Relationship between inhibition constant Ki and interaction energy?

Inhibition of an enzyme $E$ by an inhibitor $I$:

$$E + I \rightleftharpoons EI$$

The inhibition constant $K_i$ is defined as follows:

$$K_i = \frac{[E][I]}{EI}$$

where $[E]$ is the enzyme concentration, $[I]$ is the inhibitor concentration, and $[EI]$ is the enzyme-inhibitor-complex concentration. Inhibition is strong when $K_i$ is low; that is, when the equilibrium above is shifted toward the product.

Can the strength of inhibition be approximated by the relative interaction energy: $E_{Interaction} = E_{EI complex} - E_{Enzyme} - E_{Inhibitor}$? This seems like a logical and valid hypothesis to me, but I cannot find any sources confirming that this is justified. If the enzyme-inhibitor complex is very stable, I would think that the equilibrium above naturally is shifted toward the product, given a relatively fast kinetics. In the case of a commercial drug inhibiting the HIV-1 protease, for example, where it is safe to assume fast kinetics, I would model the strength of inhibition as the relative interaction energy: stronger interaction $\Rightarrow$ lower $K_i$ value.

The context is a simple theoretical modelling of a protein-ligand interaction using density functional theory.

• What do you mean by interaction energy? and what are $E_{Enzyme}$ and others? Free energy? – WYSIWYG Feb 2 '16 at 10:10

On one side of the spectrum you have competitive inhibition, which means that the $EI$ complex is functionally inactive, and cannot bind substrate. (The easiest way of getting competitive inhibition of an enzyme is to have the inhibitor bind in the active site, competing with the substrate binding. But that's not the only way. You can also have allosteric inhibitors which exhibit "competitive inhibition", if they bind the apo enzyme and by doing so prevent substrate binding.) This is the sort of inhibition you're assuming in your question, but isn't the only type of inhibition.
On the other side of the spectrum, you have uncompetitive inhibition. (N.B. uncompetitive - noncompetitive inhibition is something different.) This means that the inhibitor requires the substrate to be bound to the enzyme prior to binding itself, and it's the $ESI$ complex which is functionally inactive.
You're correct in that for pure competitive inhibition, the $K_i$ is equal to the $K_D$ of the inhibitor. The problem arises when you have one of those mixed effect or uncompetitive inhibitors, where catalytic steps and substrate concentrations get folded (partially) into the $K_i$.