Homework Assignment 1997 - 2 (Answers)

Problem 2

The Data

Time Interval (hr)

Volume of Urine Excreted (ml)

Concentration in Urine (µg/ml)

<------>

<------>

<------>

0-1

91

805

1-2

99

539

2-4

206

324

4-6

194

181

6-9

no sample

9-12

577

16

12-24

1127

5

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

Time Interval (hr)

Volume of Urine Excreted (ml)

Concentration in Urine (µg/ml)

U (mg)

tmidpoint (hr)

U/t (mg/hr)

0-1

91

805

73.4

0.5

73.4

1-2

99

539

53.3

1.5

53.3

2-4

206

324

66.8

3

33.4

4-6

194

181

35.2

5

17.6

6-9

no sample

9-12

577

16

9.3

10.5

3.1

12-24

1127

5

5.6

18

0.5

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 = 78.4 mg/hr

kel from the -ve slope of ln-linear line through the data. Note the R/E on the line at 5 hours is 17.6. NOTE: This will depend on where you put the line. Everyone may put the line in a different place.

kel = (ln 78 - ln 0.23)/20 = 0.291 hr-1

Since the Intercept = ke x Dose = 78.4 = ke x 400

ke = 78.4/400 = 0.196 hr-1

Since km = kel - ke = 0.291 - 0.196 = 0.095 hr-1

fe = ke/kel = 0.196/0.291 = 0.674

fm = km/kel = 0.095/0.291 = 0.326

Summary

Parameter

Value

Unit

kel

0.291

hr-1

ke

0.196

hr-1

km

0.095

hr-1

fe

0.674

fm

0.326