csir december 2014 - part c keys (1)

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CSIR JRF/NET Life Sciences December 2014

Booklet A

Part C

Questions + Keys + Explanations

Answer 2

Peptide Bond Hydrolysis is favorable and means Free energy always negative. Peptide bond easily hydrolysed in presence of 6N HCl.

Answer 1

Check the angle of - helix. They fall on negative scale (-, -).






Answer 4

RNA hydrolysis is a reaction in which a phosphodiester bond in the sugar-phosphate backbone of RNA is broken, cleaving the RNAmolecule. RNA is susceptible to this base-catalyzed hydrolysis because the ribose sugar in RNA has a hydroxyl group at the 2’position.[1] This feature makes RNA chemically unstable compared to DNA , which does not have this 2’ OH group and thus is notsusceptible to base-catalyzed hydrolysis.[1]

Answer 2

Presnece of disulfide bonds makes protein more stable at high temperatures.

Answer 2

Question is related to Scatchard plot.






Answer 1

Experimentally, the K m for an enzyme tends to be similar to the cellular concentration of its substrate. An enzyme that acts on asubstrate present at a very low concentration in the cell usually has a lower K m than an enzyme that acts on a substrate that is moreabundant.

Km of Isozyme A (0.1) almost similar to liver substrate concentration (0.2).

From, Biochemistry, Nelson aand Cox, Page 199

Answer 4

Both A and B statements are correct. C statements is wrong.

In the cis-Golgi a GlcNAc phosphotransferase adds a GlcNAc-1-phosphate residue onto the 6-hydroxyl group of aspecific mannose residue within theoligosaccharide. This forms a phosphodiester: Man-phosphate-GlcNAc. Once the phosphodiesterhas been formed the lysosomal enzyme will be translocated through the Golgi apparatus to the trans-Golgi.






Answer 3

If membrane already lysed, UMP fluroscence will directly measured in supernant.

Answer 1






By the mutation of deacetylase gene causes to increased production of product. Normally deacetylase represses the geneexpression but mutant deacetylase can’t repress the gene expression.

Answer 4. Protein import into mitochondria required membrane potential. Presence of Proteinase K will degrade protein outside ofmitochondria.

Answer 3

The addition of proteasomes inhibitor to G2 cells causes the cells arrest at anaphase.For metaphase to anaphase progression requires the activation of Anaphase Promoting Factor(APF), which causes ubiquitination of twotarget proteins, one is Securin and other is cyclin-B. Both Securin and cyclin-B are degraded by proteasome; hence cell enters in toanaphase.

So by inhibiting proteasome causes the cell arrest at anaphase.






Answer 2

Option 1: Wrong. They contain Pseudopeptidoglycan and 70S Ribosomes.Option 3: Wrong. They dont have Golgi bodies.Option 4: Wrong. They contain Pseudopeptidoglycan not chitinous cell wall.

Answer 1






In Tryptophan operon the structural genes are regulated by two ways

1) By trp repressor- the presence of tryptophan causes the repression of trp operon2)

Attenuation- the presence of tryptophan causes the formation of a attenuator terminator, which decouples transcription andtranslation.

In the present question the two trp codons in the leader sequence are replaced with ala codons. Then trp operon becomes sensitive toalanine concentration.But trp operon is repressed by trp repressor. Hence in the presence of tryptophan causes the low expression tryptophan synthetase.

Answer 3

Statement F: Correct.

tRNA, and it can bind to the ribosomal A site and participate in peptidebond formation. The product of this reaction, instead of beingtranslocated to the P site, dissociates from the ribosome, causing prematurechain termination.

Answer 1

Eukaryotes often have multiple origins of replication on each linear chromosome that initiate at different times (replicationtiming), with up to 100,000 present in a single human cell. Having many origins of replication helps to speed the duplication oftheir (usually) much larger store of genetic material. The segment of DNA that is copied starting from each unique replication originis called a replicon.

During the embryonic S-M cycles, entry into S phase must be controlled posttranscriptionally and must be regulated in the presenceof constitutive cyclin E kinase activity. The addition of a GI phase during embryogenesis requires that extrinsic developmental cues






Answer 1

Answer 2






Answer 2

Proapoptotic members that promote apoptosis (eg. Bax and Bak). Because Apaf-1 is mutated, apatosome copmex cold not formed. Noapotosis.

Answer 1

Answer 4

Answer 1






CSIR asking Non-Cananocal pathway and Cytoskeleton movements.

Answer 2

CSIR Asking for conditions which prevent interaction between egg and sperm.

Answer 1

A: Autonomous specificationB: Condtional Specification






Answer 4

Answer 4






Answer 3

Statement A: incorrect. Allterpenes are derived from five carbon element.Statement B: Correct.Statement C: Correct.Statement D: Incorrect.

Answer 2

Statement B: Incorrect. Cryptochromes not involved in hypocotyl length.






Answer 4

Answer 4

Statement B: Correct.

Statement C: Incorrect.






Answer

Answer 2

Answer 3






Answer 1

Answer 4

It is a frame shift mutation due to Acridine orange.

Answer 3

Stp gene is maternal and mitochondrial origin.






Answer 4

a and d having high frequency and they are closed linked.

And c are distaly linked.






Answer 1

Answer 2






Answer 1

Answer 1

Sea anemone has radial symmetry.






Answer 4

Amborella’s lineage diverged from other angiosperms around 130 million years ago, sometime after the first flowering plantappeared. Amborella has all of the defining features of a flowing plant, but at the same time it seems to have retained some“gymnospermy ” characteristics as well. For instance, it lacks the vessel elements for water conduction present in most otherflowering plants. Also, while Amborella has carpels, they are incompletely closed. This is significant because the carpel is thought tohave originated from a flat, leaf-like structure with ovules on its margins. This structure eventually rolled inward and became enfolded,creating a hollow, enclosed ovary with one or more ovules. Early angiosperms probably had carpels that were not quite fused shut but were sealed with secretions from the carpel, which is the case with Amborella.

Answer 3






Answer

Answer 3

Answer 4

Answer 3






Answer 2

Answer 4




Answer 1 or 2

Answer

csir december 2014 - part c keys (1)

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