Question

12. When the helix axis of a closed circular duplex DNA of 2310 bp is constrained to lie in a plane, the DNA has a twist (T) of 207. When released, the DNA takes up its normal twist of 10.5 bp per turn. Indicate the values of the linking number (L), writhing number (W), and the twist for both the constrained and unconstrained conformational states of this DNA circle. What is the superhelix density, ơ W/T, of both the unconstrained and constrained DNA circles.

Answer 1

L = linking number = W + T

for relaxed or unconstrained state

W= 0 and L=T

And L of relaxed DNA is = no of base pairs/ 10 = 2310/10= 231

Therefore W=0 , T=L= 231

Now we know already that L is a topological invariant and hence will not change even in constrained states.

Therefore, L= 231 for constrained state also.

Now, for constrained state T=207 (given)

L= W + T

Hence W= L - T

= 231 - 207

= + 24

Bow superhelical densities for

relaxed state = W/T= 0/231= 0

constrained state = W/T =24/231= 0.104