T19: AUC from K, dose and concentration

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC)

50.1.3.1) AUC basic concepts Normal 1


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472.83 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 4.14 mg/L. The elimination rate constant of the drug is 0.291 h-1. Determine the AUC.


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lb equals 14.23 mg × h/L kg

    We know that, AUC may be calculated from dose (`X_{0}`) and clearance (`Cl`) OR initial concentration (`C_{0}`) and the elimination rate constant (`K`).

    ` AUC = (X_{0})/(Cl) = (C_{0})/K`.

    By plugging the values of initial concentration and K, we get,

    ` AUC = (C_{0})/K = (4.14 \quad (mg)/L)/(0.291 \quad h^{-1})=14.23 \quad (mg × h)/L`. Ans.


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