Tutorials › Physiologically-Based Pharmacokinetics (PBPK) › Mixed and Noncompetitive Inhibition Models

Enzyme Inhibition and Induction · Tutorial 382 of 799

Mixed and Noncompetitive Inhibition Models

Derive the mixed-inhibition model and use its apparent kinetic parameters to interpret mixed and pure noncompetitive inhibition.

Advanced 10 min read

What You'll Learn

  • Define the inhibition constants for binding to free enzyme and enzyme–substrate complex.
  • Derive the mixed-inhibition rate equation from enzyme conservation.
  • Calculate apparent \(K_m\) and \(V_{\max}\) under mixed inhibition.
  • Identify pure noncompetitive inhibition as the special case of equal inhibitor affinity for both enzyme states.
  • Diagnose when an apparent-parameter interpretation may not fit the experimental mechanism.

Why Mixed Inhibition Changes Two Kinetic Parameters

An inhibitor that binds only free enzyme reduces the amount of enzyme available to bind substrate, producing the competitive pattern introduced in the earlier material on competitive inhibition. But some reversible inhibitors can bind both free enzyme and the enzyme–substrate complex. Binding to the complex can also remove enzyme from productive turnover. The relative strength of these two interactions determines whether inhibition changes the apparent \(K_m\), the apparent \(V_{\max}\), or both.

The standard mixed-inhibition model describes initial reaction rates at a fixed, known free inhibitor concentration. It assumes one substrate, reversible inhibitor binding, and an inactive inhibitor-bound enzyme state: neither \(EI\) nor \(ESI\) produces product. The enzyme–substrate complex \(ES\) produces product with the same catalytic rate constant as in the absence of inhibitor. The model also assumes that substrate and inhibitor concentrations do not change appreciably during the rate measurement, and that product accumulation and other reactions do not materially affect the measured initial rate.

Definition: In mixed inhibition, inhibitor \(I\) binds reversibly to free enzyme \(E\) and to enzyme–substrate complex \(ES\). Define \(K_i=[E][I]/[EI]\) and \(K_i'=[ES][I]/[ESI]\), where \(K_i\) and \(K_i'\) are the dissociation constants for the two binding interactions. Smaller values indicate stronger binding under the model. The inhibitor factors are \(\alpha=1+[I]/K_i\) and \(\alpha'=1+[I]/K_i'\), using the free inhibitor concentration \([I]\).

The unprimed constant refers to binding to free enzyme; the primed constant refers to binding to \(ES\). These labels matter: reversing them reverses the predicted effect on apparent \(K_m\). For this model, the rate equation is

$$ v=\frac{V_{\max}[S]}{\alpha K_m+\alpha'[S]} $$

Here \(v\) is the initial reaction rate, \([S]\) is substrate concentration, and \(V_{\max}\) and \(K_m\) are the corresponding parameters without inhibitor. Rearranging into Michaelis–Menten form shows which parameters are apparent:

$$ v=\frac{(V_{\max}/\alpha')[S]}{(\alpha/\alpha')K_m+[S]}, \qquad V_{\max,\mathrm{app}}=\frac{V_{\max}}{\alpha'}, \qquad K_{m,\mathrm{app}}=\frac{\alpha}{\alpha'}K_m $$

Thus, binding to \(ES\) lowers apparent \(V_{\max}\) as \(\alpha'\) increases. The apparent \(K_m\) depends on the ratio of the two inhibitor factors: it can increase, decrease, or remain unchanged. \(K_{m,\mathrm{app}}\) is a parameter describing the rate curve in the presence of inhibitor; it is not necessarily a direct measure of substrate-binding affinity.

Deriving the Rate Equation

A convenient derivation uses the usual steady-state Michaelis–Menten relationship for the uninhibited \(E\) and \(ES\) states, so that \([ES]=[E][S]/K_m\). Inhibitor binding is treated as reversible and at equilibrium for each enzyme state: \([EI]=[E][I]/K_i\) and \([ESI]=[ES][I]/K_i'\). The total enzyme concentration is the sum of all four enzyme forms:

$$ [E]_{\mathrm{T}}=[E]+[ES]+[EI]+[ESI] $$

Substitute the binding relationships and collect terms. This is the step that accounts for both ways inhibitor binding removes enzyme from productive turnover.

$$ \begin{aligned} [E]_{\mathrm{T}} &=[E]\left(1+\frac{[S]}{K_m}+\frac{[I]}{K_i} +\frac{[S]}{K_m}\frac{[I]}{K_i'}\right)\\ &=[E]\left(\alpha+\frac{[S]}{K_m}\alpha'\right) \end{aligned} $$

The rate is \(v=k_{\mathrm{cat}}[ES]\), with \(V_{\max}=k_{\mathrm{cat}}[E]_{\mathrm{T}}\). Since \([ES]=[E][S]/K_m\), replace \([E]\) using the enzyme-conservation equation:

$$ \begin{aligned} v &=k_{\mathrm{cat}}\frac{[E][S]}{K_m}\\ &=\frac{k_{\mathrm{cat}}[E]_{\mathrm{T}}[S]} {K_m\alpha+\alpha'[S]}\\ &=\frac{V_{\max}[S]}{K_m\alpha+\alpha'[S]} \end{aligned} $$

The result follows under the stated assumptions. In particular, it is not a general equation for every inhibitor that can interact with multiple enzyme states. If inhibitor-bound enzyme retains catalytic activity, if binding is effectively irreversible during the assay, or if multiple substrates or enzyme states are important, this derivation does not describe the full mechanism.

Worked Example: Mixed Inhibition Raises Apparent \(K_m\)

Suppose a hypothetical enzyme has \(V_{\max}=120\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\) and \(K_m=4.0\ \mathrm{\mu M}\). A reversible inhibitor is present at free concentration \([I]=6.0\ \mathrm{\mu M}\), with \(K_i=3.0\ \mathrm{\mu M}\) and \(K_i'=6.0\ \mathrm{\mu M}\). Calculate the apparent parameters and the rate at \([S]=6.0\ \mathrm{\mu M}\).

First calculate the inhibitor factors:

$$ \alpha=1+\frac{6.0}{3.0}=3.0, \qquad \alpha'=1+\frac{6.0}{6.0}=2.0 $$

Therefore,

$$ V_{\max,\mathrm{app}}=\frac{120}{2.0} =60\ \mathrm{nmol\,min^{-1}\,mg^{-1}}, \qquad K_{m,\mathrm{app}}=\frac{3.0}{2.0}(4.0) =6.0\ \mathrm{\mu M} $$

At \([S]=6.0\ \mathrm{\mu M}\), substitute into the inhibited rate equation:

$$ v=\frac{120(6.0)}{3.0(4.0)+2.0(6.0)} =\frac{720}{24}=30\ \mathrm{nmol\,min^{-1}\,mg^{-1}} $$

For comparison, the uninhibited rate at the same substrate concentration is \(120(6.0)/(4.0+6.0)=72\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\). In this example, inhibitor binding to both enzyme forms lowers apparent \(V_{\max}\), while stronger binding to free enzyme relative to \(ES\) makes apparent \(K_m\) larger.

Pure Noncompetitive Inhibition

Pure noncompetitive inhibition is a special case of mixed inhibition, not a separate binding scheme. It occurs when the inhibitor has equal affinity for free enzyme and enzyme–substrate complex: \(K_i=K_i'\). Consequently, \(\alpha=\alpha'\), and their ratio in the apparent \(K_m\) expression is one.

Definition: In the pure noncompetitive model, \(K_i=K_i'\). The inhibitor lowers apparent \(V_{\max}\) but leaves apparent \(K_m\) unchanged: \(V_{\max,\mathrm{app}}=V_{\max}/\alpha\) and \(K_{m,\mathrm{app}}=K_m\).

Worked Example: Pure Noncompetitive Inhibition

Consider a hypothetical assay with \(V_{\max}=90\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\), \(K_m=5.0\ \mathrm{\mu M}\), and free inhibitor concentration \([I]=8.0\ \mathrm{\mu M}\). Suppose \(K_i=K_i'=4.0\ \mathrm{\mu M}\). Find the apparent parameters and rate at \([S]=5.0\ \mathrm{\mu M}\).

The equal dissociation constants give equal inhibitor factors:

$$ \alpha=\alpha'=1+\frac{8.0}{4.0}=3.0 $$

Thus,

$$ V_{\max,\mathrm{app}}=\frac{90}{3.0} =30\ \mathrm{nmol\,min^{-1}\,mg^{-1}}, \qquad K_{m,\mathrm{app}}=\frac{3.0}{3.0}(5.0) =5.0\ \mathrm{\mu M} $$

The inhibited rate at \([S]=5.0\ \mathrm{\mu M}\) is

$$ v=\frac{30(5.0)}{5.0+5.0} =15\ \mathrm{nmol\,min^{-1}\,mg^{-1}} $$

Without inhibitor, the rate at this substrate concentration would be \(90(5.0)/(5.0+5.0)=45\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\). The rate falls, but the substrate concentration giving half of the inhibited \(V_{\max}\) remains \(5.0\ \mathrm{\mu M}\).

How the Relative Affinities Shape the Curve

The ratio \(\alpha/\alpha'\) determines the direction of the apparent \(K_m\) change. If \(\alpha>\alpha'\), the inhibitor has a larger effect on free enzyme than on \(ES\), and \(K_{m,\mathrm{app}}>K_m\). If \(\alpha<\alpha'\), its effect on \(ES\) is larger, and \(K_{m,\mathrm{app}}<K_m\). Equal factors give the pure noncompetitive case. In all of these cases, provided inhibitor binds to \(ES\) so that \(\alpha'>1\), apparent \(V_{\max}\) is below its uninhibited value.

Worked Example: Stronger Binding to the Enzyme–Substrate Complex

A hypothetical enzyme has \(V_{\max}=200\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\) and \(K_m=10.0\ \mathrm{\mu M}\). The free inhibitor concentration is \(4.0\ \mathrm{\mu M}\), \(K_i=8.0\ \mathrm{\mu M}\), and \(K_i'=2.0\ \mathrm{\mu M}\). Calculate the apparent parameters and the inhibited rate at \([S]=5.0\ \mathrm{\mu M}\).

The inhibitor factors are

$$ \alpha=1+\frac{4.0}{8.0}=1.5, \qquad \alpha'=1+\frac{4.0}{2.0}=3.0 $$

Then

$$ V_{\max,\mathrm{app}}=\frac{200}{3.0} =66.67\ \mathrm{nmol\,min^{-1}\,mg^{-1}}, \qquad K_{m,\mathrm{app}}=\frac{1.5}{3.0}(10.0) =5.0\ \mathrm{\mu M} $$

The rate at \([S]=5.0\ \mathrm{\mu M}\) is

$$ v=\frac{200(5.0)}{1.5(10.0)+3.0(5.0)} =\frac{1000}{30} =33.33\ \mathrm{nmol\,min^{-1}\,mg^{-1}} $$

Here the apparent \(K_m\) decreases because the inhibitor binds more strongly to \(ES\) than to free enzyme. This decrease does not mean the inhibitor stimulates catalysis: the apparent maximum rate falls from \(200\) to \(66.67\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\).

Comparing Inhibition Patterns and Checking Model Fit

The apparent-parameter equations help distinguish common reversible inhibition patterns. Competitive inhibition is the limit in which inhibitor binding to \(ES\) is absent, so \(\alpha'=1\): apparent \(V_{\max}\) is unchanged and apparent \(K_m\) increases. In uncompetitive inhibition, inhibitor binding to free enzyme is absent, so \(\alpha=1\): both apparent \(V_{\max}\) and apparent \(K_m\) decrease by the same factor. Mixed inhibition permits binding to both states; pure noncompetitive inhibition is the special mixed case with equal affinities.

ModelInhibitor binding patternApparent \(V_{\max}\)Apparent \(K_m\)
CompetitiveFree enzyme onlyUnchangedIncreases
UncompetitiveEnzyme–substrate complex onlyDecreasesDecreases by the same factor
MixedFree enzyme and enzyme–substrate complex, with unequal affinitiesDecreasesMay increase or decrease
Pure noncompetitiveBoth states, with equal affinitiesDecreasesUnchanged

A useful diagnostic is to measure initial rates across a range of substrate concentrations at several inhibitor concentrations, then fit the rate equation directly. Under this model, increasing inhibitor concentration should increase \(\alpha\) and \(\alpha'\) and change the rate curve in a way that can be summarized by the two apparent parameters. A systematic mismatch between observed and fitted rates, or apparent parameters that cannot be described consistently across inhibitor concentrations, is evidence that the model assumptions may not hold. Examples include time-dependent loss of active enzyme, inhibitor depletion, multiple enzyme forms, or an inhibitor that is not at rapid binding equilibrium during the assay.

The kinetic pattern is evidence about a mechanism, not proof of one by itself. Experimental noise and a narrow substrate range can make different models difficult to distinguish. Nonlinear fitting of the original rate equation is generally preferable to relying on reciprocal transformations, which alter the error structure and can give disproportionate influence to measurements at low substrate concentration. When evaluating a fitted model, inspect whether the residuals show systematic trends and whether the parameter estimates are plausible and sufficiently precise.

Common Mistake and Practical Limits

  • Wrong reasoning: “A decrease in apparent \(K_m\) means the inhibitor increases substrate affinity or activates the enzyme.” Correction: \(K_{m,\mathrm{app}}=(\alpha/\alpha')K_m\) is a ratio-dependent feature of the inhibited rate curve. In mixed inhibition, it can decrease even while \(V_{\max,\mathrm{app}}\) also decreases.
  • Wrong reasoning: “Noncompetitive inhibition always leaves \(K_m\) unchanged.” Correction: That statement applies to pure noncompetitive inhibition, where \(K_i=K_i'\). General mixed inhibition changes apparent \(K_m\) unless the two factors happen to be equal.
  • Wrong reasoning: “The inhibitor concentration in the assay is automatically the \([I]\) in the equations.” Correction: The equations use free inhibitor concentration. If the free concentration is not known or differs substantially from the nominal concentration, calculated inhibitor factors may be misleading.

These equations describe steady initial-rate behavior for a specified, reversible binding model. They do not by themselves account for changing inhibitor exposure, enzyme inactivation, induction, or other time-dependent mechanisms. They are most useful when the assay conditions support the assumptions and the observed rates across substrate and inhibitor concentrations are compatible with the model.

Check Your Understanding

Use the definitions and equations above. Each question is followed by its solution.

  1. If \(K_i=K_i'\), which apparent parameter remains unchanged, and why?
  2. Suppose \(\alpha=4\), \(\alpha'=2\), \(K_m=3.0\ \mathrm{\mu M}\), and \(V_{\max}=80\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\). Find both apparent parameters.
  3. In a mixed-inhibition model, what does \(\alpha<\alpha'\) imply about \(K_{m,\mathrm{app}}\) relative to \(K_m\)?
  4. Why can a lower apparent \(K_m\) occur without an increase in catalytic capacity?
  5. What concentration of inhibitor belongs in the definitions of \(\alpha\) and \(\alpha'\)?

Solutions. (1) Apparent \(K_m\) remains unchanged because \(\alpha/\alpha'=1\); apparent \(V_{\max}\) decreases if the inhibitor binds. (2) \(V_{\max,\mathrm{app}}=80/2=40\ \mathrm{nmol\,min^{-1}\,mg^{-1}}\), and \(K_{m,\mathrm{app}}=(4/2)(3.0)=6.0\ \mathrm{\mu M}\). (3) \(K_{m,\mathrm{app}}=(\alpha/\alpha')K_m<K_m\). (4) The decrease results from the ratio of inhibitor effects on free enzyme and \(ES\); it does not indicate an increase in \(V_{\max}\), which is reduced by the factor \(\alpha'\). (5) Use the free inhibitor concentration, not automatically the nominal total concentration added to the assay.