Question

The diffusion coefficient, D, of a rigid spherical molecule is given by the equation KT Dua 6 where n is the viscosity of the solvent, r is the radius of the sphere, k is the Boltzman constant 1.38 x 10-6 erg.K-, and T is the absolute temperature in kelvins. Assume that this protein is an unhydrated, rigid sphere with a density, p, of 1.35g.cm . Calculate the diffusion coefficient of a 100 kDa protein in a membrane that has an effective viscosity of 1 poise at 37 °C. Note that 1 poise is equal to 1 erg.s-t.cm'.) cm?.5-1 Calculate the average distance, S, traversed by this protein in Ips, 1 ms, and 1 s. S in 1 us = S in 1 ms = S in 1 ms - Sin Is=

Answer 1

Diffusion coefficient, D = kT / 6πηr

We have unhydrated rigid sphere , density = 1.35 g/cm3

Mass of protein : 100 kDa = 100*1000*1.66*10^-24 g  =1.66*10^-19 g   (1 Da = 1amu)

Volume of this sphere = 1.66*10^-19 g / 1.35 g/cm3 = 1.23*10^-19 cm3

radius 'r' = (3V / 4π)⅓ = 0.31 *10^-6 cm

η = 1poise = 1 erg/s-cm3

D = kT / 6πηr = ( 1.38*10^-16 erg/K * 310 K )/ (6π *1 erg*s /cm3 * 0.31 *10^-6 cm )

D = 7.32 *10^-9 cm² s⁻¹

Average distance(S) travelled by this :

we have following relation : t (time) ~ S² / 2D or S = (t*2D)^1/2

S (1 us ) = 1.2*10^-7 cm

S (1 ms ) = 3.9*10^-6 cm

S (1 s ) =    1.2*10^-4 cm