Protein A binds to protein B to form a non-covalent complex, AB. A cell contains

Question

Protein A binds to protein B to form a non-covalent complex, AB. A cell contains an equilibrium mixture of A, B and AB each present at a 1uM concentration.

a. What is the equilibrium constant Kd for the binding reaction?

b. What would the equilibrium constant be if each species were present at 1 nM concentration?

c. How many additional hydrogen bonds would be needed at the lower concentration (b) versus the higher concentration (a) to hold A and B together tightly enough to form the same proportion of the AB complex?

The formation of 1 hydrogen bond is accompanied by a favorable change in free energy of ~1kcal/mol, R=1.987cal/mol-K, T=310K

Explanation / Answer

A) The equilibrium constant is defined as K=(AB) / ((A)X(B)). The Brackets of AB, A, and B refer to the concentration. Therefore, if A, B, and AB are each 1 uM (10^-6 M),

Then, K will be (10^-6 / (10^-6 X 10^-6)) = 10^6 M^-1

B) In similar fashion, if A, B, and AB are each 1 nM (10^-9 M), then K will be 10^9 M^-1.

C) This example illustrates that interacting proteins that are present in cells in lower concentrations need to bind to each other with high affinities. Therefore, a significant fraction of the molecules are bound at equilibrium. In this situation, lowering the concentration by three orders of magnitude (from uM to nM) requires a change in the equilibrium constant by three orders of magnitude to maintain the AB protein complex (that corresponding to -4.3 kcal of free energy). This corresponds to about four or five extra hydrogen bonds.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.