Question

The diffusion of a molecule in a tissue is studied by measuring the uptake of labeled protein into the tissue of thickness L =149.4 um. Initially, there is no labeled protein in the tissue. At t=0, the tissue is placed in a solution with a molecular concentration of C1=1.2 uM, so the surface concentration at x=0 is maintained at C1. Assume the tissue can be treated as a semi-infinite medium. Surface area of the tissue is A = 93.9 cm2. Calculate the uptake of the labeled protein M by the tissue over 100 s.Please give your answer with a unit of pmol. Assuming the diffusion coefficient is known of D=1*10-9 cm2/s

Answer 1

According to Fick's law,

J = D A (Cf - Ci)/L

where, J = Diffusional flux, mol/s

D = Coefficient of diffusion, cm2/s

A = cross sectional area, cm2

Ci, Cf = initial and final concentrations, mol/cm3

L = length of diffusion, cm

J here represents the amount of substance that has diffused per time. In this case, it is the protein uptake.

Concentration of the protein if given in terms of molarity.

Molarity = Moles of solute / Volume of solution in litres , mol/l

Given Cf = 1.2 x 10^-6 moles/l

1litre = 10^-3 m3

1 m3 = 10^6 cm3

Therefore 1litre = 1000 cm3

Expressing Ci in terms of moles/cm3,

Cf = 1.2 x 10^-6 moles /1000 cm3

Cf = 1.2 x10^-9 moles/cm3

Since initially the tissue had no protein, Ci = 0 moles/cm3

The uptake takes place across a thickness (L) of 149.4x10^-6 m

   1m = 100cm

L = 149.4 x 10^-6 x 100 cm

   = 149.4 x 10^-4 cm

The cross-sectional area of the tissue is 93.9 cm2. Substituting the values in the equation,

J = (1 x 10^-9) x (93.9) x (1.2 x 10^-9 - 0)/(149.4 x 10^-4) (cm2/s) x (cm2) x (mol/cm3) x (1/cm)

= 0.754 x 10^-18 x 10^4    (cm4 x mol)/(cm4 x s)

= 0.754 x 10^-14 (mol/s)

The amount of protein uptake per second J = 0.754 x 10^-14 (mol/s)

In 100 seconds the protein uptake would be = 0.754 x 10^-14 x 100 (mol/s) x s

   = 0.754 x 10^-12 moles

10^-12 moles = 1 pico mole = 1pmol

In 100 seconds the protein uptake would be 0.754 pmol.