Toxicokinetic Calculations

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
• Where dose = total amount of drug in the body
, while C p,0 is the concentration of drug in
plasma at 0 hrs after injection.
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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
C p, 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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• dC p /dt = -K el. C p
• dC p /C p = -K el .dt
• Integrationof the above equation gives:
• C p = C p0 e -Kel.t
, or
• lnC p = lnC p0 - K el .t
• Between time t 1 and t 2 , we have
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• ln C p1 –lnC
p2 = k el (t 2 – t 1 )
• It follows that:
• K el = (lnC p1 – lnC p2 )/(t 2 – t 1 )
• Plotting lnC p versus time should yield a
straight line with a slope equals k el
• After plotting the curve, extrapolation should
yield C p0 .
• Finally the apparent volume can be calculated
from the relation:
• V d = Dose/C p,0
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• K el = (lnC p1 - lnC p2 )/(t 2 – t 1 )
• K el = (ln87.1 – ln4.17)/(10 – 0)
• K el = 0.304/hr
• V d = Dose/C p,0
• V d = 500/87.1 = 5.74 L
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Half life of elimination
• From first order kinetics we have:
• lnC p = lnC p0 – k el t
• lnC p - lnC p0 = -k el t
• lnC p /C p0 = -k el t
• The half life of elimination is defined as the time
required for the concentration to decrease to one half.
This means C p0 = 2C p
• Substituting in the last equation above gives:
• ln ½ = -k el t 1/2
• Or: t 1/2 = 0.693/k el
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The steps to take are:
1. Draw a line through the points (this tends to
average the data)
2. Pick any C p and t 1 on the line
3. Determine C p /2 and t 2 using the line
4. Calculate t 1/2 as (t 2 - t 1 )
• And finally calculate k el = 0.693/t 1/2
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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 life i.e. 50.0 % lost 50.0 %
2 half-lives lives i.e. 25.0 % lost 75.0 %
3 half-lives lives i.e. 12.5 % lost 87.5 %
4 half-lives lives i.e. 6.25 % lost 93.75 %
5 half-lives lives i.e. 3.125 % lost 96.875 %
6 half-lives lives i.e. 1.563 % lost 98.438 %
7 half-lives lives i.e. 0.781 % lost 99.219 %
• Thus over 95 % is lost or eliminated after 5 half-lives
lives
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Example
• If the rate of elimination of a drug is 0.3/hr,
find the half life of elimination.
• t 1/2 = 0.693/k el
• t 1/2 = 0.693/0.3
• t 1/2 = 2.31 hr
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At steady state
• To describe the plasma concentration (C(
p ) at
any time (t)(
) within a dosing interval ()(
) at
steady state:
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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. K el = 0.3/hr and
• V d =5.6 L.
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• Substitution gives:
• C p = {100 (e -0.3*2
) / 5.6(1 – e -0.3*6
)}
• = 54.88/4.67 = 11.74 mg/L
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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 ().(
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:
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• F = AUC (extravascular) /AUC IV
• S = amount of active drug/amount of total drug (salt)
administered
• Steady state:
• Input rate = elimination rate
• || ||
• Maintenance dose (MD*F/) ) = CL*C ss
• MD = CL*C ss */F
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t = t o
t’ = t
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C ss = C p0 = C pt /e -kel*t
However, now t = t’–t , since t
o = t
Rearranging gives:
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Loading dose
The time required to obtain steady-state state plasma
levels by IV infusion will be long. It is
possible to administer an intravenous loading
dose to attain the desired drug concentration
immediately and then attempt to maintain this
concentration by a continuous infusion.
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If the loading dose is to be administered orally, then
the bioavailability
term (F)(
) needs to be introduced. Thus:
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A 100 mg dose was administered every 8 h. At
steady state, two plasma concentrations are
measured:
• Sample 1 is taken at 1 h post dose: Conc 9.6
mg/L
• Sample 2 is taken pre dose: Conc 2.9 mg/L
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The volume of distribution (V(
d ) can be
calculated from either the 1 h post- or pre-dose
samples. From the 1 h post-dose sample
The following equation describes the plasma
concentration 1 h post dose at steady state:
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A patient, aged 40 years and weighing 60 kg,
receives an oral dose of Drug, 500 mg every
12 h. The patient is at steady state. A plasma
level is measured at 10 h post dose and is
reported to be 18.2 mg/L.
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Assume one-compartment kinetics, all doses
were given and F = 1. Calculate the half-life life of
the drug. Provided that : V d = 0.4 L/kg and CL
= 0.05 L/h/kg.
V d = 0.4 X 60 = 24L
CL = 0.05 X 60 = 3.0 L/h
CL = K el *Vd
3.0 = K el *24
k el = 0.125/h
t 1/2 = 0.693/K el = 0.693/0.125
t 1/2 = 5.5 h
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Example
• For a drug with : t 1/2 = 4 hr; IV dose 100 mg
every 6 hours; V d = 10 liter, calculate C pmax
and C pmin and when the plateau would be
reached?
• C
0
p = Dose/V d
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D/V d = C po
k el = 0.693/4 = 0.17 hr -1
R = e -kel
* = 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.
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Accumulation factor, R
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We can also calculate the pharmacokinetic parameters
k el and V d if we know the dose given and the plasma
concentrations at two (or more) times after an I.V.
bolus administration. This time we could use the
equation:
lnC p = ln C p0 - k el *t
If C p 2 hours = 4.5 mg/liter and Cp 6 hours = 3.7
mg/liter after a 400 mg I.V. bolus dose then:
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= - 0.0489 hr -1
thus k el
= 0.0489 hr -1
Now ln C p
2 hr = ln C p0
-k el
* t
ln 4.5 = ln C p0
- 0.0489 x 2
ln Cp 0
= 1.504 + 0.098 = 1.602
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Given k el (= 0.17 hr -1 ) and V d = (25 L) for a
given drug we can calculate the dose required,
by rapid IV bolus, to achieve a plasma
concentration of 2.4 g/ml (= 2.4 mg/L) at 6
hours.
Thus C p6 = C p0 * e -kel.t
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or 2.4 x 25 = DOSE * e -1.02 = DOSE x 0.36
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Calculation of K a
1. Method of Residuals
Starting with the equation for Cp versus time
Cp versus Time after Oral Administration this can be written as
Simplified Equation for Cp versus Time
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Semi-log plot of Cp versus Time after Oral
Administration
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Semi-log Plot of Cp versus Time Showing C late p , Slope, and
Intercept
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and plotting C p
late
versus time gives a straight line
on semi-log graph paper, with a slope (ln) = -kel and
intercept = A.
Now looking at the equation for Cp versus time
again.
Difference or Residual versus Time
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An Example Calculation Using the Method of Residuals
Example Data for the Method of Residuals
Time
(hr)
0.25
0.5
0.75
1.0
1.5
2.0
3.0
4.0
5.0
6.0
7.0
Plasma
Concentration
(mg/L)
1.91
2.98
3.54
3.80
3.84
3.62
3.04
2.49
2.04
1.67
1.37
Cp(late)
(mg/L)
5.23
4.98
4.73
4.50
4.07
3.69
Residual
[Col3 - Col2]
(mg/L)
3.32
2.00
1.19
0.70
0.23
0.07
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2. Wagner-Nelson Method
• Advantages:
i) The absorption and elimination processes can be quite similar
and still accurate determinations of ka can be made.
ii) The absorption process doesn't have to be first order. This
method can be used to investigate the absorption process.
• Disadvantages:
The major disadvantage of this method is that you need to know
the elimination rate constant, from data collected following
intravenous administration.
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• Amount absorbed = amount in body + amount eliminated
• A = X + U
• Differentiation gives:
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• Change in amount of drug in the body with
time is given by:
• dX/dt = V.dC p /dt
• Change in amount of drug eliminated is given
by:
• dU/dt = V.k el .C p
• Therefore, dA/dt = V.dC p /dt + V.k el .C p
• dA = V.dC p + V.k el .C p dt
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• Where, A is the amount absorbed up to time t
• C p = k el .AUC 0-t
• Taking this to infinity gives the maximum amount
absorbed (since C p equals zero after a time equals
infinity:
• A max = 0 + V . k el . AUC 0-
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• The amount remaining to be absorbed (A max – A), can
also be expressed as the amount remaining in the GI,
X g
• X g = (A max – A),
• Dividing by V we get:
• X g /V = {(A max /V) –(A/V)}
• If the kinetics governing the process is a 1 st order one,
then:
• (X g /V) = (X g0 /V) e -kat
• X/V = Concentration
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• It follows that:
• Plotting (A max -A)/V versus time produces a
straight line on semi-log graph paper and a
curved line on linear graph paper. From the
slope of the line on the semi-log graph paper k a
can be calculated (slope = K a )
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3. Method of inspection
• The method of Residuals and the Wagner-Nelson methods are
useful technique for determining good estimates of k a
Requirements for the Method of Inspection
• We assume that k a is much larger than k el . That is, that k a is at
least five time greater than k el . This is the same requirement as
for the Method of Residuals
• Assume that absorption is complete (i.e. greater than 95 %
complete) at the time of the peak concentration. This follows
from the first assumption
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The Method
• The first step is to estimate the time of the peak drug
concentration by inspection. If we assume that the
time of peak is approximately five time the
absorption half-life:
life:
• t peak = 5t 1/2
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• K a = (ln(
2)/t
2)/t 1/2 (absorption)
• K a = 0.693/t 1/2(absorption)
• K a = 0.693/ (t peak /5)
• K a = 0.693*5 / (t(
peak )
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Extent of absorption, F value
• the fraction of the dose which is absorbed is
termed F
• Plotting C p versus time allow the
calculation of k a and k el , as in the methods
described previously.
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• From the intercept, one can calculate the term:
• The total amount absorbed can be calculated as:
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Urine analysis
• We can do the same thing using urine data alone.
• F e is fraction excreted as the parent compound
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Clearance
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CL = K el .V d
CL = 0.693V d /t 1/2
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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, k el is 0.17 hr -1 ) and
the apparent volume of distribution, V, is 25 L
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Example
If C p after 2 hours is 4.5 mg/liter and C p after 6
hours is 3.7 mg/liter, after a 400 mg IV bolus
dose what are the values of k el and V.
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mg/L
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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, V d is 30 liter and the
elimination rate constant, k el is 0.2 hr -1
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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.
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Finding a dose necessary to achieve a
certain C p
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Plasma drug concentration after multiple IV
doses
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• C p = {100/14}{[(1- e -12*0.23*4
12*0.23*4 )/(1-e -0.23*12
)]e -0.23*3
)
• C p = 3.8 mg/L
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The following data was collected after
administration of an IV and oral dose to a
subject. Calculate C po , V d , and the
bioavailability (F), as well as other kinetic
parameters.
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IV data , Dose = 100 mg
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Thus the intercept value is 2.50 and the apparent volume of distribution is
Dose/C p0 = 100/2.5 = 40 L. From the slope k el can be estimated as 0.201 hr -
1
. The AUC (AUC = Dose/V d K el ) from the IV data is 12.6 mg.hr.L - 1 .
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Oral Dose = 250 mg
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Semi-log Plot of Drug Concentration versus Time after an Oral
dose
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• The time of peak concentration is 1.5 hour, thus the t 1/2
absorption might be estimated to be 0.3 hour:
t 1/2 absorption = 1.5/5 = 0.3hr
• and the absorption rate constant to be approximately
2.3 hr -1 .
• K a = 0.693/t 1/2 = 0.693/0.3 = 2.31
• The AUC from the oral data is 24.91 mg.hr.L -1 and
thus the bioavailability might be estimated as
• F = (24.9 x 100) / (250 x 12.6) = 0.79.
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Sample Preparation
1. Centrifugation - removing excipients and/or
macromolecules
A significant separation can be achieved at times by
simple centrifugation. Centrifugation of whole blood
after clot formation produces serum with no red blood
cells and fewer plasma proteins. Centrifugation of
whole blood pretreated with anticoagulants such as
heparin, EDTA or citrate produces plasma. Protein
can be removed from plasma by centrifugation after
the addition of trichloroacetic acid (TCA) or
acetonitrile.
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2. Extraction - using pH/pKa
pKa partitioning or lipid/aqueous solubility
differences
• Extraction can be used to remove unwanted
interfering compounds or to concentrate the
compound(s) ) of interest. The use of various organic
solvents of differing polarity and/or aqueous buffers
can provide excellent resolution based on the
solubility of the free compounds of interest and/or
their salt forms. Extraction may be used to remove
lipophilic interfering compounds or to remove the
desirable compounds into a cleaner environment.
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3. Ultrafiltration
One method of separating free drug from plasma
samples is ultrafiltration. . The sample can be forced
(by centrifugation) through a membrane filter. The
filtrate is protein free solution of the compound.
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4. Chromatography - Adsorption, partitioning, size, or charge
• Chromatography can be used in a clean-up mode or in a
sensitive analytical mode. In the clean-up mode the compound
of interest may be adsorbed tightly onto a column while the
interfering compounds are washed through it and subsequently
flushed from the column with a 'stronger' eluting solvent. In
the analytical mode the challenge is to choose the right column
(stationary phase) and mobile phase so that the compound is
eluted within a reasonable time and well resolved from other
components in the sample.
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