1. Area under the plasma concentration time curve

2. IMPORTANCE OF AUC Pharmacokinetics - measurement of bioavaibility absolute, relative Biopharmaceutics - comparison of drug products in BA/BE studies Calculation of PK parameters

3. Elimination Rate Constant, kel overall elimination rate constant describing removal of the drug by all elimination processes including excretion and metabolism

4. proportionality constant relating the rate of change drug concentration and concentration kel = - dCp/dt Cp

5. kel as slope

6. Calculation of kel

7. Unit of Kel kel = - dCp/dt Cp

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

9. Relation between Cp and time

10. Trapezoidal Rule

11. Calculation of AUC using the Trapezoidal Rule Trepezoid = Four sided figure with two parallel sides

12. We can calculate the AUC of each segment if we consider the segments to be trapezoids

13. CALCULATION OF A SEGMENT AUC 2-3 = Cp 2 + Cp 3. ( t 3 – t 2 ) 2

14. The area from the first to last data point can then be calculated by adding the areas together

15. Calculation of first segment The first segment can be calculated after determining the zero plasma concentration Cp 0 by extrapolation AUC 0-1 = Cp 0 + Cp 1. t 1 2

16. Calculation of last segment final segment can be calculated from t last to infinity

17. TOTAL AUC

18. EXAMPLE OF CALCULATION OF AUC timeconc 0????? 171 250 335 425 612 86.2 103.1 1 2 3 4 6 8 10

19. AUC

20. Calculation of kel TIME CONC ( ln ) 0 171 ( 4.26 ) 250 ( 3.91) 335 ( 3.55 ) 425 ( 3.21 ) 612 ( 2.48 ) 86.2 ( 1.82 ) 103.1 ( 1.13 ) ln(x) = 2.303. log(x)

21. ln conc vs time curve Slope = kel = - 0.34 1/time

22. Extrapolation of Cp 0 -kel.t Cp 0 = Cp / e Cp = 71 kel = 0.34 Cp 0 = 100

23. Calculation of first segment AUC 0-1 = Cp 0 + Cp 1. t1 2 = (100 + 71)/2. 1 = 85.5

24. Calculation of observed segments timeconcAUC 0100 17185.5 25060.5 33542.5 42530 61237 86.218.2 103.1 = Cp last 9.3 ??? AUC 2-3 = C p 2 + C p 3. ( t 3 – t 2 ) 2

25. Calculation of last segment = 3.1 / 0.34 = 9.11

26. AUC Total AUC timeconcAUC 0100 17185.5 25060.5 33542.5 42530 61237 86.218.2 103.19.3 ?? 9.1 = AUC last total292.1

27. Unit of AUC Conc. time mass. time / volume mg. hr / litre

28. AUC & other PK parameters AUC = dose = dose V. kel CL

29. References A First Course in Pharmacokinetics and Biopharmaceutics - David Bourne, Ph.D. Principles of Clinical Pharmacology,Elsevier-2 nd edition,Atkinson et al 2007

30. AUC Calculation using Trapezoidal Rule IV 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

31. AUC Total AUC timeconcAUC 0 14.04 23.82 32.76 41.62 60.93 80.49 100.31 ?? kel = 0.278 hr -1 total

32. Time (hr)Cp (mg/L) Δ (AUC mg.hr/L) AUC (mg.hr/L) 04.84 0.54.042.22 13.821.974.18 22.763.297.47 41.624.3811.85 60.932.5514.4 80.491.4215.82 100.310.816.62 ∞01.1217.74 Dose = 150 mgCp(0) = 4.84 mg/L kel = 0.278 hr -1 AUC (0-10 hr) = 16.62 mg.hr/L AUC (0-∞) = 17.74 mg.hr/L

34. Example: Assume that the Figure below was obtained after administration of a marketed antidiabetic drug. Calculate the AUC, Cmax Tmax, Kel and t 1 /2 ?

35. 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/mL Segment 2: 1 / 2 × w × sum of parallel sides = 1 / 2 × 1 × (2.8 + 5) = 3.9 ng · hr/mL Segment 3: 1 / 2 × w × sum of parallel sides = 1 / 2 × 2 × (5 + 13) = 18 ng · hr/mL

36. 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

37. Therefore, AUC = 1.4 + 3.9 + 18 + 30 + 32 + 24 + 11.5 + 2.5 = 123.3 ng · hr/mL C max = 17 ng/mL Tmax = 6 hours K el = (ln Y 1 − ln Y 2 )/(t 2 − t 1 ) = (ln 9 − ln 2.5)/(12 − 10) = 0.64 hours−1 t 1 / 2 = 0.693/0.64 hours = 1.08 hours