1. 2004-2005 Module 2 # 3 Pharmacokinetics

2. 2004-2005 how to make sense of the previous lesson or why do I have to learn this stuff?

3. 2004-2005 drug dosing calculations Concepts:target plasma concentration therapeutic window 3020100 15 12 9 6 3 0 time (hr) plasma conc. (mg/L) Cp tox Cp eff After distribution: Cp (free) approximates [D]

4. 2004-2005 Therapeutic window: Difference between the minimum effective concentrations (MEC) required for a desired response and one that produces an adverse effect. For some drugs it is small (only two- to three-fold difference) E.g. digoxin, theophylline, lidocaine, aminoglycosides, cyclosporine, anticonvulsants.

5. 2004-2005 the loading dose If we know the target plasma concentration, and the Vd, we can calculate the i.v.loading dose: L = C p.V d For the oral loading dose we have to take the fraction bioavailable into account (0-1) L = C p.V d /F

6. 2004-2005 Loading Dose (DL) = a dose of drug sufficient to produce a plasma concentration of drug that would fall within the therapeutic window after only one or very few doses over a very short interval. It is larger than the dose rate needed to maintain the concentration within the window and would produce toxic concentrations if given in repeated doses. Information needed to calculate a DL Volume of distribution, usually in L or L/Kg - from literature - For example: Goodman & Gilman’ Pharmacology, 2001, Table A-II-1 Desired concentration - from literature

7. 2004-2005 DL = target C p V d / F e.g. lidocaine t1/2 = 1-2 h Post MI arrythmias – life threatening – can’t wait 4-8 h – DL is used Another example: Digoxin for heart failure If only maintenanace dose is given it takes 10 days t1/2 = 61 h DL = 1.5 ng/ml x 580 / 0.7 = 1243 micg ~ 1 mg Can be given iv divided – 0.5 mg, aft 6-8 h 0.25 mg, aft 6h 0.125 mg and then 0.125 mg (to avoid overdigitalization and toxicity) Loading dose

8. Drug with low V d Volume of distribution (V d )

9. Drug with high V d high tissue binding

10. So now if we want an equation, C = -------- W V where C = concentration (plasma), W = weight of drug or dose of drug, and V = volume of solution or vol. of distribution =WV C or i.e D = Cp V d

11. Reference books say the V d of theophylline is 0.5 L per kg of body weight. What is the IV loading dose to give a serum theophylline concentration of 12 mg/L in a 62 kg man? =W V Cor LD = C p V d LD = (12 mg/L) (0.5 L/kg) (62 kg) = 372 mg we usually use mcg/mL (or ug/mL), which is the same thing as mg/L

12. Actually, theophylline for injection is avail- able only as Aminophylline (theophylline+ethylenediamine), which contains 80% theophylline. In this problem, we need to give a larger weight of Aminophylline to get the same weight of theophylline. 372 mg Theo 0.80 = 465 mg Amino This gives us a loading dose

13. Half-life ( t 1/2 ) time interval after which the concentration is half that at the beginning of the time interval has real-life meaning only in first-order kinetics!

14. 2004-2005 half life (t 1/ 2) the time taken for the C p to fall by half mathematically after i.v. bolus: Cp t = Cp 0 e -kt where k is the rate constant (in hrs -1 ) so: ln Cp t = ln Cp 0 -kt when Cp t / Cp 0 = 1/2 = t 1/ 2 = ln2/k = 0.693/k 108642 time (hr) iv bolus 100 10 Cp (mg/L) slope=-k el (k)

15. 2004-2005 most drugs follow first order kinetics: rate of change of drug in the body= -k el.Amount of Drug in the body. where k el (k) is the rate constant of elimination a few drugs (e.g. ethanol) follow 0 order kinetics: rate of change = k a few drugs (e.g. phenytoin) follow 1st order kinetics: at low conc. and 0 order at high conc.

16. Rate processes (mainly elimination) zero-order constant amount of drug eliminated per unit of time elimination rate constant has units of weight/time: K e = 50 mg/h, for example typical drug: ethanol

17. Rate processes (mainly elimination) first-order constant fraction or percent of drug eliminated per unit of time elimination rate constant has units of weight/time: K e = 0.25/h or 0.25 h -1, for example. typical drug: theophylline, most other drugs

18. first-order elimination following single IV bolus dose C p Time log C p

19. Special properties of first-order kinetics C p is 90% of the way to new steady-state level after change of dose in 3 t 1/2, 96% in 5 t 1/2 this is true regardless of initial concentration final concentration dose but only for first-order kinetics !

20. First-order is also known as linear kinetics: there is a linear relationship between dose and steady-state concentration C pss Dose

21. A patient’s peak serum phenobarbital level is 12.0 ug/mL at steady state on a dose of 60 mg/day. If we want a level of 20 ug/mL, what new dose should we order if we know that phenobarbital has linear kinetics? 12.0 ug/mL X = 60 mg/day 20 ug/mL X = 100 mg/day

22. If we change the dose to 90 mg/day, when can we get another plasma level to check our calculations? In adults, the half-life averages 100 hours. In 3 half-lives (300 h, or 12.5 days) we will be 90% of the way from 12.2 ug/mL to the new steady- state level (we hope it’s 20 ug/mL). This is close enough to get a level. Reschedule the patient to come back in 2 weeks for this level.

23. CpCp Time change dose level not OK level OK

24. 2004-2005 clearance defined as the volume of blood from which drug is irreversibly removed per unit time. ml/min can calculate: UV/P = excretion rate/plasma concentration at steady state: rate of excretion = rate in (the dose, M) M = rate of excretion = Cl drug. Cp

25. 2004-2005 Clearance  Clearance is the VOLUME of plasma that can be freed of drug per unit time, i.e., gives estimate of function of organs of elimination and rate of removal of drug from the body Rate of Elimination (mg/hr) CL = -------------------------------- = vol/time Concentration (mg/L)

26. 2004-2005 Clearance calculations From renal physiology: CL = (Ux * V) / Px L/h = ( mg/L * L/h) / mg/L Where: Ux = urine concentration of "x" (mg/L) V = urine flow rate (L/h) Px = Systemic venous plasma concn. of "x" (mg/L) CL = renal clearance (L/hr)

27. 2004-2005 Clearance Whole body clearance is simply the sum of all organ clearances CL(body) = CL(renal) + CL(hepatic) + CL(pulmonary) + CL(etc)

28. 2004-2005 relation between Cl, k and Vd the rate of drug removal (drug out) depends on the amount of drug in the body and the rate constant of excretion (k el, k) drug out = k*Amount of drug in the body (A) A = Cp*Vd drug out = k*Cp*Vd but drug out = Cl*Cp Cl*Cp = k*Cp*Vd therefore:Cl = k*Vd

29. 2004-2005 calculating clearance 3020100 100 80 60 40 20 0 plasma conc. (mg/L) time (hr) Cl drug =Dose AUC Cl drug = k.Vd = (0.693/t 1/ 2 ).Vd Cl total = Cl renal + Cl non-renal t1/2 = 0.693 k or k = 0.693 t1/2 t1/2 = 0.693 Vd Cl

30. 2004-2005  Relationship between CL(body) and Ke o Whole body clearance and the elimination rate constant are related, but emphasize different aspects of the same processes. o Elimination Rate constant (Ke) is the FRACTION of drug in the body that is eliminated each unit of time, e.g., "fraction per hour". o Calculate Ke from CL & Vd: If one knows the Vd of a drug and the CL (body), one can calculate the FRACTION of the drug in the body that is removed per unit time. This is the same fraction as the fraction of "plasma equivalent" that is completely cleared of drug per unit time. This fraction is the Ke. Ke = CL(body) / Vd (fraction/hour) CL(body) = Ke x Vd (volume/hour)

31. 2004-2005 the maintenance dose M = Cl drug.Cp for oral dosing M = Cl drug.Cp/F Maintenance Dose (DM) = The dose needed to maintain the concentration within the therapeutic window when given repeatedly at a constant interval

32. 2004-2005 Why clearance and Vd? clearance and volume of distribution are independent variables. they determine the half-life (Cl = k*Vd) t1/2 = 0.693 k or k = 0.693 t1/2 t1/2 = 0.693 Vd Cl

33. 2004-2005 bioavailability absolute bioavailability = F = AUC oral / AUC i.v. = a fraction between 0 and 1 151050 time (hrs) 150 100 50 0 plasma conc. (mg/L) iv po Some drugs with low "F" in humans (F <.5) - Alprenolol, 5-fluorouracil, Lidocaine, Morphine, Nifedipine, Propranolol, Salicylamide

34. 2004-2005 oral dosing M (average Cp) = Cl.Cp/F for drugs with short t 1/ 2, must dose frequently the time taken to reach steady state depends only on t 1/ 2 (= 4-5 x t 1/ 2) Css = F.dose CL.T

35. 2004-2005 number of half-livesDrugremaining 0100500 150250 225125 312.562.5 46.2531.25 53.125 15.625 Time to steady-state or elimination is independent of dosage Time required to reach steady state Depends on elimination rate Requires 5 elimination half-lives to reach 97% of steady state Requires 5 elimination half-lives to eliminate 97% of drug

36. 2004-2005 summary drug dosing should aim for the target plasma concentration. the volume of distribution is useful in calculating the loading dose the clearance is useful in calculating the maintenance dose the time to reach steady state depends only on the half life

37. 2004-2005 how can a drug have >100% bioavailability? (F>1.0)

38. 2004-2005

39. About the linear model.... It is based on the exponential function y = e x where e is the base of natural logarithms ( ln ) on your calculator, e x is or e x ln x inv

40. log C p Time ( C 1, t 1 ) ( C 2, t 2 ) C 2 = C 1 e - kt t C 1 = C 2 e kt C 1 /C 2 = e kt 2/1 = e kt e kt = 2 kt = 0.693 t = 0.693 / k

41. A patient’s acetaminophen level at 2:00 PM is 86.2 ug/mL and at 6:00 the same day is 27.8 ug/mL. If we know that k e of acetaminophen in this patient is 0.283 h -1, what was the level at 10:00 AM? log C p Time ( C 1, t 1 ) ( C 2, t 2 ) t C 1 = C 2 e kt C 1 = 86.2 e ( 0.283 ) ( 4 ) C 1 = 267.3 ug/mL

42. How did we know that k e of acetaminophen in this patient is 0.283 h -1 ? Either of the two previous equations can also be stated as k e = C1C1 C2C2 ln t 2 - t 1

43. If the patient’s acetaminophen level at 2:00 PM (1400) is 86.2 ug/mL and at 6:00 (1800) the same day is 27.8 ug/mL, k e = C1C1 C2C2 ln t 2 - t 1 k e = ln 86.2 ug/mL 27.8 ug/mL ( 1800 - 1400 ) h k e 0.283 h - 1

44. What is the half-life of acetaminophen in this patient? One last equation: t 1/2 = 0.693 k e t 1/2 = 0.693 0.283 h - 1 t 1/2 =2.45 h So

45. What good is all this? Well, we started with the bottom line and worked backward so that the importance of each equation might be clear. The next slide is the problem as it would occur in real life. Try working through it toward the bottom line, starting with information existing at the start of the case.

46. A patient is admitted following a suicide attempt in which he took an unknown quantity of acetaminophen. After appropriate gastric lavage and supportive ther- apy, he is being considered for therapy with the spe- cific antidote (N-acetylcysteine). This antidote is rec- ommended if the serum acetaminophen concentration 4 h after the ingestion is > 200 ug/mL. The overdose is reported to have occurred at 10 AM (1000); at 2:00 PM (1400) his serum acetaminophen level was 86.2 ug/mL and at 6:00 (1800) the same day it is 27.8 ug/mL. Should he receive N-acetylcysteine? Is this patient’s half-life consistent with the population average (2.5 h) or is there something unusual about his metabolism? Work it out.

47. A patient who is taking 400 mg of carbamazepine every 12 hours has a trough serum carbamazepine level of 4.2 ug/mL. Her seizure control is inadequate; if we want to achieve a serum drug level of 8.0 ug/mL, what should the new dose be? We know the drug usually has linear kinetics. Hint: we usually use trough levels with this drug. With linear kinetics, the trough levels are pro- portional to the dose just like peak levels are. 761 mg every 12 h; you would probably do this increase in two steps a week apart.

48. A patient weighs 88 pounds. He needs a loading dose of phenytoin for status epilepticus. If the average V d is 0.65 L/kg, what is the correct loading dose? (Although this drug has nonlinear kinetics, this fact only influences accumulation during repeated dosing; it has no effect on loading doses.) The therapeutic range of this drug is 10-20 ug/mL, and we would like to achieve a level of 12 ug/mL. 312 mg - Did you forget to convert pounds to kilograms?

49. A 76-kg patient has received an intravenous bolus of 400 mg of a drug with the following population averages: V d 2.3 L/kg, t 1/2 12 h, minimum effective concentration 1 ug/mL. How long will it be before another dose is needed? Assume linear kinetics. (Hint: first calculate k e, then calculate the drug concentration after the bolus dose. Then use the C 2 equation and try various values for the elapsed time to see the largest value that keeps C 2 above 1.0 ug/mL. [If you like math, you may want to solve this equation for t ; first take the logarithm of both sides and then solve.] I get about 14 hours.

50. First-order: steady state following repeated IV bolus dosing CpCp Time all peaks same all troughs same