Multi-compartment Models
Objectives
Compartment Modeling of Drugs
Models describing
drug concentration-time profiles
L/70 kg
50,000
General Principles of Distribution
Quinacrine
20,000
Chloroquine
10,000
5,000
1,000
500
100
50
10
5
Nortriptyline
Digoxin
Propranolol
Quinidine
Quinolones (1- 2 L/kg), Tetracycline
Phenobarbital
Phenytoin
To date, we have assumed that
Theophylline (0.45 L/kg)
distribution of drug in the
Aminoglycodises (0.25 L/kg)
ASA
body was instantaneous.
Warfarin
What if that is NOT true?
1000 mg
10 L
In this patient the drug
“instantly” distributes
throughout
the body and the
concentration is
homogeneous.
If we draw a “model”
of the compartments,
drug is loaded into
the compartment
… it rapidly distributes
(instantaneously)
The body acts as a
single tub or compartment.
and then begins
to be eliminated.
1000 mg
1000 mg
1L
10 L
10 L
In this patient the drug distributes to
In this patient the drug
“instantly” distributes to the some fluids (tissues) quickly, but from
there, drug distributes slowly
entire volume and the
concentration is homogeneous. to tissues that are poorly perfused.
It takes some time before all tissues
establish a concentration in
The body acts as a
equilibrium with other tissues.
single tub or compartment.
The body acts as if it contains
multiple compartments.
1000 mg
1L
10 L
When the body appears to have
a second compartment,
distribution into each
compartment
occurs at different rates
If we draw a “model”
of the compartments,
drug is loaded into
the first compartment
(central compartment)
… it rapidly distributes
(instantaneously)
throughout the first compartment
and then begins to distribute
into the second compartment.
1000 mg
1L
10 L
While drug is distributing
to the second compartment,
it will still be eliminated
(cleared) from the
first compartment
This means that
the drug concentration
in the second compartment
will be rising
as the concentration
in the first is falling.
1000 mg
1L
10 L
While drug is distributing
to the second compartment,
it will still be eliminated
(cleared) from the
first compartment
This means that
the drug concentration
in the second compartment
will be rising
as the concentration
in the first is falling.
But just as drug diffused
into the second compartment
it can diffuse back to the first.
2 Compartment Model:
IV Dosing
The rate constants describing drug movement are
labeled based on their movement
from one compartment to another.
k12 describes movement from compartment 1 to compartment 2.
k21 describes movement from compartment 2 to compartment 1.
k10 describes movement from compartment 1 out of the body.
2 Compartment Model:
IV Dosing
When the body appears to have
a second compartment,
distribution into the second
compartment and elimination
contribute an initial
rapid decline in concentration.
However, after some time,
distribution into the
second compartment
is complete.
The two compartments
are in equilibrium
with each other.
Although equilibrium
has been established
between the
two compartments,
concentrations in each
compartment may be different.
Equilibrium
Established
Here concentrations
in the second compartment
are in red.
Equilibrium is established
after ~4-6 hours.
After 4-6 hrs concentrations
in both compartments
decline with the same half-life.
1000 mg
When the body appears to have
a second compartment,
distribution into each
compartment
occurs at different rates.
1L
10 L
Distribution to tissues is
possibly related
to tissue blood flow.
Distribution to tissues (2)
with lower perfusion
occurs more slowly.
Distribution to tissues (1)
with greater greater blood
flow occurs more quickly
… almost instantaneously.
k12=0.3 hr-1; k21=0.4 hr-1
k12/k21= 0.75
k12=0.3 hr-1; k21=0.217 hr-1
k12/k21= 1.38
However, differences in profiles
is not solely related
to the size of k12.
Here, both profiles have the same
terminal elimination ( = 0.2 hr-1)
AND the same k12 (0.3 hr-1).
Profiles have different k21 values.
Differences in profiles are created
by different relative values of
k12 : k21.
Movement in and out
2 Compartment Model:
IV Dosing
Patients with different
body compositions,
Different proportions of
highly perfused
vs.
lean muscle
vs.
poorly perfused adipose tissue
Will have different rates and degrees
of distribution to their tissues.
Different relative values of k12:k21.
2 Compartment Model:
IV Dosing
Questions to Consider
What is the difference
between the profiles on the left?
Assume that it is the same drug
given to 3 different patients
Is the terminal half-life the same?
Is the initial volume the same?
Is the final volume the same?
Is the rate of distribution the same?
Which profile has the largest k12:k21 ratio?
Which has the smallest?
2 Compartment Model:
IV Dosing
Questions to Consider
What happens
if you change Input Rate?
(bolus  infusion)
Can the profile change from a
2-Compartment
following IV bolus
to a 1-Compartment Model
following IV Infusion?
How would you explain this?
Analysis of 2 Compartment Model:
IV Dosing
Using semi-log paper, graph the data from these two patients.
Each patient is given 1000 mg of a drug by IV bolus.
Patient 1
Time Plasma Conc
(hr)
(mg/L)
Patient 2
time Plasma Conc
(hr)
(mg/L)
0.25
0.50
1
2
3
4
6
9
12
0.25
0.5
1
2
3
4
6
9
12
94
89
79
63
50
40
25
12.5
6.25
60
41
26
18
14
11
7.0
3.5
1.8
Plotting / Graphing
Plasma Concentrations
Do we need 1 cycle or
2-cycle semi log paper?
Using semi-log paper,
graph the data from
these two patients,
each given 1000 mg
of a drug.
Patient 1
Time Plasma Conc
(hr)
(mg/L)
0.25
94
0.50
89
1
79
2
63
3
50
4
40
6
25
9
12.5
12
6.25
Graph Patient 1 Data
100
60
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Using semi-log paper,
graph the data from
these two patients,
each given a
1000 mg dose.
Patient 1
Time Plasma Conc
(hr)
(mg/L)
0.25
94
0.50
89
1
79
2
63
3
50
4
40
6
25
9
12.5
12
6.25
Graph Patient 1 Data
100
What is the Half-life?
… the Initial [ ]?
What model best
describes data?
60
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Patient 1
Time Plasma Conc
(hr)
(mg/L)
0.25
94
0.50
89
1
79
2
63
3
50
4
40
6
25
9
12.5
12
6.25
Graph Patient 1 Data
100
What is the Half-life?
… the Initial [ ]?
What model best
describes data?
60
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Patient 1
Time Plasma Conc
(hr)
(mg/L)
0.25
94
0.50
89
1
79
2
63
3
50
4
40
6
25
9
12.5
12
6.25
Graph Patient 1 Data
100
What is the Half-life? 3hr
… the Initial [ ]? …100 mg/L
What model best
describes data?
60
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Patient 1
Time Plasma Conc
(hr)
(mg/L)
0.25
94
0.50
89
1
79
2
63
3
50
4
40
6
25
9
12.5
12
6.25
Graph Patient 1 Data
100
IV Dose
(bolus)
1
60
V1
30
k10
10
6
4
2
1
0
2
4
6
8
10 12 14
In Patient 1, distribution of
drug to the body is
instantaneous (rapid). Drug
enters the blood and rapidly
distributes to tissues.
At 0.25 hr drug is
homogeneously distributed.
Graph Patient 1 Data
100
IV Dose
(bolus)
1
60
V1
30
K
10
Dose e-Kt
Ct = ---------V
6
4
2
1
0
2
4
6
8
10 12 14
Dose = 1000 mg
V=10 L
K= 0.231 hr-1
-0.231(t)
Ct = 100 e
Graph Patient 2 Data
100
60
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Using semi-log paper,
graph the data from
Patient 2,
also given a
1000 mg dose.
Patient 2
Time Plasma Conc
(hr)
(mg/L)
0.25
60
0.5
41
1
26
2
18
3
14
4
11
6
7.0
9
3.5
12
1.8
Graph Patient 2 Data
100
60
What is different
about this
patient?
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Patient 2
Time Plasma Conc
(hr)
(mg/L)
0.25
60
0.5
41
1
26
2
18
3
14
4
11
6
7.0
9
3.5
12
1.8
Graph Patient 2 Data
100
60
What is different
about this patient?
What is the Half-life?
… the Initial [ ]?
30
10
6
4
2
1
0
2
4
6
8
10 12 14
Patient 2
Time Plasma Conc
(hr)
(mg/L)
0.25
60
0.5
41
1
26
2
18
3
14
4
11
6
7.0
9
3.5
12
1.8
Graph Patient 2 Data
100
60
30
10
6
4
2
1
0
2
4
6
8
10 12 14
What is different
about this patient?
What is the Half-life? 3hr
… the Initial [ ]? ..100 mg/L
What model describes
a patient like
Patient 2 ?
Time Plasma Conc
(hr)
(mg/L)
0.25
60
0.5
41
1
26
2
18
3
14
4
11
6
7.0
9
3.5
12
1.8
Graph Patient 2 Data
100
60
30
In Patient 2, distribution of
drug to the body occurs in 2 steps.
Initial instantaneous distribution
followed by a slower phase.
This is possibly related
to organ and tissue
blood flow.
10
6
4
2
1
0
2
4
6
8
10 12 14
Drug begins to distribute,
immediately
but distribution is
not complete until ~2 hours
after the dose.
Only then is drug is in each tissue
in equilibrium with other tissues.
Graph Patient 2 Data
100
60
30
-αt
Ct = A e
10
-βt
+ Be
6
4
The appearance of the
second exponential in the
[ ] – time profile is an
indication that we have a
2-compartment model.
2
1
0
2
4
6
8
10 12 14
2 Compartment Model:
IV Dosing
Rapid distribution
[α phase]
Drug distributes from highly
perfused tissues to tissues in
2nd compartment (fat, lean tissue).
C= Ae
-αt
+ Be
-βt
2 Compartment Model:
IV Dosing
Slow Elimination
[β phase]
Distribution is complete
Pseudo-equilibrium has been
achieved, body behaving as 1C.
C= Ae
-αt
+ Be
-βt
2 Compartment Model:
IV Dosing
Slow Elimination
[β phase]
Distribution is complete
Pseudo-equilibrium has been
achieved, body behaving as 1C.
Concentration in 2nd
Compartment (in red)
Equilibrium established between
compartments at the end
of the distribution phase.
C= A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
But what are the values
A and B?
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
Terminal Phase
Slope  
B is the Zero time intercept of the
the back-extrapolated β phase
B
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
Terminal Phase
Slope = 
A is the Zero time intercept
of the Residual …
… the difference between
the back extrapolated
concentrations on the β phase
and the observed concentrations
… in the  phase.
B
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
Step 1
Method of residuals
Back extrapolate β phase
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
Step 2
Determine back-extrapolated
concentrations at same time
as observed concentrations
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
Step 3
Calculate the difference between
the observed concentration and
the back-extrapolated
concentrations at each time point
and plot these concentrations ( ).
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
Step 4
Calculate the slope
of the line of the residual.
This is done in the same way as
you would calculate K, it is the
slope of the log of concentration
vs. time. This slope is proportional
to alpha.
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 5
Zero time intercepts of
back-extrapolated β phase
and of residual (α), are
equal to B and A respectively.
α and β are “MACRO” constants
in that they are made up of the
“MICRO” constants
k12, k21 and k10.
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
Step 5
B
α and β are “MACRO” constants
in that they are made up of the
“MICRO” constants (k12, k21 and k10)
and reflect a variety of
simultaneous processes.
[α + β = k12 + k21 + k10]
(Equations can be shown in either form)
Dose * (k21-α) e -αt + -----------------------Dose * (k21-β) e -βt
C = --------------------V * ( β – α)
V * (α - β)
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 6
Begin to solve for micro
constants.
k21 = (Aβ + Bα) / ( A + B)
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 7
Begin to solve for micro
constants.
k10 = α β / k21
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 8
Begin to solve for micro
constants.
k12 = α + β - k21 – k10
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 9
Begin to solve for other
parameters
V1 can be obtained from the
initial concentration and dose
V1 = dose / C0
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 10
Begin to solve for other
parameters
Vss = V1(1 + k12/k21)
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 11
Begin to solve for other
parameters
Calculate AUC by trap. rule
or AUC = A/ α + B/β
Ct = A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Step 11
Begin to solve for other
parameters
Calculate AUC by trap. rule
or AUC = A/ α + B/β
Ct = A e
-αt
+ Be
-βt
Graph Patient 2 Data
100
-αt
Ct = A e
-βt
+ Be
60
In this [ ] – time profile the
β-phase intercept (B) is 27.8
mg/L and the α-phase
intercept (A) is 72.2 mg/L.
-αt
-βt
Ct = 72.2 e + 28.8 e
30
10
6
What would the initial
concentration be at time zero?
4
2
1
0
2
4
6
8
10 12 14
Graph Patient 2 Data
100
-αt
Ct = A e
-βt
+ Be
60
30
10
6
4
2
1
0
2
4
6
8
In this [ ] – time profile the
β-phase intercept (B) is 27.8
mg/L and the α-phase
intercept (A) is 72.2 mg/L.
-αt
-βt
Ct = 72.2 e + 28.8 e
What would the initial
concentration be at time zero?
C0=A+B
and V1 can be obtained
from the initial [ ]and dose.
V1 = Dose / C0
= 1000 mg/100 mg/L
= 10 L
10 12 14
Graph Patient 2 Data
100
-αt
Ct = A e
-βt
+ Be
60
In this [ ] – time profile the
β-phase intercept (B) is 27.8
mg/L and the α-phase
intercept (A) is 72.2 mg/L.
-αt
-βt
Ct = 72.2 e + 28.8 e
30
10
6
AUC following an IV dose
in a 2C model can be
calculated by
standard methods
4
2
1
(trapezoidal rule + kinetic method
from the last concentration: [ ]/β)
0
2
4
6
8
10 12 14
or based on the formula
AUC 0-∞ = (A/ α) + (B/ β)
Graph Patient 2 Data
100
-αt
Ct = A e
-βt
+ Be
60
-αt
-βt
Ct = 72.2 e + 28.8 e
30
A 2-C model is the result of
a drug which has an initial
small volume of distribution
(V1) which grows following
the dose to achieve its final
<larger> volume (Vss).
10
6
4
2
1
0
2
4
6
8
The micro-constants k12 and
k21 represent the transfer
constants between the central
(C-1) and peripheral (C-2)
compartments.
10 12 14
Graph Patient 2 Data
100
-αt
Ct = A e
-βt
+ Be
60
-αt
-βt
Ct = 72.2 e + 28.8 e
30
The final <larger> volume
(Vss) is a function of the
rate into the peripheral
compartment (k12) and the
rate out of the peripheral
compartment (k21).
10
6
4
Vss = V1(1 + k12/k21)
2
1
0
2
4
6
8
10 12 14
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Zero time intercepts of
back-extrapolated β phase
and of residual (α), are
equal to A and B respectively.
Terminal Phase
Slope = 
Residual
Slope = α
AUC 0-∞ = (A/ α) + (B/ β)
-αt
Ct = A e
-βt
+ Be
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Terminal Phase
Slope = 
What is the initial
time zero concentration?
Residual
Slope = α
Zero time intercepts of
back-extrapolated β phase
and of residual (α), are
equal to A and B respectively.
C= A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
Analysis of a 2C model
A
B
Terminal Phase
Slope = 
What is the initial
time zero concentration?
Residual
Slope = α
Zero time intercepts of
back-extrapolated β phase
and of residual (α), are
equal to A and B respectively.
[ ]0 = A + B
C= A e
-αt
+ Be
-βt
Analysis of 2 Compartment Model:
IV Dosing
What does the value of 
do to the
concentration – time
profile?
k12, k21 and k10 are micro constants
α and β are more complex constants and reflect a variety of simultaneous processes.
α + β = k12 + k21 + k10
α = k21*k10/β and β2 – (k12+k21+k10)*β + k21/k10 = 0
-βt
Dose
*
(k
-α)
-αt
Dose
*
(k
-β)
21
21
C = --------------------- e
+ ------------------------ e
V * ( β – α)
V * (α - β)
Analysis of 2 Compartment Model:
IV Dosing
α = 2.0
β = 0.2
k21 = 0.25
K12 = 0.35
Here there is a 4 fold difference in α,
but what else is different?
Relative size of  & 
as well as relative size of k21&k12.
α = 0.5
β = 0.2
k21 = 0.25
K12 = 0.05
Recall … Vss = V1(1 + k12/k21)
The concentration changes rapidly.
What does this imply?
-βt
Dose
*
(k
-α)
-αt
Dose
*
(k
-β)
21
21
C = --------------------- e
+ ------------------------ e
V * ( β – α)
V * (α - β)
Analysis of 2 Compartment Model:
IV Dosing
α = 2.0
β = 0.2
k21 = 0.25
K12 = 0.35
V1 = 10
V2 = 14
α = 1.0
β = 0.2
k21 = 0.25
K12 = 0.15
V1 = 10
V2 = 6
α = 0.5
β = 0.2
k21 = 0.25
K12 = 0.05
V1 = 10
V2 = 2
-βt
Dose
*
(k
-α)
-αt
Dose
*
(k
-β)
21
21
C = --------------------- e
+ ------------------------ e
V * ( β – α)
V * (α - β)
Analysis of 2 Compartment Model:
IV Dosing
α = 2.0
β = 0.2
k21 = 0.25
K12 = 0.35
V1 = 10
V2 = 14
As  increases,
the rate of change
of concentration during the
distribution phase increases
and the size of V2 increases.
Vss = V1 + V2
α = 0.5
β = 0.2
k21 = 0.25
K12 = 0.05
V1 = 10
V2 = 2
-βt
Dose
*
(k
-α)
-αt
Dose
*
(k
-β)
21
21
C = --------------------- e
+ ------------------------ e
V * ( β – α)
V * (α - β)
Summary
Compartment Models:
Describing [ ] - Time
1 compartment
with IV bolus input
[one exponential – K]
Dose e-Kt
Ct = ---------V
Summary
Compartment Models:
Describing [ ] - Time
1 compartment
with first order absorption
[2 exponentials ka and K)
ka*F*Dose e -kt - e -kat
C = -----------------V*(ka - k)
Summary
Compartment Models:
Describing [ ] - Time
2 compartment
with IV bolus input
[2 exponentials  and )
Dose * (k21-α) e -αt + -----------------------Dose * (k21-β) e -βt
C = --------------------V * ( β – α)
V * (α - β)
Summary
Compartment Models:
Describing [ ] - Time
2 compartment
with first order absorption
[3 exponentials  ,  and ka)
kaFDose * (k21-ka) -kat
C = ---------------------- e
V * (α-ka)*(β-ka)
kaFDose *(k21-α) -αt
+ ----------------------- e
V * (ka-α)*(β-α)
kaFDose *(k21-β)
-βt
+ ----------------------- e
V * (ka-β)*(α-β)
Summary
Compartment Models:
Describing [ ] - Time
3 compartment
with IV bolus input
[3 exponentials  ,  and )
C= A*e
-αt
+ B*e
-βt
+C*e
-t