Why Surface Binding Is Not the Whole Receptor Story
A surface-bound ligand can trigger receptor internalization. Once internalized, that receptor is unavailable for binding at the cell surface until it recycles. Consequently, a binding model that tracks only unbound and bound surface receptors can misrepresent how much receptor is accessible, even when it describes binding and dissociation correctly. To separate these effects, track receptors in three states: unbound at the surface, bound at the surface, and internalized.
The model below isolates internalization and recycling. It treats free ligand concentration as specified over time, and it does not track ligand after receptor internalization. This is suitable when receptor states are the question of interest and ligand availability is not materially changed by binding or uptake. It is not a model of ligand disposition or intracellular signaling.
The Three-Compartment Model
Assume ligand binds unbound surface receptor by one-site mass-action kinetics. Binding moves receptor from \(R_s\) to \(C_s\), and dissociation moves it back. Bound surface receptor internalizes at rate \(k_{\mathrm{int}}C_s\). Internalized receptor recycles to the unbound surface pool at rate \(k_{\mathrm{rec}}R_i\). The baseline model has no receptor synthesis or degradation.
The association term \(k_{\mathrm{on}}L(t)R_s\) has units of receptor amount per time: concentration\(^{-1}\) time\(^{-1}\) multiplied by ligand concentration and receptor concentration. Every term in each equation therefore has the same units as the state variable per time. The equations also specify where each receptor flux goes: internalization removes receptor from \(C_s\) and adds the same flux to \(R_i\); recycling removes receptor from \(R_i\) and adds it to \(R_s\).
This formulation assumes a well-mixed surface pool, a single receptor-binding class, and first-order internalization and recycling. Free ligand is treated as an input, not as a fourth state. The internalized pool is a lumped receptor state: ligand fate after uptake is not represented. If uptake substantially changes free ligand, or if ligand remains bound intracellularly and that distinction matters, the model needs additional ligand states and mass-balance equations.
Worked Example: Writing and Checking the Fluxes
Suppose a hypothetical cell preparation has \(R_s=60\) fmol, \(C_s=30\) fmol, and \(R_i=10\) fmol. At the current ligand concentration, let \(k_{\mathrm{on}}L=0.20\) h\(^{-1}\), \(k_{\mathrm{off}}=0.10\) h\(^{-1}\), \(k_{\mathrm{int}}=0.30\) h\(^{-1}\), and \(k_{\mathrm{rec}}=0.15\) h\(^{-1}\). Calculate the instantaneous rates of change.
For unbound surface receptor, association removes \(0.20(60)=12\) fmol/h, dissociation adds \(0.10(30)=3\) fmol/h, and recycling adds \(0.15(10)=1.5\) fmol/h. Thus,
For bound surface receptor, association adds 12 fmol/h, while dissociation and internalization remove \(0.10(30)=3\) and \(0.30(30)=9\) fmol/h:
Internalization adds 9 fmol/h to the internalized pool, and recycling removes 1.5 fmol/h:
Surface receptor is decreasing while internalized receptor is increasing. The total change is \(-7.5+0+7.5=0\) fmol/h, as required when this model has no synthesis or degradation.
Mass Balance and Steady State
Define total receptor \(R_T=R_s+C_s+R_i\). Adding the three differential equations cancels every binding, dissociation, internalization, and recycling flux:
Thus, the baseline model conserves receptor: internalization and recycling redistribute receptor among compartments but do not create or destroy it. This is a useful implementation check. If the numerical solution produces a changing \(R_T\), the equations, units, or numerical calculation should be examined.
For a steady state under constant ligand concentration, take \(L(t)=L\) and set each derivative to zero. The bound-surface equation gives a relationship that does not require the recycling equation:
The internalized-receptor equation gives \(R_i/R_s=(k_{\mathrm{int}}/k_{\mathrm{rec}})(C_s/R_s)\), provided \(k_{\mathrm{rec}}>0\). These relationships give the steady distribution when total receptor is fixed. In particular, internalization reduces the bound-to-unbound surface ratio compared with a model that has dissociation but no internalization: a bound receptor can leave the surface not only by dissociating, but also by internalizing.
The surface-fraction result follows by substituting the steady-state ratio into its definition:
The derivation requires a nonzero surface pool and a steady state under constant ligand. It does not say that surface occupancy tracks ligand instantaneously during a changing exposure. For time-varying \(L(t)\), the differential equations describe the response, including delays as receptors internalize and recycle.
Worked Example: Separating Surface Occupancy From Total Engagement
Use a hypothetical constant ligand exposure with \(k_{\mathrm{on}}L=0.20\) h\(^{-1}\), \(k_{\mathrm{off}}=0.10\) h\(^{-1}\), \(k_{\mathrm{int}}=0.30\) h\(^{-1}\), \(k_{\mathrm{rec}}=0.15\) h\(^{-1}\), and total receptor \(R_T=100\) fmol. Find the steady-state amounts and the two relevant fractions.
The steady-state ratio of bound to unbound surface receptor is \(\displaystyle C_s/R_s=0.20/(0.10+0.30)=0.5\). The internalized-to-unbound ratio is \(\displaystyle R_i/R_s=(0.30/0.15)(0.5)=1\). Therefore \(R_s:C_s:R_i=1:0.5:1\). The ratio totals 2.5 parts, so one part is \(100/2.5=40\) fmol. It follows that \(R_s=40\) fmol, \(C_s=20\) fmol, and \(R_i=40\) fmol.
The bound fraction of the surface pool is \(20/(40+20)=0.3333\), or 33.3%. The bound fraction of the total receptor pool is \(20/100=0.2000\), or 20.0%. These are different quantities: 40% of all receptors are internalized and unavailable at the surface in this model.
Now increase recycling to \(k_{\mathrm{rec}}=0.60\) h\(^{-1}\), holding the other rates and \(R_T\) fixed. The ratio \(C_s/R_s\) remains 0.5, while \(R_i/R_s=(0.30/0.60)(0.5)=0.25\). The total is now 1.75 parts, giving \(R_s=57.14\) fmol, \(C_s=28.57\) fmol, and \(R_i=14.29\) fmol. The surface-bound fraction remains \(28.57/(57.14+28.57)=0.3333\), but the fraction of all receptors bound rises to \(28.57/100=0.2857\). Faster recycling restores more receptor to the surface; it does not change the steady-state bound fraction among surface receptors under these assumptions.
Interpreting the Model and Its Alternatives
Three quantities should not be conflated: \(C_s/(R_s+C_s)\), the fraction of surface receptors bound; \(C_s/R_T\), the fraction of all receptors currently bound at the surface; and \((R_s+C_s)/R_T\), the fraction of all receptors available at the surface. Internalization can lower surface availability, while recycling can raise it. A statement about “occupancy” should specify which denominator is intended.
An equilibrium binding expression is simpler when the surface pool is effectively fixed and internalization is negligible on the timescale of interest. A two-state time-dependent binding model can represent changing ligand and binding kinetics, but not a separate internalized reservoir. The three-state model is appropriate when movement between surface and internalized receptor pools matters. If receptor synthesis or degradation is important, add explicit source or loss terms rather than treating the baseline conservation result as valid.
For example, a first-order degradation term \(k_{\mathrm{deg}}R_i\) can be added as a loss from the internalized compartment, and a synthesis rate can be added to the surface-unbound equation. Then total receptor is no longer conserved unless those sources and losses balance. The equations must reflect the biological destination of each flux; placing a degradation term in the bound-surface equation, for example, would represent a different mechanism.
The model’s compactness is also a limitation. It assumes one internalized state and one recycling route. It cannot distinguish early from late endosomes, receptor sorting, ligand dissociation within the cell, or recycling that returns receptor in a ligand-bound state. Add states only when data or a mechanistic question can support those distinctions; additional compartments introduce additional parameters that may be difficult to identify.
Diagnostics and Common Pitfalls
- Check receptor balance. With no synthesis or degradation, verify \(R_s+C_s+R_i=R_T\) throughout a numerical simulation.
- Check state values. Receptor amounts should remain nonnegative. Negative values can indicate a coding error, inconsistent units, or a numerical time step that is too large.
- Check flux directions. Internalization must subtract from \(C_s\) and add to \(R_i\); recycling must subtract from \(R_i\) and add to \(R_s\).
- Check what is measured. A surface assay and a total-cell receptor measurement do not necessarily report the same pool. Comparing either directly with \(C_s/R_T\) requires a measurement model that matches the assay.
- Check the fixed-ligand assumption. If receptor binding or uptake materially changes free ligand, prescribing \(L(t)\) independently may give inconsistent predictions.
A common mistake is to calculate \(C_s/(R_s+C_s)\) and call it the fraction of all receptors engaged. That reasoning ignores internalized receptors in the denominator. The correction is to state the pool explicitly: surface engagement uses \(R_s+C_s\), whereas the fraction of all receptors bound at the surface uses \(R_T=R_s+C_s+R_i\). In the numerical example, these are 33.3% and 20.0%, respectively.
Worked Example: A Short-Time Response to Binding and Trafficking
At time \(t\), suppose \(R_s=60\) fmol, \(C_s=30\) fmol, and \(R_i=10\) fmol, with rates as in the first example. Approximate the state after 0.10 h by one forward Euler step. This method uses the current derivative multiplied by the step length.
Using the previously calculated derivatives, \(\Delta R_s=(-7.5)(0.10)=-0.75\) fmol, \(\Delta C_s=(0)(0.10)=0\) fmol, and \(\Delta R_i=(7.5)(0.10)=0.75\) fmol. The approximations are therefore \(R_s=59.25\) fmol, \(C_s=30.00\) fmol, and \(R_i=10.75\) fmol at \(t+0.10\) h. Their sum remains 100 fmol.
Initially, the surface-bound fraction is \(30/(60+30)=0.3333\). After the step it is \(30/(59.25+30)=0.3361\). The fraction of all receptors bound at the surface remains 0.3000 over this particular step because the instantaneous derivative of \(C_s\) is zero; it need not remain constant over a longer interval. This is a local numerical approximation, not an exact trajectory. A smaller step or a suitable numerical ODE solver is needed when rates are fast relative to the step length.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- If \(k_{\mathrm{int}}=0\), what happens to the internalization flux, and which state equation loses that flux?
- At steady state, \(k_{\mathrm{on}}L=0.12\) h\(^{-1}\), \(k_{\mathrm{off}}=0.08\) h\(^{-1}\), and \(k_{\mathrm{int}}=0.10\) h\(^{-1}\). What fraction of surface receptors is bound?
- Why can faster recycling increase \(C_s/R_T\) without changing \(C_s/(R_s+C_s)\) at steady state in the baseline model?
- In the baseline model, what should happen to \(R_s+C_s+R_i\) during a time-varying ligand exposure?
Solutions. (1) The internalization flux is zero because it equals \(k_{\mathrm{int}}C_s\). The loss term in \(dC_s/dt\) and the matching gain term in \(dR_i/dt\) both disappear. (2) The surface-bound fraction is \(0.12/(0.12+0.08+0.10)=0.4000\), or 40.0%. (3) Recycling returns internalized receptors to the surface, changing the amount of receptor in the surface pool relative to total receptor. The steady-state ratio \(C_s/R_s=k_{\mathrm{on}}L/(k_{\mathrm{off}}+k_{\mathrm{int}})\) does not contain \(k_{\mathrm{rec}}\), so the bound fraction within that surface pool is unchanged. (4) The total remains constant: binding, dissociation, internalization, and recycling transfer receptor between compartments but do not add or remove it.