Question

A plasmid with 4000 bp of closed circular duplex DNA was isolated from E. coli. It has a 300-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) The plasmid remains in the Z conformation
b) The plasmid returns to the B conformation
c) The plasmid undergoes denaturation
d) The plasmid becomes linearized

Answer 1

Upon transferring the plasmid DNA back to a low salt solution after treatment with type IB topoisomerase, it returns to its B conformation, as this is the most stable form under physiological conditions.

Step-by-step explanation:

The question pertains to the behavior of a plasmid DNA upon treatment with a type IB topoisomerase and the conformational changes from B-DNA to Z-DNA and back. Initially, the plasmid is subjected to high salt conditions which induce a transition of a 300-bp CG-rich segment from the B-DNA conformation to the Z-DNA conformation. The action of a type IB topoisomerase at high salt conditions introduces negative supercoiling, which is commonly relieved by the plasmid DNA. Upon transferring the plasmid back to a low salt solution, the plasmid DNA will return to the B-DNA conformation, as this is the most energetically favorable conformation under physiological conditions. Thus, the correct answer is: b) The plasmid returns to the B conformation.

The DNA's conformational flexibility and the use of enzymes such as topoisomerases are integral to processes such as replication, transcription, and chromosome condensation in bacterial cells. These enzymes alter the topology of DNA to facilitate such processes.