Question

Question #d800d

Answer 1

`V_"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_"max"` does not change, so `1/V_"max"` is also unchanged.

But `K_"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_"max"` but doesn't change `K_"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_"max"` and `K_"m"` by the same amount, so the slope does not change but the `x`-intercept does.

The lines end up parallel to each other.