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?

a) The DNA remains in the Z conformation
b) The DNA returns to the B conformation
c) The DNA undergoes supercoiling
d) The DNA undergoes unwinding

Answer 1

A plasmid treated with a type IB topoisomerase in a high salt solution, which had undergone a transition to the Z conformation, will undergo supercoiling due to the enzyme's activity to relax over-wound or under-wound DNA. The correct option is C.

Step-by-step explanation:

When a plasmid undergoes a conformational change from the B to the Z conformation, and then is treated with a type IB topoisomerase, which is active in high salt solution, the correct outcome is that the DNA undergoes supercoiling.

Type IB topoisomerases induce changes in DNA topology by breaking and rejoining one strand of the DNA duplex, which allows for the passage of the other DNA strand through the break. This action relaxes the supercoiled DNA when it becomes over-wound or under-wound as a result of cellular processes such as replication or transcription.

In high salt conditions, which promote the Z conformation, type IB topoisomerase would work to resolve any additional supercoiling that may occur as a consequence of DNA manipulations. However, type IB topoisomerase would not necessarily cause the DNA to revert back to B-DNA conformation or unwind it completely.