978-0131569881 Solution Chapters 12-16 Part 3

subject Type Homework Help
subject Authors David F. Katz, Fan Yuan, George A. Truskey

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Figure S15.4.
15.8. The governing equation and the boundary conditions are identical to those in Problem 15.7,
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Figure S15.8.1.
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Although L increases with Deff and Deff can be increased by reducing the size of drugs, the
reduction in drug size also increases P. Thus, a better way to increase the penetration depth is to
increase the size of drugs or using large carriers of drugs.
MATLAB CODES
Problem 15.5
function main
global N s vol dr vr Deff PSV Cp f Kav Q
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for i = 1:N
vol(i) = 4/3*pi*(rm(i+1)*rm(i+1)*rm(i+1)-rm(i)*rm(i)*rm(i));
end
%Velocity
vr = zeros(N+1,1);
for i=2:N+1
x = alfa*rm(i)/R
vr(i) = K*pe*R/(e*rm(i)*rm(i)*sinh(alfa))*(x*cosh(x)-sinh(x))
end
function dcdt = func(t,c)
global N s vol dr vr Deff PSV Cp f Kav Q
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Problem 15.7
function main
global N dr s v Deff P Kav Cp0 alpha L1 L2
% Model Parameters
Deff = 1.0e-7*60e8; %cm2/s->um2/min
P = 1.5e-7*60e8; %cm/s->um2/min
Kav = 0.3;
Cp0 = 1; %uM
alpha = 0.7;
L1 = 4e-4; %1/min
L2 = 7e-6; %1/min
a = 5; %um
R = 150; %um
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Problem 15.8
function main
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global N dr s v Deff P Kav
% Model Parameters
Deff = 2.0e-7*60e8; %cm2/s->um2/min
P = 1.5e-7*60e8; %cm/s->um2/min
Kav = 0.3;
c_init = 10; % uM
a = 5; %um
R = 150; %um
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function dcdt = func(t,c)
global N dr s v Deff P Kav
Problem 15.9
function main
global N dr s v Deff P Kav k
% Model Parameters
Deff = 2.0e-7*60e8; %cm2/s->um2/min
P = 1.5e-7*60e8; %cm/s->um2/min
Kav = 0.3;
a = 5; %um
R = 150; %um
k = 0.5/60; %1/hour->1/min
c_init = 10; %uM
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%
c0 = zeros(N,1);
for i=1:N
c0(i)= c_init;
end
%
dcdt = zeros(N,1);
dcdt(1) = 1/v(1)*(-P*c(1)/Kav*s(1)+Deff*(c(2)-c(1))/dr*s(2))-k*c(1);
for i=2:N-1
iu=i-1;
id=i+1;
dcdt(i)= 1/v(i)*(Deff*(c(iu)-c(i))/dr*s(i)+Deff*(c(id)-c(i))/dr*s(id))-k*c(i);
end
dcdt(N) = 1/v(N)*(Deff*(c(N-1)-c(N))/dr*s(N))-k*c(N);
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Solution to Problems in Chapter 16, Section 16.6
16.1. The mass balance equation for the drug in the compartment is,
16.2. The mass balance equations of the drug in the central compartment and the peripheral
compartments are, respectively,
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16.3. The change in CD is due to the binding of the free drug to tissues, the release of the bound
drug, and the clearance of the free drug from the compartment, whereas the change in CB is
solely due to the reversible binding. Therefore, the mass balance equations of CD and CB are,
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16.4. The mass balance of the antibody in the tumor tissue is,

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