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Vaishaalie_6729
#1 Posted : Saturday, June 12, 2021 8:10:56 PM
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Hello,

I have a quick question about enzyme kinetics. I understand why the Km value increases for competitive inhibitors, and remains unchanged for non-competitive inhibitors. However, I am confused as to why the Km decreases for uncompetitive inhibitors. It is to my understanding that uncompetitive inhibitors only bind to the enzyme-substrate complex, therefore why is there a smaller amount of the [substrate] needed to reach 1/2 Vmax? Also, how detailed should I know enzyme kinetics for the MCAT (outside of the effects of each inhibitor on Vmax and Km)?

I would love and appreciate any clarification in this topic.

Thank You!
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INSTR_Katie_127
#2 Posted : Sunday, June 13, 2021 7:55:33 AM
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Hi there!

Yes, I can see how that would be confusing. The best way to think about uncompetitive inhibition is that when the uncompetitive inhibitor binds to the enzyme-substrate complex, it inactivates or blocks that complex from being used in the reaction. The inhibitor-bound complex effectively reduces the concentration of the ES complex. By Le Chatelier’s Principle, a shift occurs to form additional ES complex, resulting in less free enzyme and more enzyme in the forms ES and ESI (ES with inhibitor). Decreases in free enzyme correspond to an enzyme with greater affinity for its substrate. It's a bit tricky to wrap your head around, but that is how uncompetitive inhibition both decreases Vmax and increases an enzyme’s affinity for its substrate.

I will pose your question about amount of detail to the instructor in our upcoming BioBiochem session.

Hope that helps!

Katie
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