This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Repair of DNA Damage: NER, Direct Reversal”.
1. In an experiment you fuse cells from two patients suffering from XP ( Xeroderma Pigmentosum). The fused cells will be expected to agree to which of the following?
a) The resultant cells will have worsened case of XP
b) Only one of the two NER defects that is dominant will be seen
c) No XP defect in the resultant
d) Both will show a similar effect of XP as before
Explanation: As the disease Xeroderma pigmentosum is caused by different in different NER gene, such fusion will show complementation and the resultant phenotype will be normal.
2. The bulky lesions in GG-NER is detected by ___________
Explanation: XPC in association with the hHR23B is responsible for recognizing the bulky lesions and attaching the XPB and XPD.
3. Which of the following NER factors is also a subunit of a transcription factor?
Explanation: XPB and XPD are two subunits of the transcriptional factor TFIIH that acts as a helicase unwinding the region around the bulky lesion. It also marks the junctions of the ss and dsDNA where the cleavage would occur.
4. Which of the exonucleases of NER binds to which end of the lesion?
a) XPD on 3’ end and XPB on 5’ end
b) XPF on 3’ end and XPA on 5’ end
c) XPG on 3’ end and XPF on 5’ end
d) XPD on 3’ end and XPA on 5’ end
Explanation: While XPD and XPB bind to the 3’ and 5’ end respectively, they are helicases and not exonucleases. On the other hand XPF along with ERCCI that binds to 5’ end and XPG are exonucleases.
5. Which of the DNA polymerase in eukaryotes is responsible for repair synthesis of the excised strand in NER?
a) Pol epsilon
b) Pol III
c) Pol beta
d) Pol I
Explanation: Major polymerase in eukaryotes is pol delta and pol epsilon. So, they are responsible of repair synthesis of the excised out patch in NER.
6. What is the stretch of free DNA needed by NER machinery to repair the bulky lesions in TC NER?
a) 50 nt
b) 100 nt
c) 200 nt
d) 250 nt
Explanation: A stretch of free 100 nt is needed by NER machinery to excise out and replace the lesion. Thus, either the transcribing RNA polymerase then has to backtrack or it falls off.
7. Which of the factors are necessary in TC NER but not in GG NER?
b) ERCCI- XPF
Explanation: The factors CSA and CSB help the RNA polymerase to backtrack when it encounters a lesion. However, hHR23B in association with XPC is more necessary in GG NER to detect the lesion.
8. Which of the factor in prokaryotic NER acts as the helicase?
a) Urv A
b) Ruv C
d) Uvr D
Explanation: Uvr D is the factor responsible unwinding the DNA in prokaryotes and excising the sequence with the lesion. On the other hand, TFIIH is also a helicase for eukaryotes.
9. Which of the molecule in photolyase acts as an electron donor for recovery of the CPD?
c) MTHF poly Glu
Explanation: When the chromophore MTHF poly Glu receives the UV light, it passes the energy on to FADH- which in turn passes the electron to CPD. The electron helps in rearrangement of the bonds in CPD and returns them to the form of two pyrimidine residues.
10. You radiate a stock of E. Coli with UV rays and another stock is irradiated as control. The 1st stock is seen to be more resistant to alkylating agents. Which gene is expected to be triggered by this UV?
a) AlK B
b) Guanine methyl transferase
c) MTHF poly Glu
Explanation: While Alk B is also a de-methylating agent, but in E. coli the UV inducible de-alkylating agent is 8-oxo guanine methyl transferase.
11. It is good that Guanine methyl transferase is absent in human beings.
Explanation: While de-alkylating agents could lead to the direct reversal of the damages, yet it is a suicide protease. One molecule is irreversibly inactivated in correcting single methylation damage. It is wasteful, and eukaryotes have better ways of taking care of the methylation like BER.
12. What is the molecule released in 2nd step of Alk B reaction which returns the base to normal form?
b) Alpha ketoglutarate
Explanation: Alk B converts the methyl group to carbonyl group by oxidation, with the help of alpha ketoglutarate. The 1st step releases succinate while in the 2nd step the carbonyl group releases the formaldehyde.
13. What is the prokaryotic homologue of XPC?
a) RAD 51
b) RAD 10
c) RAD 23
d) RAD 50
Explanation: In eukaryotes, XPC works in conjugation with hHR23B which is simply the human homologue of RAD 23. RAd 51 is involved in DSB repair.
Sanfoundry Global Education & Learning Series – Cytogenetics.
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