T12: AUC from dose, K and V

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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The elimination rate constant of a drug is 0.339 h-1. If its volume of distribution in a patient is 121.91 L, then determine the AUC obtained if 113.38 mg of the drug is injected.


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lb equals 2.74 mg/L h kg

    We know that, `AUC = (dose)/(cl earance)`; but Cl is the product of elimination rate constant (K) and volume of distribution (V).

    ` ∴ AUC = X_{0}/ (K × V) `.

    Plugging the values in this equation we get:
    ` AUC = (113.38 \quad mg)/ (0.339 \quad h^{-1} × 121.91 \quad L ) = 2.74 \quad (mg)/L × h ` Ans


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