One turn of the helix in a B-form DNA is approximately:

One turn of the helix in a B-form DNA is approximately:
  1. $2 \mathrm{~nm}$
  2. $20 \mathrm{~nm}$
  3. $0.34 \mathrm{~nm}$
  4. $3.4 \mathrm{~nm}$

Solution

B-DNA is a helical structure with a diameter of $20 Å$ and the distance between the two base pairs is $3.4 Å$. There are 10 base pairs in each turn; hence, one turn of the helix (pitch of each helix) is approximately $34 Å$ or $3.4 \mathrm{~nm}(10=1.0 \mathrm{~nm})$. Related Theory There are three different DNA types: (i) A-DNA: It is a right-handed double helix similar to the B-DNA form. Dehydrated DNA takes an A form that protects the DNA during extreme condition such as desiccation. (ii) B-DNA: The most common DNA conformation which is a right-handed helix, normal physiological conditions. (iii) Z-DNA: Z-DNA is a left-handed DNA where the double helix winds to the left in a zig-zag pattern and play some role in gene regulation.

Asked in: NEET 2006

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