# PHAR 7633 Fall 1999

## OU HSC College of Pharmacy

### First Exam 1 October 1999

Section FOUR. Calculations This section = 12 + 20 + 16 = 48 points

Show all your work for full credit. All material not deleted or crossed-out will be considered for grading. Put labels and units on all requested graphs.

 Q 4.1 (12 points) Version A. Q 4.2 Version D. A drug is given to a group of volunteers in two different dosage forms on separate occasions according to a complete crossover design. Product A was a capsule containing 350 mg of the active drug and product B was a tablet containing 300 mg of the active drug. The average areas under the plasma concentration versus time curves were 405 and 435 mg.hr.L-1, for products A and B, respectively. i) Calculate the bioavailability of the capsule relative to the tablet. ii) Given that the absolute bioavailability of the tablet product is known to be 95 %, calculate the absolute bioavailability of the capsule dosage form. i) Fcap/Ftab (relative bioavailability) = (AUCcap/Dosecap) / (AUCtab/Dosetab) = (405/350) / (435/300) = 0.80 ii) If Ftab = 0.95 and since Fcap/Ftab = 0.80 Fcap = 0.80 x 0.95 = 0.76 (absolute bioavailability) Q 4.1 (12 points) Version B. Q 4.2 Version C. A drug is given to a group of volunteers in two different dosage forms on separate occasions according to a complete crossover design. Product A was a capsule containing 700 mg of the active drug and product B was a tablet containing 600 mg of the active drug. The average areas under the plasma concentration versus time curves were 405 and 435 mg.hr.L-1, for products A and B, respectively. i) Calculate the bioavailability of the capsule relative to the tablet. ii) Given that the absolute bioavailability of the tablet product is known to be 95 %, calculate the absolute bioavailability of the capsule dosage form. i) Fcap/Ftab (relative bioavailability) = (AUCcap/Dosecap) / (AUCtab/Dosetab) = (405/700) / (435/600) = 0.80 ii) If Ftab = 0.95 and since Fcap/Ftab = 0.80 Fcap = 0.80 x 0.95 = 0.76 (absolute bioavailability) Q 4.2 (25 points) Version A. Q 4.1 Version D. A patient (85 kg) has been given a first dose of a drug, 120 mg by IV infusion over 30 minutes, and two plasma concentrations were measured at 0.5 and 8 hours after the infusion was stopped. These results were 5.47 and 2.85 mg/L, respectively. Assume a linear one-compartment model and calculate the drug pharmacokinetic parameters, kel and V, for this patient. Calculate a suitable dosage regimen, bolus IV dose and maintenance IV infusion, to achieve and maintain a plasma concentration of 6 mg/L Rounding the dose is not required. i) kel: using the two data points: kel = (ln 5.47 - ln 2.85)/(8 - 0.5) = (1.70 - 1.05) / 7.5 = 0.0869 hr-1 ii) V: using the point at 0.5 after the infusion was stopped, i.e. 1 hour after it was started. Cp = (k0/(kel x V)) x [1 - e-kel x T] x e-kel x (t-T) V = (240/(0.0869 x 5.47)) x [1 - e-0.0869 x 0.5] x e-0.0869 x 0.5 V = 504.9 x 0.04252 x 0.9575 = 20.6 L iii) Bolus Dose: Dose = Cp0 x V = 6 x 20.6 = 124 mg iv) Maintenance (slow) infusion rate constant, k0 = Cp0 x V x kel = 6 x 20.6 x 0.0869 = 10.7 mg/hr Q 4.2 (25 points) Version B. Q 4.1 Version C. A patient (85 kg) has been given a first dose of a drug, 80 mg by IV infusion over 30 minutes, and two plasma concentrations were measured at 0.5 and 6 hours after the infusion was stopped. These results were 5.47 and 2.85 mg/L, respectively. Assume a linear one-compartment model and calculate the drug pharmacokinetic parameters, kel and V, for this patient. Calculate a suitable dosage regimen, bolus IV dose and maintenance IV infusion, to achieve and maintain a plasma concentration of 10 mg/L Rounding the dose is not required. i) kel: using the two data points: kel = (ln 5.47 - ln 2.85)/(6 - 0.5) = (1.70 - 1.05) / 5.5 = 0.1185 hr-1 ii) V: using the point at 0.5 after the infusion was stopped, i.e. 1 hour after it was started. Cp = (k0/(kel x V)) x [1 - e-kel x T] x e-kel x (t-T) V = (160/(0.1185 x 5.47)) x [1 - e-0.1185 x 0.5] x e-0.1185 x 0.5 V = 246.8 x 0.05753 x 0.9425 = 13.38 L iii) Bolus Dose: Dose = Cp0 x V = 10 x 13.38 = 134 mg iv) Maintenance (slow) infusion rate constant, k0 = Cp0 x V x kel = 10 x 13.38 x 0.1185 = 15.9 mg/hr Q 4.3 (16 points) Version A. With the dose given by IV bolus administration use the Laplace method to determine which parameters are identifiable if we can sample Cp (= Xp/Vp = Plasma concentration of unchanged drug) and U (cumulative amount of unchanged drug excreted into urine). ALSO, derive the integrated equation for U. Using Cp as data: dCp/dt = - kel x Cp s x L(Cp) - Dose/V = - kel x L(Cp) L(Cp) = (Dose/V) x (1/(s + kel)) Dose/V and thus V identifiable from intercept term; kel identifiable from 's' (slope) term Using U as data: dU/dt = ke x Cp x V s x L(U) - 0 = ke x L(Cp) x V s x L(U) = ke x V x (Dose/V) x (1/(s + kel)) L(U) = (ke x Dose)/(s x (s + kel)) ke x Dose and thus ke identifiable from intercept term Integrated Equation for U From L(U) = (ke x Dose)/(s x (s + kel)) From the denominator roots are 0 and -kel Thus: U = ke x Dose/kel - ke x Dose x e-kel*t/kel U = ke x Dose/kel x [1 - e-kel*t ] Q 4.3 (16 points) Version B. With the dose given by IV bolus administration use the Laplace method to determine which parameters are identifiable if we can sample Cp (= Xp/Vp = Plasma concentration of unchanged drug) and M1 (cumulative amount of M1 excreted into urine). ALSO, derive the integrated equation for Cp. Using Cp as data: dCp/dt = - kel x Cp s x L(Cp) - Dose/V = - kel x L(Cp) L(Cp) = (Dose/V) x (1/(s + kel)) Dose/V and thus V identifiable from intercept term; kel identifiable from 's' (slope) term Using M1 as data: dXm1/dt = km1 x Cp x V - kmu1 x Xm1 s x L(Xm1) - 0 = km1 x L(Cp) x V - kmu1 x L(Xm1) s x L(Xm1) = km1 x (Dose/V) x (1/(s + kel)) x V - kmu1 x L(Xm1) s x L(Xm1) = km1 x Dose x (1/(s + kel)) - kmu1 x L(Xm1) L(Xm1) = (km1 x Dose)/((s + kel)(s + kmu1)) Now for M1 dM1/dt = kmu1 x Xm1 s x L(M1) - 0 = kmu1 x L(Xm1) L(M1) = (kmu1 x km1 x Dose) / (s x (s + kel)(s + kmu1)) kmu1 is identifiable from the 's' (slope) term kmu1 x km1 x Dose and thus km1 are identifiable from intercept term Integrated Equation for Cp From L(Cp) = (Dose/V) x (1/(s + kel)) From the denominator root is -kel Thus: Cp = (Dose/V) x e-kel*t Q 4.3 (16 points) Version C. With the dose given by IV bolus administration use the Laplace method to determine which parameters are identifiable if we can sample Cp (= Xp/Vp = Plasma concentration of unchanged drug) and M2 (cumulative amount of M2 excreted into urine). ALSO, derive the integrated equation for Cp. Using Cp as data: dCp/dt = - kel x Cp s x L(Cp) - Dose/V = - kel x L(Cp) L(Cp) = (Dose/V) x (1/(s + kel)) Dose/V and thus V identifiable from intercept term; kel identifiable from 's' (slope) term Using M2 as data: dXm2/dt = km2 x Cp x V - kmu2 x Xm2 s x L(Xm2) - 0 = km2 x L(Cp) x V - kmu2 x L(Xm2) s x L(Xm2) = km2 x (Dose/V) x (1/(s + kel)) x V - kmu2 x L(Xm2) s x L(Xm2) = km2 x Dose x (1/(s + kel)) - kmu2 x L(Xm2) L(Xm2) = (km2 x Dose)/((s + kel)(s + kmu2)) Now for M2 dM2/dt = kmu2 x Xm2 s x L(M2) - 0 = kmu2 x L(Xm2) L(M2) = (kmu2 x km2 x Dose) / (s x (s + kel)(s + kmu2)) kmu2 is identifiable from the 's' (slope) term kmu2 x km2 x Dose and thus km2 are identifiable from intercept term Integrated Equation for Cp From L(Cp) = (Dose/V) x (1/(s + kel)) From the denominator root is -kel Thus: Cp = (Dose/V) x e-kel*t Q 4.3 (16 points) Version D. With the dose given by IV bolus administration use the Laplace method to determine which parameters are identifiable if we can sample Cp (= Xp/Vp = Plasma concentration of unchanged drug) and M1 (cumulative amount of M1 excreted into urine). ALSO, derive the integrated equation for U. Using Cp as data: dCp/dt = - kel x Cp s x L(Cp) - Dose/V = - kel x L(Cp) L(Cp) = (Dose/V) x (1/(s + kel)) Dose/V and thus V identifiable from intercept term; kel identifiable from 's' (slope) term Using M1 as data: dXm1/dt = km1 x Cp x V - kmu1 x Xm1 s x L(Xm1) - 0 = km1 x L(Cp) x V - kmu1 x L(Xm1) s x L(Xm1) = km1 x (Dose/V) x (1/(s + kel)) x V - kmu1 x L(Xm1) s x L(Xm1) = km1 x Dose x (1/(s + kel)) - kmu1 x L(Xm1) L(Xm1) = (km1 x Dose)/((s + kel)(s + kmu1)) Now for M1 dM1/dt = kmu1 x Xm1 s x L(M1) - 0 = kmu1 x L(Xm1) L(M1) = (kmu1 x km1 x Dose) / (s x (s + kel)(s + kmu1)) kmu1 is identifiable from the 's' (slope) term kmu1 x km1 x Dose and thus km1 are identifiable from intercept term Integrated Equation for U dU/dt = ke x Cp x V s x L(U) - 0 = ke x L(Cp) x V s x L(U) = ke x V x (Dose/V) x (1/(s + kel)) L(U) = (ke x Dose)/(s x (s + kel)) From the denominator roots are 0 and -kel Thus: U = ke x Dose/kel - ke x Dose x e-kel*t/kel U = ke x Dose/kel x [1 - e-kel*t ]

This file was last modified:
Copyright 1999 David W.A. Bourne