CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship
Part C
PREVIOUS SOLVED QUESTIONS – 2016 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. Hydrogen bonding between the bases in the major and minor grooves of DNA is absent.
2. Both uracil and thymine have a methyl group but at different positions.
3. The backbone dihedral angles of α-helices and β-sheets are very similar. Only the hydrogen bonding pattern is different.
4. A β-turn is formed by four amino acids. The type of β-turn is determined by the dihedral angles of the second and third amino acid.
Expln:-
There are different types of β-turns classified according to the backbone dihedral angles of the second and third residues. Type I and type II are the most common forms of β-turns.[For more details: Refer Conformation of Proteins in Simple Instant Notes].
(a) ADP + Pi (b) Dinitrophenol, an uncoupler (c) Oligomycin, an ATPase inhibitor (d) Cyanide (e) Succinate
Which of the following describes the profile appropriately?
1. I – b; II – d; III – e
2. I – a; II – d; III – c
3. I – a; II – e; III – c
4. I – a; II – c; III – b
Ans:-4
Expln:-
In Kreb’s cycle, the enzyme isocitrate dehydrogenase is inhibited by NADH and ATP but activated by ADP. Similarly, citrate synthase is inhibited by citrate and by ATP. Here at time I, the addition of ADP causes increase in mitochondrial respiration. At time II, the addition of oligomycin inhibits ATPase and thus ATP will not convert to ADP. Thus increase in ATP conc. causes, decrease in mitochondrial respiration. Dinitrophenol acts as uncoupling agent, when added to cells, they stop ATP synthesis but electron transport still continues and so oxygen is still consumed.[For more details: Refer Cellular Respiration in Simple Instant Notes].
1. A, B, C and D
2. A, B and D
Ans:- 2
Expln:- The bands of the Southern blot shows that A, B and D are the progeny of the parents M and F.[For more details: Refer methods in biology in Simple Instant Notes].
A. Eukaryotic DNA transposons excise themselves from one place in the genome and integrate into another site.
B. Retrotransposons are RNA sequences that are first reverse transcribed into cDNA and then integrate into the genome.
C. Retrotransposons move by a copy and paste mechanism through an RNA intermediate.
D. As DNA transposons move via a cut and paste mechanism, there can never be an increase in the copy number of a transposons.
Which of the statement(s) is are/true?
1. A and C
2. B and D
3. B only
4. D only
Expln:- Many transposons move by a “cut and paste” process. That is, the transposons is cut out of its location and inserted into a new location. Retrotransposons move by a ‘copy and paste’ mechanisms. Here the copy made is RNA not DNA.[For more details: Refer transposons in Simple Instant Notes].
A. DNA polymerase III and DNA ligase B. AP endonuclease and DNA glycosidase C. Mut S and Mut L D. RecA and RecF Defect in which of the above enzymes impair the process?
1. A, B and C
2. D and B
3. A and D
4. A and C
Expln:- To repair mismatched bases, the system has to know which base is the correct one. In E.coli, this is achieved by a special methylase called the “Dam methylase”, which can methylate all adenines that occur within (5′) GATC sequences. Immediately after DNA replication, the template strand has been methylated, but the[For more questions and answers refer Simple Instant Notes].
More Questions & Answers Refer Simple Instant Notes