If Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of `overset(15)N//overset(15)N: overset(15)N//overset(14):overset(14)//overset(14)N` containing DNA in the fourth generation would be

Options

A. 1 :1 :0
B. 0.044444444444444
C. 0:1 :3
D. 0:1 :7

Answer

To solve the problem regarding the ratio of DNA containing different nitrogen isotopes after four generations in Meselson and Stahl's experiment, we will follow these steps:

### Step 1: Understand the Experiment
Meselson and Stahl's experiment demonstrated the semi-conservative nature of DNA replication. They used two isotopes of nitrogen: \(^{15}N\) (heavy nitrogen) and \(^{14}N\) (light nitrogen). Initially, bacteria were grown in a medium containing \(^{15}N\), resulting in all DNA being labeled with \(^{15}N\).

### Step 2: Analyze the First Generation
In the first generation, after transferring the bacteria to a medium containing \(^{14}N\), the DNA strands consist of one \(^{15}N\) strand and one \(^{14}N\) strand. Thus, the ratio of DNA types is:
- \(^{15}N//^{15}N\): 0
- \(^{15}N//^{14}N\): 1
- \(^{14}N//^{14}N\): 0

### Step 3: Analyze the Second Generation
In the second generation, each \(^{15}N//^{14}N\) DNA strand serves as a template, leading to the formation of:
- \(^{15}N//^{14}N\): 2 (from the first generation)
- \(^{14}N//^{14}N\): 2 (newly synthesized)

The ratio becomes:
- \(^{15}N//^{15}N\): 0
- \(^{15}N//^{14}N\): 2
- \(^{14}N//^{14}N\): 2

### Step 4: Analyze the Third Generation
In the third generation, the strands will replicate again:
- Each \(^{15}N//^{14}N\) will yield one \(^{15}N//^{14}N\) and one \(^{14}N//^{14}N\).
- Each \(^{14}N//^{14}N\) will yield two \(^{14}N//^{14}N\).

Thus, the counts will be:
- \(^{15}N//^{15}N\): 0
- \(^{15}N//^{14}N\): 4 (from the second generation)
- \(^{14}N//^{14}N\): 4 (2 from \(^{15}N//^{14}N\) and 2 from \(^{14}N//^{14}N\))

### Step 5: Analyze the Fourth Generation
In the fourth generation, the replication continues:
- Each \(^{15}N//^{14}N\) will yield one \(^{15}N//^{14}N\) and one \(^{14}N//^{14}N\).
- Each \(^{14}N//^{14}N\) will yield two \(^{14}N//^{14}N\).

Thus, the counts will be:
- \(^{15}N//^{15}N\): 0
- \(^{15}N//^{14}N\): 8 (4 from the third generation)
- \(^{14}N//^{14}N\): 8 (4 from \(^{15}N//^{14}N\) and 4 from \(^{14}N//^{14}N\))

### Final Ratio Calculation
Now, we can summarize the ratios:
- \(^{15}N//^{15}N\): 0
- \(^{15}N//^{14}N\): 8
- \(^{14}N//^{14}N\): 8

Thus, the final ratio of \(^{15}N//^{15}N : ^{15}N//^{14}N : ^{14}N//^{14}N\) in the fourth generation is:
\[ 0 : 8 : 8 \]
This can be simplified to:
\[ 0 : 1 : 1 \]