One E . Coli bacterium contains a hybrid DNA with one heavy `N^(15)` Strand and one light `N^(14)` Strand. It was allowed ti 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 containing hybrid DNA strands in a medium rich in nitrogen-14 (N^14). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Condition
- We start with one E. coli bacterium that contains hybrid DNA: one heavy strand (N^15) and one light strand (N^14).

### Step 2: Replication Process
- E. coli replicates its DNA in a semi-conservative manner, meaning each strand serves as a template for the synthesis of a new strand.
- In a medium containing only N^14, any new strands synthesized will be light (N^14).

### Step 3: First Replication Cycle
- After the first replication cycle, the original hybrid DNA will yield:
- One strand will remain N^15 (heavy).
- The other strand will be newly synthesized N^14 (light).
- This results in two DNA molecules:
- One hybrid (N^15/N^14).
- One light (N^14/N^14).

### Step 4: Subsequent Replication Cycles
- E. coli divides approximately every 20 minutes. In 3 hours, it will undergo 9 cycles of replication (3 hours = 180 minutes; 180/20 = 9).
- After each cycle, the number of DNA molecules doubles:
- After 1 cycle: 2 molecules (1 hybrid, 1 light).
- After 2 cycles: 4 molecules (2 hybrids, 2 lights).
- Continuing this pattern, after 9 cycles, the total number of DNA molecules will be 2^9 = 512.

### Step 5: Determine the Proportions
- In the first cycle, we had 1 hybrid and 1 light DNA.
- In the second cycle, we will have:
- The original hybrid will produce one hybrid and one light.
- The light will produce two light strands.
- Continuing this process, after 9 cycles, we will have:
- 1 hybrid DNA molecule (N^15/N^14).
- 511 light DNA molecules (N^14/N^14).

### Step 6: Identify the DNA that Will Neither Float Nor Sink
- In cesium chloride (CsCl) density gradient centrifugation, the hybrid DNA (N^15/N^14) will be in the middle of the gradient, meaning it will neither float nor sink.
- Therefore, out of the total 512 DNA molecules, only 1 will be hybrid.

### Step 7: Calculate the Proportion
- The proportion of E. coli that will neither float nor sink is given by the ratio of hybrid DNA to total DNA:
- Proportion = Number of hybrid DNA / Total DNA = 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 out of 512**.