Therapy

currently enrolling in a pharmacology course, and i am struggling with this one question:

Acyclovir is a new potent and selective anti-viral agent.When administered to human beings,most of the drug is eliminated from the body unchanged,via glomerular filtration and tubular secretion.Due to its large therapeutic index,the maximum plasma level after administration can be 1.3mg/L without causing any side effect,whereas the clinical effective plasma concentration needs to be higher than 0.06mg/L.The bioavailability of the orally administered Acyclovir is low(20%),and dosing is required 2-5times a day.Acyclovir.For a 70kg adult,the elimination rate constant is 0.28hr and apparent volume of distribution is 56liters approx.The oral dose is considered to be formulated as 200mg or 400mg.Due to a limitation of sustained release capsule,the absorption rate constant of oral dosing form can be controlled as 0.167hr,or 0.083hr.Design a possible once-a-day oral dosing form(the best D and Ka)

This question is really more appropriate for the engineering or math sections. Consider the human body as a box. Drug is going in and drug is coming out. Your ultimate goal for this question is to figure out what the concentration of the drug is inside the box at any given time (and you will want it to be in the range of 0.06-1.3 mg/L). The only info you have are the rates at which things are going in and coming out. These won’t tell you what the concentration is, but they will tell you how it is changing over time–and you can figure out the concentration based on this since you know the starting condition.

First, what is going into the box at any given time?
(1) D * ba * exp(-ka * t) / V
where D is the dosage (either 200 or 400 mg), ba is the bioavailability (20%), ka is the absorption constant (either 0.167/hr or 0.083/hr), t is time, and V is volume of distribution (56 L).

Second, what is coming out of the box at any given time?
(2) C * ke
where C is the concentration, and ke is the elimination constant (0.28/hr).

The change in concentration with time is equal to [stuff going in] – [stuff coming out].
(3) dC/dt = [D * ba * exp(-ka * t) / V] – [C * ke]

rearrange to:
dC + ke*C*dt = D * ba * exp(-ka * t) / V * dt

If you have been assigned this problem, then you must know how to solve differential equations. Analytically, you can solve this first-order linear differential equation using an integrating factor of exp(ke*t).

(4) dC*exp(ke*t) + ke*C*dt*exp(ke*t) = D*ba*exp(-ka*t)/V*dt*exp(ke*t)

The left side of the equation (4) is the differential of C*exp(ke*t) (that’s the whole trick behind the use of integrating factors). So when you integrate both sides of the equation and you get:

(5) C*exp(ke*t) = A + Int{ D*ba*exp(-ka*t)/V *dt*exp(ke*t)}
where A is a constant of integration and Int{ denotes the integral operator.

Solve this to get
(6) C = {A + D*ba*exp[(ke-ka)*t] / V / (ke-ka)} / exp(ke*t)
You could also solve equation (3) computationally, but why bother?

You can solve for A using either the initial condition of C(0) = 0 or C(0) = 0.06. Just make sure to specify which one you used.

Now you have the function C(t). You can plot it using the two possible values for D and the two possible values of ka over a time period of 24 hours. You should find that D=200 mg and ka=0.167/hr are the only values that keep the plasma concentration within the therapeutic range.

One Response

1. grimmyTea Says:
   February 24th, 2010 at 2:54 pm

   This question is really more appropriate for the engineering or math sections. Consider the human body as a box. Drug is going in and drug is coming out. Your ultimate goal for this question is to figure out what the concentration of the drug is inside the box at any given time (and you will want it to be in the range of 0.06-1.3 mg/L). The only info you have are the rates at which things are going in and coming out. These won’t tell you what the concentration is, but they will tell you how it is changing over time–and you can figure out the concentration based on this since you know the starting condition.

   First, what is going into the box at any given time?
   (1) D * ba * exp(-ka * t) / V
   where D is the dosage (either 200 or 400 mg), ba is the bioavailability (20%), ka is the absorption constant (either 0.167/hr or 0.083/hr), t is time, and V is volume of distribution (56 L).

   Second, what is coming out of the box at any given time?
   (2) C * ke
   where C is the concentration, and ke is the elimination constant (0.28/hr).

   The change in concentration with time is equal to [stuff going in] – [stuff coming out].
   (3) dC/dt = [D * ba * exp(-ka * t) / V] – [C * ke]

   rearrange to:
   dC + ke*C*dt = D * ba * exp(-ka * t) / V * dt

   If you have been assigned this problem, then you must know how to solve differential equations. Analytically, you can solve this first-order linear differential equation using an integrating factor of exp(ke*t).

   (4) dC*exp(ke*t) + ke*C*dt*exp(ke*t) = D*ba*exp(-ka*t)/V*dt*exp(ke*t)

   The left side of the equation (4) is the differential of C*exp(ke*t) (that’s the whole trick behind the use of integrating factors). So when you integrate both sides of the equation and you get:

   (5) C*exp(ke*t) = A + Int{ D*ba*exp(-ka*t)/V *dt*exp(ke*t)}
   where A is a constant of integration and Int{ denotes the integral operator.

   Solve this to get
   (6) C = {A + D*ba*exp[(ke-ka)*t] / V / (ke-ka)} / exp(ke*t)
   You could also solve equation (3) computationally, but why bother?

   You can solve for A using either the initial condition of C(0) = 0 or C(0) = 0.06. Just make sure to specify which one you used.

   Now you have the function C(t). You can plot it using the two possible values for D and the two possible values of ka over a time period of 24 hours. You should find that D=200 mg and ka=0.167/hr are the only values that keep the plasma concentration within the therapeutic range.
   References :

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