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Pharmacokinetics Calculations

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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
  • E-mail bknanjwade_at_yahoo.co.in

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
    administration of drug.

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
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.

56
Advantages
57
Advantages
  • 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
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

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.
  • 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 .

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
bknanjwade_at_yahoo.co.in
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