Deriving TMDD Mass-Balance Equations

From Binding Reactions to State Equations

A target-mediated drug disposition (TMDD) model is a mass-balance model: each state changes according to the material entering and leaving it. The derivation is easiest to audit when binding is first written as a reaction and each flux is then assigned to the states it affects. This makes the signs in the ODEs consequences of the reaction scheme rather than choices made from memory.

Consider a well-mixed compartment in which free drug binds free target reversibly, and the resulting complex can be internalized. Drug also enters the compartment and undergoes linear elimination while free. The target is synthesized at a constant rate and turns over while free. The model tracks concentrations of free drug, free target, and complex.

Definition: Let \(L\) be free-drug concentration, \(R\) free-target concentration, and \(C\) drug–target complex concentration. The binding reaction is \(L+R \rightleftharpoons C\), with association rate constant \(k_{\mathrm{on}}\) and dissociation rate constant \(k_{\mathrm{off}}\). Complex internalization occurs at first-order rate constant \(k_{\mathrm{int}}\). Free drug is eliminated at rate constant \(k_{\mathrm{el}}\); free target turns over at rate constant \(k_{\mathrm{deg}}\) and is synthesized at concentration rate \(k_{\mathrm{syn}}\). The systemic drug input is \(u(t)\), expressed as concentration per unit time.

The state concentrations are assumed to share the same effective compartment volume, so the association flux \(k_{\mathrm{on}}LR\) has units of concentration per time. The rate constants \(k_{\mathrm{off}}, k_{\mathrm{int}}, k_{\mathrm{el}}\), and \(k_{\mathrm{deg}}\) have units of inverse time; \(k_{\mathrm{on}}\) has units of inverse concentration per time. The synthesis rate \(k_{\mathrm{syn}}\) and input \(u(t)\) both have units of concentration per time.

The assumptions are that the compartment is well mixed; binding follows mass-action kinetics; each listed first-order loss is proportional to its stated state; synthesis is constant; and internalization removes the entire complex from the modeled compartment. In this formulation, free-drug elimination acts on \(L\), target turnover acts on \(R\), and complex internalization acts on \(C\). These choices define the model; a different biological mechanism requires different flux terms.

Deriving the Three ODEs

The association flux is \(k_{\mathrm{on}}LR\). It consumes one unit of free drug and one unit of free target for every unit of complex formed. Dissociation has flux \(k_{\mathrm{off}}C\), returning one unit to each free state. Internalization has flux \(k_{\mathrm{int}}C\), removing complex from the modeled compartment. Drug input adds free drug, free-drug elimination removes free drug, target synthesis adds free target, and free-target turnover removes free target.

\begin{aligned} \frac{dL}{dt} &= u(t)-k_{\mathrm{el}}L-k_{\mathrm{on}}LR+k_{\mathrm{off}}C,\\ \frac{dR}{dt} &= k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{on}}LR+k_{\mathrm{off}}C,\\ \frac{dC}{dt} &= k_{\mathrm{on}}LR-(k_{\mathrm{off}}+k_{\mathrm{int}})C. \end{aligned}

Each equation follows by adding the fluxes that enter its state and subtracting the fluxes that leave it. The same association term appears with a negative sign in both free-state equations and a positive sign in the complex equation. Dissociation has the reverse sign pattern. Internalization appears only as a complex loss in this model because the internalized material is not represented by another state.

The initial condition must specify \(L(0)\), \(R(0)\), and \(C(0)\), with each concentration nonnegative. A baseline without drug and complex is a special case: if the free target is at steady state before dosing, then its synthesis and turnover balance, so \(k_{\mathrm{syn}}=k_{\mathrm{deg}}R(0)\). An IV bolus is represented as a jump in \(L\), not as a continuous input after the bolus. For a bolus amount \(D_{\mathrm{dose}}\) in volume \(V\), the jump is \(L(t^+)=L(t^-)+D_{\mathrm{dose}}/V\). During a constant infusion, \(u(t)\) is the infusion rate divided by \(V\).

Worked Example: Constructing the Equations From Fluxes

Suppose a hypothetical model has \(k_{\mathrm{on}}=0.02\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\), \(k_{\mathrm{off}}=0.10\ \mathrm{h}^{-1}\), and \(k_{\mathrm{int}}=0.05\ \mathrm{h}^{-1}\). Write the binding and complex-loss contributions to each state equation.

At concentrations \(L\) and \(R\), association forms complex at \(0.02LR\ \mathrm{nM/h}\), when \(L\) and \(R\) are in nM. It therefore contributes \(-0.02LR\) to \(dL/dt\), \(-0.02LR\) to \(dR/dt\), and \(+0.02LR\) to \(dC/dt\).

Dissociation returns free drug and target at \(0.10C\ \mathrm{nM/h}\). Its contributions are \(+0.10C\) to both free-state derivatives and \(-0.10C\) to the complex derivative. Internalization removes complex at \(0.05C\ \mathrm{nM/h}\), contributing another \(-0.05C\) to \(dC/dt\).

Thus the binding-related terms are \(-0.02LR+0.10C\) for each free state and \(+0.02LR-0.15C\) for the complex. The \(0.15\ \mathrm{h}^{-1}\) complex-loss coefficient is the sum \(k_{\mathrm{off}}+k_{\mathrm{int}}\); only dissociation returns material to the free states.

Mass-Balance Checks and a Short Derivation

A useful audit is to add states that contain the same molecular species. Define total drug in the compartment as free drug plus drug bound in complex, \(L_T=L+C\). Differentiating this definition gives \(dL_T/dt=dL/dt+dC/dt\). Substitution of the three ODEs cancels association and dissociation: each is an internal transfer between free and bound drug, not a source or sink of total drug.

\begin{aligned} \frac{dL_T}{dt} &=u(t)-k_{\mathrm{el}}L-k_{\mathrm{on}}LR+k_{\mathrm{off}}C +k_{\mathrm{on}}LR-(k_{\mathrm{off}}+k_{\mathrm{int}})C\\ &=u(t)-k_{\mathrm{el}}L-k_{\mathrm{int}}C. \end{aligned}

This identity is a direct consequence of the equations, not an additional model assumption. It says that total drug increases through input and decreases through free-drug elimination and complex internalization. In particular, it would be incorrect to subtract the association flux from total drug: association changes whether drug is free or bound, not whether it is present in the compartment.

Likewise, define \(R_T=R+C\), the sum of free and complexed target in the compartment. Adding the target and complex equations cancels association and dissociation and gives

\frac{dR_T}{dt}=k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C.

Target synthesis adds target, while free-target turnover and complex internalization remove it from this model. The turnover term is \(k_{\mathrm{deg}}R\), not \(k_{\mathrm{deg}}R_T\), because the specified mechanism applies turnover to free target only. Applying it to the complex as well would be a different model.

Worked Example: Checking Instantaneous Mass Balances

At one time point, let \(L=10\ \mathrm{nM}\), \(R=15\ \mathrm{nM}\), and \(C=5\ \mathrm{nM}\). Use \(k_{\mathrm{on}}=0.02\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\), \(k_{\mathrm{off}}=0.10\ \mathrm{h}^{-1}\), \(k_{\mathrm{int}}=0.05\ \mathrm{h}^{-1}\), \(k_{\mathrm{el}}=0.01\ \mathrm{h}^{-1}\), \(k_{\mathrm{deg}}=0.02\ \mathrm{h}^{-1}\), \(k_{\mathrm{syn}}=0.40\ \mathrm{nM/h}\), and \(u(t)=0.30\ \mathrm{nM/h}\). Calculate all three derivatives and check both total balances.

The association flux is \(0.02(10)(15)=3.00\ \mathrm{nM/h}\); dissociation is \(0.10(5)=0.50\ \mathrm{nM/h}\); and internalization is \(0.05(5)=0.25\ \mathrm{nM/h}\). Free-drug elimination is \(0.01(10)=0.10\ \mathrm{nM/h}\), while free-target turnover is \(0.02(15)=0.30\ \mathrm{nM/h}\).

The free-drug derivative is \(dL/dt=0.30-0.10-3.00+0.50=-2.30\ \mathrm{nM/h}\). The free-target derivative is \(dR/dt=0.40-0.30-3.00+0.50=-2.40\ \mathrm{nM/h}\). The complex derivative is \(dC/dt=3.00-(0.10+0.05)(5)=2.25\ \mathrm{nM/h}\).

For total drug, adding the free-drug and complex derivatives gives \(-2.30+2.25=-0.05\ \mathrm{nM/h}\). The independent balance gives \(u-k_{\mathrm{el}}L-k_{\mathrm{int}}C=0.30-0.10-0.25=-0.05\ \mathrm{nM/h}\), as required. For total target, \(dR/dt+dC/dt=-2.40+2.25=-0.15\ \mathrm{nM/h}\), matching \(k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C=0.40-0.30-0.25=-0.15\ \mathrm{nM/h}\).

Input, Baseline, and Model Checks

Worked Example: Converting a Dose and an Infusion to Concentration Input

A hypothetical one-compartment model has volume \(V=3.0\ \mathrm{L}\). An IV bolus of \(12\ \mathrm{mg}\) is given, followed later by an infusion of \(0.60\ \mathrm{mg/h}\). Express the bolus jump and infusion input in concentration units.

The bolus increases free-drug concentration by dose divided by volume: \(12\ \mathrm{mg}/3.0\ \mathrm{L}=4.0\ \mathrm{mg/L}\). Thus \(L(t^+)=L(t^-)+4.0\ \mathrm{mg/L}\), assuming the administered drug enters this compartment as free drug.

The infusion contributes \(0.60\ \mathrm{mg/h}/3.0\ \mathrm{L}=0.20\ \mathrm{mg\,L^{-1}h^{-1}}\), so \(u(t)=0.20\ \mathrm{mg\,L^{-1}h^{-1}}\) while the infusion runs and is zero otherwise. The bolus is an instantaneous initial-condition change; it should not also be entered as a continuing input. The input and state units agree because \(u(t)\) is concentration per time.

Worked Example: Checking a Drug-Free Target Baseline

Suppose \(k_{\mathrm{syn}}=0.72\ \mathrm{nM/h}\) and \(k_{\mathrm{deg}}=0.03\ \mathrm{h}^{-1}\). At a drug-free baseline, take \(L=0\), \(C=0\), and \(R=24\ \mathrm{nM}\). Verify that this target state is at steady state, then calculate the initial derivatives just after a free-drug concentration of \(5\ \mathrm{nM}\) is introduced. Let \(k_{\mathrm{on}}=0.01\ \mathrm{nM}^{-1}\mathrm{h}^{-1}\), \(k_{\mathrm{off}}=0.20\ \mathrm{h}^{-1}\), \(k_{\mathrm{int}}=0.10\ \mathrm{h}^{-1}\), \(k_{\mathrm{el}}=0.02\ \mathrm{h}^{-1}\), and \(u=0\).

At baseline, synthesis is \(0.72\ \mathrm{nM/h}\), and turnover is \(0.03(24)=0.72\ \mathrm{nM/h}\). With no drug or complex, \(dR/dt=0.72-0.72=0\), and \(dL/dt=dC/dt=0\).

Immediately after drug is introduced, before any complex has accumulated, the association flux is \(0.01(5)(24)=1.20\ \mathrm{nM/h}\). Therefore \(dL/dt=-0.02(5)-1.20=-1.30\ \mathrm{nM/h}\), \(dR/dt=0.72-0.72-1.20=-1.20\ \mathrm{nM/h}\), and \(dC/dt=1.20\ \mathrm{nM/h}\). The total-drug derivative is \(-1.30+1.20=-0.10\ \mathrm{nM/h}\), exactly the free-drug elimination rate; the total-target derivative is \(-1.20+1.20=0\). Binding initially redistributes drug and target, while the free-drug elimination flux reduces total drug.

Diagnostics, Alternatives, and Limitations

The mass-balance identities and units provide practical diagnostics during model construction and numerical implementation. They do not establish that the biological assumptions are true, but they reveal many bookkeeping errors before parameter fitting. Check that all states remain nonnegative, that every rate has compatible units, and that sums of state derivatives agree with the expected external sources and sinks.

Association signs are inconsistent. If association is subtracted from only one free state, or added rather than subtracted from a free state, the stoichiometry is broken. The same association flux must be removed from both \(L\) and \(R\), and added to \(C\).
Dissociation is treated as a net loss. Dissociation removes complex but returns one unit to each free state. Its contributions cancel in the total-drug and total-target balances.
Input units do not match the states. An amount-per-time infusion rate cannot be added directly to a concentration derivative. Divide by the appropriate volume, or formulate the model in amounts throughout.
Internalization is counted incorrectly. In this formulation, it removes complex from the modeled compartment. Adding its flux to free drug or free target would imply recycling that is not in the specified reaction scheme.
Negative concentrations appear. Large integration steps or inconsistent equations can produce negative numerical states even when the underlying flux model should preserve nonnegative concentrations. Check tolerances and solver behavior, and recheck the signs and units; do not silently interpret negative concentrations as biological.

A mass-balance TMDD model is more mechanistic than a single empirical clearance term: it represents binding and complex formation explicitly, so it can describe how those processes contribute to changing free drug, free target, and complex. That added structure requires more parameters and stronger assumptions. If only an aggregate exposure profile is needed, a simpler empirical disposition model may be appropriate, but it does not separately represent these states or fluxes. An amount-based formulation is an equivalent alternative when volumes differ or are explicitly tracked; convert each flux and concentration consistently rather than mixing amount and concentration equations.

The equations above do not describe spatial gradients, multiple target pools, drug distribution between tissues, recycling of internalized material, or target synthesis regulated by drug. They also do not impose equilibrium or a reduced form: all three state equations remain dynamic. If any omitted mechanism is important for the question, the reaction scheme and balances must be extended rather than inferred from these equations.

Key takeaway: Write each reaction and external flux first, then assign its stoichiometric effect to every state. In this TMDD model, association and dissociation redistribute drug and target; input, free-drug elimination, target synthesis and turnover, and complex internalization determine the total balances.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If \(u=0.5\ \mathrm{nM/h}\), \(k_{\mathrm{el}}L=0.2\ \mathrm{nM/h}\), and \(k_{\mathrm{int}}C=0.1\ \mathrm{nM/h}\), what is \(d(L+C)/dt\)?
For the specified model, which state receives the target synthesis flux, and which state has the target turnover loss?
A modeler writes \(+k_{\mathrm{on}}LR\) in both free-drug and complex equations. What is the error?
With no drug, complex, or input, what free-target concentration makes the target derivative zero when \(k_{\mathrm{syn}}=0.60\ \mathrm{nM/h}\) and \(k_{\mathrm{deg}}=0.025\ \mathrm{h}^{-1}\)?

Solutions. (1) The total-drug balance is \(u-k_{\mathrm{el}}L-k_{\mathrm{int}}C=0.5-0.2-0.1=0.2\ \mathrm{nM/h}\). (2) Synthesis enters \(R\), while the stated turnover loss is from \(R\); complex is removed by internalization. (3) Association consumes free drug, so its contribution to \(dL/dt\) must be negative. It forms complex, so its contribution to \(dC/dt\) is positive. (4) Set \(0=k_{\mathrm{syn}}-k_{\mathrm{deg}}R\), giving \(R=0.60/0.025=24.0\ \mathrm{nM}\).