This set of Cytogenetics Questions and Answers for Freshers focuses on “Post-Transcriptional Modification – 2”.
1. In an experiment, you are trying to pull down the site where guanyl transferase binds by co-immunoprecipitation. You run a gel and autoradiograph to see a band denoting GSK-RNA pol II (phosphorylated). What will you see when you perform the same experiment with overexpression of GSK in one setup and overexpression of unlabelled GSK-RNA pol II in the other setup?
a) Band is observed in both cases
b) Band in 1st case no band in 2nd
c) No band in 1st case and band in 2nd
d) No band in both cases
Explanation: There will not be any change in the intensity of the 1st band as GT is not binding to GSK tag. The 2nd case also doesn’t change the intensity of the band as GT binds to phosphorylated –GSK-RNA pol II, so no GT bound unlabelled RNA exists to compete with the labeled band and decrease its intensity.
2. The poly a tail protects the 3’ end from____________
a) 5’->3’ exonuclease
b) 3’->5’ exonuclease
Explanation: The 3’ end of the RNA is susceptible to chewing by 3’->5’ exonucleases. So, the poly A tail acts as a safeguard which even if chewed partly will protect the valuable information of the RNA. This in place of inhibition rather promotes the next steps like transcription and export.
3. Which of the following mRNA will not be polyadenylated?
a) Actin mRNA
b) Receptor mRNA
c) Canonical histone mRNA
d) Transport channel mRNA
Explanation: Most of the mRNAs are polyadenylated except the major histone mRNAs. These are the canonical mRNA that need a rapid turnover and have a unique 3’ end. However, the non-canonical mRNA produced during S phase are polyadenylated.
4. Which of the box is retained in mature mRNA?
a) TATA box
b) Pribnow box
c) G/U box
d) AU box
Explanation: The TATA box and Pribnow box are part of the promoter so they are not even transcribed into RNA. The GU box on the other hand is cleaved off during polyadenylation while the AU box is retained even after polyadenylation.
5. Which of these sequences will mark a site of polyadenylation?
Explanation: The cleavage sequence has a consensus of AAUAAA sequence towards the 5’ and a G/U box towards the 3’ end and a CA dinucleotide in the middle. The cleavage is just after 3’ to the CA.
6. Which of the following does not directly interact with DNA?
Explanation: PAP binds to CPSF and CStF by protein-protein interaction and doesn’t actually bind to the DNA. However, the other factors are cis elements binding to specific sequences.
7. CPSF and CStF bind to ____________
a) AAUAAA and G/U box respectively and bind is very strong
b) G/U box and AAUAAA respectively and bind is very strong
c) G/U box strongly but weakly AAUAAA
d) AAUAAA weakly but stabilization occurs after G/U binding
Explanation: CPSF binds to AAUAAA but the interaction is weak. This interaction is strengthened by CStF binding to G/U box and further stabilization occurs at CFI and CFII binding.
8. PAP has a processivity of ___________
a) 5-10 bases
b) 10-20 bases
c) 50-100 bases
d) 100-250 bases
Explanation: PAP has a very low processivity of about 10-20 A bases before binding of PABP II which has a very high processivity of 100 to 250 poly A bases.
9. Which experimental method can you use to detect 3’ end processing?
b) Affinity chromatography
c) Northern blot
Explanation: An affinity chromatography using oligo T sephadex beads in the column can be used to detect proper 3’ processing. If the RNA is properly processed the poly A tail will adhere to the oligo T and will not come out of the column without elution.
Sanfoundry Global Education & Learning Series – Cytogenetics.
To practice all areas of Cytogenetics for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.