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 (10 bp/turn) to the z conformation (12 bp/-ve turn). 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 change in its twist is zero
b) the change in its linking number is -55
c) its original linking number is 400
d) its final writhing number is 55 e) all answers are incorrect

Answer 1

When the plasmid is treated with a type IB topoisomerase and then transferred back to a low salt solution, c) the original linking number of the plasmid will be restored.

Step-by-step explanation:

When the plasmid is treated with a type IB topoisomerase, it will relax its supercoiled structure. This means that the change in twist will be zero (answer choice a).

The change in linking number is determined by the transition from the B conformation (10 bp/turn) to the Z conformation (12 bp/-ve turn) in the high salt solution. The difference in twist caused by this transition is 2 bp/turn. Therefore, the change in linking number is -ΔLk = ΔTw = -2.

Since the plasmid is transferred back to a low salt solution, which is the same as its original environment, its original linking number will be restored. Therefore, the original linking number is 400 (answer choice c).

The final writhing number cannot be determined based on the given information. Therefore, answer choice d is incorrect.

Therefore, the correct answer is e) all answers are incorrect, except for c).