Cytogenetics Questions and Answers – Repair of DNA Damage: Recombinational and Bypass – 2

This set of Cytogenetics MCQs focuses on “Repair of DNA Damage: Recombinational and Bypass – 2”.

1. Which factor in the resolve some drives branch migration?
a) Ruv A
b) Ruv B
c) Ruv C
d) Ruv Z
View Answer

Answer: b
Explanation: The Ruv B binds to the DNA in Holliday intermediate to two branch points. This factor helps in branch migration.

2. Choose the endonuclease that can cuts double strand of DNA.
a) Mut H
b) Ruv C
c) EcoRI
d) S1
View Answer

Answer: c
Explanation: ECoRI is a restriction endonuclease that can cut both the DNA strands generating an overhand. On the other hand Ruv C in recombination repair, Mut H in MMR and S1 endonuclease can only make single strand nicks.

3. Which factor helps in generation of chicken foot Holliday intermediate?
a) Rec A
b) Rec BCD
c) Ruv ABC
d) Rec FOR
View Answer

Answer: d
Explanation: Rec FOR is necessary for strand migration which causes the reverse fork migration during repair of bulky lesions in DNA.

4. Single stranded nicks can be repaired without any loss of genetic material by which of the following processes?
a) MMR
b) NER
c) Recombinational repair
View Answer

Answer: c
Explanation: In recombinational repair with the help of Rec BCD using the complementary strand as a template, the ssDNA breaks can not only be bypassed but also repaired effectively.

5. How many Holliday branch points are produced during DSB repair?
a) 1
b) 2
c) 3
d) None
View Answer

Answer: b
Explanation: In case of DSB there are two branch points produced. One where the 3’ overhang invades the homologous strand and other when it hybridizes with the other 3’ end.
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6. Defect in which of these genes produce an increased propensity to breast cancer?
a) MTHF poly glu
b) PKC
c) BRCA 1
d) XPB
View Answer

Answer: c
Explanation: Defects in the genes of BRCA 1 and BRCA 2 increases the propensity of breast cancer to 80%. On the other hand option a, b and d doesn’t directly play a role in breast cancer.

7. You have two circular chromosomes that are undergoing homologous recombination. The resolution of Holliday intermediate will give rise to _________________
a) 4 smaller circular chromosomes
b) 2 smaller circular chromosomes
c) One larger circular chromosome or two smaller chromosomes
d) 3 same sized circular chromosome
View Answer

Answer: c
Explanation: Resolution of Holliday intermediate will give rise to two smaller chromosomes if resolved in one way. And it can also give rise to one larger chromosome if it is resolved in the other way, but both of these have 50% chance and it doesn’t occur together.

8. What is the base incorporated by pol eta against a DNA CPD lesion?
a) AT
b) GC
c) GG
d) AA
View Answer

Answer: d
Explanation: Cyclo butane pyrimidine dimers consist of two T bases joined to form a bulky lesion. Pol eta incorporates two random A base against it.

Sanfoundry Global Education & Learning Series – Cytogenetics.


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

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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 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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