EK Bio Lecture 1 (Possible mistake?)

[orthoboy]

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[Catburr]

If E = Enzyme, S = Substrate, and P = Product; you can think of enzyme catalyzed reactions like this:

E + S <--> ES --> P

The first arrow, pointing both ways, indicates substrate binding, which is reversible. This means substrate can bind and then be either released or converted to product.

Think of a noncompetitive inhibitor as something that messes with the enzymes ability to finish the job (the ES --> P step). ES can still be formed, but we can't progress to P.

Vmax is the highest rate possible for a given situation. Throwing a noncompetitive inhibitor in basically disables the enzymes it sticks to, so Vmax drops.

Km is the equilibrium constant for the E + S <--> ES step. Since ES still forms normally after we throw the noncompetitive inhibitor in, Km doesn't change.

Hope that's useful. Good luck with your studying.

 

[orthoboy]

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[PiBond]

That makes sense, but when the noncompetitive inhibitor binds to the enzyme, does it not change it's shape? thus changing the shape of the active site? This should decrease the affinity of the substrate to that particular enzyme and lower the Km value. They said this in the audio osmosis too, which is why I think they made a mistake in the book. What do you think?

Thanks for the reply btw! Good luck with your studying as well!

Here's a slightly different way to think about it:

Noncompetitive inhibitors bind to an allosteric site on the enzyme and render the enzyme inactive (unable to form product). In doing so, we decrease our concentration of working enzyme which translates to a decrease in Vmax. However, those enzymes that didn't act with the noncompetitive inhibitor still have the same affinity for the substrate as they always have. Thus, Km for the remaining enzymes won't change.

 

[orthoboy]

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