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Correct option is (c) No change

Best EXPLANATION: The IRE, or iron response elements of the transferring gene is located at the 3’ end of the mRNA NEAR the poly A tail. Thus, deletion of a part in 5’end will have no effect at all.

The CORRECT choice is (b) Decreased

To elaborate: By turnover of mRNA the EFFECTIVE concentration of a PROTEIN can be decreased as it REDUCES the chance of translation. When already the level of iron is low, more TRANSFERRING s required, so mRNA is stabilized and its turnover is reduced.

Correct choice is (a) Increase

The best I can explain: EIF4E-BP is the active form of the FACTOR which can bind to the eIF4E and prevent TRANSLATION. When phosphorylated, this BINDING is sterically prevented. Thus, phosphorylation increases translation generating FREE eIF4E.

Right choice is (b) Release factor 1

Easy explanation: RF3.GTP binds the ribosome after the PEPTIDE is already RELEASED by the RF1. It HYDROLYSES to release the RF1 INSTEAD.

Right answer is (c) E site of 50S

The best explanation: According to hybrid model 1^st the head part SHIFTS to the NEXT site in the cycle then the foot part of tRNA bound to 30S SUBUNIT shifts ALONG with the input of energy. Thus, according to our question the head should be at E site.

The CORRECT ANSWER is (C) eIF4E

The best explanation: eIF4E recognized the 7-Methyl Guanine cap in the 5’ end of mRNA. This interacts with eIF4G and EIF3 which slides the 40S subunit on the mRNA looking for an initiation SITE.

Correct option is (a) IRES

The best EXPLANATION: IRES stands for Internal ribosomal entry site, which in no WAY is related to iron sensing. The AU rich REGIONS in IRE are actually the step loop in 3’ END of transferring that binds to IRE-BP. Also, the stem loop in 5’ end of ferritin binds IRE-BP.

Right OPTION is (C) In FRAME stop codon between two AUGs

For explanation: If there is in frame stop codon to 1^st AUG there will be an OPEN reading frame in between 1st AUG and stop codon and again another open reading frame from 2^nd AUG. This will lead to EFFECTIVELY normal translation from the 2^nd AUG.

The correct option is (d) PABP2

Explanation: PABP2 can bind the POLY a tail to the eIF4E through the MEDIATOR eIF4G. This increases the ASSOCIATION of the 40S subunit to the 5’ end of the mRNA thus increasing translation.

Right ANSWER is (b) First

Explanation: The ribosome scans the mRNA from 5’ end to the 3’ end searching for a Kozak consensus sequence to start at. Here, the 1^st AUG indeed has a purin at -3 and +4 position so it is a good INITIATION site, the ribosome will not LOOK beyond to the 2^nd AUG(WEAKER) and start initiation.

The correct OPTION is (b) Promote degradation

Explanation: The stem loop STRUCTURES in the 3’ end in the UTR of the mRNA for transferring has some stem loop structures with the loop being rich in AU bases. These LOOPS call in the protein responsible for degrading the mRNA increasing its TURNOVER.

The correct option is (c) EIF4E factors which increases translation

Explanation: Mitogens cause PHOSPHORYLATION of eIF4E-BP which binds to eIF4E and prevents translation. THUS, phosphorylation of the inhibitor will in turn PROMOTE translation.

The correct answer is (b) False

Explanation: RF1 or release FACTOR 1 is responsible for recognition of two stop CODONS- UAA and UAG, while RF2 can recognize the codon UGA.

Correct choice is (c) Deadenylation, DECAPPING, exonucleolytic degradation

The best I can EXPLAIN: The turn-over of the mRNA proceeds in the following sequence- 1^st the poly A TAIL is degraded i.e. deadenylation. Then the 5’G cap is REMOVED i.e. decapping. The mRNA is degraded by the exonucleases.

Right option is (c) RF3

For EXPLANATION I would say: The RF3 is responsible for CLEAVAGE of the PEPTIDE bond as well as for the release of the RIBOSOMAL subunits in EUKARYOTES.

Right choice is (C) RRF and EF-G

The EXPLANATION: The RRF first replaces the RF3 from the termination complex. Then, EF-G translocates the complex for the last time FREEING the mRNA, Ribosome, tRNA and the factors.

The correct choice is (d) Deacylated TRNA

For explanation I would SAY: From the A site and the P site, the tRNA moves to the E site or EXIT site. Then its tRNA is deacylated as the peptide bond is on the P site tRNA.

The correct option is (c) EMPTY

To elaborate: The EF-G factor HELPS in translocation of the ribosome such that the peptide bound tRNA shifts to the P site and A site becomes empty and AVAILABLE to BIND another CHARGED tRNA.

Correct ANSWER is (B) eIF3 and eIF6

Easy explanation: eIF1 is a multi protein complex with eight subunits; this along with eIF6 prevents the REASSOCIATION of the subunits of ribosome after translation of ONE protein is complete.

The correct answer is (B) A hairpin 52 at downstream of cap

To EXPLAIN I would say: It has been seen experimentally that a hairpin 52 at further from the cap can’t help the AUG to be recognized as initiation site even when the AUG is within it. The other OPTIONS here improve translation from the AUG involved.

The correct option is (c) Neck

Easy explanation: The 3’-OH END of the 16S rRNA is located in the neck region of the rRNA. NEAR it is the SHINE and Dalgarno sequence that BINDS the MRNA.

Right option is (d) A faint band above the original, only original and a DARK band above original

To elaborate: Rise in a band is seen only in CASE of EMSA which is not PERFORMED in this experiment. Here eIF4A acts as a helicase separating the dimer into monomer which leads to the formation of an upper band. eIF4B is a PROMOTER of helicase activity of eIF4A in absence of which there is only SLIGHT activity.

Right option is (d) eIF5B

Easiest explanation: While the eIF2B has a GEF activity, the eIF5 PROMOTES the phosphatase activity of eIF5B. This HELPS the GTP to hydrolysis to GDP thus helping in dissociation of tRNAi and initiation of TRANSLATION.

Right option is (B) Exonuclease

For explanation: NMD stands for nonsense mediated degradation which is responsible for degrading the MRNA prematurely when it has a nonsense MUTATION. Nonsense mutation can lead to a stop codon before the original stop codon. Here the only exonuclease also PARTICIPATES in normal turn over.

Correct choice is (b) Greater turnover of mRNA for CASEIN

The explanation: It has been seen in vitro that treatment with prolactin only INCREASES the PRODUCTION of casein mRNA to 3 fold, but there is an increase of 100 fold in casein level. So, this must be due to an increase in stability by the change in protein turnover RATE.

The CORRECT ANSWER is (a) Ef-Tu

For EXPLANATION I would say: EF-Tu in GTP bound form binds to amino ACYL tRNA. It helps to bring the amino acid to the ribosome, and one placed in proper position the GTP HYDROLYSES.

Correct option is (b) Phosphorylation of eIF2-alpha

The best explanation: HCR is HEME CONTROLLED Repressor protein which SENSES the lack of heme in the medium and phosphorylated eIF2-alpha. This prevents the GEF activity of eIF2B and prevents translation. This measure prevents the waste of resources.

Correct choice is (C) IF3

The best explanation: The IF3 binds to the 30S subunit and PREVENTS association of 30S and 50S subunit when the ribosome is not bound to MRNA. It also HELPS in binding t the mRNA.

Right option is (a) True

Explanation: When there is a frame shift mutation in frame initiation CODON and TERMINATION codon are transcribed. Thus, it can EVEN lead to alternate initiation and alternate protein SYNTHESIS.

Correct option is (b) 20 amino acid PER second

Easiest EXPLANATION: While the ribosome can read about 50-60 nucleotides per second, due to triplet codon the maximum number of amino acids added per second would be : (60/3) = 20 amino acid per second.

Correct choice is (d) IF-1 and IF-3

To explain: While the RELEASE factor releases the ribosome from the termination complex, in order to keep the ribosome FUNCTIONAL the 50S and 30S SUBUNITS should be SEPARATED. This is maintained by IF1 and IF3 (MAINLY by IF3) which prevents re-association.

Right OPTION is (c) -3 and +4

To explain: The CONSENSUS sequence CC(A/G)CCAUGG was DISCOVERED by Kozak, where the purine in -3 and +4 position is significant in determining which AUG should serve as INITIATION site.

The correct choice is (b) CIRCULAR

Best EXPLANATION: Polysomes are the chromosomes on with a number of ribosomes are translating the same gene in a series. These tend to appear circular as that helps in recycling the ribosomes from the poly A TAI to the initiation SITE.

Correct answer is (c) eIF2B

For explanation: The factor eIF2B is a GUANOSINE nucleotide EXCHANGE Factor or GEF. It exchanges the GDP BOUND to EIF2 to GTP so that it can then bind to the aminoacyl tRNA.

Correct OPTION is (c) eIF3 and eIF6

Explanation: While he factor eIF3 binds to the 40S subunit, the other factor eIF6 binds to the 60S subunit and prevent the association of ribosomal SUBUNITS when not bound to MRNA.

The correct OPTION is (C) f-Met-tRNAi-Met

To explain I would SAY: The initiation of translation in eukaryotes is by the codon AUG that recognizes initiation tRNA or tRNAi that binds to METHIONINE. While in prokaryotes it is by formyl met tRNA which binds to met and CONVERTS it to formyl met.

Correct answer is (C) 5

For explanation I would say: The FACTORS are- RF1-3, RRF and Ef-G RESPECTIVELY. The EF-G factor finally hydrolyses the GTP that leads to final translocation STEP freeing the ribosome.