AENaanettealyareaCH
`
`VOLUME 40
`
`|
`
`ai
`EUBs
`
`PGRfor U.S. Patent No. 9,408,862
`
`
`
`SMaLGU
`SUS
`ML)a
`CHRISTOPHER T. RHODES
`
`Grun. Exh, 1063
`
`Grun. Exh. 1063
`PGR for U.S. Patent No. 9,408,862
`
`

`

`Modern
`Pharmaceutics
`
`Second Edition, Revised and Expanded
`
`edited by
`GILBERT S. BANKER
`University of Minnesota
`Minneapolis. Minnesota
`
`CHRISTOPHER T. RHODES
`l,Jniversity of Rhode Island
`Kingston. Rhode Island
`
`MARCEL DEKKER, INC.
`
`New York and Basel
`
`

`

`Library of Congress Cataloging-in-Publication Data
`
`Modern pharmaceutics .
`
`(Drugs and the pharmaceutical sciences ; v. 40)
`Includes bibliographical references.
`1. Drugs - -Dosage forms .
`2. Biopharmaceutics .
`3. Pharmacokinetics .
`4. Pharmaceutical industry - -Quality
`control.
`I. Banker , Gilbert S.
`II. Rho des,
`Christopher T .
`III . Series .
`RS200 . M63
`1989
`615' . 1
`ISBN 0-8247-7499-X
`
`89-23365
`
`COPYRIGHT @ 1990 by MARCEL DEKKER, INC . ALL RIGHTS RESERVED
`
`Neither this book nor any part may be reproduced or transmitted in any
`form or by any means, electronic or mechanical, including photocopying,
`microfilming, and recording , . or by any information storage and retrieval
`system, without permission in writing from the publisher .
`
`MARCEL DEKKER, INC .
`270 Madison Avenue, New York, New York 10016
`
`PRINTED IN THE UNITED STATES OF AMERICA
`
`

`

`3
`Pharmacokinetics
`
`DAVID W. A. BOURNE
`University of Oklahoma, Oklahoma City , Oklahoma
`
`LEWIS W. DITTERT
`University of Kentucky, Lexington, Kentucky
`
`I .
`
`INTRODU CT ION
`
`Drug therapy is a dynamic process. When a drug product is administered,
`absorption usually proceeds over a finite time interval; and distribution,
`metabolism, and excretion of the drug and its metabolites proceed continu(cid:173)
`ously at various rates. The relative rates of these "ADME processes" deter(cid:173)
`mine the time course of the drug in the body, most important at the receptor
`sites, which are responsible for the pharmacologic action of the drug .
`The usual aim of drug therapy is to achieve and maintain effective con(cid:173)
`centrations of drug at the receptor site·. However, the body is constantly
`trying to eliminate the drug and it is therefore necessary to balance absorp(cid:173)
`tion against elimination so as to maintain the desired concentration . Often
`the receptor sites are tucked away in a specific organ or tissue of the body,
`such as the central nervous system, and it is necessary to depend on the
`blood supply to distribute the drug from the site of administration, such as
`the gastrointestinal tract, to the site of action.
`Since the body may be viewed as a very complex system of compartments,
`it might at first appear to be hopeless to try to describe the time course of
`the drug at the receptor sites in any mathematically rigorous way. The
`picture is further complicated by the fact that for many drugs, the locations
`of the receptor sites are unknown . Fortunately, ,body compavtmentfFrare
`c ~mnected by the blood s-ys-tem,;.,.,-and--distributian,,oft..,d:rmgs among the com(cid:173)
`partments -usually occurs ·much ·more -rapidly 1::han-,-absorption · or· elimination
`of the drug. The rret result -is. t hat- the body behaves as .a single -homo~
`geneous compartment with ·respect to many drugs, and the concentration of
`the drug in the blood directly reflects or is p r oportional to- the concentra(cid:173)
`tion of the drug in all organs and tissues : Thus it may never be possible
`to isolate a receptor site and determine the concentration of drug around it,
`but the concentration at the receptor site usually can be controlled if the
`blood concentration can be controlled .
`The objective of pharmacokinetics is to describe the time course of drug
`concentrations in blood in mathematical terms so that (1) the performance of
`
`91
`
`

`

`92
`
`Bourne and Dittert
`
`pharmaceutical dosage forms can be evaluated in terms of the rate and amount
`.of drug they deliver to the blood, and ( 2) the dosage regimen of a drug
`can be adjusted to produce and maintain therapeutically effective blood con(cid:173)
`centrations with little or no toxicity . , The primary objective of this chapter
`is to describe the mathematical tools needed to accomplish these aims when
`the body behaves as a single homogeneous compartment and when all phar(cid:173)
`macokinetic processes obey first - order kinetics .
`On some occasions, the body does not behave as a sirigle homogeneous
`compartment and multicompartment pharmacokinetics are required to describe
`the time course of drug concentrations.
`In other instances certain pharmaco(cid:173)
`kinetic processes may not obey first - order kinetics, and saturable or non(cid:173)
`linear models may be . required . Readers interested in such advanced topics
`are referred to a number of texts which describe these more complex phar(cid:173)
`macokinetic models in detail [ 1- 5] .
`
`II . PRINCIPLES OF FIRST - ORDER KINETICS
`A. Definition and Characteristics of First- Order Processes
`The science of kinetics deals with the mathematical description of the rate
`•of processes, for example, the appearance or disappearance of a substance.
`One of the most common types of rate processes observed in nat ure is the
`fir st- order process, in which the r ate is dependent on the concentration ·or
`amount of only one component. An example of such a process is radioactive
`decay in which the rate of deacy (i.e . , the number of radioactive decomposi(cid:173)
`tions per minute) is directly proportional to the amount of undecayed sub (cid:173)
`stance remaining . This may be written mathematically as follows :
`
`rate of radioactive decay oc [undecayed substance]
`
`or
`
`rate of radioactive decay = k(undecayed substance)
`
`(1)
`
`( 2)
`
`where k is a proportionality constant called the first-order rate constant.
`Chemical reactions usually occur through collision of at least two mole(cid:173)
`cules, very often in a solution, and the rate of the chemical reaction is
`proportional to the concentrations of all reacting molecules . For example,
`the rate of hydrolysis of an ester in an alkaline buffered solution depends
`on the concentration of both the ester and hydroxide ion:
`
`Ester + OH
`
`---? Acid- + Alcohol
`
`The rate of hydrolysis may be, expressed as follows :
`
`rate of hydrolysis ex: [Ester] · [OH- ]
`
`or
`
`rate of hydrolysis = k[Ester] · [OH - ]
`
`(3)
`
`(4)
`
`(5)
`
`where k is the proportionality constant called the second- order rate constant.
`But in a buffered system, [OH- ] is constant. Therefore, at a given
`pH, the rate of hydrolysis is dependent only on the concentration of the
`ester and may be written
`
`rate of hydrolysis(pH) = k* [Ester]
`
`(6)
`
`

`

`Pharmacokinetics
`
`93
`
`where k * is the pseudo - first - order rate constant at the pH in question .
`(The pseudo - first - order rate constant, k*, is the product of the second(cid:173)
`order rate constant and the hydroxide ion concentration : k* = k[OH"· ] . )
`Fortunately, most ADME processes behave as pseudo-first-order pro(cid:173)
`cesses - not because they are so simple, but because everything except the
`drug concentration is constant. For example, the elimination of a drug
`from the body may be written as follows :
`
`+
`
`enzymes;
`membranes;
`pH; protein
`binding;
`etc .
`
`metabolized or
`~ excreted drug
`
`(7)
`
`If everything except the concentration, of drug in the body is constant ,
`the elimination of the drug will be a pseudo - first - order process . This may
`seem to be a drastic oversimplification, but most in vivo drug processes, '
`in fact, behave as pseudo- first - order processes.
`
`B. Differential Rate Expressions
`In the previous discussion of radioactive decay it was noted that the rate
`of decay is directly proportional to the amount of undecayed substance
`remammg .
`In a solution of a radioactive substance, a similar relationship
`would hold for the concentration of undecayed substance remaining.
`If a
`solution of a radioactive substance were allowed to decay and a plot were
`constructed of the concentration remaining versus time, the plot would be
`a curve such as that shown in Fig . 1.
`In this system, the rate of decay
`might be expressed as a change in concentration per unit time , b.C I b.t,
`which corresponds to the slope of the line . But the line in Fig . 1 is
`curved, which means that the rate is constantly changing and therefore
`cannot be expressed in terms of a finite time interval .
`By resorting to differential calculus, it is possible to express the rate
`of decay in terms of an infinitesimally small change in concentration ( dC)
`over an infinitesimally small time interval ( dt) . The resulting function,
`dC/dt, is the slope of the line, and it is this function that is proportinal
`to concentration in a first - order process . Thus
`
`dC
`rate = dt = - kc
`
`(8)
`
`(The negative sign is introduced because the concentration is falling as
`time progresses . )
`Equation (8) is the differential rate expression for a first - order reac(cid:173)
`tion . The value of the rate constant, k could be calculated by determining
`the slope of the concentration ver sus time curve at any point and dividing
`by the concentration at that point. However, the slope of a curved line is
`difficult to measure accurately, and k can be determined much more easily
`using integrated rate expressions .
`
`Integrated Rate Expres sions and Working Equations
`C .
`Equation ( 8) can be rearranged and integrated as follows:
`
`dC
`C = - k dt
`
`(9)
`
`

`

`94
`
`Bourne and Dittert
`
`0
`
`0 .
`
`0
`O _
`
`a
`
`0 .
`
`0
`,(X)
`
`z o
`D~
`..,_. o _ I
`t-(0
`er:
`a:
`t(cid:173)z o wo
`u~
`Z:::I'
`D u
`
`:
`
`0
`0
`
`0
`ru
`
`g-1---~---..-- -......-- ----,r----_:_==:::::======:;:======;
`9J . 00
`12 . 00
`TIME
`
`15 . 00
`
`18 . 00
`
`21.00
`
`24 . 00
`
`3 . 00
`
`5 . 00
`
`9.00
`
`Figure 1 Plot of concentration remaining versus time for a first - order pro(cid:173)
`cess (radioactive decay) .
`
`f ~c = - k f dt
`
`lnC = - kt + constant
`
`(10)
`
`where lnC is the natural logarithm (base e) of the concentration. The
`constant in Eq. (10) can be evaluated at zero time when kt = 0 and C =
`Co, the initial concentration. Thus
`
`lnC
`
`= constant
`
`0
`and since ln x = 2 . 3 log x, Eq . (10) can be converted to common logarithms
`(base 10) as follows:.
`2. 3 log C = - kt + 2. 3 log c 0
`- kt
`log C = ~ + log c 0
`
`(11)
`
`Equation ( 11) is the integrated rate expression for a first - order process
`and can serve as a working equation for solving problems.
`It is also in
`the form of the equation of a straight line :
`
`y = mx + b
`
`

`

`Pharmacokinetics
`
`95
`
`Therefore, if log C is plotted against t, as shown in Fig . 2, the plot will
`be a straight line with a n tntercept (at t = 0) of log Co , and the slope of
`the line (m) will be - k/2.3 . Such plots are commonly used to determine
`the order of a reaction; that is, if a plot of log C versus time is a straight
`line, the reaction is assumed to be_ a first - order or pseudo- first - order
`process.
`The slope of the line and the corresponding value of k for a. plot such
`as that shown in Fig. 2 may be calculated using the following equation:
`.L
`I
`
`log c2
`log c1 -
`s~pe (m) =------(cid:173)
`tl -
`t2
`
`=
`
`k
`-2.3
`
`(12)
`
`EXAMPLE . A solution of ethyl acetate in pH 10 . 0 buffer (25°C) 1 hr
`after preparation was found to contain 3 mg/ml. Two hours after prepara(cid:173)
`tion, the solution contained 2 mg/ml. Calculate the pseudo- first - or der rate
`constant for hydrolysis of ethyl acetate at pH 10 . 0 (25°C) .
`
`= -
`
`k
`2.3
`
`slope (m)
`
`=
`
`=
`
`log 3 -
`log 2
`(1 - 2) hr
`0 . 477 - 0 . 301
`(1 - 2) hr
`- 1
`= - 0 . 176 hr
`
`k
`= - 2.3
`
`1 NTERCEPT = LOG (Cp 0 l
`
`SLDPE = -K /2.3
`
`0
`l1)
`
`N
`
`0
`0
`
`N
`
`z
`0
`;::~
`er .
`cx:::-
`1-(cid:173)z
`w
`u z g
`0 ,
`u-
`
`L'.)
`0
`....J~
`
`0
`
`0
`0
`
`9:J. 00
`
`3.00
`
`6 . 00
`
`9.00
`
`12. 00
`TIME
`
`15.00
`
`18 . 00
`
`21. 00
`
`24.00
`
`Figure 2 Plot of log ( concentration r emaining) vers us time for a first (cid:173)
`or der process.
`
`

`

`96
`
`Bourne and Dittert
`
`k = 0.176 X 2.3 hr- 1
`
`= 0.405 hr-l
`
`Note that since log C is dimensionless, the rate constant k has the dimen(cid:173)
`sions of reciprocal time (i.e., day-1, hr-1, min-1, sec-1, etc . ) .
`Another useful working equation can be obtained by rearranging Eq.
`( 11) as follows :
`
`! .!
`
`log c 0
`
`=
`
`kt
`2.3
`kt
`log C = -2 . 3
`kt
`co
`log-c = 2.3
`
`log C -
`
`(13)
`
`Equation (13) shows that since k/2.3 is a constant for a given process,
`the ratio Co/ C is determined solely by the value of t. For example, Co/ C
`will be equal to 2 after the same length of time, no matter what the value
`of the initial concentration (Co).
`EXAMPLE. For the ethyl acetate hydrolysis above (k = 0 . 405 hr- 1),
`if Co = 3 mg/ml, when would C = 1.5 mg/ml?
`
`. co
`kt
`log c = 2.3
`0 . 405 X t
`3
`=
`log 1. 5
`2 . 3
`0.405 X t
`2.3
`0,405 X t
`2.3
`
`log 2 =
`
`0 .3 01 =
`
`t = 1. 71 hr
`If Co = 1 . 5 mg/ml, when would C = 0. 75 mg/ml?
`
`kt
`co
`log -C = -2.3
`1.5
`0 .405 X t
`log 0. 75 =
`2. 3
`_ Q , 405 X t
`1
`og 2 -
`2. 3
`
`t = 1. 71 hr
`
`The time required for the concentration to fall to Co/2 is called the
`'
`half- life, and the foregoing example shows that the half- life for a first -
`order or pseudo-first- order process is a constant throughout the process;
`it also demonstrates that a first-order process theoretically never reaches
`completion, since even the lowest concentration would only fall to half its
`value in one half- life.
`
`

`

`Pharma cokinetics
`
`97
`
`For most practical purposes, a first - order process may be deemed
`"complete" if it is 95% or more complete . Table 1 shows that five half- lives
`must elapse to reach this point . Thus the elimination of a drug from the
`body may be considered to be complete after five half- lives have elapsed
`(i.e . , 97% completion), This principle becomes important, for example, in
`crossover bioavailability studies in which the subjects must be rested for
`sufficient time between each dr ug administration to ensure that "washout"
`is complete .
`The half- life of a first - order process is very important. Since it is
`often desirable to convert a half- life to a rate constant, and vice versa, a
`simple relationship between the two is very useful. The relationship may
`be derived as follows:
`
`co
`log -C
`
`kt
`= 2 . 3
`when Co/C = 2 and t = tl/2 · T hus
`
`(14)
`
`(15)
`
`log 2
`
`0 . 30
`
`ktl/2 = 0 . 69
`
`k = 0 . 69
`tl/2
`0 . 69
`=----iz--
`
`Table 1 Approach to Completeness with Increasing
`Half- Lives
`
`Number of
`half- lives
`elapsed
`
`Initial
`concentration
`remaining (%)
`
`11 Completeness 11
`of process (%)
`
`0
`1
`2
`3
`4
`5
`6
`7
`
`100 . 0
`50.0
`25 . 0
`12 . 5
`6 . 25
`3. 13
`1.56
`0 . 78
`
`0 . 0
`50.0
`75.0
`87.5
`93 . 75
`96 .8 7
`98 .44
`99.22
`
`

`

`98
`
`Bourne and Di ttert
`
`D . Examp les of Calculat ions
`Equations (13), (14), and (15) can be used to solve three types of proble ms
`involving first - order processes . These types of problems are illustrated
`in the following examples:
`1. Given the rate constant or half-life and the initial concentration,
`calculate the concentration at some time in the future .
`EXAMPLE . A p enicillin solution containing 500 units/ml has a half- life
`of 10 days . What will the concentration be in 7 days?
`
`k
`
`=
`
`=
`
`0.69
`10 day
`
`-1
`0 . 069 da y
`
`=
`
`0 . 69
`tl /2
`co
`kt
`log c = 2 . 3
`- 1
`0.069 day
`500 units/ml
`2 . 3
`C
`5oo u~itS/ml = antilog (0 . 21) = 1.62
`
`log
`
`=
`
`x 7 days
`
`= 0 . 21
`
`C = 308 . 3 units/ml
`
`2 . Given the half- life or rate constant and the initial concentration,
`calculate the time required to r each a specified lower concentration .
`EXAMPLE . A penicillin solution has a half- life of 21 days . How long
`will it take for the potency to drop to 90% of the initial potency?
`
`- 1
`0 . 69
`k = 21 days = 0 . 033 day
`co
`kt
`log - C = 2 . 3
`100%
`0,033 X
`log 90% =
`2 . 3
`
`t
`
`t = 3 . 2 days
`
`3 . Given an initial concentration a nd the concentration after a speci(cid:173)
`fied elapsed time , calculate the. rate constant or half- life .
`EXAMPLE. A penicillin solution has an initial potency of 125 mg per
`5 ml. After 1 month in a refrigera tor, the potency is found to be 100 mg/
`5 ml. Wha t is the half- life of the penicillin solution under these storage
`conditions?
`
`co
`kt
`log -C = 2 . 3
`125 mg/5 ml
`1
`og 100 mg/5 ml
`k = 0 . 0074 day - 1
`0 . 69
`= - - - - - = 93 days
`0 . 0074 day- l
`
`k X 30 day
`=----~
`2.3
`
`t
`112
`
`

`

`Pharmacokinetics
`
`99
`
`For each type of problem the following assumptions are made: (1) The
`process follows first-order kinetics, at least over the time interval and con(cid:173)
`centration range involved in the calculations; and (2) all time and concen(cid:173)
`tration values are accurate. The latter assumption is particularly critical
`in solving problems such as type 3 where a rate constant is being calculated.
`It would be unwise to rely on only two assay results at two time points to
`calculate such an important value .
`Normally, duplicate or triplicate assays would be performed at six or
`more time points throughout as much of the reaction as possible. The
`resulting mean assay values and standard deviation values would be plotted
`on semilogarithmic graph paper and a straight line carefully fitted to the
`data points . The half-life could then be determined using Eq . (14) .
`EXAMPLE. A solution of ethyl acetate in pH 9.5 buffer (25°C) was
`assayed in triplicate several times over a 20 - hr period. The data obtained
`are presented in Table 2. The results were plotted on semilogarithmic
`graph paper as shown in Fig. 3. Calculate the pseudo- first - order rate
`constant for the hydrolysis of ethyl acetate at pH 9.5 (25°C) .
`By fitting a straight line through the data points in Fig . 3 (this can
`be done by eye using a transparent straight edge) and extrapolating to
`t = O, the intercept Co is found to be 3.13 mg/ml. The half- life is the time
`at which the concentration equals 1. 57 mg/ml, and this is found by inter(cid:173)
`polation to be 2 . 4 hr. The value of k is then given by
`
`k =
`
`0 . 69
`tl/2
`
`-1
`0 . 69
`= 2 . 4 hr = 0 . 288 hr
`
`Semilogarithmic graph paper is available from all graph paper manufac(cid:173)
`turers .
`It consists of a logarithmic scale on the y axis and a cartesian
`scale on the x axis ( see Fig . 3) . On the log scale, the spatial distribution
`of lines is such that the position of each line is proportional to the log of
`the value represented by the mark . For example, plotting a concentration
`of 1. 83 mg/ ml on semilog paper is equivalent to looking up the log of 1. 83
`and plotting it on a cartesian scale. This type of graph paper is extremely
`useful for kinetic calculations because raw concentration data can be plotted
`
`Table 2 Assay of Ethyl Acetate
`
`Time
`(hr)
`
`2
`4
`6
`8
`10
`12
`16
`20
`
`Concentration
`(mg/ml) ± SD
`
`1.83 ± 0.15
`1.01 ± 0 . 09
`0 . 58 ± 0 . 07
`0 .33 ± 0.06
`0 . 18 ± 0 . 04
`0 . 10 ± 0.02
`0 . 031 ± 0 . 006
`0 . 012 ± 0.002
`
`

`

`100
`
`0
`
`Bourne a nd Dittert
`
`:::1'
`
`z
`D
`......
`t-a: ('I')
`a::
`t- (\J
`z
`w -ub
`z
`D
`U
`
`a>
`GO
`r-(cid:173)
`c.o
`If)
`
`:::1'
`
`(\J
`
`N
`
`b
`
`0.00
`
`3 . 00
`
`6 . 00
`
`9 . 00
`TI ME
`
`12 . 00
`(HaURSl
`
`15 . 00
`
`18 . 00
`
`21 . 00
`
`2 4 . 00
`
`Figure 3 Semilogarithmic plot of concentration versus time for a first (cid:173)
`order process . One standard deviation is indicated by error bars .
`
`directly without converting to logs, and concentration values can be extra(cid:173)
`polated and interpolated from the plot without converting logs to numbers .
`For example, to determine the half- life in the preceding example, the
`Co value and the time at which C = Co/2 were both read directly from the
`graph .
`If Fig . 3 had been a plot of log C ( on a cartesian scale) versus
`time, it would have been necess ary to read log Co from the graph, convert
`it to Co, divide by 2, convert back to log (Co/2), then read the half-life
`off the graph .
`If the rate constant is determined for this example using
`Eq . (12) , the slope must be calculated . To calculate the slope of the line
`it is necessary first to read C1 and C2 from the graph and then take the
`logarithm of each concentration as described in Eq . (12) .
`
`Ill. P t R9if- 0 R 0ER J?HARMAC0KINETICS·: DRUG ELIMINATION
`•FOLLOWING RAPID INTRAVENOUS INJ ECTION
`
`H wa s mentioned previously that drug elimination most often dis plays the
`char act eristics of a first - order process. Thus , if a drug is administer ed
`by · r apid intravenous (IV) injection , after mixing with the body fluids its
`rat e of elimination from the body is proportional to the amount r emaining in
`the body . Normally, the plasma concentration is used a s a measure of the
`
`

`

`Pharmacokinetics
`
`101
`
`amount of drug in the body, and a plot of plasma concentration versus time
`has the same characteristics as the plot in Fig . 1.
`A semilogarithmic plot of plasma concentration v-ersus time is a straight
`line with a slope equal to ke1I 2. 3, where kef is the overall elimination rate
`constant. The intercept at t = 0 is cg, the hypothetical plasma concentration
`after the drug is completely mixed with body fluids but before any elimina(cid:173)
`tion has occurred; a typical log plasma concentration versus time plot is
`shown in Fig. 4. This figure shows that pharmacokinetic data can also be
`expressed in terms of a half-life, called the biologic half-life, which bears
`the same relationship to kel as that shown in Eqs. (14) and (15).
`Since all the kinetic characteristics of the disappearance of a drug from
`plasma are the same as those for the pseudo-first-order disappearance of
`a substance from a solution by hydrolysis, the same working equations [Eqs.
`( 11) and ( 13)] and the same approach to solving problems can be used.
`EXAMPLE. A 100-mg dose of tetracycline was administered to a patient
`by rapid IV injection. The initial plasma concentration ( cg) was found to'
`be 10 µg/ml. After 4 hr the plasma concentration was found to be 7. 5 µg/
`ml. What is,... the biologic half-life (t112) of tetracycline in this patient?
`
`SLOPE= -KEL/2 . 3
`
`Cp0 /2
`
`'b
`
`Q)
`(Xl
`
`,....
`
`CD
`
`l/)
`
`::1'
`
`(\J
`
`z
`0
`...... -0
`I-
`a: Q)
`a: (Xl
`I- r--
`z CD
`w l/)
`u
`z
`0
`u
`a: (\J
`::E
`~'b
`_j Q)
`CL CD
`,....
`
`::1'
`
`(t')
`
`CD
`
`l/)
`
`::1'
`
`N
`
`'o
`
`T112 = BIOLOGIC HALF-LIFE
`
`0 . 00
`
`3.00
`
`6 . 00
`
`9 . 00
`
`12.00
`TIME
`
`15 . 00
`
`18 . 00
`
`21. 00
`
`24.00
`
`Figure 4 Semilogarithmetic plot of plasma concentration versus time for a
`drug administered by rapid intravenous injection.
`
`

`

`102
`
`Bourn e and Dittert
`
`c 0
`t
`k
`el
`p
`logc = u (cid:173)
`p
`
`10
`log 7. 5 =
`
`kel X 4
`2.3
`
`2. 3 X 0 , 125
`4
`
`= 0 . 072 hr - l
`0.69
`= - - - - = 9 . 6 hr
`0.072 hr - l
`
`t
`112
`
`Note that this approach involves the following assumptions: (1) t he drug
`was eliminated by a pseudo.:.first;..order process, and (2) the drug -.was
`rapidly distributed so that an ''initial plasma concentration" could be mea(cid:173)
`sur ed..-before any drug began t o leave the body . · The latter assumption
`implies that the body behaves as a single homogeneous compartment through(cid:173)
`out which the drug distributes instantaneously following IV injection.
`In
`pharmacokinetic terms, this is referred to as the one.,.compartment mode1.
`Although most drugs do not, in fact, distribute instantaneously, they do dis (cid:173)
`tribute very rapidly, and the one-compartment model can be used for many
`clinically important pharmacokinetic calculations .
`An important parameter of the one- compartment model is the apparent
`volume of the body compartment because it directly determines the relation(cid:173)
`,ship between the plasma concentration and the amount of drug in the body .
`This volume is called the appa rent volume of distribution , Vct , and it may
`be calculated using the relationship
`
`amount
`volume = concentration
`
`The easiest way to calculate V d is to use cg, the plasma concerttl'ation when
`
`distribution is complete (assumed to be ·instantaneous for a one --cOm'Pai'tment
`model) and the ·entire dose is still in the body . Thus
`
`dose
`co
`p
`
`(16)
`
`EXAMPLE. Calculate Vct for the patient in the previous example .
`
`100 mg
`= ~---,-=,,-
`10 µg/ml
`= 10 liters
`
`Note : Since 1 µg/ml = 1 mg/liter, dividing the dose in milligrams by the
`plasma concentration is micrograms per millilitet> will give V d in liters .
`The apparent volume of distribution of a drug very rarely corresponds
`to any physiologic volume , and even in cases where it does, it must never
`be construed as showing that ·the drug enters or does not enter various
`body spaces . For example, the 10- liter volume calculated in the foregoing
`
`

`

`Pharmacok i netics
`
`103
`
`example is greater than either plasma volume (about 3 liters ) or whole -
`blood volume (about 6 liters) in a standard (70- kg) man; on the other hand,
`it is less than the extracellular fluid volume ( 12 liters) or total body water
`( 41 liters ) in the same average man .
`It cannot be said that tetracycline is
`restricted to the plasma, or that it enters or does not enter red blood cells,
`or that it enters or does not enter any or all extracellular fluids .
`A discussion of all the reasons for this phenomenon is beyond the scope
`of this chapter, but a simple example will illustrate the concept.
`· Highly
`lipid soluble drugs, such as pentobarbital, are preferentially distributed
`into adipose tissue . The result is that plasma concentrations are extremely
`low after distribution is complete. When the apparent volumes of distribu(cid:173)
`tion are calculated, they are frequently found to exceed total body volume,
`occasionally by a factor of 2 or more. This would be impossible if the con(cid:173)
`centration in the entire body compartment were equal to the plasma concen(cid:173)
`'f.hus, Vct is an empirically fabricated number r e lating the concentra..,
`tration .
`tion of drug in plasma (or ·blood) with the amount of drug in the body.
`in
`the case of drugs such as pentobarbital the ratio of the concentration in
`adipose tissue to the concentration in plasma is much greater than unity,
`resulting in a large value for V d .
`In calculating V d using Eq. (16), the
`assumption is made that the drug concentration in the entire body equals
`that in plasma.
`
`IV . PHARMACOKINETIC ANALYS IS OF UR INE DATA
`
`Occasionally , it is inconvenient or impossible to assay the drug in plasma,
`If
`but it may be possible to follow the appearance of the drug in urine.
`the drug is not metabolized to any appreciable degree, the pharmacokinetic
`model may be written as shown in Scheme 1.
`
`;> DU
`
`Scheme 1
`
`A plot of cumulative amount of drug appearing in urine (Du) versus
`time will be the mirror image of a plot of amount of drug remaining in the
`body (DB) versus time. This is illustrated in Fig. 5, which shows that the
`total amount of drug recovered in urine throughout the entire study (Du)
`is equal to the dose (Di) and, at any time, the sum of drug in the body
`(DB) plus drug in urine (Du) equals the dose (Di) .
`A kinetic equation describing urine data can be developed as follows.
`
`If
`
`dDB
`dt = - kelDB
`
`then
`
`dDU
`dt =
`+kelDB
`
`

`

`Bourne and Dittert
`
`= amount recovered in urine
`
`104
`
`But
`
`Then
`
`Therefore,
`
`or
`
`---- -- ----- ---- --- ---- ------ --- ------- --
`
`\
`\ Du"' (XOJ
`\
`\
`
`\ ' ' ' ' \
`' ' \
`
`3 . 00
`
`6 . 00
`
`9.00
`
`12 . 00
`TIME
`
`15 . 00
`
`-----------------------
`18.00
`21 . 00
`24 . 00
`
`Cl
`
`Cl .
`
`Cl
`Cl
`
`::J
`Do
`Cl
`
`Cl
`(D
`
`w
`z
`..... Cl
`a:o
`::Jo
`z
`
`<D
`
`WO
`> o
`.......
`1--0
`a: (\J
`.....J
`::::,
`::c
`:::,Cl
`Uo
`
`Figure 5 Plot of cumulative amount of drug in urine, Du (solid line), and
`amount of drug in body, DB ( dashed line), versus time according to Scheme
`1.
`
`

`

`Pharmacokinetics
`
`Integration gives
`
`105
`
`Since ln(x) - 2 . 3 log(x) and D~ = 0 (there is no drug in urine when t = 0,
`
`log D~
`
`k lt
`oo
`oo
`e
`log(Du - Du) = - 2.3 + log Du
`
`(17)
`
`Equation (17) is in the form of the equation for a straight line (y =
`mx + b), where tis one variable (x), - ke1/2.3 the slope (m), log Du the
`constant (b), and log(Du - Du) the other variable (y) . Thus a plot of
`log( Du - Du) versus time is a straight line with a slope equal to - ke1/ 2. 3
`and an intercept of log Du . Since Du is the total amount excreted and Du
`is the amount excreted up to time t, Du - Du is the amount remaining to
`be excreted (ARE). A typical ARE plot is shown in Fig . 6.
`EXAMPLE . The plot in Fig . 6 was constructed using the data shown in
`Table 3 . Note that the concentration of the drug in each urine specimen is
`not the information analyzed . The total amount excreted over each time
`interval and throughout the entire study must be determined . As a result,
`the experimental details of a urinary excretion study must be very carefully
`chosen, and strict adherence to the protocol is required . Loss of a single
`urine specimen, or even an unknown part of a urine specimen, makes con(cid:173)
`struction of an ARE plot impossible .
`
`Table 3 Drug Excreted into the Urine Versus Time
`
`Time
`interval
`
`(hr)
`
`0 . 0- 0 . 5
`0 . 5- 1.0
`1.0- 1.5
`1.5 - 2 . 0
`2 . 0- 2.5
`2 . 5- 3 . 0
`3.0- 6 . 0
`6 . 0- 12 . 0
`
`anu = 75 . 3
`
`Amount
`excreted
`
`(mg)
`
`Cumulative amount
`excreted, DU a
`(mg)
`
`Doo
`u
`
`- D
`u
`
`37 . 5
`18 . 5
`10 . 0
`5 . 2
`1.8
`1. 3
`1.0
`0 . 0
`
`37 . 5
`56 . 0
`66 . 0
`71. 2
`73 . 0
`74 . 3
`75 . 3
`75 . 3
`
`37 . 8
`19 . 3
`9 . 3
`4 . 1
`2 . 3
`1.0
`0 . 0
`0 . 0
`
`. I
`
`1
`
`l
`
`

`

`Bourne and Di tte r t
`
`106
`
`'b
`
`ru
`
`-0
`
`a,
`(X)
`r--
`CJ <D
`:E If) .
`:::T'
`
`('")
`
`:::,
`D
`
`ru
`
`'b
`8
`~ al -
`(X)
`r--
`<D
`lJ')
`=r -
`
`(T)
`
`ru
`
`-;
`D
`
`HRLF-LI FE
`
`0 . 00
`
`0.50
`
`1.00
`
`I .S O
`TI ME
`
`2 . 00
`(HClURSl
`
`2 .50
`
`3 . 00
`
`3 . 50
`
`4 . 00
`
`Figure 6 Semilogarithmic plot of amount of drug remaining to be excreted
`into urine, Du - Du, versus time (ARE plot) .
`
`V . CLEARANCE RATE AS AN EXPRESSION OF
`DRUG ELIM I NAT ION RATE
`
`A clearance rate is defined as the volume of blood or plasma completely
`cleared of drug per unit time .
`It is a useful way to describe drug elimi(cid:173)
`nation because it is related to blood or plasma perfusion of various organs
`of elimination, and it can be related directly to the physiologic function of
`these organs . For example, the renal clearance rate (RCR) of a drug can
`be calculated using the following equation:
`
`t.
`
`RCR =
`
`amount excreted in urine per unit time
`plasma concentration
`
`(18)
`
`In the example plotted in Fig . 6, the amount of drug ex (cid:173)
`EXAMPLE .
`If the plasma concen(cid:173)
`creted over the 0-
`to 0 . 5 - h r interval was 37 . 5 mg .
`tration at 0 . 25 hr (the middle of the interval) was 10 µg/ml, what was the
`renal clear ance rate? From Eq. (18),
`
`RCR = 37 . 5 mg/0 . 5 hr
`10 µg/ml
`
`

`

`Pharrnacok i netics
`
`107
`
`= 7.5 liters/hr
`= 125 ml/min
`
`The glomerular filtration rate (GFR) in normal males is estimated to be
`125 ml/min, and the results of the example calculation suggest that the drug
`is cleared solely by GFR.
`If the RCR had been less than 125 ml/min, tubu(cid:173)
`lar reabsorption of the drug would have been suspected .
`If it had been
`greater than 125 ml/min, tubular secretion would have been suspected .
`Drugs can be cleared from the body by metabolism as well as renal
`excretion, and when this occurs it is impossible to measure directly the
`amount cleared. However, the total clearance rate (TCR), or total body
`clearance, of the drug can be calculated from its pharmacokinetic parameters
`using the following equation :
`
`(19)
`
`EXAMPLE . The biologic half-life of procaine in a patient was found to
`be 35 min, and its volume of distribution was estimated to be 58 liters .
`Calculate the TCR of procaine .
`
`k
`el
`
`69
`= 0
`·
`35 min
`
`TCR = kelV d
`
`= 0 . 0197 min- l
`
`= 0 . 0197 min- l x 58 liters
`= 1.14 liters/min
`
`When a drug is eliminated by both metabolism and urinary excretion, it
`is possible to calculate the metabolic clearance rate (MCR) by the difference
`between TCR and RCR:
`
`MCR = TCR - RCR
`
`(20)
`
`The RCR can be determined from urine and plasma data using Eq . (18),
`and the TCR can be determined from the pharmacokinetic parameters using
`Eq . (19) . Alternatively, the RCR can be calculated by multiplying the TCR
`by the fraction of the dose excreted unchanged into . urine, fe :
`
`RCR = f
`e
`
`TCR
`
`(21)
`
`If it is assumed that the fraction of the dose not appearing as unchanged
`drug in urine has been metabolized, the MCR can be calculated as follows :
`
`MCR = ( 1 -
`
`f ) · TCR
`e
`
`c22)
`
`EXAMPLE. Sulfadiazine in a normal volunteer was found to have a
`biologic half- life of 16 hr and a volume of distribution of 20 liters . Sixty
`percent of the dose was recovered as