Multiple Oral Dose Administration. Pedro Baratta
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1 Multiple Oral Dose Administration Pedro Baratta
2
3 However Cpmin can be more easily determined at t = 0 or t = t. Thus at t = 0 and n ->. Plot Cp Versus Time after a Single Dose showing Possible Time of Second Dose
4 This can be further simplified if we assume that the subsequent doses are given after the plasma concentration has peaked and e- ka * t is close to zero. That is the next dose is given after the absorption phase is complete. Cpmin then becomes:-
5 The relationship between loading dose and maintenance dose and thus drug accumulation during multiple dose administration can be studied by looking at the ratio between the minimum concentration at steady state and the concentration one dosing interval (t) after the first dose.
6 Which can be simplified to give: We can further simplify Equation, if we assume that ka >> kel: then (ka - kel) approximatley equal to ka and thus approximately = 1.
7 Another very useful concentration term for the calculation of oral dosing regimens is the average plasma concentration,, during the dosing interval at steady state. This term is defined as the area under the plasma concentration versus time curve during the dosing interval at steady state divided by the dosing interval.
8
9 By integrating the equation for plasma concentration at the plateau, between t = 0 and t = t gives:
10 An interesting result of this equation is that we get the same average plasma concentration whether the dose is given as a single dose every t dosing interval or is subdivided into shorter dosing intervals. For example 300 mg every 12 hours will give the same average plasma concentration as 100 mg every 4 hours. Of course, the difference between the maximum and minimum plasma concentration will be larger in the case of the less frequent dosing. For example F = 1.0; V = 30 liter; t 1/2 = 6 hours or kel = 0.693/6 = hr -1. We can now calculate the dose given every 12 hours required to achieve an average plasma concentration of 15 mg/l.
11 =
12 We could now calculate the loading dose R = e -kel * [[tau]] = e x 12 = 0.25
13 To get some idea of the fluctuations in plasma concentration we could calculate the Cpmin value.
14 Therefore the plasma concentration would probably fluctuate between 7 and 23 mg/l (very approximate) with an average concentration of about 15 mg/l. [23 = 15 + (15-7), i.e. high = average + (average - low), very approximate!].
15 As an alternative we could give half the dose, 312 mg, every 6 hours give:-
16 The Cp would be the same Thus the plasma concentration would fluctuate between about 10.4 to 20 with an average of 15 mg/l.
17 Superposition Principle Applies when all disposition processes are linear. That is, distribution, metabolism, and excretion (DME) processes are linear or first order. Thus, concentrations after multiple doses can be calculated by adding together the concentrations from each dose. For example, calculate drug concentration at 24 hours after the first dose of 200 mg. The second dose of 300 mg was given at 6 hours and the third dose of 100 mg at 18 hours.
18
19 Non-uniform dosing intervals The calculations we have looked at consider that the dosing intervals are quite uniform, however, commonly this ideal situation is not adhered to completely. Dosing three times a day may be interpreted as with meals, the plasma concentration may then look like the plot in Figure 60. The ratio between Cpmax and Cpmin is seven fold (8.2/1.1 = 7.45) in this example.
20
21 However this regimen may be acceptable if 1) the drug has a wide therapeutic index 2) there is no therapeutic disadvantage to low overnight plasma concentrations, e.g., analgesic of patient stays asleep.
22 Exercicio 1 A drug is to be given orally every six hours to achieve an average concentration of 15 mg/l. Calculate the dose (F = 0.90) required if t 1/2 = 11 hours and V = 23 L. Estimate the peak and trough concentrations after steady state is reached.
23 t 1/2 = 11 hr means the kel = 0.06 hr -1 Thus, Dose = 15 x 23 x 0.06 x 6 / 0.9 = 145 mg. Give 150 mg to achieve a of 15.5 mg/l Cp min = (150 / 23) x (0.69 / (1-0.69)) = 12.8 mg/l By rough estimation Cp max can be calculated as 18.2 mg/l ( ( ))
24 Exercicio 2 Exemplo IV Calculate an appropriate dosing regimen for the following male patient; age = 56 years, weight = 79 kg, serum creatinine = 2.0 mg/ 100 ml (mg%). With this drug the kel is a function of creatinine clearance; kel = 0.04 hr -1 with CLCr = 20 ml/min and kel = 0.12 hr -1 with CLCr = 75 ml/min. Apparent volume of distribution is calculated as 0.28 L/kg. Develop a dosing regimen to keep the peak concentration close to but below 6 µg/ml and the trough concentration below 1.0 µg/ml.
25 CL CR = (140-56) x 79 / (72 x 2.0) = 46.1 ml/min By plotting kel versus CL CR it is possible to interpolate a value of kel for the patient
26 From this graph or by calculations the kel patient = hr -1
27 R = Cp min /Cp max = 1/6 = e -kel*tau Taking ln of both sides and solving for tau gives a value of 23.0 hr. Adjusting this upwards (to avoid extending the concentration range) give a new tau value of 24 hours. The new R value is now e x 24 = Maintenance dose can be calculated from Cp max * V * (1 - R) = 6 x 0.28 x 79 x ( ) = 112 mg (probably give 100 mg which would produce a Cp max of 5.34 mg/l and Cp min of mg/l). The required loading dose can be calculated as Cp max * V = 6 x 0.28 x 79 = 133 mg (125 mg would produce a Cp max of 5.65 mg/l).
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