PK Module 3 - In-Class Activity 1

A patient received an IV bolus dose of a drug at time = 0, and resulted in the following plasma concentrations.

Determine the half-life in hours WITHOUT USING A CALCULATOR.

Answer explanation

The concentration decreased by half, going from 8 mg/L to 4 mg/L. Therefore, the time between the two measurements is equal to the half-life (9 hr - 3 hr = 6 hr),

If a drug has an elimination rate constant of 0.254 hr-1, what is the half-life in hr? Round to the nearest hundredth.

Answer explanation

t1/2 = 0.693/K

t1/2 = 0.693/0.254

t1/2 = 2.73 hr

A patient received a dose of vancomycin.  Plasma concentrations were 24 and 16 mg/L at 24 and 48 hours after infusion, respectively. Assuming first-order elimination, what is the elimination rate constant of the drug (in hr-1) in this patient? Round to the nearest thousandth.

Answer explanation

K = ln(C₀/C₁)/t  (t = time after dosage)

K = ln(24/16)/24

K = 0.017 hr-1

If the half-life for decomposition of a drug is 6 hr, how long will it take (in hr) for 500 mg of the drug to decompose to 12.5% remaining? Numerical response only.

Answer explanation

After one half-life, 50% of the drug will remain. After two half-lives, 25% of the drug will remain.

After three half-lives, 12.5% of the drug will remain. Half-life = 6 hr x 3 = 18 hr

A patient is administered a medication with an elimination rate constant of 0.25 hr-1. What percent of the drug is removed from the patient's body per hour?

(a)  

• 25%
• 2.5%
• 250%
• 0.25%

Using the equation C = C₀e^-kt, determine the plasma concentration (in mg/L) of a drug 8 hours after a peak level of 10 mg/L is observed if the elimination rate constant is 0.15 hr^-1. Round to the nearest tenth. Numerical response only. ​

Answer explanation

C = C₀e^-kt

C = 10e^-(0.15*8)

C =3.0 mg/L

The average values of clearance and volume of distribution of valproic acid, an antiepileptic drug, in the adult patient population are 0.5 L/hr and 9 L, respectively.

Calculate the K of valproic acid in hr-1. Numerical response only

Answer explanation

K = Cl /Vd

K = 0.5 L/hr / 9 L

K = 0.056 hr-1

Calculate the AUC of valproic acid following administration of a 800 mg IV dose in mg*L/hr. Numerical response only.

The average values of clearance and volume of distribution of valproic acid, an antiepileptic drug, in the adult patient population are 0.5 L/hr and 9 L, respectively.

Answer explanation

AUC = Xo/Cl

AUC = 800 mg/0.5 L/hr

AUC = 1600 mg*hr/L

Calculate the AUC of valproic acid in mg*L/hr assuming a half-life of 12 hr and an initial plasma concentration of 50 mg/L. Numerical response only. Round to the nearest whole number.

Answer explanation

K = 0.693/t1/2

K = 0.693/12 = 0.05775 hr-1

AUC = Co/K

AUC = 50/0.05775 = 866 mg*hr/L

The AUC following a 500-mg IV bolus dose of drug is 100 (mg/L)*hr.  What is the total clearance of this drug in L/hr? Round to the nearest tenth.  Numerical response only

Answer explanation

AUC = Xo/Cl, so Cl = Xo/AUC

Cl = 500/100 = 5.0 L/hr