T12: AUC from dose, K and V
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 0.226 h-1. If its volume of distribution in a patient is 132.03 L, then determine the AUC obtained if 209.93 mg of the drug is injected.
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lb equals 7.04 mg/L h kg
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We know that, `AUC = (dose)/(cl earance)`; but Cl is the product of elimination rate constant (K) and volume of distribution (V).
` ∴ AUC = X_{0}/ (K × V) `.
Plugging the values in this equation we get:
` AUC = (209.93 \quad mg)/ (0.226 \quad h^{-1} × 132.03 \quad L ) = 7.04 \quad (mg)/L × h ` Ans