Multicompartment Pharmacokinetic Models. Objectives. Multicompartment Models. 26 July Chapter 30 1

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1 Multicompartment Pharmacokinetic Models Objectives To draw schemes and write differential equations for multicompartment models To recognize and use integrated equations to calculate dosage regimens To determine parameter values using the method of residuals To calculate various V values To use the non-compartmental method of parameter estimation Multicompartment Models Rapid equilibration assumption not always true Distribution may take some finite time Body may be represented by two equilibrated compartments with distribution between the two Semi-log figure will show a distribution phase Chapter 30

2 Linear Plot One Compartment Model Dose = 0 mg; V = 2.5 L; kel = 0.5 hr - Cp = Cp 0 e -kel t One Compartment Model Semi-log Plot Cp = Cp 0 e -kel t Dose = 0 mg; V = 2.5 L; kel = 0.5 hr Drug Disposition Distribution Commonly observed when early data are collected Deviation from a single exponential line A rapid drop followed by a slower terminal phase Body can be represented by two (or more) compartments Chapter 30 2

3 Semi-log Plot Two Compartment 0 first phase second phase 0. Cp = 24 e -.5 t + 6 e t Two Compartments Central compartment rapidly perfused tissues Peripheral compartment slowly perfused tissues Scheme Two Compartment Blood Kidney Liver X Central k2 k2 kel X 2 Peripheral Fat Bone Chapter 30 3

4 Two Compartment Differential Equations Central Compartment dx = kel X dt k2 X + k2 X 2 Peripheral Compartment dx 2 dt = k2 X k2 X 2 The Differential Equations dx = k2 X dt 2 k2 X kel X dx 2 = k2 X dt k2 X 2 Take Laplace of the Equations s X X (0) = k2 X2 k2 X kel X s X2 X 2 (0) = k2 X k2 X2 Since X (0) = Dose and X 2 (0) = 0 X2 = k2 X (s + k2) Chapter 30 4

5 Substitute and Rearrange s X Dose = k2 k2 X (k2 + kel) X (s + k2) s X+(k2 + kel) X k2 k2 X = Dose (s + k2) X [ s (s + k2) + (k2 + kel) (s + k2) k2 k2] = Dose (s + k2) X [s 2 + s (k2+ k2 + kel) + k2 kel] = Dose (s + k2) Notice the similarity with [s 2 + s (α+β)+α β]=(s +α) (s +β) Getting close now If and (α + β) = k2 + k2 + kel α β= k2 kel X [s 2 + s (α+β)+α β]= Dose (s + k2) X (s +α) (s +β)= Dose (s + k2) Dose (s + k2) X = (s +α) (s +β) Back Transforming X = Dose (s+ k2) (s +α) (s +β) The denominator has a power of 2 in s and no repeat terms. Note the numerator has a power of in s Considering the denominator: (s + α) (s + β) = 0 Roots or solutions are -α and -β Chapter 30 5

6 First root is -α Root Dose (s+ k2) X = (s +α) (s +β) Dose (s+ k2) (s +α) (s+β) Dose (k2 α) e ( β α) α t Root 2 Dose (s+ k2) X = (s +α) (s +β) Second root is -β Dose (s+ k2) (s +α) (s+β) Dose (k2 β) α β ( ) e β t Putting It Together Dose (α k2) e ( α β) α t Dose (α k2) e ( α β) α t + OR Cp = A e -α t + B e -β t where Dose (k2 β) e ( β t α β) Dose (k2 β) e ( β t α β) and Dose (α k2) A = V (α β) B= Dose (k2 β) V (α β) Chapter 30 6

7 Integrated Equation α+β= kel + k2 + k2 α β= kel k2 Cp = A e -α t + B e -β t ( α,β= α+β )± ( α+β )2 4 α β 2 or ( kel + k2 + k2) ± kel + k2 + k2 α,β= ( )2 4 kel k2 2 Parameter Determination Method of Residuals Parameters A, B, α and β can be determined using the method of residuals Since α > β (if α/β > 5) e -α t approaches 0 quickly Terminal data points will be on a line Method of Residuals Cp late = B e -β t Plotted on semi-log graph paper should give a straight line Calculate β from the terminal slope Chapter 30 7

8 Terminal Slope 0 slope > β 0. Cp = 24 e -.5 t + 6 e t Terminal Half-life t /2 = ln(2)/β Biological or terminal half-life [= equivalent to ln(2)/kel with one compartment model] Residual Residual = Cp - Cp late = A e -α t 0 0. α > β slope -> α Chapter 30 8

9 Now Calculate k2, kel, k2 A β+ B α k2 = A +B kel = α β k2 k2 =α+β k2 kel Effect of k2 and k2 Ratio k2/k2 = /4 9 k2/k2 = /2 8 k2/k2 = / 7 k2/k2 = 2/ 6 k2/k2 = 4/ Time (hour) Effect of k2 and k2 Ratio Effect of k2/k2 ratio The higher the ratio greater the distribution into the peripheral compartment At the extremes low ratio - less distribution into second compartment high ratio and no early data - looks like one compartment model Chapter 30 9

10 Magnitude of k2 and k2 k2/k2 = 0.5/ k2/k2 = /0.5 8 k2/k2 = 2/ 7 k2/k2 = 4/2 6 k2/k2 = 8/4 5 k2/2 = 6/ Time (hour) Magnitude of k2 and k2 Larger values approach one compartment assumption Volume of Distribution V V Apparent volume of central compartment ( ) V = Dose A + B = Dose 0 since A+B=Cp0 Cp Use to calculate Cp 0 after an IV bolus administration Chapter 30

11 Volume of Distribution V area (= V β ) V area V area = Dose β AUC = V kel β = Clearance β Useful for dosing calculations, easy to calculate from Dose and AUC Volume of Distribution V extrap V extrap Volume extrapolated V extrap = Dose B Ignores distribution phase Volume of Distribution V ss V ss Steady state volume V ss = V k2 + k2 k2 Relates total amount in the body (at steady state) with drug concentrations in plasma or blood Chapter 30

12 Steady State Volume 0 X X Volumes of Distribution V extrap > V area > V ss > V Example Calculation Time (hr) Cp (mg/l) Cp late (mg/l) Residual (mg/l) I.V. Bolus 500 mg Chapter 30 2

13 The Plots 0 Cp Residual Calculations B = mg/l, β = (ln - ln 0.5)/2 = 2.996/2 = 0.25 hr - A = 25 mg/l, α = (ln 25 - ln 0.27)/3 = 4.528/3 =.5 hr - Cp = 25 e -.5 x t + e x t Microconstants k2 = kel = α β k2 A β+ B α = A +B = 0.6 hr = = 0.62 hr k2 =α+β k2 kel = = 0.53 hr - Chapter 30 3

14 Volumes of Distribution V = Dose A + B = = 4.3 L V area = Dose β AUC = = 34.3 L V extrap = Dose B = 500 = 50 L V ss = V k2 + k = 4.3 = 26.7 L k2 0.6 Note:V extrap > V area > V ss > V [50 > 34.3 > 26.7 > 4.3] AUC = = 58.3 mg.hr.l - Dosage Calculations Initial Concentration, Cp 0 V = Dose Cp 0 0 Dose = V Cp required To achieve a Cp 0 of 20 mg/l give 600 mg with V = 30 L If V = 6 L an IV Bolus Dose of 500 mg would result in a Cp 0 of 3.3 mg/l (=500/6) Dosage Calculation Continuous Infusion Cp ss k0 k0 = = kel V Clearance = k0 V area β k0 = Cp ss kel V If V = 5 L and kel = 0.2 hr - a k0 of 90 mg/hr is required to produce a Cp ss of 30 mg/l Chapter 30 4

15 Plasma Concentration Time to Steady State Cp can be calculated after an IV bolus dose from A, B, α and β Cp after an IV infusion somewhat more involved Time to steady state controlled by β value Can be slow with long biological half-life I.V. Bolus and Infusion I.V. Bolus and Infusion kel = 0.2 hr - ; k2 = 2 hr - ; k2 = hr - ; V = 5 L; Bolus = 450 mg; k0 = 90 mg/hr I.V. Bolus and Infusion I.V. Bolus and Infusion kel = 0.2 hr - ; k2 = 2 hr - ; k2 = hr - ; V = 5 L; Bolus = 300 or 600 mg; k0 = 90 mg/hr Chapter 30 5

16 Fast and Slow I.V. Infusion I.V. Infusion - Fast/Slow kel = 0.2 hr-; k2 = 2 hr-; k2 = hr-; V = 5 L; Infusion Rate 300 mg/hr for 4 hr then 90 mg/hr Oral Administration Scheme Drug in GI Tract ka X Central k2 X 2 Peripheral k2 kel Differential Equation dx = ka Xg + k2 X dt 2 ( k2 + kel) X Chapter 30 6

17 Semi-log Plot 0 Oral - Two Compartment with Distribution Phase A = 20 mg/l; B = 5 mg/l; C = -35 mg/l; α = hr - ; β = 0. hr - ; ka = 2 hr Semi-log Plot Oral - Two Compartment without Distribution Phase A = 20 mg/l; B = 5 mg/l; C = -35 mg/l; α =.5 hr - ; β = 0.6 hr - ; ka = hr Oral Dose Two Compartment Model Bioavailability calculations the same as for a one compartment model Use AUC comparison or Use U comparison These method work for any linear system (first order disposition) Method of residuals could be used to calculate α, β, and ka (if sufficiently different) Chapter 30 7

18 Cp Calculations Average Cp of 20 mg/l required with V = 5 L, kel = 0.5 hr -, F = 0.9, and τ = 2 hr F Dose Cp = Clearance τ = F Dose kel V τ = F Dose β V β τ Dose = = 600 mg q2h 0.9 MacKinetics Two compartment Demo Example Data - 0 mg IV Bolus Dose Chapter 30 8

19 Clinical Example Lidocaine - Rapidly attain and maintain effective concentrations (2-6 mg/l) Multiple bolus over 5 or 30 min + infusion Exponentially declining infusion Stepwise, tapering infusion k = min - t /2 = 20 min k2 = min - t /2 = 2 min k2 = min - t /2 = 30 min V = 0.49 L/kg = 34 L (70 kg patient) Evans, Schentag, and Jusko Applied Pharmacokinetics, 3rd ed., Applied Therapeutics, Vancouver, WA 992 Lidocaine - Loading Dose Usual Dose - 50 to 0 mg followed by an infusion of - 4 mg/min Dip in concentration below therapeutic range Increasing the loading dose to mg may cause toxic doses Multiple Loading dose approach Initial 75 mg followed by up to six 50 mg bolus doses to effect Ectopic ventricular beats to less than 5 per minute and no complex ventricular arrhymias Lidocaine - Usual Dose min x Chapter 30 9

20 Usual Dose - Simulation Minutes Multiple Bolus Doses Minutes Non Compartmental Analysis Linear Disposition First order elimination and distribution No assumptions about number of compartments Use t max, Cp max, AUC, AUMC New parameters: AUMC (area under moment curve) MRT (mean residence time) MAT (mean absorption time) Chapter 30 20

21 AUMC Area under the Moment Curve Use trapezoidal rule with t versus Cp t data Last segment from Cp last t last where k is slowest (last) exponential k + Cplast k 2 Non Compartmental Analysis Time Cp Cp t AUC AUMC (hr) (mg/l) (mg.hr/l) (mg.hr/l) (mg.hr 2 /L) Linear System - 0 mg IV Plot of Cp versus t IV Data Chapter 30 2

22 Plot of Cp t versus t IV Data 25 Cp t (mg.hr/l) Parameter Calculations MRT = AUMC AUC = 553 = 8.2 hr 67.4 k = MRT = - = 0.22 hr 8.2 Cl = Dose AUC = 0 =.48 L/hr 67.4 V ss = Cl MRT = =2.2 L Non Compartmental Analysis Time (hr) Cp (mg/l) Cp t (mg.hr/l) AUC (mg.hr/l) Linear System mg PO AUMC (mg.hr 2 /L) Chapter 30 22

23 Plot of Cp versus t PO Data Plot of Cp t versus t PO Data 80 Cp t (mg.hr/l) Parameter Calculations MRT( PO) = AUMC AUC = 36 = 9.08 hr 50 MAT=MRT( PO) MRT( IV) = = 0.88 hr ka = MAT = - =.4 hr 0.88 F = AUC PO Dose IV 50 0 = AUC IV Dose PO = 0.89 Chapter 30 23

24 Objectives To draw schemes and write differential equations for multicompartment models To recognize and use integrated equations to calculate dosage regimens To determine parameter values using the method of residuals To calculate various V values To use the non-compartmental method of parameter estimation Chapter 30 24