Question

E.coli has only 4.6 x $10^6$ base pairs and completes the process of replication within 18 minutes; then the average rate of polymerisation is approximately-

• A. 2000 base pairs/second
• B. 3000 base pairs/second
• C. 4000 base pairs/second
• D. 1000 base pairs/second

Answer 1

Answer: (A)

2000 base pairs/second

Answer 2

The E. coli genome has $4.6 \times 10^6$ base pairs and replication is completed in 18 minutes. First, convert the time to seconds: $18 \text{ minutes} \times 60 \text{ seconds/minute} = 1080 \text{ seconds}$. Since DNA replication is bidirectional, there are two replication forks working simultaneously. The average rate of polymerisation per fork is calculated by dividing the total number of base pairs by the product of the number of forks and the time taken: Rate = $\frac{\text{Total base pairs}}{\text{Number of forks} \times \text{Time}} = \frac{4.6 \times 10^6 \text{ bp}}{2 \times 1080 \text{ s}} = \frac{4.6 \times 10^6}{2160} \text{ bp/s} \approx 2130 \text{ bp/s}$. This value is closest to 2000 base pairs/second.