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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