If the length of E. Coli $DNA$ is $1.36 \;mm$, calculate the number of base pairs in E. Coli.
 

Given length of E. Coli $DNA$ $=1.36 \mathrm{~mm}=1.36 \times 10^{-3} \mathrm{~m} .$ Distance between two consecutive base pairs $=0.34 \mathrm{~nm}=0.34 \times 10^{-9} \mathrm{~m}$

Hence for $E .$ Coli total number of base pairs $=\frac{1.36 \times 10^{-3}}{0.34 \times 10^{-9}}=4 \times 10^{6} \mathrm{bp}$