Let's see if I have this right.

• You are using Kiapp to refer to the IC50 of the inhibitor titration, at which vi = vo/2 where vi is the rate in the presence of the inhibitor and vo is the rate in its absence.
• You are referring to the mixed linear inhibition model with a single-substrate reaction, in which the rate equation can be written:

v = Vmax[S]/{Km(1+[I]/Ki) + [S](1+[I]/alphaKi)}

and you are calling Ki the competitive inhibition constant Kic and alphaKi the noncompetitive inhibition constant Kiu.

If my algebra is correct, Kiapp = (KicKiuKm + KicKiu[S])/(KiuKm + Kic[S])

As [S] goes to zero, Kiapp approaches Kic (the y-intercept on your graph), and as [S] goes to infinity, Kiapp approaches Kiu. So, I agree with you up to that point.

I'm having trouble with Kiapp = Kic + Kiu[S]/(S1/2 + [S]) because as [S] goes to infinity Kiapp approaches Kic + Kiu instead of Kiu. Also, the form of the equation for Kiapp above does not match the form of Kiapp = Kic + Kiu[S]/(S1/2 + [S]).

Adam B Shapiro Man, you are the RG biochemisty guru! I don't think you've missed any of my Qs.

Okay, so we are in agreement up until my equation. Here's my explanation... When you increase [S] to saturate that plot, eventually there are no more free enzymes for inhibitor to bind. [S]/KM is so high relative to [I]/Ki that inhibitor doesn't have a chance to get in the active site before the substrate. So when the the plot above plateaus, there is no competitive inhibition. It's 100% noncompetitive.

I think...

I edited my answer to offer another reason why the last equation can't be correct if my equation for Kiapp for the mixed type linear inhibition single-substrate situation is correct: The equation does not have the same form as the equation you wrote containing S1/2. However, I don't think it's a problem for the approach to measuring Ki. There is no need to measure S1/2.

I agree with your statement about saturation with substrate leading to pure noncompetitive inhibition, which is why Kiapp goes to Kiu at high [S]. However, it isn't so obvious why Kiapp should approach Kic as [S] approaches zero. The rationale is that Kiu is, by definition, higher than Kic in the mixed inhibition mechanism. This mechanism is a special case of non-competitive inhibition, in which, using your nomenclature, Kic = Kiu. Therefore, the lowest Kiapp value must be Kic.

By the way, I'm not a big fan of using IC50 measurements at various substrate concentrations to measure Ki of inhibitors. This can be challenging because the range of usable substrate concentrations may not be broad enough. My preference would be to perform a global nonlinear regression of rates measured at a matrix of substrate and inhibitor concentrations to a kinetic rate equation describing the mechanism of inhibition. This is a pretty easy thing to do for single-substrate enzymes. It gives you estimates for the Km, Vmax, and inhibition constant(s) simultaneously. At the same time, it can be used to test out other kinetic mechanisms of inhibition by regression to other rate equations.

Adam B Shapiro How many concentrations of each would you start with and what would be the equation for a bimolecular reaction? I have thought about doing something similar to that,,, with like 10 concentrations of S and I. That way I could get both Michaelis Menten curves and IC50 curves.

Also, I thought Kiapp and IC50 were distinct values, IC50 being raw 50% mark in an inhibition assay, whereas Kiapp is corrected for enzyme concentration (IC50 = Kiapp +1/2 [E] ; if [E] is high enough to be relevant), and Ki is the KD at 0 substrate (which is why Ki doesn't depend on what substrate is used, I think). Then you go from Ki app to Ki like this Ki = Kiapp/(1+S/KM). It's essentially the Cheng-Prusoff equation, but their equation I don't think is right in situations where [E] is getting near the Ki of the compound.

When you say Therefore, the lowest Kiapp value must be Kic. In competitive/noncompetitive mixed inhibition, the lowest Ki app value would theoretically be at 0 substrate right? If you are sufficiently below the KM, there should be virtually no competition for the inhibitor, right?

Apart from the fact that ES complex results in product formation and EI is non-productive, the relationship between S and I, from the point of view of their interaction with the enzyme, is completely symmetric at equilibrium conditions (you should be able to interchange S and I in the equilibrium equations) and reciprocal (the effect of S over I is the same as the effect of I over S).

The equation: IC50 = Kiapp + 1/2 [E]T indicates that IC50 is just a derived parameter from the intrinsic Ki that depends on enzyme concentration ([E]T) and on experimental conditions, substrate affinity etc. (implicit in Kiapp).

Under equilibrium, or under quasi-equilibrium in the steady-state phase for enzyme activity, the reciprocal effect between S and I is expressed with the following equations:

Kmapp = Km (1+[I]/Ki)/(1+[I]/αKi)

Kiapp = Ki (1+[S]/Km)/(1+[S]/αKm)

where Ki is the intrinsic inhibition constant for I, Km is the intrinsic Michaelis constant for S, and α is the cooperative interaction constant between S and I (positive if α < 1, negative if α > 1), where a single site for S and a single site for I in E are considered, and Kxapp are the apparent constants for each reactant in the presence of the other reactant (effector) at a certain concentration.

From the first equation:

if [I]=0, Kmapp = Km

if [I] is sufficiently large, Kmapp = αKm

From the second equation:

if [S]=0, Kiapp = Ki

if [S] is sufficiently large, Kiapp = αKi

From that you can see the reciprocal effect between S and I, which would be expected if energy conservation must not be violated.

In particular, at [I] = αKi and [S] = αKm, you get:

Kmapp = 1/2 Km (1+α)

Kiapp = 1/2 Ki (1+α)

which are the arithmetic averages between intrinsic constants at zero effector and at sufficiently high effector concentration (half-way maximal effect), which are the [I]1/2 and [S]1/2 (see the picture shown in your question).

Personally, I do not like using IC50, since it is usually ill-defined, unless it is required operationally. However, as it arose in another discussion post before, any dependency of IC50 with [S] is equivalent to any dependency of Kiapp with [S], because IC50 and Kiapp differ by the constant [E]T/2.

Adrian Velazquez-Campoy Thanks for the informative reply. So... if I have this right, alphaKi is equivalent to Kiu, so I could determine Kiu by fitting the data with the rate equation to get Kic and alpha, and then Kiu=Kic * alpha? I'm going to try fitting some of my data with these equations. I'll report back.

Yes.

Given the symmetry, let's assume you have A and B interacting with a protein P.

If you have KAapp as a function of [B], then using:

KAapp = KA (1+[B]/KB)/(1+[B]/αKB)

you can estimate KA, α, and KB.

Or you can use a zero concentration of B (if possible; not if [S]=0, because then you have no activity; but it is possible in binding studies) and at sufficiently high concentration of B. In the first case you get KA and in the second case you get αKA.

Please, take a look to:

https://pubmed.ncbi.nlm.nih.gov/22052485/

(beware in that publication association constants instead of dissociation constants are considered, and α is then the inverse of the α we are talking in this post).

If they are well chosen, you can get a good result with about 6 concentrations each of S and I, but there is no harm in doing more. The result should be a well-spread group of curves covering most of the dynamic range of the effect. A little trial and error will help you identify the right range of concentrations.

If by bimolecular you mean bisubstrate, then there are several possible equations, depending upon the kinetic mechanism. Some examples are Rapid Equilibrium Random, Ordered, and Ping Pong. It is not always easy to figure out which kinetic mechanism is relevant. The equations can be found in the books by Segel or Leskovac, for example.

The 1/2[E]T part is only significant if your IC50 is close to the enzyme concentration (tight binding).

It's not quite right to say that Ki is the Kd at zero substrate. Both parameters are the same, the equilibrium dissociation constant of the inhibitor from the enzyme form to which it binds. For the competitive inhibitor, this is the dissociation constant in the absence of substrate, but for the uncompetitive inhibitor it is the dissociation constant when the enzyme is saturated with the substrate.

You have the Cheng-Prusoff relationship reversed: Kiapp (or IC50) = Ki(1 + [S]/Km) for a competitive inhibitor in a single-substrate reaction. From this you can see that IC50 is always higher than Ki for a competitive inhibitor, since you have to put some substrate into the reaction, and the IC50 increases in direct proportion to the substrate concentration. You could measure Ki by plotting IC50 v. [S] and measuring the slope, which would be Ki/Km and the intercept Ki. For a pure noncompetitive inhibitor, IC50=Ki and is not affected by substrate concentration. For a pure uncompetitive inhibitor, IC50 decreases as [S] increases, in a saturable manner.

"When you say Therefore, the lowest Kiapp value must be Kic. In competitive/noncompetitive mixed inhibition, the lowest Ki app value would theoretically be at 0 substrate right? If you are sufficiently below the KM, there should be virtually no competition for the inhibitor, right?"

Yes. In the theoretical limit, when [S] = 0, there is no ES complex for I to bind to as an uncompetitive inhibitor. There is only E. Therefore the only possible complex is EI and the inhibitor is purely competitive. The practical problem would be to get the substrate concentration close enough to zero to make the rate measurement possible. Instead, one could make a ligand binding measurement to get Kic, which might be a lot easier to do.

Hello Nathan Ponzar

I wondered how much your approach would change if, in addition to a mixed inhibition mechanism, we were faced with an enzyme that exhibits the phenomenon of substrate inhibition.

Thanks in advance!