saturable metabolism of 5FU

2 messages 2 people Latest: May 19, 2012

Re: saturable metabolism of 5FU

From: Nick Holford Date: May 19, 2012 technical
Marcus, In the case of first-order elimination of 5FU you can make an assumption of a constant fraction of total 5FU clearance describing the formation of its metabolite. But a constant fraction is not compatible with a mixed order process for formation of a metabolite. As the 5FU conc rises the fraction will decrease because the mixed order clearance to metabolite will decrease while the 5FU clearance by a first-order other process (not to the metabolite) will remain unchanged. You have to consider estimating another parameter (CLOTHER) describing the other clearance (not to the metabolite) or fixing CLOTHER to zero which assumes all of the parent is converted to metabolite. See NM-TRAN code below. I prefer to refer to the model predictions by meaningful names rather than use NONMEM's default and essentially meaningless name F. In the code below different names are used for parent and metabolite concs in the $DES and $ERROR blocks because NONMEM does not allow a variable to be created in $DES with the same name as one being created in $ERROR. I like to prefix the $ERROR name with 'D' when I refer to the same quantity in $DES. Best wishes, Nick $PK VP = THETA(1) ; volume of parent CLOTHER =THETA(2) ; clearance of parent by first-order other process VMAX =THETA(3) ; Vmax of parent to metabolite mixed order process KM =THETA(4) ; Km of parent to metabolite mixed order process VM = THETA(5) ; volume of metabolite CLM =THETA(6) ; clearance of metabolite $DES DCP =A(1)/VP ; parent i.e. 5FU DCM =A(2)/VM ; metabolite i.e. 5FU-H2 CLP2M =VMAX/(KM+DCP) ; clearance from parent to metabolite DADT(1) = - (CLP2M + CLOTHER) *DCP DADT(2) = CLP2M*DCP - CLM*DCM $ERROR CP =A(1)/VP ; This is the same as DCP in the $DES block CM =A(2)/VM ; This is the same as DCM in the $DES block IF (CMT.EQ.1) THEN Y=LOG(CP) + EPS(1) ENDIF IF (CMT.EQ.2) THEN Y=LOG(CM) + EPS(2) ENDIF
Quoted reply history
On 18/05/2012 11:23 p.m., markus joerger wrote: > dear Community, > > I am modeling 5FU --> 5FU-H2, that is potentially saturable. The linear model runs smoothly, with a fixed proportion of 85% of the parent going to the metabolite. I wonder how the coding should be if I introduce saturable metabolism from 5FU to 5FU-H2. This model does not minimize. Should the coding for saturable metabolism (again with a 85% proportion going from 5FU to 5FU-H2) look like: ? > > V1=THETA(1)*EXP(ETA(1)) > VM=THETA(2)*EXP(ETA(2)) > KM=THETA(3)*EXP(ETA(3)) > CL2=THETA(4)*EXP(ETA(4)) > V2=THETA(5)*EXP(ETA(5)) > > S1=V1/1000 > S2=V2/1000 > FM=0.85 > K20=CL2/V2 > > $DES > C1=A(1)/S1 > C2=A(2)/S2 > DADT(1)=-C1*VM/(KM+C1) > DADT(2)=C1*FM*VM/(KM+C1)-K20*A(2) > > LINEAR MODEL: > ------------------------------------------------------------------------------------------------------------------------------------ > $INPUT C=DROP ID OCC TIME AMT RATE DV CMT MDV EVID AGE SEX HEP OXA > IRI BEV WT BSA ECOG A496G DPD4 DPD5 CCL > > $DATA 08088_6.CSV IGNORE=C > $SUBROUTINES ADVAN5 > $MODEL COMP=(CENTRAL, DEFDOSE) > COMP=(5FUH2) > > $ABBREVIATED DERIV2=NOCOMMON > > $PK > FLAG1=0 > FLAG2=0 > IF(OCC.EQ.1) FLAG1=1 > IF(OCC.EQ.2) FLAG2=2 > > TVCL1=THETA(1)*THETA(5)**SEX > CL1=TVCL1*EXP(ETA(1)+FLAG1*ETA(2)+FLAG2*ETA(3)) > V1=THETA(2)*EXP(ETA(4)) > TVCL2=THETA(3)*THETA(6)**SEX > CL2=TVCL2*EXP(ETA(5)) > V2=THETA(4)*EXP(ETA(6)) > > S1=V1/1000 > S2=V2/1000 > FM=0.85 > K10=(1-FM)*CL1/V1 > K12=FM*CL1/V1 > K20=CL2/V2 > > $ERROR CALLFL=0 > IPRE=-3 > IF(F.GT.0) IPRE=LOG(F) > PROP=0 > IF(CMT.EQ.1)PROP=EPS(1) > IF(CMT.EQ.2)PROP=EPS(2) > Y=LOG(F)+PROP > W=LOG(F) > IRES=DV-IPRE > IWRES=IRES/W > > $THETA > (0, 152) ;CL1 > (0, 43.3) ;V1 > (0, 144) ;CL2 > (0, 122) ;V2 > (0,1) ;SEX > CL-5FU > (0,1) ;SEX > CL-5FU-H2 > > $OMEGA > 0.04 ;IIV CL1 > $OMEGA BLOCK(1) > 0.004 ;IOV OCC1; CL-5FU > $OMEGA BLOCK(1) SAME ;IOV OCC2; CL-5FU > $OMEGA > 0.04 ;IIV V1 > 0.04 ;IIV CL2 > 0.04 ;IIV V2 > > $SIGMA > 0.329 ;EPS COMP1 > 0.136 ;EPS COMP2 > > -- > Markus Joerger MD PhD > Associate Professor > Medical Oncology&Clinical Pharmacology > Cantonal Hospital > 9007 St. Gallen > Switzerland > [email protected] <mailto:[email protected]> > [email protected] <mailto:[email protected]> > Phone: +41-765591070 > Fax: +41-714946325 -- Nick Holford, Professor Clinical Pharmacology First World Conference on Pharmacometrics, 5-7 September 2012 Seoul, Korea http://www.go-wcop.org Dept Pharmacology& Clinical Pharmacology, Bldg 505 Room 202D University of Auckland,85 Park Rd,Private Bag 92019,Auckland,New Zealand tel:+64(9)923-6730 fax:+64(9)373-7090 mobile:+64(21)46 23 53 email: [email protected] http://www.fmhs.auckland.ac.nz/sms/pharmacology/holford

RE: saturable metabolism of 5FU

From: Martin Bergstrand Date: May 19, 2012 technical
Dear Markus, First of all it is contradictory to both assume a fix fraction (FM) of substance converted to 5FU-H2 and at the same say that hat pathway is saturable. Given that the alternative pathway isn't saturable in the exact same way the fraction FM will be dependent on the concentration of 5FU. If you leave the FM parameter out of the equation for the linear model you will have the exact same model fit but instead of estimating CL of 5FU-H2 you will estimate a parameter that is equal to CL of 5FU-H2 over FM. This way you do not need to make any assumption about FM. ;[1] Non saturable elimination of 5FU and formation of 5FU-H2 K10 = 0 K12 = CL1/V1 K20 = CL2/V2 This model can be compare to a few alternative models that corresponds to different hypothesis regarding saturable metabolism. These models are not strictly nested hierarchical models but in my mind at least you should still be able to judge them as if they were that (when KM>>C1 the models are highly similar). ;[2] Saturable elimination of 5FU --> 5FU-H2 (alternative elimination path) C1 = A(1)/S1 DADT(1) = -C1*VM/(KM+C1) DADT(2) = C1*VM/(KM+C1)-K20*A(2) ;[3] Saturable metabolism of 5FU --> 5FU-H2 and an alternative non-saturable elimination path for 5FU. CL1A = THETA(6)*EXP(ETA(6)) ; Alternative elimination path CL for 5FU K10 = CL1A/V C1 = A(1)/S1 DADT(1) = -C1*VM/(KM+C1) -K10*A(1) ; Same as: -C1*VM/(KM+C1) -CL1A*C1 DADT(2) = C1*VM/(KM+C1)-K20*A(2) ; FM = (C1*VM/(KM+C1))/(CL1A*C1) ; FM as a function of C1 ;[4] Non-saturable metabolism of 5FU --> 5FU-H2 and an alternative saturable elimination path for 5FU. CL1M = THETA(6)*EXP(ETA(6)) ; CL1M K10 = CL1M/V C1 = A(1)/S1 DADT(1) = -C1*VM/(KM+C1) -K10*A(1) ; Same as: -C1*VM/(KM+C1) -CL1M*C1 DADT(2) = K10*A(1)-K20*A(2) ; FM = (CL1M*C1)/(C1*VM/(KM+C1)) ; FM as a function of C1 ;[5] Non-saturable metabolism of 5FU --> 5FU-H2 and an saturable elimination of 5FU-H2. C2 = A(2)/S2 DADT(1) = -K10*A(1) DADT(2) = K10*A(1) -C2*VM/(KM+C2) The above scenarios [2-5] can further be combined to account for more than I saturable elimination path. For running the differential eq. models I would recommend looking into ADVAN13 that in my experience is the best performing differential eq. solver in NONMEM and settings for TOL, SIGL and SIGN (see NONMEM 7 userguide). Kind regards, Martin Bergstrand, PhD Pharmacometrics Research Group Dept of Pharmaceutical Biosciences Uppsala University Sweden [email protected] Visiting scientist: Mahidol-Oxford Tropical Medicine Research Unit, Bangkok, Thailand Phone: +66 8 9796 7611
Quoted reply history
From: [email protected] [mailto:[email protected]] On Behalf Of markus joerger Sent: den 19 maj 2012 04:23 To: [email protected] Subject: [NMusers] saturable metabolism of 5FU dear Community, I am modeling 5FU --> 5FU-H2, that is potentially saturable. The linear model runs smoothly, with a fixed proportion of 85% of the parent going to the metabolite. I wonder how the coding should be if I introduce saturable metabolism from 5FU to 5FU-H2. This model does not minimize. Should the coding for saturable metabolism (again with a 85% proportion going from 5FU to 5FU-H2) look like: ? V1=THETA(1)*EXP(ETA(1)) VM=THETA(2)*EXP(ETA(2)) KM=THETA(3)*EXP(ETA(3)) CL2=THETA(4)*EXP(ETA(4)) V2=THETA(5)*EXP(ETA(5)) S1=V1/1000 S2=V2/1000 FM=0.85 K20=CL2/V2 $DES C1=A(1)/S1 C2=A(2)/S2 DADT(1)=-C1*VM/(KM+C1) DADT(2)=C1*FM*VM/(KM+C1)-K20*A(2) LINEAR MODEL: ---------------------------------------------------------------------------- -------------------------------------------------------- $INPUT C=DROP ID OCC TIME AMT RATE DV CMT MDV EVID AGE SEX HEP OXA IRI BEV WT BSA ECOG A496G DPD4 DPD5 CCL $DATA 08088_6.CSV IGNORE=C $SUBROUTINES ADVAN5 $MODEL COMP=(CENTRAL, DEFDOSE) COMP=(5FUH2) $ABBREVIATED DERIV2=NOCOMMON $PK FLAG1=0 FLAG2=0 IF(OCC.EQ.1) FLAG1=1 IF(OCC.EQ.2) FLAG2=2 TVCL1=THETA(1)*THETA(5)**SEX CL1=TVCL1*EXP(ETA(1)+FLAG1*ETA(2)+FLAG2*ETA(3)) V1=THETA(2)*EXP(ETA(4)) TVCL2=THETA(3)*THETA(6)**SEX CL2=TVCL2*EXP(ETA(5)) V2=THETA(4)*EXP(ETA(6)) S1=V1/1000 S2=V2/1000 FM=0.85 K10=(1-FM)*CL1/V1 K12=FM*CL1/V1 K20=CL2/V2 $ERROR CALLFL=0 IPRE=-3 IF(F.GT.0) IPRE=LOG(F) PROP=0 IF(CMT.EQ.1)PROP=EPS(1) IF(CMT.EQ.2)PROP=EPS(2) Y=LOG(F)+PROP W=LOG(F) IRES=DV-IPRE IWRES=IRES/W $THETA (0, 152) ;CL1 (0, 43.3) ;V1 (0, 144) ;CL2 (0, 122) ;V2 (0,1) ;SEX > CL-5FU (0,1) ;SEX > CL-5FU-H2 $OMEGA 0.04 ;IIV CL1 $OMEGA BLOCK(1) 0.004 ;IOV OCC1; CL-5FU $OMEGA BLOCK(1) SAME ;IOV OCC2; CL-5FU $OMEGA 0.04 ;IIV V1 0.04 ;IIV CL2 0.04 ;IIV V2 $SIGMA 0.329 ;EPS COMP1 0.136 ;EPS COMP2 -- Markus Joerger MD PhD Associate Professor Medical Oncology&Clinical Pharmacology Cantonal Hospital 9007 St. Gallen Switzerland [email protected] [email protected] Phone: +41-765591070 Fax: +41-714946325