If the length of e coli DNA is 1.36mm can you calculate the number of base pairs in ecoli
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Answered by
1
base pair =
1.36× 10^-3
---------------------
3.4 × 10^-10
= 4× 10^-6
1.36× 10^-3
---------------------
3.4 × 10^-10
= 4× 10^-6
Answered by
1
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.
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