If a normal DNA molecule replicates in an N^{ | vedclass

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(C) The replication of $DNA$ is semi-conservative.In the $1^{st}$ generation,all $DNA$ molecules are hybrid $(N^{14}-N^{15})$.In the $2^{nd}$ generati

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(C) The replication of $DNA$ is semi-conservative.In the $1^{st}$ generation,all $DNA$ molecules are hybrid $(N^{14}-N^{15})$.In the $2^{nd}$ generation,there are $2$ hybrid and $2$ heavy $(N^{15}-N^{15})$ molecules.In subsequent generations,the number of hybrid molecules remains constant at $2$,while the number of heavy molecules increases.Since the medium is $N^{15}$ (heavy),no new light $(N^{14})$ $DNA$ strands are formed after the first replication.Therefore,after the $1^{st}$ generation,the percentage of light $DNA$ is $0\%$.

List-$I$	List-$II$
$a$. Ligase	Joins short segments of $\text{DNA}$ together.
$b$. $\text{DNA}$ polymerase	Cuts $\text{RNA}$ at specific $\text{RNA}$ sequence.
$c$. Helicase	Breaks the hydrogen bonds between complementary pairs during $\text{DNA}$ replication.

If a normal $DNA$ molecule replicates in an $N^{15}$ medium for consecutive generations,what is the percentage of light $DNA$ in the $4^{th}$ generation (in $\%$)?

Similar Questions

$DNA$ multiplication is called:

New $DNA$ synthesis occurs in the $P$ direction,and the $DNA$ polymerase enzyme reads the template in the $Q$ direction.
$\quad \quad \quad \quad P \quad \quad \quad \quad \quad Q$

Which of the following diagrams correctly represents the process of $DNA$ replication at the replication fork?

Semi-conservative replication of $DNA$ was first demonstrated in

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