Introduction :• Pharmacokinetic Parameters:

Elimination rate constant Biological Half life Rate constant of absorption Apparent volume of distributions Area under the curve

1

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

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

=ln2.0 – ln 10.0

7.0 hr – 0 hr=

0.6931 – 2.3026

7.0 hr

=- 1.6095

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 , therefore

t½ =0.693

Ke

0.2299/hr

0.693= 3.01 hr=

5

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 AUC0-7.0 , then determine AUC7.0-∞

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

AUC 7.0- ∞ = Ct

Ke

=2.0mcg/ml

0.2299/hr= 8.7 mcg. hr/ml

AUC0-∞ = C0

Ke

=10mcg / ml

0.2299/ hr= 43.55 mcg.hr /ml

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Volume of distribution :

Vd = Dose

Co

= 250 mg

10 mcg / ml

= 250 mg

10 mg/L

= 25 L

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25 LITRES

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.

250 mgintravenous

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 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.

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Rate constants

K12 = a + b – K21 – K10

K12 = 1.3 /hr + 0.13 /hr – 0.2554 / hr – 0.6617 / hr

K12 = 0.5433 / hr

K 21 =Ab + Ba

B + A=

1.625 + 1.95

1.5 + 12.5=

3.575

14.0= 0.2554/hr

K10 = abK21

=(1.3/hr) ( o.13/hr)

0.2554/hr=

0.169/hr

0.2554/hr=0.6617/hr

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Apprent volume of distribution

B + A

250 mg

Vd =Vc (K12 + K21)

K21

Vt = Vd – Vc = 55.843 L – 17.857 L = 37.986 L

Vc =D

=14 mcg ml

= 17.857 L

Vd =(17.857 L) (0.5433 / hr+ 0.2554 / hr)

0.2554 / hr= 14.2624 L

0.2554= 55.843 L

Vt =(Vc) (K12)

K21=

(17.857 L) (0.5433 / hr)0.2554 /hr

= 37. 986 L

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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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250 mgintravenous

injection

17.857 L

0.2554/hr

0.5433/hr

37.986 L

0.6617 /hr

Schematic of the two compartment model

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• 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

Example -3

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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

b = - slope

= - ln 5.734 - ln 4.343

(18 - 20) hr= - 0.2778

2 hr = 0.139/ 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

=

19

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 20

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,

B = (F)(D)(a)

(Vd)(a-b)

(1)(500mg)(0.71/hr)70 mcg/ml=

(Vd)(0.71/hr – 0.139 /hr)

(Vd) = 621.72 mg

70mcg / ml = 8.88 L

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Area under the curve

AUC = B A-

b a

70 mcg/ml -

0.139/hr 0.71/hr

70 mcg/ml

AUC = 503.597 mcg – hr/ml – 98.592 mcg – hr /ml

= 405.005 mcg – hr/ml

=

26

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 LITRS500 mg0.71 /hr

0.139 / hr

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• 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

Example -3.1

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• 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.

t max = ln Ka – ln Ke

Ka – Ke

ln 0.71 /hr – ln 0.139 /hr

0.71 /hr – 0.139 /hrt max =

t max = – 0.3425 – (- 1.9733)

0.571 /hr

1.6308

0..571 /hr= = 2.856 hr

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• 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

Example -3.2

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• B = 70 mcg /ml, Ka = 0.71 /hr, Ke = 0.139 /hr, and tmax = t’ = 2.856 hr

C max =

C max =

C max =

C max =

B (e-bt’ – e-at’)

(70 mcg/ml) (e-(0.139/hr))(2.856 hr) – e-(0.71/hr)(2.856hr))

(70 mcg/ml) (0.623 – 0.1316)

(70 mcg/ml) (0.5407)

= 37.85 mcg ml

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• 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

Example -4

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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

= -

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 = (0.3 mcg/ml) e(0.15/hr)(28hr)

B = (0.3 mcg/ml) (66.6863)

B = 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 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/ml36

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

=

37

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

= 5.1988hr

= 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 lag-time. The lag-time (L) is determined using equation

ln 40 mcg/ml – ln 0.20 mcg/ml

0.65 – 0.15

= 0.6930.5 / hr

= 1.386 hr

L = ln A – ln B

a – b

=

39

• 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

0.65 /hr – 0.15 /hrt max =

t max = 1.4663

0.5 /hr= 4.319 hr

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C max (L) =

C max (L) =

C max (L) =

C max (L) =

Be-bt” – Ae-at”)

(20 mcg/ml) (e-(0.15/hr))(4.319 hr) – e-(0.65 /hr)(4.319 hr))

(20 mcg/ml) (0.5253) – (40 mcg /ml) (0.0604)

10.463 mcg/ml – 2.415 mcg /ml

= 8.048 mcg /mlC max (L)

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Pharmacokinetics and

Pharmacodynamics Parameters

42

Measurement of bioavailability

• Pharmacokinetic methods ( indirect ) 1. Blood analysis

2. Urinary excretion data

• Pharmacodynamic methods ( direct ) 1. Acute pharmacological response

2. Therapeutic response

43

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.

44

45

Parameters determined

• 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.

Pharmacokinetic parameters

Pharmacodynamics parameters

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• 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

48

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

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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 9.2

3 14.9

4 10.3

5 7.3

6 2.2

AUC = ½ (5.5 +9.2) (2-1) + ½ (9.2+14.9) (3-2) + ½ (14.9+10.3) (4-3) ½ (10.3+ 7.1)(5-4) + ½ (7.1 +2.2) (6-5)

AUC = 45.35 mcg/ml hr50

Parameters determinedParameters determined

AUC = ½ (5.5 +9.2) (2-1) + ½ (9.2+14.9) (3-2) + ½ (14.9+10.3) (4-3) ½ (1AUC = ½ (5.5 +9.2) (2-1) + ½ (9.2+14.9) (3-2) + ½ (14.9+10.3) (4-3) ½ (10.3+ 7.1)(5-4) + ½ (7.1 +2.2) (6-5)

AUC = 45.35 mcg/ml hr

• 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%

51

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

52

Parameters determined

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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55

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

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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%

57

Advantages

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.

58

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 AB AE

Ka Kc

61

62

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

63

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

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

64

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

65

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)

66

The method of ‘Residuals’

Wagner Nelson MethodASSUMPTIONS• 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 .

67

• 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.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

68

Wagner Nelson Method

69

Wagner Nelson Method