One approach is to use the population values for phenytoin. With this method we would use the population values of Vm = 7 mg/kg/day and Km = 5 mg/L. Aiming at 15 mg/L for with a patient weight of 80 kg, the equation
Equation XX-3 Dosing Rate versus Average Cp>
Probably better to start out low since toxicity is more probable above 20 mg/L.
That is after giving a continuous dose regimen to steady state, measure plasma concentration and adjust dose. For example if after 420 mg/day, is 20 mg/L then a downward adjustment would be necessary. If we assume that the Km is close to the average value of 5 mg/L we can estimate Vm from the equation above
thus a new dose rate can be calculated
approximately 400 mg/day. Note: A reduction in dose of 20 mg/day (5 %) is calculated to give a 5 mg/L change (25 %) in . Another approach can be describe using the 'graph' (nomogram) shown below.
In Figure XX-6, Line A represents Cpss = 8 mg/L on 300 mg/day (70 kg = 4.3 mg/kg/day). Line B was drawn to achieve a new Cpss = 15 mg/L with a dose of 5.2 mg/kg/day (= 364 mg/day)
The graph is used by plotting the line described by the current Cpss and R on the graph, marking a point in the middle of the contour. From that point draw a line to the desired Cpss, the value on the vertical axis gives the required dose rate, R.
Figure 93, Nomogram for Phenytoin Dosing
If we already have two plasma concentrations after two dose rates we can solve the equation
using simultaneous equations.
With = 8.0 mg/L and = 27.0 mg/L for R1 = 225 mg/day and R2 = 300 mg/day
225 * Km + 225 * 8 = 8 * Vm (1)
300 * Km + 300 * 27 = 27 * Vm (2)
or multiplying (1) x 300
300 * 225 * Km + 300 * 225 * 8 = 300 * 8 * Vm (3)
and multiplying (2) x 225
300 * 225 * Km + 300 * 225 * 27 = 225 * 27 * Vm (4)
subtracting (4) - (3)
300 * 225 * (27 - 8) = (225 * 27 - 300 * 8) * Vm
With these Vm and Km values we can now calculate the next dosing regimen to try.
There are a number of graphical methods which have been described for when you have data from more than two dosing intervals. Basically these rely on converting the equations mentioned above into a straight line form which can be plotted to give the Vm and Km as a function of the intercept and/or slope.
Want more practice with this type of problem!
Copyright 2001 David W.A. Bourne