3
$\begingroup$

Suppose there is a solution of enzyme, its substrate, a competitive inhibitor, and a suicide inhibitor. The reaction rate constant for the reaction of the suicide inhibitor with the enzyme is known. The Ki for the competitive inhibitor is also known. The occupation of the enzyme's active site by the competitive inhibitor prevents deactivation of the enzyme by the suicide inhibitor. The initial concentrations of the inhibitors are also known. How would the presence of the competitive inhibitor affect the observed reaction rate constant for the inactivation of the enzyme by the suicide inhibitor in the following cases:

  1. When the enzyme has been inactivated by the suicide inhibitor, the competitive inhibitor cannot enter the enzyme's active site.
  2. When the enzyme has been inactivated by the suicide inhibitor, a molecule of the competitive inhibitor can still enter the enzyme's active.
  3. A molecule of the competitive inhibitor can still enter an inactivated enzyme's active site, but at a lower probability compared to that of the active enzyme.

Is there an equation to determine the observed inhibition reaction rate constant based on the actual rate constant, the Ki of the competitive inhibitor, and the concentrations of the enzyme and the inhibitors.

$\endgroup$
2
  • $\begingroup$ Hello! Still interested in an answer, @user73910 ? $\endgroup$ Aug 14, 2023 at 12:57
  • $\begingroup$ @MetalStorm yes $\endgroup$
    – user73910
    Aug 15, 2023 at 5:05

1 Answer 1

1
$\begingroup$

Yes, there are many such models you could make.

A general, difficult approach

You could make a very general model that has lots and lots of unknown parameters, and try to numerically fit observed kinetics data to this "full" model. But you'll probably need a lot of data and you'll need an accurate model. Fitting large (>=3 or 4) parameters by numerical integration of ODEs is not fun; you need to carefully consider which of your parameters are independent. You'd probably want to non-dimensionalize your ODE model to be sure the parameterization you have chosen is actually fully identifiable from your data.

A specific approach, with lots more assumptions

You could also make much simpler models by making some simplifying assumptions.

$$\ce{E + S <=>[K_d] (ES)}$$ $$\ce{(ES) ->[k_f] E + P}$$ $$\ce{E + I <=>[K_i] (EI)}$$ $$\ce{E + C ->[k_{cov}] C}$$

One of the simplest sets of assumptions might be:

  1. Covalent inhibition by covalent inhibitor $\ce{C}$ is slow relative to the equilibration of free enzyme with substrate and with non-covalent, reversible inhibitor, and also slower than the reaction of (ES) to form product.

  2. Non-covalent reversible inhibition by $\ce{I}$ happens only to free enzyme. Once an enzyme is inactivated by $\ce{C}$, nothing else can happen to it. (This violates your third bullet point in your question, but the goal here is simplicity.)

  3. Usual quasi-equilibrium assumptions about $\ce{ES}$. These can often be replaced by less stringent steady-state approximations about $\ce{ES}$, but at the expense of harder math.

  4. You are using a large excess of covalent inhibitor relative to the initial total amount of enzyme.

  5. Usual assumptions about a large excess of substrate relative to initial total enzyme.

I think with these assumptions, you can write a rate law for the formation of product like:

$\frac{d P}{d t} = k_f E_{tot} \frac{1}{(1 +\frac{I}{K_i})} \frac{S}{(K_M + S)}$

This is the usual equation for competitive inhibition, but now $E_{tot}$ is a time-dependent variable that has first-order depletion kinetics. So you could substitute in:

$\frac{d P}{d t} = k_f E_0 e^{-k_{cov}t} \frac{S}{(\frac{I}{K_i} + 1)K_d + S}$

This equation is itself already complicated enough! But it is a starting point. By going through the math of competitive inhibition you could see if you could relax some of the assumptions I've made. The hardest one to relax will be #1; if covalent inhibition is not slow relative to all other reaction steps, everything gets considerably more complicated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.