Problem $1$:

The simple Michaelis$-$Menten model does not deal with all aspects of enzyme$-$catalyzed reactions. The

model must be modified to treat the phenomena of inhibition.

The inhibition process in general may be represented by the following six$-$step scheme, in which I is the

inhibitor, EI is a binary enzyme$-$inhibitor complex, and EIS is a ternary enzyme$-$inhibitor$-$substrate

complex.

$E+S{?}_{k_{-1}}^{k_1}ES$

$E+I{?}_{k_{-2}}^{k_2}EI$

$ES+I{?}_{k_{-3}}^{k_3}$ EIS

$EI+S{?}_{k_{-4}}^{k_4}$EIS

$ES{k}_{r1}E+P$

EIS$k_{r2}EI+P$

In steps $(1)$ and $(2),S$ and I compete for $($sites on$)$ E to form the binary complexes ES and EI$.$ In steps $(3)$

and $(4),$ the ternary complex EIS is formed from the binary complexes. In steps $(5)$ and $(6),$ ES and EIS

form the product $P$; if EIS is inactive, step $(6)$ is ignored.

Treatment of the full six$-$step kinetic scheme above with the PSSH leads to very cumbersome expressions

for $E$

$EI,$ etc., such that it would be better to use a numerical solution. However, these can be greatly

simplified if we assume $(1)$ the first four steps are at equilibrium, and $(2)k_{r1}=k_{r2}.$ With these

assumptions, show that:

$r_P=\frac{V_{\text{m}\text{a}\text{x}}[S]}{[S]+K_M\frac{1+\frac{\text{I}}{K_2}}{1+\frac{\text{I}}{K_3}}}$

where $K_2=\frac{k_{-2}}{k_2}($dissociation constant for EI$)$

and $,K_3=\frac{k_{-3}}{k_3}($dissociation constant for EIS$)$