### 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 0.339 h-1. If its volume of distribution in a patient is 121.91 L, then determine the AUC obtained if 113.38 mg of the drug is injected. Click on the button below to see the answer and explanations

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lb equals 2.74 mg/L h kg

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 = (113.38 \quad mg)/ (0.339 \quad h^{-1} × 121.91 \quad L ) = 2.74 \quad (mg)/L × h  Ans

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