Extravascular administration: monitoring drug in urine.

Extravascular administration: monitoring

drug in urine

Methods to computePK 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

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

KaK

Ke

KaK

e

KaK

KrKaFXXu

tt KaKo

K

FXKX r 0

u

uX

tt KaKuu eKeKa

KKa

XXuX

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

tKuu eKa

KKa

XXuX

303.2

loglogKt

KKa

KaXXuX u

u

XuX u log

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


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

Sigma-Minus Method: 1- Calculate cumulative amount of drug eliminated

Time interval (hr)

Volume (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

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

uu XX

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

uu XX

0

0.5

1

1.5

2

2.5

0 5 10 15 20 25 30

time (endpoint of the interval, hr)

log

(AR

E)

Terminal phase (straight line)

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

(AR

E)


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 22.0303.20937.0

303.2

hrK

slopeK

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:

XKt

Xur

KatKt eeKKa

KaFXoK

t

Xu r

)(

KteKKa

KaFXoK

t

Xu r

)(

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)(loglog

Kt

KKa

KaFXoK

t

Xu r

K

FXKX r 0

u

uX

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


Time interval (hr)

Volume (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

The rate method: 1- Calculate amount of drug eliminated

Time interval (hr)

Volume (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

The rate method: 2- Calculate the change in time

Time interval (hr)

Volume (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

The rate method: 3- Calculate the rate of urinary excretion

Time interval (hr)

Volume (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

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

(d

Xu

/dt)

Terminal phase (straight line)

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

(d

Xu

/dt)


The plot of dXu/dt vs. mid point of urine collection time gives a line with a slope equal to –K/2.303

1 2.0303.20874.0

303.2

hrK

slopeK

Extravascular administration: monitoring drug in urine.

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