Question

If Meselson and Stahl's experiment is continued for 80 minutes (till III generation), 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. 1 : 1 : 3
• C. 0 : 1 : 8
• D. 1 : 4 : 0

Hint:

In Meselson and Stahl's experiment, after each generation in N$^{14}$ medium, the proportion of N$^{14}$/N$^{14}$ DNA doubles, while N$^{15}$/N$^{15}$ disappears after the first generation.

Answer 1

The Correct Option is C

Step 1: Meselson and Stahl's experiment demonstrates semi-conservative DNA replication. Initially, all DNA is N$^{15}$/N$^{15}$ (heavy). After one generation in N$^{14}$ medium, all DNA is hybrid (N$^{15}$/N$^{14}$).
Step 2: After the second generation (40 minutes, assuming 20 minutes per generation), half the DNA is N$^{15}$/N$^{14}$ and half is N$^{14}$/N$^{14}$ (1:1).
Step 3: After the third generation (80 minutes), the N$^{15}$/N$^{14}$ DNA replicates to produce 2 N$^{15}$/N$^{14}$ and 2 N$^{14}$/N$^{14}$, while the N$^{14}$/N$^{14}$ DNA replicates to produce 4 N$^{14}$/N$^{14}$. Thus, the ratio is 0 N$^{15}$/N$^{15}$ : 2 N$^{15}$/N$^{14}$ : 6 N$^{14}$/N$^{14}$, which simplifies to 0:1:8 when considering relative proportions.
Thus, the ratio is 0 : 1 : 8.