T20: Mixed types from T01-T20

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

Total tried:       Correct:       Wrong:

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

Notice: Undefined variable: input_value1 in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 220
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.

Notice: Undefined index: TOTALTRY in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 675