CBSE 2025, Set 2

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). $1:4:0$

Edvaya · Accepted Answer

The correct answer is $0:1:3$. **SOLUTION** In Meselson and Stahl's experiment, *E. coli* divides every 20 minutes. * **Generation I (20 min):** 100% hybrid DNA ($N^{15}/N^{14}$). Ratio is $0:2:0$ (simplifies to $0:1:0$). * **Generation II (40 min):** 50% hybrid, 50% light DNA. Ratio is $0:2:2$ (simplifies to $0:1:1$). * **Generation III (60 min):** 2 molecules of hybrid DNA, 6 molecules of light DNA. Ratio is $0:2:6$, which simplifies to **$0:1:3$**. *(Note: The question text mentions "80 minutes" alongside "III generation". 80 minutes would mathematically be the IV generation with a ratio of 0:1:7. However, the official CBSE marking scheme strictly adheres to the "III generation" ratio of $0:1:3$ as the intended answer).*

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Biology > Reproduction And Genetics > Molecular Basis Of Inheritance

Keep Practicing

Edvaya Target

Current Question

Community Stats

Track Your Progress

Quick Actions

We Value Your Privacy