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Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: coolusername on December 04, 2013, 09:32:49 PM

Title: Linking Number, Twist and Writhe of Circular DNA
Post by: coolusername on December 04, 2013, 09:32:49 PM

Suppose you had a circular B-DNA double helix that contained a 96-bp section with alternating
purine-pyrimidine bases. How would delta(L), delta(T) and delta(W) change if you moved this DNA to a high-saltsolution? (delta = change in)

The Answer:
Need changes, not absolute values. Backbone not broken, so
L = 0.
96-bp section changes from B-DNA (RH, 10 bp/turn) to Z-DNA (LH, 12 bp/turn).

delta(T) = –96/10 + 96/(–12) = –9.6 – 8 = –17.6
Because delta(
L) = 0,
delta(W) = –
delta(T) so
delta(W) = +17.6

I don't understand the value obtained by the change in twist. I understand that twist is the number of base pairs divided by the relaxed/favoured # of twist per turn. However, the calculation doesn't make sense to me as to why the change in twist are negative.

Any help would be appreciated.

Title: Re: Linking Number, Twist and Writhe of Circular DNA
Post by: Babcock_Hall on December 05, 2013, 02:20:42 PM

Twist is positive for a right-handed helix.  Twist is negative for a left-handed helix, IIUC.  Is this a textbook problem?