### T20: Mixed types from T01-T20

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 0.066 h-1. It is parenterally administered to a patient weighing 73 Kg. If the volume of distribution of the drug is 0.53 L/Kg, and the AUC was measured to be 50.45 mg × h/L, then determine the amount of dose the patient received. Click on the button below to see the answer and explanations

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lb equals 128.838 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=

(50.45 \quad mg × h)/L × 0.066/h × (0.53 \quad L)/(Kg) × 73 \quad Kg

∴ dose= 128.84 \quad mg. Ans.

Clarification: The AUC shown in this question (50.45) has been rounded from its more precise value (50.454545454545), that is why the displayed answer looks slightly distorted.

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