Vmax and Km in competitive versus non-competitive inhibition

[medicalschoolislife]

Hi everyone,

I was having trouble understanding the Kaplan explanation for why Vmax is unchanged in competitive inhibition, but it decreases with non-competitive inhibition.

I searched on Google, and found this explanation:

"Vmax is the maximum velocity of the enzyme. Competitive inhibitors can only bind to E and not to ES. They increase Km by interfering with the binding of the substrate, but they do not affect Vmax because the inhibitor does not change the catalysis in ES because it cannot bind to ES."

If there is only a limited amount of enzyme, then wouldn't taking up a bunch of the active sites by the competitive inhibitor cause Vmax to decrease? This would happen since there's less enzyme to bind the actual substrate.

In non-competitive inhibition, the inhibitor can bind to enzyme-substrate complexes, so I understand why Vmax decreases there. But I don't see why the same wouldn't happen for competitive inhibition, since a certain amount of the enzyme is also out of commission by being inhibited by the competitive inhibitor.

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[Zenabi90]

Great question!

I tend to think about things like this in terms of food.

Lay out a buffet table: 10 plates of food laid out. You and your crew arrive, 10 deep, and y'all starving. So y'all grab one at random and get all the food you want ASAP. You get to eat a plate each, at your maximum velocity or Vmax. 10 plates of food knocked down per meal no problem.

Tomorrow, same deal BUT 20 plates of food laid out, and 10 of them are fake and plastic. You and your crew arrive, 10 deep, hungry again. Y'all grab one at random and try to eat. Some of you ended up with the plastic food, so when you sit down to eat, bam disappointment. Those guys gotta get up, get other plates of food, come back and sit down again to eat. Maybe it happens again. Bad luck. The point is it took MORE plates and more time to get real food, so your Km went up! But each of y'all still managed to combine for 10 plates of food per meal, so Vmax isn't changed.

Next day, same deal. But now we have 100 plates of food, and only 10 of them are fake and plastic. Basically we pumped a ton of substrates in. You and your crew arrive, 10 deep, always hungry. Y'all grab one at random and try to eat. This time it takes much less attempts to find real food. You've maxed out the substrate to overcome the competition provided by the plastic food. You all managed to get real food, and your crew puts away 10 plates of food again. Vmax still the same cuz y'all can still put away 10 plates of food per meal, and because with a high enough substrate concentration, the inhibition provided by a competitive inhibitor can be overcome. The point is even if you got fake food, you can still go back and get real food, and still get your grub on.

Now non-competitive inhibition, lets redo the scenario.
20 plates of food, BUT 10 are poisoned. Your crew comes in, grabs plates at random. 3 of them die with the first bite. Now you only manage to clear 7 plates of real food PLUS you lost three of your crew. Vmax has dropped from 10 plates of food per meal to 7 plates of food per meal. You come back tomorrow, 20 plates of food but none of them are poisoned. Y'all eat with reckless abandon but your Vmax is still stuck at 7 plates of food per meal. So with the noncompetitive inhibition of poison, Vmax has dropped, and won't go back until you bring in more crew members, aka synthesize more enzymes.

 

[btuck]

Lol I just had a question about this on my first med school exam. It's important!