Different ka and f? [Power / Sample Size]

posted by d_labes  – Berlin, Germany, 2012-11-28 13:10 (4547 d 14:38 ago) – Posting: # 9614
Views: 16,790

Dear Helmut,

❝ Example (one-compartment, no lag time, ka 0.6 h-1 (R) and 3.905 h-1 (T), f 1 (R) and 0.85552 (T); linear trapezoidal rule for simplicity):

❝ ...

......................       T/R

❝ ——————————————————————————————————

❝ AUCt     748.96   762.32   101.78%

❝ AUC∞     839.85   839.93   100.01%

❝ ...

❝ Kel        0.10     0.10     ← identical elimination


Different f but ratio of AUC ~ 1 with the same elimination :confused:.

This seems to me like the discovery of a new fundamental law in pharmacokinetics :cool: or the other way round like a violation of fundamental pharmacokinetic relationships.

AFAIK is the one-compartiment AUC formula
    AUCpo = f*C0/kel with C0=D/V
where D is the dose, V the volume of the central compartment. If kel, D and V are the same you can't get the same AUC if your f = fraction absorbed is different.

BTW1: Your data are from f=1 for both curves as a recalculation via the Bateman function easily shows (R code):
ct <- function(time,f,c0,ka,ke){ f*c0*(ka/(ka-ke))*(exp(-ke*time)-exp(-ka*time)) }
time <- c(0.0, 0.25, 0.5, 0.75, 1, 1.5, 2, 2.5, 3,  3.5, 4, 6, 9, 12, 16, 24)
round(ct(time, f=1, c0=83.35, ka=0.6, ke=0.1),2)
round(ct(time, f=1, c0=83.35, ka=3.905, ke=0.1),2)


BTW2: c0 is an educated guess. A non-linear fit gives slight different values for both curves.

BTW3: This doesn't invalidate your demonstration of different ratios for the different PK metrics, of course.

Regards,

Detlew

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