Question

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ex of activity? in a single intra exactly, what 3. A new drug was given in a single intravenous dose of 200 mg to an 80-kg adult male patient. After 6 hours, the plasma drug concentration of drug was 1.5 mg/100 mL of plasma. Assum- ing that the apparent V, is 10% of body weight, compute the total amount of drug in the body fluids after 6 hours. What is the half-life of this drug? A new antibiotic drug was given in a single intravenous bolus of 4 mg/kg to five healthy male adults ranging in age from 23 to 38 years (average weight 75 kg). The pharmacokinetics of the plasma drug concentration-time curve for this drug fits a one-compartment model. The equation of the curve that best fits the data is C = 78e-0.461 Determine the following (assume units of ug/mL for C and hour for t): a. What is the t,,? b. What is the V.?

Answer 1

3)200 mg dose was administered.

Vd = 10/100 * 80 = 8 L

So initial concentration = 200/8 = 25 mg/L = 2.5 mg/100 ml

After 6 hours, concentration of the drug = 1.5 mg/100 ml

A= Ao e^(-kt)

1.5 = 2.5 e^(-6k)

Solving for k and then putting t1/2 = 0.693/k

We get t1/2 = 8.11 hours = 8 hours 6 minutes (approximately)

Hence half life of the drug =8 hours 6 minutes

Total amount of drug in body after 6 hours = 8000 ml * 1.5 mg/100 ml = 120 mg

4)
78 0 e-o. -0.46€ A = o e-k (p= So, initial concentration of the droug (A) = 78 k= 0.46 So; *1/2 = 0.693 = 0.693 ~ 1.5 hawes - (a). 0.46

Dose administered = 4 malung Vd = = 4x75 = 300 mg . Dose Plasma concentration 300 mg (mg/L = ug/m 78 mg IL 3.85 litre - - - - - (6)

(c) After 4 hours; epe 78 é 0.4604 = 12:48 ug/ml . (d) Drug left = 12.480 3.85 = 48.05 mg

(e) ud is less than blood volume. So; it is mostly confined to extracellular component (protein bound in blood). If it had been sequestered inside cell, Vd would have been much higher - (6) 2 = 78 e-0467 =) t = 7.96 hours
So the dose should be administered after 8 hours.