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(D) In $E. coli$ mutants lacking $DNA$ ligase,the joining of $Okazaki$ fragments on the lagging strand is impaired. During $DNA$ replication,the leadi

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Graph showing a single peak at low molecular weight.

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(D) In $E. coli$ mutants lacking $DNA$ ligase,the joining of $Okazaki$ fragments on the lagging strand is impaired. During $DNA$ replication,the leading strand is synthesized continuously,while the lagging strand is synthesized as short $Okazaki$ fragments. Normally,$DNA$ ligase joins these fragments into a long,high-molecular-weight strand. In the absence of functional $DNA$ ligase,the newly synthesized $DNA$ remains as short,low-molecular-weight fragments. When these strands are separated by denaturation and subjected to density gradient centrifugation,the short fragments will migrate to the region corresponding to low molecular weight. Therefore,the radioactive signal will show a peak at the low molecular weight position,as represented in graph $(d)$.

Similar Questions

Which bond does $DNA$ helicase break in $DNA$?

During $DNA$ replication, the two strands separate. Each of these acts as a template for the synthesis of a new strand. This mode of replication is called:

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

$DNA$ replication requires:

Which of the following did Meselson and Stahl use in the experiment to prove semiconservative $DNA$ replication?

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