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In a 3.2 Kbp long piece of DNA , 820 a...

In a 3.2 Kbp long piece of DNA , 820 adenine bases were found. What would be number of cyosine bases

A

780

B

1560

C

740

D

1480

Text Solution

Verified by Experts

The correct Answer is:
A
780 chargaff's equivalency rule
`(A+G)/(T+C)=1`
A=T
G = C
if A = 820 then t= 820
A + T = 820 + 820 = 1640
32kbp = 3200 bp
G + C = 3200 - 1640
G + C = 1560
So' Cytosine `=(1560)/(2)`
Cytosine = 780
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