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Length of one loop (turn) of B-DNA :
A
3.4 nm
B
0.34 nm
C
20 nm
D
10 nm
Detailed Solution
In the Watson–Crick (B-form) double helix, one complete turn (pitch) of the helix contains 10 base pairs.
The distance between adjacent base pairs is 0.34 nm, so one turn is $10 \times 0.34 = 3.4$ nm (34 Å).
The diameter of the helix is 2 nm.
So the length of one turn of B-DNA is 3.4 nm.
The distance between adjacent base pairs is 0.34 nm, so one turn is $10 \times 0.34 = 3.4$ nm (34 Å).
The diameter of the helix is 2 nm.
So the length of one turn of B-DNA is 3.4 nm.