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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 (-, -).
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Answer 4 RNA hydrolysis is a reaction in which a phosphodiester bond in the sugar-phosphate backbone of RNA is broken, cleaving the RNA molecule. 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 not susceptible 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.
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Answer 1 Experimentally, the Km for an enzyme tends to be similar to the cellular concentration of its substrate. An enzyme that acts on a substrate present at a very low concentration in the cell usually has a lower Km than an enzyme that acts on a substrate that is more abundant. 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 a specific mannose residue within theoligosaccharide. This forms a phosphodiester: Man-phosphate-GlcNAc. Once the phosphodiester has been formed the lysosomal enzyme will be translocated through the Golgi apparatus to the trans-Golgi.
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Answer 3 If membrane already lysed, UMP fluroscence will directly measured in supernant.
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By the mutation of deacetylase gene causes to increased production of product. Normally deacetylase represses the gene expression 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 of mitochondria.
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 two target proteins, one is Securin and other is cyclin-B. Both Securin and cyclin-B are degraded by proteasome; hence cell enters in to anaphase. So by inhibiting proteasome causes the cell arrest at anaphase.
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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
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In Tryptophan operon the structural genes are regulated by two ways 1) 2)
By trp repressor- the presence of tryptophan causes the repression of trp operon Attenuation- the presence of tryptophan causes the formation of a attenuator terminator, which decouples transcription and translation.
In the present question the two trp codons in the leader sequence are replaced with ala codons. Then trp operon becomes sensitive to alanine 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 being translocated 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 (replication timing), with up to 100,000 present in a single human cell. Having many origins of replication helps to speed the duplication of their (usually) much larger store of genetic material. The segment of DNA that is copied starting from each unique replication origin is called a replicon. During the embryonic S-M cycles, entry into S phase must be controlled posttranscriptionally and must be regulated in the presence of constitutive cyclin E kinase activity. The addition of a GI phase during embryogenesis requires that extrinsic developmental cues helpBIOTECH Academy Hyderabad | CSIR JRF/NET, GATE, MSc Entrance | Study Material for CSIR JRF /NET 2015 Contact: 040 64611970 /9652956019 |
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influence the onset of S phase. In Drosophila, this is mediated at least in part through cyclin E. During the S-G cycle that produces polytene cells, entry into S phase can no longer have the completion of mitosis as a prerequisite. In addition to differential control of the onset of S phase, the actual parameters of S phase are altered during development. By parameters of S phase, we mean the intrinsic properties of DNA replication. The parameters of DNA replication changed in modified cell cycles include replication origin usage and activation, the rate of replication fork movement, and the block to rereplication. By pulse-labeling cells with nucleotides prior to this treatment, it is possible to directly examine sites of DNA synthesis. Originally used with radioactively labeled nucleotides to accurately measure the rate of replication fork progression, more recent studies have used this approach to compare the efficiency of origin firing between early and late replicating regions.
Answer 4 Nortern blot for RNA and Westernblot for protein. Absence of bands in control for both means regulatory mechanism operating at tranxription and translation also. Alternative splicing is a regulated process during gene expression that results in a single gene coding for multiple proteins
Answer 1
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Answer 1
Answer 2 helpBIOTECH Academy Hyderabad | CSIR JRF/NET, GATE, MSc Entrance | Study Material for CSIR JRF /NET 2015 Contact: 040 64611970 /9652956019 |
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Answer 2
Answer 2
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Answer 2 Proapoptotic members that promote apoptosis (eg. Bax and Bak). Because Apaf-1 is mutated, apatosome copmex cold not formed. No apotosis.
Answer 1
Answer 4
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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 specification B: Condtional Specification
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Answer 4
Answer 4
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Answer 4
Answer 2 IPT gene required for Cytokinin biosynthesis. Cytokinin delays senescence.
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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.
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Answer 4
Answer 4 Statement B: Correct.
Statement C: Incorrect.
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Answer 3 Statement C: Incorrect. Galactoce is not osmolyte.
Answer
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Answer
Answer 4
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Answer
Answer 2
Answer 3
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Answer 1
Answer 4 It is a frame shift mutation due to Acridine orange.
Answer 3 Stp gene is maternal and mitochondrial origin.
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Answer 4 a and d having high frequency and they are closed linked. And c are distaly linked.
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Answer 1
Answer 2
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Answer 1
Answer 1 Sea anemone has radial symmetry.
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Answer 4 Amborella’s lineage diverged from other angiosperms around 130 million years ago, sometime after the first flowering plant appeared. 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 other flowering plants. Also, while Amborella has carpels, they are incompletely closed. This is significant because the carpel is thought to have 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
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Answer 2
Answer 1 Euglenids only contain paraxial rods in flagella.
Answer 2 It is k-slected life history.
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Answer 4
Answer 1 Here, the carrying capacity of species 1 (K1) is higher than the carrying capacity of species 2 divided by the competition coefficient (K2/a21), and the carrying capacity of species 2 (K2) is higher than the carrying capacity of species 1 divided by the competition coefficient (K1/a12). Below both isoclines and above both isoclines the populations increase or decrease as in the first two scenarios, and there is an unstable equilibrium point (closed circle) where the isoclines intersect. For points above the dashed pink line (species 2 isocline) and below the solid yellow line (species 1 isocline), the outcome is the same as in the first scenario: competitive exclusion of species 2 by species 1. On the other hand, for points above the solid yellow line (species 1 isocline) and below the dashed pink line (species 2 isocline), the outcome is the same as in the second scenario: competitive exclusion of species 1 by species 2. The two stable equilibrium points are again represented by open circles. In this scenario, the outcome depends on the initial abundances of the two species.
Answer 1
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Answer
Answer 3
Answer 4
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Cost is higher than benefit.
Answer 1
Answer 3
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Answer 3
Answer 4
Answer 3
Answer 1
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Answer 2
Answer 4
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Answer 4
Answer
Answer 4
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Answer 1 or 2
Answer
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