71
Basic Pharmacokinetics REV. 99.4.25 9-1 Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/ CHAPTER 9 Clearance OBJECTIVES 1. Given patient information regarding organ function, the student will calculate (III) changes in clearance and other pharmacokinetic parameters inherent in compro- mised patients. 2. Determine the total clearance based on Dose and AUC. 3. Determine clearance of an organ based on dose, AUC, and fraction of drug elimi- nated by the organ 4. Determine change in clearance due to functional changes in an organ. 5. Determine change in clearance due to change in blood flow through an organ. 6. Prepare a professional consult (V) and justify (VI) modifications in drug therapy based on clearance of a drug.

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Page 1: Clearance

Basic Pharmacokinetics REV. 99.4.25 9-1Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

CHAPTER 9 Clearance

OBJECTIVES

1. Given patient information regarding organ function, the student will calculate (III) changes in clearance and other pharmacokinetic parameters inherent in compro-mised patients.

2. Determine the total clearance based on Dose and AUC.

3. Determine clearance of an organ based on dose, AUC, and fraction of drug elimi-nated by the organ

4. Determine change in clearance due to functional changes in an organ.

5. Determine change in clearance due to change in blood flow through an organ.

6. Prepare a professional consult (V) and justify (VI) modifications in drug therapy based on clearance of a drug.

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Basic Pharmacokinetics REV. 99.4.25 9-2Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

9.1 Equations

(EQ 9-1)

(EQ 9-2)

(EQ 9-3)

(EQ 9-4)

(EQ 9-5)

(EQ 9-6)

(EQ 9-7)

(EQ 9-8)

(EQ 9-9)

(EQ 9-10)

(EQ 9-11)

(EQ 9-12)

(EQ 9-13)

(EQ 9-14)

(EQ 9-15)

Cl Rate of EliminationSerum Concentration---------------------------------------------------=

Cltotf D⋅ osAUC

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

Clr Cltot Fraction of drug that is renally eliminated( )⋅=

ClH Cltot Fraction of drug that is hepatically eliminated( )⋅=

Qr 0.0191 Lmin kg⋅-------------------- renal blood perfusion 70kg 60min

hr---------⋅ ⋅ 80 L

hr-----blood≈=

QH 0.0238 Lmin kg⋅-------------------- hepatic blood perfusion 70kg 60min

hr---------⋅ ⋅ 100 L

hr-----blood≈=

Er Clr( ) Qr⁄=

EH ClH( ) QH⁄=

fu Clint⋅ Q Cl⋅Q Cl–----------------=

Clint

Q Cl⋅Q Cl–----------------

fu

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

Fi

fu∗ Cl∗int⋅fu Clint⋅

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

FRQ∗Q-------=

FCl

Fi FR⋅FR Er Fi FR–( )+-----------------------------------------=

FCltot

Cl∗tot

Cltot

-------------- k∗ V∗⋅k V⋅

----------------Cl∗H Cl∗r+

ClH Clr+-----------------------------= = =

0.80 FClto t≤ 1.20≤

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Basic Pharmacokinetics REV. 99.4.25 9-3Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

9.2 Definitions and Terms

Clearance: The hypothetical volume of a fluid from which a substance istotally and irreversibly removed per unit time.

Dimensions:

Examples of fluids: blood, serum, plasma, bile, gut contents, CSF.

Systemic or Total Body Clearance: Removal process is elimination

(excretion and metabolism). Fluid is usually plasma or serum (rarely blood).

Renal Clearance: Removal process is urinary excretion of

unchanged drug. Fluid is usually plasma or serum (rarely blood).

Metabolic Clearance: Removal process is metabolism. Fluid is usu-

ally blood (rarely plasma or serum).

Hepatic Clearance: This is when the liver is the metabolic

organ.

Creatinine Clearance: This is applied to endogenous creatinine.

It is used to monitor renal function, and thus is a valuable parameter for calculatingdosage regimens in elderly patients or those suffering from renal dysfunction. Cre-atinine

Value for normal males: 117 ± 20 ml/minValue for normal females: 108 ± 20 ml/min

Inulin Clearance: This is for inulin, and yields the glomerularfiltration rate.

Value for normal males: 124.5 ± 9.7 ml/minValue for normal females: 108.8 ± 13.5 ml/min

L3

T⁄

Cl( ), Cltot( )

Clr( )

Clm( )

ClH( ) Clm

Clcr( ) Clr

t1 2⁄ 231min=

Clinulin Clr

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Basic Pharmacokinetics REV. 99.4.25 9-4Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

9.3 Measurement of Creatinine Clearance

The mass of endogenous creatinine excreted into the urine collected over a given

time interval is determined. The mean serum creatinine concentration over that interval is calculated from sample determinations; this should be the con-

centration halfway through the interval. In practice, and, as is rela-tively constant, the serum sample is taken at any convenient time.

Let “a” be a volume of serum having a creatinine concentration of . The mass

of creatinine in the serum will be . If this creatinine is totally and irrevers-ibly removed from the serum to the urine in the time interval, , then

(EQ 9-16)

Thus, (EQ 9-17)

The volume of serum from which this creatinine is removed in unit time is ;this is the definition of clearance. Hence,

(EQ 9-18)

Siersbaek-Neilson et al. report a value of 11.1 for in 149 males (aged20-99). The value of decreased with age from 16.53 per Kg body

weight (age 20-29) to 6.53 per Kg body weight (age 90-99). For a 25 year

old 70Kg male, equation 9-18 yields

∆t( ) Cs( )cr

∆t 24hr= Cs( )cr

Cs( )cr

a Cs( )cr⋅

∆t

a Cs( )cr⋅ ∆Xu=

a∆Xu

Cs( )cr

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

a ∆t⁄

Clcra∆t-----

∆Xu ∆t⁄( )T

Cs( )cr

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

µg ml⁄ Cs( )cr

∆Xu ∆t⁄( )T

µg min⁄

µg min⁄

Clcr 104.2 mlmin---------=

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Basic Pharmacokinetics REV. 99.4.25 9-5Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

9.4 Model Correlations

Although intrinsically model independent, clearance can also be related to com-partmental models.

9.4.1 RENAL CLEARANCE

The plasma renal clearance of a drug may be measured analogously to creatinineclearance:

(EQ 9-19)

The practical versions of and states:

(EQ 9-20)

Comparing equation 9-19 and equation 9-20,

(EQ 9-21)

This relates clearance to model parameters. What is the slope of a plot of

against ?

Note that if (males), it may indicate active secretion of the druginto the kidney tubules. If (females), it may indicate reabsorp-

tion of the drug from the kidney tubules.

9.4.2 SYSTEMIC CLEARANCE AND METABOLIC CLEARANCE

How could you measure ? and

By analogy,

(EQ 9-22)

and , so .

Consequently, fractional changes in clearance,

Clr

∆Xu ∆t⁄( )T

Cp

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

∆Xu ∆t⁄( )T

kuX kuVCp= =

Clr kuV=

∆Xu ∆t⁄( )T

Cp

Clr 117 20ml min⁄±>

Clr 108 20ml min⁄±<

Cl Clm

Cltot KV 0.693Vt1 2⁄

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

K ku km+= Cltot Clr Clm+=

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Basic Pharmacokinetics REV. 99.4.25 9-6Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

(EQ 9-23)

where is new or altered variable. Hepatic function and renal function are not apriori connected, although some physiological functional changes might result insimilar changes in clearance of both organs. We can see from equation 9-12 thatchanges in total body clearance can result in changes in either K, V, or both. Theconsequences of that will be discussed in the section on dosage regimens.

9.4.3 USE IN PHARMACOKINETIC EQUATIONS

Systemic Clearance (Cl) can be used in many equations where the drug is removedby elimination (renal excretion and metabolism). If renal excretion is the onlyremoval process, use l if metabolism, use . Some examples:

Intravenous infusion:

Oral and Intravenous Bolus:

This equation becomes a means of calculating Cl from plasma data.

Dosage Regimen:

These examples are all model-independent expressions, which are very useful incalculating dosage regimens. The importance of clearance terms rests on their abil-ity to account for variations in both and / or V simultaneously, as both these

parameters can change in disease states and with age.

FCl tot

Clh°

Clr°

+

Clh Clr+------------------------- K

°V

°⋅K V⋅

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

Clr Clm

Cp( )ss

QCl------=

Clf Xa( )

0

Cp td

0

∫--------------- fD

Cp td

0

∫--------------- fD

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

Cp( )ssfD

τ Cl⋅-------------=

t1 2⁄

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Basic Pharmacokinetics REV. 99.4.25 9-7Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

9.5 Physiological Factors Affecting Clearance

9.5.1 INTRINSIC CLEARANCE

Intuitively, it may be recognized that two factors will affect the clearance of a drug:

1. The rate at which blood is presented to the eliminating organ.2. The intrinsic ability of the eliminating organ to clear the drug.

Mathematically, a hyperbolic equation has been derived to illustrate the relativeeffect of these factors. (Note: this is one model of clearance. There are several oth-ers which also illustrate the effect of these factors.)

Liver Drug Metabolism

(EQ 9-24)

Where is the rate of blood flow through the liver (assumed 23.8 ml/min/Kgbody weight in normal adult),

fu is the fraction unbound of the drug, and

is the intrinsic hepatic clearance of the drug.

If there were no physiological limits to the rate of blood flow , hence equa-tion 9-24 becomes

(EQ 9-25)

This equation provides a definition for intrinsic clearance, namely the clearance ofa drug were there to be no physiological limits on the rate of blood flow throughthe clearing organ.

Cl int( )

ClH

QH fu⋅ ClH( )⋅ int

QH fu ClH( )⋅ int+------------------------------------------=

QH

ClH( )int

QH ∞→( )

ClH ClH( )int

=

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Basic Pharmacokinetics REV. 99.4.25 9-8Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Kidney drug excretion By analogy for excretion of unchanged drug by the kidney:

(EQ 9-26)

Where is the rate of blood flow through the kidney (assumed 19.1 ml/min per

Kg body weight in normal adults), and

is the intrinsic renal clearance of the drug.

Note that the value of (assumed 1.75 ml.min per Kg body weight in normaladults) is about 9% of .

9.5.2 EXTRACTION RATIO (E)

This is defined as “the ratio of the clearance of a drug compared to the rate of blood flow through the clearing organ.” As such, it indicates what fraction of the drug in the blood is cleared (extracted) on each passage through the clearing organ. Note: when using clearance to calculate extraction ratio, blood flow must be used.

Drug metabolism by the liver

(EQ 9-27)

Where is the steady-state hepatic extraction ratio.

By comparison with equation 9-24,

(EQ 9-28)

Thus, the range of values of is from zero, when , to one, when

or . For example, propanolol has , yielding

and in normal adult males.

Clr

Qr fu Clr( )int

⋅ ⋅

Qr fu Clr( )⋅ int+---------------------------------------=

Qr

Clr( )int

Clcr

Qr

EH

ClH

QH

---------=

EH

EH

fu ClH( )⋅ int

QH fu ClH( )⋅ int+------------------------------------------=

EH ClH( )int

0=

QH 0= ClH( )int

QH» EH 0.75=

ClH17.9ml

min Kg body weight⋅-----------------------------------------------------= ClH( )

int71.4ml

min Kg body weight⋅-----------------------------------------------------=

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Basic Pharmacokinetics REV. 99.4.25 9-9Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Kidney excretion of unchanged drug

(EQ 9-29)

(EQ 9-30)

where is the steady-state renal extraction ratio.

Thus the range of values is from zero, when , to one, when or

. For example, digoxin has , yielding

and in normal adult males. In this case, note that

Er

Clr

Qr

--------=

Er

fu Clr( )⋅int

Qr fu Clr( )⋅ int+---------------------------------------=

Er

Clr( )int

0= Qr 0=

Clr( )int

Qr» Er 0.09= Clr1.72ml

min Kg body weight⋅-----------------------------------------------------=

Clr( )int

1.89mlmin Kg body weight⋅-----------------------------------------------------=

Clr Clcr≈

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Basic Pharmacokinetics REV. 99.4.25 9-10Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

9.6 Hepatic Function and Clearance

9.6.1 ALTERATIONS IN HEPATIC BLOOD FLOW

For a given drug, equation 9-24 predicts that alterations in the hepatic blood perfu-sion rate will cause a change in drug clearance, assuming the intrinsic hepatic clearance is unaltered. A general equation may be derived relating the ratio of hepatic clearances at two blood perfusion rates to the fractional change in perfu-sion rate and the extraction ratio of the drug.

(EQ 9-31)

where denotes normal hepatic clearance,

denotes altered hepatic clearance

is the new flow rate over the old flow rate, the fractional change in blood

perfusion rate, and

is the hepatic extraction ratio under normal conditions.

The equation predicts that, for any given decrease in blood perfusion rate, drugshaving a large normal extraction ratio will experience a proportionally greaterreduction in clearance than drugs having a small normal extraction ratio.

Liver blood flow can be reduced by congestive heart failure, for example. Theintrinsic hepatic clearance can be represented by the inherent activity of theenzymes responsible for drug metabolism.

9.6.2 ALTERATIONS IN HEPATIC INTRINSIC CLEARANCE

For any given drug, equation 9-24 predicts that alterations in the intrinsic hepatic clearance will cause a change in drug clearance, assuming the blood flow rate is unchanged. A general equation may be derived relating the ratio of hepatic clear-ance at two intrinsic hepatic clearances to the fractional change in intrinsic hepatic and the extraction ratio of the drug.

Cl∗H

ClH

------------FR

FR EH 1 FR–( )+-----------------------------------------=

ClH

Cl∗H

FR

QH∗

QH----------=

EH

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(EQ 9-32)

where is the fractional change in fraction unbound times the frac-

tional change in intrinsic hepatic clearance.

The equation predicts that, for any given decrease in intrinsic hepatic clearance,drugs having a small normal extraction ratio will experience a proportionatelygreater reduction in clearance than drugs having a large normal extraction ratio.

The intrinsic hepatic clearance of a drug can be reduced by cirrhosis or increasedby enzyme inducers, such as phenobarbitol.

9.6.3 TABULATED OR GRAPHICAL ALTERATIONS

A table or graph of clearance changes when the hepatic blood flow (but not the intrinsic hepatic clearance) is altered shows that drugs having a low extraction ratio need little adjustment in dosage. Even if the hepatic blood flow were halved , the hepatic clearance is still 91% of its normal value. Con-

versely, dosage adjustment is necessary for drugs having a high extraction ratio and predominantly eliminated by hepatic metabolism (e.g., propanolol).

A table or graph of clearance changes when the intrinsic hepatic clearance (but notthe hepatic blood flow) is altered shows that drugs having a high extraction ratio

need little adjustment in dosage. Even if the intrinsic hepatic clearance

were halved , the hepatic clearance is still 91% of its normal value. Con-versely, dosage adjustment is necessary for drugs having a low extraction ratio andpredominately eliminated by hepatic metabolism (e.g., phenylbutazone).

Cl∗H

ClH

------------Fi

1 EH Fi 1–( )+-----------------------------------=

Fi

fu∗ Clint( )∗⋅fu Clint( )⋅

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

EH 0.1=( )

FR 0.5=( )

EH 0.9=( )

Fi 0.5=( )

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9.7 Renal Function and Clearance

Approximately 25% of cardiac output goes to the kidneys or approximately 735 ml/min of plasma is presented to the kidneys of a 70 kg man (19.1 mL/min/kg x 70 kg). Approximately 125 ml/min (1.8 mL/min/kg) of that goes to the glomeruli for filtration (Glomerular Filtration Rate, GFR). Unbound drug is filtered into the proximal renal tubule at this point. The remaining plasma (as blood) is shunted around the tubule in the arterioles adjascent to the proximal tubule where drug may be actively secreted from the arteriol into the proximal tubule or actively reabsorbed in the opposite direction. As the blood flows down the vessels adjas-cent to the loop of Henle, the drug may be also passively reabsorbed into the blood vessel as the water in the urine is being reabsorbed and the urine is being concen-trated.

This leads to some interesting possibilities:

1. . It is likely that the drug is filtered only, in this case, . It is also possible that

secretion and reabsortion balance and cancel each orther out but are still occurring. The actual clearance of the drug may be low as the drug may be bound to plasma protiens or red blood cells.

11. . Net active secretion is infered in this case. These active mechanisms are non-

specific and consequently, drugs actively secreted compete with each other. Secretion, if it occurs, occurs on the unbound drug and thus is also effected by changes in free fraction. In cases where secretion is very rapid and as a consequence, virtually all of the drug is removed by the single pass through the kidney (Er ~1), the disssociation of the drug from the protien or out of the red blood cells is not a hinderance. Some reabsorption may occur but it is less than secre-tion.

12. . Net active reabsorbtion is infered in this case. Active reabsorption occurs for

many exogenous compounds, including glucose and vitamins. For many compounds, reabsorp-tion is passive, occurring only as a consequence of the concentration gradient produced as water is removed from the urine as is proceeds down the renal tubule. Since the membrane is lipoidal in nature, polar compounds, ionized acids and ionized bases are less likely to be reabsorbed. Thus changing the pH of the urine would result in changing the reabsorption characteristics of weakly acidic or basic drugs.

For low molecular weight drugs (<2,000 dalton) , filtration always occurs. Active secretion, active reabsorption and passive reabsorption may occur.

It has been found that renal blood flow is little affected by changes in blood flow elsewhere. However, in chronic renal dysfunction there are two effects which exhibit a parallel decline. One is a decrease in glomerular filtration rate (GFR), as measured by , and the other is the net secretion of drugs into the kidney

tubules. Note that p-amino hippurate (PAH) clearance measures the sum of both effects.

Clr fu GFR⋅=

Clr fu GFR⋅>

Clr fu GFR⋅<

Clcr

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For any given drug, equation 9-26 predicts that alterations in both the renal bloodperfusion rate (as manifest by the GFR) and the intrinsic renal clearance will causea change in drug clearance. A general equation may be derived relating the ratio ofrenal clearances at two different blood flow rates and two different intrinsic renalclearances.

(EQ 9-33)

where is the fractional change in drug unbound times the fractional change inintrinsic renal clearance.

In this case,

(EQ 9-34)

where is the fractional change in blood flow rate (or GFR).

Thus, equation 9-33 shows that the renal clearance of a drug is reduced by a con-stant fraction, independent of the renal extraction ratio . This fractionaldecrease can be estimated by changes in creatinine clearance:

(EQ 9-35)

where is the altered creatinine clearance

Substituting, equation 9-33 through equation 9-35, we get:

(EQ 9-36)

This equation shows why, in cases of chronic renal dysfunction, a change in themeasured creatinine clearance indicated a likely change in drug renal clearance.Hence, dosage adjustments are made on this basis, particularly for drugs predomi-nantly eliminated by renal filtration (e.g., gentamicin, digoxin).

Cl∗r

Clr

----------- Fi=

Fi

Fi

fu∗

fu

------- F⋅R

=

FR

Er( )

FRCl∗cr

Clcr

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

Cl∗cr

Cl∗r

Clr

-----------Cl∗cr

Clcr

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

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9.8 General Equations for Changes in Clearance

For each clearing organ,

(EQ 9-37)

When a drug has a high extraction ratio, , then equation 9-26 becomes

(EQ 9-38)

and when a drug has a low extraction ratio, , then equation 9-26 becomes

(EQ 9-39)

Thus, the clearance of drugs with a high extraction ratio are more effected by phys-iological changes in flow of blood to the clearing organ, while drugs with a lowextraction ratio are more effected by physiological changes in the function of theorgan.

9.8.1 PLASMA/BLOOD RATIO

Calculation of Extraction Ratio requires measurement in whole blood by defini-tion. Since most clinical measurements are done in plasma, knowledge of the plasma/blood ratio is necessary. Blood in made up of plasma and red blood cells (RBCs). Thus the amount of drug in the blood is made up of the amount of drug in the plasma and the amount of drug in the RBCs.

(EQ 9-40)

where

and b = blood, p = plasma, rbc = red blood cell

If we define the ratio of the concentration of the drug in the RBCs to the concentra-tion of the free drug in plasma as

(EQ 9-41)

and

(EQ 9-42)

FClClCl------

° FI FR⋅FR E+ FI FR–( )----------------------------------------= =

E 1≈

FCl FR≈

E 0≈

FCl FI≈

Cb Vb⋅ Cp Vp⋅ Crbc Vrbc⋅+=

Cx Vx⋅ AMOUNTx=

ρCrbc

fu Cp⋅---------------=

Vrbc H Vb⋅=

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Basic Pharmacokinetics REV. 99.4.25 9-15Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

(EQ 9-43)

Pluging equation 9-41 thru equation 9-43 into equation 9-40 results in:

(EQ 9-44)

Rearranging and simplifying results in:

(EQ 9-45)

(EQ 9-46)

Thus equation 9-46 determines the affinity of the drug for the RBCs. Drugs with ahigh affinity for the RBCs should result in a smaller volume of distribution.

For drugs that are primarily filtered by the glomeruli, the renal extraction ratio is:

(EQ 9-47)

Putting equation 9-45 into equation 9-47 results in:

(EQ 9-48)

Thus:

1. If the the ratio of to calculated Erf from equation 9-48 is one, it is likely that the drug is

filtered only.

13. If the the ratio of to calculated Erf from equation 9-48 is greater than one, active secretion

is infered in this case.

14. If the the ratio of to calculated Erf from equation 9-48 is less than one, active reabsorb-

tion is infered in this case.

Vp 1 H–( ) Vb⋅=

Cb Vb⋅ 1 H–( ) Vb Cp⋅ fu ρ H Vb Cp⋅ ⋅ ⋅ ⋅+⋅=

Cb

Cp

------ 1 H fu ρ 1–⋅( )⋅+=

ρH 1– Cb Cp⁄( )+

fu H⋅-----------------------------------------=

ErfRate of filtration

Rate of presentation------------------------------------------------

GFR fu Cp⋅ ⋅Qr Cb⋅

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

Erf

GFR fu⋅Qr 1 H fu ρ 1–⋅( )⋅+( )⋅-----------------------------------------------------------=

Clr

Qr

--------

Clr

Qr--------

Clr

Qr--------

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9.8.2 HALF LIFE AND ELIMINATION RATE CONSTANT IN RELATIONSHIP TO CLEARANCE

The elimination rate constant is related to the volume of distribution and the total body clearance by equation 9-22 above which when rewritten yields:

(EQ 9-49)

(EQ 9-50)

where b = blood and pwu = unbound plasma water. Clearance of drug from blood(Clb) is useful in considering drug extraction in the eliminating organs. Volumeand clearance terms based on unbound drug concentration are particularly usefulin therapeutics, because only the unbound drug is thought to cause the therapeuticaction.

9.8.3 EFFECTS OF ALTERATIONS IN PROTEIN BINDING ON CLEARANCE

Protein binding of drugs may be altered in disease states and by interferance bind-ing by other drugs on the protein. These changes in binding effect Fi , the frac-tional change in intrinsic clearance in both renal and hepatic clearances eventhough the actual intrinsic clearance, the indicator of organ function, may be unef-fected as shown in equation 9-11:

where . Thus, a change in protein binding will cause a proportionalchange in Fi of both clearances. There is more discussion in the chapter on proteinbinding.

K Rate of EliminationAmount in the body------------------------------------------------ Mass( ) Time( )⁄

Mass---------------------------------------- Cl

V------ Volume( ) Time( )⁄

Volume----------------------------------------------= = = =

K ClV------

Clb

Vb

--------Clpwu

Vpwu

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

Fi

fu∗ Cl∗int⋅fu Clint⋅

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

Cl∗int Clint=

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9.9 Problems

For each of the problems, do questions A through R now and S through W after completing chapter 10. In doing questions S through W, please try to obtain a

plasma concentration of free drug within 120 % of normal and 80 % of

.

Cpss

maxfree

Cpss

minfree

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Basic Pharmacokinetics REV. 99.4.25 9-18Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Acebutolol (Problem 9 - 1)

Piquette-Miller, M., et. al., “Effect of aging on the pharmacokinetics of acebutolol enantiomers”, Journal of Clinical Pharmacol-ogy, Vol. 32, (1992), p. 148 - 156. Kukes, VG; Gneushev ET; Mamedov TS; Gneusheva IA; “Acebutolol and diacetolol: thier bind-ing to plasma and erythrocytes and secretion with saliva.” Farmakol-Toksikol. 1991 Jan-Feb; 54(1) Acebutolol is a beta-adrenergic blocking agent which is often used in the treatment of hypertension.(Use Qr = 72 L/hr;

Qh = 90 L/hr)

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

PROBLEM TABLE 9 - 1. Acebutolol

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 200 400

B f 0.4

C fu 0.867

D 1.93

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 5.5

H %Clr 40

I %Clnr 60

J AUC (mg/L*hr) 3.97

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 12

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

maxfree

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 1. Answers for Acebutolol

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 200 400 BID 400 BID 400 BID 400 BID 200 QID

FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 1

FRH 1 1 1 1 0.5 1

FIH 1 1 1 1 1 0.5

B f 0.4

C fu 0.867

D 1.93

E Vd (L) 160

F k (hr-1) 0.126 0.101 0.102 0.12 0.091

G T 1/2 (hr) 5.5 6.87 6.77 5.89 7.64

H %Clr 40 25 26.1 42.9 55.5

I %Clnr 60 75 73.9 57.1 44.6

J AUC (mg/L*hr) 3.97 9.93 9.78 8.51 5.5

K (L/hr) 20.2 16.1 16.4 18.8 14.5

L (L/hr) 12.1 12.1 12.1 10.7 6.5

M (L/hr) 8.1 4.03 4.27 8.1 8.1

L 0.126

O 0.112

P (L/hr) 16

Q (L/hr) 10.5

R FCL 1 0.8 0.81 0.93 0.72

S (hr) 12 12 12 12 6

T N 2.18 1.75 1.77 2.04 0.785

U

1.11 1.24 1.23 1.15 1.03

V 0.57 0.72 0.71 0.61 0.80

W 0.25 0.37 0.36 0.28 0.6

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

maxfree

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-20Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Bisoprolol (Problem 9 - 2)

Kirch, W., et. al., “Pharmacokinetics of bisoprolol during repeated oral administration to healthy volunteers and patients with kid-ney or liver disease”, Clinical Pharmacokinetics, Vol. 13, (1987), p. 110 - 117.Bisoprolol (comes as 5 and 10 mg tablets) is a - selective adrenergic antagonist. It is used in the treatment of hyperten-sion and angina pectoris.(Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 2. Bisoprolol

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 10 10 TID

B f 0.7

C fu 1

D 1

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 10

H %Clr 50

I %Clnr 50

J AUC (mg/L*hr) 0.661

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

maxfree

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 2. Answers for Bisoprolol

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 10 10 10 10 10 10

FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 1

FRH 1 1 1 1 0.5 1

FIH 1 1 1 1 1 0.5

B f 0.9

C fu 0.7

D

E Vd (L) 152.8

F k (hr-1) 0.0693 0.052 0.0526 0.067 0.0525

G T 1/2 (hr) 10 13.3 13.2 10.3 13.2

H %Clr 50 33.3 34 51.3 66

I %Clnr 50 66.7 66 48.7 34

J AUC (mg/L*hr) 0.661 0.85 0.87 0.69 0.87

K (L/hr) 10.6 7.94 8.0 10.3 8.0

L (L/hr) 5.3 5.3 5.3 5.0 2.7

M (L/hr) 5.3 2.65 2.8 5.3 5.3

L 0.055

O 0.074

P (L/hr) 5.6

Q (L/hr) 5.7

R FCL 1 0.75 0.76 0.97 0.76

S (hr) 8 12 12 8 12

T N 0.8 0.9 0.9 0.78 0.91

U

0.108 0.099 0.098 0.11 0.098

V 0.083 0.073 0.073 0.085 0.073

W 0.062 0.052 0.052 0.064 0.052

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

maxfree

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Cefonicid (Problem 9 - 3)

Fillastre, J., et. al., “Pharmacokinetics of cefonicid in uraemic patients”, Journal of Antimicrobial Chemotherapy, Vol. 18, (1986), p. 203 - 211.Cefonicid is a beta-lactamase resistant cephalosporin which is useful in treating many infections caused by Gram-posi-tive and Gram-negative organisms. Cefonicid is 80% renally excreted.(Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 3. Cefonicid

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 1000 1000 TID

B f 1

C fu 0.06

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 5.3

H %Clr 80

I %Clnr 20

J AUC (mg/L*hr) 654

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-23Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

TABLE 9 - 3. Answers for Cefonicid

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 1000 1000 1000 1000 1000 1000

FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 1

FRH 1 1 1 1 0.5 1

FIH 1 1 1 1 1 0.5

B f 1

C fu 0.06

D

E Vd (L) 11.7

F k (hr-1) 0.131 0.078 0.079 0.131 0.118

G T 1/2 (hr) 5.3 8.8 8.8 5.3 5.9

H %Clr 80 66.7 66.9 80 88.9

I %Clnr 20 33.3 33.1 20 11.1

J AUC (mg/L*hr) 654 1090 1083 654 727

K (L/hr) 1.53 0.917 0.93 1.5 1.4

L (L/hr) 0.3 0.31 0.31 0.3 0.15

M (L/hr) 1.22 0.612 0.62 1.22 1.22

L 0.0032

O 0.017

P (L/hr) 5.11

Q (L/hr) 20.7

R FCL 1 0.6 0.6 1 0.9

S (hr) 8 12 12 8 8

T N 1.51 1.4 1.4 1.51 1.36

U

7.9 8.4 8.4 7.9 8.4

V 4.9 5.5 5.4 4.9 5.4

W 2.8 3.3 3.3 2.8 3.3

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Cefpirome (Problem 9 - 4)

Lameire, N., et. al., “Single-dose pharmacokinetics of cefpirome in patients with renal impairment”, Clinical Pharmacology and Therapeutics, Vol. 52, (1992), p. 24 - 30.Cefpirome is a third-generation, broad-spectrum cephalosporin which is useful against many cephalosporin-resistantorganisms.

PROBLEM TABLE 9 - 4. Cefpirome

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 2000 IV 2000 TID

B f 1

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 2.6

H %Clr 85

I %Clnr 15

J AUC (mg/L*hr) 342

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 4. Answers for Cefpirome

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Cefprozil (Problem 9 - 5)

Shyu, W., et. al., “Pharmacokinetics of cefprozil in healthy subjects and patients with renal impairment”, Journal of Clinical Phar-macology, Vol. 31, (1991), p. 362 - 371.Cefprozil is a broad-spectrum oral cephalosporin.

PROBLEM TABLE 9 - 5. Cefprozil

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 1000

B f 0.95

C fu 0.7

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 1.2

H %Clr 75

I %Clnr 25

J AUC (mg/L*hr) 58.1

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 5. Answers for Cefprozil

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Chloramphenicol (Problem 9 - 6)

Ambrose, P., “Clinical pharmacokinetics of chloramphenicol and chloramphenicol succinate”, Clinical Pharmacokinetics, Vol. 9, (1984), p. 222 - 238.Chloramphenicol succinate is a prodrug which is converted in vivo to the active form, chloramphenicol.

PROBLEM TABLE 9 - 6. Chloramphenicol

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 1000

B f 1

C fu 0.4

D

E Vd (L/kg) 2.8

F k (hr-1)

G T 1/2 (hr) 0.6

H %Clr 30

I %Clnr 70

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-29Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

TABLE 9 - 6. Answers for Chloramphenicol

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-30Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Enalapril (Problem 9 - 7)

Ohnishi, A., et. al., “Kinetics and dynamics of enalapril in patients with liver cirrhosis”, Clinical Pharmacology and Therapeutics, Vol. 45, (1989), p. 657 - 665.Enalapril is an ACE inhibitor which is a prodrug that is metabolized in the liver to the active metabolite, enalaprilat.

PROBLEM TABLE 9 - 7. Enalapril

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 10 mg po 10 BID1

B f 0.65

C fu 0.55

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 0.63

H %Clr 27

I %Clnr 73

J AUC (mg/L*hr) 0.123

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-31Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

TABLE 9 - 7. Answers for Enalapril

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Enoxacin (Problem 9 - 8)

Somogyi, A., and Bochner, F., “The absorption and disposition of enoxacin in healthy subjects”, Journal of Clinical Pharmacol-ogy, Vol. 28, (1988), p. 707 - 713.Enoxacin is a fluorinated quinolone which is used to treat infections caused by gram-negative organisms andPseudomonoas aeruginosa.

PROBLEM TABLE 9 - 8. Enoxacin

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 400 200 bid

B f 0.9

C fu 0.8

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 7.75

H %Clr 60

I %Clnr 40

J AUC (mg/L*hr) 15.61

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 12

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-33Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

TABLE 9 - 8. Answers for Enoxacin

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Enprofylline (Problem 9 - 9)

Nadai, M., et. al, “Dose-dependent pharmacokinetics of enprofylline and its renal handling in rats”, Journal of Pharmaceutical Sciences, Vol. 80, No. 7, (1991), p. 648 - 651 Enprogylline is a xanthine bronchodilator which is more potent than theophylline.

PROBLEM TABLE 9 - 9. Enprofylline

TABLE 9 - 9. Answers for Enprofylline

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg/kg) 2.5 2.5 TID

B f 1

C fu 0.4

D

E Vd (L/kg)

F k (hr-1)

G T 1/2 (hr) 0.36

H %Clr 90

I %Clnr 10

J AUC (mg/L*hr) 3.6

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Erythromycin (Problem 9 - 10)

Welling and Creig (JPS 67, 1057-9,1978).Erythromycin is a macrolide antibiotic.(Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 10. Erythromycin

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 300 300 TID

B f .8

C fu 0.06 0.06 0.12 0.06 0.18

D

E Vd (L) 57 57 100 57 150

F k (hr-1)

G T 1/2 (hr)

H %Clr 10

I %Clnr 90

J AUC (mg/L*hr)

K (L/hr) 16.5

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 8

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 10. Answers for Erythromycin

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 300 300 300 200 300 200

FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 3

FRH 1 1 1 1 0.5 1

FIH 1 1 1 2 1 0.5

B f .8

C fu 0.06 0.06 0.12 0.06 0.18

D

E Vd (L) 57 57 100 57 150

F k (hr-1) 0.255 0.242 0.235 0.227 0.075

G T 1/2 (hr) 2.71 2.85 2.92 3.05 9.27

H %Clr 10 5.3 3.1 11.2 37

I %Clnr 90 94.7 96.9 88.8 63

J AUC (mg/L*hr) 14.5 17.4 6.7 18.5 14.2

K (L/hr) 16.5 13.8 23.7 13.0 11

L (L/hr) 13.1 13.1 23.4 11.5 7.02

M (L/hr) 1.45 0.73 0.73 1.45 4.19

L 0.136

O 0.020

P (L/hr) 252

Q (L/hr) 24.7

R FCL 1 0.95 1.63 0.89 0.77

S (hr) 8 8 8 8 24

T N 2.94 2.79 2.74 2.63 2.59

U

0.29 0.30 0.23 0.30 0.23

V 0.12 0.13 0.10 0.14 0.11

W 0.038 0.042 0.034 0.049 0.038

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-38Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Fleroxacin (Problem 9 - 11)

Singlas, E., et. al., “Disposition of fleroxacind, a new trifluoroquinolone, and its metabolites - pharmacokinetics in renal failure and influence of haemodialysis”, Clinical Pharmacokinetics, Vol. 19, No. 1, (1990), p. 67 - 79.Fleroxacin is a trifluorinated quinolone with activity against a variety of gram-negative and gram-positive organisms.

PROBLEM TABLE 9 - 11. Fleroxacin

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 400 200 tid

B f 0.95

C fu 0.5

D 1.45

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 14

H %Clr 65

I %Clnr 35

J AUC (mg/L*hr) 92

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 11. Answers for Fleroxacin

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Fosinopril (Problem 9 - 12)

Hui, K., et. al., “Pharmacokinetics of fosinopril in patients with various degrees of renal function”, Journal of Clinical Pharmacol-ogy and Therapeutics, Vol. 49, No. 4, (1991), p. 457 - 466.Fosinopril is an angiotensin converting enzyme inhibitor which is a prodrug that is metabolized to active form, fosino-prilat.

PROBLEM TABLE 9 - 12. Fosinopril

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 7.5 IV 20 po qd

B f 1 (Oral 0.36)

C fu 0.01

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 7

H %Clr 0

I %Clnr 100

J AUC (mg/L*hr) 5.1

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 24

T N

U

V

W

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 12. Answers for Fosinopril

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Glutathione (Problem 9 - 13)

Mulders, T., et. al., “Characterization of glutathione conjugation in humans: stereoselectivity in plasma elimination pharmacoki-netics and urinary excretion of (R)- and (S)-2-bromoisovalerylurea in healthy volunteers”, Clinical Pharmacology and Therapeu-tics, Vol. 53, (1993), p. 49 - 58.This study explored the pharmacokinetics of glutathione.

PROBLEM TABLE 9 - 13. Glutathione

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 600 500 tid

B f 1

C fu 1

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 4.4

H %Clr 35

I %Clnr 65

J AUC (mg/L*hr) 276

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 13. Answers for Glutathione

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Guanadrel (Problem 9 - 14)

Halstenson, C., et. al., “Disposition of guanadrel in subjects with normal and impaired renal function”, Journal of Clinical Phar-macology, Vol. 29, (1989), p. 128 - 132.Guanadrel is adrenergic blocker used in the treatment of hypertension.

PROBLEM TABLE 9 - 14. Guanadrel

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 25 mg po 25 BID

B f 1

C fu 0.8

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 3.7

H %Clr 40

I %Clnr 60

J AUC (mg/L*hr) 0.234

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 12

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 14. Answers for Guanadrel

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Monoxidine (Problem 9 - 15)

Kirch, W., Hutt, H., and Plänitz, V., “The influence of renal function on clinical pharmacokinetics of monoxidine”, Clinical Phar-macokinetics, Vol. 15, (1988), p. 245 - 253.Monoxidine is a centrally acting antihypertensive agent which stimulates -adrenergic receptors.

PROBLEM TABLE 9 - 15. Monoxidine

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

α2

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 0.25 IV 0.25 tid

B f 1

C fu 1

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 2.75

H %Clr 95

I %Clnr 5

J AUC (mg/L*hr) 0.05

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 15. Answers for Monoxidine

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Nalmefene (Problem 9 - 16)

Dixon, R., et. al., “Nalmefence: safety and kinetics after single and multiple oral doses of a new opiod antagonist”, Journal of Clinical Pharmacology, Vol. 27, (1987), p. 233 - 239.Nalmefene is a pure opiod antagonist which is currently being investigated for use.

PROBLEM TABLE 9 - 16. Nalmefene

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 20 IV

B f 0.6 oral

C fu 1

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 9.8

H %Clr 70

I %Clnr 30

J AUC (mg/L*hr) 0.3

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 16. Answers for Nalmefene

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Nitrendipine (Problem 9 - 17)

Dylewicz, P, et. al, “Bioavailability and elimination of nitrendipine in liver disease”, European Journal of Clinical Pharmacology, Vol, 32, (1987), p. 563 - 568.Nitrendipine is a calcium antagonist

PROBLEM TABLE 9 - 17. Nitrendipine

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 5 IV 25 po BID

B f 1 (oral 0.2)

C fu 0.05

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 11.7

H %Clr 0.1

I %Clnr 99.9

J AUC (mg/L*hr)

K (L/hr) 90

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 17. Answers for Nitrendipine

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Ofloxacin (Problem 9 - 18)

Lamerire, N., et. al., “Ofloxacin pharmacokinetics in chronic renal failure and dialysis”, Clinical Pharmacokinetics, Vol. 21, No. 4, (1995), p. 357 - 371.

PROBLEM TABLE 9 - 18. Ofloxacin

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 300 300

B f 0.93

C fu 0.74

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 6

H %Clr 97

I %Clnr 3

J AUC (mg/L*hr) 28.47

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 12

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 18. Answers for Ofloxacin

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Omeprazole (Problem 9 - 19)

(Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 19. Omeprazole

TABLE 9 - 19. Answers for Omeprazole

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 10 10 BID

B f 0.5 0.75

C fu 0.04 0.06 0.08

D

E Vd (L) 30

F k (hr-1)

G T 1/2 (hr) 2

H %Clr 0

I %Clnr 100

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 12

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 10 10 10 5 10 5

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FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 2

FRH 1 1 1 1 0.5 1

FIH 1 1 1 1.5 1 0.5

B f 0.5 0.75

C fu 0.04 0.06 0.08

D

E Vd (L) 30

F k (hr-1) 0.347 0.347 0.493 0.31 0.184

G T 1/2 (hr) 2 2 1.41 2.2 3.8

H %Clr 0

I %Clnr 100

J AUC (mg/L*hr) 0.48 0.48 0.338 0.53 0.68

K (L/hr) 10.4 10.4 14.8 9.38 5.5

L (L/hr) 10.4 10.4 14.8 9.38 5.5

M (L/hr) 0 0 0 0 0

L 0.108

O 0

P (L/hr) 291.5

Q (L/hr) 0

R FCL 1 1 1.42 0.90 0.53

S (hr) 6 6 8 6 12

T N 3 3 2.83 2.71 3.2

U

0.0076 0.0076 0.0051 0.0079 0.011

V 0.0032 0.0032 0.0025 0.0035 0.0045

W 0.00095 0.00095 0.00099 0.0012 0.0012

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Piperacillin (Problem 9 - 20)

Johnson, C., et. al., “Single-dose pharmacokinetics of piperacillin and tazobactam in patients with renal disease”, Clinical Phar-macology and Therapeutics, Vol. 51, (1992), p. 32 - 41.Piperacillin is a beta-lactam antibiotic.

PROBLEM TABLE 9 - 20. Piperacillin

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 3000 2000 qid

B f 1

C fu 0.82

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 0.95

H %Clr 75

I %Clnr 25

J AUC (mg/L*hr) 276

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 6

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 20. Answers for Piperacillin

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Piroxicam (Problem 9 - 21)

Boudinot, S., Funderburg, E., and Boudinot, F., “Effects of age on the pharmacokinetics of piroxicam in rats”, Journal of Pharma-ceutical Sciences, Vol. 82, No. 3, (1993), p. 254 - 257.Piroxicam is a nonsteroidal anti-inflammatory drug (NSAID) commonly used in the treatment of arthritis.

PROBLEM TABLE 9 - 21. Piroxicam

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 70 20 qd

B f 1

C fu 0.007

D

E Vd (L) 9

F k (hr-1)

G T 1/2 (hr) 50

H %Clr 5

I %Clnr 95

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 21. Answers for Piroxicam

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-60Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Quinidine (Problem 9 - 22)

Quinidine sulfate is used to treat ventricular and supraventricular arrythmias and is available in 200 and 300 mg tablets.It is known to bind to -acid glycoprotein (AAG), which is an acute phase reactant. AAG rises in trauma, inflamation,malignancy and stress and falls in hepatic disease, nephrotic syndrome and malneurtrition for example.(Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 22. Quinidine

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

α

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg/kg) 10 10 TID

B f 0.7

C fu 0.2 0.10 0.25 0.15 0.3

D s 0.83

E Vd (L/kg) 2.6 2.0 2.75 2.2 3.0

F k (hr-1)

G T 1/2 (hr) 6.4

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr/kg)

L (L/hr/kg) 0.20

M (L/hr/kg) 0.056

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 8

T N

U

MTC

1.7

V

W MEC

0.37

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 22. Answers for Quinidine

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 10 10 10 10 10 5

FRR 1 1 0.5 1 1 1

FIR 1 1 0.25 0.5 0.75 1.5

FRH 1 1 1 1 0.5 1

FIH 1 1 0.5 1.25 0.75 0.5

B f 0.7

C fu 0.20 0.10 0.25 0.15 0.30

D s 0.83

E Vd (L) 2.6 2.0 2.75 2.2 3.0

F k (hr-1) 0.098 0.057 0.101 0.087 0.061

G T 1/2 (hr) 7 12.1 6.9 7.9 11.3

H %Clr 22 12.3 10 22 46

I %Clnr 78 87.7 90 78 54

J AUC (mg/L*hr) 27.3 61.3 20.6 36.5 19.0

K (L/hr) 0.256 0.114 0.28 0.19 0.18

L (L/hr) 0.20 0.10 0.25 0.15 0.1

M (L/hr) 0.056 0.014 0.028 0.04 0.08

L 0.0021

O 0.00078

P (L/hr) 1.0

Q (L/hr) 0.28

R FCL 1 0.45 1.1 0.75 0.72

S (hr) 8 8 8 8 8

T N 1.3 0.66 1.16 1.0 0.71

U

MTC

1.7

0.82 0.79 0.95 0.79 0.75

V 0.57 0.63 0.65 0.56 0.59

W MEC

0.3

0.37 0.5 0.44 0.39 0.46

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Tazobactam (Problem 9 - 23)

Johnson, C., et. al., “Single-dose pharmacokinetics of piperacillin and tazobactam in patients with renal disease”, Clinical Phar-macology and Therapeutics, Vol. 51, (1992), p. 32 - 41.Tazobactam is an irreversible beta-lactamase inhibitor.

PROBLEM TABLE 9 - 23. Tazobactam

TABLE 9 - 23. Answers for Tazobactam

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 375 IV 375 qid

B f 1

C fu 0.96

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 0.89

H %Clr 68

I %Clnr 32

J AUC (mg/L*hr) 30.3

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr) 6

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-64Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

Theophylline (Problem 9 - 24)

Wagner, J., “Theophylline - pooled Michaelis-Menten parameters and implications”, Clinical Pharmacokinetics, Vol. 10, (1985), p. 432 - 442.

PROBLEM TABLE 9 - 24. Theophylline

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-65Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

TABLE 9 - 24. Answers for Theophylline

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Tolrestat (Problem 9 - 25)

Troy, S., et. al., “The effect of renal disease on tolrestat pharmacokinetics”, Clinical Pharmacology and Therapeutics, Vol. 51, (1992), p. 271 - 277.Tolrestat is an aldose reductase inhibitor used in the treatment of diabetic neuropathy, diabetic nephropathy, and dia-betic retinopathy. (Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 25. Tolrestat

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 200 200 tid

B f 0.8

C fu

D

E Vd (L)

F k (hr-1)

G T 1/2 (hr) 10.6

H %Clr 25

I %Clnr40 bile 35 metab.

J AUC (mg/L*hr) 86

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Basic Pharmacokinetics REV. 99.4.25 9-67Copyright © 1996-1999 Michael C. Makoid All Rights Reserved http://kiwi.creighton.edu/pkinbook/

TABLE 9 - 25. Answers for Tolrestat

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 200 200 TID 200 TID 200 TID 200 TID 200 BID

FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 1

FRH 1 1 1 1 0.5 1

FIH 1 1 1 1 1 0.5

B f 0.8

C fu 1

E Vd (L) 28.5

F k (hr-1) 0.0654 0.057 0.057 0.065 0.041

G T 1/2 (hr) 10.6 12.1 12.1 10.7 16.9

H %Clr 25 14.3 14.3 25.3 39.8

I %Clnr 75 85.7 85.7 74.7 60.2

J AUC (mg/L*hr) 86 98.3 98.2 87 137

K (L/hr) 1.86 1.63 1.63 1.84 1.17

L (L/hr) 1.4 1.4 1.4 1.4 0.7

M (L/hr) 0.46 0.23 0.47 0.47 0.47

L 0.0155

O 0.0065

P (L/hr) 1.42

Q (L/hr) 0.47

R FCL 1 0.88 0.88 0.99 0.63

S (hr) 8 8 8 8 12

T N 0.75 0.66 0.66 0.75 0.71

U

13.8 15.3 15.3 13.9 14.4

V 10.8 12.3 12.3 10.9 11.4

W 8.2 9.6 9.7 8.3 8.8

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Vancomycin (Problem 9 - 26)

Macais, W., Meuller, B., and Scarim, S., “Vancomycin pharmacokinetics in acute renal failure; preservation of nonrenal clear-ance”, Clinical Pharmacology and Therapeutics, Vol., 50, (1991), p. 688 - 694.Vancomycin is a glycopeptide antibiotic used in the treatment of infections caused by Gram-positive organisms.

PROBLEM TABLE 9 - 26. Vancomycin

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 1000 500 QID

B f 1

C fu

D 0.44

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr 70

I %Clnr 30

J AUC (mg/L*hr) 543

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

ρ

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 26. Answers for Vancomycin

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg)

B f

C fu

E Vd (L)

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr)

M (L/hr)

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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Xipamide (Problem 9 - 27)

Knauf, H, et. al., “Xipamide disposition in liver cirrhosis”, Clinical Pharmacology and Therapeutics, Vol. 48, No. 6, (1990), p. 328 - 632.Xipamide is a diuretic that has been used in the treatment of congestive heart failure, hypertension, advanced renal fail-ure, and hepatic edema.(Use Qr = 72 L/hr; Qh = 90 L/hr)

PROBLEM TABLE 9 - 27. Xipamide

Problem Submitted By: Maya Leicht AHFS 00:00.00Problem Reviewed By: Vicki Long GPI: 0000000000

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 40 40 QID

B f 0.8

C fu 0.01 0.01 0.02 0.01 0.025

E Vd (L) 21

F k (hr-1)

G T 1/2 (hr)

H %Clr

I %Clnr

J AUC (mg/L*hr)

K (L/hr)

L (L/hr) 1.38

M (L/hr) 0.72

L

O

P (L/hr)

Q (L/hr)

R FCL 1

S (hr)

T N

U

V

W

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------

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TABLE 9 - 27. Answers for Xipamide

Patient Condition Normal Normal FRR=0.5 FIR=0.5 FRH=0.5 FIH=0.5

A Dose(mg) 40 40 QID 40 QID 20 QID 40 QID 20 QID

FRR 1 1 0.5 1 1 1

FIR 1 1 0.5 0.5 1 2.5

FRH 1 1 1 1 0.5 1

FIH 1 1 1 2 1 0.5

B f 0.8

C fu 0.01 0.01 0.02 0.01 0.025

E Vd (L) 21

F k (hr-1) 0.1 0.083 0.146 0.1 0.118

G T 1/2 (hr) 6.9 8.4 4.7 7.0 5.9

H %Clr 34 21 12 35 72

I %Clnr 66 79 88 65 28

J AUC (mg/L*hr) 15.2 18.4 7.8 15.4 6.5

K (L/hr) 2.1 1.74 3.1 2.1 2.5

L (L/hr) 1.38 1.38 2.71 1.36 0.7

M (L/hr) 0.72 0.36 0.36 0.72 1.8

L 0.0144

O 0.01

P (L/hr) 140

Q (L/hr) 73

R FCL 1 0.83 1.46 1 1.18

S (hr) 6 6 4 6 6

T N 0.87 0.71 0.847 0.86 1

U

0.034 0.039 0.034 0.034 0.038

V 0.025 0.031 0.026 0.025 0.027

W 0.019 0.024 0.019 0.019 0.019

Cltot

Clh

Clr

Eh

Er

Clhint

Clrint

τ

Cpss

max free

µgmL--------

Cpss

avgfree

µgmL--------

Cpss

minfree

µgmL--------