In equlibrium dialysis a twochamber setup is used with each chamber serperated b
ID: 907132 • Letter: I
Question
In equlibrium dialysis a twochamber setup is used with each chamber serperated by a semipermeable membrane. A protein is assed to one side, and a diffusible ligand is added to the other side. The protein can not pass through the membrane. A leucine-binding protein of 35000 MW is added to one side (0.5mg/ml) and on the other side an equal volume of buffer containing 4 x 10^-5 M leucine labelled with 14C is added. After allowing it to equlibrate it was found that the side containing protein had 2.3x10^-5 M leucine. The other compartment had 1.7x10^-5 M leucine.
What is the concentration of leucine bound to protein? What is the concentration of free protein adter equalibration? from this data calculate a Kd for leucine for this protein
Explanation / Answer
Equilibrium constants
An equilibrium constant, designated by a upper case K, is the ratio of the equilibrium
concentrations of reaction products to reactants or vice versa.
For the bimolecular reaction, A+B <=> AB, we can define an equilibrium dissociation
constant (Kd), which are reciprocally related, as shown below:
Kd = [A][B] / [AB]
For bimolecular reactions, the units of Kd are concentration (M, mM, mM, etc.)
For this protein, A = 2.3x10^-5 M and B = 1.7x10^-5 M substitute these values in Kd.
Then Kd = [2.3x10^-5] [1.7x10^-5] / [2.3x10^-5 + 1.7x10^-5]
Therefore, [Kd = 9.775 x 10^-6 M]