Homework Assignment 1995 - 2 (Answers)

Problem 2

The Data

Table 1. Volume and Concentration of Unchanged Drug found in Urine after an i.v. Bolus Dose of 400 mg.
Time Interval (hr) Volume of Urine Excreted (ml) Concentration in Urine (µg/ml) <------> <------> <------>
0-1 65 480


1-2 64 390


2-4 121 315


4-6 130 205


6-9 no sample



9-12 175 75


12-24 750 25


NOTE: Missing data means that the Amount Remaining to be Excreted method cannot be used. Thus, need to plot R/E versus midpoint time

Complete the Table

Table 1. Volume and Concentration of Unchanged Drug found in Urine after an i.v. Bolus Dose of 400 mg.
Time Interval(hr) Volume of Urine Collected (ml) Concentration in Urine (µg/ml) U (mg) tmidpoint (hr) U/t (mg/hr)
0-1 65 480 31.2 0.5 31.2
1-2 64 390 25.0 1.5 25.0
2-4 121 315 38.1 3 19.1
4-6 130 205 26.7 5 13.3
6-9 no sample



9-12 175 75 13.1 10.5 3.5
12-24 750 25 18.8 18 1.5

Graphing the Data

Plot U/t versus tmidpoint on semi-log graph paper.

The Parameters

Intercept by inspection after putting a line through the data on semi-log graph paper = 31.8 mg/hr

kel from the -ve slope of ln-linear line through the data. Note the R/E on the line at 18 hours is 1.26. NOTE:: This value and the value for R/E will depend on where you put the line. Everyone may put the line in a different place.

kel = (ln 31.8-ln 1.26)/18 = 0.179 hr-1

Since the Intercept = ke x Dose = 31.8 = ke x 400

ke = 31.8/400 = 0.080 hr-1

Since km = kel - ke = 0.179 - 0.080 = 0.099 hr-1

fe = ke/kel = 0.080/0.179 = 0.447

fm = km/kel = 0.099/0.179 = 0.553

Summary

Parameter Value Unit
kel 0.179 hr-1
ke 0.099 hr-1
km 0.099 hr-1
fe 0.447
fm 0.553