Cytogenetics Questions and Answers – Post-Transcriptional Modification – 3

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This set of Cytogenetics Interview Questions and Answers for freshers focuses on “Post-Transcriptional Modification – 3”.

1. IN case of splicing the attacking group is _________________
a) ATP
b) GTP
c) A
d) G
View Answer

Answer: c
Explanation: Splicing requires a branch point A that can attach the 5’ end with the intron making a lariat structure. This doesn’t require energy molecules like ATP.
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2. In the case of group 1 RNA, which one of the following can’t attack?
a) GMP
b) GDP
c) GTP
d) ATP
View Answer

Answer: d
Explanation: It is seen that in the case of group II auto splicing RNA the attack is made by the 3’-OH end of the ribose attached to the G base. It is immaterial whether its GMP, GDP or GTP. However, A base is not seen to attack in that case.

3. The 3’- OH end of the branch point A makes the attack in splicing of an intron.
a) True
b) False
View Answer

Answer: b
Explanation: The 2’-OH of A base is seen to make this attack on the 5’ end of the intron at the phosphodiester bond while the original 3’-OH end of the A is still in the bonding with the phosphate. This makes a lariat structure.
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4. Choose the wrong answer.
a) The A base saves the 5’-P of the intron by forming bond which saves requirement of ATP
b) The 5’-P attacks the 3’-OH of the intron
c) In self splicing group I G base makes this lariet structure
d) Group II self-splicing RNA splices better when proteins are added
View Answer

Answer: c
Explanation: The G base attacking the 5’ end of the intron doesn’t form a bond with the 2’OH but with the 3’-OH end. Thus, in place of a leriet like structure a linear structure is seen.

5. Which of the following shows incorrect matching?
a) Group I : G base
b) U1 : Exon
c) U2AF : U2
d) U6 : U2
View Answer

Answer: d
Explanation: In the spliceosome complex we see that U1 is replaced by U6 and not U2. U2AF on the other hand is replaced by U2. Other matches but d are correct.
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6. 5’______ and 3’ ____ is the sequence for two ends of an intron.
a) GA TC
b) GU AG
c) AG TA
d) GA AT
View Answer

Answer: b
Explanation: The 5’ end of the splice site has the sequence GU, the 3’ end has the sequence AG. These help to locate the intron but more determining factors exist.

7. In the cytoplasm we can find ___________________ U rich snRNA.
a) U3, U4 and U6
b) U4, U6
c) U4, U5, U6
d) U3, U5
View Answer

Answer: b
Explanation: U4 and U6 snRNA are found in the cytoplasm. They can associate to form a common RNP with the other proteins when they can move to the nucleus. In the nucleus, it interacts with U5 to form the tripartite complex.
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8. Which of the following is retained in the C complex of splicosome?
a) U5
b) U4
c) U2AF
d) U1
View Answer

Answer: a
Explanation: The C complex is the final structure before the attack. Then U6 has replaced U1 and U2 has replaced U2AF and U4 dissociates. Only U5 is still there stabilizing the structure by binding the two ends to close.

9. Proteins rich in _____________ and ______________ help in splicing.
a) A and G
b) R and S
c) F and G
d) S and R
View Answer

Answer: d
Explanation: The proteins rich in Serine (S) and Arginine (R) are needed for proper splicing. FG rich proteins on the other hand are seen in the nucleoporin complex.
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10. The SR proteins bind to the __________________
a) Introns to be skipped
b) Introns not to be skipped
c) Exons to be skipped
d) Exons not to be skipped
View Answer

Answer: d
Explanation: Binding of SR proteins on the exons ensure that the exon is not skipped. It doesn’t bind to intron and in some weak exons it helps in skipping the exon thus helping in alternate splicing.

11. SR proteins can interacts with__________________
a) U1
b) U2AF
c) U1 and U2AF
d) None
View Answer

Answer: c
Explanation: SR protein interacts with U1 and helps it to recognize the 5’ end. It also interacts with the U2AF that helps to bind it to the branch point.

12. Intermediate sequences in the intron are unimportant for splicing.
a) True
b) False
View Answer

Answer: a
Explanation: Splicing requires the 5’ end GU and the 3’ end AG with a branch point A. How big is the intron or the intermediate sequences play no role in splicing.

13. Which of the following is the proper sequence of assembly of factors on the branch poinr A?
a) U1-> U2-> U6
b) U2AF -> U2->BBP
c) BBP->U2AF->U2
d) BBP->U1->U2
View Answer

Answer: c
Explanation: BBP stands for branch point binding proteins. In the branch site A 1st the U2AF binds followed by BBP then it is replaced by U2. Then U2 can interact with U6 that replaced U1.

14. ESE bound to SR proteins ______________ splicing and ESE bound to hnRNA ______________ splicing.
a) Activate, activate
b) Activate, inactivate
c) Inactivate, activate
d) Inactivate, inactivate
View Answer

Answer: b
Explanation: Binding of SR proteins to ESE or exon splicing enhancers promotes splicing where that exon is retained. hnRNA is a competitive inhibitor of SR proteins that tend to bind ESE as well, thus it inactivates.

15. U rich RNA that helps in splicing are______________
a) sn RNA
b) snoRNA
c) hnRNA
d) mRNA
View Answer

Answer: a
Explanation: The U rich NA are the small nuclear or snRNA. The snoRNA are needed for ribosomal rRNA processing in the nucleolus. hnRNA is heterogeneous nuclear RNA are mRNA is the RNA that is spliced by snRNA.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter