A bacteriophage T2 particle contains within its head a double-stranded B-from DNA of molecular weight `1.2 xx 10^8` Da. Assume that the head of a T2 Phage particle is of 210 nm in length and the average molecular weight of a nucleotide is 330 Da. The length of the T2 genome is in the range of

Options

A. `6 xx 10^5 "to "6.4 xx 10^5 `nm
B. ` 40 xx 10^4 " to" 41 xx 10^4 `nm
C. `1.8 xx 10^5 " to "2 xx 10^5 ` nm
D. `6 xx 10^4 " to " 6.4 xx 10^4 ` nm

Answer

To find the length of the T2 bacteriophage genome, we will follow these steps:

### Step 1: Determine the number of base pairs in the DNA
Given:
- Molecular weight of the double-stranded DNA = \(1.2 \times 10^8\) Da
- Average molecular weight of a nucleotide = 330 Da

To find the number of base pairs (bp), we can use the formula:
\[ \text{Number of base pairs} = \frac{\text{Molecular weight of DNA}}{\text{Molecular weight of one nucleotide}} \]

Substituting the values:
\[ \text{Number of base pairs} = \frac{1.2 \times 10^8 \text{ Da}}{330 \text{ Da}} \approx 3.636 \times 10^5 \text{ bp} \]

### Step 2: Calculate the length of the DNA
Since the DNA is double-stranded, we need to consider the length of the single strand. Each base pair contributes approximately 0.34 nm to the length of the DNA.

To find the total length of the single-stranded DNA, we first calculate the number of nucleotides in one strand:
\[ \text{Number of nucleotides in one strand} = \frac{\text{Number of base pairs}}{2} \approx \frac{3.636 \times 10^5}{2} \approx 1.818 \times 10^5 \]

Now, we can calculate the total length of the DNA:
\[ \text{Length of DNA} = \text{Number of nucleotides} \times \text{Distance between nucleotides} \]
\[ \text{Length of DNA} = 1.818 \times 10^5 \text{ nucleotides} \times 0.34 \text{ nm} \approx 6.18 \times 10^4 \text{ nm} \]

### Step 3: Final Result
The length of the T2 genome is approximately \(6.18 \times 10^4\) nm.

### Conclusion
The correct answer is \(6.18 \times 10^4\) nm.

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