1. Rolling circle mode of replication is ________
a) Conservative
b) Non Conservative
c) Semi – Conservative
d) Dispersed
Explanation: 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.
2. 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
Explanation: 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.
3. 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.
Explanation: 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.
4. 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
Explanation: 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.
5. 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.
Explanation: 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.
6. In case of eukaryotes replication initiates at ________
a) TATA
b) CpG islets
c) AUG
d) ARS
Explanation: Replication begins at ARS – Autonomous Replicating Sequences. While TATA and CpG islets are characteristics of promoter elements which controls transcription. Translation initiates at AUG codon
7. Here are names of some factors necessary for prokaryotic replication. Which of these or their homologue is unnecessary for eukaryotes?
a) Dna G
b) Dna b
c) Beta clamp
d) SSB
Explanation: DNA G (gyrase) is not needed for eukaryotes. While Dna b homologue MCM, Beta clamp homologue PCNA and SSB homologue RPA are necessary.
8. You wanted to find out the importance of the DNA polymerases. So, you mutated the polymerases one by one and checked for survival. In which of these mutant cases will you see least survival rate?
a) Alpha
b) Beta
c) Delta
d) Eta
Explanation: Alpha has primase activity. Without a primer DNA synthesis can’t initiate thus Alpha polymerase is indispensible. While the beta is mainly needed for repair mechanisms it doesn’t make much difference. However, delta and eta are main polymerases for eukaryotes and their mutation will have major effects as well.
9. You design a circular ssDNA with a labeled RNA primer (alpha-P32 labeling). You add polymerase epsilon and required enzymes for replication to the fragment. What will you expect to see on autoradiography after gel electrophoresis?
a) Single labeled band with some labeling at the bottom region
b) Two labeled band one much lower than the other with labeling at the bottom region
c) No band but labeling at the bottom
d) Two labeled band one much lower than the other with no labeling at the bottom
Explanation: Alpha-P32 incorporated the label in the entire RNA primer fragment. As polymerization will take place the 5’->3’ exonuclease activity of pol eta will produce labeled oligomers of that RNA. One band will correspond to unreplicated DNA, the lower one to oligomer. No labeled monomer is produced.
10. You design a plasmid with a gene for Histidine synthesis and a fragment containing ARS. You introduce it yeast culture and try to grow then in Histidine deficit media. What will be your observation?
a) No growth
b) One or two colonies
c) Rapid growth with many colonies
d) Normal growth with countable colonies
Explanation: ARS promotes replication of the yeast. Even if the plasmid was initially taken up by few cells due to rapid replication brought about by ARS the number of colony massively increases. If the same experiment was done with any random stretch of DNA none or few colonies would be seen