Transcript Lecture 6

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.
Example:
 After an IV bolus dose of 500 mg, the following data
was collected: Find the elimination rate and the
apparent volume of distribution.
Time (hr)
1
2
3
4
6
8
10
Cp, mg/L
72
51
33
20
14
9
4
Solution:
 First, we should be familiar with the first order kinetics
where:
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dCp/dt = -Kel. Cp
dCp/Cp = -Kel.dt
Integrationof the above equation gives:
Cp = Cp0 e-Kel.t , or
lnCp = lnCp0 - Kel.t
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Between time t1 and t2, we have
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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
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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
Half life of elimination
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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
The steps to take are:
1.
Draw a line through the points (this tends to
average the data)
2.
Pick any Cp and t1 on the line
3.
Determine Cp/2 and t2 using the line
4.
Calculate t1/2 as (t2 - t1)
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And finally calculate kel = 0.693/t1/2
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Cp/2
Cp/4
Cp/8
Cp/16
Cp/32
Cp/64
Cp/128
in
in
in
in
in
in
in
1 half-life i.e. 50.0 % lost 50.0 %
2 half-lives i.e. 25.0 % lost 75.0 %
3 half-lives i.e. 12.5 % lost 87.5 %
4 half-lives i.e. 6.25 % lost 93.75 %
5 half-lives i.e. 3.125 % lost 96.875 %
6 half-lives i.e. 1.563 % lost 98.438 %
7 half-lives i.e. 0.781 % lost 99.219 %
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Thus over 95 % is lost or eliminated after 5 half-lives
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
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t1/2 = 2.31 hr
Clearance
At t = 0, e-kel*t = 1 and at t = ∞, e-kel*t = 0 , Therefore:
Or, V = Dose/(AUC * kel)
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In the same mannar,
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From these equations and since
CL=Doseiv/AUCiv ,
and since AUC0-a=Dose/(Vd*Kel)
It Turns out that CL = Vd*Kel
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AUC by Trapezoidal Approach
The area of a trapezoid is calculated as:
A = ½ sum of the two parallel sides * height
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However, look at the AUC where we have
plasma concentrations usually taken within
some interval, say from 1-10 hrs:
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It is evident that the overall AUC value should
involve the trapezoids from 0-1 and from 10infinity:
Example Calculation of AUC
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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.
Time (hr)
Cp (mg/L)
0
0.5
5.42
1
4.61
2
3.28
4
1.28
6
0.65
8
0.32
10
0.14
∞
Δ (AUC mg.hr/L)
AUC (mg.hr/L)
Solution:
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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
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
Time (hr)
Cp (mg/L)
Δ (AUC mg.hr/L)
AUC (mg.hr/L)
0
6.65
Segment
Summation
0.5
5.42
3.02
3.02
1
4.61
2.51
5.52
2
3.28
3.95
9.47
4
1.28
4.56
14.03
6
0.65
1.93
15.96
8
0.32
0.97
16.93
10
0.14
0.46
17.39
∞
0
0.36
17.75
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
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CL = rate of elimination/plasma concentration
Rate of elimination = change of amount per
time = d(amount)/dt
Therefore CL = {d(amount)/dt}/Cp
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Taking in consideration that:
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Example
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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
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.
mg/L
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
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.
Finding a dose necessary to achieve a
certain Cp
Plasma drug concentration after multiple IV
doses
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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 = ∞
Example calculation of plasma drug
concentration after multiple IV doses
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Cp = {100/14}{[(1- e-12*0.23*4)/(1-e-0.23*12)]e-0.23*3)
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Cp = 3.8 mg/L
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:
For IV Route
t = to
t’ = t
Css = Cp0 = Cpt/e-kel*t
However, now t = t’ – t , since t
o=t
Rearranging gives:
For Non-IV Routes
From multiple doses to steady state
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We have the equation for multiple doses
where:
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When n is infinity then the value e-nKelt = 0, the
equation then becomes:
Therefore, at steady state
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The plasma concentration (Cp) at any time (t)
within a dosing interval (t) at steady state is
represented by the equation:
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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.
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Substitution gives:
Cp = {100 (e-0.3*2) / 5.6(1 – e-0.3*6)}
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= 54.88/4.67 = 11.74 mg/L
Cpmax and Cpmin
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At steady state we have:
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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:
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Immediately after many doses, after t = 0, we have:
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Immediately before many doses, after t = t , we have:
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Therefore,
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An example may be helpful: t1/2 = 4 hr; IV dose 100
mg every 6 hours; V = 10 liter
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.