1 / 52

# Toxicokinetic Calculations - PowerPoint PPT Presentation

Toxicokinetic Calculations. Extent of distribution The parameter that reflects the extent of distribution is the apparent volume of distribution, V d , where: V d = Dose/C p,0

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Toxicokinetic Calculations' - koen

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Extent of distribution

• The parameter that reflects the extent of distribution is the apparent volume of distribution, Vd, where:

• Vd = Dose/Cp,0

• Where dose = total amount of drug in the body , while Cp,0 is the concentration of drug in plasma at 0 hrs after injection.

• After an IV bolus dose of 500 mg, the following data was collected: Find the elimination rate and the apparent volume of distribution.

Solution:

• First, we should be familiar with the first order kinetics where:

• dCp/dt = -Kel. Cp

• dCp/Cp = -Kel.dt

• Integrationof the above equation gives:

• Cp = Cp0 e-Kel.t , or

• lnCp = lnCp0 - Kel.t

• Between time t1 and t2, we have

• ln Cp1– lnCp2 = kel (t2– t1)

• It follows that:

• Kel = (lnCp1– lnCp2)/(t2– t1)

• Plotting lnCp versus time should yield a straight line with a slope equals kel

• After plotting the curve, extrapolation should yield Cp0.

• Finally the apparent volume can be calculated from the relation:

• Vd = Dose/Cp,0

• Kel = (lnCp1 - lnCp2)/(t2– t1)

• Kel = (ln87.1 – ln4.17)/(10 – 0)

• Kel = 0.304/hr

• Vd = Dose/Cp,0

• Vd = 500/87.1 = 5.74 L

• From first order kinetics we have:

• lnCp = lnCp0– kelt

• lnCp - lnCp0 = -kelt

• lnCp/Cp0 = -kelt

• The half life of elimination is defined as the time required for the concentration to decrease to one half. This means Cp0 = 2Cp

• Substituting in the last equation above gives:

• ln ½ = -kelt1/2

• Or: t1/2 = 0.693/kel

• Draw a line through the points (this tends to average the data)

• Pick any Cp and t1 on the line

• Determine Cp/2 and t2 using the line

• Calculate t1/2 as (t2 - t1)

• And finally calculate kel = 0.693/t1/2

• Cp/2 in 1 half-life i.e. 50.0 % lost 50.0 %Cp/4 in 2 half-lives i.e. 25.0 % lost 75.0 %Cp/8 in 3 half-lives i.e. 12.5 % lost 87.5 %Cp/16 in 4 half-lives i.e. 6.25 % lost 93.75 %Cp/32 in 5 half-lives i.e. 3.125 % lost 96.875 %Cp/64 in 6 half-lives i.e. 1.563 % lost 98.438 %Cp/128 in 7 half-lives i.e. 0.781 % lost 99.219 %

• Thus over 95 % is lost or eliminated after 5 half-lives

• If the rate of elimination of a drug is 0.3/hr, find the half life of elimination.

• t1/2 = 0.693/kel

• t1/2 = 0.693/0.3

• t1/2 = 2.31 hr

At t = 0, e-kel*t = 1 and at t = ∞, e-kel*t = 0 , Therefore:

Or, V = Dose/(AUC * kel)

• In the same mannar,

• From these equations and since CL=Doseiv/AUCiv ,

and since AUC0-a=Dose/(Vd*Kel)

• It Turns out that CL = Vd*Kel

A = ½ sum of the two parallel sides * height

Example Calculation of AUC trapezoids from 0-1 and from 10-infinity:

• A dose of 250 mg was administered to healthy volunteer. Seven blood samples were collected at 0.5, 1, 2, 4, 6, 8, 10 hours. Plasma was separated from each blood sample and analyzed for drug concentration. The collected data are shown in the table below. Use these data with the trapezoidal rule shown in the related equations to calculate each AUC segment including the last segment.

Solution: trapezoids from 0-1 and from 10-infinity:

• First we plot the data with ln Cp against time

• Extrapolate to y axis and find Cpo

• From the slope find Kel

• Start calculating Δ(AUC mg.hr/L) segments using the equation

For the first segment and then go on for the other ones using the same equation

4. Find trapezoids from 0-1 and from 10-infinity:

5. Find AUC(0-∞)

The following table summarizes the results:

• Dose = 250 mg

• Cpo = 6.65 mg/L

• kel = 0.386 hr-1

• AUC(0-10 hr) = 17.39 mg.hr/L

• AUC(0-∞) = 17.75 mg.hr/L

For the first segment we have: {(6.65+5.42)/2}*(0.5 trapezoids from 0-1 and from 10-infinity:– 0) = 3.02

For the second segment we have: {(5.42+4.61)/2}*(1 – 0.5) = 2.51

Example trapezoids from 0-1 and from 10-infinity:

• What IV bolus dose is required to achieve a plasma concentration of 2.4 µg/ml (2.4 mg/L) at 6 hours after the dose is administered. The elimination rate constant, kel is 0.17 hr-1) and the apparent volume of distribution, V, is 25 L

Example trapezoids from 0-1 and from 10-infinity:

If Cp after2 hours is 4.5 mg/liter and Cp after6 hours is 3.7 mg/liter, after a 400 mg IV bolus dose what are the values of kel and V.

mg/L trapezoids from 0-1 and from 10-infinity:

Example trapezoids from 0-1 and from 10-infinity:

What is the concentration of a drug 0, 2 and 4 hours after a dose of 500 mg. Known pharmacokinetic parameters are apparent volume of distribution, Vd is 30 liter and the elimination rate constant, kel is 0.2 hr-1

Make Predictions trapezoids from 0-1 and from 10-infinity:Once we have a model and parameter values we can use this information to make predictions. For example we can determine the dose required to achieve a certain drug concentration.

Finding a dose necessary to achieve a certain C trapezoids from 0-1 and from 10-infinity:p

Plasma drug concentration after multiple IV doses trapezoids from 0-1 and from 10-infinity:

• The anticipated plasma concentration is meant to be before the steady state is reached. The equation used for such a calculation is:

The steady state is reached when the number of doses exceeds 5 half lives, but is surely attainable when n = ∞

• C multiple IV dosesp = {100/14}{[(1- e-12*0.23*4)/(1-e-0.23*12)]e-0.23*3)

• Cp = 3.8 mg/L

Steady state from first principles multiple IV doses

At steady state the rate of drug administration is equal to the rate of drug elimination. Mathematically the rate of drug administration can be stated in terms of the dose (D) and dosing interval (t). It is always important to include the salt factor (S) and the bioavailability (F). The rate of drug elimination will be the clearance of the plasma concentration at steady state:

For IV Route multiple IV doses

t = t multiple IV doseso

t’ = t

C multiple IV dosesss = Cp0 = Cpt/e-kel*t

However, now t = t’– t , since to = t

Rearranging gives:

For Non-IV Routes multiple IV doses

From multiple doses to steady state multiple IV doses

• We have the equation for multiple doses where:

• When n is infinity then the value e-nKelt = 0, the equation then becomes:

Therefore, at steady state multiple IV doses

• The plasma concentration (Cp) at any time (t) within a dosing interval (t) at steady state is represented by the equation:

• Example: multiple IV doses

• Calculate the concentration of drug in plasma 2 hrs after the last dose of a series of doses (6hrs interval and 100 mg each) that brought the patient to a steady state. Kel = 0.3/hr and

• Vd =5.6 L.

• Substitution gives: multiple IV doses

• Cp = {100 (e-0.3*2) / 5.6(1 – e-0.3*6)}

• = 54.88/4.67 = 11.74 mg/L

C multiple IV dosespmax and Cpmin

• At steady state we have:

• Now, the maximum plasma concentration for each dose administration occurs at t = 0, while the minimum plasma concentration at steady state occurs at t = t , back to equation of Cpt at steady state

• Applying the conditions for t = 0 and t = t and taking in consideration that e-Kelt = 1 when t = 0, and e-Kelt = e-Keltwhen t = t , we get:

• Immediately after many doses, after t = 0, we have: administration occurs at t = 0, while the minimum plasma concentration at steady state occurs at t =

• Immediately before many doses, after t = t , we have:

• Therefore,

• An example may be helpful: t1/2 = 4 hr; IV dose 100 mg every 6 hours; V = 10 liter

Dose/V administration occurs at t = 0, while the minimum plasma concentration at steady state occurs at t = d = Cpo1 = 100/10 = 10 mg/L

Where Cpo1 is the plasma concentration at zero time after the first dose

kel = 0.693/4 = 0.17 hr-1

R = e-kel *t = e-0.17 x 6 = 0.35

therefore

Therefore the plasma concentration will fluctuate between 15.5 and 5.4 mg/liter during each dosing interval when the plateau is reached.