250+ TOP MCQs on DNA Replication Prokaryotes and Answers

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Cytogenetics Multiple Choice Questions on “DNA Replication Prokaryotes”.

1. You grow a bacterial culture in a media containing N15 and transfer it to a media with N14. After two rounds of replication you perform a CsCl density gradient centrifugation of the DNA. How many bands will you observe what will be their intensity?
A. One, very intense
B. Two, equally intense
C. Three with middle one more intense than upper and lower
D. Three equally intense
Answer: B
Clarification: In semi conservative replication initially the DNA was N15-N15. After 1st round of replication both strands were N14-N15, making only a single band as they have same density. In the next round we will get N14N14, and N14-N15 making two bands which are equally intense for having 1:1 concentration.

2. What will be the fate of a Seq A mutant E. Coli?
A. DNA replication will occur less frequently
B. Replication will be error prone
C. Replication will not occur
D. Uncontrolled replication
Answer: B
Clarification: Seq A ( Seq standing for sequestering) blocks the newly produced DNA stand from Methylase. This plays an important role in Mis-Match repair. In its absence Mis-match repair will be non functional producing erroneous DNA.

3. At which end are the new DNA bases added?
A. 5’ triphosphate end
B. 3’ triphosphate end
C. 5’ OH end
D. 3’ OH end
Answer: D
Clarification: DNA replication proceeds from 5’ -> 3’ direction with new bases being added to 3’-OH end. 5’-P is at the beginning of DNA Strand.

4. How many prokaryotic DNA polymerases have 5’->3’ proofreading activity?
A. 1
B. 2
C. 3
D. 4
Answer: A
Clarification: Only DNA polymerase I has 5’->3’ proofreading activity. Other polymerases like pol II and III have 3’ -> 5’ proofreading activity. There are mainly 3 prokaryotic DNA polymerases.

5. Which is the most processive of prokaryotic DNA polymerases?
A. pol I
B. pol II
C. pol III
D. klenow fragment

Answer: C
Clarification: DNA pol III holoenzyme has a processivity of >500,000 which is highest among prokaryotic polymerases. Pol I,II and Klenow fragment has no beta clamp resulting in their lesser processivity.

6. Bacterial polymerases are slower than eukaryotic polymerases.
A. True
B. False
Answer: B
Clarification: Prokaryotic polymerases can add about 50,000 bases per minute while eukaryotic polymerases can add only 2000 bases per minute. This slower rate is compensated by more number of origins of replication.

7. In an experiment you take a DNA in vitro and attempt to replicate it. Which combination will you add to your DNA to get the maximum replication product?
A. Dna a, Dna b, HU, SSB, topoisomerase I, polymerase I
B. Dna a, Dna b, HU, SSB, polymerase III
C. Dna a, Dna b, Dna c, HU ,SSB polymerase I
D. Dna a, Dna b, Dna c, HU, SSB, polymerase III
Answer: C
Clarification: DNA pol I has primase activity which is mandatory for starting DNA replication. As pol III lacks primase activity it will not be able to initiate replication. Also in vitro(specially in open DNA stranD. topoisomerase doesn’t have much role to play but Dna c is a must for replication as it loads the helicase.

8. In an experiment you use DNA pol I – Klenow fragment. When all other requisites for replication are added, then what will be the effect on the newly replicated DNA? Consider leading strand only.
A. No difference from intact DNA pol I replication
B. Replication will be slower
C. Replication will be error prone
D. DNA produced will be shorter
Answer: D
Clarification: Klenow fragment provides the 5’ -> 3’ exonuclease activity used to remove the primer. In its absence the primer remains and hence DNA produced is shorter (i.e. original length-length of primer).

9. You take a circular ssDNA and to it you attach a small labeled complimentary fragment. You add different reagents and try to get free labeled probe. Which of this reagent will give you your desired result?
A. Dna b
B. Dna c
C. Dna G
D. Dna a

Answer: A
Clarification: Dna b has helicase activity which can separate two complementary DNA strands, this will break the link between labeled fragment and circular DNA rendering it free. Dna G mainly has primase activity while Dna a and Dna c are required for replication initiation which don’t give the desired result.

10. Which of this subunit is not a part of core DNA polymerase?
A. Alpha
B. Beta
C. Theta
D. Eta
Answer: B
Clarification: Beta is the clamp which helps to provide polymerase with higher processivity, however even in its absence DNA replication will take place although processivity of polymerase would be slow. Alpha is the centre for polymerase activity and it along with theta and eta forms the core polymerase.

11. Rolling circle mode of replication is ________
A. Conservative
B. Non Conservative
C. Semi – Conservative
D. Dispersed
Answer: C
Clarification: One cycle of replication results in an inner parent circular strand and outer daughter circular strand making the process semi-conservative. It would have been conservative if resultant parent and daughter strand would be on different molecules and dispersed if both strands would have some part of parent and daughter.

12. What constitutes Primosome?
A. Dna a, Dna b, Dna c, Dna G
B. Dna b, Dna G
C. Dna c, Dna b
D. Dna a, Dna c
Answer: B
Clarification: Dna b, the helicase, and Dna G, the primase, constitutes the primosome necessary for initiation. Dna a mainly generates a strain allowing Dna c to load the helicase Dna b.

13. If we mutate the DNA ligase and observe the length of the replicated strands in different time slots after replication initiation, what will we observe?
A. The DNA will gradually increase in length till it is fully replicated
B. Small fragments of DNA will be obtained increasing in number with time
C. Mixture of small and long Fragments of definite length from the start whose concentration simply increases with time
D. At first small fragment, then two separate bands showing long fragment with increasing length and short fragments of definite length.

Answer: D
Clarification: The small fragments are the Okazaki fragments of lagging strand while the longer is the leading stand DNA. While leading strand grows in length the Okazaki fragments in absence of ligation to each other results in small fragments of definite length.

14. Acyclovir is a drug used to treat viral infection by impairing its replication. Why will it not effect bacterial replication as well?
A. Bacteria under viral attack don’t replicate
B. Viral polymerase binds to it and thus can’t perform its function
C. Virus uses it in a polymerization
D. Cellular mechanism deactivates it
Answer: C
Clarification: Acyclovir is a GMP analogue that cellular mechanism activates to GTP analogue, which the virus uses as a substrate while polymerizing. However, as it lacks 3’-OH it terminates the replication process. Thus, cellular processes activate this drug, yet its specificity to viral polymerase protects the cell.

15. DNA replication in the two strands proceed in opposite direction as they are aligned oppositely with respect to 3’ and 5’ ends
( 5’——————————-3’
3’——————————-5’).
In this context which of the following is true.
A. The two arms of the DNA Pol are exactly same with same orientation
B. The two arms of the DNA Pol are exactly same with opposite orientation
C. The two arms of the DNA Pol have different catalytic mechanism i.e. one polymerizes 3’ -> 5’ other 5’ -> 3’
D. The two arms are isomers i.e. they have different arrangement of the subunits.
Answer: A
Clarification: The two arms have the core subunits and beta clamp in the same orientation. The DNA strand loops through the lagging strand synthesizing arm to allow polymerization in the same direction (say right to left). However, both arms would polymerize from 5’->3’ direction only.

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