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Target-Mediated Drug Disposition in NLME · Tutorial 352 of 797

Parameterizing Binding, Complex Turnover, and Clearance

Learn to parameterize TMDD binding, target turnover, volume, and clearance consistently, and to transform the equations correctly when concentration units change.

Advanced 12 min read

What You'll Learn

  • Identify the units and roles of \(k_{\mathrm{on}}\), \(k_{\mathrm{off}}\), \(k_{\mathrm{int}}\), target turnover, volume, and systemic clearance.
  • Write a full TMDD model using free-drug, free-target, and complex concentrations.
  • Relate target synthesis and degradation rates to a baseline target concentration and target half-life.
  • Convert between clearance and the first-order elimination rate \(CL/V\).
  • Transform every relevant parameter and input term when changing concentration units.
  • Diagnose common unit and parameterization errors in fitted TMDD models.

Why Parameterization and Units Matter

A TMDD model can have the right biological structure and still produce incorrect estimates if its parameters do not match the state definitions and units. A binding rate expressed per molar concentration cannot be inserted unchanged when concentrations are represented in nanomolar units. Likewise, systemic clearance \(CL\) is not itself the first-order rate constant for concentration decline: in a one-compartment model that rate is \(CL/V\).

This tutorial uses the full mass-action model with free drug concentration \(C\), free target concentration \(R\), and drug–target complex concentration \(B\). The species are expressed in compatible molar units, with one drug molecule binding one target site. If drug and target are initially recorded in different units, convert them to compatible units before applying these equations. The input \(u(t)\) is an amount rate into the compartment; a bolus dose can instead be represented as an instantaneous concentration change.

Definition: \(k_{\mathrm{on}}\) is the association rate constant, with units concentration\(^{-1}\) time\(^{-1}\). \(k_{\mathrm{off}}\) is the dissociation rate constant, and \(k_{\mathrm{int}}\) is the complex internalization or removal rate constant; both have units time\(^{-1}\). Target synthesis \(k_{\mathrm{syn}}\) has units concentration/time, while free-target degradation \(k_{\mathrm{deg}}\) has units time\(^{-1}\). \(V\) is the apparent distribution volume, and \(CL\) is systemic clearance, with units volume/time.

The corresponding full model is:

$$ \begin{aligned} \frac{dC}{dt} &= \frac{u(t)}{V}-\frac{CL}{V}C-k_{\mathrm{on}}CR+k_{\mathrm{off}}B,\\ \frac{dR}{dt} &= k_{\mathrm{syn}}-k_{\mathrm{deg}}R-k_{\mathrm{on}}CR+k_{\mathrm{off}}B,\\ \frac{dB}{dt} &= k_{\mathrm{on}}CR-(k_{\mathrm{off}}+k_{\mathrm{int}})B. \end{aligned} $$

The association term removes free drug and free target while forming complex; dissociation returns both free species. Complex internalization removes \(B\), and systemic elimination removes free drug at rate \((CL/V)C\). In this model, systemic clearance acts on free drug, while drug in complex is lost through the complex-removal process. These are model assumptions, not universal biological facts; a different mechanism requires different equations.

Rate Constants, Turnover, and Affinity

The product \(k_{\mathrm{on}}CR\) must have concentration/time units. For example, if \(C\) and \(R\) are in nM and time is measured in hours, \(k_{\mathrm{on}}\) is in nM\(^{-1}\)h\(^{-1}\), and \(k_{\mathrm{on}}CR\) is in nM/h. Both \(k_{\mathrm{off}}B\) and \(k_{\mathrm{int}}B\) also have concentration/time units. This unit check is a quick way to detect a missing concentration conversion or an incorrectly entered rate constant.

For the reversible binding reaction by itself, the equilibrium dissociation constant is \(K_D=k_{\mathrm{off}}/k_{\mathrm{on}}\), with concentration units. This ratio describes the binding reaction’s equilibrium scale when internalization is not part of that reaction. It does not replace the separate rate constants in the full dynamic model: two parameter sets can share a ratio but imply different binding and dissociation time courses.

Target turnover can be parameterized directly by \(k_{\mathrm{syn}}\) and \(k_{\mathrm{deg}}\). A useful alternative uses a baseline free-target concentration \(R_0\) and \(k_{\mathrm{deg}}\), setting \(k_{\mathrm{syn}}=k_{\mathrm{deg}}R_0\). This identity follows when synthesis and first-order degradation balance in the absence of drug and complex. The next tutorial addresses the use of baseline target equilibrium in setting initial conditions; here, the identity is a parameterization relationship.

A target half-life \(t_{1/2,R}\) under first-order degradation is related to \(k_{\mathrm{deg}}\) by \(k_{\mathrm{deg}}=\ln(2)/t_{1/2,R}\). It is the half-life of free target under the specified degradation process, not necessarily the half-life of total target in a system where binding and complex internalization also affect target amounts.

Worked Example: Parameterizing Target Turnover

Suppose a hypothetical model uses a baseline free-target concentration \(R_0=20\) nM and a free-target degradation half-life of 24 h. Find \(k_{\mathrm{deg}}\) and the corresponding synthesis rate \(k_{\mathrm{syn}}\).

From first-order degradation, \(k_{\mathrm{deg}}=\ln(2)/t_{1/2,R}\). Substituting the stated half-life gives:

$$ k_{\mathrm{deg}}=\frac{0.6931}{24\ \mathrm{h}} =0.02888\ \mathrm{h}^{-1}. $$

Using \(k_{\mathrm{syn}}=k_{\mathrm{deg}}R_0\):

$$ k_{\mathrm{syn}} =(0.02888\ \mathrm{h}^{-1})(20\ \mathrm{nM}) =0.5776\ \mathrm{nM/h}. $$

Thus the turnover parameterization is \(k_{\mathrm{deg}}=0.02888\) h\(^{-1}\) and \(k_{\mathrm{syn}}=0.5776\) nM/h. Their units differ because one is a fractional loss rate and the other is an amount of target concentration supplied per unit time.

Volume and Systemic Clearance

In the free-drug equation, \(CL\) and \(V\) enter through \(CL/V\). Since \(CL\) has units volume/time and \(V\) has units volume, \(CL/V\) has units time\(^{-1}\). For a one-compartment model with no input or binding, \(dC/dt=-(CL/V)C\), so concentration declines exponentially at rate \(CL/V\). This is why a fitted elimination rate \(k_{\mathrm{el}}\) can be translated to clearance by \(CL=k_{\mathrm{el}}V\), provided the model uses this same one-compartment elimination structure.

Worked Example: Converting Clearance to a Concentration-Decline Rate

Consider a hypothetical model with \(CL=3.0\) L/h, \(V=30\) L, and free drug concentration \(C=4.0\) mg/L. Find the first-order elimination rate and the elimination contribution to \(dC/dt\).

The rate constant is:

$$ \frac{CL}{V} =\frac{3.0\ \mathrm{L/h}}{30\ \mathrm{L}} =0.100\ \mathrm{h}^{-1}. $$

The elimination contribution is negative:

$$ -\frac{CL}{V}C =-(0.100\ \mathrm{h}^{-1})(4.0\ \mathrm{mg/L}) =-0.400\ \mathrm{mg/(L\cdot h)}. $$

If an amount input of 12 mg/h enters the same compartment, its concentration-rate contribution is \(u/V=(12\ \mathrm{mg/h})/(30\ \mathrm{L})=0.400\) mg/(L·h). With elimination as the only other process, the two contributions cancel at this instant. Binding could change the net derivative, but it does not change the unit conversion for either input or clearance.

Changing Concentration Units Without Changing the Model

Changing concentration units changes the numerical values of some parameters, even though the physical process is unchanged. Suppose every concentration state is rescaled by a positive factor \(q\), so \(C'=qC\), \(R'=qR\), and \(B'=qB\), with time units unchanged. For example, conversion from mol/L to nmol/L uses \(q=10^9\). The derivatives must then satisfy \(dC'/dt=q\,dC/dt\), with the same relationship for \(R\) and \(B\).

The transformed parameters follow by requiring every term in each equation to scale by \(q\). In particular, \(k'_{\mathrm{on}}=k_{\mathrm{on}}/q\), because \(k'_{\mathrm{on}}C'R'=(k_{\mathrm{on}}/q)(qC)(qR)=qk_{\mathrm{on}}CR\). The first-order rates \(k_{\mathrm{off}}, k_{\mathrm{int}}, k_{\mathrm{deg}}\), and \(CL/V\) are unchanged. The synthesis rate becomes \(k'_{\mathrm{syn}}=qk_{\mathrm{syn}}\).

The input term must transform too. If \(a(t)=u(t)/V\) is the concentration-rate input, then \(a'(t)=q\,a(t)\). Equivalently, when \(V\) retains its numerical value in the same volume unit, the input amount rate must be expressed in the amount unit corresponding to the new concentration unit. Leaving \(u(t)/V\) numerically unchanged while scaling \(C\), \(R\), and \(B\) would not produce the same trajectories.

Derivation: With \(C'=qC\), \(R'=qR\), and \(B'=qB\), require \(dC'/dt=q\,dC/dt\). The association term then requires \(k'_{\mathrm{on}}C'R'=qk_{\mathrm{on}}CR\), giving \(k'_{\mathrm{on}}=k_{\mathrm{on}}/q\). The input must satisfy \(u'(t)/V'=q\,u(t)/V\); for an unchanged numerical volume, this is \(u'(t)/V=q\,u(t)/V\). The remaining terms scale correctly with unchanged first-order rates, while synthesis must scale as \(k'_{\mathrm{syn}}=qk_{\mathrm{syn}}\). The target and complex equations obey the same scaling, so all three state derivatives are multiplied by \(q\).

Worked Example: Converting a TMDD Equation From Molar to Nanomolar Units

Use hypothetical values \(C=2.0\times10^{-9}\) M, \(R=1.0\times10^{-8}\) M, and \(B=1.0\times10^{-9}\) M. Let \(k_{\mathrm{on}}=5.0\times10^7\) M\(^{-1}\)h\(^{-1}\), \(k_{\mathrm{off}}=0.20\) h\(^{-1}\), \(CL/V=0.030\) h\(^{-1}\), and \(u/V=0.40\times10^{-9}\) M/h. Calculate \(dC/dt\), then express the same terms in nM and verify the converted derivative.

In molar units, the association contribution is \((5.0\times10^7)(2.0\times10^{-9})(1.0\times10^{-8})=1.0\times10^{-9}\) M/h. The elimination contribution is \(-(0.030)(2.0\times10^{-9})=-0.060\times10^{-9}\) M/h, and dissociation contributes \((0.20)(1.0\times10^{-9})=0.20\times10^{-9}\) M/h. Including the stated input:

$$ \frac{dC}{dt} =(0.40-0.060-1.00+0.20)\times10^{-9} =-0.460\times10^{-9}\ \mathrm{M/h}. $$

For nM units, \(q=10^9\): \(C'=2.0\) nM, \(R'=10.0\) nM, and \(B'=1.0\) nM. The transformed association constant is \(k'_{\mathrm{on}}=(5.0\times10^7)/10^9=0.050\) nM\(^{-1}\)h\(^{-1}\). The input concentration rate is \(u'/V=0.40\) nM/h, not \(0.40\times10^{-9}\) nM/h. The transformed derivative is therefore:

$$ \frac{dC'}{dt} =0.40-(0.030)(2.0)-(0.050)(2.0)(10.0)+(0.20)(1.0) =-0.460\ \mathrm{nM/h}. $$

Because \(1\) nM equals \(10^{-9}\) M, this is precisely \(10^9\) times the derivative in M/h, as required by \(C'=10^9C\). The physical trajectory is unchanged; only its numerical concentration scale and associated parameter values differ.

Practical Checks, Alternatives, and Limitations

Before fitting, write down the units of every state, parameter, and input. Then check each right-hand-side term for concentration/time units. A useful model comparison is between parameterizing elimination with \(CL\) and \(V\), or directly with \(k_{\mathrm{el}}=CL/V\). The former has a direct clearance interpretation and can be useful when volume is separately meaningful; the latter is a compact parameterization of concentration decline. If only \(CL/V\) is informed by the data, reporting separate \(CL\) and \(V\) estimates without adequate information about both can imply more precision than the model supports.

Similarly, target turnover can be fitted as \(k_{\mathrm{syn}}\) and \(k_{\mathrm{deg}}\), or expressed using \(R_0\) and \(k_{\mathrm{deg}}\) through \(k_{\mathrm{syn}}=k_{\mathrm{deg}}R_0\). The second form can make a baseline target scale easier to communicate; it does not add information by itself. Binding can be reported using \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\), or using \(k_{\mathrm{on}}\) and their ratio \(K_D\), but the dynamic equations still require a consistent representation of the rate processes.

Common warning signs include a fitted \(k_{\mathrm{on}}\) changing by a factor of \(10^9\) after a concentration-unit conversion without the reciprocal conversion being applied, or \(k_{\mathrm{syn}}\) retaining the same numerical value even though its concentration unit changed. Inspect units and transformed equations before interpreting a striking parameter estimate. A poor fit can also result from an incorrect structural assumption—such as treating complex internalization as free-drug clearance—rather than from an optimizer failure.

  • Wrong: “All concentrations were converted to nM, so I can keep \(k_{\mathrm{on}}\) and the input term \(u/V\) unchanged.” Correction: divide \(k_{\mathrm{on}}\) by the concentration scale factor and multiply the concentration-rate input by it; also scale synthesis and verify every equation term.
  • Wrong: “Clearance is the rate constant in the concentration equation.” Correction: the rate constant is \(CL/V\); \(CL\) alone has units volume/time.
  • Wrong: “The target half-life is always \(\ln(2)/k_{\mathrm{deg}}\) for total target.” Correction: that relationship describes first-order degradation of free target in this parameterization; binding and complex removal affect the broader system dynamics.

These parameterizations do not guarantee that every parameter can be estimated precisely from a given data set. The full model specifies the meaning and units of the parameters, but data informativeness and numerical behavior must still be assessed during fitting.

Check Your Understanding

Answer each question, then compare with the solutions.

  1. If concentrations are converted from M to nM, by what factor does the numerical value of \(k_{\mathrm{on}}\) change?
  2. A model has \(CL=2.4\) L/h and \(V=12\) L. What is \(CL/V\), including units?
  3. For \(R_0=15\) nM and \(k_{\mathrm{deg}}=0.040\) h\(^{-1}\), calculate the turnover-based \(k_{\mathrm{syn}}\).
  4. Why must \(u/V\) change numerically when the concentration states are multiplied by \(q\), if volume is unchanged?

Solutions. (1) The scale factor is \(q=10^9\), so \(k'_{\mathrm{on}}=k_{\mathrm{on}}/10^9\) when expressed per nM rather than per M. (2) \(CL/V=(2.4\ \mathrm{L/h})/(12\ \mathrm{L})=0.200\ \mathrm{h}^{-1}\). (3) \(k_{\mathrm{syn}}=k_{\mathrm{deg}}R_0=(0.040\ \mathrm{h}^{-1})(15\ \mathrm{nM})=0.600\ \mathrm{nM/h}\). (4) Each state derivative must also be multiplied by \(q\); therefore the input concentration-rate term must become \(q(u/V)\), or the equation would not represent the same scaled trajectory.