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Site of
Action
Dosage Effects
Plasma
Concen.
Pharmacokinetics Pharmacodynamics
Dose
PlasmaConcentration
0 1 2 3 4 5 6 7 8 9
0
2
4
6
8
10
12
TOXIC RANGE
THERAPEUTIC RANGE
SUB-THERAPEUTIC
DISPOSITION OF DRUGS
The disposition of chemicals entering the body (from C.D. Klaassen, Casarett and Doull’s Toxicology, 5th
ed., New York: McGraw-Hill, 1996).
Bound Free Free Bound
LOCUS OF ACTION
“RECEPTORS”
TISSUE
RESERVOIRS
SYSTEMIC
CIRCULATION
Free Drug
Bound Drug
ABSORPTION EXCRETION
BIOTRANSFORMATION
Plasma concentration vs. time profile of a
single dose of a drug ingested orally
0
2
4
6
8
10
12
14
0 5 10 15 20
TIME (hours)
PlasmaConcentration
Dose
PlasmaConcentration
0 1 2 3 4 5 6 7 8 9
0
2
4
6
8
10
12
TOXIC RANGE
THERAPEUTIC RANGE
SUB-THERAPEUTIC
Bound Free Free Bound
LOCUS OF ACTION
“RECEPTORS”
TISSUE
RESERVOIRS
SYSTEMIC
CIRCULATION
Free Drug
Bound Drug
ABSORPTION EXCRETION
BIOTRANSFORMATION
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 to
destruction in gastric acid and liver metabolism
First Pass Effect
Bioavailability
Dose
Destroyed
in gut
Not
absorbed
Destroyed
by gut wall
Destroyed
by liver
to
systemic
circulation
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 a
therapeutic effect by one route but a
toxic or no effect by another.
Drugs appear to distribute in the body as if it
were a single compartment. The magnitude of
the drug’s distribution is given by the apparent
volume 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
Pharmacokinetics
Drug L/Kg L/70 kg
Sulfisoxazole 0.16 11.2
Phenytoin 0.63 44.1
Phenobarbital 0.55 38.5
Diazepam 2.4 168
Digoxin 7 490
Examples of apparent Vd’s for
some drugs
K I D N E Y
f ilt r a t io n
s e c r e t io n
( r e a b s o r p t i o n )
L I V E R
m e t a b o lis m
s e c r e t io n
L U N G S
e x h a la t io n
O T H E R S
m o t h e r 's m ilk
s w e a t, s a liv a e t c .
E l i m i n a t i o n
o f d r u g s f r o m t h e b o d y
M
A
J
O
R
M
I
N
O
R
Elimination by the Kidney
• Excretion - major
1) 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
Elimination by the Liver
• Metabolism - major
1) Phase I and II reactions
2) Function: change a lipid soluble to more
water soluble molecule to excrete in kidney
3) Possibility of active metabolites with
same or different properties as parent
molecule
• Biliary Secretion – active transport, 4 categories
The enterohepatic shunt
Portal circulation
Liver
gall bladder
Gut
Bile
duct
Drug
Biotransformation;
glucuronide
produced
Bile formation
Hydrolysis by
beta glucuronidase
Dose
PlasmaConcentration
0 1 2 3 4 5 6 7 8 9
0
2
4
6
8
10
12
TOXIC RANGE
THERAPEUTIC RANGE
SUB-THERAPEUTIC
0
2
4
6
8
10
12
14
0 5 10 15 20
TIME (hours)
Plasmaconcentration
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
Elimination
• Zero order: constant rate of elimination
irrespective of plasma concentration.
• First order: rate of elimination proportional to
plasma concentration. Constant Fraction of drug
eliminated per unit time.
Rate of elimination ∝ Amount
Rate of elimination = K x Amount
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 must
ethanol be taken now?
at 10 ml/h of pure ethanol, or 20 ml/h of drink.
Rarely Done DRUNKENNES Coma Death
0
2
4
6
8
10
12
14
0 5 10 15 20
TIME (hours)
Plasmaconcentration
Ct = C0 . e – Kel •t
lnCt = lnC0 – Kel • t
logCt = logC0 – Kel •t
2.3
DC/dt = – k•C
y = b – a.x
First Order Elimination
dA/dt A∝ DA/dt = – k•A
Time
PlasmaConcentration
0 1 2 3 4 5 6
1
10
100
1000
10000
First Order Elimination
logCt = logC0 - Kel . t
2.303
Time
PlasmaConcentration
0 1 2 3 4 5 6
1
10
100
1000
10000
C0
Distribution equilibrium
Elimination only
Distribution and Elimination
Plasma Concentration Profile
after a Single I.V. Injection
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
lnCt = lnC0 – Kel.t
When Ct = ½ 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
PRINCIPLEPRINCIPLE
Elimination of drugs from the
body usually follows first order
kinetics with a characteristic
half-life (t1/2) and fractional rate
constant (Kel).
First Order Elimination
• Clearance: volume of plasma cleared of drug
per 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.
IN
OUT
Blood Flow = Q
CA CV
BLOOD
B
LO
O
D
ELIMINATED
Rate of Elimination = QCA – QCV = Q(CA-CV)
Liver Clearance = Q(CA-CV)/CA = Q x EF
SIMILARLY FOR
OTHER ORGANS
Renal Clearance = Ux•V/Px
Total Body Clearance = CLliver + CLkidney + CLlungs + CLx
•
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
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
Multiple dosing
• On continuous steady administration of a drug,
plasma concentration will rise fast at first then more
slowly and reach a plateau, where:
rate of administration = rate of elimination
ie. 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.
Constant Rate of Administration (i.v.)
0
1
2
3
4
5
6
7
0 5 10 15 20 25 30
Time
PlasmaConcentration
Repeated doses –
Maintenance dose
Therapeutic
level
Single dose –
Loading dose
Concentration due to a single dose
Concentration due to
repeated doses
The time to reach steady
state is ~4 t1/2’s
Pharmacokinetic parameters
• Volume of distribution Vd = DOSE / C0
• Plasma clearance Cl = Kel .Vd
• 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
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
0
10
20
30
40
50
60
70
0 2 4 6 8 10
Plasmaconcentration
Time (hours)
Bioavailability (AUC)o
(AUC)iv
=
i.v. route
oral route
Daily Dose (mg/kg)
PlasmaDrug
Concentration(mg/L)
0 5 10 15
0
10
20
30
40
50
60
Variability in Pharmacokinetics
PRINCIPLEPRINCIPLE
The absorption, distribution and
elimination of a drug are qualitatively
similar in all individuals. However, for
several reasons, the quantitative aspects
may differ considerably. Each person
must be considered individually and
doses adjusted accordingly.

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Pharmacokinetics

  • 2. Dose PlasmaConcentration 0 1 2 3 4 5 6 7 8 9 0 2 4 6 8 10 12 TOXIC RANGE THERAPEUTIC RANGE SUB-THERAPEUTIC
  • 3. DISPOSITION OF DRUGS The disposition of chemicals entering the body (from C.D. Klaassen, Casarett and Doull’s Toxicology, 5th ed., New York: McGraw-Hill, 1996).
  • 4. Bound Free Free Bound LOCUS OF ACTION “RECEPTORS” TISSUE RESERVOIRS SYSTEMIC CIRCULATION Free Drug Bound Drug ABSORPTION EXCRETION BIOTRANSFORMATION
  • 5. Plasma concentration vs. time profile of a single dose of a drug ingested orally 0 2 4 6 8 10 12 14 0 5 10 15 20 TIME (hours) PlasmaConcentration
  • 6. Dose PlasmaConcentration 0 1 2 3 4 5 6 7 8 9 0 2 4 6 8 10 12 TOXIC RANGE THERAPEUTIC RANGE SUB-THERAPEUTIC
  • 7. Bound Free Free Bound LOCUS OF ACTION “RECEPTORS” TISSUE RESERVOIRS SYSTEMIC CIRCULATION Free Drug Bound Drug ABSORPTION EXCRETION BIOTRANSFORMATION
  • 8. 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 to destruction in gastric acid and liver metabolism First Pass Effect
  • 9. Bioavailability Dose Destroyed in gut Not absorbed Destroyed by gut wall Destroyed by liver to systemic circulation
  • 10. 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 a therapeutic effect by one route but a toxic or no effect by another.
  • 11. Drugs appear to distribute in the body as if it were a single compartment. The magnitude of the drug’s distribution is given by the apparent volume 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
  • 13. Drug L/Kg L/70 kg Sulfisoxazole 0.16 11.2 Phenytoin 0.63 44.1 Phenobarbital 0.55 38.5 Diazepam 2.4 168 Digoxin 7 490 Examples of apparent Vd’s for some drugs
  • 14. K I D N E Y f ilt r a t io n s e c r e t io n ( r e a b s o r p t i o n ) L I V E R m e t a b o lis m s e c r e t io n L U N G S e x h a la t io n O T H E R S m o t h e r 's m ilk s w e a t, s a liv a e t c . E l i m i n a t i o n o f d r u g s f r o m t h e b o d y M A J O R M I N O R
  • 15. Elimination by the Kidney • Excretion - major 1) 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
  • 16. Elimination by the Liver • Metabolism - major 1) Phase I and II reactions 2) Function: change a lipid soluble to more water soluble molecule to excrete in kidney 3) Possibility of active metabolites with same or different properties as parent molecule • Biliary Secretion – active transport, 4 categories
  • 17. The enterohepatic shunt Portal circulation Liver gall bladder Gut Bile duct Drug Biotransformation; glucuronide produced Bile formation Hydrolysis by beta glucuronidase
  • 18. Dose PlasmaConcentration 0 1 2 3 4 5 6 7 8 9 0 2 4 6 8 10 12 TOXIC RANGE THERAPEUTIC RANGE SUB-THERAPEUTIC
  • 19. 0 2 4 6 8 10 12 14 0 5 10 15 20 TIME (hours) Plasmaconcentration 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
  • 20. Elimination • Zero order: constant rate of elimination irrespective of plasma concentration. • First order: rate of elimination proportional to plasma concentration. Constant Fraction of drug eliminated per unit time. Rate of elimination ∝ Amount Rate of elimination = K x Amount
  • 21. 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 must ethanol be taken now? at 10 ml/h of pure ethanol, or 20 ml/h of drink. Rarely Done DRUNKENNES Coma Death
  • 22. 0 2 4 6 8 10 12 14 0 5 10 15 20 TIME (hours) Plasmaconcentration Ct = C0 . e – Kel •t lnCt = lnC0 – Kel • t logCt = logC0 – Kel •t 2.3 DC/dt = – k•C y = b – a.x First Order Elimination dA/dt A∝ DA/dt = – k•A
  • 23. Time PlasmaConcentration 0 1 2 3 4 5 6 1 10 100 1000 10000 First Order Elimination logCt = logC0 - Kel . t 2.303
  • 24. Time PlasmaConcentration 0 1 2 3 4 5 6 1 10 100 1000 10000 C0 Distribution equilibrium Elimination only Distribution and Elimination Plasma Concentration Profile after a Single I.V. Injection
  • 25. 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
  • 26. lnCt = lnC0 – Kel.t When Ct = ½ 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
  • 27. PRINCIPLEPRINCIPLE Elimination of drugs from the body usually follows first order kinetics with a characteristic half-life (t1/2) and fractional rate constant (Kel).
  • 28. First Order Elimination • Clearance: volume of plasma cleared of drug per 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.
  • 29. IN OUT Blood Flow = Q CA CV BLOOD B LO O D ELIMINATED Rate of Elimination = QCA – QCV = Q(CA-CV) Liver Clearance = Q(CA-CV)/CA = Q x EF SIMILARLY FOR OTHER ORGANS Renal Clearance = Ux•V/Px Total Body Clearance = CLliver + CLkidney + CLlungs + CLx •
  • 30. 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
  • 31. 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
  • 32. Multiple dosing • On continuous steady administration of a drug, plasma concentration will rise fast at first then more slowly and reach a plateau, where: rate of administration = rate of elimination ie. 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.
  • 33. Constant Rate of Administration (i.v.)
  • 34. 0 1 2 3 4 5 6 7 0 5 10 15 20 25 30 Time PlasmaConcentration Repeated doses – Maintenance dose Therapeutic level Single dose – Loading dose
  • 35. Concentration due to a single dose Concentration due to repeated doses The time to reach steady state is ~4 t1/2’s
  • 36. Pharmacokinetic parameters • Volume of distribution Vd = DOSE / C0 • Plasma clearance Cl = Kel .Vd • 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
  • 37. 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
  • 38. 0 10 20 30 40 50 60 70 0 2 4 6 8 10 Plasmaconcentration Time (hours) Bioavailability (AUC)o (AUC)iv = i.v. route oral route
  • 39. Daily Dose (mg/kg) PlasmaDrug Concentration(mg/L) 0 5 10 15 0 10 20 30 40 50 60 Variability in Pharmacokinetics
  • 40. PRINCIPLEPRINCIPLE The absorption, distribution and elimination of a drug are qualitatively similar in all individuals. However, for several reasons, the quantitative aspects may differ considerably. Each person must be considered individually and doses adjusted accordingly.