A 340 text{ Å} long segment of DNA molecule h | vedclass

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Question

(C) The total length of the $DNA$ segment is $340 	ext{ Å}$.Since the distance between two consecutive base pairs in a $DNA$ helix is $3.4 	ext{ Å}$

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) The total length of the $DNA$ segment is $340 	ext{ Å}$.Since the distance between two consecutive base pairs in a $DNA$ helix is $3.4 	ext{ Å}$,the total number of base pairs is $340 	ext{ Å} / 3.4 	ext{ Å} = 100 	ext{ base pairs}$.This means there are $200$ total nitrogenous bases ($100$ on each strand).According to Chargaff's rule,the number of adenine $(A)$ equals the number of thymine $(T)$,and the number of guanine $(G)$ equals the number of cytosine $(C)$.Given that there are $20$ thymine bases,there must be $20$ adenine bases.The total number of $A + T$ bases is $20 + 20 = 40$.The remaining bases are $G + C = 200 - 40 = 160$.Since $G = C$,the number of guanine bases is $160 / 2 = 80$.

$A$ $340 \text{ Å}$ long segment of $DNA$ molecule has $20$ thymine nitrogenous bases. What will be the number of guanine nitrogen bases in the same segment?

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