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and ρ is the tissue mass density, which is approximately equal to 1 g/ml. Note that Ct is defined
as the number of moles per unit tissue volume whereas Ca and CaG are defined as the number of
moles per unit volume of the interstitial space. The volume fraction of the interstitial space is φ.
To solve Equation (S16.4.1), we need to determine how Ct depends on Ca. Since the binding
between the antibody and the tumor-associated antigen is at equilibrium,
Equations (S16.4.6) through (S16.4.8) are solved numerically using the MATLAB. (The
MATLAB code is attached at the end of this chapter.) The concentration profiles of the drug are
shown in Fig. S16.4.1.
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Figure S16.4.1
16.5. The mass balance equations in the three compartments are,
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16.6. With the alternative approach to modeling the process of methotrexate transport in the
hepatic duct, the only change in the PBPK analysis is to replace Equations (16.4.10a, b, and c) by
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16.7. The concentration distributions of methotrexate in a mouse, a rat, a dog, and a monkey are
determined by solving the equations for its concentrations in different organs described in
Section 16.4.2. The solutions are obtained numerically using the MATLAB. (The MATLAB
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16.8. The concentration distribution of methotrexate in a mouse is determined by the same
MATLAB code as that for Problem 16.7. The only difference is the values of model constants.
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The shapes of concentration profiles in Panel (a) are similar to those in Panel (b); and the
concentrations during the first few minutes are similar between the two panels in corresponding
organs. However, the concentrations in all compartments in Panel (b) are significantly lower than
those in Panel (a) in later time points, due to the higher value of predicted clearance rate in the
kidney. The AUC during the first 6 hours is listed in the table.
Model
Plasma
Muscle
Kidney
Liver
Gut Lumen
* The unit of AUC is µg/ml×min.
MATLAB CODES
Problem 16.4
function main
global cp0 alpha lamda1 lamda2 k ke ka fi g0 ro
cp0 = 100; % nM
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alpha = 0.459;
lamda1 = 1.11e-2; % 1/min
lamda2 = 1.38e-4; % 1/min
k = 0.5; % ul/min/g
ke = 0.8; % ul/min/g
ka = 1; %1/nM
fi = 0.243;
g0 = 100; % nM
ro = 1e-3; %g/ul
function dcdt = func(t,c)
global cp0 alpha lamda1 lamda2 k ke ka fi g0 ro
cp = cp0*(alpha*exp(-lamda1*t)+(1-alpha)*exp(-lamda2*t));
dcdt = ro*(k*cp/fi-ke*c)/(1+g0*ka/((1+ka*c)^2));
Problem 16.6
% Physiologically Based Pharmacokinetic Analysis of Methotrexate
clear all;
global para n ct m td
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%Initialization
para = zeros(22,1);
%Plasma flow rate (ml/min)
para(7) = 420; %muscle
para(8) = 700; %kidney
para(9) = 800; %liver
para(10) = 700; %G.I. tract
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c0(1) = para(21)*1000/para(1); % microgram/ml
function y = f16_6(c,t)
global para n ct m td
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Problem 16.7
% Physiologically Based Pharmacokinetic Analysis of Methotrexate
clear all;
global para n ct
h = 0.1; % delta t (min)
tend = 400; % last time point (min)
N = fix(tend/h)+1; % total steps in time
%Initialization
para = zeros(22,1);
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para(5) = 2100; %G.I. tract
para(6) = 2100; %gut lumen
%Plasma flow rate (ml/min)
para(7) = 420; %muscle
para(8) = 700; %kidney
para(9) = 800; %liver
para(10) = 700; %G.I. tract
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function y = f16_7(c,t)
global para n ct
y = zeros(12,1);
y(1) = (para(7)*c(2)/para(11) + para(8)*c(3)/para(12) + para(9)*c(4)/para(13)...
- (para(9)+para(8)+para(7))*c(1))/para(1);
y(2) = para(7)*(c(1) - c(2)/para(11))/para(2);
y(3) = (para(8)*(c(1) - c(3)/para(12)) - para(15)*c(3)/para(12)) / para(3);
r = para(16) * c(4) / para(13);
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