Equilibria Question

Question

Upon conducting an electrophoretic mobility shift assay with DNA, 30% of your DNA is bound by a protein (P) when the total protein concentration is 2.0 μM and the total DNA concentration is 1 nM. Assuming only one protein molecule binds to one DNA (D) molecule, what is dissociation equilibrium constant expression for the binding event? What is the dissociation constant for the 1:1 interaction? Calculate the association constant for the complex if the fraction of bound DNA and the total protein and DNA concentrations are the same as above, but two proteins cooperatively bind one DNA.

Can anyone help me with the bolded parts above? I am very confused on how to solve this problem! I have tried using an ICE table, but I don't understand what the question really means...

Answer 1

P + D <==> PD

Kd = dissociation constant = [P][D]/[PD] = 1/Ka

[P] = 2 uM = 2000 nM

[D] = 1 nM - 0.3 nM = 0.7 nM

[PD] = 0.3 nM

Kd = (2000 nM)(0.7 nM)/0.3 nM

Kd = 4667 nM

If you are looking to use an ICE table, for the association, it would look something like this:

P + D <===> PD

2000....1..............0......Initial

-0.3....-0.3.........+0.3....Change

2000-0.3...1-0.3...0.3...Equilibrium

Ka = [PD]/[P][D] = 0.3/(2000)(0.7) = 0.0002143

Kd = 1/Ka = 1/0.0002143 = 4666 nM