Why Free and Total Target Can Tell Different Stories
Antibody binding changes how much target is free and how much is bound in an antibody–target complex. A measurement that detects only unbound target can therefore change even when the amount of target in the system has not changed. Conversely, a measurement that detects both free and complexed target can rise as binding redistributes target into a form that the assay also counts. To interpret either measurement, the model must distinguish target turnover from binding.
We will use a minimal model in which target is synthesized continuously, free target is degraded, and antibody binding is reversible. The free antibody concentration is treated as prescribed rather than calculated from a pharmacokinetic model. This isolates the target balance and the distinction between free and total target.
Concentrations of \(R\), \(C\), and \(D\) have the same units, such as nM. The units of \(k_{\mathrm{syn}}\) are concentration per time; \(k_{\mathrm{deg}}\), \(k_{\mathrm{off}}\), and \(k_{\mathrm{on}}D\) have units of inverse time. \(K_D\) has units of concentration. We assume one-to-one binding, constant synthesis, first-order degradation of free target, and no separate loss of complex in this minimal model.
Mass Balance and Baseline Target
With \(D\) specified as a function of time, the target equations are
The binding terms remove free target when a complex forms and return it when a complex dissociates. Adding the equations cancels these internal transfers:
This balance is the central accounting result: reversible binding alone does not create or destroy total target. In this model, total target changes through synthesis and degradation of free target. The equation is not \(dR_{\mathrm{tot}}/dt=k_{\mathrm{syn}}-k_{\mathrm{deg}}R_{\mathrm{tot}}\), because that would also apply the specified degradation rate to the bound target, contrary to the model assumptions.
Before treatment, \(D=0\) and \(C=0\). At baseline equilibrium, synthesis balances free-target degradation. Write the baseline concentration as \(R_0\):
This baseline relation is useful for checking units and model inputs: a synthesis rate divided by a first-order rate constant gives a concentration. It also provides a reference for assessing treatment-associated changes.
Worked Example: Calculating Baseline Target
Suppose a hypothetical system has \(k_{\mathrm{syn}}=120\) nM/day and \(k_{\mathrm{deg}}=0.400\) day\(^{-1}\). Calculate its untreated equilibrium target concentration.
At baseline, \(R_0=k_{\mathrm{syn}}/k_{\mathrm{deg}}\). Substituting the values gives
The units reduce to nM, and \(0.400\ \mathrm{day}^{-1}\times300\ \mathrm{nM}=120\ \mathrm{nM/day}\), so degradation exactly balances synthesis. These values describe a hypothetical model, not a measured biological system.
Binding Equilibrium and the Free–Total Distinction
At binding equilibrium, formation and dissociation of complex balance:
For a constant, externally prescribed free-antibody concentration \(D\), a steady state of the full target model also requires \(dR_{\mathrm{tot}}/dt=0\). The mass balance then gives \(R=R_0\). Thus,
These steady-state results have a specific interpretation. Under the stated assumptions, antibody binding increases the steady-state total target by adding a bound pool, but does not change the steady-state free-target concentration. This does not mean that free target cannot change after treatment: binding can rapidly redistribute target, producing a transient fall in free target while total target is initially conserved. Synthesis can then replenish free target over time.
The steady-state bound fraction of total target follows from the same equations:
This fraction concerns the target pool, not the fraction of antibody bound. It is 0.5 when \(D=K_D\), and approaches 1 as \(D\) becomes much larger than \(K_D\), provided the assumptions of the model remain appropriate.
Worked Example: Steady-State Free and Total Target Under Treatment
Continue with the hypothetical baseline \(R_0=300\) nM. Suppose \(K_D=20\) nM and free antibody is held at \(D=80\) nM. Calculate complex and total target at steady state.
The binding-equilibrium relation gives
Therefore,
Free target remains 300 nM at this model steady state, while total target is 1500 nM. The fraction of total target that is complexed is \(1200/1500=0.800\), or 80.0%. The conclusion depends on the steady-state and no-complex-loss assumptions; it is not a general prediction that antibody treatment raises total target in every biological system.
Time Course After a Change in Antibody
The full differential equations describe binding and turnover on their respective timescales. A useful approximation applies when binding and dissociation are fast relative to target turnover, so that binding remains near equilibrium as free antibody changes. This is the rapid-binding-equilibrium approximation. It is an assumption about timescales, not a consequence of target turnover alone.
Under this approximation, \(C=DR/K_D\), so \(R_{\mathrm{tot}}=R(1+D/K_D)\). Combining this relation with the target mass balance gives
For constant \(D\), the factor multiplying \(R\) is constant. The equation reduces to
Its steady state is \(R_0=k_{\mathrm{syn}}/k_{\mathrm{deg}}\), as before. If \(D\) steps from zero to a constant value at time zero, rapid binding redistributes the pre-existing target immediately: the total target does not jump, while \(R\) and \(C\) do. Thereafter, target synthesis and free-target degradation move the system toward its new steady state.
Worked Example: Free and Total Target During Turnover
Use \(k_{\mathrm{syn}}=120\) nM/day, \(k_{\mathrm{deg}}=0.400\) day\(^{-1}\), and \(R_0=300\) nM. At time zero, free antibody steps to \(D=80\) nM, with \(K_D=20\) nM. Assume rapid binding equilibrium, constant free antibody after the step, and no complex loss. Find free and total target immediately after redistribution and after 7 days.
Here \(1+D/K_D=1+80/20=5\). Immediately before dosing, total target is 300 nM. Rapid redistribution preserves that total at the instant of the step, so \(5R(0^+)=300\) nM. Thus \(R(0^+)=60.0\) nM and \(C(0^+)=300-60=240\) nM.
For later times, the constant-\(D\) equation becomes
The solution approaches 300 nM, with time constant \(1/0.0800=12.5\) days. Applying the initial value \(R(0^+)=60.0\) nM gives
where \(t\) is measured in days. At \(t=7\) days, \(e^{-0.56}=0.5712\), so
The free concentration has risen from 60.0 nM immediately after redistribution toward its 300 nM steady state. Total target has also risen from 300 nM toward 1500 nM as synthesis replenishes free target, some of which is then bound. The instantaneous redistribution and the subsequent turnover are distinct processes.
Reading Assay Results and Checking the Model
A free-target assay is intended to quantify unbound target, \(R\). A total-target assay is intended to quantify both \(R\) and \(C\), typically after a measurement procedure that detects the complexed form as well. Assay selectivity and sample handling matter: the model's \(R\) and \(R_{\mathrm{tot}}\) are defined quantities, while an actual assay may not measure them perfectly.
Worked Example: Interpreting Two Target Measurements
At steady state in the hypothetical setting with \(D=40\) nM and \(K_D=20\) nM, suppose a free-target assay reports 300 nM and a total-target assay reports 900 nM. Assuming the assays measure the model quantities, calculate the complex concentration and the bound fraction.
By definition, \(R_{\mathrm{tot}}=R+C\), so
The bound fraction is \(C/R_{\mathrm{tot}}=600/900=0.667\), or 66.7%. The model also predicts \(C=DR/K_D=(40)(300)/20=600\) nM and \(R_{\mathrm{tot}}=900\) nM, consistent with the hypothetical measurements. This agreement checks the arithmetic and the equilibrium relation; it does not establish that the model assumptions hold in an actual system.
A useful diagnostic is to compare free and total measurements with the expected baseline and with each other. For example, a total-target value below the measured free-target value is inconsistent with the definition \(R_{\mathrm{tot}}=R+C\), unless measurement error or differing assay definitions explain it. A measured steady-state free target that differs substantially from the baseline prediction may indicate that assumptions such as unchanged synthesis, free-target degradation, or externally fixed free antibody are not adequate.
The full kinetic equations are the main alternative to the rapid-binding-equilibrium approximation. They retain finite association and dissociation rates and are appropriate when binding is not fast relative to turnover or when early-time behavior matters. The approximation is simpler for interpreting slow target turnover, but may misrepresent the initial transient if binding is not sufficiently fast. Both approaches here assume free antibody is known; a coupled drug-disposition model would be needed if free antibody changes in response to binding or clearance.
Common Pitfall and Model Limitations
Wrong reasoning: “Antibody binding lowers free target, so it must lower total target by the same amount.” This treats binding as target destruction. In the model, association transfers target from \(R\) to \(C\), and dissociation transfers it back; adding the balances shows that these terms cancel. The correction is to calculate \(R_{\mathrm{tot}}=R+C\) and account separately for synthesis and degradation.
The conclusion is conditional. This minimal model assumes that complex is not independently removed and that degradation acts only on free target. If complex is lost, target may be removed while bound; if target synthesis or degradation changes with treatment, the baseline turnover balance may change as well. Those mechanisms require different equations and can produce different free- and total-target responses. Model evaluation should therefore ask whether observed time courses and assay definitions are compatible with the assumed turnover and binding processes, rather than treating the equations as universal.
Check Your Understanding
Try each question, then check the solution immediately below it.
- If \(k_{\mathrm{syn}}=90\) nM/day and \(k_{\mathrm{deg}}=0.300\) day\(^{-1}\), what is baseline free target?
- At binding equilibrium, \(D=30\) nM, \(K_D=10\) nM, and \(R=250\) nM. Calculate \(C\), \(R_{\mathrm{tot}}\), and the bound fraction.
- Why does the target mass-balance equation contain no association or dissociation terms?
- Under the model assumptions, what does it mean if a steady-state total-target assay is higher than baseline while a free-target assay is unchanged?
Solutions. (1) \(R_0=k_{\mathrm{syn}}/k_{\mathrm{deg}}=90/0.300=300\) nM. (2) \(C=DR/K_D=(30)(250)/10=750\) nM; \(R_{\mathrm{tot}}=250+750=1000\) nM; the bound fraction is \(750/1000=0.750\), or 75.0%. (3) Association removes free target and adds the same amount to complex, while dissociation does the reverse; adding the two state equations cancels both transfers. (4) It is consistent with target being redistributed into a bound pool while free target returns to its turnover baseline; total target includes both forms and is not, by definition, a free-target measurement.