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6 Homework PHA 5127 Fall 2004 1.) The extent to which a drug is absorbed partially determines its: a.) b.) c.) d.) half-life bioavailability volume of distribution elimination rate Use the following information to answer questions 2 and 3. A patient is to be started on a continuous infusion of an antibiotic. The desired average steady state concentration (Css) is 5.0 g/ml. The population pharmacokinetic values for its half-life and volume of distribution are 2.0 hours and 1.5 L/kg respectively. The patient weighs 45 kg. 2.) What would be the rate of infusion (k0) to achieve a Css of 5.0 g/ml? If the Css (desired) is 5.0 g/ml what should the loading dose be? k = 0.693/2 hr = 0.35 hr-1 V = 1.5 L/kg*45 kg = 67.5 L Cl = 0.35 hr-1 * 67.5 L = 23.6 L/hr 5.0 g/ml = 5000 g/L k0 = Css * Cl 5000 g/L * 23.6 L/hr = 118000 g/hr or 118 mg/hr Loading dose = Css *V 5000 g/L* 67.5 L = 337500 g or 337.5 mg 3.) After dosing, the patient s Css is measured to be 7.5 g/ml and not 5 g/ml. Possible reasons that could account for this higher than expected concentration are: a.) Increased clearance in the patient compared to the population value. b.) Decreased clearance in the patient compared to the population value. c.) Decreased bioavailability (F) of the drug in the patient compared the to population value. d.) Both a and c e.) Both b and c 4.) T.M., a 45 year old male weighing 65 kg, is to be given 200 mg of a drug i.v. over 1 hour every 12 hours. If the patient is assumed to have an average ke of 0.35 hr-1 and a Vd of 0.31 L/kg, how long will it take to reach steady state? Calculate the peak and trough plasma concentrations at steady state. 5 half lives will reach steady state. t1/2 = 0.693/0.35 hr-1 = 2 hr 5*2 hr = 10 hr k0 = 200 mg/hr V = 65 kg * 0.31 L/kg = 20 L k = 0.35 hr-1 t = 1 hr, T = 12 hr C max( ss ) = k 0 (1 e k *t ) 200(1 e 0.35*1 ) = 8.6 mg/L = V * k (1 e k * ) 20 * 0.35(1 e 0.35*12 ) ) Ctrough = C max * e k *( t ) = 8.6 * e 0.35*(12 1) = 0.18 mg/L 5.) Calculate the appropriate maintenance dose, including the most appropriate dosing interval for a patient. The Vd is 15 L and the ke is 0.5 hr-1. The desired Cpeak is 16 mg/Land the Ctrough is 0.5 mg/L and 1 hr is assumed to be the drug administration time. = 1/-k [ln Ctrough- ln Cpeak] + t = 1/-0.5 [ ln 0.5 ln 16] + 1 = 7.93 hr 8 hr C max(ss ) = k 0 (1 e k *t ) = 16 mg / L V * k (1 e k * ) V= 15 L, k = 0.5 hr-1, t = 1 hr, T = 8 hr, solve for k0 k0 = 299.4 mg/hr 300 mg/hr infusion over 1 hour every 8 hours.

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