Answer to Practice Problem #1 Noncompartmental Pharmacokinetics Tutorial:
The first step is to plot the data on a semi-log plot:
From the graph we can determine the C0 to be 8.57 mg/L The next step is to calculate the AUC and AUMC:
AUC AUMC
Time (hr) |
Concentration (mg/L) |
Area
(mg-hr/L) |
C x t
(mg/L)(hr) |
Area (mg-hr2/L) |
0
|
8.57
|
- |
0 |
- |
|
0.25 |
8.21 |
2.10 |
2.05 |
0.26 |
|
0.5 |
7.87 |
2.01 |
3.94 |
0.75 |
|
1.0 |
7.23 |
3.78 |
7.23 |
2.79 |
|
3.0 |
5.15 |
12.38 |
15.45 |
22.68 |
|
6.0 |
3.09 |
12.36 |
18.54 |
50.99 |
|
12.0 |
1.11 |
12.60 |
13.32 |
95.58 |
|
18.0 |
0.4 |
4.53 |
7.20 |
61.56 |
|
|
Total |
49.76 |
|
234.61 |
Thus, AUC0-18 = 49.76 mg-hr/L, while AUMC0-18 = 234.61 mg-hr2/L. However, to calculate the pharmacokinetic parameters of interest, we need to have these values from zero to infinity. This requires using the plot above to determine the elimination rate constant.
Slope
= -0.074
k =
-slope x 2.303
k = -(-0.074) x 2.303 = 0.170 hr-1
To calculate AUC and AUMC,
AUC0-infinity = AUC0-18 + (C18)/k = 49.76 mg-hr/L + (0.4 mg/L)/(0.170 hr-1) = 49.76 mg-hr/L +2.35 mg-hr/L = 52.11mg-hr/L
AUMC0-infinity = AUMC0-18 + (C18)/k2 + (t18) (C18)/k = 234.61 mg-hr2/L + (0.4 mg/L)/(0.170 hr-1)2 + (18 hr)(0.4 mg/L)/(0.170 hr-1)
AUMC0-infinity = 234.61 mg-hr2/L + 13.84 mg-hr2/L + 42.35 mg-hr2/L = 290.8 mg-hr2/L
Once AUC and AUMC are calculated, we can determine the kinetic parameters of interest:
CL = Div/AUC = 300 mg/52.11 mg-hr/L = 5.8 L/hr
Vss = DivAUMC/AUC2 = (300 mg)(290.8 mg-hr2/L)/(52.11 mg-hr/L)2 = (87240 mg2-hr2/L)/(2715.5 mg2-hr2/L2) = 32.1 L
MRT = AUMC/AUC = (290.8 mg-hr2/L)/ (52.11mg-hr/L) = 5.5 hr
Last revised 7/13/05
ã 2005 -
Craig K. Svensson, Pharm.D., Ph.D.
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