Question d800d

1 Answer
Apr 7, 2015

${V}_{\text{max}}$ is unaffected by competitive inhibition.

A Lineweaver-Burk plot can distinguish among competitive, non-competitive, and uncompetitive enzyme inhibition.

1/V = (K_"m" + [S])/(V_"max"[S]) = K_"m"/V_"max" 1/([S]) + 1/V_"max"#

Competitive Inhibition

A competitive inhibitor has a similar shape to the substrate. It binds reversibly at the active site, where it competes with substrate for binding.

${V}_{\text{max}}$ does not change, so $\frac{1}{V} _ \text{max}$ is also unchanged.

But ${K}_{\text{m}}$ changes, so there are different slopes and $x$-intercepts for the two processes.

Non-competitive inhibition

The shape of a non-competitive inhibitor is different from that of the substrate.

It binds to another site on the enzyme or enzyme-substrate and prevents the formation of products.

It causes a decrease in ${V}_{\text{max}}$ but doesn't change ${K}_{\text{m}}$, so the plot has the same $x$-intercept as the uninhibited reaction, but a different slope.

Uncompetitive inhibition

The shape of an uncompetitive inhibitor is different from that of the substrate.

It cannot bind to the active site. It can bind to the enzyme-substrate complex only after it has been formed.

This reduces both ${V}_{\text{max}}$ and ${K}_{\text{m}}$ by the same amount, so the slope does not change but the $x$-intercept does.

The lines end up parallel to each other.