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Hi, All:
Rate constant (k) is often used in the in vitro studies as a initial parameter
to further calculate other important parameters such as intrinsic clearance
(CLint) in metabolic stability study and inactive parameters in the inhibition
studies (KI and kinact in time-dependent inhibition study, where kobs is
used to
describe rate constant), which is calculated from the slope of the initial
linear phase of the plots. For example: In the metabolic stability study, k is
determined with a linear regression analysis of the logarithm of remaining
percent (C%) of parent drug against incubation-time. While in the
time-dependent inhibition study, the logarithm of the enzymatic activity
is
firstly plotted against the pre-incubation time, then the apparent inactivation
rate constant (kobs) can be calculated from the slope of the initial linear
phase. My question is when we plot these lines to obtain slope values,
is it
necessary to set the intercept to a constant value such as 0 since when
t=0, the
initial value of y should also be 0 based on the equation of y=ln(C0)-kt? We
all know that the data quality can influence the intercept values, if we
don't
set the intercept values when doing linear regression, the intercept will be
changed a lot with different data set especially when Microsoft excel was used
as the tool, which obviously conflicts with the equation and at the same
time,
the R2 is still possibly good enough (>0.9). I have reviewed a lot of paper and
surprisingly noticed that no one set the intercept to a constant value when
calculating k so I guess that if the intercept was not constrained, a criterion
is necessary to judge if the regression is acceptable as well as the calculated
slope is correct. I will be appreciated that if anyone can share your
experiences on this kind of data processing.
Many thanks!
Regards,
Jian Wang
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Hi,
My advice is to use what you already know to be true without doubt. The
intercept must be zero so fix it to zero when doing the regression.
If you find that the intercept is importantly different from zero then
you need to think very carefully about the methodology you are using and
find the problem.
R^2 is a waste of time. It is not a reliable measure of goodness of
fit.
Nick
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Hi Nick,
Which statistical method provides the best measure of goodness of fit
for clinical PK data?
Thanks,
Mike
Michael A. Jones, Pharm.D.
Informatics Pharmacist - Clinical Decision Support
University of Colorado Hospital
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Mike,
Visual predictive check is my gold standard for showing the model fits the data.
Nick
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The following message was posted to: PharmPK
Thanks Nick,
Are there any statistical methods that can be automated that would be as
good as visual inspection of the curve and data?
Thanks,
Mike
Michael A. Jones, Pharm.D.
Informatics Pharmacist - Clinical Decision Support
University of Colorado Hospital
[I once had a statistics professor 'offer' to leave the dissertation
committee of one of my graduate students (a week before the defense)
unless we removed R^2 from standard curve plots ;-) A visual check of
the data and best-fit line is always an important part of the process
but as far as automating the fit the use of a number of statistics could
be useful including R^2 in a controlled fashion. I've recently seen it
used to automatic the process of selecting the number of points to use
in the determination of the terminal half-life or rate constant.
However, it didn't work very well and I feel that I did a (much) better
job looking at each plot individually. This can especially important
with sparse data sets. - db]
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Mike,
Anybody with common sense will always look at the data and the fit. Statistics have a limited role. They can be helpful in guiding the search for an appropriate model eg. I use the extended least squares log-likelihood. The final decision rests with the human brain via the eyes.
Nick
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That makes sense. So my takeaway from this is; even with automated
methods we still must provide a sufficiently granular plot for the
end-user to inspect.
Thanks again Nick,
Mike
Michael A. Jones, Pharm.D.
Informatics Pharmacist - Clinical Decision Support
University of Colorado Hospital
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My question is when we plot these lines to obtain slope values is it
necessary to set the intercept to a constant value such as 0 since when
t=0
the initial value of y should also be 0 based on the equation of y=ln(C0)-kt?
-In any kinetic exp it is often difficult to get concn data for t=0 unless you
have a cell-free incubation. Hence=2C one uses t= 2 or 3 min to obtain 1st
concn value which is close to initial conc. The y-axis is usually Ln (Concn).
Parnali Chatterjee
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Parnali,
Calculation of the terminal rate constant by linear regression involves
estimating both the slope and intercept. The slope value will be
negative because the concentrations are declining as time increases.
Thus the slope value is equal to -1 * terminal rate constant. The
intercept is not fixed to zero, and the intercept value may be useful
depending on the pharmacokinetic profile. If you have a IV bolus dose,
and the drug exhibits 1-compartment kinetics, then the intercept value
corresponds to C(0), or the "concentration at time=0". More
appropriately the intercept value can be used to calculate the volume
of distribution using the following equation: V = Dose/C(0).
Good luck in your analysis.
Nathan S. Teuscher, PhD
Founder, PK/PD Associates
www.learnpkpd.com
nathan.-at-.learnpkpd.com
Twitter: www.twitter.com/learnpkpd
Facebook: www.facebook.com/LearnPKPD
YouTube Channel: www.youtube.com/learnpkpd
[I think Nathan means, semi-log linear regression, that is linear regression of the log/ln of Cp versus time.
For example http://www.boomer.org/c/php/pk0202a.php
and http://www.boomer.org/c/php/pk0504a.php
- db]
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Hi all,
It seems the discussion is biased from what it was originally asked. I think
Jian is talking about the in vitro metabolism assay, so theratically at
time zero, the drug should be of no loss and the remaining should be 100%. From this
point, I agree with Nick to set the intercept.
However, actaully calculating k is a process of linear model fitting. in terms
of linear model fitting, manully setting the intercept should be better avoided,
because it will change the model variance. That's why it is generally not fixing
the intercept.
However, regarding to this question itself, if the method is robust and the
result is reliable, there may be an intercept but it will be not statistically
significantly different from the ideal (theratical) value (for example, 0 or
100, depended on what kind of model equation). Assuming the theratical value is
100, always the modeling intercept result will range around 100. But the
significane (p value) of the linear model parameters is more meaningful, for
this intercept it is better to be not significant. However, if the modeling
intercept is significantly different from the ideal (theratical) value, such as
p<0.05, it implicates the data has some problems. You need to refine the
experiment. But for early projects or screening purpose, I think manually fixing
the intercept of this kind of questionable data to generate a k is reasonable.
Thanks
Yi
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