Question

1.38 milimeter DNA is present in E. coli. How many base pair will be present in it?

• A. $4\times {10}^{6}bp$
• B. $4\times {10}^{7}bp$
• C. $4\times {10}^{8}bp$
• D. $4\times {10}^{5}bp$

Answer 1

The correct option is A $4\times {10}^{6}bp$

The distance between two base pairs is 0.34 X 10${}^{-9}$ m. The total length of DNA in E. coli is 1.38 millimeter.

Total length of DNA = the total number of base pairs x distance between two consecutive base pairs

Hence, total number of base pairs = total length of DNA / distance between two consecutive base pairs

= 1.38 / 0.34 X 10${}^{-9}$ = 4 X 10${}^6$ base pairs.

Thus, the correct answer is '4 x 10${}^6$ bp'.