If the DNA of E. coli is 1.36 mm long,how ma | vedclass

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(D) The distance between two consecutive base pairs in $DNA$ is $0.34 \ nm$ or $0.34 	imes 10^{-6} \ mm$.Total length of $DNA = 1.36 \ mm$.Number of

Vedclass · Accepted Answer

$4 	imes 10^{6} \ bp$

Vedclass · Answer

(D) The distance between two consecutive base pairs in $DNA$ is $0.34 \ nm$ or $0.34 	imes 10^{-6} \ mm$.Total length of $DNA = 1.36 \ mm$.Number of base pairs = $\frac{	ext{Total length}}{	ext{Distance between two base pairs}}$.Number of base pairs = $\frac{1.36 \ mm}{0.34 	imes 10^{-6} \ mm/bp} = 4 	imes 10^{6} \ bp$.Therefore,the correct option is $D$.

If the $DNA$ of $E. coli$ is $1.36 \ mm$ long,how many base pairs $(bp)$ does it contain?

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