V max and Km and noncomp inhibitor

[dray5150]

Could somebody explain why a noncompetitive inhibitor would decrese Vmax and keep Km constant.

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[xanthomondo]

Could somebody explain why a noncompetitive inhibitor would decrese Vmax and keep Km constant.

So the two different types of inhibition are 1) competitive inhibition and 2) allosteric (or noncompetitive) inhibition. Km = the concentration of substrate that is needed to achieve 1/2 Vmax

Imagine A+E -> B+E which is inhibited by I, A is the substrate and E is the enzyme

If I is a competitive inhibitor, I competes with A for the active site of E. This means that I can (by chance) bump into the enzyme and block A's conversion of B. So if you have 1 molecule of A and 100000 molecules of I, you probably won't get A converted to B. However, if you had 100000 molecules of A and 1 molecule of I, you won't see a change in Vmax (the As just outnumber the Is too much). If you had the concentrations equal: 100 molecules of A and 100 molecules of I you would have a reaction but it would be much slower than what you'd have without I because of competition for the active site of the enzyme. Thus, Vm is unchanged and Km is changed.

It's the same enzyme, same binding ability, just the inhibitor blocks the substrate from getting to the active site. This just requires more substrate molecules to get to Vmax (which is a change in Km).


Now, if I is a noncompeitive inhibitor, it will not touch the active site, rather it binds a regulatory site which changes the conformation of the enzyme. This change in the enzyme will change the active site and make it less likely to bind the substrate. So here there is no direct competition, but I essentially changes the enzyme and makes it slower. Since this "new", slower enzyme's affinity is changed, you can flood the enzyme with A and it will reach a Vmax, but a Vmax that is less than what is observed without I. However, since there is no competition for the active site (only A binds!) you still find that the same saturation level will produce 1/2Vmax as without inhibitor, just 1/2Vmax will be slower.

So the figure shows that its the same enzyme, just the inhibited version is slower (because the active site is changed). It will still show the same saturation kinetics though because there is no competition for the active site.

Hope this makes some sense

 

[Calheesh]

Very well put xanthomondo

 

[dray5150]

Very good explanation!!! Thanks so much!!

 

[Hibrida]

Can someone describe the images of the graphs? I have broken images and I just see the boxed red - X. Thanks for the great answer!

 

[xanthomondo]

haha glad I could help

sorry about the images, they worked when I checked but I guess the web site they come from doesnt allow bandwith stealing so i uploaded them to photobucket