E. coli has only 4 . 6 × 10 6 base pairs and completes the process of replication within 18 minutes; then…

Given,

Number of base pairs $=4.6\mathrm{x}{10}^6=4600000$

Time $=$$18$ minutes $=18 \mathrm{x} 60 =1080$ seconds

The average rate of polymerisation can be calculated using the formula:

$\mathrm{Rate} \mathrm{of} \mathrm{polymerisation}=\frac{\mathrm{Number} \mathrm{of} \mathrm{base} \mathrm{pairs}}{\mathrm{Time}}$

                                             =$=\frac{4600000}{1080}=4259 \mathrm{bp}/\mathrm{s}$

It is approximately $4000 \mathrm{base} \mathrm{pairs}/\mathrm{second}$.