Question

Consider a protein that has a diffusion constant in water of \(D=10^{-6} \mathrm{~cm}^{2} \mathrm{~s}^{-1}\). (a) What is the time that it takes the protein to diffuse a root-mean-square distance of \(10 \mu \mathrm{m}\) in a threedimensional space? This distance is about equal to the radius of a typical cell. (b) If a cell is spherical with radius \(r=10 \mu \mathrm{m}\), and if the protein concentration outside the cell is \(1 \mu \mathrm{M}\) (micromolar), what is the number of protein molecules per unit time that bombard the cell at the diffusion-limited rate?

Short answer

The time it takes is approximately 1.67 seconds. The flux of protein molecules is 7.57 \times 10^9 molecules/s.

Step-by-step solution

Understand the problem

We need to calculate the time it takes for a protein to diffuse a given distance in a three-dimensional space and then find the number of protein molecules per unit time that bombard a cell surface given the diffusion-limited rate.

Use the diffusion equation

The root-mean-square (rms) distance for diffusion in three dimensions is given by the equation \[ \text{rms} = \sqrt{6Dt}\] where D is the diffusion constant, t is the time, and rms is the root-mean-square distance.

Solve for time

Rearrange the equation to solve for t: \[ t = \frac{\text{rms}^2}{6D}\] Plug in the given values: rms = 10 µm (\times 10^{-4} cm to convert to cm), and D = 10^{-6} cm^2/s. \[ t = \frac{(10 \times 10^{-4} \text{cm})^{2}}{6 \times 10^{-6} \text{cm}^{2}/\text{s}} \approx 1.67 \text{seconds}\]

Find the flux of protein molecules

To find the number of protein molecules per unit time (the flux), use the formula: \[ J = 4\pi DrC\] where J is the flux, r is the radius of the cell, D is the diffusion constant, and C is the concentration of the protein. Plug in the given values: \[ J = 4 \pi \times 10^{-6} \frac{cm^{2}}{s} \times 10 \times 10^{-4} \text{cm} \times 1 \times 10^{-6} \frac{mol}{L}\] Convert the concentration to molecules/cm\textsuperscript{3} using Avogadro’s number (6.022 \times 10^{23} mol^{-1}): \[ 1 \times 10^{-6} \frac{mol}{L} = 6.022 \times 10^{17} \frac{molecules}{cm^{3}}\] So \[ J = 4 \pi \times 10^{-6} \frac{cm^{2}}{s} \times 10^{-3} \text{cm} \times 6.022 \times 10^{17} \frac{molecules}{cm^{3}} = 7.57 \times 10^9 \frac{molecules}{s}\]

Key concepts

Diffusion Constant

The diffusion constant, often denoted as D, is a fundamental parameter in the study of diffusion processes. It quantifies how quickly molecules spread through a medium. Consider it as the 'speed' at which particles move on average. A higher diffusion constant means faster diffusion.

For proteins in biological systems, the diffusion constant can be influenced by several factors:

• Temperature - Typically, diffusion rates increase with temperature.
• Medium - Proteins may diffuse differently in water compared to cytoplasm due to viscosity differences.
• Molecular Size - Larger proteins tend to move more slowly.
  
  In the given problem, the diffusion constant for the protein in water is given as \(D = 10^{-6} \text{ cm}^2/\text{s}\). This value indicates that the protein spreads at a moderate pace suitable for many biological processes.

Root-Mean-Square Distance

The root-mean-square (rms) distance is a measure of the average distance a diffusing particle will travel over a given period. In three dimensions, this is calculated using the formula:

\[\text{rms} = \sqrt{6Dt} \]

Here, D is the diffusion constant, and t is the time. For example, if you want to find how far a protein will diffuse in a specific time, you use this equation. Conversely, if you know the distance and want to find the time, you rearrange it to:

\[t = \frac{\text{rms}^2}{6D}\]



In our problem, we wanted to know how long it takes for a protein to diffuse a distance of \(10 \mu \text{m}\). Substituting the given values, we found it takes approximately 1.67 seconds.

The rms distance concept is crucial in understanding how effectively proteins move and interact within cellular environments.

Diffusion-Limited Rate

The diffusion-limited rate refers to the rate at which molecules collide with a target due to diffusion alone, without any other active forces driving the movement. In biological contexts, this is often relevant for processes such as ligand-receptor binding or nutrient uptake by cells.

The number of protein molecules hitting a cell per unit time can be calculated using:

\[J = 4\pi DrC\]

Here,

• J is the flux or number of molecules per unit time.
• D is the diffusion constant.
• r is the radius of the cell.
• C is the concentration of the protein outside the cell.
  
  In the problem, substituting the values:
  
  \[J = 4\pi \times 10^{-6} \text{ cm}^2/\text{s} \times 10^{-3} \text{ cm} \times 6.022 \times 10^{17} \frac{molecules}{cm^{3}} \approx 7.57 \times 10^9 \frac{molecules}{s}\]
  
  This calculation tells us that, on average, about 7.57 billion protein molecules bombard the cell surface every second.

Protein Concentration

Protein concentration is key in determining the likelihood of molecular interactions and is often measured in micromolar (\(\mu M\)) concentrations in biological studies. It represents the number of molecules in a given volume.

For instance, an external protein concentration of \(1 \mu M\) indicates there is one micromole of protein per liter of solution. This concentration can be translated into a number of molecules using Avogadro’s number (\(6.022 \times 10^{23} \text{ mol}^{-1}\)).

For our exercise, we converted the concentration to molecules per cubic centimeter:

• \(1 \mu M = 1 \times 10^{-6} \text{ mol}/L\)
• \(1 \text{ mol} = 6.022 \times 10^{23} \text{ molecules}\)

Thus providing \(6.022 \times 10^{17} \text{ molecules}/cm^3\).

Understanding protein concentration helps predict the rate of diffusion and molecular collisions which are fundamental for processes such as enzyme reactions, cellular signaling, and nutrient transport.