# PharmPK Discussion - Naive question on ke and t1/2

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• On 16 Jan 2010 at 09:24:23, Gilberto De Nucci (denucci.at.gilbertodenucci.com) sent the message
`The following message was posted to: PharmPKIn 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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• On 20 Jan 2010 at 12:16:38, veeravalli vijay bhaskar (veeravalli.bhaskar.at.gmail.com) sent the message
`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*KelHope I answered your query, Any doubts do revert back.Regards,Vijay,G7 Synergon Pvt Ltd.`
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• On 22 Jan 2010 at 16:36:17, "J.H. Proost" (j.h.proost.at.rug.nl) sent the message
`The following message was posted to: PharmPKDear 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 theamount 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 notcorrect; it means indeed that at each time point the decrease of plasmaconcentration and amount in the body is 0.231 times the plasma concentrationand amount in the body, respectively. In equations:dC/dt = k * CordA/dt = k * A(please note from the discussion 'A question of clearance' that thisnotations are not mechanistically based, but a re-parametrization of theequations dA/dt = CL * C and A = V * C but I use the rate constant equationshere 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 timesthe initial C and A. Since C and A are decreasing over time, the rate ofdecline dC/dt and dA/dt are also decreasing over time. So, after one hourthere 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% hasbeen 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 meanthat 231% is removed after 1 hour. After one hour, C = C0 * exp(-2.31) = C0* 0.099.best regards,Hans ProostJohannes H. ProostDept. of Pharmacokinetics, Toxicology and TargetingUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlands[There may be something useful onhttp://www.boomer.org/c/p3/c02/c0209.htmlwhich discusses Euler's method, note the effect of step-size - db]`
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• On 22 Jan 2010 at 17:13:28, Peter W Mullen (pmullen.-at-.kemic.com) sent the message
`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,PeterPeter W. Mullen, PhD, FCSFSKEMIC BIORESEARCH (www.kemic.com)KentvilleNova Scotia, B4N 4H8Canada[Check out the wikipedia entryhttp://en.wikipedia.org/wiki/Rate_equationthe 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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• On 23 Jan 2010 at 00:15:56, "Wang, Yaning" (Yaning.Wang.-a-.fda.hhs.gov) sent the message
`The following message was posted to: PharmPKPeter:I think the interpretation of ke as "a rate of approximately 23% perhour" may come from dA/dt=ke*A. Move A on the right side to the leftside: dA/A/dt=ke or %/dt=ke. Therefore ke can be approximatelyinterpreted as the fraction eliminated per unit time. This approximationbecomes more and more precise when you use smaller and smaller timescale, such as minute, second or millisecond (as a result, ke becomessmaller and smaller). It is the same concept as hazard in survivalanalysis. Depending on the time scale, hazard can be even greater than1, which can always be converted to a smaller number (<<1) by changingthe time scale so that "proportion of people dying in a unit timeinterval" can be used to explain the hazard concept."The contents of this message are mine personally and do not necessarilyreflect any position of the Government or the Food and DrugAdministration."Yaning Wang, Ph.D.Team Leader, PharmacometricsOffice of Clinical PharmacologyOffice of Translational ScienceCenter for Drug Evaluation and ResearchU.S. Food and Drug Administration`
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• On 23 Jan 2010 at 13:31:41, "Wang, Yaning" (Yaning.Wang.aaa.fda.hhs.gov) sent the message
`The following message was posted to: PharmPKHi Gilberto,You wrote:"In the book of Rowland, it shows that ke=0.693/t1/2. It is given anexample, stating that if the half-life is 3h, ke would 0.231 h-1 (whichmeans a rate of approximately 23% per hour, which makes sense). What isthe interpretation of a t1/2 of one hour?"You can think of "t1/2 of one hour" as t1/2 of 60 minutes and itscorresponding ke will be 0.693/60=0.0116 min-1, which means a rate ofapproximately 1.16% per minute."The contents of this message are mine personally and do not necessarilyreflect any position of the Government or the Food and DrugAdministration."Yaning Wang, Ph.D.Team Leader, PharmacometricsOffice of Clinical PharmacologyOffice of Translational ScienceCenter for Drug Evaluation and ResearchU.S. Food and Drug Administration`
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• On 24 Jan 2010 at 23:27:26, Nick Holford (n.holford.-a-.auckland.ac.nz) sent the message
`The following message was posted to: PharmPKYaning,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, asI 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. Sowhat possible meaning can we attach to a rate of 1664% per day? Itimplies that more than 100% of the drug is eliminated per day which isobviously impossible. Rate constants cannot be interpreted sensibly inthe way you have tried to use in your example.You ran into a similar problem in an earlier posting when you tried toexplain 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 timeinterval". The hazard is the event rate at an instant in time. It hasthe same dimensions as a PK rate constant i.e. 1/time.  The numericalvalue  depends on the  chosen time unit e.g. 24 per day, or 1 per houror 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 thepopulation to die. In the same way for PK ln(2)/rate constant will tellyou long it takes for half of the drug to be eliminated.Nick--Nick Holford, Professor Clinical PharmacologyDept Pharmacology & Clinical PharmacologyUniversity of Auckland,85 Park Rd,Private Bag 92019,Auckland,New Zealand`
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• On 24 Jan 2010 at 15:07:56, Peter W Mullen (pmullen.aaa.kemic.com) sent the message
`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,PeterPeter W. Mullen, PhD, FCSFSKEMIC BIORESEARCH (www.kemic.com)KentvilleNova Scotia, B4N 4H8Canada`
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• On 24 Jan 2010 at 23:35:38, "Wang, Yaning" (Yaning.Wang.at.fda.hhs.gov) sent the message
`The following message was posted to: PharmPKNick:Your wrote:"By analogy this means the rate is 69.3% per hour or 1664% per day. Itimplies that more than 100% of the drug is eliminated per day which isobviously impossible"As I explained earlier, it is an approximation and the approximation ismore 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 askyou a similar question: when V is 20L and CL is 10L/hr, what is theexplanation of CL of 2400L/day? Whatever answer you come up with can beused to answer 1664% per day question too.I actually found "fraction eliminated per unit time" a very useful andintuitively appealing explanation of rate constant when I convert it toa smaller time scale. When I first learned survival analysis, I foundhazard a very difficult concept to be translated into a simple concepteveryone can understand. When I finally linked Ke with hazard, I felt"proportion of people dying in a unit time interval (among the peoplesurviving up to the starting time point of the interval, to bestatistically exact)" a very easy explanation of hazard. In fact, thedefinition of hazard starts from this concept and converts to a limitconcept (when dt goes to infinitely small). My favorite example todemonstrate this concept when I teach survival analysis is: at a certaintime point, if there are 100 subjects alive and 10 people die in thenext month, the hazard during that month is approximately 0.1 month-1;if another 10 subjects die in the following month, the hazard during thefollowing month is 1/9 month-1. I am glad that you noticed hazard has aunit 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 thehazard in the unit of month-1 rather than decade-1 even though you canconvert numerically. For those repeatable event, hazard>1 has ameaningful explanation. But can you translate a hazard of 2 per year fornon-repeatable event (such as death) into something a layman canunderstand?"The contents of this message are mine personally and do not necessarilyreflect any position of the Government or the Food and DrugAdministration."Yaning Wang, Ph.D.Team Leader, PharmacometricsOffice of Clinical PharmacologyOffice of Translational ScienceCenter for Drug Evaluation and ResearchU.S. Food and Drug Administration`
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• On 25 Jan 2010 at 10:16:25, (Ted.Parton.aaa.ucb.com) sent the message
`The following message was posted to: PharmPKDear 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 therate constant is converted to the % change in C per unit time, theresult is the INSTANTANEOUS RATE of change; it is the tangent to thecurve of log(concentration) vs time plot, and as it is the rate, not therate constant, it keeps changing.A better analogy for this is in radioactivity. The radioactive half-lifeis a familiar concept, and those who use short half-life isotopes suchas 32P or PET isotopes 11C and 18F know that corrections have to be madeto the specific activity as the batch of isotope ages. The half-life isconstant (Geiger counters usually average over a few seconds) as long asthe number of radionuclei is large enough for statistically smoothchange, but when the number of radionuclei is countable, then the rateof disintegration becomes slow and the time between events becomesextremely variable; this is more like a survival curve.For more complicated functions, the concept of half-life is rarelyuseful and often extremely misleading, which is why it is so often thesubject of discussion in this group. I am not aware of a constanthalf-life for human life - the probability of death is not constant inmy experience, though some causes might be approximately constant for atime, 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 thecommoner mistakes.Regards.Ted`
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• On 25 Jan 2010 at 15:20:14, "J.H. Proost" (j.h.proost.-a-.rug.nl) sent the message
`The following message was posted to: PharmPKDear 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 itavoids the question of 'fraction'. Clearance is not a fraction (as I'vstated in the beginning of our 'A question of clearance' discussion). It isthe capacity of the system to eliminate the drug. The analogy with afiltering system of a swimming pool can be helpful, also to understand (yes,I must take this opportunity!) that half-life and rate constant aredepending on the volume and filtering capacity.best regards,Hans ProostJohannes H. ProostDept. of Pharmacokinetics, Toxicology and TargetingUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlands`
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• On 25 Jan 2010 at 22:53:09, "Wang, Yaning" (Yaning.Wang.-at-.fda.hhs.gov) sent the message
`The following message was posted to: PharmPKHans: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 weavoid the question of "fraction"? Maybe I should have put the questionin a simpler and more straightforward way: when V is 20L and CL is10L/hr, does that mean after 1 hour, 50% of drug is eliminated; after 2hours, no drug is left?"The contents of this message are mine personally and do not necessarilyreflect any position of the Government or the Food and DrugAdministration."Yaning Wang, Ph.D.Team Leader, PharmacometricsOffice of Clinical PharmacologyOffice of Translational ScienceCenter for Drug Evaluation and ResearchU.S. Food and Drug Administration`
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• On 25 Jan 2010 at 23:05:24, "Wang, Yaning" (Yaning.Wang.aaa.fda.hhs.gov) sent the message
`The following message was posted to: PharmPKDear 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 thecurve of log(concentration) vs time plot, and as it is the rate, not therate constant, it keeps changing"Could you explain why % change in C per unit time (or the tangent to thecurve 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 necessarilyreflect any position of the Government or the Food and DrugAdministration."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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• On 25 Jan 2010 at 20:58:54, "Walt Woltosz" (walt.aaa.simulations-plus.com) sent the message
`The following message was posted to: PharmPKI 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 Kewill have diminishing effects as the remaining amount/concentration isreduced, which is why we see exponential functions as reasonableapproximations in many cases. Only through integrating the differentialequation will you get the answer at any future time point. You can't treateither one as linear, and that's why you can't say that if CL = 10 L/h andVolume = 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 constantamount vs time if the concentration stayed the same. But the concentrationis diminishing (assuming no further dosing), so clearing 10 L/h when theconcentration is, say 100 ng/mL, is not clearing the same amount per unittime when the concentration drops to 99 ng/mL. The same goes for Ke - it isa rate constant/coefficient that is being applied to a diminishing amount orconcentration, so the fraction cleared per unit time is the fraction of whatis left at that instant, and will be different in the next instant.Clearing a hypothetical volume per unit time that has a changingconcentration is no easier or more difficult for me to visualize thanclearing a fraction per unit time. They both apply to the instantaneousamount or concentration at that point in time.Perhaps I'm stating the obvious, but it seems from some of the posts that itmay not be.Walt WoltoszChairman & CEOSimulations Plus, Inc. (NASDAQ: SLP)42505 10th Street WestLancaster, CA  93534-7059U.S.A.http://www.simulations-plus.com`
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• On 26 Jan 2010 at 09:19:10, "J.H.Proost" (J.H.Proost.-at-.rug.nl) sent the message
`The following message was posted to: PharmPKDear 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 CLof 2400 L/day. I don't see any problem here, as I explained in mymessage, and we don't need fractions to view the situation in terms ofV and CL (of course, a fraction is involved in the calculation ofT1/2). If you say '1664% per day', some people may not understand, andyou need an explanation, as a result of the 'fraction' (a % is afraction, 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 issimple and straightforward to get the right answer. And again, thereis a problem if you start with rate constants (the result) instead ofstarting with the basic parameters: the capacity of the swimming pool(volume) and filtering system (clearance).best regards,HansJohannes H. ProostDept. of Pharmacokinetics, Toxicology and TargetingUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlands`
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• On 26 Jan 2010 at 09:25:40, (Ted.Parton.-a-.ucb.com) sent the message
`The following message was posted to: PharmPKDear Yaning,Good question - I got that wrong and fell into my own hole. In thelog(concentration)-time plot, the line is straight and the slope isconstant -k. I should have considered the concentration-time plot wherethe slope (rate) is -k*C0*exp(-k*t), the tangent to the curve, and itchanges with time while it is negatively proportional to theconcentration.So the answer to the original question is that the % rate of change isan 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 andnot -[dC/dt]/C0; the former is constant whereas the latter decreasesexponentially over time.Thanks for clearing that up.Ted`
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• On 26 Jan 2010 at 09:35:56, "Lu, Chuang" (Chuang.Lu.aaa.MPI.com) sent the message
`The following message was posted to: PharmPKI 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]ChuangChuang Lu, Ph.D.Associate DirectorDMPK, Drug Safety and DispositionMillennium, The Takeda Oncology Company35 Landsdowne StreetCambridge, MA 02139`
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• On 26 Jan 2010 at 09:57:13, "Wang, Yaning" (Yaning.Wang.at.fda.hhs.gov) sent the message
`The following message was posted to: PharmPKDear Walt:Thank you for pointing out "the fraction cleared per unit time is thefraction of what is left at that instant". Even though I demonstratedthis in my survival example for hazard, I did not explicitly mentionthis for ke. Basically, the denominator for the fraction is not theinitial amount/concentration of the drug, but the remainingamount/concentration at that moment. This is why hazard is also calledthe conditional instantaneous failure rate, where conditional means thedenominator is changing over time, just like the remainingamount/concentration for ke. I thought it is obvious from the definitiondA/dt=-ke*A converted to dA/A/dt=-ke where dA/A is the fractioneliminated during dt relative to A, the remaining amount at that time,not A0. This "conditional" fraction per unit time is constant over timein a first-order elimination process. When A is getting smaller overtime, dA over a unit time is also getting smaller, therefore keeping the"conditional" fraction eliminated per unit time constant. I think whenpeople say it will be different or changing over time, they most likelyare thinking about dA/A0/dt, where the denominator for the fraction isthe initial amount. That should be obvious from dA/A0/dt=-ke*A/A0 whereA is exponentially declining over time. This concept corresponds to theunconditional failure rate in survival analysis, which is NOT constantover time for a first-order process.I find "conditional" fraction per unit time to explain ke is very easyto visualize because I can easily link it to the hazard concept insurvival analysis, where everyone can understand the fraction ofsubjects that die this month relative to the total number of subjects atthe beginning of this month, the fraction of subjects that die nextmonth relative to the total number of subjects at the beginning of nextmonth, and so on. This fraction is constant in a first-order process (oran exponential distribution), which means that even though the totalnumber of subjects keeps declining, the number of subjects that dieevery month is also declining proportionally, making the ratio constant."The contents of this message are mine personally and do not necessarilyreflect any position of the Government or the Food and DrugAdministration."Yaning Wang, Ph.D.Team Leader, PharmacometricsOffice of Clinical PharmacologyOffice of Translational ScienceCenter for Drug Evaluation and ResearchU.S. Food and Drug Administration`
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• On 28 Jan 2010 at 15:54:24, "J.H.Proost" (J.H.Proost.aaa.rug.nl) sent the message
`The following message was posted to: PharmPKDear 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 drugelimination dA/dt, i.e. the amount of drug eliminated per unit oftime, is at any time proportional to the drug concentration, being thedriving force for both metabolism and excretion; in formula:dA/dt = CL * CDue to drug elimination the concentration decreases, so the rate ofelimination also decreases over time. Why bother the audience withirrelevant and unnecesssary explanations of the interpretation of k?(even the 'rate constant' experts were discussing about this!).best regards,Hans ProostJohannes H. ProostDept. of Pharmacokinetics, Toxicology and TargetingUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlands`
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• On 29 Jan 2010 at 11:03:21, Angusmdmclean.-at-.aol.com sent the message
`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 eliminationCan 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.htmlFig 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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• On 29 Jan 2010 at 14:01:09, Peter W Mullen (pmullen.-at-.kemic.com) sent the message
`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,PeterPeter W. Mullen, PhD, FCSFSKEMIC BIORESEARCH (www.kemic.com)KentvilleNova Scotia, B4N 4H8Canada`
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• On 29 Jan 2010 at 18:19:12, (Ted.Parton.-a-.ucb.com) sent the message
`The following message was posted to: PharmPKDear Angus,David has very helpfully annotated some diagrams on boomer[http://www.boomer.org/c/p4/c04/c0403.htmlFig 4.3.2 and 4.3.3] and yes, they DO HELP enormously. See how thetangents (rates) in Figure 4.3.2 are all different and only touch thecurve for an instant. They could have the units ng/mL/min or nM/hr orany 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 transformthe axis in the graphing software rather than transform theconcentration numbers or formula. If we do the latter, the equationbecomes    ln(C) = ln(C0) - k*t    from which it is easy to see thatthis is a straight line with (constant) slope -k. Log values aredimensionless, so this slope has the units   1/(unit of time).I still maintain that it is unfortunate that this is called the rateconstant because it gets confused with rate, and it gives the impressionthat the rate is constant, which it isn't. So, the result is that Idon't agree with your final sentence regarding average k values. kreally is constant from t=0 to t=infinity,  but the rate is NEVERCONSTANT. Rate = -k*C0*exp(-k*t). If you try to average this, youranswer will vary with the time interval that you use (and thestart-time) and if you extrapolate to infinity, the average is zero!So to summarise, for this first order or exponential decay - the rateconstant is constant and the rate isn't ... ever.Happy plotting.Ted`
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• On 29 Jan 2010 at 14:39:47, Angusmdmclean.at.aol.com sent the message
`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.htmlFig 8.2.3 - db]Do you agree?Angus`
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• On 29 Jan 2010 at 17:21:31, "Walt Woltosz" (walt.at.simulations-plus.com) sent the message
`The following message was posted to: PharmPKAngus,David was right - I simply must reply.Ka is never a constant. It can sometimes be successfully approximated as aconstant, but only for drugs like propranolol that are highly permeable andhighly soluble.  I'll call them "very Class 1". It's drugs like these thatgive the illusion of a constant Ka. For that matter, CL and Ke are neverconstant, but are approximated as constants often enough to lead some tothink you can (almost) always treat them as such. For students, suchapproximations provide a way to begin to understand oral absorption andpharmacokinetics. However, students need to learn that there's much more toit than that before they go out in to the world and apply themselves toreal-world problems.[I try ;-)http://www.boomer.org/c/p4/ja/Fig31a/Fig31a.htmlhttp://www.boomer.org/c/p4/ja/Fig31b/Fig31b.html- db]Most of the drugs we see today in development are challenging in variousways (I guess that's why we're called in to consult - we don't see the easyones that can be done the easy way). Simple constant Ka or constant CL or Kemodels would provide very misleading results for these challenging cases.Best regards,WaltWalt WoltoszChairman & CEOSimulations Plus, Inc. (NASDAQ: SLP)42505 10th Street WestLancaster, CA  93534-7059U.S.A.`
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• On 1 Feb 2010 at 13:40:32, "J.H. Proost" (j.h.proost.-at-.rug.nl) sent the message
`The following message was posted to: PharmPKDear 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 exponentialfunctions is probably the best way to make simple thing complex and tointroduce 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 rateconstant!). Here Angus is making simple things quite complex.best regards.Hans ProostJohannes H. ProostDept. of Pharmacokinetics, Toxicology and TargetingUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlands`
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• On 1 Feb 2010 at 14:16:11, Peter W Mullen (pmullen.-a-.kemic.com) sent the message
`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,PeterPeter W. Mullen, PhD, FCSFSKEMIC BIORESEARCH (www.kemic.com)KentvilleNova Scotia, B4N 4H8Canada`
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• On 1 Feb 2010 at 22:10:29, Gilberto De Nucci (denucci.-at-.gilbertodenucci.com) sent the message
`The following message was posted to: PharmPKIt 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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• On 2 Feb 2010 at 09:12:48, "J.H.Proost" (J.H.Proost.-at-.rug.nl) sent the message
`The following message was posted to: PharmPKDear 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,HansJohannes H. ProostDept. of Pharmacokinetics, Toxicology and TargetingUniversity Centre for PharmacyAntonius Deusinglaan 19713 AV Groningen, The Netherlands`
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• On 2 Feb 2010 at 18:12:25, (Ted.Parton.-a-.ucb.com) sent the message
`The following message was posted to: PharmPKDear Angus,I am not sure where you want to go with this, but I can make a fewobservations. First, it is proving very difficult to get everythingright. For example, David's text & diagrams - excellent though they are- contain the occasional confusion of parameters [sorry, David]. InFigure 8.2.3 (http://www.boomer.org/c/p4/c08/c0802.html), the Y axis isconcentration but the labels on the tangents refer to amount as do theequations in the text. While this does not change the visualappreciation of the shape of the curve, it is a bit confusing,particularly as there are those who balance equations in amount (what wedose) 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 canbe used, eg in PBPK simulations and small numbers of subjects, and thereare other cases where a massive compromise is called for because thelarge 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 forlarge studies, 1 or 2 compartments with zero or first order absorptionare as complicated as can be comfortably managed on affordablecomputers. 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 appropriatenessof model such as AIC. And I agree that it is good to simulate using themost informed model possible, but it is not possible to parameterise themodel from any realistic number of data points - usually there are moreparameters than timepoints!Finally, to get to Angus's point, it is difficult to read the curve ofan absorption "phase" because it is (1) asymptotic in an upwarddirection and does not simplify on log transformation (natural orotherwise) and (2) the blood concentration curve is a composite ofabsorption and elimination except when t=0 (and ignoring the probablelag). You can do the subtraction of the observed values from theback-extrapolated terminal phase and get the absorption exponential thatway. Is that what you meant?Best regards.Ted`
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• On 4 Feb 2010 at 12:22:38, Angusmdmclean.at.aol.com sent the message
`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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• On 9 Feb 2010 at 13:12:09, "Walt Woltosz" (walt.-a-.simulations-plus.com) sent the message
`The following message was posted to: PharmPKTed,You said:"I am not a Pop-PK modeller, but I believe that for large studies, 1 or 2compartments with zero or first order absorptionare as complicated as can be comfortably managed on affordable computers. Iagree with Walt that no drug has first order absorption, but a useful modelmight very reasonably have first order absorption, and this might bedetermined objectively by a measure of appropriateness of model such as AIC.And I agree that it is good to simulate using the most informed modelpossible, but it is not possible to parameterise the model from anyrealistic number of data points - usually there are moreparameters than timepoints!"A full-blown PBPK model for midazolam including the 9-compartment ACAT modelfor concentration-gradient-based gut absorption, saturable gut and livermetabolism, and a 17-compartment PBPK model runs at the rate of about 4sec/simulation in GastroPlus - on my laptop! Thus, a 1,000 patient VirtualTrial would take about 4,000 seconds. So a bit over an hour on an affordablelaptop would provide a very reasonable run time with a model that is asmechanistic as you can get today. By the way, the model uses in vitro Vmaxand Km for 3A4 and all default GastroPlus physiological settings, includingexpression levels of 3A4 in gut and liver, and calculated Kp's using ourmodified Rodgers method. The prediction is quite close to the observed datausing no human inputs and no fitted parameters.So not only is speed not an issue, but neither is parameterization.Best regards,WaltWalt WoltoszChairman & CEOSimulations Plus, Inc. (NASDAQ: SLP)42505 10th Street WestLancaster, CA  93534-7059U.S.A.http://www.simulations-plus.com`
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• On 10 Feb 2010 at 06:08:23, (Ted.Parton.at.ucb.com) sent the message
`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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