Inhibitors

[happyfellow]

---

Members don't see this ad.

I've always been confused on why Km is kept constant during non-competitive inhibition and why Km increases during competitive inhibition.

Please first listen to my current understanding on the matter:
I know that non-competitive inhibitors bind to an allosteric site (not the active site) and thus change the conformation of the enzyme. I understand that this renders the enzyme unable to bind more substrate and thus reduces our Vmax because our "turnover rate" has decreased. However, if we've effectively "disabled" our enzyme by changing its shape then why in the heck does the Km stay the same??? Shouldn't the binding affinity for substrate change as well since the shape of our beloved enzyme has changed?? This idea has eluded me far too long.

For competitive inhibition I understand that the inhibitor competes with the substrate directly for the active site. However, this can be overcome by merely adding more substrate so our Vmax doesn't change. I also know that our enzyme has not changed so it still has the same "liking" or "affinity" for the substrate that it always had because nothing about the enzyme has been altered, right? Therefore, why in the world does Km change??? We haven't altered the enzyme in any way to make it have less of a liking for the substrate.

I would great APPRECIATE any clarification. This has been driving me nuts for a while.

-Happy

 

[MDub1]

Km is the concentration of substrate required for the reaction to run at 1/2 vmax. With a competitive inhibitor you have to add more substrate to reach the 1/2 vmax thus the Km increases. With non-competitive inhibition, the structure of the enzyme does change, decreasing the affinity of the enzyme for the substrate but the vmax decreases as well, meaning that the Km remains unchanged.

In other words, with a competitive inhibitor the capacity doesn't change but the concentration required to reach 1/2 capacity increases (because it's competing for the active site). With a non-competitive inhibitor the capacity decreases but the concentration to reach 1/2 capacity doesn't change (because there is no competitor for the active site). Hope this helps.

 

[unique135]

Short version:
Km has nothing to do with the altering of enzyme. Importantly, Km is ratio of rate constants so think in terms of equilibrium (due to steady-state assumption). Inhibitors bind and alter enzyme causing to change these constants, ultimately affecting Km.
High Km is represented with lower affinity (when Km = k-1/k1). However, in other cases, Km does not always directly represent as an affinity measure of an enzyme for its substrate.

Lol, organizing in my mind took more time, so here’s the longer version:

There are more types of inhibitors under reversible inhibition
a. Competitive Inhibitors
b. Uncompetitive Inhibitors
c. Mixed Competitive Inhibitors
Non-competitive Inhibitors (special case of Mixed Competitive Inhibitors)


In competitive inhibition, inhibitors directly bind to free enzymes at an active site. The shape of a substrate-bound enzyme will change, not of the free enzyme. Of course, a substrate-bound enzyme will not bind to a substrate. In presence of inhibitors, substrate will have to compete with inhibitors for binding with enzymes.
*E + S <---------(k1/k-1)-------------> ES
*If the free enzymes (E) are removed by Inhibitor (I), then according to Le Chatelier’s principle, the above equation will shift towards left. Less of enzyme-substrate complex (ES) will be formed and more of E + S will be formed. You should be able to see how the reaction is inhibited. When comparing to reaction without inhibitor where Km = when Vo = 1/2Vmax, substrate concentration, in this case, increases such that they can effectively compete and win over the inhibitors. From equation, such increase in substrate concentration will increase Km (which represents decrease in affinity), and notice Vmax is not affected here. Using Le Chatelier’s principle, increase in can achieve Vmax.

In uncompetitive inhibition, inhibitors bind to enzyme-substrate complex, not the free enzyme at a different site (Notice: This is (may) NOT be allosteric site. Specifically, allosteric site is completely different topic). The shape of the free enzyme has nothing to do with inhibition. How is substrate affinity of an enzyme affected if shape has nothing to do?
E + S <---------(k1/k-1)-------------> *ES
*Inhibitors, blocking at enzyme-substrate complex (ES), will suck up the enzyme-substrate complex. Again according to Le Chatelier’s principle, inhibitors will shift the equilibrium but this time towards right. In this case, more ES will be formed. This means that Km has decreased (representing higher affinity). This works well with high . In this case, Vmax also decreases. It would take longer for product formation since catalysis has been affected.

In mixed inhibition, inhibitors bind to free enzyme or enzyme-substrate complex mainly at a different site. Combining above two effects here will give reduced Vmax and variable Km (not sure if it would be variable). I think all depends on where inhibitor binds better (E or ES). If inhibitor binds better at E, then Km should increase; if it binds better at ES, then Km should decrease.
Non-competition (special case of mixed Inhibition): Inhibitors bind equally well, their dissociation constants with E or ES are equal, and thus there is no effect on Km. Vmax does decrease due to same reason as uncompetitive inhibition. With this case, I can see why Km is a play of rate constants.

Phew! I hope this clears the confusion. I would appreciate if someone verifies it as well.

 

[Techmed07]

Short version:
Km has nothing to do with the altering of enzyme. Importantly, Km is ratio of rate constants so think in terms of equilibrium (due to steady-state assumption). Inhibitors bind and alter enzyme causing to change these constants, ultimately affecting Km.
High Km is represented with lower affinity (when Km = k-1/k1). However, in other cases, Km does not always directly represent as an affinity measure of an enzyme for its substrate.

Lol, organizing in my mind took more time, so here’s the longer version:

There are more types of inhibitors under reversible inhibition
a. Competitive Inhibitors
b. Uncompetitive Inhibitors
c. Mixed Competitive Inhibitors
Non-competitive Inhibitors (special case of Mixed Competitive Inhibitors)


In competitive inhibition, inhibitors directly bind to free enzymes at an active site. The shape of a substrate-bound enzyme will change, not of the free enzyme. Of course, a substrate-bound enzyme will not bind to a substrate. In presence of inhibitors, substrate will have to compete with inhibitors for binding with enzymes.
*E + S <---------(k1/k-1)-------------> ES
*If the free enzymes (E) are removed by Inhibitor (I), then according to Le Chatelier’s principle, the above equation will shift towards left. Less of enzyme-substrate complex (ES) will be formed and more of E + S will be formed. You should be able to see how the reaction is inhibited. When comparing to reaction without inhibitor where Km = when Vo = 1/2Vmax, substrate concentration, in this case, increases such that they can effectively compete and win over the inhibitors. From equation, such increase in substrate concentration will increase Km (which represents decrease in affinity), and notice Vmax is not affected here. Using Le Chatelier’s principle, increase in can achieve Vmax.

In uncompetitive inhibition, inhibitors bind to enzyme-substrate complex, not the free enzyme at a different site (Notice: This is (may) NOT be allosteric site. Specifically, allosteric site is completely different topic). The shape of the free enzyme has nothing to do with inhibition. How is substrate affinity of an enzyme affected if shape has nothing to do?
E + S <---------(k1/k-1)-------------> *ES
*Inhibitors, blocking at enzyme-substrate complex (ES), will suck up the enzyme-substrate complex. Again according to Le Chatelier’s principle, inhibitors will shift the equilibrium but this time towards right. In this case, more ES will be formed. This means that Km has decreased (representing higher affinity). This works well with high . In this case, Vmax also decreases. It would take longer for product formation since catalysis has been affected.

In mixed inhibition, inhibitors bind to free enzyme or enzyme-substrate complex mainly at a different site. Combining above two effects here will give reduced Vmax and variable Km (not sure if it would be variable). I think all depends on where inhibitor binds better (E or ES). If inhibitor binds better at E, then Km should increase; if it binds better at ES, then Km should decrease.
Non-competition (special case of mixed Inhibition): Inhibitors bind equally well, their dissociation constants with E or ES are equal, and thus there is no effect on Km. Vmax does decrease due to same reason as uncompetitive inhibition. With this case, I can see why Km is a play of rate constants.

Phew! I hope this clears the confusion. I would appreciate if someone verifies it as well.[/QUOTE




I think you got it all right man, I'll like to add some clarification and perhaps venture away from interwinding this concept with general chemistry (ex: Le Chatelier Principle).


Competitive Inhibitor can be easily "defeated" with increase substrate concentration. This simply equates to a higher Km, which equals the substrate concentration at 1/2Vmax. The Vmax in this senario will not be changed because in short the active site is not being effected (just being proccupied) and with proper understand of the equation that gives us Vmax we can also deduce that substrate concentration has no effect on Vmax, Vmax=[E][Kcat]. Thus, Km is increased while Vmax is not effected.


Noncompetitive Inhibitor- In this case, the inclusion of substrates in the reaction will not have any affect on the reaction, because the inhibitor is bound in different place than the active site. The inhibitor has a drastic effect on the shape of the enzyme and thus resulting in a change in the catalytic site and in short effecting Kcat. Thus, with noncompetitive Km is not affected however the Vmax is drastically decreased. This can be deduced by using this simple equation Vmax= [Kcat][E]


Allosteric Enzymes- Allosteric enzymes are classically multisubunit enzymes and they do not adhere to Michealis-Menton enzyme kinetics. So no Vmax,Km etc etc.

 

[doc615]

unique135,as you say in uncompetitive inhibition " inhibitors will shift the equilibrium but this time towards right. In this case, more ES will be formed. This means that Km has decreased (representing higher affinity)".km being the reciprocal of the equilibrium constant for the formation of ES, i dont think it will be affected by equilibrium shifts. can u please clarify to this

 

[MedPR]

Other "short" versions aside, here is a real short version.

Km depends only on available enzymes.



Explanation:
Non-competitive inhibitors decrease the number of available enzymes (decrease Vmax), but Km depends only on the ones that can actually participate with the substrate. In other words, the ones that are not being bound by a non-competitive inhibitor.