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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

$DNA$ is synthesized from viral $RNA$ using the enzyme:

$A$ $DNA$ molecule containing radioactive thymidine is allowed to replicate in a non-radioactive environment. What will be the exact number of $DNA$ molecules containing radioactive thymidine after three replication cycles?

Which one of the following correctly represents the manner of replication of $DNA$?

$A$: Kornberg enzyme is associated with the removal of primers and thymine dimers.
$R$: $DNA$ polymerase $I$ performs exonuclease activity in $5^{\prime} \rightarrow 3^{\prime}$ and $3^{\prime} \rightarrow 5^{\prime}$ directions.

$DNA$ repairing is done by

Vedclass Test Series

Exam Paper Generator

Online Exam Module