CSIR NET Life Sciences December 2015 Question Paper with Answer Key (Part-1)

NET December 2015 Answer Key

Previous Year Question Papers of CSIR – UGC /JRF/NET: Life Science Examination, December 2015 (II)
(Original Question Paper with Answer Key, Explanations and References)

PART: B (Questions 21 – 40)

(21). The ionic strength of a 0.2 M Na2HPO4 solution will be:

a.       0.2 M
b.      0.3 M
c.       0.6 M
d.      0.8 M

Ans. (c)

Ionic strength of a solution is the function of all ions present in a solution.

(22). A cell line deficient in salvage pathway for nucleotide biosynthesis was fed with medium containing 15N labelled amino acids. Purines were then extracted. Treatment with which one of the following amino acids is likely to produce 15N labelled purines?

a.       Aspartic acid
b.      Glycine
c.       Glutamine
d.      Aspartamine

Ans. (c)

(23). Enzymes accelerate a reaction by which one of the following strategies?

a.       Decreasing energy required to form the transition state.
b.      Increasing kinetic energy of the substrate.
c.       Increasing the free energy difference between substrate and the product.
d.      Increasing the turn over number of enzymes.

Ans. (a)

Transition state is the state of substrate with highest energy; they are least stable than products and other substrate molecules. They have the highest energy along the reaction coordinate and they have the highest free energy. Enzymes accelerate the chemical reactions by lowering the energy required to reach the transition state of a substrate.

(24). Coupling of the reaction centers of oxidative phosphorylation is achieved by which one of the following?

a.       Making a complex of all four reaction centers.
b.      Locating all four complexes in the inner membrane.
c.       Ubiquinones and cytochrome C.
d.      Pumping of protons.

Ans. (c)

(25). The genome of a bacterium is composed of a single DNA molecule which is 109 bp long. How many moles of genomic DNA is present in the bacterium? [Consider Avogadro No = 6 X 1023]

a.       1/6 X 10-23
b.      1/6 X 10-13
c.       6 X 1014
d.      6 X 1023

Ans. (a)

26.  It takes 40 minutes for a typical E. coli cell to completely replicate its chromosome. Simultaneous to the ongoing replication, 20 minutes of a fresh round of replication is completed before the cell divides. What would be the generation time of E. coli growing at 37oC in complex medium?

a.       20 Minutes
b.      40 Minutes
c.       60 Minutes
d.      30 Minutes

Ans. (a)

Generation time: the rate of exponential growth of a bacterial culture in optimum condition. In simple term it is the time required to double the population. Generation time (G) is defined as the time (t) per generation (n = number of generations) [ G=t/n]. Generation time of E. coli is 17 minutes. So in the options provided the best answer is option (a).

27.  Glycophorin having one highly hydrophobic domain is able to span a phospholipid bilayer membrane only

a.       Once
b.      Twice
c.       Thrice
d.      Four times

Ans. (a)

Glycophorins are the membrane proteins of RBC, they facilitate the transport of sugar molecules through the membrane. They are highly glycosylated with many sialic acid residues.

28.  Given below are events in the cell cycle.

(a) Phosphorylation of lamin A, B, C
(b) Phosphorylation of Rb (Retinoblastoma protein)
(c) Polyubiquitination of securin
(d) Association of inner nuclear membrane proteins and nuclear pore complex proteins with chromosomes.

Which one of the following reflects the correct sequence of events in the mammalian cell cycle?

a.      a → b → c → d
b.      b → c → d → a
c.       c → a → b → d
d.      b → a → c → d

Ans. (d)

Dephosphorylated Rb (a tumor suppressor gene) inhibits cell cycle. When Rb is phosphorylated, it becomes inactive and cells proceed to cell division.

Phosphorylation of lamin activates the nuclear membrane disorganization.

Securin activate separase, a protease, which degrade the cohesion proteins that links the sister chromatids during cell division. Once the cohesion proteins are completely degraded and the sister chromatids are separated, the securing should be inactivated. In the phosphorylated state of securing, the APC (anaphase promoting complex) cannot initiate the onset of anaphase. Thus, in order to initiate anaphase, the APC complex ubiquitinates sequrin and target it for degradation.

29.  Which one of the following chemicals is a DNA intercalator?

a.       5-bromouracil
b.      Ethyl methane sulfonate
c.       Acridine orange
d.      UV

Ans. (c)

Intercalation is a reversible inclusion of molecules to another molecule which have a layered structural organization. 5-bromouracil, ethyl methane sulfonate, acridine orange and UV are DNA damaging agents. Among these, the acridine organge intercalate to double stranded DNA. Apart from its mutation potential, acridine orange is a nucleic acid specific fluorescent dye which can be used for the detection and labeling of nucleic acid samples in molecular biology experiments. Ethidium bromide (EtBr) is another DNA intercalating agent which is also used as fluorescent tag for DNA and RNA.

30.  An antibiotic that resembles the 3’end of a charged tRNA molecule is:

a.       Streptomycin
b.      Sparsomycin
c.       Puromycin
d.      Tetracycline

Ans. (c)

Puromycin is an antibiotic obtained from Streptomyces alboniger. Puromycin cause premature chain termination in protein synthesis. Part of the molecule resembles the 3′ end of the aminoacylated tRNA. Puromycin enters the A site and transfers to the growing chain, causing the formation of a puromycylated nascent chain. This results in premature of release of polypeptide from the ribosome

31.  α-Amanitin is a fungal toxin which inhibits eukaryotic RNA polymerases. The three eukaryotic RNA polymerases show differential sensitivity to this toxin. Which one of the following order (higher to lower) is correct in respect of sensitivity towards α-amanitin?


Ans. (b)

Alpha amanitin is a cyclic peptide of 8 amino acid produced by a mushroom Amanita. It inhibits the transcription process of protein synthesis by inhibiting RNA polymerase II which synthesize the mRNA. Amanitin can also inhibit other RNA polymerases such as RNA polymerase I and RNA polymerase III which synthesis rRNA and tRNA respectively. The sensitivity of α-amanitin towards RNA Polymerase enzyme shows differential sensitivity as follows RNA POL II > RNA POL III > RNA POL I. RNA Pol I is least inhibited by α-amanitin.

32.  . In eukaryotic replication, helicase loading occurs at all replicators during

a.       G0 phase
b.      G1 phase
c.       S phase
d.      G2 phase

Ans. (b)

Helicase loading is carefully regulated to control the location and frequency of DNA replication initiation. In eukaryotic cells, helicase loading is tightly restricted to the G1 phase of the cell cycle. This is to ensure that no origin of replication is initiated more than once per cell cycle.

33.  Cytotoxic T cells express

a.       CD8 marker and are class II MHC restricted
b.      CD4 marker and are class I MHC restricted
c.       CD4 marker and are class II MHC restricted
d.      CD8 marker and are class I MHC restricted

Ans. (d)

34.  The mutation in an oncogene falls under which of the following classes?

a.       Loss of function mutation
b.      Frame shift mutation
c.       Gain of function mutation
d.      Dominant negative mutation

Ans. (c)

Los of function mutation: gene lost its function due to mutation

Frame shift mutation: the reading frame of mRNA changes due to point mutation

Gain of function mutation: give enhanced or new activities to a protein

Dominant negative mutation: produce an altered gene product which acts antagonistically to the gene product of its wild allele. Usually these mutations will be dominant.

Oncogenes are tumour producing genes. Usually these genes are very essential genes for the normal functioning of the cells and we call them as proto-oncogenes. Proto-oncogenes code for proteins that help to regulate cell growth and differentiation. Mutations of proto-oncogenes cause the formation of oncogenes.

35.  Which of the following is NOT a cell adhesion protein?

a.       Cadherin
b.      Selectin
c.       Immunoglobulin (Ig) superfamily
d.      Laminin

Ans. (d)

Lamins are fibrous proteins; they are present in the nucleus as nuclear lamins. Nuclear lamins interact with membrane-associated proteins to form the nuclear lamina on the interior of the nuclear envelope.

Cadherins: they are calcium dependent transmembrane adhesion proteins forming adherence junctions.

Selectin: are a type of lectins that functions as adhesion proteins

36.  Which of the following is NOT a second messenger?

a.       Cyclic GMP
b.      Diacylglycerol
c.       Inositol triphosphate
d.      Phosphatidyl inositol

Ans. (d)

37.  In chick, development of wing feather, thigh feather and claws depends on epithelial specificity conferred by induction from mesenchymal components from different sources of the dermins. This may be attributed to?

a.       Autocrine interaction
b.      Regional specificity of induction
c.       Receptor activation by hormones
d.      Inactivation of genetic interactions

Ans. (b)

38.  Alveolar cells of the lung arise from which one of the following layer(s)?

a.       Mesoderm
b.      Endoderm
c.       Ectoderm
d.      Both ectoderm and endoderm

Ans. (b)

39.  Migration of individual cells from the surface into the embryo’s interior is termed as

a.       Ingression
b.      Involution
c.       Invagination
d.      Delamination

Ans. (a)

40.  Floral organ development is controlled by overlapping expression of ‘A’ class, ‘B’ class and ‘C’ class genes in different whorls. In an Arabidopsis mutant, the flowers had sepals, sepals, carpels and carpels in the four whorls. Mutation in which one of the following is the cause for the mutant phenotype?

a.       ‘A’ class gene alone
b.      ‘B’ class gene alone
c.       ‘A’ and ‘B’ class genes
d.      ‘C’ class gene alone

Ans. (b).

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