A DNA fragment contains 3.2 kilobases (kb). I | vedclass

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(C) $1$. Total number of bases in the $DNA$ fragment = $3.2 \ kb = 3.2 	imes 1000 = 3200$ bases.$2$. According to Chargaff's rule,the total number of

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$780$

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(C) $1$. Total number of bases in the $DNA$ fragment = $3.2 \ kb = 3.2 	imes 1000 = 3200$ bases.$2$. According to Chargaff's rule,the total number of purines equals the total number of pyrimidines $(A + G = T + C)$.$3$. In a double-stranded $DNA$,$A = T$ and $G = C$. Therefore,$A + G = 3200 / 2 = 1600$.$4$. Given that the number of adenine $(A)$ molecules = $820$.$5$. Since $A = T$,the number of thymine $(T)$ molecules = $820$.$6$. Using the equation $A + G = 1600$,we get $820 + G = 1600$.$7$. $G = 1600 - 820 = 780$.$8$. Since $G = C$,the number of cytosine $(C)$ molecules = $780$.

$A$ $DNA$ fragment contains $3.2$ kilobases $(kb)$. If the $DNA$ fragment contains $820$ adenine molecules,calculate the number of cytosine molecules.

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