Medical Pharmacology Topics   

Pharmacokinetics: Dosing

Dosages are planned to achieve a steady-state plasma concentration where ther rate of administration equals the rate of elimination.

Steady-state Kinetics

The steady-state kinetics of a drug in the body yield a time course graph approaching an increasing rectangular hyperbola.

A maintenance dose can be obtained from the drug clearance and desired steady-state plasma concentration:

elimination  =  administration

        Cl Pss = dose  x   1
                     time       F

Þ dose = Cl Pss t F

The minimum, maximum and average steady-state concentrations can also be calculated:

Pss mean  =  dose F  =  dose F t1/2
                     t Cl           t Vd 0.693 

Pss max  =        dose F    
                    (1 - Pt/P0) Vd

Pss min  =  (Pss max) (Pt/P0)

The steady-state concentration should be achieved in about four half-times. To achieve Pss faster, one loading dose may be given before starting the maintenance dosage:

loading dose  =  Vd Pss
                             F

If the loading dose is very close to a toxic dose, it should be divided into 2-3 doses.

Fluctuation Allowance

The plasma concentration of a drug will fall by 50% if it is given at intervals equal to the elimination half-time. But in most cases this is not desirable, plasma concentration should fluctuate around a narrower range. The timing of dosing can be calculated using the first-order equation. For example, if the plasma concentration should not fall more than 10%:

ln (9/10)  =  -ke t   Þ   t = 0.105 / ke

As long as the total doses within a half-life interval are the same, steady-state concentration will be achieved, but a narrower range of fluctuation will come frm smaller doses at shorter time intervals. <more, see notes>

Impaired Renal Function

There is a direct correlation between the elimination half-life for drugs excreted largely by the kidney and glumerular filtrartion rate (GFR). If a patient has a GFR half of normal (normal ~ 120), the elimination half-life will be about double. The same calculation may be applied to the rrenal component of a drug that is eliminated by both the kidneys and the liver (there is no similar calculation yet available for the hepatic component):

t1/2  =  0.693
             ke renal  +  ke hepatic

Clearance will also be reduced, for example by 50% in a patient with half the normal GFR.

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