Introduction to Pharmacometrics and Quantitative Pharmacology with an Emphasis on Physiologically Based Pharmacokinetics

Abstract

Pharmacometrics, enabled by advanced pharmacostatistical modeling and simulation technique, has gained significant importance as a tool for model-based drug development and for pharmacotherapy. This introductory chapter provides an overview of the models and mathematical framework that are commonly used in the field. Both pharmacokinetic and pharmacodynamic models are discussed; the physiologically based pharmacokinetic (PBPK) approach is emphasized through illustrations and examples. We have shown how systemic and tissue clearances are related through mathematical proofs using the PBPK approach. As for the pharmacodynamic section, models for continuous and noncontinuous responses were explained, as well as a disease progression model. We also provided select examples of systems pharmacology models that investigate physiological and biochemical processes. This chapter also discusses the software tools and their specific application. As pharmacometrics and quantitative pharmacology has evolved as a discipline, more and more different types of statistical and mathematical models have been utilized. The advancement of computing software and systems facilitate these complex computations.

Appendices

Appendix

Derivation of Eq. 1.55

Implicit in the assumption that the concentration of drug in the system that is being cleared is constant or at steady state, we assumed that the rate of change in the drug concentration is 0 and thus set all the left-hand sides in the list of equations in Eq. 1.53 to 0.

$$ 0={{V}_{1}}\frac{d{{C}_{1(vein)}}}{dt}={{Q}_{1}}\cdot {{C}_{0}}-{{Q}_{1}}\cdot {{C}_{1}}+{{K}_{I}}(t) $$

(1.112)

$$ 0={{V}_{2}}\frac{d{{C}_{2(artery)}}}{dt}={{Q}_{1}}\cdot {{C}_{3o}}-{{Q}_{1}}\cdot {{C}_{2}}$$

(1.113)

$$ 0={{V}_{3}}\frac{d{{C}_{3(lung)}}}{dt}={{Q}_{1}}\cdot {{C}_{1}}-{{Q}_{1}}\cdot {{C}_{3o}}-{{r}_{3}}({{C}_{3}}) $$

(1.114)

$$ 0={{V}_{4}}\frac{d{{C}_{4(liver)}}}{dt}={{Q}_{7}}\cdot {{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}})\cdot {{C}_{2}}-{{Q}_{4}}\cdot {{C}_{4o}}-{{r}_{4}}({{C}_{4}}) $$

(1.115)

$$ 0={{V}_{5}}\frac{d{{C}_{5(kidney)}}}{dt}={{Q}_{5}}\cdot {{C}_{2}}-{{Q}_{5}}\cdot {{C}_{5o}}-{{r}_{5}}({{C}_{5}}) $$

(1.116)

$$ 0={{V}_{6}}\frac{d{{C}_{6(rest\text{ }of\text{ }body)}}}{dt}={{Q}_{6}}\cdot {{C}_{2}}-{{Q}_{6}}\cdot {{C}_{6o}}$$

(1.117)

$$ 0={{V}_{7}}\frac{d{{C}_{7(GI\text{ }tract)}}}{dt}={{Q}_{7}}\cdot {{C}_{2}}-{{Q}_{7}}\cdot {{C}_{7o}}-{{r}_{7}}({{C}_{7}}). $$

(1.118)

From Eq. 1.112, we integrate

$$ \int_{0}^{\infty }{{{Q}_{1}}\cdot {{C}_{1}}dt=}\int_{0}^{\infty }{({{Q}_{1}}\cdot {{C}_{0}}+{{K}_{I}}(t) )dt}. $$

(1.119)

By applying Eq. 1.54 and substituting ${{Q}_{1}}\cdot {{C}_{0}}$ with ${{Q}_{4}}\cdot {{C}_{4o}}+{{Q}_{5}}\cdot {{C}_{5o}}+{{Q}_{6}}\cdot {{C}_{6o}}$ to Eq. 1.119, the resulting expression is

$$ {{Q}_{1}}\int_{0}^{\infty }{{{C}_{1}}dt=}\int_{0}^{\infty }{({{Q}_{4}}\cdot {{C}_{4o}}+{{Q}_{5}}\cdot {{C}_{5o}}+{{Q}_{6}}\cdot {{C}_{6o}})dt+}\int_{0}^{\infty }{{{K}_{I}}\text{(t)}dt}. $$

(1.120)

Rearrangement of Eq. 1.120 by adding and subtracting $({{Q}_{7}}\cdot {{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}}){{C}_{2}}+{{Q}_{5}}\cdot {{C}_{5o}}+{{Q}_{6}}\cdot {{C}_{2}})$ to Eq. 1.120 such that the net result is 0 will yield the following expression:

$$ {{Q}_{1}}\int_{0}^{\infty }{{{C}_{1}}dt=}\int_{0}^{\infty }{[({{Q}_{7}}\cdot {{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}}){{C}_{2}}+{{Q}_{5}}\cdot {{C}_{5o}}+{{Q}_{6}}\cdot {{C}_{2}})-({{Q}_{7}}\cdot {{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}}){{C}_{2}}+{{Q}_{5}}\cdot {{C}_{5o}}+{{Q}_{6}}\cdot {{C}_{2}}) ]dt+}\int_{0}^{\infty }{({{Q}_{4}}\cdot {{C}_{4o}}+{{Q}_{5}}\cdot {{C}_{5o}}+{{Q}_{6}}\cdot {{C}_{6o}})dt+}\int_{0}^{\infty }{{{K}_{I}}(t)dt}. $$

(1.121)

From Eq. 1.117, $\int_{0}^{\infty }{({{Q}_{6}}\cdot {{C}_{2}}-{{Q}_{6}}\cdot {{C}_{6o}})dt=0}$. Given that ${{Q}_{7}}\cdot {{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}}){{C}_{2}}$ represents the total amount of drug entering the liver from the artery and the gut which is also C_p, wherein${{Q}_{4}}\cdot {{C}_{p}}={{Q}_{7}}\cdot {{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}}){{C}_{2}}$, we can then simplify Eq. 1.121 as follows:

$$ {{Q}_{1}}\int_{0}^{\infty }{{{C}_{1}}dt=}\int_{0}^{\infty }{({{Q}_{4}}+{{Q}_{5}}+{{Q}_{6}}){{C}_{2}}dt-}\left\{\int_{0}^{\infty }{{{Q}_{4}}({{C}_{p}}-{{C}_{4o}})dt+}\int_{0}^{\infty }{{{Q}_{5}}({{C}_{2}}-{{C}_{5o}})dt+}\int_{0}^{\infty }{{{Q}_{7}}({{C}_{2}}-{{C}_{7o}})dt} \right\}+\int_{0}^{\infty }{{{K}_{I}}\text{(t)}dt}. $$

(1.122)

Furthermore, $\int_{0}^{\infty }{({{Q}_{4}}+{{Q}_{5}}+{{Q}_{6}}){{C}_{2}}dt={{Q}_{1}}}\int_{0}^{\infty }{{{C}_{2}}dt}$, since ${{Q}_{1}}={{Q}_{4}}+{{Q}_{5}}+{{Q}_{6}}$ and from Eq. 1.113, ${{Q}_{1}}\underset{0}{\overset{\infty }{\mathop \int }}\,{{C}_{3o}}dt$, then Eq. 1.122 can be rewritten as Eq. 1.55:

$$ \int_{0}^{\infty }{{{Q}_{1}}({{C}_{1}}-{{C}_{3o}})dt+}\int_{0}^{\infty }{{{Q}_{4}}({{C}_{p}}-{{C}_{4o}})dt+}\int_{0}^{\infty }{{{Q}_{5}}({{C}_{2}}-{{C}_{5o}})dt+}\int_{0}^{\infty }{{{Q}_{7}}({{C}_{2}}-{{C}_{7o}})dt=}\int_{0}^{\infty }{{{K}_{I}}(t)dt}. $$

Derivation of Eq. 1.60

Using the expression for mean clearance, the rearrangement of Eqs. 1.55 through 1.59 will result in the following:

$$ {{\overline{CL}}_{systemic}}={{\overline{CL}}_{3}}+\left(\frac{\mathop{\int }_{0}^{\infty }{{C}_{p}}dt}{\mathop{\int }_{0}^{\infty }{{C}_{1}}dt} \right){{\overline{CL}}_{4}}+\left(\frac{\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}{\mathop{\int }_{0}^{\infty }{{C}_{1}}dt} \right)({{\overline{CL}}_{5}}+{{\overline{CL}}_{7}}). $$

(1.123)

We set ${{Q}_{4}}\cdot {{C}_{p}}={{Q}_{7}}{{C}_{7o}}+({{Q}_{4}}-{{Q}_{7}}){{C}_{2}}$ to ${{Q}_{4}}({{C}_{2}}-{{C}_{p}})={{Q}_{7}}({{C}_{2}}-{{C}_{7o}})$. Then integrate the latter expression from 0 to infinity, ${{Q}_{4}}\int_{0}^{\infty }{({{C}_{2}}-{{C}_{p}})dt={{Q}_{7}}}\int_{0}^{\infty }{({{C}_{2}}-{{C}_{7o}})dt}$ and divide both sides of the equation by $\int_{0}^{\infty }{{{C}_{2}}dt}$ to obtain $\frac{{{Q}_{4}}\mathop{\int }_{0}^{\infty }({{C}_{2}}-{{C}_{p}})dt}{\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}=\frac{{{Q}_{7}}\mathop{\int }_{0}^{\infty }({{C}_{2}}-{{C}_{7o}})dt}{\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}$. The previous equation can be rewritten as $1-\frac{\mathop{\int }_{0}^{\infty }{{C}_{p}}dt}{\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}=\frac{{{Q}_{7}}\mathop{\int }_{0}^{\infty }({{C}_{2}}-{{C}_{7o}})dt}{{{Q}_{4}}\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}$. The ratio consisting the integrating components on the left side of the equation is the expression for the extraction ratio for tissue compartment 7 for the GI tract, ${{\overline{E}}_{7}}=\frac{\mathop{\int }_{0}^{\infty }({{C}_{2}}-{{C}_{7o}})\text{d}t}{\mathop{\int }_{0}^{\infty }{{C}_{2}}\text{d}t}$. Therefore,

$$ \frac{\mathop{\int }_{0}^{\infty }{{C}_{p}}dt}{\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}=1-\frac{{{Q}_{7}}}{{{Q}_{4}}}{{\overline{E}}_{7}}$$

(1.124)

From Eq. 1.113 above, ${{Q}_{1}}\int_{0}^{\infty }{{{C}_{2}}dt={{Q}_{1}}}\int_{0}^{\infty }{{{C}_{3}}dt}$, and from Eq. 1.57, ${{\overline{CL}}_{3}}=\frac{{{Q}_{1}}\mathop{\int }_{0}^{\infty }({{C}_{1}}-{{C}_{3o}})dt}{\mathop{\int }_{0}^{\infty }{{C}_{1}}dt}=\frac{{{Q}_{1}}\mathop{\int }_{0}^{\infty }({{C}_{1}}-{{C}_{2}})dt}{\mathop{\int }_{0}^{\infty }{{C}_{1}}dt}$. Thus,

$$ \frac{\mathop{\int }_{0}^{\infty }{{C}_{2}}dt}{\mathop{\int }_{0}^{\infty }{{C}_{1}}dt}=1-\frac{{{\overline{CL}}_{3}}}{{{Q}_{1}}}=1-{{\overline{E}}_{lung}}$$

(1.125)

In the following step, multiply Eq. 1.124 by A125:

$$ \frac{\mathop{\int }_{0}^{\infty }{{C}_{p}}dt}{\mathop{\int }_{0}^{\infty }{{C}_{1}}dt}=\left(1-\frac{{{\overline{CL}}_{3}}}{{{Q}_{1}}}\right)\left(1-\frac{{{\overline{CL}}_{7}}}{{{Q}_{4}}}\right). $$

(1.126)

Since ${{\overline{CL}}_{7}}={{Q}_{7}}\cdot {{\overline{E}}_{7}}$. In the following step, we substitute Eqs. 1.126 and 1.125 to Eq. 1.123, ${{\overline{CL}}_{systemic}}={{\overline{CL}}_{3}}+\left(1-\frac{{{\overline{CL}}_{3}}}{{{Q}_{1}}}\right)\left(1-\frac{{{\overline{CL}}_{7}}}{{{Q}_{4}}}\right){{\overline{CL}}_{4}}+\left(1-\frac{{{\overline{CL}}_{3}}}{{{Q}_{1}}}\right)({{\overline{CL}}_{5}}+{{\overline{CL}}_{7}})={{\overline{CL}}_{3}}+(1-{{\overline{E}}_{3}})[{{\overline{CL}}_{4}}+(1-{{\overline{E}}_{4}}){{\overline{CL}}_{7}}+{{\overline{CL}}_{5}}].$