CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship
Some Important Questions Collected from the Memory of Students and Probable Questions and Answers
Previous Year Questions & Answers –Part B. June 2017
A.In proteins the amino acids that can undergo oxidation are Cys and Met.
B.A tetrasaccharide composed of alternate L and D isomers will not be optically active.
C.The ΔG (Kcal/mol) values for Keqof 0.1, 0.01 and 0.001 are 1.36, 2.72 and 4.09, respectively. It can be concluded that the relationship between ΔG and Keq is parabolic.
D.The oxidation states of Fe in haemoglobin is +2. In cytochrome C, the oxidation states of Fe can be +2 or +3
E.InDNA, the sugar and bases are planar.
F.High-energy bonds hydrolyze with large negative Δ
Choose the combination with ONLY ONE WRONG statement.
1.A, E, F
2. B, C, D
3. C, D, E
4. A, B, C
Expln:- Unlike the nucleotide bases, which are planar and rigid, the sugar ring is easily bent and twisted into various conformations.
The five nitrogen bases are all planar molecules meaning that they are fairly flat and rigid. The N atom in each molecule is the point of attachment for a sugar molecule.
The DNA molecule’s stability and rigidity is due to stacking interactions between adjacent bases. These stacking interactions are a form of Van der Waals interaction.[Base stacking : Two or more bases are positioned with the planes of their rings parallel like a stack of coins].
A noncovalent attractive force between two aromatic rings is known as Pi stacking. In addition to hydrogen bond, Pi bond stabilizes the DNA.
Amino acids that can undergo oxidation include methionine, cystein, histidine, tryptophan and tyrosine.
Among these AAs, cysteine is more prone to oxidation by ROS because of its high nucleotiphilic property. The reactivity of Cyst with ROS is due to the presence of thiol group. In the oxidized form Cyst forms disulphide bond, which are primary covalent cross-link found in proteins, and which stabilize the native conformation of a protein. Methione is among the most hydrophobic of the amino acids. This means that most of the methione residues in globular proteins are found in the interior hydrophobic core. In membrane-spanning protein domains, methionine is found to interact with the lipid bilayer. In some proteins, a fraction of the methionine residues are somewhat surface exposed. These are susceptible to oxidation to methione sulfoxide residues.
1.N, N-dimethylguanosine, pseudouridine, 2’O-methyluridine
2.2-thiouridine, dihydrouridine, N-isopentenyladenine
3.5-methyldeoxycytosine, 5-thiouridine, pseudouridine
4.Dihydrouridine, 4-thiouridine, 2’O-methyluridine
Expln:- N, N-dimethylguanosine :- The most prominent case of dual methylation is found in the majority of eukaryotic tRNAs.
Pseudouridine (psi- ψ) is an isomer of the nucleoside uridine in which the uracil is attached via a carbon-carbon instead of a nitrogen-carbon glycosidic bond.
2’O-methylation is a common nucleoside modification of RNA, where a methyl group is added to the 2’ hydroxyl of the ribose moiety of a nucleoside producing a methoxy group. 2’O-methylated nucleosides are mostly found in ribosomal RNA and small nuclear RNA.
2-thiouridine :– A modified nucleobase found in tRNAs that is known to stabilize U:A pairs and modestly destabilize U:G Wobble pairs.
Dihydrouridine (DHU) :- is a pyramidine which is the result of adding two hydrogen atoms to a uridine, making it a fully saturated pyramidine ring with no remaining double bonds.
N-isopentenyl adenine :- is a naturally occurring cytokinin that regulates cell division, development and nutrient processing in plants.
4-thiouridine:- is a photoreactive (crosslinking) uridine analogue that upon phosphorylation to 4-thioUTP may be incorporated into RNA structures.
5-methyl deoxycytosine:- pairs with deoxyguanine, and substitutes for deoxycytosine. It is particularly useful in studies like anti-sense, DNA methylation studies, strong-binding PCR primers etc.
2’O-methyluridine is a uridine analogue, used for preservation of antiviral nucleoside derivatives as inhibitors of subgenomic hepatitis C virus RNA replication.
Answer is 3 , because the uridine analogue that is incorporated into RNA structure is 4- thiouridine; but not 5-thiouridine.
A.The dihedral angles of an amino acid X in Acetyl-X-NMethyl amide in the Ramachandran plot, occur in very small but equal areas in the left and right quadrants. It can be concluded that X is not one of the 20-coded amino acids.
B.The dihedral angles of a 20-residue peptide are represented in the Ramachandran plot. It is possible to conclude that the peptide does not have a proline.
C.Two proteins can have a similar fold even if they do not share significant similarity in their primary structure.
D.On denaturation of a protein by urea, the interactions that would be disrupted are ionic bonds and van der Waal’s interaction but not disulfide bonds.
Choose the combination with ALL CORRECT answers:
1.A, B , C
4. A, B, D
Expln:- Proline has cyclic side chain. Therefore rotation around bond is constrained by its inclusion in the pyrrolidine ring. The phi values are restricted to angles around -60 and hence proline is the most conformationally restricted AA residue.
A.Fructose 2,6-biphosphate is an allosteric inhibitor of phosphofructokinase 1.
B.The TCA cycle intermediates, succinate and oxaloacetate can both be derived from amino acids.
C.A diet rich in cysteine can compensate for a methionine deficient diet in humans.
D.dTTP for DNA synthesis can be obtained from UTP.
E.In the fatty acid biosynthetic pathway the carbon atom from HCO3- in the synthesis of malonyl CoA is not incorporated into palmitic acid.
Choose the option that represents the combination of all the CORRECT statements
1.A, B, C and E
2. B, D and E
3. A, D and E
4. Only B and C
Expln:-Fructose 2,6-biphosphate is a potent stimulator of phosphofructokinase and an inhibitor of fructose 1,6 biphosphatase.
Methionine is required to make cysteine, glutathionine and taurine.
A.Capping protects the mRNA from degradation by 5’-exoribonucleases.
B.During capping, the α-phosphate is released from the 5’end of the nascent mRNA.
C.Phosphorylation mediated conformational change in carboxyl terminal domain (CTD) of RNA Pol II enables its binding with capping enzymes.
D.During capping, a 5’-5’ triphosphate bond is formed between the β -phosphate of the nascent mRNA and α -phosphate of GTP.
Which of the above statement(s) is/are INCORRECT ?
2. B only
3. A and B
4. C and D
Expln:-The beta phosphate of the RNA transcript displaces a pyrophosphate group at the 5’ position of the GTP molecule as seen in the diagram
More Questions & Answers Refer Simple Instant Notes