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MH SET 2013: 17 February, Paper III, Life Sciences

  • Date: 17 February 2013
  • Paper code: 17FEB_Life_Science_PIII
  • Language: Bilingual
  • 22 pages
MH SET 2013 question paper, 17 February 2013, Paper III, Life Sciences, page 1 of 22
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FEB - 34313/III
                                Life Sciences
                                   Paper III
Time Allowed : 2½ Hours]                           [Maximum Marks : 150
Note : This paper contains Seventy Five (75) multiple choice questions, each
       question carrying Two (2) marks. Attempt All questions.

1.   Thar desert stretches to Indian           3.   Dendrocronology     deals     with
     states like :                                  determination of age of a tree

     (A) Rajasthan,        Uttarakhand,             with :

         Madhya Pradesh                             (1) Diffused porous wood
     (B) Rajasthan, Haryana, NCR
                                                    (2) Ring porous wood
     (C) Rajasthan, Andhra Pradesh,
                                                    (3) Heart ring porous wood
         Jharkhand
                                                    (4) Sap wood
     (D) Rajasthan, Punjab, Gujrat
                                                    (A) 1 and 3
2.   Which of the following sequences
     indicates trend in the evolution of            (B) 2 and 3

     thallus in algae ?                             (C) 1 and 4

     (A) Unicellular → Heterotrichous               (D) 2 and 4
         → Colonial → Filamentous
                                               4.   The immediate products of meiosis
     (B) Unicellular → Filamentous →
                                                    in megaspore mother cell are :
         Heterotrichous → Colonial
                                                    (A) Eggs
     (C) Unicellular      → Colonial   →
                                                    (B) Megaspores
         Filamentous → Heterotrichous

     (D) Unicellular → Filamentous →                (C) Embryo sacs

         Colonial → Heterotrichous                  (D) Female gametophytes

                                           1                                    [P.T.O.
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5.   ‘Reserpine’ a therapeutically active         7.   Lolium is a long day plant, which

     compound used for hypertension and
                                                       of the following light treatments will
     epilepsy is obtained from :
                                                       result in flowering in this plant ?
     (A) Roots of Rauwolfia serpentina
                                                       (A) Red—Far Red
     (B) Roots of Glycyrrhiza glabra

                                                       (B) Far Red—Red—Far Red
     (C) Leaves      of        Podophyllum

          emodi                                        (C) Far Red

     (D) Rhizome          of     Dryopteris
                                                       (D) Red—Far Red—Red
          filixmas
                                                  8.   Allopolyploids eventually give rise to
6.   In plants first step in the synthesis
                                                       amphidiploids. These can be easily
     of   sulfur     containing    organic

     compounds is reduction of sulfate.                recognised      by     looking     at

     Which of the following amino acids                their :

     is synthesized in this process ?
                                                       (A) Morphological characters
     (A) Methionine
                                                       (B) Karyotype
     (B) Cysteine
                                                       (C) Fertility characters
     (C) Glutamine

                                                       (D) Meiotic analysis
     (D) Glycine

                                              2
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9.   In animal kingdom, there are               11. If gonadotropin releasing hormone

     animals who show true and false                was to be secreted in large amounts

     body   segmentation.      Animals              and at a constant rate rather than
                                                    in a pulsatile manner, one of the
     characterized by the presence of
                                                    following would happen :
     coelom and segmentation of the body
                                                    (A) Secretion      of    luteinising
     belong to the phylum :
                                                        hormone would increase at first
     (A) Rotifera
                                                        and then decrease
     (B) Annelida                                   (B) Secretion      of    luteinising

     (C) Platyhelminthes                                hormone        would    increase
                                                        indefinitely
     (D) Echinodermata
                                                    (C) Testosterone secretion in a
10. During last few centuries, our earth
                                                        male would be continuously
     has seen considerable extinction of
                                                        high
     fauna and flora. However, scientists
                                                    (D) Ostradiol secretion in a female
     came across some species, thought
                                                        would be continuously high
     to have been extinct by chance. The
                                                12. Prevention of polyspermy of an egg
     living fossil Latimeria chalumnae is
                                                    predominantly contributed by a
     an example of :
                                                    process is called :
     (A) Lung fish                                  (A) Acrosome reaction

     (B) Placoderm                                  (B) Cortical granule breakdown

     (C) Lobe-finned fish                           (C) Membrane receptor activation

     (D) Xiphosuran                                 (D) Inactivation of sperm

                                            3                                    [P.T.O.
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13. In amphibians, the embryonic gene                      16. Completion of the life cycle of the
    expression does not begin during
                                                               malarial parasite Plasmodium
    cleavage in early embryos. The
    maternal mRNAs are translated.                             requires two hosts because :
    Whereas at later stages, there is
                                                               (A) After passing through the pre-
    induction of embryonic gene
    expression. This phenomenon is                                 erythrocytic and erythrocytic
    called :                                                       cycles in its human host, the
      (A) Maternal-paternal transition                             parasite requires low ambient
      (B) Midblastula transition                                   temperature      for    further
      (C) Late embryonic switch                                    development. Therefore it

      (D) Embryonic reprogramming                                  enters the second host, the

14. .........................mimicry is exhibited by               mosquito.
    harmless colubrid snakes which
                                                               (B) There are too many parasites in
    resemble the poisonous snakes of the
    genus Elaps in having beautiful                                the human blood. Therefore, a
    body colouration of alternating red                            transfer to the second host
    and black bands.                                               becomes necessary.
      (A) Protective
                                                               (C) The infected person starts
      (B) Müllerian
                                                                   taking      medicines    which
      (C) Warning
                                                                   prevents further development.
      (D) Batesian
                                                                   Hence a transfer to a second
15. A period of arrested development in                            host becomes necessary.
    the life cycle of some insect pests is
    known as :                                                 (D) The person develops resistance
      (A) Ecdysis                                                  to the parasite by the time the
      (B) Diastasis                                                erythrocytic    cycle   starts.

      (C) Diapause                                                 Therefore another host is

      (D) Pupa                                                     required.

                                                       4
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                                                   18. Exposure of an organism to
17. Ability to taste phenylthiocarbamide
                                                      carcinogens      often   results    in
    (PTC) in human beings is controlled               mutagenesis.      This      leads   to
                                                      alterations in important growth
    by a single gene, the tasters being               regulatory genes. Carcinogens are
                                                      therefore identified by :
    either TT or Tt and non-tasters
                                                      (A) Ame’s test
    being tt. Suppose in a population of
                                                      (B) Commet assay

    human beings, different genotypes                 (C) Complementation test

    are represented by ratio 0.50 TT :                (D) Fluctuation test

                                                   19. State which of the following
    0.20 Tt : 0.30 tt (The gene
                                                      statements is correct in case of

    frequencies will be T = 0.60, t = 0.40).          population analysis

                                                      (A) The rare allele is represented
    If population attains equilibrium,
                                                          predominantly in heterozygotes
                                                          particularly as it becomes rarer
    one would expect the genotypic
                                                      (B) The rare allele is always under
    ratios as :
                                                          selection

                                                      (C) Equilibrium values in any
    (A) 0.50 TT : 0.20 Tt : 0.30 tt
                                                          population cannot be reached in

    (B) 0.36 TT : 0.48 Tt : 0.16 tt                       a single generation of random
                                                          mating
    (C) 0.30 TT : 0.40 Tt : 0.30 tt                   (D) The maximum frequency of
                                                          heterozygotes cannot exceed
    (D) 0.40 TT : 0.40 Tt : 0.20 tt                       0.75

                                               5                                    [P.T.O.
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20. The    frequency     with     which         22. In mammals, sex is determined by

    spontaneous mutations arise in a
                                                    presence     or     absence   of   y
    normal diploid population is :
                                                    chromosome. The search for sex
    (A) 10–8
                                                    determining genes has led to
    (B) 10–2
                                                    identification of Hy, bkm, zfy as
    (C) 10–12
                                                    testes determining gene. Recently
    (D) 10–3
                                                    identified Sry gene, a putative sex

21. Activation of formation of different
                                                    determining gene has one of the
    kinds of microfilament like actin
                                                    following properties assigned :
    network and actin bundles (stress

    fibers) are caused by cell signalling           (A) DNA binding protein expressed

    mediated by a particular protein                    in genital ridge at 8.5 pc

    family member. Which of the
                                                    (B) Zn finger protein expressed in
    following is involved in formation of
                                                        testes
    stress fibers ?

    (A) cdc 42                                      (C) Homeo-domain protein in


    (B) Rac                                             genital ridge


    (C) Rho                                         (D) Leads to sex transformation

    (D) Both Rho and Rac                                from males to females

                                            6
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23. During G2 phase of cell cycle, cyclin       24. Lampbrush chromosomes in frog
    B is synthesized and it forms a
                                                    oocytes     exhibit   an   index   of
    complex with CDK. This complex,
    called MPF, is crucial for the cells            amplification of :
    to enter mitosis. However, there is
    an extensive regulation of activity             (A) rRNA

    of MPF. There are kinases and
                                                    (B) mRNA
    phosphatases which take part in it.

    For conversion of an inactive MPF               (C) tRNA
    to an active MPF that initiates
                                                    (D) snoRNA
    mitosis, there are the following
    possibilities :
                                                25. Principle       of      holocoenotic
    (1) Phosphorylation of CDK by
                                                    environment means :
         Weel kinase

    (2) Phosphorylation of CDK by                   (A) Every factor affects every other
         CAK
                                                        factor in an ecosystem
    (3) Dephosphorylation of CDK
         inhibitory site by CDC 25                  (B) Only one factor affects every

    (4) Dephosphorylation by PTEN                       other factor in an ecosystem

    Which of the following is correct ?
                                                    (C) Only one factor affects only one
    (A) 1 and 2
                                                        other factor in an ecosystem
    (B) 1 and 3
                                                    (D) All the factors affect only one
    (C) 2 and 3

    (D) 2 and 4                                         factor in an ecosystem

                                            7                                    [P.T.O.
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26. Last few years humans have seen               28. A student examined two different
    rapid degradation of ecosystems.                  groups of cells and made the
    This has led global pressure on
                                                      following observations :
    conservation of fauna and flora. In
                                                      Trait            Cell I          Cell II
    this direction the trees and shrubs
    are planted on all unused and fallow              Cell wall        Present         Present

    land to provide fuelwood, fodder etc.
                                                      Ribosomes        Present         Present
    thereby reducing pressure on
                                                      Nucleus          Absent          Present
    existing forests. For example,
    unused farmlands, community land,                 Ability to do    Present         Absent

    road and rail sides etc. are planted
                                                      photosynthesis
    with suitable indigenous and/or
    exotic tree species. This is known as :           Respiration      Present         Present


    (A) Forest plantations                            These observations support which of

    (B) Social forestry                               the following conclusions :

    (C) Topiary                                       (A) Cell I is more complex in its

    (D) Conservation forestry                                 organisation than cell II

27. Typha angustata grows abundantly                  (B) Cell I is a prokaryote

    in/on :                                           (C) The ancestors of cell type II
    (A) Mountain plateaus                                     appeared earlier in the fossil

    (B) Wetlands                                              record than the ancestor of

    (C) Grasslands                                            cell I

    (D) Arid deserts                                  (D) Both cell I and II are plant cells

                                              8
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                                                31. The stimulation of microbial
29. Viruses generally infect only one or
                                                    decomposition of recalcitrant
    a few particular species. This is
                                                    compounds by the addition of
    called :                                        microorganisms specific for its
                                                    degradation is termed as :
    (A) Host restriction
                                                    (A) Biostimulation
    (B) Host range
                                                    (B) Bioaugmentation
    (C) Cell specificity                            (C) Biotransformation

    (D) Viral range                                 (D) Bioconversion
                                                32. Two statements I and II are given
30. E.coli is used as a faecal pollution
                                                    below :
    indicator. The water giving positive
                                                    Statement I : Some vaccines when
    results for the presence of E.coli is
                                                    administered protect the recepient
    declared non-potable or unfit for               from the specific infection for long
    drinking. Which of the following is             periods.

    the most correct explanation for the            Statement II : Antibodies induced by
                                                    some antigens are stable enough to
    above strategy ?
                                                    persist in body for long periods.
    (A) Faecal E.coli is extremely                  Which of the following options is
                                                    correct ?
         pathogenic
                                                    (A) Both the statements are
    (B) Presence of E.coli also means                   incorrect
         the presence of pathogens
                                                    (B) Both the statements are correct
    (C) E.coli and enteropathogens both                 and statement II is the cause
                                                        of statement I
         enter water through faecal
                                                    (C) Statement I is correct, but the
         matter
                                                        statement II is incorrect
    (D) E.coli grows much faster than               (D) Statement II is correct, but
         most enteropathogens                           statement I is incorrect

                                            9                                   [P.T.O.
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33. In a single substrate reaction,              34. Although pancreas is the site of

    competitive inhibitor was added to               storage of proteolytic enzymes, these

    examine the effect on Km and Vmax.               enzymes do not act on pancreatic

                                                     tissues because :
    Which of the following equations

    will be correct ?                                (A) They are inactivated


        1     1       [I]                          (B) They are not released from
    (A) V = V    1 +       
                      K ESI 
             max 
                                                         pancreas

                    K ES      [I]   1
                +        1 +      +                (C) They are in the form of
                    Vmax     K EI  [S]
                                                         proenzymes
        1   1   K
    (B)   =    + ES
        V Vmax  Vmax                                 (D) Their substrate is absent in

                               [I]  1                  pancreas
                          1 +      
                              K EI  [S]
                                                 35. A sigmoidal plot of substrate
          1   1       [I] 
    (C)     =    1 +                               concentration       verses   velocity
          V Vmax     K EI 
                                                     indicates :
                    K ES     [I]   1
                +        1 +     +
                    Vmax  K EI  [S]                (A) Michaelis-Menten kinetics

          1   1       [I]                          (B) Competitive inhibition
    (D)     =    1 +       
          V Vmax     K ESI 
                                                     (C) Non-competitive inhibition
                             K ES   1
                           +      .
                             Vmax [S]                (D) Co-operative binding

                                            10
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36. The disaccharide that does not react
                                                38. Which of the following sugar pucker
    with Benedict’s reagent is :
                                                    favours Z-conformation of DNA ?
    (A) Sucrose
                                                    (A) 2'–0 H steric hindrance
    (B) Lactose

    (C) Maltose                                     (B) Anti–G and C 2'–endosugar

    (D) Isomaltose                                  (C) Syn G and C 3'–endosugar
37. An amino acid contains no ionizable
                                                    (D) G 4 – DNA quartet
    groups in its side chain (R). It is
    titrated from pH 0 to 14. Which of          39. The sequence in which the following
    the following ionizable state is not
                                                    electron carriers are organised in
    observed during the entire titration
                                                    electron transport chain is :
    in the pH range 0 to 14 ?

    (A)
          +
              NH3 − CH − COO−                       (A) Ubiquinone, cytochrome a + a3,
                    |
                    R                                   NADH dehydrogenase


          +                                         (B) Cytochrome a + a3, ubiquinone,
    (B)       NH3 − CH − COOH
                    |
                                                        NADH dehydrogenase
                    R

                                                    (C) NADH          dehydrogenase,
    (C) NH2 − CH − COO−
              |
                                                        ubiquinone, cytochrome a + a3
              R

                                                    (D) NADH          dehydrogenase,
    (D) NH2 − CH − COOH
              |
              R                                         cytochrome a + a3, ubiquinone

                                           11                                  [P.T.O.
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                                                  42. The tertiary structure of proteins is
40. Maple syrup disease is a result of
                                                      stabilized by various interactions.
     genetic defect in metabolism of :                Which is the weakest interaction of

                                                      the following ?
     (A) Sulphur containing amino acids
                                                      (A) Hydrogen bond

     (B) Branched chain amino acids                   (B) Hydrophobic interactions

     (C) Aromatic amino acids                         (C) van der Waals interactions

                                                      (D) Covalent bond
     (D) Dicarboxylic amino acids
                                                  43. Hibernating animals can maintain

41. Which of the following tissues utilize            their body temperature during

                                                      hybernation period. This is made
     ketone body acetoacetate as a source
                                                      possible by :

     of   energy    during    prolonged               (A) Maintaining the BMR at low

                                                           level
     starvation ?
                                                      (B) Coupling              oxidative
     (A) Brain                                             phosphorylation       to   ATP

                                                           synthesis
     (B) Liver
                                                      (C) Uncoupling            oxidative

     (C) Skeletal muscle                                   phosphorylation from ATP

                                                           synthesis
     (D) Lung
                                                      (D) Catabolising muscle proteins

                                             12
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44. There are many compounds that are             46. Gene   density     in   eukaryotic

                                                     organisms is consistently lower and
    known       to   be    inhibitors   of
                                                     more variable than in prokaryotes.
    mitochondrial electron transport
                                                     Compared to lower eukaryotes like
    chain (ETC). Which of the following
                                                     the yeast S. cerevisiae ( 500 genes/
    is not an inhibitor of ETC ?                     Mb), the human genome has 50 fold

    (A) Rotenone                                     lower gene density. The eukaryotic

                                                     genes are also much larger in size.
    (B) Antimycin A
                                                     This is because :
    (C) Sodium azide
                                                     (A) eukaryotic genome has high

    (D) Dinitrophenol                                    proportion of repeats and

                                                         satellite DNAs and genes have
45. Coding strand in ds-DNA is :
                                                         introns
    (A) identical     in    sequence    to
                                                     (B) eukaryotic genome is diploid
        transcribed RNA                                  and has to cross over; also genes

                                                         are polycistronic
    (B) complementary in sequence to

        transcribed RNA                              (C) Prokaryotic genome has more

                                                         non-coding DNA and interrupted
    (C) identical to sense strand of
                                                         genes
        DNA
                                                     (D) eukaryotic      genome      has
    (D) identical in sequence to ORF                     telomeres and centromeres and

        alone                                            genes are clustered as familites

                                             13                                   [P.T.O.
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47. Non-homologous         nucleotide           49. In an experiment to determine
                                                   genetic    code,   synthetic     RNA
   sequences in two DNA molecules
                                                   molecules with different composition
   could be visualised using :                     of A and C nucleotide were used,
                                                   viz. (A) poly (AC) (5 : 1) and (B) poly
   (A) Heteroduplex technique                      (AC) (1 : 5). Lysine was incorporated
                                                   in case of A with very high
   (B) Fluorescent in-situ hybridisation
                                                   frequency and proline in case of B.
                                                   Lysine incorporation was least in B.
   (C) Immunoblotting technique
                                                   The codon assignment could be

   (D) Southern        blotting    and             concluded as :

                                                   (A) AAA lysine; CCC proline
       hybridisation
                                                   (B) AAC lysine; CCA proline
48. RNA type C tumor viruses are
                                                   (C) ACA lysine; CAC proline
   negative strand viruses and amplify             (D) CAA lysine; ACC proline

   copy number of their genome by               50. Eukaryotic RNAs (transcripts)
                                                   are much longer than the final
   using an enzyme encoded by them,
                                                   mRNAs. Splicing is a characteristic
   which is :                                      feature of eukaryotic processing.
                                                   This process involves :
   (A) RNA replicase
                                                   (A) ribozymes         and       lariat
                                                       formation
   (B) RNA polymerase
                                                   (B) heteroduplex formation
   (C) RNA       dependent        DNA
                                                   (C) end joining complexes and
       polymerase                                      rRNA

                                                   (D) adaptor      RNA     and    micro
   (D) RNA helicase
                                                       RNA

                                           14
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         51. In an ipozya situation of a lac              52. Which of the following is not a

             operon, where β-galactosidase is                 feature of transcription factors ?

             generally inducible by lactose,
                                                              (A) glycosylation domain
             diploids    were     created   in   a
                                                              (B) zinc finger motif
             transduction experiment. In an

                     /          situation which of            (C) helix-turn-helix


             the following statements is valid ?              (D) leucine zipper

             (A) beta galactosidase will be               53. For precise quantification of
ipozya
  ozya
                 constitutively expressed
                                                              differential gene expression at

             (B) beta galactosidase will not be               mRNA level for a gene ‘X’ which of

                 constitutively expressed
                                                              the following methods will be used ?

             (C) beta galactosidase will not be
                                                              (A) Western blot
                 expressed
                                                              (B) Real time PCR
             (D) beta galactosidase will not be
                                                              (C) RT-PCR
                 synthesised due to catabolite

                 repression                                   (D) Northern blot


                                                     15                                   [P.T.O.
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                                                  55. Cancer is said to be a multistep
54. Modified     nucleotides    provide
                                                      process.          Oncogenes              and
   additional level of information over               antioncogenes are involved in the
                                                      process of carcinogenesis. The
   and above genetic code. They
                                                      reason for cancer phenotype is :
   influence     genotype-phenotype
                                                      (A) Oncogenes acquire a dominant

   correlation. This is due to the ability                 activating mutation while
                                                           antioncogenes get inactivated
   of such modified nucleotides to
                                                      (B) Oncogenes are activated and

   influence :                                             duplicated and translocated

                                                      (C) Oncogene is inactivated and
   (A) DNA protein interactions and
                                                           antioncogene is activated

       DNA conformations                              (D) Loss of heterozygosity for
                                                           oncogenes
   (B) DNA       bending,      promoter
                                                  56. If a woman desires to conceive after

       stability       and     rate    of             the age of 35, she needs genetic
                                                      counselling. This is particularly
       transcription                                  required because of incidence
                                                      of...........................in the babies born
   (C) RNA splicing and termination
                                                      to such mothers.

       of transcription                               (A) Klinefelters syndrome

                                                      (B) Still birth
   (D) RNA editing, transcription and
                                                      (C) Down’s syndrome
       protein folding
                                                      (D) Abortion

                                             16
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57. CI 857 is a mutation in lambda               58. Chromosome walking technique

    phage wherein the repressor protein              relies on :

    is temperature sensitive and gets in-
                                                     (A) Genomic libraries
    activated at 45ºC. Which of the
                                                     (B) cDNA libraries
    following is true for this mutant of

    λ phage ?                                        (C) RFLP


    (A) This mutant will not produce                 (D) Real time PCR

        plaques in plate assay on E.coli         59. During natural transformation in

        host at 37ºC or 45ºC
                                                     bacteria, a fragment of single strand

    (B) Mutant will produce plaques at               enters recipient cell. Upon its

        37ºC but not at 45ºC                         interaction with recipient, a

    (C) Mutant will produce turbid                   structure that results is termed as :

        plaques at 37ºC and clear
                                                     (A) Merozygote
        plaques at 45ºC
                                                     (B) Exogenote
    (D) Mutant will produce clear
                                                     (C) Endogenote
        plaques at 37ºC and turbid

        plaques at 45ºC                              (D) Zygote


                                            17                                    [P.T.O.
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60. Given genome size of an organism           62. Transgenic      animals      can   be
                                                   utilised to produce recombinant
    is 4 × 105 bP and its G + C content
                                                   therapeutic proteins such as
    is 50%. Complete digestion of                  insulin and growth hormones. In
                                                   order to produce genetically
    this genomic DNA by Bam H1                     identical individuals which can
    (GG AT CC) will yield an average               be used for generation of animal
                                                   farms for biopharming the
    fragment size of :                             following strategies can be
                                                   utilised.
    (A) 1.5 kb
                                                   (A) embryo transfer
    (B) 4 kb                                       (B) artificial insemination
                                                   (C) in-vitro fertilisation
    (C) 2 kb
                                                   (D) Cloning of animals
    (D) 10 kb                                  63. In barley plants cross was carried
                                                   out between Hordeum vulgare and
61. DNA foot-printing technique is
                                                   H. bulbosum. Immature embryos
    useful in :                                    were cultured in-vitro on M. S.
                                                   medium to overcome natural
    (A) Detection of genomic variations            abortion. 35% of the embryos grown
                                                   in-vitro showed development of
    (B) Detection of predisposition of             haploid plants with Vulgare genome.
                                                   This phenomenon has occurred due
        genetic diseases
                                                   to :
    (C) Detection of protein-DNA                   (A) distant hybridization

        interactions                               (B) genotypic variation
                                                   (C) somaclonal variation
    (D) Prenatal sex determination
                                                   (D) phenotypic variation

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64. For cloning hybridomas, the cells               66. Suppose the Na + concentration
    need to be distributed through serial
                                                        outside a cell is 100 mM and inside
    dilutions to the extent of a single cell
                                                        it   is    10   mM.    The   lactose
    deposited in a well in the 96-well cell
    culture plates. To ensure that such                 concentration on both sides of the
    a cell grows and propagates the                     membrane is 10 mM. If there is an
    following is required :
                                                        integral membrane protein for the
    (A) grow them with addition of
                                                        symport of Na+ and lactose, which
         growth factors
                                                        of the following results would you
    (B) grow them in the presence of
         high serum concentration                       expect ?

    (C) plate them on a feeder-layer of                 (A) No change, as ATP hydrolysis
         uv-irradiated fibroblast cells
                                                             would be necessary to transport
    (D) grow them along with myeloma
                                                             lactose
         cells

65. Release of Ca++ from the ER in a                    (B) Na + will enter but lactose

    fertilised egg is triggered by :                         concentration will not change

    (A) Map kinase pathway                                   because it is at equilibrium

    (B) Innositol triphosphate (IP 3)
                                                        (C) Na+ will enter and lactose will
         pathway
                                                             leave the cell
    (C) Wnt signalling pathway
                                                        (D) Both lactose and Na+ will enter
    (D) C-AMP-dependent signalling
         pathway                                             the cell


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67. The most suitable ligand for                  70. DNA sequence determination
    purification of glycoproteins from a
                                                      classically involves sequencing by
    given mixture of proteins is :
                                                      synthesis or degradation which is
    (A) Heparin
                                                      used in Sanger or Maxam Gibert
    (B) Concanavalin A
                                                      method.     In    the   NGS   (next
    (C) Avidin
                                                      generation sequencing) which of the
    (D) 5'-AMP
                                                      following methods is used ?
68. Which of the following microscopic
    techniques is useful to observe three-            (A) Radioactivity

    dimensional structure of the cell ?
                                                      (B) Fluorescence
    (A) SEM
                                                      (C) Luminiscence
    (B) Light microscopy
                                                      (D) Electrical conductance
    (C) Fluorescence microscopy

    (D) TEM                                       71. Which of the following molecules

69. You have been asked to measure                    has the highest molar extinction

    DNA synthesis, which radioisotope                 coefficient ?
    is of better choice ?
                                                      (A) GFP
    (A) 3H Thymidine
                                                      (B) Fluorescein
    (B) 32 P
                                                      (C) Rhodamine
    (C) 14C ATP

    (D) 15N cytosine                                  (D) Chlorophyll

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72. Leukemia or      blood cancer is            74. Which of the following enzymatic
    uncontrolled proliferation of an
                                                    reaction produces chromatogenic
    abnormal clone of hematopoietic
    cells. The diagnosis is generally               product ?
    based on detection of abnormal cells
                                                    (A) Phosphatase     using    para-
    in blood and bone marrow. One of
    the most powerful confirmatory                      nitrophenyl phosphate
    approach is immunophenotyping of
    the cells. The immunophenotyping                (B) β-galactosidase using X-gal

    analysis can be carried out by :
                                                    (C) Cytochrome     oxidase    with
    (A) Enzyme linked immunosorbant
                                                        cytochrome C
        assay

    (B) Radioimmunoassay                            (D) Alcohol dehydrogenase with

    (C) Flow cytometry                                  ethanol
    (D) Immunoprecipitation
                                                75. The resemblance between groups
73. In a HPLC separation of a sample,
                                                    and within groups of relatives can
    it is suspected that two known
    molecules could be present in a                 be determined by applying which
    single peak. How can you verify
                                                    test ?
    this ?

    (A) Changing the elution gradient               (A) Correlation

    (B) Increasing the run time
                                                    (B) Covariances
    (C) Spiking     with    suspected
        molecules                                   (C) ANOVA

    (D) Changing the mobile phase                   (D) Regression

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ROUGH WORK




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Source: Savitribai Phule Pune University. Question paper published by Savitribai Phule Pune University.