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GATE 2017-2018 :: GATE Biotechnology

GATE Biotechnology - Practice Test Paper 1
GATE Biotechnology - Practice Test Paper 2
  1. 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
    2. A.
    S only
    B.
    P and R only
    C.
    Q and S only
    D.
    Q only

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  2. Match the entries in Group I with the enzymes in Group 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
    3. 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

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  3. Match the herbicides in Group I with the target enzymes in 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
    4. 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

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  4. 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

    5. A.
    substrate inhibition
    B.
    uncompetitive inhibition
    C.
    mixed inhibition
    D.
    competitive inhibition

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  5. Match the entries in Group I with 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
    6. 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

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  6. 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.
    7. 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

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  7. Match the entries in the Group I with the entries in Group II.
    Group I                  Group II
    P. Threading    1. Gene duplication
    Q. FASTA           2. Fold prediction
    R. Profile           3. HMM
    S. Paralogs       4. k-tuple
    8. A.
    P-2, Q-1, R-3, S-4
    B.
    P-2, Q-4, R-3, S-1
    C.
    P-3, Q-4, R-2, S-1
    D.
    P-1, Q-4, R-3, S-2

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  8. The Laplace transform of f(t) = 2t + 6 is
    9. A.
    1/s + 2/s2
    B.
    3/s - 6/s2
    C.
    6/s + 2/s2
    D.
    -6/s + 2/s2

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  9. The solution of the following set of equations is
    x + 2y + 3z = 20
    7x + 3y + z = 13
    x + 6y + 2z = 0
    10. A.
    x = -2, y = 2, z = 8
    B.
    x = 2, y = -3, z = 8
    C.
    x = 2, y = 3, z = -8
    D.
    x = 8, y = 2, z = -3

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  10. The solution to dy/dx + y cot x = cosec x is
    11. A.
    y = (c+x) cot x
    B.
    y = (c+x) cosec x
    C.
    y = (c + x) cosec x cot x
    D.
    y = (c+x) cosec x/cot x

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