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Methods of solving rapid binding target-mediated drug disposition model for two drugs competing for the same receptor

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Abstract

The target-mediated drug disposition (TMDD) model has been adopted to describe pharmacokinetics for two drugs competing for the same receptor. A rapid binding assumption introduces total receptor and total drug concentrations while free drug concentrations C A and C B are calculated from the equilibrium (Gaddum) equations. The Gaddum equations are polynomials in C A and C B of second degree that have explicit solutions involving complex numbers. The aim of this study was to develop numerical methods to solve the rapid binding TMDD model for two drugs competing for the same receptor that can be implemented in pharmacokinetic software. Algebra, calculus, and computer simulations were used to develop algorithms and investigate properties of solutions to the TMDD model with two drugs competitively binding to the same receptor. A general rapid binding approximation of the TMDD model for two drugs competing for the same receptor has been proposed. The explicit solutions to the equilibrium equations employ complex numbers, which cannot be easily solved by pharmacokinetic software. Numerical bisection algorithm and differential representation were developed to solve the system instead of obtaining an explicit solution. The numerical solutions were validated by MATLAB 7.2 solver for polynomial roots. The applicability of these algorithms was demonstrated by simulating concentration–time profiles resulting from exogenous and endogenous IgG competing for the neonatal Fc receptor (FcRn), and darbepoetin competing with endogenous erythropoietin for the erythropoietin receptor. These models were implemented in ADAPT 5 and Phoenix WinNonlin 6.0, respectively.

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Acknowledgments

This work was supported by the Laboratory for Protein Therapeutics at the University at Buffalo, and Grant GM 57980 from the National Institute of Health.

Author information

Correspondence to Wojciech Krzyzanski.

Electronic supplementary material

Appendices

Appendix 1

Existence of the unique solution of Eqs. 27 and 28

Let x and y denote the RC A and RC B divided by R tot :

$$ x = \frac{{RC_{A} }}{{R_{tot} }}\; {\text{and}}\;y = \frac{{RC_{B} }}{{R_{tot} }} $$
(103)

Then Eqs. 27 and 28 can be expressed in the following form:

$$ (1 - x - y)(a_{A} - x) = k_{A} x $$
(104)
$$ (1 - x - y)(a_{B} - y) = k_{B} y $$
(105)

Note that because of definitions in Eqs. 103 and 42, x and y satisfy the following relationships:

$$ 0 < x < a_{A} ,0 < y < a_{B} , {\text{ and }} x + y < 1 $$
(106)

Existence Theorem

Let k A , k B , a A , a B  > 0. If kA ≠ kB, then there exist exactly three solutions to Eqs. 104 and 105: (x 1, y 1), (x 2, y 2), (x 3, y 3) such that

  1. (a)

    If k B  > k A , then:

$$ 0 < x_{1} < a_{A} < x_{2} < x_{3} ,y_{3} < 0 < y_{1} < a_{B} < y_{2} ,\;{\text{and}}\;x_{1} + y_{1} < 1 $$
(107)
  1. (b)

    If k B  < k A , then:

$$ x_{3} < 0 < x_{1} < a_{A} < x_{2} {\text{ and }} 0 < y_{1} < a_{B} < y_{2} < y_{3} ,\;{\text{and}}\;x_{1} + y_{1} < 1 $$
(108)

If k A  = k B , then there exist exactly two solutions to Eqs. 104 and 105: (x 1, y 1), (x 2, y 2) such that:

$$ 0 < x_{1} < a_{A} < x_{2} {\text{ and }} 0 < y_{1} < a_{B} < y_{2} ,\;{\text{and}}\;x_{1} + y_{1} < 1 $$
(109)

Proof of Existence Theorem is based on the observation that the solutions of Eqs. 104 and 105 can be geometrically interpreted as intersections of the following curves:

$$ y = 1 - x + \frac{{k_{A} x}}{{x - a_{A} }} $$
(110)
$$ x = 1 - y + \frac{{k_{B} y}}{{y - a_{B} }} $$
(111)

The curves of Eqs. 110 and 111 are transformed Eqs. 104 and 105, respectively. The asymptotes of Eq. 110 are:

$$ x = a_{A} \; {\text{and}}\;y = 1 + k_{A} - x $$
(112)

whereas the asymptotes for Eq. 111 are:

$$ y = a_{B} \; {\text{and}}\;x = 1 + k_{B} - y $$
(113)

If k B  > k A , the examination of the monotonicity of Eqs. 110 and 111 and the horizontal and vertical asymptotes imply that there are two intersection points (x 1, y 1) and (x 2, y 2) such that:

$$ 0 < x_{1} < a_{A} < x_{2} \;{\text{and}}\;0 < y_{1} < a_{B} < y_{2} $$
(114)

as shown in Fig. 8. k B  > k A implies that the diagonal asymptote for Eq. 110 is below the diagonal asymptote for Eq. 111. Consequently, there is a third intersection point (x 3, y 3) such that:

$$ x_{2} < x_{3} \;{\text{and}}\;y_{3} < 0 $$
(115)
Fig. 8
figure8

Graphical representation of solutions of the equilibrium equations when K DA  ≠ K DB (upper panel) and K DA  = K DB (lower panel). The equilibrium equations Eqs. 27 and 28 are equivalent to a system of two hyperbolic equations represented by the solid lines. Their intersection coordinates (x 1, y 1), (x 2, y 2), (x 3, y 3) (upper panel), and (x 1, y 2), (x 2, y 2) (lower panel) are all possible solutions of the system. The dashed lines represent the asymptotes for the hyperbolas. In case K DA  = K DB (equivalent to k A  = k B ) the diagonal asymptotes collapse to a single one reducing the number of solutions to two. The vertical and horizontal asymptotes intersect the axes at aA and aB, respectively. Only the solution (x 1, y 1) is inside the rectangle of vertices defined by 0, a A , and a B

If k B  < k A , the positions of the diagonal asymptotes reverses and the third intersection (x 3, y 3) satisfies the following:

$$ x_{3} < 0\;{\text{and}}\;y_{2} < y_{3} $$
(116)

If k B  = k A , then the existence of (x 1, y 1) and (x 2, y 2) satisfying Eq. 109 is a consequence of the same argument. Since the diagonal asymptotes collapse into one (see Fig. 8), there is no third intersection point. A formal proof of Existence Theorem not referring to a geometric interpretation of Eqs. 104 and 105 presented in Fig. 8 follows below.

Define the functions:

$$ f(x) = 1 - x - \frac{{k_{A} x}}{{a_{A} - x}},x \ne a_{A} \;{\text{and}}\;g(y) = 1 - y - \frac{{k_{B} y}}{{a_{B} - y}},\;y \ne a_{B} $$
(117)

Since the derivatives are negative:

$$ \frac{df}{dx}(x) = - 1 - \frac{{k_{A} a_{A} }}{{(a_{A} - x)^{2} }} < 0\;{\text{and}}\;\frac{dg}{dy}(y) = - 1 - \frac{{k_{B} a_{B} }}{{(a_{B} - B)^{2} }} < 0 $$
(118)

both functions are strictly decreasing. Because a discontinuity at x = a A , there are two solutions to the equation:

$$ f\left( x \right) = 0 $$
(119)
$$ x_{a} = \frac{1}{2}\left( {1 + a_{A} + k_{A} - \sqrt {(1 + a_{A} + k_{A} )^{2} - 4a_{A} } } \right) $$
(120)

and

$$ x_{b} = \frac{1}{2}\left( {1 + a_{A} + k_{A} + \sqrt {(1 + a_{A} + k_{A} )^{2} - 4a_{A} } } \right) $$
(121)

One can verify by direct calculation that:

$$ 0 < x_{a} < \min \{ 1,a_{A} \} \le \max \{ 1,a_{A} \} < x_{b} $$
(122)

Similarly, there are two solutions to the equation:

$$ g\left( y \right) = 0 $$
(123)
$$ y_{a} = \frac{1}{2}\left( {1 + a_{B} + k_{B} - \sqrt {(1 + a_{B} + k_{B} )^{2} - 4a_{B} } } \right) $$
(124)

and

$$ y_{b} = \frac{1}{2}\left( {1 + a_{B} + k_{B} + \sqrt {(1 + a_{B} + k_{B} )^{2} - 4a_{B} } } \right) $$
(125)

Additionally

$$ 0 < y_{a} < \min \{ 1,a_{B} \} \le \max \{ 1,a_{B} \} < y_{b} $$
(126)

Because function g(y) is strictly decreasing and continuous in the intervals (−∞, a B ) and (a B , ∞), there exist inverse functions h 1(x) and h 2(x), respectively, such that:

$$ g\left( {h_{1} \left( x \right)} \right) = x\;{\text{and}}\;g\left( {h_{2} \left( x \right)} \right) = x, - \infty < {\text{ x }} < \infty , $$
(127)

To show existence of (x 1, y 1) consider a new function F 1(x) = f(x) − h 1(x) defined on the interval 0  x  x a . Then Eq. 126 implies that:

$$ F_{ 1} \left( 0 \right) = { 1} - y_{a} > 0 $$
(128)

As an inverse to a decreasing function h 1(x) is also decreasing and Eq. 122 implies h 1(x a ) > h 1(1) = 0, and consequently

$$ F_{ 1} \left( {x_{a} } \right) = 0 -h_{ 1} \left( {x_{a} } \right) < 0 $$
(129)

Because the function F 1(x) is continuous, and it changes the sign at the ends of the interval [0, x a ], the intermediate value theorem guarantees there exists a 0 < x 1 < x a such that:

$$ F_{ 1} \left( {x_{1} } \right) = 0 $$
(130)

Let y 1 = h 1(x 1). Then Eqs. 130, 127, and 117 imply that (x 1, y 1) is a solution to Eqs. 104 and 105. Since x a  < a A , then x 1 < a A . This and Eq. 104 also yields that x1 + y1 < 1.

To show existence of (x 2, y 2) consider a new function F 2(x) = f(x) − h 2(x) defined on the interval a A  < x  x b . By definition h 2(x b ) > a B  > 0, and consequently

$$ F_{2} \left( {x_{b} } \right) = 0- h_{2} \left( {x_{b} } \right) < 0 $$
(131)

From Eq. 117 it follows that:

$$ \mathop {\lim }\limits_{{x \to a_{A} + }} F_{2} (x) = + \infty $$
(132)

The intermediate value theorem implies there exists a a A  < x 2 < x b such that:

$$ F_{ 2} \left( {x_{ 2} } \right) = 0 $$
(133)

Let y 2 = h 2(x 2). By definition of h 2(x), y 2 > a B . Also, Eqs. 133, 127, and 117 imply that (x 2, y 2) is a solution to Eq. 104.

To show existence of (x 3, y 3) for the case k A  < k B consider a new function F 3(x) = F(x) − h 1(x) defined on the interval x b   x < ∞. The function h 1(x) is decreasing and Eq. 112 implies h 1(x b ) < h 1(1) = 0. Hence

$$ F_{ 3} \left( {x_{b} } \right) = 0 - h_{ 1} \left( {x_{b} } \right) > 0 $$
(134)

Eq. 117 implies that:

$$ f(x) - 1 + x - k_{A} \to 0\;{\text{as}}\;x \to \infty ,\; {\text{and}} \; g(y) - 1 + y - k_{B} \to 0\;{\text{as}}\;y \to - \infty $$
(135)

As a decreasing function h 1(x) → −∞ as x  ∞. Consequently, Eq. 135 implies that:

$$ g(h_{1} (x)) - 1 + h_{1} (x) - k_{B} \to 0 \;{\text{as}}\;x \to \infty $$
(136)

Hence and from Eq. 127

$$ h_{1} (x) - 1 + x - k_{B} \to 0\; {\text{as}}\;x \to \infty $$
(137)

Thus

$$ F_{3} (x) \to k_{A} - k_{B} < 0\;{\text{as}}\;x \to \infty $$
(138)

The function F 3(x) changes its sign at the ends of the interval x b   x < ∞. The intermediate value theorem implies that there exists x b  < x 3 < ∞ such that:

$$ F_{ 3} \left( {x_{ 3} } \right) = 0 $$
(139)

Since x 2 < x b , then x 2 < x 3. Let y 3 = h 1(x 3), then y 3 < h 1(1) = 0. Also Eqs. 137, 127, and 117 imply that (x 3, y 3) is a solution to Eq. 104.

A similar argument holds to show existence of (x 3, y 3) for the case k A  > k B . Consider a function F 4(x) = f(x) − h 2(x) defined on the interval −∞ < x  0. Eqs. 117, 123, and 126 imply that:

$$ F_{ 4} \left( 0 \right) = { 1 }-y_{b} < 0 $$
(140)

The same derivations as above lead to:

$$ F_{4} (x) \to k_{A} - k_{B} > 0\; {\text{as x}} \to \infty $$
(141)

The intermediate value theorem implies that there exists x 3 < 0 such that:

$$ F_{ 4} \left( {x_{ 3} } \right) = 0 $$
(142)

Let y 3 = h 2(x 3). Since h 2(x) is decreasing y 3 > h 2(0) = y b  > y 2, and Eq. 117 implies that y 3 > y 2. Also Eqs. 137, 127, and 117 imply that (x 3, y 3) is a solution to Eq. 104.

To show uniqueness of (x 1, y 1), (x 2, y 2), (x 3, y 3) for k A  ≠ k B one can notice that x 1, x 2, and x 3 pairwise distinct. There are also roots of a polynomial obtained from Eqs. 104 and 105 as follows. One can calculate from Eq. 104 the term:

$$ y(a_{A} - x) = (1 - x)(a_{A} - x) - k_{A} x $$
(143)

To enforce this term in Eq. 105 multiply both sides by (a A -x)2:

$$ ((1 - x)(a_{A} - x) - y(a_{A} - x))(a_{B} (a_{A} - x) - y(a_{A} - x)) = k_{B} y(a_{A} - x)^{2} $$
(144)

Eq. 143 implies that:

$$ k_{A} x = (1 - x)(a_{A} - x) - y(a_{A} - x) $$
(145)

Substituting Eqs. 143 and 145 into A38 yields:

$$ k_{A} x(a_{B} (a_{A} - x) - (1 - x)(a_{A} - x) + k_{A} x) = k_{B} (a_{A} - x)((1 - x)(a_{A} - x) - k_{A} x) $$
(146)

which can further transformed to

$$ k_{A} a_{B} x(a_{A} - x) - \left( {k_{A} x + k_{B} (a_{A} - x)} \right)\left( {(1 - x)(a_{A} - x) + k_{A} x} \right) = 0 $$
(147)

A leading term of the polynomial in Eq. 147 is (k B  − k A )x 3. Therefore Eq. 147 is a cubic equation with three distinct roots x 1, x 2, and x 3. If (x*, y*) is a solution to Eqs. 104 and 105, then x* must be a solution to Eq. 147 and hence x* = x i for some i = 1, 2, 3. Eq. 104 implies that x* ≠ a A and Eq. 143 yields:

$$ y^{\ast} = \frac{{(1 - x^{\ast} )(a_{A} - x^{\ast} ) - k_{A} x^{\ast} }}{{a_{A} - x^{\ast} }} $$
(148)

Since (x i , yi) is a solution to Eq. 104 as well y i can be expressed by the right hand side of Eq. 148 with x i substituted for x* and hence y* = y i .

To show uniqueness of (x 1, y 1) and (x 2, y 2) for k A  = k B one can notice that then the polynomial equation Eq. 147 reduces to a quadratic equation since the highest order term (k B  − k A )x 3 vanishes. This quadratic equation has two distinct roots x 1 and x 2. If (x*, y*) is a solution to Eqs. 104 and 105 then x* must be also a solution to the quadratic equation Eq. 147, and consequently x* = x i for some i = 1, 2. Then y* = y i by the same argument as above. This completes proof of Existence Theorem.

Derivation of Eqs. 40, 43, and 57

For the case k A  ≠ k B , to solve Eqs. 104 and 105 one can add them side by side:

$$ (1 - x - y)(a_{A} + a_{B} - x - y) = k_{A} x + k_{B} y $$
(149)

Multiplying Eq. 104 by kB and Eq. 105 by kA followed and adding equation side by side yields:

$$ (1 - x - y)(k_{B} a_{A} + k_{A} a_{B} - k_{B} x - k_{A} y) = k_{A} k_{B} (x + y) $$
(150)

Since

$$ k_{B} x + k_{A} y = (k_{A} + k_{B} )(x + y) - (k_{A} x + k_{B} y) $$
(151)

Eq. 151 can be substituted in Eq. 150:

$$ (1 - x - y)(k_{B} a_{A} + k_{A} a_{B} + k_{A} x + k_{B} y - (k_{A} + k_{B} )(x + y)) = k_{A} k_{B} (x + y) $$
(152)

With introducing a new variable:

$$ z = x + y $$
(153)

Eq. 152 becomes:

$$ (1 - z)(k_{B} a_{A} + k_{A} a_{B} + k_{A} x + k_{B} y - (k_{A} + k_{B} )z) = k_{A} k_{B} z $$
(154)

The right hand side of Eq. 149 coincides with a term in Eq. 154 that can be replaced by the left hand side of Eq. 149, resulting in:

$$ (1 - z)(k_{B} a_{A} + k_{A} a_{B} + (1 - z)(a_{A} + a_{B} - z) - (k_{A} + k_{B} )z) = k_{A} k_{B} z $$
(155)

The only unknown in Eq. 50 is z and ordering the terms by the power of z produces a cubic equation Eq. 43.

If k A  = k B , then Eq. 149 assumes the following form:

$$ (1 - z)(a_{A} + a_{B} - z) = k_{A} z $$
(156)

Rearranging terms in Eq. 156 and ordering them by the power of z yields Eq. 57. In order to express C A and C B in terms of z, one should notice that according to Eq. 153:

$$ C_{Atot} - C_{A} + C_{Btot} - C_{B} = R_{tot} z $$
(157)

Upon substitution of Eq. 157 to the Eqs. 27 and 28 they reduce to:

$$ R_{tot} (1 - z)C_{A} = K_{DA} (C_{Atot} - C_{A} ) $$
(158)
$$ R_{tot} (1 - z)C_{B} = K_{DB} (C_{Btot} - C_{B} ) $$
(159)

Solving Eq. 158 for C A and Eq. 159 for C B results in Eq. 40.

Appendix 2

Derivation of Eqs. 63 and 64

Differentiating both sides of Eqs. 27 and 28 leads to:

$$ \begin{aligned} & (R_{tot} - C_{Atot} - C_{Btot} + 2C_{A} + C_{B} + K_{DA} )\frac{{dC_{A} }}{dt} + C_{A} \frac{{dC_{B} }}{dt} \\& \quad = K_{DA} \frac{{dC_{Atot} }}{dt} - C_{A} \left(\frac{{dR_{tot} }}{dt} - \frac{{dC_{Atot} }}{dt} - \frac{{dC_{Btot} }}{dt}\right) \\ \end{aligned} $$
(160)
$$ \begin{aligned} & C_{B} \frac{{dC_{A} }}{dt} + (R_{tot} - C_{Atot} - C_{Btot} + C_{A} + 2C_{B} + K_{DB} )\frac{{dC_{B} }}{dt} \\ &\quad = K_{DB} \frac{{dC_{Btot} }}{dt} - C_{B} \left(\frac{{dR_{tot} }}{dt} - \frac{{dC_{Atot} }}{dt} - \frac{{dC_{Btot} }}{dt}\right) \\ \end{aligned} $$
(161)

Rearranging terms in Eqs. 160 and 161 so that dC A /dt and dC B /dt are unknowns leads to:

$$ a\frac{{dC_{A} }}{dt} + b\frac{{dC_{B} }}{dt} = e $$
(162)
$$ c\frac{{dC_{A} }}{dt} + d\frac{{dC_{B} }}{dt} = f $$
(163)

where

$$ a = R_{tot} - C_{Atot} - C_{Btot} + 2C_{A} + C_{B} + K_{DA} $$
(164)
$$ b = C_{A} $$
(165)
$$ c = C_{B} $$
(166)
$$ d = R_{tot} - C_{Atot} - C_{Btot} + C_{A} + 2C_{B} + K_{DB} $$
(167)
$$ e = K_{DA} \frac{{dC_{Atot} }}{dt} - C_{A} \left(\frac{{dR_{tot} }}{dt} - \frac{{dC_{Atot} }}{dt} - \frac{{dC_{Btot} }}{dt}\right) $$
(168)
$$ f = K_{DB} \frac{{dC_{Btot} }}{dt} - C_{B} \left(\frac{{dR_{tot} }}{dt} - \frac{{dC_{Atot} }}{dt} - \frac{{dC_{Btot} }}{dt}\right) $$
(169)

The system of two linear equations Eqs. 162 and 163 has a unique solution defined by the Cramer’s rule [17] if

$$ ad - bc \ne 0 $$
(170)

Then

$$ \frac{{dC_{A} }}{dt} = \frac{ed - bf}{ad - bc} $$
(171)
$$ \frac{{dC_{B} }}{dt} = \frac{af - ec}{ad - bc} $$
(172)

Replacing the derivatives in Eqs. 168 and 169 by the right hand sides of differential equations Eqs. 22, 24, and 26 and using the variables CP Atot , CP Btot , ans RP tot defined by Eqs. 6769 one can notice that:

$$ a = E + C_{A} $$
(173)
$$ d = F + C_{B} $$
(174)
$$ e = K_{DA} CP_{Atot} - C_{A} (RP_{tot} - CP_{Atot} - CP_{Btot} ) $$
(175)
$$ f = K_{DB} CP_{Btot} - C_{B} (RP_{tot} - CP_{Atot} - CP_{Btot} ) $$
(176)

Upon performing the calculations ad-bf, af-ec, and ad-bc become equal to the numerators and denominators of the ratios in Eqs. 63 and 64. The conditions in Eq. 29 imply that the left hand sides of the equilibrium equations are positive, and consequently E > 0 and F > 0. This guaranties that:

$$ ad - bc = EF + FC_{A} + EC_{B} > 0 $$
(177)

and the condition Eq. 170 is satisfied.

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Yan, X., Chen, Y. & Krzyzanski, W. Methods of solving rapid binding target-mediated drug disposition model for two drugs competing for the same receptor. J Pharmacokinet Pharmacodyn 39, 543–560 (2012). https://doi.org/10.1007/s10928-012-9267-z

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Keywords

  • Target-mediated drug disposition
  • Gaddum equation
  • Erythropoietin
  • FcRn
  • Therapeutic antibody