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Dose
PlasmaC
oncentration
0 1 2 3 4 5 6 7 8 90
2
4
6
8
10
12
TOXIC RANGE
THERAPEUTIC RANGE
SUB-THERAPEUTIC
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DISPOSITION OF DRUGS
The disposition of chemicals entering the body (from C.D. Klaassen, Casarett and DoullsToxicology, 5th ed., New York: McGraw-Hill, 1996).
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Bound Free Free Bound
LOCUS OF ACTION
RECEPTORS
TISSUE
RESERVOIRS
SYSTEMICCIRCULATION
Free Drug
Bound Drug
ABSORPTION EXCRETION
BIOTRANSFORMATION
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Dose
PlasmaC
oncentration
0 1 2 3 4 5 6 7 8 90
2
4
6
8
10
12
TOXIC RANGE
THERAPEUTIC RANGE
SUB-THERAPEUTIC
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Bound Free Free Bound
LOCUS OF ACTION
RECEPTORS
TISSUE
RESERVOIRS
SYSTEMICCIRCULATION
Free Drug
Bound Drug
ABSORPTION EXCRETION
BIOTRANSFORMATION
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Bioavailability
Definition: the fraction of the administered
dose reaching the systemic circulation
for i.v.: 100%
for non i.v.: ranges from 0 to 100%
e.g. lidocaine bioavailability 35% due todestruction in gastric acid and liver metabolism
First Pass Effect
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Bioavailability
Dose
Destroyed
in gut
Not
absorbed
Destroyed
by gut wall
Destroyed
by liver
to
systemic
circulation
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PRINCIPLEPRINCIPLE
For drugs taken by routes other than the
i.v. route, the extent of absorption and
the bioavailability must be understood in
order to determine what dose will induce
the desired therapeutic effect. It will also
explain why the same dose may cause atherapeutic effect by one route but a
toxic or no effect by another.
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Drugs appear to distribute in the body as if itwere a single compartment. The magnitude of
the drugs distribution is given by the apparentvolume of distribution (Vd).
Vd = Amount of drug in body Concentration in Plasma
PRINCIPLEPRINCIPLE
(Apparent) Volume of Distribution:
Volume into which a drug appears to distribute with
a concentration equal to its plasma concentration
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Drug L/Kg L/70 kg
Sulfisoxazole 0.16 11.2Phenytoin 0.63 44.1
Phenobarbital 0.55 38.5
Diazepam 2.4 168
Digoxin 7 490
Examples of apparent Vds for
some drugs
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K I D N E
f i l t r a t i os e c r e t i( r e a b s o r p t i
L I V E R
m e t a b os e c r e t i
L U N Ge x h a l a
O T H Em o t h e r '
s w e a t , s
E l i m i n a t i o
o f d r u g s f r o
M
A
J
O
R
MI
N
O
R
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Elimination by the Kidney
Excretion - major1) glomerular filtration
glomerular structure, size constraints,protein binding
2) tubular reabsorption/secretion
-acidification/alkalinization,
- active transport, competitive/saturable,
organic acids/bases -protein binding
Metabolism - minor
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Elimination by the Liver
Metabolism - major1) Phase I and II reactions
2) Function: change a lipid soluble to morewater soluble molecule to excrete in kidney
3) Possibility of active metabolites withsame or different properties as parent
molecule
Biliary Secretion active transport, 4 categories
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The enterohepatic shunt
Portal circulation
Liver
gall bladder
Gut
Bile
duct
Drug
Biotransformation;
glucuronide
produced
Bile formation
Hydrolysis by
beta glucuronidase
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Dose
PlasmaC
oncentration
0 1 2 3 4 5 6 7 8 90
2
4
6
8
10
12
TOXIC RANGE
THERAPEUTIC RANGE
SUB-THERAPEUTIC
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0
2
4
6
8
10
12
14
0 5 10 15 20
TIME (hours)
Plasm
aco
ncentratio
n
Influence of Variations in Relative Rates of
Absorption and Elimination on Plasma
Concentration of an Orally Administered Drug
Ka/Ke=10
Ka/Ke=0.1
Ka/Ke=0.01
Ka/Ke=1
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Elimination
Zero order: constant rate of eliminationirrespective of plasma concentration.
First order: rate of elimination proportional toplasma concentration. ConstantFraction of drugeliminated per unit time.
Rate of elimination Amount
Rate of elimination = Kx Amount
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Zero Order Elimination
Pharmacokinetics of Ethanol Ethanol is distributed in total body water.
Mild intoxication at 1 mg/ml in plasma.
How much should be ingested to reach it?Answer: 42 g or 56 ml of pure ethanol (VdxC)
Or 120 ml of a strong alcoholic drink like whiskey
Ethanol has a constant elimination rate = 10 ml/h
To maintain mild intoxication, at what rate mustethanol be taken now?
at 10 ml/h of pure ethanol, or 20 ml/h of drink.
Rarely Done DRUNKENNES Coma Death
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Time
Plasm
aC
oncentratio
n
0 1 2 3 4 5 61
10
100
1000
10000
First Order Elimination
logCt = logC0 - Kel . t
2.303
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lnCt = lnCo Kel .t
When t = 0, C = C0, i.e., the concentration at
time zero when distribution is complete and
elimination has not started yet. Use this value
and the dose to calculate Vd.
Vd = Dose/C0
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lnCt = lnC0 Kel .t
When Ct = C0, then Kel .t = 0.693. This is the
time for the plasma concentration to reach half
the original, i.e., the half-life of elimination.
t1/2 = 0.693/Kel
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PRINCIPLEPRINCIPLE
Elimination of drugs from the
body usually follows first order
kinetics with a characteristic
half-life (t1/2) and fractional rateconstant (Kel).
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First Order Elimination
Clearance:volume of plasma cleared of drugper unit time.
Clearance = Rate of elimination plasma conc. Half-life of elimination:time for plasma
conc. to decrease by half.
Useful in estimating:
- time to reach steady state concentration.- time for plasma concentration to fall after
dosing is stopped.
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IN
OUT
Blood Flow = Q
CA CV
BLOOD
BLO
OD
ELIMINATED
Rate of Elimination = QCA QCV = Q(CA-CV)
Liver Clearance = Q(CA-CV)/CA = Q x EFSIMILARLY FOR
OTHER ORGANS
Renal Clearance = UxV/Px
Total Body Clearance = CLliver
+ CLkidney
+ CLlungs
+ CLx
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Rate of elimination = Kel x Amount in body
Rate of elimination = CL x Plasma Concentration
Therefore,
Kel x Amount = CL x Concentration
Kel = CL/Vd
0.693/t1/2 = CL/Vd
t1/2 = 0.693 x Vd/CL
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PRINCIPLEPRINCIPLE
The half-life of elimination of a drug (and
its residence in the body) depends on its
clearance and its volume of distribution
t1/2 is proportional to Vd
t1/2 is inversely proportional to CL
t1/2 = 0.693 x Vd/CL
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Multiple dosing
On continuous steady administration of a drug, plasmaconcentration will rise fast at first then more slowly andreach a plateau, where:
rate ofadministration = rate of eliminationie.steady state is reached.
Therefore, at steady state:
Dose (Rate of Administration) = clearance x plasma conc.
Or
If you aim at a target plasma level and you know the
clearance, you can calculate the dose required.
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Constant Rate of Administration (i.v.)
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0
1
2
3
4
5
6
7
0 5 10 15 20 25 30
Time
Plasm
aConcen
tration
Repeated doses
Maintenance dose
Therapeutic
level
Single dose
Loading dose
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Concentration due to a single dose
Concentration due to
repeated doses
The time to reach steady
state is ~4 t1/2s
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Pharmacokinetic parameters
Volume of distribution Vd= DOSE / C
0
Plasma clearance Cl = Kel.V
d
plasma half-life t1/2= 0.693 / Kel
Bioavailability (AUC)x/ (AUC)
iv
Get equation of regression line; from it get Kel , C0 , and AUC
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But C x dt = small area under the curve. For total
amount eliminated (which is the total given, or the
dose, if i.v.), add all the small areas = AUC.
Dose = CL x AUC and Dose x F = CL x AUC
dC/dt = CL x C
dC = CL x C x dt
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0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
0 2 4 6 8 1 0
Plasm
aco
nce
ntrati
on
Time (hours)
Bioavailability (AUC)o(AUC)iv
=
i.v. route
oral route
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Daily Dose (mg/kg)
PlasmaD
rug
Concentrat io
n(m
g/L
)
0 5 10 150
10
20
30
40
50
60
Variability in Pharmacokinetics
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PRINCIPLEPRINCIPLE
The absorption, distribution and
elimination of a drug are qualitativelysimilar in all individuals. However, for
several reasons, the quantitative aspects
may differ considerably. Each person
must be considered individually and
doses adjusted accordingly.