Pharmacokinetics Problems PDF

Practical - 2 Pharmacokinetic problems A subject in whom the renal clearance of inulin is 120 ml/min is given a drug, the clearance of which is found to be 18 ml/min. If the drug is 40% plasma protein bound, how much filtered drug must be reabsorbed in the renal tubules?  The formula to use is Cl...

Practical - 2 Pharmacokinetic problems A subject in whom the renal clearance of inulin is 120 ml/min is given a drug, the clearance of which is found to be 18 ml/min. If the drug is 40% plasma protein bound, how much filtered drug must be reabsorbed in the renal tubules?  The formula to use is Cl = ff x GFR. The drug is 40% protein bound so the (free fraction) ff = 60%. 120 mL/min x 60% = 72 mL/min theoretical clearance of the drug. Since only 18 ml/min was actually cleared, there must have been tubular reabsorption of the drug. 72 - 1 8 = 54 mL/min of reabsorbed drug A patient was given a 200 mg dose of a drug IV, and 100 mg was eliminated during the first two hours. If the drug follows first-order elimination kinetics, how much of the drug will remain 6 hours after its administration?  One half of the dose is eliminated in the first two hours so its elimination half-life equals two hours. With the passage of each half-life the amount in the body (or in the blood) will decrease to 50% of a former level. Thus, at 6 hours after administration, three half-lives have passed: 1 )200 mg to 100 mg, 2) 100 mg to 50 mg, 3 ) 50 mg to 25 mg. If a drug is known to be distributed into total body water, what dose (mg) is needed to obtain an initial plasma level of 5 mg/L in a patient weighing 70 kg?  This is a "loading dose" question. The equation for loading dose or the volume of distribution equation can be used (LD = Vd x CP). Since the patient weighs 70 kg and 60% of body weight is water, he has 42L (70L x 60%) of total body water. LD = 42 L x 5 mg/L = 210 mg. A 500-mg dose of a drug has therapeutic efficacy for 6 h. If the half-life of the drug is 8 h, for how long would a 1-g dose be effective?  The fact that the drug has therapeutic efficacy for 6 h has no direct relationship to its half-life-it simply means that the drug is above its minimal effective concentration for 6 h. Doubling the dose (to 1 g) means that the drug level will be above the minimum for a longer period. Because the elimination half-life is 8 h, 500 mg of the drug will remain in the body 8 h after a dose of 1 g. Thus, the total duration of effectiveness must be 8 +6 = 14 h. Pharmacokinetic characteristics of propranolol include V d = 300 L/70 kg, Cl = 700 mL/min, and oral bioavailability f = 0.25. What is the dose needed to achieve a plasma level equivalent to a steady-state level of 20 μg/L? At 6 h after IV administration of bolus dose, the plasma level of a drug is 5mg/L. If the Vd = 10 L and the elimination half-life = 3 h, what was the dose administered?  At 6 h after IV injection (which corresponds to two half-lives of the drug) , the plasma level is 5 mg/L. Extrapolating back to zero time, "doubling" plasma level for each half-life results in an initial plasma level at zero time (C o) = 5 mg/L x 2 x 2 = 20 mg/L. Dose = Co x Vd = 20 mg/L x 10 L = 200 mg Question 1 Mr X is admitted to the hospital with cough, shortness of breath, and fever. History, physical examination, and culture of the sputum lead to a diagnosis of pneumonia due to gram-negative bacteria. The antibiotic tobramycin is administered. The clearance and Vd of tobramycin in Mr X are 0.16 L/min and 40 L, respectively. 17. What maintenance dose should be administered intravenously every 6 h to eventually obtain average steady-state plasma concentrations of 8 mg/L? A. 1.28 mg B. 6.4 mg C. 30.72 mg D. 320 mg E. 460.8 mg Maintainence dose Maintenance dose = Clearance X Desired concentration Bioavailability = 0.16 L/min X 8mg/L = 1.28 mg/min 1 We have to calculate maintainence dose to be given every 6 hours (6hours = 60 min x 6= 360 minutes) = 1.28 mg/min x 360 minutes = 460.8 mg Question 2 If you wish to give Mr. X an intravenous loading dose to achieve the therapeutic plasma concentration of 8 mg/L rapidly, how much should be given? 1.28 mg 10 mg 115.2 mg 320 mg Loading dose = Vd X desired plasma concentration Bioavailability = 40 L x 8 mg/L 1 = 320 mg Question 3 A continuous intravenous infusion of lidocaine, 2.9 mg/min, is started at 8 AM. The average pharmacokinetic parameters of lidocaine are: V d 77 L; clearance 145 mL/min; half-life 1.4 h. What is the expected steady-state plasma concentration? 2 mg/L 7 mg/L 20 mg/L 74 mg/l 223 mg/L Css The drug is being administered continuously and the steady state concentration (Cpss) for a continuously administered drug is given by Dosage = Plasma levelss (Cpss )x Clearance 2.9 mg/min = Cpss x 145 mL/minute Cpss= 2.9mg/min 145mL/min =0.02mg/mL = 20 mg/L Question 4 The pharmacokinetics of theophylline include the following average parameters: Vd 35 L; CL 48 mL/min; half-life 8 h. If an intravenous infusion of theophylline is started at a rate of 0.48 mg/min, how long would it take to reach 93.75% of the final steady-state concentration? (A) 48 min (B) 7.4 h (C) 8 h (D) 24 h (E) 32 h  The approach of the drug plasma concentration to steady state concentration during continuous infusion follows a stereotypical curve that rises rapidly at first and gradually reaches a plateau. It reaches 50% of steady state at 1 half-life, 75% at 2 half-lives, 87.5% at 3, 93.75% at 4 Question 5 A single intravenous bolus of Aminoglycoside (100 mg) was administered to a 70 Kg man. The figure above shows the concentration of Aminoglycoside in plasma as a function of time. Aminoglycosides does not bind to plasma proteins.  What is the half life of drug X in this man? 1 Hour 2 hours 3 hours 4 hours The initial redistribution phase(rapid fall in conc.) should be skipped for calculating the half life use a scale, the concentration of 4 mcg/mL at 2 hours, and 2 mg/mL at 5 hours So half life = 3 hours  What is the Elimination constant of the drug X 0.173 0.231 0.346 0.693 693 t ½ = 0.693 so Ke = 0.693 = 0.231 Ke t½ The table below gives the amount of a drug X remaining inside the body following the passage of different amounts of time Time in hours Drug amount remaining inside the body Amount of drug eliminated (mg) (mg) 0 500 - 1 450 50 2 405 45 3 364.5 40.5 4 328.05 36.45 5 295.25 32.80 6 262.45 29.525 7 232.925 26.245  What is the type of elimination for drug X and why? First order kinetics as constant fraction of drug is eliminated What is the elimination rate constant of the drug X? 0.10 , 50/500 = 0.10 Calculate the half life of drug X. Half life is the time necessary for the concentration of drug in plasma to decrease by half With 500mg/L being the initial plasma drug concentration , t1/2 should there fore be time required for plasma drug concentration to be lowered to 250 mg/L (can be considered if student directly gives answer with logic) between 6 and 7 T1/2 = 0.693/Ke (1 marks) Where Ke = elimination rate constant Half life = 0.693/0.10 = 6.93 hours