Q: A spherical protein of diameter R dimerizes and rearranges to form another spherical shape equal to the sum of the original monomer volumes. What will be the percent increase or decrease in (a) the diffusion coefficient and (b) the sedimentation?

Solution

(a) Diffusion coefficient

D = RT / Nf

After dimerization, the only variable is the viscosity coefficient f. It changes from f₀ to f' ,where f = 6phR, R is the radius of the sphere. To calculate the radius of a sphere, we need the relation between radius and volume, V = (4/3)pR³,

V' = 2V₀

i.e.,

(4/3)pR'³ = 2[(4/3)pR₀³]

that is

R' = (2)1/3R₀

thus

f' = (2)^1/3f0

D' = (2)^-1/3D₀

DD = 1- (2)^-1/3 = 29.3% (decrease)

(b) Sedimentation coefficient

s = M(1-Vr) / Nf

From question (a), we know that f' = (2)^1/3f0, and here the mass M is doubled due to dimerization. Hence,

s' = 2M(1-Vr) / Nf'

i.e.,

s' = 2M(1-Vr) / N(2)^1/3f ₀ = 2^2/3 s₀

Ds = 2^2/3-1 = 58.7% (increase)