T16: Mixed types from T12-T18

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC) AUC basic concepts Normal 1

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The elimination rate constant of a drug is 0.089 h-1. If its volume of distribution in a patient is 49.68 L, then determine the quantity of drug to be administered as an IV push which will produce an AUC of 26.07 mg × h/L.

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lb equals 115.26 mg kg

    We know that, `AUC = (dose)/(cl earance)`

    `⇒ dose=AUC × cl earance`.
    However, Cl is the product of elimination rate constant (K) and volume of distribution (V).

    ` ∴ X_{0} = AUC × K × V`.

    Plugging the values in this equation we get:
    ` X_{0} = 26.07 \quad (mg × h)/L × 0.089 \quad h^{-1} × 49.68 \quad L`

    `= 115.26 \quad mg` Ans