pharmacokinetics new2

8/8/2019 Pharmacokinetics New2

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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.

pharmacokinetics new2

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