Why Target Turnover Belongs in a TMDD Model
A target is not necessarily a fixed pool waiting to bind drug. Cells or tissues may continually synthesize target, while free target is degraded or otherwise removed. During drug exposure, binding redistributes target between its free form and drug-target complex; if the complex is internalized, that process can also remove target. These processes determine how much target is available to bind drug and how much drug can be removed through target-mediated disposition.
The key distinction is between the drug-free target steady state and the target balance during exposure. In the absence of drug, synthesis and degradation determine the baseline target concentration. During exposure, the target balance also reflects binding, dissociation, and complex internalization. The resulting target concentration can therefore differ from baseline even when synthesis and degradation rates do not change.
Full TMDD Equations With Target Synthesis and Turnover
Under the standard full TMDD structure, free drug can undergo linear elimination at rate constant \(k_{\mathrm{el}}\), bind free target, or be released from complex. The equations below assume a well-mixed compartment, 1:1 reversible binding, constant rate parameters, constant basal target synthesis, first-order degradation of free target, and internalization of complex. For this model, internalization removes both the drug and target in the complex; there is no complex recycling. Drug input \(u(t)\) is expressed as a concentration input rate.
The target equation includes synthesis and degradation whether or not drug is present. Binding removes free target from \(R\), while dissociation returns it; internalization removes target while it is in \(C\). The complex equation accounts for formation by binding and loss by dissociation or internalization. In the free-drug equation, binding reduces \(L\) and dissociation returns drug to \(L\); internalization is represented as loss from \(C\), rather than as an extra loss term in the \(L\) equation.
Worked Example: Drug-Free Target Steady State
Suppose a hypothetical target has \(k_{\mathrm{syn}}=12.0\ \mathrm{nM/day}\) and \(k_{\mathrm{deg}}=0.300\ \mathrm{day}^{-1}\). Find its drug-free steady-state concentration.
With no drug, \(L=0\) and \(C=0\). The target equation reduces to
At steady state, \(dR/dt=0\), so synthesis equals degradation:
The units reduce to concentration because \((\mathrm{nM/day})/(\mathrm{day}^{-1})=\mathrm{nM}\). At this concentration, degradation is \((0.300\ \mathrm{day}^{-1})(40.0\ \mathrm{nM})=12.0\ \mathrm{nM/day}\), matching synthesis. This result assumes the target has reached its drug-free steady state; it is not an estimate of target concentration during drug exposure.
Deriving the Drug-Free Steady State
The result follows directly from the target balance. When drug is absent, the binding and complex terms vanish, leaving a first-order differential equation. Its solution, for initial concentration \(R(0)\), is
Since \(k_{\mathrm{deg}}>0\), the exponential term tends to zero as time increases. Thus \(R(t)\) approaches \(k_{\mathrm{syn}}/k_{\mathrm{deg}}\), which is the unique stable drug-free steady state. The approach rate is governed by \(k_{\mathrm{deg}}\); the corresponding half-life for return toward baseline is \(\ln(2)/k_{\mathrm{deg}}\). For example, if \(k_{\mathrm{deg}}=0.300\ \mathrm{day}^{-1}\), that half-life is \(0.6931/0.300=2.31\) days. This is a target turnover timescale, not a drug half-life.
This derivation also clarifies the model assumption: \(k_{\mathrm{syn}}\) is a constant rate, not a rate proportional to the current target concentration. If synthesis changes with target abundance, feedback or another regulatory mechanism is needed, and the simple ratio \(k_{\mathrm{syn}}/k_{\mathrm{deg}}\) need not apply.
How Turnover Changes the Complex During Exposure
To isolate the target balance during exposure, consider a constant free drug concentration \(L\) maintained by an external clamp. Assume the target and complex have reached a steady state under this fixed \(L\). This is a steady state of the target subsystem, not a claim that an administered drug concentration will itself remain constant or that the whole drug system is at steady state.
At steady state, the complex equation gives
Here \(K_{\mathrm{ss}}\) is the concentration scale implied by the steady-state complex balance; it is not necessarily the equilibrium dissociation constant. Substituting the steady-state complex relation into the target equation, or adding the \(R\) and \(C\) equations first, yields
Therefore,
Adding the equations cancels binding and dissociation because those processes only move target between free and bound forms. The total target \(R+C\) is changed by synthesis, degradation of free target, and internalization of complex. If \(k_{\mathrm{int}}=0\), target is not removed from the complex in this model: \(R\) remains \(k_{\mathrm{syn}}/k_{\mathrm{deg}}\), while complex can add to the total target pool. This limiting case is a useful check on the algebra.
Worked Example: Target and Complex at Fixed Drug Concentration
For a hypothetical system, let \(k_{\mathrm{syn}}=12.0\ \mathrm{nM/day}\), \(k_{\mathrm{deg}}=0.300\ \mathrm{day}^{-1}\), \(k_{\mathrm{on}}=0.0200\ \mathrm{nM}^{-1}\mathrm{day}^{-1}\), \(k_{\mathrm{off}}=0.100\ \mathrm{day}^{-1}\), and \(k_{\mathrm{int}}=0.200\ \mathrm{day}^{-1}\). Suppose free drug is clamped at \(L=30.0\ \mathrm{nM}\). Calculate the steady-state free target and complex.
First calculate the steady-state concentration scale and the complex-to-free-target ratio:
Then use the target balance:
Thus \(C=2.00R=34.29\ \mathrm{nM}\). Check the target balance using these displayed rounded concentrations: free-target degradation is \((0.300)(17.14)=5.142\ \mathrm{nM/day}\), and complex internalization removes target at \((0.200)(34.29)=6.858\ \mathrm{nM/day}\). Their sum is \(12.000\ \mathrm{nM/day}\), consistent with synthesis to the shown precision. The internalization-mediated drug loss rate is also \(k_{\mathrm{int}}C=6.858\ \mathrm{nM/day}\) in this clamped-concentration example.
Turnover, Complex Formation, and Drug Disposition
The complex internalization flux, \(k_{\mathrm{int}}C\), is a drug-loss flux as well as a target-loss flux under the assumptions above. Its magnitude depends on the amount of complex, which in turn depends on target availability and binding kinetics. Target turnover can therefore change drug disposition indirectly: it alters the supply of target available to form complex and the balance between free-target degradation and complex removal.
Worked Example: Comparing Turnover Rates at the Same Baseline Target
Compare two hypothetical systems with the same drug-free target concentration \(R_0=40.0\ \mathrm{nM}\), the same \(K_{\mathrm{ss}}=15.0\ \mathrm{nM}\), the same \(k_{\mathrm{int}}=0.200\ \mathrm{day}^{-1}\), and the same clamped \(L=30.0\ \mathrm{nM}\). System A has \(k_{\mathrm{deg}}=0.100\ \mathrm{day}^{-1}\); system B has \(k_{\mathrm{deg}}=0.400\ \mathrm{day}^{-1}\). Set synthesis in each system to maintain the stated baseline, so \(k_{\mathrm{syn}}=k_{\mathrm{deg}}R_0\).
Both systems have \(C/R=L/K_{\mathrm{ss}}=2.00\). System A has \(k_{\mathrm{syn}}=(0.100)(40.0)=4.00\ \mathrm{nM/day}\), giving
System B has \(k_{\mathrm{syn}}=(0.400)(40.0)=16.0\ \mathrm{nM/day}\), giving
The corresponding internalization-mediated drug-loss fluxes are \(k_{\mathrm{int}}C_A=(0.200)(16.0)=3.20\ \mathrm{nM/day}\) and \(k_{\mathrm{int}}C_B=(0.200)(40.0)=8.00\ \mathrm{nM/day}\). The higher-turnover system has greater complex abundance and internalization flux in this comparison, despite having the same drug-free target concentration. This is because its synthesis rate is also higher, replenishing target as complex is internalized. The result depends on the stated assumptions and fixed drug concentration; it is not a universal prediction for every dosing regimen.
This comparison explains why baseline target abundance alone may not characterize target-mediated drug disposition. Two systems can share the same \(R_0\) but have different \(k_{\mathrm{deg}}\) and \(k_{\mathrm{syn}}\), and therefore different dynamic responses under drug exposure. In a full concentration-time model, \(L\) is not clamped: target turnover changes complex formation and can consequently change the time course of free drug. The fixed-\(L\) calculation isolates that mechanism without claiming to solve the overall drug exposure profile.
Assumptions, Diagnostics, and Alternatives
The turnover equations are appropriate only when their biological and mathematical assumptions are plausible. Synthesis is treated as constant, free-target degradation as first order, and rate parameters as time-invariant. The equations also assume the specified complex fate: internalization removes complex-bound drug and target, with no recycling. If evidence supports target-dependent synthesis, multiple target pools, or a different complex fate, the model structure must be modified rather than forcing those processes into \(k_{\mathrm{syn}}\) or \(k_{\mathrm{deg}}\).
Several checks help detect a misspecified or incorrectly implemented turnover model:
- Drug-free baseline check. In the absence of drug, the model should approach \(R_0=k_{\mathrm{syn}}/k_{\mathrm{deg}}\). A simulated baseline that does not approach this value suggests a coding, unit, or initialization error.
- Rate and unit check. \(k_{\mathrm{syn}}\) must have concentration/time units, whereas \(k_{\mathrm{deg}}\), \(k_{\mathrm{off}}\), and \(k_{\mathrm{int}}\) have inverse-time units. The ratio \(k_{\mathrm{syn}}/k_{\mathrm{deg}}\) must be a concentration.
- Mass-balance check. Adding the free and complex target equations must give \(d(R+C)/dt=k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C\). Binding and dissociation should cancel.
- Boundary check. With nonnegative initial states and nonnegative rates, the model should not generate negative concentrations. A negative numerical solution may indicate integration problems or a coding error.
- Baseline interpretation check. If measured target changes substantially over time without drug, constant synthesis and first-order degradation may be inadequate, or the system may not have been at steady state when sampled.
A fixed-target approximation treats target availability as unchanged and can be reasonable when target turnover is negligible over the observation period or drug has little effect on the target pool. It cannot represent target depletion or replenishment. Reduced TMDD approaches can simplify binding dynamics under their own assumptions, but they do not remove the need for a coherent target turnover balance when turnover affects the data. The full differential-equation model is more flexible, but requires sufficient information to estimate its rates and can be harder to fit. In particular, similar baseline concentrations can arise from different synthesis and degradation rates, so baseline data alone do not distinguish turnover speed.
Common Mistake
A related error is to count binding as a net loss of total target. Binding moves target from \(R\) to \(C\); it does not remove target from \(R+C\). Only the specified degradation and internalization terms remove target in this model. Keeping that distinction clear prevents incorrect mass balances and helps interpret which mechanism is changing drug exposure.
Check Your Understanding
Answer each question, then compare your work with the solutions.
- A hypothetical target has \(k_{\mathrm{syn}}=6.00\ \mathrm{nM/day}\) and \(k_{\mathrm{deg}}=0.150\ \mathrm{day}^{-1}\). What is its drug-free steady-state concentration?
- Why do binding and dissociation cancel when the free and complex target equations are added?
- At fixed \(L=15.0\ \mathrm{nM}\), let \(K_{\mathrm{ss}}=10.0\ \mathrm{nM}\), \(k_{\mathrm{int}}=0.100\ \mathrm{day}^{-1}\), \(k_{\mathrm{deg}}=0.200\ \mathrm{day}^{-1}\), and \(k_{\mathrm{syn}}=8.00\ \mathrm{nM/day}\). Calculate \(R\), \(C\), and the internalization flux.
- Two systems have the same drug-free target concentration but different target turnover rates. Explain why their complex concentrations during drug exposure need not be equal.
Solutions. (1) \(R_0=k_{\mathrm{syn}}/k_{\mathrm{deg}}=(6.00\ \mathrm{nM/day})/(0.150\ \mathrm{day}^{-1})=40.0\ \mathrm{nM}\). (2) Each binding event decreases \(R\) by the same amount that it increases \(C\), and each dissociation event does the reverse; the net change in \(R+C\) from either process is zero. (3) \(C/R=L/K_{\mathrm{ss}}=15.0/10.0=1.50\). Thus \(R=8.00/[0.200+(0.100)(1.50)]=8.00/0.350=22.86\ \mathrm{nM}\), and \(C=(1.50)(22.86)=34.29\ \mathrm{nM}\). The internalization flux is \((0.100)(34.29)=3.429\ \mathrm{nM/day}\). The target balance checks: \((0.200)(22.86)+(0.100)(34.29)=4.572+3.429=8.001\ \mathrm{nM/day}\), approximately \(8.00\ \mathrm{nM/day}\) after rounding. (4) Equal baseline concentration fixes the ratio \(k_{\mathrm{syn}}/k_{\mathrm{deg}}\), not the individual rates. Different turnover rates imply different synthesis rates if baseline is held constant, which can change target replenishment, complex abundance, and internalization-mediated drug loss during exposure.