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

$(4 \times 10^6 \; BP)$ The length of $DNA$ is calculated by multiplying the number of base pairs by the distance between two consecutive base pairs.
Given:
Length of $E. \; coli$ $DNA$ $= 1.36 \; mm = 1.36 \times 10^{-3} \; m$.
Distance between two consecutive base pairs $= 0.34 \; nm = 0.34 \times 10^{-9} \; m$.
Number of base pairs $= \frac{\text{Total length of DNA}}{\text{Distance between two consecutive base pairs}}$.
Number of base pairs $= \frac{1.36 \times 10^{-3} \; m}{0.34 \times 10^{-9} \; m} = 4 \times 10^{6} \; bp$.