Question

The human genome contains about 3 billion base pairs. During the first cell division after fertilization of a human embryo, S phase is approximately three hours long. Assuming an average DNA polymerase rate of 50 nucleotides/second over the entire S phase, what is the minimum number of origins of replication you would expect to find in the human genome?

Answer 1

Answer:

5556

Explanation:

If a DNA polymerase synthesizes in average 50 nucleotides/second, that means that in three hours (10800 seconds) it synthesizes about 540000 nucleotides.

However, if the human genome is composed of 3000000000 (3 billion) base pairs (nucleotids), the minimum number of DNA polymerases (working in the same number of origins of replication) to finish the duplication of all the genome in three hours is 5555,5. (3000000000/540000). As we know there is no half polymerase, so we round to 5556.

5556 molecules of DNA polymerases acting on 5556  origins of replication are needed.

Answer 2

Answer:

5555 replication origins.

Explanation:

The human genome contains about 3000 million base pairs, all of them are copied during the S phase by the DNA polymerase. In this case, you are supposing that the DNA polymerase can copy 50 nucleotides/second, so all of this work is done in 10800 seconds, if you multiply 50 nuc/sec by 10800 sec, we obtain that a single replication origin can copy 540000 nucleotides in three hours. Now if you divide 3000 millions by 540000 you will obtain the aproximate number of replication origins in human cells, that is 5555.