# Pharmacokinetics Calculations - PowerPoint PPT Presentation

PPT – Pharmacokinetics Calculations PowerPoint presentation | free to download - id: 3e10a8-OGU5N The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## Pharmacokinetics Calculations

Description:

### Pharmacokinetics Calculations Prof. Dr. Basavaraj K. Nanjwade M. Pharm., Ph. D Department of Pharmaceutics KLE University s College of Pharmacy – PowerPoint PPT presentation

Number of Views:567
Avg rating:3.0/5.0
Slides: 72
Provided by: apiNingC81
Category:
Tags:
Transcript and Presenter's Notes

Title: Pharmacokinetics Calculations

1
Pharmacokinetics 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

2
Introduction
• Pharmacokinetic Parameters
• Elimination rate constant
• Biological Half life
• Rate constant of absorption
• Apparent volume of distributions
• Area under the curve

3
Example 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.

4
Solution
• Graph

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
• Biological half-life

7
• 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

8
Total area under curve
• This is a two step method, first determine
AUC0-7.0 , then determine AUC7.0-8

Adding this value to AUC 0-7.0, we
have AUC0-infi AUC0-7.0 AUC7.0-8
34.85mcg.hr/ml 8.7mcg.hr/ml 43.50mcg.hr/ml
9
• Volume of distribution

10
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.

25 LITRES
11
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 e1.3t.
• Draw a schematic of the pharmacokinetic model,
assuming concentration is in mcg / ml and time is
in hours.

12
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 y-intercept
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.

13
Rate constants
K12 a b K21 K10
K12 1.3 /hr 0.13 /hr 0.2554 / hr 0.6617
/ hr
K12 0.5433 / hr
14
Apprent volume of distribution
250 mg
B A
Vc (K12 K21)
Vd
K21
Vt Vd Vc 55.843 L 17.857 L
37.986 L
15
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.

16
Schematic of the two compartment model
17.857 L
0.2554/hr
17
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.

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
18
Graph
19
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
• b - slope

ln 5.734 - ln 4.343
0.2778
0.139/ hr
-
-
2 hr
(18 - 20) hr
20
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

21
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

22
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

23
Graph
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

25
Rate 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
26
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,

27
Area 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
28
Description of the model
• Schematic shows that a 500 mg of the dose of the
drug was administered by an extravascular route.
The first-order 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
29
Example -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
first-order rate constant of absorption, Ka
0.71/hr and the first order rate constant
elimination, Ke 0.139/hr.

31
Example -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 (e-bt e-at)
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
33
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.

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
34
graph
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 y-intercept, b, of this straight line is
determined using the first-order 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 y-intercept
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 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
38
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

39
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

ln 40 mcg/ml ln 0.221 mcg/ml
Ka a - slope -
0 hr 8 hr
5.198

8hr
0.65/ hr
40
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 lag-time. The lag-time (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
lag-time, 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
42
Be-bt Ae-at)
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
43
Pharmacokinetics and Pharmacodynamics Parameters
44
Measurement of bioavailability
• Pharmacokinetic methods ( indirect )
• 1. Blood analysis
• 2. Urinary excretion data
• Pharmacodynamic methods ( direct )
• 1. Acute pharmacological response
• 2. Therapeutic response

45
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

46
(No Transcript)
47
Parameters 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
48
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

49
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

50
Parameters determined
• Relative bioavailability
• F rel ( AUC) drug . (Dose) standard
• (AUC) standard .(Dose) drug
• Absolute bioavailability
• Fab (AUC)drug . (Dose) IV
• (AUC)IV . (Dose) drug

51
From 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) (2-1) ½ (9.214.9) (3-2) ½
(14.910.3) (4-3) ½ (10.3 7.1)(5-4) ½ (7.1
2.2) (6-5) AUC 45.35 mcg/ml hr
AUC ½ (5.5 9.2) (2-1) ½ (9.214.9) (3-2) ½
(14.910.3) (4-3) ½ (1AUC ½ (5.5 9.2) (2-1)
½ (9.214.9) (3-2) ½ (14.910.3) (4-3) ½ (10.3
7.1)(5-4) ½ (7.1 2.2) (6-5) AUC 45.35 mcg/ml
hr
52
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

53
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

54
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.

55
• 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.

56
57
• 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

58
• 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
• Absolute bioavailability (cumulative amt.of
drug excreted)sample

• (cumulative amt.of drug excreted)IV
• 280
• 400
• 0.7 or 70

59
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.

60
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.

61
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.

62
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 AB
AE

Ka
Kc
63
The method of Residuals
64
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 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
65
The 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
66
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

67
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 gt 3)

68
Wagner Nelson Method
• ASSUMPTIONS
• The body behaves as a single homogenous
compartment.
• Drug elimination obeys 1st order kinetics.
• 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 .

69
Wagner 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

70
Wagner Nelson Method
71
• Thank you..

Cell No 0091 974243100 E-mail