Q1: Drug A is a small and hydrophilic compound that distributes to extracellular fluids only. It has a volume of distribution of 5.6 L in a healthy 70-kg person . The free fraction of the drug in plasma is 0.1. What is the free fraction of the drug in the tissues? fu = 0.1 VT = 13 L fuT = 0.5 1 1

Drug A (100 mg) is intravenously injected to a patient Drug A (100 mg) is intravenously injected to a patient. The AUC of plasma drug concentration vs. time curve is 20 mg/ml·h. Drug A is eliminated via hepatic metabolism and renal excretion only, and the fraction of the unchanged drug excreted in urine is 0.3. What is the total body clearance (CLT), hepatic clearance (CLH) and renal clearance (CLR) of drug A? CLT = Dose/AUC = 100 X 103 mg / (20 mg/ml·h) = 5 L/h CLR = fe · CL = 0.3 X 5 L/h = 1.5 L/h CLH = CLT - CLR = 3.5 L/h EH= CLH/QH = (3.5 L/h)/(1.35 L/min) = (3.5 L/h)/(81 L/h) = 0.043

In an average 70-kg adult, Drug B has a hepatic blood clearance of 1 In an average 70-kg adult, Drug B has a hepatic blood clearance of 1.2 L/min and is 95% bound to plasma protein. What is the new hepatic blood clearance of drug B, (a) if the plasma protein binding of the drug is decreased to 90%? (b) if the hepatic blood flow is decreased to 1.2 L/min? a. CLH = 1.2 L/min, fu = 0.05, Q = 1.35 L/min EH= CLH/QH = (1.2 L/min)/(1.35 L/min) = 0.89 Drug B is a high extraction ratio drug. The hepatic clearance of drug A is insensitive to changes in plasma protein binding. fu’=0.1 CLH’≈ 1.2 L/min b. CLH’ = QH’  EH = 1.2 L/min x 0.89 = 1.07 L/min

CLH = 10 ml/min, fu = 0.05, Q = 1.35 L/min In an average 70-kg adult, Drug C has a hepatic blood clearance of 10 ml/min and is 95% bound to the plasma protein. What is the new hepatic blood clearance of drug C (a) if the plasma protein binding of the drug is decreased to 90% ? (b) if the hepatic blood flow is decreased to 1.2 L/min? a. CLH = 10 ml/min, fu = 0.05, Q = 1.35 L/min EH= CLH/Q = (10 ml/min)/(1.35 L/min) = 0.007 Drug C is a low extraction ratio drug. CLint ≈ CLH/fu = (10 ml/min)/0.05 = 200 ml/min fu’ = 0.1 CLH’ ≈ fu’ CLint = 0.1 X 200 ml/min = 20 ml/min b. CLH’ = QH’  EH = 1.2 L/min x 0.007 = 8.4 ml/min 4

In an average 70-kg adult, Drug D is completely absorbed into the intestinal epithelium following oral administration and does not undergo intestinal metabolism. The oral bioavailability of Drug D is 25%. If the protein binding of the drug is decreased from 99% to 98%, what is the new oral bioavailability? F = 0.25, fu = 0.01 EH = 1-F = 0.75 Drug D is a high extraction ratio drug. fu’=0.02, fu’/fu = 2, i.e. the free fraction increases by 2 fold F’ = F/2 = 12.5% i.e. the new bioavailability decreases by a half.

Drug E is a low extraction ratio drug In an average 70-kg adult, Drug E is completely absorbed into the intestinal epithelium following oral administration and does not undergo intestinal metabolism. The oral bioavailability of Drug E is 75%. If the protein binding of the drug is decreased from 99% to 98%, what is the new oral bioavailability? F = 0.75, fu = 0.01 EH = 1-F = 0.25 Drug E is a low extraction ratio drug Even though fu’=0.02, fu’/fu = 2, i.e. the free fraction increases by 2 fold F’ = F = 75% i.e. the bioavailability is insensitive to changes in fu 6

CLT = Ro/Css = (100 mg/h)/(20mg/L) =5 L/h Drug F is administered via intravenous infusion to a patient at a rate of 100 mg/h for 10 hours. The steady-state plasma drug concentration is 20 mg/L, a total of 300 mg of the drug is excreted unchanged in the urine. Drug F is only eliminated via renal excretion and hepatic metabolism. What is the total clearance (CLT), renal clearance (CLR) and hepatic clearance (CLH)? CLT = Ro/Css = (100 mg/h)/(20mg/L) =5 L/h fe = excreted unchanged in the urine/IV dose = 300 mg / (100 mg/h X 10 h) = 0.3 CLR = fe · CLT = 0.3 X 5 L/h = 1.5 L/h CLH = CLT - CLR = 3.5 L/h 7

In an average 70-kg adult, intravenous Drug G is eliminated by renal excretion only. When Drug G is given as i.v. infusion at 1 mg/min, an steady-state plasma concentration of 10 mg/L (fu =0.1) is achieved. a) What is the renal clearance of Drug G? CLR= CLT = Ro/Css = (1mg/min)/(10 mg/L) = 0.1 L/min = 100 ml/min CLR = fu GFR + (CLtubular secretion – CLtubular reabsorption) fu GFR = 0.1 X 120 ml/min =12 ml/min CLR > fu GFR, there is a net tubular secretion of 88 ml/min

b) If the plasma protein binding of Drug G is decreased to 80%, what is the new renal clearance of Drug G? Assume no saturation in tubular secretion or reabsorption. CLR = fu GFR + (CLtubular secretion – CLtubular reabsorption) = 0.2 X 120 ml/min + 88 ml/min = 112 ml/min