Clearance Determinations Arthur G Roberts Routes of Elimination

Clearance Determinations Arthur G. Roberts

Routes of Elimination

Zero versus First-Order Elimination

Zero Order

First Order

1 st Order Elimination rate of renal excretion = Krenal * [Drug] rate of metabolism = Kmetabolism * [Drug] rate of biliary excretion = Kbiliary * [Drug] rate of other = Kother * [Drug] Overall elimination = sum of all the rates Overall elimination = (Krenal+Kmetabolism+Kbiliary+Kother)*[Drug]

Clearance, Elimination and Excretion • Clearance (Renal) (V/time, L/hr) • Elimination (mass) (a. k. a. Extraction) • Mechanisms for either – biotransformation (e. g. Clint) – excretion

Clearance • efficiency for kidney or liver removal • L/hour • rapid or slow? – rel. large clearance and small volume distribution – rel. small clearance and large volume distribution • rate of elimination (extraction) = Clearance * [Drug]plasma

Clearance Terms Vein Artery [Drug]in (mg/L) [Drug]out (mg/L) Q (L/h) Rate of Drug in = [Drug]in * Q Rate of Drug out = [Drug]out * Q Extraction Rate = ([Drug]in-[Drug]out ) * Q Extraction Ratio ([Drug]in-[Drug]out)/[Drug]in) Clearance = E * Q

Example Kidneys Continuous IV infusion 1 L/(kg*hr) Rate Drug in, Rate Drug out, Extraction rate, extraction ratio, Clearance, Elimination Rate

Clearances

Clearance is additive sometimes Clh

Elimination Rate Constant and Half Life Versus Clearance • elimination rate = - K * Drug doseplasma = -Cl * [Drug]plasma • K = Clearance (Cl)/Volume Distribution (Vd) • T 1/2 = (ln 2)/K = (ln 2 * Vd)/Cl

[Drug]plasma Total Body Clearance

Renal Clearance Proximal Tubule Distal Tubule Collecting Duct Loop of Henle

Glomerular Filtration • • • Glomerulus Q at 1. 2 L/min. or 72 L/hour prefer neutral 4 -8 nm anionic albumin (7 nm) not filtered insulin (5 k. D) filtered Normal glomerulus filtration rate 10% Q 7. 2 L/ hour

Renal Clearance • Neutral and small drug (< 8 nm) – not bound to plasma proteins • Glomerular Filtration Rate (GFR) = Clearance (Cl. GF) – bound to plasma proteins • fraction unbound (fu) * GFR = Cl. GF Example: GFR = 9 L/hour fu = 30% CLGF = ?

Tubular Secretion

Tubular Reabsorption • Lipophilicity • p. Ka (acid and base forms)

Renal Clearance • Renal Clearance (Clr) = Cl. GF + Clearance from Tubular Secretion (Cl. TS) – Tubular Reabsorption (TR) • Clr = fu * GFR + Clts - TR

Renal Clearance Example GFR = 7. 2 L/hr fraction bound = 80% Clr = 1. 8 L/hr What are the relative values for active tubular secretion (Cl. TS) and tubular reabsorption (TR)?

Renal Clearance [drug]plasma

Clr Example L mg/L Period Volume [Drug]urine Period [Drug]plasma 0 -1 0. 2 15 0. 240 1 -3 0. 180 19. 4 2 0. 142 3 -5 0. 140 12. 8 4 0. 071 5 -10 0. 4 3. 5 7. 5 0. 021 hours mg/L Average Excretion Rate (mg/hour) = (Volume * [Drug]urine)/time 3 mg/hour = ( 0. 2 *15)/1

Clr Example hours mg/hours mg/L Period Avg. Excretion Rate Period [Drug]plasma 0 -1 3 0. 5 0. 240 1 -3 1. 746 2 0. 142 3 -5 0. 896 4 0. 071 5 -10 0. 280 7. 5 0. 021 slope = (3 -0. 280)/(0. 240 -0. 21) slope = 2. 72/0. 219=12. 4 L/hour=Clr

Single-Point Determination of Renal Clearance Under Steady-State Conditions 55 year old 65 kg female Measurement Amount Collection Period 24 hours Volume Urine Collected 1. 05 L Urinary Creatine Concentration 1140 mg/L 400 -3000 mg/L Plasma Creatine Concentration 10 mg/L 6 -12 mg/L What is the average Clr? Normal

Hepatic Clearance Mechanisms • Biotransformation • Excretion

Restrictive and Non-restrictive Clearance • Non-restrictive Eh > 0. 7 • Restrictive Eh < 0. 3

Non-Restrictive Hepatic Clearance: High Eh

Restrictive Hepatic Clearance: Low Eh

Blood and Plasma Hepatic Clearance plasma = protein and water

Example • • • Clplasma hepatic = 54 L /hr Qhb = 78 L/hr fu=1 ([drug]plasma/[drug]blood) = 0. 3 What is the extraction ratio? Does the drug undergo restrictive or nonrestrictive clearance?

End of Clearance Determinations