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Example Calculation of Urine Analysis Plots

with Excretion by Elimination Only

Example: Consider a drug given by IV administration, assuming all of the dose is excreted as unchanged drug in urine.

Table 12.6.1. Example Data Analysis of Drug in Urine Data

Time Interval (hr)	Urine Volume (ml)	Urine Concentration (mg/ml)	Amount Excreted ΔU (mg)	Cumulative Amount Excreted U (mg)	Midpoint Time t_midpt (hr)	Rate of Excretion ΔU/Δt (mg/hr)	A.R.E. (mg)
0 - 0.5	30	3.43	103	103			397
					0.25	206
0.5 -1	35	2.34	82	185			315
					0.75	164
1 - 2	60	1.95	117	302			198
					1.5	117
2 - 4	140	0.85	119	421			79
					3	59.5
4 - 8	224	0.30	67	488			12
					6	16.8
8 - 12	250	0.04	10	498			2
					10	2.5
12 - 24	800	0.0025	2	500			-
					18	0.17
24 - ∞	-	-	-	500

After an IV dose of 500 mg, total urine samples were collected and assayed for drug concentration. Thus the data collected is the volume of urine collected and the drug concentration in urine at the end of each interval. These are the data in columns 1, 2 and 3 above.

Multiplying column 2 by column 3 gives ΔU, column 4, the amount excreted during the interval.
Accumulating the ΔU amounts in column 4 gives the cumulative amount excreted, U, up to the end of the current time interval. The total amount excreted, U^∞ is found at the bottom of column 5, that is 500 mg.
The rate of excretion, ΔU/Δt, is calculated by dividing the ΔU amount by the length of the time interval, Δt.
A.R.E. in the last column is calculated by subtracting the value for U in each row from U^∞.

Figure 12.6.1 Linear Plot of Cumulative Amount Excreted versus Time

The plot shows U rapidly increasing at first then leveling off to U^∞. U^∞ = DOSE for this set of data. Notice that U^∞/2 (250 mg) is excreted in about 1.5 hours which gives an estimate of the elimination half-life. Otherwise this plot is a fairly qualitative representation of the data.

Rate of Excretion and ARE Plots on Semi-log Graph

Figure 12.6.2 Semi-log Plot of Rate of Excretion and A.R.E. versus Time

Calculation Using Rate of Excretion Data

Figure 12.6.2 provides a semi-log plot of ΔU/Δt versus t _midpoint (blue squares). As you can see this gives a reasonably straight line plot.

From the equation before we have:-

Intercept = ln (DOSE • kel)

Equation for Slope = -kel

kel = 0.440 hr^-1 (and t_1/2 = 1.58 hr)

This plot can be used to estimate kel and t_1/2. 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.

Calculation Using A.R.E. Data

The A.R.E. data are plotted as red circles on the semi-log graph in Figure 12.6.2 above. From these data we can calculate the slope and thus kel.

kel = 0.464 hr^-1 (and t_1/2 = 1.49 hr)

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.

This file was last modified: Friday 23 Jan 2004 at 10:23 AM