Question

The length of one turn of the helix in a $B$- form DNA is approximately-

• A. 0.34 nm
• B. 20 nm
• C. 2 nm
• D. 3.4 nm

Answer 1

Answer: (D)

$B$-form of DNA shows right handed coiling which contains $10$ base pairs in each turn. One complete turn of DNA occupies $34 \,\mathring{A}$ distance so in $nm$ it will $3.4\, nm$ because
$1 nm =10^{-9} m $
$1 \,\mathring{A}=10^{-10} m$
$34 \,\mathring{A}=34 \times 10^{-10} m $
or $=3.4 \times 10^{-9} m $
So, $=3.4 \,nm$.