Question

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

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

Answer 1

Answer: (C)

No. of base pair $\frac{\text{Total length }}{\text{Length between 2 bases }}$
$=\frac{1.36\times 10^{-3}}{0.34\times 10^{-9}}$
$=4\times 10^6$