Bolus Dosing

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Bolus Dosing

Bolus Dosing

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  • 1. I.V. Bolus Dosing CHAPTER 4 Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich OBJECTIVES For an IV one compartment model plasma and urine: 1. Given patient drug and/or metabolite concentration, amount, and/or rate vs. time profiles, the student will calculate (III) the relevant pharmacokinetic parameters available from IV plasma, urine or other excreta data: e.g. V d, K, k m, k r, AUC, AUMC, CL, MRT, t 1 ⁄ 2 2. The student will provide professional communication regarding the pharmacoki- netic parameters obtained to patients and other health professionals. 3. The student will be able to utilize computer programs for simulations and data analysis. 4-1 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 2. I.V. Bolus Dosing 4.1 I.V. Bolus dosing of Parent compound 4.1.1 PLASMA ln C p = – K ⋅ t + ln C p 0 Valid equations: (EQ 4-1) (Obtained from the ln X = – K ⋅ t + ln X0 LaPlace transforms (EQ 4-2) derived from the appropriate models – Kt derived from the C p = C p0 e (EQ 4-3) pharmacokinetic descriptions of the drug) D- C p 0 = ----- (EQ 4-4) Vd t ½ = 0.693 ------------ - (EQ 4-5) K ( ∞) ( Cp n + Cp n + 1 ) Cp last Cp dt = Σ  ------------------------------------- ⋅ ∆t + -------------- ∫ AUC =   (EQ 4-6) 2 K 0 ( ∞) t ( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 ) ( t l ast ⋅ Cp last ) Cp l ast ∑  ------------------------------------------------------------------- ⋅ ∆t + --------------- + ---------------------------------- ∫ t ⋅ C p dt = - AUMC = (EQ 4-7)   2 2 K K 0 0 MRT = AUMC ----------------- - (EQ 4-8) AUC Cl = K ⋅ Vd (EQ 4-9) Utilization: Can you determine the • You should be able to plot a data set Concentration vs. time on semilog yielding a straight line slope and intercept from with slope = – K and an intercept of C p0 . a graph? Plot the data in table 4 -1.on semi-log graph paper. Extrapo- Nifedipine 25 mg IV bolus TABLE 4-1 late the line back to time = 0 to get Cp0. Find the Cp half life. Calculate the Time (hr) (mcg/L) elimination rate con- 2 139 stant. 4 65.6 6 31.1 8 14.6 FIGURE 4-1. 4-2 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 3. I.V. Bolus Dosing Does your Graph look Nifedipine IV Bolus (25 mg IV Bolus) FIGURE 4-1 like this? 100 10 3 Concentration (mic/L) Cp0 = 295 mic/L -K1 = -0.375 hr -1 Concentration (ng/mL) 10 2 1.85 hr 50 10 1 0 2 4 6 8 Time (hr) Time (hours) • You should be able to determine K. A plot of the data in TABLE 4-1 results in FIGURE 4-1 dy Remember from high school algebra, the slope of any straight line is the rise over the run, ----- , - dx In the case of semi-log graphs dy is the difference in the logarithms of the concentrations. Thus, using the rules of logarithms, when two logs are subtracted, the numbers themselves are divided. i.e. ln ( C1 ) – ln ( C2 ) = ln  ------ . Thus if we are judicious in the concentrations that we C1 -  C2 take, we can set the rise to a constant number. So, if we take any two concentrations such that one concentration is half of the other (In FIGURE 4-1 above, we took 100 and 50), the time it takes for the concentration to halve is the half life (in the graph above, 1.85 hr). Then 0.693 0.693 –1 K = ------------ = --------------- = 0.375 hr - - 1.85hr t½ • You should be able to determine V d :. To do this, extrapolate the line to t = 0 . The value of Cp mic when t = 0 is C p0 (in the graph above, C p0 = 295 -------- which is equal to D ⁄ V d for an IV bolus - L dose only. Dose 25mg 1000mic Thus, Cp 0 = ------------ , V d = Dose = ------------------ ⋅ -------------------- = 85L - - ------------ - Vd mg Cp 0 295mic ----------------- - L The volume of distribution is a mathematical construct. It is merely the proportionality constant between two knowns - the C p0 which results from a given D 0 . It is, however, useful because it is patient specific and therefore can be used to predict how the patient will treat a subsequent dose of the same drug. You should be able to obtain the volume of distribution from graphical analysis of the data. Pay attention to the units! Make sure that they are consistent on both sides of the equation. NOTE: the volume of distribution is not necessarily any physiological space. For example the approximate volume of distribution of digoxin is about 600 L If that were a physiological space and I were all water, that would mean that I would weigh about 1320 pounds. I’m a little overweight (I prefer to think that I’m underheight), but REALLY! • Given any three of the variables of the IV bolus equation, either by direct information (the vol- ume of distribution is such and such) or by graphical data analysis, you should be able to find the fourth. 4-3 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 4. I.V. Bolus Dosing • You should be able to calculate Area Under the Curve (AUC) from IV Bolus data (Time vs. Cp). From the above data in TABLE 4-1 the AUC is calculated using (EQ 4-6): (∞) Cpn + Cp n + 1 Cp = Σ  --------------------------------  + --------l which in this case is: ∫ Cp dt - AUC =   ∆t K 0  Cp o + Cp 1 Cp l ast  Cp 1 + Cp 2 Cp2 + Cp 3 Cp 3 + Cp last Σ  ------------------------- ⋅ ∆t 1 + ------------------------- ⋅ ∆t 2 + ------------------------- ⋅ ∆t 3 + ------------------------------ ⋅ ∆t last + --------------  - - - - - 2 2 2 2  K1   295 + 139 14.6 mcg 139 + 65.6 65.6 + 31.1 31.1 + 14.6 Σ  ----------------------- ⋅ 2 + ------------------------- ⋅ 2 + -------------------------- ⋅ 2 + -------------------------- ⋅ 2 + ------------ ---------- hr or - - - - 2 2 2 2 0.375  L  mcg mcg Σ { 434 + 204.6 + 96.7 + 45.7 + 38.9 } --------- hr = 819.9 ---------- hr . In tabular format, the AUC calculation - L L is shown in TABLE 4-2. TABLE 4-2 AUC t t AUC AUC TIME Cp t–1 0 0 295 2 139 434.0 434.0 4 65.6 204.6 638.6 6 31.1 96.7 735.3 8 14.6 45.7 781.0 ∞ 0 38.9 819.9 The AUC of a plot of plasma concentration vs. time, in linear pharmacokinetics, is a number which is proportional to the dose of the drug which gets into systemic circulation. The propor- tionality constant, as before, is the volume of distribution. It is useful as a tool to compare the amount of drug obtained by the body from different routes of administration or from the same route of administration by dosage forms made by different manufacturers (calculate bioavail- ability in subsequent discussions). The AUC of a plot of Rate of Excretion of a drug vs. time, in linear pharmacokinetics, is the mass of drug excreted into the urine, directly. • You should be able to calculate the AUMC from IV Bolus data (Time vs. Cp). The equation for AUMC is equation 4-7: ( ∞) t ( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 ) ( t l ast ⋅ Cp last ) Cp l ast ∑  ------------------------------------------------------------------- ⋅ ∆t + --------------- + ---------------------------------- ∫ t ⋅ C p dt = - which in the AUMC =   2 2 K K 0 0 data given in TABLE 4-1 is: T0 ⋅ C po + T1 ⋅ C p1 T1 ⋅ C p1 + T2 ⋅ C p2 T 2 ⋅ C p 2 + T3 ⋅ C p3 Σ ---------------------------------------------- ⋅ ∆t 1 + ---------------------------------------------- ⋅ ∆t 2 + ---------------------------------------------- ⋅ ∆t 3 + - - - 2 2 2 4-4 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 5. I.V. Bolus Dosing T 3 ⋅ C p 3 + T last ⋅ C p last T last ⋅ C p l ast Cplast --------------------------------------------------------- ⋅ ∆t l ast + ------------------------------ + -------------- and thus, - - - 2 2 K K  0 ⋅ 295 + 2 ⋅ 139 4 ⋅ 65.6 + 6 ⋅ 31.1 mcg 2 2 ⋅ 139 + 4 ⋅ 65.6 Σ  -------------------------------------- ⋅ 2 + ---------------------------------------- ⋅ 2 + ----------------------------------------- ⋅ 2 --------- hr + - - - 2 2 2 L   6 ⋅ 31.1 + 8 ⋅ 14.6 14.6  mcg 2 8 ⋅ 14.6  ----------------------------------------- ⋅ 2 + ----------------- + ---------------  ---------- hr or - - - 2 0.375  L 2  0.375 mcg 2 Σ { 278 + 540.4 + 449 + 303.4 + 311.47 + 103.82 } = 1986.1 --------- hr - L Thus in tabular format the AUMC for data given in TABLE 4-1 is TABLE 4-3 below. TABLE 4-3 AUMC t AUMC AUMC t TIME Cp Cp*T 0 0 295 0 2 139 278 278.0 278.0 4 65.6 262.4 540.4 818.4 6 31.1 186.6 449.0 1267.4 8 14.6 116.8 303.4 1570.8 ∞ 0 0 415.3 1986.1 The AUMC is the Area Under the first Moment Curve. A plot of T*Cp vs. T is the first moment curve. The time function buried in this plot, the Mean Residence Time (MRT), can be extracted using equation 4-8 below. It is the geometric mean time that the molecules of drug stay in the body. It has utility in the fact that, as drug moves from the dosage form into solution in the gut, from solution in the gut into the body, and from the body out, each process is cumulatively additive. That means if we can physically separate each of these processes in turn, we can calculate the MRT of each process. The MRT of each process is the the inverse of the rate constant for that process. • You should be able to calculate MRT from IV Bolus data (Time vs. Cp) using equation 4-8 AUMC 1986.1 MRT = ----------------- = --------------- = 2.42 - - AUC 819.9 Since there is only the process of elimination (no release of the drug from the dosage form, no absorption), the MRT is the inverse of the elimination rate constant, K. Thus MRT = 1/K. 4-5 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 6. I.V. Bolus Dosing IV Bolus Flow Chart 4-1 K X MRT(IV) = 1/K Suppose the drug were given in a solution. Then the drug would have to be absorbed and then eliminated. Since the MRTs are additive, the MRT of the oral solution would be made up of the MRTs of the two processes, thus: Oral Solution Flow Chart 4-2 Ka K Xa X MRT(os) = MAT(os)+MRT(IV) MRT(os) = 1/Ka + 1/K Consequently, if a drug has to be released from a dosage form for the drug to get into solution which is subsequently absorbed, a tablet for example, the MRT of the tablet will consist of the MRT(IV) and the MAT(os) and the Mean Dissolution Time (MDT), thus: Tablet Flow Chart 4-3 Kd Ka K Xd Xa X MRT(tab) = MDT + MAT(os) + MRT(IV) MRT(tab) = 1/Kd + 1/Ka + 1/K MRT(tab) = MAT(tab) + MRT(IV) MRT(tab) = 1/Ka (apparent) + 1/K 4-6 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 7. I.V. Bolus Dosing Normally, we don’t have information from the oral solution, just IV and tablet. So in that case the information obtained about absorption from the tablet is bundled together into an apparent absorption rate constant consisting of both dissolution and absorption. It should be apparent that this is a reasonably easily utilized and powerful tool used to obtain pharmacokinetic parameters. 4-7 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 8. I.V. Bolus Dosing 4.1.2 IV BOLUS, PARENT COMPOUND, PLASMA PROBLEMS 1. K = –slope from equation 4-3 Equations used in this section: ln 2 2. t 1 ⁄ 2 = ------- equation 4-5 - K 1 MRT = --- ( estimate ) MRT = AUMC equation 4-8 - 3. ----------------- - K AUC 4. Cp 0 = the y-intercept of the line from equation 4-3 Cp 0 ∞ ∫0 Cp dt 5. Estimate for AUC = AUC = --------- which is K (∞) ( Cp n + Cp n + 1 ) Cp last Cp dt = Σ  ------------------------------------- ( ∆t ) + -------------- ∫ AUC =   2 K 0 Trapezoidal rule applied to equation 4-6 Estimate for AUMC = AUMC = AUC ⋅ MRT from equation 4-8 6. ( ∞) t ( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 ) ( t last ⋅ Cp l ast ) Cp last ∑  ------------------------------------------------------------------- ⋅ ∆tn + --------------- + ---------------------------------- ∫ - AUMC Cp dt =   2 2 K K 0 0 from equation 4-7 V d = Dose from equation 4-4 7. ------------ - Cp 0 Cp 0 Dose Cl = K1 ⋅ V d = ----------- ⋅ ------------ = Dose - - ------------ 8. - AUC Cp 0 AUC 4-8 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 9. I.V. Bolus Dosing Acyclovir (Problem 4 - 1) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals De Miranda and Burnette, “Metabolic Fate and Pharmacokinetics of the Acyclovir Prodrug Valaciclovir in Cynomolgus Mon- keys”, Drug Metabolism and Disposition (1994): 55-59. Acyclovir is an antiviral drug used in the treatment of herpes simplex, varicella zoster, and in suppressive therapy. In this study, three male cynomolgus monkeys were each given a 10 mg ⁄ kg intravenous dose. The monkeys weighed an average of 3.35 kg each. Blood samples were collected and the following data was obtained: Acyclovir PROBLEM TABLE 4 - 1. Serum concentration ( µg ⁄ mL ) Time (hours) 0.167 26.0 0.300 23.0 0.500 19.0 0.75 16.0 1.0 12.0 1.5 7.0 2.0 5.0 From the data presented in the Preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-9 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 10. I.V. Bolus Dosing (Problem 4 - 1) Acyclovir: 2 10 CONCENTRATION (MIC/ML) 1 10 100 0.0 0.5 1.0 1.5 2.0 TIME (HR) –1 k = 0.93hr 1. t½ = 0.75hr . 2. MRT = 1.08hr . 3. ( C p )0 = 30.4ug ⁄ mL . 4. AUC = 32.75ug ⁄ mL ⋅ hr . 5. 2 AUMC = 35.2ug ⁄ mL ⋅ hr . 6. Vd = 1.1L 7. Cl = 1.02L ⁄ hr . 8. 4-10 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 11. I.V. Bolus Dosing Aluminum (Problem 4 - 2) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Xu, Pai, and Melethil, quot;Kinetics of Aluminum in Rats. II: Dose-Dependent Urinary and Biliary Excretionquot;, Journal of Pharmaceu- tical Sciences, Oct 1991, p 946 - 951. A study by Xu, Pai, and Melethil establishes the pharmacokinetics of Aluminum in Rats. In this study, four rats with an average weight of 375g, were given an IV bolus dose of aluminum (1 mg/kg). Blood samples were taken at various intervals and the following data was obtained: Aluminum PROBLEM TABLE 4 - 2. ng Serum concentration, ------- - mL Time (hours) 0.4 19000 0.6 18000 1.4 15000 1.6 14500 2.3 12500 3.0 10500 4.0 8500 5.0 6500 6.0 5000 8.0 3250 10.0 2000 12.0 1250 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-11 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 12. I.V. Bolus Dosing (Problem 4 - 2) Aluminum: 5 10 CONCENTRATION (NG/ML) 4 10 3 10 0 2 4 6 8 10 12 TIME (HR) –1 k = 0.234hr 1. t½ = 3hr . 2. MRT = 4.3hr . 3. ( C p )0 = 21000ng ⁄ mL . 4. AUC = 89285ng ⁄ mL ⋅ hr . 5. 2 AUMC = 383926ng ⁄ mL ⋅ hr . 6. Vd = 17.86mL 7. Cl = 4.18mL ⁄ hr . 8. 4-12 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 13. I.V. Bolus Dosing Amgen (Problem 4 - 3) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Salmonson, Danielson, and Wikstrom, quot;The pharmacokinetics of recombinant human erythropoetin after intravenous and subcuta- neous administration to healthy subjectsquot;, Br. F. clin. Pharmac. (1990), p 709- 713. Amgen (r-Epo) is a form of recombinant erythropoetin. Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produce erythropoetin; therefore, r-Epo is being investigated for use in these patients in order to treat the anemia that results from the lack of erythropoetin. In a study by Salmonson et al, six healthy volunteers were used to demonstrate that both IV and subcutaneous administration of erythropoetin have similar effects in the treatment of anemia due to chronic renal failure. The six volunteers were each given a 50 U/kg intravenous dose of Amgen. The average weight of the six volunteers was 79 kg. Blood samples were drawn at various times and the data obtained is summarized below: Amgen PROBLEM TABLE 4 - 3. mU Serum concentration, -------- - mL Time (hours) 2 700 4 600 6 400 8 300 12 150 24 40 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-13 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 14. I.V. Bolus Dosing (Problem 4 - 3) Amgen: 103 CONCENTRATION (MU/ML) 102 Con (mU/mL) 101 0 5 10 15 20 25 TIME (HR) –1 k = 0.134hr 1. t½ = 5.2hr . 2. MRT = 7.46hr . 3. ( C p )0 = 900mU ⁄ mL . 4. AUC = 6945mU ⁄ mL ⋅ hr . 5. 2 AUMC = 49600 mU ⁄ mL ⋅ hr . 6. Vd = 4.44L 7. Cl = 0.6L ⁄ hr . 8. 4-14 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 15. I.V. Bolus Dosing Atrial Naturetic Peptide (ANP) (Problem 4 - 4) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Brier and Harding, quot;Pharmacokinetics and Pharmacodynamics of Atrial Naturetic Peptide after Bolus and Infusion Administra- tion in the Isolated Perfused Rat Kidneyquot;, The Journal of Pharmacology and Experimental Therapeutics (1989), p 372 - 377. A study by Brier and Harding a dose of 45 ng was given by IV bolus to rats. Samples of blood were taken at various intervals throughout the length of the study and the following data was obtained: Atrial Naturetic Peptide (ANP) PROBLEM TABLE 4 - 4. pg Serum concentration, ------- - mL Time (minutes) 3 380 10 280 20 170 30 130 40 100 50 70 60 50 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-15 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 16. I.V. Bolus Dosing (Problem 4 - 4) Atrial Naturetic Peptide (ANP): 103 Con CONCENTRATION (PG/ML) 102 (pg/mL) 101 0 10 20 30 40 50 60 Time (min) –1 k = 0.0345min 1. t½ = 20.09min . 2. MRT = 28.95min . 3. ( C p )0 = 386.6pg ⁄ mL . 4. AUC = 11206.4pg ⁄ mL ⋅ min . 5. 2 AUMC = 324425.4pg ⁄ mL ⋅ min . 6. Vd = 116.4mL 7. Cl = 4.02mL ⁄ min . 8. 4-16 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 17. I.V. Bolus Dosing Aztreonam (Problem 4 - 5) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Cuzzolim et al., quot;Pharmacokinetics and Renal Tolerance of Aztreonam in Premature Infantsquot;, Antimicrobial Agents and Chemo- therapy (Sept. 1991), p. 1726 - 1928. Aztreonam is a monolactam structure which is active against aerobic, gram-negative bacilli. The pharmacokinetic parameters of Aztreonam were established in a study presented in by Cuzzolim et al in which Aztreonam (100 mg/ kg) was administered intravenously to 30 premature infants over 3 minutes every 12 hours. The group of neonates had an average weight of 1639.6g. The following set of data was obtained: Aztreonam PROBLEM TABLE 4 - 5. µg Serum concentration, ------- - mL Time (minutes) 1 40.50 2 34.99 3 29.99 4 23.88 5 22.20 6 19.44 7 16.55 8 14.99 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-17 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 18. I.V. Bolus Dosing (Problem 4 - 5) Aztreonam: 10 2 CONCENTRATION (UG/ML) Con (ug/mL) 10 1 0 2 4 6 8 TIME (MIN) –1 k = 0.144min 1. t½ = 4.81min . 2. MRT = 6.94min . 3. ( C p )0 = 45.75ug ⁄ mL . 4. AUC = 317.7ug ⁄ mL ⋅ min . 5. 2 AUMC = 2204.8ug ⁄ mL ⋅ min . 6. Vd = 3.58L 7. Cl = 0.516L ⁄ min . 8. 4-18 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 19. I.V. Bolus Dosing Recombinant Bovine Placental Lactogen (Problem 4 - 6) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Byatt, et. al., quot;Serum half-life and in-vivo actions of recombinant bovine placental lactogen in the dairy cowquot;, Journal of Endocri- nology (1992), p. 185 - 193. Bovine placental lactogen (bPL) is a hormone similar to growth hormone and prolactin. It binds to both prolactin and growth hormone receptors in the rabbit and stimulates lactogenesis in the rabbit. In a study by Byatt, et. al., four cows (2 pregnant and 2 nonpregnant) were given IV bolus injections of 4 mg and the following data was obtained: Recombinant Bovine Placental Lactogen PROBLEM TABLE 4 - 6. µg Serum concentration ------ L Time (minutes) 3.8 117 6.8 72 12.0 43 16.0 27 20.0 18 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-19 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 20. I.V. Bolus Dosing (Problem 4 - 6) Recombinant Bovine Placental Lactogen: 103 CONCENTRATION (MIC/L) 102 Con (ug/L) 101 0 5 10 15 20 Time (min) –1 k = 0.113min 1. t½ = 6.13min . 2. MRT = 8.85min . 3. ( C p )0 = 167.8ug ⁄ L . 4. AUC = 1484.9ug ⁄ L ⋅ min . 5. 2 AUMC = 13141.1ug ⁄ L ⋅ min . 6. Vd = 23.84L 7. Cl = 2.69L ⁄ min . 8. 4-20 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 21. I.V. Bolus Dosing Caffeine (Problem 4 - 7) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Dorrbecker et. al., quot;Caffeine and Paraxanthine Pharmacokinetics in the Rabbit: Concentration and Product Inhibition Effects.quot;, Journal of Pharmacokinetics and Biopharmaceutics (1987), p.117 - 131. This study examines the pharmacokinetics of caffeine in the rabbit. In this study type I New Zealand White rabbits were given an 8 mg intravenous dose of caffeine. Blood samples were taken and the following data was obtained: Caffeine PROBLEM TABLE 4 - 7. µg Serum concentration ------- - mL Time (minutes) 12 3.75 40 2.80 65 2.12 90 1.55 125 1.23 173 0.72 243 0.37 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-21 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 22. I.V. Bolus Dosing (Problem 4 - 7) Caffeine: CONCENTRATION (MIC/ML) Caffeine 101 100 10-1 Con (ug/L) 0 50 100 150 200 250 Time (min) –1 k = 0.00997min 1. t½ = 69.51min . 2. MRT = 100.3min . 3. ( C p )0 = 4.105ug ⁄ mL . 4. AUC = 411.7ug ⁄ mL ⋅ min . 5. 2 AUMC = 41293.5ug ⁄ mL ⋅ min . 6. Vd = 1.95L 7. Cl = 19.44mL ⁄ min . 8. 4-22 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 23. I.V. Bolus Dosing Ceftazidime (Problem 4 - 8) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Demotes-Mainard, et. al., quot;Pharmacokinetics of Intravenous and Intraperitoneal Ceftazidime in Chronic Ambulatory Peritoneal Dyialysisquot;, Journal of Clinical Pharmacology (1993), p. 475 - 479. Ceftazidime is a third generation cephalosporin which is administered parenterally. In this study, eight patients with chronic renal failure were each given 1 g of ceftazidime intravenously. Both blood samples were taken the data obtained from the study is summarized in the following table: Ceftazidime PROBLEM TABLE 4 - 8. mg Serum concentration ------ - L Time (hours) 1 50 2 45 4 38 24 21 36 14 48 11 60 8 72 4 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-23 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 24. I.V. Bolus Dosing (Problem 4 - 8) Ceftazidime: 102 CONCENTRATION (MG/L) 101 Con (mg/L) 100 0 20 40 60 80 Time (hours) –1 k = 0.0324hr 1. t½ = 21.39hr . 2. MRT = 30.86hr . 3. ( C p )0 = 47.57mg ⁄ L . 4. AUC = 1468.2mg ⁄ L ⋅ hr . 5. 2 AUMC = 45308.6mg ⁄ L ⋅ hr . 6. Vd = 21.02L 7. Cl = 0.681L ⁄ hr . 8. 4-24 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 25. I.V. Bolus Dosing Ciprofloxacin (Problem 4 - 9) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Lettieri, et. al., quot;Pharmacokinetic Profiles of Ciprofloxacin after Single Intravenous and Oral Dosesquot;, Antimicrobial Agents and Chemotherapy (May 1992), p. 993 -996. Ciprofloxacin is a fluoroquinolone antibiotic which is used in the treatment of infections of the urinary tract, lower res- piratory tract, skin, bone, and joint. In this study, twelve healthy, male volunteers were each given 300 mg intravenous doses of Ciprofloxacin. Blood and urine samples were collected at various times throughout the day and the following data was collected: Ciprofloxacin PROBLEM TABLE 4 - 9. mg Serum concentration ------ - L Time (hours) 2 1.20 3 0.85 4 0.70 6 0.50 8 0.35 10 0.25 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-25 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 26. I.V. Bolus Dosing (Problem 4 - 9) Ciprofloxacin: 101 CONCENTRATION (MG/L) 100 Con (mg/L) 10-1 0 2 4 6 8 10 Time (hours) –1 k = 0.1875hr 1. t½ = 3.7hr . 2. MRT = 5.33hr . 3. ( C p )0 = 1.57mg ⁄ L . 4. AUC = 8.395mg ⁄ L ⋅ hr . 5. 2 AUMC = 44.74mg ⁄ L ⋅ hr . 6. Vd = 190.6L 7. Cl = 35.74L ⁄ hr . 8. 4-26 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 27. I.V. Bolus Dosing The effect of Probenecid on Diprophylline (DPP) (Problem 4 - 10) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Nadai et al, quot;Pharmacokinetics and the Effect of Probenecid on the Renal Excretion Mechanism of Diprophyllinequot;, Journal of Pharmaceutical Sciences (Oct 1992), p. 1024 - 1027. Diprophylline is used as a bronchodilator. A study by Nadai et al was designed to determine whether or not coadmin- istration of Diprophylline with Probenecid affected the pharmacokinetic parameters of Diprophylline. In this study, male rats (average weight: 300 g) were given 60 mg/kg of Diprophylline intravenously and a 3 mg/kg loading dose of Probenecid followed by a continuous infusion of 0.217 mg/min/kg of Probenecid. The following set of data was obtained for Diprophylline (DPP): The effect of Probenecid on Diprophylline (DPP) PROBLEM TABLE 4 - 10. µg Serum concentration ------- - mL Time (minutes) 16 40.00 31 27.00 60 13.00 91 6.50 122 3.50 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-27 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 28. I.V. Bolus Dosing (Problem 4 - 10) The effect of probenecid on diprophylline (DPP): 102 CONCENTRATION (MIC/ML) 101 Con (ug/mL) 100 0 20 40 60 80 100 Time (min) –1 k = 0.023min 1. t½ = 30.13min . 2. MRT = 43.48min . 3. ( C p )0 = 55.13ug ⁄ mL . 4. AUC = 2396.96ug ⁄ mL ⋅ min . 5. 2 AUMC = 104219.8ug ⁄ mL ⋅ min . 6. Vd = 326.5mL 7. Cl = 7.5mL ⁄ min . 8. 4-28 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 29. I.V. Bolus Dosing Epoetin (Problem 4 - 11) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals MacDougall et. al., quot;Clinical Pharmacokinetics of Epoetin (Recombinant Human Erythropoetinquot;, Clinical Pharmacokinetics (1991), p 99 - 110. Epoetin is recombinant human erythropoetin. Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produce erythropo- etin; therefore, Epoetin is used in these patients to treat the anemia that results from the lack of erythropoetin. Epoetin has also been used in the treatment of anemias resulting from AIDS. malignant disease, prematurity, rheumatoid arthri- tis, sickle-cell anemia, and myelosplastic syndrome. In a study by Macdougall et al, eight patients who were on perito- neal dialysis (CAPD) were given an IV bolus dose of 120 U/kg which decayed monoexponentially from a peak of 3959 U/L to 558 U/L at 24 hours. The following data was obtained: Epoetin PROBLEM TABLE 4 - 11. U Serum concentration --- - L Time (hours) 0.0 4000 0.5 3800 1.0 3600 2.0 3300 3.0 3000 4.0 2550 5.0 2350 6.0 2150 7.0 1900 From the data presented in the preceding table and assuming that the patient weighs 65 kg, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-29 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 30. I.V. Bolus Dosing (Problem 4 - 11) Epoetin: 104 CONCENTRATION (U/L) Con (U/L) 103 0 1 2 3 4 5 6 7 Time (hours) –1 k = 0.107 hr 1. t½ = 6.5 hr . 2. MRT = 9.38 hr . 3. ( C p )0 = 4023 Units/L . 4. Units ⋅ hr AUC = 37775 ----------------------- . - 5. L 2 Units ⋅ hr - AUMC = 354697 -------------------------- . 6. L Vd = 1.9 L 7. L- Cl = 0.2065 ---- . 8. hr 4-30 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 31. I.V. Bolus Dosing Famotidine (Problem 4 - 12) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Kraus, et. al., quot;Famotidine--Pharmacokinetic Properties and Suppression of Acid Secretion in Pediatric Patients Following Car- diac Surgeryquot;, Clinical Pharmacokinetics (1990), p 77 - 80. Famotidine is a histamine H2-receptor antagonist. The study by Kraus, et. al., focuses on the kinetics of famotidine in children. In the study, ten children with normal kidney function and a body weight ranging from 14 - 25 kg, were each given a single intravenous 0.3 mg/kg dose of famotidine. Blood and urine samples were taken providing the following data: Famotidine PROBLEM TABLE 4 - 12. µg Serum concentration ------ L Time (hours) 0.33 300 0.50 250 1.00 225 4.00 125 8.00 70 12.00 40 16.00 15 From the data presented in the preceding table, determine the following assuming that the patient weighs 17.2 kg: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-31 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 32. I.V. Bolus Dosing (Problem 4 - 12) Famotidine: 10 3 ConCONCENTRATION (MIC/L) 10 2 (ug/mL) 10 1 0 5 10 15 20 Time (hours) –1 k = 0.17 hr 1. t½ = 3.9 hr . 2. MRT = 5.7 hr . 3. µg ( C p )0 = 285 ------ . 4. L µg ⋅ hr AUC = 1600 ---------------- . - 5. L 2 µg ⋅ hr AUMC = 9000 ------------------ . 6. L Vd = 18 L 7. Cl = 3.2L . 8. 4-32 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 33. I.V. Bolus Dosing Ganciclovir (Problem 4 - 13) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Trang, et. al., quot;Linear single-dose pharmacokinetics of ganciclovir in newborns with congenital cytomegalovirus infectionsquot;, Clin- ical Pharmacology and Therapeutics (1993), p. 15 - 21. Ganciclovir (mw: 255.23) is used against the human herpes viruses, cytomegalovirus retinitis, and cytomegalovirus infections of the gastrointestinal tract. In this study, twenty-seven newborns with cytomegalovirus disease were given 4 mg/kg of ganciclovir intravenously over one hour. Blood samples were taken and the data obtained is summarized in the following table: Ganciclovir PROBLEM TABLE 4 - 13. Time (hours) Serum concentration 1.50 4.50 2.00 4.00 3.00 3.06 4.00 2.40 6.00 1.45 8.00 0.87 From the data presented in the preceding table and assuming the patient weighs 3.6 kg, determine the following : Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-33 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 34. I.V. Bolus Dosing (Problem 4 - 13) Ganciclovir: 10 CONCENTRATION (MICMOLE/L) 10 10 0 2 4 6 8 TIME (HR) –1 k = 0.288hr 1. t½ = 2.4hr . 2. MRT = 3.5hr . 3. µmole ( C p )0 = 23 --------------- . - 4. mL µmole ⋅ hr AUC = 80 ------------------------- . - 5. mL 2 µmole ⋅ hr - AUMC = 280 ---------------------------- . 6. mL mg 1000µg 4 ------ ⋅ 3.6kg ⋅ ------------------ - - Dose ------------------------------------------------------------ kg mg - Vd = ------------ = - = 2.45L 7. µmole µg - Cp 0 --------------- ⋅ 255.23 --------------- - 23 µmole L L- Cl = 0.7 ---- . 8. hr 4-34 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 35. I.V. Bolus Dosing Imipenem (Problem 4 - 14) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Heikkila, Renkonen, and Erkkola, quot;Pharmacokinetics and Transplacental Passage of Imipenem During Pregnancyquot;, Antimicrobial Agents and Chemotherapy (Dec. 1992), p 2652 - 2655. Imipenem is a beta-lactam antibiotic which is used in combination with cilastin and is active against a broad spectrum of bacteria. The pharmacokinetics of Imipenem in pregnant women is established in this study. Twenty women (six of which were non-pregnant controls) were given a single intravenous dose of 500 mg of imipenem-cilastin (1:1). Blood samples were taken at various intervals and the data obtained is summarized in the following table: Imipenem PROBLEM TABLE 4 - 14. mg Serum concentration ------ - L Time (minutes) 10 27.00 15 23.50 30 15.50 45 9.50 60 6.50 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-35 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 36. I.V. Bolus Dosing (Problem 4 - 14) Imipenem: 2 10 CONCENTRATION (MG/L) 1 10 0 10 0 10 20 30 40 50 60 TIME (MIN) –1 k = 0.029 min 1. t½ = 24 min . 2. MRT = 34.5 min . 3. mg ( C p )0 = 36.2 ------ . - 4. L mg ⋅ min AUC = 1250 --------------------- . 5. L 2 mg ⋅ min - AUMC = 43125 ----------------------- . 6. L Dose 500mg Vd = ------------ = ------------------ = 13.8L - 7. Cp 0 mg 36.2 ------ - L L- Cl = 0.4 -------- . 8. min 4-36 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 37. I.V. Bolus Dosing Methylprednisolone (Problem 4 - 15) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Patel, et. al., quot;Pharmacokinetics of High Dose Methylprednisolone and Use in Hematological Malignanciesquot;, Hematological Oncology (1993), p. 89 - 96. Methylprednisolone is a corticosteriod that has been used in combination chemotherapy for the treatment of hemato- logical malignancy, myeloma, and acute lymphoblastic leukemia. In a study by Patel et. al., eight patients were given 1.5 gram intravenous doses of methylprednisolone from which the following data was obtained: Methylprednisolone PROBLEM TABLE 4 - 15. µg Serum concentration ------- - mL Time (hours) 0.5 19.29 1.0 17.56 1.8 15.10 4.0 9.98 5.8 7.10 8.0 4.70 12.0 2.21 18.0 0.71 24.0 0.23 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-37 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 38. I.V. Bolus Dosing (Problem 4 - 15) Methylprednisolone: 102 CONCENTRATION (MIC/ML) 101 100 Con (ug/mL) 10-1 0 5 10 15 20 25 Time (hours) –1 k = 0.188 hr 1. t½ = 3.69hr . 2. MRT = 5.3hr . 3. µg ( C p )0 = 21.2 ------- . - 4. mL µg ⋅ hr AUC = 112.5 ---------------- . - 5. mL 2 µg ⋅ hr AUMC = 598.4 ------------------ . 6. mL Vd = 71L 7. L- Cl = 13.3 ---- . 8. hr 4-38 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 39. I.V. Bolus Dosing Omeprazole (Problem 4 - 16) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Anderson, et. al., quot;Pharmacokinetics of [14C] Omeprazole in Patients with Liver Cirrhosisquot;, Clinical Pharmacokinetics (1993), p. 71 - 78. Omeprazole (mw: 345.42) is a gastric proton-pump inhibitor which decreases gastric acid secretion. It is effective in the treatment of ulcers and esophageal reflux. In normal patients 80% of the omeprazole dose is excreted as metabo- lites in the urine and the remainder is excreted in the feces. In the study by Anderson, et. al., eight patients with liver cirrhosis were given 20 mg, IV bolus doses of omeprazole. The patients had a mean body weight of 70 kg. Both blood were taken at various intervals throughout the study and the following data was obtained: Omeprazole PROBLEM TABLE 4 - 16. ρmole Serum concentration --------------- - mL Time (hours) 0.75 3.49 1.00 3.25 2.00 2.46 3.00 1.86 4.00 1.40 5.00 1.06 6.00 0.80 7.00 0.61 8.00 0.46 10.00 0.26 12.00 0.15 From the data presented in the preceding table, determine the following : Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-39 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 40. I.V. Bolus Dosing (Problem 4 - 16) Omeprazole: 10 1 ConCONCENTRATION (PICOMOLE/ML) 10 0 (umol/mL) 10-1 0 2 4 6 8 10 12 Time (hours) –1 k = 0.280hr 1. t½ = 2.5hr . 2. MRT = 3.57hr . 3. ρmole ( C p )0 = 4.3 --------------- . - 4. mL ρmole ⋅ hr AUC = 15.4 ------------------------- . - 5. mL 2 ρmole ⋅ hr - AUMC = 55 ---------------------------- . 6. mL Dose 20mg Vd = ------------ = ------------------------------------------------------------------------------------------------------------ = 13465L - 7. ρmole mmole - 345.42mg 1000mL Cp 0 4.3 --------------- ⋅ ------------------------ ⋅ ----------------------- ⋅ ------------------- - - - mL 10 9 ρmole mmole L L- Cl = 3.9 ---- . 8. hr 4-40 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 41. I.V. Bolus Dosing Pentachlorophenol (Problem 4 - 17) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Reigner, Rigod, and Tozer, quot;Absorption, Bioavailability, and Serum Protein Binding of Pentachlorophenol in the B6C3F1 Mousequot;, Pharmaceutical Research (1992), p 1053 - 1057. Pentachlorophenol (PCP) is a general biocide. That is, it is an insecticide, fungicide, bactericide, herbicide, algaecide, and molluskicide, that is used as a wood preservative. Extensive exposure to PCP can be fatal. In a study by Reigner et al, six mice (average weight: 27 g) were given 15 mg/kg of PCP by intravenous bolus. Blood samples were taken at various intervals from which the following data was obtained: Pentachlorophenol PROBLEM TABLE 4 - 17. µg Serum concentration ------- - mL Time (hours) 0.083 38.00 4.000 22.00 8.000 14.00 12.000 7.90 24.000 1.30 28.000 0.75 32.000 0.60 36.000 0.40 From the data presented in the preceding table, determine the following : Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-41 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 42. I.V. Bolus Dosing (Problem 4 - 17) Pentachlorphenol: CONCENTRATION (MIC/ML) 102 101 100 Con (ug/mL) 10-1 0 10 20 30 40 Time (hours) –1 k = 0.134 hr 1. t½ = 5.2hr . 2. MRT = 7.5hr . 3. µg ( C p )0 = 35.6 ------- . - 4. mL µg ⋅ hr AUC = 281 ---------------- . - 5. mL 2 µg ⋅ hr - AUMC = 2100 ------------------ . 6. mL Vd = 11.4mL 7. ml Cl = 1.5 ----- . - 8. hr 4-42 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 43. I.V. Bolus Dosing 9-(2-phophonylmethoxyethyl) adenine (Problem 4 - 18) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Naesens, Balzarini, and Clercq, quot;Pharmacokinetics in Mice of the Anti-Retrovirus Agent 9-(2-phophonylmethoxyethyl) adeninequot;, Drug Metabolism and Disposition (1992), p. 747- 752. 9-(2-phophonylmethoxyethyl) adenine (PEMA) is an anti-retrovirus (anti-HIV) agent. The pharmacokinetics of PEMA in mice were established in a study by . In this study there were three different PEMA doses given: 25 mg/kg, 100 mg/kg, and 500 mg/kg. Each of these doses was injected intravenously into male mice. The data obtained from study using the 25 mg/kg dose is summarized in the following table: 9-(2-phophonylmethoxyethyl) adenine PROBLEM TABLE 4 - 18. µg Serum concentration ------- - mL Time (minutes) 2.0 90.3 2.9 83.9 5.6 67.3 8.9 51.5 10.5 45.2 13.5 35.4 15.0 31.3 20.0 20.9 24.0 15.1 59.6 0.9 From the data presented in the preceding table, determine the following. (Assume that the mouse weighs 200g.) Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-43 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 44. I.V. Bolus Dosing (Problem 4 - 18) Pema: 10 2 CONCENTRATION (MIC/ML) 10 1 10 0 Con (ug/mL) 10 -1 0 10 20 30 40 50 60 Time (min) –1 k = 0.08min 1. t½ = 8.67min . 2. MRT = 12.5min . 3. µg ( C p )0 = 105 ------- . - 4. mL µg ⋅ hr AUC = 1300 ---------------- . - 5. mL 2 µg ⋅ hr - AUMC = 16250 ------------------ . 6. mL Vd = 47.6ml 7. mL- Cl = 3.8 -------- . 8. min 4-44 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 45. I.V. Bolus Dosing Thioperamide (Problem 4 - 19) Problem Submitted By: Maya Lyte AHFS 12:34.56 Antivirals Problem Reviewed By: Vicki Long GPI: 1234567890 Antivirals Sakurai, et. al., quot;The Disposition of Thioperamide, a Histamine H3-Antagonist, in Ratsquot;, J. Pharm. Pharmacol. (1994), p. 209 - 212. Thioperamide is a histamine (H3) receptor-antagonist. In a study by Sakurai et al, rats were given 10 mg/kg intrave- nous injections of Thioperamide. The following data was obtained from the study: Thioperamide PROBLEM TABLE 4 - 19. µg Serum concentration ------- - mL Time (minutes) 3.7 3.1 7.5 2.8 13 2.4 45 1.1 60 0.74 120 0.16 From the data presented in the preceding table, determine the following: Find the elimination rate constant, k . 1. Find the half life, t ½ . 2. Find MRT . 3. Find ( C p )0 . 4. Find the Area Under the Curve, AUC . 5. Find the area under the first moment curve, AUMC . 6. Find the volume of distribution, V d 7. Find the clearance, Cl . 8. 4-45 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 46. I.V. Bolus Dosing (Problem 4 - 19) thioperamide: 101 ConCONCENTRATION (MIC/ML) 100 (ug/mL) 10-1 0 20 40 60 80 100 120 Time (min) –1 k = 0.0254min 1. t½ = 27.3min . 2. MRT = 39.4min . 3. µg ( C p )0 = 3.39 ------- . - 4. mL µg ⋅ min AUC = 133.5 -------------------- . - 5. mL 2 µg ⋅ min - AUMC = 5256 ---------------------- . 6. mL mg 10 ------ - Dose ------------------------------------------------------------------- kg L- Vd = ------------ = - - = 2.95 ----- 7. µg Cp 0 mg - 1000mL kg 3.39 ------- ⋅ ------------------ ⋅ ------------------- - - mL 1000µg L L- 1000ml mL - –1 ⋅ 2.95 ----- ⋅ ----------------- = 75 ------------------- . - Cl = 0.0254min 8. min ⋅ kg kg L 4-46 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 47. I.V. Bolus Dosing Cocaine (Problem 4 - 20) Khan,vM. et. al. “Determination of pharmacokinetics of cocaine in sheep by liquid chromatography” J. Pharm. Sci. 76:1 (39-43) Jan 1987 4-47 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 48. I.V. Bolus Dosing 4.1.3 URINE From the Laplace Transform of a drug given by IV bolus we find that : ku ( –K ⋅ t ) Xu = ---- ⋅ X0 ⋅ ( 1 – e ) - (EQ 4-10) K where Xu is the cumulative amount of drug in the urine at time t. Rearranging, we get: ku ( Xu )∞ – Xu =  ---- ⋅ X 0 ⋅ e – Kt -  K (EQ 4-11) ku ( X u ) ∞ = ---- ⋅ X 0 . where the amount of drug that shows up in the urine at infinite time, - K Thus a plot of ( Xu )∞ – X u vs. time on semi-log paper would result in a straight line with a slope of -K and an intercept of ( X u ) ∞ .. and we can get ku from the intercept ( X u )∞ k u = K ⋅ ------------- and the slope. Rearranging the intercept equation, we get This method - X0 of obtaining pharmacokinetic parameters is known as the Amount Remaining to be Excreted (ARE) method. TABLE 4-4 Enalapril urinary excretion data from 5 mg IV Bolus Cumulative ∞ Enalapril in urine X – X u mg u Time (hr) (mg) 1 0.41 0.59 2 0.65 0.35 3 0.80 0.20 4 0.88 0.12 6 0.96 0.04 ∞ 1.0 ------ 4-48 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 49. I.V. Bolus Dosing Utilizations: A.R.E. Cumulative Enalapril in urine FIGURE 4-2. Method 0 10 0.2 Xu(inf) - Xu 0.1 -1 10 1.3 hr half life -2 10 0 1 2 3 4 5 6 Hours • You should be able to transform a data set containing amount of drug in the urine vs. time into cumulative amount of drug in the urine vs. time and plot the ARE. (Amount Remaining to be { ( Xu )∞ – Xu ( cum ) } vs. time on semi-log yielding a straight line with a slope of Excreted -> ku ⋅ X0 –1 ( Xu ) ∞ = --------------- = 1.0 mg – K = – 0.533 hr and an intercept of K • You should be able to determine the elimination rate constant, K1, from cumulative urinary excretion data. (Calculate the slope of the graph on SL paper.) • You should be able to determine the excretion rate constant, ku, from cumulation urinary excre- tion data. (Divide the intercept of the graph by X0 and multiply by K1. ( X u )∞ –1 1.0 mg –1 k u = K ⋅ ------------- = 0.53 hr ⋅ ---------------- = 0.106 hr ) - - X0 5.0 mg • You should be able to determine k m . K = k u + k m • You should be able to calculate percent metabolized or excreted from a data set. Thus, km ku ----- ⋅ 100 and percent excreted unchanged = ---- ⋅ 100 assuming - - Percent metabolized = K K K = k u + km A second method is to plot the rate at which the drug shows up in the urine over time. Again, using the LaPlace transforms, we find that: dX u –K t –K t -------- = k u ⋅ X0 ⋅ e = R0 ⋅ e - (EQ 4-12) dt Thus, a plot of the rate of excretion vs. time results in a straight line on semi-log Utilization: Rate of excretion method paper with a slope of -K1 and an intercept, R0 , of kuX0 . Rearranging the intercept 4-49 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 50. I.V. Bolus Dosing R0 equation yields k u = ----- . In real data, we don’t have the instantaneous excretion - X0 ∆X u dX u rate , but the average excretion rate, --------- , over a much larger interval. What - ∆t dt that means to our calculations is that over the interval of data collection, the total amount of drug collected divided by the total time interval is the average rate. In the beginning of the interval the rate was faster than at the end of the interval. So the average rate must have occurred in the middle of the interval. Thus equation 4- 12 which is the instantaneous rate can be rewritten to ∆Xu –K t –K t mid mid --------- = k u ⋅ X 0 ⋅ e = R0 ⋅ e - (EQ 4-13) ∆t TABLE 4-5 Enalapril Urinary Rate Data ∆X u Enalapril in --------- - urine ∆X u ,(mg) ∆t ∆t Interval (hr) t(mid) 0-1 0.5 1 0.41 0.41 1-2 1.5 1 0.24 0.24 2-3 2.5 1 missed sample ? 3-4 3.5 1 0.08 0.08 4-6 5 2 0.08 0.04 • You should be able to transform a data set containing amount of drug in the urine vs. time inter- ∆X u t ,(t mid the time of the midpoint of the interval), on semilog val into Average Rate, --------- , vs. - ∆t – K and an intercept of k u ⋅ X 0 . as shown below. yielding a straight line with a slope of -1 0 10 R0 = 0.53 mg/hr Urinary Excretion Rate (mg/hr) -1 -2 10 1.3 hr half life -2 10 0 1 2 3 4 5 T (Mid) • You should be able to determine k u extrapolate the line to t = 0 . The value of Rate (at t = 0 ), R0, = k r ⋅ X0 = 0.53 ( mg ⁄ hr ) which when divided by X 0 .is kr. 4-50 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 51. I.V. Bolus Dosing R 0.53mg/hr –1 0 Thus, ----- = ------------------------ = 0.106hr - - 5mg X0 • You should be able to determine k m . K = k u + k m • You should be able to calculate percent metabolized or excreted from a data set. The rate equation is superior clinically because the ARE method requires collec- tion of all of the urine which is usually only possible when you have a catheterized patient while the Rate Method does not. (People don’t urinate on command, and your data could be in the toilet, literally.) ∞ An additional advantage of the rate equations is that the has the units of AUC 0 mass, which gives the total amount of drug excreted into the urine directly. Thus: ∞ R 0 0.53 mg/hr = ----- = -------------------------- = 1 mg - AUC –1 K 0 0.53 hr AN INTERESTING OBSERVATION: If you look at the LaPlace Transform of the rate equation for any terminal compartment, you would see that the resulting equa- tion is that of the previous compartment times the rate constant through which the drug entered the terminal compartment. Thus, the rate of drug showing up in the urine (terminal compartment) is: dX u –K t –K t -------- = k u ⋅ X 0 ⋅ e = R0 ⋅ e - dt where ku is the rate constant through which the drug entered the urine and dX –K t ------ = X 0 ⋅ e is the equation of the previous compartment. - dt 4-51 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 52. I.V. Bolus Dosing 4.2 Metabolite 4.2.1 PLASMA Remember, the LaPlace Transform of the metabolite data yielded ( km ⋅ Xo ) ( km ⋅ Xo ) –Kt – K 1t – K 1t – Kt X m = --------------------- ⋅ ( e – e ) or Xm = --------------------- ⋅ ( e – e ) depending on - - ( K1 – K ) ( K – K1 ) which rate constant that we arbitrarily assigned to be K, the summation of all the ways that the drug is removed from the body and K1, the summation of all the ways that the metabolite is removed from the body. When we begin to manipulate the data, we know that we have a curve with two different exponents in it. (If they were the same, the equation would be different.) We don’t know which is bigger, K1 or K, but we can rewrite the equation to simply reflect Klarge and Ksmall, know- ing that one is K1 and the other is K but not which is which. If we, then, devided both sides of the equation by Vdm, the volume of distribution of the metabolite, we would get : – ( K l arg e ⋅ t )   X 0   – ( Ksmall ⋅ t )   km C pm =  -------------------------------------   ---------   e  (EQ 4-14) - - –e  K l arg e – K small  V dm   • You should be able to plot a data set of plasma concentration of metabolite vs. time on semi-log Utilization: Curve Stripping paper yielding a bi-exponential curve. –k t –k t – Kt l arg e small → 0 as t → ∞ . And e → 0 faster than e → 0 . So, at some long e –K t –K t –K t l arg e small l arg e time, t, e «e e . In fact is small enough to be ignored. Thus at long time, t, the equation becomes : –( K ⋅ t) km X0 C pm =  ----------------------------------   --------   e small  - -  K l arg e – K small  V dm   (EQ 4-15) So that the plot of the terminal portion of the graph would yield a straight line with a slope of km X0  ----------------------------------   --------  - - -Ksmall and an intercept of I =  K l arg e – K small  Vdm • You should be able to obtain the slope of the terminal portion of the curve, the negative of which would be the smaller of the two rate constants, K small , (either the summation of all the K , or the summation of all the ways that the metabolite is ways that the drug is eliminated, K1 ). eliminated, • Subtracting the two previous equations yields 4-52 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 53. I.V. Bolus Dosing –( K ⋅ t) km X0 C pm – C pm =  ----------------------------------  --------   e big  - -  K l arg e – K small  Vdm   (EQ 4-16) which is a straight line on semi-log paper with a slope of -kbig and an intercept of km X0 I =  -----------------------------------   --------- . Note: we can get the larger of the two rate constants from this -  K l arg e – K small  V dm method. TABLE 4-6 Drug Metabolite (1) (2) (3) (4) (5) Cp – Cpm Cpm Cpm Time (hr) (mcg/L) Cpm1 (mcg/L) 0 0 181.2 181.2 0.5 24.7 175 150.3 1 44.4 168.9 124.5 2 139 71.8 157.5 85.7 4 65.6 96.5 136.9 40.4 6 31.1 100 119 19 8 14.6 94.7 12 76.5 24 34 In the above data Cp vs. Time is the plasma profile of the drug from Table 4-1 on page 2 and Cpm1 vs. Time is the plasma profile of the metabolite. A plot of Cp vs. Time yielded a straight 0.693 –1 line with a slope,(-K) of -0.375 hr-1, K = --------------------- = 0.375 hr and and intercept of 295 mic/ - –1 1.85 hr L, 4-53 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 54. I.V. Bolus Dosing Figure 4-1 on page 3 (column 2 vs. 1 in Table 4-6 on page 53) 10 3 Cpo = 295 mic/L Concentration (mic/L) 100 10 2 Concentration (ng/mL) 50 1.85 hr 10 1 0 2 4 6 8 Time (hours) Time (hr) while a plot of Cpm1 vs. Time( Figure 4-3 on page 54) yields a biexponential plot with a termi- nal slope of 0.07 hr-1 , k small = 0.693 and extrapolating the terminal line back to time = 0 ------------ - 10 hr yields 181 mic/L. Nifedipine Metabolite (column 3 vs. 1 in Table 4-6 on page 53) FIGURE 4-3. ) Nifedipine IV bolus - Metabolite 103 mic Cpm0 = 181 -------- - L Concentration (mic/L) 80 102 40 10 hr 101 0 4 8 12 16 20 24 Time (hours) 4-54 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 55. I.V. Bolus Dosing • You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to t = 0 ) terminal line (column 4 vs. 1 in Table 4-6 on page 53) and the observed data (at early times) (column 3 vs. 1 in Table 4-6 on page 53) yield- ing a straight line with the slope of the line equal to the negative of the other (larger) rate con- stant (column 5 vs. 1 in Table 4-6 on page 53). First you would fill in the Cpm column (column 4 in Table 4-6 on page 53) by computing Cpm – k small t for various values of time i.e Cpm = Cpm 0 ⋅ e where – k small is the terminal slope of the graph. Then Cpm – Cpm (column 5 in Table 4-6 on page 53) would be column 4 - column 3. Then a plot of Cpm – Cpm vs. time (column 5 vs. 1 in Table 4-6 on page 53) is shown below. Curve strip of Nifedipine Metabolite data FIGURE 4-4. 3 10 Intercept 2 Column 5 100 102 1 1.85 hr 50 Half life 1 10 0 1 2 3 4 5 6 Time (hr) In this case, the slope of the stripped line line is -0.375 hr-1 and the intercept is 0.181.2 mic/L. The slope of -0.375 hr-1 should not be surprising as the plot of the data in Figure 4-3 on page 54 resulted in a terminal slope of -.07 hr-1 . Since the data set yielded a bi-exponential plot, sepa- rating out the exponents could only yield K (0.375 hr-1) or K1 as determined by our Laplace Transform information. Thus, the terminal slope could be either -K1 or -K. Since it was obvi- ously not -K, it had to be -K1. Thus the other rate constant obtained by stripping has to be K. You can determine which slope is which rate constant if you have any data regarding intact drug (i e. either plasma or urine time profiles of intact drug) as the slope of any of those profiles is –K . always • You should be able to determine V dm if you have any urine data regarding intact drug (i.e. urine time profiles of intact drug) as the intercept of those profiles allow for the solution of k m . Thus the intercept, I, of the extrapolated line of equation 4-14 could be rearranged to contain 1000 mic –1 0.375hr ⋅ 25mg ⋅ ---------------------- km ⋅ X0 mg - = ----------------------------------------------- = ------------------------------------------------------------------------- = 170 L . only one unknown variable, V dm ( K l arg e – K small ) ⋅ I mic –1 ( 0.375 – 0.07 ) hr ⋅ 181.2 -------- - L 4-55 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 56. I.V. Bolus Dosing • You should be able to determine the rate constants using MRT calculations. Utilization: MRT Calculations In a caternary chain, each compartment contributes its MRT to the overall MRT of the drug, thus: IV Bolus Flow Chart 4-4 K X MRT(IV) = 1/K Suppose the drug were given by IV bolus. Then the drug would have to be metabolized and the metabolite eliminated. Since the MRTs are additive, the overall MRT of the metabolite would be made up of the MRTs of the two processes, thus: Metabolite Flow Chart 4-5 km kmu X Xm MRT(met) = MRT(elim)+MRT(IV) MRT(met) = 1/K1 + 1/K Thus, using the data from Table 4-3 on page 5 the MRT(IV)Trap is MRT = AUMC = 1986.1 = 2.42 hr or about MRT = AUMC = 2100 = 2.67 hr using calculus. ----------------- - --------------- - ----------------- - ----------- - AUC 819.9 AUC 787 And using the data from columns 1 and 3 from Table 4-6 on page 53 the MRT(met) using calcu- lus is MRT = AUMC = 36000 = 17 hr. ----------------- - -------------- - AUC 2116 MRT(elim) = MRT(met) - MRT(IV) = 17 hr - 2.67 hr = 14.33 hr = 1/K2. Thus K2 = 0.07 hr-1. 4-56 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf
  • 57. I.V. Bolus Dosing 4.2.2 URINE Valid equations: k mu ⋅ k m X0  – K small t – K l arg e t  dXmu ------------ = --------------------------------------- ⋅  e  - - –e (EQ 4-17) ( K l arg e – K small )   dt as in the previous urinary rate equation, clinically we work with the average rate Utilization: over a definite interval which results in rewriting equation 4-17 as: k mu ⋅ k m X 0 ∆Xmu  – Ksmall tmid – K l arg e t mid  ------------ = --------------------------------------- ⋅  e  - - –e (EQ 4-18) ( K l arg e – K small )  ∆t  • You should be able to plot a data set of rate of metabolite excreted vs. time (mid) on semi-log paper yielding a bi-exponential curve. • You should be able to obtain the slope of the terminal portion of the curve, the negative of K1 or K ). which would be the smaller of the two rate constants (either • You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to t = 0 ) terminal line and the observed data (at early times) yielding a straight line with the slope of the line equal to the negative of the other (larger) K1 or K ). rate constant (either • You should be able to utilize MRT calculations to obtain K1 and K . • You should be able to determine which slope is which rate constant if you have any data regard- ing intact drug (i.e. either plasma or urine time profiles of intact drug) as the slope of any of –K . those profiles is always By this time, it should be apparent that data which fits the same shape curve (mono-exponential, bi-exponential, etc.) are treated the same way. When the curves are evaluated, the slopes and intercepts are obtained in the same manner. The only difference is what those slopes and intercepts represent. These represen- tations come from the equations which come from the LaPlace Transforms which come from our picture of the pharmacokinetic description of the drug. Please refer back to the section on graphical analysis in the Chapter 1, Math review for a interpretation of slopes and intercepts of the various graphs. Temporarily, please refer to exam section 1, chapter 14 for problems for this sec- tion (until problems can be generated) as well as additional problems for the previ- ous sections. 4-57 Basic Pharmacokinetics REV. 00.1.27 Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf