Download Absorption

PK-Fourth
ou t year
yea
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
ao
one-compartment
e co pa t e t model
ode with
t first-order
st o de
absorption and elimination.
Consider the concentration versus time
profile in the following slide following a
single oral dose.
١
Extravascular Administration
Scheme of the Model
٢
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 =
٣
[
KaFXo
e − Kt − e − Kat
Vd ( Ka − K )
]
Pharmacokinetic Parameters
Back projection or stripping technique
(Method of Residuals)
• K,
K Ka
• Half-lives
• AUC
• Clearance
• Volume
• Tmax and Cmax
Method of Residuals
٤
٥
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
AUC =
KaFXo
1 ⎤
⎡ 1
−
Vd ( Ka − K ) ⎢⎣ K
Ka ⎥⎦
AUC =
FXo
KVd
Cl =
FXo
AUC
FXo
Vd =
K . AUC
٦
]
٧
Normal or Flip-Flop??
Pharmacokinetic parameters of oral absorption
Tmax, Cmax and AUC
¾ These three parameter are important for evaluation of an
extravascular dosage forms and evaluation of bioequivalence of
drugs.
d
¾ Tmax : indicator of how fast a drug is absorbed, if the rate of
absorption is fast, Tmax is short
¾ Cmax represents the highest plasma concentration that can be
achieved with a single dose.
¾ AUC: a parameter that covers the entire period of sampling and it
represents the extent of absorption and how complete is the
absorption
¾ AUC and Cmax are dependent on the dose, as the dose increases the
AUC and Cmax increases
٨
Pharmacokinetic parameters of oral absorption
¾ Effect of ka and k on Cmax, tmax, and AUC
¾ Changes in ka and k may affect t max, C max, and AUC
¾ If the values for ka and k are reversed, then the same t max is
obtained but the C max and AUC are different
obtained,
different.
¾ 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).
¾ In contrast,
contrast when the absorption rate constant is kept at 0.3
0 3 hr
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).
٩
Elimination rate constant is kept constant and the ka is
increased; tmax becomes shorter, the C max increases
but the AUC remains constant.
١٠
¾when the absorption rate constant is kept constant,
and k increases, then elimination rate increases
¾The tmax decreases ,the C max decreases and
th AUC decreases.
the
d
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
A
K=0.1
١١
K=0.4
K=0.6
7
8
9
10
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
١٢
K=0.4
K=0.6
7
8
9
10
١٣
Bioavailability
¾ Bioavailability and area under plasma conc. time curve are
important factors in absorption kinetics
¾ The completeness of absorption is of primary importance
in therapeutic situations.
¾ 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
١٤
F*dose =clearance* AUC
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=
١٥
AUC IV
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.
١٦
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.
١٧
١٨
١٩
٢٠
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
٢١
=
KrKaFXo
(Ka − K)
[e − Kt ]
(4)
dXu
dt
log
dXu
dt
=
KrKaFXo
− K)
(Ka
= 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
Δ Xu
KrKaFXo
= log
Δt
(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
d
drawn
b
between llog(∆Xu/
(∆X / ∆t)
∆ ) versus time
i
will
ill represent the
h absorption
b
i rate
constant (slope = -Ka/2.303) and the residual represent the elimination rate
constant (slope = -K/2.303).
٢٢
log(
ΔXu
) residual
Δt
= log
KrKaFXo
(Ka − K)
−
Ka
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
Since the amount of the drug to be excreted at
infinity may be expressed as:
Xu ∞ =
KrFXo
K
[
Xu∞
(Xu − Xu)=
Kae−Kt − Ke−Kat
Ka− K
∞
(Xu
٢٣
∞
]
[ ]
Xu ∞ K a − Kt
− Xu) =
e
Ka − K
]
(3)
⎤
⎥
⎥⎦
(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.
y collection of sufficient number of urine samples
p
during
g the
* Practically,
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.
٢٤
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.
٢٥
٢٦
Case 2
In a controlled, crossover bioavailability study
of oral dosage forms, formulation A,
containing 250 mg of a drug, was tested
against
i
fformulation
l i
B
B, containing
i i
1 0 mg off
150
the same drug. The drug is extensively
metabolized in the body to one major
metabolite. Urinary recovery of the metabolite
was 210 mg
g 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
d is
i eliminated
li i t d ffrom th
the b
body
d iin equall
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 )?
٢٧
Dose
400
Bioavailable
Not abs
200
200
In urine
100
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
g 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
٢٨
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
y excretion rate against
g
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 iff Vd =22L
3- Find Tmax ,Cmax, and
4- If we double the dose what is the new Tmax
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/ml.
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
urine.
1-Calculate the renal clearance, (Vd = 18.9L and
F=1) and find Tmax ,Cmax
2-If we changed the dose to 600 mg what is the new
Tmax ,Cmax
٢٩