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. First, we should be familiar with the first order kinetics where:
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: 10 8 6 4 3 2 1
Time (hr) 9 14 20 33 51 72
Cp, mg/L

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

6. 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

7. 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

9. 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

10. Thus over 95 % is lost or eliminated after 5 half-lives
Cp/2 in 1 half-life i.e % lost 50.0 % Cp/4 in 2 half-lives i.e % lost 75.0 % Cp/8 in 3 half-lives i.e % lost 87.5 % Cp/16 in 4 half-lives i.e % lost % Cp/32 in 5 half-lives i.e % lost % Cp/64 in 6 half-lives i.e % lost % Cp/128 in 7 half-lives i.e % lost %
Thus over 95 % is lost or eliminated after 5 half-lives

12. 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

13. Clearance

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

15. 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

16. AUC by Trapezoidal Approach

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

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

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

20. 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.

21. AUC (mg.hr/L)
Δ (AUC mg.hr/L)
Cp (mg/L)
Time (hr)
  0  
5.42
  0.5  
4.61
  1  
3.28
  2  
1.28
  4  
0.65
  6  
0.32
  8  
0.14
  10  
  ∞  

22. 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

23. 4. Find
5. Find AUC(0-∞)
The following table summarizes the results:
Dose = 250 mg
Cpo = 6.65 mg/L
kel = hr-1
AUC(0-10 hr) = mg.hr/L
AUC(0-∞) = mg.hr/L

24. For the first segment we have: {(6.65+5.42)/2}*(0.5 – 0) = 3.02
AUC (mg.hr/L)
Δ (AUC mg.hr/L)
Cp (mg/L)
Time (hr)
Summation 
  Segment
6.65
3.02
5.42
0.5
5.52
2.51
4.61 1
9.47
3.95
3.28 2
14.03
4.56
1.28 4
15.96
1.93
0.65 6
16.93
0.97
0.32 8
17.39
0.46
0.14 10
17.75
0.36 ∞
For the first segment we have: {( )/2}*(0.5 – 0) = 3.02
For the second segment we have: {( )/2}*(1 – 0.5) = 2.51

25. 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:

27. 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

29. Example
If Cp after 2 hours is 4.5 mg/liter and Cp after 6 hours is 3.7 mg/liter, after a 400 mg IV bolus dose what are the values of kel and V.

30. mg/L

31. 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

33. Make Predictions 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.

34. Finding a dose necessary to achieve a certain Cp

36. 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 = ∞

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

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

39. 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:

40. For IV Route

41. t = to
t’ = t

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

43. For Non-IV Routes

44. 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:

45. 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:

46. 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.

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

48. Cpmax and Cpmin
At steady state we have:

49. 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-Kelt when t = t , we get:

51. Immediately after many doses, after t = 0, we have:
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

52. Dose/Vd = 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.