Part B-Dec 2013

CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship

Part B


This time CSIR does not allow candidates to carry questions with them. We have collected maximum questions from our candidates (memory based).

1. The interaction energy between atom A and B is ~ 400 kjmol-1.
The type of interaction between them is

1.   pi – pi
2.   covalent
3.   ion-dipole
4.   hydrogen bond

Ans: 2
Expln:- Interaction energy between
1. Pi-Pi = 50-60 KJ/mol
2. Covalent = 400 KJ/mol.
3. ion-dipole = 1-1.5 KJ/mol
4. Hydrogen bond = 12-16 KJ/mol
5. < KJ/mol – London Dispersion forces between non-polar molecules
6. 2-0.5 KJ/mol – Dipole-Dipole attraction.

2. one of the following bases has the largest hydrogen bonding possibility?

1. Adenine
2. Guanine
3. Cytosine
4. Uracil

Ans:- 2

Expln:- In the given choice, Guanine has the largest hydrogen bonding possibility, because it contains the highest number of electron donor sites.

3. Enzymes help to lower the activation energies of reactions by



1.   covalent interaction with substrates
2.   binding only with the solvent molecules
3.   changing reaction equilibria
4.   forming weak interactions with substrates.

Ans:- 4

Expln:- Indicates a continuous change in the conformation and shape of an enzyme in response to substrate binding. This makes the enzyme catalytic which results in the lowering of the activation energy barrier causing an increase in the overall rate of the reaction. In other words, when a substrate binds to an enzyme, it will change the conformation of the enzyme. This forms a transitional intermediate which lowers the activation energy and allows the reactants to proceed towards the product at a faster rate.


4. Glucose residues in amylose are linked by

1. β 1 → 4
2. α 1 → 4
3. α 1 → 6
4. β 1 → 6

Ans:- 2

Expln:- The structure of starch includes two chemical substances, viz., amylase and amylopectin and these two substances are again made up of a number of glucose units joined through the glycosidic linkages like maltose ie., amylose consists of glucose residue with repeating maltose units, while amylopectin consists of glucose units in α -glycosidic linkage. In its structure amylose shows α -1,4 glucoside linkage, while amylopectin in addition to α -1,4 glucoside linkage has α -1,6 glucoside linkage in side chains. Thus the chain of amylopectine is branched.


5.  In the lysogenic state of λ phage

1.   both CI and Cro are on
2.   both CI and Cro are off
3.   CI is on while Cro is off
4.   CI is off while Cro is on


Ans:- 3

Expln:- Choice between lysogeny and lytic cycle:- There is simultaneous production of cro-protein repressor (by cro gene) and the lambda repressor protein (by cI gene). Since both are repressor proteins, they can block the transcription by each other and there is a race between the production of Cro protein and lambda repressor protein.

If the lambda repressor protein (product of cI gene) win the race, it blocks, the action of the left promoter and the right promoter. Blocking of the left promoter prevents the transcription of gene N. Blocking of the right promoter prevents the formation of cro-protein and the transcription of the O, P and Q genes. Products of these genes are necessary for late stage of lytic cycle.

6.  The “Uvr ABC” repair mechanism is involved in repairing

1.   missing bases
2.   strand break
3.   Cross linked strands
4.   DNA damage caused by “bulky” chemical adducts

Ans:- 4

Expln:- The nucleotide excision repair process uses a set of proteins called UvrA, UvrB, UvrC, and UvrD. “Uvr” stands for “ultraviolet light repair” of pyrimidine dimers, but remember that this system is also used to repair other types of DNA Damage. In addition, DNA polymerase and DNA ligase are used to complete the repair.

In the first step, the protein trimer (two Uvr molecules and one UvrB molecule) binds to the DNA molecule. This trimer moves along the DNA, scanning for damage that distorts the double helix. UvrD, a helicase, then binds to the site and separates the segment of damaged strand from the rest of the DNA molecule. UvrB, C, and D are then all released from the damaged region.

This leaves a gap that is filled with the proper nucleotides by DNA polymerase and then sealed by DNA ligase.

7. The cylindrical channels in gap junctions are made of
1.   connexin
2.   collagen
3.   fibronectin
4.   N-CAM

Ans:- 1

Expln:- Gap junctions are the most widespread cell junctions found in large numbers in most animal tissues. These cell junctions are usually 2 to 4 nm. Gap junctions are constructed of transmembrane proteins that form specialised structures called connexons. Each connexon is formed of a ring of six identical protein subunits called connexins. The subunits are arranged around a central channel. Each connexin is a transmembrane protein. The connexons are about 7 nm in diameter. A continuous channel is formed by the alignment of two connexons of the adjacent cell membranes. The channel is about 2 nm in diameter and permit the passage of substances having molecular weight of upto 1000 daltons. These allow the passage of most sugars, amino acids, nucleotides, vitamins, steroid hormones, cyclic AMP and various ions. Thus the adjacent cells can share these substances through these channels. Such a connecting system is important especially in the embryonic cells before the establishment of circulatory system. The connexons from each cell surface keep the connecting plasma membranes at a distance from each other (about 5 nm) and hence the term gap junction. Each gap junction can contain a cluster of upto several hundred connexons.

The permeability of gap junction channels is influenced by calcium ions. Ca2+ usually inhibits the coupling of the adjacent cells by way of connexons. In electrically excitable cells, the gap junctions permit transmission of electrical activity without the involvement of neurotransmitters.

8. The cell death pathway in C. eleganscan be schematically represented as :
Ced 9 —-> ced 4 —-> ced 3
Based on the above, which one of the following statements is TRUE ?

1.   A loss-of-function allele of ced 9 would lead to survival of cells that normally die.
2.   A loss-of function allele of ced 9 would lead to excessive cell death.
3.   A gain-of-function allele of ced 9 would lead to excessive cell death.
4.   Neither loss or gain-of-function of ced9would make any change to the cell death pathway.

Ans:- 2

Expln:- Programmed cell death in C.elegans is controlled through the interaction between ced-3, the gene that promotes cell death, and ced-9 the gene that promotes cell survival. Examining the effects of over expressing cell-death-related genes in specific C.elegans neurons that normally live, it was demonstrated that the cell-death genes ced-3 (a caspase), ced-4 (an Apaf-1 homolog), and ced-9 (a BCL-2 homolog) all can act cell autonomously to control programmed cell death.

9.Which of the following mechanisms is NOT involved in providing photoprotection to plants ?
1.   Degradation of D1 protein
2.   Zeaxanthin formation
3.   Photolysis of water
4.   Thermal dissipation

Ans:- 3

Expln:- Photoprotection is the mechanism that nature has developed to minimize the damage caused by UV radiation. Photoinhibition of photosystem II leads to loss of PS II electron transfer activity. PS II is continuously repaired via degradation and synthesis of the D1 protein.D1 protein synthesis can be blocked by Lincomycin. The photoinhibited PS II centres are continuously repaired via degradation and synthesis of the D1 protein.

Zeaxanthin formation: In xanthophyll cycle there is the enzymatic removal of epoxy groups from xanthophylls like violoxanthin, antheraxanthin and diadinoxanthin to form de-epoxidised xanthophylls like zeaxanthin, diatoxanthin etc. These de-epoxed xanthophylls stimulates energy dissipation by non-photochemical quenching.

Thermal dissipation: Under conditions of excess sunlight, the light harvesting complex (LHC) is switched into a photoprotected quenched (to suppress) state in which the harmful absorbed energy is dissipated as heat by the process of non-photochemical quenching of chlorophyll fluorescence.

Light energy that has been absorbed by a leaf will excite electrons in chlorophyll molecules. Energy in photosystem II can be converted to chemical energy to drive photosynthesis (photochemistry). If photochemistry is inefficient, excess energy can damage the leaf. Energy can be emitted (known as energy quenching) in the form of heat (called non-photochemical quenching) or emitted as chlorophyll fluorescence. These three processes are in competition, so fluorescence yield is high when less energy is emitted as heat or used in photochemistry. Therefore by measuring the amount of chlorophyll fluorescence, the efficiency of photochemistry and non-photochemical quenching can be assessed.

10.Which of the following plant hormones can mimic the det 1 mutation, causing de-etiolation and chloroplast development in dark ?
1.   Cytokinin
2.   Gibberellin
3.   Auxin
4.   Ethylene

Ans:- 3

Expln:- Cytokinins are also known to greatly enhance conversion of etioplasts into chloroplasts when etiolated seedlings after treatment with cytokinins are exposed to light. In such cases, the chloroplasts develop extensive grana and chlorophylls and the rate of synthesis of photosynthetic enzymes is much greater in comparison to those etiolated seedlings which are illuminated without cytokinin treatment.

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