Medical Pharmacology Chapter 2: Pharmacokinetic Problems Set
Practice Questions and Explanations
Choose the correct answer for each question.
Explanation for Practice Question Answers
Question #1: t1/2 = ln 2 /kel = 0.693/kel where t1/2 is the elimination half-life (units=time) ANSWER:
kel = 0.693/15 hours = 0.0462 / hour Review: kel = km + kex where kel = drug elimination rate constant km = elimination rate constant due to metabolism kex = elimination rate constant due to excretion Question #2:
Css = F · (D/t)/(kel * Vd ) F = 0.7 (fraction absorbed) D =300 mg t= 24 h (dosing interval, i.e. once a day)
kel = 0.0462 / hour (elimination rate constant) Vd = 40 L (apparent volume of distribution)
Css = 0.7 · (300 mg/24 hours) / (0.0462/hour · 40,000 ml) ANSWER:
Css = 0.0047 mg/ml or 4.7 μg/ml at 24 hours
Question #3:
Css = F · (D/t)/(kel * Vd ) F = 0.7 (fraction absorbed) D =300 mg t= 6 h (dosing interval, i.e. 4 times a day)
kel = 0.0462 / hour (elimination rate constant) Vd = 40 L (apparent volume of distribution)
Css = 0.7 · (300 mg/6 hours) / (0.0462/hour · 40,000 ml) ANSWER: Css = 0.019 mg/ml or 19
μg/ml at 6 hours
Question #4: Css = F · (D/t)/(kel
*Vd) F = 0.7 (fraction absorbed) D =300 mg t= 12 h (dosing interval, i.e. 2 times a day) kel = 0.0462 / hour (elimination rate constant)
Vd = 40 L (apparent volume of distribution) Css = 0.7 · (300 mg/12 hours) / (0.0462/hour · 40,000 ml) ANSWER:
Css = 0.0095 mg/ml or 9.5 μg/ml at 12 hours
Question #5: Css = F · (D/t)/(kel
* Vd ) F = 0.7 (fraction absorbed) D =300 mg t= 48 h (dosing interval, i.e. once every 2 days)
kel = 0.0462 / hour (elimination rate constant) Vd = 40 L (apparent volume of distribution)
Css = 0.7 · (300 mg/48 hours) / (0.0462/hour · 40,000 ml) ANSWER:
Css = 0.0024 mg/ml or 2.4μg/ml at 12 hours Note what happens to
Css as we shorten or lengthen the dosage interval (qualitative % quantitative effects) Dose = 300 mg of Drug X Time (hours) Css (μg/ml) 6 19 12 9.5 24 4.7 48 2.4
Question #6: Css = F · (D/t)/(kel . Vd ) F = 0.7 (fraction absorbed) D =150 mg t= 24 h (dosing interval, i.e. once a day)
kel = 0.0462 / hour (elimination rate constant) Vd = 40 L (apparent volume of distribution) Css= 0.7 · (150 mg/24 hours) / (0.0462/hour · 40,000 ml) ANSWER:
Css = 0.0024 mg/ml or 2.4 μg/ml at 24 hours
Question #7: Css = F · (D/t)/(kel . Vd ) F = 0.7 (fraction absorbed) D =600 mg t= 24 h (dosing interval, i.e. once a day) kel= 0.0462 / hour (elimination rate constant) Vd = 40 L (apparent volume of distribution) Css = 0.7 · (600 mg/24 hours) / (0.0462/hour · 40,000 ml) ANSWER:
Css = 0.0095 mg/ml or 9.5 μg/ml at 24 hours T is constant (once every 24 hours) Dosage (mg)
Css(μg/ml) 150 2.4 300 4.7 600 9.5 Questions and answers based on material presented by Dr. Edward Flynn, Department of Pharmacology, New Jersey School of Medicine and Dentistry used with permission
Consider the following information about drug X: D = 300 mg; T = 24 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the ke (elimination rate constant)?
0.133/hr
0.933/hr
0.046/hr
1.44/hr
0.693/hr
Consider the following information about drug X: D = 300 mg; T = 24 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the Css (concentration at steady-state) for drug X?
0.0462 ug/ml at 24 hours
4.7 ug/ml at 24 hours
2.4 ug/ml at 24 hours
6.76 ug/ml at 24 hours
4.73 mg/ml at 24 hours
Consider the following information about drug X: D = 300 mg; T =6 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the Css (concentration at steady-state) for drug X? (note dosing interval is now 6 hours)
4.7 ug/ml at 6 hours
2.4 ug/ml at 6 hours
19 ug/ml at 6 hours
9.5 ug/ml at 6 hours
none of the above
Consider the following information about drug X: D = 300 mg; T =12 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the Css (concentration at steady-state) for drug X? (note dosing interval is now 12 hours)
9.5 ug/ml at 12 hours
2.4 ug/ml at 12 hours
190 ug/ml at 12 hours
95 ug/ml at 12 hours
4.7 /ug/ml at 12 hours
Consider the following information about drug X: D = 300 mg; T =48 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the Css (concentration at steady-state) for drug X? (note dosing interval is now 48 hours)
16 ug/ml
4.7 ug/ml
19 ug/ml
2.4 ug/ml
9.5 ug/ml
Consider the following information about drug X: D =150mg; T =24 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the Css (concentration at steady-state) for drug X?
2.4 ug/ml
4.7 ug/ml
9.5 ug/ml
19 ug/ml
none of the above
Consider the following information about drug X: D = 600 mg; T =24 h; F = 0.70 (70% absorption);Vd = 40 L; half-life = 15 hours: What is the Css (concentration at steady-state) for drug X?