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- 1. Site of Action Dosage Effects Plasma Concen. PharmacokineticsPharmacodynamics
- 2. Dose Plasma Concentration 0 1 23 4 5 6 7 8 9 0 2 4 6 8 10 12 TOXIC RANGE THERAPEUTIC RANGE SUB-THERAPEUTIC
- 3. DISPOSITION OF DRUGS Thedisposition of chemicals entering the body (from C.D. Klaassen, Casarett and Doull’s Toxicology, 5th ed., New York: McGraw-Hill, 1996).
- 4. Bound Free FreeBound 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) Plasma Concentration
- 6. Dose Plasma Concentration 0 1 23 4 5 6 7 8 9 0 2 4 6 8 10 12 TOXIC RANGE THERAPEUTIC RANGE SUB-THERAPEUTIC
- 7. Bound Free FreeBound LOCUS OF ACTION “RECEPTORS” TISSUE RESERVOIRS SYSTEMIC CIRCULATION Free Drug Bound Drug ABSORPTION EXCRETION BIOTRANSFORMATION
- 8. Bioavailability Definition: the fractionof 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 gutwall Destroyed by liver to systemic circulation
- 10. PRINCIPLE PRINCIPLE For drugs takenby 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 todistribute 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 PRINCIPLE PRINCIPLE (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/70kg 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. KIDNEY filtration secretion (reabsorption) LIVER m etabolism secretion LUNGS exhalation OTHERS m other'sm ilk sweat, saliva etc. Elimination of drugs from the body M A J O R M I N O R
- 15. Elimination by theKidney • 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 theLiver • 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 Portalcirculation Liver gall bladder Gut Bile duct Drug Biotransformation; glucuronide produced Bile formation Hydrolysis by beta glucuronidase
- 18. Dose Plasma Concentration 0 1 23 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 1015 20 TIME (hours) Plasma concentration 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 Pharmacokineticsof 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 1015 20 TIME (hours) Plasma concentration 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 Plasma Concentration 0 1 23 4 5 6 1 10 100 1000 10000 First Order Elimination logCt = logC0 - Kel . t 2.303
- 24. Time P la sm a C o nc e ntra tio n 0 1 23 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. PRINCIPLE PRINCIPLE Elimination of drugsfrom 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 L O 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. PRINCIPLE PRINCIPLE The half-life ofelimination 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 • Oncontinuous 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 ofAdministration (i.v.)
- 34. 0 1 2 3 4 5 6 7 0 5 1015 20 25 30 Time Plasma Concentration Repeated doses – Maintenance dose Therapeutic level Single dose – Loading dose
- 35. Concentration due toa single dose Concentration due to repeated doses The time to reach steady state is ~4 t1/2’s
- 36. Pharmacokinetic parameters • Volumeof 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 xdt = 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 46 8 10 Plasma concentration Time (hours) Bioavailability (AUC)o (AUC)iv = i.v. route oral route
- 39. Daily Dose (mg/kg) Plasma Drug Concentration (mg/L) 05 10 15 0 10 20 30 40 50 60 Variability in Pharmacokinetics
- 40. PRINCIPLE PRINCIPLE The absorption, distributionand 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.