What a competitive inhibitor really does to Km
Two inhibitors, two different lies your intuition tells you. The Michaelis-Menten curve settles the argument.
A competitive inhibitor is in the tube. What happens to Vmax and Km?
Vmax is unchanged. Km rises. The inhibitor competes for the active site, but it doesn't cap the top speed.
That can't be right, if something is blocking the active site, surely the maximum rate has to drop?
That's the intuition the exam preys on. A competitive inhibitor only sits where substrate would. Flood the tube with enough substrate and it outcompetes the inhibitor at every site, so the enzyme still reaches its original Vmax. It just takes more substrate to get there, which is precisely what a higher apparent Km means.
Define Km for me, since the whole answer hinges on it.
Km is the substrate concentration at which velocity is half of Vmax. It's an inverse proxy for affinity, low Km, tight binding. It is not a speed, which is the second trap people fall into.
Now swap in a pure noncompetitive inhibitor. Same two questions.
Mirror image. Vmax falls, Km is unchanged. It binds an allosteric site whether or not substrate is present, so you can't out-add it. Fewer functional enzymes, lower ceiling, but the ones still working bind substrate exactly as before.
I hand you a Lineweaver-Burk plot with no labels. How do you tell the two apart?
Look at the intercepts. Competitive shares the y-intercept, same 1/Vmax, with a steeper slope. Noncompetitive shares the x-intercept, same −1/Km, but lifts the y-intercept because Vmax dropped. The point they pivot around tells you which quantity was spared.
↑ answer it in your head first ↑
Traps
- ⚠ Thinking a competitive inhibitor lowers Vmax. It doesn't, enough substrate outcompetes it and Vmax is restored.
- ⚠ Reading Km as "how fast the enzyme goes." Km is a substrate concentration, an inverse proxy for affinity.
- ⚠ Assuming all inhibitors can be overcome by adding substrate. Only competitive inhibition is surmountable that way.