Question

Calculate the diffusive force (J) experienced by potassium in a cell where the concentration gradient is 35 mM, the membrane width is 1 cm, solubility of potassium across the membrane is 1 and the diffusive area (number of channels available) is 10. Assume that the cell is at 1 degree.

Answer 1

Answer:

Diffusive force J = 9590

Explanation:

From the given information:

Concentration gradient C₁ - C₂ = 35 mM = 35 × 10⁻³ M

membrane width Δ x = 1 cm = 10⁻² m

Solubility S = 1

diffusive area A = 10

Temperature T = 1° C = (273 + 1 )K = 274 K

Using the expression:

$\text {diffusive force J} = \dfrac{C_1-C_2}{\Delta x }\times T \times S \times A$

$J = \dfrac{35 \times 10^{-3}}{10^{-2} }\times 274 \times 1 \times 10$

$J = 3.5 \times 274 \times 1 \times 10$

Diffusive force J = 9590