Title: Pharmacokinetics Calculations
1Pharmacokinetics Calculations
 Prof. Dr. Basavaraj K. Nanjwade M. Pharm., Ph. D
 Department of Pharmaceutics
 KLE Universitys College of Pharmacy
 BELGAUM 590010, Karnataka, India
 Cell No 0091 974243100
 Email bknanjwade_at_yahoo.co.in
2Introduction
 Pharmacokinetic Parameters
 Elimination rate constant
 Biological Half life
 Rate constant of absorption
 Apparent volume of distributions
 Area under the curve
3Example 1
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
 The plasma concentration after the 250 mg
intravenous bolus dose of an antibiotic is given
below. Plot the data and describe the
pharmacokinetic model.
4Solution
5 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
ln2.0 ln 10.0
ln y2 ln y1
0.6931 2.3026
Slope
7.0 hr
7.0 hr 0 hr
x2 x1
 1.6095
 0.2299/ hr
7.0 hr
Therefore Ke  slope  (0.2299/hr)
0.2299/hr
6 7 Area under curve
 Area from 0 to 7.0 hours

 AUC07.0 by trapezoidal rule 34.85 mcg.hr/ml
 AUC07.0 by counting squares 34.85 mcg.hr/ml
 AUC07.0 by Cutting and Weighing 34.85
mcg.hr/ml
8Total area under curve
 This is a two step method, first determine
AUC07.0 , then determine AUC7.08
Adding this value to AUC 07.0, we
have AUC0infi AUC07.0 AUC7.08
34.85mcg.hr/ml 8.7mcg.hr/ml 43.50mcg.hr/ml
9 10Description 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
halflife is 3.01 hr.
25 LITRES
11Example 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 1.5e0.13t 12.5 e1.3t.
 Draw a schematic of the pharmacokinetic model,
assuming concentration is in mcg / ml and time is
in hours.
12Solution
 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.
13Rate constants
K12 a b K21 K10
K12 1.3 /hr 0.13 /hr 0.2554 / hr 0.6617
/ hr
K12 0.5433 / hr
14Apprent volume of distribution
250 mg
B A
Vc (K12 K21)
Vd
K21
Vt Vd Vc 55.843 L 17.857 L
37.986 L
15Schematic 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 firstorder rate constant of transfer of the
from the central compartment into the tissue
compartment is 0.5433 /hr and the firstorder
rate constant of transfer of drug from the tissue
compartment in to the central compartment is
0.2554 / hr. the firstorder rate constant of
elimination of drug from the central compartment
is 0.6617 / hr.
16Schematic of the two compartment model
17.857 L
0.2554/hr
17Example 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.
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
18Graph
19Solution
 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 yintercept B. if
natural log are used the rate constant of
elimination (b) negative slope of this straight
line.  Therefore
 b  slope

ln 5.734  ln 4.343
0.2778
0.139/ hr


2 hr
(18  20) hr
20Biological 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
21The Yintercept, B
 The Yintercept 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
22Feathering 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
23Graph
24 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
25Rate constant of absorption
 The rate constant of absorption is obtained from
the slope of the straight line which represent
absorption as follows
ln 70 ln 0.1
Ka a  slope 
0 hr 9.22 hr
6.5511

 9.22 hr
 0.71/ hr
26Apparent 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,
27Area under the curve
70 mcg/ml
B
70 mcg/ml
A


AUC
a
b
0.139/hr
0.71/hr
AUC 503.597 mcg hr/ml 98.592 mcg hr
/ml 405.005 mcg hr/ml
28Description 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.
8.88 LITRS
29Example 3.1
 From the data given Calculate the time when
administered drug dose reaches its maximum
concentration in the plasma.
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
30 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.
31Example 3.2
 From the data given Calculate the maximum
concentration of the drug in plasma attained
after the administration of the dose.
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
32 B 70 mcg /ml, Ka 0.71 /hr, Ke 0.139 /hr,
and tmax t 2.856 hr
B (ebt eat)
C max
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)
37.85 mcg ml
33Example 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.
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
34graph
35 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

ln 0.546 mcg/ml  ln 0.300 mcg/ml

(24  28) hr
0.5988
0.15/ hr

4 hr
36 The yintercept, b, of this straight line is
determined using the firstorder rate equation
B Ct e bt (0.3 mcg/ml) e(0.15/hr)(28hr)
B (0.3 mcg/ml) (66.6863)
B 20 mcg/ml
37 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 yintercept
B. To feather the first concentration point, the
concentration at 2 hr on the straight line having
y intercept B is subtracted from the data
concentration at 2 hr.
C Bebt (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
38Biological 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
39Rate constant of absorption
 The rate constant of absorption is obtained from
the slope of the straight line having the
yintercept A. It is calculated as follows
ln 40 mcg/ml ln 0.221 mcg/ml
Ka a  slope 
0 hr 8 hr
5.198
8hr
0.65/ hr
40LagTime
 Since the value of the yintercept A is not equal
to the value of the yintercept B, the dosage
from exhibits lagtime. The lagtime (L) is
determined using equation
ln A ln B
L
a b
ln 40 mcg/ml ln 0.20 mcg/ml
0.65 0.15
0.693
0.5 / hr
1.386 hr
41 The equation for calculating the time of maximum
concntration of drug in plasma in presence of
lagtime, tmax (L), is
ln A ln B ln a ln b
t max
a  b
ln 40 ln 20 ln 0.65 ln 0.15
t max
0.65 /hr 0.15 /hr
1.4663
t max
4.319 hr
0.5 /hr
42Bebt Aeat)
C max (L)
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)
8.048 mcg /ml
43Pharmacokinetics and Pharmacodynamics Parameters
44Measurement of bioavailability
 Pharmacokinetic methods ( indirect )
 1. Blood analysis
 2. Urinary excretion data
 Pharmacodynamic methods ( direct )
 1. Acute pharmacological response
 2. Therapeutic response
45Blood 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.
46(No Transcript)
47Parameters determined
Pharmacokinetic parameters
 Peak Plasma Concentration (Cmax)
 Time of Peak concentration (tmax).
 Area Under Curve (AUC)
 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.
Pharmacodynamics parameters
48Parameters 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
49Parameters determined
 3. Trapezoidal method

 AUC ½ ( C1 C2) (t2 t1) ½ (C2 C3) (t3
t2) .  ½ (C n1 C n ) (tn tn1 )

 C Concentration
 t time
 subscript sample number
 AUC Area Under Curve
50Parameters determined
 Relative bioavailability
 F rel ( AUC) drug . (Dose) standard
 (AUC) standard .(Dose) drug
 Absolute bioavailability
 Fab (AUC)drug . (Dose) IV
 (AUC)IV . (Dose) drug
51From the following blood data obtained after the
oral administration of 50mg of drug A. calculate
the AUC?
Parameters determined
Time in hr Plasma drug con in mcg/ml
1 5.5
2 9.2
3 14.9
4 10.3
5 7.3
6 2.2
AUC ½ (5.5 9.2) (21) ½ (9.214.9) (32) ½
(14.910.3) (43) ½ (10.3 7.1)(54) ½ (7.1
2.2) (65) AUC 45.35 mcg/ml hr
AUC ½ (5.5 9.2) (21) ½ (9.214.9) (32) ½
(14.910.3) (43) ½ (1AUC ½ (5.5 9.2) (21)
½ (9.214.9) (32) ½ (14.910.3) (43) ½ (10.3
7.1)(54) ½ (7.1 2.2) (65) AUC 45.35 mcg/ml
hr
52Parameters 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
53Parameters 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
54Urinary 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.
55Advantages
 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.
56Advantages
57Advantages
 Bioavailability is determined by.
 F ( U ) oral . D IV
 (U ) IV . D oral
 U Cumulative amt of unchanged drug
excreted in  urine
 D IV IV dose
 D oral oral dose
 F absolute bioavailability
58Advantages
 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
59Acute 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. 
60Therapeutic 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.
61Rate 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.
62The 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 AB
AE
Ka
Kc
63The method of Residuals
64The 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 e kEt e Kat
 Vd ( Ka KE )
 If Ka F X0 / Vd ( Ka kE ) A, a hybrid
constant then,  C A e kEt A e Kat

1
2
65The method of Residuals
 During the elimination phase, when the absorption
is almost over, Ka gt gt 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
 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.
3
4
66The method of Residuals
 Back extrapolation of this straight line to zero
yields yintercept 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

67The 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 gt 3)
68Wagner 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 doesnt 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 .
69Wagner Nelson Method
 The amount of drug in the body X the amt of
drug eliminated in the body XE thus  XA X XE
 If the amt of drug in the body is X V.dCthe
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
70Wagner Nelson Method
71Cell No 0091 974243100 Email
bknanjwade_at_yahoo.co.in