Question

If Meselson and Stahl’s experiment is continued for 80 minutes (till III generation), then what would be the ratio of DNA containing \( N^{15}/N^{15} : N^{15}/N^{14} : N^{14}/N^{14} \) in the medium?

• A. 1:1:0
• B. 0:1:3
• C. 0:1:8
• D. 1:4:0

Hint:

In semi-conservative DNA replication, after each replication cycle, the amount of DNA with one strand labeled \( N^{15} \) and the other labeled \( N^{14} \) increases, while the amount of DNA with both strands labeled \( N^{15} \) decreases.

Answer 1

The Correct Option is C

Step 1: Meselson and Stahl’s experiment is a landmark experiment that demonstrated the semi-conservative replication of DNA. The experiment used nitrogen isotopes \( N^{15} \) (heavy nitrogen) and \( N^{14} \) (light nitrogen) to label the DNA strands.
Step 2: In the experiment:
- The parental DNA was initially grown in a medium containing \( N^{15} \).
- After one round of DNA replication in \( N^{14} \), the newly synthesized strands contained \( N^{14} \), while the parental strands retained the \( N^{15} \) label.
- In subsequent generations, the proportion of DNA containing only \( N^{14} \) increases, while the amount of DNA with both \( N^{15} \) and \( N^{14} \) decreases.
Step 3: Analysis of DNA after 80 minutes (till III generation):
- In the first generation, the DNA will be of the type \( N^{15}/N^{14} \) (one strand of \( N^{15} \) and the other strand of \( N^{14} \)).
- In the second generation, DNA of type \( N^{14}/N^{14} \) will appear along with the DNA of type \( N^{15}/N^{14} \).
- In the third generation, the ratio will be dominated by \( N^{14}/N^{14} \) and \( N^{15}/N^{14} \) in a 1:8 ratio, with the DNA of type \( N^{15}/N^{15} \) being absent.
Step 4: Conclusion:
Thus, the ratio of DNA containing \( N^{15}/N^{15} : N^{15}/N^{14} : N^{14}/N^{14} \) in the medium after 80 minutes will be 0:1:8.
Hence, the correct answer is option (C).