One E . Coli bacterium contains a hybrid DNA with one heavy `N^(15)` Strand and one light `N^(14)` Strand. It was allowed to replicate for 3 hours in a medium containing `N^(14)` What proportion of E. coli which will neither float nor sink in the test tube on ultra - centrifugation with CsCl ?

Options

A. one out of `2^8`
B. one out of `2^9`
C. one out of `2^10`
D. one out of `2^7`

Answer

To solve the problem, we need to analyze the replication of E. coli bacteria that initially contain a hybrid DNA strand composed of one heavy (N^15) strand and one light (N^14) strand. The bacteria are then allowed to replicate in a medium containing only N^14 for 3 hours.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- We start with one E. coli bacterium that has a hybrid DNA: one strand is heavy (N^15) and the other is light (N^14).

2. **Replication Time**:
- The bacterium divides every 20 minutes. In 3 hours (180 minutes), the number of divisions can be calculated as:
\[ \text{Number of divisions} = \frac{180 \text{ minutes}}{20 \text{ minutes/division}} = 9 \text{ divisions} \]

3. **Calculating the Total Number of Bacteria**:
- After 9 divisions, the total number of bacteria can be calculated using the formula \(2^n\), where \(n\) is the number of divisions:
\[ \text{Total bacteria} = 2^9 = 512 \]

4. **Analyzing the DNA Composition After Replication**:
- During replication in the N^14 medium, the heavy strand (N^15) will remain unchanged, while the new strands synthesized will be light (N^14).
- The possible combinations of DNA after replication will be:
- **1 hybrid DNA (N^15/N^14)**: This is the original bacterium.
- **1 heavy (N^15/N^14)**: This will occur in the first generation after replication.
- **2 light (N^14/N^14)**: These will be produced in subsequent generations.

5. **Identifying the Proportion of Hybrid DNA**:
- After 9 cycles of replication, the only bacterium that retains the hybrid DNA (N^15/N^14) is the original one. All other bacteria will either have two light strands (N^14/N^14) or the heavy strand paired with a light strand (N^15/N^14).
- Therefore, only 1 out of 512 bacteria will have the hybrid DNA.

6. **Calculating the Proportion**:
- The proportion of E. coli that will neither float nor sink in the test tube (i.e., the hybrid DNA) is:
\[ \text{Proportion} = \frac{1 \text{ hybrid}}{512 \text{ total}} = \frac{1}{512} \]

### Final Answer:
The proportion of E. coli that will neither float nor sink in the test tube on ultracentrifugation with CsCl is **1/512**.