Chapter 24, Problem 1P

Summary Introduction

To determine: The length of the DNA and compare it with the length of the T2 head.

Introduction:

In the life cycle of a virus, “genome packaging” is a basic process. Virus Bacteriophage T2, is also termed as phage. In, the Virus Bacteriophage T2, during the packaging of DNA, weight of DNA molecule is ${\text{120 × 10}}^{\text{6}}$ found in a head about 210 nm long. The molecular weight of a nucleotide pair is 650 in virus bacteriophage.

Explanation of Solution

Explanation:

The nucleotide pairs of T2 DNA molecule can be calculated as:

$\begin{array}{c}\text{Nucleotide pairs}=\frac{\text{Molecular}\text{weight}\text{of}\text{DNA}}{\text{Molecular}\text{weight}\text{of}\text{a}\text{nucleotide}\text{pair}}\\ =\frac{120\times {10}^6}{650}\\ =184615.38\\ =1.8\times {10}^5\text{base}\text{pairs}\end{array}$

Each base pair occupied length is3.4 Å.

Thus, the length of T2 DNA is:

$\begin{array}{c}\text{Length}\text{of}\text{T-DNA}=\text{Total}\text{length}\text{of}\text{nucleotide}\text{pairs}\text{×Length}\text{of}\text{each}\text{base}\text{pair}\\ =1.8\times {10}^5\times 3.4\\ =612000\\ =\text{6}\text{.1}\text{×}{\text{10}}^{\text{5}}\stackrel{\text{o}}{\text{A}}\text{/6}\text{.1}\text{×}{\text{10}}^{\text{4}}\text{nm}\end{array}$

The length of the DNA molecule can be compared with the length of head of the virus is:

$\begin{array}{c}\text{Length of DNA molecule}=\frac{\text{length}\text{of}\text{the}\text{T2}\text{DNA}\text{molecule}}{\text{length}\text{of}\text{the}\text{DNA}\text{in}\text{head}}\\ =\frac{6.1\times {10}^4}{210}\\ =290.476\\ =290\end{array}$

Thus, the DNA length is $6.1\times {10}^4\text{nm}\text{.}$

In comparison to the length of the T2 phage head it is 290 times longer.

Conclusion

Conclusion:

Thus, the length of the DNA is $\underset{\_}{6.1\times {10}^4\text{nm}}$ and when compared to the head length of T2 bacteriphage, it is $\underset{\_}{290}$ times longer.