From Processes to Differential Equations
A full target-mediated drug disposition (TMDD) model represents binding and disposition through separate states for free drug, free target, and drug-target complex. The central modeling task is to account for every process exactly once: a process that removes material from one state may add it to another, or may remove it from the modeled system altogether. Writing the equations from this accounting prevents sign errors and clarifies what the model assumes.
We derive a one-compartment model with reversible 1:1 binding. Concentrations are measured in the same well-mixed effective volume, which is constant. This makes it possible to express all three states in concentration units and to use a concentration-rate input. The equations describe concentrations over time; they do not assume that binding is at equilibrium.
The units provide an immediate check. \(D\), \(R\), and \(C\) have concentration units, such as nM. The association constant has units concentration\(^{-1}\) time\(^{-1}\), so \(k_{\mathrm{on}}DR\) has concentration per time. The first-order rate constants have units time\(^{-1}\), and \(k_{\mathrm{syn}}\) and \(u(t)\) have concentration per time.
The rate of association is \(k_{\mathrm{on}}DR\): it increases with both free drug and free target. The dissociation rate is \(k_{\mathrm{off}}C\). Association removes one concentration unit of drug and one of target from their free states while adding one unit of complex; dissociation does the reverse. The remaining processes act on the state they are specified to affect.
Constructing the Full Model
For each state, add its rates of entry and subtract its rates of exit. Drug input adds free drug. Nonspecific elimination removes free drug. Association removes free drug, while dissociation returns drug from the complex to the free-drug state. Thus the free-drug equation is
Free target is supplied by synthesis and removed by its own first-order loss process. Binding removes free target, and dissociation returns it. The free-target equation is
The complex is formed by association and lost by dissociation or internalization. Therefore,
Together, these equations are the full TMDD mass-balance system under the stated assumptions. Each term has units of concentration per time. In particular, \(k_{\mathrm{on}}DR\) is a rate, not a concentration, and it must appear with a negative sign in both free-state equations and a positive sign in the complex equation.
The process-by-process accounting can also be represented by the state changes caused by one unit of each process rate. The entries show the change in \((D,R,C)\) per concentration unit of process extent.
| Process | Rate | Change in \((D,R,C)\) |
|---|---|---|
| Drug input | \(u(t)\) | \((+1,0,0)\) |
| Nonspecific free-drug elimination | \(k_{\mathrm{el}}D\) | \((-1,0,0)\) |
| Free-target synthesis | \(k_{\mathrm{syn}}\) | \((0,+1,0)\) |
| Free-target loss | \(k_{\mathrm{deg}}R\) | \((0,-1,0)\) |
| Association | \(k_{\mathrm{on}}DR\) | \((-1,-1,+1)\) |
| Dissociation | \(k_{\mathrm{off}}C\) | \((+1,+1,-1)\) |
| Complex internalization | \(k_{\mathrm{int}}C\) | \((0,0,-1)\) |
Worked Example: Calculating Instantaneous State Changes
Consider hypothetical concentrations \(D=2.00\) nM, \(R=0.800\) nM, and \(C=0.200\) nM. Let \(k_{\mathrm{on}}=0.500\) nM\(^{-1}\)h\(^{-1}\), \(k_{\mathrm{off}}=0.300\) h\(^{-1}\), \(k_{\mathrm{el}}=0.100\) h\(^{-1}\), \(k_{\mathrm{deg}}=0.0500\) h\(^{-1}\), \(k_{\mathrm{int}}=0.200\) h\(^{-1}\), \(k_{\mathrm{syn}}=0.0400\) nM/h, and \(u(t)=0\). Find the instantaneous derivatives.
First calculate the association rate:
The free-drug derivative is
The free-target derivative is
The complex derivative is
At this instant, both free states are decreasing and the complex is increasing. These are local rates of change, not predictions that the same rates continue over time: as the states change, the association, elimination, and other rates change too.
Mass-Balance Checks
The equations can be checked by summing states that contain the same molecular species. The total drug concentration represented in the compartment is \(D+C\), because each complex contains one drug molecule. Add the free-drug and complex equations. Association and dissociation cancel: each only transfers drug between free and bound states. The result is
This balance says that drug in the modeled compartment changes through input, nonspecific elimination of free drug, and internalization of complex. If the binding terms do not cancel in this sum, a binding term has likely been omitted or counted with the wrong sign. The equation also makes clear that nonspecific elimination is specified here for free drug, not for complex.
Likewise, the total target concentration represented in the compartment is \(R+C\). Adding the free-target and complex equations gives
Binding and dissociation cancel because they redistribute target between its free and complexed forms. Target synthesis adds free target; free-target loss and complex internalization remove target from these modeled states. These sums are accounting identities for the specified model, not claims that total drug or target is conserved: input, elimination, synthesis, and internalization change the totals.
For a short justification of why the association and dissociation terms must cancel, consider association over a small interval of duration \(\Delta t\). If the association rate is \(v=k_{\mathrm{on}}DR\), then approximately \(v\Delta t\) concentration units of drug and target leave the free states and \(v\Delta t\) units of complex form. The change in represented total drug is \(-v\Delta t+v\Delta t=0\), and the same is true for represented total target. Dissociation has the reverse state changes and also contributes zero to each total. Dividing these changes by \(\Delta t\) gives the cancellation in the differential balances.
Worked Example: Checking the Drug and Target Balances
Use the hypothetical state and rate values from the previous example. The sum of the free-drug and complex derivatives is
The total-drug balance gives the same result:
For target, the state derivatives sum to \(-0.740+0.700=-0.0400\) nM/h. The target balance independently gives
Both checks agree. Notice that the association rate does not appear in either total balance; association changes the distribution among states, not the represented totals.
Initial Conditions, Inputs, and Scope
The differential equations require initial values \(D(0)\), \(R(0)\), and \(C(0)\), as well as a specified input function \(u(t)\). For a bolus dose administered into this compartment, it is usually clearer to represent the dose as an instantaneous change in the initial free-drug concentration than as an ordinary continuous input rate. If the dose amount is \(A\) and the effective volume is \(V\), the concentration jump is \(A/V\), so \(D(0^+)=D(0^-)+A/V\), with no instantaneous change in \(R\) or \(C\) under this convention. An infusion, by contrast, can be represented by a nonzero \(u(t)\) during its duration. The input units must match the concentration-per-time units in the differential equations.
Worked Example: A Bolus Followed by Binding
Suppose a hypothetical 500 nmol bolus enters a compartment with effective volume 5.00 L. Initially, before the dose, \(D=0\), \(R=10.0\) nM, and \(C=0\). Let \(k_{\mathrm{on}}=0.0200\) nM\(^{-1}\)h\(^{-1}\), \(k_{\mathrm{el}}=0.100\) h\(^{-1}\), \(k_{\mathrm{deg}}=0.0500\) h\(^{-1}\), \(k_{\mathrm{int}}=0.200\) h\(^{-1}\), and \(k_{\mathrm{syn}}=0.500\) nM/h. Take \(u(t)=0\) after the bolus.
The dose concentration is
Immediately after dosing, association occurs at rate
Since \(C=0\), there is no initial dissociation or internalization. The initial free-drug rate is
For free target, synthesis and loss initially cancel: \(0.500-(0.0500)(10.0)=0\). Binding therefore gives \(\frac{dR}{dt}=-20.0\) nM/h, while \(\frac{dC}{dt}=+20.0\) nM/h. Their sum is zero, as required by the target balance at that instant: \(k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C=0.500-0.500-0=0\). The total-drug derivative is \(-30.0+20.0=-10.0\) nM/h, matching \(-k_{\mathrm{el}}D-k_{\mathrm{int}}C=-10.0\) nM/h. Thus the fast initial decrease in free drug includes transfer into complex, while the decrease in total drug at that instant comes only from nonspecific elimination.
Assumptions, Diagnostics, and Alternatives
The equations are appropriate only when their process assumptions match the biological and experimental setting. They specify one well-mixed compartment; 1:1 reversible binding; constant rate parameters; constant synthesis of free target; first-order free-target loss; irreversible complex internalization; and nonspecific elimination of free drug. They also assume the tracked states are concentrations in a shared, constant effective volume. A model with multiple compartments, different effective volumes, additional binding partners, complex recycling, or elimination of bound drug needs additional or altered terms.
Several checks are useful before fitting or simulating the model:
- Check dimensions. Every right-hand-side term must have concentration-per-time units. A missing concentration factor in the association term, for example, will fail this check.
- Check signs and destinations. Association must reduce \(D\) and \(R\) while increasing \(C\); dissociation must do the reverse. Internalization removes \(C\), not free \(D\) or \(R\) directly under these assumptions.
- Check the summed balances. Adding the equations must cancel both binding terms. A failure indicates inconsistent stoichiometry or a transcription error.
- Check state behavior. With nonnegative parameters, nonnegative inputs, and nonnegative initial states, a state at zero should not have a negative derivative caused by the model: at \(D=0\), its derivative is \(u+k_{\mathrm{off}}C\); at \(R=0\), it is \(k_{\mathrm{syn}}+k_{\mathrm{off}}C\); and at \(C=0\), it is \(k_{\mathrm{on}}DR\). Simulated negative concentrations therefore suggest a numerical or implementation problem, or a model formulation outside these assumptions.
A common mistake is to subtract association from free drug and free target but forget to add it to complex, or to add dissociation to the complex instead of subtracting it. That error can create or destroy represented drug or target during binding. The correction is to write the state changes for each process first and then sum the derivatives to verify the total balances.
A simple linear pharmacokinetic model may be a reasonable alternative when target binding does not materially affect the observed drug disposition or when the data do not support separate target-related states; it does not explicitly represent complex formation. An algebraic occupancy description can summarize bound fraction under its own assumptions, but it does not by itself account for the changing drug and target amounts in these differential equations. Reduced TMDD formulations can be useful when their approximations are justified, but they should not be substituted for the full state balance without checking those conditions. The full model’s strength is mechanistic accounting; its limitations include greater data and parameter requirements and sensitivity to incorrect assumptions about the biological processes.
Check Your Understanding
Answer each question using the process accounting and balances above.
- If \(k_{\mathrm{on}}DR=0.60\) nM/h, what are the association contributions to \(dD/dt\), \(dR/dt\), and \(dC/dt\)?
- Why does \(k_{\mathrm{off}}C\) have a positive sign in the free-drug and free-target equations but a negative sign in the complex equation?
- Write the total-drug balance for this model and identify which terms represent loss from the modeled total.
- If \(C=0\), \(D=0\), and \(u=0\), what is \(dD/dt\)? Explain why.
Solutions. 1. The contributions are \(-0.60\), \(-0.60\), and \(+0.60\) nM/h, respectively, because one drug and one target form each complex. 2. Dissociation converts complex into one free drug and one free target, so it removes complex and adds to both free states. 3. \(\frac{d(D+C)}{dt}=u-k_{\mathrm{el}}D-k_{\mathrm{int}}C\); nonspecific elimination and complex internalization remove drug from the modeled total, while input adds it. 4. The derivative is zero: there is no free drug to eliminate, no complex to dissociate, no association when \(D=0\), and no input.