Question

Assume that a viral DNA molecule is a 50 - \(\mu\) m-long circular strand with a uniform \(20-\AA\) 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

Answer: No, the DNA molecule cannot fit inside the viral head, given the dimensions provided and assuming complete flexibility of the molecule.

Step-by-step solution

Convert given dimensions to appropriate units

First, convert the given dimensions of the DNA molecule and the viral head into meters, as it will help us in calculating the volumes easily. Length of DNA molecule: \(50\,\mu m = 50 × 10^{-6}\,m\) Diameter of DNA molecule: \(20\,\text{\AA} = 20 × 10^{-10}\,m\) Diameter of viral head: \(0.08\,\mu m = 0.08 × 10^{-6}\,m\)

Calculate the volume of the DNA molecule

The DNA molecule is given as a circular strand. To calculate the volume of the DNA molecule, we'll consider it as a cylinder with length = 50 \(\mu m\) and diameter = 20 \(\text{\AA}\). Volume of the DNA molecule (cylinder) can be found using the formula: \(V_\text{DNA} = \pi r_\text{DNA}^2 h_\text{DNA}\) where \(r_\text{DNA}\) is the radius of DNA molecule, \(h_\text{DNA}\) is the length of DNA molecule. Radius: \(r_\text{DNA} = \frac{d_\text{DNA}}{2} = \frac{20 × 10^{-10}\,\text{m}}{2} = 10^{-10}\,\text{m}\) Length: \(h_\text{DNA} = 50 × 10^{-6}\,\text{m}\) Now, substitute these values and calculate the volume of the DNA molecule: \(V_\text{DNA} = \pi (10^{-10}\,\text{m})^2 \times (50 × 10^{-6}\,\text{m})\) \(V_\text{DNA} \approx 1.57 × 10^{-21}\,\text{m}^3\)

Calculate the volume of the viral head

The viral head is given as a sphere with a diameter of 0.08 \(\mu m\). To calculate the volume of the viral head, we'll use the formula for the volume of a sphere: \(V_\text{head} = \frac{4}{3}\pi r_\text{head}^3\) where \(r_\text{head}\) is the radius of the viral head. Radius: \(r_\text{head} = \frac{d_\text{head}}{2} = \frac{0.08 × 10^{-6}\,\text{m}}{2} = 4 × 10^{-8}\,\text{m}\) Now, substitute this value and calculate the volume of the viral head: \(V_\text{head} = \frac{4}{3}\pi (4 × 10^{-8}\,\text{m})^3\) \(V_\text{head} \approx 2.68 × 10^{-22}\,\text{m}^3\)

Compare the volumes of the DNA molecule and the viral head

Now that we have the volume of the DNA molecule and the viral head, we can compare them to see if the DNA molecule can fit inside the viral head: \(V_\text{DNA} \approx 1.57 × 10^{-21}\,\text{m}^3\) \(V_\text{head} \approx 2.68 × 10^{-22}\,\text{m}^3\) Since \(V_\text{DNA} > V_\text{head}\), the DNA molecule will not fit inside the viral head, given the dimensions provided and assuming complete flexibility of the molecule.

Key concepts

Viral DNA Structure

The structure of viral DNA can affect the way it's packaged into the viral head, also known as the capsid. Typically, viruses contain either DNA or RNA, and those that have DNA can possess either single-stranded or double-stranded DNA. In our exercise, we consider viral DNA as a circular strand, which is a common form for many bacteriophages, viruses that infect bacteria.

Viral DNA must be compactly packed to fit within the limited space of the capsid. Special proteins, such as histones in eukaryotes, are often involved in this packaging process in more complex organisms. However, in many viruses, the DNA is tightly wound and folded into intricate structures to ensure it can fit inside the capsid. The length and diameter of the DNA, as well as the internal volume of the viral capsid, are crucial factors in determining how and if the DNA can be successfully packaged.

DNA Molecule Volume Calculation

Calculating the volume of a DNA molecule can be intriguing due to its microscopic size. To find the volume, we need to model the DNA as a geometrical object. In the textbook exercise, the DNA is approximated by a cylinder with a given length and diameter. Using the formula for the volume of a cylinder, \( V = \pi r^2h \), where \( r \) is the radius and \( h \) is the height (length of DNA), we obtain the volume it occupies.

The calculation involves dimensional analysis and unit conversion, which are essential skills in scientific computations. To handle the extreme smallness of the DNA dimensions, the calculations are done using metric prefixes like micrometers (\( \mu m \)) and angstroms (\( \text{\AA} \)), which need to be converted into meters to have consistent units, enabling us to determine the volume accurately. This step-by-step quantitative analysis is vital in assessing the fit of DNA within a viral head.

Spherical Viral Head Volume Calculation

The volume of a viral head can be calculated using the formula for the volume of a sphere, \( V = \frac{4}{3}\pi r^3 \). Here, the viral head is assumed to be a perfect sphere, and the challenge is to determine whether the viral DNA will fit inside this spherical space.

The calculation requires first determining the radius of the sphere, which is half the diameter, and then applying it to the formula. Such calculations are common in virology and microbiology when researchers estimate how genetic material can be encapsulated within viruses. Converting the diameter into meters allows for determining the volume in cubic meters, which is a standard SI unit for volume, thus facilitating the comparison with the DNA volume. Understanding the volumes involved is critical in assessing the capacity of the viral capsid and the viability of the virus's structure.