Question

If the length of E.coli DNA is 1.36 mm, then calculate no of base pairs in E.coli?

Answer 1

1.36 mm=0.36 cm
lwgth of 10 base pair is 34angstrom
so o.36*10^10/34=and
and*9.8 or 10=ans

Answer 2

Distance between two consecutive base pair = 0.34 nm

Total length of given DNA = 1.36mm  

∴ total no. of base pair =  1.36mm /0.34nm

⇒ 4 × 10⁶ Base pair.