Chapter 5 : Basic Pharmacokinetics Flashcards
Directions: Each question, statement, or incomplete
statement in this section can be correctly answered or
completed by one of the suggested answers or phrases.
Choose the best answer.
1. Creatinine clearance is used as a measurement of
(A) renal excretion rate.
(B) glomerular fi ltration rate (GFR).
(C) active renal secretion.
(D) passive renal absorption.
(E) drug metabolism rate.
- Th e answer is B [see I.E.2.a].
A substance that is used to measure the GFR must be
fi ltered but not reabsorbed or actively secreted. Al-
though inulin clearance gives an accurate measure-
ment of GFR, creatinine clearance is generally used
because no exogenous drug must be given. However,
creatinine formation depends on muscle mass and
muscle metabolism, which may change with age and
various disease conditions
For questions 2–5: A new cephalosporin antibiotic was
given at a dose of 5 mg/kg by a single intravenous bolus
injection to a 58-year-old man who weighed 75 kg.
Th e antibiotic follows the pharmacokinetics of a one-
compartment model and has an elimination half-life of
2hrs. Th e apparent volume of distribution is 0.28 L/kg, and
the drug is 35% bound to plasma proteins.
2. What is the initial plasma drug concentration ( CP
0 ) in
this patient?
(A) 0.24 mg/L
(B) 1.80 mg/L
(C) 17.9 mg/L
(D) 56.0 mg/L
(E) 1339 mg/L
(C) 17.9 mg/L
0.28 x75:21
5mg/kg X 75:375
375/21:17.9
VD is needed to estimate the amount of drug in the body relative to the concentration of
drug in the plasma, as shown in the following:
VD x Cp = D
- What is the predicted plasma drug concentration (Cp)
at 8 hr aft er the dose?
(A) 0.73 mg/L
(B) 1.11 mg/L
(C) 2.64 mg/L
(D) 4.02 mg/L
(E) 15.10 mg/L
- Th e answer is B [see I.A.3.b.(1); I.B.1 ].
(B) 1.11 mg/L
- How much drug remains in the patient’s body (DB)
8hrs aft er the dose?
(A) 15.3 mg
(B) 23.3 mg
(C) 84.4 mg
(D) 100.0 mg
(E) 112.0 mg
(B) 23.3 mg
- How long aft er the dose is exactly 75% of the drug
eliminated from the patient’s body?
(A) 2 hrs
(B) 4 hrs
(C) 6 hrs
(D) 8 hrs
(E) 10 hr
- Th e answer is B [see I.B.1.a.(2)].
Substituting the data for this patient in the equation for
the initial plasma drug concentration ( CP
0
) gives
C P
0 -
_
D0
VD
-
_
5 mg
0.28 L/kg -
17.9 mg/L
To obtain the patient’s plasma drug concentration
( Cp ) 8 hrs aft er the dose, the following calculation is
performed:
Cp -
C P
0 e
kt
k -
_0.693
t½
-
_0.963
2
-
0.347 hr1
Cp -
17.9e
(0.347)(8)
Cp -
(17.9)(0.0623) -
1.11 mg/L
Th e amount of drug in the patient’s body at 8 hrs is
calculated as follows:
DB -
CpVD -
(1.11)(0.28)(75) -
23.3 mg
For any fi rst-order elimination process, 50% of the
initial amount of drug is eliminated at the end of the
fi rst half-life, and 50% of the remaining drug (i.e., 75%
of the original amount) is eliminated at the end of the
second half-life. Because the drug in the current case
has an elimination half-life ( t½ ) of 2 hrs, 75% of the
dose is eliminated in two half-lives or 4 hrs.
For questions 6–11: A 35-year-old man who weighs
70kg and has normal renal function needs an intravenous
infusion of the antibiotic carbenicillin. Th e desired steady-
state plasma drug concentration is 15 mg/dL. Th e physician
wants the antibiotic to be infused into the patient for 10hrs.
Carbenicillin has an elimination half-life (t½) of 1 hr and an
apparent volume distribution (VD) of 9 L in this patient.
6. Assuming that no loading dose was given, what rate of
intravenous infusion is recommended for this patient?
(A) 93.6 mg/hr
(B) 135.0 mg/hr
(C) 468.0 mg/hr
(D) 936.0 mg/hr
(E) 1350.0 mg/hr
- Th e answer is D [see I.B.3.e.(3)].
(D) 936.0 mg/hr
- Assuming that no loading intravenous dose was
given, how long aft er the initiation of the intravenous
infusion would the plasma drug concentration reach
95% of the theoretic steady-state drug concentration?
(A) 1.0 hrs
(B) 3.3 hrs
(C) 4.3 hrs
(D) 6.6 hrs
(E) 10.0 hrs
- Th e answer is C [see I.B.3.c].
(C) 4.3 hrs
- What is the recommended loading dose?
(A) 93.6 mg
(B) 135.0 mg
(C) 468.0 mg
(D) 936.0 mg
(E) 1350.0 mg
- Th e answer is E [see I.B.3.f.(2)].
(E) 1350.0 mg
- To infuse the antibiotic as a solution containing 10-g
drug in 500 mL 5% dextrose, how many milliliters per
hour of the solution would be infused into the patient?
(A) 10.0 mL/hr
(B) 46.8 mL/hr
(C) 100.0 mL/hr
(D) 936.0 mL/hr
(E) 1141.0 mL/hr
- Th e answer is B [ see I.B.3.e.(3)].
(B) 46.8 mL/hr
- What is the total body clearance rate for carbenicillin
in this patient?
(A) 100 mL/hr
(B) 936 mL/hr
(C) 4862 mL/hr
(D) 6237 mL/hr
(E) 9000 mL/hr
- Th e answer is D [see I.B.3.e.(3); I.E.1.a].
(D) 6237 mL/hr
- If the patient’s renal clearance for carbenicillin is 86 mL/
min, what is the hepatic clearance for carbenicillin?
(A) 108 mL/hr
(B) 1077 mL/hr
(C) 3840 mL/hr
(D) 5160 mL/hr
(E) 6844 mL/hr
- Th e answer is B [see I.E.4.a ].
(B) 1077 mL/hr
Th e equation for the plasma concentration at steady
state ( Css ) provides the formula for calculating the rate
of an intravenous infusion ( R ). Th e equation is
Css -
_R
kVD
where k is the fi rst-order elimination rate constant and
VD is the apparent volume of distribution. Rearranging
the equation and substituting the data for this patient
give the following calculations:
R -
CsskVD -
_
15 mg
100 mL
_
0.693
1 hr 9000 mL
R -
936 mg/hr
Th e time it takes for an infused drug to reach the
Css depends on the elimination half-life of the drug.
Th e time required to reach 95% of the Css is equal to
4.3 times the half-life, whereas the time required to
reach 99% of the Css is equal to 6.6 times the half-life.
Because the half-life in the current case is 1 hr, the time
to reach 95% of the Css is 4.3 1 hr or 4.3 hrs.
Th e loading dose ( DL ) is calculated as follows:
DL -
CssVD -
_
15 mg
100 mL 9000 mL -
1350 mg
Th e answer to question 6 shows that the infusion rate
should be 936 mg/hr. Th erefore, if a drug solution con-
taining 10 g in 500 mL is used, the required infusion
rate is
_
936 mg
1 hr
_
500 mL
10000 mg -
46.8 mL/hrs
- Th e earliest evidence that a drug is stored in tissue is
(A) an increase in plasma protein binding.
(B) a large apparent volume of distribution (VD).
(C) a decrease in the rate of formation of metabolites
by the liver.
(D) an increase in the number of side eff ects
produced by the drug.
(E) a decrease in the amount of free drug excreted in
the urine.
- Th e answer is B [see I.B.1.b.(1)].
A large apparent volume of distribution (VD) is an early
sign that a drug is not concentrated in the plasma but
is distributed widely in tissue. An increase in plasma
protein binding suggests that the drug is located in the
plasma rather than in tissue. A decrease in hepatic me-
tabolism, an increase in side eff ects, or a decrease in uri-
nary excretion of free drug is caused by a decrease in drug
elimination.
- Th e intensity of the pharmacologic action of a drug is
most dependent on the
(A) concentration of the drug at the receptor site.
(B) elimination half-life (t½) of the drug.
(C) onset time of the drug aft er oral administration.
(D) minimum toxic concentration (MTC) of the drug
in plasma.
(E) minimum eff ective concentration (MEC) of the
drug in the body.
- Th e answer is A [see I.A.5.d.(3)].
(A) concentration of the drug at the receptor site.
As more drug is concentrated at the receptor site, more
receptors interact with the drug to produce a pharma-
cologic eff ect. Th e intensity of the response increases
until it reaches a maximum. When all of the available
receptors are occupied by drug molecules, additional
drug does not produce a more intense response.
