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Receptor Theory Behind Concentration-Effect Curves · Tutorial 69 of 598

Fractional Occupancy Under Reversible One-Site Binding

Derive and apply the equilibrium relationship between free agonist concentration, dissociation constant, and fractional receptor occupancy.

Advanced 9 min read

What You'll Learn

  • Derive fractional occupancy from reversible binding at equilibrium
  • Define the dissociation constant using association and dissociation rate constants
  • Calculate occupancy or the concentration needed for a target occupancy
  • Infer a dissociation constant from an equilibrium occupancy measurement
  • Identify when ligand depletion makes total concentration an unsuitable substitute for free concentration
  • Diagnose failures of the equilibrium, one-site, and free-concentration assumptions

Why Fractional Occupancy Has a Simple Form

A receptor-binding model connects a ligand concentration to the fraction of receptors occupied by ligand. Under a particular set of assumptions, this relationship is compact: occupancy increases with free agonist concentration and approaches, but does not exceed, one. The simplicity of the equation is useful, but also makes it important to know exactly what concentration it uses and what conditions allow it to describe a binding system.

This tutorial derives that equation for reversible binding at equilibrium to one binding site per receptor. It concerns binding, not the downstream response produced by occupied receptors. In particular, an occupancy fraction is not automatically a response fraction.

Binding States and the Dissociation Constant

Represent an unoccupied receptor as \(R\), free agonist as \(A\), and the receptor–agonist complex as \(RA\). Binding is reversible: receptors and agonist can associate to form \(RA\), and \(RA\) can dissociate. Let \(k_{\mathrm{on}}\) be the association rate constant and \(k_{\mathrm{off}}\) the dissociation rate constant. Under the mass-action description, the association rate is proportional to the concentrations of free receptor and free agonist, while the dissociation rate is proportional to the concentration of complex.

Definition: The equilibrium dissociation constant is \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\). For reversible one-site binding at equilibrium, it is also the ratio of the product of free agonist concentration and free receptor concentration to the receptor–agonist complex concentration. \(K_D\) has concentration units. A smaller \(K_D\) means that a lower free agonist concentration is associated with a given fractional occupancy under this model.

Write \(C\) for the free agonist concentration, \(R\) for the concentration of unoccupied receptor, and \(RA\) for the concentration of occupied receptor. Let \(R_{\mathrm{tot}}\) be the total concentration of receptors, occupied and unoccupied. Fractional occupancy, denoted \(f\), is the fraction of this total receptor concentration that is in the complex:

$$ f=\frac{[RA]}{R_{\mathrm{tot}}} =\frac{[RA]}{[R]+[RA]}. $$

Here the brackets denote concentrations. The expression for \(f\) assumes that the receptor population being counted consists of these two states: unoccupied and occupied by the agonist.

Deriving the Equilibrium Occupancy Equation

At equilibrium, the forward association rate and reverse dissociation rate are equal. This is a statement about the rates of the two opposing processes; it does not mean that binding and unbinding have stopped.

$$ k_{\mathrm{on}}\,C[R]=k_{\mathrm{off}}[RA]. $$

Rearrange this equality and use the definition of \(K_D\). Then substitute receptor mass balance, \(R_{\mathrm{tot}}=[R]+[RA]\), into the definition of fractional occupancy. The complete chain is:

$$ \begin{aligned} K_D &=\frac{k_{\mathrm{off}}}{k_{\mathrm{on}}} =\frac{C[R]}{[RA]},\\ K_D[RA]&=C[R],\\ [R]&=\frac{K_D[RA]}{C},\\ R_{\mathrm{tot}} &=[RA]+\frac{K_D[RA]}{C} =[RA]\left(1+\frac{K_D}{C}\right),\\ f &=\frac{[RA]}{R_{\mathrm{tot}}} =\frac{1}{1+K_D/C} =\frac{C}{K_D+C}. \end{aligned} $$

Thus, under the stated assumptions, fractional occupancy is a function of the ratio between free agonist concentration and \(K_D\). At \(C=K_D\), the formula gives \(f=K_D/(K_D+K_D)=0.5\). If \(C\) is much smaller than \(K_D\), then \(f\) is approximately \(C/K_D\); if \(C\) is much larger than \(K_D\), then \(f\) approaches one.

Key result: For reversible binding at equilibrium to one class of independent sites, with free agonist concentration \(C\), the fractional occupancy is \(f=C/(K_D+C)\). The half-occupancy concentration equals \(K_D\) only under this model and when \(C\) means free agonist concentration.

Assumptions Behind the Result

The derivation is short because its assumptions are specific. They should be considered before applying the formula to a measured concentration or a receptor system.

  • Equilibrium has been reached. The association and dissociation rates balance at the time occupancy is interpreted. A transient measurement during approach to equilibrium need not follow the equilibrium equation.
  • Binding is reversible and follows mass action. Association and dissociation can be represented by the rate terms \(k_{\mathrm{on}}C[R]\) and \(k_{\mathrm{off}}[RA]\), with a constant ratio defining \(K_D\).
  • There is one binding site per receptor in the model. Each receptor is either unoccupied or occupied by one agonist molecule. The equation does not represent multiple binding sites or interactions among sites.
  • The receptor sites are treated as one uniform class. They share the same binding constants; the calculation does not average over distinct receptor populations with different affinities.
  • \(C\) is the free concentration at the binding site. The concentration available to bind is not necessarily the same as the total amount of agonist divided by the sample volume. This distinction matters when binding removes a substantial fraction of agonist from solution.
  • The concentration scale is appropriate for mass action. In ordinary dilute solutions, concentration is commonly used as a practical proxy for chemical activity. Strong non-ideal solution conditions may require a more appropriate activity description.

The equation also presumes that the receptor total is counted consistently and that other processes do not introduce additional receptor states into this two-state binding description. These conditions are reasons to assess the model against the experimental system rather than treating the formula as a universal law.

Worked Example: Calculating Occupancy at a Given Free Concentration

Suppose a hypothetical in-vitro system has \(K_D=4.00\ \mathrm{nM}\), and the free agonist concentration at equilibrium is \(C=12.0\ \mathrm{nM}\). Calculate the fractional occupancy.

Assume reversible one-site binding has reached equilibrium, the receptors form one uniform class, and the stated \(12.0\ \mathrm{nM}\) is the free concentration at the binding site. These assumptions allow use of the derived equation:

$$ f=\frac{C}{K_D+C} =\frac{12.0\ \mathrm{nM}}{4.00\ \mathrm{nM}+12.0\ \mathrm{nM}} =\frac{12.0}{16.0} =0.750. $$

The predicted fractional occupancy is \(0.750\), or \(75.0\%\). The units cancel in the ratio. This result describes the fraction of receptors bound in the model; it does not specify the magnitude of any biological response.

Using the Relationship in Both Directions

The equation can be rearranged to answer two common questions: what free concentration corresponds to a target occupancy, and what \(K_D\) is implied by an occupancy observed at a known free concentration. For a target occupancy \(f\) strictly between zero and one, rearrangement gives:

$$ f=\frac{C}{K_D+C} \quad\Longrightarrow\quad fK_D+fC=C \quad\Longrightarrow\quad C=\frac{f}{1-f}K_D. $$

For an occupancy measurement with known \(C>0\), the corresponding rearrangement is:

$$ f(K_D+C)=C \quad\Longrightarrow\quad fK_D=C(1-f) \quad\Longrightarrow\quad K_D=C\frac{1-f}{f}. $$

These rearrangements require the same binding assumptions as the original equation. They are not independent models that repair a failure of equilibrium or a mismatch between total and free concentration.

Worked Example: Finding the Concentration for a Target Occupancy

In a hypothetical system with \(K_D=6.00\ \mathrm{nM}\), find the free agonist concentration required for \(90.0\%\) occupancy under the equilibrium one-site model.

The target fraction is \(f=0.900\), which is between zero and one. Assume the same equilibrium, uniform-site, and free-concentration conditions used in the derivation. Substitute into the rearranged equation:

$$ C=\frac{f}{1-f}K_D =\frac{0.900}{1-0.900}(6.00\ \mathrm{nM}) =\frac{0.900}{0.100}(6.00\ \mathrm{nM}) =9(6.00\ \mathrm{nM}) =54.0\ \mathrm{nM}. $$

The model therefore predicts \(90.0\%\) occupancy at a free agonist concentration of \(54.0\ \mathrm{nM}\). Substitution back into the original equation verifies the result: \(54.0/(6.00+54.0)=0.900\).

Worked Example: Inferring a Dissociation Constant

A hypothetical equilibrium binding measurement reports \(f=0.400\) at a free agonist concentration of \(C=5.00\ \mathrm{nM}\). Under the one-site model, calculate the implied \(K_D\).

Treat the measured fraction as an equilibrium occupancy, and assume that the free concentration is known and the receptor sites form one uniform class. Because \(f\) is between zero and one, rearrange the equation and substitute:

$$ K_D=C\frac{1-f}{f} =(5.00\ \mathrm{nM})\frac{1-0.400}{0.400} =(5.00\ \mathrm{nM})\frac{0.600}{0.400} =7.50\ \mathrm{nM}. $$

The implied dissociation constant is \(7.50\ \mathrm{nM}\). This calculation uses one hypothetical measurement to illustrate the algebra; it does not establish the precision of an estimate from experimental data or show that the one-site assumptions fit those data.

Free Concentration Is Not Always Total Concentration

A frequent application error is to substitute total agonist concentration into the occupancy formula without checking whether it represents the free concentration. If receptor binding consumes only a negligible fraction of agonist, free and total concentrations may be close enough for a particular application. If binding substantially depletes the free agonist pool, they differ, and the equation must use the free concentration.

For illustration, consider a hypothetical closed sample with total agonist concentration \(10.0\ \mathrm{nM}\), total receptor concentration \(10.0\ \mathrm{nM}\), and \(K_D=2.00\ \mathrm{nM}\). At equilibrium, let \(x\) be the complex concentration. Then the free agonist and free receptor concentrations are both \(10.0-x\ \mathrm{nM}\), by mass balance. The equilibrium definition gives:

$$ 2.00=\frac{(10.0-x)^2}{x}, \qquad x^2-22.0x+100=0. $$

The quadratic solutions are \(x=(22.0\pm\sqrt{84})/2\), or approximately \(15.58\ \mathrm{nM}\) and \(6.417\ \mathrm{nM}\). The first is impossible because it exceeds the total receptor concentration. Therefore \(x=6.417\ \mathrm{nM}\), the free agonist concentration is \(10.0-6.417=3.583\ \mathrm{nM}\), and the fractional occupancy is \(6.417/10.0=0.642\). Using total agonist concentration as though it were free would instead predict \(10.0/(2.00+10.0)=0.833\), which is inconsistent with the mass balance in this example.

Diagnostics, Alternatives, and Common Pitfalls

A calculation can be arithmetically correct and still be a poor description of the binding system. Check whether the measurement was taken after binding had time to equilibrate; whether estimates change with incubation time; whether free agonist is measured or credibly approximated; and whether one uniform site class is plausible. A systematic mismatch between observed and predicted occupancy across concentrations can signal that one or more assumptions do not describe the experiment. A single occupancy point cannot diagnose all such failures.

When equilibrium is not a defensible approximation, a kinetic description is the main alternative. For the same reversible one-site process, the time-dependent complex concentration can be represented by:

$$ \frac{d[RA]}{dt} =k_{\mathrm{on}}C[R]-k_{\mathrm{off}}[RA]. $$

This equation describes change over time; with a constant free concentration and the stated mass-action assumptions, its equilibrium condition is the zero-rate condition used in the derivation above. Applying a kinetic model requires information about time and rate parameters, whereas the equilibrium equation directly relates concentration and occupancy at equilibrium. Neither approach automatically corrects an incorrect free-concentration value or an unsuitable one-site representation.

  • Pitfall: interpreting \(K_D\) as an agonist concentration that always gives half occupancy. The correction is that \(K_D\) gives half occupancy when the concentration inserted is the free agonist concentration and the equilibrium one-site model applies.
  • Pitfall: assuming any stated concentration is free concentration. The correction is to distinguish free from total agonist and, when depletion matters, use mass balance to determine free concentration.
  • Pitfall: treating equilibrium binding as a response model. The correction is to interpret \(f\) as the fraction of receptors occupied, not as a percentage of maximal effect.
  • Pitfall: treating a mismatch as a rounding issue without checking assumptions. The correction is to check equilibration, concentration measurement, and whether the receptor population is adequately represented by one uniform binding class.

The equilibrium formula is valuable because it makes the concentration–occupancy relationship transparent and identifies the half-occupancy concentration under its assumptions. Its limitations follow from those same assumptions: it does not, by itself, describe binding before equilibrium, distinguish receptor subpopulations, or turn occupancy into a functional effect.

Check Your Understanding

Answer each question using the equilibrium one-site model where its assumptions are stated. Short solutions follow the questions.

  1. If \(K_D=3.00\ \mathrm{nM}\) and free agonist concentration is \(3.00\ \mathrm{nM}\), what is the predicted fractional occupancy?
  2. For \(K_D=8.00\ \mathrm{nM}\), what free concentration gives \(75.0\%\) occupancy?
  3. At \(C=4.00\ \mathrm{nM}\), occupancy is \(0.200\). What \(K_D\) is implied by the one-site equilibrium equation?
  4. Why can using total agonist concentration overstate predicted occupancy in a sample with substantial receptor binding?
  5. What does it mean physically that the association and dissociation rates are equal at equilibrium?

Solutions. (1) \(f=3.00/(3.00+3.00)=0.500\), or \(50.0\%\). (2) \(C=[0.750/(1-0.750)](8.00\ \mathrm{nM})=3(8.00\ \mathrm{nM})=24.0\ \mathrm{nM}\). (3) \(K_D=C(1-f)/f=(4.00\ \mathrm{nM})(0.800/0.200)=16.0\ \mathrm{nM}\). (4) Receptor binding can remove agonist from the free pool, so total concentration can exceed free concentration; substituting the larger total concentration into \(C/(K_D+C)\) can overpredict occupancy. (5) Binding and unbinding continue, but their rates balance, so the complex concentration has no net change at that equilibrium condition.

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