Question

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.

Short answer

Yes, the DNA molecule can fit into the viral head, as its volume (approximately \(157 \times 10^3 \textrm{ nm}^3\)) is less than the volume of the viral head (approximately \(2.56 \times 10^5 \textrm{ nm}^3\)), assuming the molecule has complete flexibility.

Step-by-step solution

Calculate the volume of the DNA molecule

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: \(V_\textrm{DNA} = \pi (1 \textrm{ nm})^2 (50 \times 10^3 \textrm{ nm}) \approx 157 \times 10^3 \pi \textrm{ nm}^3\)

Calculate the volume of the viral head

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: \(V_\textrm{viral\ head} = \frac{4}{3}\pi (40 \textrm{ nm})^3 \approx 2.56 \times 10^5 \pi \textrm{ nm}^3\)

Compare the volumes of the DNA molecule and the viral head

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.

Key concepts

Viral DNA

Viral DNA is a type of genetic material found in viruses, the tiny entities responsible for various infections. Unlike cells in our bodies which house DNA in a structured nucleus, viral DNA can often be found as a singular circular or linear strand within a protective protein shell known as a capsid. This DNA carries all the instructions a virus needs to hijack a host cell and replicate itself. Due to its compact nature, viral DNA must efficiently pack a wealth of information into a tiny space, often resulting in fascinating structural challenges.

In our exercise, the viral DNA is described as a 50 μm long circular strand with a diameter of 2 nm. This strand can be visualized like an extremely thin hose rolled up to fit into a very tiny area, balancing the need to pack tightly while allowing for possible structural changes such as bending or rearranging during viral replication. This characteristic helps to efficiently store genetic instructions that the virus uses to take over host cell functions. Understanding viral DNA gives insight into how viruses succeed in infecting host organisms, showcasing the incredibly precise balance of space and function in genetic materials.

Volume Calculation

In molecular biology, volume calculation is key when determining whether molecules or particles can fit within specific environments. The task of finding out if a viral DNA fits into its viral head involves calculating and comparing volumes.

The DNA is approximated as a long cylindrical strand, where the volume of the DNA is calculated using the formula for a cylinder: \[ V_\text{cylinder} = \pi r_\text{cylinder}^2 h_\text{cylinder} \]where \( r_\text{cylinder} \) is the cylinder's radius, and \( h_\text{cylinder} \) is its height (or length). In this case:

• Diameter of DNA = 2 nm, thus Radius = 1 nm
• Length of DNA = 50 \m = 50,000 nm

The DNA's volume calculated is approximately \( 157 \times 10^3 \pi \) nm³.

On the other hand, the viral head is treated as a sphere, calculated by:\[V_\text{sphere} = \frac{4}{3}\pi r_\text{sphere}^3\]where the diameter equals 0.08 μm, giving us:

• Radius of sphere = 40 nm

Resulting in a viral head volume of \( 2.56 \times 10^5 \pi \) nm³. The comparison shows the viral DNA, having a smaller volume, fits inside the spherical head, confirming the theoretical conclusion.

DNA Flexibility

DNA flexibility is an essential feature, especially significant in the structural dynamics of viral DNA packing. Viral DNA often needs to fold, bend, or even wrap in compact forms to accommodate the confines of a viral head, a task that relies heavily on its inherent flexibility.

DNA flexibility describes how the strand can undergo these physical manipulations without breaking. It allows the genetic material to twist or bend into different shapes, facilitating the capacity of viral particles to store large molecule folds tightly packed in small spaces. For instance, the ability of the 50 μm circular DNA strand to fit inside a viral sphere demonstrates not merely mathematical fitting but the practical flexibility that the strand assumes while being packed.

Understanding DNA flexibility reveals more than just a physical trait; it opens up insight into viral infection mechanisms and propagation traits. This knowledge is crucial for developing anti-viral strategies and understanding viral resistance and evolution. In scientific experiments and real-world applications, such knowledge drives the development of treatments and interventions to combat viral infections.