If a hybrid DNA (^{14}N-^{15}N) is allowed to | vedclass

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(D) Initially,the $DNA$ is hybrid $(^{14}N-^{15}N)$.After the first replication in a medium with $^{15}N$: Two $DNA$ molecules are formed,one hybrid $

Vedclass · Accepted Answer

$0:3:1$

Vedclass · Answer

(D) Initially,the $DNA$ is hybrid $(^{14}N-^{15}N)$.After the first replication in a medium with $^{15}N$: Two $DNA$ molecules are formed,one hybrid $(^{14}N-^{15}N)$ and one heavy $(^{15}N-^{15}N)$.After the second replication: $A$ total of four $DNA$ molecules are formed.The hybrid $(^{14}N-^{15}N)$ molecule produces one hybrid $(^{14}N-^{15}N)$ and one heavy $(^{15}N-^{15}N)$ molecule.The heavy $(^{15}N-^{15}N)$ molecule produces two heavy $(^{15}N-^{15}N)$ molecules.Thus,the final composition is $0$ light,$3$ heavy,and $1$ hybrid $DNA$ molecules.Therefore,the correct proportion is $0:3:1$.

If a hybrid $DNA$ $(^{14}N-^{15}N)$ is allowed to undergo replication twice in a medium containing nucleotides with $^{15}N$ base,then what proportion of light,heavy,and hybrid $DNA$ will be obtained respectively?

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