Why Repeated Dosing Changes the TMDD Profile
With repeated antibody doses, the next dose arrives while some drug from earlier doses may remain. That residual drug can be free, bound to target, or both. Meanwhile, target is synthesized, redistributed between free and bound states, and removed through turnover and complex internalization. The observed accumulation and trough therefore depend on more than the dose and the antibody’s nonspecific elimination rate.
A useful starting point is to track the three concentrations separately: free drug \(C\), free target \(R\), and drug–target complex \(B\). Their sum \(C+B\) is total drug concentration for a one-to-one complex, while \(R+B\) is total target concentration. These totals are not interchangeable with free concentrations: binding can shift drug out of the free pool without immediately removing it from the system. The earlier material on free drug, total drug, and complex assays establishes why the distinction matters for measurement as well as modeling.
Repeated-Dose TMDD Mass Balance
For a one-compartment model with one-to-one binding, let \(k_{\mathrm{on}}\) be the association rate constant, \(k_{\mathrm{off}}\) the dissociation rate constant, and \(k_{\mathrm{int}}\) the rate constant for complex internalization. Let \(k_{\mathrm{el}}\) be the first-order elimination rate constant for free drug. Let \(k_{\mathrm{syn}}\) be the target synthesis rate and \(k_{\mathrm{deg}}\) the first-order degradation rate constant for free target. Concentrations are measured in consistent molar units; rate constants have units of inverse time, except \(k_{\mathrm{on}}\), which has units of inverse concentration per time.
Between doses, these equations describe the system. For an intravenous bolus of dose \(D\) into a compartment of volume \(V\), the concentration jump is \(D/V\): \(C(t_j^+)=C(t_j^-)+D/V\) at dose time \(t_j\). Target and complex concentrations do not jump instantaneously under this bolus convention. Between boluses, all three states evolve continuously. Without drug, the free-target baseline is \(R_0=k_{\mathrm{syn}}/k_{\mathrm{deg}}\).
Adding the free-drug and complex equations gives an especially useful mass-balance result:
Binding and dissociation cancel from this total-drug equation: they move drug between free and complex states but do not themselves remove it. Elimination and complex internalization do remove drug. Thus a complex can contribute to measured total drug while simultaneously being a route of drug loss. Adding the target equations likewise gives \(d(R+B)/dt=k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}B\). Target abundance need not recover at a single fixed rate when free and bound target have different fates.
The model assumes well-mixed concentrations, mass-action binding, one-to-one complex stoichiometry, and the specified turnover and elimination pathways. It also assumes that dose input and assay definitions are represented consistently. If distribution between tissues, time-varying target synthesis, or additional complex fates matter, those processes need additional states or terms rather than an unexamined adjustment to an apparent half-life.
A Baseline: Linear Accumulation
A useful comparison is a linear one-compartment model with instantaneous boluses and first-order elimination, but no target binding. If each dose raises concentration by \(\Delta C=D/V\), and doses are separated by \(\tau\), then the residual concentration from one interval is multiplied by \(q=e^{-k\tau}\), where \(k\) is the elimination rate constant. At steady state, the pre-dose trough \(C_{\mathrm{trough,ss}}\) is followed by a dose jump and then exponential decline back to that trough.
The corresponding steady-state peak is \(\Delta C/(1-e^{-k\tau})\). Relative to the concentration increment from one dose, the peak accumulation ratio is \(1/(1-e^{-k\tau})\). This result follows from a geometric series: each preceding dose contributes a fraction \(q, q^2, q^3,\ldots\) of its immediate post-dose increment. It depends on linear, time-invariant elimination; it is not a general TMDD accumulation formula.
Worked Example: Linear Steady-State Peak and Trough
Suppose a hypothetical antibody regimen produces an immediate concentration increment of \(10.0\ \text{mg/L}\) per bolus, with \(k=0.100\ \text{day}^{-1}\) and \(\tau=14.0\ \text{days}\). Assume first-order elimination and no target-mediated binding.
First calculate \(q=e^{-k\tau}=e^{-1.40}=0.2466\). The peak accumulation ratio is \(1/(1-q)=1/(1-0.2466)=1.327\). The steady-state peak is therefore \(10.0(1.327)=13.27\ \text{mg/L}\).
The steady-state trough is the peak multiplied by the fraction remaining over 14 days: \(13.27(0.2466)=3.272\ \text{mg/L}\). Equivalently, \(10.0/(e^{1.40}-1)=10.0/(4.055-1)=3.273\ \text{mg/L}\), with the small difference due to rounding. Thus the approximate peak and trough are \(13.27\) and \(3.27\ \text{mg/L}\), respectively. The assumptions justify using the linear recurrence; these values should not be transferred directly to a saturable TMDD model.
Target Recovery and Residual Complex
Target turnover affects repeated-dose profiles because the target available for binding can change between doses. The full mass-balance model should be used when free and bound target have distinct turnover or internalization pathways. A simpler recovery calculation is sometimes useful as a clearly stated approximation: suppose a target pool \(T\) returns toward its baseline \(T_0\) by first-order turnover, with \(dT/dt=k_{\mathrm{deg}}(T_0-T)\). This approximation treats the pool’s net recovery as first order; it is not a replacement for the full free-target/complex equations.
Solving the linear differential equation gives \(T(t)=T_0+[T(0)-T_0]e^{-k_{\mathrm{deg}}t}\). If binding is maintained at a fixed ratio \(B/R=2\), then \(T=R+B=3R\), so \(R=T/3\) and \(B=2T/3\). That fixed-ratio condition is an additional simplifying assumption, not a general consequence of target turnover.
Worked Example: Target Recovery Under a First-Order Approximation
In a hypothetical calculation, let the baseline target pool be \(T_0=60.0\ \text{nM}\), the post-perturbation pool be \(T(0)=30.0\ \text{nM}\), and \(k_{\mathrm{deg}}=0.1667\ \text{day}^{-1}\). Assume the first-order recovery approximation and a maintained binding ratio \(B/R=2\). Find the pool and its free and bound components after 7.00 days.
Substitution into the recovery solution gives \(T(7)=60.0+(30.0-60.0)e^{-(0.1667)(7.00)}\ \text{nM}\). The exponent is \(1.1669\), so \(T(7)=60.0-30.0e^{-1.1669}\approx50.66\ \text{nM}\).
Since \(T=3R\), \(R(7)=50.66/3\approx16.89\ \text{nM}\). The complex contains the remaining two-thirds: \(B(7)=2(50.66)/3\approx33.77\ \text{nM}\). The components sum to \(16.89+33.77=50.66\ \text{nM}\), up to rounding. These values describe the stated recovery approximation and fixed binding ratio, not a general prediction from the full TMDD equations.
Rapid Binding Can Redistribute Drug After a Dose
When binding is fast relative to the time scale of interest, a quasi-equilibrium snapshot can help interpret an immediate post-dose redistribution. Let \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\) be the equilibrium dissociation constant. For a fixed total target pool \(T=R+B\), equilibrium binding satisfies \(K_D=CR/B\), giving \(B=TC/(K_D+C)\). This relationship describes binding at equilibrium; by itself it does not describe elimination, target turnover, or the trough reached after an interval.
Worked Example: Redistribution After a Bolus
Immediately before a hypothetical bolus, let free drug be \(C=4.00\ \text{nM}\), complex be \(B=30.0\ \text{nM}\), and total target be \(T=60.0\ \text{nM}\). Let \(K_D=4.00\ \text{nM}\). A bolus adds \(20.0\ \text{nM}\) to free drug. Assume binding re-equilibrates rapidly, while elimination and target turnover are negligible during this short redistribution.
Before the bolus, \(C+B=4.00+30.0=34.0\ \text{nM}\). The bolus raises the conserved drug total during redistribution to \(34.0+20.0=54.0\ \text{nM}\). At the new equilibrium, \(B=60.0C/(4.00+C)\), and \(C+B=54.0\). Therefore, \(C+60.0C/(4.00+C)=54.0\). Multiplying through by \(4.00+C\) yields \(C^2+10.0C-216=0\).
The positive root is \(C=(-10.0+\sqrt{964})/2\approx10.52\ \text{nM}\); the negative root is physically impossible for a concentration. The complex is \(B=54.0-10.52\approx43.48\ \text{nM}\). Thus, after redistribution, free drug is about \(10.52\ \text{nM}\) and complex is about \(43.48\ \text{nM}\), summing to the \(54.0\ \text{nM}\) drug total. This is an immediate equilibrium calculation, not a steady-state trough prediction: subsequent elimination, dissociation, internalization, and target synthesis alter all three states.
Interpreting Accumulation, Diagnostics, and Limitations
In the full TMDD system, each dose can encounter a different amount of free target and residual complex. At low free-drug concentrations, binding may remove a substantial fraction of drug from the free pool; at higher concentrations, available target can become occupied and the incremental fraction bound may change. As target turns over, new free target can become available for binding. These mechanisms can make successive peaks and troughs differ from the geometric pattern predicted by linear superposition. The direction and size of the change depend on the balance among binding, dissociation, target synthesis, degradation, complex internalization, and free-drug elimination.
A practical analysis should inspect state profiles, not just a single reported concentration. Check that the dose jump is applied to the intended state, that \(C+B\) changes by the administered concentration increment at the instant of a bolus, and that \(R+B\) changes continuously unless the model explicitly includes an instantaneous target input or removal. Between doses, verify that binding and dissociation cancel in the total-drug balance and that complex internalization reduces total drug. Compare the simulated trough at successive dosing times: a periodic steady state is indicated when the relevant state values repeat from one interval to the next, not merely when one concentration appears visually flat.
The main alternatives answer different questions. A linear model gives a transparent accumulation ratio when target-mediated processes are negligible or adequately approximated by constant clearance. A full TMDD ODE model is appropriate when target, complex, and drug states change on relevant time scales. A quasi-steady-state or quasi-equilibrium reduction can reduce computation when its timescale assumptions are justified, but it may miss fast transients after a bolus or target recovery between doses. The equilibrium binding calculation above is narrower still: it is a snapshot, not an interval model.
Common warning signs include fitting a linear accumulation ratio to visibly dose-dependent profiles, treating total drug as if it were free drug, assuming target remains at baseline throughout a dosing interval, or interpreting a high complex concentration as evidence that the same amount of free antibody is available. Another error is to infer a trough from the binding equation alone. Binding equilibrium does not specify how much drug is removed over the interval; that requires the elimination and internalization terms and the evolving target state. Sparse sampling, uncertain target turnover, or an assay that combines free and bound forms can also make distinct mechanisms difficult to distinguish. Model predictions should therefore be checked against the measured analyte definition and the timing of samples relative to dose.
Check Your Understanding
Answer each question, then compare with the short solutions below.
- In the mass-balance equations, why does binding not appear as a net loss term in \(d(C+B)/dt\), while complex internalization does?
- For the linear model in the worked example, what fraction of a post-dose concentration increment remains after 14 days?
- Under the first-order target recovery approximation, what happens to \(T(t)\) as \(t\) increases without bound?
- Why is the equilibrium calculation after a bolus insufficient to predict the next trough?
Solutions. (1) Association subtracts drug from \(C\) and adds the same amount to \(B\); dissociation does the reverse, so both cancel in the sum. Internalization removes complex and therefore removes drug from \(C+B\). (2) The remaining fraction is \(e^{-k\tau}=e^{-1.40}\approx0.2466\), or about \(24.66\%\). (3) Since \(e^{-k_{\mathrm{deg}}t}\) tends to zero for positive \(k_{\mathrm{deg}}\), \(T(t)\) approaches \(T_0\). (4) The equilibrium calculation omits interval-time elimination, internalization, and target turnover; those processes determine how the states evolve before the next dose.