A completely relaxed 500bp circular DNA molecule has a linking number (L) = 50.

Question

A completely relaxed 500bp circular DNA molecule has a linking number (L) = 50. DNA gyrase, a type II topoisomerases, hydrolyzes two molecules of ATP while adding left-handed supercoils to this molecule. What is the linking number after this occurs?

3) A completely relaxed 500 bp circular DNA molecule has a mber (L) = 50. DNA gyrase, a type II topoisomerase, hydrol o molecules of ATP while adding left-handed supercoils to t olecule. What is the linking number after this occurs? A) 46 B) 48 C) 52 D) 54 4) Fill in the blanks: is more likely to interact with site I while is ely to interact with site II. A) Photolyase; a histone B) A restriction enzyme; MutL2Mu DNAP polymerase; a restrictio enzyme D) MutL2MutS2: DNAP polymeras

Explanation / Answer

Solution 3:

The answer is- D) 54

Left handed supercoil means- positive supercoiling. That means DNA gyrase type II topoisomerase will introduce supercoil to the relaxed DNA.

In relaxed state it has 50 turns. So after the action of DNA gyrase type II topoisomerase the linking number would become = 50 + 4 = 54.

Because DNA gyrase type II topoisomerase can add 2 turns at a time. And here 2 molecules of ATP have been utilized. So 4 turns will be added and the linking number will become 54.

Solution 4:

Option a is correct because at this site photolyase help in reparing of DNA and histone help in binding of DNA in uniform structure.

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