CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship
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).
(2) 3.5 m
(3) 2.2 m
(4) 4.0 m
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
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.
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.
2. increased phosphorylation of CREB protein.
3. no change in transcription level
4. increased GTPase activity of G α subunit.
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.
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.
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)
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.
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