Part C-June 2013

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

Part C

PREVIOUS SOLVED QUESTIONS – 2013 June CSIR – NET EXAM

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

1.     The diploid genome of a species comprises 6.4 x 109 bp and fits into a nucleus that is 6 μm in diameter. If base pairs occur at intervals of 0.34 nm along the DNA helix, what is the total length of DNA in a resting cell ?
(1) 3.0 m
(2) 3.5 m
(3) 2.2 m
(4) 4.0 m

Ans: 3
Expln:- Length of DNA double helix can be calculated by multiplying the total number of base pair with distance between two consecutive base pairs.
Here the total number of base pair is = 6.4 x 109bp.
Distance between two consecutive base pair = 0.34 nm.
To convert 0.34 nm to metre multiply this with 10-9        ie; 0.34 x 10-9
Therefore Total length of DNA = 6.4 x 109 x 0.34 x 10-9 = 2.2 m

2. You are working with an in vitro eukaryotic transcription system, which produced both capped and uncapped mRNAs. You incubated these mRNAs with mammalian cell nuclear extract and then quantified the different products as shown below. Which of the following graphs correctly represents the expected result?
2013_june_c_1
Ans:- 1

Expln:- The first step in mRNA processing is the addition of a Cap. The cap structure binds two proteins in the nucleus, the CBP 20 and CBP 80 proteins (CBP = Cap Binding Proteins) to form a Cap Binding Complex, CBC. The CBC influences splicing of the pre mRNA and it may be involved in transport of the pre-mRNA to the cytoplasm.

In this question answer is 1, because in the diagram the uncapped DNA does not undergoes splicing and polyadenylation.

Ans:- 2
Expln:- The outer membrane contains many copies of a transport protein called porin which forms large aqueous channels through the lipid bilayer. This membrane thus resembles a sieve that is permeable to all molecules of 5000 daltons or less, including small proteins. Such molecules can enter the intermembrane space, but most of them cannot pass the impermeable inner membrane. Thus, whereas the intermembrane space is chemically equivalent to the cytosol with respect to the small molecules it contains, the matrix contains a highly selected set of these molecules.

3. In cells having G protein coupled receptor, inhibition of protein kinase A by siRNA technology led to diminished transcription of androgen binding protein (ABP) and CREB protein. Addition of cAMP, which is a second messenger, will lead to
1.   increased transcription of ABP
2.   increased phosphorylation of CREB protein.
3.   no change in transcription level
4.   increased GTPase activity of G α subunit.

Ans:- 3

Expln:- When protein kinase A is inhibited in cells having GPCR (G protein coupled receptor), transcription of androgen binding protein (ABP) and CRER protein are diminished. While, addition of cAMP will lead to no change in transcription level.

When hormone like epinephrine binds to β- adrenergic receptors on the liver cell, G proteins on the inner side of the cell membrane are activated. Each G protein is composed of 3 subunits and the binding of epinephrine to its receptor protein causes one of the G protein units to dissociate from the other two.

The G protein subunits which dissociates from the others carries a GDP, which is replaced by GTP when the subunit is activated. The activated G protein subunit then diffuses within the plasma membrane until it encounters adenyle cyclase, a membrane enzyme that is inactive until it interacts with the G protein subunit. When activated by the G protein subunit, adenyle cyclase catalyses the formation of cAMP from ATP. The cAMP formed at the inner surface of the membrane diffuses within the cytoplasm, where it binds to and activates protein kinase-A an enzyme that adds phosphate groups to specific cellular proteins.

In liver cells, protein kinase-A phosphorylates and thereby activates another enzyme called phosphorylase, which converts glycogen into glucose-6-phosphate. The glucose-6-phosphate is then converted to glucose. Through this multistep mechanism, hormone like epinephrine causes the liver to secrete glucose into the blood during the fight – or – flight reactions.

It is clear from the above descripton that, cAMP activates proteinkinase. But if the protein kinase get inhibited, then there is no change in transcription level.

4. Cells undergo apoptosis by two distinct and inter-connected pathways: extrinsic and intrinsic. Extrinsic pathway is activated by extracellular ligand binding to cell surface death receptors. Whenever an apoptotic stimulus activates intrinsic pathway, the proapoptotic Bax and Bak proteins become activated and induce the release of cytochrome C from mitochondria leading to caspase cascade activation resulting in apoptosis. In cell A, cytochrome C is introduced by microinjection whereas in cell B, cytochrome C is introduced by microinjection but Bax and Bak are inactivated. What will be the most appropriate apoptotic response type in both cells ?
2013_june_c_2
Ans:- 1
Expln:- Two distinct pathways of apoptosis are intrinsic or mitochondrial pathway and extrinsic or death receptor pathway.

Intrinsic pathway:- The outer membrane of mitochondria display the protein Bcl-2 on their surface. Bcl-2 inhibits apoptosis. Internal damage to the cell causes the proapoptotic proteins Bax, Bak, Bad etc. and they migrate to the surface of the mitochondrion and bind to Bcl-2-blocking its protective effect. As a result holes are formed in the outer mitochondrial membrane, causing cytochrome c to leak out. Cytochrome c binds to the protein Apaf-1 (apoptotic protease activating factor-1). This complex aggregate to form apoptosomes using ATP. Apoptosomes activates caspase-9. (Caspase 9 is one of a family of over a dozen caspases). Caspases-9 cleaves and activates other caspases (caspases-3 and -7). The activation of these “executioner” caspases creates an expanding cascade of proteolytic activity which leads to phagocytosis of the cell.

So in the given question, cell A has cytochrome c and and hence can bind with Apaf 1 and can for apoptosomes which causes apoptosis. Similarly in cell B, eventhough Bat and Bak are inactivated, cytochrome c is introduced; hence can lead to apoptosis. So answer is 1.


5.    According to the current model of alternative oxidase regulation, the following factors cause induction of alternative oxidase:

1.   Significant increase in the ubiquitin pool in the cytosol.
2.   presence of α -keto acids (like pyruvate and glyoxylate).
3.   cold stress
4.   increase in cytosolic ATP concentration

Which one of the following combinations of above statements is true ?
(1)   (A) and (D)      (2)   (B) and (C)      (3)   (A) and (B)      (4)   ( A) and (C)

Ans:- 3

Expln:- Plants, some fungi, and protists contain a cyanide-resistant, alternative mitochondrial respiratory pathway. This pathway branches at the ubiquinone pool and consists of an alternative oxidase encoded by the nuclear gene Aox 1. Alternative oxidase expression is influenced by stress stimuli – cold, oxidative stress, pathogen attack-and by factors constricting electron flow through the cytochrome pathway of respiration.

More Questions & Answers Refer Simple Instant Notes