Bioequivalence and Bioavailability Forum

Dear Helmut!

❝ Reach for the stars, even if you have to stand on a cactus. Susan Longacre

Uno puede estar mirando las estrellas y al mismo tiempo verse la punta de las pestañas (Julio Cortázar)

I am very grateful for your answers and the references provided! I've found some of them and will search for others although for some reason I have strong doubts that our local library has books on pharmacokinetics on german printed in 50th

❝ Not only that. As a rule of thumb at \(\small{MRT}\) ~⅔ of the drug is eliminated. It is very useful comparing PK models with different compartments. The slowest t_½ might be misleading (see there, slides 24–28). There is a big problem with it. To get a reliable estimate of AUC one has to cover 95% (!) of AUC_0–∞ (note that I’m not taking about BE but hard-core PK). For AUMC is should be 99%. I’m quoting Les Benet. Don’t blame me.

I was wondering from where such a rule of thumb was going and integrated the area for simple exponential elimination. It turns out that at MRT (1-exp(-1))~0,632 of the drug is eliminated for IV and slightly lower for EV (so the rule of pinky is 0,632 versus the rule of thumb (2/3=0,(6))

As for physics there exists inaccuracy in the considerations on the slide "Excursion to Hydrodynamics" . "Same proportions is emptied in the same time interval" is true only when you are solving school problems with a pool. Exactly the unexpired volume leaked depends on the form of the vessel. For the cylindric vessel for example water height and thus the volume is proportional to t². If you want to have a constant proportion you need a vessel with a form of parabola x⁴ that is clepsydra or consider Mariotte’s bottle.

❝ What I learned: The variability of VRT sucks. Not surprising cause we have \(\small{t^2}\) and \(\small{C^2}\) in it.

I've calculated C_c for several real studies according to simple linear trapezoidal rule:
$$C_c=\frac{1}{3}\frac{\sum\limits_{i}(t_{i+1}-t_i)(C^{2}_{i}+C^{2}_{i+1}+C_{i}\cdot C_{i+1})}{\sum\limits_{i}(t_{i+1}-t_{i})(C_{i+1}+C_i)}\tag{7}.$$ Although it has C² in it, it's variability was always lower than that of C_max, but I should've check it more carefully.

Dear ElMaestro!

❝ I think F may be in its own right also included on your list of crackpot ideas from the odd sock drawer? PMDA have a sentence about it in their guidance. "If F can be calculated by deconvolution, F may be used instead of AUC"

Thank you! I will definitely add it to my collection of weird PK parameters! Need to know more about deconvolution...

Dear mittyri!

❝ By the way I couldn't follow Yamaoka's logic regarding that magic cut-off errors. How did they find it?

I am puzzled with the same question. How did they calculated the time to reach 5% of C_max?
I slightly modified the Helmut's considerations on the article of Scheerans et al. (2008)
Let us consider a one-compartment model with first-order absorbtion of the form:
$$C=\frac{A}{k_a-k_e}(\textrm{e}^{-k_et}-\textrm{e}^{-k_at}) \tag{6}$$
then residual area (1-AUC_0-t/AUC_0-inf) should be as follows:
$$AUC_{resid}(x,t)=\frac{x\textrm{e}^{-t\cdot k_e}-\textrm{e}^{-x\cdot k_e*t}}{x-1},\quad
\textrm{where}\qquad x=\frac{k_a}{k_e}.$$ Let n define the ratio of t to T_1/2,e, then
$$AUC_{resid}(x,n)=\frac{x\cdot2^{-n}-2^{-nx}}{x-1}\sim \frac{x\cdot 2^{-n}}{x-1}\qquad for\qquad nx>>1. \tag{8}$$
In order to estimate the duration of sampling to achieve specific AUC_resid we can use the simplifyed formula
$$n=\textrm{log}_2\left(\frac{x}{(x-1)AUC_{resid}}\right) \tag{9},$$
for example for x=2 and AUC_resid=1% the duration should be n=7.64 T_1/2, for AUC_resid=20% the duration should be n=3.32 T_1/2 (the exact value is 3.24.)
In particular, $$AUC_{resid}(T_{1/2},x)=\frac{x-2^{1-x}}{2(x-1)};\quad AUC_{resid}(T_{max},x)=\frac{x^{\frac{2-x}{1-x}}-x^{\frac{x}{1-x}}}{(x-1)};\quad AUC_{resid}(2T_{max},x)=\frac{x^{\frac{3-x}{1-x}}-x^{\frac{2x}{1-x}}}{(x-1)}. $$
AUC_resid(T_max,x) is a monotone function of x limited from 2/e (0.736) to 1;
AUC_resid(2T_max,x) is a monotone function of x limited from 3/e³ (0.406) to 1.