Half Life Steady State Derivation Continuous Infusion
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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.
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Figure 2-1
Graphic Jump Location
Using compartment models, the body can be represented as a series of discrete sections. The simplest model is the one-compartment model which depicts the body as one large container where drug distribution between blood and tissues occurs instantaneously. Drug is introduced into the compartment by infusion (ko ), absorption (ka ), or IV bolus; distributes immediately into a volume of distribution(V); and is removed from the body via metabolism and elimination via the elimination rate constant (ke). The simplest multicompartment model is a two-compartment model which represents the body as a central compartment into which drug is administered and a peripheral compartment into which drug distributes. The central compartment (1) is composed of blood and tissues which equilibrate rapidly with blood. The peripheral compartment (2) represents tissues that equilibrate slowly with blood. Rate constants (k12 , k21 ) represent the transfer between compartments and elimination from the body (k10 ).
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Intravenous Bolus Equation
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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– k e t, 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.
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Figure 2-2
Graphic Jump Location
The solid line shows the serum concentration/time graph for a drug that follows one-compartment model pharmacokinetics after intravenous bolus administration. Drug distribution occurs instantaneously, and serum concentrations decline in a straight line on semilogarithmic axes. The dashed line represents the serum concentration/time plot for a drug that follows two-compartment model pharmacokinetics after an intravenous bolus is given. Immediately after the dose is given, serum concentrations decline rapidly. This portion of the curve is known as the distribution phase. During the distribution phase, drug is distributing between blood and tissues and is removed from the body via hepatic metabolism and renal elimination. Later, serum concentrations decline more slowly during the elimination phase. During the elimination phase, drug is primarily being removed from the body.
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For example, a patient is given a theophylline loading 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– k e t = (400 mg/30 L)e– (0.115 h–1 )(4 h) = 8.4 mg/L.
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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– k e t = (1000 mg/50 L)e– (0.077 h–1 )(12 h) = 7.9 mg/L.
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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.
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Figure 2-3
Graphic Jump Location
Phenobarbital concentrations are plotted on semilogarithmic axes, and a straight line is drawn connecting the concentrations. Half-life(t1/2) is determined by measuring the time needed for serum concentrations to decline by 1/2 (i.e., from 12.6 mg/L to 6.3 mg/L), and is converted to the elimination rate constant (ke = 0.693/t1/2 = 0.693/4d = 0.173d– 1). The concentration/time line can be extrapolated to the concentration axis to derive the concentration at time zero (C0 = 15 mg/L) and used to compute the volume of distribution (V = D/C0).
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Figure 2-4
Graphic Jump Location
For a one-compartment model, the body can be thought of as a beaker containing fluid. If 600 mg of phenobarbital is added to a beaker of unknown volume and the resulting concentration is 15 mg/L, the volume can be computed by taking the quotient of the amount placed into the beaker and the concentration: V = D/C0 = 600 mg/(15 mg/L) = 40 L.
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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– k e t, 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– k e t = (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.
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Continuous and Intermittent Intravenous Infusion Equations
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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– k e t) = [k0/(keV)](1 – e– k e t), 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).
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Figure 2-5
Graphic Jump Location
If a drug is given as a continuous intravenous infusion, serum concentrations increase until a steady-state concentration (Css) is achieved in 5–7 half-lives. The steady-stateconcentration is determined by the quotient of the infusion rate (k0) and drug clearance (Cl): Css = k0/Cl. When the infusion is discontinued, serum concentrations decline in a straight line if the graph is plotted on semilogarithmic axes. When using log10 graph paper, the elimination rate constant (ke) can be computed using the following formula: slope = –ke/2.303.
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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– k e t postinfusion, where ke is the elimination rate constant and tpostinfusion is the postinfusion time (tpostinfusion = 0 at end of infusion and increases from that point).
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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– k e t) = [(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– k e t postinfusion = (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– k e t postinfusion = (10.8 mg/L)e– (0.139 h–1 )(6 h) = 4.7 mg/L.
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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– k e t) = [(100 mg/1 h)/ (0.231 h– 1· 20 L)](1 – e– (0.231 h–1 )(1 h)) = 4.5 mg/L.
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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.
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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.
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Figure 2-6
Graphic Jump Location
Tobramycin concentrations are plotted on semilogarithmic axes, and a straight line is drawn connecting the concentrations. Half-life (t1/2) is determined by measuring the time needed for serum concentrations to decline by 1/2 (i.e., from 6.2 mg/L to 3.1 mg/L), and is converted to the elimination rate constant (ke = 0.693/t1/2 = 0.693/2 h = 0.347 h– 1). Volume of distribution is computed using the equation given in the text.
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The volume of distribution (V) can be computed using the following equation4:
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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:
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Extravascular Equation
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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– k e t – e– k a t), 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).
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Figure 2-7
Graphic Jump Location
Serum concentration/time curves for extravascular drug administration for agents following a one-compartment pharmacokinetics. The absorption rate constant (ka) controls how quickly the drug enters the body. A large absorption rate constant allows drug to enter the body quickly while a small elimination rate constant permits drug to enter the body more slowly. The solid line shows the concentration/time curve on semilogarithmic axes for an elimination rate constant equal to 2 h– 1. The dashed and dotted lines depict serum concentration/time plots for elimination rate constants of 0.5 h– 1 and 0.2 h– 1, respectively.
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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:
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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– k e t, 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– k e t = [(0.90 · 24 mEq)/60 L] e– (0.058 h–1 )(12 h) = 0.18 mEq/L.
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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.
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Figure 2-8
Graphic Jump Location
Valproic acid concentrations are plotted on semilogarithmic axes, and a straight line is drawn connecting the concentrations. Half-life(t1/2) is determined by measuring the time needed for serum concentrations to decline by 1/2 (i.e., from 51.9 mg/L to 26 mg/L), and is converted to the elimination rate constant (ke = 0.693/t1/2 = 0.693/14 h = 0.0495 h– 1). The concentration/ time line can be extrapolated to the concentration axis to derive the concentration at time zero (C0 = 70 mg/L) and used to compute the hybrid constant volume of distribution/bioavailability fraction (V/F = D/C0).
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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– k e t, 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– k e t = (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.
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Multiple-Dose and Steady-State Equations
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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– nk i τ)/(1 – e– k i τ), 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– k i τ), 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.
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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– k e t, 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– k e t/(1 – e– k e τ)], 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.
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Table Graphic Jump Location
Table 2-1 Single-Dose, Multiple-Dose, and Steady-State One-Compartment Model Equations
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Table 2-1 Single-Dose, Multiple-Dose, and Steady-State One-Compartment Model Equations
| Route of Administration | Single Dose | Multiple Dose | Steady State |
|---|---|---|---|
| Intravenous bolus | C = (D/V)e– ket | C = (D/V)e– ket[(1 – e– nke τ)/(1 – e– ke τ)] | C = (D/V)[e– ket/(1 – e– ke τ)] |
| Continuous intravenous infusion | C = [k0/(keV)](1 – e– ket) | N/A | Css = k0/Cl = k0/(keV) |
| Intermittent intravenous infusion | C = [k0/(keV)](1 – e– ket ′) | C = [k0/(keV)](1 – e– ket ′) [(1 – e– nke τ)/(1 – e– ke τ)] | C = [k0/(keV)][(1 – e– ket ′)/(1 – e– ke τ)] |
| Extravascular (postabsorption, postdistribution) | C = [(FD)/V]e– ket | C = [(FD)/V]e– ket[(1 – e– nke τ)/(1 – e– ke τ)] | C = (FD/V)[e– ket/(1 – e– ke τ)] |
| Average steady-state concentration (any route of administration) | N/A | N/A | Css = [F(D/τ)]/Cl |
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The following are examples of steady-state one compartment model equations for intravenous, intermittent intravenous infusions, and extravascular routes of administration:
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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– k e t/(1 – e– k e τ)] = (100 mg/ 75 L)[e– (0.116 d–1 )(0.96 d)/(1 – e– (0.116 d–1 )(1 d))]= 10.9 mg/L.
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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– k e t ′)/(1 – e– k e τ)] = [(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.
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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– k e t/(1 – e– k e τ)] = [(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.
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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:
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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.
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Table Graphic Jump Location
Table 2-2 Single-Dose, Multiple-Dose, and Steady-State Pharmacokinetic Constant Computations Utilizing a One Compartment Model
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Table 2-2 Single-Dose, Multiple-Dose, and Steady-State Pharmacokinetic Constant Computations Utilizing a One Compartment Model
| Route of Administration | Single Dose | Multiple Dose | Steady State |
|---|---|---|---|
| Intravenous bolus | ke = – (ln C1 – ln C2)/(t1 – t2) | ke = – (ln C1 – ln C2)/(t1 – t2) | ke = – (ln C1 – ln C2)/(t1 – t2) |
| t1/2 = 0.693/ke | t1/2 = 0.693/ke | t1/2 = 0.693/ke | |
| V = D/C0 | V = D/(C0 – Cpredose) | V = D/(C0 – Cpredose) | |
| Cl = keV | Cl = keV | Cl = keV | |
| Continuous intravenous infusion | N/A | N/A | Cl = k0/Css |
| Intermittent intravenous infusion | ke = – (ln C1 – ln C2)/(t1 – t2) | ke = – (ln C1 – ln C2)/(t1 – t2) | ke = – (ln C1 – ln C2)/(t1 – t2) |
| t1/2 = 0.693/ke | t1/2 = 0.693/ke | t1/2 = 0.693/ke | |
| V = [k0(1 – e– ket ′)]/{ke[Cmax – (Cpredosee– ket ′)]} | V = [k0(1 – e– ket ′)]/{ke[Cmax – (Cpredosee– ket ′)]} | V = [k0(1 – e– ket ′)]/{ke[Cmax – (Cpredosee– ket ′)]} | |
| Cl = keV | Cl = keV | Cl = keV | |
| Extravascular (postabsorption, postdistribution) | ke = – (ln C1 – ln C2)/(t1 – t2) | ke = – (ln C1 – ln C2)/(t1 – t2) | ke = – (ln C1 – ln C2)/(t1 – t2) |
| t1/2 = 0.693/ke | t1/2 = 0.693/ke | t1/2 = 0.693/ke | |
| V/F = D/C0 | V/F = D/(C0 – Cpredose) | V/F = D/(C0 – Cpredose) | |
| Cl/F = ke(V/F) | Cl/F = ke(V/F) | Cl/F = ke(V/F) | |
| Average steady-state concentration (any route of administration) | N/A | N/A | Cl/F = (D/τ)/Css |
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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:
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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– k e t = (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.
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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:
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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:
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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– k e t = (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.
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Average Steady-State Concentration Equation
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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:
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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.
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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.
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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.
Source: https://accesspharmacy.mhmedical.com/Content.aspx?bookid=510§ionid=40843069
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