Duration of Action of a drug treatment

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  • Thread starter Master1022
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  • #1
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Homework Statement:
A single dose of 10 mg of the drug with 200 ml water immediately after a low fat meal consisting of fruit juice, skim milk, cereal and toast with jelly. The concentration of the drug (in units of ##\mu##g ##l^{-1}##) in the blood following treatment A is given by the following equation:
[tex] C_p (t) = 91.417 \left( e^{-0.4t} - e^{-3.455t} \right) [/tex]
Knowing that the minimal effective blood concentration (MEC) of the drug is 50 ##\mu##g per litre blood, calculate the duration of action for treatment A assuming that the time it takes to initially get above the MEC is 20 minutes.
Relevant Equations:
Concentration
Hi,

I was attempting the following question, but didn't understand how to make further progress with it.

Question:
A single dose of 10 mg of the drug with 200 ml water immediately after a low fat meal consisting of fruit juice, skim milk, cereal and toast with jelly. The concentration of the drug (in units of ##\mu##g ##l^{-1}##) in the blood following treatment A is given by the following equation:
[tex] C_p (t) = 91.417 \left( e^{-0.4t} - e^{-3.455t} \right) [/tex]
Knowing that the minimal effective blood concentration (MEC) of the drug is 50 ##\mu##g per litre blood, calculate the duration of action for treatment A assuming that the time it takes to initially get above the MEC is 20 minutes.

Attempt:
1) What is the duration of action?
Is it the time which the drug is effective? In this case, that would be when the concentration is above 50 ##\mu##g per liter of blood.

2) How can I solve this question by hand?
This problem was set to be done without the use of a computer or advanced calculator techniques...

If I let ## C_p (t) = 50 ##, we get:
[tex] 50 = 91.417 \left( e^{-0.4t} - e^{-3.455t} \right) [/tex]
which is non-linear and I don't understand how to proceed. Perhaps we can make an assumption that the second term is significantly smaller than the first by now - thus we can ignore it to yield:

[tex] 50 \approx 91.417 \left( e^{-0.4t} \right) \rightarrow T = -ln(\frac{50}{91.417})/0.4 = 1.508... \text{hrs} [/tex]

Then we can subtract the time taken for the drug to first become effective: ## \text{Duration} = 1.508 - 1/3 = 1.17518... \text{hours} ##

Does this approach seem correct?
 

Answers and Replies

  • #2
162
58
Try graphing the original function and then the other two where you remove one of the exponential terms and see if one of them seems like a good approximation in a certain interval. :)
 
  • #3
501
92
Try graphing the original function and then the other two where you remove one of the exponential terms and see if one of them seems like a good approximation in a certain interval. :)
Thanks @Mayhem ! After plotting the graphs, it seems like the approximation above was a fairly good one. I think this is probably how it was intended for us to solve. Was my understanding of the term 'duration of action' correct?
 
  • #4
162
58
Thanks @Mayhem ! After plotting the graphs, it seems like the approximation above was a fairly good one. I think this is probably how it was intended for us to solve. Was my understanding of the term 'duration of action' correct?
Yes, as when it gets below 50 mmg/l it is no longer considered to be active.
 
  • #5
501
92
Hmm, I don't think they are the same. However, the threshold of 50 is quoted as a concentration, so I feel it is okay to use the equation provided.
 
  • #6
epenguin
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A somewhat academic question which no one in practice would be calculating for such a problem, but meant I guess to get you familiar with behaviour corresponding to exponential formulae.

I don't think you can 'solve' the equation mathematically for the required t.
But what you can do mathematically that I think gives a reasonable answer, is solve mathematically (fairly elementary calculus) to find the t at which the circulating drug concentration is maximum, and find this maximum concentration. Then with that, call it C', as starting concentration find the time to decay by the slower exponential to 50 ug/L - forget the original (and ridiculously precise) 91.::: now. Still have to subtract the time to reach the 50 ug/L concentration. Not all that different from what you did. Bit of a fudge any way. (But maybe the errors tell you something.)
 
  • #7
501
92
A somewhat academic question which no one in practice would be calculating for such a problem, but meant I guess to get you familiar with behaviour corresponding to exponential formulae.

I don't think you can 'solve' the equation mathematically for the required t.
But what you can do mathematically that I think gives a reasonable answer, is solve mathematically (fairly elementary calculus) to find the t at which the circulating drug concentration is maximum, and find this maximum concentration. Then with that, call it C', as starting concentration find the time to decay by the slower exponential to 50 ug/L - forget the original (and ridiculously precise) 91.::: now. Still have to subtract the time to reach the 50 ug/L concentration. Not all that different from what you did. Bit of a fudge any way. (But maybe the errors tell you something.)
Thanks @epenguin ! That's an interesting method and I will give it a go.
 
  • #8
epenguin
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Just to say that I hope you realise this equation is not out of the blue but the solution ##y(t)## to the differential equation system
$$\dfrac {dx\left(t\right) }{dt}=-k_{1}x(t),\dfrac {dy\left(t\right) }{dt}=k_{2}x(t)-k_{3}y(t)$$
corresponding to a simple two-compartment throughput model where ##x(t), y(t)## represent molarities of the drug in the two compartments (which could be stomach and circulating blood) and ##x(0)>0, y(0)=0##. ##k_{1}, k_{2}, k_{3} > 0## are kinetic constants, which are unpredictable except the first two are related by: ##k_{1}V_{1}=k_{2}V_{2}##, where the ##V## are the volumes of the compartments (mass conservation). Around ##k_{3}## there maybe reasonable physiological considerations to be made, e.g. there would be a maximum possible if the elimination was purely by excretion. In other cases there are other possible mechanisms like enzymatic destruction of the drug.

With or without that I think - can you see? - that your method for estimating the time would overestimate it, and mine would underestimate it. To have an estimate of both sides has even an appearance of a useful exercise!

I don't know whether you are able to solve the d.e. system. If you have done anything like it before (linear ordinary differential equations) you would do yourself some good if you solve them yourself.
 
  • #9
501
92
Just to say that I hope you realise this equation is not out of the blue but the solution ##y(t)## to the differential equation system
$$\dfrac {dx\left(t\right) }{dt}=-k_{1}x(t),\dfrac {dy\left(t\right) }{dt}=k_{2}x(t)-k_{3}y(t)$$
corresponding to a simple two-compartment throughput model where ##x(t), y(t)## represent molarities of the drug in the two compartments (which could be stomach and circulating blood) and ##x(0)>0, y(0)=0##. ##k_{1}, k_{2}, k_{3} > 0## are kinetic constants, which are unpredictable except the first two are related by: ##k_{1}V_{1}=k_{2}V_{2}##, where the ##V## are the volumes of the compartments (mass conservation). Around ##k_{3}## there maybe reasonable physiological considerations to be made, e.g. there would be a maximum possible if the elimination was purely by excretion. In other cases there are other possible mechanisms like enzymatic destruction of the drug.

With or without that I think - can you see? - that your method for estimating the time would overestimate it, and mine would underestimate it. To have an estimate of both sides has even an appearance of a useful exercise!

I don't know whether you are able to solve the d.e. system. If you have done anything like it before (linear ordinary differential equations) you would do yourself some good if you solve them yourself.
Thanks @epenguin ! Yes, I have seen the O.D.E derivation via the compartment model before. I think this question just wanted to start with that equation so as to not ask us to spend time writing it out, taking Laplace transforms, etc...

Thanks for providing a bit more of the biological interpretation to something I have only really considered through a mathematical/engineering lens!
 

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