Part C-Dec 2013

CSIR-UGC National Eligibility Test (NET) for Junior Research Fellowship and Lecturer-ship

Part C

PREVIOUS SOLVED QUESTIONS – 2013 December 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. Tryptic digest of a heptapeptide[built from 3 lysine (K), 2 alanine (A), 1 tyrosine (Y) and 1-phenylalanine (F)] yielded tri and tetrapeptide. Which of the following is the correct sequence of the heptapeptide ?
The type of interaction between them is

1.   KAYAKFK
2.   YKAAFKK
3.   KYKAAKF
4.   KYAAKFK

Ans: 3
Expln:- Trypsin cleaves peptide chains mainly at the carbonyl side of lysine and arginine, except when either is followed by proline.

2. When cells enter mitosis, their existingarray of cytoplasmic microtubules has to be rapidly broken down and replaced with the mitotic spindle, which pulls the chromosomes into the daughter cells. The enzyme Katanin is activated during the onset of mitosis and chops microtubules into short pieces. The possible fate of the microtubule fragments created by Katanin will be

1. depolymerization
2. aggregation
3. degradation
4. translocation

Ans:- 1

Expln:- Katanin is a microtubule – severing (severe = to divide)AAA protein (AAA = ATPases associated with diverse cellular Activities). The severing of microtubules by katanin is regulated by nucleotide exchange factors, which can exchange ADP with ATP, protective microtubule-associated proteins (MAPs). Katanin-mediated microtubule severing is an important step in mitosis and meiosis. Katanin is responsible for severing microtubules at the mitotic spindles when disassembly is required to segregate sister chromatids during anaphase.

3. A bacterial strain can use carbohydrates and hydrocarbons as growth substrates. The strain uses glucose following a minimal lag period after inoculation, regardless of the other carbohydrates and hydrocarbons in the growth medium. The following observations were also made.

A)  In the absence of glucose, lactose is used after a lag period of about three times as long as the lag period for glucose utilization.
B)  The presence of hydrocarbons does not affect the lag period for the utilization of lactose.
C)  The utilization pattern for all hydrocarbons is similar to that of lactose.
D)  Branched hydrocarbons are not immediately utilized if straight chain hydrocarbons are initially present.
Which one of the following specific regulatory mechanisms is consistent with the above observations related to carbohydrate and hydrocarbon utilization ?

1.   Diauxie
2.   End point repression
3.   Catabolite repression
4.   Transcription attenuation

Ans:- 1

Expln:- Diauxic growth: In a culture medium containing two carbon sources, bacteria such as E.coli displays a growth curve, called diauxic. Under this conditions, if glucose and lactose are supplemented in medium having E.coli, first E.coli will utilize glucose and after it is exhausted lactose will be utilized. In between a short lag period is there. This led us to conclude that E.coli preferentially utilizes certain carbon source.

4.Cell cycle is regulated by various cyclins and cyclin dependent kinases (CDK). On receiving mitotic stimuli, cyclin D, the first cyclin expressed, binds with existing CDK4 to form the active cyclin D-CDK4 complex. This in turn phosphorylates retinoblastoma protein (Rb) which activates E2f to further activate the transcription of various downstream cyclins. In a particular cell type there is a mutation in Rb such that it cannot be phosphorylated. What will be the correct expression pattern of cylin E in these cells after mitotic stimulation ?

2013_dec_c_1

Ans:- 3

Expln:-A rising level of G, cyclins, D cyclins bind to Cdk4 and signal the cell to prepare the chromosome for replication. Then the S phase cyclin A, binds to Cdk2 and prepares the cell to duplicate its DNA. The other S phase cyclin E, binds to Cdk2 and phosphorylate Rb. Here, since Rb get mutated, cyclin E can not be expressed.

5. AminoacyltRNAsynthetases face two important challenges :
i) they must recognize the correct set of tRNAs for a particular amino acid.
ii) they must charge all of these isoacceptingtRNAs with the correct amino acid.
Both of these processes are carried out with high fidelity by the following possible mechanisms :
A. The discrimination ability resides predominantly at the acceptor stem of the tRNAs.
B. The specificity is contributed by the anticodon loop in tRNAs.
C. The specificity is embedded in the amino acyl synthetase at the ‘N’ terminus
D. The specificity is contributed by the variable loop of the tRNA.
Which of the following is corrected ?

1.   A and B
2.   A and C
3.   B and C
4.   A and D

Ans:- 1

Expln:-   Aminoacyl tRNA synthetase

Aminoacyl tRNA synthetase catalyze the formation of “charged” transfer RNA. To begin the process, a specific aminoacyl tRNA synthetase binds a particular amino acid and also a molecule of ATP at its active site. A covalent bond is formed between AMP and the carboxyl group of the amino acid, and phosphate is released from the enzyme. A specific tRNA that has an anticodon that corresponds to the amino acid then binds to the enzyme.

The nucleotide sequence in the anticodon region as well as other parts of the tRNA molecule, such as the acceptor stem, are important for recognition between the tRNA and the aminoacyl tRNA synthetase. The bond is broken between the amino acid and AMP, and the AMP is released. At the same time, a covalent bond is formed between the amino acid and the 3` end of the tRNA. Now the tRNA called a “charged” tRNA molecule. The charged tRNA is then released from the enzyme and is now ready to be used in the process of translation on a ribosome.

6.During heat shock, mammalian cells shut down global protein synthesis while inducing heat shock proteins (Hsps). The possible molecular regulation(s) that could explain the phenomenon are :

A.   mRNA of all proteins, except those of Hsps, undergoes degradation during heat shock.
B.   Cap – dependent translation of most mRNAs is affected during heat shock due to denaturation of cap binding protein, elF – 4E.
C.   Translation initiation of Hsp mRNAs takes places through their internal ribosome entry sites (IRES)
D.   Hsp mRNAs are abundant during heat shock and thus they compete out other mRNAs for ribosome binding and translation.
Which of the following sets is correct ?

1.   A and D
2.   B and C
3.   C and D
4.   A and D

Ans:- 2

Expln:-   Eukaryotic translation initiation may be cap-dependent or cap-independent.

Cap-dependent initiation: In cap-dependent initiation, translation can be initiated only at the 5` end of the mRNA molecule, since 5` cap recognition is required for the assembly of the initiation complex. The location for these sites is often in the 5`UTR. The protein factors bind the 40S subunit of ribosome. The eukaryotic Initiation Factor 3 (eIF3) is associated with the small ribosomal subunit, and plays a role in keeping the large ribosomal subunit from prematurely binding. eIF3 also interacts with the eIF4F complex, which consists of three other initiation factors: eIF4A, eIF4E, and eIF4G. eIF4G is a scaffolding protein that directly associates with both eIF3 and the other two components. eIF4E is the cap-binding protein.

Cap-independent initiation: The best- studied example of the cap-independent mode of translation initiation in eukaryotes is the Internal Ribosome Entry Site (IRES) approach. What differentiates cap-independent translation from cap-dependent translation is that cap-independent translation does not require the ribosome to start scanning from the 5` end of the mRNA cap until the start codon. The ribosome can be trafficked to the start site by ITAFs (IRES trans-acting factors) bypassing the need to scan from the 5` UTR. This method of translation has been recently discovered, and has found to be important in conditions that require the translation of specific mRNAs, despite cellular stress or the inability to translate most mRNAs. Examples include factors responding to apoptosis, stress-induced responses.

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