Psycho-Babble Medication Thread 347588

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Re: Chemist...question » chess

Posted by chemist on May 16, 2004, at 23:14:38

In reply to Chemist...question, posted by chess on May 16, 2004, at 21:37:06

> Chemist
>
> What's the difference between the half-life of a drug and its duration of action?
>
> Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
>
> Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?

hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist

 

Re: Chemist...question

Posted by poppi on May 17, 2004, at 2:47:46

In reply to Re: Chemist...question » chess, posted by chemist on May 16, 2004, at 23:14:38

Well, I don't know about him or her

But you just confused the heck out of me. But then, that was what you intended! Correcto??

popi

 

Re: Chemist...questionChemist

Posted by chess on May 17, 2004, at 7:45:24

In reply to Re: Chemist...question » chess, posted by chemist on May 16, 2004, at 23:14:38

chemist,

so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?


> > Chemist
> >
> > What's the difference between the half-life of a drug and its duration of action?
> >
> > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> >
> > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
>
> hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist

 

Re: Chemist...question » poppi

Posted by chemist on May 17, 2004, at 11:58:09

In reply to Re: Chemist...question, posted by poppi on May 17, 2004, at 2:47:46

> Well, I don't know about him or her
>
> But you just confused the heck out of me. But then, that was what you intended! Correcto??
>
> popi

no, i did not intend to confuse, only tried to answer the question.....all the best, chemist

 

Re: Chemist...questionChemist » chess

Posted by chemist on May 17, 2004, at 12:09:32

In reply to Re: Chemist...questionChemist, posted by chess on May 17, 2004, at 7:45:24

> chemist,
>
> so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?
>
>
**** the deal is: if you were to measure for the presence of klonopin in your body - and i mean all throughout your body - after a dose, then somewhere beteen 18 and 50 hours, the number (let's say you took 1 mg) would be 0.5 mg. the time between when you took the 1 mg and the peak amount found in your blood is 2 hours. so, for the next 6 to 10 hours, there is enough drug in your system to do what it needs to do to the receptor it targets, which is becoming desensitized over time. now, there is still klonopin in your body. the plasma-bound fraction results immediately after ingestion and some hepatic metabolism, and then it rides along into your brain. but now we are talking 12 hours later, and frankly, there isn't much of the `mobile'' drug piggybacking on proteins in your blood to keep the anxiolytic effect at full-strength. but there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction. but this is why you take another dose 12 hours later, to boost it back up. if you stop taking klonopin (slowly!), then after a week or so your body would probably be quite literally clear of any and all traces of the drug. but 2 or 3 days into your taper, you could still find it lingering around in fatty tissues, maybe some in the blood, etc......all the best, chemist
>
>
> > > Chemist
> > >
> > > What's the difference between the half-life of a drug and its duration of action?
> > >
> > > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> > >
> > > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
> >
> > hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist
>
>

 

Re: Chemist...questionChemist » chemist

Posted by chess on May 18, 2004, at 0:04:28

In reply to Re: Chemist...questionChemist » chess, posted by chemist on May 17, 2004, at 12:09:32

chemist, I think I got it, but please correct me if I'm wrong with the following example ...

accepting 8 hours as being the duration of action of klonopin, and 18 hours being the half-life of klonopin, is the following example correct?:

6AM: I take a 1mg tab of Klonopin
8AM-2PM : there is 1mg in my blood
2PM: amount in my blood starts to fall below 1mg
12midnight: there is .5mg in my blood

> > chemist,
> > so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?
> >
> >
> **** the deal is: if you were to measure for the presence of klonopin in your body - and i mean all throughout your body - after a dose, then somewhere beteen 18 and 50 hours, the number (let's say you took 1 mg) would be 0.5 mg. the time between when you took the 1 mg and the peak amount found in your blood is 2 hours. so, for the next 6 to 10 hours, there is enough drug in your system to do what it needs to do to the receptor it targets, which is becoming desensitized over time. now, there is still klonopin in your body. the plasma-bound fraction results immediately after ingestion and some hepatic metabolism, and then it rides along into your brain. but now we are talking 12 hours later, and frankly, there isn't much of the `mobile'' drug piggybacking on proteins in your blood to keep the anxiolytic effect at full-strength. but there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction. but this is why you take another dose 12 hours later, to boost it back up. if you stop taking klonopin (slowly!), then after a week or so your body would probably be quite literally clear of any and all traces of the drug. but 2 or 3 days into your taper, you could still find it lingering around in fatty tissues, maybe some in the blood, etc......all the best, chemist
> >
> >
> > > > Chemist
> > > >
> > > > What's the difference between the half-life of a drug and its duration of action?
> > > >
> > > > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> > > >
> > > > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
> > >
> > > hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist
> >
> >
>

 

Re: Chemist...questionChemist » chess

Posted by chemist on May 18, 2004, at 0:27:12

In reply to Re: Chemist...questionChemist » chemist, posted by chess on May 18, 2004, at 0:04:28

> chemist, I think I got it, but please correct me if I'm wrong with the following example ...
>
> accepting 8 hours as being the duration of action of klonopin, and 18 hours being the half-life of klonopin, is the following example correct?:
>
> 6AM: I take a 1mg tab of Klonopin
> 8AM-2PM : there is 1mg in my blood
> 2PM: amount in my blood starts to fall below 1mg
> 12midnight: there is .5mg in my blood
>
>
*** almost. given an 85% plasma-bound concentration, at midnight, you are looking more likely at 0.425 mg plasma-bound....but this is splitting hairs, you are on the right track! all the best, and please let me know if this helps or if there are data that indicate otherwise....all the best, chemist***
>
> > > chemist,
> > > so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?
> > >
> > >
> > **** the deal is: if you were to measure for the presence of klonopin in your body - and i mean all throughout your body - after a dose, then somewhere beteen 18 and 50 hours, the number (let's say you took 1 mg) would be 0.5 mg. the time between when you took the 1 mg and the peak amount found in your blood is 2 hours. so, for the next 6 to 10 hours, there is enough drug in your system to do what it needs to do to the receptor it targets, which is becoming desensitized over time. now, there is still klonopin in your body. the plasma-bound fraction results immediately after ingestion and some hepatic metabolism, and then it rides along into your brain. but now we are talking 12 hours later, and frankly, there isn't much of the `mobile'' drug piggybacking on proteins in your blood to keep the anxiolytic effect at full-strength. but there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction. but this is why you take another dose 12 hours later, to boost it back up. if you stop taking klonopin (slowly!), then after a week or so your body would probably be quite literally clear of any and all traces of the drug. but 2 or 3 days into your taper, you could still find it lingering around in fatty tissues, maybe some in the blood, etc......all the best, chemist
> > >
> > >
> > > > > Chemist
> > > > >
> > > > > What's the difference between the half-life of a drug and its duration of action?
> > > > >
> > > > > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> > > > >
> > > > > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
> > > >
> > > > hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist
> > >
> > >
> >
>
>

 

Re: Chemist...questionChemist » chemist

Posted by chess on May 18, 2004, at 1:44:37

In reply to Re: Chemist...questionChemist » chess, posted by chemist on May 18, 2004, at 0:27:12

chemist,

before asking more questions, let me say THANK YOU!, and also say that you are more than helpful!

question ... if I take 1mg klonopin at 6AM, and around midnight there is around .5mg still in me, is that .5mg still affecting gaba receptors in my brain? Or is that .5mg at midnight of inert quality and having no affect on my brain gaba receptors?

question ... when you said “12 hours after taking klonopin there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction”, what do you mean by "less-accessible"?


>>
>>
>>
>>chemist, I think I got it, but please correct me if I'm wrong with the following example ...
> > accepting 8 hours as being the duration of action of klonopin, and 18 hours being the half-life of klonopin, is the following example correct?:
> > 6AM: I take a 1mg tab of Klonopin
> > 8AM-2PM : there is 1mg in my blood
> > 2PM: amount in my blood starts to fall below 1mg
> > 12midnight: there is .5mg in my blood
> *** almost. given an 85% plasma-bound concentration, at midnight, you are looking more likely at 0.425 mg plasma-bound....but this is splitting hairs, you are on the right track! all the best, and please let me know if this helps or if there are data that indicate otherwise....all the best, chemist***
> > > > chemist,
> > > > so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?
> > > **** the deal is: if you were to measure for the presence of klonopin in your body - and i mean all throughout your body - after a dose, then somewhere beteen 18 and 50 hours, the number (let's say you took 1 mg) would be 0.5 mg. the time between when you took the 1 mg and the peak amount found in your blood is 2 hours. so, for the next 6 to 10 hours, there is enough drug in your system to do what it needs to do to the receptor it targets, which is becoming desensitized over time. now, there is still klonopin in your body. the plasma-bound fraction results immediately after ingestion and some hepatic metabolism, and then it rides along into your brain. but now we are talking 12 hours later, and frankly, there isn't much of the `mobile'' drug piggybacking on proteins in your blood to keep the anxiolytic effect at full-strength. but there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction. but this is why you take another dose 12 hours later, to boost it back up. if you stop taking klonopin (slowly!), then after a week or so your body would probably be quite literally clear of any and all traces of the drug. but 2 or 3 days into your taper, you could still find it lingering around in fatty tissues, maybe some in the blood, etc......all the best, chemist
> > > > > > Chemist
> > > > > > What's the difference between the half-life of a drug and its duration of action?
> > > > > > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> > > > > > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
> > > > >
> > > > > hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist

 

Re: Chemist...questionChemist » chess

Posted by chemist on May 18, 2004, at 14:15:30

In reply to Re: Chemist...questionChemist » chemist, posted by chess on May 18, 2004, at 1:44:37

well, thank you...the ~ 0.5 mg klonopin still in you at midnight will be hitting your GABA receptor BUT not with the heft of the plasma-bound fraction, and will include active metabolites of klonopin more than the parent. the plasma-bound fraction is more accessible in that it crosses the blood/brain barrier, and the hydrophobicity of klonopin (relatively low solubility in water) means it ``prefers'' to partition into something other than water, such as a protein matrix that does contain chemical groups that in turn make the whole mess soluble and thus transport is enhanced. the non-plasma-bound fraction is the stuff that partitions into your fatty tissues, and it's harder to remove the drug from these depositories than from the adsorbed/incorporated goods in your serum.....hope this helps, all the best, chemist

> chemist,
>
> before asking more questions, let me say THANK YOU!, and also say that you are more than helpful!
>
> question ... if I take 1mg klonopin at 6AM, and around midnight there is around .5mg still in me, is that .5mg still affecting gaba receptors in my brain? Or is that .5mg at midnight of inert quality and having no affect on my brain gaba receptors?
>
> question ... when you said “12 hours after taking klonopin there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction”, what do you mean by "less-accessible"?
>
>
>
>
> >>
> >>
> >>
> >>chemist, I think I got it, but please correct me if I'm wrong with the following example ...
> > > accepting 8 hours as being the duration of action of klonopin, and 18 hours being the half-life of klonopin, is the following example correct?:
> > > 6AM: I take a 1mg tab of Klonopin
> > > 8AM-2PM : there is 1mg in my blood
> > > 2PM: amount in my blood starts to fall below 1mg
> > > 12midnight: there is .5mg in my blood
> > *** almost. given an 85% plasma-bound concentration, at midnight, you are looking more likely at 0.425 mg plasma-bound....but this is splitting hairs, you are on the right track! all the best, and please let me know if this helps or if there are data that indicate otherwise....all the best, chemist***
> > > > > chemist,
> > > > > so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?
> > > > **** the deal is: if you were to measure for the presence of klonopin in your body - and i mean all throughout your body - after a dose, then somewhere beteen 18 and 50 hours, the number (let's say you took 1 mg) would be 0.5 mg. the time between when you took the 1 mg and the peak amount found in your blood is 2 hours. so, for the next 6 to 10 hours, there is enough drug in your system to do what it needs to do to the receptor it targets, which is becoming desensitized over time. now, there is still klonopin in your body. the plasma-bound fraction results immediately after ingestion and some hepatic metabolism, and then it rides along into your brain. but now we are talking 12 hours later, and frankly, there isn't much of the `mobile'' drug piggybacking on proteins in your blood to keep the anxiolytic effect at full-strength. but there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction. but this is why you take another dose 12 hours later, to boost it back up. if you stop taking klonopin (slowly!), then after a week or so your body would probably be quite literally clear of any and all traces of the drug. but 2 or 3 days into your taper, you could still find it lingering around in fatty tissues, maybe some in the blood, etc......all the best, chemist
> > > > > > > Chemist
> > > > > > > What's the difference between the half-life of a drug and its duration of action?
> > > > > > > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> > > > > > > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
> > > > > >
> > > > > > hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist
>

 

Re: Chemist...questionChemist » chemist

Posted by chess on May 18, 2004, at 14:39:19

In reply to Re: Chemist...questionChemist » chess, posted by chemist on May 18, 2004, at 14:15:30

> well, thank you...the ~ 0.5 mg klonopin still in you at midnight will be hitting your GABA receptor BUT not with the heft of the plasma-bound fraction
-----does that mean that fewer gaba receptors are being hit as time goes by, or does that mean that klonopin is still hitting the same amount of gaba receptors but just not as powerfully as before?


, and will include active metabolites of klonopin more than the parent. the plasma-bound fraction is more accessible in that it crosses the blood/brain barrier,
-----does the blood-brain barrier allow hydrophilic meds or lipophilic meds to pass through it?


and the hydrophobicity of klonopin (relatively low solubility in water) means it ``prefers'' to partition into something other than water
-----does that mean that klonopin is lipophilic?






, such as a protein matrix that does contain chemical groups that in turn make the whole mess soluble and thus transport is enhanced. the non-plasma-bound fraction is the stuff that partitions into your fatty tissues, and it's harder to remove the drug from these depositories than from the adsorbed/incorporated goods in your serum.....hope this helps, all the best, chemist
> > chemist,
> > before asking more questions, let me say THANK YOU!, and also say that you are more than helpful!
> > question ... if I take 1mg klonopin at 6AM, and around midnight there is around .5mg still in me, is that .5mg still affecting gaba receptors in my brain? Or is that .5mg at midnight of inert quality and having no affect on my brain gaba receptors?
> > question ... when you said “12 hours after taking klonopin there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction”, what do you mean by "less-accessible"?
> > >>
> > >>chemist, I think I got it, but please correct me if I'm wrong with the following example ...
> > > > accepting 8 hours as being the duration of action of klonopin, and 18 hours being the half-life of klonopin, is the following example correct?:
> > > > 6AM: I take a 1mg tab of Klonopin
> > > > 8AM-2PM : there is 1mg in my blood
> > > > 2PM: amount in my blood starts to fall below 1mg
> > > > 12midnight: there is .5mg in my blood
> > > *** almost. given an 85% plasma-bound concentration, at midnight, you are looking more likely at 0.425 mg plasma-bound....but this is splitting hairs, you are on the right track! all the best, and please let me know if this helps or if there are data that indicate otherwise....all the best, chemist***
> > > > > > chemist,
> > > > > > so then what you're saying is that klonopin lasts 18-50 hours in the body BUT it is only bioavailable to get into the brain and work therapeutically for only 8-12 hours BECAUSE after 12 hours it begins to bind to plasma in the body and thus is no longer bioavailable to get into the brain?
> > > > > **** the deal is: if you were to measure for the presence of klonopin in your body - and i mean all throughout your body - after a dose, then somewhere beteen 18 and 50 hours, the number (let's say you took 1 mg) would be 0.5 mg. the time between when you took the 1 mg and the peak amount found in your blood is 2 hours. so, for the next 6 to 10 hours, there is enough drug in your system to do what it needs to do to the receptor it targets, which is becoming desensitized over time. now, there is still klonopin in your body. the plasma-bound fraction results immediately after ingestion and some hepatic metabolism, and then it rides along into your brain. but now we are talking 12 hours later, and frankly, there isn't much of the `mobile'' drug piggybacking on proteins in your blood to keep the anxiolytic effect at full-strength. but there is still klonopin in your body, it's just less-accessible than the plasma-bound (and mostly used up by now) fraction. but this is why you take another dose 12 hours later, to boost it back up. if you stop taking klonopin (slowly!), then after a week or so your body would probably be quite literally clear of any and all traces of the drug. but 2 or 3 days into your taper, you could still find it lingering around in fatty tissues, maybe some in the blood, etc......all the best, chemist
> > > > > > > > Chemist
> > > > > > > > What's the difference between the half-life of a drug and its duration of action?
> > > > > > > > Like klonopin for example. I've read that its half-life is 18-50 hours, but that its duration of action is 8-12 hours.
> > > > > > > > Is that why you have to take klonopin more than once a day even though it has a half-life of 18-50 hours, because even though it lasts in the body for 18-50 hours it is only working therapeutically for the first 8-12 hours?
> > > > > > >
> > > > > > > hi chess....half-life of elimination is the time at which one-half of the parent compound can be found to reside in your system: more accurately, it is determined (usually) via excretion of drug in urine. the duration of action has to do with a measure called (in general) bioavailability. when you take a dose of klonopin, the time to what is called maximum area under (the) plasma concentration curve (AUC) [t__{max} AUC] is what we want to talk about. since the drug is being metabolized, excreted, and, as in the case of klonopin, highly plasma-bound, we can expect that klonopin will partition into fatty tissue in your body (this comes into play later) and that the onset of action - and duration - is determined by the ``balance'' of bioavailability vs. elimination. these factors - and others, but these are the major ones (in my experience) that address your question. so: you dose. the dose gose to your gut, where some degradation of the parent occurs. the kidneys excrete parent + metabolites and this is a (usually) first-order process, i.e., the change in concentration of the parent is linear in time or, at best, a pseudo-first-order kinetic process). now, it turns out that the plasma binding is not the only way to go, because much of the parent (and metabolites) have partitioned into your fatty tissue (like likes like, i.e., oil and oil are miscible, oil and water are only slightly so). so the AUC business must be taken into account with plasma binding, which is how the parent gets to your active site. turns out that the higher plasma binding, the lower the potential for unchanged parent to reach the receptor. but this is an issue iff (not a typo: if and only if) the plasma protein interacts with the drug, thus making the unbound fraction more potent due to retention of potency. so: the elimination includes perfusion from the fatty tissues, and this is what make withdrawl, well, withdrawl. it is also a boost if you can perfuse the fatty-bound drug into your system, e.g., with a vasodilator. the onset of action is how quickly the drug starts doing it's thing; and the duration of action is a balancing act between how much parent is partitioned into adipose, plasma, and is being eliminated. please let me know if this helps or hinders....all the best, chemist
> >
>

 

Re: Chemist...questionChemist » chess

Posted by chemist on May 18, 2004, at 15:59:43

In reply to Re: Chemist...questionChemist » chemist, posted by chess on May 18, 2004, at 14:39:19

you asked if K is hitting fewer GABA receptors with respect to time or if the same receptors were being hit but with lower ``strength:'' it is the latter that is closest to the correct answer. the presence of a more lipophilic derivative of K (the result of miking a nitro group into a primary amino group in the 7 position) - which is pharmacologically active - means that this metabolite will stay around in the brain longer, and i am not aware of any change in specificity for the BZD site in the GABA type A receptor.

the blood/brain barrier is where hydrophilic moieties hit the lipophilic wall, and the answer to whether hydrophilic and lipophilic compounds cross is yes, but you can see that a hydrophilic drug will be transported more quickly to the BBB than a hydrophobic one BUT we have to consider that if a lipophilic drug hitches a ride by being partitioned into a carrier that is itself hydrophilic, then the lipophilic drug can reach the BBB quite quickly and cross quite quickly (this is why, in prn dosing, diazepam has a quicker onset than does lorazepam, due to the extreme hydrophobicity of diazepam and the albeit slight hydrophilicity of lorazepam, not to mention the slightly increased hydrophilicity of nordiazepam and oxazepam, 2 of the active metabolites of diazepam). and yes, clonazepam is lipophilic on the whole, but the presence of the nitro group (which becomes an animo group upon metabolism) does make it somewhat hydrophilic....hope this helps, all the best, chemist

 

Re: Chemist...questionChemist » chemist

Posted by chess on May 19, 2004, at 7:01:18

In reply to Re: Chemist...questionChemist » chess, posted by chemist on May 18, 2004, at 15:59:43

> you asked if K is hitting fewer GABA receptors with respect to time or if the same receptors were being hit but with lower ``strength:'' it is the latter that is closest to the correct answer. the presence of a more lipophilic derivative of K (the result of miking a nitro group into a primary amino group in the 7 position) - which is pharmacologically active - means that this metabolite will stay around in the brain longer, and i am not aware of any change in specificity for the BZD site in the GABA type A receptor.
-----that's interesting, i was expecting fewer gaba receptors being hit to have been the answer.
so then why is the K metabolite less potent than the parent K?

 

Re: Chemist...questionChemist » chess

Posted by chemist on May 19, 2004, at 13:25:55

In reply to Re: Chemist...questionChemist » chemist, posted by chess on May 19, 2004, at 7:01:18

> > you asked if K is hitting fewer GABA receptors with respect to time or if the same receptors were being hit but with lower ``strength:'' it is the latter that is closest to the correct answer. the presence of a more lipophilic derivative of K (the result of miking a nitro group into a primary amino group in the 7 position) - which is pharmacologically active - means that this metabolite will stay around in the brain longer, and i am not aware of any change in specificity for the BZD site in the GABA type A receptor.
> -----that's interesting, i was expecting fewer gaba receptors being hit to have been the answer.
> so then why is the K metabolite less potent than the parent K?
>
>
actually, the affinity is only for the A type GABA receptor, so in any case, there is only one receptor involved....the interaction at the benzodiazepine binding site in re: the amino vs. the nitro group has to do with higher binding affinity for the parent due to direct hydrogen bonding or water-mediated hydrogen bonding with the higly diffuse and electron-rich number of lone-pairs, in contrast to the amino, which is less likely to bind as tightly. thus, it will be eliminated faster than the parent.....hope this helps, all the best, chemist

 

Re: Chemist...questionChemist » chemist

Posted by chess on May 19, 2004, at 22:56:52

In reply to Re: Chemist...questionChemist » chess, posted by chemist on May 19, 2004, at 13:25:55

so then K becomes less potent as time goes by because it is being changed from a higher-affinity-binding nitro molecule to a lower-affinity-binding amino molecule?

> > > why is the K metabolite less potent than the parent K?
> actually, the affinity is only for the A type GABA receptor, so in any case, there is only one receptor involved....the interaction at the benzodiazepine binding site in re: the amino vs. the nitro group has to do with higher binding affinity for the parent due to direct hydrogen bonding or water-mediated hydrogen bonding with the higly diffuse and electron-rich number of lone-pairs, in contrast to the amino, which is less likely to bind as tightly. thus, it will be eliminated faster than the parent.....hope this helps, all the best, chemist

 

Re: Chemist...questionChemist » chess

Posted by chemist on May 20, 2004, at 0:34:20

In reply to Re: Chemist...questionChemist » chemist, posted by chess on May 19, 2004, at 22:56:52

hello there...you have it! all the best, chemist

> so then K becomes less potent as time goes by because it is being changed from a higher-affinity-binding nitro molecule to a lower-affinity-binding amino molecule?
>
> > > > why is the K metabolite less potent than the parent K?
> > actually, the affinity is only for the A type GABA receptor, so in any case, there is only one receptor involved....the interaction at the benzodiazepine binding site in re: the amino vs. the nitro group has to do with higher binding affinity for the parent due to direct hydrogen bonding or water-mediated hydrogen bonding with the higly diffuse and electron-rich number of lone-pairs, in contrast to the amino, which is less likely to bind as tightly. thus, it will be eliminated faster than the parent.....hope this helps, all the best, chemist
>
>

 

Thank You Chemist ! (nm) » chemist

Posted by chess on May 20, 2004, at 1:59:37

In reply to Re: Chemist...questionChemist » chess, posted by chemist on May 20, 2004, at 0:34:20

 

you're welcom, and one last bit... » chess

Posted by chemist on May 20, 2004, at 12:04:21

In reply to Thank You Chemist ! (nm) » chemist, posted by chess on May 20, 2004, at 1:59:37

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

 

Re: you're welcom, and one last bit...

Posted by chess on May 20, 2004, at 23:05:22

In reply to you're welcom, and one last bit... » chess, posted by chemist on May 20, 2004, at 12:04:21

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

 

Re: you're welcom, and one last bit... » chess

Posted by chemist on May 20, 2004, at 23:32:17

In reply to Re: you're welcom, and one last bit..., posted by chess on May 20, 2004, at 23:05:22

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
>
>

 

Re: you're welcom, and one last bit... » chemist

Posted by chess on May 21, 2004, at 2:38:21

In reply to Re: you're welcom, and one last bit... » chess, posted by chemist on May 20, 2004, at 23:32:17

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
> >
> >
>

 

Re: you're welcom, and one last bit... » chess

Posted by chemist on May 21, 2004, at 14:46:35

In reply to Re: you're welcom, and one last bit... » chemist, posted by chess on May 21, 2004, at 2:38:21

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
> > >
> > >
> >
>
>

 

Re: you're welcom, and one last bit... » chemist

Posted by chess on May 21, 2004, at 21:13:29

In reply to Re: you're welcom, and one last bit... » chess, posted by chemist on May 21, 2004, at 14:46:35

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
> > > >
> > > >
> > >
> >
> >
>

 

where is this going???? » chess

Posted by chemist on May 21, 2004, at 22:17:44

In reply to Re: you're welcom, and one last bit... » chemist, posted by chess on May 21, 2004, at 21:13:29

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
> > > > >
> > > > >
> > > >
> > >
> > >
> >
>
>

 

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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Re: where is this going???? » chess

Posted by chemist on May 22, 2004, at 13:45:49

In reply to Re: where is this going???? » chemist, posted by chess on May 22, 2004, at 7:37:05

hi there...in my estimation, the more K you take at one time, the longer it lingers in your system. this is mainly because the rate at which it is metabolized is fixed. there is a second factor that has to do with desensitizing the receptor, but the main reason why a higher dose lasts longer is that you can only eliminate just so much of K over a given period of time. thus, the more you put into your system, the more lingers around. and given the high affinity for plasma binding, you would expect that elimination via other conduits (such as renal) would be low, especially for the parent. the time for AUC_{max} will be the same, but the trail-off will be longer for a higher dose....does this answer your question? all the best, chemist


> 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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