Question

If the lenth of e.coli DNA is 1.36mm, can u dcalculate the number of base pair e.coli ?

Answer 1

《Answer》

=> Let's The Number Is x

so,

1.36mm = x X 0.34 X 10^-9m

=>x = 4 X 10^6 bp

So the number of base pairs in given strand of E coli = 4 X 10^6bp.

《Hope Helped》

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.