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

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

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

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?

“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

Absorption

“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)

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

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

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

Modified-release preparation

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

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

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

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) ??

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

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

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

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?

Protein binding

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

C normal binding = 5 mg/L (1 – 0.1)

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

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

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

Salicylate: increasing Vd with incr. dose

pH and andVd

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

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

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 ?

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

Dose was 150 mg IV... Vd = ?

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

Digoxin OD in a child

Lithium

Elimination: Can you say “haff-life”?

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

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

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)

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

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

“They reported the CLEARANCE was really good 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

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

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

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

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

What happens in OD?  Saturation of normal routes of elimination  “zero-order kinetics” First-order Half-life = 1 hour HoursLevel 00: : : :007.5 First-order Half-life = 1 hour HoursLevel 00: : : :007.5 Zero-order Elim. = 30 mg/L/hr HoursLevel 00: : : :00120 Zero-order Elim. = 30 mg/L/hr HoursLevel 00: : : :00120

Nonlinear kinetics

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?

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

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)

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

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

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

Extraction ratio = = 2/3 Cl HD = flow rate x ER = 300 x 2/3 = 200 mL/min

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

OK, what’s really the half-life? Cl t 1/2 = Vd Cl intrinsic = 1500 mL/min (90 L/hr) C total = 90 L/hr + 12 L/hr (HD) = 102 t 1/2 = 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

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

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

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