CSIR UGC JRF NET Life Sciences Part B Questions and Answers June 2015

11. Which one of the following enzymes is NOT a part of pyruvate dehydrogenase enzyme complex in glycolysis pathway?
a. Pyruvate dehydrogenase
b. Dihydrolipoyl transferase
c. Dihydrolypoyl dehydrogenase
d. Dihydrolypoyl oxidase
Ans: d. Dihydrolypoyl oxidase
Explanation: Pyruvate dehydrogenase enzyme complex (PDC) is a complex of three enzymes that convert pyruvate into acetyl-CoA by a process called pyruvate decarboxylation. The complex consist of Pyruvate dehydrogenase (E1), Dihydrolipoyl transferase (E2)and Dihydrolypoyl dehydrogenase (E3).

12. What phenotype would you predict for a mutant mouse lacking one of the genes required for site-specific recombination in lymphocytes?
a. Decrease in T cell counts
b. Immunodeficient
c. Increase in T cell count
d. Increase in B cell count
Ans: b. Immunodeficient
Explanation: Site-specific recombination is a type of genetic recombination in which DNA strand exchange takes place between segments possessing only a limited degree of sequence homology. Mutation in genes required for site specific recombination results in reduced or absence of mature T and B cells. Immature cells are still present resulting in immune deficient state.
2015 csir

13. A 1% (w/v) solution of a sugar polymer is digested by an enzyme (20µg, MW=200,000). The rate of monomer sugar (MW=400) liberated was determined to have a maximal initial velocity of 10 mg formed/min. the turnover number (min-1) will be
a. 5 x 104
b. 2.5 x 10-2
c. 4.0 x 10-6
d. 2.5 x 105
Ans: d. 2.5 x 105
Explanation: Kcat is the turn over number  and is calculated as Kcat
Vmax=max velocity; Et= total enzyme concentration.
In the question we should convert the concentration terms to molarity.
Vmax= 10x10-3/400= 0.25x10-4;
Et= 20x10-6/200000=10-10.
Kcat= 0.25x 10-4/10-10=2.5x105/min.

14. Beating of cilia is regulated by
a. actin
b. Myosin
c. cofilin
d. Nexin
Ans: d. Nexin
Explanation: Nexin is highly extensible protein responsible for the beating of cilia. Other options are related to microfilaments. (Refer Cilia vs Flagella)

15. Which one of the following events NEVER activates the G-protein coupled receptor for sequestering Ca2+ release?
a. Interaction of bindin to sperm receptors.
b. Activation of Frizzled by Wnt.
c. Cortical reaction blocking polyspermy
d. DNA synthesis and nuclear envelop breakdown.
Ans: d. DNA synthesis and nuclear envelop breakdown.
Explanation: G-protein coupled receptor ((GPCRs) form a very large group cell surface receptors that are coupled to signal transducting trimeric G proteins. Here, all other options would result in increased concentration of Ca2+ except for DNA synthesis and nuclear envelop breakdown which might occur due to activation of a cAMP dependent kinase.

16. Hydra shows morphallactic regeneration and involves which one of the following signal transduction pathway in its axis formation?
a. Wnt/β-catenin pathway
b. Retinoic acid pathway
c. FGF pathway
d. Delta- Notch pathway
Ans: a. Wnt/β-catenin pathway
Explanation: When a hydra is cut in half, the half containing the head will regenerate a new basal disc, and the half containing the basal disc will regenerate a new head. Moreover, if a hydra is cut into several portions, the middle portions will regenerate both heads and basal discs at their appropriate ends. No cell division is required for this to happen, and the result is a small hydra. This regeneration is morphallactic.
Reference: Developmental Biology by Gilbert S F
Recent detailed analyses of the activation of the Wnt–β-catenin pathway in bisected Hydra shows that the route taken to regenerate a structure as complex as the head varies dramatically according to the level of the amputation.
Galliot, Brigitte, and Simona Chera. "The Hydra model: disclosing an apoptosis-driven generator of Wnt-based regeneration." Trends in cell biology 20.9 (2010): 514-523.

17. The main difference between normal and transformed cells are
a. Immortality and contact inhibition
b. Shorter generation time and cell mobility
c. Apotopsis and tumour suppressor gene hyper function
d. Inactivation of oncogenes and shorter cell cycle duration
Ans: a. Immortality and contact inhibition
Explantion: Apart from check point regulation, one of the crucial mechanism that prevents uncontrolled cell division is contact inhibition, a hall mark character of normal cells. Once transformed, the cells become immortal or cancerous due to the absence of ‘contact inhibition’ mechanism. (Refer: Characteristics of cancer cells)

18. Following are three single stranded DNA sequences that form secondary structures.
(a) A T T G A G C G A T C A A T
(b) A T T G A G C G A T A T C A A T
(c) A G GG A G C G A T C CC T
Based on their stability, which one is correct?
a. (a) = (b)=(c)
b. (c)>(a)>(b)
c. (b)>(c)=(a)
d. (b)>(c)>(a)
Ans: b. (c)>(a)>(b)
Explanation: Here, C has the highest GC pairing. So, it shows the highest stability. Three H-bonds are involved in GC pairing. Therefore, more the number of G and C, the more will be the stability.
option (c) has the maximum number of G and C
Option (a) and (b) has same number of G and C, but option (a) is more stable as the total percentage of GC bond is more in option (a) as option (b) sequence length is slightly longer.
For More..Refer  DNA double helix model, Chargaff’s rule, Stability of DNA and RNA
19. The key determinant of the plane of cytokinesis in mammalian cells is the position of
a. Chromosomes
b. Central spindle
c. Centrioles
d. Pre- prophase band
Ans: b. Central spindle
Explanation: Mitotic spindle, more specifically the astral spindle, determines the the site of the cleavage furrow in mammalian cell. (Read more: Cytokinesis in Animal cell vs Cytokinesis in Plant cell)

20. Dark- grown seedlings display ‘triple response’ when exposed to ethylene. Which one of the following is NOT a part of ‘triple response’?
a. Decrease in epicotyl elongation
b. Rapid unfolding and expansion of leaves
c. Thickening of shoot.
d. Horizontal growth of epicotyl.
Ans: b. 2. Rapid unfolding and expansion of leaves
Triple response of seedlings: The hormone (ethylene) prevents the seedling stem and root from elongating. It induces the stem and root to swell radially, thereby increasing in thickness. Together these responses strengthen the seedlings stem and root. The seedling stem bends so that embryonic leaves and the delicate meristem grow horizontally rather than vertically. The bend portion of the stem then pushes through the soil. This bent portions forms as the result of an imbalance of auxin across the stem axis. Ethylene drives this auxin imbalance.


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  2. thank u for sharing qsn & ans..

  3. These explanations are so valuable to us. Thank you for the effort you put into these.

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