If the length of E.coli DNA is 1.36 mm ,what is the number of base pair in E.coli
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the distance between two consecutive base pair is 0.34 (multiply)*10power-9
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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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