martin ★★ Austria, 2017-05-08 21:50 (2538 d 13:56 ago) Posting: # 17322 Views: 3,845 |
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Dear all, I would like to open again a new thread regarding endogenous compounds in BE assessments. Simple case: - assuming no circadian rhythm / temporarily constant - baseline subtraction method (e.g. median of at several pre-dose samples) Several guidelines mention that a negative pre-dose corrected value should be set to zero. However, what about subsequent levels? I think setting also all concentrations after the first negative to zero would make some sense as otherwise we just have “noise” in the AUC (i.e. increase variability) and it’s rather impossible to get a reliable estimate for lambda_z (for calculation of AUC0-inf). I think this should be also in-line with guidelines as the method for baseline subtraction needs to be specified a-priori. Best regards and looking forward to thoughts from more experienced members in this forum Martin PS.: premise of stable endogenous could be verified by looking at adjusted concentrations after the first negative (inclusive) adjusted concentration: should be symmetrically distributed around zero. |
Helmut ★★★ Vienna, Austria, 2017-05-08 23:10 (2538 d 12:35 ago) @ martin Posting: # 17324 Views: 3,257 |
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Hi Martin, ❝ Several guidelines mention that a negative pre-dose corrected value should be set to zero. However, what about subsequent levels? ❝ I think setting also all concentrations after the first negative to zero would make some sense […] this should be also in-line with guidelines as the method for baseline subtraction needs to be specified a-priori. Well, this is exactly what I have done for ages. never got a deficiency letter. ❝ PS.: premise of stable endogenous could be verified by looking at adjusted concentrations after the first negative (inclusive) adjusted concentration: should be symmetrically distributed around zero. Really? You explained that to me so may times; excuse my walnut-sized brain. Baseline (C0) = lognormal. Post-dose (C1) = lognormal. C0–C1 = normal?
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martin ★★ Austria, 2017-05-08 23:19 (2538 d 12:26 ago) @ Helmut Posting: # 17325 Views: 3,231 |
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Hi Helmut, Thank you very much for your response - much appreciated. Regarding distributions - thank you for the reminder & sorry for not being clear We look at the end of the profile assuming C0=C1 this would lead to
C0 <- rlnorm(n=1e6, meanlog=log(1)-0.5*log(0.3^2+1), sdlog=sqrt(log(0.3^2+1))) best regards Martin |