Bolus Dosing

CHAPTER 4 I.V. Bolus Dosing

Author: Michael Makoid and John CobbyReviewer: 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.

2. The student will provide professional communication regarding the pharmacoki-netic parameters obtained to patients and other health professionals.

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

Vd K km kr AUC AUMC CL MRT t1 2⁄,,,,, , , ,

Basic Pharmacokinetics REV. 00.1.27 4-1Copyright © 1996-2000 Michael C. Makoid All Rights Reserved http://pharmacy.creighton.edu/pha443/pdf

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)

(EQ 4-1)

(EQ 4-2)

(EQ 4-3)

(EQ 4-4)

(EQ 4-5)

(EQ 4-6)

(EQ 4-7)

(EQ 4-8)

(EQ 4-9)

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

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

FIGURE 4-1.

Cpln K t Cp0ln+⋅–=

Xln K t X0ln+⋅–=

Cp Cp0eKt–

=

Cp0DVd

------=

t½0.693

K-------------=

AUC Cp td

0

∞( )

∫= ΣCpn Cpn 1++( )

2------------------------------------- ∆t⋅

Cplast

K--------------+=

AUMC t C⋅ p tdtn Cpn⋅( ) tn 1+ Cpn 1+⋅( )+

2-------------------------------------------------------------------- ∆t⋅

0

t

∑Cplast

K2

---------------tlast Cplast⋅( )

K----------------------------------+ +=

0

∞( )

∫=

MRT AUMCAUC

------------------=

Cl K Vd⋅=

TABLE 4-1 Nifedipine 25 mg IV bolus

Time (hr)Cp

(mcg/L)

2 139

4 65.6

6 31.1

8 14.6

K– Cp0


I.V. Bolus Dosing

Does your Graph look like this?

FIGURE 4-1 Nifedipine IV Bolus (25 mg IV Bolus)

• You should be able to determine K. A plot of the data in TABLE 4-1 results in FIGURE 4-1

Remember from high school algebra, the slope of any straight line is the rise over the run, ,

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. . Thus if we are judicious in the concentrations that we

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

• You should be able to determine :. To do this, extrapolate the line to . The value of

when is (in the graph above, which is equal to for an IV bolus

dose only.

Thus,

The volume of distribution is a mathematical construct. It is merely the proportionality constant between two knowns - the which results from a given . 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 vol-ume of distribution is such and such) or by graphical data analysis, you should be able to find the fourth.

0 2 4 6 8

Time (hours)

101

102

103

Co

ncen

tratio

n (

ng

/mL)

-K1 = -0.375 hr -1Cp0 = 295 mic/L

Co

nce

ntr

atio

n (

mic

/L)

Time (hr)

1.85 hr

100

50

dydx------

C1( )ln C2( )ln–C1C2-------

ln=

K 0.693t½

-------------0.693

1.85hr---------------- 0.375 hr

1–== =

Vd t 0= Cp

t 0= Cp0 Cp0 295micL

---------= D Vd⁄

Cp0Dose

Vd------------- Vd

DoseCp0

------------- 25mg295mic

L------------------------------------=

1000micmg

--------------------- 85L=⋅=,=

Cp0 D0


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):

which in this case is:

or

. In tabular format, the AUC calculation

is shown in TABLE 4-2.

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 propor-tionality 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 bioavail-ability 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:

which in the

data given in TABLE 4-1 is:

+

TABLE 4-2 AUC

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

AUC Cp td

0

∞( )

∫= ΣCpn Cpn 1++

∆t---------------------------------

Cpl

K--------+=

ΣCpo Cp1+

2-------------------------- ∆t1

Cp1 Cp2+

2-------------------------- ∆t2

Cp2 Cp3+

2-------------------------- ∆t3

Cp3 Cplast+

2------------------------------- ∆tlast

Cplast

K1---------------+⋅+⋅+⋅+⋅

Σ 295 139+2

------------------------ 2⋅ 139 65.6+2

------------------------- 2⋅ 65.6 31.1+2

--------------------------- 2⋅ 31.1 14.6+2

--------------------------- 2⋅ 14.60.375-------------+ + + +

mcg

L----------hr

Σ 434 204.6 96.7 45.7 38.9+ + + +{ }mcgL

----------hr 819.9mcg

L----------hr=

AUCt 1–t

AUC0t

∞

AUMC t C⋅ p tdtn Cpn⋅( ) tn 1+ Cpn 1+⋅( )+

2-------------------------------------------------------------------- ∆t⋅

0

t

∑Cplast

K2

---------------tlast Cplast⋅( )

K----------------------------------+ +=

0

∞( )

∫=

ΣT0 C⋅ po T1 C⋅ p1+

2----------------------------------------------- ∆t1

T1 C⋅ p1 T2 C⋅ p2+

2----------------------------------------------- ∆t2

T2 C⋅ p2 T3 C⋅ p3+

2----------------------------------------------- ∆t3⋅+⋅+⋅


I.V. Bolus Dosing

and thus,

+

or

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

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

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.

TABLE 4-3 AUMC

TIME Cp Cp*T

0 295 0

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

T3 C⋅ p3 Tlast C⋅ plast+

2---------------------------------------------------------- ∆tlast

Tlast C⋅ plast

K-------------------------------

Cplast

K2

---------------+ +⋅

Σ 0 295⋅ 2 139⋅+2

--------------------------------------- 2⋅ 2 139⋅ 4 65.6⋅+2

---------------------------------------- 2⋅ 4 65.6⋅ 6 31.1⋅+2

------------------------------------------ 2⋅+ + mcg

L----------hr

2

6 31.1⋅ 8 14.6⋅+2

------------------------------------------ 2⋅ 8 14.6⋅0.375

------------------ 14.6

0.3752

----------------+ + mcg

L----------hr

2

Σ 278 540.4 449 303.4 311.47 103.82+ + + + +{ } 1986.1mcg

L----------hr

2=

AUMCtAUMC

0t

∞

MRT AUMCAUC

------------------= 1986.1819.9

---------------- 2.42= =


I.V. Bolus Dosing

Flow Chart 4-1 IV Bolus

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

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

XK

MRT(IV) = 1/K

KKaXa

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

X

X XaXd Kd Ka K

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


I.V. Bolus Dosing

ase rent

tain

Normally, we don’t have information from the oral solution, just IV and tablet. So in that cthe information obtained about absorption from the tablet is bundled together into an appaabsorption rate constant consisting of both dissolution and absorption.

It should be apparent that this is a reasonably easily utilized and powerful tool used to obpharmacokinetic parameters.


I.V. Bolus Dosing

4.1.2 IV BOLUS, PARENT COMPOUND, PLASMA PROBLEMS

Equations used in this section:

1. from equation 4-3

2. equation 4-5

3. equation 4-8

4. = the y-intercept of the line from equation 4-3

5. Estimate for AUC = which is

Trapezoidal rule applied to equation 4-6

6. Estimate for AUMC = from equation 4-8

from equation 4-7

7. from equation 4-4

8.

K slope–=

t1 2⁄2ln

K--------=

MRT1K---- estimate( )= MRT AUMC

AUC------------------=

Cp0

AUCCp0

K---------= Cp td

0

∞∫

AUC Cp td

0

∞( )

∫= ΣCpn Cpn 1++( )

2-------------------------------------

∆t( )Cplast

K--------------+=

AUMC AUC MRT⋅=

AUMC Cp tdtn Cpn⋅( ) tn 1+ Cpn 1+⋅( )+

2-------------------------------------------------------------------- ∆tn⋅

0

t

∑Cplast

K2

---------------tlast Cplast⋅( )

K----------------------------------+ +=

0

∞( )

∫

VdDoseCp0

-------------=

Cl K1 Vd

Cp0

AUC------------ Dose

Cp0-------------⋅=⋅ Dose

AUC-------------==


I.V. Bolus Dosing

n-

Acyclovir (Problem 4 - 1)

De Miranda and Burnette, “Metabolic Fate and Pharmacokinetics of the Acyclovir Prodrug Valaciclovir in Cynomolgus Mokeys”, 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. Inthis study, three male cynomolgus monkeys were each given a 10 intravenous dose. The monkeys weighed anaverage of 3.35 each. Blood samples were collected and the following data was obtained:

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

1. Find the elimination rate constant, .

2. Find the half life, .

3. Find .

4. Find .

5. Find the Area Under the Curve, .

6. Find the area under the first moment curve, .

7. Find the volume of distribution,

8. Find the clearance, .

Problem Submitted By: Maya Lyte AHFS 12:34.56 AntiviralsProblem Reviewed By: Vicki Long GPI: 1234567890 Antivirals

PROBLEM TABLE 4 - 1. Acyclovir

Time (hours)

Serum concentration

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

mg kg⁄kg

µg mL⁄( )

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 1) Acyclovir:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0.5 1.0 1.5 2.010

10

10

0.0

0

1

2

CO

NC

EN

TR

AT

ION

(M

IC/M

L)

TIME (HR)

k 0.93hr1–

=

t½ 0.75hr=

MRT 1.08hr=

Cp( )0 30.4ug mL⁄=

AUC 32.75ug mL hr⋅⁄=

AUMC 35.2ug mL⁄ hr2⋅=

Vd 1.1L=

Cl 1.02L hr⁄=


I.V. Bolus Dosing

Aluminum (Problem 4 - 2)

Xu, Pai, and Melethil, "Kinetics of Aluminum in Rats. II: Dose-Dependent Urinary and Biliary Excretion", Journal of Pharmaceu-tical 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 anaverage weight of 375g, were given an IV bolus dose of aluminum (1 mg/kg). Blood samples were taken at variousintervals and the following data was obtained:

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



3. Find .

4. Find .






PROBLEM TABLE 4 - 2. Aluminum

Time (hours)Serum concentration,

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

ngmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 2) Aluminum:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 2 4 6 8 1210

10

10

10

TIME (HR)

CO

NC

EN

TR

AT

ION

(N

G/M

L)

3

4

5

k 0.234hr1–

=

t½ 3hr=

MRT 4.3hr=

Cp( )0 21000ng mL⁄=

AUC 89285ng mL hr⋅⁄=

AUMC 383926ng mL hr2⋅⁄=

Vd 17.86mL=

Cl 4.18mL hr⁄=


I.V. Bolus Dosing

Amgen (Problem 4 - 3)

Salmonson, Danielson, and Wikstrom, "The pharmacokinetics of recombinant human erythropoetin after intravenous and subcuta-neous 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 andused in the production of red blood cells. The kidneys of patients who have end-stage renal failure cannot produceerythropoetin; therefore, r-Epo is being investigated for use in these patients in order to treat the anemia that resultsfrom the lack of erythropoetin. In a study by Salmonson et al, six healthy volunteers were used to demonstrate thatboth IV and subcutaneous administration of erythropoetin have similar effects in the treatment of anemia due tochronic renal failure. The six volunteers were each given a 50 U/kg intravenous dose of Amgen. The average weightof the six volunteers was 79 kg. Blood samples were drawn at various times and the data obtained is summarizedbelow:




3. Find .

4. Find .






PROBLEM TABLE 4 - 3. Amgen

Time (hours)Serum concentration,

2 700

4 600

6 400

8 300

12 150

24 40

mUmL---------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 3) Amgen:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 5 10 15 20 25

101

102

103

Con (

mU/m

L)

TIME (HR)

CO

NC

EN

TR

AT

ION

(M

U/M

L)

k 0.134hr1–

=

t½ 5.2hr=

MRT 7.46hr=

Cp( )0 900mU mL⁄=

AUC 6945mU mL hr⋅⁄=

AUMC 49600= mU mL hr2⋅⁄

Vd 4.44L=

Cl 0.6L hr⁄=


I.V. Bolus Dosing

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

Brier and Harding, "Pharmacokinetics and Pharmacodynamics of Atrial Naturetic Peptide after Bolus and Infusion Administra-tion 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 variousintervals throughout the length of the study and the following data was obtained:




3. Find .

4. Find .






PROBLEM TABLE 4 - 4. Atrial Naturetic Peptide (ANP)

Time (minutes)Serum concentration,

3 380

10 280

20 170

30 130

40 100

50 70

60 50

pgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

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

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 10 20 30 40 50 60

Time (min)

101

102

103

Con

(pg/

mL)

C

ON

CE

NT

RA

TIO

N (

PG

/ML

)

k 0.0345min1–

=

t½ 20.09min=

MRT 28.95min=

Cp( )0 386.6pg mL⁄=

AUC 11206.4pg mL⁄ min⋅=

AUMC 324425.4pg mL⁄ min2⋅=

Vd 116.4mL=

Cl 4.02mL min⁄=


I.V. Bolus Dosing

Aztreonam (Problem 4 - 5)

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

Aztreonam is a monolactam structure which is active against aerobic, gram-negative bacilli. The pharmacokineticparameters 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 anaverage weight of 1639.6g. The following set of data was obtained:




3. Find .

4. Find .






PROBLEM TABLE 4 - 5. Aztreonam

Time (minutes)Serum concentration,

1 40.50

2 34.99

3 29.99

4 23.88

5 22.20

6 19.44

7 16.55

8 14.99

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 5) Aztreonam:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 2 4 6 8

101

102

Con

(ug/

mL)

TIME (MIN)

CO

NC

EN

TR

AT

ION

(U

G/M

L)

k 0.144min1–

=

t½ 4.81min=

MRT 6.94min=

Cp( )0 45.75ug mL⁄=

AUC 317.7ug mL min⋅⁄=

AUMC 2204.8ug mL min⋅⁄ 2=

Vd 3.58L=

Cl 0.516L min⁄=


I.V. Bolus Dosing

Recombinant Bovine Placental Lactogen (Problem 4 - 6)

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

Bovine placental lactogen (bPL) is a hormone similar to growth hormone and prolactin. It binds to both prolactin andgrowth 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:




3. Find .

4. Find .






PROBLEM TABLE 4 - 6. Recombinant Bovine Placental Lactogen

Time (minutes)Serum concentration

3.8 117

6.8 72

12.0 43

16.0 27

20.0 18

µgL

------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 6) Recombinant Bovine Placental Lactogen:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 5 10 15 20

Time (min)

101

102

103

Con

(ug/L

) C

ON

CE

NT

RA

TIO

N (

MIC

/L)

k 0.113min1–

=

t½ 6.13min=

MRT 8.85min=

Cp( )0

167.8ug L⁄=

AUC 1484.9ug L min⋅⁄=

AUMC 13141.1ug L min⋅⁄ 2=

Vd 23.84L=

Cl 2.69L min⁄=


I.V. Bolus Dosing

Caffeine (Problem 4 - 7)

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 rabbitswere given an 8 mg intravenous dose of caffeine. Blood samples were taken and the following data was obtained:




3. Find .

4. Find .






PROBLEM TABLE 4 - 7. Caffeine


12 3.75

40 2.80

65 2.12

90 1.55

125 1.23

173 0.72

243 0.37

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 7) Caffeine:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 50 100 150 200 250

Time (min)

10-1

100

101

Con (

ug/L)

Caffeine

CO

NC

EN

TR

AT

ION

(M

IC/M

L)

k 0.00997min1–

=

t½ 69.51min=

MRT 100.3min=

Cp( )0 4.105ug mL⁄=

AUC 411.7ug mL⁄ min⋅=

AUMC 41293.5ug mL⁄ min2⋅=

Vd 1.95L=

Cl 19.44mL min⁄=


I.V. Bolus Dosing

Ceftazidime (Problem 4 - 8)

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 withchronic renal failure were each given 1 g of ceftazidime intravenously. Both blood samples were taken the dataobtained from the study is summarized in the following table:




3. Find .

4. Find .






PROBLEM TABLE 4 - 8. Ceftazidime

Time (hours)Serum concentration

1 50

2 45

4 38

24 21

36 14

48 11

60 8

72 4

mgL

-------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 8) Ceftazidime:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 20 40 60 80

Time (hours)

100

101

102

Con

(mg/

L)

CO

NC

EN

TR

AT

ION

(M

G/L

)

k 0.0324hr1–

=

t½ 21.39hr=

MRT 30.86hr=

Cp( )0 47.57mg L⁄=

AUC 1468.2mg L hr⋅⁄=

AUMC 45308.6mg L hr⋅⁄ 2=

Vd 21.02L=

Cl 0.681L hr⁄=


I.V. Bolus Dosing

Ciprofloxacin (Problem 4 - 9)

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 res-piratory tract, skin, bone, and joint. In this study, twelve healthy, male volunteers were each given 300 mg intravenousdoses of Ciprofloxacin. Blood and urine samples were collected at various times throughout the day and the followingdata was collected:




3. Find .

4. Find .






PROBLEM TABLE 4 - 9. Ciprofloxacin


2 1.20

3 0.85

4 0.70

6 0.50

8 0.35

10 0.25

mgL

-------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 9) Ciprofloxacin:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 2 4 6 8 10

Time (hours)

10-1

100

101

Con (

mg/L)

C

ON

CE

NT

RA

TIO

N (

MG

/L)

k 0.1875hr1–

=

t½ 3.7hr=

MRT 5.33hr=

Cp( )0

1.57mg L⁄=

AUC 8.395mg L hr⋅⁄=

AUMC 44.74mg L hr⋅⁄ 2=

Vd 190.6L=

Cl 35.74L hr⁄=


I.V. Bolus Dosing

The effect of Probenecid on Diprophylline (DPP) (Problem 4 - 10)

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 coadmin-istration 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 ofProbenecid followed by a continuous infusion of 0.217 mg/min/kg of Probenecid. The following set of data wasobtained for Diprophylline (DPP):




3. Find .

4. Find .






PROBLEM TABLE 4 - 10. The effect of Probenecid on Diprophylline (DPP)


16 40.00

31 27.00

60 13.00

91 6.50

122 3.50

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

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

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 20 40 60 80 100

Time (min)

100

101

102

Con (

ug/m

L) C

ON

CE

NT

RA

TIO

N (

MIC

/ML

)

k 0.023min1–

=

t½ 30.13min=

MRT 43.48min=

Cp( )0 55.13ug mL⁄=

AUC 2396.96ug mL min⋅⁄=

AUMC 104219.8ug mL min⋅⁄ 2=

Vd 326.5mL=

Cl 7.5mL min⁄=


I.V. Bolus Dosing

Epoetin (Problem 4 - 11)

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

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



3. Find .

4. Find .






PROBLEM TABLE 4 - 11. Epoetin


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

UL----

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 11) Epoetin:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 1 2 3 4 5 6 7

Time (hours)

103

104

Con (

U/L)

C

ON

CE

NT

RA

TIO

N (

U/L

)

k 0.107 hr1–

=

t½ 6.5 hr=

MRT 9.38 hr=

Cp( )0 4023 Units/L=

AUC 37775Units hr⋅

L------------------------=

AUMC 354697Units hr

2⋅L

---------------------------=

Vd 1.9 L=

Cl 0.2065Lhr-----=


I.V. Bolus Dosing

Famotidine (Problem 4 - 12)

Kraus, et. al., "Famotidine--Pharmacokinetic Properties and Suppression of Acid Secretion in Pediatric Patients Following Car-diac 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 inchildren. In the study, ten children with normal kidney function and a body weight ranging from 14 - 25 kg, were eachgiven a single intravenous 0.3 mg/kg dose of famotidine. Blood and urine samples were taken providing the followingdata:

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



3. Find .

4. Find .






PROBLEM TABLE 4 - 12. Famotidine


0.33 300

0.50 250

1.00 225

4.00 125

8.00 70

12.00 40

16.00 15

µgL

------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 12) Famotidine:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 5 10 15 20

Time (hours)

101

102

103

Con

(ug/

mL)

C

ON

CE

NT

RA

TIO

N (

MIC

/L)

k 0.17 hr1–

=

t½ 3.9 hr=

MRT 5.7 hr=

Cp( )0

285µgL

------=

AUC 1600µg hr⋅

L-----------------=

AUMC 9000µg hr

2⋅L

------------------=

Vd 18 L=

Cl 3.2L=


I.V. Bolus Dosing

Ganciclovir (Problem 4 - 13)

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

Ganciclovir (mw: 255.23) is used against the human herpes viruses, cytomegalovirus retinitis, and cytomegalovirusinfections of the gastrointestinal tract. In this study, twenty-seven newborns with cytomegalovirus disease were given4 mg/kg of ganciclovir intravenously over one hour. Blood samples were taken and the data obtained is summarized inthe following table:

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



3. Find .

4. Find .






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

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 13) Ganciclovir:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 2 4 6 8

10

10

10

CO

NC

EN

TR

AT

ION

(M

ICM

OL

E/L

)

TIME (HR)

k 0.288hr1–

=

t½ 2.4hr=

MRT 3.5hr=

Cp( )0

23µmole

mL----------------=

AUC 80µmole hr⋅

mL--------------------------=

AUMC 280µmole hr

2⋅mL

-----------------------------=

VdDoseCp0

-------------4

mgkg------- 3.6kg

1000µgmg

-------------------⋅⋅

23µmole

L---------------- 255.23

µgµmole----------------⋅

------------------------------------------------------------- 2.45L===

Cl 0.7Lhr-----=


I.V. Bolus Dosing

Imipenem (Problem 4 - 14)

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 spectrumof bacteria. The pharmacokinetics of Imipenem in pregnant women is established in this study. Twenty women (six ofwhich were non-pregnant controls) were given a single intravenous dose of 500 mg of imipenem-cilastin (1:1). Bloodsamples were taken at various intervals and the data obtained is summarized in the following table:




3. Find .

4. Find .






PROBLEM TABLE 4 - 14. Imipenem


10 27.00

15 23.50

30 15.50

45 9.50

60 6.50

mgL

-------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 14) Imipenem:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 10 20 30 40 50 60

10

10

10

TIME (MIN)

CO

NC

EN

TR

AT

ION

(M

G/L

)

1

2

0

k 0.029 min1–

=

t½ 24 min=

MRT 34.5 min=

Cp( )0

36.2mgL

-------=

AUC 1250mg min⋅

L---------------------=

AUMC 43125mg min

2⋅L

------------------------=

VdDoseCp0

------------- 500mg

36.2mgL

------------------------- 13.8L===

Cl 0.4L

min---------=


I.V. Bolus Dosing

Methylprednisolone (Problem 4 - 15)

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 hemato-logical malignancy, myeloma, and acute lymphoblastic leukemia. In a study by Patel et. al., eight patients were given1.5 gram intravenous doses of methylprednisolone from which the following data was obtained:




3. Find .

4. Find .






PROBLEM TABLE 4 - 15. Methylprednisolone


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

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 15) Methylprednisolone:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 5 10 15 20 25

Time (hours)

10-1

100

101

102

Con (

ug/m

L) C

ON

CE

NT

RA

TIO

N (

MIC

/ML

)

k 0.188 hr1–

=

t½ 3.69hr=

MRT 5.3hr=

Cp( )0

21.2 µgmL--------=

AUC 112.5µg hr⋅

mL-----------------=

AUMC 598.4µg hr

2⋅mL

------------------=

Vd 71L=

Cl 13.3Lhr-----=


I.V. Bolus Dosing

Omeprazole (Problem 4 - 16)

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 inthe treatment of ulcers and esophageal reflux. In normal patients 80% of the omeprazole dose is excreted as metabo-lites in the urine and the remainder is excreted in the feces. In the study by Anderson, et. al., eight patients with livercirrhosis were given 20 mg, IV bolus doses of omeprazole. The patients had a mean body weight of 70 kg. Both bloodwere taken at various intervals throughout the study and the following data was obtained:

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



3. Find .

4. Find .






PROBLEM TABLE 4 - 16. Omeprazole


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

ρmolemL

----------------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 16) Omeprazole:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 2 4 6 8 10 12

Time (hours)

10-1

100

101

Con

(um

ol/m

L)

CO

NC

EN

TR

AT

ION

(P

ICO

MO

LE

/ML

)

k 0.280hr1–

=

t½ 2.5hr=

MRT 3.57hr=

Cp( )0 4.3ρmole

mL----------------=

AUC 15.4ρmole hr⋅

mL--------------------------=

AUMC 55ρmole hr

2⋅mL

-----------------------------=

VdDoseCp0

------------- 20mg

4.3ρmole

mL---------------- mmole

109ρmole

------------------------- 345.42mgmmole

------------------------ 1000mLL

--------------------⋅ ⋅ ⋅------------------------------------------------------------------------------------------------------------ 13465L===

Cl 3.9Lhr-----=


I.V. Bolus Dosing

Pentachlorophenol (Problem 4 - 17)

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 Reigneret al, six mice (average weight: 27 g) were given 15 mg/kg of PCP by intravenous bolus. Blood samples were taken atvarious intervals from which the following data was obtained:

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



3. Find .

4. Find .






PROBLEM TABLE 4 - 17. Pentachlorophenol


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

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 17) Pentachlorphenol:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 10 20 30 40

Time (hours)

10-1

100

101

102

Con (

ug/m

L) C

ON

CE

NT

RA

TIO

N (

MIC

/ML

)

k 0.134 hr1–

=

t½ 5.2hr=

MRT 7.5hr=

Cp( )0 35.6µgmL--------=

AUC 281µg hr⋅

mL-----------------=

AUMC 2100µg hr

2⋅mL

-------------------=

Vd 11.4mL=

Cl 1.5mlhr------=


I.V. Bolus Dosing

9-(2-phophonylmethoxyethyl) adenine (Problem 4 - 18)

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 ofPEMA 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 fromstudy using the 25 mg/kg dose is summarized in the following table:

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



3. Find .

4. Find .






PROBLEM TABLE 4 - 18. 9-(2-phophonylmethoxyethyl) adenine


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

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 18) Pema:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 10 20 30 40 50 60

Time (min)

10-1

100

101

102

Con

(ug/

mL)

C

ON

CE

NT

RA

TIO

N (

MIC

/ML

)

k 0.08min1–

=

t½ 8.67min=

MRT 12.5min=

Cp( )0 105µgmL--------=

AUC 1300µg hr⋅

mL-----------------=

AUMC 16250µg hr

2⋅mL

-------------------=

Vd 47.6ml=

Cl 3.8mLmin---------=


I.V. Bolus Dosing

Thioperamide (Problem 4 - 19)

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 intrave-nous injections of Thioperamide. The following data was obtained from the study:




3. Find .

4. Find .






PROBLEM TABLE 4 - 19. Thioperamide


3.7 3.1

7.5 2.8

13 2.4

45 1.1

60 0.74

120 0.16

µgmL--------

k

t½

MRT

Cp( )0

AUC

AUMC

Vd

Cl


I.V. Bolus Dosing

(Problem 4 - 19) thioperamide:

1.

2. .

3. .

4. .

5. .

6. .

7.

8. .

0 20 40 60 80 100 120

Time (min)

10-1

100

101

Con

(ug/

mL)

C

ON

CE

NT

RA

TIO

N (

MIC

/ML

)

k 0.0254min1–

=

t½ 27.3min=

MRT 39.4min=

Cp( )0

3.39µgmL--------=

AUC 133.5µg min⋅

mL---------------------=

AUMC 5256µg min

2⋅mL

-----------------------=

VdDoseCp0

-------------10

mgkg-------

3.39µgmL-------- mg

1000µg------------------- 1000mL

L--------------------⋅ ⋅

-------------------------------------------------------------------- 2.95Lkg------===

Cl 0.0254min1–

2.95Lkg------ 1000ml

L------------------ 75

mLmin kg⋅--------------------=⋅ ⋅=


I.V. Bolus Dosing

9-43)

Cocaine (Problem 4 - 20)

Khan,vM. et. al. “Determination of pharmacokinetics of cocaine in sheep by liquid chromatography” J. Pharm. Sci. 76:1 (3Jan 1987


I.V. Bolus Dosing

4.1.3 URINE

From the Laplace Transform of a drug given by IV bolus we find that :

(EQ 4-10)

where Xu is the cumulative amount of drug in the urine at time t. Rearranging, weget:

(EQ 4-11)

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

Thus a plot of vs. time on semi-log paper would result in a straight line

with a slope of -K and an intercept of .. and we can get ku from the intercept

and the slope. Rearranging the intercept equation, we get This method

of obtaining pharmacokinetic parameters is known as the Amount Remaining to beExcreted (ARE) method.

TABLE 4-4 Enalapril urinary excretion data from 5 mg IV Bolus

Time (hr)

Cumulative Enalapril in urine

(mg) 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 ------

Xu

ku

K----- X0 1 e

K t⋅–( )–( )⋅ ⋅=

Xu( )∞ Xu–ku

K-----

X0 eKt–⋅ ⋅=

Xu( )∞

ku

K----- X0⋅=

Xu( )∞ Xu–

Xu( )∞

ku K=Xu( )∞X0

--------------⋅

X∞u

Xu–

∞


I.V. Bolus Dosing

Utilizations: A.R.E. Method

FIGURE 4-2. Cumulative Enalapril in urine

• 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 -> vs. time on semi-log yielding a straight line with a slope of

and an intercept of

• 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 excre-tion data. (Divide the intercept of the graph by X0 and multiply by K1.

)

• You should be able to determine .

• You should be able to calculate percent metabolized or excreted from a data set. Thus,

Percent metabolized = and percent excreted unchanged = assuming

A second method is to plot the rate at which the drug shows up in the urine overtime. Again, using the LaPlace transforms, we find that:

(EQ 4-12)

Utilization: Rate of excretion method

Thus, a plot of the rate of excretion vs. time results in a straight line on semi-logpaper with a slope of -K1 and an intercept, R0 , of kuX0 . Rearranging the intercept

0 1 2 3 4 5 6

10

10

10

Xu(

inf)

- X

u

0

-1

-2

Hours

1.3 hr

half life

0.2

0.1

Xu( )∞ Xu cum( )–{ }

K 0.533 hr1–

–=– Xu( )∞

ku X0⋅K

--------------- 1.0 mg==

ku K=Xu( )∞X0

-------------- 0.53 hr1– 1.0 mg

5.0 mg----------------- 0.106 hr

1–=⋅=⋅

km K ku km+=

km

K------ 100⋅

ku

K----- 100⋅

K ku km+=

dXu

dt--------- ku X0 e

K t–R0 e

K t–⋅=⋅ ⋅=


I.V. Bolus Dosing

etion

What

total. Inl. So

tion 4-

equation yields . In real data, we don’t have the instantaneous excr

rate , but the average excretion rate, , over a much larger interval.

that means to our calculations is that over the interval of data collection, theamount of drug collected divided by the total time interval is the average ratethe beginning of the interval the rate was faster than at the end of the intervathe average rate must have occurred in the middle of the interval. Thus equa12 which is the instantaneous rate can be rewritten to

(EQ 4-13)

TABLE 4-5 Enalapril Urinary Rate Data

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

val into Average Rate, , vs. ,(t mid the time of the midpoint of the interval), on semilog

yielding a straight line with a slope of and an intercept of . as shown below.

• You should be able to determine extrapolate the line to . The value of Rate (at

), R0, = which when divided by .is kr.

Interval (hr) t(mid)

Enalapril in urine ,(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

ku

R0

X0------=

td

dXu ∆Xu

∆t----------

∆Xu

∆t---------- ku X0 e

K tmid–R0 e

K tmid–⋅=⋅ ⋅=

∆t ∆Xu

∆Xu

∆ t----------

∆Xu

∆ t---------- t

K– ku X0⋅

0 1 2 3 4 510

-210

-110

T (Mid)

Uri

nary

Exc

reti

on R

ate

(mg/

hr)

-1

-2

0

1.3 hr

half life

R0 = 0.53 mg/hr

ku t 0=

t 0= kr X0 0.53 mg hr⁄( )=⋅ X0


I.V. Bolus Dosing

, and

its of

us:

heequa-h thein the

and

Thus,

• You should be able to determine .

• 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 collec-tion of all of the urine which is usually only possible when you have a catheterizedpatient while the Rate Method does not. (People don’t urinate on commandyour data could be in the toilet, literally.)

An additional advantage of the rate equations is that the has the un

mass, which gives the total amount of drug excreted into the urine directly. Th

AN INTERESTING OBSERVATION: If you look at the LaPlace Transform of trate equation for any terminal compartment, you would see that the resulting tion is that of the previous compartment times the rate constant through whicdrug entered the terminal compartment. Thus, the rate of drug showing up urine (terminal compartment) is:

where ku is the rate constant through which the drug entered the urine

is the equation of the previous compartment.

R0

X0------

0.53mg/hr5mg

------------------------- 0.106hr1–

==

km K ku km+=

AUC∞0

AUC∞0

R0

K------ 0.53 mg/hr

0.53 hr1–

-------------------------- 1 mg===

dXu

dt--------- ku X0 e

K t–R0 e

K t–⋅=⋅ ⋅=

dXdt------- X0 e

K t–⋅=


I.V. Bolus Dosing

ger,

idedlite,

4.2 Metabolite

4.2.1 PLASMA

Remember, the LaPlace Transform of the metabolite data yielded

or depending on

which rate constant that we arbitrarily assigned to be K, the summation of all theways that the drug is removed from the body and K1, the summation of all theways that the metabolite is removed from the body. When we begin to manipulatethe data, we know that we have a curve with two different exponents in it. (If theywere the same, the equation would be different.) We don’t know which is bigK1 or K, but we can rewrite the equation to simply reflect Klarge and Ksmall, know-ing that one is K1 and the other is K but not which is which. If we, then, devboth sides of the equation by Vdm, the volume of distribution of the metabowe would get :

(EQ 4-14)

Utilization: Curve Stripping

• 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.

as . And faster than . So, at some long

time, t, . In fact is small enough to be ignored. Thus at long time, t, the equation becomes :

(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 =

• 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 the summation of all the

ways that the drug is eliminated, , or the summation of all the ways that the metabolite is

eliminated, ).

• Subtracting the two previous equations yields

Xm

km X⋅ o( )K1 K–( )

---------------------- eK– t

eK– 1t

–( )⋅= Xm

km X⋅ o( )K K1–( )

---------------------- eK– 1t

eKt–

–( )⋅=

Cpm

km

Kl earg Ksmall–--------------------------------------

X0

Vdm----------

eKsmall t⋅( )–

e–

Kl earg t⋅( )–

=

eKt–

0→ t ∞→ ekl earg t–

0→ eksmallt–

0→

eKl earg t–

eKsmallt–

« eKl earg t–

Cpm

km

Kl earg Ksmall–-----------------------------------

X0

Vdm

--------- e

Ksmall t⋅( )–

=

km

Kl earg Ksmall–-----------------------------------

X0

Vdm

---------

Ksmall

K

K1


I.V. Bolus Dosing

(EQ 4-16)

which is a straight line on semi-log paper with a slope of -kbig and an intercept of

. Note: we can get the larger of the two rate constants from this

method.TABLE 4-6

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, and and intercept of 295 mic/

L,

Drug Metabolite

(1) (2) (3) (4) (5)

Time (hr)Cp

(mcg/L) Cpm1 (mcg/L)

0 0 181.2 181.2

0.5 24.7 175 150.3

1 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

Cpm Cpm–km

Kl earg Ksmall–-----------------------------------

X0

Vdm

--------- e

Kbig t⋅( )–

=

Ikm

Kl earg Ksmall–------------------------------------

=X0

Vdm

---------

Cpm Cpm Cpm–

K0.693

1.85 hr1–

---------------------- 0.375 hr1–

==


I.V. Bolus Dosing

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

while a plot of Cpm1 vs. Time( Figure 4-3 on page 54) yields a biexponential plot with a termi-

nal slope of 0.07 hr-1 , and extrapolating the terminal line back to time = 0

yields 181 mic/L.

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

0 2 4 6 8

Time (hours)

101

102

103

Con

cent

ratio

n (n

g/m

L)

Cpo = 295 mic/L

Co

nce

ntr

atio

n (

mic

/L)

Time (hr)

50

100

1.85 hr

ksmall0.69310 hr-------------=

Nifedipine IV bolus - Metabolite

Time (hours)

)

0 4 8 12 16 20 24101

102

103

Con

cent

ratio

n (m

ic/L

)

10 hr

80

40

Cpm0 181mic

L---------=


I.V. Bolus Dosing

• You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to ) 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) yield-ing a straight line with the slope of the line equal to the negative of the other (larger) rate con-stant (column 5 vs. 1 in Table 4-6 on page 53).

First you would fill in the column (column 4 in Table 4-6 on page 53) by computing

for various values of time i.e where is the terminal slope of the

graph. Then (column 5 in Table 4-6 on page 53) would be column 4 - column 3.

Then a plot of 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

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, sepa-

rating 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 obvi-ously 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 .

• You should be able to determine 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 .

Thus the intercept, I, of the extrapolated line of equation 4-14 could be rearranged to contain

only one unknown variable, .

t 0=

Cpm Cpm

Cpm Cpm0 eksmallt–

⋅= ksmall–

Cpm Cpm–

Cpm Cpm–

0 1 2 3 4 5 6

10

1

10

2

10

Time (hr)

1

2

3

Col

umn

5

1.85 hr

Intercept

100

50

Half life

K–

Vdm

km

Vdm

km X0⋅Kl earg Ksmall–( ) I⋅

-----------------------------------------------0.375hr

1–25mg

1000 micmg

----------------------⋅⋅

0.375 0.07–( ) hr1–

181.2micL

---------⋅-------------------------------------------------------------------------- 170 L===


I.V. Bolus Dosing

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

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

Thus, using the data from Table 4-3 on page 5 the MRT(IV)Trap is

hr or about hr using calculus.

And using the data from columns 1 and 3 from Table 4-6 on page 53 the MRT(met) using calcu-

lus is hr.

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

XK

MRT(IV) = 1/K

kmukmXmX

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

MRT(met) = 1/K1 + 1/K

MRT AUMCAUC

------------------= 1986.1819.9

---------------- 2.42= = MRT AUMCAUC

------------------= 2100787

------------ 2.67= =

MRT AUMCAUC

------------------= 360002116--------------- 17= =


I.V. Bolus Dosing

4.2.2 URINE

Valid equations:

(EQ 4-17)

Utilization: as in the previous urinary rate equation, clinically we work with the average rateover a definite interval which results in rewriting equation 4-17 as:

(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 or ).

• You should be able to feather (curve strip) the other rate constant out of the data by plotting the difference between the extrapolated (to ) 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 or ).

• You should be able to utilize MRT calculations to obtain and .

• You should be able to determine which slope is which rate constant if you have any data regard-ing intact drug (i.e. either plasma or urine time profiles of intact drug) as the slope of any of

those profiles is always .

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 thecurves are evaluated, the slopes and intercepts are obtained in the same manner.The only difference is what those slopes and intercepts represent. These represen-tations come from the equations which come from the LaPlace Transforms whichcome from our picture of the pharmacokinetic description of the drug. Pleaserefer back to the section on graphical analysis in the Chapter 1, Math review for ainterpretation of slopes and intercepts of the various graphs.

Temporarily, please refer to exam section 1, chapter 14 for problems for this sec-tion (until problems can be generated) as well as additional problems for the previ-ous sections.

dXmu

dt-------------

kmu kmX0⋅Kl earg Ksmall–( )

---------------------------------------- eKsmallt–

eKl earg t–

–

⋅=

∆Xmu

∆t-------------

kmu kmX0⋅Kl earg Ksmall–( )

---------------------------------------- eKsmalltmid–

eKl earg tmid–

–

⋅=

K1 K

t 0=

K1 K

K1 K

K–


Bolus Dosing

Health & Medicine

k0 t t n cp n

cp 2cp2

vdmg cp

cp dt auc

cp half life

cp lastcp dt

t c p dt

bolus data time