Why Write the Full State Equations?
When drug binds a target, the free-drug concentration can change for two different reasons: drug can be cleared from the system, or it can move temporarily into a bound complex. These processes are not interchangeable. A mechanistic model therefore tracks free drug, free target, and drug–target complex as separate states and accounts explicitly for each reaction that moves material between them.
The equations below describe a standard one-to-one mass-action TMDD model. They specify how the states change over time; they do not, by themselves, determine which concentrations an assay measures or how those measurements are modeled.
The standard model assumes a well-mixed compartment of fixed volume, one drug molecule binding one target molecule, and constant rate parameters over the modeled period. The association reaction follows mass action, so its rate is proportional to both free concentrations. Dissociation, internalization, free-target degradation, and systemic drug clearance are first-order in their respective states. Target production is zero-order. The model also assumes that internalization removes both drug and target in the complex, and that systemic clearance acts on free drug rather than directly on complex.
From Reactions to Differential Equations
Write each process as a reaction or input, then record its effect on the three states. The association flux is \(k_{\mathrm{on}}DR\), with concentration-per-time units. Dissociation returns one drug and one target to their free states, at flux \(k_{\mathrm{off}}C\). Internalization removes complex at flux \(k_{\mathrm{int}}C\). The remaining processes are production of free target, degradation of free target, and clearance of free drug.
Let \(u_D(t)\) denote an external drug input expressed as concentration per unit time. For an infusion into a fixed-volume compartment, this is the amount infusion rate divided by volume. A bolus is instead represented as a jump in \(D\), not as an ordinary finite rate acting over time.
Each term has a direct state-based interpretation. Association decreases both \(D\) and \(R\) and increases \(C\); dissociation does the reverse. Internalization decreases \(C\), while target production and degradation affect free \(R\), and systemic clearance decreases free \(D\). The model does not add an internalized state because internalized material is outside the three-state system being represented.
Worked Applications
Worked Example: Calculate the Instantaneous State Changes
Suppose a hypothetical system has \(D=100\) nM, \(R=20\) nM, and \(C=5\) nM. Let \(k_{\mathrm{on}}=0.010\) nM\(^{-1}\)h\(^{-1}\), \(k_{\mathrm{off}}=0.20\) h\(^{-1}\), \(k_{\mathrm{int}}=0.10\) h\(^{-1}\), \(k_{\mathrm{el}}=0.030\) h\(^{-1}\), \(k_{\mathrm{syn}}=2.0\) nM/h, and \(k_{\mathrm{deg}}=0.050\) h\(^{-1}\). There is no infusion at this instant, so \(u_D(t)=0\). Find each derivative.
First calculate the association flux:
The dissociation flux is \(k_{\mathrm{off}}C=(0.20)(5)=1.0\) nM/h. Free-drug clearance is \(k_{\mathrm{el}}D=(0.030)(100)=3.0\) nM/h. Thus
For free target, degradation is \(k_{\mathrm{deg}}R=(0.050)(20)=1.0\) nM/h, so
Complex internalization is \(k_{\mathrm{int}}C=(0.10)(5)=0.50\) nM/h. Therefore
At this instant, complex is accumulating, while both free states are decreasing. These are instantaneous slopes, not predictions that the same rates persist for an hour; the fluxes change as the state concentrations change.
Worked Example: Represent a Bolus and Check the Post-Dose Slope
Consider a hypothetical one-compartment system with volume \(5.0\) L. A \(500\) nmol intravenous bolus is given when \(D=10\) nM, \(R=20\) nM, and \(C=0\) nM. Immediately before dosing, all other processes are finite-rate processes; suppose \(k_{\mathrm{el}}=0.010\) h\(^{-1}\) and \(k_{\mathrm{on}}=0.0020\) nM\(^{-1}\)h\(^{-1}\).
The bolus raises the drug concentration by dose divided by volume. Since \(500\) nmol in \(5.0\) L gives \(100\) nmol/L, and \(1\) nmol/L equals \(1\) nM, the immediate post-dose drug concentration is \(110\) nM. The bolus does not instantaneously change \(R\) or \(C\), so they remain \(20\) nM and \(0\) nM.
Evaluate the post-dose derivative using the state values just after the jump. There is no continuing infusion, and dissociation contributes zero because \(C=0\). Association is
Clearance is \((0.010)(110)=1.1\) nM/h. Hence \(dD/dt=-1.1-4.4=-5.5\) nM/h immediately after the bolus. The instantaneous derivatives of target and complex are not fully determined by this example because their production, degradation, dissociation, and internalization parameters were not specified. The bolus is a state jump; it should not be treated as a sustained input of \(100\) nM/h.
A Mass-Balance Check
A useful derivation follows by adding the free-drug and complex equations. This sum tracks drug moiety in the modeled states: each molecule of \(D\) or \(C\) contains one drug molecule. Association and dissociation redistribute drug between the two states, so their terms must cancel in the sum.
The cancellation is a structural check: binding alone cannot create or destroy drug. Drug leaves the modeled free-plus-complex pool through systemic clearance of free drug and internalization of complex. In the first worked example, the derivative of this pool is \(-22+18.5=-3.5\) nM/h. The same result follows from \(-3.0-0.50=-3.5\) nM/h, exactly the free-drug clearance plus complex internalization losses.
A parallel check can be made for target moiety by adding the equations for \(R\) and \(C\). Binding and dissociation cancel there too, leaving \(d(R+C)/dt=k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C\). This reflects the stated assumption that degradation applies to free target and internalization removes complex.
Assumptions, Diagnostics, and Alternatives
Before using these equations, check that state concentrations and parameter units are compatible. If \(D\), \(R\), and \(C\) are in nM and time is in hours, then \(k_{\mathrm{on}}\) must be in nM\(^{-1}\)h\(^{-1}\), while the first-order rate constants must be in h\(^{-1}\), and \(k_{\mathrm{syn}}\) in nM/h. Every term in each derivative must have concentration-per-time units. If amount states are used instead, the association expression and all input and clearance terms must be formulated consistently for that representation.
The equations also imply nonnegative forward fluxes when states and rate constants are nonnegative. At a zero boundary, the derivative should not point toward a negative concentration: for example, at \(D=0\), \(dD/dt=u_D+k_{\mathrm{off}}C\), which is nonnegative for nonnegative input and complex. At \(R=0\), \(dR/dt=k_{\mathrm{syn}}+k_{\mathrm{off}}C\), also nonnegative. At \(C=0\), \(dC/dt=k_{\mathrm{on}}DR\), nonnegative. A numerical solution that becomes substantially negative is therefore a warning about numerical integration, coding, or incompatible inputs—not a biological prediction.
Several additional checks help locate errors:
- Association must have a negative sign in both free-state equations and a positive sign in the complex equation.
- Dissociation must have a positive sign in both free-state equations and a negative sign in the complex equation.
- Internalization must reduce complex. Under the stated model, it does not return free drug or free target.
- Systemic clearance must reduce free drug, and target production must increase free target.
- The association and dissociation terms must cancel from both \(d(D+C)/dt\) and \(d(R+C)/dt\).
The full state model is appropriate when binding and complex turnover are represented as distinct processes. A simpler empirical concentration-dependent clearance model may be easier to fit, but it does not explicitly represent target or complex dynamics and should not be interpreted as if it did. A target turnover model without explicit complex formation can describe changing target abundance but omits the mass-action binding mechanism. Choosing among such formulations depends on the scientific question and the information in the data; the full equations do not guarantee that every rate parameter can be estimated precisely.
Check Your Understanding
Answer each question, then compare with the solutions below.
- If \(k_{\mathrm{on}}DR=6\) nM/h, which derivatives receive a \(+6\) nM/h contribution and which receive a \(-6\) nM/h contribution?
- Why does a bolus change \(D\) discontinuously rather than appear as a constant infusion term in the differential equation?
- Suppose \(D=0\), \(C=4\) nM, and \(k_{\mathrm{off}}=0.25\) h\(^{-1}\), with no input. What are the clearance and binding-related contributions to \(dD/dt\)?
- State the expression for \(d(D+C)/dt\) and explain why the association and dissociation terms do not appear in it.
Solutions. (1) Association contributes \(-6\) nM/h to \(dD/dt\), \(-6\) nM/h to \(dR/dt\), and \(+6\) nM/h to \(dC/dt\), because one free drug and one free target form one complex. (2) A bolus delivers a finite amount over a negligible time interval in the model, so it is represented by an instantaneous state increment, dose divided by volume. An infusion delivers drug continuously and is represented by a rate \(u_D(t)\). (3) Clearance contributes zero because \(k_{\mathrm{el}}D=0\). Dissociation contributes \(k_{\mathrm{off}}C=(0.25)(4)=1.0\) nM/h to free drug; association is zero because \(D=0\). Thus the stated contributions give \(dD/dt=+1.0\) nM/h. (4) \(d(D+C)/dt=u_D(t)-k_{\mathrm{el}}D-k_{\mathrm{int}}C\). Association and dissociation only transfer drug between free and complex states, so their negative and positive contributions cancel when the two derivatives are added.