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# Toxicokinetic Calculations

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1. Toxicokinetic Calculations 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.

2. Example: • 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:

3. 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

4. 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

5. 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

6. Half life of elimination • 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

7. The steps to take are: • 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

8. 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

9. Example • 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

10. Clearance

11. At t = 0, e-kel*t = 1 and at t = ∞, e-kel*t = 0 , Therefore: Or, V = Dose/(AUC * kel)

12. 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

13. AUC by Trapezoidal Approach

14. The area of a trapezoid is calculated as: A = ½ sum of the two parallel sides * height

15. However, look at the AUC where we have plasma concentrations usually taken within some interval, say from 1-10 hrs:

16. It is evident that the overall AUC value should involve the trapezoids from 0-1 and from 10-infinity:

17. Example Calculation of AUC • 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.

18. Solution: • 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

19. 4. Find 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

20. For the first segment we have: {(6.65+5.42)/2}*(0.5 – 0) = 3.02 For the second segment we have: {(5.42+4.61)/2}*(1 – 0.5) = 2.51

21. CL = rate of elimination/plasma concentration • Rate of elimination = change of amount per time = d(amount)/dt • Therefore CL = {d(amount)/dt}/Cp • Taking in consideration that:

22. Example • 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

23. Example 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.

24. mg/L

25. Example 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

26. Make PredictionsOnce 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.

27. Finding a dose necessary to achieve a certain Cp

28. Plasma drug concentration after multiple IV doses • 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 = ∞

29. Example calculation of plasma drug concentration after multiple IV doses

30. Cp = {100/14}{[(1- e-12*0.23*4)/(1-e-0.23*12)]e-0.23*3) • Cp = 3.8 mg/L

31. Steady state from first principles 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:

32. For IV Route

33. t = to t’ = t

34. Css = Cp0 = Cpt/e-kel*t However, now t = t’– t , since to = t Rearranging gives:

35. For Non-IV Routes

36. From multiple doses to steady state • We have the equation for multiple doses where: • When n is infinity then the value e-nKelt = 0, the equation then becomes:

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

38. Example: • 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.

39. Substitution gives: • Cp = {100 (e-0.3*2) / 5.6(1 – e-0.3*6)} • = 54.88/4.67 = 11.74 mg/L

40. Cpmax and Cpmin • At steady state we have:

41. 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: