Read the given passage and answer the questions that follow :-
In prokaryotes such as E. coli ,though they do not have a defined nucleus the DNA is not scattered throughout the cell. The DNA is held with some proteins in a specific region. in eukaryotes this organisation is much more complex .There is a set of positively charged basic proteins called histones. Histones are organised to form a unit of 8 molecules called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer.

If the length of E. coli DNA is 1.36 mm what will be the number of base pairs in E. coli ?

4 x 10⁶

Here it is given that, the length of the E.coli DNA is 1.36mm. We know the length of the DNA can be calculated by multiplying the number of base pairs into the distance between the basepairs. Thus, from this we can get, Number of base pairs = Length of DNA/ Distance between the base pairs =1.36 / 0.34 X 10⁻⁹ = 4 X 10⁶ base pairs.