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Table 5.6.1. Example Data Analysis of Drug in Urine Data
Cumulative Amount Excreted
Rate of Excretion
|0 - 2||50||1.666||83.3||83.3||116.8|
|2 - 4||46||1.069||49.2||132.5||67.7|
|4 - 6||48||0.592||28.4||160.9||39.2|
|6 - 8||49||0.335||16.4||177.3||22.8|
|8 - 10||46||0.210||9.7||187||13.2|
|10 - 12||48||0.116||5.6||192.5||7.6|
|12 - 18||134||0.047||6.3||198.8||1.3|
|18 - 24||144||0.009||1.3||200.1||-|
|24 - ∞||-||-||-||200.1|
Figure 5.6.1 Linear Plot of Cumulative Amount Excreted versus Time
The plot in Figure 5.6.1 shows U rapidly increasing at first then leveling off to U∞ (= 200 mg). NOTE: U∞ ≠ DOSE for this set of data. Notice that U∞/2 (100 mg) is excreted in about 3 hours which gives an estimate of the elimination half-life. Otherwise this plot is a qualitative representation of the data.
Figure 5.6.2 Semi-log Plot of Rate of Excretion versus Time
The plots in Figures 5.6.2 and 5.6.3 (below) are more useful for calculating parameter values. A straight line can be drawn through the data on each semi-log plot. The elimination rate constant, kel, can be determined from the slope of this line and ke or fe determined from the intercept.
Figure 5.6.2 provides a semi-log plot of ΔU/Δt versus t midpoint. As you can see this gives a reasonably straight line plot.
Estimating the intercept value to be 53 mg/hr and if the line crosses the axis at 23.6 hr where the rate of excretion is 0.1 mg/hr a value for kel can be estimated.
and ke can be determined from the intercept.
Thus fe = ke/kel = 0.177/0.266 = 0.665.
This plot can be used to estimate kel, ke and fe. A disadvantage of this type of plot is that the error present in "real" data can obscure the straight line and lead to results which lack precision. Also it can be difficult to collect frequent, accurately timed urine samples. This is especially true when the elimination half-life is small.
Figure 5.6.3 Analysis of Urine Data - Rate of Excretion
Figure 5.6.4 Semi-log Plot of A.R.E. versus Time
The A.R.E. data are plotted as red circles on Figure 5.6.3 above. Estimating the intercept value to be 210 mg and if the line crosses the axis at 19 hr where the A.R.E. is 1 mg a value for kel can be estimated.
and fe can be determined from the intercept.
Thus ke = fe • kel = 0.7 x 0.281 = 0.197 hr-1
One disadvantage of this approach is that the errors are cumulative, with collection interval, and the total error is incorporated into the U∞ values and therefore into each A.R.E. value. Another problem is that total (all) urine collections are necessary. One missed sample means errors in all the results calculated.
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