Structural Biochemistry/Enzyme/Noncompetitive Inhibitor

General information
Noncompetitive inhibitor can bind to an enzyme with or without a substrate at different places at the same time. It changes the conformation of an enzyme, but it does not change the efficiency of binding or the Km. A noncompetitive inhibitor binds to the enzyme away from the active site, altering the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively. Most of the time, the inhibitor is reversible. However, this inhibition decreases the turnover number, meaning the rate of reaction decreases. As the inhibitor binds to the enzyme and the enzyme-substrate complex, it reduces the concentration of enzyme available for proper catalysis. Fewer functional enzymes leads to fewer available active sites and thus a smaller Vmax. Unlike competitive inhibition, raising [S] (substrate concentration) is pointless with noncompetitive inhibition. A noncompetitive inhibitor binds to a different site that is not the active site of the enzyme and changes the structure of the enzyme; therefore, it blocks the enzyme from converting substrate to product, but it still allows the substrate to bind. Thus, it decreases the rate of the chemical reaction of enzyme and substrate, which can not be changed by increasing concentration of substrate; the binding decreases Vmax and has no change on the Km of the chemical reaction.

E + I − (through a substrate) → ES + I → E + P

ES + I ⇌ ESI → NR (no reaction)

where E is enzyme, I is inhibitor, ES is enzyme-substrate complex, P is product. ESI is the molecule after the inhibitor is bound to the enzyme-substrate complex. ESI cannot form any products, so the later reaction is not allowed (or, no reaction).





Based on the Michaelis-Menten Model, KM, the concentration of the substrate when the velocity is the half of the maximum velocity (or half of the substrates at maximum velocity), remains same, but the maximum velocity is decreased.



The picture shows a double-reciprocal plot of V0 and [S]. The x-intercept is equal to -1/Km while the y-intercept is 1/Vmax. The slope of the line is Km/Vmax. Thus, the plot shows that there is no change in Km and Vmax is decreased. In relation to the original plot, the x intercept stays constant while the y intercept increases along with the slope.