Please explain this question on non-competitive and competitive enzyme inhibitor

[zut212]

Please explain this question on non-competitive and competitive enzyme inhibitors

According to Wikipeida:
"In competitive inhibition (CI), the substrate and inhibitor cannot bind to the enzyme at the same time, as shown in the figure on the left. This usually results from the inhibitor having an affinity for the active site of an enzyme where the substrate also binds; the substrate and inhibitor compete for access to the enzyme's active site. This type of inhibition can be overcome by sufficiently high concentrations of substrate, i.e., by out-competing the inhibitor. Competitive inhibitors are often similar in structure to the real substrate.

Non-competitive inhibition (NCI) is a form of mixed inhibition where the binding of the inhibitor to the enzyme reduces its activity but does not affect the binding of substrate. As a result, the extent of inhibition depends only on the concentration of the inhibitor."

CI can be overcome by "sufficiently high concentrations of substrate". So another words, if the CI concentration is sufficiently low compared to that of the substrate, than it will be overcome. Here, we see that the RELATIVE concentrations of the CI and the substrate affect the rate speed.

NCI inhibition "depends only on the concentration of the inhibitor." However, they really mean to say the "RELATIVE CONCENTRATION" to the substrate, in my opinion.

So how are the rates not affected by the relative concentration of the inhibitors for both CI and NCI?

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[SN2ed]

Thread moved. These types of questions are best suited to the Study Q&A forum.

 

[StIGMA]

NCI inhibition does not depend on the amount of substrate - it depends on the amount of enzyme. Ie: the ratio of NCI to enzyme matters, but not NCI to substrate (because they are not binding the same thing).

Make sense?

 

[PiBond]

I'll take a stab but it's a tough topic because of word choice, in particular the word "rate"

The Vmax is only affected in non-competitive inhibition. thus, the overall reaction velocity for the population of the enzymes as a whole is decreased (since the NCI has knocked a few of them out of commission) but the individual enzymes that haven't been affected by the non competitive inhibitor still catalyze at their same rate. So, if you have 100 enzymes and 10 NCI's, you end up having only 90 functional enzymes and thus your Vmax has decreased. But those 90 enzymes are still functioning normally and thus continue to catalyze their specific reaction at their normal rate, if that makes sense.

For the competitive inihbitor situation, the enzyme hasn't been changed by our inhibitor and thus our Vmax is not diminished. When the enzymes come in contact with their substrates they still catalyze their respective reaction at the same velocity. However, because a competitive inhibitor exists they have an apparet lower affinity (higher Km) for the substrate and thus take more of it to attain their max velocity (Vmax).

hope that helps

 

[PltEnthusiast]

Rate of reaction depends only on the concentration of the reactant (or, substrate in this case)

Competitive inhibitors bind to the binding site of the enzyme (affect Km). Thus, if there is more of the substrate, they can beat the CI and occupy that binding site. Thus, rate of reaction can be increased to the original by increasing the conc. of the substrate.

Non comp. inh. bind to a site other than the binding site (don't affect Km). Thus, no matter how much substrate is there, they will lower the rate of reaction by attaching to the enzymes and leaving them of no use.

 

[docelh]

"[NCI]... does not affect the binding of substrate."

This statement sounds incorrect to me. While NCI does not bind at the activity site, it certainly can affect the enzyme's binding to a substrate.

 

[StIGMA]

"[NCI]... does not affect the binding of substrate."

This statement sounds incorrect to me. While NCI does not bind at the activE site, it certainly can affect the enzyme's binding to a substrate.

Agreed. These are textbook scenarios and are used to teach concepts. Practically, Km is altered.

 

[ioud]

I might be wrong on this,
but I think the change in Km for competitive inhibitors and the change in Vmax for the noncompetitive inhibitors are only for reversible inhibition.
For irreversible inhibition, competitive and noncompetitive inhibitors could affect more than one factor and this is more complicated. I don't think we go into detail about irreversible inhibition that much.

 

[UCB05]

"[NCI]... does not affect the binding of substrate."

This statement sounds incorrect to me. While NCI does not bind at the activity site, it certainly can affect the enzyme's binding to a substrate.

You're thinking of normal mixed inhibition, in which binding of the inhibitor can affect binding to a substrate, e.g. allosteric inhibition. Noncompetitive inhibition is a subset of mixed inhibition where this is not the case. NCIs only affect the activity of the enzyme and not its binding to a substrate.

 

[JP2740]

I'll take a stab but it's a tough topic because of word choice, in particular the word "rate"

The Vmax is only affected in non-competitive inhibition. thus, the overall reaction velocity for the population of the enzymes as a whole is decreased (since the NCI has knocked a few of them out of commission) but the individual enzymes that haven't been affected by the non competitive inhibitor still catalyze at their same rate. So, if you have 100 enzymes and 10 NCI's, you end up having only 90 functional enzymes and thus your Vmax has decreased. But those 90 enzymes are still functioning normally and thus continue to catalyze their specific reaction at their normal rate, if that makes sense.

For the competitive inihbitor situation, the enzyme hasn't been changed by our inhibitor and thus our Vmax is not diminished. When the enzymes come in contact with their substrates they still catalyze their respective reaction at the same velocity. However, because a competitive inhibitor exists they have an apparet lower affinity (higher Km) for the substrate and thus take more of it to attain their max velocity (Vmax).

hope that helps

How would you fit in uncompetitive inhibition into all of this? Thank you, your explanation is helpful.