Second Semester Exam - 1995 - Answers
Section FOUR
Question 1.
- kel = ln2/t1/2 = 0.0365 hr-1
- V = 162 L (= 1.9 x 85]
- Since Cpaverage = F * Dose / kel * V * tau
- Doserequired = Cpaverage * Dose / kel * V * tau/ F = 8 x 0.0365 x 162 x 24 / 0.75 = 1508 mg which leads to 1500 mg (= 6 x 250 mg tablets)
- Cpaverage = 7.96 mg/L (= 7.96 microgram/ml)
- Cpmin = (F*Dose/V) * (R/(1-R)) = 4.97 mg/L
Question 2.
See Chapter 16
- CLCr = (140-age)*Wt/(72 * SerumCR) = (140-56) x 79 / (72 x 1.8) = 51.2 ml/min
- kel = 0.021 + 0.00275 x 51.2 = 0.162 hr-1
- R = Cpmin/Cpmax = 1/8 = 0.125 = e-kel*tau
- tau = lnR/-kel = 12.9 hr
- Possible new tau values include 13, 15, or 16 hours.
- Using 15 hours the new R value is 0.0880
- Dose (maintenance dose) = Cpmax * V * (1 - R) = 8 x 0.21 x 79 x (1 - 0.0880) = 121 mg or 120 mg every 15 hours). Values with rounded doses: Cpmax = 7.93 mg/L and Cpmin = 0.70 mg/L
- Using 13 hours the new R value is 0.122
- Dose (maintenance dose) = Cpmax * V * (1 - R) = 8 x 0.21 x 79 x (1 - 0.122) = 117 mg or 110 mg every 13 hours). Values with rounded doses: Cpmax = 7.55 mg/L and Cpmin = 0.92 mg/L
- Using 16 hours the new R value is 0.0749
- Dose (maintenance dose) = Cpmax * V * (1 - R) = 8 x 0.21 x 79 x (1 - 0.0749) = 123 mg or 120 mg every 16 hours). Values with rounded doses: Cpmax = 7.82 mg/L and Cpmin = 0.59 mg/L
- Loading dose = Cpmax x V = 8 x 0.21 x 79 = 133 mg or 130 mg. Value with rounded dose: Cpmax = 7.84 mg/L
Question 3.
- kel = ln 13.3 - ln 1.85 / 14 = 0.141 hr-1
- ka = ln 10 - ln 0.43 / (1.75 - 0.165) = 2.00 hr-1. NOTE: ka/kel = 14
- V/F = Dose * ka / (A * (ka - kel)) = 200 x 2.00 / (13.3 x (2.00 - 0.141) = 16.2 L