Deriving Fractional Inhibition From Unbound Concentration

From Unbound Concentration to Inhibition

A concentration at the target is not itself an effect. A mechanistic model needs a relationship that maps that concentration to a change in target activity. For a reversible inhibitor, the simplest starting point is equilibrium binding: as unbound drug concentration rises, a larger fraction of target molecules is occupied. If occupied target is inactive, occupancy gives fractional inhibition directly. More flexible concentration-response models allow for incomplete inhibition and a non-unit slope.

The concentration used here is $C_u$, the unbound drug concentration at the target. It is not automatically the total plasma concentration or the total tissue concentration. The earlier material on unbound drug concentration at the intracellular target establishes why the concentration at the site of action is the relevant exposure input. This tutorial treats $C_u$ as known at the time of interest and focuses on the reversible concentration-response relationship.

Definition: Let $A_0$ be target activity in the absence of inhibitor and $A(C_u)$ be target activity at unbound concentration $C_u$. Fractional inhibition is $$ I(C_u)=\frac{A_0-A(C_u)}{A_0}. $$ Thus $I=0$ means no reduction from baseline, and $I=1$ means complete inhibition relative to baseline. The residual fraction of baseline activity is $1-I(C_u)$.

This definition presumes that baseline activity is positive and that the measured activity can be meaningfully compared with that baseline. In a concentration-response model with maximal fractional inhibition $I_{\max}$, $0\leq I_{\max}\leq 1$. If activity is measured on a scale where the no-drug baseline is normalized to one, predicted activity is $1-I(C_u)$. If the baseline is $A_0$ in activity units, predicted activity is $A_0[1-I(C_u)]$.

Deriving the Simple Reversible Occupancy Relationship

Consider a one-site binding model in which free target $T$ and unbound inhibitor $D$ reversibly form an inactive complex $TD$. At equilibrium, the dissociation constant $K_d$ is defined as the product of the free concentrations divided by the complex concentration. Treating $C_u$ as the free inhibitor concentration at the target, the fraction of total target bound is obtained directly from this equilibrium relationship.

T+D \rightleftharpoons TD, \qquad K_d=\frac{[T]C_u}{[TD]}, \qquad \theta=\frac{[TD]}{[T]+[TD]}.

Rearrange the definition of $K_d$ to get $[TD]/[T]=C_u/K_d$. Substituting this ratio into the expression for $\theta$ gives the occupancy equation:

\theta =\frac{[TD]/[T]}{1+[TD]/[T]} =\frac{C_u/K_d}{1+C_u/K_d} =\frac{C_u}{K_d+C_u}.

If binding and unbinding are at equilibrium, the unbound concentration is the free concentration available to bind, and bound target is fully inactive while unbound target retains baseline activity, then the fraction of target inhibited equals $\theta$. The model therefore predicts $I(C_u)=C_u/(K_d+C_u)$. At $C_u=K_d$, half the target is occupied and inhibited. This special case has complete maximal inhibition and a unit Hill slope.

Key result: For one-site reversible binding with complete inhibition upon binding, $$ I(C_u)=\frac{C_u}{K_d+C_u}. $$ This is a special case of a broader Hill concentration-response model: $I_{\max}=1$, $IC_{50}=K_d$, and $h=1$.

The equality between $IC_{50}$ and $K_d$ is not universal. It follows here from the specified simple binding assumptions and the definition of inhibition as target occupancy. In more general concentration-response models, $IC_{50}$ is the concentration producing half of the model’s maximal inhibition. It is a potency parameter for the modeled response, not automatically a binding constant.

The Hill Concentration-Response Model

A useful model for reversible concentration-response behavior is the Hill equation. It separates three features of the predicted inhibition: the concentration scale, the steepness of the response, and the maximal inhibition. The Hill slope $h$ allows the curve to be shallower or steeper than the one-site occupancy curve; in a phenomenological model, it should not by itself be interpreted as the number of binding sites.

Definition: The Hill model for fractional inhibition is $$ I(C_u)=I_{\max}\frac{C_u^h}{IC_{50}^h+C_u^h}, $$ where $IC_{50}>0$ is the unbound concentration giving half of $I_{\max}$, $h>0$ is the Hill slope, and $I_{\max}$ is the maximal fractional inhibition. At $C_u=IC_{50}$, $I(C_u)=I_{\max}/2$.

The equation can be understood in two stages. First, the Hill term $C_u^h/(IC_{50}^h+C_u^h)$ ranges from zero toward one as concentration rises. It is the fraction of the model’s attainable effect. Multiplying by $I_{\max}$ scales that fraction to the maximum reduction in baseline activity. Consequently, a drug with $I_{\max}=0.7$ cannot inhibit more than 70% of baseline activity in this model, no matter how large $C_u$ becomes.

Justification and Useful Consequences

For the one-site model, the fractional occupancy odds are $\theta/(1-\theta)$. Substituting the occupancy expression shows that these odds equal $C_u/K_d$. More generally, the Hill curve describes a response whose odds on the normalized effect scale are proportional to a power of concentration. Define the normalized effect $q=I/I_{\max}$. For $0<q<1$, the model implies:

q=\frac{C_u^h}{IC_{50}^h+C_u^h}, \qquad \frac{q}{1-q} =\left(\frac{C_u}{IC_{50}}\right)^h.

Solving the second expression for concentration gives a useful inverse relationship. It applies only to a desired inhibition strictly between zero and $I_{\max}$; the model approaches its maximum asymptotically rather than reaching it at a finite concentration.

C_u =IC_{50}\left(\frac{I}{I_{\max}-I}\right)^{1/h}, \qquad 0<I<I_{\max}.

The model also has clear limiting behavior. As $C_u$ approaches zero, $I(C_u)$ approaches zero. As $C_u$ grows without bound, $I(C_u)$ approaches $I_{\max}$. If $I_{\max}=1$ and $h=1$, the Hill equation reduces to the one-site occupancy form when $IC_{50}=K_d$. These checks are useful both when interpreting a fitted model and when implementing it computationally.

Worked Example: One-Site Inhibition at and Above IC50

Suppose a hypothetical reversible inhibitor is described by complete inhibition at occupied target, with $K_d=IC_{50}=40$ nM, $I_{\max}=1$, and $h=1$. Calculate fractional inhibition and residual activity at $C_u=40$ nM and $C_u=120$ nM. The one-site model assumes equilibrium binding, a free target-site concentration input, and inactive bound target; these are stipulated model conditions in this example.

At 40 nM, substitute into the occupancy equation:

I(40)=\frac{40}{40+40} =\frac{40}{80} =0.500.

Thus inhibition is 50.0% of baseline activity and residual activity is $1-0.500=0.500$, or 50.0% of baseline.

At 120 nM:

I(120)=\frac{120}{40+120} =\frac{120}{160} =0.750.

Inhibition is 75.0%, and residual activity is $1-0.750=0.250$, or 25.0% of baseline. Tripling the concentration from $IC_{50}$ does not triple inhibition: the response approaches its maximum rather than increasing in direct proportion to concentration.

Worked Example: Partial Maximal Inhibition and a Steeper Slope

A hypothetical inhibitor has $I_{\max}=0.80$, $IC_{50}=25$ nM, and $h=2$. Predict inhibition and residual activity at $C_u=50$ nM. Assume the fitted Hill model is appropriate over this concentration range and that $C_u$ is the unbound concentration at the target.

Substituting the values into the model gives:

I(50) =0.80\frac{50^2}{25^2+50^2} =0.80\frac{2500}{625+2500} =0.80(0.800) =0.640.

The predicted inhibition is 64.0% of baseline activity. The residual fraction is $1-0.640=0.360$, so predicted activity is 36.0% of baseline. The Hill term is 0.800, meaning the response has reached 80.0% of its attainable maximum; it does not mean 80.0% inhibition, because the attainable maximum is only 0.80.

At $C_u=IC_{50}=25$ nM, the model would predict $I=I_{\max}/2=0.400$, not 0.500. This illustrates why the model defines $IC_{50}$ as the concentration giving half of maximal inhibition, especially when $I_{\max}<1$.

Worked Example: Finding Concentration for a Desired Inhibition

For a hypothetical compound, let $I_{\max}=0.80$, $IC_{50}=30$ nM, and $h=1.5$. Find the unbound target concentration predicted to produce 60.0% fractional inhibition. Assume the desired effect is within the model’s attainable range and that its parameters apply at the time and in the system being modeled.

Here $I=0.60$, which is below $I_{\max}=0.80$, so the inverse equation can be used:

C_u =30\left(\frac{0.60}{0.80-0.60}\right)^{1/1.5} =30(3)^{2/3} \approx 30(2.080) \approx 62.40\text{ nM}.

The model therefore predicts 60.0% inhibition at approximately 62.4 nM unbound concentration. This is a model-based concentration prediction, not a guarantee that a particular administered dose will produce that target-site concentration.

Interpreting Parameters and Checking the Model

Each parameter changes the curve in a distinct way, although their effects should be interpreted together. A smaller $IC_{50}$ shifts the concentration-response curve toward lower concentrations: less unbound drug is required to reach a given fraction of $I_{\max}$. A larger $IC_{50}$ indicates lower potency in this modeled system. Changing $IC_{50}$ does not change the limiting maximum when $I_{\max}$ is held fixed.

The Hill slope controls how rapidly the response changes around $IC_{50}$. For $h=1$, the rise is the standard hyperbolic occupancy shape. For $h>1$, the transition around $IC_{50}$ is steeper; for $0<h<1$, it is shallower. $I_{\max}$ sets the upper limit. A lower $I_{\max}$ means that increasing exposure cannot produce as large a reduction in baseline activity under this model.

A practical diagnostic is to compare observations with the model across the concentration range rather than checking a single predicted value. Look for systematic discrepancies, such as consistent underprediction at low concentrations, overprediction near the plateau, or an observed plateau that differs from the specified $I_{\max}$. Also check whether residual activity remains nonnegative, whether the observed response lies within the allowed range, and whether the input is truly unbound target-site concentration. A fitted curve that appears adequate only because data are sparse near the transition or plateau may leave potency, slope, or maximal inhibition poorly determined.

Confirm the response definition.
Determine what activity is used as $A_0$, and verify that the reported effect is a reduction relative to that baseline.

Confirm the concentration input.
Use unbound concentration at the target, with compatible concentration units for $C_u$ and $IC_{50}$.

Calculate and check limits.
Apply the Hill equation, verify $0\leq I(C_u)\leq I_{\max}\leq1$, then calculate residual activity as $1-I(C_u)$.

The simple occupancy model is appropriate when equilibrium binding is a reasonable approximation and the occupied target is functionally inactive. A fitted Hill model is more flexible, but that flexibility does not establish a binding mechanism. Alternative approaches include the mechanistic one-site occupancy equation when its assumptions are supported, or a more detailed model when the target response is not adequately represented by an instantaneous equilibrium curve. A direct linear relationship may be a local approximation over a narrow concentration range, but it does not naturally represent a saturating effect. A step-like threshold model may be useful for a decision rule, but it is not a smooth reversible concentration-response relationship.

Assumptions and Limitations

The concentration-response model is a static relationship: for a given $C_u$, it returns one predicted inhibition. It assumes that any binding and functional equilibration needed for that relationship are sufficiently fast, or that the model is being applied at equilibrium. If concentration changes faster than the target response can follow, a single-valued curve may not describe the time course. The earlier material on effect-site equilibration addresses delayed equilibration between an exposure compartment and the site of action; the Hill equation here does not itself represent that delay.

The simple one-site derivation also assumes a reversible binding equilibrium, a concentration input that represents free inhibitor available at the target, and a direct link between bound fraction and inhibition. If bound target retains activity, if multiple processes contribute to measured activity, or if the maximal effect is incomplete, the equality between occupancy and inhibition does not hold. The Hill model can represent incomplete maximum and non-unit slope, but its parameters remain descriptive unless additional mechanistic evidence supports a particular interpretation.

Finally, concentration-response parameters are conditional on the system and response definition in which they are determined. A potency value measured under one set of assay conditions should not be assumed to apply unchanged in another context. The model predicts inhibition given $C_u$; it does not by itself calculate drug exposure from dose, account for changing substrate conditions, or translate target inhibition into a downstream pathway response.

Common Mistakes

Using total concentration as though it were target-site unbound concentration. The model input is $C_u$; substituting a different exposure measure requires justification.
Calling $IC_{50}$ the concentration for 50% absolute inhibition when $I_{\max}<1$. It gives half of maximal inhibition, so the predicted inhibition at $IC_{50}$ is $I_{\max}/2$.
Interpreting $h$ as a count of binding sites. A non-unit Hill slope describes curve steepness in this model and does not, on its own, establish a binding stoichiometry.
Assuming $IC_{50}=K_d$ in every model. That equality follows for the simple one-site, complete-inhibition occupancy model, not as a general identity.
Reporting inhibition without residual activity or a baseline. State the effect relative to the defined $A_0$, and use $1-I$ for the residual fraction of baseline.

Key takeaway: The Hill model maps unbound target-site concentration to fractional inhibition while keeping potency ($IC_{50}$), curve steepness ($h$), and maximal inhibition ($I_{\max}$) distinct. Its prediction is meaningful only when the concentration, response definition, and assumptions match the modeled system.

Check Your Understanding

Try each question, then compare with the solutions below.

A complete-inhibition one-site model has $K_d=IC_{50}=20$ nM and $h=1$. What inhibition and residual activity does it predict at $C_u=20$ nM?
A Hill model has $I_{\max}=0.60$, $IC_{50}=10$ nM, and $h=1$. What inhibition does it predict at $C_u=10$ nM? Explain why this is not 50.0% inhibition.
For a model with $I_{\max}=0.90$, $IC_{50}=15$ nM, and $h=2$, calculate inhibition at $C_u=30$ nM.
Can a model with $I_{\max}=0.70$ predict 80.0% fractional inhibition at any finite concentration? Explain.
State two assumptions needed to identify the one-site occupancy fraction with fractional inhibition.

Solutions. (1) $I=20/(20+20)=0.500$, so inhibition is 50.0% and residual activity is 50.0% of baseline. (2) At $IC_{50}$, inhibition is $I_{\max}/2=0.60/2=0.300$, or 30.0%; $IC_{50}$ marks half the maximum, not necessarily 50.0% absolute inhibition. (3) $I=0.90[30^2/(15^2+30^2)]=0.90(900/1125)=0.720$, or 72.0%. (4) No. The model’s inhibition approaches $I_{\max}=0.70$ as concentration increases, so 80.0% exceeds its maximum. (5) Examples include reversible binding at equilibrium with free inhibitor concentration represented by $C_u$, and complete inactivity of bound target with unbound target retaining baseline activity.