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❘❙❚■ Chapter 13 | Bioequivalence TrialsTable 1. Anagrelide concentration and calculation of AUC.Samples T i(h) C i(ng/mL) (C i+ C i–1) / 2 T i+ T i–1AUC i(ng.h/mL)1 0.00
Clinical Trials: A Practical Guide ■❚❙❘λ and T 1/2For calculating the remaining PK parameters, a statistical model must beestablished. In general, there is no special parametric PK model that fits the‘plasma concentration over time’ profile for the whole time interval. Nevertheless,it is empirically accepted that a single exponential model can be fitted to describethe ‘plasma concentration over time’ profile during the so-called terminal orelimination phase, and that this declining exponential curve will continue evenbeyond the observation interval. In other words, it is assumed that thedisappearance of the drug molecules follows the most simple linear onecompartmentalmodel (treating the body as a single compartment) during theelimination phase, but not for the entire interval. Based on the above assumption,the other PK parameters can be derived.In most cases, elimination of the drug is a first-order process (ie, the rate of drugelimination is directly proportional to the concentration of the drug), and a logtransformation makes it possible to draw a straight line through data from theelimination phase. The slope of this regression line in the elimination phase isequivalent to the elimination rate constant [1,2].A log transformation of the concentration values in Table 1 is plotted in Figure 2,from which the elimination phase and rate constant can be determined. It iscrucially important to select the correct starting point for the elimination phase –ie, where elimination is no longer influenced by absorption and after which thetransformed concentration values tend to be linear. This time point is the firstpoint of the elimination phase, while the last data point measured is the last pointof the elimination phase. From the anagrelide data in Figure 2, it is obvious thatthe elimination phase is from 2 to 8 hours and that the log concentration data arevery close to the fitted regression line:Log(C i) = 2.25 – 0.58 × T iwith R 2 = 0.9998This gives λ = 0.58 (the loss of drug activity per hour). R 2 is a measurement of thegoodness-of-fit of the regression line (a value close to 1 means that the regressionline is a very good fit to the data), 2.25 is the intercept, and –0.58 is the slope ofthe fitted regression line.As elimination is a first-order process, by simply dividing 0.693 (ln 2) by theestimated λ value, the half-life (T 1/2) can be calculated [1,2]:T 1/2= ln 2 / λ = 0.693 / 0.58 = 1.19123
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❘❙❚■ ContributorsStephen L
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Clinical Trials: A Practical Guide
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■■❚❙❘Chapter 1RandomizedC
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❘❙❚■ Chapter 2 | Uncontroll
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■■❚❙❘Chapter 3Protocol De
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■■❚❙❘Chapter 4EndpointsAm
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■■❚❙❘Chapter 6Source and
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■■❚❙❘Chapter 19Comparison
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■■❚❙❘Chapter 20Comparison
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■■❚❙❘Chapter 21Analysis o
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■■❚❙❘Part IVSpecial Trial
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■■❚❙❘Chapter 22Intention-
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■■❚❙❘Chapter 23Subgroup A
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■■❚❙❘Chapter 24Regression
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■■❚❙❘Chapter 25Adjustment
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■■❚❙❘Chapter 33Trial Prof
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■■❚❙❘Chapter 34Presenting
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■■❚❙❘Chapter 35Use of Tab
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■■❚❙❘GlossaryGlossary453
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■■❚❙❘IndexIndex467
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