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 $k_{\mathrm{cat}}$; $EI$ 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.

An inhibition constant sets an inhibitor concentration scale in an enzyme kinetics model. In the rapid-binding construction under reversible enzyme inhibition, $K_i=[E]I/[EI]$ and $K_i^{\prime }=[ES]I/[ESI]$ are dissociation constants. Smaller values favour the corresponding bound population at fixed free inhibitor concentration. More generally, fitted inhibition constants need not be true binding dissociation constants; a mechanism must justify that interpretation.^[ref]

For example, choosing $I=K_i$ makes $[EI]=[E]$ in this model, but does not imply that half of all enzyme is $EI$: $ES$ and $ESI$ also contribute to the total population.

In the model of reversible enzyme inhibition, competitive inhibition makes substrate binding and inhibitor binding mutually exclusive: $EI$ is allowed but $ESI$ is absent. Set ${\alpha }^{\prime }=1$, equivalently take $K_i^{\prime }\to \infty$. ThenFor the chosen $I=30\mu \mathrm{M}$, $\alpha =4$: the half-limiting substrate concentration rises from $20$ to $80\mu \mathrm{M}$, while the limiting reaction rate remains $100\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. At $s=200\mu \mathrm{M}$ the inhibited reaction rate is $500/7\approx 71.43\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, compared with the uninhibited $1000/11\approx 90.91\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. “Overcome by substrate” means that their ratio tends to one as $s\to \infty$, not that any finite addition removes the difference.

A competitive inhibitor is an enzyme inhibitor that produces competitive inhibition relative to a specified substrate. In the usual single-substrate kinetic description, it reduces the low-substrate response while leaving the limiting rate unchanged. A common molecular model has the inhibitor bind free enzyme and prevent formation of the enzyme-substrate complex; the kinetic classification alone does not prove a particular binding location.^[ref]

For the simple reversible model at fixed free inhibitor concentration $[I]$,Here $K_i$ is the inhibition constant for this model and $K_M$ the uninhibited Michaelis constant. Take illustrative values $V_{\max }=12\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, $K_M=2\mu \mathrm{M}$, $K_i=5\mu \mathrm{M}$ and $[I]=10\mu \mathrm{M}$. Then $\alpha =3$ and the apparent Michaelis constant is6micromolar. At $[S]=2\mu \mathrm{M}$, the inhibited rate is3micromolar per minute, compared with6without inhibitor.

As substrate concentration grows, both rates approach the same limiting value in this model. That is a limit, not an assertion that a particular finite substrate addition removes all inhibition. Characterizing a candidate therefore requires rates over suitable substrate and inhibitor ranges with controlled enzyme amount; one reduced rate does not establish that the inhibitor is competitive.^[ref]

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.

Set $K_i=K_i^{\prime }$, so that $\alpha ={\alpha }^{\prime }$. The reversible enzyme inhibition equation reduces toAt the chosen $I=30\mu \mathrm{M}$, every reaction rate is one quarter of its uninhibited value. The plateau falls to $25\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, but half that plateau still occurs at $s=20\mu \mathrm{M}$. Increasing substrate cannot restore the uninhibited plateau. An unchanged apparent Michaelis constant here expresses equality of the two binding effects, rather than an absence of inhibitor binding to the enzyme-substrate complex.

The complementary limit of the branching model permits inhibitor binding only to the enzyme-substrate complex: set $K_i\to \infty$, hence $\alpha =1$, with finite $K_i^{\prime }$. Then both $V_{\max ,\mathrm{app}}$ and $K_{M,\mathrm{app}}$ decrease by ${\alpha }^{\prime }$, while their ratio stays fixed. This limit helps distinguish preferential binding to $ES$ in mixed inhibition from exclusive binding to $ES$ in uncompetitive inhibition.