Transcript Multicompartment Models

Multicompartment Models
Prof. Dr. Henny Lucida, Apt
These models were developed to explain
the observation that, after a rapid IV
injection, the Cp vs t curve does not
decline linearly as a single, first order
process.
The drug distributes at various rates into
different tissue groups. After drug
equilibration within the peripheral tissues,
the Cp vs t curve reflects first-order
elimination of the drug from the body
General Groupin of Tissues
according to Blood Supply
Blood Supply Tissue Group
% BW
Highly
perfused
9
Slowly
perfused
Heart, brain, hepatic-portal
system, kidney and
endocrine glands
Skin and muscle
Adipose (fat) tissue &
marrow
Bone, ligaments, tendons,
cartilage, teeth and hair
50
19
22
Two-Compartment open model
• The plasma drug concentration declines
biexponentially as the sum of two firstorder processes– distribution and
elimination.
• The drug distributes into 2 comp: the
central comp and the tissue (peripheral)
comps. Drug transfer between the two
comps is assumed to take place by firstorder process.
Basic Equations
C p  Ae
 at
 Be
 bt
D0 a  k 21 
A
V p a  b 
D0 k 21  b 
B
V p a  b 
a  b  k12  k 21  k
a  b  k 21  k
Residual Method
Example:
100 mg of a drug was administered by
rapid IV injection to a 70-kg, healthy adult
male. Blood samples were taken
periodically after the administration of
drug, and the plasma fraction of each
sample was assayed for drug. The
following data were obtained:
Time
(hr)
0.25
Cp (mg/mL)
Cp (mg/mL)
43.00
Time
(hr)
4.0
0.5
32.00
8.0
2.80
1.0
20.00
12.0
1.20
1.5
14.00
16.0
0.52
2.0
11.00
6.50
Step1. Transform data to log Cp
Time
(hr)
0.25
Log Cp
Log Cp
1.6335
Time
(hr)
4.0
0.5
1.5051
8.0
0.4472
1.0
1.3010
12.0
0.0792
1.5
1.1461
16.0
-0.2840
2.0
1.0414
0.8129
Plot log Cp vs time
a
b
Step 2. Determine the linear regression of line b
(after 4 hr), these data represent only the
elimination process thus the slopeof line is equal
to K
C p  Be  bt
bt
logC p  logB 
2.303
The linear regression:
Y1=1.1785 – 0.0915 X1 (r=-0.9999)
Slope = -0.0915; b = 0.2107/hr
B=antilog 1.1785= 15.0834 mg/mL
Equation for b : Cp = 15.08 e-0.21t
Step 3. The ka is determined residually by subtracting the data point at
< 4 hr with the fitted data using linear regression of b.
At these data point, the absorption process is predominant, then the
slope would be the ka.
This step would be easily solved by using a semi-logarithmic paper.
Then the linear regression of the subtracted data is determined:
Y2 = 1.6727 – 0.8055 X2
The slope = -0.8055, then a or ka = 1.8551/hr
The intercept = 1.6727, then A or the C0 = antilog 1.6727 = 47.0652
mg/mL.
Equation for a : Cp = 47.06 e-1.85t
Thus the profile of plasma conc based on two compartment model is
Cp = 47.06 e-1.85t + 15.08 e-0.21t
Then: k, k12 and k21 can be calculated.
Problems
A drug was administered by rapid iv injection
into a 70-kg adult male. Blood samples
were withdrawn over a 7-hour period and
assayed for intact drug. The results are
tabulated below. Using the method of
residuals, calculate the values for
intercepts A and B and slopes a, b, k, k12
and k21.
Time (hr)
Cp (mg/mL)
Time (hr)
Cp (mg/mL)
0.00
70.0
2.5
14.3
0.25
53.8
3.0
12.6
0.50
43.3
4.0
10.5
0.75
35.0
5.0
9.0
1.00
29.1
6.0
8.0
1.50
21.2
7.0
7.0
2.00
17.0