Back to the Top
Hi to all,
does anybody know for which reason the metabolite to parent
drug ratio could increase when increasing the iv administered dose of a
parent drug?
For example when administering 100 mg of the parent drug, metabolite
to parent drug percentage ratio is 5%; when administering 200 mg of
the parent drug, the percentage ratio increases to 12%.
Thank you in advance
Federico Pea, MD
Back to the Top
[A few replies - db]
Reply-To: "Stephen Duffull"
From: "Stephen Duffull"
To:
Subject: Re: PharmPK Increasing AUC metabolite/AUC parent drug %ratio
Date: Fri, 8 Oct 1999 09:40:29 +0100
Organization: University of Manchester
X-Priority: 3
If you haven't suddenly developed a better assay since the
last experiment then have you considered non-linear PK?
Steve
=====================
Stephen Duffull
School of Pharmacy
University of Manchester
Manchester, M13 9PL, UK
Ph +44 161 275 2355
Fax +44 161 275 2396
---
Date: Fri, 8 Oct 1999 02:24:50 -0700 (MST)
X-Sender: ml11439.at.pop.goodnet.com
To: PharmPK.-a-.boomer.org
From: ml11439.aaa.goodnet.com (Michael J. Leibold)
Subject: Re: PharmPK Increasing AUC metabolite/AUC parent drug %ratio
Hello Dr.Pea,
If the drug in question undergoes saturable protein binding, the
free fraction may increase with increasing doses. This increase in
free fraction could allow a greater percentage of the administered
drug to be metabolized, thus increasing the metabolite to parent drug
ratio recovered in the urine or plasma. This is basically the idea, but
this intuitively simple concept becomes more complex when examined
mathematically. The following mathematical treatment will find that
a relative shift from renal tubular secretion to hepatic metabolism
must occur for there to be an increase in the fraction metabolized as
a result of in increase in free fraction secondary to concentration-
dependent protein binding.
Concentration dependent protein binding is described mathematically as:
Fu= 1/[1+ K*P]
Fu= fraction unbound
K= affinity constant for protein binding
P= concentration of free protein binding sites
Molar Concentrations of Plasma Proteins
Protein Molar Concentrations
albumin 5-7.5 x 10-4
AAG 0.9-2.2 x 10-5
Concentration-dependent protein binding occurs with all protein bound
drugs, as the fraction unbound increases with increased drug concentration
according the above equation (i.e. P decreases). The therapeutic drug
concentrations can be used to predict if concentration-dependent protein
binding occurs in the therapeutic range. For example, the 10-20mg/l
phenytoin therapeutic range corresponds to a 0.4-0.8 x 10-4 molar
concentration which
is well below the normal 6 x 10-4 molar concentration of albumin, and
concentration-dependent protein binding does not occur in the therapeutic
range. However, concentration-dependent protein bnding does occur with
salicylate
with a therapeutic range of 7-22 x 10-4 molar.
For low hepatic extraction ratio drugs (drugs with low intrinsic clearance
relative to hepatic blood flow such as warfarin), the plasma concentration
is related to protein binding by the following equation:
Css free= FbKo/Cl
Css free= plasma steady state concentration of unbound drug
Fb= free fraction or fraction unbound to plasma proteins
Cl= systemic clearance
For restricted drugs or low hepatic extraction ratio drugs, clearance
is approximately equal to FbCli, which is the product of the fraction
unbound and intrinsic clearance. So that Css is obtained from the following
equation:
Css = Ko/FbCli
where FbCLi~= Cl or systemic clearance
and Css free= Ko/Cli (unaffected by protein binding)
So saturable protein binding would predict an increase in clearance with
increased Fb with increased dosages. In the following equation, the fraction
metabolized (Fm) is expressed as a ratio of Clh (hepatic clearance)to
Cls (systemic clearance):
Fm= Clh/Cls= Clh/[Clh+ Clr]
Where, Clh = hepatic clearance
Clr= renal clearance
Cls= systemic clearance
Fm= fraction metabolized
If the drug has a low hepatic extraction ratio and is excreted solely
by glomerular filtration, the above equation reduces to:
Fm= (FbCLii)/[FbCLi + FbClr(1-FR)]
= Cli/[Cli +Clr(1-FR)]
Where, FR= fracton of drug filtered that is reabsorbed
In the above case, the fraction metabolized is unaffected by protein
binding, since the Clr is similarly affected by the change in Fb. However,
in the case of the low extraction ratio drugs that undergo renal tubular
secretion as well as filtration, the above equation becomes:
Fm= Cli/[Cli + Clr(1-FR) + QkCli(k)(1-FR)/[Qk +FbCli(k)]]
Where, Qk= renal blood flow
Cli(k)= intrinsic renal secratory clearance
Fb= fraction unbound
Thus, in this case the Fm increases as the Fb increases. That is, the
increase in free fraction results in an increase in the fraction
metabolized, as
more drug is metabolized relative to that undergoing tubular excretion.
In the differential equation describing the concentration of the
metabolite in the above case, the Kf for the metabolite would increase
as the Fb increases:
dM/dt= KfX -KmM
Kf= formation constant for metabolite
Km= elimination constant for metabolite
X= parent drug
M= metabolite
The concentration of the metabolite would increase with Kf by the
following equation:
Cm= KfXo(e-kmt - e-ket)/[Vm(Ke-Km)]
So it would seem mathematically, that an increase in free fraction
could result in an increase in the concentration of the metabolite by
increasing the Kf or rate of formation. However, a the same time,
mathematical examination requires that there be an interplay between
renal tubular secretion and hepatic metabolism whereby more parent
drug is routed towards metabolism.
I hope this was of some help!! The above pharmacokinetic equations
are available in both editions of "Pharmacokinetics" by Gibaldi and
Perrier (Marcel Dekker).
Mike Leibold, PharmD, RPh
ML11439.-a-.goodnet.com
---
X-Sender: flerner.aaa.mail.intramed.net.ar
Date: Fri, 08 Oct 1999 06:53:57 -0300
To: PharmPK.-a-.boomer.org
From: Federico Ezequiel Lerner
Subject: Re: PharmPK Increasing AUC metabolite/AUC parent drug %ratio
Dear Federico:
Maybe, if the the metabolite elimination process is enzymatic pathway. Its
a phase 0 kinetics
Regards
Federico
---
Date: Fri, 08 Oct 1999 06:58:17 -0400
From: "Edward F. O'Connor"
Reply-To: efoconnor.aaa.snet.net
Organization: conn chem
X-Accept-Language: en
To: PharmPK.-at-.boomer.org
Subject: Increasing AUC metabolite/AUC parent drug % ratio
It may be due to the accuracy of the bioanalytical method. The higher dose
may be expected to be less of a challenge to the analytical method than the
lower dose, both for the parent and the metabolite but more importantly for
the metabolite. What accuracy and precision do you have for the metabolite
method at the levels arising from the low level? Are they the same as those
for the metabolite at the high range? Is the method linear ? Are you
accepting less than ELOQ values in the assay?
Ed O'Connor
Microtest
---
From: "Ram Chandra Gupta"
To:
Subject: Re: PharmPK Increasing AUC metabolite/AUC parent drug %ratio
Date: Fri, 8 Oct 1999 19:48:17 -0400
X-Priority: 3
Dear Federico Pea,
The reason could be that at 100 mg dose particular CYP isoform is active and
200 mg dose acts as inducer for other CYP isoform. As net result the ratio
changes. Does the compound has stereo chemical center? have you checked
the formation of another metabolite?
R. C. Gupta
Back to the Top
> From: ml11439.at.goodnet.com (Michael J. Leibold)
> Subject: Re: PharmPK Increasing AUC metabolite/AUC parent drug %ratio
>
> If the drug in question undergoes saturable protein binding, the
> free fraction may increase with increasing doses. This increase in
> free fraction could allow a greater percentage of the administered
> drug to be metabolized,
This is a common misunderstanding which arises from confusing=20
clearance based on total drug concentrations with drug elimination=20
mechanisms. Unbound drug clearance is more clearly linked with drug=20
elimination.
Plasma protein binding has no influence on unbound drug clearance for=20
drugs with low extraction ratios. Drugs with high extraction ratios=20
are expected to have *decreased* unbound drug clearance when the free=20
fraction increases because of decreased drug delivery to the=20
extracting organ. See these textbook chapters for details:
Holford NHG and Benet LZ. Pharmacokinetics and pharmacodynamics:=20
Rational dose selection & the time course of drug action. Chapter 3=20
in Basic and Clinical Pharmacology. 7th Edition. Edited by B Katzung.=20
1997; Lange Medical Publications, Palo Alto, CA.
Holford NHG, Tett S. Therapeutic Drug Monitoring. Chapter in Avery=92s=20
Drug Treatment (4th Edition) Eds. Speight T, Holford NHG. 1997; Adis=20
Intl, Auckland
--
Nick Holford, Dept Pharmacology & Clinical Pharmacology
University of Auckland, Private Bag 92019, Auckland, New Zealand
email:n.holford.aaa.auckland.ac.nz tel:+64(9)373-7599x6730 fax:373-7556
http://www.phm.auckland.ac.nz/Staff/NHolford/nholford.html
Back to the Top
Hello Dr.Holford,
My answer was based on a section in Gibaldi& Perrier's
"Pharmacokinetics", where equations for the fraction metabolized
are derived. The proof is in the equation and its derivation.
In the case of the low extraction ratio drugs that undergo renal
tubular secretion as well as filtration, the following equation applies
as it appears in "Pharmacokinetics":
Fm= Cli/[Cli + Clr(1-FR) + QkCli(k)(1-FR)/[Qk +FbCli(k)]]
Where, Qk= renal blood flow
Cli(k)= intrinsic renal secretory clearance
Fb= fraction unbound
Cli= intrinsic hepatic clearance
Clr= renal filtration clearance
FR= fraction reabsorbed
This equation is derived from the simpler expression:
Fm= Clh/Cls= Clh/[Clh + Clr]
In in this mathematical treatment the Fm increases as the
Fb increases. That is, the increase in free fraction results in
an increase in the fraction metabolized, as more drug is metabolized
relative to that undergoing tubular excretion. Or, in mathematical
terms, the numerator of the above equation increases relative to
the denominator with an increase in Fb.
So, the point I was trying to make is that there is a possible
pharmacokinetic explanation in "Pharmacokinetics" for the phenomenon
described by Dr.Pea. The derivation indicates that this increase in fraction
metabolized would only occur with low extraction drugs which undergo
renal tubular secretion as well as hepatic metabolism.
However, "Pharmacokinetics" agrees with you in that the clearance
of unbound portion of low extraction drugs is unaffected by protein binding.
Css(total) = Ko/FbCli
where FbCLi~= Cl or systemic clearance which increases with Fb
In contrast, the Css free portion is unaffected by protein binding:
Css free= Ko/Cli (unaffected by protein binding)
This also agrees with your chapter in "Basic and Clinical Pharmacology"
7th edition. However, the Fm or fraction metabolized equation has a more
complex derivation as it appears in "Pharmacokinetics", and could theoretically
at least, answer Dr.Pea's question. That is, according to the Fm equation in
"Pharmacokinetics", there is a mechanism whereby an increase in Fm could occur
with an increase in Fb.
Mike Leibold, PharmD, RPh
ML11439.aaa.goodnet.com
Back to the Top
Michael,
Please accept my apologies for implying your proposal was based on a
misunderstanding of the effect of protein binding on clearance. Indeed
you are correct that an increase in the formation of metabolite is
possible due to changes in plasma protein binding. The mechanism is
quite subtle and I had not appreciated it until I read your remarks more
carefully.
An increase in metabolite formation requires not only that there is a
decrease in the fraction unbound but also that the metabolite is formed
by a low extraction ratio clearance process and that there is clearance
of the parent by another route with a high extraction ratio. When the
fraction unbound increases the unbound clearance due to the low
extraction ratio process (formation of metabolite) does not change but
the unbound organ clearance of the parent due to the high extraction
process will decrease:
CLo = Q*CLi*Fu/(Q+CLi*Fu)
CLou = CLo/Fu
where CLo=organ clearance of whole blood, CLi=intrinsic clearance of
unbound drug, Fu=fraction unbound, Q=organ blood flow, CLou=organ
clearance of unbound drug
e.g. if CLi=100, Q=60 and Fu increases from 0.1 to 0.33 then CLou will
decrease from 85 to 64. Assuming that the metabolite is formed by a
parallel low extraction ratio process then the formation of the
metabolite will increase because of the decrease in clearance of the
parent by the high extraction ratio process. The fractional formation of
metabolite, Fm, is given by:
Fm = CLmu/(CLmu+CLou)
where CLmu is the unbound clearance of parent to metabolite. An increase
in Fm will be most clearly seen if CLmu is negligible in relation to
CLou. In that case metabolite formation will be inversely proportional
to Clou. In the example above, Fm would increase from 1/85 to 1/64 i.e.
33%.
The following equation is not correct as you have written it. The Cli(k)
term in the numerator of of the expression for renal tubular secretion
should be multiplied by Fb. Note that Fb as defined by Gibaldi and
Perrier rather confusingly refers to the Fraction Unbound. It has the
same meaning as Fu in the expressions I have used above for CLo and
CLou.
> Fm= Cli/[Cli + Clr(1-FR) + QkCli(k)(1-FR)/[Qk +FbCli(k)]]
>
> Where, Qk= renal blood flow
> Cli(k)= intrinsic renal secretory clearance
> Fb= fraction unbound
> Cli= intrinsic hepatic clearance
> Clr= renal filtration clearance
> FR= fraction reabsorbed
--
Nick Holford, Dept Pharmacology & Clinical Pharmacology
University of Auckland, Private Bag 92019, Auckland, New Zealand
email:n.holford.aaa.auckland.ac.nz tel:+64(9)373-7599x6730 fax:373-7556
http://www.phm.auckland.ac.nz/Staff/NHolford/nholford.htm
Back to the Top
[Two more comments - db]
Date: Fri, 15 Oct 1999 09:57:27 +0200
From: furlanut
Subject: Increasing AUC metabolite/AUC parent drug % ratio
To: Multiple recepient PK
X-Accept-Language: it
Dear all,
thank you for the many answers to the question of
increasing AUC metabolite/AUC parent drug % ratio.
Federico Pea, MD
---
Date: Fri, 15 Oct 1999 01:25:09 -0700 (MST)
X-Sender: ml11439.aaa.pop.goodnet.com
To: PharmPK.at.boomer.org
From: ml11439.at.goodnet.com (Michael J. Leibold)
Subject: Re: PharmPK Re: Increasing AUC metabolite/AUC parent drug % ratio
Dr.Holford,
Thanks for your reply! The equation I emailed you is the
correct form! The Fb in the numerator of the renal secretory
clearance expression was factored out as follows:
Fm= FbCli/[FbCLi + FbClr(1-FR) + FbCli(k)(1-FR)/[Qk+FbCli(k)]]
Dividing the numerator and denominator by Fb results in the
expression with Fb only in the denominator of the renal secretory
clearance expresssion:
Fm= Cli/[Cli + Clr(1-FR) + QkCli(k)(1-FR)/[Qk +FbCli(k)]]
Where, Qk= renal blood flow
Cli(k)= intrinsic renal secretory clearance
Fb= fraction unbound
Cli= intrinsic hepatic clearance
Clr= renal filtration clearance
FR= fraction reabsorbed
This is the equation by which the text predicts an increase
in free fraction of a low extraction drug which undergoes renal
tubular secretion as well as renal filtration clearance in addition
to its low extraction hepatic metabolism. The drug must be low
extraction since:
Cl= QFbCli/[Q+ FbCli] reduces to:
Cl= FbCli for low extraction drugs since Q>>Cli.
The Cll= FbCli fits into the above equation, and this is the
requirement for the low extraction drug. The parent drug must be
a low hepatic extraction drug which undergoes renal filtration and
renal tubular secretion as well, according to the text. The renal
tubular clearance can be high or low extraction.
Mike Leibold, PharmD, RPh
ML11439.-a-.goodnet.com
Back to the Top
> From: ml11439.aaa.goodnet.com (Michael J. Leibold)
> Thanks for your reply! The equation I emailed you is the
> correct form!
My apologies are due to you again. You are quite correct. All the
clearance terms refer to unbound clearance and thus the expression for
the renal tubular clearance must indeed be expressed as an unbound
clearance.
> The parent drug must be
> a low hepatic extraction drug which undergoes renal filtration and
> renal tubular secretion as well, according to the text. The renal
> tubular clearance can be high or low extraction.
At risk of being shown to be wrong for a third time on this issue (!) I
will attempt to disagree with the above paragraph. I cannot see any
reason for specifically including any renal components of clearance.
Sufficient requirements for an increase in Fm due to saturable plasma
protein binding are:
1. The metabolite is formed by a low extraction ratio process
2. The parent is also eliminated by a high extraction ratio process
The first requirement means that the metabolite formation clearance is
not influenced by changes in protein binding. The second requirement
means that the overall clearance of the parent will decrease when the
unbound fraction increases. If metabolite formation clearance is not
changed and overall clearance is decreased then Fm must increase.
There is no need for renal filtration of the parent. There must be a
high extraction elimination mechanism of the parent. If you want to use
a renal mechanism for the high extraction process then this must be
tubular secretion because renal filtration clearance is always low
extraction.
--
Nick Holford, Dept Pharmacology & Clinical Pharmacology
University of Auckland, Private Bag 92019, Auckland, New Zealand
email:n.holford.aaa.auckland.ac.nz tel:+64(9)373-7599x6730 fax:373-7556
http://www.phm.auckland.ac.nz/Staff/NHolford/nholford.htm
Back to the Top
[Two (?) replies - db]
Date: Thu, 21 Oct 1999 01:08:28 -0700 (MST)
X-Sender: ml11439.at.pop.goodnet.com
To: PharmPK.at.boomer.org
From: ml11439.at.goodnet.com (Michael J. Leibold)
Subject: Re: PharmPK Increasing AUC metabolite/AUC parent drug % ratio
Dr. Pea,
Along with the possible concentration-dependent protein effect and
an increase in Fm, another possiblity I should mention is the accumulation
of a metabolite which undergoes renal clearance. In the case of meperidine,
its use in patients with renal insufficiency can result in the accumulation
of normeperidine, a toxic metabolite which can cause excitability and seizures.
So that the AUC of the metabolite can increase as the its renal clearance
decreases with decreasing renal function.
This later mechanism is probably more clinically significant than the
previously described increase in Fm. Although increased Fm is also very likely,
the renal accumulation of an active metabolite is probably more significant,
probably has more clinical consequence.
Actually, I just happen to look up the Fm equation in a textbook, and then
wrote the previous emails.
Mike Leibold,PharmD, RPh
ML11439.at.goodnet.com
---
Date: Thu, 21 Oct 1999 00:52:01 -0700 (MST)
X-Sender: ml11439.aaa.pop.goodnet.com
To: PharmPK.aaa.boomer.org
From: ml11439.-a-.goodnet.com (Michael J. Leibold)
Subject: Increasing AUC metabolite/AUC parent drug % ratio
Dr.Holford,
Thanks again for your reply! I agree that renal filtration
clearance need not be included in the expression, that is, for there
to be an affect on Fm. But, renal filtration clearance is likely to
occur along with renal tubular secretion clearance, the later being
necessary for an increase in Fm with an increase in Fb.
I think your concept of a low extraction metabolism of the parent
drug occuring along with a high extraction parent drug elimination
process, is also a mechanism by which an increase in Fm could occur with
an increase in Fb.
1) Low extraction parent drug metabolism
Cl= FbCli [equation 1]
Css= Ko/FbCli
2) High extraction process where Cl= Q since Cli>>>Q
Cl= Q [equation 2]
Css= Ko/Q
3) Both equations are derived from:
Cl= QFbCli/[Q+ FbCli]
a) Low extraction drugs: Q>>FbCli and Cl= FbCli
b) High extraction drugs: Cli>>Q and Cl= Q
Equation 1 predicts that there would be an increase in clearance
with an increase in Fb for a low extraction drug. Equation 2 predicts
that there would be no change in the clearance of the parent drug
by the high extraction process with an increase in Fb. As a result,
there would be an increase in Fm since there would be an increase in the
relative clearance of the parent drug by the low extraction metabolic
pathway.
The Fm equation for parallel metabolic and renal secretory
clearance also has a high extraction process form for the renal
secretory clearance expression:
Fm= Cli/[Cli + Clr(1-FR) + (Qk/Fb)(1-FR)]
The Fb factoring in this equation also predicts the same increase
in Fm with an increase in Fb as a result of a relative decrease in
a high extraction renal secretory elimination process in favor of an
increased low extraction metabolic clearance.
The above equation could be modified for the situation you propose:
Fm= Cli/[Cli + Q/Fb] [equation 3]
This is where the factoring of Fb from the numerator and denominator
results in the same relative increase in the low extraction Cli, and a
consequent increase in Fm. So, I think that this last equation describes
the general low/high extraction interplay that might occur with increase
in Fb which could increase the Fm of the low extraction metabolism versus
the high extraction elimination process (hepatic or renal).
I think this basically agrees with what you are saying, although the
dynamics of the Fb in equations 1, 2 and 3 are a little different.
Mike Leibold, PharmD, RPh
ML11439.aaa.goodnet.com
PharmPK Discussion List Archive Index page
Copyright 1995-2010 David W. A. Bourne (david@boomer.org)