Question

the whole genome of E. Coli has 4.6*10^6 bp and is replicated within 38 minutes. The rate of DNA polymerisation in this bacteria is approximately (1)4000 bp/sec
(2)2000 bp/sec
(3)1500 bp/sec
(4)200 bp/sec

Answer 1

Explanation:

here 38 minutes : 38×60 s

: 2280 s

so rate of DNA polymerisation in bacteria will be :

4.6×10^6/2280

= 2000bp/sec( approx.)

Answer 2

Given: The genome of E.Coli bacteria has $4.6 \times 10^{6}$ bp. It is replicated within 38 minutes.

To Find: The rate of DNA polymerization in this bacteria.

Solution:

OPTION (2) 2000 bp/sec is the right answer.

Explanation:

We know that, the rate of DNA polymerization in bacteria = $\frac{Number\ of\ base\ pairs}{time\ (in\ seonds)}$

Substituting the values in the above equation we get, the rate of DNA polymerization in bacteria = $\frac{4.6 \times 10^{6}}{38 \times 60}$

⇒ 2000 bp/sec (approx.)

Hence, the rate of DNA polymerization in this bacteria is approximately 2000 bp/sec

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