++
When medications are administered to humans, the body acts as
if it is a series of compartments1 (Figure 2-1). In many
cases, the drug distributes from the blood into the tissues quickly,
and a pseudoequilibrium of drug movement between blood and tissues
is established rapidly. When this occurs, a one-compartment model
can be used to describe the serum concentrations of a drug.2,3 In
some clinical situations, it is possible to use a one-compartment
model to compute doses for a drug even if drug distribution takes
time to complete.4,5 In this case, drug serum concentrations
are not obtained in a patient until after the distribution phase
is over.
+++
Intravenous
Bolus Equation
++
When a drug is given as an intravenous bolus and the drug distributes
from the blood into the tissues quickly, the serum concentrations
often decline in a straight line when plotted on semilogarithmic
axes (Figure 2-2). In this case, a one-compartment model intravenous
bolus equation can be used: C = (D/V)e–ket,
where t is the time after the intravenous bolus was given (t = 0
at the time the dose was administered), C is the concentration at
time = t, V is the volume of distribution,
and ke is the elimination rate constant. Most drugs given
intravenously cannot be given as an actual intravenous bolus because
of side effects related to rapid injection. A short infusion of
5–30 minutes can avoid these types of adverse effects,
and if the intravenous infusion time is very short compared to the
half-life of the drug so that a large amount of drug is not eliminated
during the infusion time, intravenous bolus equations can still
be used.
++
++
For example, a patient is given a theophyllineloading dose of
400 mg intravenously over 20 minutes. Because the patient received
theophylline during previous hospitalizations, it is known that
the volume of distribution is 30 L, the elimination rate constant
equals 0.116 h–1, and the half-life
(t1/2) is 6 hours (t1/2 = 0.693/ke = 0.693/0.115
h–1 = 6
h). To compute the expected theophylline concentration 4 hours after
the dose was given, a one-compartment model intravenous bolus equation
can be used: C = (D/V)e–ket = (400
mg/30 L)e–(0.115 h–1)(4
h) = 8.4 mg/L.
++
If drug distribution is not rapid, it is still possible to use
a one compartment model intravenous bolus equation if the duration
of the distribution phase and infusion time is small compared to
the half-life of the drug and only a small amount of drug is eliminated
during the infusion and distribution phases.6 The strategy
used in this situation is to infuse the medication and wait for
the distribution phase to be over before obtaining serum concentrations
in the patient. For instance, vancomycin must be infused slowly
over 1 hour in order to avoid hypotension and red flushing around
the head and neck areas. Additionally, vancomycin distributes slowly
to tissues with a 1/2–1 hour
distribution phase. Because the half-life of vancomycin in patients
with normal renal function is approximately 8 hours, a one compartment
model intravenous bolus equation can be used to compute concentrations
in the postinfusion, postdistribution phase without a large amount
of error. As an example of this approach, a patient is given an
intravenous dose of vancomycin 1000 mg. Since the patient has received
this drug before, it is known that the volume of distribution equals
50 L, the elimination rate constant is 0.077 h–1,
and the half-life equals 9 h (t1/2 = 0.693/ke = 0.693/0.077
h–1 = 9
h). To calculate the expected vancomycin concentration 12 hours
after the dose was given, a one compartment model intravenous bolus
equation can be used: C = (D/V)e–ket = (1000
mg/50 L)e–(0.077 h–1)(12
h) = 7.9 mg/L.
++
Pharmacokinetic parameters for patients can also be computed
for use in the equations. If two or more serum concentrations are
obtained after an intravenous bolus dose, the elimination rate constant,
half-life and volume of distribution can be calculated (Figure 2-3).
For example, a patient was given an intravenous loading dose of
phenobarbital 600 mg over a period of about an hour. One day and
four days after the dose was administered phenobarbital serum concentrations
were 12.6 mg/L and 7.5 mg/L, respectively. By
plotting the serum concentration/time data on semilogarithmic
axes, the time it takes for serum concentrations to decrease by
one-half can be determined and is equal to 4 days. The elimination
rate constant can be computed using the following relationship:
ke = 0.693/t1/2 = 0.693/4
d = 0.173 d–1.
The concentration/time line can be extrapolated to the
y-axis where time = 0. Since this was
the first dose of phenobarbital and the predose concentration was
zero, the extrapolated concentration at time = 0
(C0 = 15 mg/L in this
case) can be used to calculate the volume of distribution (Figure
2-4): V = D/C0 = 600
mg/ (15 mg/L) = 40 L.
++
++
++
Alternatively, these parameters could be obtained by calculation
without plotting the concentrations. The elimination rate constant
can be computed using the following equation: ke = –(ln
C1 – ln C2)/(t1 – t2),
where t1 and C1 are the first time/concentration
pair and t2 and C2 are the second time/concentration
pair; ke = –[ln
(12.6 mg/L) – ln (7.5 mg/L)]/(1
d – 4 d) = 0.173 d–1.
The elimination rate constant can be converted into the half-life
using the following equation: t1/2 = 0.693/ke = 0.693/0.173
d–1 = 4
d. The volume of distribution can be calculated by dividing the
dose by the serum concentration at time = 0.
The serum concentration at time = zero
(C0) can be computed using a variation of the intravenous
bolus equation: C0 = C/e–ket,
where t and C are a time/concentration pair that occur
after the intravenous bolus dose. Either phenobarbital concentration
can be used to compute C0. In this case, the time/concentration
pair on day 1 will be used (time = 1 d,
concentration = 12.6 mg/L): C0 = C/e–ket = (12.6
mg/L)/e–(0.173 d–1)(1
d) = 15.0 mg/L. The volume
of distribution (V) is then computed: V = D/C0 = 600
mg/(15 mg/L) = 40 L.
+++
Continuous and
Intermittent Intravenous Infusion Equations
++
Some drugs are administered using a continuous intravenous infusion,
and if the infusion is discontinued the serum concentration/time
profile decreases in a straight line when graphed on a semilogarithmic
axes (Figure 2-5). In this case, a one compartment model intravenous
infusion equation can be used to compute concentrations (C) while
the infusion is running: C = (k0/Cl)(1 – e–ket) = [k0/(keV)](1 – e–ket),
where k0 is the drug infusion rate (in amount per unit
time, such as mg/h or μg/min),
Cl is the drug clearance (since Cl = keV,
this substitution was made in the second version of the equation),
ke is the elimination rate constant, and t is the time
that the infusion has been running. If the infusion is allowed to
continue until steady state is achieved, the steady-state concentration
(Css) can be calculated easily: Css = k0/Cl = k0/(keV).
++
++
If the infusion is stopped, postinfusion serum concentrations
(Cpostinfusion) can be computed by calculating the concentration
when the infusion ended (Cend) using the appropriate equation
in the preceding paragraph, and the following equation: Cpostinfusion = Cende–ketpostinfusion,
where ke is the elimination rate constant and tpostinfusion is
the postinfusion time (tpostinfusion = 0
at end of infusion and increases from that point).
++
For example, a patient is administered 60 mg/h of theophylline.
It is known from previous hospital admissions that the patient has
the following pharmacokinetic parameters for theophylline: V = 40
L and ke = 0.139 h–1.
The serum concentration of theophylline in this patient after receiving
the drug for 8 hours and at steady state can be calculated: C = [k0/(keV)](1 – e–ket) = [(60
mg/h)/(0.139 h–1 · 40
L)](1 – e–(0.139
h–1)(8 h)) = 7.2
mg/L; Css = k0/(keV) = (60
mg/h)/(0.139 h–1 · 40
L) = 10.8 mg/L. It is possible
to compute the theophylline serum concentration 6 hours after the
infusion stopped in either circumstance. If the infusion only ran
for 8 hours, the serum concentration 6 hours after the infusion
stopped would be: Cpostinfusion = Cende–ketpostinfusion = (7.2
mg/L)e–(0.139 h–1)(6
h) = 3.1 mg/L. If the infusion
ran until steady state was achieved, the serum concentration 6 hours
after the infusion ended would be: Cpostinfusion = Cende–ketpostinfusion = (10.8
mg/L)e–(0.139 h–1)(6
h) = 4.7 mg/L.
++
Even if serum concentrations exhibit a distribution phase after
the drug infusion has ended, it is still possible to use one compartment
model intravenous infusion equations for the drug without a large
amount of error.4,5 The strategy used in this instance
is to infuse the medication and wait for the distribution phase
to be over before measuring serum drug concentrations in the patient.
For example, gentamicin, tobramycin, and amikacin are usually infused
over one-half hour. When administered this way, these aminoglycoside
antibiotics have distribution phases that last about one-half hour.
Using this strategy, aminoglycoside serum concentrations are obtained
no sooner than one-half hour after a 30-minute infusion in order
to avoid the distribution phase. If aminoglycosides are infused
over 1 hour, the distribution phase is very short and serum concentrations
can be obtained immediately. For example, a patient is given an
intravenous infusion of gentamicin 100 mg over 60 minutes. Because
the patient received gentamicin before, it is known that the volume
of distribution is 20 L, the elimination rate constant equals 0.231
h–1, and the half-life equals 3
h (t1/2 = 0.693/ke = 0.693/0.231
h–1 = 3
h). To compute the gentamicin concentration at the end of infusion,
a one compartment model intravenous infusion equation can be employed:
C = [k0/(keV)](1 – e–ket) = [(100
mg/1 h)/ (0.231 h–1· 20
L)](1 – e–(0.231
h–1)(1 h)) = 4.5
mg/L.
++
Pharmacokinetic constants can also be calculated for use in the
equations. If a steady-state concentration is obtained after a continuous
intravenous infusion has been running uninterrupted for 3–5
half-lives, the drug clearance (Cl) can be calculated by rearranging
the steady-state infusion formula: Cl = k0/Css.
For example, a patient receiving procainamide via intravenous infusion
(k0 = 5 mg/min) has
a steady-state procainamide concentration measured as 8 mg/L.
Procainamide clearance can be computed using the following expression:
Cl = k0/Css = (5
mg/min)/(8 mg/L) = 0.625
L/min.
++
If the infusion did not run until steady state was achieved,
it is still possible to compute pharmacokinetic parameters from
postinfusion concentrations. In the following example, a patient
was given a single 120-mg dose of tobramycin as a 60-minute infusion,
and concentrations at the end of infusion (6.2 mg/L) and
4 hours after the infusion ended (1.6 mg/L) were obtained.
By plotting the serum concentration/time information on
semilogarithmic axes, the half-life can be determined by measuring
the time it takes for serum concentrations to decline by one-half
(Figure 2-6), and equals 2 hours in this case. The elimination rate
constant (ke) can be calculated using the following formula:
ke = 0.693/t1/2 = 0.693/2
h = 0.347 h–1.
Alternatively, the elimination rate constant can be calculated without
plotting the concentrations using the following equation: ke = –(ln
C1 – ln C2)/(t1 – t2),
where t1 and C1 are the first time/concentration
pair and t2 and C2 are the second time/concentration
pair; ke = –[ln
(6.2 mg/L) – ln (1.6 mg/L)]/(1
h – 5 h) = 0.339 h–1 (note
the slight difference in ke is due to rounding errors).
The elimination rate constant can be converted into the half-life
using the following equation: t1/2 = 0.693/ke = 0.693/0.339
h–1 = 2
h.
++
++
The volume of distribution (V) can be computed using the following
equation4:
++
++
where k0 is the infusion rate, ke is the elimination
rate constant, t′ = infusion
time, Cmax is the maximum concentration at the end of infusion,
and Cpredose is the predose concentration. In this example,
the volume of distribution is:
++
+++
Extravascular
Equation
++
When a drug is administered extravascularly (e.g., orally, intramuscularly,
subcutaneously, transdermally, etc.), absorption into the systemic
vascular system must take place (Figure 2-7). If serum concentrations
decrease in a straight line when plotted on semilogarithmic axes
after drug absorption is complete, a one compartment model extravascular
equation can be used to describe the serum concentration/time
curve: C = {(FkaD)/[V(ka – ke)]}(e–ket – e–kat),
where t is the time after the extravascular dose was given (t = 0
at the time the dose was administered), C is the concentration at
time = t, F is the bioavailability fraction,
ka is the absorption rate constant, D is the dose, V is
the volume of distribution, and ke is the elimination rate
constant. The absorption rate constant describes how quickly drug
is absorbed with a large number indicating fast absorption and a
small number indicating slow absorption (Figure 2-7).
++
++
An example of the use of this equation would be a patient that
is administered 500 mg of oral procainamide as a capsule. It is
known from prior clinic visits that the patient has a half-life
equal to 4 hours, an elimination rate constant of 0.173 h–1 (ke = 0.693/t1/2 = 0.693/4
h = 0.173 h–1),
and a volume of distribution of 175 L. The capsule that is administered
to the patient has an absorption rate constant equal to 2 h–1,
and an oral bioavailability fraction of 0.85. The procainamide serum
concentration 4 hours after a single dose would be equal to:
++
++
If the serum concentration/time curve displays a distribution
phase, it is still possible to use one compartment model equations
after an extravascular dose is administered. In order to do this,
serum concentrations are obtained only in the postdistribution phase.
Since the absorption rate constant is also hard to measure in patients,
it is also desirable to avoid drawing drug serum concentrations
during the absorption phase in clinical situations. When only postabsorption,
postdistribution serum concentrations are obtained for a drug that
is administered extravascularly, the equation simplifies to: C = [(FD)/V]e–ket,
where C is the concentration at any postabsorption, postdistribution
time; F is the bioavailability fraction; D is the dose; V is the
volume of distribution; ke is the elimination rate constant;
and t is any postabsorption, postdistribution time. This approach
works very well when the extravascular dose is rapidly absorbed
and not a sustained- or extended-release dosage form. An example would
be a patient receiving 24 mEq of lithium ion as lithium carbonate
capsules. From previous clinic visits, it is known that the patient
has a volume of distribution of 60 L and an elimination rate constant
equal to 0.058 h–1. The bioavailability
of the capsule is known to be 0.90. The serum lithium concentration
12 hours after a single dose would be: C = [(FD)/V]e–ket = [(0.90 · 24
mEq)/60 L] e–(0.058
h–1)(12 h) = 0.18
mEq/L.
++
Pharmacokinetic constants can also be calculated and used in
these equations. If two or more postabsorption, postdistribution
serum concentrations are obtained after an extravascular dose, the
volume of distribution, elimination rate constant, and half-life
can be computed (Figure 2-8). For example, a patient is given an
oral dose of valproic acid 750 mg as capsules. Six and twenty-four
hours after the dose, the valproic acid serum concentrations are
51.9 mg/L and 21.3 mg/L, respectively. After graphing
the serum concentration/time data on semilogarithmic axes,
the time it takes for serum concentrations to decrease by one-half
can be measured and equals 14 hours. The elimination rate constant is
calculated using the following equation: ke = 0.693/t1/2 = 0.693/14
h = 0.0495 h–1.
The concentration/time line can be extrapolated to the
y-axis where time = 0. Since this was
the first dose of valproic acid, the extrapolated concentration
at time = 0 (C0 = 70
mg/L) is used to estimate the hybrid volume of distribution/bioavailability
(V/F) parameter: V/F = D/C0 = 750
mg/70 L = 10.7 L. Even though
the absolute volume of distribution and bioavailability cannot be
computed without the administration of intravenous drug, the hybrid
constant can be used in extravascular equations in place of V/F.
++
++
An alternative approach is to directly calculate the parameters
without plotting the concentrations. The elimination rate constant
(ke) is computed using the following relationship: ke = – (ln
C1 – ln C2)/(t1 – t2),
where C1 is the first concentration at time = t1,
and C2 is the second concentration at time = t2;
ke = – [ln
(51.9 mg/L) – ln (21.3 mg/L)]/ (6
h – 24 h) = 0.0495 h–1.
The elimination rate constant can be translated into the half-life
using the following equation: t1/2 = 0.693/ke = 0.693/0.0495
h–1 = 14
h. The hybrid constant volume of distribution/bioavailability
(V/F) is computed by taking the quotient of the dose and
the extrapolated serum concentration at time = 0.
The extrapolated serum concentration at time = zero
(C0) is calculated using a variation of the intravenous
bolus equation: C0 = C/e–ket,
where t and C are a time/concentration pair that occur
after administration of the extravascular dose in the postabsorption
and postdistribution phases. Either valproic acid concentration
can be used to compute C0. In this situation, the time/concentration
pair at 24 hours will be used (time = 24
hours, concentration = 21.3 mg/L):
C0 = C/e–ket = (21.3
mg/L)/e–(0.0495 h–1)(24
h) = 70 mg/L. The hybrid
volume of distribution/bioavailability constant (V/F)
is then computed: V/F = D/C0 = 750
mg/ (70 mg/L) = 10.7
L.
+++
Multiple-Dose
and Steady-State Equations
++
In most cases, medications are administered to patients as multiple
doses, and drug serum concentrations for therapeutic drug monitoring
are not obtained until steady state is achieved. For these reasons,
multiple dose equations that reflect steady-state conditions are
usually more useful in clinical settings than single dose equations.
Fortunately, it is simple to convert single dose compartment model
equations to their multiple dose and steady-state counterparts.7 In
order to change a single dose equation to the multiple dose version,
it is necessary to multiply each exponential term in the equation
by the multiple dosing factor: (1 – e–nkiτ)/(1 – e–kiτ),
where n is the number of doses administered, ki is the
rate constant found in the exponential of the single dose equation,
and τ is the dosage interval. At steady state,
the number of doses (n) is large, the exponential term in the numerator
of the multiple dosing factor (–nkiτ)
becomes a large negative number, and the exponent approaches zero.
Therefore, the steady-state version of the multiple dosing factor
becomes the following: 1/(1 – e–kiτ), where
ki is the rate constant found in the exponential of the
single dose equation and τ is the dosage interval.
Whenever the multiple dosing factor is used to change a single dose
equation to the multiple dose or steady-state versions, the time
variable in the equation resets to zero at the beginning of each
dosage interval.
++
As an example of the conversion of a single dose equation to
the steady-state variant, the one compartment model intravenous
bolus equation is: C = (D/V)e–ket,
where C is the concentration at time = t,
D is the dose, V is the volume of distribution, ke is the
elimination rate constant, and t is time after the dose is administered.
Since there is only one exponential in the equation, the multiple
dosing factor at steady state is multiplied into the expression
at only one place, substituting the elimination rate constant (ke)
for the rate constant in the multiple dosing factor: C = (D/V)[e–ket/(1 – e–keτ)],
where C is the steady-state concentration at any postdose time (t)
after the dose (D) is given, V is the volume of distribution, ke is
the elimination rate constant, and τ is the dosage
interval. Table 2-1 lists the one compartment model equations for
the different routes of administration under single dose, multiple
dose, and steady-state conditions.
++
++
The following are examples of steady-state one compartment model
equations for intravenous, intermittent intravenous infusions, and
extravascular routes of administration:
++
Intravenous bolus. A patient with
tonic-clonic seizures is given phenobarbital 100 mg intravenously
daily until steady-state occurs. Pharmacokinetic constants for phenobarbital
in the patient are: ke = 0.116
d–1, V = 75
L. The steady-state concentration 23 hours [(23 h)/(24
h/d) = 0.96 d] after
the last dose equals: C = (D/V)[e–ket/(1 – e–keτ)] = (100
mg/ 75 L)[e–(0.116
d–1)(0.96 d)/(1 – e–(0.116
d–1)(1 d))]= 10.9
mg/L.
++
Intermittent intravenous infusion. A
patient with gram-negative pneumonia is administered tobramycin
140 mg every 8 hours until steady state is achieved. Pharmacokinetic
parameters for tobramycin in the patient are: V = 16
L, ke = 0.30 h–1.
The steady-state concentration immediately after a 1 hour infusion
equals: C = [k0/(keV)][(1 – e–ket′)/(1 – e–keτ)] = [(140
mg/h)/(0.30 h–1· 16
L)][(1 – e(–0.30
h–1·1
h))/(1 – e(–0.30
h–1·8
h))] = 8.3 mg/L.
++
Extravascular. A patient with an
arrhythmia is administered 250 mg of quinidine orally (as 300 mg
quinidine sulfate tablets) every six hours until steady state occurs.
Pharmacokinetic constants for quinidine in the patient are:V = 180
L, ke = 0.0693 h–1,
F = 0.7. The postabsorption, postdistribution
steady-state concentration just before the next dose (t = 6
h) equals: C = (FD/V)[e–ket/(1 – e–keτ)] = [(0.7 · 250
mg)/180 L][e(–0.0693
h–1·6
h)/ (1 – e(–0.0693
h–1·6
h))] = 1.9 mg/L.
++
It is also possible to compute pharmacokinetic parameters under
multiple dose and steady-state conditions. Table 2-2 lists the methods
to compute pharmacokinetic constants using a one compartment model
for different routes of administration under single-dose, multiple-dose,
and steady-state conditions. The main difference between single-dose
and multiple-dose calculations is in the computation of the volume
of distribution. When a single dose of medication is given, the
predose concentration is assumed to be zero. However, when multiple
doses are given, the predose concentration is not usually zero,
and the volume of distribution equation (V) needs to have the baseline, predose
concentration(Cpredose) subtracted from the extrapolated
drug concentration at time = 0 (C0)
for the intravenous bolus (V = D/[C0 – Cpredose],
where D is dose) and extravascular (V/F = D/ [C0 – Cpredose],
where F is the bioavailability fraction and D is dose) cases. In
the case of intermittent intravenous infusions, the volume of distribution
equation already has a parameter for the predose concentration in
it4:
++
++
where k0 is the infusion rate, ke is the elimination
rate constant, t′ = infusion
time, Cmax is the maximum concentration at the end of infusion,
and Cpredose is the predose concentration. For each route
of administration, the elimination rate constant (ke) is
computed using the same equation as the single dose situation: ke = – (ln
C1 – ln C2)/(t1 – t2),
where C1 is the first concentration at time = t1,
and C2 is the second concentration at time = t2.
++
++
The following are examples of multiple dose and steady-state
computations of pharmacokinetic parameters using a one compartment
model for intravenous, intermittent intravenous infusions, and extravascular
routes of administration:
++
Intravenous bolus. A patient receiving
theophylline 300 mg intravenously every 6 hours has a predose concentration
equal to 2.5 mg/L and postdose concentrations of 9.2 mg/L
one hour and 4.5 mg/L five hours after the second dose
is given. The patient has an elimination rate constant (ke)
equal to: ke = – (ln
C1 – ln C2)/(t1 – t2) = – [(ln
9.2 mg/L) – (ln 4.5 mg/L)]/(1
h – 5 h) = 0.179 h–1.
The volume of distribution (V) of theophylline for the patient is:
C0 = C/e–ket = (9.2
mg/L)/e(–0.179
h–1)(1 h) = 11.0
mg/L and V = D/[C0 – Cpredose] = (300
mg)/(11.0 mg/L – 2.5 mg/L) = 35.3
L.
++
Intermittent intravenous infusion. A
patient is prescribed gentamicin 100 mg infused over 60 minutes
every 12 hours. A predose steady-state concentration (Cpredose)
is drawn and equals 2.5 mg/L. After the 1-hour infusion,
a steady-state maximum concentration (Cmax) is obtained
and equals 7.9 mg/L. Since the patient is at steady state,
it can be assumed that all predose steady-state concentrations are
equal. Because of this the predose steady-state concentration 12
hours after the dose can also be considered equal to 2.5 mg/L
and used to compute the elimination rate constant (ke)
of gentamicin for the patient: ke = – (ln
C1 – ln C2)/(t1 – t2) = – [(ln
7.9 mg/L) – (ln 2.5 mg/L)]/(1
h – 12 h) = 0.105 h–1.
The volume of distribution (V) of gentamicin for the patient is:
++
++
where k0 is the infusion rate, ke is the elimination
rate constant, t′ = infusion
time, Cmax is the maximum concentration at the end of infusion,
and Cpredose is the predose concentration. In this example,
volume of distribution is:
++
++
Extravascular. A patient is given
procainamide capsules 750 mg every 6 hours. The following concentrations
are obtained before and after the second dose: Cpredose = 1.1
mg/L, concentrations 2 hours and 6 hours postdose equal
4.6 mg/L and 2.9 mg/L. The patient has an elimination rate
constant (ke) equal to: ke = – (ln
C1 – ln C2)/(t1 – t2) = – [(ln
4.6 mg/L) – (ln 2.9 mg/L)]/(2
h – 6 h) = 0.115 h–1.
The hybrid volume of distribution/bioavailability constant
(V/F) of procainamide for the patient is: C0 = C/e–ket = (2.9
mg/L)/e(–0.115
h–1)(6 h) = 5.8
mg/L and V/F = D/[C0 – Cpredose] = (750
mg)/(5.8 mg/L – 1.1 mg/L) = 160
L.
+++
Average Steady-State
Concentration Equation
++
A very useful and easy equation can be used to compute the average
steady-state concentration (Css) of a drug: Css = [F(D/τ)]/Cl,
where F is the bioavailability fraction, D is the dose, τ is
the dosage interval, and Cl is the drug clearance.8 This
equation works for any single or multiple compartment model, and
because of this it is deemed a model-independent equation. The steady-state
concentration computed by this equation is the concentration that
would have occurred if the dose, adjusted for bioavailability, was
given as a continuous intravenous infusion. For example, 600 mg
of theophylline tablets given orally every 12 hours (F = 1.0)
would be equivalent to a 50 mg/h (600 mg/12 h = 50
mg/h) continuous intravenous infusion of theophylline.
The average steady-state concentration equation is very useful when
the half-life of the drug is long compared to the dosage interval
or if a sustained-release dosage form is used. Examples of both
situations follow:
++
Long half-life compared to dosage interval. A
patient is administered 250 μg of digoxin tablets
daily for heart failure until steady state. The pharmacokinetic
constants for digoxin in the patient are: F = 0.7,
Cl = 120 L/d. The average steady-state
concentration would equal: Css = [F(D/τ)]/Cl = [0.7(250 μg/d)]/(120
L/d) = 1.5 μg/L.
++
Sustained-release dosage form. A
patient is given 1500 mg of procainamide sustained-release tablets
every 12 hours until steady state for the treatment of an arrhythmia.
The pharmacokinetic parameters for procainamide in the patient are:
F = 0.85, Cl = 30
L/h. The average steady-state concentration would be: Css = [F(D/τ)]/Cl = [0.85(1500
mg/12 h)]/ (30 L/h) = 3.5
mg/L.
++
If an average steady-state concentration (Css) is known for a
drug, the hybrid pharmacokinetic constant clearance/bioavailability
(Cl/F) can be computed: Cl/F = (D/τ)/Css,
where D is dose and τ is the dosage interval. For
example, a patient receiving 600 mg of sustained-release theophylline
every 12 hours has a steady-state concentration equal to 11.2 mg/L.
The clearance/bioavailability constant for theophylline
in this patient would equal: Cl/F = (D/τ)/Css = (600
mg/12 h)/11.2 mg/L = 4.5
L/h.