Transcript 02_One compartment IV Bolus

One Compartment
Open Model
IV bolus
Dr Mohammad Issa
1
One compartment
2
More than one compartment
3
One compartment open model

The one-compartment open model offers the simplest
way to describe the process of drug distribution and
elimination in the body.

This model assumes that the drug can enter or leave the
body (ie, the model is "open"), and the body acts like a
single, uniform compartment.

The simplest route of drug administration from a
modeling perspective is a rapid intravenous injection (IV
bolus).
4
One compartment open model

The simplest kinetic model that describes drug
disposition in the body is to consider that the
drug is injected all at once into a box, or
compartment, and that the drug distributes
instantaneously and homogenously throughout
the compartment.

Drug elimination also occurs from the
compartment immediately after injection.
5
One compartment:
6
Properties of a Pharmacokinetic
Compartment
1.
Kinetic homogeneity. A compartment contains tissues
that can be grouped according to similar kinetic
properties to the drug allowing for rapid distribution
between tissues
2.
Although tissues within a compartment are kinetically
homogeneous, drug concentrations within a
compartment may have different concentrations of drug
depending on the partitioning and binding properties of
the drug.
3.
Within each compartment, distribution is immediate and
rapidly reversible.
4.
Compartments are interconnected by first-order rate
constants. Input rate constants may be zero order
7
One compartment:
IV bolus
administration
(dose = X0)
Drug amount in the
Body (X)
Elimination process
Elimination rate
constant (K)
Based on the assumption of first order elimination process:
eliminatio n rate  K  X
8
One compartment open model
Drug Conc (C)
D  K t
C
e
Vd
Time
C= concentration
D= dose
Vd: Volume of
distribution
K: elimination rate
constant
t: time
9
How to distinguish one comp?
log (C)
Plotting log(C) vs. time yields a
straight line
Time
10
Fundamental parameters in one
compartment
Apparent Volume of Distribution (Vd)
 Elimination rate constant (K)
 Elimination half life (t1/2)
 Clearance (Cl)

11
Apparent Volume of Distribution
(Vd)
100 mg
C= 10 mg/L
C= 1 mg/L
Vd= 10 L
Vd= 100 L
12
Apparent Volume of Distribution
(Vd)

In general, drug equilibrates rapidly in the body. When
plasma or any other biologic compartment is sampled
and analyzed for drug content, the results are usually
reported in units of concentration instead of amount

Each individual tissue in the body may contain a different
concentration of drug due to differences in drug affinity
for that tissue. Therefore, the amount of drug in a given
location can be related to its concentration by a
proportionality constant that reflects the volume of fluid
the drug is dissolved in

The volume of distribution represents a volume that must
be considered in estimating the amount of drug in the
body from the concentration of drug found in the
13
sampling compartment
The real Volume of Distribution has physiological
meaning and is related to body water
Total body water 42 L
Plasma
Interstitial
fluid
Intracellular
fluid
Plasma volume 4 L
Interstitial fluid volume 10 L
Intracellular fluid volume 28 L
14
Apparent Volume of Distribution

Drugs which binds selectively to plasma proteins, e.g.
Warfarin have apparent volume of distribution smaller
than their real volume of distribution

Drugs which binds selectively to extravascular tissues,
e.g. Chloroquines have apparent volume of distribution
larger than their real volume of distribution. The Vd of
such drugs is always greater than 42 L (Total body
water)
15
Apparent Volume of Distribution
Lipid solubility of drug
 Degree of plasma protein binding
 Affinity for different tissue proteins
 Fat : lean body mass
 Disease like Congestive Heart Failure
(CHF), uremia, cirrhosis

16
Apparent Volume of Distribution

In general, drug equilibrates rapidly in the body. When
plasma or any other biologic compartment is sampled
and analyzed for drug content, the results are usually
reported in units of concentration instead of amount

Each individual tissue in the body may contain a different
concentration of drug due to differences in drug affinity
for that tissue. Therefore, the amount of drug in a given
location can be related to its concentration by a
proportionality constant that reflects the volume of fluid
the drug is dissolved in

The volume of distribution represents a volume that must
be considered in estimating the amount of drug in the
body from the concentration of drug found in the
17
sampling compartment
Apparent Volume of Distribution:
Mathematics

In order to determine the apparent volume of distribution
of a drug, it is necessary to have plasma/serum
concentration versus time data
X0
dose
Vd 

initial conc. C 0
18
Apparent volume of distribution
estimation
1.
Plot log(C) vs. time
2.
Plot the best-fit line
3.
Extrapolate to the Y-axis intercept (to estimate
initial concentration, C0)
4.
Estimate Vd:
X0
dose
Vd 

initial conc. C 0
19
1- Plot log(C) vs. time
7
Log (Conc)
6.8
6.6
6.4
6.2
6
5.8
0
1
2
3
4
5
6
Time
20
2- Plot the best-fit line
7
Log (Conc)
6.8
6.6
6.4
6.2
6
5.8
0
1
2
3
4
5
6
Time
21
3-Extrapolate to the Y-axis intercept
(to estimate C0)
7
Log (Conc)
6.8
6.6
6.4
C0
6.2
6
5.8
0
1
2
3
4
5
6
Time
22
4-Estimate Vd
7
X0
dose
Vd 

initial conc. C 0
Log (Conc)
6.8
6.6
6.4
Log(C0)
6.2
6
5.8
0
1
2
3
4
5
6
Time
23
The Extent of Distribution and Vd in
a 70 kg Normal Man
Vd, L
% Body
Weight
Extent of Distribution
5
7
Only in plasma
5-20
7-28
In extracellular fluids
20-40
28-56
In total body fluids.
>56
In deep tissues; bound to
peripheral tissues
>40
24
Elimination rate constant (K)

Elimination rate constant represents the fraction
of drug removed per unit of time

K has a unit of reciprocal of time (e.g. minute-1,
hour-1, and day-1)

With first-order elimination, the rate of
elimination is directly proportional to the serum
drug concentration
25
Elimination rate constant estimation
1.
Plot log(C) vs. time
2.
Plot the best-fit line
3.
Calculate the slope using two points on
the best-fit line
Estimate K: K   Slope  2.303
4.
26
1- Plot log(C) vs. time
6.8
6.7
Log (Conc)
6.6
6.5
6.4
6.3
6.2
6.1
6
0
1
2
3
4
5
6
Time
27
2- Plot the best-fit line
6.8
6.7
Log (Conc)
6.6
6.5
6.4
6.3
6.2
6.1
6
0
1
2
3
4
5
6
Time
28
3- Calculate the slope using two
points on the best-fit lin
6.8
log( C1 )  log( C2 )
Slope 
t1  t 2
6.7
Log (Conc)
6.6
6.5
6.4
(Log(C1), t1)
6.3
6.2
(Log(C2), t2)
6.1
6
0
1
2
3
4
5
6
Time
29
4- Estimate K
6.8
K   Slope  2.303
6.7
Log (Conc)
6.6
6.5
6.4
6.3
6.2
6.1
6
0
1
2
3
4
5
6
Time
30
Elimination half life (t1/2)

The elimination half life is sometimes
called ‘‘biological half-life’’ of a drug

The elimination half life is defined as the
time (h, min, day, etc.) at which the mass
(or amount) of unchanged drug becomes
half (or 50%) of the initial mass of drug
31
Elimination half life (t1/2) estimation

Two methods:
 From
the value of K:
t1/ 2
 Directly
0.693

K
from Conc vs. time plot
Select a concentration on the best fit line (C1)
 Look for the time that is needed to get to 50% of
C1  half-life

32
Clearance (Cl)

Clearance is a measure of the removal of drug
from the body

Plasma drug concentrations are affected by the
rate at which drug is administered, the volume in
which it distributes, and its clearance

A drug’s clearance and the volume of distribution
determine its half life
33
Clearance (Cl)

Clearance (expressed as volume/time) describes the
removal of drug from a volume of plasma in a given unit
of time (drug loss from the body)

Clearance does not indicate the amount of drug being
removed. It indicates the volume of plasma (or blood)
from which the drug is completely removed, or cleared,
in a given time period.

Figures in the following two slides represent two ways of
thinking about drug clearance:

In the first Figure, the amount of drug (the number of dots)
decreases but fills the same volume, resulting in a lower
concentration
 Another way of viewing the same decrease would be to
calculate the volume that would be drug-free if the
concentration were held constant as resented in the
34
second Figure
Clearance (Cl)
the amount of drug (the number of
dots) decreases but fills the same
volume, resulting in a lower
concentration
35
Clearance (Cl)
36
Clearance (Cl)

The most general definition of clearance is that it is ‘‘a
proportionality constant describing the relationship
between a substance’s rate of elimination (amount per
unit time) at a given time and its corresponding
concentration in an appropriate fluid at that time.’’

Clearance can also be defined as ‘‘the hypothetical
volume of blood (plasma or serum) or other biological
fluids from which the drug is totally and irreversibly
removed per unit time.’’
37
Clearance (Cl) estimation

For ALL LINEAR pharmacokinetics (including
one compartment) , clearance is calculated
using:
dose
Cl 
AUC
where AUC is the area under the concentration
curve (it will be discussed later)
38
Clearance (Cl) estimation

For One compartment pharmacokinetics ,
clearance is calculated using:
Cl  K Vd
39
Clearance (Cl)

Drugs can be cleared from the body by
different pathways, or organs, including
hepatic biotransformation and renal and
biliary excretion. Total body clearance of a
drug is the sum of all the clearances by
various mechanisms.
Clt  Clr  Clh  Clother
(Cl t , Clr and Clh  total, renal, and hepatic Cl)
40
Elimination rate

The elimination rate at any time can be
calculated using:
 Elimination
rate = K*X(t)
OR
 Elimination rate = Cl*C(t)
where

X(t) is the amount of drug in the body at time t,
 C(t) is the concntration of drug at time t
41
Area Under the Conc. Time Curve
(AUC) calculation

Two methods:
 Model
dependent: can be used only for one
compartment IV bolus
 Model independent: Can be used for any drug
with any route of administration
42
AUC calculation: Model dependent

With one compartment model, first-order
elimination, and intravenous drug
administration, the AUC can be calculated
using:
Dose C0
AUC 

K  Vd K
43
AUC calculation: Model
independent
1200
1000
800
600
400
200
0
0
2
4
6
8
10
12
44
AUC calculation: Model
independent
1200
1- Divide the area into different parts
based on the observed concentration
points (parts 1-5)
1000
800
600
1
400
2
200
3
4
5
0
0
2
4
6
8
10
12
45
AUC calculation: Model
independent
1200
2- Calculate the area for each part of the
parts 1,2,3 and 4 (until the last observed
concentration) using trapezoidal rule
1000
800
600
1
400
2
200
3
4
5
0
0
2
4
6
8
10
12
46
Trapezoidal rule
(Trapezoid = ‫)شبه المنحرف‬
C1
C2
C 2  C1
area 
 (t 2  t1 )
2
where C = concentration
t = time
t1
t2
47
AUC calculation: Model
independent
3- For part 5 (area between the last
observed concentration and infinity)
use the following equation:
C*
C*
area 
K
48
AUC calculation: Model
independent
1200
4- The total AUC (from zero to infinity) is the
sum of the areas of parts: 1,2,3,4, and 5
1000
AUC 0  AUC1  AUC 2  AUC 3  AUC 4  AUC 5
800
600
1
400
2
200
3
4
5
0
0
2
4
6
8
10
12
49
Fraction of the dose remaining

Fraction of dose remainig (F = X(t)/X0) is
given by the following equation:
 K t
Amount at time t X 0  e
F

dose
X0
e
 K t
since t1/2= 0.693/k, the equation can be
represented as:
t
t
0.693
 1  t1 / 2
t1 / 2
Fe
 
2
50
Time to get to certain conc.

Time to get to certain concentration (C*) is
given by:
C*  C 0  e
 K t
e
 C0 
K  t  ln 

 C *
 C0 
ln 

C
*


t
K
 K t
C0

C*
51
Applications of one compartment
model

Case 1: Predicting Plasma Concentrations

Case 2: Duration of Action

Case 3:Value of a Dose to Give a Desired
Initial Plasma Concentration
52
Case 1: Predicting Plasma
Concentrations

A 20-mg dose of a drug was
administered as an intravenous bolus
injection. The drug has the following
pharmacokinetic parameters: k = 0.1 h−1
and Vd = 20 L
Calculate the initial concentration (C0 )
2. Calculate the plasma concentration at 3 h
1.
53
Case 1: Predicting Plasma
Concentrations
1.
Calculate the initial concentration (C0 )
dose 20 mg
C0 

 1 mg/L
Vd
20 L
2.
Calculate the plasma conc. at 3 h
C  C0  e
 K t
 1 e
-(0.1)(3)
 0.74 mg/L
54
Case 2: Duration of Action

The duration of action of a drug may be
considered to be the length of time the
plasma concentration spends above the
MEC. Its determination is best illustrated
by example 2.
55
Case 2: Duration of Action

Continuing with the
drug used in Example
1, if the therapeutic
range is between 5
and 0.3 mg/L, how
long are the plasma
concentrations in the
therapeutic range?
56
Case 2: Duration of Action

As indicated in the diagram (previous
slide) C0 =1 mg/L. Thus, at time zero the
plasma concentration is in the therapeutic
range. The plasma concentration will
remain therapeutic until it falls to the MEC
(0.3 mg/L). At what time does this occur?
 C0 
ln 

C
*


t
K
 1 
 ln 
 / 0.1  12.0 hr
 0. 3 
57
Case 3: Value of a Dose to Give a Desired
Initial Plasma Concentration

Continuing with the drug used in Examples
1 and 2, If the initial Cp of 1 mg/L is
unsatisfactory, Calculate a dose to provide
an initial plasma concentration of 5 mg/L.
dose
dose  C0 Vd
C0 
Vd
mg
dose  5
 20 L  100 mg
L
58
Examples
59
Example 1

Ten mg metoclopramide was administered
intravenously to a 72 kg patient. The
minimum plasma concentration required to
cause significant enhancement of gastric
emptying is 50 ng/mL. The following
plasma concentrations were observed
after analysis of the specimen.
60
Example 1
Time (hr)
Conc. (ng/ml)
1
90.0
2
68.0
4
40.0
6
21.5
8
12.0
10
7.0
61
Example 1

Calculate the biological half-life of the drug
elimination (t½), the overall elimination
rate constant (K), the volume (Vd), the
coefficient of distribution and the duration
of action (td)
62
Example 1
2.5
2
log (Conc)
y = -0.1243x + 2.0832
2
R = 0.9995
1.5
1
0.5
0
0
2
4
6
time (hr)
8
10
12
63
Example 1

The elimination rate constant can be
obtained from the slope:
K  Slope  2.303
 (0.1243)  (2.303)  0.286 hr

1
Another way to calculate the slope is
using:
log(C1) - log(C2)
Slope 
t1 - t2
64
Example 1

Another way to calculate the slope (if you do not
have the ability to do regression) is using:
log(C1) - log(C2)
Slope 
t1 - t2
where (C1,t1) and (C2,t2) are two different conc.
time points

It is important to note that the first method for
calculating the slope is more accurate
65
Example 1

Calculating the slope using the second method:
log(40) - log(21.5)
Slope 
 -0.13481
4-6

Note that the value of the slope is similar to the
value estimated using the first method (-0.1243)
66
Example 1

the biological half-life of the drug
elimination (t½):
t 0.5

0.693 0.693


 2.42 hr
K
0.286
The volume of distribution (Vd):
dose
10
10
Vd 
 2.0832 
C0
10
121.12
mg 10 6 ng
L
 0.083

 3  83 L
ng/ml mg 10 ml
67
Example 1
Intercept = log (C0)
C0= 10intecept
2.5
2
log (Conc)
y = -0.1243x + 2.0832
2
R = 0.9995
1.5
1
y = -0.1243x + 2.0832
R2 = 0.9995
0.5
0
0
2
4
6
time (hr)
8
10
12
68
Example 1

the coefficient of distribution = Vd/wt
=83 L/ 72 kg= 1.15 L/kg

the duration of action (td). td is the time
needed for the conc. To get to 50 ng/ml :
 C0 
ln 

C
*


t
K
 121.12 
ln 

50



0.286
 3.1 hr
69
Example 2

An adult male patient was given the first dose of an
antibiotic at 6:00 AM. At 12:00 noon the plasma level of the
drug was measured and reported as 5 µg/ml. The drug is
known to follow the one compartment model with a half-life
of 6 hours. The recommended dosage regimen of this drug
is 250 mg q.i.d. the minimum inhibitory concentration is 3
µg/ml. Calculate the following:









Apparent volume of distribution
Expected plasma concentration at 10 AM.
Duration of action of the first dose
Total body clearance
Fraction of the dose in the body 5 hours after the injection
Total amount in the body 5 hours after the injection
Cumulative amount eliminated 5 hours after the injection
Total amount in the body immediately after injection of a second dose at
12:00 noon
Duration of action of first dose only if dose administered at 6:00 AM was 500
70
mg.
Example 2
 Elimination
K  0.693
 Initial
rate constant:
t 0.5
 0.693  0.116 hr 1
6
concentration:
The
conc. at 12:00 noon (6 hrs after the
first dose) is 5 µg/ml:
C (t  5)  C0  e
 k t
C (t  5)
5
 C0 
 0.1166  10 ug/ml
 k t
e
e
71
Example 2

Apparent volume of distribution:
C(t=6hrs)= 5 ug/ml. Since the half life is 6 hrs, C0
= 10 ug/ml.
X0
VD 

C0

250 mg
 25000 ml  25 L
-3
μg 10 mg
10

ml
μg
Expected plasma concentration at 10 AM
K  0.693 /t 0 .5  0.693 / 6  0.1155 hr -1
C(t  4)  C0  e
 K t
 6.3 μg/ml
72
Example 2

Duration of action of the first dose
C 
ln  0 
t   C *

K
 10 
ln  
 3
0.1155
 10.42 hr
Total body clearance
Cl  K  VD  2.89 L/hr

Fraction of the dose in the body 5 hours after the
injection 5
 1 6
F     0.56
2
73
Example 2

Total amount in the body 5 hours after the
injection = (0.56)(250 mg) = 140 mg

Cumulative amount eliminated 5 hours after
the injection = dose – amount in the body =
250 – 140 = 110 mg
Total amount in the body immediately after
injection of a second dose at 12:00 noon
Total amount = amount from the first dose +
amount from the second dose = 125 + 250 =
375 mg

74
Example 2

Duration of action of first dose only if dose
administered at 6:00 AM was 500 mg
t d  10.42 hr  6 hr  16.42 hrs

Note that 6 hrs (one t0.5) is needed for the
amount in the body to decline from 500 mg
to 250 mg
75
Example 3

The therapeutic range of a drug is 20-200 mg/L.
After an intravenous bolus injection of 1.0 gm
followed by regression analysis of the
concentration of the drug in plasma (in units of
mg/L) versus time (in hours), the following linear
equation was obtained
log Cp  2  0.1t

Calculate the following
 Duration of action
 Total body clearance
 Rate of elimination at
2 hours
76
Example 3

From the equation:
log Cp  2  0.1t  log( C0 )  slope  t
The following were estimated:
C0  10 2  100 mg/L
K  Slope  2.303  (0.1)  (2.303)  0.23 hr 1
VD 
X 0 1000 mg

 10 L
C 0 100 mg/L
77
Example 3

Duration of action:
C 
ln  0 
td   C * 



K
 100 
ln 

  20 
0.23
 7 hr
Total body clearance= K∙Vd=(0.23)(10) =2.3 L/hr
Rate of elimination at 2 hours:
Elimination rate = Cl*C(t=2) = 2.3*63 =145 mg/hr
log(Cp(t  2))  2  (0.1)(2)  1.8
 Cp(t  2)  101.8  63 mg/L
78
79