Nonlinear pharmacokinetics

5 Nonlinear pharmacokinetics 5 Introduction 33 5 Capacity-limited metabolism 35 53 Estimation of Michaelis Menten parameters(v max andk m ) 37 55 Time to reach a given fraction of steady state 56 Example: calculation of parameters for phenytoin 33 333 54 Relationship between the area under the plasma concentration versus time curve and the administered dose 33 Objectives Upon completion of this chapter, you will have the ability to: perform pharmacokinetic calculations for drugs that have partially saturated their metabolic sites (capacity-limited metabolism) obtain values of the Michaelis Menten elimination parameters from plasma drug concentration data, either graphically or by equation estimate the daily dosing rate necessary to attain a target steady-state plasma drug concentration calculate the steady-state plasma drug concentration that will be attained from a given daily dosing rate estimate the time necessary to reach 9% of steady-state plasma drug concentrations calculate target phenytoin concentrations for a patient with hypoalbuminemia 5 Introduction Pharmacokinetic parameters, such as elimination half life (t/), the elimination rate constant (K), the apparent volume of distribution (V), and the systemic clearance (Cl) of most drugs are not expected to change when different doses are administered and/or when the drug is administered via different routes as a single dose or multiple doses The kinetics of these drugs is described as linear, or dose-independent, pharmacokinetics and is characterized by the first-order process The term linear simply means that plasma concentration at a given time at steady state and the area under the plasma concentration versus time curve (AUC) will both be directly proportional to the dose administered, as illustrated in Fig 5 For some drugs, however, the above situation may not apply For example, when the daily dose of phenytoin is increased by 5% in a patient from 3 mg to 45 mg, the average steady-state plasma concentration, (C p ) ss, may increase by as much as -fold This dramatic increase in the concentration (greater than

34 Basic Pharmacokinetics (a) (C p ) ss (mg L - ) 9 8 7 6 5 4 3 3 4 5 6 Dose (mg) 7 8 9 (b) AUC (mg h L ) 9 8 7 6 5 4 3 3 4 5 Dose (mg) 6 7 8 9 Figure5 Relationshipbetweentheplasmaconcentration (C p )atagiventimeatsteadystate(a)andtheareaunderthe plasma concentration versus time curve(auc)(b) against the administered dose for a drug that exhibits dose-independent pharmacokinetics directly proportional) is attributed to the nonlinear kinetics of phenytoin For drugs that exhibit nonlinear or dosedependent kinetics, the fundamental pharmacokinetic parameters such as clearance, the apparent volume of distribution, and the elimination half life may vary depending on the administered dose This is because one or more of the kinetic processes (absorption, distribution and/or elimination) of the drug may be occurring via a mechanism other than simple first-order kinetics For these drugs, therefore, the relationship between the AUC or the plasma concentration at a given time at steady state and the administered dose is not linear (Fig 5) Furthermore, administration of different doses of these drugs may not result in parallel plasma concentration versus time profiles expected for drugs with linear pharmacokinetics (Fig 53) For drugs with nonlinear metabolism, the initial rate of decline in the plasma concentrations of high doses of drug may be less than proportional to the plasma concentration; by comparison, after the administration of lower doses, the rate of decline will be proportional to plasma concentration and the proportionality constant will be K (Fig 54) This means that the rate of elimination is not directly proportional to the plasma concentration for these drugs The reason for this nonlinearity is explained as follows: (a) (C p ) ss (mg L - ) 7 6 5 4 3 (b) 6 4 AUC (mg h L ) 8 6 4 4 6 8 Dose (mg) 4 6 8 Dose (mg) Figure5 Relationshipbetweentheplasmaconcentration (C p )atagiventimeatsteadystate(a)andtheareaunderthe plasma concentration versus time curve(auc)(b) against the administered dose for a drug that exhibits dose-dependent pharmacokinetics

Nonlinear pharmacokinetics 35 Cp (mg L - ) 4 6 8 4 6 Time (h) been shown that the antibacterial agent dicloxacillin has saturable active secretion in the kidneys, resulting in a decrease in renal clearance as dose is increased Both phenytoin and ethanol have saturable metabolism, which means that an increase in dose results in a decrease in hepatic clearance and a more than proportional increase in AUC In the remainder of this chapter, nonlinearity in metabolism, which is one of the most common sources of nonlinearity, will be discussed Figure 53 The relationship between plasma concentration (C p )andtimefollowingtheadministrationof different doses of a drug that exhibits dose-dependent elimination pharmacokinetics 5 Capacity-limited metabolism Capacity-limited metabolism is also called saturable metabolism, Michaelis Menten kinetics, or mixedorder kinetics The process of enzymatic metabolism of drugs may be explained by the relationship depicted below Elimination rate/c p 8 6 4 5 5 5 3 Time (h) Figure54 Plotofeliminationrate (dc p /dt)normalizedfor plasmadrugconcentration (C p )versustimeearlyafteradose of drug, when drug levels are high, dose-dependent elimination kinetics may apply In this case, the elimination rate is less than proportional to plasma drug concentration When plasma drug levels have declined sufficiently(after about5hinthisfigure),theeliminationrateisdirectly proportionaltoc p,withproportionalityconstantk (horizontal section) Nonlinearity may arise at any one of the various pharmacokinetic steps, such as absorption, distribution and/or elimination For example, the extent of absorption of amoxicillin decreases with an increase in dose For distribution, plasma protein binding of disopyramide is saturable at the therapeutic concentration, resulting in an increase in the volume of distribution with an increase in dose of the drug As for nonlinearity in renal excretion, it has Enzyme + Substrate (drug) Enzyme drug complex Enzyme + Metabolite First the drug interacts with the enzyme to produce a drug enzyme intermediate Then the intermediate complex is further processed to produce a metabolite, with release of the enzyme The released enzyme is recycled back to react with more drug molecules According to the principles of Michaelis Menten kinetics, the rate of drug metabolism changes as a function of drug concentration, as illustrated in Fig 55 Based on this relationship, at very low drug concentration, the concentration of available enzymes is much greater than the number of drug molecules or the drug concentration Therefore, when the concentration of drug is increased, going from left to right in Fig 55, the rate of metabolism is also increased proportionally (linear elimination kinetics) However, after a certain point, as the drug plasma concentration increases, the rate of metabolism increases less than proportionally The other extreme occurs when the concentration of drug is very high relative to the concentration of available enzyme molecules Under this condition, all of the enzyme molecules are saturated with the drug molecules and, when concentration is increased further, there will be no change in the rate of metabolism of the drug In other words, the maximum rate of metabolism (V max ) has been achieved

36 Basic Pharmacokinetics Elimination rate as fraction of V max 5 Michaelis Menten region First-order (linear) region Zero-order region Drug concentration or mass of drug in body Figure 55 Relationship between elimination rate and the plasma concentration of a drug that exhibits dose-dependent pharmacokineticsathighdrugconcentrations,wheresaturationoccurs,theeliminationrateapproachesitsmaximum,v max The rate of metabolism, or the rate of elimination if metabolism is the only pathway of elimination, is defined by the Michaelis Menten equation: Metabolism rate = V maxc K m + C (5) where V max is the maximum rate (mg h ) of metabolism; K m is the Michaelis Menten constant (mg L ), and C is the drug concentration (mg L ) The maximum rate of metabolism (ie, V max ) is dependent on the amount or concentration of enzyme available for metabolism of the drug; K m is the concentration of the drug that results in a metabolic rate equal to one half of V max (V max /) In addition, K m is inversely related to the affinity of the drug for the metabolizing enzymes (the higher the affinity, the lower the K m value) The unit for the maximum rate of metabolism is the unit of elimination rate and is normally expressed as amount per unit time (eg, mg h ) However, in some instances, it may be expressed as concentration per unit time (eg, mg L h ) Equation (5) describes the relationship between the metabolism (or elimination) rate and the concentration over the entire range of concentrations However, different regions of the Michaelis Menten curve (Fig 55) can be examined with regard to drug concentrations At one extreme, the drug concentration may be much smaller than K m In this case, the concentration term may be deleted from the denominator of Eq (5), yielding Metabolism rate = V maxc K m (5) Because both V max and K m are constants, the metabolism rate is proportional to the drug concentration and a constant (ie, first-order process) in this region: where Metabolism rate = KC (53) K = V max K m and where the units of K are mg L h mg L = h Equation (53) is analogous to the classical first-order rate equation ( dx/dt = KX) At the other extreme, the drug concentrations are much higher than K m ; therefore, the term K m may be deleted from the denominator of Eq (5): Metabolism rate = V maxc C = V max (54) Equation (54) is analogous to the zero-order equation ( dx/dt = K ) Equation (54) shows that, when the drug concentration is much higher than K m, the rate of metabolism is a constant (V max ), regardless of drug concentration This situation is similar to

Nonlinear pharmacokinetics 37 zero-order kinetics; at drug concentrations around the K m, a mixed order is observed, which is defined by the Michaelis Menten equation (Eq 5) 53 Estimation of Michaelis Menten parameters (V max andk m ) Estimation of Michaelis Menten parameters from administration of a single dose Following the administration of a drug as an intravenous solution, drug plasma concentration is measured at various times This gives a set of concentration versus time data (Fig 56) Use concentration versus time data and follow the following steps to obtain the information necessary to determine V max and K m Determine dc p /dt (rate; units of mg L h ): for example, (C p ) (C p ) t t = C p t Determine the midpoint concentration (ie, average concentration; units of mg L ): (C p ) + (C p ) (a) 4 C p (mg L ) 3 Zero-order region Michaelis Menten region First-order region 5 3 45 6 75 Time (h) 9 5 35 5 (b) Zero-order region Michaelis Menten region First-order region C p (mg L ) 5 3 45 6 75 Time (h) 9 5 35 5 Figure56 Plasmaconcentration (C p )versustimeprofilefollowingtheadministrationofanintravenousbolusdoseofadrug that exhibits the characteristics of dose-dependent pharmacokinetics(a) Rectilinear plot;(b) semilogarithmic plot

38 Basic Pharmacokinetics The practical expression of the Michaelis Menten equation becomes: ( ) Cp t t = V max( C p ) t K m + ( C p ) t (55) There are two ways to linearize Eq (55), which will then enable determination of V max and K m Lineweaver Burke plot Take the reciprocal of Eq (55) ( ) K m ) = V max ( C p ) t ( Cp t t The plot of / ( ) Cp t t + V max (56) against /( C p ) t straight line (Fig 57) On this plot, the intercept is /V max So V max = /Intercept yields a The slope of the plot is given by K m /V max Thus using two corresponding sets of values on the line for x and y, the slope will equal (Y Y )/(X X ) and will have units of (h L mg )/ (L mg ) = h (ie, units of time) The calculated slope then equals K m /V max Slope (h) V max (mg L h ) = K m (mg L ) A second way of linearizing Eq(56) ( ) = ( Cp t t K m V max ( C p ) t ) + V max Each side of Eq (56) is multiplied by ( C p ) t : ( C p ) t ) ( Cp t t = K m V max + ( C p ) t V max (57) ( Cp t The plot of ( C p ) t / versus ( C p ) t will yield a t straight line (Fig 58) The slope of this line is found as before, (Y Y )/(X X ), and will have units of (h)/(mg L ) = h L mg The slope is /V max So V max = /Slope (units of mg L h ) ) The intercept is K m /V max Intercept (h) V max (mg L h ) = K m (mg L ) Estimation of Michaelis Menten parameters following administration of multiple doses If the drug is administered on a multiple-dosing basis, then the rate of metabolism (or elimination) at steady state will be a function of the steady-state plasma concentration; ie, (C p ) ss : 8 7 (h mg L) ( C p / t) t Intercept = /V max /Cp (L mg ) Slope = K m /V max (C p ) t (h) ( C p / t) t 6 5 4 3 Slope = /V max Intercept = K m /V max 5 5 5 3 35 (C p ) t (mg L ) Figure 57 Lineweaver Burke plot to estimate the fundamental pharmacokinetic parameters of a drug that exhibitsnonlinearkineticsk m,michaelis Mentenconstant; V max,maximumvelocity Figure 58 Wolf s plot to estimate the fundamental pharmacokinetic parameters of a drug that exhibits nonlinear kinetics Abbreviations as in Fig 57

Nonlinear pharmacokinetics 39 Elimination rate or metabolism rate = dx dt Intercept = V max = V max(c p ) ss K m + (C p ) ss Note that at steady state, the rate of elimination is equal to the drug dosing rate (R): R (mg day ) Slope = K m R = V max(c p ) ss K m + (C p ) ss (58) In order to estimate V max and K m, Eq (58) must first be linearized The following approach may be used Equation (58) is rearranged as follows R[K m + (C p ) ss ] = V max (C p ) ss RK m + R(C p ) ss = V max (C p ) ss R(C p ) ss = V max (C p ) ss RK m R = [V max (C p ) ss RK m ]/(C p ) ss R = V max [K m R/(C p ) ss ] (59) Equation (59) indicates that a plot of R versus R/(C p ) ss will be linear, with a slope of K m and a y-axis intercept of V max To construct such a plot and estimate the values of Michaelis Menten parameters (ie, V max and K m ), one needs at least two sets of doses (three or four is ideal) along with their corresponding steady-state plasma concentration values CalculationofK m fromtwosteady-state drug concentrations arising from two infusion rates The equation derived above (Eq 59) can be used to derive two steady-state drug concentrations (eg, for a drug such as phenytoin, which undergoes capacity-limited metabolism) arising from two infusion rates: R = V max [K m R/(C p ) ss ] This equation forms the basis for the linear plot of R versus R/(C p ) ss, as seen in Fig 59, with a slope of K m and a y-axis intercept of V max In practice, two pairs of infusion-rate steady-state drug concentration data will suffice to define the straight line of this plot R/(C p ) ss (L day ) Figure59 Theuseofdosingrate (R)andthe corresponding steady-state plasma concentrations to obtain the pharmacokinetic parameters for a drug that exhibits nonlinear kinetics Abbreviations as in Fig 57 and evaluate V max in a patient The two rates will be R = V max [K m R /(C p ) ss ] (5) R = V max [K m R /(C p ) ss ] (5) Now, the slope of the line connecting the points [R, R /(C p ) ss ] and [R, R /(C p ) ss ] will equal Y/ X, or R R R (C p ) ss R (C p ) ss Substituting the values of R and R from Eqs (5) and (5) into the numerator of the above expression yields V max K m R (C p ) ss V max + K m R (C p ) ss R (C p ) ss R (C p ) ss = K m R R (C p ) ss + K m (C p ) ss R (C p ) ss R (C p ) ss Upon simplification, this slope equals [ ] K R m (C p ) ss R (C p ) ss R (C p ) ss R (C p ) ss = K m Once we have the value of K m = slope, we can obtain the patient s V max by rearrangement of Eq (58), as follows: V max = R[K m + (C p ) ss ] (C p ) ss

33 Basic Pharmacokinetics Example: estimation of dosing rate(r) and steady-stateplasmaconcentration (C p ) ss for phenytoin from Michaelis Menten parameters If the Michaelis Menten parameters are known in a patient or are reported in the literature, the dosing rate necessary to obtain a desired steady-state plasma concentration can be estimated For example, for phenytoin, the estimated values of K m and V max are 65 mg L and 548 mg day, respectively It is desired to obtain a steady-state plasma concentration of 5 mg L The dosing rate can be determined from Eq (58): R = V max (C p ) ss K m + (C p ) ss R = 548 mg day 5 mg L 65 mg L + 5 mg L = 38 mg day Alternatively, if the Michaelis Menten parameters and the administered dosing rate are known, the steady-state plasma concentration can be predicted Equation (58) may be rearranged to solve for (C p ) ss : (C p ) ss = K m R V max R (5) In the earlier example, if a dose of 4 mg day is administered, (C p ) ss = 65 mg L 4 mg day 548 mg day 4 mg day (C p ) ss = 756 mg L Using Eq (5) and administered daily doses of phenytoin of,, 3, and 45 mg, the steady-state plasma concentration can be calculated for each dose For mg dose: (C p ) ss = 65 mg L mg day 548 mg day mg day (C p ) ss = 45 mg L For mg dose: (C p ) ss = 65 mg L mg day 548 mg day mg day (C p ) ss = 373 mg L For 3 mg dose: (C p ) ss = 65 mg L 3 mg day 548 mg day 3 mg day (C p ) ss = 786 mg L For 45 mg dose: (C p ) ss = 65 mg L 454 mg day 548 mg day 45 mg day (C p ) ss = 984 mg L What (C p ) ss would be achieved by a 548 mg day dose? Construct a plot of (C p ) ss versus administered dose 54 Relationship between the area under the plasma concentration versus time curve and the administered dose For a single intravenous bolus eliminated by a first-order process: (AUC) = C p dt = (Dose) VK (53) whereas for nonlinear kinetics, capacity-limited kinetics, or Michaelis Menten kinetics: (AUC) High doses = C p dt = (C p) V max [ (Cp ) + K m ] (54) At high single doses or when (C p ) K m, the value of K m in Eq (54) is negligible (AUC) = = C p dt = (C p) (V max ) (X ) (V max )(V) (55) where V is the apparent volume of distribution and X is the dose administered

Nonlinear pharmacokinetics 33 Equation (55) shows that, for a single, relatively high dose of phenytoin, if you double the dose, the AUC will increase by four times, while tripling the dose will increase plasma concentration nine times In other words, (AUC) is proportional to the square of the dose Therefore, a relatively modest increase in the dose may produce a dramatic increase in the total AUC The effect of increasing the size of doses of phenytoin that will be given in a multiple-dosing regimen can have even more dramatic effects As mentioned in the introduction to this chapter, increasing the daily dose of phenytoin to a size that is large (compared with V max ) can cause steady-state phenytoin concentrations to skyrocket In fact, daily doses equal to or greater than a patient s maximum rate of metabolism (V max ) will cause phenytoin concentrations to increase without an upper limit! These assertions can be validated by use of Eq (5) Low doses At low doses: When (C p ) K m, or (C p ) / K m, the value of (C p ) / in Eq (54) is ignored because it is much smaller than the value of K m (AUC) = C p dt = (C p) K m V max Moreover, V max /K m = K Therefore, K m /V max = /K and (C p ) = Dose/V, which yields Eq (53): (AUC) = Dose VK For example, for phenytoin, the reported values of the apparent volume of distribution and V max are 5 L and 548 mg day, respectively Determine the (AUC) following the daily administration of,, 3, and 45 mg doses of phenytoin Remember that, for phenytoin, the literature average K m = 65 mg L V max = 548 mg day /5 L = 96 mg L day This situation does not qualify for use of the high-dose equation (Eq 55) Neither are the plasma drug concentrations obtained small enough to warrant use of the low-dose (linear kinetic) equation (Eq 53) This leaves the requirement to use the general equation for Michaelis Menten elimination kinetics (Eq 54): (AUC) = For mg dose: C p dt = (C p) V max [ (Cp ) + K m ] (C p ) = Dose/V = mg/5 L = mg L (AUC) mg L = C p dt = 96 mg L day [ ] mg L + 65 mg L For mg dose: = 37 mg L day (C p ) = Dose/V = mg/5 L = 4 mg L (AUC) 4 mg L = C p dt = 96 mg L day [ ] 4 mg L + 65 mg L For 3 mg dose: = 3 mg L day (C p ) = Dose/V = 3 mg/5 L = 6 mg L (AUC) 6 mg L = C p dt = 96 mg L day [ ] 6 mg L + 65 mg L For 45 mg dose: = 5 mg L day (C p ) = Dose/V = 45 mg/5 L = 9 mg L (AUC) 9 mg L = C p dt = 96 mg L day [ ] 9 mg L + 65 mg L = 93 mg L day Using these data, a plot of (AUC) against the administered dose can be constructed From the plot, make an observation of the relationship

33 Basic Pharmacokinetics 55 Timetoreachagivenfraction of steady state In nonlinear elimination pharmacokinetics, the time required to reach a given fraction of the steady-state concentration varies with the rate of drug administration and depends upon the values of V max and K m For a constant rate of infusion or input, the rate of change of a drug in the body, V(dC p /dt), is the difference between the rate drug in and the rate drug out: V dc p dt = R V maxc p K m + C p = RK m + RC p V max C p K m + C p Collecting terms with their appropriate differentials: K m + C p RK m + (R V max )C p dc p = V dt Multiplying numerator and denominator of the left side by and splitting terms: K m RK m + (V max R)C p dc p + = V dt C p RK m + (V max R)C p dc p Integrating from (C p ) to (C p ) t : K m (Cp) t + = V RK m + (V max R)C p dc p (C p) (Cp) t C p dc p (C p) RK m + (V max R)C p t dt Performing the integrations (using Selby (97), integrals 7 and 3 for the first two terms): K m V max R ln RK m + (V max R)(C p )t RK m + (V max R)(C p ) ( (Cp )t V max R (C p) V max R RK m (V max R) ln RK m + (V max R)(C p ) t RK m + (V max R)(C p ) = V t ) Multiplying through by V max R: K m ln RK m + (V max R)(C p ) t RK m + (V max R)(C p ) ( (C p ) t (C p ) which equals RK m (V max R) ln RK ) m + (V max R)(C p ) t RK m + (V max R)(C p ) = (V max R) t V + K m ln RK m + (V max R)(C p ) RK m + (V max R)(C p ) t ( (C p ) t (C p ) +RK m (V max R) ln RK ) m + (V max R)(C p ) RK m + (V max R)(C p ) t = (V max R) t V Multiplying the first term by unity in the form of (V max R)/(V max R) and expanding yields K m V max V max R ln RK m + (V max R)(C p ) RK m + (V max R)(C p ) t K mr V max R ln RK m + (V max R)(C p ) RK m + (V max R)(C p ) t ( + (C p ) t + (C p ) + +RK m (V max R) ln RK ) m + (V max R)(C p ) RK m + (V max R)(C p ) t = (V max R) t V Cancellation of terms yields K m V max V max R ln RK m + (V max R)(C p ) RK m + (V max R)(C p ) t + [ (C p ) t ] + (C p ) = (V max R) t V Multiplying both numerator and denominator of the logarithmic expression by yields K m V max V max R ln +RK m (V max R)(C p ) +RK m (V max R)(C p ) t

Nonlinear pharmacokinetics 333 + [ (C p ) t ] + (C p ) = (V max R) t V Rearranging to solve for t, the time to go from (C p ) to (C p ) t, yields K m V max V (V max R) ln RK m (V max R)(C p ) RK m (V max R)(C p ) t V(C p) t V max R + V(C p) V max R = t (56) If the starting condition is (C p ) =, then K m V max V (V max R) ln RK m RK m (V max R)(C p ) t V(C p) t V max R = t (57) If the objective is to determine how much time it will take to reach a given fraction of the drug concentration that will be achieved at steady state (ie, (C p ) t = f ss (C p ) ss ), then Eq (5) needs to be modified as follows: and (C p ) t = f ssk m R V max R (58) (V max R)(C p ) t = f ss K m R (59) Now the expression in Eq (58) for (C p ) t and the expression in Eq (59) for (V max R)(C p ) t can both be substituted into Eq (57): K m V max V (V max R) ln RK m RK m f ss RK m Cancellation yields f ssrk m V (V max R) = t f ss (5) K m V max V (V max R) ln f ssrk m V f ss (V max R) = t f ss (5) Combining common factors gives the expression for time to go from an initial concentration of zero to the concentration achieved at a particular fraction of steady state: ( ) K m V (V max R) V max ln f ss R = t fss f ss For the commonly used f ss = 9, we obtain K m V (V max R) (V max ln() 9R) = K m V (V max R) (33V max 9R) = t 9 Dimensional analysis The dimensions will be (mg L )(L) (mg day ) (33 [mg day ] 9 [mg day ]) mg = (mg day = time (days) ) (5) (53) It is clear from Eq (53) that the time required to reach 9% of steady state for drugs with nonlinear kinetics is affected by the dosing rate (R) as well as the values of V max, K m, and the apparent volume of distribution (V) of the drug For a given drug, however, the time required to attain a given fraction of the steady-state concentration is determined by the chosen rate of drug administration (R), the other parameters being constant for a particular drug administered to a particular patient 56 Example: calculation of parameters for phenytoin Thetimerequiredtoattain9%ofthe true steady-state plasma concentration for phenytoin Let us assume that we are interested in determining the time required to attain 9% of the true steady-state plasma concentration for phenytoin, which is administered at different rates, where V = 5 L, V max = 5 mg day and K m = 4µg ml (= 4 mg L ) Using Eq (53), the time required to attain 9% of the steady-state concentration can be determined for various daily doses

334 Basic Pharmacokinetics For mg dose: (4 mg L )(5 L) (4 mg day ) (33[5 mg day ] 9[ mg day ]) = 33 days For mg dose: (4 mg L )(5 L) (3 mg day ) (33[5 mg day ] 9[ mg day ]) = 6 days For 3 mg dose: (4 mg L )(5 L) ( mg day ) (33[5 mg day ] 9[3 mg day ]) = 44 days For 4 mg dose: (4 mg L )(5 L) ( mg day ) (33[5 mg day ] 9[4 mg day ]) = 58 days What plasma phenytoin concentrations would be achieved at the times and doses above? The answer is that slight modifications of the steady-state phenytoin concentration can be made as follows (using Eq 5): 9(C p ) ss = 9K m R/(V max R) (54) From this a table can be constructed (Table 5) What if K m was equal to 57 mg L instead of 4 mg L? The answer is that, since K m appears only in the numerator of Eq (53), there is a direct proportion between K m and the time to reach 9% of the steady-state phenytoin concentration Similarly, the value 9(C p ) ss is also directly proportional to K m Therefore, each of the above values can be multiplied by the factor (57/4), resulting in the figures given in Table 5 Alternative equation to calculate the time to reach a given fraction of steady state If V max is not known but the desired steady-state drug (eg, phenytoin) concentration is known, a different equation must be used to calculate the time required to reach a given fraction of steady state Equation (58) can be easily rearranged to the Table 5 Time (t 9 ) to 9% of steady-state plasma concentration (C p ) ss level as a function of dailydose(r);k m =4mgL R (mgday ) t 9 (days) 9(C p ) ss (mgl ) 33 9 6 4 3 44 5 4 58 44 Table 5 Time (t 9 ) to 9% of steady-state plasma concentration (C p ) ss level as a function of dailydose(r);k m = 57mgL R (mgday ) t 9 (days) 9(C p ) ss (mgl ) 9 8 38 34 3 68 77 4 5 5 following expression: Thus, V max = RK m (C p ) ss + R (55) V max R = RK m (C p ) ss (56) Substitution of the above expressions for V max and (V max R) into Eq (5) results in t f ss = V(C [ ( ) p) ss (Cp ) ss ln f ss R K m f ss ] + ln (57) f ss For f ss = 9, this equals t 9 = V(C p) ss R ( 43 (C ) p) ss + 33 (58) K m The time required for plasma phenytoin concentration to decline from an initial value (C p ) toaparticularvalue (C p ) t Rate of metabolism, or more generally elimination, is the decrease of plasma drug concentration over time Therefore, Eq (5) can be rewritten as dc p dt = V maxc p K m + C p

Nonlinear pharmacokinetics 335 Collecting terms with their appropriate differentials: dc p C p (K m + C p ) = V max dt Expanding: dc p K m C p dc p = V max dt Taking integrals for t = to t = t: (Cp) (Cp) dc p K m dc p (C p) t (C p) t C p = V max t dt Performing the integration: (C p ) (C p ) t + K m ln (C p) (C p ) t = V max t (59) In this case V max would have units of concentration per unit time For V max expressed in units of mass per unit time, the equation would be (C p ) (C p ) t + K m ln (C p) = V max t (53) (C p ) t V Free and total plasma phenytoin concentrations Only unbound (free) drug is capable of exerting pharmacological (or toxic) effect The following example depicts the adjustments required in the case of an atypical free fraction of phenytoin in the blood Problem A patient with hypoalbuminemia has a phenytoin free fraction of Calculate a reasonable target total (free + bound) plasma phenytoin concentration in this patient Solution The average percentage phenytoin bound to plasma proteins is 89%, which corresponds to a percentage free equal to 89 = % Expressed as a decimal, the nominal free fraction of phenytoin is This is termed the fraction unbound in the plasma (f up ) A reasonable total phenytoin concentration at steady state is in the middle of the therapeutic range: 5 µg ml Since unbound drug concentration is equal to the free fraction times the total drug concentration, we define: (C p ) free = f up C p (53) Therefore, a patient with a nominal free fraction of phenytoin will have a target free phenytoin concentration of () (5µg ml ) = 65µg ml Target total plasma concentrations will be different for different free fractions; however, the target free plasma phenytoin concentration remains 65µg ml That is, this value is invariant, regardless of the patient s free phenytoin fraction This fact can be used when Eq (53) is rearranged to solve for the target total phenytoin concentration in the patient in this example: Target (C p ) patient = Target (C p) free (f up ) patient = 65µg ml = 75µg ml Notice that, in this problem, the patient s free fraction is twice normal; so, her total plasma phenytoin concentration requirement is exactly one-half of the usual 5µg ml

Problem set 8 Problems for Chapter 5 Calculations Question Patient AB, a 4-year-old woman weighing 67 kg, presented to the hospital emergency room with epileptiform seizures A phenytoin assay shows no phenytoin in the plasma In the absence of individual pharmacokinetic parameters, the literature average values should be used: maximum reaction rate (V max ) = 75 mg kg day Michaelis Menten constant (K m ) = 57 mg L volume of distribution (V) = 64 L kg salt value for sodium phenytoin = 96 absolute oral bioavailability (F PO ) = 98 (a) (b) (c) Suggest a loading dose of intravenous sodium phenytoin, (D L ) IV, for this patient Base the desired plasma concentration, (C p ) desired, on a maximum concentration (C p ) max of mg L The patient has been stabilized by a series of short intravenous infusions of sodium phenytoin For a target average phenytoin concentration of 5 mg L, calculate a daily oral maintenance dose (D o ) of sodium phenytoin for this patient If sodium phenytoin capsules are available in mg and 3 mg strengths, what combination of these will closely approximate (d) (e) the daily dose of sodium phenytoin calculated in part (b)? One month later, the patient s regimen is to be reviewed Steady-state plasma concentrations from two different dosage rates have shown that this patient s V max equals mg kg day and her K m equals 6 mg L Calculate a new daily oral maintenance dose of sodium phenytoin What intravenous dosing rate in milligrams per day of sodium phenytoin would equal the oral dosing regimen in part (d)? Question Patient BC, the identical twin of patient AB in question, also weighing 67 kg but with a history of hypoalbuminemia, presents to the hospital with epileptiform seizures A phenytoin assay shows no phenytoin in the plasma Patient BC is stabilized by a series of short intravenous infusions of sodium phenytoin Although her seizures have been stabilized, the patient exhibits a plasma total phenytoin concentration of only 75 mg L, prompting the determination of a free (unbound) phenytoin concentration The free phenytoin concentration is 65 mg L In the absence of individual pharmacokinetic parameters, employ literature average values: maximum reaction rate (V max ) = 75 mg kg day

338 Basic Pharmacokinetics Michaelis Menten constant (K m ) = 57 mg L volume of distribution (V) = 64 L kg salt value for sodium phenytoin = 96 absolute bioavailability (F PO ) = 98 (a) (b) (c) (d) What is the free fraction of phenytoin in patient BC? Does the total (bound and free) concentration of 75 mg L represent a subtherapeutic concentration in this patient? To what total phenytoin concentration, (C p ) total, would this value of 75 mg L in patient BC correspond in a patient with normal serum albumin and a normal free fraction of phenytoin? Calculate a daily oral maintenance dose of sodium phenytoin for patient BC using the literature average values 4 Michaelis Menten elimination kinetics is due to saturation of metabolic sites; while zero-order elimination kinetics is due to partial saturation of these sites (A) (B) True False 5 For a drug eliminated by nonlinear kinetics, slight errors in the calculation of the optimal daily dose can cause disproportionate changes in the drug s average steady-state plasma drug concentration and corresponding changes in therapeutic or toxic effect (A) (B) Answers True False Answers to calculations Multiple choice and true/false questions Answer the following statements by selecting A for true and B for false or appropriate answer for multiple choice and numerical value questions A daily oral dose of a drug eliminated by Michaelis Menten kinetics will reach practical steady-state equilibrium in about 43 drug elimination half lives Question answer (a) (b) (D L ) IV = (64 L kg )(67 kg)( mg L mg L ) (96)() = 936 mg Daily oral maintenance dosage: (A) (B) True False Doubling the daily dose of a drug eliminated by Michaelis Menten kinetics will likely more than double the drug s average steady-state plasma drug concentration (A) (B) True False 3 Doubling the intravenous bolus dose of a drug eliminated by Michaelis Menten kinetics would more than double the area under the plasma drug level versus time curve but would not affect the initial plasma drug concentration, (C p ) (A) (B) True False (c) (d) D o τ = V max (C p ) [K m + (C p ) ](SF PO ) (75 mg kg )(67 kg)(5 mg L ) = (57 mg L + 5 mg L )(96)(98) = 46 mg D o = 46 mg sodium phenytoin daily Four capsules of mg Revised daily oral maintenance dose: ( Do τ ) PO = ( mg kg )(67 kg)(5 mg L ) (6 mg L + 5 mg L )(96)(98) = 533 mg D o = 533 mg sodium phenytoin daily

Problem set 8 339 (e) Daily intravenous dose equivalent to oral dose of 533 mg: ( ) Do = (533 mg)(98) = 5 mg τ IV D IV = 5 mg sodium phenytoin daily Question answer (a) (b) (c) The free fraction is given by (C p ) free (C p ) total = 65 mg L = 75 mg L In fact a free phenytoin concentration of 65 mg L is in the midrange of the therapeutic window for phenytoin and, as such, represents a useful target Since the normal free fraction of phenytoin is, the total phenytoin concentration in a normal person would be (C p ) total = 75 mg L = 5 mg L (d) To calculate a daily dose using the Michaelis Menten equation, the target (C p ) total of 75 mg L is not used Instead, the value used is the (C p ) total to which the observed concentration of 75 mg L would correspond if this patient had a normal free fraction The adjusted total phenytoin concentration, calculated above in (c), equals 5 mg L This adjusted value is used in the Michaelis Menten equation This reflects the fact that more phenytoin is available for elimination when the free fraction is higher than normal In patient BC, a dose based on a target total phenytoin concentration of 5 mg L should achieve a steady-state concentration: ( ) (C p ) total = (5 mg L ) = 75 mg L and a therapeutic free phenytoin concentration, (C p ) free = 75 mg L = 65 mg L Now, we can calculate the daily oral dose: X o τ = V max(c p ) adjusted [K m + (C p ) ](SF PO ) (75 mg kg )(67 kg)(5 mg L ) = (57 mg L + 5 mg L )(96)(98) = 46 mg Notice that this is the same daily dose as used in patient AB above, the identical twin with normal serum albumin and a normal phenytoin free fraction of We expect that 4 mg day of oral sodium phenytoin will produce a (C p ) total of 75 mg L and a (C p ) free of 65 mg L in patient BC Answers to multiple choice and true/false questions B; A; 3 A; 4 B; 5 A