Question

Assume DNA replication proceeds at a rate of 100. base pairs per second in human cells and origins of replication occur every 300. kbp. Assume also that DNA polymerase III is highly processive and only 2 molecules of DNA polymerase III are needed per replication fork. How long would it take to replicate the entire diploid human genome? How many molecules of DNA polymerase does each cell need to carry out this task?

Answer 1

The replication of DNA takes place at different frequencies on the basis of the kind of enzyme monitoring the process. There are five distinct enzymes that catalyze the reaction, these are, alpha, beta, gamma, delta, and epsilon. They all possess distinct rates of replication on the basis of the number of nucleotides.

If DNA replication progresses at a rate of 100 base pairs per second and the origin of replication is found at every 300-kilo base pairs. A diploid human genome comprises 6 × 10^9 base pairs of DNA. The number of replication forks can be determined by using the formula:

N = 6 × 10^9 nt/genome / 300 × 10^3 bp/origin

= 2 × 10^4 origins (2 replication forks / 1 origin)

= 4 × 10^4 replication forks

Here, N is the number of replication forks, nt are nucleotides. Thus, the number of replication forks found in the given length of the genome is 4 × 10^4 replication forks.

It has been assumed that the replication of DNA proceeds at a rate of 100 base pairs per second, then the rate of replication can be determined by:

t = 6 × 10^9 bp / 4 × 10^4 replication forks × 100 bp/sec

t = 1500 seconds

It has been assumed that there are two DNA polymerase molecules taking part per replication fork, then the number of DNA polymerases needed for 4 × 10^4 replication forks are as follows:

4 × 10^4 replication forks × 2 DNA polymerase/replication fork

= 8 × 10^4 DNA polymerase.

Thus, the cell would need a sum of 80,000 DNA polymerase molecules for the completion of replication in 4 × 10^4 replication forks.