Question

If the length of $E.coli$ DNA is 1.36 ~mm\), then how many base pairs are present in $E.coli$\(?

• A. $4.6\times {10}^6$
• B. $4.0\times {10}^6$
• C. $3.6\times {10}^6$
• D. $3.6\times {10}^4$

Answer 1

The correct option is B $4.0\times {10}^6$

Base pairs present in $E.coli$
One turn of DNA helix has 10 nucleotides or 10 base pairs
Length of one complete turn of the DNA double helix is $34Å=3.4nm$
Length between two adjacent base pairs$=\left(3.4/10\right)nm=0.34nm=0.34x10-9m$
Length of $E.coli$ DNA $=1.36mm=1.36x10-3m$
Number of base pairs in \9E. coli =\) (Total length of DNA/ Length between adjacent base pairs) $=(1.36x10-3m)/(0.34x10-9m)=4x106basepairs.$
Hence, the correct answer is option (C).
Final answer
(C) $4.0\times {10}^6$