Assume that a viral BNA molecule is a 50 - \(\mu \mathrm{m}\) -long circular strand of a uniform 2-nm diameter. If this molecule is contained in a viral head that is a 0.08 - \(\mu\) m-diameter sphere, will the DNA molecule fit into the viral head, assuming complete flexibility of the molecule? Justify your answer mathematically.

Given that the DNA molecule is a circular strand of uniform diameter, we can find its volume using the formula for the volume of a cylinder. The formula for the volume of a cylinder is \(V_\textrm{cylinder} = \pi r_\textrm{cylinder}^2h_\textrm{cylinder}\), where \(r_\textrm{cylinder}\) is the radius, and \(h_\textrm{cylinder}\) is the height. Since the DNA molecule is a circular strand of 50 μm length and 2 nm diameter, we can consider it as a long cylinder bent in a circle. Here, \(h_\textrm{cylinder} = 50\textrm{ μm} = 50 \times 10^3 \textrm{ nm}\) and the radius \(r_\textrm{cylinder} = \frac{2 \textrm{ nm}}{2} = 1 \textrm{ nm}\). Using the given information, let's calculate the volume of the DNA molecule: \(V_\textrm{DNA} = \pi (1 \textrm{ nm})^2 (50 \times 10^3 \textrm{ nm}) \approx 157 \times 10^3 \pi \textrm{ nm}^3\)

The viral head is a sphere with a diameter of 0.08 μm. To calculate the volume, we will use the formula for the volume of a sphere, which is \(V_\textrm{sphere} = \frac{4}{3}\pi r_\textrm{sphere}^3\), where \(r_\textrm{sphere}\) is the radius of the sphere. In this case, the radius of the viral head is half the diameter, so \(r_\textrm{sphere} = \frac{0.08 \textrm{ μm}}{2} = 0.04 \textrm{ μm} = 40 \textrm{ nm}\). Now, let's calculate the volume of the viral head: \(V_\textrm{viral\ head} = \frac{4}{3}\pi (40 \textrm{ nm})^3 \approx 2.56 \times 10^5 \pi \textrm{ nm}^3\)

Now that we have calculated the volumes of the DNA molecule and the viral head, we can compare them to determine if the DNA molecule will fit into the viral head. To do this, we will check if the volume of the DNA molecule is less than or equal to the volume of the viral head: \(V_\textrm{DNA} \leq V_\textrm{viral\ head}\) \(157 \times 10^3 \pi \textrm{ nm}^3 \leq 2.56 \times 10^5 \pi \textrm{ nm}^3\) Dividing both sides by \(\pi\) and comparing the numerical values: \(157 \times 10^3 \textrm{ nm}^3 \leq 2.56 \times 10^5 \textrm{ nm}^3\) As we can see, the inequality holds because \(157 \times 10^3\) is less than \(2.56 \times 10^5\). Therefore, the DNA molecule will fit into the viral head, assuming complete flexibility of the molecule.