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Uncompetitive inhibition, also known as anti-competitive inhibition, takes place when an enzyme inhibitor binds only to the complex formed between the enzyme and the substrate (the E-S complex).

While uncompetitive inhibition requires that an enzyme-substrate complex must be formed, non-competitive inhibition can occur with or without the substrate present.

Uncompetitive inhibition is distinguished from competitive inhibition by two observations: first uncompetitive inhibition cannot be reversed by increasing [S] and second as shown in the Lineweaver–Burk plot yields parallel rather than intersecting lines.

This behavior is found in the inhibition of acetylcholinesterase by tertiary amines (R3N). Such compounds bind to the enzyme in its various forms, but the acyl-intermediate-amine complex cannot break down into enzyme plus product.

Mechanism
This reduction in the effective concentration of the E-S complex increases the enzyme's apparent affinity for the substrate through Le Chatelier's principle (Km is lowered) and decreases the maximum enzyme activity (Vmax), as it takes longer for the substrate or product to leave the active site. Uncompetitive inhibition works best when substrate concentration is high. An uncompetitive inhibitor need not resemble the substrate of the reaction it is inhibiting.

Mathematical definition
The Lineweaver–Burk equation states that:
 * $$\ \frac{1}{v}=\frac{K_m}{V_{max}[S]} + {1 \over V_\max}$$

Where v is the initial reaction velocity, Km is the Michaelis–Menten constant, Vmax is the maximum reaction velocity, and [S] is the concentration of the substrate.

The Lineweaver–Burk plot for an uncompetitive inhibitor produces a line parallel to the original enzyme-substrate plot, but with a higher y-intercept, due to the presence of an inhibition term $$\ \frac{[I]}{K_i} $$:
 * $$\ \frac{1}{v}=\frac{K_m}{V_{max}[S]}+\frac{1+\frac{[I]}{K_i}}{V_{max}}$$

Where [I] is the concentration of the inhibitor and Ki is an inhibition constant characteristic of the inhibitor.

The Michaelis-Menten equation is altered to:
 * $$\ {V_0}=\frac{V_{max}[S]}{K_m+\alpha'[S]}$$

where
 * $$\ \alpha'=1+\frac{[I]}{K'_I}$$ and $$\ K'=\frac{[ES][I]}{[ESI]}$$

As described by the equation above, at high concentrations of substrate, V0 approaches Vmax/α'. Thus, an uncompetitive inhibitor lowers the measured Vmax. Apparent Km also decreases, because [S] required to reach one-half Vmax decreases by the factor α'.