USMLE Biochemistry Practice Questions
USMLE Step 1, Biochemistry Multiple Choice Questions, (Mcqs) Page 3. Mcqs are obtaining from reliable sources, Uworld and others.
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USMLE Step 1, Biochemistry 10 Questions.
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- Question 1 of 10
1. Question
A 29-year-old woman with a long-standing history of asthma and eczema presents with watery, itchy eyes and a stuffy nose of 3 days’ duration. The woman states that her symptoms are similar to those she experiences during the spring. Her heart rate is 82/min, blood pressure is 117/80 mm Hg, respiratory rate is 14/min, and oxygen saturation is 96%. This patient’s symptoms are due to the activation of which of the following receptors?
CorrectThe correct answer is D. The patient’s symptoms
are consistent with those of seasonal allergies.
The patient has experienced asthma, eczema,
and allergies and most likely suffers from
atopy. Seasonal allergies are a result of histamine1-
receptor activation, which results in
pruritus, bronchoconstriction, and increased
nasal and bronchial mucus production. Seasonal
allergy symptoms can be treated with antihistamines
such as loratadine, which are histamine1-
antagonists.IncorrectThe correct answer is D. The patient’s symptoms
are consistent with those of seasonal allergies.
The patient has experienced asthma, eczema,
and allergies and most likely suffers from
atopy. Seasonal allergies are a result of histamine1-
receptor activation, which results in
pruritus, bronchoconstriction, and increased
nasal and bronchial mucus production. Seasonal
allergy symptoms can be treated with antihistamines
such as loratadine, which are histamine1-
antagonists. - Question 2 of 10
2. Question
An 8-month-old boy is brought to the pediatrician by his parents because he has recently lost the ability to crawl or hold his toys. On examination the patient is tachypneic and breathing with considerable effort; the liver is palpable fi ve fi ngerwidths below the right costal margin. X-ray of the chest reveals cardiomegaly. He has a diffi cult time sitting upright and cannot squeeze the physician’s fi ngers or the ring of his pacifi er with any noticeable force. Despite a number of interventions, the child’s symptoms continue to worsen until his death 2 weeks later. On autopsy, it is likely that this patient’s cells will contain an accumulation of which of the following substances?
CorrectThe correct answer is B. This patient has
Pompe’s disease, a glycogen storage disorder.
Pompe’s disease is an autosomal recessive disease
that is characterized by a defi ciency or defect
in lysosomal α-1,4-glucosidase. This enzyme
is necessary for the dissolution of the
polymer linkages in glycogen. In its absence,
glycogen accumulates to toxic levels in both
the cytoplasm and lysosomes.IncorrectThe correct answer is B. This patient has
Pompe’s disease, a glycogen storage disorder.
Pompe’s disease is an autosomal recessive disease
that is characterized by a defi ciency or defect
in lysosomal α-1,4-glucosidase. This enzyme
is necessary for the dissolution of the
polymer linkages in glycogen. In its absence,
glycogen accumulates to toxic levels in both
the cytoplasm and lysosomes. - Question 3 of 10
3. Question
After consumption of a carbohydrate-rich meal, the liver continues to convert glucose to glucose- 6-phosphate. The liver’s ability to continue this processing of high levels of glucose is important in minimizing increases in blood glucose after eating. What is the best explanation for the liver’s ability to continue this conversion after eating a carbohydrate-rich meal?
CorrectThe correct answer is B. Glucokinase is found
in liver and pancreatic β cells. It catalyzes the
initial step of glycolysis, phosphorylation of
glucose to glucose-6-phosphate, which is catalyzed
by hexokinase in other tissues. Both enzymes
are found in the liver. Glucokinase has
a higher Michaelis-Menten constant (Km) and
a higher maximum reaction rate (Vmax) than
hexokinase; therefore, glucokinase has lower
affi nity for glucose and a higher capacity to
make glucose into glucose-6-phosphate. At low
glucose levels, hexokinase, with its higher affi
nity for glucose, processes glucose to glucose-
6-phosphate. At higher glucose levels, hexokinase
is overwhelmed (operating at its low Vmax)
and suffi cient substrate is available for glucokinase
to process the excess glucose even with its
low affi nity. Glucokinase thus helps to handle
large increases in glucose from the gut.IncorrectThe correct answer is B. Glucokinase is found
in liver and pancreatic β cells. It catalyzes the
initial step of glycolysis, phosphorylation of
glucose to glucose-6-phosphate, which is catalyzed
by hexokinase in other tissues. Both enzymes
are found in the liver. Glucokinase has
a higher Michaelis-Menten constant (Km) and
a higher maximum reaction rate (Vmax) than
hexokinase; therefore, glucokinase has lower
affi nity for glucose and a higher capacity to
make glucose into glucose-6-phosphate. At low
glucose levels, hexokinase, with its higher affi
nity for glucose, processes glucose to glucose-
6-phosphate. At higher glucose levels, hexokinase
is overwhelmed (operating at its low Vmax)
and suffi cient substrate is available for glucokinase
to process the excess glucose even with its
low affi nity. Glucokinase thus helps to handle
large increases in glucose from the gut. - Question 4 of 10
4. Question
A 30-year-old man is diagnosed with type I familial dyslipidemia. He has had recent laboratory studies showing elevated triglycerides and normal cholesterol levels. Which of the following explains the pathophysiology of this disease?
CorrectThe correct answer is C. Type I dyslipidemia
is caused by a defi ciency of lipoprotein lipase.
This enzyme exists in capillary walls of adipose
and muscle tissue and cleaves triglycerides into
free fatty acids and glycerol. The enzyme is activated
by apolipoprotein C-II, which is found
on VLDL cholesterol and chylomicrons. Type
I dyslipidemia is characterized by an accumulation
of triglyceride-rich lipoproteins in the
plasma. It can also occur with an alteration in
apolipoprotein C-II.IncorrectThe correct answer is C. Type I dyslipidemia
is caused by a defi ciency of lipoprotein lipase.
This enzyme exists in capillary walls of adipose
and muscle tissue and cleaves triglycerides into
free fatty acids and glycerol. The enzyme is activated
by apolipoprotein C-II, which is found
on VLDL cholesterol and chylomicrons. Type
I dyslipidemia is characterized by an accumulation
of triglyceride-rich lipoproteins in the
plasma. It can also occur with an alteration in
apolipoprotein C-II. - Question 5 of 10
5. Question
A 36-year-old woman returned from a trip to Japan 2 days ago. Yesterday she started experiencing left calf pain. The woman is afebrile with a heart rate of 82/min, a blood pressure of 129/86 mm Hg, and a respiratory rate of 14/ min. On examination, the patient’s calf pain is found to intensify on dorsifl exion of the left foot. Suspecting a deep venous thrombosis (DVT), the physician performs a venous ultrasound, which confi rms the diagnosis. Which of the following factors is most responsible in the formation of the DVT?
CorrectThe correct answer is D. This patient’s long
airplane fl ight placed her at an increased risk
of developing DVT. In addition, the patient
has a positive Homans’ sign (calf pain on dorsifl
exion), which is seen in some patients with a
DVT. Virchow’s triad refers to three factors that
predispose a patient to developing a venous
thrombosis: (1) local trauma to the vessel wall,
(2) stasis, and (3) hypercoagulability. Thromboxane
A2 stimulates platelet aggregation and
vasoconstriction and will be elevated at the site
of a clot.IncorrectThe correct answer is D. This patient’s long
airplane fl ight placed her at an increased risk
of developing DVT. In addition, the patient
has a positive Homans’ sign (calf pain on dorsifl
exion), which is seen in some patients with a
DVT. Virchow’s triad refers to three factors that
predispose a patient to developing a venous
thrombosis: (1) local trauma to the vessel wall,
(2) stasis, and (3) hypercoagulability. Thromboxane
A2 stimulates platelet aggregation and
vasoconstriction and will be elevated at the site
of a clot. - Question 6 of 10
6. Question
A 53-year-old man comes to his physician because he noticed blood in his urine and has been having some low back pain. Physical examination reveals palpable fl ank masses felt bilaterally as well as mild hypertension. CT of the abdomen is shown in the image. Which of the following conditions is also associated with this disorder?
CorrectThe correct answer is B. This patient has adult
polycystic kidney disease, an autosomal dominant
condition characterized by massive bilateral
cysts in the kidneys (as shown in the image).
Cysts can also sometimes be found in the
liver. The renal cysts eventually lead to endstage
renal disease. In addition to the cysts, patients
are more prone to mitral valve prolapse
as well as berry aneurysms, which can rupture
and lead to strokes.IncorrectThe correct answer is B. This patient has adult
polycystic kidney disease, an autosomal dominant
condition characterized by massive bilateral
cysts in the kidneys (as shown in the image).
Cysts can also sometimes be found in the
liver. The renal cysts eventually lead to endstage
renal disease. In addition to the cysts, patients
are more prone to mitral valve prolapse
as well as berry aneurysms, which can rupture
and lead to strokes. - Question 7 of 10
7. Question
The bacterial ribosome and its involvement in protein synthesis pathways hold great importance to the function of various antibiotics that antagonize various steps in the process. Which of the following is true of the 50S ribosomal subunit in bacterial protein synthesis?
CorrectThe correct answer is C. The 50S ribosomal
subunit is the larger of the two ribosomal macromolecules.
Its main function in translation is
the creation of a new peptide bond between
the incoming aminoacyl-tRNA and the growing
peptide chain. Essential for this purpose is
the 23S rRNA molecule, which holds the peptidyl
transferase enzyme activity. Linezolid is
an antibiotic that is active against gram-positive
bacteria including enterococci and staphylococci.
The linezolid mechanism of action is
inhibition of protein synthesis by binding to
the 23S rRNA and interacting with the initiation
complex. Linezolid resistance has been
reported in staphylococci and enterococci.
Since bacteria carry several copies of the 23S
rRNA gene, the number of rRNA genes mutated
seems to be a signifi cant determinant of
resistance.IncorrectThe correct answer is C. The 50S ribosomal
subunit is the larger of the two ribosomal macromolecules.
Its main function in translation is
the creation of a new peptide bond between
the incoming aminoacyl-tRNA and the growing
peptide chain. Essential for this purpose is
the 23S rRNA molecule, which holds the peptidyl
transferase enzyme activity. Linezolid is
an antibiotic that is active against gram-positive
bacteria including enterococci and staphylococci.
The linezolid mechanism of action is
inhibition of protein synthesis by binding to
the 23S rRNA and interacting with the initiation
complex. Linezolid resistance has been
reported in staphylococci and enterococci.
Since bacteria carry several copies of the 23S
rRNA gene, the number of rRNA genes mutated
seems to be a signifi cant determinant of
resistance. - Question 8 of 10
8. Question
A researcher investigating mechanisms of intracellular signaling develops a mouse model using recombinant DNA technology to generate a gene knockout. The F2 progeny exhibit dwarfi sm, hypogonadism, and hypothyroidism with low levels of follicle-stimulating hormone, luteinizing hormone, and thyroid-stimulating hormone but normal levels of ACTH. In addition, the female progeny exhibit impaired milk secretion. Which of the following is the second- messenger molecule for the intracellular signaling system that is most likely impaired in this mouse model?
CorrectThe correct answer is C. The knockout mouse
model described exhibits a phenotype consistent
with impaired function of all of the hypothalamic
hormones with the exception of corticotropin-
releasing hormone (CRH). The hypothalamus
produces a number of important
hormones, including trophic hormone-releasing
hormone, that act on the anterior pituitary;
these include thyrotropin-releasing hormone
and gonadotropin-releasing hormone as well as
hormones to be stored and released in the posterior
pituitary, such as ADH and oxytocin. All of
the hypothalamic hormones exert their actions
on target cells via a phospholipase C (PLC) intracellular
signaling cascade with inositol trisphosphate
(IP3) as the second messenger, with
the exception of CRH, which acts via adenylate
cyclase-cAMP. Thus, the PLC-IP3 signaling system
is the most likely target of the gene knockout
described in the vignette. Important nonhypothalamic
hormones that also act via this
signaling system include angiotensin II and α1-
adrenergic agonists.IncorrectThe correct answer is C. The knockout mouse
model described exhibits a phenotype consistent
with impaired function of all of the hypothalamic
hormones with the exception of corticotropin-
releasing hormone (CRH). The hypothalamus
produces a number of important
hormones, including trophic hormone-releasing
hormone, that act on the anterior pituitary;
these include thyrotropin-releasing hormone
and gonadotropin-releasing hormone as well as
hormones to be stored and released in the posterior
pituitary, such as ADH and oxytocin. All of
the hypothalamic hormones exert their actions
on target cells via a phospholipase C (PLC) intracellular
signaling cascade with inositol trisphosphate
(IP3) as the second messenger, with
the exception of CRH, which acts via adenylate
cyclase-cAMP. Thus, the PLC-IP3 signaling system
is the most likely target of the gene knockout
described in the vignette. Important nonhypothalamic
hormones that also act via this
signaling system include angiotensin II and α1-
adrenergic agonists. - Question 9 of 10
9. Question
Hemoglobin consists of 2 α subunits and 2 β subunits. Each α unit is bound to a β unit by strong hydrophobic bonds. The β units and α units are each bound to one another by weaker polar bonds that are somewhat mobile. This change in quaternary structure changes the affi nity for oxygen that hemoglobin exhibits as it shifts between its taut (T) form and relaxed form. At a given partial pressure of oxygen, which of the following will increase the amount of T hemoglobin?
CorrectThe correct answer is A. Hemoglobin carries
oxygen better when it is in the relaxed (R) form,
in which it has a higher affi nity for oxygen. As a
result, the oxygen dissociation curve shifts to the
left and decreased unloading of oxygen results.
Conversely, tightening hemoglobin decreases its
affi nity for oxygen. Hemoglobin in the T form is
stabilized by all the processes that result in increased
oxygen unloading. Binding of CO2 stabilizes
the T form, which decreases oxygen affi
nity.IncorrectThe correct answer is A. Hemoglobin carries
oxygen better when it is in the relaxed (R) form,
in which it has a higher affi nity for oxygen. As a
result, the oxygen dissociation curve shifts to the
left and decreased unloading of oxygen results.
Conversely, tightening hemoglobin decreases its
affi nity for oxygen. Hemoglobin in the T form is
stabilized by all the processes that result in increased
oxygen unloading. Binding of CO2 stabilizes
the T form, which decreases oxygen affi
nity. - Question 10 of 10
10. Question
A group of scientists have recently discovered a new drug for treating hypercholesterolemia. In vitro studies with a hepatocyte cell line have revealed that the drug increases the number of LDL cholesterol receptors by acting in a manner similar to steroids. What is the mechanism by which this drug is acting on hepatocytes?
CorrectThe correct answer is D. Like steroid hormones,
this drug is binding to a cytosolic receptor
to form a hormone-receptor complex. The
complex is then transported to the nucleus,
where it acts on DNA to increase the transcription
of LDL cholesterol receptor mRNA, which
can be translated to LDL receptors.IncorrectThe correct answer is D. Like steroid hormones,
this drug is binding to a cytosolic receptor
to form a hormone-receptor complex. The
complex is then transported to the nucleus,
where it acts on DNA to increase the transcription
of LDL cholesterol receptor mRNA, which
can be translated to LDL receptors.