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Re: where is this going???? chemist

Posted by chess on May 22, 2004, at 7:37:05

In reply to where is this going???? chess, posted by chemist on May 21, 2004, at 22:17:44

i sincerely don't understand why taking more of K qd would mean a longer duration of action since it is the same compound that is being degraded.
so to be more exact with my questioning...
if i took 1mg K qd at 6AM then how long would K's duration of action be?
if i took 2mg K qd at 6AM then how long would K's duration of action be?

thanks again for your all your responses! they have been very helpful to myself and others i have shared them with. you're knowledge is envied and comforting to many, and i apologize if i have been redundant in my eagerness to understand what i am putting in my brain and what it is doing to me

> i am curious about the length of this thread and the redundancies...i am wondering what the point of continuing this dialogue is: we have covered metabolites, half-lives, partitioning, and who knows what else. where is this going, if i might ask? i am confused as to the point of this discourse. if you take 2 mg K instead of 1 mg, then the duration of action will be longer, but not in a linear manner. again, please clarify what it is you are after, as it is not clear to me that the issues have not already been covered. what *exactly* do you want to know? all the best, chemist
> > let's compare "the duration of action of K from taking 1mg K once a day" versus "the duration of action of K from taking 2mg K once a day"
> >
> > > how are you proposing the dosing? qd, bid, etc? all the best, chemist
> > >
> > > > why and how would taking more K also increase the duration of action of K?
> > > >
> > > > > yup. all the best, chemist
> > > > >
> > > > > > i would assume it's well accepted that taking 2mg of K instead of 1mg of K would produce a greater tranquilizing effect.

> > > > > > question: would taking more K also increase the duration of action of K?
> > > > > >
> > > > > > > it occurred to me that i should clarify the half-life business a little...there have been statements (mine, others) that have been made along the lines of ``if the elimination half-life is 30 hours, then if you took 1 mg klonopin, at 30 hours there would be 0.5 mg in your system, and at 60 hours 0.25 mg, and so forth.'' this is not entirely accurate. assuming that the elimination can be described by first-order kinetics, we need to know k, the rate constant, which will be in units of inverse time. going through the math, you end up with the usual expression (i am assuming the activity coefficient for klonopin - K - to be unity, thus [K] == {K}) for t_{1/2} = (ln 2)/k. this yields (assuming t_{1/2} == 30 hours) k = 0.023 hr^{-1}. for the time at which one quarter of your initial dose remains, t_{1/4} = (ln 4)/k = 60.26 hours. the decay is exponential, and my colleagues would berate me if they ever found out that i had not clarified....thus, t_{1/10} = 100.11 hours, and t_{1/1000} = 300.34 hours, which means that 1 microgram of klonopin - which is 3 orders of magnitude of the lowest detection limit i am aware of for the metabolite (i.e., part per trillion) - will be present at 1.8 weeks after the initial 1 mg dose....all the best, chemist
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