic material between bacterial cells uncoupled with cell division.

11)  Phylum proteobacteria is subdivided into -, -, -, - and -proteobacteria based on

(A) G+C content (B) 23S rRNA sequences (C)tRNA sequences (D)16S rRNA sequences

SOL: simple guess, 16S rRNA sequences makes small sub unit of ribosomes, phylum divided based on it,

12 Which one of the following is an ABC transporter?

(A)multidrug resistance protein (B) acetylcholine receptor (C)bacteriorhodopsin (D) ATP synthase

SOL:  refer rathnaswamy SIR module1 cellbiology 30.

13)The catalytic efficiency for an enzyme is definedas (A) kcat (B)  (C)

SOL: Kcat/Km

14) 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

SOL: doono

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

SOL: DNA polymeraseI removes RNA primers synthesized,

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

SOL: PAM 40 or BLOSSUM 80 , is used for less divergent alignments.

17) If        2 2 1 1 P ,        2 2 2 1 Q and        1 3 3 0 R ,which one of the following statements is TRUE?

(A) PQ = PR (B) QR= RP (C) QP = RP (D) PQ = QR

SOL: calculate  by calci 😛

18)SOL: ans is D (ie) 1

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?

(A) ½ sons and ½ daughters (B) all daughters and no sons (C) all sons and no daughters (D) ¼ daughters and ¼ sons

SOL:

X Y
X’ X’X X’Y
X XX XY

Half sons and half daughters,

20) One of theeigen values  of  XXXX is

(A) 2 (B) 4 (C) 6 (D) 8

SOL: simple maths

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 _________

SOL:  (1.5*5)-5/5= 0.5

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-1h-1) after 3h of steady state operation will be ___________

SOL: Productivity is DX=0.6*30=18.

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. _________

SOL:  N2=N1(D1/D2)

=450(1/10)=45.

24)  Calculate the percentage sequence identity for the pairwise alignment given below. _______

H E L L O –             Y E L L O W

SOL :      66.5

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). ___________

SOL: Yp/s=0.215/0.25=0.86.

26)  Match the commercial microbial sources in Group I with the products in Group II.

A B
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)   (B) P-3,Q-1,R-4,S-2

(C)   (C) P-1,Q-3,R-2,S-4

(D)    (D) P-1,Q-3,R-4,S-2

SOL:  B, Principles Of Fermentation technology pg;106

27)

A B
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

SOL:  C,

Lets start up with  S . easy to guess that answer for S  is 1, as Q name suggest answer must be some where near to polarity (ie) 4.

And for P answer is 3, so combine all and guess answer,

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.

SOL: A

Immobilization of plant cells can enhance secondary metabolite production during bioreactor cultivation by protecting the plant cells from shear forces in the bioreactor,

29)

A B
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

SOL:  Methano bacteria needs pseudomurine in cell wall for surviving in harsh conditions, clostridium is a genus of bacteria capable of producing endospores,spirillum id a bipolar flagella.

30)

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

SOL: D

Sulfonamide is used up in folic acid biosynthesis, close looking at answers we come to a conclusion that quinolones have something to do with  topoisomerase , next look for erythromycin, as it antibiotic it must inhibit either protein synthesis or PPDG layer, Cephalosporins are bactericidal and have the same mode of action as other beta-lactam antibiotics (such as penicillins) but are less susceptible to penicillinases. Cephalosporins disrupt the synthesis of the peptidoglycan layer of bacterial cell walls.and finally answer is D.

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About YANAMALA VIJAY RAJ

Mtech in Clinical Eng Jointly offered by Indian institute of technology Madras& Christian medical college Vellore& Sree chitra tirunal institute for medical sciences and technology Trivandrum.
This entry was posted in GATE and tagged . Bookmark the permalink.

4 Responses to GATE BT 2013 solutions (1-30)

  1. koorishma says:

    hey thank u so much….

  2. Dhanya says:

    Answer to question number 14 may be obtained as follows:
    Allotriploid plants contain two genomes of one species and one genome from a different species.
    In this question the allotriploid may be obtained by either getting 2 haploid genomes from species 1 i.e 36 chromosomes + a haploid set from species 2 i.e 28/2. so chromosome number = 36+14=50 or
    a haploid set from species 1 (i.e 36/2=18) + 2 haploid sets from species 2 = 18+28=46
    so the answer is 46 or 50

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