This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Control of Translation”.
1. Mitogens can control translation efficiency by acting on _____________ factor which ___________
a) eIF4 factors which increases translation
b) eIF4A factors which decreases translation
c) eIF4E factors which increases translation
d) eIF4G factors which decreases 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.
2. In an experiment with the effect of insulin on the protein synthesis, which of the factors doesn’t play a role?
Explanation: During insulin signalling, the receptor which has an intrinsic tyrosine kinase property activates mTOR, which phosphorylates and inhibits PHAS1 thus releasing eIF4E for increasing translation.
3. Which of the following acts as a GAP during translation?
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.
4. In a synergistic model of initiation which of the following factor doesn’t directly lead to initiation?
d) Protein X
Explanation: In a synergistic model of initiation both the poly a tail and the G cap plays a role. So, PAB1 binds to the poly A tail and eIF4E binds to the 5’-G cap with the eIF4G acting as a linker. But protein X plays a role in IRES mode of initiation not this.
5. When the intracellular iron level is low, the turnover of transferrin mRNA is________
c) Kept normal
d) Doesn’t effect
Explanation: 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.
6. The production of casein by rat mammary gland tissue reduces 100 fold in the absence of prolactin hormone. This reduction is due to_____________
a) Binding of eIF4E by PHAS 1
b) Greater turnover of mRNA for casein
c) Lesser synthesis of casein mRNA
d) Development of secondary structure on mRNA preventing translation
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.
7. In an experiment if you take a stretch of hybridized RNA, and add only eIF4A, eIF4B and eIF4a and eIF4B together what will happen to the band obtained by that fragment under SDS.
a) Slight rise, no rise, appearance of two bands one on top of original
b) No rise in all cases but appearance of a 2nd band on top of 1st in 3rd case
c) Slight rise, slightly more rise, and highest rise
d) A faint band above the original, only original and a dark band above original
Explanation: 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.
8. Translation of which sigma factor is dependent on a genetic thermostat?
a) Sigma 70
b) Sigma 32
c) Sigma E
d) Sigma S
Explanation: When the temperature is favorable the sigma 32 gene is hidden from initiation by complex mRNA structure. When the temperature rises, this complex opens up stabilizing the mRNA of sigma 32 thus leading to increase in its translation. This is like a genetic thermostat.
9. In an experiment you delete a part of the 5’ UTR of transferring mRNA, what will be its effect on iron sensitivity?
a) It will be able to sense iron changes better
b) It will not be able to sense the iron changes
c) No change
d) It will have leaky sensitivity
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.
10. The Stem loop structures at the 5’ UTR of Ferritin is not ______________
a) Binding to IRE-BP
b) Regulation ferritin level
c) Degraded when unbound
d) Acting as iron sensor
Explanation: The 5’ UTR of the Ferritin can also bind to IRE-BP when iron concentration is low. In that case the translation is prevented. But it is not AU rich so, it is not degraded.
11. Which of these is not an NMD factor?
d) Ski 7
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.
12. Which of the following is the correct sequence of turnover?
a) Exonucleolytic degradation, decapping, deadenylation
b) Exonucleolytic degradation, deadenylation and decapping
c) Deadenylation, decapping, exonucleolytic degradation
d) Decapping, deadenylation and exonucleolytic degradation
Explanation: The turn-over of the mRNA proceeds in the following sequence- 1st 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.
Sanfoundry Global Education & Learning Series – Cytogenetics.
To practice all areas of Cytogenetics, here is complete set of 1000+ Multiple Choice Questions and Answers.