Noncompetitive inhibitor and rate on reaction

[Buono]

Can someone explain how a noncompetitive inhibitor changes Vmax while Km is same?

I understand how a competitive inhibitor (binds directly to active site) affects the rate and how increasing the concentration of substrate nullifies the effect on rate by just adding more substrate so there is basically no more "competition" and turnover can commence at its maximum rate, but with a higher Km since higher concentration needed to reach Vmax, a higher concentration is needed to reach 1/2 of Vmax which is Km

But how does a noncompetitive inhibitor affect the reaction rate and not affect Km? Is it just by cancelling out the catalyzing ability of the enzymes themselves so the product cannot be made even if the substrate binds? Therefore the reaction would still progress, just slower because there are less "active" enzymes? But if substrates are still binding to enzymes that are inhibited, shouldnt the concentration at 1/2 of Vmax need to increase since some of the substrate molecules are effectively being "distracted" by inhibited active sites producing no product? I don't understand why Km would stay the same here.

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[AwayFromReality]

I believe that Km refers to 1/2 the concentration of substrate required to take up all possible active sites that are available (saturate the enzyme). The ONLY time you can think of Km as 1/2 the substrate concentration required to reach Vmax is when there are no non competitive inhibitors present (since, in the case of no inhibitor or only competitive inhibitor, maximum turnover occurs only when substrates fully saturate the enzyme)

Noncompetitive inhibitors usually don't modify the active site but rather the enzyme so that it can no longer produce the product. Thus the concentration of substrate that is required to saturate the enzyme is still the same therefore Km is still the same but since the enzymes are being inhibited, the maximum turnover rate is reduced therefore Vmax is reduced.

 

[amy_k]

Noncompetitive inhibitors do prevent the substrate binding to the active site of the enzyme as a result of the change in enzyme structure. The reason that Km does not change is because the enzyme cannot become saturated with substrate no matter how much you increase the substrate concentration, the substrate cannot bind and thus the reaction velocity (and Vmax) decreases but Km stays the same.

 

[AwayFromReality]

Noncompetitive inhibitors do prevent the substrate binding to the active site of the enzyme as a result of the change in enzyme structure. The reason that Km does not change is because the enzyme cannot become saturated with substrate no matter how much you increase the substrate concentration, the substrate cannot bind and thus the reaction velocity (and Vmax) decreases but Km stays the same.

All the text I've read and some research I just did online state that Noncompetitive inhibitors do not prevent the substrate from binding but rather only prevents product formation

 

[Buzzin]

AWAYFROMREALITY has it perfectly correct, it jus took me a while to put all the pieces together bc he stated it so briefly yet effectively. I agree that noncompetitive inhibitors don't prevent enzyme and substrate from binding rather it prevents the product from forming.

the REASON this is so complicated has to do with the DEFINITION of Km, which is a measure of affinity or strength of binding between enzyme and substrate. Km is directly related to SUBSTRATE concentration at 1/2Vmax but it does not EQUAL this at all times.

The lower the Km, the greater the affinity (so the lower the concentration of substrate needed to achieve the given rate).

In competitive inhibition, the inhibition of product can be UNDONE by increasing substrate concentration BECAUSE the substrate and inhibitor are competing for the same active site, therefore increasing the substrate concentration diminishes the effect of the inhibitor completely using strength in numbers. Vmax stays the same, but at the cost of Km since more substrate molecules are needed to reach the same rate Vmax, the efficiency of the enzyme suffers in terms of substrate concentration, which is the perspective of Km.

BUT in NONcompetitive inhibition, the effect can NOT be undone, no matter how many substrate molecules are added by increasing concentration since the substrate and inhibitor are binding at different locations. The inhibitor has NO EFFECT on the binding of substrate and enzyme. The enzymes that are catalyzing the reaction are already working at full capacity (saturated) so Km is the same, Vmax is lowered bc there are less of them.

 

[amy_k]

Yes... the conformational change induced by the noncompetitive inhibitor prevent the substrate binding properly to the active site which is why the active site cannot function correctly and thus the enzyme cannot catalyze the reaction.

 

[amy_k]

See this image http://science.halleyhosting.com/sci/ibbio/chem/notes/chpt8/allostericnotes.gif

 

[AwayFromReality]

Yes... the conformational change induced by the noncompetitive inhibitor prevent the substrate binding properly to the active site which is why the active site cannot function correctly and thus the enzyme cannot catalyze the reaction.

From Wikipedia: "It [Non-competitive inhibtion] differs from competitive inhibition in that the binding of the inhibitor does not prevent binding of substrate, and vice versa, it simply prevents product formation for a limited time". Source: http://en.wikipedia.org/wiki/Non-competitive_inhibition

From chem.wisc.edu: "[Non competitive inhibition]. This changes the enzyme's three-dimensional structure so that its active site can still bind substrate with the usual affinity."
Source:http://www.chem.wisc.edu/deptfiles/genchem/netorial/modules/biomolecules/modules/enzymes/enzyme6.htm

Although I did find a couple other websites (aside from yours) that did say it prevents substrate from binding which contradicts what I learned and have posted above.

With some more research, I found this site: http://www.chemguide.co.uk/organicprops/aminoacids/enzymes3.html If you scroll down to noncompetitve inhibition it mentions two types. "pure" and "other". It seems "pure" inhibitors don't modify the active site therefore allowing the substrate to bind whereas "other" does prevent substrate from binding. It also mentions that "pure" is not as common in nature.

I'm gonna use the "pure" explanation when it comes to enzyme kinetics because that helps me personally visualize the graph and the effects of Km vs Vmax. I don't think the MCAT is gonna specifically ask us whether noncompetitive inhibitors prevent substrate from binding since there could be two possible answers.