Protein A binds to protein B to form a complex, AB. At equilibrium in a cell the concentrations of A, B, and AB are all at 1 μM.

A. Referring to Figure 3−19, calculate the equilibrium constant for the reaction A + B ↔ AB.

B. What would the equilibrium constant be if A, B, and AB were each present in equilibrium at the much lower concentrations of 1nM each?

C. How many extra hydrogen bonds would be needed to hold A and B together at this lower concentration so that a similar proportion of the molecules are found in the AB complex?

Transcribed Image Text:

association rate A + B association rate constant association A B dissociation rate= [AB] [A] [B] X association rate = kon [A] [B] = AB concentration concentration of A of B dissociation dissociation rate constant dissociation rate = koff [AB] x A + B AT EQUILIBRIUM: association rate = dissociation rate Kon [A] [B] = Koff [AB] Kon Koff concentration of AB = K = equilibrium constant