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The following message was posted to: PharmPK
In the book of Rowland, it shows that ke=0.693/t1/2. It is given an example, stating that if the half-life is 3h, ke would 0.231 h-1 (which means a rate of approximately 23% per hour, which makes sense). What is the interpretation of a t1/2 of one hour?
G De Nucci
[Sorry for the delay in sending out messages over the weekend. I had a hard time getting wifi to work in everywhere I tried except B&N. Qwest, McDonald's, DIA let me done ;-) Never mind, mission accomplished!! My daughter's wedding at Chief Hosa Lodge (in the Denver foothills, in January) was a wonderful success. - db]
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Dear Nucci,
Its based on a simple mathematical calculation. According to example given by you, when half life is 1 hour, then the elimination rate will be 69.3% per hour. But you shouldn't get confused because elimination rate apart from half life in the equation used for calculation, also depends upon clearance and volume of distribution. This is where you will get the difference while correlating half life and elimination rate constant.
CL= Vd*Kel
Hope I answered your query, Any doubts do revert back.
Regards,
Vijay,
G7 Synergon Pvt Ltd.
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The following message was posted to: PharmPK
Dear Gilberto,
You wrote:
> In the book of Rowland, it shows that ke=0.693/t1/2. It is given an
> example, stating that if the half-life is 3h, ke would 0.231 h-1 (which
> means a rate of approximately 23% per hour, which makes sense). What is
> the interpretation of a t1/2 of one hour?
A half-life of 1 hour implies that after 1 hour, the concentration, and the
amount in the body (assuming a one-compartment model) has decreased by 50%.
The interpretation that k = 0.231 h-1 means a rate of 23.1% per hour is not
correct; it means indeed that at each time point the decrease of plasma
concentration and amount in the body is 0.231 times the plasma concentration
and amount in the body, respectively. In equations:
dC/dt = k * C
or
dA/dt = k * A
(please note from the discussion 'A question of clearance' that this
notations are not mechanistically based, but a re-parametrization of the
equations dA/dt = CL * C and A = V * C but I use the rate constant equations
here to explain the meaning of the rate constant).
So, if k=0.231 h-1, the initial rate of decline of C and A is 0.231 times
the initial C and A. Since C and A are decreasing over time, the rate of
decline dC/dt and dA/dt are also decreasing over time. So, after one hour
there is not 23.1% eliminated, but less than that.
Upon integration of the above equation for C one gets:
C = C0 * exp(-k*t)
For k=0.231 h-1 and t=1, C = C0 * exp(-0.231) = C0 * 0.7937. So, 20.63% has
been eliminated (not 23.1%).
For a half-life of 1 hour, k = ln(2) / 1 = 0.693 h-1. After one hour, C = C0
* exp(-0.693) = C0 * 0.5 (as expected!).
For a half-life of 0.3 hour, k = ln(2) / 0.3 = 2.31 h-1. This does not mean
that 231% is removed after 1 hour. After one hour, C = C0 * exp(-2.31) = C0
* 0.099.
best regards,
Hans Proost
Johannes H. Proost
Dept. of Pharmacokinetics, Toxicology and Targeting
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
[There may be something useful on
http://www.boomer.org/c/p3/c02/c0209.html
which discusses Euler's method, note the effect of step-size - db]
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Hi Gilberto,
You wrote:
"In the book of Rowland, it shows that ke=0.693/t1/2. It is given an example, stating that if the half-life is 3h, ke would 0.231 h-1 (which means a rate of approximately 23% per hour, which makes sense). What is the interpretation of a t1/2 of one hour?"
I don't know where it started but it seems to be a common misconception that ke refers to the fraction of drug eliminated per unit time.
Based on the definition of half-life, when T1/2 = 1 h, the fraction eliminated after one hour is obviously 0.5 (not 0.693).
This can be readily seen (in the case of a one compartment model with linear elimination kinetics) by examining the monoexponential equation describing the concentration-time relationship, namely Ct = C0 * exp(-ke*t). From this equation, it is obvious (divide through by C0) that the fraction remaining (FR) at any time = exp(-ke*t). Thus, when T1/2 = 1 h, ke = 0.693/h and the FR = exp(-0.693*1) = 0.5.
If FR = exp(-ke*t), then the *fraction eliminated*, FE = 1 - FR.
To take another example, say T1/2 = 0.1 h. (In this instance ke = 6.93/h but, of course, this does not mean that 693 % of drug would be eliminated after one hour!) The actual FE (in percentage terms) after one hour would be 99.9%. i.e. FE = 1 - FR = 1 - exp(-6.93*1) = 1 - 0.000978 = 0.999 (consistent with the fact that one hour represents 10 half-lives in this case).
Best regards,
Peter
Peter W. Mullen, PhD, FCSFS
KEMIC BIORESEARCH (www.kemic.com)
Kentville
Nova Scotia, B4N 4H8
Canada
[Check out the wikipedia entry
http://en.wikipedia.org/wiki/Rate_equation
the first-order section under 'An alternative view of first order kinetics'
"kinetic constant must represent the fraction of the population of reactant present that will breakdown in a given time period and the fraction must be less than one" - db]
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The following message was posted to: PharmPK
Peter:
I think the interpretation of ke as "a rate of approximately 23% per
hour" may come from dA/dt=ke*A. Move A on the right side to the left
side: dA/A/dt=ke or %/dt=ke. Therefore ke can be approximately
interpreted as the fraction eliminated per unit time. This approximation
becomes more and more precise when you use smaller and smaller time
scale, such as minute, second or millisecond (as a result, ke becomes
smaller and smaller). It is the same concept as hazard in survival
analysis. Depending on the time scale, hazard can be even greater than
1, which can always be converted to a smaller number (<<1) by changing
the time scale so that "proportion of people dying in a unit time
interval" can be used to explain the hazard concept.
"The contents of this message are mine personally and do not necessarily
reflect any position of the Government or the Food and Drug
Administration."
Yaning Wang, Ph.D.
Team Leader, Pharmacometrics
Office of Clinical Pharmacology
Office of Translational Science
Center for Drug Evaluation and Research
U.S. Food and Drug Administration
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The following message was posted to: PharmPK
Hi Gilberto,
You wrote:
"In the book of Rowland, it shows that ke=0.693/t1/2. It is given an
example, stating that if the half-life is 3h, ke would 0.231 h-1 (which
means a rate of approximately 23% per hour, which makes sense). What is
the interpretation of a t1/2 of one hour?"
You can think of "t1/2 of one hour" as t1/2 of 60 minutes and its
corresponding ke will be 0.693/60=0.0116 min-1, which means a rate of
approximately 1.16% per minute.
"The contents of this message are mine personally and do not necessarily
reflect any position of the Government or the Food and Drug
Administration."
Yaning Wang, Ph.D.
Team Leader, Pharmacometrics
Office of Clinical Pharmacology
Office of Translational Science
Center for Drug Evaluation and Research
U.S. Food and Drug Administration
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The following message was posted to: PharmPK
Yaning,
You have found another good reason not to use rate constants in PK --
because trying to explain what they mean in simple terms is hard and, as
I think you show below, such an explanation may not make much sense.
You wrote:
> You can think of "t1/2 of one hour" as t1/2 of 60 minutes and its
> corresponding ke will be 0.693/60=0.0116 min-1, which means a rate of
> approximately 1.16% per minute.
>
By analogy this means the rate is 69.3% per hour or 1664% per day. So
what possible meaning can we attach to a rate of 1664% per day? It
implies that more than 100% of the drug is eliminated per day which is
obviously impossible. Rate constants cannot be interpreted sensibly in
the way you have tried to use in your example.
You ran into a similar problem in an earlier posting when you tried to
explain the hazard concept:
> Depending on the time scale, hazard can be even greater than
> 1, which can always be converted to a smaller number (<<1) by changing
> the time scale so that "proportion of people dying in a unit time
> interval" can be used to explain the hazard concept.
The hazard is NOT the "proportion of people dying in a unit time
interval". The hazard is the event rate at an instant in time. It has
the same dimensions as a PK rate constant i.e. 1/time. The numerical
value depends on the chosen time unit e.g. 24 per day, or 1 per hour
or 1/3600 per second, all represent the same hazard.
IF the hazard is constant (a rather unlikely situation for death events)
then ln(2)/hazard will tell you how long it takes for half of the
population to die. In the same way for PK ln(2)/rate constant will tell
you long it takes for half of the drug to be eliminated.
Nick
--
Nick Holford, Professor Clinical Pharmacology
Dept Pharmacology & Clinical Pharmacology
University of Auckland,85 Park Rd,Private Bag 92019,Auckland,New Zealand
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Hi Gilberto (and Yaning),
I wish to revise the statement, "it seems to be a common misconception that ke refers to the fraction of drug eliminated per unit time" in my previous posting. I misspoke. I should have said that "ke CAN be interpreted as approximating the fraction of drug eliminated per unit time (providing it is of relatively low numerical value)". In this regard, simply changing the time units as pointed out by Yaning is the obvious way to ensure that this definition of ke works (e.g. if the value of ke per hour approaches 1.0 convert to per minute by dividing by 60).
Thus, in my example, it was just a matter of converting the T1/2 value of 0.1 h to 6 min in which case ke = 0.1155/min indicating a loss rate of approximately 11.6% per minute which is in reasonable agreement with the actual value (per min.) one would obtain using my fraction eliminated calculation (i.e. FE = 0.109/min or ca. 11% /min).
Best wishes,
Peter
Peter W. Mullen, PhD, FCSFS
KEMIC BIORESEARCH (www.kemic.com)
Kentville
Nova Scotia, B4N 4H8
Canada
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The following message was posted to: PharmPK
Nick:
Your wrote:
"By analogy this means the rate is 69.3% per hour or 1664% per day. It
implies that more than 100% of the drug is eliminated per day which is
obviously impossible"
As I explained earlier, it is an approximation and the approximation is
more precise when the time scale is small and therefore ke is <<1.
Obviously, you can not apply this approximation to a value >1. I can ask
you a similar question: when V is 20L and CL is 10L/hr, what is the
explanation of CL of 2400L/day? Whatever answer you come up with can be
used to answer 1664% per day question too.
I actually found "fraction eliminated per unit time" a very useful and
intuitively appealing explanation of rate constant when I convert it to
a smaller time scale. When I first learned survival analysis, I found
hazard a very difficult concept to be translated into a simple concept
everyone can understand. When I finally linked Ke with hazard, I felt
"proportion of people dying in a unit time interval (among the people
surviving up to the starting time point of the interval, to be
statistically exact)" a very easy explanation of hazard. In fact, the
definition of hazard starts from this concept and converts to a limit
concept (when dt goes to infinitely small). My favorite example to
demonstrate this concept when I teach survival analysis is: at a certain
time point, if there are 100 subjects alive and 10 people die in the
next month, the hazard during that month is approximately 0.1 month-1;
if another 10 subjects die in the following month, the hazard during the
following month is 1/9 month-1. I am glad that you noticed hazard has a
unit of 1/time just like ke. I guess most people don't realize that.
When all the subjects die within a year, you may want to express the
hazard in the unit of month-1 rather than decade-1 even though you can
convert numerically. For those repeatable event, hazard>1 has a
meaningful explanation. But can you translate a hazard of 2 per year for
non-repeatable event (such as death) into something a layman can
understand?
"The contents of this message are mine personally and do not necessarily
reflect any position of the Government or the Food and Drug
Administration."
Yaning Wang, Ph.D.
Team Leader, Pharmacometrics
Office of Clinical Pharmacology
Office of Translational Science
Center for Drug Evaluation and Research
U.S. Food and Drug Administration
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The following message was posted to: PharmPK
Dear friends,
For a first order rate function such as C = C0*exp(-k*t), k, the "rate"
= the differential: dC/dt = -k*C. It might be helpful if we called k the
"rate coefficient" rather than the "rate constant" because clearly the
"rate" is not constant, but proportional to the concentration. When the
rate constant is converted to the % change in C per unit time, the
result is the INSTANTANEOUS RATE of change; it is the tangent to the
curve of log(concentration) vs time plot, and as it is the rate, not the
rate constant, it keeps changing.
A better analogy for this is in radioactivity. The radioactive half-life
is a familiar concept, and those who use short half-life isotopes such
as 32P or PET isotopes 11C and 18F know that corrections have to be made
to the specific activity as the batch of isotope ages. The half-life is
constant (Geiger counters usually average over a few seconds) as long as
the number of radionuclei is large enough for statistically smooth
change, but when the number of radionuclei is countable, then the rate
of disintegration becomes slow and the time between events becomes
extremely variable; this is more like a survival curve.
For more complicated functions, the concept of half-life is rarely
useful and often extremely misleading, which is why it is so often the
subject of discussion in this group. I am not aware of a constant
half-life for human life - the probability of death is not constant in
my experience, though some causes might be approximately constant for a
time, eg death by road traffic accident between age 30 and age 50.
I think that the confusion between RATE and RATE CONSTANT is one of the
commoner mistakes.
Regards.
Ted
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The following message was posted to: PharmPK
Dear Yaning,
Again we are talking on clearance and rate constants.
> a similar question: when V is 20L and CL is 10L/hr, what is the
> explanation of CL of 2400L/day? Whatever answer you come up
> with can be used to answer 1664% per day question too.
What is the problem here? (apart from the obvious numerical mistake (10 * 24
= 240, or 100 * 24 = 2400). One of the nice things of clearance is that it
avoids the question of 'fraction'. Clearance is not a fraction (as I'v
stated in the beginning of our 'A question of clearance' discussion). It is
the capacity of the system to eliminate the drug. The analogy with a
filtering system of a swimming pool can be helpful, also to understand (yes,
I must take this opportunity!) that half-life and rate constant are
depending on the volume and filtering capacity.
best regards,
Hans Proost
Johannes H. Proost
Dept. of Pharmacokinetics, Toxicology and Targeting
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
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The following message was posted to: PharmPK
Hans:
You wrote:
"One of the nice things of clearance is that it avoids the question of
'fraction'"
We are talking about T1/2 here, time to 50% loss of the drug. How can we
avoid the question of "fraction"? Maybe I should have put the question
in a simpler and more straightforward way: when V is 20L and CL is
10L/hr, does that mean after 1 hour, 50% of drug is eliminated; after 2
hours, no drug is left?
"The contents of this message are mine personally and do not necessarily
reflect any position of the Government or the Food and Drug
Administration."
Yaning Wang, Ph.D.
Team Leader, Pharmacometrics
Office of Clinical Pharmacology
Office of Translational Science
Center for Drug Evaluation and Research
U.S. Food and Drug Administration
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The following message was posted to: PharmPK
Dear Ted:
You wrote:
"When the rate constant is converted to the % change in C per unit time,
the result is the INSTANTANEOUS RATE of change; it is the tangent to the
curve of log(concentration) vs time plot, and as it is the rate, not the
rate constant, it keeps changing"
Could you explain why % change in C per unit time (or the tangent to the
curve of log(concentration) vs time) keeps changing in your example C C0*exp(-k*t)?
"The contents of this message are mine personally and do not necessarily
reflect any position of the Government or the Food and Drug
Administration."
Yaning Wang, Ph.D. Team Leader, Pharmacometrics Office of Clinical Pharmacology Office of Translational Science Center for Drug Evaluation and Research U.S. Food and Drug Administration
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The following message was posted to: PharmPK
I have been following this thread with great interest, and often amusement.
Like the fly walking half way to the wall with each step, both CL and Ke
will have diminishing effects as the remaining amount/concentration is
reduced, which is why we see exponential functions as reasonable
approximations in many cases. Only through integrating the differential
equation will you get the answer at any future time point. You can't treat
either one as linear, and that's why you can't say that if CL = 10 L/h and
Volume = 20 L that it will all be cleared in 2 hr.
CL of 10 L/h simply says that you're instantaneously clearing the amount
(volume*concentration) calculated at that rate. It would only be a constant
amount vs time if the concentration stayed the same. But the concentration
is diminishing (assuming no further dosing), so clearing 10 L/h when the
concentration is, say 100 ng/mL, is not clearing the same amount per unit
time when the concentration drops to 99 ng/mL. The same goes for Ke - it is
a rate constant/coefficient that is being applied to a diminishing amount or
concentration, so the fraction cleared per unit time is the fraction of what
is left at that instant, and will be different in the next instant.
Clearing a hypothetical volume per unit time that has a changing
concentration is no easier or more difficult for me to visualize than
clearing a fraction per unit time. They both apply to the instantaneous
amount or concentration at that point in time.
Perhaps I'm stating the obvious, but it seems from some of the posts that it
may not be.
Walt Woltosz
Chairman & CEO
Simulations Plus, Inc. (NASDAQ: SLP)
42505 10th Street West
Lancaster, CA 93534-7059
U.S.A.
http://www.simulations-plus.com
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The following message was posted to: PharmPK
Dear Yaning,
Your wrote:
> We are talking about T1/2 here, time to 50% loss of the drug.
> How can we avoid the question of "fraction"?
My comment was not on T1/2, but on your question with respect to a CL
of 2400 L/day. I don't see any problem here, as I explained in my
message, and we don't need fractions to view the situation in terms of
V and CL (of course, a fraction is involved in the calculation of
T1/2). If you say '1664% per day', some people may not understand, and
you need an explanation, as a result of the 'fraction' (a % is a
fraction, isn't it?).
> when V is 20L and CL is 10L/hr, does that mean after 1 hour, 50%
> of drug is eliminated; after 2 hours, no drug is left?
No. Again the analogy with the filtering of the swimming pool is
simple and straightforward to get the right answer. And again, there
is a problem if you start with rate constants (the result) instead of
starting with the basic parameters: the capacity of the swimming pool
(volume) and filtering system (clearance).
best regards,
Hans
Johannes H. Proost
Dept. of Pharmacokinetics, Toxicology and Targeting
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
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The following message was posted to: PharmPK
Dear Yaning,
Good question - I got that wrong and fell into my own hole. In the
log(concentration)-time plot, the line is straight and the slope is
constant -k. I should have considered the concentration-time plot where
the slope (rate) is -k*C0*exp(-k*t), the tangent to the curve, and it
changes with time while it is negatively proportional to the
concentration.
So the answer to the original question is that the % rate of change is
an instantaneous constant, and its value depends on the units of time.
The reason that it is misleading is that the % ratio is -[dC/dt]/C and
not -[dC/dt]/C0; the former is constant whereas the latter decreases
exponentially over time.
Thanks for clearing that up.
Ted
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The following message was posted to: PharmPK
I have been watching this discussion for a while, some questions became too silly especially they were from established people (excuses me if these are not the right words). Such as connect intrinsic clearance to in vivo clearance without acknowledge the underline assumptions, confuse about rate from the rate constant, etc.
In the comment of "Could you explain why % change in C per unit time (or the tangent to the curve of log(concentration) vs time) keeps changing in your example C C0*exp(-k*t)?", if the author thinks C(t2)=C(t1)*exp(-k(t2-t1)), then the question should never be asked. Co here never meant absolutely time 0, it is the beginning time of your integration. In a fixed time interval (C1-C2)/C1 is constant for first order reaction.
[Only if the time interval is very small - db]
Chuang
Chuang Lu, Ph.D.
Associate Director
DMPK, Drug Safety and Disposition
Millennium, The Takeda Oncology Company
35 Landsdowne Street
Cambridge, MA 02139
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The following message was posted to: PharmPK
Dear Walt:
Thank you for pointing out "the fraction cleared per unit time is the
fraction of what is left at that instant". Even though I demonstrated
this in my survival example for hazard, I did not explicitly mention
this for ke. Basically, the denominator for the fraction is not the
initial amount/concentration of the drug, but the remaining
amount/concentration at that moment. This is why hazard is also called
the conditional instantaneous failure rate, where conditional means the
denominator is changing over time, just like the remaining
amount/concentration for ke. I thought it is obvious from the definition
dA/dt=-ke*A converted to dA/A/dt=-ke where dA/A is the fraction
eliminated during dt relative to A, the remaining amount at that time,
not A0. This "conditional" fraction per unit time is constant over time
in a first-order elimination process. When A is getting smaller over
time, dA over a unit time is also getting smaller, therefore keeping the
"conditional" fraction eliminated per unit time constant. I think when
people say it will be different or changing over time, they most likely
are thinking about dA/A0/dt, where the denominator for the fraction is
the initial amount. That should be obvious from dA/A0/dt=-ke*A/A0 where
A is exponentially declining over time. This concept corresponds to the
unconditional failure rate in survival analysis, which is NOT constant
over time for a first-order process.
I find "conditional" fraction per unit time to explain ke is very easy
to visualize because I can easily link it to the hazard concept in
survival analysis, where everyone can understand the fraction of
subjects that die this month relative to the total number of subjects at
the beginning of this month, the fraction of subjects that die next
month relative to the total number of subjects at the beginning of next
month, and so on. This fraction is constant in a first-order process (or
an exponential distribution), which means that even though the total
number of subjects keeps declining, the number of subjects that die
every month is also declining proportionally, making the ratio constant.
"The contents of this message are mine personally and do not necessarily
reflect any position of the Government or the Food and Drug
Administration."
Yaning Wang, Ph.D.
Team Leader, Pharmacometrics
Office of Clinical Pharmacology
Office of Translational Science
Center for Drug Evaluation and Research
U.S. Food and Drug Administration
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The following message was posted to: PharmPK
Dear Yaning,
You wrote:
> I find "conditional" fraction per unit time to explain ke is very
> easy to visualize because I can easily link it to the hazard
> concept in survival analysis, where everyone can understand the
> fraction of subjects that die this month relative to the total
> number of subjects at the beginning of this month, the fraction
> of subjects that die next month relative to the total number of
> subjects at the beginning of next month, and so on. This fraction
> is constant in a first-order process (or an exponential
> distribution), which means that even though the total number of
> subjects keeps declining, the number of subjects that die
> every month is also declining proportionally, making the ratio
> constant.
You will understand that I prefer to explain that the rate of drug
elimination dA/dt, i.e. the amount of drug eliminated per unit of
time, is at any time proportional to the drug concentration, being the
driving force for both metabolism and excretion; in formula:
dA/dt = CL * C
Due to drug elimination the concentration decreases, so the rate of
elimination also decreases over time. Why bother the audience with
irrelevant and unnecesssary explanations of the interpretation of k?
(even the 'rate constant' experts were discussing about this!).
best regards,
Hans Proost
Johannes H. Proost
Dept. of Pharmacokinetics, Toxicology and Targeting
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
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TED:
Thank you for your clarification (see original message below). it is correct to say for the log(concentration)-time plot, the line is straight and the slope is constant -k (units per hour or minutes) for one compartment model with first order elimination
Can you comment on below:
For the concentration time plot the k parameter referred to in the "slope (rate) is -k*C0*exp(-k*t) the tangent to the curve" varies with time and is perhaps better referred to as an instantaneous rate coefficient (units per hour or minutes)
[Does http://www.boomer.org/c/p4/c04/c0403.html
Fig 4.3.2 and 4.3.3 help? - db]
Then for the first order elimination the first k from the Ln transformed plot is an average value of the individual different k values from the above exp expression.
Do you agree?
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Dear Angus,
In your message to Ted, you wrote:
"it is correct to say for the log(concentration)-time plot, the line is straight and the slope is constant -k (units per hour or minutes) for one compartment model with first order elimination"
Although seemingly rather pedantic of me to point out, it is important to remember that for a *log*(concentration)-time plot (i.e. using "ordinary", base 10 logarithms), the slope is actually -k/2.303. {The relevant equation, assuming first-order drug elimination from a one compartment model, being: logCt = logC0 - (k/2.303)* t.}
Later you state, "...the first k from the Ln transformed plot is an average value of the individual different k values...". I'm afraid I don't understand what is meant by this statement; however, the slope of a *Ln*(concentration)-time plot (i.e. using "natural", base e logarithms), is just -k. {The equation being: lnCt = lnC0 - k*t .}
Best wishes,
Peter
Peter W. Mullen, PhD, FCSFS
KEMIC BIORESEARCH (www.kemic.com)
Kentville
Nova Scotia, B4N 4H8
Canada
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The following message was posted to: PharmPK
Dear Angus,
David has very helpfully annotated some diagrams on boomer
[http://www.boomer.org/c/p4/c04/c0403.html
Fig 4.3.2 and 4.3.3] and yes, they DO HELP enormously. See how the
tangents (rates) in Figure 4.3.2 are all different and only touch the
curve for an instant. They could have the units ng/mL/min or nM/hr or
any combination of concentration /time.
Let me have one more go at clarification. The concentration curve is C
= C0*exp(-k*t) and when log-linear plotting this, I usually transform
the axis in the graphing software rather than transform the
concentration numbers or formula. If we do the latter, the equation
becomes ln(C) = ln(C0) - k*t from which it is easy to see that
this is a straight line with (constant) slope -k. Log values are
dimensionless, so this slope has the units 1/(unit of time).
I still maintain that it is unfortunate that this is called the rate
constant because it gets confused with rate, and it gives the impression
that the rate is constant, which it isn't. So, the result is that I
don't agree with your final sentence regarding average k values. k
really is constant from t=0 to t=infinity, but the rate is NEVER
CONSTANT. Rate = -k*C0*exp(-k*t). If you try to average this, your
answer will vary with the time interval that you use (and the
start-time) and if you extrapolate to infinity, the average is zero!
So to summarise, for this first order or exponential decay - the rate
constant is constant and the rate isn't ... ever.
Happy plotting.
Ted
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Elimination terminology:
Thanks Ted, David and Peter. David your graphical material is excellent. Agreed; I think the term "rate" constant is a little confusing and of course the standard procedure for elimination rate constant evaluation is the regression graph of the Ln transformed plasma concentrations (with respect to time of sample) over the duration of the elimination phase to give the best fit value of elimination rate constant (-k).
Now where this takes me is where I want to go:
Absorption Phase
to the absorption phase, in case of a drug with observed first order rate absorption , then analogous considerations apply. In other words the absorption rate constant (ka) is a constant over the duration time of the absorption phase, but the absorption rate at any instant in time changes.
Then analogous equations apply for evaluation of Ka.
[We may get Walt upset but I do have http://www.boomer.org/c/p4/c08/c0802.html
Fig 8.2.3 - db]
Do you agree?
Angus
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Angus,
David was right - I simply must reply.
Ka is never a constant. It can sometimes be successfully approximated as a
constant, but only for drugs like propranolol that are highly permeable and
highly soluble. I'll call them "very Class 1". It's drugs like these that
give the illusion of a constant Ka. For that matter, CL and Ke are never
constant, but are approximated as constants often enough to lead some to
think you can (almost) always treat them as such. For students, such
approximations provide a way to begin to understand oral absorption and
pharmacokinetics. However, students need to learn that there's much more to
it than that before they go out in to the world and apply themselves to
real-world problems.
[I try ;-)
http://www.boomer.org/c/p4/ja/Fig31a/Fig31a.html
http://www.boomer.org/c/p4/ja/Fig31b/Fig31b.html
- db]
Most of the drugs we see today in development are challenging in various
ways (I guess that's why we're called in to consult - we don't see the easy
ones that can be done the easy way). Simple constant Ka or constant CL or Ke
models would provide very misleading results for these challenging cases.
Best regards,
Walt
Walt Woltosz
Chairman & CEO
Simulations Plus, Inc. (NASDAQ: SLP)
42505 10th Street West
Lancaster, CA 93534-7059
U.S.A.
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Dear Peter,
You wrote to Angus:
> Although seemingly rather pedantic of me to point out, it is important to
> remember that for a *log*(concentration)-time plot (i.e. using "ordinary",
> base 10 logarithms), the slope is actually -k/2.303. {The relevant
> equation, assuming first-order drug elimination from a one compartment
> model, being: logCt = logC0 - (k/2.303)* t.}
Why do you use base 10 logarithms? Using base 10 logarithms in exponential
functions is probably the best way to make simple thing complex and to
introduce mistakes.
> Later you state, "...the first k from the Ln transformed plot is an
> average value of the individual different k values...". I'm afraid I
> don't understand what is meant by this statement; however, the slope of a
> *Ln*(concentration)-time plot (i.e. using "natural", base e logarithms),
> is just -k. {The equation being: lnCt = lnC0 - k*t .}
I fully agree (including the lack of understanding and the use of a rate
constant!). Here Angus is making simple things quite complex.
best regards.
Hans Proost
Johannes H. Proost
Dept. of Pharmacokinetics, Toxicology and Targeting
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
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Hi Hans,
You wrote:
"...Using base 10 logarithms in exponential functions is probably the best way to make simple thing complex and to introduce mistakes."
I agree. Concerning base10 logs, I was merely *emphasizing* the difference between these and natural logs (lns). It is not uncommon, amongst the uninitiated (and even in text books) to see confusion between the two types of logs, especially when just the word 'log' is used. [In my teaching experience, I have also observed that many students of pharmacokinetics today are unfamiliar with plotting concentration-time data on common "semi-log graph paper" (which, of course, is based on log10).]
IMHO, it's imperative that there be no misunderstanding regarding the important and *essential* skill of determining rate constants! :-)
Regards,
Peter
Peter W. Mullen, PhD, FCSFS
KEMIC BIORESEARCH (www.kemic.com)
Kentville
Nova Scotia, B4N 4H8
Canada
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It is just a matter of units. ke should be expressed in minutes rather than hours. If t 1/2= 1h, ke (expressed in minutes) is 1.15% per minute, which is exactly the elimination rate (it works for any half-life, at least I tried up to t1/2=24h).
P.S. - my apologies to Malcolm, my thanks to Peter.
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Dear Peter,
Good to hear that we agree at several points!
>I have also observed that many >students of pharmacokinetics today are unfamiliar with plotting >concentration-time data on common "semi-log graph paper" (which, of >course, is based on log10).]
I don't think this is correct. How do you conclude that 'semi-log graph paper' is based on log10? Using base e (or base 2, or base pi) would result in exactly the same division of number along the axis, only different by some arbitrary scaling factor (also needed for base 10).
[I was interested to see this too. A few years ago I 'researched' the question and there didn't seem to be a consensus on log versus ln. With a new semester started I seem to be moving through this material in class. ;-) I decided to use the log for the 'raw' slope but use ln to calculate kel (described as: from the slope of the line). With a calculator, once you have points from the line you can easily estimate kel using the ln button. - db]
You are probably pointing at the decades (10^0, 10^1, 10^2) indicated along the axis.
[Ah, but are they really decades?... decade = period of ten (years). I've grappled with that one in class too. Last week I started calling them 'powers of ten'. Seemed to go over OK, maybe - db]
This is, however, the result of our decimal system of presenting numbers, and has nothing to do with the base of the logarithmic values.
> IMHO, it's imperative that there be no misunderstanding regarding >the important and *essential* skill of determining rate constants!
Please note: I agree!
best regards,
Hans
Johannes H. Proost
Dept. of Pharmacokinetics, Toxicology and Targeting
University Centre for Pharmacy
Antonius Deusinglaan 1
9713 AV Groningen, The Netherlands
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Dear Angus,
I am not sure where you want to go with this, but I can make a few
observations. First, it is proving very difficult to get everything
right. For example, David's text & diagrams - excellent though they are
- contain the occasional confusion of parameters [sorry, David]. In
Figure 8.2.3 (http://www.boomer.org/c/p4/c08/c0802.html), the Y axis is
concentration but the labels on the tangents refer to amount as do the
equations in the text. While this does not change the visual
appreciation of the shape of the curve, it is a bit confusing,
particularly as there are those who balance equations in amount (what we
dose) and others who use concentration (what we quantify).
[Maybe my slight adjustments will help; Ted, thanks for the observation - db]
To respond to Walt, I say that there are cases where the best models can
be used, eg in PBPK simulations and small numbers of subjects, and there
are other cases where a massive compromise is called for because the
large number of simultaneous differential equations are too many to use
- eg Pop-PK(/PD). I am not a Pop-PK modeller, but I believe that for
large studies, 1 or 2 compartments with zero or first order absorption
are as complicated as can be comfortably managed on affordable
computers. I agree with Walt that no drug has first order absorption,
but a useful model might very reasonably have first order absorption,
and this might be determined objectively by a measure of appropriateness
of model such as AIC. And I agree that it is good to simulate using the
most informed model possible, but it is not possible to parameterise the
model from any realistic number of data points - usually there are more
parameters than timepoints!
Finally, to get to Angus's point, it is difficult to read the curve of
an absorption "phase" because it is (1) asymptotic in an upward
direction and does not simplify on log transformation (natural or
otherwise) and (2) the blood concentration curve is a composite of
absorption and elimination except when t=0 (and ignoring the probable
lag). You can do the subtraction of the observed values from the
back-extrapolated terminal phase and get the absorption exponential that
way. Is that what you meant?
Best regards.
Ted
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Ted; Thanks. yes that is what I am thinking of. For the absorption phase the plot of the plasma concentrations (as is or Ln transformed) represents the resultant of absorption and elimination processes, whereas the plot of the elimination phase (usually looked at as Ln transformed) relates to elimination processes.
I find it useful to look at the Ln transformed plasma concentrations from both phases.
I looked again at David's page. http://www.boomer.org/c/p4/c08/c0802.html
[I did make some changes - db]
The diff. equation for and integral form of the equation for drug remaining to be absorbed (Xg) is shown prior to considering corresponding graphical presentation of the integral form.
On the other hand the diff. equation for amount drug absorbed (Xp) in mg is shown, but the corresponding integral form of the equation for amount of drug in the body (Xp) is not shown. The graph shown is a representation of what the integral equation would look like so the concentration units will be on the y axis (as mass/unit volume) I agree it is not necessary to show the integral form to highlight the point made. The changes made are helpful. Presumably the integral form of the equation is elsewhere in the course.
[On the next page http://www.boomer.org/c/p4/c08/c0803.html
- db]
Angus
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Ted,
You said:
"I am not a Pop-PK modeller, but I believe that for large studies, 1 or 2
compartments with zero or first order absorption
are as complicated as can be comfortably managed on affordable computers. I
agree with Walt that no drug has first order absorption, but a useful model
might very reasonably have first order absorption, and this might be
determined objectively by a measure of appropriateness of model such as AIC.
And I agree that it is good to simulate using the most informed model
possible, but it is not possible to parameterise the model from any
realistic number of data points - usually there are more
parameters than timepoints!"
A full-blown PBPK model for midazolam including the 9-compartment ACAT model
for concentration-gradient-based gut absorption, saturable gut and liver
metabolism, and a 17-compartment PBPK model runs at the rate of about 4
sec/simulation in GastroPlus - on my laptop! Thus, a 1,000 patient Virtual
Trial would take about 4,000 seconds. So a bit over an hour on an affordable
laptop would provide a very reasonable run time with a model that is as
mechanistic as you can get today. By the way, the model uses in vitro Vmax
and Km for 3A4 and all default GastroPlus physiological settings, including
expression levels of 3A4 in gut and liver, and calculated Kp's using our
modified Rodgers method. The prediction is quite close to the observed data
using no human inputs and no fitted parameters.
So not only is speed not an issue, but neither is parameterization.
Best regards,
Walt
Walt Woltosz
Chairman & CEO
Simulations Plus, Inc. (NASDAQ: SLP)
42505 10th Street West
Lancaster, CA 93534-7059
U.S.A.
http://www.simulations-plus.com
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Walt,
that's great. I love to simulate, too, but how would I analyse the data from a 1000 real patient trial?
Best regards.
Ted
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