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

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

Vmax is the maximal rate of reaction (k2 * Total enzyme concentration), Ki is the dissociation constant of the inhibitor (k-3/k3) and Km is (k-1+k2)/k1

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?

• @tomd point (2) was my mistake (typo). OP posted a picture (which was quite low-res). I edited the question to show reactions using LaTeX. Corrected it. – WYSIWYG Mar 3 '17 at 6:24
• Yup, sorry - I meant I got k-1/k1 * I/Ki rather than (k-1 + k3)/k1 * I/Ki. Also, yes, the system is describing a noncompetitive reaction scheme. Should I edit the question to correct these? – Bobbybobbobbo Mar 3 '17 at 16:36
• Actually, k-1/k1 is the Michaelis constant for a rapid equilibrium simulation of equation 1. Is that significant? – Bobbybobbobbo Mar 3 '17 at 16:39