LIPINCOTT pharmacokinetics Flashcards

1
Q

1.1 An 18-year-old female patient is brought to the
emergency department due to drug overdose. Which
of the following routes of administration is the most
desirable for administering the antidote for the drug
overdose?
A. Intramuscular.
B. Subcutaneous.
C. Transdermal.
D. Oral.
E. Intravenous.

A

Correct answer = E. The intravenous route of administration
is the most desirable because it results in achievement of
therapeutic plasma levels of the antidote rapidly.

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2
Q

1.2 Chlorothiazide is a weakly acidic drug with a pKa of 6.5.
If administered orally, at which of the following sites of
absorption will the drug be able to readily pass through
the membrane?
A. Mouth (pH approximately 7.0).
B. Stomach (pH of 2.5).
C. Duodenum (pH approximately 6.1).
D. Jejunum (pH approximately 8.0).
E. Ileum (pH approximately 7.0).

A

Correct answer = B. Because chlorothiazide is a weakly
acidic drug (pKa = 6.5), it will be predominantly in nonionized
form in the stomach (pH of 2.5). For weak acids,
the nonionized form will permeate through cell membrane
readily.

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3
Q

1.3 Which of the following types of drugs will have maximum
oral bioavailability?
A. Drugs with high first-pass metabolism.
B. Highly hydrophilic drugs.
C. Largely hydrophobic, yet soluble in aqueous
solutions.
D. Chemically unstable drugs.
E. Drugs that are P-glycoprotein substrates.

A

Correct answer = C. Highly hydrophilic drugs have poor oral
bioavailability, because they are poorly absorbed due to their
inability to cross the lipid-rich cell membranes. Highly lipophilic
(hydrophobic) drugs also have poor oral bioavailability,
because they are poorly absorbed due their insolubility in
aqueous stomach fluids and therefore cannot gain access to
the surface of cells. Therefore, drugs that are largely hydrophobic,
yet have aqueous solubility have greater oral bioavailability
because they are readily absorbed.

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4
Q

1.4 Which of the following is true about the blood–brain
barrier?
A. Endothelial cells of the blood–brain barrier have slit
junctions.
B. Ionized or polar drugs can cross the blood–brain
barrier easily.
C. Drugs cannot cross the blood–brain barrier through
specific transporters.
D. Lipid-soluble drugs readily cross the blood–brain
barrier.
E. The capillary structure of the blood–brain barrier is
similar to that of the liver and spleen.

A

Correct answer = D. Lipid-soluble drugs readily cross the
blood–brain barrier because they can dissolve easily in the
membrane of endothelial cells. Ionized or polar drugs generally
fail to cross the blood–brain barrier because they are
unable to pass through the endothelial cells, which do not
have slit junctions.

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5
Q

1.5 A 40-year-old male patient (70 kg) was recently
diagnosed with infection involving methicillin-resistant
S. aureus. He received 2000 mg of vancomycin as an
IV loading dose. The peak plasma concentration of
vancomycin was reported to be 28.5 mg/L. The apparent
volume of distribution is:
A. 1 L/kg.
B. 10 L/kg.
C. 7 L/kg.
D. 70 L/kg.
E. 14 L/kg.

A

Correct answer = A. Vd = dose/C = 2000 mg/28.5 mg/L =
70.1 L. Because the patient is 70 kg, the apparent volume
of distribution in L/kg will be approximately 1 L/kg
(70.1 L/70 kg).

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6
Q

1.6 A 65-year-old female patient (60 kg) with a history of
ischemic stroke was prescribed clopidogrel for stroke
prevention. She was hospitalized again after 6 months
due to recurrent ischemic stroke. Which of the following
is a likely reason she did not respond to clopidogrel
therapy? She is a:
A. Poor CYP2D6 metabolizer.
B. Fast CYP1A2 metabolizer.
C. Poor CYP2E1 metabolizer.
D. Fast CYP3A4 metabolizer.
E. Poor CYP2C19 metabolizer.

A

Correct answer = E. Clopidogrel is a prodrug, and it is activated
by CYP2C19, which is a cytochrome P450 (CYP450)
enzyme. Thus, patients who are poor CYP2C19 metabolizers
have a higher incidence of cardiovascular events
(for example, stroke or myocardial infarction) when taking
clopidogrel.

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7
Q

1.7 Which of the following phase II metabolic reactions
makes phase I metabolites readily excretable in urine?
A. Oxidation.
B. Reduction.
C. Glucuronidation.
D. Hydrolysis.
E. Alcohol dehydrogenation

A

Correct answer = C. Many phase I metabolites are too lipophilic
to be retained in the kidney tubules. A subsequent
phase II conjugation reaction with an endogenous substrate,
such as glucuronic acid, results in more watersoluble
conjugates that excrete readily in urine.

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8
Q

1.8 Alkalization of urine by giving bicarbonate is used to
treat patients presenting with phenobarbital (weak acid)
overdose. Which of the following best describes the
rationale for alkalization of urine in this setting?
A. To reduce tubular reabsorption of phenobarbital.
B. To decrease ionization of phenobarbital.
C. To increase glomerular filtration of phenobarbital.
D. To decrease proximal tubular secretion.
E. To increase tubular reabsorption of phenobarbital.

A

Correct answer = A. As a general rule, weak acid drugs
such as phenobarbital can be eliminated faster by alkalization
of the urine. Bicarbonate alkalizes urine and keeps
phenobarbital ionized, thus decreasing its reabsorption.

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9
Q

1.9 A drug with a half-life of 10 hours is administered by
continuous intravenous infusion. Which of the following
best approximates the time for the drug to reach steady
state?
A. 10 hours.
B. 20 hours.
C. 33 hours.
D. 40 hours.
E. 60 hours.

A

Correct answer = D. A drug will reach steady state in about
four to five half-lives. Thus, for this drug with a half-life of
10 hours, the approximate time to reach steady state will
be 40 hours.

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10
Q

1.10 A 55-year-old male patient (70 kg) is going to be treated
with an experimental drug, Drug X, for an irregular
heart rhythm. If the Vd is 1 L/kg and the desired steadystate
plasma concentration is 2.5 mg/L, which of the
following is the most appropriate intravenous loading
dose for Drug X?
A. 175 mg.
B. 70 mg.
C. 28 mg.
D. 10 mg.
E. 1 mg.

A

Correct answer = A. For IV infusion, Loading dose =
(Vd) × (desired steady-state plasma concentration). The Vd
in this case corrected to the patient’s weight is 70 L. Thus,
Loading dose = 70 L × 2.5 mg/L = 175 mg.

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