area under the plasma concentration time curve. importance of auc pharmacokinetics - measurement of...

Area under the plasma concentration time curve

IMPORTANCE OF AUC

Pharmacokinetics -measurement of bioavaibility

absolute , relative

Biopharmaceutics - comparison of drug products in BA/BE studies

Calculation of PK parameters

Elimination Rate Constant, kel

overall elimination rate constant describing removal of the drug by all elimination processes including excretion and metabolism

proportionality constant relating the rate of change drug concentration and concentration

kel = - dCp/dt Cp

kel as slope

Calculation of kel

Unit of Kel kel = - dCp/dt

Cp

Drug kel, 1 / hour Acetaminophen 0.28 Diazepam 0.021 Digoxin 0.017 Gentamicin 0.35 Lidocaine 0.43 Theophylline 0.063

Relation between Cp and time

Trapezoidal Rule

Calculation of AUC using the Trapezoidal Rule

Trepezoid= Four sided figure with two

parallel sides

We can calculate the AUC of each segment if we consider

the segments to be trapezoids

CALCULATION OF A SEGMENT

AUC2-3 = Cp2 + Cp3 . ( t3 – t2 )

2

The area from the first to last data point can then be calculated by adding the

areas together

Calculation of first segment

• The first segment can be calculated after determining the zero plasma concentration Cp0 by extrapolation

AUC0-1 = Cp0+ Cp1 . t1

2

Calculation of last segment

final segment can be calculated from

tlast to infinity

TOTAL AUC

EXAMPLE OF CALCULATION OF

AUCtime conc

0 ?????

1 71

2 50

3 35

4 25

6 12

8 6.2

10 3.1

1

2

3

4

6

8

10

AUC

Calculation of kel

TIME CONC ( ln )

0

1 71 ( 4.26 )

2 50 ( 3.91)

3 35 ( 3.55 )

4 25 ( 3.21 )

6 12 ( 2.48 )

8 6.2 ( 1.82 )

10 3.1 ( 1.13 )

ln(x) = 2.303 . log(x)

ln conc vs time curve

• Slope = kel = - 0.34 1/time

00.5

11.5

22.5

33.5

44.5

5

0 2 4 6 8 10 12

ln co

nc

time

Extrapolation of Cp0

-kel.tCp0 = Cp / e

Cp = 71kel = 0.34

Cp0 = 100

Calculation of first segment

AUC0-1 = Cp0 + Cp1 . t1

2

= (100 + 71)/2 . 1

= 85.5

Calculation of observed segments

time conc AUC

0 100

1 71 85.5

2 50 60.5

3 35 42.5

4 25 30

6 12 37

8 6.2 18.2

10 3.1 = Cplast 9.3

??? ??? ???

AUC2-3 = Cp2 + Cp3 . ( t3 – t2 ) 2

Calculation of last segment

= 3.1 / 0.34

= 9.11

Total AUCAUCtime conc AUC

0 100

1 71 85.5

2 50 60.5

3 35 42.5

4 25 30

6 12 37

8 6.2 18.2

10 3.1 9.3

?? ?? 9.1 = AUClast

total 292.1

Unit of AUC

• Conc . time

• mass . time / volume

• mg . hr / litre

AUC & other PK parameters

AUC = dose = dose

V . kel CL

References

A First Course in Pharmacokinetics and Biopharmaceutics

- David Bourne, Ph.D.

Principles of Clinical Pharmacology ,Elsevier-2nd edition ,Atkinson et al 2007

AUC Calculation using Trapezoidal RuleIV Bolus - Linear One Compartment

• A dose of 150 mg was administered to healthy volunteer. Seven blood samples were collected at 0.5, 1, 2, 4, 6, 8, 10 hours. Plasma was separated from each blood sample and analyzed for drug concentration.

• The collected data are shown in the table below.

• kel = 0.278 hr-1

Total AUCAUCtime conc AUC

0

1 4.04

2 3.82

3 2.76

4 1.62

6 0.93

8 0.49

10 0.31

?? ??

kel = 0.278 hr-1 total

Time (hr) Cp (mg/L)Δ (AUC mg.hr/L)

AUC (mg.hr/L)

0 4.84    

0.5 4.04 2.22 2.22

1 3.82 1.97 4.18

2 2.76 3.29 7.47

4 1.62 4.38 11.85

6 0.93 2.55 14.4

8 0.49 1.42 15.82

10 0.31 0.8 16.62

∞ 0 1.12 17.74

Dose = 150 mgCp(0) = 4.84 mg/Lkel = 0.278 hr-1

AUC(0-10 hr) = 16.62 mg.hr/LAUC(0-∞) = 17.74 mg.hr/L

Example:

Assume that the Figure below was obtained after administration of a marketed antidiabetic drug. Calculate the AUC, Cmax

Tmax, Kel and t1/2?

As shown in Figure, the AUC is divided into eight segments.

Segments 1 and 8 are triangles and rest of the segments are trapezoids.

The AUC is calculated for all the segments individually. The total AUC is then calculated by adding up the AUC values of all the individual segments.

The calculations are shown as follows:Segment 1: (w × h)/2 = (1 × 2.8)/2 = 1.4 ng·hr/mLSegment 2:1/2× w × sum of parallel sides =1/2× 1 × (2.8 + 5)

= 3.9 ng·hr/mLSegment 3:1/2× w × sum of parallel sides =1/2× 2 × (5 + 13)

= 18 ng·hr/mL

Segment 5:1/2× w× sum of parallel sides =1/2× 2 × (17 + 15) = 32 ng·hr/mL

Segment 6:1/2× w × sum of parallel sides =1/2× 2 × (15 + 9) =

24 ng·hr/mL

Segment 7:1/2× w × sum of parallel sides =1/2× 2 × (9 + 2.5) =

11.5 ng·hr/mL

Segment 8: (w × h)/2 = (2 × 2.5)/2 = 2.5 ng·hr/mL

Therefore, AUC = 1.4 + 3.9 + 18 + 30 + 32 + 24 + 11.5+ 2.5 = 123.3 ng·hr/mL

Therefore, AUC = 1.4 + 3.9 + 18 + 30 + 32 + 24 + 11.5+ 2.5 = 123.3 ng·hr/mL

Cmax= 17 ng/mL

Tmax = 6 hours

Kel= (ln Y1− ln Y2)/(t2− t1) = (ln 9 − ln 2.5)/(12 − 10)

= 0.64 hours−1

t1/2= 0.693/0.64 hours = 1.08 hours