Cytogenetics Questions and Answers – Repair of DNA Damage: Recombinational and Bypass – 1


This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Repair of DNA Damage: Recombinational and Bypass – 1”.

1. Homologous recombination can’t occur in which stage?
a) G2
b) Bacterial fission
c) G1
d) S
View Answer

Answer: c
Explanation: For homologous recombination to take place there must be two sister chromatids with same or similar composition. This is available only after DNA replication, thus it can’t take place in G1 phase before replication.

2. The DSB are indistinguishable from normal chromosomes of shorter length.
a) True
b) False
View Answer

Answer: b
Explanation: DSB is in the endonuclear region. So the ends created lack telomere. DNA ends produced by synthesis in vitro have 5’ tri-phosphate and that in between strands has 5’ monophosphate, DSB can thus be distinguished from normal chromosomal ends.

3. What is the role of Rec BCD in DSB repair?
a) It helps in strand invasion
b) It chews off the flap
c) It helps to generate 3’ overhang
d) It helps by aligning the two homologous chromosomes
View Answer

Answer: c
Explanation: Rec BCD acts as an exonuclease that selectively chews off the 5’ end of DNA to generate a 3’ overhang. This is necessary for strand invasion by this 3’ overhang which brings about recombination repair.
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4. Rec BCD can chew 3’ end of the DSB till it reaches _____________ sequence.
b) Chi
c) Phi
d) Eta
View Answer

Answer: b
Explanation: Initially the exonuclease Rec BCD can chew up both the 3’ and 5’ end of the DSB till it reaches the specific chi sequence in the 3’ strand. Then it stops chewing the 3’ end and increases its activity in the 5’ end.

5. Which of the following will be a marker of DSB?
a) H2A
b) Gamma-P H2AX
c) H5G
d) Alpha-P H3BX
View Answer

Answer: b
Explanation: In DSB the canonical bases are replaced by the non-canonical bases, which it also phosphorylated by ATM kinase. H2AX is one such non-cannonical base, while there is no H3BX or H5G. H2AX is phosphorylated at gamma not alpha.
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6. Which of these is not a part of the tri protein complex in homologous DSB repair?
a) MRE11
b) NBS1
c) RAD 51
d) RAD 50
View Answer

Answer: c
Explanation: The MRN complex or the tri protein complex that can recognize the DSB consists of MRN11, NBS1 and RAD 50. Rad 51 is a eukaryotic homologue of Rec A which guides the strand to invade the other homologue.

7. What is the structure of Rec A filament?
a) Cylindrical
b) Straight
c) Like rolled beta sheet
d) Alpha helical
View Answer

Answer: d
Explanation: Rec A factors join together to form right handed alpha helical structure around the single strand of DNA. This has 6.2 Rec A per turn.

8. Which of these factors is not responsible for the assembly of Rec A oh the DNA?
a) Rec B
b) Rec F
c) Rec R
d) Rec O
View Answer

Answer: b
Explanation: Rec B is a pard of Rec BCD complex that helps to generate the 3’ overhang. On the other hand, Rec FOR complex is responsible for assembling Rec A on the DNA stand and guides its invasion.

9. Which of the following is not true about ATM kinase?
a) It is guided to the DSB site by MRN complex
b) It exits as inactive dimer
c) It is activated to monomer by another factor of DSB repair
d) It phosphorylates BRCA1 and BRCA2
View Answer

Answer: c
Explanation: ATM kinase is activated by autocatalysis. It however is a major kinase which phosphorylates the H2AX, BRCA 1 and 2, 53BP1 and many other factors in DSB repair.

10. In DSB repair in G1 phase, which of these factors would be seen which is absent in other phases?
a) RAD 51
b) MRN
c) RAD 23
d) Ku 70/80
View Answer

Answer: d
Explanation: In G1 phase we have non homologous end joining. In this phase Ku 80/90 hetero dimer senses are repairs the DSB in absence of a homologous chromosome. RAD 51 and MRN are also seen in homologous repair in S or G2 phase, and RAD 23 is needed for NER.

11. Which of the following is most flexible ligase?
a) Ligase I
c) Ligase IV
d) Ligase X
View Answer

Answer: c
Explanation: Ligase IV has exceptional flexibility as it can even ligate dissimilar DNA ends. It can also ligate the overhand with a complex structure. This is necessary for the immediate repair of the NHEJ breaks.

12. Which of the following has both kinase and nuclease property?
a) Pol Beta
b) Ku-PKC
c) Ku- pol gamma
d) Pol delta
View Answer

Answer: b
Explanation: Ku PKC has the property of both kinase as well as nuclease. Thus, it can work in non-homologous DSB without the help of additional nucleases.

13. Which of the following is not true about NHEJ?
a) In most of the cases, it is associated with loss of genetic material
b) Ku is present at a concentration as high as 400000 per cell
c) It strand invasion looks for micro-homology regions
d) The 3’overhang chewed up to chi sequence invades to search for homology
View Answer

Answer: d
Explanation: In NHEJ to reduce the loss of genetic information and for a quicker search of homology domains the 3’ over hand is only 4 nt long. This increases the probability of fining micro homology region as well as prevents loss.

14. Which of the following is not a sensor of DSB?
a) Ku
b) P21
c) H2AX
d) P53
View Answer

Answer: b
Explanation: P53 is associated with DSB as ATM kinase phosphorylates 53BP1 that on activation codes for P53. On the other hand P21 is not associated with DSB. Ku and H2AX the non-canonical base is also associated with DSB repair.

Sanfoundry Global Education & Learning Series – Cytogenetics.

To practice all areas of Cytogenetics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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