Question

(4 pts) A sample of cells has a total receptor concentration of 25 mM. Ninety percent of the receptors have bound ligand and the concentration of free ligand is 125 PM. What is the Kd for the receptor-ligand interaction?

Answer 1

Binding of ligand to receptor protein can be represented by the chemical equilibrium

P + L <=^Kf_Kr==> PL where P = protein , L = Ligand

K_f = forward rate , K_r = reverse rate

The apparant dissociation constant

K_d = K_r / K_f = [P] [L] / [PL] -------(1)  

=> [PL] = [P] [L] / K_d

the fraction of protein receptors occupied by ligand = (occupied protein / total protein) = [PL] / ([P] + [PL])

= ([P] [L] / K_d) / ([P] + ( [P] [L] / K_d))

= [P] [L] / (K_d[P] + [P] [L])

= [L] / (K_d + [L]) ----------(2)

Now when 90% of receptors have bound to ligands then θ = 0.9 & given [L] = 125μM = 0.125 mM

=> 0.9 = 0.125 / (K_d + 0.125)

=> 0.9 K_d + 0.1125 = 0.125

=> K_d = (0.0125 / 0.9)

=> K_d = 0.0139