extravascular urin

1. Extravascular
administration: monitoring
drug in urine

2. Methods to compute
PK parameters from urinary data
1. the ‘‘amount remaining to be excreted’’
method (ARE); also known as the sigma-
minus method
2. The rate of excretion method

3. Sigma-Minus Method

Amount of unchanged or excreted drug in urine (Xu)
is given by:
the cumulative amount of drug excreted in the urine at
t = ∞ is:
Substitution for and rearrangement yields:















)
(
1
Ka
K
Ke
Ka
K
e
Ka
K
KrKaFX
Xu
t
t Ka
K
o
K
FX
K
X r 0
u 


u
X
 
t
t Ka
K
u
u e
K
e
Ka
K
Ka
X
Xu
X 










4. Sigma-Minus Method
 Generally, when Ka>>K, at certain time point the
absorption process ends (become negligible) as we
referred earlier by the terminal phase. Previous Eqn.
become:
 Taking the logarithm, we get:
 Thus the plot of vs. end of the time
interval gives a line with a slope equal to –K/2.303
 
t
K
u
u e
Ka
K
Ka
X
Xu
X 






  303
.
2
log
log
Kt
K
Ka
Ka
X
Xu
X u
u 













 
Xu
Xu 

log

5. Sigma-Minus Method: Example
 An oral tablet with a strength of 500 mg of a
drug was administered. The drug is one that
is partially eliminated by urinary excretion of
unchanged drug following one-compartment
model distribution and first-order elimination.
 Using the urinary data presented in the
following table, calculate elimination rate
constant

6. Sigma-Minus Method: Example
Time
interval (hr)
Volume
(ml)
Concentration
(mg/ml)
0-2 50 1.272
2-4 25 2.972
4-8 100 0.915
8-16 200 0.280
16-24 150 0.075
24-32 200 0.011

7. Sigma-Minus Method:
1- Calculate cumulative amount of drug
eliminated
Time
interval (hr)
Volume
(ml)
Concentration
(mg/ml)
Amount
(mg)
Cumulative
Amount
(mg)
0-2 50 1.272 63.60 63.6
2-4 25 2.972 74.30 137.9
4-8 100 0.915 91.50 229.4
8-16 200 0.280 56.00 285.4
16-24 150 0.075 11.25 296.7
24-32 200 0.011 2.20 298.9

8. Sigma-Minus Method:
2- Calculate amount remaining to be excreted
(ARE)
Time
interval (hr)
Amount
(mg)
Cumulative
Amount
(mg)
ARE
(mg)
0-2 63.60 63.6 235.3
2-4 74.30 137.9 161.0
4-8 91.50 229.4 69.5
8-16 56.00 285.4 13.5
16-24 11.25 296.7 2.2
24-32 2.20 298.9 0
 
u
u X
X 


9. Sigma-Minus Method:
3- Plot time (end of interval) vs. log(ARE)
Time
(hr)
ARE
(mg)
2 235.3
4 161.0
8 69.5
16 13.5
24 2.2
32 0
 
u
u X
X 

0
0.5
1
1.5
2
2.5
0 5 10 15 20 25 30
time (endpoint of the interval, hr)
log(ARE)
Terminal phase
(straight line)

10. Sigma-Minus Method:
4- draw the best fit line to the linear portion of the
curve (terminal phase)
y = -0.0937x + 2.6045
R2
= 0.9991
0
0.5
1
1.5
2
2.5
0 5 10 15 20 25 30
time (endpoint of the interval, hr)
log(ARE)

11. Sigma-Minus Method: Example
 The plot of log(ARE) vs. end of the time
interval point of urine collection time gives a
line with a slope equal to –K/2.303
1
22
.
0
303
.
2
0937
.
0
303
.
2







hr
K
slope
K

12. The rate of excretion method
substituting the value of X from previous lecture (oral
equation), we get:
Generally, when Ka>>K, at certain time point the
absorption process ends (become negligible) as we
referred earlier by the terminal phase. Previous Eqn.
become:
X
K
t
Xu
r 



 
Kat
Kt
e
e
K
Ka
KaFXo
K
t
Xu r 






)
(
 
Kt
e
K
Ka
KaFXo
K
t
Xu r 




)
(

13. The rate of excretion method
Taking the logarithm, we get:
Thus the plot of dXu/dt vs. mid point of urine
collection time gives a line with a slope equal
to –K/2.303
The total amount to be eliminated ( ) is:
303
.
2
)
(
log
log
Kt
K
Ka
KaFXo
K
t
Xu r



















K
FX
K
X r 0
u 


u
X

14. The rate of excretion method:
Example
 An oral tablet with a strength of 500 mg of a
drug was administered. The drug is one that
is partially eliminated by urinary excretion of
unchanged drug following one-compartment
model distribution and first-order elimination.
 Using the urinary data presented in the
following table, calculate elimination rate
constant

15. The rate of excretion method:
Example
Time
interval (hr)
Volume
(ml)
Concentration
(mg/ml)
0-2 50 1.272
2-4 25 2.972
4-8 100 0.915
8-16 200 0.280
16-24 150 0.075
24-32 200 0.011

16. The rate method:
1- Calculate amount of drug eliminated
Time
interval (hr)
Volume
(ml)
Concentration
(mg/ml)
Amount
(mg)
0-2 50 1.272 63.60
2-4 25 2.972 74.30
4-8 100 0.915 91.50
8-16 200 0.280 56.00
16-24 150 0.075 11.25
24-32 200 0.011 2.20

17. The rate method:
2- Calculate the change in time
Time
interval (hr)
Volume
(ml)
Concentration
(mg/ml)
Amount
(mg)
Δt
(hr)
0-2 50 1.272 63.60 2
2-4 25 2.972 74.30 2
4-8 100 0.915 91.50 4
8-16 200 0.280 56.00 8
16-24 150 0.075 11.25 8
24-32 200 0.011 2.20 8

18. The rate method:
3- Calculate the rate of urinary excretion
Time
interval (hr)
Volume
(ml)
Concentration
(mg/ml)
Amount
(mg)
Δt
(hr)
mg/hr
0-2 50 1.272 63.60 2 31.80
2-4 25 2.972 74.30 2 37.15
4-8 100 0.915 91.50 4 22.88
8-16 200 0.280 56.00 8 7.00
16-24 150 0.075 11.25 8 1.41
24-32 200 0.011 2.20 8 0.28








t
Xu

19. The rate method:
4- Plot time (mid of interval) vs. log(dXu/dt)
Time
(h)
mg/hr
1 31.80
3 37.15
6 22.88
12 7.00
20 1.41
28 0.28








t
Xu
-1
-0.5
0
0.5
1
1.5
2
0 5 10 15 20 25 30
time (midpoint of the interval, hr)
log
(dXu/dt)
Terminal phase
(straight line)

20. The rate method:
5- draw the best fit line to the linear portion of
the curve (terminal phase)
y = -0.0874x + 1.8946
R2
= 1
-1
-0.5
0
0.5
1
1.5
2
0 5 10 15 20 25 30
time (midpoint of the interval, hr)
log
(dXu/dt)

21. The rate of excretion method:
Example
 The plot of dXu/dt vs. mid point of urine
collection time gives a line with a slope equal
to –K/2.303
1
2
.
0
303
.
2
0874
.
0
303
.
2







hr
K
slope
K