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