For the following problem, you should refer to Chapter 24.2, which discusses sup

Question

For the following problem, you should refer to Chapter 24.2, which discusses supercoiling and topoisomerases. Assume that B DNA has 10.5 bp/turn. A) What is the linking number of a 5775 bp DNA circular duplex DNA molecule with a nick in one strand? B) What is the linking number if the nick is sealed? C) What is the linking number of the molecule in part B after 10 rounds of enzymatic turnover by the enzyme DNA gyrase in the presence of ATP? (See your textbook for how DNA gyrase changes DNA topology.) D) What is the linking number of the molecule from part C if it undergoes 4 rounds of enzymatic turnover using the enzyme E. coli topoisomerase I. E) What is the superhelical density of the DNA molecule in part D? (See equations 24-1 and 24-2.) F) Illustrate what would happen if you ran an agarose gel on the DNA from part b, part c and part d.

Explanation / Answer

Linking number, Lk = No. of base pairs of DNA / Twist

Given, No. of base pairs of DNA = 5775

and Twist = 10.5 bp/turn

so, Lk = 5775/10.5 = 550 turns

A) If one strand is nicked, then the linking no. is decreased by 1 unit.

so, Lk' = Lk - 1 = 550 -1 = 549 turns

B) If nick is sealed, then the DNA is relaxed, so the linking no.remains unchanged.

i.e. Lk" = Lk = 550 turns

C) When DNA gyrase acts in presence of ATP, then the linking no.is decreased by 2 units.

i.e. Lk* = Lk - 2 = 550-2 = 548 turns

D) Toposisomerase-I nicks one strand of DNA, then after relaxation, the nicked (cut) strand is resealed. In this process, the linking no.is increased by 1 unit.

i.e. Lk^ = Lk* + 1 = 548 + 1 = 549 turns

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