Quasi-Equilibrium Pharmacokinetic Model for Drugs Exhibiting Target-Mediated Drug Disposition

Purpose

The aim of this study is to derive and evaluate an equilibrium model of a previously developed general pharmacokinetic model for drugs exhibiting target-mediated drug disposition (TMDD).

Methods

A quasi-equilibrium solution to the system of ordinary differential equations that describe the kinetics of TMDD was obtained. Computer simulations of the equilibrium model were carried out to generate plasma concentration-time profiles resulting from a large range of intravenous bolus doses. Additionally, the final model was fitted to previously published pharmacokinetic profiles of leukemia inhibitory factor (LIF), a cytokine that seems to exhibit TMDD, following intravenous administration of 12.5, 25, 100, 250, 500, or 750 μg/kg in sheep.

Results

Simulations show that pharmacokinetic profiles display steeper distribution phases for lower doses and similar terminal disposition phases, but with slight underestimation at early time points as theoretically expected. The final model well-described LIF pharmacokinetics, and the final parameters, which were estimated with relatively good precision, were in good agreement with literature values.

Conclusions

An equilibrium model of TMDD is developed that recapitulates the essential features of the full general model and eliminates the need for estimating drug-binding microconstants that are often difficult or impossible to identify from typical in vivo pharmacokinetic data.

Appendices

Appendix A

Quasi-equilibrium equations for the TMDD model

Equations(7a) and (7b) define the variables C_tot and R_tot as the concentration sums of free and bound complexes. To derive the differential equations describing them, similar operations should be performed on the equations for the TMDD model. Adding Eq. (1) to Eq. (4) will result in:

$${{\text{d}}C_{{{\text{tot}}}} } \mathord{\left/ {\vphantom {{{\text{d}}C_{{{\text{tot}}}} } {{\text{d}}t}}} \right. \kern-\nulldelimiterspace} {{\text{d}}t} = {\text{In}}{\left( t \right)} - k_{{\operatorname{int} }} {\text{RC}} - {\left( {k_{{{\text{el}}}} + k_{{{\text{pt}}}} } \right)}C + {k_{{{\text{tp}}}} \cdot A_{{\text{T}}} } \mathord{\left/ {\vphantom {{k_{{{\text{tp}}}} \cdot A_{{\text{T}}} } {V_{{\text{c}}} }}} \right. \kern-\nulldelimiterspace} {V_{{\text{c}}} }$$

(A1)

The RC variable can be calculated from Eq. (7a), such that:

$${\text{RC}} = C_{{{\text{tot}}}} - C$$

(A2)

After substituting Eq. (A2) for RC in Eq. (A1), it can be rearranged to Eq. (8). Similarly, to obtain the differential equation for R_tot, one needs to add Eqs. (3) and (4) and eliminate the variables RC and R using Eq. (A2):

$$R{\text{ = }}R_{{{\text{tot}}}} - {\text{RC = }}R_{{{\text{tot}}}} - {\text{(}}C_{{{\text{tot}}}} - {\text{C)}}$$

(A3)

This procedure results in differential equations for C_tot and R_tot that contain the variable C which in turn can be calculated from the quasi-equilibrium Eq. (6) after substituting Eqs. (A2) and (A3) for RC and R:

$$\frac{{C{\left[ {R_{{{\text{tot}}}} - {\left( {C_{{{\text{tot}}}} - C} \right)}} \right]}}} {{C_{{{\text{tot}}}} - C}} = K_{{\text{D}}} $$

(A4)

Equation (A4) can be multiplied side by side by the denominator of the left-hand side and rearranged to the quadratic equation for C:

$$C^{{\text{2}}} - {\text{(}}C_{{{\text{tot}}}} - R_{{{\text{tot}}}} - K_{{\text{D}}} {\text{)}}C - K_{{\text{D}}} C_{{{\text{tot}}}} {\text{ = 0}}$$

(A5)

Because K_D > 0 and C_tot ≥ 0, this equation has exactly one nonnegative solution:

$$C = 1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2{\left[ {{\left( {C_{{{\text{tot}}}} - R_{{{\text{tot}}}} - K_{{\text{D}}} } \right)} + {\sqrt {{\left( {C_{{{\text{tot}}}} - R_{{{\text{tot}}}} - K_{{\text{D}}} } \right)}^{2} + 4K_{{\text{D}}} C_{{{\text{tot}}}} } }} \right]}$$

(A6)

which is Eq. (11). Equation (A6) is algebraically equivalent to Eq. (A4) under assumption C ≥ 0. The differential equation for A_T in the equilibrium model is identical to Eq. (2).

Appendix B

Steady-state equations for the quasi-equilibrium TMDD model

To derive the steady-state equations for the quasi-equilibrium model, we will assume that the input rate is constant [In(t) ≡ In] and the nonspecific tissue-binding site (A_T) is present (k_tp, k_pt > 0). The derivations will hold for receptor turnover with the synthesis rate k_syn > 0 and degradation or internalization (k_int + k_deg > 0). Also, we consider the model with at least one clearance mechanism (k_el + k_int > 0) and the binding process always present (K_D > 0). At steady state, there is no change in the model variables, and therefore, the time derivatives in Eqs. (8)–(10) can be set to 0 resulting in a system of four algebraic equations with four unknowns (C_ss, A_Tss, C_totss, and R_totss):

$$0 = {\text{In}} - k_{{\operatorname{int} }} C_{{{\text{totss}}}} - {\left( {k_{{{\text{el}}}} + k_{{{\text{pt}}}} - k_{{\operatorname{int} }} } \right)}C_{{{\text{ss}}}} + {k_{{{\text{tp}}}} \cdot A_{{{\text{Tss}}}} } \mathord{\left/ {\vphantom {{k_{{{\text{tp}}}} \cdot A_{{{\text{Tss}}}} } {V_{{\text{c}}} }}} \right. \kern-\nulldelimiterspace} {V_{{\text{c}}} }$$

(B1)

$$0 = k_{{{\text{pt}}}} C_{{{\text{ss}}}} V_{{\text{c}}} - k_{{{\text{tp}}}} A_{{{\text{Tss}}}} $$

(B2)

$$0 = k_{{{\text{syn}}}} - {\left( {k_{{\operatorname{int} }} - k_{{\deg }} } \right)}{\left( {C_{{{\text{totss}}}} - C_{{{\text{ss}}}} } \right)} - k_{{\deg }} R_{{{\text{totss}}}} $$

(B3)

$$C_{{{\text{ss}}}} = 1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2{\left[ {{\left( {C_{{{\text{totss}}}} - R_{{{\text{totss}}}} - K_{{\text{D}}} } \right)} + {\sqrt {{\left( {C_{{{\text{totss}}}} - R_{{{\text{totss}}}} - K_{{\text{D}}} } \right)}^{2} + 4K_{{\text{D}}} C_{{{\text{totss}}}} } }} \right]}$$

(B4)

Equation (B4) is equivalent to the equilibrium Eq. (6) that can be rewritten as follows:

$$C_{{{\text{totss}}}} = C_{{{\text{ss}}}} + {R_{{{\text{totss}}}} \cdot C_{{{\text{ss}}}} } \mathord{\left/ {\vphantom {{R_{{{\text{totss}}}} \cdot C_{{{\text{ss}}}} } {{\left( {K_{{\text{D}}} + C_{{{\text{ss}}}} } \right)}}}} \right. \kern-\nulldelimiterspace} {{\left( {K_{{\text{D}}} + C_{{{\text{ss}}}} } \right)}}$$

(B5)

where the relationships defined by Eqs. (7a) and (7b) have been exploited. Equation (B2) allows one to express A_Tss as a function of C_ss:

$$A_{{{\text{Tss}}}} = k_{{{\text{pt}}}} V_{{{\text{ss}}}} \cdot {C_{{{\text{ss}}}} } \mathord{\left/ {\vphantom {{C_{{{\text{ss}}}} } {k_{{{\text{tp}}}} }}} \right. \kern-\nulldelimiterspace} {k_{{{\text{tp}}}} }$$

(B6)

Dividing Eq. (B2) by V_c and adding it to Eq. (B1), we cancel off the A_T term and obtain:

$$k_{{\operatorname{int} }} {\left( {C_{{{\text{totss}}}} - C_{{{\text{ss}}}} } \right)} + k_{{{\text{el}}}} C_{{{\text{ss}}}} = {\text{In}}$$

(B7)

If In = 0, and because C_totss ≥ C_ss and k_syn > 0, then Eqs. (B3), (B5), and (B7) imply that k_deg > 0, and:

$$C_{{{\text{ss}}}} = C_{{{\text{totss}}}} = 0\,{\text{and}}\,R_{{{\text{totss}}}} = {k_{{{\text{syn}}}} } \mathord{\left/ {\vphantom {{k_{{{\text{syn}}}} } {k_{{\deg }} }}} \right. \kern-\nulldelimiterspace} {k_{{\deg }} }$$

(B8)

Later, we will assume that In > 0. Let us first consider the case where receptors are eliminated through the turnover process (k_deg > 0). Equation (B5) can be used to calculate C_totss − C_ss and substitute for this term in Eq. (B3) yielding:

$$R_{{{\text{totss}}}} = \frac{{k_{{{\text{syn}}}} {\left( {K_{{\text{D}}} + C_{{{\text{ss}}}} } \right)}}} {{k_{{\operatorname{int} }} C_{{{\text{ss}}}} + k_{{\deg }} K_{{\text{D}}} }}$$

(B9)

The only unknown now is C_ss. If k_int = 0, then C_ss can be easily calculated from Eq. (B7):

$$C_{{{\text{ss}}}} = {{\text{In}}} \mathord{\left/ {\vphantom {{{\text{In}}} {k_{{{\text{el}}}} }}} \right. \kern-\nulldelimiterspace} {k_{{{\text{el}}}} }$$

(B10)

If k_int > 0, then calculation of the steady states becomes more troublesome. C_ss can be determined from Eq. (B3) if Eq. (B5) is used to replace R_totss and Eq. (B7) to substitute for C_totss − C_ss resulting in the following equation for C_ss:

$$0 = k_{{{\text{syn}}}} - {\left( {k_{{\operatorname{int} }} - k_{{\deg }} } \right)}{\left( {{\text{In}} - k_{{{\text{el}}}} C_{{{\text{ss}}}} } \right)}/k_{{\operatorname{int} }} - k_{{\deg }} {\left( {{\text{In}} - k_{{{\text{el}}}} C_{{{\text{ss}}}} } \right)}{\left( {K_{{\text{D}}} + C_{{{\text{ss}}}} } \right)}/{\left( {k_{{\operatorname{int} }} C_{{{\text{ss}}}} } \right)}$$

(B11)

After multiplying both sides of Eq. (B11) by k_intC_ss, it can be rearranged to the equation for C_ss:

$$k_{{{\text{el}}}} k_{{\operatorname{int} }} C^{2}_{{{\text{ss}}}} + {\left[ {k_{{\operatorname{int} }} {\left( {k_{{{\text{syn}}}} - {\text{In}}} \right)} + k_{{\deg }} k_{{{\text{el}}}} K_{{\text{D}}} } \right]}C_{{{\text{ss}}}} - k_{{\deg }} K_{{\text{D}}} {\text{In}} = 0$$

(B12)

If k_el = 0, then the positive solution exists only if k_syn > In:

$$C_{{{\text{ss}}}} = \frac{{k_{{\deg }} K_{{\text{D}}} {\text{In}}}} {{{\left( {k_{{{\text{syn}}}} - {\text{In}}} \right)}k_{{\operatorname{int} }} }}$$

(B13)

If k_el > 0, then the positive solution is:

$$C_{{{\text{ss}}}} = \frac{{{\text{In}}k_{{\operatorname{int} }} - k_{{\deg }} k_{{{\text{el}}}} K_{{\text{D}}} - k_{{{\text{syn}}}} k_{{\operatorname{int} }} + {\sqrt {{\left( {{\text{In}}k_{{\operatorname{int} }} - k_{{\deg }} k_{{{\text{el}}}} K_{{\text{D}}} - k_{{{\text{syn}}}} k_{{\operatorname{int} }} } \right)}^{2} + 4k_{{{\text{el}}}} k_{{\operatorname{int} }} k_{{\deg }} {\text{In}}K_{{\text{D}}} } }}} {{2k_{{{\text{el}}}} k_{{\operatorname{int} }} }}$$

(B14)

Let us now consider the case k_deg = 0. Then the only clearance mechanism for bound receptors is the internalization process. If k_int > 0, then Eq. (B3) implies:

$$C_{{{\text{totss}}}} - C_{{{\text{ss}}}} = {k_{{{\text{syn}}}} } \mathord{\left/ {\vphantom {{k_{{{\text{syn}}}} } {k_{{\operatorname{int} }} }}} \right. \kern-\nulldelimiterspace} {k_{{\operatorname{int} }} }$$

(B15)

and from Eq. (B7), it follows that for k_el > 0 and In > k_syn:

$$C_{{{\text{ss}}}} = {{\left( {{\text{In}} - k_{{{\text{syn}}}} } \right)}} \mathord{\left/ {\vphantom {{{\left( {{\text{In}} - k_{{{\text{syn}}}} } \right)}} {k_{{{\text{el}}}} }}} \right. \kern-\nulldelimiterspace} {k_{{{\text{el}}}} }$$

(B16)

R_totss can be determined from Eq. (B8). If k_el = 0, then after adding Eqs. (B3) and (B7), one can conclude that k_syn = In, thus implying:

$$C_{{{\text{tot}}}} + {A_{{\text{T}}} } \mathord{\left/ {\vphantom {{A_{{\text{T}}} } {V_{{\text{c}}} }}} \right. \kern-\nulldelimiterspace} {V_{{\text{c}}} } - R_{{{\text{tot}}}} = {\text{constant}}$$

(B17)

Hence,

$$R_{{{\text{totss}}}} {\text{ = }}C_{{{\text{totss}}}} {\text{ + }}k_{{{\text{pt}}}} \cdot {C_{{{\text{ss}}}} } \mathord{\left/ {\vphantom {{C_{{{\text{ss}}}} } {k_{{{\text{pt}}}} }}} \right. \kern-\nulldelimiterspace} {k_{{{\text{pt}}}} }{\text{ + constant}}$$

(B18)

Inserting R_totss from Eq. (B18) and C_totss described by Eq.(B15) into Eq. (B5) yields an equation for C_ss:

$${k_{{{\text{syn}}}} } \mathord{\left/ {\vphantom {{k_{{{\text{syn}}}} } {k_{{\operatorname{int} }} }}} \right. \kern-\nulldelimiterspace} {k_{{\operatorname{int} }} }{\text{ = (}}C_{{{\text{ss}}}} {\text{ + }}k_{{{\text{syn}}}} {\text{/}}k_{{\operatorname{int} }} {\text{ + }}k_{{{\text{pt}}}} \cdot {C_{{{\text{ss}}}} } \mathord{\left/ {\vphantom {{C_{{{\text{ss}}}} } {k_{{{\text{pt}}}} }}} \right. \kern-\nulldelimiterspace} {k_{{{\text{pt}}}} }{\text{ + constant)}} \cdot {C_{{{\text{ss}}}} } \mathord{\left/ {\vphantom {{C_{{{\text{ss}}}} } {{\left( {K_{{\text{D}}} {\text{ + }}C_{{{\text{ss}}}} } \right)}}}} \right. \kern-\nulldelimiterspace} {{\left( {K_{{\text{D}}} {\text{ + }}C_{{{\text{ss}}}} } \right)}}$$

(B19)

which can be transformed to the following quadratic equation:

$${\left( {{\text{1 + }}{k_{{{\text{pt}}}} } \mathord{\left/ {\vphantom {{k_{{{\text{pt}}}} } {k_{{{\text{tp}}}} }}} \right. \kern-\nulldelimiterspace} {k_{{{\text{tp}}}} }} \right)}C^{2}_{{{\text{ss}}}} {\text{ + constant}} \cdot C_{{{\text{ss}}}} {\text{ - }}K_{{\text{D}}} \cdot {k_{{{\text{syn}}}} } \mathord{\left/ {\vphantom {{k_{{{\text{syn}}}} } {k_{{\operatorname{int} }} }}} \right. \kern-\nulldelimiterspace} {k_{{\operatorname{int} }} }{\text{ = 0}}$$

(B20)

Equation (B21) has only one nonnegative solution:

$$C_{{{\text{ss}}}} = \frac{1} {{2{\left( {1 + \frac{{k_{{{\text{pt}}}} }} {{k_{{{\text{tp}}}} }}} \right)}}}{\left( {{\text{constant}} + {\sqrt {{\text{constant}}^{2} + 4\frac{{k_{{{\text{syn}}}} K_{{\text{D}}} }} {{k_{{\operatorname{int} }} }}{\left( {1 + \frac{{k_{{{\text{pt}}}} }} {{k_{{{\text{tp}}}} }}} \right)}} }} \right)}$$

(B21)

This completes derivations of the steady-state solutions forthe quasi-equilibrium model, which are summarized in Table II.

Table II Steady-State Solutions for the Quasi-EquilibriumTMDD Model^a

Appendix C

Derivation of Wagner nonlinear tissue binding [Eq. (16)]

For derivation of Eq. (16), the best form of the equilibrium Eq. (6) is:

$$C_{{{\text{tot}}}} = C + R_{{{\text{tot}}}} C/(K_{{\text{D}}} + C)$$

(C1)

where R_tot is constant. One can differentiate both sides of Eq.(C1) and obtain:

$$\frac{{{\text{d}}C_{{{\text{tot}}}} }} {{{\text{d}}t}} = \frac{{{\text{d}}C}} {{{\text{d}}t}} + \frac{{R_{{{\text{tot}}}} K_{{\text{D}}} }} {{(K_{{\text{D}}} + C)^{2} }}\frac{{{\text{d}}C}} {{{\text{d}}t}}$$

(C2)

Equation (8), with k_int = 0, can be substituted in the left-hand side of Eq. (C2) resulting in:

$${\text{In}}{\left( t \right)} - {\left( {k_{{{\text{el}}}} + k_{{{\text{pt}}}} } \right)}C + {k_{{{\text{tp}}}} \cdot A_{{\text{T}}} } \mathord{\left/ {\vphantom {{k_{{{\text{tp}}}} \cdot A_{{\text{T}}} } {V_{{\text{c}}} }}} \right. \kern-\nulldelimiterspace} {V_{{\text{c}}} } = {\left( {1 + \frac{{R_{{{\text{tot}}}} K_{{\text{D}}} }} {{(K_{{\text{D}}} + C)^{2} }}} \right)}\frac{{{\text{d}}C}} {{{\text{d}}t}}$$

(C3)

Finally, one can solve Eq. (C3) for dC/dt and arrive at Eq. (16).