Question:
If Meselson and Stahl's experiment on the mode of DNA replication is continued up to the sixth generation in bacteria, the ratio of Heavy strands 15N/15N: Hybrid15N/14N: light strand 14N/14N containing DNA in the sixth generation would be-
If Meselson and Stahl's experiment on the mode of DNA replication is continued up to the sixth generation in bacteria, the ratio of Heavy strands 15N/15N: Hybrid15N/14N: light strand 14N/14N containing DNA in the sixth generation would be-
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For any generation $n$, the ratio of Hybrid : Light molecules simplifies to $1 : (2^{n-1} - 1)$. For $n=6$, this is $1 : (2^5 - 1) = 1 : (32 - 1) = 1 : 31$.
Updated On: Aug 13, 2026
- 1:1:1
- 0:1:7
- 0:1:15
- 0:1:31
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The Meselson-Stahl experiment elegantly demonstrated the semi-conservative mode of DNA replication.
In semi-conservative replication, each original parental strand is conserved and serves as a template for the synthesis of a new complementary strand.
Key Formula or Approach:
If we start with one double-stranded DNA molecule where both strands are labeled with heavy nitrogen (\(^{15}\text{N}/^{15}\text{N}\)), and replication is carried out in a light nitrogen (\(^{14}\text{N}\)) medium:
The total number of double-stranded DNA molecules at generation \(n\) is given by \(2^n\).
The number of heavy DNA molecules (\(^{15}\text{N}/^{15}\text{N}\)) is \(0\) for any generation \(n \geq 1\).
The number of hybrid DNA molecules (\(^{15}\text{N}/^{14}\text{N}\)) always remains constant at \(2\) because there are only two original heavy (\(^{15}\text{N}\)) strands.
The number of light DNA molecules (\(^{14}\text{N}/^{14}\text{N}\)) at generation \(n\) is \(2^n - 2\).
Step 2: Detailed Explanation:
Let us calculate the distribution of DNA molecules for the sixth generation (\(n = 6\)):
Total number of DNA molecules: \[ 2^6 = 64 \]
Distribution of these 64 molecules:
- Heavy DNA (\(^{15}\text{N}/^{15}\text{N}\)): \(0\)
- Hybrid DNA (\(^{15}\text{N}/^{14}\text{N}\)): \(2\)
- Light DNA (\(^{14}\text{N}/^{14}\text{N}\)): \[ 64 - 2 = 62 \]
Now, let us write down the ratio of Heavy : Hybrid : Light DNA: \[ 0 : 2 : 62 \]
Simplifying the ratio by dividing each component by 2: \[ 0 : 1 : 31 \]
Step 3: Final Answer:
Therefore, the ratio of heavy, hybrid, and light DNA in the sixth generation is 0:1:
The Meselson-Stahl experiment elegantly demonstrated the semi-conservative mode of DNA replication.
In semi-conservative replication, each original parental strand is conserved and serves as a template for the synthesis of a new complementary strand.
Key Formula or Approach:
If we start with one double-stranded DNA molecule where both strands are labeled with heavy nitrogen (\(^{15}\text{N}/^{15}\text{N}\)), and replication is carried out in a light nitrogen (\(^{14}\text{N}\)) medium:
The total number of double-stranded DNA molecules at generation \(n\) is given by \(2^n\).
The number of heavy DNA molecules (\(^{15}\text{N}/^{15}\text{N}\)) is \(0\) for any generation \(n \geq 1\).
The number of hybrid DNA molecules (\(^{15}\text{N}/^{14}\text{N}\)) always remains constant at \(2\) because there are only two original heavy (\(^{15}\text{N}\)) strands.
The number of light DNA molecules (\(^{14}\text{N}/^{14}\text{N}\)) at generation \(n\) is \(2^n - 2\).
Step 2: Detailed Explanation:
Let us calculate the distribution of DNA molecules for the sixth generation (\(n = 6\)):
Total number of DNA molecules: \[ 2^6 = 64 \]
Distribution of these 64 molecules:
- Heavy DNA (\(^{15}\text{N}/^{15}\text{N}\)): \(0\)
- Hybrid DNA (\(^{15}\text{N}/^{14}\text{N}\)): \(2\)
- Light DNA (\(^{14}\text{N}/^{14}\text{N}\)): \[ 64 - 2 = 62 \]
Now, let us write down the ratio of Heavy : Hybrid : Light DNA: \[ 0 : 2 : 62 \]
Simplifying the ratio by dividing each component by 2: \[ 0 : 1 : 31 \]
Step 3: Final Answer:
Therefore, the ratio of heavy, hybrid, and light DNA in the sixth generation is 0:1:
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