Pharmacokinetics of Anti-hepcidin Monoclonal Antibody Ab 12B9m and Hepcidin in Cynomolgus Monkeys

Abstract

Hepcidin is a key regulator responsible for systemic iron homeostasis. A semi-mechanistic PK model for hepcidin and a fully human anti-hepcidin monoclonal antibody (Ab 12B9m) was developed to describe their total (free + bound) serum concentration-time data after single and multiple weekly intravenous or subcutaneous doses of Ab 12B9m. The model was based on target mediated drug disposition and the IgG–FcRn interaction concepts published previously. Both total Ab 12B9m and total hepcidin exhibited nonlinear kinetics due to saturable Fc–FcRn interaction. Ab 12B9m showed a limited volume of distribution and negligible linear elimination from serum. The nonlinear elimination of Ab 12B9m was attributed to the endosomal degradation of Ab 12B9m that was not bound to the FcRn receptor. The terminal half-life, assumed to be the same for free and total serum Ab 12B9m, was estimated to be 16.5 days. The subcutaneous absorption of Ab 12B9m was described with a first-order absorption rate constant k _a of 0.0278 h⁻¹, with 86% bioavailability. The model suggested a rapid hepcidin clearance of approximately 800 mL h⁻¹ kg⁻¹. Only the highest-tested Ab 12B9m dose of 300 mg kg⁻¹ week⁻¹ was able to maintain free hepcidin level below the baseline during the dosing intervals. Free Ab 12B9m and free hepcidin concentrations were simulated, and their PK profiles were nonlinear as affected by their binding to each other. Additionally, the total amount of FcRn receptor involved in Ab 12B9m recycling at a given time was calculated empirically, and the temporal changes in the free FcRn levels upon Ab 12B9m administration were inferred.

APPENDIX

Derivation of Eqs. 9–16

The full (non-reduced) TMDD PK model as presented in Fig. 1 encompasses differential equations describing Ab_E, FcAb_E, AbH_E, and FcAbH_E variables:

$$ \frac{{{\text{dA}}{{\text{b}}_{\rm{E}}}}}{{{\hbox{d}}t}} = {k_{\rm{up}}} \cdot {\hbox{Ab}} - {k_{\deg }} \cdot {\hbox{A}}{{\hbox{b}}_{\rm{E}}} - {k_{\rm{onA}}} \cdot {\hbox{Fc}} \cdot {\hbox{A}}{{\hbox{b}}_{\rm{E}}}/{V_{\rm{E}}} + {k_{\rm{offA}}} \cdot {\hbox{FcA}}{{\hbox{b}}_{\rm{E}}} $$

(23)

$$ \frac{{{\text{dFcA}}{{\text{b}}_{\rm{E}}}}}{{{\hbox{d}}t}} = {k_{\rm{onA}}} \cdot {\hbox{Fc}} \cdot {\hbox{A}}{{\hbox{b}}_{\rm{E}}}/{V_{\rm{E}}} - ({k_{\rm{offA}}} + {k_{\rm{R}}}) \cdot {\hbox{FcA}}{{\hbox{b}}_{\rm{E}}} $$

(24)

$$ \frac{{{\text{dAb}}{{\text{H}}_{\rm{E}}}}}{{{\hbox{d}}t}} = {k_{\rm{up}}} \cdot {\hbox{AbH}} - {k_{\deg }} \cdot {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}} - {k_{\rm{onA}}} \cdot {\hbox{Fc}} \cdot {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}}/{V_{\rm{E}}} + {k_{\rm{offA}}} \cdot {\hbox{FcAb}}{{\hbox{H}}_{\rm{E}}} $$

(25)

$$ \frac{{{\text{dFcAb}}{{\text{H}}_{\rm{E}}}}}{{{\hbox{d}}t}} = {k_{\rm{onA}}} \cdot {\hbox{Fc}} \cdot {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}}/{V_{\rm{E}}} - ({k_{\rm{offA}}} + {k_{\rm{R}}}) \cdot {\hbox{FcAb}}{{\hbox{H}}_{\rm{E}}} $$

(26)

Let E denote the amount of free and H25 bound Ab 12B9m in the endosomal compartment:

$$ E = {\hbox{A}}{{\hbox{b}}_{\rm{E}}} + {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}} $$

(27)

Similarly, let FcE denote the amount of the complexes FcRn–Ab 12B9m and FcRn–Ab 12B9m–H25 in the endosomal compartment:

$$ {\hbox{FcE}} = {\hbox{FcA}}{{\hbox{b}}_{\rm{E}}} + {\hbox{FcAb}}{{\hbox{H}}_{\rm{E}}} $$

(28)

Adding Eqs. 23 and 25, and Eqs. 24 and 26 results in the following differential equations for E and FcE:

$$ \frac{{{\text{d}}E}}{{{\hbox{d}}t}} = {k_{\rm{up}}} \cdot ({\hbox{Ab}} + {\hbox{AbH}}) - {k_{\deg }} \cdot E - {k_{\rm{onA}}} \cdot {\hbox{Fc}} \cdot E/{V_{\rm{E}}} + {k_{\rm{offA}}} \cdot {\hbox{Fc}}E $$

(29)

$$ \frac{{{\text{dFc}}E}}{{{\hbox{d}}t}} = {k_{\rm{onA}}} \cdot {\hbox{Fc}} \cdot E/{V_{\rm{E}}} - ({k_{\rm{offA}}} + {k_{\rm{R}}}) \cdot {\hbox{Fc}}E $$

(30)

Consequently, the combined species E bind to the FcRn receptor as if it is a single drug. The equilibrium assumptions Eq. 1a, b can be written as

$$ {\hbox{Fc}} \cdot {\hbox{A}}{{\hbox{b}}_{\rm{E}}}/{V_{\rm{E}}} = {K_{\rm{DA}}} \cdot {\hbox{FcA}}{{\hbox{b}}_{\rm{E}}}\;{\hbox{and}}\;{\hbox{Fc}} \cdot {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}}/{V_{\rm{E}}} = {K_{\rm{DA}}} \cdot {\hbox{FcAb}}{{\hbox{H}}_{\rm{E}}} $$

(31a, b)

Adding Eq. 31a, b and dividing by FcE results in

$$ \frac{{{\text{Fc}} \cdot E/{V_{\rm{E}}}}}{{{\hbox{Fc}}E}} = {K_{\rm{DA}}} $$

(32)

which means that the equilibrium assumption applies also to the combined species E. Repeating the argument for the quasi-equilibrium TMDD model (43), one can conclude that

$$ E = \frac{1}{2}\left( {{E_{\rm{tot}}} - {\hbox{F}}{{\hbox{c}}_{\rm{tot}}} - {K_{\rm{DA}}} \cdot {V_{\rm{E}}} + \sqrt {{{{({E_{\rm{tot}}} - {\hbox{F}}{{\hbox{c}}_{\rm{tot}}} - {K_{\rm{DA}}} \cdot {V_{\rm{E}}})}^2} + 4 \cdot {E_{\rm{tot}}} \cdot {K_{\rm{DA}}} \cdot {V_{\rm{E}}}}} } \right) $$

(33)

and

$$ \frac{{{\text{d}}{E_{\rm{tot}}}}}{{{\hbox{d}}t}} = {k_{\rm{up}}} \cdot ({\hbox{Ab}} + {\hbox{AbH}}) - {k_{\deg }} \cdot E - {k_{\rm{R}}} \cdot {\hbox{Fc}}E $$

(34)

where

$$ {E_{\rm{tot}}} = E + {\hbox{Fc}}E $$

(35)

Notice that Eq. 33 is identical to Eq. 14, since Eq. 34 is the sum of Eqs. 9 and 10. In order to derive Eqs. 15 and 16, one needs to re-write Eq. 2 to the following form

$$ {\hbox{Fc}} = {\hbox{F}}{{\hbox{c}}_{\rm{tot}}} - {\hbox{Fc}}E $$

(36)

and substitute it to Eq. 31a, b. These will constitute a system of two linear equations with unknowns FcAb_E and FcAbH_E that can be easily solved resulting in Eqs. 15 and 16. Finally, Eqs. 11a, b and 15 imply

$$ {\hbox{A}}{{\hbox{b}}_{\rm{Etot}}} = {\hbox{A}}{{\hbox{b}}_{\rm{E}}} + \frac{{{\text{F}}{{\text{c}}_{\rm{tot}}} \times {\text{A}}{{\text{b}}_{\rm{E}}}}}{{{K_{\rm{DA}}} \cdot {V_{\rm{E}}} + {\hbox{A}}{{\hbox{b}}_{\rm{E}}} + {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}}}} $$

(37)

Multiplication of Eq. 37 by the denominator yields

$$ {\hbox{A}}{{\hbox{b}}_{\rm{Etot}}} \cdot ({K_{\rm{DA}}} \cdot {V_{\rm{E}}} + {\hbox{A}}{{\hbox{b}}_{\rm{E}}} + {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}}) = {\hbox{A}}{{\hbox{b}}_{\rm{E}}} \cdot ({K_{\rm{DA}}} \cdot {V_{\rm{E}}} + {\hbox{F}}{{\hbox{c}}_{\rm{tot}}} + {\hbox{A}}{{\hbox{b}}_{\rm{E}}} + {\hbox{Ab}}{{\hbox{H}}_{\rm{E}}}) $$

(38)

and Eq. 15 follows. Equation 16 can be derived in a similar way.

Equations 32 and 33 imply that

$$ {\hbox{Fc}}E = \frac{{{\text{F}}{{\text{c}}_{\rm{tot}}} \cdot E}}{{{K_{\rm{DA}}} \cdot {V_{\rm{E}}} + E}} $$

(39)