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One Compartment Open ModelIV bolus
Dr Mohammad Issa
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One compartment
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More than one compartment
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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).
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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.
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One compartment:
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Properties of a Pharmacokinetic Compartment1. 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
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One compartment:
Drug amount in the Body (X)
IV bolus administration
(dose = X0)
Elimination process
Elimination rate constant (K)
Based on the assumption of first order elimination process:
XK raten eliminatio
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C= concentration
D= dose
Vd: Volume of distribution
K: elimination rate constant
t: time
One compartment open modelD
rug
Co
nc
(C)
Time
tKeVd
DC
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How to distinguish one comp?lo
g (
C)
Time
Plotting log(C) vs. time yields a straight line
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Fundamental parameters in one compartment Apparent Volume of Distribution (Vd) Elimination rate constant (K) Elimination half life (t1/2) Clearance (Cl)
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Apparent Volume of Distribution (Vd)
100 mg
C= 10 mg/L C= 1 mg/L
Vd= 10 L Vd= 100 L
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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 sampling compartment
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The real Volume of Distribution has physiological meaning and is related to body water
Plasma
Interstitial fluid
Total body water 42 L
Intracellular fluid
Plasma volume 4 L
Interstitial fluid volume 10 L
Intracellular fluid volume 28 L
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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)
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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
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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 sampling compartment
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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
0
0
C
X
conc. initial
doseVd
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Apparent volume of distribution estimation1. 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:
0
0
C
X
conc. initial
doseVd
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1- Plot log(C) vs. time
5.8
6
6.2
6.4
6.6
6.8
7
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
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2- Plot the best-fit line
5.8
6
6.2
6.4
6.6
6.8
7
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
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3-Extrapolate to the Y-axis intercept (to estimate C0)
5.8
6
6.2
6.4
6.6
6.8
7
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
C0
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4-Estimate Vd
5.8
6
6.2
6.4
6.6
6.8
7
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
Log(C0)
0
0
C
X
conc. initial
doseVd
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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.
>40 >56In deep tissues; bound to peripheral tissues
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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
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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
4. Estimate K: 303.2 SlopeK
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1- Plot log(C) vs. time
6
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
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2- Plot the best-fit line
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6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
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3- Calculate the slope using two points on the best-fit lin
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6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
(Log(C1), t1)
(Log(C2), t2)
21
21 )log()log(
tt
CCSlope
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4- Estimate K
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6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
0 1 2 3 4 5 6
Lo
g (
Co
nc)
Time
303.2 SlopeK
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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
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Elimination half life (t1/2) estimation
Two methods:From the value of K:
Directly 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
Kt
693.02/1
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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
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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 second Figure
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Clearance (Cl)
the amount of drug (the number of dots) decreases but fills the same volume, resulting in a lower concentration
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Clearance (Cl)
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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.’’
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Clearance (Cl) estimation
For ALL LINEAR pharmacokinetics (including one compartment) , clearance is calculated using:
where AUC is the area under the concentration curve (it will be discussed later)
AUC
doseCl
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Clearance (Cl) estimation
For One compartment pharmacokinetics , clearance is calculated using:
VdKCl
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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.
Cl) hepatic and renal,total,Cl and Cl,(Cl
ClClClCl
hrt
otherhrt
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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
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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
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AUC calculation: Model dependent
With one compartment model, first-order elimination, and intravenous drug administration, the AUC can be calculated using:
K
C
VdK
DoseAUC 0
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0
200
400
600
800
1000
1200
0 2 4 6 8 10 12
AUC calculation: Model independent
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0
200
400
600
800
1000
1200
0 2 4 6 8 10 12
AUC calculation: Model independent
1
23
4 5
1- Divide the area into different parts based on the observed concentration points (parts 1-5)
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0
200
400
600
800
1000
1200
0 2 4 6 8 10 12
AUC calculation: Model independent
1
23
4 5
2- Calculate the area for each part of the parts 1,2,3 and 4 (until the last observed concentration) using trapezoidal rule
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Trapezoidal rule(Trapezoid = المنحرف (شبه
C1
C2
t1 t2
)(2
C area 12
12 ttC
where C = concentration
t = time
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AUC calculation: Model independent
3- For part 5 (area between the last observed concentration and infinity) use the following equation:
C*K
C area
*
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0
200
400
600
800
1000
1200
0 2 4 6 8 10 12
AUC calculation: Model independent
1
23
4 5
4- The total AUC (from zero to infinity) is the sum of the areas of parts: 1,2,3,4, and 5
543210 AUCAUCAUCAUCAUC AUC
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Fraction of the dose remaining
Fraction of dose remainig (F = X(t)/X0) is given by the following equation:
since t1/2= 0.693/k, the equation can be represented as:
tKtK
ee
0
0
X
X
dose
tat timeAmount F
2/12/1
2
1F
693.0 t
t
t
t
e
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Time to get to certain conc.
Time to get to certain concentration (C*) is given by:
tKeC 0C **
C 0
Ce tK
*
Cln 0
CtK
KCt
*C
ln
0
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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
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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
1. Calculate the initial concentration (C0 )
2. Calculate the plasma concentration at 3 h
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1. Calculate the initial concentration (C0 )
2. Calculate the plasma conc. at 3 h
Case 1: Predicting Plasma Concentrations
mg/L 1L 20
mg 20
Vd
dose C0
mg/L 0.74e1 eC C (3)-(0.1)tK0
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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.
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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?
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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?
hr 0.121.0/3.0
1ln*
Cln
0
KCt
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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.
Vd
dose C0 Vd 0C dose
mg 100 L 20L
mg 5 dose
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Examples
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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.
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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
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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)
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Example 1
y = -0.1243x + 2.0832
R2 = 0.9995
0
0.5
1
1.5
2
2.5
0 2 4 6 8 10 12
time (hr)
log
(C
on
c)
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Example 1
The elimination rate constant can be obtained from the slope:
Another way to calculate the slope is using:
1hr 0.286(2.303)(0.1243)
2.303SlopeK
t2-t1
log(C2)-log(C1) Slope
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Example 1
Another way to calculate the slope (if you do not have the ability to do regression) is using:
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
t2-t1
log(C2)-log(C1) Slope
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Example 1
Calculating the slope using the second method:
Note that the value of the slope is similar to the
value estimated using the first method (-0.1243)
-0.134816-4
log(21.5)-log(40) Slope
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Example 1
the biological half-life of the drug elimination (t½):
The volume of distribution (Vd):
hr 2.420.286
0.693
K
0.693t0.5
L 83ml10
L
mg
ng10
ng/ml
mg 0.083
121.12
10
10
10
C
doseVd
3
6
2.08320
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Example 1
y = -0.1243x + 2.0832
R2 = 0.9995
y = -0.1243x + 2.0832
R2 = 0.9995
0
0.5
1
1.5
2
2.5
0 2 4 6 8 10 12
time (hr)
log
(C
on
c)
Intercept = log (C0)
C0= 10intecept
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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 :
hrKCt 1.3286.0
50121.12
ln *
Cln
0
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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 mg.
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Example 2
Elimination rate constant:
Initial concentration: The conc. at 12:00 noon (6 hrs after the
first dose) is 5 µg/ml:
1
0.5hr 0.1166
0.693t
0.693K
ug/ml 105)5(
)5(
6116.00
0
ee
tCC
eCtC
tk
tk
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Example 2
Apparent volume of distribution: C(t=6hrs)= 5 ug/ml. Since the half life is 6 hrs, C0
= 10 ug/ml.
Expected plasma concentration at 10 AM
L 25ml 25000
μg
mg10
ml
μg 10
mg 250
C
XV
3-0
0D
150 11550669306930 -. hr././t.K
μg/ml 6.3 eC4)C(t tK0
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Example 2
Duration of action of the first dose
Total body clearance
Fraction of the dose in the body 5 hours after the injection
hr 42.101155.03
10ln *
Cln
0
KCt
L/hr 89.2 DVKCl
56.02
1F
6
5
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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
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Example 2
Duration of action of first dose only if dose administered at 6:00 AM was 500 mg
Note that 6 hrs (one t0.5) is needed for the amount in the body to decline from 500 mg to 250 mg
hrs 16.42 hr 6 hr 42.10 dt
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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
Calculate the following Duration of action Total body clearance Rate of elimination at 2 hours
tCp 1.02log
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Example 3
From the equation:
The following were estimated:
tslopeCtCp )log(1.02log 0
mg/L 1001020 C
1hr 0.23(2.303)(0.1) 2.303SlopeK
L 10mg/L 100
mg 1000
C
XV
0
0D
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Example 3
Duration of action:
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
hr 723.020
100ln *
Cln
0
KCtd
mg/L 63102)Cp(t
1.8(0.1)(2)22))log(Cp(t1.8
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