Pharmacokinetics Rate Method - PDF

The rate method (Example) Time interval (h) Volume (mL) Concentration (mg/mL) 0-1 200 0.200 1-2 50 0.400 2-3 50 0.200 3-4 100 0.050 4-5 25 0.100 5-6 125 0.010 6-12 133 0.005 10 The rate method: 1- Calculate amount of drug eliminated Time interval (h) Volume (mL) Concentrat...

The rate method (Example) Time interval (h) Volume (mL) Concentration (mg/mL) 0-1 200 0.200 1-2 50 0.400 2-3 50 0.200 3-4 100 0.050 4-5 25 0.100 5-6 125 0.010 6-12 133 0.005 10 The rate method: 1- Calculate amount of drug eliminated Time interval (h) Volume (mL) Concentration (mg/mL) Drug amount in the urine (mg) 0-1 200 0.200 40 1-2 50 0.400 20 2-3 50 0.200 10 3-4 100 0.050 5 4-5 25 0.100 2.5 5-6 125 0.010 1.25 6-12 133 0.005 0.665 Amount = volume*conc 11 The rate method: 2- Calculate the change in time Time interval (h) Volume (mL) Concentration (mg/mL) Drug amount in the urine (mg) Δt (hr) 0-1 200 0.200 40 1 1-2 50 0.400 20 1 2-3 50 0.200 10 1 3-4 100 0.050 5 1 4-5 25 0.100 2.5 1 5-6 125 0.010 1.25 1 6-12 133 0.005 0.665 6 12 The rate method: 3- Calculate the rate of urinary excretion Time interval (h) Volume (mL) 0-1 200 1-2  Du     t  Drug amount in the urine (mg) Δt (hr) 0.200 40 1 40 50 0.400 20 1 20 2-3 50 0.200 10 1 10 3-4 100 0.050 5 1 5 4-5 25 0.100 2.5 1 2.5 5-6 125 0.010 1.25 1 1.25 6-12 133 0.005 0.665 6 0.111 Concentration (mg/mL) (mg/hr) 13 The rate method: 4- Plot time (mid of interval) vs. (dDu/dt) Time (h)  Du     t  (mg/hr) 0.5 40 1.5 20 2.5 10 3.5 5 4.5 2.5 5.5 1.25 9 0.111 ∆𝐷𝑈 = 𝑘𝑒 ∙ 𝐷𝐵0 ∙ 𝑒 −𝑘𝑡 ∆𝑡 Intercept Slope 14 The rate method: 5- Estimate PK parameters from the values of the slope and the intercept • The urinary excretion rate is described according to the following Equation: ∆𝐷𝑈 = 56.5 ∙ 𝑒 −0.693𝑡 ∆𝑡 • Hence, 𝑘 = 0.693 𝑘𝑒 ∙ 𝐷𝐵0 = 56.5 𝑘𝑒 = 56.5 𝐷𝐵0 56.5 = = 0.471 ℎ𝑟 −1 120 15 Additional calculations • Fraction of drug renally excreted = 𝑘𝑒 𝑘 • First order metabolism rate constant: 𝑘𝑚 = 𝑘 − 𝑘𝑒 • Cumulative amount excreted in the urine at t=∞: 𝐷𝑈∞ 𝑘𝑒 = ∙ 𝐷𝐵0 𝑘 16 Example 1 continued • From previous Example, calculate the following: • Fraction of drug renally excreted = 𝑘𝑒 𝑘 = 0.471 0.693 = 0.68 (This implies that 68% of the IV Bolus dose is renally excreted) • First order metabolism rate constant: 𝑘𝑚 = 𝑘 − 𝑘𝑒 = 0.693 − 0.471 = 0.222 ℎ𝑟 −1 • Cumulative amount excreted in the urine at t=∞: 𝐷𝑈∞ 𝑘𝑒 0.471 0 = ∙ 𝐷𝐵 = ∙ 120 = 81.6 𝑚𝑔 𝑘 0.693 17 General comments on the use of urinary data in PK analysis • Urine collection is a non-invasive technique • It is, perhaps, a more convenient method of sample collection, and sample size is generally not a problem. The sampling time, however, reflects drug in urine collected over a period of time, rather than a drug concentration at a discrete time • Urinary data allows direct measurement of bioavailability, both absolute and relative, without the need of fitting the data to a mathematical model • Volume of Distribution cannot be calculated without plasma concentrations 18 Cumulative amount eliminated in the urine • Cumulative amount eliminated in the urine at infinity (𝐷𝑈∞ ) is calculated according to the following: 𝑘𝑒 0 ∞ 𝐷𝑈 = 𝐷𝐵 𝑘 • A drug with 40% renal elimination, an IV bolus dose of 100 mg is expected to result in a cumulative amount eliminated in the urine of 40 mg 19 Example 2 • A new drug was given in a single intravenous dose of 200 mg to an adult male patient. Assuming an elimination half-life of 6 hours, with 30% renal excretion. • What is the excretion rate constant? • What is the Cumulative amount eliminated in the urine at infinity? 20 Example 2 • Elimination rate constant (k) is calculated according to: 0.693 0.693 𝑘= = = 0.116 ℎ𝑟 −1 𝑡1ൗ 6 2 • Excretion rate constant (ke): 𝑘𝑒 = 0.3 ∙ 𝑘 = 0.3 ∙ 0.116 ℎ𝑟 −1 = 0.035ℎ𝑟 −1 • What is the Cumulative amount eliminated in the urine at infinity? 𝑘𝑒 0 ∞ 𝐷𝑈 = 𝐷𝐵 = 0.3 ∙ 200 = 60 𝑚𝑔 𝑘 • An IV bolus dose of 200 mg is expected to result in a cumulative amount eliminated in the urine of 60 mg 21 Chapter 3 One-Compartment IV bolus Part 1 Practice Problems 1 Example 1 • Ten mg metoclopramide was administered intravenously to a 72 kg patient. The minimum plasma concentration required to cause significant enhancement of gastric emptying is 50 ng/mL. The following plasma concentrations were observed after analysis of the specimen. 2 Example 1 Time (hr) Conc. (ng/ml) 1 90.0 2 68.0 4 40.0 6 21.5 8 12.0 10 7.0 3 Example 1 • Calculate the biological half-life of the drug elimination (t½), the overall elimination rate constant (K), the volume (Vd), and the duration of action 4 Example 1 5 Example 1 • The elimination rate constant can be obtained from the equation: k = 0.286 hr 6 −1 Example 1 • Another way to calculate the slope (if you do not have the ability to do regression) is using: log 𝐶𝑃2 − log 𝐶𝑃1 𝑆𝑙𝑜𝑝𝑒 = 𝑡2 − 𝑡1 where (𝐶𝑃1 ,t1) and (𝐶𝑃2 ,t2) are two different conc. time points • It is important to note that the first method for calculating the slope is more accurate 7 Example 1 • Calculating the slope using the second method: log(40) - log(21.5) Slope = = -0.13481 4-6 𝑘 = −2.3 ∙ 𝑆𝑙𝑜𝑝𝑒 = −2.3 ∙ −0.13481 = 0.310 ℎ𝑟 −1 • Note that the value of k is similar to the value estimated using the first method (0.286) 8 Example 1 • the biological half-life of the drug elimination (t½): t 0.5 0.693 0.693 = = = 2.42 hr K 0.286 • The volume of distribution (Vd): dose 10 mg Vd = 0 = CP 121.1 ng/ml 6 mg 10 ng L = 0.083   3 = 83 L ng/ml mg 10 ml 9 Example 1 • the duration of action (td). td is the time needed for the conc. To get to 50 ng/ml : C   ln C *  t= 0 P K  121.1  ln  50   = 10 0.286 = 3.1 hr Example 2 • An adult male patient was given the first dose of an antibiotic at 6:00 AM. At 12:00 noon the plasma level of the drug was measured and reported as 5 µg/ml. The drug is known to follow the one compartment model with a half-life of 6 hours. The recommended dosage regimen of this drug is 250 mg q.i.d. the minimum inhibitory concentration is 3 µg/ml. Calculate the following: – – – – – – – – – Apparent volume of distribution Expected plasma concentration at 10 AM. Duration of action of the first dose Total body clearance Fraction of the dose in the body 5 hours after the injection Total amount in the body 5 hours after the injection Cumulative amount eliminated 5 hours after the injection Total amount in the body immediately after injection of a second dose at 12:00 noon Duration of action of first dose only if dose administered at 6:00 AM was 500 mg. 11 Example 2 • Elimination rate constant: K = 0.693 t 0.5 = 0.693 = 0.116 hr −1 6 • Initial concentration: – The conc. at 12:00 noon (6 hrs after the first dose) is 5 µg/ml: C (t = 5) = C 0  e − k t P P C P (t = 5) 5 C = = (−0.116 )(6 ) = 10 ug/ml = 10 mg/L − k t e e 0 P 12 Example 2 • Note: The initial concentration can be calculated without the use of calculator. Since the half-life is 6 hrs and the Conc. 6 hours after the dose is 5 µg/ml, the initial plasma conc. is 10 µg/ml. 13 Example 2 • Apparent volume of distribution: The initial plasma concentration is 10 mg/L. D 0B 250 mg VD = 0 = = 25 L C P 10 mg L • Expected plasma concentration at 10 AM (4 hrs after dose adminstration) C P = C P0 e − K t = 10e −0.1164 = 6.29 mg/L 14 Example 2 • Duration of action of the first dose  C 0P  ln *  CP   Duration of Action = • Total body clearance k  10  ln  = 3 0.116 = 10.4 hr Cl = K  VD = 2.89 L/hr • Fraction of the dose in the body 5 hours after the injection 5 6 1 F =   = 0.56 2 15 Example 2 • Total amount in the body 5 hours after the injection = (0.56)(250 mg) = 140 mg • Cumulative amount eliminated 5 hours after the injection = dose – amount in the body = 250 – 140 = 110 mg • Total amount in the body immediately after injection of a second dose at 12:00 noon Total amount = amount from the first dose + amount from the second dose = 125 + 250 = 375 mg 16 Example 2 • Duration of action of first dose only if dose administered at 6:00 AM was 500 mg t d = 10.42 hr + 6 hr = 16.42 hrs • Note that 6 hrs (one t0.5) is needed for the amount in the body to decline from 500 mg to 250 mg 17 Example 3 • The therapeutic range of a drug is 20-200 mg/L. After an intravenous bolus injection of 1.0 gm followed by regression analysis of the concentration of the drug in plasma (in units of mg/L) versus time (in hours), the following linear equation was obtained log Cp = 2 − 0.1t • Calculate the following – Duration of action – Total body clearance – Rate of elimination at 2 hours 18 Example 3 • From the equation: k log Cp = 2 − 0.1t = log(C ) − t 2.3 0 P The following were estimated: C P0 = 10 2 = 100 mg/L k = −Slope  2.3 = (0.1)  (2.3) = 0.23 hr −1 D 0B 1000 mg VD = 0 = = 10 L C P 100 mg/L 19 Example 3 • Duration of action:  C 0P  ln *  C Duration of Action =  P  k  100  ln  =  20  0.23 = 7 hr • Total body clearance= k∙VD=(0.23)(10) =2.3 L/hr • Rate of elimination at 2 hours: • Elimination rate = Cl*CP(t=2) = 2.3*63 =145 mg/hr log(Cp(t = 2)) = 2 − (0.1)(2) = 1.8  Cp(t = 2) = 101.8 = 63 mg/L 20 Example 4 • A 70-kg volunteer is given an intravenous dose of an antibiotic, and serum drug concentrations were determined at 2 hours and 5 hours after administration. The drug concentrations were 1.2 and 0.3 µg/mL, respectively. What is the biologic half-life for this drug, assuming first-order elimination kinetics? 21 Example 4: First approach • Calculating the slope: log(0.3) - log(1.2) Slope = = -0.201 5-2 𝑘 = −2.3 ∙ 𝑆𝑙𝑜𝑝𝑒 = −2.3 ∙ −0.201 = 0.462 ℎ𝑟 −1 t 0.5 0.693 0.693 = = = 1.5 hr k 0.462 22 Example 4: Second approach 1.2 µg/ml 1 halflife 0.6 µg/ml 2 hours 1 halflife 0.3 µg/ml 5 hours 3 hours = 2 half-lifes Half-life = 1.5 hour 23 Example 5 • A new drug was given in a single intravenous dose of 200 mg to an 80-kg adult male patient. After 6 hours, the plasma drug concentration of drug was 15 µg/mL. Assuming that the apparent V D is 10% of body weight, compute the total amount of drug in the body fluids after 6 hours. What is the half-life of this drug? 24 Example 5 • The total amount of drug in the body fluids after 6 hours 𝑉𝐷 = 10% ∙ 𝑤𝑒𝑖𝑔ℎ𝑡 = 8 𝐿 𝑚𝑔 𝐷𝐵 𝑡 = 6 ℎ𝑟𝑠 = 𝐶𝑃 𝑡 = 6 ℎ𝑟𝑠 ∙ 𝑉𝐷 = 15 ∙ 8 𝐿 = 120 mg 𝐿 25 Example 5 • What is the half-life of this drug? • Calculating the slope (using D B): log(120) - log(200) Slope = = -0.0370 6-0 𝑘 = −2.3 ∙ 𝑆𝑙𝑜𝑝𝑒 = −2.3 ∙ −0.0370 = 0.0851 ℎ𝑟 −1 t 0.5 0.693 0.693 = = = 8.14 hr k 0.0851 26 Example 6 • A 50-kg woman was given a single IV dose of an antibacterial drug at a dose level of 6 mg/kg. Blood samples were taken at various time intervals. The concentration of the drug (Cp) was determined in the plasma fraction of each blood sample and the following data were obtained: Time CP (μg/mL) 0.25 8.21 0.5 7.87 1 7.23 3 5.15 6 3.09 12 1.11 18 0.4 27 Example 6 a) What are the values for VD, k, and t1/2 for this drug? b) This antibacterial agent is not effective at a plasma concentration of less than 2 µg/mL. What is the duration of activity for this drug? c) How long would it take for 99.9% of this drug to be eliminated? d) If the dose of the antibiotic was doubled exactly, what would be the increase in duration of activity? 28 Example 6 29 Example 6 • The elimination rate constant can be obtained from the equation: k = 0.17 hr 30 −1 Example 6 • Another way to calculate the slope (if you do not have the ability to do regression) is using: log 𝐶𝑃2 − log 𝐶𝑃1 𝑆𝑙𝑜𝑝𝑒 = 𝑡2 − 𝑡1 where (𝐶𝑃1 ,t1) and (𝐶𝑃2 ,t2) are two different conc. time points • It is important to note that the first method for calculating the slope is more accurate 31 Example 6 • Calculating the slope using the second method: log(3.09) - log(5.15) Slope = = -0.0740 6-3 𝑘 = −2.3 ∙ 𝑆𝑙𝑜𝑝𝑒 = −2.3 ∙ −0.074 = 0.170 ℎ𝑟 −1 • Note that the value of k is similar to the value estimated using the first method 32 Example 6 • the biological half-life of the drug elimination (t½): t 0.5 0.693 0.693 = = = 4.08 hr K 0.17 • The volume of distribution (Vd): dose 6 * 50 Vd = 0 = CP 8.57 g/ml 300 mg = = 35 L 8.57 mg/L 33 Example 6 • Duration of action:  C 0P  ln *  C Duration of Action =  P  k  8.57  ln  =  2  34 0.17 = 8.56 hr Example 6 • How long would it take for 99.9% of this drug to be eliminated? 𝐶𝑃 = 𝐶𝑃0𝑒 −𝑘𝑡 (𝐶 = 0.001 ∙ 𝐶0 ) 0.001 ∙ 𝐶0 = 𝐶0 𝑒 −𝑘𝑡 0.001 = 𝑒 −𝑘𝑡 ln 0.001 = ln 𝑒 −𝑘𝑡 6.908 6.908 𝑡= = = 40.6 ℎ𝑟𝑠 𝑘 0.17 35 −6.908 = −𝑘𝑡 Example 6 • If the dose of the antibiotic was doubled exactly, what would be the increase in duration of activity? – Please refer to Example 2 36 Example 7 • A new antibiotic drug was given in a single intravenous bolus of 4 mg/kg to 5 healthy male adults ranging in age from 23 to 38 years (average weight 75 kg). The pharmacokinetics of the plasma drug concentration–time curve for this drug fits a one-compartment model. The equation of the curve that best fits the data is 𝐶𝑃 = 78 𝑒 −0.46𝑡 37 Example 7 • Determine the following (assume units of mg/mL for Cp and hours for t): A. B. C. D. E. What is the t1/2? What is the VD? What is the plasma level of the drug after 4 hours? How much drug is left in the body after 4 hours? Predict what body water compartment this drug might occupy and explain why you made this prediction. F. Assuming the drug is no longer effective when levels decline to less than 2 mg/mL, when should you administer the next dose? 38