PHAR 7632 Spring 1999
Biopharmaceutics
OU HSC College of Pharmacy
First Semester Exam - Answers
18 February 1999
Section FOUR Calculations This section = 49 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 (24 points) A 300 mg dose of a drug was given by IV bolus injection and the data below were collected. Plot the data on semi-log graph paper, draw the ‘best’ line and determine Cp0, kel, and total AUC using the trapezoidal rule.
Cp(0) = 15.1 mg/L
Cp(14) = 1.49 mg/L
kel = [ln(15.1) - ln(1.49)]/14 = 2.316/14 = 0.165 hr-1
|
Time (hr) |
Concentration (mg/L) |
delta AUC (mg.hr/L) |
AUC (mg.hr/L) |
|
|
15.1 |
||||
|
1 |
13.5 |
14.3 |
14.3 |
|
|
2 |
10.2 |
11.85 |
26.15 |
|
|
4 |
8.0 |
18.2 |
44.35 |
|
|
8 |
3.6 |
23.2 |
67.55 |
|
|
12 |
2.2 |
11.6 |
79.15 |
|
|
14 |
1.4 |
3.6 |
82.75 |
|
| Infinity | kel = 0.165 hr-1 | 8.46 | 91.21 |
Q 4.2 (25 points) Determine the analytical equation for X2 as a function of time using the Laplace transform method. The dose is put into the GI, component 1, at time zero. The two rate constants, ka and kel, are first order.
The diagram is shown below:
dX(1)/dt = -ka x X1
s x L(X(1)) - X(1)(0) = -ka x L(X(1))
Since X(1)(0) = Dose
L(X(1)) = Dose/(s + ka)
dX(2)/dt = ka x X1 - kel x X(2)
s x L(X(2)) - X(2)(0) = ka x L(X(1)) - kel x L(X(2))
Since X(2)(0) = 0
L(X(2) x (s + kel) = ka x L(X(1)) = ka x Dose/(s+ka)
L(X(2)) = ka x Dose / ((s + kel)(s + ka))
From the table
X(2) = (ka x Dose / (ka - kel)) x (exp(-kel x t) - exp(-ka x t))