RE: enzyme turnover
There are no measurement in A(3), therefore there are no units. The
steady-state amount can be any arbitrary value, as all changes are
relative. Set A_0(3)=1, and KENZ=KIN=KOUT, and KENZ/KENZ=1. KENZ
determines the half-life, and is the only quantity the can be estimated
without direct data.
If the code isn't working, the most likely case is that there is a
clearance process that is not mediated by the destructible enzyme, and
therefore the amount going into the metabolite compartment may depend on
the dose administration schedule. There are a lot of possible problems
in parent-metabolite models which can be difficult to unravel due to the
uncertain input of metabolite. The metabolite could require more than
one compartment, or first-pass formation (possibly saturable) of
metabolite may need to be included in the model.
Best regards, James
James G Wright PhD
Scientist
Wright Dose Ltd
Tel: 44 (0) 772 5636914
Quoted reply history
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Magnusson, Mats
Sent: 15 February 2008 11:26
To: Khaled Nm; [email protected]
Subject: RE: [NMusers] enzyme turnover
Khaled,
I agree with John, you will in this case need a separate production rate
term (RIN) and state that A_0(3) = RIN / KENZ, and then let CL24 change
by A(3)/A_0(3). Depending on the mechanism of the inhibition, you might
consider adding -KIRR*C2*A(3) (after being transformed to only masses or
concentrations) to DADT(2).
/Mats
Mats Magnusson, PhD
Pharmacometrician
Pfizer Global Research & Development
Sandwich, Kent
CT13 9NJ
UK
Khaled,
You seem need another parameter. KENZ cannot be both a creation and a
death rate for A(3) (i.e. [mg/h] and [1/h]). Also, it is confusing the
way you use both masses and concentrations in your interaction terms. Do
you intend CL24 to be L h-1 mg-1?
John
John C Lukas
Strategic Consulting Services
Pharsight Corp.
line: + 33 492 726 495
cell: + 33 626 496 777
_____
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Khaled Nm
Sent: Wednesday, February 13, 2008 2:08 PM
To: [email protected]
Subject: [NMusers] enzyme turnover
Dear nmuser,
I had analyzed data set of a CYP- inhibitor agent using NONMEM ADVAN 6.
A two-comp. model was used to describe the parent and its metabolite
data sets sid by side to a hypothetical enzyme compartment. All
estimates are fine except the KENZ (enzyme turnover rate was about (2
hr-1) which is 100-fold the published value. my code is
$DES
C2=A(2)/V2 ; for parent
C4=A(4)/V4 ; for metab
DADT(1)= -KA*A(1)
DADT(2)=KA*A(1) - CL24*A(3)*C2
DADT(3)= KENZ - KENZ*A(3) - KIRR *C2 *A(3)
DADT(4)=CL24*A(3)*C2 - CL40*C4
If i use the Emax and IC50 mode, the result is similar. what is wrong
with my code ?
All suggesion are wellcome.
Regards
Khaled
_____
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