E.coli has only 4.6 × 10 6 base pairs and completes the process of replication within 18 minutes, then the average rate of polymerisation is approximately- (1) 2000 base pairs/second (2) 3000 base pairs/second (3) 4000 base pairs/second (4) 1000 base pairs/second

Options

A. 2000 base pairs/second
B. 3000 base pairs/second
C. 4000 base pairs/second
D. 1000 base pairs/second

Answer

To find the average rate of polymerization of E. coli during DNA replication, we can follow these steps:

### Step 1: Identify the total number of base pairs
The total number of base pairs in E. coli is given as:
\[ 4.6 \times 10^6 \text{ base pairs} \]

### Step 2: Convert the time taken for replication from minutes to seconds
The time taken for replication is given as 18 minutes. To convert this into seconds:
\[ \text{Time in seconds} = 18 \text{ minutes} \times 60 \text{ seconds/minute} = 1080 \text{ seconds} \]

### Step 3: Calculate the average rate of polymerization
The average rate of polymerization can be calculated using the formula:
\[ \text{Average Rate} = \frac{\text{Total Base Pairs}}{\text{Time in seconds}} \]
Substituting the values we have:
\[ \text{Average Rate} = \frac{4.6 \times 10^6 \text{ base pairs}}{1080 \text{ seconds}} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \text{Average Rate} = \frac{4.6 \times 10^6}{1080} \approx 4259.26 \text{ base pairs/second} \]

### Step 5: Round the result
Rounding \( 4259.26 \) to the nearest thousand gives us approximately:
\[ \text{Average Rate} \approx 4000 \text{ base pairs/second} \]

### Conclusion
Thus, the average rate of polymerization during DNA replication in E. coli is approximately:
\[ \text{Answer: } 4000 \text{ base pairs/second} \]