Welcome to the NCERT Solutions for Class 12 Biology - Chapter Molecular Basis of Inheritance. This page offers a step-by-step solution to the specific question from Exercise 0, Question 9: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
Replication was allowed to take place in the presence of radioactive deoxynucleotides precursors in E.coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result?
In above case which is shown in the question, as E.coli bacteria is a mutant for DNA ligase (which helps in binding) enzyme, it will lead to no further joining of Okazaki fragments on lagging strand.
This will results into the formation of both high molecular weight fragments on the leading strand and low molecular weight fragments on the lagging strand. Hence, only the graph (a) could be the appropriate result after centrifugation.
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