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: The Meselson and Stahl experiment fundamentally established DNA's semi-conservative replication mechanism. This experiment utilized nitrogen isotopes, specifically \( N^{15} \) (heavy) and \( N^{14} \) (light), for DNA strand labeling.

Step 2: The experimental procedure involved:
- Initial cultivation of parental DNA in a \( N^{15} \) medium.
- After one replication cycle in \( N^{14} \), new strands incorporated \( N^{14} \), while parental strands retained their \( N^{15} \) label.
- Subsequent generations showed an increasing proportion of \( N^{14} \) only DNA and a decreasing proportion of DNA containing both \( N^{15} \) and \( N^{14} \).

Step 3: DNA analysis after 80 minutes (up to the third generation) revealed:
- First generation DNA: exclusively \( N^{15}/N^{14} \) type (one \( N^{15} \) strand, one \( N^{14} \) strand).
- Second generation DNA: appearance of \( N^{14}/N^{14} \) type alongside the existing \( N^{15}/N^{14} \) type.
- Third generation DNA: dominated by \( N^{14}/N^{14} \) and \( N^{15}/N^{14} \) in an 8:1 ratio, with no \( N^{15}/N^{15} \) DNA present.

Step 4: Conclusion:
Therefore, the ratio of \( N^{15}/N^{15} : N^{15}/N^{14} : N^{14}/N^{14} \) DNA after 80 minutes is determined to be 0:1:8.
Option (C) represents the correct answer.