RE: Determining metabolite clearance fraction

From: "Leonid Gibiansky" leonidg@metrumrg.com Subject: RE: [NMusers] Determining metabolite clearance fraction Date: Mon, November 29, 2004 11:19 am Paul, The way how it is written in the message, the system is not identifiable: the code should ensure that FM1+FM2+FM3 = 1. You may try to use DADT(2)=A(1)*KA-A(2)*K; EQ FOR PARENT rather than DADT(2)=A(1)*KA-A(2)*K*(FM1+FM2+FM3) but you may end up with FM1+FM2 > 1. In the other variant, if you use FM3=1-(FM1+FM2) (is this how it was coded? FM3=1-FM1+FM2 is incorrect) you must ensure that FM3 > 0. I would do the following: FM1=THETA(3)/(1+THETA(3)+THETA(5) ) FM2=THETA(5)/(1+THETA(3)+THETA(5) ) ; FM3=1/(1+THETA(3)+THETA(5) ) do not put this into the code, this is implicit DADT(2)=A(1)*KA-A(2)*K; use FM1+FM2+MF3=1 DADT(3)=A(2)*K*FM1-A(3)*KM1; EQ FOR 4OH METABOLITE "COMPARTMENT" DADT(4)=A(2)*K*FM2-A(4)*KM2; EQ FOR NDESTAM METABOLITE "COMPARTMENT" Alternative is to use K23=THETA(.) K24=THETA(.) K20=THETA(.) ; direct (or via other metabolites) elimination DADT(2)=A(1)*KA-A(2)*(K23+K24+K20); use FM1+FM2+MF3=1 DADT(3)=A(2)*K23-A(3)*KM1; EQ FOR 4OH METABOLITE "COMPARTMENT" DADT(4)=A(2)*K24-A(4)*KM2; EQ FOR NDESTAM METABOLITE "COMPARTMENT" The drug effect at week 12 and random effects can be added to all or some of K23,K24,K20. Fractions F1, F2, F3 can be re-computed from Kij values. This should solve the problem of F1+F2+F3=1. Leonid