Transcript 5-absorption kinetic..

CHAPTER 7
ABSORPTION KINETICS
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ABSORPTION
GIT
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ABSORPTION FROM GIT
Oral Dosage Forms
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Advantages of Oral Drugs
 Convenient, portable, no pain
 Easy to take
 Cheap, no need for sterilization
 Compact, multi-dose bottles
 Automated machines producing
tablets in large quantities
 Variety- fast release, enteric coated,
capsules, slow release, …..
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ABSORPTION
Definition: is the net transfer
of drug from the site of
absorption into the circulating
fluids of the body.
For Oral Absorption
1- Cross the epithelium of the
GIT and entering the blood
via capillaries
2- Passing through the hepatoportal system intact into the
systemic circulation
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ABSORPTION
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Effect of Food on Drug Absorption
Propranolol
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Effect of Diseases on Drug Absorption
Diseases that cause changes in:
 Intestinal blood flow
 GI motility
 Stomach emptying time
 Gastric and intestinal pH
 Permeability of the gut wall
 Bile and digestive enzyme secretion
 Alteration of normal GI flora
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ABSORPTION KINETICS
Plasma Concentration-Time Curve
Cmax
Cp
Absorption
Elimination
Phase
Phase
Tmax
Time
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First-Order Absorption
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Absorption
Zero-Order Absorption: is seen with controlled
release dosage forms as well as with poorly soluble
drugs. The rate of input is constant.
First-Order Absorption: is seen with the majority of
extravascular administration (oral, IM, SC, rectal,
ect..) Most PK models assume first-order absorption
unless otherwise stated.
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One Compartment Model for First-Order
Absorption and First-Order Elimination
Gastrointestinal, Percutaneous, Subcutaneous,
Intramuscular, Ocular, Nasal, Pulmonary, Sublingual,…
Drug in dosage
form
Release
Drug particles
In body fluid
Dissolution
Drug in
solution
Central
Compartment
Absorption
(Plasma)
ka
kel
Elimination
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COMPARTMENTAL MODEL
One compartment model with Extravascular
Administration(dosage form parameter and
drug patient parameter)
Drug in
Dosage
GIT
ka
Central
Compartment
kel
Route of Administration: Oral, IM, SC, Rectal, ect…
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First-Order Absorption Model
Rate of change = rate of input – rate of output
dDB
 Fka DGI  k el DB
dt
dDB
 Fka D0 e  k a t  kel DB
dt
Integrated Equation:
Fka D0
 k el t
kat
Cp 
(e
e )
Vd (k a  kel )
C p  A(e
 kel t
e
 kat
)
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The Residual Method
The rising phase is not log-linear because absorption
and elimination are occurring simultaneously
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The Residual Method
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The Residual Method
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The Residual Method
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Cmax and tmax
The time needed to reach Cmax is tmax
t max
ln( k a  k el )

k a  k el
At the Cmax the rate of drug absorbed is
equal to the rate of drug eliminated
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Lag Time
The time delay prior to the commencement of
first-order drug absorption is known as lag time
Cp
Lag time
Time
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FLIP-FLOP of ka and kel
In a few cases, the kel obtained from oral
absorption data does not agree with that
obtained after i.v. bolus injection. For
example, the kel calculated after i.v. bolus
injection of a drug was 1.72 hr -1, whereas
the kel calculated after oral administration
was 0.7 hr -1. When ka was obtained by the
method of residuals, the rather surprising
result was that the ka was 1.72 hr -1
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FLIP-FLOP of ka and kel
Drugs observed to have flip-flop
characteristics are drugs with fast elimination
(kel > ka)
The chance for flip-flop of ka and kel is
greater for drugs that have a kel > 0.69 hr-1
The flip-flop problem also often arises when
evaluating controlled-release products
The only way to be certain of the estimates
is to compare the kel calculated after oral
administration with the kel from intravenous
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data.
FLIP-FLOP of ka and kel
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Effect of size of the dose of a drug on the peak
concentration and time of peak concentration
The time of peak conc is the same for all doses
A >B >C
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Effect of altering ka on Cmax and Tmax
The faster the absorption the higher is the Cmax and the
shorter is the Tmax
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Effect of altering kel on Cmax and Tmax
The faster the elimination the lower is the Cmax and the
shorter is the Tmax
ka= 0.5 hr-1
kel= 0.02 hr-1
ka= 0.5 hr-1
kel= 0.2 hr-1
Cp
ka= 0.5 hr-1
kel= 20 hr-1
Time
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Equations
C p  A (e
Fk a D0
A
Vd (k a  k el )
F .Dose
AUC 
Cl
 kel t
e
 kat
tmax
)
ln( ka / kel )

ka  kel
t1 / 2
0.693

k el
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