= Reversible enzyme inhibition
Construct a minimal model with one <substrate> and no appreciable accumulated <reaction product>. Only the <enzyme-substrate complex> $ES$ produces product, with catalytic <rate constant> $k_{\rm cat}$; $EI$ and $ESI$ are inactive. Write $s=[S]$ and $I=[\mathrm I]$ for free <molar concentrations>, and $E_T$ for total <enzyme> <molar concentration>. Assume a suitable <quasi-steady-state approximation>, rapid inhibitor binding and release relative to turnover, and negligible depletion of free inhibitor. In this construction the population relations are
$$
[ES]=[E]\frac{s}{K_M},\qquad
[EI]=[E]\frac{I}{K_i},\qquad
[ESI]=[ES]\frac{I}{K_i'},\qquad
\alpha=1+\frac{I}{K_i},\quad \alpha'=1+\frac{I}{K_i'}.
$$
Here $K_M$ is the uninhibited <Michaelis constant>, and $K_i,K_i'$ are the <inhibition constants> for the two indicated binding branches. The symbol $I$ denotes a free <concentration>, not an amount added to the tube. Define $V_{\max}=k_{\rm cat}E_T$. Conservation of <enzyme> gives an original short derivation:
$$
\begin{aligned}
E_T&=[E]+[EI]+[ES]+[ESI]
=[E]\left(\alpha+\alpha'\frac{s}{K_M}\right),\\
v&=k_{\rm cat}[ES]
=\frac{V_{\max}s}{\alpha K_M+\alpha's}
=\frac{(V_{\max}/\alpha')s}{(\alpha/\alpha')K_M+s}.
\end{aligned}
$$
Thus fitting a <Michaelis-Menten equation> at a fixed $I$ would return the apparent parameters
$$
V_{\max,\rm app}=\frac{V_{\max}}{\alpha'},\qquad
K_{M,\rm app}=\frac{\alpha}{\alpha'}K_M.
$$
“Apparent” means conditional on that inhibitor <concentration>. The uninhibited <Michaelis constant> has not been redefined. <Dimensional analysis> checks the denominator: both $\alpha K_M$ and $\alpha's$ have units of <molar concentration>; the resulting $v$ has units of <reaction rate>.
\Image[/enzyme-inhibition-rate-curves.png]
{title=Chosen <competitive inhibition>, <pure noncompetitive inhibition> and <mixed inhibition> curves}
{description=Original calculations, not measured rates or fits to <trypsin>. Every panel uses $V_{\max}=100\,\mathrm{\mu M\,min^{-1}}$, $K_M=20\,\mathrm{\mu M}$ and $K_i=10\,\mathrm{\mu M}$, with free $I=0,10,30\,\mathrm{\mu M}$. <Competitive inhibition> uses $\alpha'=1$; <pure noncompetitive inhibition> uses $K_i'=10\,\mathrm{\mu M}$; the unequal-constant <mixed inhibition> example uses $K_i'=30\,\mathrm{\mu M}$. Dots mark $s=20\,\mathrm{\mu M}$. Dotted horizontal lines are limiting rates, approached rather than reached at finite $s$.}
The shared numerical choices make the mechanisms comparable. At $I=0$ and $s=20\,\mathrm{\mu M}$, all three models give $v=50\,\mathrm{\mu M\,min^{-1}}$. At $I=10\,\mathrm{\mu M}$, the respective <reaction rates> are $100/3$, $25$, and $30\,\mathrm{\mu M\,min^{-1}}$; at $I=30\,\mathrm{\mu M}$ they are $20$, $12.5$, and $50/3\,\mathrm{\mu M\,min^{-1}}$. These are independently chosen teaching models, not measured <inhibition constants> for a named <enzyme>.
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