Challenges in Pharmacology Modelling

Abstract

The increased emphasis on mechanistic and quantitative approaches in pharmacology presents new challenges to the design and analysis of mathematical models used in drug discovery. In this paper we present three of such challenges: (i) How to proceed when exposure data are scarce or plainly absent? Response versus time data for different doses are then the only source of information. (ii) The advent of biological (biologics) therapeutics has resulted in a new class of models which involve the interaction of drug and target. (iii) How to determine the impact of proteins at the target site on drug efficacy? We illustrate these challenges with three case studies.

Appendix A: Derivation of the expression for \(R_n(t)\) in (2.11)

Putting $S(A_{b})=1+S_\mathrm{max}$ in the the system (2.4) we obtain the model

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{dR_{1}}{dt} &{}= k_\mathrm{in}(1+S_\mathrm{max})-k_\mathrm{out} R_{1}, \\ &{}............. \\ \displaystyle \frac{dR_{i}}{dt} &{}= k_\mathrm{out}(R_{i-1}-R_{i}), \qquad i=2,3,\dots n. \end{array}\right. \end{aligned}$$

(5.1)

We shall derive an explicit solution of this problem.

For convenience we put $R_{i} = R_{0}(1+x_{i})$ and $\tau = k_\mathrm{out}\, t$. Then the system (5.1) can be rewritten as

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \frac{dx_{1}}{d\tau } &{}= S_\mathrm{max}- x_{1}, \\ &{}............. \\ \displaystyle \frac{dx_{i}}{d\tau } &{}= x_{i-1}-x_{i}, \qquad i=2,3,\dots n. \end{array}\right. \end{aligned}$$

(5.2)

This system can be solved sequentially. Since $x_{i}(0)=0$ for $i = 1, \dots . n$ we obtain

$$\begin{aligned} x_{1}(\tau )&= S_\mathrm{max}e^{-\tau }(e^{\tau } - 1), \nonumber \\ x_{2}(\tau )&= S_\mathrm{max}e^{-\tau }(e^{\tau } - 1-t), \nonumber \\ x_{3}(\tau )&= S_\mathrm{max}e^{-\tau }\left( e^{\tau } - 1-\tau - \frac{1}{2}\tau ^{2}\right) , \end{aligned}$$

(5.3)

and for general $n \ge 1$:

$$\begin{aligned} x_n(\tau ) = S_\mathrm{max}\left( 1 - e^{-\tau }\sum _{m=0}^{n-1} \frac{\tau ^m}{m!}\right) . \end{aligned}$$

(5.4)

Thus, in terms of the original variables we obtain the following expression for the responses:

$$\begin{aligned} R_n(t) = R_{0}\left\{ 1+ S_\mathrm{max}\left( 1 - e^{-k_\mathrm{out}t}\sum _{m=0}^{n-1} \frac{(k_\mathrm{out}t)^m}{m!}\right) \right\} . \end{aligned}$$

(5.5)