Mixed inhibition

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.