Steady-State Mass Balance for TMDD

Why Steady-State Balances Matter

In a target-mediated drug disposition (TMDD) model, drug binds target, the complex may dissociate, and the complex may be removed. At steady state, the concentrations do not change, but binding and removal need not stop. Instead, the rates of formation and loss balance. Writing those balances makes clear which pathways actually eliminate drug and target, and which only redistribute them between free and bound forms.

The focus here is a full, one-to-one TMDD model under a constant drug input. The resulting balances are useful as a check on numerical solutions and as a way to understand what a proposed mechanism permits. They are not a dosing-regimen calculation: they describe concentrations only when the stated constant-input system has a finite steady state.

Model and Assumptions

Definition: Let $D$ be the free-drug concentration, $R$ the free-target concentration, and $C$ the concentration of the one-to-one drug–target complex. Drug enters at a constant rate $q$, measured as concentration per unit time. Free drug is eliminated by first-order clearance with rate constant $k_{\mathrm{el}}$. Target is synthesized at rate $k_{\mathrm{syn}}$ and lost from the free state at rate $k_{\mathrm{deg}}R$. Drug and target associate with rate constant $k_{\mathrm{on}}$, dissociate with rate constant $k_{\mathrm{off}}$, and the complex is internalized and removed with rate constant $k_{\mathrm{int}}$. This model assumes that internalization removes both drug and target and that internalized target is not recycled.

Concentrations $D$, $R$, and $C$ have units of amount per volume. The rate constants $k_{\mathrm{el}}$, $k_{\mathrm{deg}}$, $k_{\mathrm{off}}$, and $k_{\mathrm{int}}$ have units of inverse time; $k_{\mathrm{on}}$ has units of concentration$^{-1}$ time$^{-1}$. Both $q$ and $k_{\mathrm{syn}}$ have units of concentration per time. The model treats the input as constant, uses a fixed volume, and includes no additional drug or target loss pathways.

Under these assumptions, the full concentration balances are:

\begin{aligned} \frac{dD}{dt} &= q-k_{\mathrm{el}}D-k_{\mathrm{on}}DR+k_{\mathrm{off}}C,\\ \frac{dR}{dt} &= k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{on}}DR+k_{\mathrm{off}}C,\\ \frac{dC}{dt} &= k_{\mathrm{on}}DR-(k_{\mathrm{off}}+k_{\mathrm{int}})C. \end{aligned}

Association removes free drug and free target at the same rate that it forms complex; dissociation reverses those changes. Internalization removes complex. At steady state, each derivative is zero. That condition does not mean every individual process is zero: association, dissociation, and internalization can all continue while their combined rates balance.

Deriving the Steady-State Mass Balances

Adding the free-drug and complex equations cancels the association and dissociation terms. The sum $D+C$ is total drug in the modeled free and complex forms, so its balance is input minus the two drug-loss pathways:

\frac{d(D+C)}{dt}=q-k_{\mathrm{el}}D-k_{\mathrm{int}}C.

Similarly, adding the free-target and complex equations gives the total-target balance:

\frac{d(R+C)}{dt}=k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C.

Setting these total balances to zero gives the central result. The drug input is balanced by nonspecific elimination of free drug and removal of drug in internalized complex. Target synthesis is balanced by loss of free target and loss of target in internalized complex.

Theorem: At a finite steady state in this model, $$ q=k_{\mathrm{el}}D+k_{\mathrm{int}}C,\qquad k_{\mathrm{syn}}=k_{\mathrm{deg}}R+k_{\mathrm{int}}C. $$ These are mass balances, not binding-equilibrium assumptions. In particular, complex formation by itself is not a net elimination pathway: it removes drug from the free pool but does not remove drug from the modeled system unless the complex is removed.

The complex equation supplies the additional steady-state relation:

k_{\mathrm{on}}DR=(k_{\mathrm{off}}+k_{\mathrm{int}})C, \qquad C=\frac{DR}{K_{\mathrm{ss}}}, \qquad K_{\mathrm{ss}}=\frac{k_{\mathrm{off}}+k_{\mathrm{int}}}{k_{\mathrm{on}}}.

Here $K_{\mathrm{ss}}$ has units of concentration and is the binding scale implied by a steady complex balance. It is not generally the equilibrium dissociation constant $k_{\mathrm{off}}/k_{\mathrm{on}}$: the two coincide only when $k_{\mathrm{int}}=0$.

When $k_{\mathrm{el}}>0$, $k_{\mathrm{deg}}>0$, and $k_{\mathrm{int}}>0$, the total balances express free concentrations in terms of the complex:

D=\frac{q-k_{\mathrm{int}}C}{k_{\mathrm{el}}}, \qquad R=\frac{k_{\mathrm{syn}}-k_{\mathrm{int}}C}{k_{\mathrm{deg}}}.

Substituting these expressions into $C=DR/K_{\mathrm{ss}}$ yields one equation for the steady-state complex concentration:

C=\frac{(q-k_{\mathrm{int}}C)(k_{\mathrm{syn}}-k_{\mathrm{int}}C)} {k_{\mathrm{el}}k_{\mathrm{deg}}K_{\mathrm{ss}}}.

This equation also provides a physical-feasibility check. For positive input and synthesis, any solution must satisfy $0\leq C\leq \min(q/k_{\mathrm{int}},\,k_{\mathrm{syn}}/k_{\mathrm{int}})$, so that the implied free concentrations are nonnegative. Under the positive-rate assumptions above, the right-hand side decreases over this feasible interval as $C$ increases, while the left-hand side increases. At $C=0$, the right-hand side is positive; at the upper endpoint, it is zero. Therefore the two sides cross exactly once within the interval, giving one physically feasible steady-state complex concentration.

Worked Steady-State Calculations

Worked Example: Solving for All Three Concentrations

Consider a hypothetical system with $q=10$ nM/h, $k_{\mathrm{el}}=0.50$ h$^{-1}$, $k_{\mathrm{syn}}=6.0$ nM/h, $k_{\mathrm{deg}}=0.20$ h$^{-1}$, $k_{\mathrm{on}}=0.010$ nM$^{-1}$h$^{-1}$, $k_{\mathrm{off}}=0.10$ h$^{-1}$, and $k_{\mathrm{int}}=0.20$ h$^{-1}$. Find the steady-state concentrations.

First calculate the steady-state binding scale:

K_{\mathrm{ss}}=\frac{0.10+0.20}{0.010}=30\ \text{nM}.

Substitution into the scalar equation gives:

C=\frac{(10-0.20C)(6.0-0.20C)} {(0.50)(0.20)(30)} =\frac{60-3.2C+0.040C^2}{3}.

Rearrange to a quadratic:

$$ 0.01333C^2-2.0667C+20=0. $$

The roots are approximately $C=10.37$ nM and $C=144.6$ nM. The feasibility bound is $C\leq\min(10/0.20,6.0/0.20)=30$ nM, so only $C=10.37$ nM is physically possible. The free concentrations follow from the mass balances:

D=\frac{10-(0.20)(10.37)}{0.50}=15.85\ \text{nM}, \qquad R=\frac{6.0-(0.20)(10.37)}{0.20}=19.63\ \text{nM}.

Check both balances. Drug loss is $(0.50)(15.85)+(0.20)(10.37)=7.925+2.074=9.999$ nM/h, matching the 10 nM/h input to rounding. Target loss is $(0.20)(19.63)+(0.20)(10.37)=3.926+2.074=6.000$ nM/h, matching synthesis. The binding relation also checks: $DR/K_{\mathrm{ss}}=(15.85)(19.63)/30\approx10.37$ nM.

Worked Example: No Complex Internalization

Suppose, hypothetically, $q=4.0$ nM/h, $k_{\mathrm{el}}=0.20$ h$^{-1}$, $k_{\mathrm{syn}}=3.0$ nM/h, $k_{\mathrm{deg}}=0.30$ h$^{-1}$, $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$, $k_{\mathrm{off}}=0.10$ h$^{-1}$, and $k_{\mathrm{int}}=0$. Find the steady state.

With no complex removal, the total balances reduce to $q=k_{\mathrm{el}}D$ and $k_{\mathrm{syn}}=k_{\mathrm{deg}}R$. Thus:

D=\frac{4.0}{0.20}=20.0\ \text{nM}, \qquad R=\frac{3.0}{0.30}=10.0\ \text{nM}.

The complex steady-state relation uses $K_{\mathrm{ss}}=k_{\mathrm{off}}/k_{\mathrm{on}}=0.10/0.020=5.0$ nM, giving:

C=\frac{DR}{K_{\mathrm{ss}}} =\frac{(20.0)(10.0)}{5.0}=40.0\ \text{nM}.

The total drug concentration is $D+C=60.0$ nM, but the drug input is still balanced entirely by free-drug elimination: $(0.20)(20.0)=4.0$ nM/h. Binding changes the distribution between free and complexed drug; without complex removal, it does not contribute to net drug elimination. Likewise, target synthesis is balanced by free-target loss, not by complex loss.

Worked Example: A Steady State Supported Only by Target-Mediated Removal

For a hypothetical limiting case, let $q=2.0$ nM/h, $k_{\mathrm{el}}=0$, $k_{\mathrm{syn}}=5.0$ nM/h, $k_{\mathrm{deg}}=0.25$ h$^{-1}$, $k_{\mathrm{on}}=0.020$ nM$^{-1}$h$^{-1}$, $k_{\mathrm{off}}=0.10$ h$^{-1}$, and $k_{\mathrm{int}}=0.50$ h$^{-1}$. Assume no nonspecific drug elimination.

The drug balance requires $q=k_{\mathrm{int}}C$, so:

C=\frac{2.0}{0.50}=4.0\ \text{nM}.

The target balance then gives:

R=\frac{k_{\mathrm{syn}}-k_{\mathrm{int}}C}{k_{\mathrm{deg}}} =\frac{5.0-(0.50)(4.0)}{0.25} =12.0\ \text{nM}.

Here $K_{\mathrm{ss}}=(0.10+0.50)/0.020=30$ nM. From $C=DR/K_{\mathrm{ss}}$, the free drug concentration is:

D=\frac{CK_{\mathrm{ss}}}{R} =\frac{(4.0)(30)}{12.0}=10.0\ \text{nM}.

The drug input is exactly matched by complex removal, $(0.50)(4.0)=2.0$ nM/h. Target loss is $(0.25)(12.0)+(0.50)(4.0)=3.0+2.0=5.0$ nM/h, matching target synthesis. This finite solution is possible because the input is below the target synthesis rate, leaving positive free target to support complex formation.

Reading the Balances and Checking a Model

The balances identify the controlling pathways directly. If complex internalization is absent, free-drug elimination must carry the entire input. If nonspecific clearance is absent, complex internalization must carry it. With both pathways present, the input is divided between them according to the steady-state concentrations; neither pathway’s share can be inferred from its rate constant alone.

The target balance has a related implication. Under the no-recycling assumption, internalization removes target as well as drug. Therefore the internalized complex flux cannot exceed target synthesis at steady state. In the limiting case $k_{\mathrm{el}}=0$, the drug balance requires $C=q/k_{\mathrm{int}}$, while the target balance requires $q\leq k_{\mathrm{syn}}$ for a nonnegative free-target concentration. If $q>k_{\mathrm{syn}}$, these balances cannot produce a finite nonnegative steady state. At $q=k_{\mathrm{syn}}$, they imply $R=0$; with positive $C$ and finite $K_{\mathrm{ss}}$, the binding relation cannot hold at a finite $D$. Thus this boundary also fails to yield a finite steady state under the stated model.

For a numerical or fitted model, verify that all concentrations are nonnegative, the complex lies within the bounds implied by the mass balances, and the binding relation holds. Then compute the two residuals:

q-k_{\mathrm{el}}D-k_{\mathrm{int}}C,\qquad k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{int}}C.

Both should be close to zero relative to the scale of the rates. A negative concentration, an infeasible quadratic root, or a substantial balance residual indicates an algebraic error, a numerical convergence problem, or a mismatch between the equations being solved and the assumed mechanism. A solution to the total balances alone is not sufficient: it must also satisfy the complex balance.

How This Relates to Other TMDD Approaches

The full steady-state calculation retains free drug, free target, and complex, and enforces each mechanism’s balance. It is therefore a direct check on the full TMDD model rather than a substitute for simulating its time course. A quasi-steady-state reduction or rapid-equilibrium approximation simplifies the dynamic description by making additional assumptions about complex behavior. Those approximations can be useful, but their reduced relations should not be confused with the mass-balance identities derived here.

A Michaelis–Menten approximation can summarize saturable target-mediated disposition without explicitly tracking all three concentrations. That can be convenient when its assumptions are appropriate, but it does not by itself report the free-target and complex concentrations needed to check target turnover. Conversely, the full steady-state balances are limited to constant input and the specific pathways stated above. Target recycling, additional complex degradation routes, time-varying input, or other elimination processes require modifying the balances before using them.

Common Mistake

Common pitfall: Treating association as drug elimination and subtracting $k_{\mathrm{on}}DR$ from the total-drug balance. Association only transfers drug from the free pool into complex. When the free-drug and complex equations are added, association and dissociation cancel. Net drug loss through the target pathway occurs only when complex is removed, at rate $k_{\mathrm{int}}C$ in this model.

A second warning is to use $k_{\mathrm{off}}/k_{\mathrm{on}}$ as the complex relation when $k_{\mathrm{int}}>0$. At steady state, complex formation balances both dissociation and internalization, so the appropriate scale is $(k_{\mathrm{off}}+k_{\mathrm{int}})/k_{\mathrm{on}}$. Using the equilibrium dissociation constant in that setting generally gives a different, inconsistent complex concentration.

Check Your Understanding

Answer each question, then compare with the concise solutions below.

Why does adding the free-drug and complex equations remove the association and dissociation terms?
For $q=3.0$ nM/h, $k_{\mathrm{el}}=0.30$ h$^{-1}$, and $k_{\mathrm{int}}=0$, what is the steady-state free-drug concentration?
What is $K_{\mathrm{ss}}$ if $k_{\mathrm{off}}=0.12$ h$^{-1}$, $k_{\mathrm{int}}=0.18$ h$^{-1}$, and $k_{\mathrm{on}}=0.015$ nM$^{-1}$h$^{-1}$?
If $k_{\mathrm{el}}=0$ and the input is $q=7.0$ nM/h, what minimum condition on $k_{\mathrm{syn}}$ is required by the steady-state target balance for a nonnegative free-target concentration?
A calculated complex concentration makes $q-k_{\mathrm{el}}D-k_{\mathrm{int}}C$ substantially positive. What does that indicate?

Solutions. (1) Association and dissociation move drug between $D$ and $C$, so their equal-and-opposite terms cancel when those pools are added. (2) With $k_{\mathrm{int}}=0$, $D=q/k_{\mathrm{el}}=3.0/0.30=10.0$ nM. (3) $K_{\mathrm{ss}}=(0.12+0.18)/0.015=20.0$ nM. (4) The drug balance requires $C=q/k_{\mathrm{int}}$, so the target balance requires $k_{\mathrm{syn}}\geq k_{\mathrm{int}}C=q=7.0$ nM/h. Equality implies zero free target and does not give a finite binding-consistent solution for positive complex; a finite solution requires $k_{\mathrm{syn}}>7.0$ nM/h. (5) The drug mass balance is not satisfied; check the calculation, the steady-state solution, and whether the model includes an unaccounted pathway.