steady state for modified release [Regulatives / Guidelines]

posted by Helmut Homepage – Vienna, Austria, 2019-09-23 12:04 (1732 d 08:54 ago) – Posting: # 20639
Views: 4,652

Hi Kelen,

❝ I would like to ask help for understand the objetive of the steady state study when you are comparing a MR formulation with a IR formulation.

I don’t know the current state of affairs in Brazil. I don’t speak Portuguese and the English site is practically useless. Sorry.

❝ The objective would be compare the effect of the accumulation?

That a drug accumulates is trivial. The question is only, how much and whether the drug follows linear PK. That’s already established by the originator. A study in a paired design; SD → saturation → (pseudo-) steady state.If you think about a generic drug, no. Requirements according to the guidelines:

❝ Is there any reason to conduct the study only with de highest dose?

EMA again: In general the highest dose because – if linear PK is established – the other strengths can be waived. Makes sense. “Perfect” dose-proportionality (slope β of the power model = 1) is rare. Metabolic saturation is more common than induction. Furthermore, the GL asks for the “most sensitive strength”. Might be difficult to know beforehand. If case of nonlinear PK it might be the lowest strength as well.

Another trap (EMA GL Section 5.1.1):

Fluctuation in drug concentrations should be studied following repeated dosing. Unless otherwise justified, the modified release product should produce similar or less fluctuations as the immediate release product.

(my emphasis)
That’s a one-sided test (‘non-superiority’) and not the two-sided test for equivalence. For the same variability and T/R-ratio <1 you need a lower sample size. Example:

sampleN.TOST(CV = 0.2, theta0 = 0.95, theta1 = 0.80, theta2 = 1.25)

+++++++++++ Equivalence test - TOST +++++++++++
            Sample size estimation
Study design: 2x2 crossover
log-transformed data (multiplicative model)

alpha = 0.05, target power = 0.8
BE margins = 0.8 ... 1.25
True ratio = 0.95,  CV = 0.2

Sample size (total)
 n     power

20   0.834680

sampleN.noninf(CV = 0.2, theta0 = 0.95, margin = 1.25)

++++++++++++ Non-inferiority test +++++++++++++
            Sample size estimation
Study design: 2x2 crossover
log-transformed data (multiplicative model)

alpha = 0.025, target power = 0.8
Non-inf. margin = 1.25
True ratio = 0.95,  CV = 0.2

Sample size (total)
 n     power

12   0.863279

Only if the T/R-ratio is >1 it will get nasty. Try

sampleN.TOST(CV = 0.2, theta0 = 1.1, theta1 = 0.80, theta2 = 1.25)
sampleN.noninf(CV = 0.2, theta0 = 1.1, margin = 1.25)

Some people decided to ignore the comparison of %PTF and opt for the so-called “bracketing approach” instead, where Cmin is assessed for ‘non-inferiority’ (surrogate for efficacy) and Cmax for ‘non-superiority’ (surrogate for safety).

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