University:
Michigan State University
Course:
MED 609 | Hematology Clerkship
Academic year:

2021
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142

Pages:

47
Author:

Vallesewp7

ONE-COMPARTMENT OPEN MODEL: INTRAVENOUS BOLUS ADMINISTRATION: The one-compartment open model offers the simplest way to describe the process of drug distribution and elimination in the body. This model assumes that the drug can leave the body (i.e., the model is "open"), and the body acts like a single, uniform compartment (kinetic). The simplest route of drug administration from a modeling perspective is a rapid intravenous injection (IV bolus). The simplest kinetic model that describes drug disposition in the body is to consider that the drug is injected all at once into a box, or compartment, and that the drug distributes instantaneously and homogenously (kinetically) throughout the compartment. Drug elimination also occurs from the compartment immediately after injection. One-Comp. Open Linear Model Assumptions • Rapid Mixing – drug is mixed instantaneously in blood or plasma. • One compartment – drug in the blood (plasma) is in rapid equilibrium with drug in the extravascular tissues. • Linear Model – drug elimination follows first order kinetics. Changes in the plasma drug concentration reflect changes in drug concentrations in other tissues. However, the plasma drug concentration does not equal the concentration at other sites but rather indicates how it changes with time. Generally, if the plasma concentration of a drug is decreasing, the concentration in tissues will also decrease. Figure 1.3 is a simplified plot of the drug concentration versus time profile following an intravenous drug dose and illustrates the property of kinetic homogeneity. The property of kinetic homogeneity is important for the assumptions made in clinical pharmacokinetics. It is the foundation on which all therapeutic and toxic plasma drug concentrations are established. That is, when studying concentrations of a drug in plasma, we assume that these plasma concentrations directly relate to concentrations in tissues where the disease process is to be modified by the drug (e.g., the central nervous system in Parkinson's disease or bone in osteomyelitis). This assumption, however, may not be true for all drugs. Linear Kinetics (First order) Elimination rate or change in concentration is proportional to the amount available for elimination. Glumerular Filtration } Passive -Tubular secretion -Biliary secretion -Biotransformation } Involve enzymatic processes (active) At therapeutic Levels most drugs do not reach the saturation levels of those enzymes. Accordingly the whole elimination process can be approximated very well by first order kinetics. 1-Comp. Model: IV Bolus Dosing Xt: the amount of drug remained in the compartment K: first-order elimination rate constant (OVERALL) (unit = time-1) Rate of elimnation  dX   KX dt X  Xo e C  Co e  Kt Vd  Xt  Kt Ct ln Cp  ln Co  Kt logCp  logCo  Kt 2.303 Apparent Volume of Distribution (Vd) This apparent volume of distribution is not a physiological volume. It won't be lower than blood or plasma volume but it can be much larger than body volume for some drugs. It is a mathematical factor relating the amount of drug in the body and the concentration of drug in the measured compartment, usually plasma: Vd = AMOUNT of drug in the body CONCENTRATION in plasma Factors affecting drug distribution: Rate of distribution - Membrane permeability - Lipid Solubility - pH - pKa (pH-partition theory for ionizable molecules) - Blood perfusion of organs and tissues Extent of Distribution - Plasma protein binding - Intracellular binding Diagram IV-3. Apparent Volume of Distribution Volume of Distribution Erythropoietin Warfarin 5L 8L 0.07 L/kg* 0.12 L/kg* Phenytoin Digoxin Amiodarone Chloroquine Quinacrine 45 L 500 L 5000 L 15000 L 35000 L 0.63 L /kg* 7 L /kg* 70 L /kg* 215 L/kg* 500 L/kg* * Distribution Coefficient Body water Intra-vascular Interstitial Intra-cellular Distribution - Evan’s Blue Intra-vascular space only Distribution - Ethanol All water Distribution - Quinacrine Concentration into cells Volume of distribution A measure of the tendency of a drug to move out of the blood plasma to some other site. Volume of Distribution Erythropoietin Warfarin Phenytoin Digoxin Amiodarone Chloroquine Quinacrine 5L 8L 45 L 500 L 5000 L 15000 L 35000 L * Distribution Coefficient 0.07 L/kg* 0.12 L/kg* 0.63 L /kg* 7 L /kg* 70 L /kg* 215 L/kg* 500 L/kg* Patterns of Vd Relationship Between the Extent of Distribution and Vd in a 70 kg Normal Man % Vd, L Body Weight 5 7 5-20 7-28 20-40 28-56 >40 >56 Extent of Distribution Only in plasma In extracellular fluids In total body fluids. In deep tissues; bound to peripheral tissues Conc (mg/L) 1000 y = 149.95e-0.203x 100 10 0 2 4 Time (hr) 6 Determination of K Cp versus Time Data Time (hr) 1 2 4 8 12 Cp (mg/L) 20 15 6.8 3.2 1.3 Elimination half-life (t1/2) Definition: Elimination half-life is the time it takes the drug concentration in the blood to decline to one half of its initial value. It is a secondary parameter : The elimination half-life is dependent on the ratio of CL and VD. Unit : time (min, h, day) Half-life Area Under the Conc. Time Curve AUC • Model dependent Approach According to the equation (model) AUC  Xo Xo Co   K .Vd Cl K 2) Model-independent Approach Trapezoidal Rule Area between t2 and t3 Total Area Clearance (CL) Definition : Clearance of a drug is the ratio of the rate of elimination by all routs to the concentration of drug in plasma. CL = Rate of elimination [mg / h ] C in plasma [mg /L ] Unit: Volume/Time [L/h] or adjusted for body weight [l/h/kg] Cl  KVd Clearance (Cl) is the most important pharmacokinetic parameter because it determines the maintenance dose (MD) that is required to obtain a given steadystate serum concentration (Css): MD = Css The definition of clearance is the volume of serum or blood completely cleared of the drug per unit time. Thus, the dimension of clearance is volume per unit time, such as L/h or mL/min. The liver is most often the organ responsible for drug metabolism while in most cases the kidney is responsible for drug excretion. The gastrointestinal wall, lung, and kidney can also metabolize some drugs, and some medications are eliminated unchanged in the bile Clearance (CL) Clearance has an additive character: it is the sum of clearances in all eliminating organs CL = CLRENAL + CLHEPATIC +CLpulmonary . . . . other renal + nonrenal 500/40=12.5 L 20/1.5=13.3 L Case 1 Ten mg metoclopramide were 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. Time (h) CP (ng/ml) 1 2 4 6 8 10 90 68 40 21.5 12 7 1) Plot the metoclopramide concentration-time data and draw a compartmental scheme showing the number of compartment involved. 2) Write the equation describing the disposition kinetics of the drug. 3) Calculate the biological half-life of the drug elimination (t½), the overall elimination rate-constant (K), the volume (Vd), the coefficient of distribution and the duration of action (td). 4) Comment on the extent of metoclopramide distribution in the body. 120 100 80 60 40 20 0 0 1 2 3 4 5 6 7 8 9 10 Time (hr) 1000 100 10 1 0 1 2 3 4 5 Time (hr) 6 7 8 9 10 Metoclopramide 10 mg IV dose AUC 424.95 ng.hr/ml Vd 83.33 L Half-life 2.465 hr K 0.28 hr-1 Cmax (Co) 119.5 ng/ml Cl 23.532 L/hr Case 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 g/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 g/ml. Calculate the following: 1)Apparent volume of distribution 1) Vd=Xo/Co=250/10=25L 2)Expected plasma concentration at 10 AM. 2) Ct=4 = 6.3 mg/L 3. Duration of action of the first dose 3  10e 0.1155*t  t  10.4hr 4. Total body clearance Cl=K*Vd=2.89 L/hr 5. Fraction of the dose in the body 5 hours after the injection X  e 0.1155*5  0.56 Xo 6) Total amount in the body 5 hours after the injection X  250e 0.1155*5  140mg 7) Exponential and logarithmic equation (pharmacokinetic model) C  10e 0.1155*t LnC  2.303  0.1155 * t LogC  1 0.0502 * t 8) Total amount in the body immediately after injection of a second dose at 12:00 noon X  250  125  375mg 9) Duration of action of first dose only if dose administered at 6:00 AM was 500 mg. 3  20e 0.1155*t  t  16.4hr Case 3 A general anesthetic has a volume of distribution of 15 L and a minimum effective concentration of 2 μg/mL (the drug is effective as long as the drug concentration is above 2 μg/mL). After administration of 120 mg of the drug as an IV bolus dose to a patient the drug produced anesthetic effect for 6 h. a. Calculate the half-life of this drug. 3 hrs a. Calculate the minimum effective concentration for the drug if the dose was 400mg. 2 mg/L a. Calculate the expected duration of effect if an IV bolus dose of 240mg was administered. 9 hrs a. Calculate the lowest dose that will produce an effect for 3 h. 60 mg a. Calculate the expected duration of effect if 20mg was given as an IV bolus dose. Zero 20/15=1.33 mg/L. less than the MEC. Case 4 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 Co=100 mg/L Vd=10L K=2.303*0.1 K=0.2303 hr-1 Calculate the following 1) Duration of action (7 hr) 2) Total body clearance (2.303 L/hr) 3) Rate of elimination at 2 hours (dX/dt=K*X) 3) Rate of elimination at 2 hours dX  KX  0.2303 * X t  2 dt 0.2303*2 X t  2  1000 * e  630.9mg dX  145.3mg / hr dt Case 5 Drug X has a therapeutic range of 15-80 mg/L. After an intravenous bolus injection of 500 mg of drug X, the concentration of the drug in plasma (in units of µg/ml) versus time (in hours), were described by the following equation Co=50 mg/L 0.12 t Ct  50 Vd=10L K=0.12 hr-1 Calculate the following e 1) Duration of action after the 500-mg dose.(10 hr) 2) Amount eliminated at 2 hours (106.7 mg) 3) Rate of elimination at 2 hours. (0.12x393.3=47.2 mg/hr) Case 6 1) The plasma concentration–time profile after a single IV dose of 300mg of a drug was back extrapolated and the y-intercept was 7.5mg/L. a. Calculate the Vd of this drug. (40 L) b. b. Calculate the dose that should achieve an initial drug concentration of 12mg/L. (480mg) Case 7 After IV bolus administration of 450mg of a drug, the initial drug concentration was found to be 15mg/L. a. Calculate the Vd of this drug in this patient.(30L) b. What is the IV bolus dose required to achieve an initial drug concentration of 20mg/L? (600mg) c. If the patient receives a single IV bolus dose of 2g of the drug, calculate the expected initial drug concentration after this large dose. (66.7 mg/L) Evaluation of Drug Kinetics By the Utilization of Urinary Excretion Data One comp IV-Bolus Clearance may be applied to any organ that is involved in drug elimination from the body. As long as first-order elimination processes are involved, clearance represents the sum of the clearances for each drug-eliminating organ Knr K Kr dXu/dt = KrX Urine Rate Method • log(dXu/dt) = log(KrXo)-Kt/2.303 log Xu Kt  log KrXo  t 2.303 Example After administration of 1000 mg of cefazolin, urine samples were collected and the following data were obtained Time interval 0-1 Volume (ml) 65 Conc (mg/ml) 5.1 1-3 3-5 5-8 8-12 114 140 225 180 3.0 1.0 0.3 0.1 Rate Method Time Volume Conc. Amount ∆t interval (ml) (mg/ml) hr 0-1 65 5.1 331.5 1 ∆Xu/ ∆t T-mid 331.5 0.5 1-3 114 3.0 342 2 171 2 3-5 140 1.0 140 2 70 4 5-8 225 0.3 67.5 3 22.5 6.5 8-12 180 0.1 18 4 4.5 10 KrXo 1000 100 Slope (Ln or Log) 10 1 0 1 2 3 4 5 6 7 8 Time K=0.45hr-1, t0.5=1.54 hr, Kr=0.41hr-1, Knr=0.04hr-1 9 10 Sigma-Minus Method Xu  KrXo (1  e  Kt ) K Xu  Xu  (1  e  Kt ) log( Xu   Xu )  log Xu   Kt 2.303 Sigma-Minus Method Time Volu Conc Amount Xu ARE T-end interval (ml) (mg/ml) 0-1 65 5.1 331.5 331.5 567.5 1 1-3 114 3.0 342 673.5 225.5 3 3-5 140 1.0 140 813.5 85.5 5 5-8 225 0.3 67.5 881 18 8 8-12 180 0.1 18 899 0 Sigma-Minus Method 1000 Xu∞ 100 Observed Predicted 10 0 1 2 3 4 5 6 7 8 9 10 Time(hr) Practice Problem A single IV dose of an antibiotic was given to a 50-kg woman at a dose level of 20 mg/kg. Urine and blood samples were removed periodically and assayed for parent drug. The following data were obtained. Determine the different PK parameters: Problems in Obtaining Valid Urinary Excretion Data. Certain factors can make it difficult to obtain valid urinary excretion data. Some of these factors are as follows: 1. A significant fraction of the unchanged drug must be excreted in the urine. 2. The assay technique must be specific for the unchanged drug and must not include interference due to drug metabolites that have similar chemical structures. 3. Frequent sampling is necessary for a good curve description. 4. Urine samples should be collected periodically until almost all of the drug is excreted. A graph of the cumulative drug excreted versus time will yield a curve that approaches an asymptote at "infinite" time (∞). In practice, approximately seven elimination half-lives are needed for 99% of the drug to be eliminated. 5. Variations in urinary pH and volume may cause significant variation in urinary excretion rates. 6. Subjects should be carefully instructed as to the necessity of giving a complete urine specimen (ie, completely emptying the bladder). Comparison of the Rate and the Sigma-Minus Methods The rate method does not require knowledge of Xu∞, and the loss of one urine specimen does not invalidate the entire urinary drug excretion study. The sigma-minus method requires an accurate determination of Xu∞, which requires the collection of urine until urinary drug excretion is complete. A small error in the assessment of Xu∞ introduces an error in terms of curvature of the plot, because each point is based on log (Xu∞ – Xu) versus time. Fluctuations in the rate of drug elimination and experimental errors including incomplete bladder emptying for a collection period cause appreciable departure from linearity using the rate method, whereas the accuracy of the sigma-minus method is less affected. Lastly, the renal drug excretion rate (Kr) constant may be obtained from the rate method but not from the sigma-minus method !!!!. ٧٩ ٨٠ ٨١ ٨٢ ٨٣ ٨٤ ٨٥ ٨٦ ٨٧ ٨٨ ٨٩ ٩٠ ٩١ ٩٢ ٩٣ ٩٤

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