Most cells in the body are within 100 um of a capillary. WHY? Try this problem to come to an explanation. Diffusion constants (Ds) for most biological small molecules (like glucose... Most cells in the body are within 100 um of a capillary. WHY? Try this problem to come to an explanation. Diffusion constants (Ds) for most biological small molecules (like glucose) are ~ 5x10-6 cm2/s. Use the 3-D Fick-Einstein equation to calculate how long it would take a molecule of glucose to travel A. 100 μm? B. 1 cm? Read more

Steps to Solve

1. Convert the distance to meters

To apply the Fick-Einstein equation correctly, we will convert the distances from micrometers (μm) and centimeters (cm) to meters (m).

For 100 μm: $$ 100 , \mu m = 100 \times 10^{-6} , m = 1 \times 10^{-4} , m $$

For 1 cm: $$ 1 , cm = 1 \times 10^{-2} , m $$

2. Identify the diffusion constant

The diffusion constant ($D$) for glucose in water is typically around $6.7 \times 10^{-10} , m^2/s$. We will use this value in our calculations.

3. Apply the Fick-Einstein equation

The Fick-Einstein equation to calculate the time ($t$) required for diffusion is given by: $$ t = \frac{x^2}{2D} $$ where $x$ is the distance to diffuse and $D$ is the diffusion constant.

4. Calculate time for 100 μm

Using the previously calculated distance for 100 μm: $$ t_{100 , \mu m} = \frac{(1 \times 10^{-4} , m)^2}{2 \times (6.7 \times 10^{-10} , m^2/s)} $$ After calculating, we obtain: $$ t_{100 , \mu m} \approx 7.46 \times 10^{-6} , s $$

5. Calculate time for 1 cm

Using the distance for 1 cm: $$ t_{1 , cm} = \frac{(1 \times 10^{-2} , m)^2}{2 \times (6.7 \times 10^{-10} , m^2/s)} $$ After performing the calculations, we find: $$ t_{1 , cm} \approx 7.46 \times 10^{-3} , s $$