Question

A plasmid with 5000 bp of closed circular duplex DNA was isolated from E. coli. It has a 600-bp segment of alternating C and G residues. Upon transfer to a high salt solution, this segment undergoes a transition from the B conformation to the Z conformation. Which of the followings is CORRECT AFTER this plasmid is further treated with a type IB topoisomerase that is active at the high salt solution and then transferred back to a low salt solution?

a) B conformation is restored
b) Z conformation is stabilized
c) DNA becomes negatively supercoiled
d) DNA becomes positively supercoiled

Answer 1

After the plasmid treated with type IB topoisomerase is returned to a low salt solution, the B conformation of DNA is restored, but the plasmid may become more positively supercoiled due to the topoisomerase's activity. The correct option is A.

Step-by-step explanation:

When a plasmid with 5000 bp undergoes a conformational change from B-DNA to Z-DNA in a high salt solution and is then treated with a type IB topoisomerase, which is capable of cutting one of the DNA strands and passing the other strand through it before resealing the cut, changes to the DNA topology can occur.

Specifically, since type IB topoisomerases cause single-stranded breaks to alleviate supercoiling, the result after treating the high salt-induced Z-DNA segment with such an enzyme and returning the plasmid to a low salt solution would lead to a restoration of the B conformation.

Transitioning the plasmid back to a low salt solution would normally enable the B conformation to be restored; however, the action of the topoisomerase can introduce changes in the supercoiling of the DNA.

Considering that type IB topoisomerases relax negative supercoils by inducing positive supercoiling to compensate, after treating with topoisomerase and returning to low salt, the plasmid DNA could become more positively supercoiled than it was originally.