Selection between Michaelis–Menten and target-mediated drug disposition pharmacokinetic models

Abstract

Target-mediated drug disposition (TMDD) models have been applied to describe the pharmacokinetics of drugs whose distribution and/or clearance are affected by its target due to high binding affinity and limited capacity. The Michaelis–Menten (M–M) model has also been frequently used to describe the pharmacokinetics of such drugs. The purpose of this study is to investigate conditions for equivalence between M–M and TMDD pharmacokinetic models and provide guidelines for selection between these two approaches. Theoretical derivations were used to determine conditions under which M–M and TMDD pharmacokinetic models are equivalent. Computer simulations and model fitting were conducted to demonstrate these conditions. Typical M–M and TMDD profiles were simulated based on literature data for an anti-CD4 monoclonal antibody (TRX1) and phenytoin administered intravenously. Both models were fitted to data and goodness of fit criteria were evaluated for model selection. A case study of recombinant human erythropoietin was conducted to qualify results. A rapid binding TMDD model is equivalent to the M–M model if total target density R _tot is constant, and R _tot K _D /(K _D  + C) ² ≪ 1 where K _D represents the dissociation constant and C is the free drug concentration. Under these conditions, M–M parameters are defined as: V _max  = k _int R _tot V _c and K _m  = K _D where k _int represents an internalization rate constant, and V _c is the volume of the central compartment. R _tot is constant if and only if k _int  = k _deg, where k _deg is a degradation rate constant. If the TMDD model predictions are not sensitive to k _int or k _deg parameters, the condition of R _tot K _D /(K _D  + C) ² ≪ 1 alone can preserve the equivalence between rapid binding TMDD and M–M models. The model selection process for drugs that exhibit TMDD should involve a full mechanistic model as well as reduced models. The best model should adequately describe the data and have a minimal set of parameters estimated with acceptable precision.

Appendices

Appendix 1

Derivation of Wagner equation (Eq. 19)

The Eq. 12a, b can be rearranged to the following form:

$$ C_{tot} = C + {\frac{{R_{tot} C}}{{K_{D} + C}}} $$

(A1)

Differentiating both sides of Eq. A1 gives:

$$ {\frac{{dC_{tot} }}{dt}} = {\frac{dC}{dt}} + {\frac{d}{dt}}\left( {{\frac{{R_{tot} C}}{{K_{D} + C}}}} \right) = {\frac{dC}{dt}} + {\frac{{R_{tot} K_{D} }}{{(K_{D} + C)^{2} }}}{\frac{dC}{dt}} + {\frac{C}{{K_{D} + C}}}\,{\frac{{dR_{tot} }}{dt}} $$

(A2)

Solving above equation for dC/dt results in:

$$ {\frac{dC}{dt}} = {\frac{{{\frac{{dC_{tot} }}{dt}} - {\frac{C}{{K_{D} + C}}}{\frac{{dR_{tot} }}{dt}}}}{{1 + {\frac{{R_{tot} K_{D} }}{{(K_{D} + C)^{2} }}}}}} $$

(A3)

Substitution of dC _tot /dt from Eq. 13 into Eq. A3 and further replacement of C _tot with Eq. A1 yield:

$$ {\frac{dC}{dt}} = {\frac{{In(t) - {\frac{{\left( {k_{int} R{}_{tot} + {\frac{{dR_{tot} }}{dt}}} \right)C}}{{K_{D} + C}}} - (k_{el} + k_{pt} ) C + k_{tp} A_{T} /V_{c} }}{{1 + {\frac{{R_{tot} K_{D} }}{{(K_{D} + C)^{2} }}}}}} $$

(A4)

If R _tot is constant, then dR _tot /dt = 0, and one obtains Eq. 19.

Appendix 2

Theorem: for the TMDD model, R _tot is constant if and only if k _int  = k _deg

Proof

• (a) Proof of the sufficient condition

At steady state, from Eq. 14 it can be reached:

$$ k_{syn} - k_{deg } R_{0} - k_{int} RC_{0} = 0 $$

(A5)

Hence,

$$ k_{syn} = k_{deg } R_{0} + k_{int} RC_{0} $$

(A6)

If R _tot is constant, Eq. 14 implies:

$$ k_{syn} - k_{deg } R - k_{int} RC = 0 $$

(A7)

Replacing k _syn with Eq. A6, one can obtain:

$$ k_{deg } (R_{0} - R) + k_{int} (RC_{0} - RC) = 0 $$

(A8)

From Eq. 12a, b, it can be concluded:

$$ R_{tot} = RC + R;\,R_{tot0} = RC_{0} + R_{0} $$

(A9, 10)

That R _tot is constant implies:

$$ R_{tot} = R_{tot0} $$

(A11)

Substitution of R _tot and R _tot0 from Eqs. A9 and A10 to Eq. A11 results in:

$$ RC + R = RC_{0} + R{}_{0} $$

(A12)

Rearrangement of Eq. A12 gives:

$$ R_{0} - R = RC - RC_{0} $$

(A13)

Replacing R ₀ –R in Eq. A8 with Eq. A13 yields:

$$ (k_{deg } - k_{int} )(RC_{0} - RC) = 0 $$

(A14)

Since $RC_{0} - RC \ne 0$ after drug administration, the Eq. A14 implies:

$$ k_{int} = k_{deg } $$

(A15)

• (b) Proof of the necessary condition

If $k_{int} = k_{deg } ,$ then Eq. 14 implies:

$$ {\frac{{dR_{tot} }}{dt}} = - k_{deg } R_{tot} + k_{syn} $$

(A16)

Replacing k _syn in Eq. A16 with Eq. A6 yields:

$$ {\frac{{dR_{tot} }}{dt}} = - k_{deg } R_{tot} + k_{deg } R_{0} + k_{int} RC_{0} $$

(A17)

Substitution of k _int with k _deg gives:

$$ {\frac{{dR_{tot} }}{dt}} = - k_{deg } R_{tot} + k_{deg } (R_{0} + RC_{0} ) $$

(A18)

Given $R_{tot0} = RC_{0} + R_{0} ,$ Eq. A18 reduces to:

$$ {\frac{{dR_{tot} }}{dt}} = - k_{deg } R_{tot} + k_{deg } R_{tot0} $$

(A19)

Since Eq. A19 starts from initial condition, it can be concluded that:

$$ {\frac{{dR_{tot} }}{dt}} = 0 $$

(A20)

End of proof.