PCEUT532_Final_AnswerKey_2009

PCEUT532_Final_AnswerKey_2009 - PHARMACEUTICS 532 Winter...

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PHARMACEUTICS 532 Winter 2009 Final March 17 th Name_____________Exam Key_________________ Student #___________________________________
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Page 2 of 12 Questions Points 1. 15 2. 10 3. 3 4. 2 5. 5 6. 3 7. 4 8. 5 9. 5 10. 10 11. 3 12. 8 13. 12 14. 14 15. 2 16. 2 17. 2 18. 3 19. 3 20. 3 21. 3 22. 3 23. 10
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Page 3 of 12 1. DT has been receiving carbamazepine 500 mg twice daily for 1 year. His average steady state plasma concentration of carbamazepine for this dose is 8 mg/L. CBZ-epoxide is an active metabolite of carbamazepine. Urine was collected for one dosage interval (12 hrs) and the following amounts in the urine were determined. Carbamazepine (CBZ) 5 mg CBZ epoxide 30 mg CBZ dihydrodiol 150 mg CBZ diol -glucuronide 75 mg CBZ N-glucuronide 5 mg The elimination half-life of carbamazepine is 20 hours. When CBZ-epoxide is administered intravenously itself, the elimination half-life of CBZ-epoxide is 10 hours. CYP3A epoxide hydrolase UGT CBZ CBZ epoxide CBZ-dihydrodiol CBZ- diol-glucuronide UGT CBZ-N-Gluc urine urine urine Using the information provided above, answer the following questions. a. The concentration time curve of CBZ-epoxide is described by formation rate limited pharmacokinetics after carbamazepine is administered. (2 pts). b. When carbamazepine therapy was first started in DT, the time to maximum pharmacologic effect was reached how long after his first dose? (3 pts) 3 -5 T1/2 of CBZ or 60-100 hours . c. Calculate the CYP3A4 dependent formation clearance of carbamazepine epoxide. Assume that absorption is complete (F=1.0) and that the carbamazepine metabolites have been corrected for molar weight. (10 pts) Fm = CBZ-epoxide + CBZ-dihydrodiol + CBZ-diol glucuronide = 30 mg + 150 mg + 75 mg Dose 500 mg Fm = 255 mg = 0.51 500 mg Css,ave = F*D/ τ therefore Cl = F*D/ τ = 1.0*500 mg/12 hr = 5.21 L/hr Cl Css,ave 8 mg/L Fm = Cl f therefore Cl f = Fm * Cl = 0.51 * 5.21 L/hr = 2.66 L/hr Cl If use D/ τ as 1000 mg/D then Cl f = 63.84 L/D
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Page 4 of 12 2. Fill in the Blanks or circle the correct answer. (1 pt each, 10 pts total) A. The incidence of the slow acetylator phenotype (NAT2) in the Asian population is (greater, less , or the same) as the incidence in the Caucasians and African- American population. B. The incidence of the poor metabolizer phenotype of CYP2D6 in the Caucasian population is (greater , less, or the same ) as found in the Asian population. C.
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This note was uploaded on 01/31/2010 for the course PHCOL 402 taught by Professor Storm,d during the Spring '08 term at University of Washington.

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PCEUT532_Final_AnswerKey_2009 - PHARMACEUTICS 532 Winter...

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