1. Which nucleotide is present in the 5’ cap?
a) ADP
b) GDP
c) CDP
d) UDP
Explanation: The 5’ cap has a guanine residue which makes Gpp(pN)n where N is any nucleotide. This is also called the 5’ G cap. Other bases like A, T, C or U don’t participate in capping.
2. Which is the 1st enzyme in capping?
a) Guanyltransferase
b) RNA 5’ triphosphatase
c) N7G methytransferase
d) Guanul transferase
Explanation: 1st the 5’ triphosphate group is removed by the phosphatase. Then the Gpp is added to it making it the G cap. Lastly the methyl transferase transfers the methyl group to the N7 position of the cap guanine
3. The post transcriptional modifications are done after the transcript is completely transcribed. State whether the statement is true or false.
a) true
b) false
Explanation: The modifications are carried out as the transcript is produced. Like when the transcript is 25-30 nucleotides long it is capped and there has also been evidence showing that splicing also takes place during transcription.
4. CBP acts as a(an)_______________
a) Enzyme
b) Adaptor
c) Catalyst
d) Suicide enzyme
Explanation: CBP acts as a catalyst which concentrates the enzyme activity at the pol II transcripts so that they are selectively capped at their 5’ ends.
5. In an experiment you try to hybridize a Dna single strand with a mature RNA. You observe loops being formed. These loops have ____________
a) Dna
b) RNA
c) HIstone octamer
d) Histone H1
Explanation: Due to splicing the mature mRNA is shorter than the DNA with the introns in DNA looping out when they hybridize. Thus the loops have DNA.
6. Which of the following corresponding to the same gene will you expect to be the shortest?
a) DNA
b) hnRNA
c) cDNA
d) mRNA
Explanation: cDNA is made from reverse transcription of mRNA so it lacks the excess modifications present in mRNA. And as mRNA has only exons unlike hnRNA or DNA, cDNA would be the shortest of the four relating to the same gene.
7. Splicing concensus sequence is ___________
a) Exon/GU–intron–AG/exon
b) Exon/UG–intron–AT/exon
c) Exon/GU–intron–GA/exon
d) Exon/AU–intron–CG/exon
Explanation: It has been seen that splicing occurs at this consensus sequence. However, as there could be many more motief similar to this within the intron there are additional consensus sequences.
8. In the consensus sequence 5′-AG/GUAAGU–intron–YNCURAC–YnNAG/G-3′ Y could be ______
a) Base A
b) Base G
c) Base C
d) Base T
Explanation: Y could be any pyrimidine that rules out A and G as base. Now we know that RNA doesn’t have T, so the only pyrimidine possible from the given combination is C
9. 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.
10. The poly a tail protects the 3’ end from____________
a) 5’->3’ exonuclease
b) 3’->5’ exonuclease
c) Translation
d) Export
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.