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Receptor Theory Behind Concentration-Effect Curves · Tutorial 75 of 598

Competition Between Full and Partial Agonists

Learn to calculate the combined response of competing full and partial agonists and identify when the partial agonist lowers the response.

Advanced 12 min read

What You'll Learn

  • Express each agonist concentration relative to its dissociation constant.
  • Calculate receptor occupancy when two agonists compete for one receptor population.
  • Combine occupancy and operational efficacy to calculate response.
  • Determine when increasing partial-agonist concentration raises or lowers response.
  • Solve for the partial-agonist concentration needed to reach a specified response.
  • Recognize assumptions and diagnostics for the shared-receptor equilibrium model.

Why Competition Can Lower a Response

A partial agonist can activate a receptor, but it produces less response than a full agonist in the same system at sufficiently high concentrations. If both drugs compete for the same receptor population, increasing the partial agonist can displace some full agonist. The resulting response depends on both occupancy and efficacy: displacement lowers response only when the partial agonist’s contribution is less than the response it replaces.

The calculations below use a shared-receptor equilibrium model with an operational transducer. This makes the distinction between who occupies receptors and how strongly each occupied receptor contributes to response explicit. It also accounts for a nonlinear relationship between receptor stimulus and measured response.

Definition: Let \(A\) be a full agonist and \(B\) a partial agonist competing reversibly for one receptor population. Let \([A]\) and \([B]\) be their free concentrations, and \(K_A\) and \(K_B\) their equilibrium dissociation constants. Define \(x=[A]/K_A\) and \(y=[B]/K_B\). Under equilibrium, mutually exclusive binding, and a single shared receptor population, the fractions occupied by \(A\), \(B\), and neither ligand are respectively \(x/(1+x+y)\), \(y/(1+x+y)\), and \(1/(1+x+y)\).

The ratios \(x\) and \(y\) are dimensionless. A ligand’s concentration relative to its own \(K\) determines its contribution to occupancy: for example, \([A]=K_A\) gives \(x=1\). The two ligands need not have the same affinity; their concentrations enter the competition through separate ratios.

From Occupancy to Response

Let \(\tau_A\) and \(\tau_B\) denote the operational efficacy parameters of the two agonists in the system. They summarize ligand efficacy together with features of the receptor system, such as receptor abundance. For a partial agonist, \(0<\tau_B<\tau_A\) in the system being modeled. The combined stimulus \(S\) is the efficacy-weighted sum of the occupied fractions:

$$ S=\tau_A\frac{x}{1+x+y}+\tau_B\frac{y}{1+x+y} =\frac{\tau_Ax+\tau_By}{1+x+y}. $$

Assume the common operational transducer maps stimulus to response as \(E=E_0+E_{\max}S/(1+S)\), where \(E_0\) is the response with no agonist and \(E_{\max}\) is the system’s response scale. The normalized response above baseline is therefore \(S/(1+S)\). Because this mapping increases with \(S\), comparisons of responses can be made by comparing stimuli.

Model assumptions: The ligands bind at equilibrium and compete for the same mutually exclusive binding site; \(K_A\) and \(K_B\) are constant; the concentrations used are free concentrations at the receptor; occupied receptors contribute additively to stimulus through \(\tau_A\) and \(\tau_B\); and both ligands act through the same operational transducer. The model does not represent time-dependent binding, multiple receptor populations, or concentration-dependent changes in transduction.

This response model is not the claim that response is always proportional to occupancy. A receptor system with amplification or reserve can respond nonlinearly to stimulus. If the response is known to be linear in stimulus over the range studied, a linear stimulus-to-response mapping may be more appropriate; the occupancy calculation remains the same, but the response calculation changes.

Worked Example: Calculating Response as Partial Agonist Concentration Rises

Suppose hypothetical concentrations are \([A]=8\) nM and \([B]=0\), 10, or 90 nM. Let \(K_A=2\) nM, \(K_B=10\) nM, \(\tau_A=8\), and \(\tau_B=1\). Calculate the receptor fractions and normalized response \(r=(E-E_0)/E_{\max}\).

First, \(x=8/2=4\). At \([B]=0\), \(y=0\), so the fraction occupied by \(A\) is \(4/(1+4)=0.8000\). Thus \(S=8(0.8000)=6.400\), and

$$ r=\frac{S}{1+S}=\frac{6.400}{7.400}=0.8649. $$

At \([B]=10\) nM, \(y=10/10=1\), and the shared denominator is \(1+4+1=6\). The fractions occupied by \(A\), occupied by \(B\), and unoccupied are \(4/6\), \(1/6\), and \(1/6\). Therefore,

$$ S=8\left(\frac{4}{6}\right)+1\left(\frac{1}{6}\right) =\frac{33}{6}=5.500, \qquad r=\frac{5.500}{6.500}=0.8462. $$

At \([B]=90\) nM, \(y=9\) and the denominator is \(14\). The corresponding stimulus and response are

$$ S=8\left(\frac{4}{14}\right)+1\left(\frac{9}{14}\right) =\frac{41}{14}=2.929, \qquad r=\frac{2.929}{3.929}=\frac{41}{55}=0.7455. $$

In this example, adding more partial agonist reduces the response from 0.8649 to 0.7455 of the response scale above baseline. The partial agonist displaces the more efficacious full agonist, and its own efficacy is not enough to maintain the initial stimulus.

When Does Adding the Partial Agonist Reduce Response?

At a fixed full-agonist concentration, the stimulus with no partial agonist is \(S_A=\tau_Ax/(1+x)\). To determine what happens as \(y\) increases, differentiate the combined stimulus with respect to \(y\):

$$ \frac{\partial S}{\partial y} = \frac{\tau_B(1+x)-\tau_Ax}{(1+x+y)^2} = \frac{(1+x)(\tau_B-S_A)}{(1+x+y)^2}. $$

The denominator is positive, as is \(1+x\). Thus the sign of the derivative is the sign of \(\tau_B-S_A\). If \(S_A>\tau_B\), increasing partial-agonist concentration lowers stimulus and therefore lowers response. If \(S_A<\tau_B\), it raises stimulus and response. If \(S_A=\tau_B\), stimulus is unchanged as \(y\) varies. This result follows because the competing ligand shifts occupancy toward a ligand whose operational efficacy is either below or above the stimulus level it replaces.

As \(y\) becomes very large, nearly all receptors are occupied by \(B\), so \(S\) approaches \(\tau_B\) and the normalized response approaches \(\tau_B/(1+\tau_B)\). This limit does not mean that every addition of \(B\) must lower response: if the full agonist’s initial stimulus is below \(\tau_B\), the response rises toward the partial agonist’s single-ligand limit instead.

Worked Example: A Partial Agonist Can Initially Increase Response

Consider hypothetical values \(K_A=10\) nM, \([A]=1\) nM, \(\tau_A=8\), \(K_B=5\) nM, and \(\tau_B=1\). Here \(x=0.1\), and with no \(B\), \(S_A=8(0.1)/(1.1)=0.7273\). Since \(S_A<\tau_B=1\), the derivative result predicts that adding \(B\) initially raises response.

At \([B]=5\) nM, \(y=1\). Substitution gives

$$ S=\frac{8(0.1)+1(1)}{1+0.1+1} =\frac{1.8}{2.1}=0.8571, \qquad r=\frac{0.8571}{1.8571}=0.4615. $$

With no \(B\), \(r=0.7273/1.7273=0.4211\). Thus the normalized response increases from 0.4211 to 0.4615. As \(B\) becomes very high, \(S\) approaches 1 and \(r\) approaches \(1/(1+1)=0.5000\). The partial agonist is still less efficacious than \(A\), but at this low concentration of \(A\), its efficacy exceeds the stimulus produced by the full agonist alone.

Solving for a Concentration That Gives a Target Response

The model can also be rearranged to find the partial-agonist concentration associated with a target response. First convert the target normalized response \(r\) to stimulus using \(S=r/(1-r)\). Then solve the stimulus equation for \(y\):

$$ S=\frac{\tau_Ax+\tau_By}{1+x+y} \quad\Longrightarrow\quad y=\frac{\tau_Ax-S(1+x)}{S-\tau_B}, \qquad [B]=K_By. $$

This solution is meaningful only if it gives \(y\geq 0\), and the target must be compatible with the range of stimulus attainable by varying \(B\) at the fixed \(A\) concentration. In particular, when \(S=\tau_B\), the rearrangement has a zero denominator: that stimulus is the limiting value as \(y\) grows without bound, unless the response is already constant at that value.

Worked Example: Finding the Concentration for a Specified Response

Use \(x=4\), \(\tau_A=8\), \(\tau_B=1\), and \(K_B=10\) nM. Find the \(B\) concentration that reduces the normalized response to 0.8000.

Convert the target response to stimulus: \(S=r/(1-r)=0.8000/0.2000=4\). Substitute into the rearranged equation:

$$ y=\frac{8(4)-4(1+4)}{4-1} =\frac{32-20}{3}=4. $$

Since \(y=[B]/K_B\), \([B]=10(4)=40\) nM. Checking directly, the denominator is \(1+4+4=9\), and \(S=(8(4)+1(4))/9=36/9=4\). The normalized response is \(4/(1+4)=0.8000\), as required.

Assumptions, Diagnostics, and Alternatives

The calculations are useful only to the extent that the assumptions match the experiment. Equilibrium may be doubtful if the assay is read before binding has stabilized. The use of free concentrations may be inappropriate if ligand depletion is substantial. A single shared receptor population may be inadequate if the data show clearly distinct response phases or different binding behavior. These concerns call for checking the experimental design and model fit rather than treating a calculated occupancy as directly observed.

A practical diagnostic is to compare observed responses over a concentration matrix that varies both agonists. Systematic residual patterns across combinations of \([A]\) and \([B]\) can indicate that the shared transducer or binding assumptions are inadequate. It is also useful to compare the predicted limiting responses when either ligand is present alone with the observed single-ligand concentration-response data. A model that reproduces one agonist alone but consistently misses mixtures may have an incorrect efficacy-weighted competition structure.

The main alternatives differ in what they assume about response generation. A linear occupancy-response model makes response proportional to efficacy-weighted occupancy; it can be suitable over a range where transduction is approximately linear. The operational transducer used here allows a nonlinear response to stimulus. A simple sum of the two drugs’ responses is generally not an appropriate mixture calculation under shared-site competition, because the drugs cannot each occupy the same receptor at once. A competitive antagonist is a different limiting case: it has no activating efficacy in this model, whereas a partial agonist has \(\tau_B>0\).

ApproachResponse assumptionKey limitation
Shared-site occupancy with operational transducerEfficacy-weighted occupancy produces a common stimulus, then a nonlinear responseRequires the stated equilibrium, shared-site, and common-transducer assumptions
Linear occupancy-response approximationResponse changes in direct proportion to efficacy-weighted occupancyMay not capture amplification or response saturation
Adding single-drug responsesEach drug’s response is calculated separately and summedIgnores mutually exclusive receptor occupancy and can overpredict mixtures

A common mistake is to assume that a partial agonist must always lower response because it is less efficacious than a full agonist. That reasoning compares the ligands’ efficacy parameters but ignores the initial stimulus. The correct comparison at fixed \(A\) is \(S_A\) versus \(\tau_B\): adding \(B\) lowers response when \(S_A>\tau_B\), and raises it when \(S_A<\tau_B\). The phrase “partial agonist” describes its efficacy relative to the full agonist in the system; it does not determine the direction of every mixture response.

Key takeaway: Competition changes both occupancy and response. Calculate the dimensionless concentration ratios, form the efficacy-weighted shared-site stimulus, and pass that stimulus through the stated response model. Whether the partial agonist reduces response depends on the full agonist’s starting stimulus relative to the partial agonist’s operational efficacy.

Check Your Understanding

Use the shared-site model in this tutorial. The solutions are included below each question.

  1. If \(x=2\) and \(y=3\), what fraction of receptors is occupied by \(B\)?
    Solution: The denominator is \(1+2+3=6\), so the fraction occupied by \(B\) is \(3/6=0.5000\).
  2. At fixed \(x\), what condition determines whether increasing \(y\) lowers response?
    Solution: It lowers response when \(S_A=\tau_Ax/(1+x)>\tau_B\), because \(\partial S/\partial y\) then is negative and the response function increases with stimulus.
  3. Let \(x=1\), \(\tau_A=6\), and \(\tau_B=1\). Does adding \(B\) raise or lower response from the no-\(B\) condition?
    Solution: \(S_A=6(1)/(1+1)=3\). Since \(3>1\), increasing \(B\) lowers response.
  4. For \(S=2\), \(x=1\), \(\tau_A=6\), and \(\tau_B=1\), calculate \(y\).
    Solution: \(y=[6(1)-2(1+1)]/(2-1)=(6-4)/1=2\). Thus the required partial-agonist concentration is \([B]=2K_B\).
  5. Why is adding the two single-drug responses not generally valid for these mixtures?
    Solution: Under mutually exclusive shared-site binding, the agonists compete for the same receptors. Their occupied fractions are determined jointly by \(1+x+y\), so responses calculated separately cannot simply be added as if each ligand acted on an independent receptor population.
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