Tutorials › Pharmacokinetic-Pharmacodynamic (PK/PD) Modeling › Interpreting Selectivity Across Multiple Targets

Target Binding, Occupancy and Engagement · Tutorial 240 of 598

Interpreting Selectivity Across Multiple Targets

Learn to compare target affinity and fractional occupancy at matched free concentrations, and to interpret how apparent selectivity changes across the concentration range.

Advanced 11 min read

What You'll Learn

  • Define an affinity selectivity ratio and interpret its direction.
  • Calculate fractional occupancy for multiple targets at the same free concentration.
  • Distinguish a constant affinity or occupancy-odds ratio from a concentration-dependent occupancy difference.
  • Derive the concentration at which the occupancy difference between two targets is greatest.
  • Check whether a concentration range can meet specified on-target and off-target occupancy thresholds.
  • Identify assumptions and diagnostics that matter when comparing predicted target selectivity.

Why Selectivity Depends on Concentration

A compound can bind one target more tightly than another and still occupy both targets substantially at sufficiently high free concentration. Conversely, at very low concentration, occupancy of both targets may be small even when their affinities differ greatly. Thus, an affinity comparison and an occupancy comparison answer related but distinct questions: affinity describes binding strength, while occupancy describes the predicted fraction of a target bound at a specified concentration.

To make a meaningful occupancy comparison, evaluate each target at the same free drug concentration. Comparing occupancies at different concentrations can make a target appear more or less selective simply because the exposure differs. The comparison here uses a simple equilibrium model; it does not describe competition among ligands or more complex binding mechanisms.

Definition: For target \(j\), let \(K_{d,j}\) be the equilibrium dissociation constant and let \(C\) be the free drug concentration at equilibrium. Under a reversible, one-to-one binding model, the fractional occupancy is $$ \theta_j(C)=\frac{C}{K_{d,j}+C}. $$ The affinity selectivity ratio of target A relative to target B is defined here as \(S_{A:B}=K_{d,B}/K_{d,A}\). If \(S_{A:B}>1\), target A has higher affinity than target B because its \(K_d\) is lower. The ratio is dimensionless, and its direction must be stated whenever it is reported.

This model assumes a single class of independent binding sites, reversible binding at equilibrium, and a known free concentration that is effectively the same for both targets. It also assumes that binding does not appreciably deplete free drug and that the reported \(K_d\) values refer to comparable conditions. Under these assumptions, lower \(K_d\) means higher affinity, but occupancy also depends on \(C\).

Affinity Ratios and Occupancy Comparisons

A useful way to separate affinity selectivity from fractional occupancy is to compare the occupancy odds, \(\theta_j/(1-\theta_j)\). Rearranging the one-site equation gives \(\theta_j/(1-\theta_j)=C/K_{d,j}\). Therefore, when both targets are evaluated at the same positive free concentration,

$$ \frac{\theta_A/(1-\theta_A)}{\theta_B/(1-\theta_B)} = \frac{K_{d,B}}{K_{d,A}} = S_{A:B}. $$

The ratio of occupancy odds equals the affinity selectivity ratio and does not vary with concentration in this model. The ratio of fractional occupancies, \(\theta_A/\theta_B\), does vary with concentration, as does their difference, \(\theta_A-\theta_B\). At low concentrations, occupancy is approximately proportional to concentration divided by \(K_d\), so the fractional-occupancy ratio approaches the affinity ratio. At high concentrations, both occupancies approach one, so their ratio approaches one and their difference approaches zero. These patterns explain why a strong affinity difference need not look like a similarly large difference in fractional occupancy at every concentration.

Worked Example: Translating Affinity Selectivity Into Occupancy

Suppose two targets have hypothetical dissociation constants \(K_{d,A}=10\) nM and \(K_{d,B}=100\) nM. Calculate the affinity selectivity ratio and compare predicted occupancy at a matched free concentration of 10 nM.

The ratio is \(S_{A:B}=K_{d,B}/K_{d,A}=100/10=10\). Thus, target A has tenfold higher affinity than target B under the stated model and conditions.

At \(C=10\) nM, target A occupancy is \(\theta_A=10/(10+10)=0.500\), or 50.0%. Target B occupancy is \(\theta_B=10/(100+10)=0.09091\), or 9.09%. The occupancy difference is \(0.500-0.09091=0.4091\), or 40.9 percentage points; the fractional-occupancy ratio is \(0.500/0.09091=5.50\), not 10.

The odds are \(0.500/(1-0.500)=1\) for A and \(0.09091/(1-0.09091)=0.1000\) for B. Their ratio is \(1/0.1000=10\), matching the affinity selectivity ratio. The example illustrates why the odds ratio, not the fractional-occupancy ratio, is constant across concentration for this model.

Where Is the Occupancy Difference Largest?

For two targets where \(K_{d,A}<K_{d,B}\), the occupancy difference at a common concentration is

$$ \Delta(C)=\theta_A(C)-\theta_B(C) =\frac{C(K_{d,B}-K_{d,A})}{(C+K_{d,A})(C+K_{d,B})}. $$

This difference is zero at \(C=0\), and tends toward zero as \(C\) becomes very large. Its maximum occurs between those extremes. Differentiating with respect to \(C\), and writing \(a=K_{d,A}\) and \(b=K_{d,B}\), gives

$$ \frac{d\Delta}{dC} =(b-a)\frac{ab-C^2}{(C+a)^2(C+b)^2}. $$

Because \(b-a>0\) and the denominator is positive, the derivative is positive when \(C^2<ab\), zero when \(C^2=ab\), and negative when \(C^2>ab\). Consequently, the maximum occurs at \(C=\sqrt{K_{d,A}K_{d,B}}\), the geometric mean of the two dissociation constants. Substitution into \(\Delta(C)\) gives

$$ \Delta_{\max} =\frac{\sqrt{K_{d,B}}-\sqrt{K_{d,A}}} {\sqrt{K_{d,B}}+\sqrt{K_{d,A}}}. $$

This result concerns the absolute difference in fractional occupancy, measured in percentage points. It does not mean that the drug is selectively binding only one target at that concentration; it identifies where the model predicts the largest separation in occupancy between the two targets.

Worked Example: How the Occupancy Difference Changes Across Concentrations

Use the hypothetical \(K_{d,A}=10\) nM and \(K_{d,B}=100\) nM values from the previous calculation. Compare occupancy at 1, 10, 100, and 1,000 nM free drug, and locate the greatest predicted occupancy difference.

At 1 nM, \(\theta_A=1/(10+1)=0.09091\) and \(\theta_B=1/(100+1)=0.009901\). The difference is \(0.08101\), or 8.10 percentage points.

At 10 nM, the occupancies are \(0.500\) and \(10/110=0.09091\), giving a difference of \(0.4091\), or 40.9 percentage points. At 100 nM, they are \(100/110=0.9091\) and \(100/200=0.500\), again a difference of \(0.4091\), or 40.9 percentage points. At 1,000 nM, they are \(1000/1010=0.9901\) and \(1000/1100=0.9091\), a difference of \(0.08101\), or 8.10 percentage points.

The maximum occurs at \(C=\sqrt{10\times100}=\sqrt{1000}=31.62\) nM. At this concentration, \(\theta_A=31.62/(31.62+10)=0.7597\) and \(\theta_B=31.62/(31.62+100)=0.2403\). The difference is \(0.5194\), or 51.9 percentage points. The same result follows from \(\Delta_{\max}=(\sqrt{100}-\sqrt{10})/(\sqrt{100}+\sqrt{10})=0.5195\), with the small rounding discrepancy due to displayed values.

The affinity ratio remains 10 at each concentration, but the fractional-occupancy ratio and occupancy difference change. At high concentration the targets are both nearly saturated, so their occupancies look similar despite the unchanged affinity ratio.

Using Occupancy Thresholds to Assess a Selective Range

A practical question is whether any concentration can achieve a specified occupancy on a desired target while keeping occupancy of another target below a specified limit. For example, under the one-site model, \(\theta_A\geq0.90\) requires \(C\geq9K_{d,A}\): this follows by solving \(C/(K_{d,A}+C)\geq0.90\). Likewise, \(\theta_B\leq0.10\) requires \(C\leq K_{d,B}/9\). Both conditions can hold only if \(9K_{d,A}\leq K_{d,B}/9\), or \(K_{d,B}/K_{d,A}\geq81\).

Worked Example: Checking Whether an Occupancy Window Exists

Suppose a hypothetical compound has \(K_{d,A}=2\) nM for the desired target and \(K_{d,B}=100\) nM for a second target. Determine whether there is a free concentration that gives at least 90% occupancy of A while keeping B at or below 10%.

For A, the requirement is \(C\geq9(2)=18\) nM. For B, it is \(C\leq100/9=11.11\) nM. No concentration satisfies both inequalities because \(18>11.11\). Equivalently, the affinity ratio is \(100/2=50\), which is below the required ratio of 81.

At the minimum concentration that reaches 90% occupancy of A, \(C=18\) nM, occupancy of B is \(18/(100+18)=0.1525\), or 15.3%. The 90%/10% target pair is therefore not attainable under this model, even though A has 50-fold higher affinity. This calculation is a binding-model prediction, not evidence about clinical efficacy or safety.

Assumptions, Diagnostics, and Alternatives

The equations are only as useful as their assumptions and inputs. A selectivity comparison should state the targets, the \(K_d\) values and their measurement conditions, the matched free concentration, and whether occupancy is measured or model-predicted. Check that concentrations and \(K_d\) values use the same units. If total concentration is substituted for free concentration, the calculation can be misleading when protein binding, nonspecific binding, or target binding makes the free fraction materially different or target-dependent.

Look for signs that the simple model is not adequate. Occupancy observations that systematically depart from \(C/(K_d+C)\), or fitted affinity estimates that change substantially with the concentration range, may indicate model misspecification, non-equilibrium behavior, heterogeneity, or cooperative binding. Differences between assay conditions, uncertain free concentrations, and uncertainty in fitted \(K_d\) values can also change the predicted selectivity. In such cases, compare like with like, examine uncertainty, and use a binding model appropriate to the evidence rather than treating the ratio as an exact property of every setting.

The affinity ratio is a compact way to rank binding strength, while matched-concentration occupancy is more directly interpretable for a particular exposure. Occupancy difference is useful when the practical concern is separation in percentage points; the occupancy-odds ratio is useful when relating the comparison directly to the affinity ratio under the one-site model. These summaries are not interchangeable. A plot of each target’s occupancy against log free concentration can reveal where both are unsaturated, where their occupancy difference is largest, and where both approach saturation.

The independent-target equation also does not account for other ligands competing for the same binding site. If competition or ligand depletion is important, \(C/(K_d+C)\) may not represent occupancy, and the simple cross-target comparison can fail. That is a reason to use a model that includes the relevant binding processes, not a reason to interpret a simple-model affinity ratio as an observed occupancy ratio. Nor does fractional occupancy by itself establish a functional response: biological effects can depend on target abundance, signaling, and other mechanisms.

Common Pitfall

Incorrect reasoning: “Target A has tenfold higher affinity, so it will have tenfold higher fractional occupancy at every concentration.” This confuses the affinity ratio with the fractional-occupancy ratio. Under the stated model, the ratio of occupancy odds equals the affinity ratio; fractional occupancies themselves change nonlinearly with concentration and both approach one at high concentration.

Correction: Specify the matched free concentration, calculate each occupancy separately, and report the comparison that answers the question. State the direction of the \(K_d\) ratio, and do not infer an occupancy difference from the affinity ratio alone.

Key takeaway: For reversible one-to-one binding at equilibrium, lower \(K_d\) indicates higher affinity, and the affinity ratio equals the ratio of occupancy odds at matched free concentration. The fractional occupancies and their difference depend on concentration; in particular, the occupancy difference between two targets is greatest at the geometric mean of their \(K_d\) values.

Check Your Understanding

Answer each question using the one-site equilibrium model, and then compare your reasoning with the solutions.

  1. Targets A and B have \(K_d\) values of 4 nM and 40 nM. Which has higher affinity, and what is \(S_{A:B}\)?
  2. For the targets in question 1, calculate both occupancies at a matched free concentration of 4 nM. Give the occupancy difference in percentage points.
  3. At what concentration is the occupancy difference between targets with \(K_{d,A}=4\) nM and \(K_{d,B}=40\) nM greatest?
  4. Could these targets reach at least 90% occupancy of A while maintaining no more than 10% occupancy of B? Show the concentration constraints.

Solutions. (1) A has higher affinity because its \(K_d\) is lower. \(S_{A:B}=40/4=10\). (2) \(\theta_A=4/(4+4)=0.500\), or 50.0%; \(\theta_B=4/(40+4)=0.09091\), or 9.09%. The difference is \(0.500-0.09091=0.4091\), or 40.9 percentage points. (3) The greatest difference occurs at \(C=\sqrt{4\times40}=\sqrt{160}=12.65\) nM. (4) A requires \(C\geq9(4)=36\) nM, while B requires \(C\leq40/9=4.444\) nM. Since \(36>4.444\), no concentration meets both requirements under this model.

Clinical Biostats