T20: Mixed types from T12-T20

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 0.131 h-1. It is parenterally administered to a patient weighing 163 Kg. If the volume of distribution of the drug is 0.69 L/Kg, and the AUC was measured to be 29.62 mg × h/L, then determine the amount of dose the patient received.

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lb equals 436.384 mg kg

    We know that:
    `AUC=(dose)/(cl earance)=(dose)/(K × V)`.

    Rearranging the equation: `dose=AUC × K × V`.

    However, the total volume of distribution (V) will be equal to the volume of distribution per Kg body weight times the weight of the patient in Kg.

    `∴ dose=`

    `(29.62 \quad mg × h)/L × 0.131/h × (0.69 \quad L)/(Kg) × 163 \quad Kg`

    `∴ dose= 436.38 \quad mg`. Ans.

    Clarification: The AUC shown in this question (29.62) has been rounded from its more precise value (29.618320610687), that is why the displayed answer looks slightly distorted.

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