Mixed Inhibition - v vs S - Biology

Mixed Inhibition - v vs S - Biology

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Mixed Inhibition - v vs S

No, you cannot assume that an allosteric inhibitor is non-competitive. "Allosteric" refers to the location of binding, whereas terms like competitive, non-competitive, uncompetitive, or mixed inhibition refer to how (or if) the binding of the inhibitor affects binding with the substrate.

The Wikipedia page on competitive inhibition is a reasonable source for this question.

A competitive inhibitor is one that prevents binding of the normal substrate(s) at the active site. It is possible to achieve this either by literally getting in the way by binding in the active site or by binding elsewhere, causing a conformational change that prevents binding the active site. Both cases are considered competitive inhibition because of how they influence substrate binding.

It may be technically correct to say that an allosteric inhibitor can be (or is) a “competitive inhibitor”, but, in terms of communication and education, I would advise a more measured approach. Most students encounter the terms “competitive inhibitor” and “non-competitive inhibitor” in the context of reactions exhibiting Michaelis–Menten kinetics. Their mental image of such an inhibitor will therefore be of the type:

In discussing allosteric inhibitors I would therefore focus on the different mental model, below, in which substrate (or positive effector) and negative effector (use of this different term being deliberate) have opposing effects on the equilibrium between tense and relaxed states:

I would then go on to say something to the effect that because the relative concentrations of substrate and negative effector determine the rate of the reaction “the negative effector indirectly competes with the substrate”.

First of all, I take your point about the previous post being too technical. On this occasion, there will be no equations and no diagrams.

When dealing with reversible enzyme inhibition, we are mainly concerned with the meaning of four terms: competitive, uncompetitive, non-competitive and mixed. As I said above, the literature is very confusing to the point of making relatively simple concepts appear very complex, and the confusion often arises from the meaning of the term 'non-competitive'.

First things first: reversible inhibitors are best classified by their effects on two kinetic constants rather than by the mechanism that gives rise to a particular inhibition pattern. The two kinetic constants are kcat and kcat/ Km. (So why not kcat and Km? This is discussed in a previous post, but the short answer is that it is simpler: Km is often a very complex kinetic constant). We can distinguish two 'limiting' cases. (i) A competitive inhibitor effects kcat/ Km but not kcat and (ii) an uncompetitive inhibitor effects kcat but not kcat / Km. (iii) We can also now distinguish a third case where both kinetic constants are affected. Let's call this one non-competitive inhibition.

The confusion in the literature arises because some authorities call non-competitive inhibition as defined above mixed inhibition, and (to really screw things up), they consider non-competitive inhibition a special case of mixed inhibition where both kinetic constants are changed to the same extent! For the purposes of this answer, I am not going there as I think what you mean by non-competitive is as defined in (iii) above. In any event, this is how Cleland, the leading authority in this field, defines non-competitive inhibition. In other words, 'mixed' inhibition is banished from this answer from this point on.

We could be a little bit more adventurous here and say that a competitive inhibitor only effects binding of the substrate to the enzyme and that an uncompetitive inhibitor effects only the catalytic step. Or, we could say that a competitive inhibitor effects only the apparent second order rate constant for the combination of enzyme with substrate (which the enzymologists call kcat/ Km) and that an uncompetitive inhibitor effects only the apparent first order rate constant (which the enzymologists term kcat). But that is becoming too abstract. But perhaps you can see where I am going? There are only two limiting cases, one affecting substrate binding and one affecting catalysis, and a combination of both. It really is as simple as that.

Having got the nomenclature mess out of the way, I will now attempt to answer your question: if an inhibitor does not bind to the active site can we conclude it is a non-competitive inhibitor?

IMO, the answer is an emphatic no. An allosteric inhibitor (one which binds to a site other than the active site) may be competitive, uncompetitive or non-competitive.

As Bryan Krause put it , a competitive inhibitor prevents binding of the normal substrate(s) at the active site, but there is no requirement that the inhibitor bind to the active site. Binding of the inhibitor at the allosteric site may cause a conformational change that prevents substrate binding, for example.

The same argument applies to uncompetitive inhibition. A simple mechanism that gives rise to uncompetitive inibition is where the inhibitor cannot bind to the 'free' enzyme but binds to the enzyme-substrate complex. And of course binding to the allosteric site might not be possible unless substrate is bound.

C2 . Competitive Inhibition

Reversible Competitive inhibition occurs when substrate (S) and inhibitor (I) both bind to the same site on the enzyme. In effect, they compete for the active site and bind in a mutually exclusive fashion. This is illustrated in the chemical equations and molecular cartoon below.

There is another type of inhibition that would give the same kinetic data. If S and I bound to different sites, and S bound to E and produced a conformational change in E such that I could not bind (and vice versa), then the binding of S and I would be mutually exclusive. This is called allosteric competitive inhibition. Inhibition studies are usually done at several fixed and non-saturating concentrations of I and varying S concentrations.

The key kinetic parameters to understand are Vm and Km. Let us assume for ease of equation derivation that I binds reversibly, and with rapid equilibrium to E, with a dissociation constant Kis. The "s" in the subscript "is" indicates that the slope of the 1/v vs 1/S Lineweaver-Burk plot changes while the y intercept stays constant. Kis is also named Kic where the subscript "c" stands for competitive inhibition constant.

A look at the top mechanism shows that even in the presence of I, as S increases to infinity, all E is converted to ES. That is, there is no free E to which I could bind. Now remember that Vm= kcatEo. Under these condition, ES = Eo hence v = Vm. Vm is not changed. However, the apparent Km, Kmapp, will change. We can use LaChatelier's principle to understand this. If I binds to E alone, and not ES, it will shift the equilibrium of E + S <==> ES to the left, which would have the affect of increasing the Km app (i.e. it would appear that the affinity of E and S has decreased.). The double reciprocal plot (Lineweaver Burk plot) offers a great way to visualize the inhibition. In the presence of I, Vm does not change, but Km appears to increase. Therefore, 1/Km, the x-intercept on the plot will get smaller, and closer to 0. Therefore the plots will consists of a series of lines, with the same y intercept (1/Vm), and the x intecepts (-1/Km) closer and closer to the 0 as I increases. These intersecting plots are the hallmark of competitive inhibition.

Note that in the first three inhibition models discussed in this section, the Lineweaver-Burk plots are linear in the presence and absence of inhibitor. This suggests that plots of v vs S in each case would be hyperbolic and conform to the usual form of the Michaelis Menton equation, each with potentially different apparent Vm and Km values.

An equation, shown in the figure above, can be derived which shows the effect of the competitive inhibitor on the velocity of the reaction. The only change is that the Km term is multiplied by the factor 1+I/Kis. Hence Kmapp = Km(1+I/Kis). This shows that the apparent Km does increase as we predicted. Kis is the inhibitor dissociation constant in which the inhibitor affects the slope of the double reciprocal plot.

Wolfram Mathematica CDF Player - Competitive Inhibition v vs S (free plugin required)

4/6/14Wolfram Mathematica CDF Player - Competitive Inhibition - Lineweaver Burk(free plugin required)

If the data was plotted as vo vs log S, the plots would be sigmoidal, as we saw for plots of ML vs log L in Chapter 5B. In the case of competitive inhibitor, the plot of vo vs log S in the presence of different fixed concentrations of inhibitor would consist of a series of sigmoidal curves, each with the same Vm, but with different apparent Km values (where Kmapp = Km(1+I/Kis), progressively shifted to the right. Enyzme kinetic data is rarely plotted this way, but simple binding data for the M + L < == > ML equilibrium, in the presence of different inhibitor concentrations is .

Reconsider our discussion of the simple binding equilibrium, M + L <==> ML. When we wished to know how much is bound, or the fractional saturation, as a function of the log L, we considered three examples.

  1. L = 0.01 Kd (i.e. L << Kd), which implies that Kd = 100L. Then Y = L/[Kd+L] = L/[100L + L] ≈1/100. This implies that irrespective of the actual [L], if L = 0.01 Kd, then Y ≈0.01.
  2. L = 100 Kd (i.e. L >> Kd), which implies that Kd = L/100. Then Y = L/[Kd+L] = L/[(L/100) + L] = 100L/101L ≈ 1. This implies that irrespective of the actual [L], if L = 100 Kd, then Y ≈1.
  3. L = Kd, then Y = 0.5

These scenarios show that if L varies over 4 orders of magnitude (0.01Kd < Kd < 100Kd), or, in log terms, from
-2 + log Kd < log Kd < 2 + log Kd), irrespective of the magnitude of the Kd, that Y varies from approximately 0 - 1.

In other words, Y varies from 0-1 when L varies from log Kd by +2. Hence, plots of Y vs log L for a series of binding reactions of increasingly higher Kd (lower affinity) would reveal a series of identical sigmoidal curves shifted progressively to the right, as shown below.

The same would be true of vo vs S in the presence of different concentration of a competitive inhibitor, for initial flux, Jo vs ligand outside, in the presence of a competitive inhibitor, or ML vs L (or Y vs L) in the presence of a competitive inhibitor.

Wolfram Mathematica CDF Player - Competitive Inhibition v vs logS (free plugin required)

In many ways plots of v0 vs lnS are easier to visually interpret than plots of v0 vs S . As noted for simple binding plots, textbook illustrations of hyperbolas are often misdrawn, showing curves that level off too quickly as a function of [S] as compared to plots of v0 vs lnS, in which it is easy to see if saturation has been achieved. In addition, as the curves above show, multiple complete plots of v0 vs lnS at varying fixed inhibitor concentration or for variant enzyme forms (different isoforms, site-specific mutants) over a broad range of lnS can be made which facilitates comparisons of the experimental kinetics under these different conditions. This is especially true if Km values differ widely.

Now that you are more familiar with binding, flux, and enzyme kinetics curves, in the presence and absence of inhibitors, you should be able to apply the above analysis to inhibition curves where the binding, initial flux, or the initial velocity is plotted at varying competitive inhibitor concentration at different fixed concentration nonsaturating concentrations of ligand or substrate. Consider the activity of an enzyme. Lets say that at some reasonable concentration of substrate (not infinite), the enzyme is approximately 100% active. If a competitive inhibitor is added, the activity of the enzyme would drop until at saturating (infinite) I, no activity would remain. Graphs showing this are shown below.

Figure: Inhibition of Enzyme Activity - % Activity vs log [Inhibitor]

A special case of competitive inhibition: the specificity constant: In the previous chapter, the specificity constant was defined as kcat/KM which we also described as the second order rate constant associated with the bimolecular reaction of E and S when S << KM. It also describes how good an enzyme is in differentiating between different substrates. If has enzyme encounters two substrates, one can be considered to be a competitive inhibitor of the other. The following derivation shows that the ratio of initial velocities for two competing substrates at the same concentration is equal to the ratio of their kcat/KM values.

How Do Enzymes Work?

A stretched out spring. (Photo Credit : Pixabay)

Think of enzymes as a spring. While in use, the spring stretches and changes its shape, but once the work is done, it returns to its original shape. Enzymes work in a similar way. During a chemical reaction, the enzyme may change its composition or form to promote the reaction, and once the task is done, it returns to its original state, leaving no permanent change in the enzyme.

C4. Noncompetitive and Mixed Inhibition

Reversible noncompetitive inhibition occurs when I binds to both E and ES. We will look at only the special case in which the dissociation constants of I for E and ES are the same. This is called noncompetiive inhibition. It is quite rare as it would be difficult to imagine a large inhibitor which inhibits the turnover of bound substrate having no effect on binding of S to E. However covalent interaction of protons with both E and ES can lead to noncompetitive inhibition. In the more general case, the Kd's are different, and the inhibition is called mixed. Since inhibition occurs, we will hypothesize that ESI can not form product. It is a dead end complex which has only one fate, to return to ES or EI. This is illustrated in the chemical equations and in the molecular cartoon below.

Let us assume for ease of equation derivation that I binds reversibly to E with a dissociation constant of Kis (as we denoted for competitive inhibition) and to ES with a dissociation constant Kii (as we noted for uncompetitive inhibition). Assume for noncompetitive inhibition that Kis = Kii. A look at the top mechanism shows that in the presence of I, as S increases to infinity, not all of E is converted to ES. That is, there is a finite amount of ESI, even at infinite S. Now remember that Vm = kcatEo if and only if all E is in the form ES . Under these conditions, the apparent Vm, Vmapp is less than the real Vm without inhibitor. In contrast, the apparent Km, Kmapp, will not change since I binds to both E and ES with the same affinity, and hence will not perturb that equilibrium, as deduced from LaChatelier's principle. The double reciprocal plot (Lineweaver Burk plot) offers a great way to visualize the inhibition. In the presence of I, just Vm will decrease. Therefore, -1/Km, the x-intercept will stay the same, and 1/Vm will get more positive. Therefore the plots will consists of a series of lines intersecting on the x axis, which is the hallmark of noncompetitive inhibition. You should be able to figure out how the plots would appear if Kis is different from Kii (mixed inhibition).
An equation, shown in the diagram above can be derived which shows the effect of the noncompetitive inhibitor on the velocity of the reaction. In the denominator, Km is multiplied by 1+I/Kis, and S by 1+I/Kii. We would like to rearrange this equation to show how Km and Vm are affected by the inhibitor, not S, which obviously isn't. Rearranging the equation as shown above shows that Kmapp = Km(1+I/Kis)/(1+I/Kii) = Km when Kis=Kii, and Vmapp = Vm/(1+I/Kii). This shows that the Km is unchanged and Vm decreases as we predicted. The plot shows a series of lines intersecting on the x axis. Both the slope and the y intercept are changed, which are reflected in the names of the two dissociation constants, Kis and Kii. Note that if I is zero, Kmapp = Km and Vmapp = Vm. Sometimes the Kis and Kii inhibition dissociations constants are referred to as Kc and Ku (competitive and uncompetitive inhibition dissociation constants.

Mixed (and non-)competitive inhibition (as shown by mechanism above) differ from competitive and uncompetiive inhibition in that the inhibitor binding is not simply a dead end reaction in which the inhibitor can only dissociate in a single reverse step. In the above equilibrium, S can dissociate from ESI to form EI so the system may not be at equilibrium. With dead end steps, no flux of reactants occurs through the dead end complex so the equilibrium for the dead end step is not perturbed.

Other mechanisms can commonly give mixed inhibition. For example, the product released in a ping pong mechanism (discussed in the next chapter) can give mixed inhibition.

If P, acting as a product inhibitor, can bind to two different forms of the enzyme (E' and also E), it will act as an mixed inhibitor.

4/26/13Wolfram Mathematica CDF Player - Mixed Inhibition v vs S curves Kis and Kii called Kc and Ku (start sliders at high values) (free plugin required)

I nteractive SageMath MIxed Inhibition

4/26/13Wolfram Mathematica CDF Player - Lineweaver-Burk plots for Mixed Inhibition v vs S curves (start sliders at high values) (free plugin required). Note where the inhibited and inhibited curves intersect at different values of Kis and Kii (in the graph termed Kc and Ku). When Kis = Kii, the inhibition is noncompetitive.

I nteractive SageMath Mixed Inhibition ( Red graph + Inhibitor, Blue graph - Inhibitor, Green Axes

If you can apply LeChatilier's principle, you should be able to draw the Lineweaver-Burk plots for any scenario of inhibition or even the opposite case, enzyme activation!

Archived version of full Chapter 6C: Enzyme Inhibition

Biochemistry Online by Henry Jakubowski is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.

What's the difference between a mixed inhibitor that preference for the enzyme-substrate complex vs. Uncompetitive inhibitors

Mixed inhibitors bind to an allosteric site and can bind preferentially to either the enzyme or enzyme substrate conformation. If it binds to the enzyme substrate preferentially then it lowers Km and V max. A uncompetitive inhibitor binds only to the enzyme substrate complex at an allosteric site but it loses Km and Vmax. I'm confused. What's the difference between these two?

Mixed inhibition has different affinities for the E and E•S complex, with one of the two being preferential. Noncompetitive inhibition is a special case of mixed inhibition. with noncompetitive inhibitors you have equal affinities for the E and E•S complex. These differences are largely due to any conformational changes of the enzyme in allosteric sites following substrate binding.

With noncompetitive inhibition we do not affect affinity for the substrate at all (considering the affinities are the exact same for E and E•S). So the effect of the inhibitor is to halt some conformational change that is essential for turnover of the substrate into product, thereby decreasing the measured Vmax but not affecting Km.

With mixed inhibition this can go both ways. If we have a higher affinity for the enzyme alone (no substrate), we will see a conformational change in the enzyme and specifically in the active site that will make it more difficult/impossible for the substrate to bind. This would essentially turn off the enzymes with these inhibitors bound(reducing the amount of functional enzymes left), thereby reducing Vmax. We would also see a increase in apparent Km because it looks like we need more substrate to reach Vmax because the conformational change in the active site reduces affinity for the substrate.

The other side of mixed inhibition is if we have a higher affinity for the E•S complex. Here, we again will cause a conformational change in the enzyme that decrease Vmax because it will inhibit some essential conformational change for enzyme turnover. In this case, we already have the substrate bound so the apparent Km will decrease in the same way it does with uncompetitive inhibition. Binding inhibitor to the E•S complex gives us E•S•I complex (still binding allosterically though). Think of le chatelier’s. E + S -> E•S + I -> E•S•I. Reducing E•S by formation of E•S•I will cause equilibrium to shift from E to E•S, which makes it look like E has a higher affinity for S, ergo lower Km.

Watch the video: mixed inhibition and non competitive inhibition (August 2022).