PHA 5127Case Study 3 PHA 5127 Assume GFR is 130 mL min-1 and urine flow is 1.5 mL min-1 . For the following situations, indicate whether the drug is: - only filtered - filtered and reabsorbed through passive diffusion - filtered and actively secreted 1a) A drug with fu =0.04 and a ClREN = 40 mL min-1 filtered and actively secreted b) A drug with fu =0.20 and a ClREN = 26 mL min-1 only filtered c) A drug with fu =0.30 and a ClREN = 30 mL min-1 filtered and reabsorbed through passive diffusion 2.) A drug is eliminated through glomerular filtration (no other clearance mechanism is observed). The drug is 50% bound to plasma proteins. Glomerular filtration rate is normal (130 ml/min). No active renal secretion and passive or active reabsorption after renal filtration is observed. The volume of distribution is 50L. What is the clearance of the drug? What is the elimination rate constant of the drug? CL = GFR * fu = 130 ml/min *0.5 = 65 ml/min CL = ke*V 65 ml/min = ke* 50,000 mL ke = 0.0013 min-1 3.) Complete the following statements for a low molecular weight drug which does not bind to plasma proteins. A. The maximum value which renal clearance can approach that of Renal Blood Flow B. If this drug is hydrophilic and not interacting with transporters the largest possible renal clearance is that of GFR C. If this drug is lipophilic, neutral drug and not interacting with transporters the smallest possible renal clearance is that of Urine Flow D. Assuming that the lipophilic drug is a base, renal clearance can be reduced by increasing pH 4.) An investigational new drug is eliminated entirely by liver (hepatic) metabolism, with a clearance of 1.40 L/min in subjects with an average liver blood flow of 1.50 L/min. What would be its expected clearance in a congestive heart failure patient with a liver blood flow of 1.10 L/min but no change in hepatic enzyme activity?CLH = QH *E E = CLH/QH = 1.4 L/min/1.5 L/min = 0.933 For CHF patient: QH = 1.10 L/min CLH = QH*E = 1.10 L/min * 0.933 = 1.03 L/min 5.) J.P is a 40 y.o. man, height 5’6” with a serum creatinine level of 1.3 mg/dL. S.G is a 20 y.o. female, height 5’2”, weight with a serum creatinine level of 1.5 mg/dL. Determined the creatinine clearance for each person. What can be deduced about the GFR for each of these people? For J.P.: Clcreat(male) = ((140-age)*IBW)/(72*Cpcreat) IBW(male) = 50kg + 2.4 kg for every inch above 5 ft IBW for JP: 50 kg + 2.4(6) = 64.4 kg Creat Clearance = ((140-40)(64.4)/(72*1.3) = 68.8 ml/min For S.G.: Clcreat(female) = ((140-age)*IBW)/(85*Cpcreat) IBW(female) = 45.5 kg + 2.3 for every inch over 5 ft IBW for S.G.: 45.5 kg + 2.3(2) = 50.1 kg Creat Clearance = ((140-20)(50.1)/(85*1.5) = 47.1 ml/min Creatinine is eliminated from the body via glomular filtration and shown no protein binding. Thus, calculating creatinine clearance is a means of obtaining the GFR. Since the creatinine clearance was very low for these patients (normal value is 125 ml/min), it may be deduced that there is some kidney
View Full Document
Unlocking...