Study Guide
One Compartment Model with First Order Absorption
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University:
Michigan State University -
Course:
PHM 430 | Human Pharmacology Academic year:
2020
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Views:
308
Pages:
35
Author:
crawniadxfo
One Compartment Model with
First order Absorption
Single extravascular dose
The plasma concentration–time
profile of a large number of drugs
can be described by a onecompartment model with first-order
absorption and elimination.
Consider the concentration versus
time profile in the following slide
following a single oral dose. The time delay prior to the commencement of first-order drug
absorption is known as lag time.
Extravascular Administration
Scheme of the Model The Model
Assuming first-order absorption and firstorder elimination, the rate of change of
amount of drug (X) in the body is described
by
dX
KaXa KX
dt
Cp
KaFXo
e Kt e Kat
Vd ( Ka K )
Pharmacokinetic Parameters
Back projection or stripping technique
(Method of Residuals)
• K, Ka
• Half-lives
• AUC
• Clearance
• Volume
• Tmax and Cmax Method of Residuals AUC
KaFXo
1
1
Vd ( Ka K ) K
Ka
AUC
FXo
KVd
Cl
Vd
FXo
AUC
FXo
K . AUC tp
Cp max
2 . 303
Ka
log
( Ka K )
K
KaFXo
e Kt max e Kat max
Vd ( Ka K )
FXo Kt max
C max
Vd e
Calculation of F, bioavailability
FXooral
AUCoral
KVd
AUCIV
since KVd = Cl
Xoiv
KVd
Cl IV
Cloral
FXooral
AUCoral
Xo iv
AUC IV
FXooral
Xoiv
AUCoral AUCIV
F
Xo iv AUC oral If the two doses were equal
*
Xo oral AUC IV
AUC oral
F
AUC IV Normal or Flip-Flop??
Dr. Mutasim Al-Ghazawi
18 Sustained Release Products
Dr. Mutasim Al-Ghazawi
19
Sustained Release Products
Dr. Mutasim Al-Ghazawi
20 Pharmacokinetic parameters of oral absorption
Tmax, Cmax and AUC
These three parameters are important for evaluation of an
extravascular dosage forms and evaluation of bioequivalence of
drugs.
Tmax : indicator of how fast a drug is absorbed, if the rate of
absorption is fast, Tmax is short
Cmax represents the highest plasma concentration that can be
achieved with a single dose.
AUC: a parameter that covers the entire period of sampling and it
represents the extent of absorption and how complete is the
absorption
AUC and Cmax are dependent on the dose, as the dose increases the
AUC and Cmax increases
Pharmacokinetic parameters of oral absorption
Effect of ka and k on Cmax, tmax, and AUC
Changes in ka and k may affect t max, C max, and AUC
If the values for ka and k are reversed, then the same t max is
obtained, but the C max and AUC are different.
If the elimination rate constant is kept at 0.1 /hr and the ka changes
from 0.2 to 0.6 hr– 1 (absorption rate increases), then the tmax
becomes shorter (from 6.93 to 3.58 hr), the C max increases (from
5.00 to 6.99 µg/mL), but the AUC remains constant (100 µg hr/mL).
In contrast, when the absorption rate constant is kept at 0.3 hr– 1
and k changes from 0.1 to 0.5 hr– 1 (elimination rate increases), then
the tmax decreases (from 5.49 to 2.55 hr), the C max decreases
(from 5.77 to 2.79 µg/mL), and the AUC decreases (from 100 to 20
µg hr/mL). Dose=100 mg, F=1, Vd= 10 L
Elimination rate constant is kept constant and the ka is
increased; tmax becomes shorter, the C max increases
but the AUC remains constant. when the absorption rate constant is kept constant,
and k increases, then elimination rate increases
The tmax decreases ,the C max decreases and
the AUC decreases. Effect of K on Tmax and Cmax
F=1, Dose=500, Ka=1.0 hr-1
40
35
30
25
20
15
10
5
0
0
1
2
3
4
5
6
7
8
9
10
A
K=0.1
K=0.4
K=0.6
Effect of K on Tmax and Cmax
F=1, Dose=500, Ka=1.0 hr-1
100.0
10.0
1.0
0.1
0
1
2
3
4
5
6
A
K=0.1
K=0.4
K=0.6
7
8
9
10 Bioavailability
Bioavailability and area under plasma conc. time curve are
important factors in absorption kinetics
The completeness of absorption is of primary importance
in therapeutic situations.
The bioavailability, F, is proportional to the total area
under plasma-time curve (AUC) irrespective to its shape.
Total amount eliminated =clearance* AUC
But the Total amount eliminated is the amount absorbed,
F*dose
F*dose =clearance* AUC Pharmacokinetics 475
Dr. M. Al-Ghazawi
Case 1
A 500-mg dose of the sulfonamide sulfamethoxazole
is administered as an oral tablet to a human subject.
Eighty percent of the drug is absorbed, and the
balance is excreted unchanged in feces. The drug
distributes into an apparently homogeneous body
volume of 12 L, and has an absorption half-life of 15
min and overall elimination half-life of 12 h.
1) Calculate the following:
(i) AUC0→∞,(ii) tmax and (iii) C max.
2) Recalculate the values in Problem 1 if all
parameter values remained unchanged, but the
elimination half-life was increased to 18 h. Case 2
• A patient received a single dose of 500
mg erythromycin in the form of a tablet
that is known to have 80%
bioavailability. Calculate
• the time to reach the maximum
concentration,
• the maximum concentration,
• AUC and Clearance after this single
dose
If K is 0.2 hr-1, Ka is 1.3 hr-1, and Vd is 40
liters. Determination of elimination rate constant from urinary drug excretion data:
Approach I: The urinary excretion rate:
Urinary drug excretion data may also be used for calculation of the first-order
elimination rate constant. The rate of drug excretion after a single oral dose of
dXu
dt
drug is given by
KrX
(1)
Considering the amount of the drug in the body after oral administration is given by the
equation:
X
KaFXo
(Ka K)
e Kt
e Kat
and substituting for X in equation 1 as given by equation (2)
dXu
dt
KrKaFXo
(Ka K)
e Kt
e Kat
(2)
(3)
This equation suggests that if Ka >> K, the term e-Kat will approach zero after enough time,
while e-Kt still assumes a finite value, then equation 3 reduces to
dXu
dt
KrKaFXo
(Ka K)
e Kt
(4) dXu
dt
log
dXu
dt
KrKaFXo
(Ka
K)
log
KrKaFXo
(Ka K)
e Kt
Kt
2.303
(4)
(5)
Because the rate of urinary drug excretion, dxu/dt, cannot be
determined directly for any given time point, an average rate of
urinary drug excretion (∆X/∆t) is obtained and this value is plotted
against the midpoint of the collection period for each urine sample.
log
Xu
KrKaFXo
log
t
(Ka K)
Kt *
2.303
(5)
This means that plots of log(∆Xu/ ∆t) versus time will yield a curvilinear profile
with the terminal phase representing the elimination rate constant (slope = K/2.303) and the residual line representing the absorption rate constant (slope
= - Ka/2.303).
By contrast, drugs exhibiting flip-flop kinetics, the terminal segment of the plot
drawn between log(∆Xu/ ∆t) versus time will represent the absorption rate
constant (slope = -Ka/2.303) and the residual represent the elimination rate
constant (slope = -K/2.303). log(
ΔXu
) residual
Δt
log
KrKaFXo
Ka
(Ka K)
2.303 Approach II: Drug remaining to be excreted (Sigma Minus Method):
This represent an alternative approach to the rate approach.
It makes use of the amount of drug remaining to be excreted.
dXu
dt
Xu
KrKaFXo
(Ka K)
KrKaFXo
K
e Kt
e Kat
1
e Kt
Ke K a t
(K Ka)
Ka(K Ka)
Ka
(3)
Since the amount of the drug to be excreted at infinity may be
expressed as:
Xu
KrFXo
K
Xu
(Xu Xu)
KaeKt KeKat
Ka K
(Xu
Xu K a Kt
Xu)
e
Ka K
(Xu Xu)
Xu
K a e Kt Ke K at
Ka K
(Xu Xu)
Xu K a Kt
e
Ka K
* Potting the logarithm of the amount remaining to be eliminated ((Xu - Xu )
versus time will yield a curvilinear plot with terminal segment representing
elimination rate constant (slope = -K/2.303)
* The residual line representing the absorption rate constant
(slope = -Ka/2.303) and vice versa in flip-flop.
* Practically, collection of sufficient number of urine samples during the
absorption phase to enable a pharmacokinetics analysis of this phase is generally
not possible unless the absorption half-life is long.
A serious shortcoming for this approach is manifested by the need to collect
the urine for a long period of time (seven t1/2), and also that samples cannot be
skipped. Case 1
The following information is available for ampicillin: 90% is excreted
unchanged and a 250 mg IV bolus dose yields an AUC of 11 mic/mL*hr.
The following blood level profile has been reported for two brands of
ampicillin which were given as 500 mg oral capsules. Case 2
In a controlled, crossover bioavailability study
of oral dosage forms, formulation A,
containing 250 mg of a drug, was tested
against formulation B, containing 150 mg of
the same drug. The drug is extensively
metabolized in the body to one major
metabolite. Urinary recovery of the metabolite
was 210 mg from formulation A and 130 from
formulation B.
Calculate the ratio of systemic availability
from formulation A to availability from
formulation B. Case 3
After 400 mg of a drug administered orally, 50% is
absorbed into the circulation. The drug has a
first- order absorption rate constant of 2.0 h-1
and is eliminated from the body in equal
proportions by metabolism and by excretion in
unchanged form in urine. The overall biological
half-life is 3.46 h. If a sigma-minus plot was
constructed from the urinary excretion data,
what would be values of
(i) the rate constant from the slope of the terminal
and residual lines, and
(ii)the intercept of each of the two lines on the
ordinate ( t = 0 )?
Dose
400
Bioavailable
200
In urine
100
Not abs
200
Metabolite
100 Case 4
An oral dose of 100 mg of drug X in a solution
form was administered to a 70 Kg subject.
Assuming the administered dose is
completely absorbed, and knowing that
absorption half-life =0.2hr, the elimination
half-life =1.5hr, the y-intercept=3.9mg/ml in
the conc.time profile.
Calculate the amount of drug in body at Tmax
And if 76% of the drug is excreted unchanged
find Kr & renal clearance
Case 5
Following oral administration of a dose in solution
form, an elimination half-life of 4hrs & an absorption
half-life of 0.5 hrs were obtained by plotting the
urinary excretion rate against tmid for urine
collection. The intercept of the elimination phase
extension was 38.5 mg/hr. Assuming complete
absorption of the drug,
1- what was the administered dose if 30% of the
available dose is excreted unchanged in urine?
2- What is the renal clearance if Vd =22L
3- Find Tmax ,Cmax, and
4- If we double the dose what is the new Tmax log(
ΔXu
KrKaFXo
K
) terminal log
Δt
(Ka K)
2.303
KrKaFXo
38 . 5
(Ka K)
Ka=1.386 hr-1, K=0.1733 hr-1
Kr=0.3 * 1.1733=0.052 hr-1
Xo≈650 mg
6- A patient receives a drug dose of 400 mg orally,
the drug is 100% bioavailable, 25% is metabolized
and 75% is excreted as unchanged drug by the
kidneys. If K=0.2/hr, Ka=1.5/hr, AUC=105.8 mg.hr/L.
Find Vd, total clearance, Cmax ,Xu∞.
7- Using the sigma-minus method of constructing
urine data after the administration of 400 mg oral
dose we found that K=0.2/hr and Ka=1.5/hr, and that
75% of the dose is excreted unchanged in urine.
1-Calculate the renal clearance, (Vd = 18.9L and
F=1) and find tmax ,Cmax
2-If we change the dose to 600 mg what is the new
tmax, Cmax
One Compartment Model with First Order Absorption
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