Re: The double peak

Dear Chenguang, You wrote: "The ka of the first peak is greater than that of the second, and the ke of the first is also greater than the second." You seem to mix absorption and disposition here. I assume you mean that the plasma concentrations decline faster after the second peak than after the first. One reason for this observation would be that the two absorption processes are not of the same order. I.e. absorption might be more like a first-order process for the first peak and more like a zero-order process after the second peak. This might be caused by an absorption window for the 2nd peak. In your model, you could dose a bolus into the gut compartment with or without a short lag-time and then you dose a zero order infusion with a longer lag-time into your central compartment. This should work and does not require differential equations. ADVAN4 and TRANS3 or 4 will work and will run much faster than the ADVAN6 model you have. Unless you have a drug which reaches concentrations in the liver during the absorption which are high enough to inhibit systemic metabolism, disposition should be independent from the absorption process. I would suggest you use your favorite simulation tool and try if you can generate profiles of the same shape as the ones you are seeing in your study to find out, if a candidate model may be appropriate. Hope it helps. Best regards Jurgen ----------------------------------------------- Jurgen Bulitta, PhD, Post-doctoral Fellow Pharmacometrics, University at Buffalo, NY, USA Phone: +1 716 645 2855 ext. 281, [EMAIL PROTECTED] ----------------------------------------------- -----Ursprüngliche Nachricht----- Von: "Chenguang Wang" <[EMAIL PROTECTED]> Gesendet: 16.11.07 16:35:45 An: [email protected] Betreff: [NMusers] The double peak Dear Nonmem users, I am working with a double peaks PK model. I refered to a recent question about this topic and tried the double absorption compartment model suggested by somebody. Howerver, in my real data, the Ka and Ke for each peak are different. The ka of the first peak is greater than that of the second, and the ke of the first is also greater than the second. Therefore, the profile has a sharp peak at first, then a flat peak after that. If I use the double absorption compartment model, it can give me two different Ka, but can't give me two different Ke in different time. Then I wanted to add a periphal compartment to solve this problem. My idea is after the admin, part of drug(F1) gets into the plasma quickly(Ka1) then quickly goes to the periphal. So there will be a big Kcp (from central to periphal). In the same time, drug eliminates to outside from plasma in a speed of Ke, which is very small. After a period of time, the drug in periphal begins to flow back to central in a same speed of Kcp( Kpc=Kcp). This will retain the drug concentration in the plasma and prevent the fast drop after the second peak in the c-t profile. Also, a sencond part of drug(F2) will be absorbed with a slower Ka2 after a proper lag time. Finally, the drug will eliminate outside in the speed of Ke, which is very slow. In order to do this, I use lag time between periphal and central. I am not sure if it is permitted by NONMEM, but at least I didn't find error message when I did simulation. Also, I don't know if my model can really work as I expected. But when I did the estimation, it crashed. Would someone have some good idea about this or can tell me if there is any wrong in my script? Thank you in advance! My model is : $DATA ****.csv $SUBROUTINES ADVAN6 TRANS1 TOL=3 $MODEL NCOMP=4 COMP=(ABS1) COMP=(ABS2) COMP=(CENTRAL) COMP=(PERI) $PK K13 =THETA(1) K23 =THETA(2) KE =THETA(3)*EXP(ETA(1)) ALAG2=THETA(4)*EXP(ETA(2)) F1 = THETA(5) F2 = 1- F1 K34 =THETA(6)*EXP(ETA(3)) K43 =THETA(7)*EXP(ETA(4)) ALAG4=THETA(8)*EXP(ETA(5)) $DES DADT(1)=-K13*A(1) DADT(2)= -K23*A(2) DADT(3)= K13*A(1)+K23*A(2)-KE*A(3)-K34*A(3)+K43*A(4) DADT(4)= K34*A(3)-K43*A(4) $ERROR IPRED= F Y = IPRED * (1+ERR(1)) + ERR(2) IRES = DV - F IWRES = IRES/IPRED $THETA 1 FIX ; K13 0.6 FIX ; K23 (0.0001,0.001,0.01) ; KE (20,40,60) ; ALAG2 0.7 FIX ; F1 (0.5,0.75,1) ; K34 (0.5,0.75,1) ;K43 (20,40,60) ;ALAG4 $OMEGA 0.2 0.1 0.1 0.1 0.1 $SIGMA 0.1 0.1 $ESTIMATION METHOD=1 PRINT=1 MAXEVAL=9999 NOABORT SIGDIGITS=3 POSTHOC INTERACTION MSFO=msfo.outputfile ................ Maurice