Pharmacokinetics Calculations
Prof. Dr. Basavaraj K. Nanjwade M. Pharm., Ph. D
Department of Pharmaceutics
KLE University’s College of Pharmacy
BELGAUM- 590010, Karnataka, India
Cell No: 0091 974243100
E-mail: bknanjwade@yahoo.co.in
26-11-2010
KLECOP, Nipani
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Introduction :
• Pharmacokinetic Parameters:
Elimination rate constant
Biological Half life
Rate constant of absorption
Apparent volume of distributions
Area under the curve
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Example : 1
• The plasma
concentration after the
250 mg intravenous
bolus dose of an
antibiotic is given below.
Plot the data and
describe the
pharmacokinetic model.
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Time (hrs)
Conc (mcg/ml)
1.0
8.0
2.0
6.3
3.0
4.9
4.0
4.0
5.0-
3.2
6.0
2.5
7.0
1.9
3
Solution :
• Graph
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Elimination rate constant :
Suppose we choose the following two points to determine
the slope of the straight line :
• x1= 0 hr, y1 = 10.0 mcg/ml, and x2 = 7.0hr, y2 = 2.0
mcg/ml. then
Slope =
ln y2 – ln y1
x2 – x1
=
- 1.6095
7.0 hr
ln2.0 – ln 10.0
=
7.0 hr – 0 hr
=
0.6931 – 2.3026
7.0 hr
= - 0.2299/ hr
Therefore Ke = - slope = - (-0.2299/hr) = 0.2299/hr
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• Biological half-life :
Ke = 0.693
t1/2
,
0.693
t½ =
Ke
=
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therefore
0.693
0.2299/hr
= 3.01 hr
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• Area under curve :
Area from 0 to 7.0 hours –
AUC0-7.0 by trapezoidal rule = 34.85 mcg.hr/ml
AUC0-7.0 by counting squares = 34.85 mcg.hr/ml
AUC0-7.0 by Cutting and Weighing = 34.85 mcg.hr/ml
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Total area under curve :
• This is a two step method, first determine
then determine AUC7.0-∞
AUC 7.0- ∞ =
Ct
=
Ke
2.0mcg/ml
AUC0-7.0 ,
= 8.7 mcg. hr/ml
0.2299/hr
Adding this value to AUC 0-7.0, we have
AUC0-infi = AUC0-7.0 + AUC7.0-∞
= 34.85mcg.hr/ml + 8.7mcg.hr/ml
= 43.50mcg.hr/ml
AUC0-∞ =
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C0
Ke
=
10mcg / ml
0.2299/ hr
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= 43.55 mcg.hr /ml
8
• Volume of distribution :
Vd =
Dose
Co
250 mg
=
10 mcg / ml
=
250 mg
10 mg/L
= 25 L
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Description of model :
• It shoes that a 250mg dose is administered
intravenously. The apparent volume of
distribution is 25 L and the rate constant of
elimination (Ke) is 0.2299 / hr. since biological
half-life is 3.01 hr.
intravenous
250 mg
25 LITRES
injection
0.2299 / hr
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Example 2
• The plasma concentration versus time data
following the administration of a single 250 mg
rapid intravenous bolus dose of a drug is
represented by the biexponential equation;
C = 1.5e-0.13t + 12.5 e–1.3t.
Draw a schematic of the pharmacokinetic model,
assuming concentration is in mcg / ml and time
is in hours.
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Solution
• From the biexponential equation, the following parameters
of the two compartment pharmacokinetic model are
deduced : b= 0.13/hr (because the smallest hybrid rate
constant always b), and B = 1.5 mcg/ml (because B is yintercept corresponding to b). therefore a must be equal to
1.3/hr, and A = 12.5mcg/ml.
In order to draw a schematic of the pharmacokinetic
model, the following parameters need to be calculated:
rate constants K10, K12, K21,, and apparent volumes of
distribution Vc, and Vt.
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Rate constants
K 21 =
1.625 + 1.95
Ab + Ba
=
B+A
ab
K10 =
K21
1.5 + 12.5
3.575
=
14.0
= 0.2554/hr
0.169/hr
(1.3/hr) ( o.13/hr)
=
=
=0.6617/hr
0.2554/hr
0.2554/hr
K12 = a + b – K21 – K10
K12 = 1.3 /hr + 0.13 /hr – 0.2554 / hr – 0.6617 / hr
K12 = 0.5433 / hr
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Apprent volume of distribution
D
Vc =
B+A
250 mg
=
= 17.857 L
14 mcg ml
Vc (K12 + K21)
Vd =
K21
(17.857 L) (0.5433 / hr+ 0.2554 / hr) 14.2624 L = 55.843 L
=
Vd =
0.2554 / hr
0.2554
Vt = Vd – Vc
(Vc) (K12)
Vt =
K21
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= 55.843 L – 17.857 L
= 37.986 L
(17.857 L) (0.5433 / hr)
= 37. 986 L
=
0.2554 /hr
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Schematic representation :
This schematic shows that the 250 mg dose can was
given intravenously. The apparent volume of the central
and tissue compartment are 17.857 L and 37.986 L,
respectively.
The first-order rate constant of transfer of the from the
central compartment into the tissue compartment is
0.5433 /hr and the first-order rate constant of transfer of
drug from the tissue compartment in to the central
compartment is 0.2554 / hr. the first-order rate constant
of elimination of drug from the central compartment is
0.6617 / hr.
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Schematic of the two compartment model
0.5433/hr
250 mg
intravenous
17.857 L
37.986 L
0.2554/hr
injection
0.6617
/hr
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Example -3
• The following data
were obtained when a
500 mg dose of an
antibiotic was given
orally. Calculate the
pharmacokinetic
parameters,
assuming 100% of
the administered dose
was absorbed.
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TIME (Hr)
Concentration (mcg/ml)
1
26.501
2
36.091
3
37.512
4
36.055
5
32.924
6
29.413
8
22.784
16
7.571
18
5.734
20
4.343
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Graph
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Solution:
Elimination rate constant:
The rate constant of elimination is calculated
from the terminal linear portion of plasma
profile.
To determine it, we need to calculate slope of
the straight line having y-intercept = B. if
natural log are used the rate constant of
elimination (b) = negative slope of this straight
line.
Therefore
ln 5.734 - ln 4.343
0.2778 = 0.139/ hr
=
=
b = - slope
2 hr
(18 - 20) hr
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Biological half life:
The biological half life (t1/2) is determined using the
equation
t1/2 = 0.693/b
=
0.693
0.139/hr
= 4.98 hr
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The Y-intercept, B
The Y-intercept of this straight line is B and is
determined using the first order equation
ln Ct = ln B – bt
Which upon rearrangement gives
ln B = ln Ct + bt
= ln 4.343 + (0.139/hr)(20hr)
=1.4686 + 2.78
= 4.2486
B = Inverse ln 4.2486
= 70.0 mcg/ml
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Feathering the curve :
• To obtain the straight line which represents absorption
phase, the technique of feathering or the method of
residuals is used. for example, to feather the first plasma
conc. point at 1 hr, the plasma conc. at 1 hr on the
straight line having the y intercept = B is subtracted from
the plasma conc. data provided in the data.
• ln Ct = ln B - bt
ln Ct = ln 70 – (0.139) (1)
= 4.2485 – 0.139
= 4.1095
Ct = inverse ln 4.109 = 60.916 mcg/ml
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Graph
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• The residual conc. at 1 hr is obtained by subtracting from
this concentration at 1 hr provided in the data.
therefore the residual concentration at 1hr is,
1 hr = 60.916 – 26.501
= 34.415mcg/ml
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Rate constant of absorption
• The rate constant of absorption is obtained from the
slope of the straight line which represent absorption as
follows;
Ka = a = - slope
=-
ln 70 – ln 0.1
0 hr – 9.22 hr
=-
6.5511
- 9.22 hr
= - 0.71/ hr
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Apparent volume of distribution
• Since 100% of the administstered dose was absorbed,
F = 1. substittuting the values of B= 70mcg/ml, D= 50mg,
a = Ka = 0.71/hr, b Ke = 0.139/hr,
(F)(D)(a)
B=
(Vd)(a-b)
(1)(500mg)(0.71/hr)
70 mcg/ml=
(Vd)(0.71/hr – 0.139 /hr)
(Vd) =
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621.72 mg
70mcg / ml = 8.88 L
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Area under the curve
-
b
A
a
=
70 mcg/ml
0.139/hr
-
AUC =
B
70 mcg/ml
0.71/hr
AUC = 503.597 mcg – hr/ml – 98.592 mcg – hr /ml
= 405.005 mcg – hr/ml
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Description of the model
• Schematic shows that a 500 mg of the dose of the drug
was administered by an extravascular route. The firstorder rate constant of absorption is 0.71/hr and the first
order rate constant elimination is 0.139/hr. the apparent
volume of the central compartment is 8.88 L.
0.71 /hr
500 mg
8.88 LITRS
0.139 / hr
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Example -3.1
• From the data given
Calculate the time
when administered
drug dose reaches its
maximum
concentration in the
plasma.
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TIME (Hr)
Concentration (mcg/ml)
1
26.501
2
36.091
3
37.512
4
36.055
5
32.924
6
29.413
8
22.784
16
7.571
18
5.734
20
4.343
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• From the pharmacokinetic parameters found, the firstorder rate constant of absorption, Ka = 0.71/hr and the
first order rate constant elimination, Ke = 0.139/hr.
t max =
ln Ka – ln Ke
Ka – Ke
ln 0.71 /hr – ln 0.139 /hr
t max =
0.71 /hr – 0.139 /hr
t max =
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– 0.3425 – (- 1.9733)
=
0.571 /hr
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1.6308
0..571 /hr
= 2.856 hr
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Example -3.2
• From the data given
Calculate the
maximum
concentration of the
drug in plasma
attained after the
administration of the
dose.
26-11-2010
TIME (Hr)
Concentration (mcg/ml)
1
26.501
2
36.091
3
37.512
4
36.055
5
32.924
6
29.413
8
22.784
16
7.571
18
5.734
20
4.343
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• B = 70 mcg /ml, Ka = 0.71 /hr, Ke = 0.139 /hr,
and tmax = t’ = 2.856 hr
C max = B (e-bt’ – e-at’)
C max = (70 mcg/ml) (e-(0.139/hr))(2.856 hr) – e-(0.71/hr)(2.856hr))
C max =
(70 mcg/ml) (0.623 – 0.1316)
C max =
(70 mcg/ml) (0.5407)
=
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37.85 mcg ml
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Example -4
• The following data
were obtained when a
500 mg dose of an
antibiotic was given
orally calculate the
pharmacokinetic
parameters,
assuming 100% of
the administered dose
was absorbed.
26-11-2010
TIME (Hr)
Concentration (mcg/ml)
2
3.915
4
8.005
6
7.321
8
5.803
10
4.403
16
1.814
18
1.344
20
0.996
24
0.546
28
0.300
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graph
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Elimination rate constant:
The rate constant of elimination (b) is
calculated using the terminal two points of the
plasma profile as follows;
Therefore
b = - slope
=-
=26-11-2010
ln 0.546 mcg/ml - ln 0.300 mcg/ml
(24 - 28) hr
0.5988
4 hr
= 0.15/ hr
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• The y-intercept, b, of this straight line is determined
using the first-order rate equation :
B = Ct e bt
B
B
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= (0.3 mcg/ml) e(0.15/hr)(28hr)
= (0.3 mcg/ml) (66.6863)
= 20 mcg/ml
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• To obtained the straight line which represents absorption
phase, the technique of feathering is used. The plasma
profile is feathered with respect to the straight line having
y-intercept = B. To feather the first concentration point,
the concentration at 2 hr on the straight line having yintercept = B is subtracted from the data concentration at
2 hr.
C = Be-bt = (20 mcg/ml)e-(1.5/hr)(2hr)
C = (20 mcg/ml) (0.7408)
= 14.816 mcg/ml
Therefore, residual concentration at 2 hr is :
14.816 mcg/ml – 3.915 mcg/ml = 10.901 mcg/ml
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Biological half life:
The biological half life (t1/2) is determined using the
equation
t1/2 = 0.693/b
=
0.693
0.15/hr
= 4.62hr
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Rate constant of absorption
• The rate constant of absorption is obtained from the
slope of the straight line having the y-intercept = A. It is
calculated as follows;
Ka = a = - slope
=-
=
ln 40 mcg/ml – ln 0.221 mcg/ml
0 hr – 8 hr
5.198
8hr
= 0.65/ hr
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Lag-Time
• Since the value of the y-intercept A is not equal to the
value of the y-intercept B, the dosage from exhibits lagtime. The lag-time (L) is determined using equation
ln A – ln B
L =
=
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a–b
ln 40 mcg/ml – ln 0.20 mcg/ml
0.65 – 0.15
=
0.693
0.5 / hr
=
1.386 hr
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• The equation for calculating the time of maximum
concntration of drug in plasma in presence of lag-time,
tmax (L), is
t max =
ln A – ln B + ln a – ln b
a-b
ln 40 – ln 20 + ln 0.65 – ln 0.15
t max =
0.65 /hr – 0.15 /hr
t max =
1.4663
= 4.319 hr
0.5 /hr
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C max (L) = Be-bt” – Ae-at”)
C max (L) = (20 mcg/ml) (e-(0.15/hr))(4.319 hr) – e-(0.65 /hr)(4.319 hr))
C max (L) = (20 mcg/ml) (0.5253) – (40 mcg /ml) (0.0604)
C max (L) = 10.463 mcg/ml – 2.415 mcg /ml
C max (L) =
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8.048 mcg /ml
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Pharmacokinetics
and
Pharmacodynamics Parameters
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Measurement of
bioavailability
• Pharmacokinetic methods ( indirect )
1. Blood analysis
2. Urinary excretion data
• Pharmacodynamic methods ( direct )
1. Acute pharmacological response
2. Therapeutic response
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Blood analysis
• Plasma level time studies or The plasma concentration – time
curve or blood level curve.
• A direct relationship exists concentration of drug at the site of
action & concentration of drug in the plasma.
• Serial blood samples are taken after drug administration &
analyzed for drug concentration.
• A typical blood level curve obtained after oral administration of
drug.
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Parameters determined
Pharmacokinetic parameters
• Peak Plasma Concentration (Cmax)
• Time of Peak concentration (tmax).
• Area Under Curve (AUC)
•
•
•
•
•
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Pharmacodynamics parameters
Minimum Effective Concentration (MEC) / Minimum
Inhibitory Concentration (MIC).
Maximum Safe Concentration (MSC) / Maximum Safe
Dose (MSD).
Duration of action
Onset of action.
Intensity of action.
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Parameters determined
•
AUC or Extent of absorption can be measured by 3
methods…
1.Planimeter
Instrument for mechanically measuring the area
2. Cut & weigh method
AUC is cut & weighed on analytical balance. The weight
obtained is converted to proper unit by dividing it by the wt
of a unit area of same paper.
3. Trapezoidal method
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Parameters determined
3. Trapezoidal method
AUC = ½ ( C1 + C2) (t2 – t1) + ½ (C2 + C3) (t3 – t2) +…….
½ (C n-1 + C n ) (tn – tn-1 )
C = Concentration
t = time
subscript= sample number
AUC = Area Under Curve
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Parameters determined
Relative bioavailability
F rel =
( AUC) drug . (Dose) standard
(AUC) standard .(Dose) drug
Absolute bioavailability
Fab =
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(AUC)drug . (Dose) IV
(AUC)IV . (Dose) drug
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Parameters determined
•From the following blood data obtained after the oral administration of 50mg of
drug A. calculate the AUC?
Time in hr
Plasma drug con in mcg/ml
1
5.5
2
3
9.2
14.9
4
5
10.3
7.3
6
2.2
AUC
= ½½(5.5
(5.5+9.2)
+9.2)
(2-1)
+ ½ (9.2+14.9)
(3-2)
+ ½ (14.9+10.3)
AUC =
(2-1)
+ ½ (9.2+14.9)
(3-2) + ½
(14.9+10.3)
(4-3) ½ (1AUC(4-3)
= ½
+9.2)7.1)(5-4)
(2-1) + ½ (9.2+14.9)
½ (14.9+10.3) (4-3) ½ (10.3+ 7.1)(5-4) + ½
½(5.5
(10.3+
+ ½ (7.1 (3-2)
+2.2)+ (6-5)
(7.1 +2.2) (6-5)
AUC
= 45.35
45.35mcg/ml
mcg/ml
AUC =
hr hr
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Parameters determined
• The AUC of a new sustained release diclofenac sodium
developed in the lab after giving in a dose of 100mg was
found to be 250.30 mcg/ml hr.The AUC of the standard
marketed sustained release tablets of the same at the
same dose was found to be 261.35 mcg/ml hr. what is
the the relative bioavailability of he same drug.
F rel = 250.30 X 100
261.35 X 100
= 0.9577 or 95.77%
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Parameters determined
• The AUC of salbutamol sulphate from a 10 mg
IV dose was found to be 94.6mcg/ml hr.when
the same dose was given orally, the AUC was
found to be 60.5 mcg/ml hr. What is the
absolute bioavailability of the drug?
Fabs = 60.5 X 10
94.6 X 10
Fabs = 0.6395 or 63.95
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Urinary excretion data
• The method of determination bioavailability
provided that the active ingredient is excreted
unchanged in the significant quantity of urine.
• The cumulative amount of active drug excreted in
urine is directly proportional to extent of systemic
drug absorption.
• The rate of drug excretion is directly proportional
to rate of systemic drug absorption.
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Advantages
• Useful when there is lack of sufficiently sensitive analytical
techniques to measure concentration of drug in plasma.
• Noninvasive method therefore better subject compliance.
• Convenience of collecting urine samples in comparison to
drawing of blood periodically.
• If any case the urine drug concentration is low, assaying of
larger sample volume is relatively more.
• Direct measurement of bioavailability, both absolute &
relative is possible without the necessity of fitting the data
to the mathematical model.
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Advantages
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Advantages
• Bioavailability is determined by….
F= (U
(U
) oral . D IV
) IV . D oral
U
= Cumulative amt of unchanged drug excreted in
urine
D IV = IV dose
D oral = oral dose
F = absolute bioavailability
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Advantages
• When drug A was administered IV to a group of volunteers, 80%
of the 500mg dose was recovered unchanged in the urine. When
the same drug was administered to the same volunteers
orally.280 mg was recovered unchanged in urine. What is the
absolute bioavailability of Drug A following oral administration.
Absolute bioavailability = (cumulative amt.of drug excreted)sample
(cumulative amt.of drug excreted)IV
=
280
400
= 0.7 or 70%
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Acute pharmacological response
• Bioavailability can be determined from the acute
pharmacologic effect – time curve as well as from dose
response graph.
• DISADVANTAGE is that pharmacological response tends to
more variable & accurate correlation between the measured
response & drug available from the formulation is difficult.
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Therapeutic response
• This method is based on the observing the clinical response to a
drug formulation given to a patients suffering from disease for
which it is intended to be used.
• Ex …for anti inflammatory drugs, the reduction in the
inflammation is determined.
• The major DRAWBACK is …quantification of observed response
is too improper to allow for reasonable assessment of relative
bioavailability between two dosage forms of a same drug.
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Rate of Absorption
• AUC/dose gives an average extent of bioavailability.
• The rate of absorption is usually also important
for the onset of drug action.
• The time of peak plasma concentration is used
often as a measure of the rate of drug absorption.
• The peak plasma concentration is also an
important parameter - for keeping the drug
concentration within the therapeutic window.
• Absorption can be characterized by evaluating the absorption rate
constant Ka from the plasma concentration –time data.
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The method of ‘Residuals’
• Also called as Feathering or peeling or stripping.
ASSUMPTIONS
• Absorption & elimination process follows 1st order kinetics.
• Absorption from the dosage form is complete.
• Ka is at least five times larger than Ke
• Kinetic model is
AG
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Ka
AB
Kc
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62
The method of ‘Residuals’
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The method of ‘Residuals’
• This technique is used to resolve a multiexponential
curve into its individual components.
• For a drug that follows one compartment kinetics &
administered e.v, the concentration of drug in plasma is
expressed by…
C=
Ka F X0
Vd ( Ka –KE )
[ e –kEt – e –Kat ]
1
If Ka F X0 / Vd ( Ka – kE ) = A, a hybrid constant then,
C=Ae
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–kEt
-A e –Kat
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64
The method of ‘Residuals’
• During the elimination phase, when the absorption is
almost over, Ka > > KE & the value of second exponential e
–Kat approaches zero whereas the 1st exponential e –kEt
retains some finite value….at this time the equation is
C = A e –kEt
in log form
log C = log A – KEt/ 2.303
3
4
Where C is the back extrapolated plasma concentration
value.
A plot of log C versus t yields a biexponential curve with a
terminal linear phase having slope KE/ 2.303.
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The method of ‘Residuals’
• Back extrapolation of this straight line to zero yields
y-intercept equal to log A.
• Subtraction of true plasma concentration values that
is equation 2 from the extrapolated plasma
concentration values that is equation 3 yields a series
of residual concentration values Cr
( C - C ) = Cr = A e –Kat ,
in log form the equation is :
log Cr = log A - Kat/ 2.303
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The method of ‘Residuals’
• A plot of logCr versus t yields a straight line with
slope –Ka / 2.303 & y intercept log A.
• Absorption half life can be computed from Ka using
the relation 0.693/Ka …thus the method of residual
enables resolution of the biexponential plasma level
time curve into its exponential components.
• The technique works best when the difference
between Ka & KE is large ( Ka >= 3)
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Wagner Nelson Method
ASSUMPTIONS
• The body behaves as a single homogenous compartment.
• Drug elimination obeys 1st order kinetics.
DISADVANTAGES
• The absorption & elimination processes can be quite similar &
still accurate determination of Ka can not be made.
• The absorption process doesn’t have to be 1st order.
• The kinetics of absorption may be zero order, mixed order,
mixed zero order & 1st order or even more complex.
• This method involves determination of Ka from percent
absorbed –time plot & does not require the assumption of zero .
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Wagner Nelson Method
• The amount of drug in the body X & the amt of drug eliminated
in the body XE thus :
XA = X + X E
If the amt of drug in the body is X = V.dC…the amt of drug
eliminated at any time t can be calculated as ..
XE = KE Vd ( AUC)to
Substitution of values of X & XE in above equation…
XA = Vd C + KE Vd ( AUC)to
from this equation we can get the value for drug absorbed in to
the systemic circulation from time zero to
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Wagner Nelson Method
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Thank you…..
Cell No: 0091 974243100
E-mail: bknanjwade@yahoo.co.in
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