How to derive rate of product formation for an allosterically inhibited enzyme?

Question

I'm working through Mathematical Modelling in Systems Biology: An Introduction by Brian Ingalls, and in Chapter 3 there's an example of a simple allosterically inhibited enzyme with reaction scheme shown here:

$$\ce{S + E<=>[k_1][k_{-1}] ES->[k_2] E + P}\\[2em] \ce{I + E<=>[k_3][k_{-3}] EI}\\[2em] \ce{I + ES<=>[k_3][k_{-3}] ESI}\\[2em] \ce{S + EI<=>[k_1][k_{-1}] ESI} $$

where E is the free enzyme, I is the allosteric inhibitor, S is the substrate, P is the product, and ES, ESI, EI etc. are the various complexes between them. The book goes on to state that the rate of formation of P is

$$\frac{V_{max}}{1+I/K_I}\frac{S}{K_M +S}$$

where

$V_\text{max}$ is the maximal rate of reaction ($k_2$ * Total enzyme concentration), $K_I$ is the dissociation constant of the inhibitor $k_{-3}/k_3$ and $K_M$ is $(k_{-1}+k_2)/k_1$

No proof for this is shown, although the book says the solution is assuming a quasi-steady state of the complexes ES, EI and ESI. I've attempted to work it out, and arrived at almost, the same equation, except I have I/Ki * (k-1+k3)/k1 instead of I/Ki * KM. I have no idea how to get to the equation present by the book. Can someone explain how their equation was derived?

Answer 1

If you apply quasi-steady state approximation for the complexes then you will get three equations for each of the complexes. Then you have to use the conservation law for enzyme to obtain the equation with respect to just substrate and inhibitor. If you try solving these algebraic equations, it gets very complex and difficult to solve by hand. However, in this system of chemical reactions, you can notice one thing: the binding of the inhibitor is not affected by the binding of substrate. In other words, the amount of enzyme inhibited will not depend on the amount of substrate present. So, in the absence of substrate: $$\frac{d}{dt}(EI)=k_3I.E - k_{-3}EI = 0$$

So the total uninhibited enzyme ($E_u$) would be: $$E_u=\frac{E_0}{1+\frac{I}{K_I}}$$

When the substrate is present, the amount of total enzyme available for catalysis would be $E_u$. Now, apply the Michaelis-Menten equation for uninhibited enzyme but substitute total enzyme ($E_0$) with total available enzyme ($E_u$):

$$V=\frac{k_2S\left(\frac{E_0}{1+\frac{I}{K_I}}\right)}{K_M+S}$$

From here you can get the equation given in the book.

The above logic works only if the rate of inhibitor binding is much faster than that of substrate binding. If not, the dynamics of product formation will depend on the different reaction rates and this simplistic equation would not be valid.