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

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The elimination half life of a drug is 6.93 h. It is administered as an IV bolus to a 133 Kg patient. If the volume of distribution of the drug is 0.29 L/Kg, then calculate the clearance of the drug in this patient's body.

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lb equals 3.857 L/h kg

    We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).

    ` Cl = K × V `.

    From the relationship of `t_{½}=0.693/K`, we can calculate K as `0.693/t_{½}=0.693/(6.93 h)= 0.1 \quad h^{-1}`.

    Also, the total volume of distribution will be the product of volume per Kg body weight and the weight of the patient. Thus,
    ` V = 0.29 \quad L/(Kg) × 133 \quad Kg =38.57 \quad L`.

    Plugging the values in the first equation we get:
    ` Cl = 0.1 \quad h^{-1} × 38.57 \quad L = 3.86 \quad L/h ` Ans