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USMLE Biochemistry Quiz 1
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USMLE Biochemistry Free quiz
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Question 1 of 10
1. Question
A 6-year-old boy presents to his pediatrician with skin lesions all over his body. For several years he has been very sensitive to sunlight. Neither the boy’s parents nor his siblings have the same skin lesions or sun sensitivity. Biopsies of several of the boy’s lesions reveal squamous cell carcinoma. Which mutation would one expect to see in this patient’s DNA?
Correct
The correct answer is G. This patient has xeroderma
pigmentosa, an autosomal recessive
disease characterized by a defect in excision repair.
This disease results in an inability to
repair thymidine dimers that can form in the
presence of ultraviolet light. This can lead to
the development of skin cancer and photosensitivity.Incorrect
The correct answer is G. This patient has xeroderma
pigmentosa, an autosomal recessive
disease characterized by a defect in excision repair.
This disease results in an inability to
repair thymidine dimers that can form in the
presence of ultraviolet light. This can lead to
the development of skin cancer and photosensitivity. -
Question 2 of 10
2. Question
A metabolic process is pictured in the image below. Which intermediate in this process inhibits the rate-limiting enzyme of glycolysis and activates the rate-limiting enzyme of fatty acid synthesis?
Correct
The correct answer is B. Citrate, formed from
oxaloacetate and acetyl CoA by the enzyme
citrate synthase, inhibits phosphofructokinase
and allosterically activates acetyl CoA carboxylase.
Citrate synthase regenerates a molecule of
CoA and is an important regulator of the tricarboxylic
acid cycle. It is inhibited by adenosine
triphosphate.Incorrect
The correct answer is B. Citrate, formed from
oxaloacetate and acetyl CoA by the enzyme
citrate synthase, inhibits phosphofructokinase
and allosterically activates acetyl CoA carboxylase.
Citrate synthase regenerates a molecule of
CoA and is an important regulator of the tricarboxylic
acid cycle. It is inhibited by adenosine
triphosphate. -
Question 3 of 10
3. Question
A 32-year-old develops polyuria. Her nephrologist monitors her closely on a water deprivation test, and she continues to have increased urine output. A diagnosis of diabetes insipidus is made. The nephrologist orders an antidiuretic hormone level and determines that it is inappropriately elevated. Which of the following is the site of pathology in this patient?
Correct
The correct answer is A. The adenohypophysis
is not related to the pathology of diabetes insipidus.Incorrect
The correct answer is A. The adenohypophysis
is not related to the pathology of diabetes insipidus. -
Question 4 of 10
4. Question
A 35-year-old man presents to the physician with arthritic pain in both knees along with back pain. He states that the pain has been present for months. In an effort to obtain relief, he has taken only aspirin, but this has been of little benefi t. The patient is afebrile, and his slightly swollen knee joints are neither hot nor tender to palpation; however, the pain does restrict his motion. The cartilage of his ears appears slightly darker than normal. No tophi are present. A urine specimen is taken for analysis of uric acid content and turns black in the laboratory while standing. A defect in which of the following is the most likely underlying cause of the patient’s condition?
Correct
The correct answer is B. The patient has
alkaptonuria, a condition corresponding to the
one described in the stem. A defi ciency of the
enzyme homogentisic acid oxidase leads to deposition
of homogentisic acid in the joints and
cartilage, giving them a dark color (ochronosis)
and resulting in degenerative changes. Classically,
the urine of these patients turns black on
contact with air or when the urine is made alkaline.
The associated defect is on chromosome
3.Incorrect
The correct answer is B. The patient has
alkaptonuria, a condition corresponding to the
one described in the stem. A defi ciency of the
enzyme homogentisic acid oxidase leads to deposition
of homogentisic acid in the joints and
cartilage, giving them a dark color (ochronosis)
and resulting in degenerative changes. Classically,
the urine of these patients turns black on
contact with air or when the urine is made alkaline.
The associated defect is on chromosome
3. -
Question 5 of 10
5. Question
A patient who is a carrier of sickle cell trait presents to the clinic. The single base-pair mutation for sickle cell anemia destroys the MstII restriction enzyme recognition site represented by an asterisk in the image. The restriction enzyme- binding sites are shown as arrows on the map. DNA from this patient is treated with MstII and run on an electrophoresis gel. The DNA is then hybridized with a labeled probe that binds to the normal gene in the position shown on the map. In the Southern blot shown in the image, which lane represents the patient?
Correct
The correct answer is B. Lane B represents
the Southern blot of a heterozygous carrier of
sickle cell anemia. The β-A-globin gene results
in a 1.15-kb fragment of DNA cut by the MstII
restriction enzyme. The β-S-globin gene results
in a 1.35-kb band because the single basepair
mutation responsible for sickle cell anemia
eliminates an MstII restriction site.Incorrect
The correct answer is B. Lane B represents
the Southern blot of a heterozygous carrier of
sickle cell anemia. The β-A-globin gene results
in a 1.15-kb fragment of DNA cut by the MstII
restriction enzyme. The β-S-globin gene results
in a 1.35-kb band because the single basepair
mutation responsible for sickle cell anemia
eliminates an MstII restriction site. -
Question 6 of 10
6. Question
Phosphatidylcholine is a major component of red blood cell membranes, myelin, surfactant, and cholesterol. Phosphatidylcholine is synthesized through phosphorylation of choline obtained from the diet or with reused choline derived from phospholipid turnover. De novo synthesis requires an addition of three methyl groups, transferred from an amino acid. Without the turnover component, defi ciency in which essential amino acid would make dietary choline essential for phosphatidylcholine synthesis?
Correct
The correct answer is C. The key to answering
this question correctly is an understanding that
phosphatidylcholine is formed by donation of
methyl groups. Methionine is the only amino
acid listed that can donate methyl groups. Its
activated form, S-adenosyl-L-methionine, is a
very common methyl group donor.Incorrect
The correct answer is C. The key to answering
this question correctly is an understanding that
phosphatidylcholine is formed by donation of
methyl groups. Methionine is the only amino
acid listed that can donate methyl groups. Its
activated form, S-adenosyl-L-methionine, is a
very common methyl group donor. -
Question 7 of 10
7. Question
A DNA fragment is added to four different tubes along with DNA polymerase, a radiolabeled primer, and the adenine, thymine, cytosine, and guanine deoxynucleotides. Each tube also contains one of the four bases as dideoxynucleotides. The four tubes are then run on electrophoresis gel and visualized by autoradiography. Which of the following laboratory techniques does this describe?
Correct
The correct answer is E. This question describes
sequencing. Sequencing is a laboratory
technique that utilizes dideoxynucleotides to
randomly terminate growing strands of DNA.
Gel electrophoresis is used to separate the varying
lengths of DNA. The DNA sequence can
then be read based on the position of the bands
on the gel.Incorrect
The correct answer is E. This question describes
sequencing. Sequencing is a laboratory
technique that utilizes dideoxynucleotides to
randomly terminate growing strands of DNA.
Gel electrophoresis is used to separate the varying
lengths of DNA. The DNA sequence can
then be read based on the position of the bands
on the gel. -
Question 8 of 10
8. Question
A DNA segment is treated with restriction enzymes, pipetted into a well of polyacrylamide gel, and subjected to an electric fi eld. Next, the gel is stained with ethidium bromide and visualized under ultraviolet lights. What laboratory technique does this describe?
Correct
The correct answer is B. This question describes
gel electrophoresis. Gel electrophoresis
uses an electric fi eld to separate molecules
based on their sizes. The negatively charged DNA migrates in the electric fi eld toward the
positive end. Smaller fragments move more
rapidly through the gel. Bands of DNA can be
visualized by staining the gel with dyes such as
ethidium bromide.Incorrect
The correct answer is B. This question describes
gel electrophoresis. Gel electrophoresis
uses an electric fi eld to separate molecules
based on their sizes. The negatively charged DNA migrates in the electric fi eld toward the
positive end. Smaller fragments move more
rapidly through the gel. Bands of DNA can be
visualized by staining the gel with dyes such as
ethidium bromide. -
Question 9 of 10
9. Question
Increases in intracellular calcium can be especially detrimental to the cell. Therefore, calcium homeostasis is very tightly regulated not only across the cell membrane but also through the additional work of sequestration in endoplasmic reticulum and mitochondria. In which of the following ways does increased intracellular calcium concentration cause the most cell damage?
Correct
The correct answer is A. Calcium is maintained
in high concentrations outside the cell
and within discrete compartments within the
cell (ie., mitochondria). Free intracellular calcium
can activate several enzymes whose cumulative
effect is to induce signifi cant cell injury.
A few important enzyme classes include
ATPases, which decrease ATP supply; phospholipases,
which decrease membrane stability;
endonucleases, which induce DNA damage;
and several proteases.Incorrect
The correct answer is A. Calcium is maintained
in high concentrations outside the cell
and within discrete compartments within the
cell (ie., mitochondria). Free intracellular calcium
can activate several enzymes whose cumulative
effect is to induce signifi cant cell injury.
A few important enzyme classes include
ATPases, which decrease ATP supply; phospholipases,
which decrease membrane stability;
endonucleases, which induce DNA damage;
and several proteases. -
Question 10 of 10
10. Question
A scientist working in a research laboratory has been examining different agonists of serotonin receptor 1B (5-HT1B), a G-protein-coupled receptor. Compound A has a much higher affi nity for 5-HT1B than compound B. Both compounds have a higher affi nity for the receptor than serotonin. Which of the following describes the relationship between compound A and compound B when considering the guanine- nucleotide exchange activity of 5-HT1B?
Correct
The correct answer is B. G-protein-coupled
receptors exist in an equilibrium between their
active and inactive states that depends on
whether ligand is present and the affi nity of ligand
for the receptor. When active, these receptors
catalyze guanine-nucleotide exchange
(GTP for GDP) of their associated G-proteins.
The Michaelis-Menten constant (Km) for any
enzyme-catalyzed reaction is inversely proportional
to the affi nity of the enzyme for its substrate.
Therefore, the Km with compound A
will be lower than that with compound B because
compound A has a higher affi nity for theIncorrect
The correct answer is B. G-protein-coupled
receptors exist in an equilibrium between their
active and inactive states that depends on
whether ligand is present and the affi nity of ligand
for the receptor. When active, these receptors
catalyze guanine-nucleotide exchange
(GTP for GDP) of their associated G-proteins.
The Michaelis-Menten constant (Km) for any
enzyme-catalyzed reaction is inversely proportional
to the affi nity of the enzyme for its substrate.
Therefore, the Km with compound A
will be lower than that with compound B because
compound A has a higher affi nity for the
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