Uncompetitive inhibitors

[basophilic]

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Uncompetitive inhibitors can bind to enzyme-substrate complexes and reduce Km and hence enhance affinity. But why does this cause a decrease in reaction rate? Doesn't high affinity generally mean high activity?
I don't see how holding the substrate tighter would decrease the enzyme's efficiency. The reasoning seems to be that the enzyme clasps the substrate or product so hard that the substrate/product can't leave; but you would want the substrate to be held tightly. Also, product would likely have a different 3D shape, so it should still be able to escape, no?
Also, textbook cites Le Chatelier's principle: formation of ESI removes the available number of ES, leading to fewer product molecules for the given enzyme concentration, thereby decreasing Vmax. But I can also use LC principle to argue that removing ES from system can drive the Step 1 equilibrium (E + S <-- --> ES) more to the right, so that little/none of the available ES reverts back to E + S.

 

[chemist16]

My biochem is a bit rusty so someone correct me if I'm wrong. Uncompetitive inhibition, by definition, is an inhibitor binding to an enzyme-substrate complex so that said complex cannot go on to form product (hence "inhibitor"). Therefore, it must decrease the reaction rate. You can imagine it as an E-S forming, which causes some geometric change in the enzyme that allows for the inhibitor to bind, forming the E-S-I complex. Now the inhibitor is keeping the E-S-I in place. For the enzyme to work, it has to convert the substrate to some product, which would require a change in geometry again so that you can regenerate the enzyme. But the inhibitor prevents this from happening and so the E-S-I is "stuck" and cannot go on to form product.

Formation of an E-S-I complex removes E-S and therefore causes more E-S complexes to form - that's what they mean by enhancing enzyme affinity for the substrate. Other uncompetitive inhibitor molecules can bind to these newly-formed E-S complexes until the inhibitor concentration is so low that you start to have E-S complexes forming and going on to product. But since you've removed so many E-S complexes via the inhibitor, you get a lot less product than you would expect with no inhibitor (since the amount of enzyme you have is some finite number). Therefore, you have fewer product molecules for a given enzyme concentration.

 

[basophilic]

Oh so the high affinity is not for the enzyme in the ESI but for the enzyme that's in the ES, which will be catalyzed more readily. However, overall Vmax decreases since fewer catalyses are taking place

 

[chemist16]

Not exactly. Your reasoning for Vmax is sound - you have fewer catalytic turnovers for the amount of catalyst you loaded. But the high affinity is for free enzyme, meaning that when you add an inhibitor, that makes free enzyme much more likely to bind the substrate to make the ES (because ES is removed by the inhibitor thus pushing equilibrium to the right).