This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Post-Transcriptional Modification – 1”.
1. If the events in post transcriptional modification are listed as-
i) Polyadenylation ii) capping iii) splicing
What would be the correct sequence?
Explanation: Post transcriptional modification is very strictly synchronized process and no next step takes place unless the preceding one is complete. The sequence is 1st adding cap, then Polyadenylation and then splicing.
2. Which RNA transcript would be capped by the capping enzymes in vitro mixture of RNA?
a) Pol II transcripts
b) Pol I and Pol III transcripts
c) Pol I transcripts
d) Pol I, Pol II and Pol III transcripts
Explanation: In vivo, only pol II transcripts are capped, however in vitro when the different pol transcripts are mixed with the capping enzymes all of them are capped showing no specificity.
3. Which nucleotide is present in the 5’ cap?
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.
4. Which is the 1st enzyme in capping?
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.
5. The post transcriptional modifications are done after the transcript is completely transcribed. State whether the statement is true or 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.
6. CBP acts as a(an)_______________
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.
7. In an experiment you try to hybridize a Dna single strand with a mature RNA. You observe loops being formed. These loops have ____________
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.
8. Which of the following corresponding to the same gene will you expect to be the shortest?
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.
9. Splicing concensus sequence is ___________
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.
10. 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.
Sanfoundry Global Education & Learning Series – Cytogenetics.
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