1. Toxicokinetics is not rocket science Kent R. Olson, MD Medical Director, SF Division California Poison Control System

2. ... a stepwise approach to complicate simple kinetics concepts and freak out fellows

3. Dr. Bart’s blackboard fun™ presents... Kinetics for DUMMIES ! 1. Absorption 2. Distribution 2. Distribution 3. Elimination

4. Case 1  Biff says he drank “2 beers”  His serum ethanol = 0.28 gm/dL  Possible Questions:  How big is Biff?  How big are his beers?  How honest is Biff about his beers?

5. “2 beers” = ? EtOH  Assume:  Pint-sized: 500 mL each  6.8% EtOH v/v  EtOH ~ 0.7 g/mL Calculation: 1000mL x 6.8mL EtOH/100mL beer x 0.7 g/mL 1000mL x 6.8mL EtOH/100mL beer x 0.7 g/mL = 47.6 g EtOH = 47.6 g EtOH

6. Absorption

7. “First-pass effect”  Removal of drug after ingestion, by:  Enzymes in the gut wall  Uptake by the liver  Vomiting, AC, WBI, etc  Ethanol: first-pass removal ~ 6-7% Biff’s absorbed dose: down to 44.5 g (47.6 x 93.5% = 44.5)

8. EtOH FP effect modified by:  Gastric emptying time  Food  Medications (eg, Reglan, ranitidine)  Gender  Age Most rapid and complete EtOH absorption: older female empty stomach on metoclopramide Most rapid and complete EtOH absorption: older female empty stomach on metoclopramide

9. Some drugs w/ high FP effect a.k.a. “low bioavailability”  Propranolol  Cyclosporine  Morphine  Desipramine & other TCAs  Implications:  FP removal can be saturated in an OD  Greater proportion of drug will reach the systemic circulation

10. Other absorption issues:  Delayed or altered absorption  Massive OD  GI motility altered by drug effect  Anticholinergics  Opioids  Solubility  Modified-release preparations

11. Modified-release preparation

12. APAP (mg/L) Poss. Toxic Prob. Toxic hrs Serum APAP level Note: co-ingestion of Nyquil plus up to 44 g Tylenol ER Ref: Bizovi K et al: J Toxicol Clin Toxicol 1995; 33:510 Tylenol “Extended Relief” ingestion

13. Volume of Distribution (Vd)  Where the drug goes  Vd = = mg/kg / mg/L = L/kg  Total body water = 0.7 L/kg or ~ 50 L  ECF = 0.25 L/kg or about 15 L in adult  Plasma = 0.07 L/kg or ~ 5 L  For EtOH: Vd ~ 0.7 L/kg amount in body Cp

14. Vd for some common drugs Large Vd:  camphor  antidepressants  digoxin  opioids  phencyclidine  phenothiazines Small Vd:  alcohols  lithium  phenobarbital  phenytoin  salicylate  valproic acid

15. Back to Biff’s beers...  How big is Biff? If Vd = amount in body, then Cp 0.7 L/kg x Biff (kg) = 44.5 g 0.28 g/dL and Biff = 22.7 kg (50 lb) ??

16. Practice Question:  Boff ingested the contents of his mother’s old Rx of theophylline  What is the highest possible serum concentration he could achieve?  Boff weighs 80 kg  Vd theophylline 0.5 L/kg  Bottle had # 20 pills 300 mg Theo-Dur

17. Cp = dose / Vd Max dose = 20 x 300 = 6000 mg Max dose = 20 x 300 = 6000 mg Vd = 0.5 L/kg x 80 kg = 40 L Vd = 0.5 L/kg x 80 kg = 40 L Max Cp = 6000 mg = 150 mg/L Max Cp = 6000 mg = 150 mg/L 40 L 40 L

18. Try this on your own: How many vials of Digoxin-Fab would be needed to neutralize a digoxin serum concentration of 4 ng/mL? (assuming equilibrium) How many vials of Digoxin-Fab would be needed to neutralize a digoxin serum concentration of 4 ng/mL? (assuming equilibrium)  Vd = 6 L/kg  50 kg elderly woman  Each vial binds ~ 0.5 mg digoxin

19. Question:  Joe has a serum phenytoin level of 10 mg/L w/ serum albumin 4.4 gm/dL  Josette has a serum phenytoin level of 5 mg/L w/ albumin 2.2 gm/dL  What do they have in common?

20. Protein binding

22. C normal binding = C’C’ (1 – fu) + fu P ’ P normal fu = fraction unbound

23. C normal binding = 5 mg/L (1 – 0.1) + 0.1 2.2 4.4

24. C normal binding = = 9.09 mg/L 5 mg/L 0.55

25. Some drugs w/ high Pr binding  Carbamazepinefu = 0.2  Phenytoin0.1  Salicylic acid0.16  Valproic acid0.15  Warfarin0.03  Note: Pr binding can be saturated in OD, resulting in greater free fraction

26. Effect of saturated Pr binding Plasma protein bound drug Free drug Drug in tissues Plasma proteins SATURATED Free drug Drug in tissues

27. Salicylate: increasing Vd with incr. dose

28. pH and andVd

29. Salicylate is a Weak Acid ( pK a 3.5 ) TISSUES (pH 6.8) BLOOD (pH 7.4) URINE (pH variable) SH H + + S - SH SH AcidosisAlkalosis

30. Log = pKa – pH OR... OR... Remember Henderson-Hasselbalch? protonated/unprotonated = 10 pKa -pH protonated species unprotonated species

31. Question:  What is the proportion of salicylate in the non-ionized (protonated) state compared with the ionized (non- protonated) state in urine with:  pH = 3.5 ?  pH = 7.5 ?

32. Answer:  pH 3.5 Protonated / nonprotonated = 10 3.5-3.5 Salicylic acid / salicylate = 10 0 = 1 Ratio = 1:1  pH 7.5 Protonated / nonprotonated = 10 3.5-7.5 Salicylic acid / salicylate = 10 -4 Ratio = 1:10,000

33. Dose was 150 mg IV... Vd = ?

34. Cp at t = 0 ~ 7.5 mg/L Vd = dose / Cp = 150 / 7.5 = 20 L

36. Digoxin OD in a child

38. Lithium

39. Elimination: Can you say “haff-life”?

40. 2 half-lives 1 half-life Half-life = the time it takes for the Cp to drop in half Half-life = the time it takes for the Cp to drop in half

41. No. of half-livesIncrementPercent of maximum 150%50% 225%75% 312.5%87.5% 46.25%93.75% 53.125%96.875% 61.5625%98.4375% No. of half-livesIncrementPercent of maximum 150%50% 225%75% 312.5%87.5% 46.25%93.75% 53.125%96.875% 61.5625%98.4375%

42. K = slope of Slope = the proportion of drug elimination per unit time (natural log graph) Slope = the proportion of drug elimination per unit time (natural log graph)

43. What is Clearance? (Cl) VOLUME per unit TIMEcleared of the drug VOLUME per unit TIMEcleared of the drug units = mL/min or L/hr

44. Clearance calculation:  If the reported Cl is 200 mL/min, What is the Half-life? How much drug is gone after 2 hours?

45. “They reported the CLEARANCE was really good - - - 200 mL/min...”  But, Cl is expressed in mL/min... (NOT mg/min or gm/hr or tons/day)  Total drug elimination depends on drug concentration: mcg/mL x mL/min = mg/min

46. Now try again:  Cl is 200 mL/min  Drug concentration is 1000 ng/mL

47. Cl x Cp = 200 mL/min x 1000 ng/mL = 200,000 ng/min = 200 mcg/min = 0.2 milligrams/minute !

48. What is the relationship between Cl and Vd? Slope = Cl Vd Vd Slope = Cl Vd Vd Cl t 1/2 =.693 Vd 0.693 Vd

49. First-order kinetics Elimination is LINEAR when plotted on semi-log graph Elimination is LINEAR when plotted on semi-log graph a.k.a. “concentration- dependent” kinetics

50. What happens in OD?  Saturation of normal routes of elimination  “zero-order kinetics” First-order Half-life = 1 hour HoursLevel 00:0060 01:0030 02:0015 03:007.5 First-order Half-life = 1 hour HoursLevel 00:0060 01:0030 02:0015 03:007.5 Zero-order Elim. = 30 mg/L/hr HoursLevel 00:00210 01:00180 02:00150 03:00120 Zero-order Elim. = 30 mg/L/hr HoursLevel 00:00210 01:00180 02:00150 03:00120

52. Nonlinear kinetics

55. C ss = K m x dose rate V m – dose rate V m = maximum rate of metabolism K m = C p at which the rate of metabolism is ½ of maximum V m = maximum rate of metabolism K m = C p at which the rate of metabolism is ½ of maximum What happens when the dose = Vm?

56. Other factors affecting elimination rate in OD  Continued absorption from the GUT  Combined effect on Cp vs time plot makes it appear that half-life prolonged  Hepatic  Decreased hepatic blood flow  Liver damage  Renal  Oliguria due to hypotension  Acute renal failure

57. Clearance can be:  Metabolic (Cl M )  Renal (Cl R )  Lungs (Cl L )  etc (Cl etc )  and even hemodialysis (Cl HD ) and other extracorporeal methods  Total Cl = Cl M + Cl R + Cl L + Cl etc + Cl HD “Intrinsic clearance” (by the body)

58. Extracorporeal removal Blood from patient ARTERY or VEIN Return to patient Hemodialysis

59. Question:  What is the hemodialysis clearance of Endital™, a new (but not very effective) antidepressant?  Vd = 40 L/kg  Cp = 1000 ng/mL  C out = 340 ng/mL  Dialysis flow rate = 300 mL/min

60. Hemodialysis Cl Flow rates = 250-350 mL/min (w/catheter) Extraction ratio = Clearance = Flow rate x Extraction ratio C in C out C in – C out C in

61. Extraction ratio 1000 - 340 1000 = = 2/3 Cl HD = flow rate x ER = 300 x 2/3 = 200 mL/min

62. What’s the “half-life” on HD? Cl t 1/2 = 0.693 Vd = 0.693 x 40 L/kg x 100 kg / 12 L/hr = 231 hours !

63. OK, what’s really the half-life? Cl t 1/2 = 0.693 Vd Cl intrinsic = 1500 mL/min (90 L/hr) C total = 90 L/hr + 12 L/hr (HD) = 102 t 1/2 = 0.693 x 4000 L / 102 L/hr = 27.2 hours Need to use Cl total not Cl HD Need to use Cl total not Cl HD

64. Continuous Renal Replacement Therapy (or vein) (CVVH) (CVVHD) (or vein) Rate of ultrafiltrate production = up to 3 L/hr Clearance = x CpCp CuCu C ultrafiltrate C plasma Volume/time of ultrafiltrate

65. Beckmann U et al: JTCT 2001; 39:393-7

66. Estimate for Lithium  Usual renal Cl 25-35 mL/min  Hemodialysis adds 100-150 mL/min  But only for 3-4 hours at a time  Rebound between dialysis sessions  CVVH adds 20-35 mL/min  But can be provided continuously  Volume cleared ~ 50L/day vs 36 L/day w/ 4 hours of HD  No rebound