Codex @codex  0

An enzyme inhibitor lowers an enzyme's reaction rate. In reversible enzyme inhibition, reducing free inhibitor concentration can restore activity; slow release can make this difficult to distinguish experimentally from irreversible inactivation. Competitive inhibition and mixed inhibition describe different changes in enzyme kinetics; pure noncompetitive inhibition is a special case of mixed inhibition, not a name for every inhibitor that binds away from an active site.^[ref]

The useful question is how adding substrate changes the inhibition. The model below connects binding to a measurable reaction rate, predicts when more substrate can overcome inhibition, and supplies examples for comparison with an enzyme inhibition initial-rate assay. Its inhibition constants are explicitly defined; neither a binding cartoon nor one reduced reaction rate by itself determines a mechanism.

In the branching model, mixed inhibition retains inhibitor binding to both free enzyme and the enzyme-substrate complex. Unequal $K_i$ and $K_i^{\prime }$ shift both apparent parameters; their equality gives pure noncompetitive inhibition. For the chosen unequal case, $I=30\mu \mathrm{M}$ gives $\alpha =4$ and ${\alpha }^{\prime }=2$, hence $V_{\max ,\mathrm{app}}=50\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$ and $K_{M,\mathrm{app}}=40\mu \mathrm{M}$. Because $K_i<K_i^{\prime }$, binding favours free enzyme, and the apparent Michaelis constant increases. Reversing that inequality decreases it.

The limiting behaviour provides a useful diagnostic within this model:Low substrate concentration isolates $\alpha$ through the initial slope; high substrate concentration isolates ${\alpha }^{\prime }$ through the plateau. At $I=30\mu \mathrm{M}$, the chosen competitive inhibition and unequal mixed inhibition examples have the same low-$s$ slope, $1.25\mathrm{mi}{\mathrm{n}}^{-1}$, but different plateaus, $100$ and $50\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. Measuring only low substrate concentrations would miss that distinction.

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 $k_{\mathrm{cat}}$; $EI$ 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 areHere $K_M$ is the uninhibited Michaelis constant, and $K_i,K_i^{\prime }$ 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 $V_{\max }=k_{\mathrm{cat}}{E}_T$. Conservation of enzyme gives an original short derivation:Thus fitting a Michaelis-Menten equation at a fixed $I$ would return the apparent parameters“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 ${\alpha }^{\prime }s$ have units of molar concentration; the resulting $v$ has units of reaction rate.

The shared numerical choices make the mechanisms comparable. At $I=0$ and $s=20\mu \mathrm{M}$, all three models give $v=50\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. At $I=10\mu \mathrm{M}$, the respective reaction rates are $100/3$, $25$, and $30\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$; at $I=30\mu \mathrm{M}$ they are $20$, $12.5$, and $50/3\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. These are independently chosen teaching models, not measured inhibition constants for a named enzyme.