Basic Pharmacokinetics - Pter 4: Iv Bolus

CHAPTER 4

I.V. Bolus Dosing

Author: Michael Makoid and John Cobby Reviewer: Phillip Vuchetich

OBJECTIVES For an IV one compartment model plasma and urine: 1.

Given patient drug and/or metabolite concentration, amount, and/or rate vs. time profiles, the student will calculate (III) the relevant pharmacokinetic parameters available from IV plasma, urine or other excreta data: e.g.

V d, K, k m, k r, AUC, AUMC, CL, MRT, t 1 ⁄ 2 2.

The student will provide professional communication regarding the pharmacokinetic parameters obtained to patients and other health professionals.

3.

The student will be able to utilize computer programs for simulations and data analysis.

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4-1

I.V. Bolus Dosing

4.1 I.V. Bolus dosing of Parent compound 4.1.1

PLASMA

Valid equations: (Obtained from the LaPlace transforms derived from the appropriate models derived from the pharmacokinetic descriptions of the drug)

ln C p = – K ⋅ t + ln C p 0

(EQ 4-1)

ln X = – K ⋅ t + ln X0

(EQ 4-2)

C p = C p0 e

( ∞)

AUC =

∫

– Kt

(EQ 4-3)

DC p 0 = ----Vd

(EQ 4-4)

t ½ = 0.693 ------------K

(EQ 4-5)

( Cp n + Cp n + 1 ) Cp last Cp dt = Σ  ------------------------------------⋅ ∆t + -------------2 K

(EQ 4-6)

0 ( ∞)

AUMC =

∫

t

t ⋅ C p dt =

0

Utilization: Can you determine the slope and intercept from a graph? Plot the data in table 4 -1.on semi-log graph paper. Extrapolate the line back to time = 0 to get Cp0. Find the half life. Calculate the elimination rate constant.

( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 )

Cp l ast

( t l ast ⋅ Cp last )

- ⋅ ∆t + --------------- + ---------------------------------∑  ------------------------------------------------------------------ 2 2 K K

(EQ 4-7)

0

MRT = AUMC -----------------AUC

(EQ 4-8)

Cl = K ⋅ Vd

(EQ 4-9)

• You should be able to plot a data set Concentration vs. time on semilog yielding a straight line with slope = – K and an intercept of C p0 .

TABLE 4-1

Nifedipine 25 mg IV bolus

Time (hr)

Cp (mcg/L)

2

139

4

65.6

6

31.1

8

14.6

FIGURE 4-1.

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4-2

I.V. Bolus Dosing

FIGURE 4-1

100 Concentration (mic/L)

Does your Graph look like this?

Nifedipine IV Bolus (25 mg IV Bolus)

10 3

Cp0 = 295 mic/L

Concentration (ng/mL)

-K1 = -0.375 hr -1 10 2

1.85 hr

50

10 1 0

2

4

6

8

Time (hr) Time (hours)

• You should be able to determine K. A plot of the data in TABLE 4-1 results in FIGURE 4-1 dy Remember from high school algebra, the slope of any straight line is the rise over the run, ------ , dx

In the case of semi-log graphs dy is the difference in the logarithms of the concentrations. Thus, using the rules of logarithms, when two logs are subtracted, the numbers themselves are divided. i.e. ln ( C1 ) – ln ( C2 ) = ln  ------- . Thus if we are judicious in the concentrations that we C2 C1

take, we can set the rise to a constant number. So, if we take any two concentrations such that one concentration is half of the other (In FIGURE 4-1 above, we took 100 and 50), the time it takes for the concentration to halve is the half life (in the graph above, 1.85 hr). Then 0.693 –1 0.693 K = ------------- = ---------------- = 0.375 hr 1.85hr t½

• You should be able to determine V d :. To do this, extrapolate the line to t = 0 . The value of Cp mic L

when t = 0 is C p0 (in the graph above, C p0 = 295 --------- which is equal to D ⁄ V d for an IV bolus dose only. Dose 25mg ⋅ 1000mic --------------------- = 85L Thus, Cp 0 = ------------- , V d = Dose ------------- = -----------------Vd

Cp 0

295mic -----------------L

mg

The volume of distribution is a mathematical construct. It is merely the proportionality constant between two knowns - the C p0 which results from a given D 0 . It is, however, useful because it is patient specific and therefore can be used to predict how the patient will treat a subsequent dose of the same drug. You should be able to obtain the volume of distribution from graphical analysis of the data. Pay attention to the units! Make sure that they are consistent on both sides of the equation. NOTE: the volume of distribution is not necessarily any physiological space. For example the approximate volume of distribution of digoxin is about 600 L If that were a physiological space and I were all water, that would mean that I would weigh about 1320 pounds. I’m a little overweight (I prefer to think that I’m underheight), but REALLY!

• Given any three of the variables of the IV bolus equation, either by direct information (the volume of distribution is such and such) or by graphical data analysis, you should be able to find the fourth.

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4-3

I.V. Bolus Dosing

• You should be able to calculate Area Under the Curve (AUC) from IV Bolus data (Time vs. Cp). From the above data in TABLE 4-1 the AUC is calculated using (EQ 4-6): (∞)

AUC =

∫ Cp dt

Cpn + Cp n + 1 Cp = Σ  --------------------------------- + --------l which in this case is:   ∆t K

0

Cp 1 + Cp 2 Cp2 + Cp 3 Cp 3 + Cp last Cp l ast   Cp o + Cp 1 - Σ  -------------------------- ⋅ ∆t 1 + -------------------------- ⋅ ∆t 2 + -------------------------- ⋅ ∆t 3 + ------------------------------- ⋅ ∆t last + -------------2 2 2 2 K1    295 + 139 31.1 + 14.6 65.6 + 31.1 139 + 65.6 14.6 mcg Σ  ------------------------ ⋅ 2 + ------------------------- ⋅ 2 + --------------------------- ⋅ 2 + --------------------------- ⋅ 2 + ------------- ---------- hr or 2 2 2 2 0.375   L mcg mcg Σ { 434 + 204.6 + 96.7 + 45.7 + 38.9 } ---------- hr = 819.9 ---------- hr . In tabular format, the AUC calculation L L

is shown in TABLE 4-2.

TABLE 4-2 AUC t t–1

AUC

AUC

t 0

TIME

Cp

0

295

2

139

434.0

434.0

4

65.6

204.6

638.6

6

31.1

96.7

735.3

8

14.6

45.7

781.0

∞

0

38.9

819.9

The AUC of a plot of plasma concentration vs. time, in linear pharmacokinetics, is a number which is proportional to the dose of the drug which gets into systemic circulation. The proportionality constant, as before, is the volume of distribution. It is useful as a tool to compare the amount of drug obtained by the body from different routes of administration or from the same route of administration by dosage forms made by different manufacturers (calculate bioavailability in subsequent discussions). The AUC of a plot of Rate of Excretion of a drug vs. time, in linear pharmacokinetics, is the mass of drug excreted into the urine, directly.

• You should be able to calculate the AUMC from IV Bolus data (Time vs. Cp). The equation for AUMC is equation 4-7: ( ∞)

AUMC =

∫

t

t ⋅ C p dt =

0

( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 )

Cp l ast

( t l ast ⋅ Cp last )

- ⋅ ∆t + --------------- + ---------------------------------∑  ------------------------------------------------------------------ 2 2 K K

which in the

0

data given in TABLE 4-1 is: T0 ⋅ C po + T1 ⋅ C p1 T1 ⋅ C p1 + T2 ⋅ C p2 T 2 ⋅ C p 2 + T3 ⋅ C p3 Σ ----------------------------------------------- ⋅ ∆t 1 + ----------------------------------------------- ⋅ ∆t 2 + ----------------------------------------------- ⋅ ∆t 3 + 2 2 2

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I.V. Bolus Dosing

T 3 ⋅ C p 3 + T last ⋅ C p last T last ⋅ C p l ast Cplast ---------------------------------------------------------- ⋅ ∆t l ast + ------------------------------ + --------------- and thus, 2 2 K K

 0 ⋅ 295 + 2 ⋅ 139 2 ⋅ 139 + 4 ⋅ 65.6 4 ⋅ 65.6 + 6 ⋅ 31.1 mcg 2 Σ  --------------------------------------- ⋅ 2 + ---------------------------------------- ⋅ 2 + ------------------------------------------ ⋅ 2 ---------- hr + 2 2 2   L  6 ⋅ 31.1 + 8 ⋅ 14.6 ⋅ 14.6 14.6  mcg 2 - ⋅ 2 + 8----------------- + ----------------  ---------- hr or  ----------------------------------------2 2 0.375  0.375  L

mcg 2 Σ { 278 + 540.4 + 449 + 303.4 + 311.47 + 103.82 } = 1986.1 ---------- hr L

Thus in tabular format the AUMC for data given in TABLE 4-1 is TABLE 4-3 below. TABLE 4-3 AUMC

TIME

Cp

Cp*T

0

295

0

AUMC t

AUMC

2

139

278

278.0

278.0

4

65.6

262.4

540.4

818.4

6

31.1

186.6

449.0

1267.4

8

14.6

116.8

303.4

1570.8

∞

0

0

415.3

1986.1

t 0

The AUMC is the Area Under the first Moment Curve. A plot of T*Cp vs. T is the first moment curve. The time function buried in this plot, the Mean Residence Time (MRT), can be extracted using equation 4-8 below. It is the geometric mean time that the molecules of drug stay in the body. It has utility in the fact that, as drug moves from the dosage form into solution in the gut, from solution in the gut into the body, and from the body out, each process is cumulatively additive. That means if we can physically separate each of these processes in turn, we can calculate the MRT of each process. The MRT of each process is the the inverse of the rate constant for that process.

• You should be able to calculate MRT from IV Bolus data (Time vs. Cp) using equation 4-8 AUMC 1986.1 MRT = ------------------ = ---------------- = 2.42 AUC 819.9

Since there is only the process of elimination (no release of the drug from the dosage form, no absorption), the MRT is the inverse of the elimination rate constant, K. Thus MRT = 1/K.

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4-5

I.V. Bolus Dosing

Flow Chart 4-1

IV Bolus

X

K

MRT(IV) = 1/K Suppose the drug were given in a solution. Then the drug would have to be absorbed and then eliminated. Since the MRTs are additive, the MRT of the oral solution would be made up of the MRTs of the two processes, thus: Flow Chart 4-2

Oral Solution

Xa

Ka

X

K

MRT(os) = MAT(os)+MRT(IV) MRT(os) = 1/Ka + 1/K

Consequently, if a drug has to be released from a dosage form for the drug to get into solution which is subsequently absorbed, a tablet for example, the MRT of the tablet will consist of the MRT(IV) and the MAT(os) and the Mean Dissolution Time (MDT), thus: Flow Chart 4-3

Tablet

Xd

Kd

Xa

Ka

MRT(tab) = MDT + MAT(os) MRT(tab) = 1/Kd + 1/Ka MRT(tab) = MAT(tab) MRT(tab) = 1/Ka (apparent)

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X

K

+ MRT(IV) + 1/K + MRT(IV) + 1/K

4-6

I.V. Bolus Dosing

Normally, we don’t have information from the oral solution, just IV and tablet. So in that case the information obtained about absorption from the tablet is bundled together into an apparent absorption rate constant consisting of both dissolution and absorption. It should be apparent that this is a reasonably easily utilized and powerful tool used to obtain pharmacokinetic parameters.

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4-7

I.V. Bolus Dosing

4.1.2

IV BOLUS, PARENT COMPOUND, PLASMA PROBLEMS

Equations used in this section:

1.

K = –slope from equation 4-3

2.

ln 2- equation 4-5 t 1 ⁄ 2 = ------K

3.

1 MRT = ---- ( estimate ) MRT = AUMC ------------------ equation 4-8 K AUC

4.

Cp 0 = the y-intercept of the line from equation 4-3

5.

Cp Estimate for AUC = AUC = ---------0 which is K (∞)

AUC =

∫

∞

∫0 Cp dt

( Cp n + Cp n + 1 ) Cp last Cp dt = Σ  ------------------------------------( ∆t ) + ------------- 2 K

0

Trapezoidal rule applied to equation 4-6 6.

Estimate for AUMC = AUMC = AUC ⋅ MRT from equation 4-8 ( ∞)

AUMC

∫

t

Cp dt =

0

( t n ⋅ Cp n ) + ( t n + 1 ⋅ Cp n + 1 )

Cp last

( t last ⋅ Cp l ast )

- ⋅ ∆t n + --------------- + ---------------------------------∑  ------------------------------------------------------------------ 2 2 K K 0

from equation 4-7 7.

V d = Dose ------------- from equation 4-4 Cp 0

8.

Cp 0 Dose Cl = K1 ⋅ V d = ------------ ⋅ ------------- = Dose ------------AUC Cp 0 AUC

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4-8

I.V. Bolus Dosing

Acyclovir Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 1)

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De Miranda and Burnette, “Metabolic Fate and Pharmacokinetics of the Acyclovir Prodrug Valaciclovir in Cynomolgus Monkeys”, Drug Metabolism and Disposition (1994): 55-59.

Acyclovir is an antiviral drug used in the treatment of herpes simplex, varicella zoster, and in suppressive therapy. In this study, three male cynomolgus monkeys were each given a 10 mg ⁄ kg intravenous dose. The monkeys weighed an average of 3.35 kg each. Blood samples were collected and the following data was obtained: PROBLEM TABLE 4 - 1.

Acyclovir Serum concentration

Time (hours)

( µg ⁄ mL )

0.167

26.0

0.300

23.0

0.500

19.0

0.75

16.0

1.0

12.0

1.5

7.0

2.0

5.0

From the data presented in the Preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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4-9

I.V. Bolus Dosing

(Problem 4 - 1) Acyclovir:

2

CONCENTRATION (MIC/ML)

10

1

10

100 0.0

0.5

1.0

1.5

2.0

TIME (HR) –1

1.

k = 0.93hr

2.

t½ = 0.75hr .

3.

MRT = 1.08hr .

4.

( C p )0 = 30.4ug ⁄ mL .

5.

AUC = 32.75ug ⁄ mL ⋅ hr .

6.

AUMC = 35.2ug ⁄ mL ⋅ hr .

7.

Vd = 1.1L

8.

Cl = 1.02L ⁄ hr .

2

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I.V. Bolus Dosing

Aluminum

(Problem 4 - 2)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

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Xu, Pai, and Melethil, "Kinetics of Aluminum in Rats. II: Dose-Dependent Urinary and Biliary Excretion", Journal of Pharmaceutical Sciences, Oct 1991, p 946 - 951.

A study by Xu, Pai, and Melethil establishes the pharmacokinetics of Aluminum in Rats. In this study, four rats with an average weight of 375g, were given an IV bolus dose of aluminum (1 mg/kg). Blood samples were taken at various intervals and the following data was obtained: PROBLEM TABLE 4 - 2.

Aluminum

Time (hours)

ng Serum concentration, -------mL

0.4

19000

0.6

18000

1.4

15000

1.6

14500

2.3

12500

3.0

10500

4.0

8500

5.0

6500

6.0

5000

8.0

3250

10.0

2000

12.0

1250

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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I.V. Bolus Dosing

(Problem 4 - 2) Aluminum:

5

CONCENTRATION (NG/ML)

10

4

10

3

10

0

2

4

6

8

10

12

TIME (HR)

–1

1.

k = 0.234hr

2.

t½ = 3hr .

3.

MRT = 4.3hr .

4.

( C p )0 = 21000ng ⁄ mL .

5.

AUC = 89285ng ⁄ mL ⋅ hr .

6.

AUMC = 383926ng ⁄ mL ⋅ hr .

7.

Vd = 17.86mL

8.

Cl = 4.18mL ⁄ hr .

2

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4-12

I.V. Bolus Dosing

Amgen

(Problem 4 - 3)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

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Salmonson, Danielson, and Wikstrom, "The pharmacokinetics of recombinant human erythropoetin after intravenous and subcutaneous administration to healthy subjects", Br. F. clin. Pharmac. (1990), p 709- 713.

Amgen (r-Epo) is a form of recombinant erythropoetin. Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produce erythropoetin; therefore, r-Epo is being investigated for use in these patients in order to treat the anemia that results from the lack of erythropoetin. In a study by Salmonson et al, six healthy volunteers were used to demonstrate that both IV and subcutaneous administration of erythropoetin have similar effects in the treatment of anemia due to chronic renal failure. The six volunteers were each given a 50 U/kg intravenous dose of Amgen. The average weight of the six volunteers was 79 kg. Blood samples were drawn at various times and the data obtained is summarized below: PROBLEM TABLE 4 - 3.

Amgen

Time (hours)

mU Serum concentration, --------mL

2

700

4

600

6

400

8

300

12

150

24

40

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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I.V. Bolus Dosing

(Problem 4 - 3) Amgen:

Con (mU/mL)

CONCENTRATION (MU/ML)

103

102

101 0

5

10

15

20

25

TIME (HR)

–1

1.

k = 0.134hr

2.

t½ = 5.2hr .

3.

MRT = 7.46hr .

4.

( C p )0 = 900mU ⁄ mL .

5.

AUC = 6945mU ⁄ mL ⋅ hr .

6.

AUMC = 49600 mU ⁄ mL ⋅ hr .

7.

Vd = 4.44L

8.

Cl = 0.6L ⁄ hr .

2

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I.V. Bolus Dosing

Atrial Naturetic Peptide (ANP) Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 4)

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Brier and Harding, "Pharmacokinetics and Pharmacodynamics of Atrial Naturetic Peptide after Bolus and Infusion Administration in the Isolated Perfused Rat Kidney", The Journal of Pharmacology and Experimental Therapeutics (1989), p 372 - 377.

A study by Brier and Harding a dose of 45 ng was given by IV bolus to rats. Samples of blood were taken at various intervals throughout the length of the study and the following data was obtained: PROBLEM TABLE 4 - 4.

Atrial Naturetic Peptide (ANP)

Time (minutes)

pg Serum concentration, -------mL

3

380

10

280

20

170

30

130

40

100

50

70

60

50

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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I.V. Bolus Dosing

(PG/ML) Con CONCENTRATION (pg/mL)

(Problem 4 - 4) Atrial Naturetic Peptide (ANP):

103

102

101 0 10 20 30 40 50 60

Time (min) –1

1.

k = 0.0345min

2.

t½ = 20.09min .

3.

MRT = 28.95min .

4.

( C p )0 = 386.6pg ⁄ mL .

5.

AUC = 11206.4pg ⁄ mL ⋅ min .

6.

AUMC = 324425.4pg ⁄ mL ⋅ min .

7.

Vd = 116.4mL

8.

Cl = 4.02mL ⁄ min .

2

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4-16

I.V. Bolus Dosing

Aztreonam

(Problem 4 - 5)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

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Cuzzolim et al., "Pharmacokinetics and Renal Tolerance of Aztreonam in Premature Infants", Antimicrobial Agents and Chemotherapy (Sept. 1991), p. 1726 - 1928.

Aztreonam is a monolactam structure which is active against aerobic, gram-negative bacilli. The pharmacokinetic parameters of Aztreonam were established in a study presented in by Cuzzolim et al in which Aztreonam (100 mg/ kg) was administered intravenously to 30 premature infants over 3 minutes every 12 hours. The group of neonates had an average weight of 1639.6g. The following set of data was obtained: PROBLEM TABLE 4 - 5.

Aztreonam

Time (minutes)

µg Serum concentration, -------mL

1

40.50

2

34.99

3

29.99

4

23.88

5

22.20

6

19.44

7

16.55

8

14.99

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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4-17

I.V. Bolus Dosing

Con (ug/mL) CONCENTRATION (UG/ML)

(Problem 4 - 5) Aztreonam:

10 2

10 1 0

2

4

TIME (MIN)

6

8

–1

1.

k = 0.144min

2.

t½ = 4.81min .

3.

MRT = 6.94min .

4.

( C p )0 = 45.75ug ⁄ mL .

5.

AUC = 317.7ug ⁄ mL ⋅ min .

6.

AUMC = 2204.8ug ⁄ mL ⋅ min .

7.

Vd = 3.58L

8.

Cl = 0.516L ⁄ min .

2

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4-18

I.V. Bolus Dosing

Recombinant Bovine Placental Lactogen Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 6)

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Byatt, et. al., "Serum half-life and in-vivo actions of recombinant bovine placental lactogen in the dairy cow", Journal of Endocrinology (1992), p. 185 - 193.

Bovine placental lactogen (bPL) is a hormone similar to growth hormone and prolactin. It binds to both prolactin and growth hormone receptors in the rabbit and stimulates lactogenesis in the rabbit. In a study by Byatt, et. al., four cows (2 pregnant and 2 nonpregnant) were given IV bolus injections of 4 mg and the following data was obtained: PROBLEM TABLE 4 - 6.

Recombinant Bovine Placental Lactogen

Time (minutes)

µg Serum concentration -----L

3.8

117

6.8

72

12.0

43

16.0

27

20.0

18

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-19

I.V. Bolus Dosing

(Problem 4 - 6) Recombinant Bovine Placental Lactogen:

(MIC/L) ConCONCENTRATION (ug/L)

103

102

101 0

5

10

15

20

Time (min) –1

1.

k = 0.113min

2.

t½ = 6.13min .

3.

MRT = 8.85min .

4.

( C p )0 = 167.8ug ⁄ L .

5.

AUC = 1484.9ug ⁄ L ⋅ min .

6.

AUMC = 13141.1ug ⁄ L ⋅ min .

7.

Vd = 23.84L

8.

Cl = 2.69L ⁄ min .

2

Basic Pharmacokinetics

REV. 00.1.27

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4-20

I.V. Bolus Dosing

Caffeine

(Problem 4 - 7)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Dorrbecker et. al., "Caffeine and Paraxanthine Pharmacokinetics in the Rabbit: Concentration and Product Inhibition Effects.", Journal of Pharmacokinetics and Biopharmaceutics (1987), p.117 - 131.

This study examines the pharmacokinetics of caffeine in the rabbit. In this study type I New Zealand White rabbits were given an 8 mg intravenous dose of caffeine. Blood samples were taken and the following data was obtained: PROBLEM TABLE 4 - 7.

Caffeine

Time (minutes)

µg Serum concentration -------mL

12

3.75

40

2.80

65

2.12

90

1.55

125

1.23

173

0.72

243

0.37

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-21

I.V. Bolus Dosing

Con (ug/L)

CONCENTRATION (MIC/ML)

(Problem 4 - 7) Caffeine:

Caffeine 101 100 10-1 0

50 100 150 200 250

Time (min)

–1

1.

k = 0.00997min

2.

t½ = 69.51min .

3.

MRT = 100.3min .

4.

( C p )0 = 4.105ug ⁄ mL .

5.

AUC = 411.7ug ⁄ mL ⋅ min .

6.

AUMC = 41293.5ug ⁄ mL ⋅ min .

7.

Vd = 1.95L

8.

Cl = 19.44mL ⁄ min .

2

Basic Pharmacokinetics

REV. 00.1.27

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4-22

I.V. Bolus Dosing

Ceftazidime

(Problem 4 - 8)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Demotes-Mainard, et. al., "Pharmacokinetics of Intravenous and Intraperitoneal Ceftazidime in Chronic Ambulatory Peritoneal Dyialysis", Journal of Clinical Pharmacology (1993), p. 475 - 479.

Ceftazidime is a third generation cephalosporin which is administered parenterally. In this study, eight patients with chronic renal failure were each given 1 g of ceftazidime intravenously. Both blood samples were taken the data obtained from the study is summarized in the following table:

PROBLEM TABLE 4 - 8.

Ceftazidime

Time (hours)

mg Serum concentration ------L

1

50

2

45

4

38

24

21

36

14

48

11

60

8

72

4

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-23

I.V. Bolus Dosing

CONCENTRATION (MG/L) Con (mg/L)

(Problem 4 - 8) Ceftazidime:

102 101 100 0

20

40

60

80

Time (hours) –1

1.

k = 0.0324hr

2.

t½ = 21.39hr .

3.

MRT = 30.86hr .

4.

( C p )0 = 47.57mg ⁄ L .

5.

AUC = 1468.2mg ⁄ L ⋅ hr .

6.

AUMC = 45308.6mg ⁄ L ⋅ hr .

7.

Vd = 21.02L

8.

Cl = 0.681L ⁄ hr .

2

Basic Pharmacokinetics

REV. 00.1.27

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4-24

I.V. Bolus Dosing

Ciprofloxacin Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 9)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Lettieri, et. al., "Pharmacokinetic Profiles of Ciprofloxacin after Single Intravenous and Oral Doses", Antimicrobial Agents and Chemotherapy (May 1992), p. 993 -996.

Ciprofloxacin is a fluoroquinolone antibiotic which is used in the treatment of infections of the urinary tract, lower respiratory tract, skin, bone, and joint. In this study, twelve healthy, male volunteers were each given 300 mg intravenous doses of Ciprofloxacin. Blood and urine samples were collected at various times throughout the day and the following data was collected: PROBLEM TABLE 4 - 9.

Ciprofloxacin

Time (hours)

mg Serum concentration ------L

2

1.20

3

0.85

4

0.70

6

0.50

8

0.35

10

0.25

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-25

I.V. Bolus Dosing

CONCENTRATION (MG/L) Con (mg/L)

(Problem 4 - 9) Ciprofloxacin:

101 100 10-1 0

2

4

6

8

10

Time (hours) –1

1.

k = 0.1875hr

2.

t½ = 3.7hr .

3.

MRT = 5.33hr .

4.

( C p )0 = 1.57mg ⁄ L .

5.

AUC = 8.395mg ⁄ L ⋅ hr .

6.

AUMC = 44.74mg ⁄ L ⋅ hr .

7.

Vd = 190.6L

8.

Cl = 35.74L ⁄ hr .

2

Basic Pharmacokinetics

REV. 00.1.27

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4-26

I.V. Bolus Dosing

The effect of Probenecid on Diprophylline (DPP) Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 10)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Nadai et al, "Pharmacokinetics and the Effect of Probenecid on the Renal Excretion Mechanism of Diprophylline", Journal of Pharmaceutical Sciences (Oct 1992), p. 1024 - 1027.

Diprophylline is used as a bronchodilator. A study by Nadai et al was designed to determine whether or not coadministration of Diprophylline with Probenecid affected the pharmacokinetic parameters of Diprophylline. In this study, male rats (average weight: 300 g) were given 60 mg/kg of Diprophylline intravenously and a 3 mg/kg loading dose of Probenecid followed by a continuous infusion of 0.217 mg/min/kg of Probenecid. The following set of data was obtained for Diprophylline (DPP):

PROBLEM TABLE 4 - 10.

The effect of Probenecid on Diprophylline (DPP)

Time (minutes)

µg Serum concentration -------mL

16

40.00

31

27.00

60

13.00

91

6.50

122

3.50

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-27

I.V. Bolus Dosing

CONCENTRATION (MIC/ML) Con (ug/mL)

(Problem 4 - 10) The effect of probenecid on diprophylline (DPP):

102

101

100 0

20

40

60

80

100

Time (min) –1

1.

k = 0.023min

2.

t½ = 30.13min .

3.

MRT = 43.48min .

4.

( C p )0 = 55.13ug ⁄ mL .

5.

AUC = 2396.96ug ⁄ mL ⋅ min .

6.

AUMC = 104219.8ug ⁄ mL ⋅ min .

7.

Vd = 326.5mL

8.

Cl = 7.5mL ⁄ min .

2

Basic Pharmacokinetics

REV. 00.1.27

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4-28

I.V. Bolus Dosing

Epoetin

(Problem 4 - 11)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

MacDougall et. al., "Clinical Pharmacokinetics of Epoetin (Recombinant Human Erythropoetin", Clinical Pharmacokinetics (1991), p 99 - 110.

Epoetin is recombinant human erythropoetin. Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produce erythropoetin; therefore, Epoetin is used in these patients to treat the anemia that results from the lack of erythropoetin. Epoetin has also been used in the treatment of anemias resulting from AIDS. malignant disease, prematurity, rheumatoid arthritis, sickle-cell anemia, and myelosplastic syndrome. In a study by Macdougall et al, eight patients who were on peritoneal dialysis (CAPD) were given an IV bolus dose of 120 U/kg which decayed monoexponentially from a peak of 3959 U/L to 558 U/L at 24 hours. The following data was obtained: PROBLEM TABLE 4 - 11.

Epoetin

Time (hours)

U Serum concentration ---L

0.0

4000

0.5

3800

1.0

3600

2.0

3300

3.0

3000

4.0

2550

5.0

2350

6.0

2150

7.0

1900

From the data presented in the preceding table and assuming that the patient weighs 65 kg, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-29

I.V. Bolus Dosing

Con (U/L) CONCENTRATION (U/L)

(Problem 4 - 11) Epoetin:

104

103 0

1

2

3

4

5

6

7

Time (hours) –1

1.

k = 0.107 hr

2.

t½ = 6.5 hr .

3.

MRT = 9.38 hr .

4.

( C p )0 = 4023 Units/L .

5.

Units ⋅ hr AUC = 37775 ------------------------ . L

6.

Units ⋅ hr AUMC = 354697 --------------------------- . L

7.

Vd = 1.9 L

8.

L Cl = 0.2065 ----- . hr

2

Basic Pharmacokinetics

REV. 00.1.27

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4-30

I.V. Bolus Dosing

Famotidine

(Problem 4 - 12)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Kraus, et. al., "Famotidine--Pharmacokinetic Properties and Suppression of Acid Secretion in Pediatric Patients Following Cardiac Surgery", Clinical Pharmacokinetics (1990), p 77 - 80.

Famotidine is a histamine H2-receptor antagonist. The study by Kraus, et. al., focuses on the kinetics of famotidine in children. In the study, ten children with normal kidney function and a body weight ranging from 14 - 25 kg, were each given a single intravenous 0.3 mg/kg dose of famotidine. Blood and urine samples were taken providing the following data: PROBLEM TABLE 4 - 12.

Famotidine

Time (hours)

µg Serum concentration -----L

0.33

300

0.50

250

1.00

225

4.00

125

8.00

70

12.00

40

16.00

15

From the data presented in the preceding table, determine the following assuming that the patient weighs 17.2 kg: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-31

I.V. Bolus Dosing

(Problem 4 - 12) Famotidine:

(MIC/L) ConCONCENTRATION (ug/mL)

10 3

10 2

10 1 0

5

10

15

20

Time (hours) –1

1.

k = 0.17 hr

2.

t½ = 3.9 hr .

3.

MRT = 5.7 hr .

4.

µg ( C p )0 = 285 ------ . L

5.

µg ⋅ hr AUC = 1600 ----------------- . L

6.

µg ⋅ hr AUMC = 9000 ------------------ . L

7.

Vd = 18 L

8.

Cl = 3.2L .

2

Basic Pharmacokinetics

REV. 00.1.27

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4-32

I.V. Bolus Dosing

Ganciclovir Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 13)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Trang, et. al., "Linear single-dose pharmacokinetics of ganciclovir in newborns with congenital cytomegalovirus infections", Clinical Pharmacology and Therapeutics (1993), p. 15 - 21.

Ganciclovir (mw: 255.23) is used against the human herpes viruses, cytomegalovirus retinitis, and cytomegalovirus infections of the gastrointestinal tract. In this study, twenty-seven newborns with cytomegalovirus disease were given 4 mg/kg of ganciclovir intravenously over one hour. Blood samples were taken and the data obtained is summarized in the following table:

PROBLEM TABLE 4 - 13.

Ganciclovir

Time (hours)

Serum concentration

1.50

4.50

2.00

4.00

3.00

3.06

4.00

2.40

6.00

1.45

8.00

0.87

From the data presented in the preceding table and assuming the patient weighs 3.6 kg, determine the following : 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-33

I.V. Bolus Dosing

(Problem 4 - 13) Ganciclovir:

CONCENTRATION (MICMOLE/L)

10

10

10 0

2

4

6

8

TIME (HR) –1

1.

k = 0.288hr

2.

t½ = 2.4hr .

3.

MRT = 3.5hr .

4.

µmole ( C p )0 = 23 ---------------- . mL

5.

µmole ⋅ hr AUC = 80 -------------------------- . mL

6.

µmole ⋅ hr AUMC = 280 ----------------------------- . mL

7.

mg 1000µg 4 ------- ⋅ 3.6kg ⋅ ------------------Dose kg mg Vd = ------------- = ------------------------------------------------------------- = 2.45L Cp 0 µmole µg 23 ---------------- ⋅ 255.23 ---------------L µmole

8.

L Cl = 0.7 ----- . hr

2

Basic Pharmacokinetics

REV. 00.1.27

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4-34

I.V. Bolus Dosing

Imipenem

(Problem 4 - 14)

Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Heikkila, Renkonen, and Erkkola, "Pharmacokinetics and Transplacental Passage of Imipenem During Pregnancy", Antimicrobial Agents and Chemotherapy (Dec. 1992), p 2652 - 2655.

Imipenem is a beta-lactam antibiotic which is used in combination with cilastin and is active against a broad spectrum of bacteria. The pharmacokinetics of Imipenem in pregnant women is established in this study. Twenty women (six of which were non-pregnant controls) were given a single intravenous dose of 500 mg of imipenem-cilastin (1:1). Blood samples were taken at various intervals and the data obtained is summarized in the following table:

PROBLEM TABLE 4 - 14.

Imipenem

Time (minutes)

mg Serum concentration ------L

10

27.00

15

23.50

30

15.50

45

9.50

60

6.50

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-35

I.V. Bolus Dosing

(Problem 4 - 14) Imipenem: 2

CONCENTRATION (MG/L)

10

1

10

0

10

0

10

20

30

40

50

60

TIME (MIN) –1

1.

k = 0.029 min

2.

t½ = 24 min .

3.

MRT = 34.5 min .

4.

mg ( C p )0 = 36.2 ------- . L

5.

mg ⋅ min AUC = 1250 --------------------- . L

6.

mg ⋅ min AUMC = 43125 ------------------------ . L

2

7.

8.

Dose 500mg Vd = ------------- = ------------------ = 13.8L Cp 0 mg 36.2 ------L L Cl = 0.4 --------- . min

Basic Pharmacokinetics

REV. 00.1.27

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4-36

I.V. Bolus Dosing

Methylprednisolone Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 15)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Patel, et. al., "Pharmacokinetics of High Dose Methylprednisolone and Use in Hematological Malignancies", Hematological Oncology (1993), p. 89 - 96.

Methylprednisolone is a corticosteriod that has been used in combination chemotherapy for the treatment of hematological malignancy, myeloma, and acute lymphoblastic leukemia. In a study by Patel et. al., eight patients were given 1.5 gram intravenous doses of methylprednisolone from which the following data was obtained:

PROBLEM TABLE 4 - 15.

Methylprednisolone

Time (hours)

µg Serum concentration -------mL

0.5

19.29

1.0

17.56

1.8

15.10

4.0

9.98

5.8

7.10

8.0

4.70

12.0

2.21

18.0

0.71

24.0

0.23

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-37

I.V. Bolus Dosing

(Problem 4 - 15) Methylprednisolone:

CONCENTRATION (MIC/ML)

102 101

Con (ug/mL)

100 10-1 0

5

10

15

20

25

Time (hours) –1

1.

k = 0.188 hr

2.

t½ = 3.69hr .

3.

MRT = 5.3hr .

4.

µg ( C p )0 = 21.2 -------- . mL

5.

µg ⋅ hr AUC = 112.5 ----------------- . mL

6.

µg ⋅ hr AUMC = 598.4 ------------------ . mL

7.

Vd = 71L

8.

L Cl = 13.3 ----- . hr

2

Basic Pharmacokinetics

REV. 00.1.27

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4-38

I.V. Bolus Dosing

Omeprazole Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 16)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Anderson, et. al., "Pharmacokinetics of [14C] Omeprazole in Patients with Liver Cirrhosis", Clinical Pharmacokinetics (1993), p. 71 - 78.

Omeprazole (mw: 345.42) is a gastric proton-pump inhibitor which decreases gastric acid secretion. It is effective in the treatment of ulcers and esophageal reflux. In normal patients 80% of the omeprazole dose is excreted as metabolites in the urine and the remainder is excreted in the feces. In the study by Anderson, et. al., eight patients with liver cirrhosis were given 20 mg, IV bolus doses of omeprazole. The patients had a mean body weight of 70 kg. Both blood were taken at various intervals throughout the study and the following data was obtained:

PROBLEM TABLE 4 - 16.

Omeprazole

Time (hours)

ρmole Serum concentration ---------------mL

0.75

3.49

1.00

3.25

2.00

2.46

3.00

1.86

4.00

1.40

5.00

1.06

6.00

0.80

7.00

0.61

8.00

0.46

10.00

0.26

12.00

0.15

From the data presented in the preceding table, determine the following : 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

Basic Pharmacokinetics

REV. 00.1.27

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4-39

I.V. Bolus Dosing

ConCONCENTRATION (umol/mL) (PICOMOLE/ML)

(Problem 4 - 16) Omeprazole:

10 1 10 0 10-1 0

2

4

6

8

10 12

Time (hours)

–1

1.

k = 0.280hr

2.

t½ = 2.5hr .

3.

MRT = 3.57hr .

4.

ρmole ( C p )0 = 4.3 ---------------- . mL

5.

ρmole ⋅ hr AUC = 15.4 -------------------------- . mL

6.

ρmole ⋅ hr AUMC = 55 ----------------------------- . mL

2

7.

8.

Dose 20mg Vd = ------------- = ------------------------------------------------------------------------------------------------------------ = 13465L Cp 0 ρmole mmole - ⋅ 345.42mg ------------------------ ⋅ 1000mL -------------------4.3 ---------------- ⋅ -----------------------9 mL 10 ρmole mmole L L Cl = 3.9 ----- . hr

Basic Pharmacokinetics

REV. 00.1.27

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4-40

I.V. Bolus Dosing

Pentachlorophenol Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 17)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Reigner, Rigod, and Tozer, "Absorption, Bioavailability, and Serum Protein Binding of Pentachlorophenol in the B6C3F1 Mouse", Pharmaceutical Research (1992), p 1053 - 1057.

Pentachlorophenol (PCP) is a general biocide. That is, it is an insecticide, fungicide, bactericide, herbicide, algaecide, and molluskicide, that is used as a wood preservative. Extensive exposure to PCP can be fatal. In a study by Reigner et al, six mice (average weight: 27 g) were given 15 mg/kg of PCP by intravenous bolus. Blood samples were taken at various intervals from which the following data was obtained: PROBLEM TABLE 4 - 17.

Pentachlorophenol

Time (hours)

µg Serum concentration -------mL

0.083

38.00

4.000

22.00

8.000

14.00

12.000

7.90

24.000

1.30

28.000

0.75

32.000

0.60

36.000

0.40

From the data presented in the preceding table, determine the following : 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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I.V. Bolus Dosing

CONCENTRATION (MIC/ML)

(Problem 4 - 17) Pentachlorphenol:

102 101

Con (ug/mL)

100 10-1 0

10

20

30

40

Time (hours) –1

1.

k = 0.134 hr

2.

t½ = 5.2hr .

3.

MRT = 7.5hr .

4.

µg ( C p )0 = 35.6 -------- . mL

5.

µg ⋅ hr AUC = 281 ----------------- . mL

6.

µg ⋅ hr AUMC = 2100 ------------------- . mL

7.

Vd = 11.4mL

8.

ml Cl = 1.5 ------ . hr

2

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9-(2-phophonylmethoxyethyl) adenine Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 18)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Naesens, Balzarini, and Clercq, "Pharmacokinetics in Mice of the Anti-Retrovirus Agent 9-(2-phophonylmethoxyethyl) adenine", Drug Metabolism and Disposition (1992), p. 747- 752.

9-(2-phophonylmethoxyethyl) adenine (PEMA) is an anti-retrovirus (anti-HIV) agent. The pharmacokinetics of PEMA in mice were established in a study by . In this study there were three different PEMA doses given: 25 mg/kg, 100 mg/kg, and 500 mg/kg. Each of these doses was injected intravenously into male mice. The data obtained from study using the 25 mg/kg dose is summarized in the following table:

PROBLEM TABLE 4 - 18.

9-(2-phophonylmethoxyethyl) adenine

Time (minutes)

µg Serum concentration -------mL

2.0

90.3

2.9

83.9

5.6

67.3

8.9

51.5

10.5

45.2

13.5

35.4

15.0

31.3

20.0

20.9

24.0

15.1

59.6

0.9

From the data presented in the preceding table, determine the following. (Assume that the mouse weighs 200g.) 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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I.V. Bolus Dosing

(Problem 4 - 18) Pema:

(MIC/ML) ConCONCENTRATION (ug/mL)

10 2 10 1 10 0 10 -1 0

10 20 30 40 50 60

Time (min) –1

1.

k = 0.08min

2.

t½ = 8.67min .

3.

MRT = 12.5min .

4.

µg ( C p )0 = 105 -------- . mL

5.

µg ⋅ hr AUC = 1300 ----------------- . mL

6.

µg ⋅ hr AUMC = 16250 ------------------- . mL

7.

Vd = 47.6ml

8.

mL Cl = 3.8 --------- . min

2

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Thioperamide Problem Submitted By: Maya Lyte Problem Reviewed By: Vicki Long

(Problem 4 - 19)

AHFS 12:34.56 Antivirals GPI: 1234567890 Antivirals

Sakurai, et. al., "The Disposition of Thioperamide, a Histamine H3-Antagonist, in Rats", J. Pharm. Pharmacol. (1994), p. 209 212.

Thioperamide is a histamine (H3) receptor-antagonist. In a study by Sakurai et al, rats were given 10 mg/kg intravenous injections of Thioperamide. The following data was obtained from the study: PROBLEM TABLE 4 - 19.

Thioperamide

Time (minutes)

µg Serum concentration -------mL

3.7

3.1

7.5

2.8

13

2.4

45

1.1

60

0.74

120

0.16

From the data presented in the preceding table, determine the following: 1.

Find the elimination rate constant, k .

2.

Find the half life, t ½ .

3.

Find MRT .

4.

Find ( C p )0 .

5.

Find the Area Under the Curve, AUC .

6.

Find the area under the first moment curve, AUMC .

7.

Find the volume of distribution, V d

8.

Find the clearance, Cl .

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(MIC/ML) ConCONCENTRATION (ug/mL)

(Problem 4 - 19) thioperamide:

101

100

10-1 0

20

40

60

80 100 120

Time (min) –1

1.

k = 0.0254min

2.

t½ = 27.3min .

3.

MRT = 39.4min .

4.

µg ( C p )0 = 3.39 -------- . mL

5.

µg ⋅ min AUC = 133.5 --------------------- . mL

6.

µg ⋅ min AUMC = 5256 ----------------------- . mL

7.

mg 10 ------Dose kg L Vd = ------------- = -------------------------------------------------------------------- = 2.95 -----Cp 0 µg mg 1000mL kg 3.39 -------- ⋅ ------------------- ⋅ -------------------mL 1000µg L

8.

Cl = 0.0254min

2

–1

L 1000ml mL ⋅ 2.95 ------ ⋅ ------------------ = 75 -------------------- . kg L min ⋅ kg

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Cocaine

(Problem 4 - 20)

Khan,vM. et. al. “Determination of pharmacokinetics of cocaine in sheep by liquid chromatography” J. Pharm. Sci. 76:1 (39-43) Jan 1987

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4.1.3

URINE From the Laplace Transform of a drug given by IV bolus we find that : k ( –K ⋅ t ) Xu = ----u- ⋅ X0 ⋅ ( 1 – e ) K

(EQ 4-10)

where Xu is the cumulative amount of drug in the urine at time t. Rearranging, we get: k – Kt ( Xu )∞ – Xu =  ----u- ⋅ X 0 ⋅ e  K

(EQ 4-11)

where the amount of drug that shows up in the urine at infinite time,

ku ( X u ) ∞ = ----- ⋅ X 0 . K

Thus a plot of ( Xu )∞ – X u vs. time on semi-log paper would result in a straight line with a slope of -K and an intercept of ( X u ) ∞ .. and we can get ku from the intercept and the slope. Rearranging the intercept equation, we get

( X u )∞ k u = K ⋅ -------------X0

This method

of obtaining pharmacokinetic parameters is known as the Amount Remaining to be Excreted (ARE) method. TABLE 4-4 Enalapril

urinary excretion data from 5 mg IV Bolus

Time (hr)

Cumulative Enalapril in urine (mg)

1

0.41

0.59

2

0.65

0.35

3

0.80

0.20

4

0.88

0.12

6

0.96

0.04

∞

1.0

------

X

∞ u

– X u mg

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Utilizations: A.R.E. Method

FIGURE 4-2.

10

Cumulative Enalapril in urine

0

Xu(inf) - Xu

0.2 0.1 -1

10

1.3 hr

half life

-2

10

0

1

2

3

4

5

6

Hours

• You should be able to transform a data set containing amount of drug in the urine vs. time into cumulative amount of drug in the urine vs. time and plot the ARE. (Amount Remaining to be Excreted ->

{ ( Xu )∞ – Xu ( cum ) } vs. time on semi-log yielding a straight line with a slope of

– K = – 0.533 hr

–1

and an intercept of

ku ⋅ X0 ( Xu ) ∞ = --------------- = 1.0 mg K

• You should be able to determine the elimination rate constant, K1, from cumulative urinary excretion data. (Calculate the slope of the graph on SL paper.)

• You should be able to determine the excretion rate constant, ku, from cumulation urinary excretion data. (Divide the intercept of the graph by X0 and multiply by K1. ( X u )∞ –1 1.0 mg –1 k u = K ⋅ -------------- = 0.53 hr ⋅ ----------------- = 0.106 hr ) 5.0 mg X0

• You should be able to determine k m . K = k u + k m • You should be able to calculate percent metabolized or excreted from a data set. Thus, Percent metabolized =

km k ------ ⋅ 100 and percent excreted unchanged = ----u- ⋅ 100 assuming K K

K = k u + km A second method is to plot the rate at which the drug shows up in the urine over time. Again, using the LaPlace transforms, we find that:

Utilization: Rate of excretion method

dX u –K t –K t --------- = k u ⋅ X0 ⋅ e = R0 ⋅ e (EQ 4-12) dt Thus, a plot of the rate of excretion vs. time results in a straight line on semi-log paper with a slope of -K1 and an intercept, R0 , of kuX0 . Rearranging the intercept

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equation yields rate

dX u dt

R k u = -----0- . X0

In real data, we don’t have the instantaneous excretion

, but the average excretion rate,

∆X ---------u- , ∆t

over a much larger interval. What

that means to our calculations is that over the interval of data collection, the total amount of drug collected divided by the total time interval is the average rate. In the beginning of the interval the rate was faster than at the end of the interval. So the average rate must have occurred in the middle of the interval. Thus equation 412 which is the instantaneous rate can be rewritten to –K t –K t ∆Xu mid mid ---------- = k u ⋅ X 0 ⋅ e = R0 ⋅ e ∆t TABLE 4-5 Enalapril

(EQ 4-13)

Urinary Rate Data ∆X ---------u∆t

Interval (hr)

t(mid)

∆t

Enalapril in urine ∆X u ,(mg)

0-1

0.5

1

0.41

0.41

1-2

1.5

1

0.24

0.24

2-3

2.5

1

missed sample

?

3-4

3.5

1

0.08

0.08

4-6

5

2

0.08

0.04

• You should be able to transform a data set containing amount of drug in the urine vs. time inter∆X

val into Average Rate, ---------u- , vs.

t ,(t mid the time of the midpoint of the interval), on semilog

∆t

yielding a straight line with a slope of

– K and an intercept of k u ⋅ X 0 . as shown below.

-1 0

Urinary Excretion Rate (mg/hr)

10

R0 = 0.53 mg/hr

-2 -1

10

1.3 hr

half life -2

10

0

1

2

3

4

5

T (Mid)

• You should be able to determine k u extrapolate the line to t = 0 . The value of Rate (at

t = 0 ), R0, = k r ⋅ X0 = 0.53 ( mg ⁄ hr ) which when divided by X 0 .is kr.

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I.V. Bolus Dosing

R

0.53mg/hr –1 Thus, -----0- = ------------------------- = 0.106hr X0

5mg

• You should be able to determine k m . K = k u + k m • You should be able to calculate percent metabolized or excreted from a data set.

The rate equation is superior clinically because the ARE method requires collection of all of the urine which is usually only possible when you have a catheterized patient while the Rate Method does not. (People don’t urinate on command, and your data could be in the toilet, literally.) An additional advantage of the rate equations is that the

AUC

∞ 0

has the units of

mass, which gives the total amount of drug excreted into the urine directly. Thus: AUC

∞

R 0 0.53 mg/hr = ------ = -------------------------= 1 mg –1 K 0 0.53 hr

AN INTERESTING OBSERVATION: If you look at the LaPlace Transform of the rate equation for any terminal compartment, you would see that the resulting equation is that of the previous compartment times the rate constant through which the drug entered the terminal compartment. Thus, the rate of drug showing up in the urine (terminal compartment) is: dX –K t –K t = R0 ⋅ e --------u- = k u ⋅ X 0 ⋅ e dt

where ku is the rate constant through which the drug entered the urine and –K t dX ------- = X 0 ⋅ e dt

is the equation of the previous compartment.

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4.2 Metabolite 4.2.1

PLASMA Remember, the LaPlace Transform of the metabolite data yielded ( km ⋅ Xo ) ( km ⋅ Xo ) –Kt – K 1t – K 1t – Kt X m = ---------------------- ⋅ ( e – e ) or Xm = ---------------------- ⋅ ( e – e ) depending on ( K1 – K ) ( K – K1 ) which rate constant that we arbitrarily assigned to be K, the summation of all the ways that the drug is removed from the body and K1, the summation of all the ways that the metabolite is removed from the body. When we begin to manipulate the data, we know that we have a curve with two different exponents in it. (If they were the same, the equation would be different.) We don’t know which is bigger, K1 or K, but we can rewrite the equation to simply reflect Klarge and Ksmall, knowing that one is K1 and the other is K but not which is which. If we, then, devided both sides of the equation by Vdm, the volume of distribution of the metabolite, we would get :

km    X 0   – ( Ksmall ⋅ t ) C pm =  --------------------------------------  ----------  e –e  K l arg e – K small  V dm  Utilization: Curve Stripping

– ( K l arg e ⋅ t )

  (EQ 4-14) 

• You should be able to plot a data set of plasma concentration of metabolite vs. time on semi-log paper yielding a bi-exponential curve.

e

– Kt

→ 0 as t → ∞ . And e –K

l arg e

t

–K

small

–k

l arg e

t

→ 0 faster than e

t

–K

time, t, e . In fact «e time, t, the equation becomes :

e

l arg e

–k

small

t

→ 0 . So, at some long

t

is small enough to be ignored. Thus at long

km X0   –( Ksmall ⋅ t ) -  -------- e C pm =  --------------------------------- K l arg e – K small  V dm 

(EQ 4-15)

So that the plot of the terminal portion of the graph would yield a straight line with a slope of -Ksmall and an intercept of I =

km X0   ----------------------------------  ------- K l arg e – K small  Vdm

• You should be able to obtain the slope of the terminal portion of the curve, the negative of which would be the smaller of the two rate constants, K small , (either the summation of all the ways that the drug is eliminated, eliminated,

K , or the summation of all the ways that the metabolite is

K1 ).

• Subtracting the two previous equations yields Basic Pharmacokinetics

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km X 0   –( Kbig ⋅ t ) -  -------- e C pm – C pm =  --------------------------------- K l arg e – K small  Vdm 

(EQ 4-16)

which is a straight line on semi-log paper with a slope of -kbig and an intercept of X0 km I =  ------------------------------------  --------- . Note: we can get the larger of the two rate constants from this  K l arg e – K small  V dm

method. TABLE 4-6

Drug

Metabolite

(1)

(2)

(3)

(4)

Time (hr)

Cp (mcg/L)

Cpm1 (mcg/L)

Cpm

0

0

181.2

181.2

0.5

24.7

175

150.3

1

(5)

Cpm

– Cpm

44.4

168.9

124.5

2

139

71.8

157.5

85.7

4

65.6

96.5

136.9

40.4

6

31.1

100

119

19

8

14.6

94.7

12

76.5

24

34

In the above data Cp vs. Time is the plasma profile of the drug from Table 4-1 on page 2 and Cpm1 vs. Time is the plasma profile of the metabolite. A plot of Cp vs. Time yielded a straight line with a slope,(-K) of -0.375 hr-1,

0.693 –1 K = ---------------------- = 0.375 hr and and intercept of 295 mic/ –1 1.85 hr

L,

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Figure 4-1 on page 3 (column 2 vs. 1 in Table 4-6 on page 53)

10 3

Concentration (mic/L)

Cpo = 295 mic/L

100

Concentration (ng/mL)

10 2

50

1.85 hr 10 1 0

2

4

6

8

Time (hours)

Time (hr)

while a plot of Cpm1 vs. Time( Figure 4-3 on page 54) yields a biexponential plot with a termi------------- and extrapolating the terminal line back to time = 0 nal slope of 0.07 hr-1 , k small = 0.693 10 hr

yields 181 mic/L. FIGURE 4-3.

Nifedipine Metabolite (column 3 vs. 1 in Table 4-6 on page 53)

)

Nifedipine IV bolus - Metabolite

Concentration (mic/L)

103

mic Cpm0 = 181 --------L

80 102

40 10 hr 101 0

4

8

12

16

20

24

Time (hours)

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• You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to t = 0 ) terminal line (column 4 vs. 1 in Table 4-6 on page 53) and the observed data (at early times) (column 3 vs. 1 in Table 4-6 on page 53) yielding a straight line with the slope of the line equal to the negative of the other (larger) rate constant (column 5 vs. 1 in Table 4-6 on page 53). First you would fill in the Cpm column (column 4 in Table 4-6 on page 53) by computing Cpm for various values of time i.e Cpm = Cpm 0 ⋅ e

– k small t

where – k small is the terminal slope of the

graph. Then Cpm – Cpm (column 5 in Table 4-6 on page 53) would be column 4 - column 3. Then a plot of Cpm – Cpm vs. time (column 5 vs. 1 in Table 4-6 on page 53) is shown below. FIGURE 4-4.

Curve strip of Nifedipine Metabolite data

3

10

Intercept Column 5

2

100 102

1

1.85 hr

50

Half life 1

10

0

1

2

3

4

5

6

Time (hr)

In this case, the slope of the stripped line line is -0.375 hr-1 and the intercept is 0.181.2 mic/L. The slope of -0.375 hr-1 should not be surprising as the plot of the data in Figure 4-3 on page 54 resulted in a terminal slope of -.07 hr-1 . Since the data set yielded a bi-exponential plot, separating out the exponents could only yield K (0.375 hr-1) or K1 as determined by our Laplace Transform information. Thus, the terminal slope could be either -K1 or -K. Since it was obviously not -K, it had to be -K1. Thus the other rate constant obtained by stripping has to be K. You can determine which slope is which rate constant if you have any data regarding intact drug (i e. either plasma or urine time profiles of intact drug) as the slope of any of those profiles is always

–K .

• You should be able to determine V dm if you have any urine data regarding intact drug (i.e. urine time profiles of intact drug) as the intercept of those profiles allow for the solution of k m . Thus the intercept, I, of the extrapolated line of equation 4-14 could be rearranged to contain only one unknown variable, V dm

1000 mic –1 0.375hr ⋅ 25mg ⋅ ---------------------km ⋅ X0 mg = ----------------------------------------------- = -------------------------------------------------------------------------- = 170 L . mic ( K l arg e – K small ) ⋅ I –1 ( 0.375 – 0.07 ) hr ⋅ 181.2 --------L

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Utilization: MRT Calculations

• You should be able to determine the rate constants using MRT calculations. In a caternary chain, each compartment contributes its MRT to the overall MRT of the drug, thus: Flow Chart 4-4

IV Bolus

K

X

MRT(IV) = 1/K Suppose the drug were given by IV bolus. Then the drug would have to be metabolized and the metabolite eliminated. Since the MRTs are additive, the overall MRT of the metabolite would be made up of the MRTs of the two processes, thus: Flow Chart 4-5

Metabolite km

kmu

X

Xm

MRT(met) = MRT(elim)+MRT(IV) MRT(met) =

1/K1

+ 1/K

Thus, using the data from Table 4-3 on page 5 the MRT(IV)Trap is MRT = AUMC ------------------ = 1986.1 ---------------- = 2.42 hr or about MRT = AUMC ------------------ = 2100 ------------ = 2.67 hr using calculus. AUC 819.9 AUC 787

And using the data from columns 1 and 3 from Table 4-6 on page 53 the MRT(met) using calcu--------------- = 17 hr. ------------------ = 36000 lus is MRT = AUMC AUC

2116

MRT(elim) = MRT(met) - MRT(IV) = 17 hr - 2.67 hr = 14.33 hr = 1/K2. Thus K2 = 0.07 hr-1.

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4.2.2

URINE Valid equations: k mu ⋅ k m X0 dXmu  – K small t – K l arg e t  –e ------------- = ---------------------------------------- ⋅  e  ( K l arg e – K small )  dt 

Utilization:

(EQ 4-17)

as in the previous urinary rate equation, clinically we work with the average rate over a definite interval which results in rewriting equation 4-17 as: k mu ⋅ k m X 0 ∆Xmu  – Ksmall tmid – K l arg e t mid  - ⋅ e ------------- = --------------------------------------–e  ( K l arg e – K small )  ∆t 

(EQ 4-18)

• You should be able to plot a data set of rate of metabolite excreted vs. time (mid) on semi-log paper yielding a bi-exponential curve.

• You should be able to obtain the slope of the terminal portion of the curve, the negative of which would be the smaller of the two rate constants (either

K1 or K ).

• You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to t = 0 ) terminal line and the observed data (at early times) yielding a straight line with the slope of the line equal to the negative of the other (larger) rate constant (either

K1 or K ).

• You should be able to utilize MRT calculations to obtain K1 and K . • You should be able to determine which slope is which rate constant if you have any data regarding intact drug (i.e. either plasma or urine time profiles of intact drug) as the slope of any of those profiles is always

–K .

By this time, it should be apparent that data which fits the same shape curve (mono-exponential, bi-exponential, etc.) are treated the same way. When the curves are evaluated, the slopes and intercepts are obtained in the same manner. The only difference is what those slopes and intercepts represent. These representations come from the equations which come from the LaPlace Transforms which come from our picture of the pharmacokinetic description of the drug. Please refer back to the section on graphical analysis in the Chapter 1, Math review for a interpretation of slopes and intercepts of the various graphs. Temporarily, please refer to exam section 1, chapter 14 for problems for this section (until problems can be generated) as well as additional problems for the previous sections.

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