Kaplan Biochem Flashcards

1
Q
1. Which of the following is the 3-carbon end-product of fatty acid metabolism?
A. Acetoacetate
B. Acetone
C. ß-Hydroxybutyrate
D. Diacylglycerol
E. Monoacylglycerol
A

The correct answer is choice B. The theme of this question is lipid metabolism. Ketone bodies
[acetoacetate (choice A), acetone (choice B), and ß-hydroxybutyrate (choice C)] are lipid
breakdown products. The levels of these molecules can increase in poorly controlled diabetes mellitus
or starvation. Acetone is the 3-carbon fragment of this group.
Monoacylglycerols (choice E) and diacylglycerols (choice D) are neutral fats made from glycerol with
one or two fatty acids esterified to the glycerol, respectively.

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2
Q
  1. Folate plays a role in single-carbon unit transfer in the synthesis of nucleotides. Which of
    the following nucleotides require folate for synthesis?
    A. Adenosine, cytosine, and uracil
    B. Adenosine, guanine, and thymidine
    C. Adenosine, guanine, and uracil
    D. Cytosine, thymidine, and uracil
    E. Guanine, thymidine, and uracil
A

The correct answer is choice B. Folate is involved in the transfer of carbons 2 and 8 of the purine
nucleus (affecting adenosine and guanosine) and the 5-methyl group of thymidine. This means that
folate is required for synthesis of 3 of the 4 nucleic acid bases of DNA and 2 of the 4 nucleic acid
bases in RNA. It is thus no wonder that folate deficiency affects so many tissues with high mitotic rate.
Megaloblastic changes analogous to those seen in erythrocytes and their precursors can also be seen
in other cells produced by bone marrow (neutrophil, eosinophil, basophil, and macrophage and
megakaryocyte lines) and in epithelia throughout the body, including skin, mucous membranes such
as the mouth and vagina (where the changes can be seen on pap smear), stomach, intestinal linings,
and cells from lung or liver. Similar megaloblastic changes are observed throughout the body when
cobalamin (vitamin B12 ) deficiency is present, since cobalamin plays a role in methionine synthesis,
which is the source of the one-carbon unit “active-formate.” Cobalamin is also invovled in the
conversion of methylmalonic acid to succinic acid and is required to maintain the integrity of nerve
cells via an unknown biochemical pathway.

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3
Q
3. A patient is taking hormone supplements. The hormone binds to a receptor in the cell
membrane, which activates tyrosine kinase activity. Which of the following hormones is this
patient taking?
A. Calcitrol
B. Thyroxine
C. Retinoic acid
D. Insulin
E. Protein synthesis
A

The correct answer is D.Insulin, a water-soluble hormone, binds to receptors in the cell membrane.
Receptor tyrosine kinase is activated, leading to protein phosphorylation. Tyrosine kinase receptors
are also involved in signaling by growth factors like PDGF (platelet derived growth factors) and EGF
(epidermal growth factor). Choice A, B and C - are all lipid soluble harmones, which diffuse through
the cell membrane, and bind to their respective receptors inside the cell. The hormone-receptor
complex then binds to hormone receptor elements in DNA, which then results in the observed
hormone effect.

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4
Q
  1. A 5-year-old child has blue-tinged sclera, hearing loss, and small, slightly blue, misshapen
    teeth. Radiologic studies confirm the presence of numerous fractures of various ages. No
    significant degree of bruising is seen over sites of recent fracture. The disease this child most
    likely has is related to abnormal metabolism involving which of the following substances?
    A. Collagen
    B. Glycogen
    C. Mucopolysaccharides
    D. Purines
    E. Tyrosine
A

The correct answer is A. The suspected disease is osteogenesis imperfecta, which is a rare genetic
disorder that occurs in both recessive and dominant forms. The clinical presentation, depending on
the specific form, varies from death in utero, to that described in the question stem, to very mild
disease with only a modest increase in bone fragility. Spontaneous fractures occur in utero or during
childhood. The different types all have defects in the synthesis of type I collagen, often with
insufficient or abnormal pro-1(1) or pro-2(1) chains. These deficits produce an unstable collagen triple
helix that is not as strong as normal collagen. Less severe mutations on type I collagen genes are
common, resulting in collagen disarray and predisposing to hypogonadal or idiopathic osteoporosis.
Defective glycogen (choice B) metabolism is associated with the various glycogen storage diseases,
such as von Gierke disease and Pompe disease. These diseases tend to present with profound
hypoglycemia, hepatomegaly, or muscle weakness.
Defective mucopolysaccharide (choice C) metabolism is associated with the mucopolysaccharidoses,
such as Hurler and Hunter syndromes. These diseases tend to present with abnormal facies
(“gargoylism”), deformed (“gibbus”) back, claw hand, and stiff joints.
Abnormalities of purine metabolism (choice D) are present in gout, which presents with joint
inflammation and often involves the great toe. The net result is due to chronic hyperuricemia.
Abnormalities of tyrosine metabolism (choice E) are associated with phenylketonuria (pale hair and
skin, mental retardation, musty smelling urine), albinism (pale hair, skin, increased skin cancer),
cretinism (decreased T3 and T4), tyrosinosis (liver and kidney disease), and alkaptonuria (chronic
arthritis and urine that turns black upon standing).

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5
Q
  1. An IgG2 molecule is composed of which of the following?
    A. One alpha, one gamma2, and two kappa chains
    B. One gamma1 chain and two kappa chains
    C. Two gamma1 chains and one kappa and one lambda chain
    D. Two gamma1 chains and two kappa chains
    E. Two gamma2 chains and two kappa chains
A
The correct answer is E. IgG molecules contain two gamma heavy chains of a given subtype and
two light chains (either kappa or lambda). The 2 in IgG2 indicates the subclass to which the molecule
belongs. IgG2 contains two gamma2 chains (since a given B cell can only form one type of heavy
2
chain). The IgG molecule will contain either two kappa chains or two lambda chains, but never one of
each (choice C). The standard serum concentration is 12 mg/mL.
A given cell produces immunoglobulin molecules with a single type of heavy chain (compare with
choice A). IgG molecules with gamma1 chains (choices B, C, and D) would be of the IgG1 subclass.
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6
Q
6. What is the lipase enzyme that degrades stored triacylglycerols in adipocytes?
A. Gastric lipase
B. Pancreatic lipase
C. Lipoprotein lipase
D. Hormone-sensitive lipase
A

The correct answer is D.Hormone-sensitive lipase is found in and degrades stored triacylglcerols in
adipocytes.
Gastric lipase (choice A) originates in the stomach, where it degrades dietary triacylglycerols.
Pancreatic lipase (choice B) is found in pancreas and degrades dietary triacylglycerols in the small
intestine.
Lipoprotein lipase (choice C) is found in extrahepatic tissues and acts on surface endothelial-cell
lining. The capillaries degrade triacylglycerols circulating in chylomicrons or VLDL.

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7
Q
  1. A 50-year-old man presents with headache, nausea, memory loss, abdominal pain, dark
    colored lines in gums, neuropathy, and anemia. Which of the following enzyme/s is/are
    inhibited in this patient?
    A. ALA dehydrase and ferrochelatase
    B. ALA synthase
    C. Uroporphyrinogen I synthase
    D. UDP-glucuronyl transferase
A

The correct answer is A. This is a case of lead poisoning. Lead inhibits ALA dehydrase and
ferrochelatase, the enzymes of the heme synthesis pathway. Other signs and symptoms include
coarse basophilic stippling of erythrocytes, lead deposits in abdomen, a gingiva, and epiphysis of
bone.
Choice B - ALA synthase is the rate-limiting enzyme of heme synthesis and is repressed by heme.
Choice C - is the enzyme of heme synthesis; its absence leads to acute intermittent porphyria.
Choice D - UDP-glucuronyl transferase is the enzyme of heme degradation, and is involved in the
formation of the conjugated form of bilirubin.

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8
Q
  1. Which of the following metabolic alterations would most likely be present in a chronic
    alcoholic compared to a non-drinker?
    A. Fatty acid oxidation is stimulated
    B. Gluconeogenesis is stimulated
    C. Glycerophosphate dehydrogenase is stimulated
    D. The ratio of lactate to pyruvate is decreased
    E. The ratio of NADH to NAD+ is increased
A

The correct answer is E. The principal route of metabolism of ethanol is via alcohol dehydrogenase,
which uses hydrogen from ethanol to form NADH from NAD+, markedly increasing the ratio of NADH
to NAD+. The relative excess of NADH has a number of effects, including inhibiting, rather than
stimulating fatty acid oxidation (choice A); inhibiting gluconeogenesis rather than stimulating it
(choice B); inhibiting, rather than stimulating (choice C) glycerophosphate dehydrogenase; and
favoring the formation of lactate rather than pyruvate from glycolysis (thereby increasing, rather than
decreasing the lactate/pyruvate ratio; choice D).
Alcoholism is a syndrome consisting of two phases: problem drinking and alcohol addiction. Alcohol
addiction is defined as a physiologic dependence as manifested by evidence of withdrawal when
intake is interrupted. As a side note, thiamine is commonly deficient in chronic alcoholics. Remember,
thiamine is a necessary ketolase enzyme cofactor.

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9
Q
9. A young adult with albinism is found to be at increased risk for skin cancer. Which of the
following is the precursor of melanin?
A. Tryptophan
B. Tyrosine
C. Arginine
D. Glycine
A

The correct answer is choice B. Albinism is caused by congenital deficiency of tyrosinase. It results
in an inability to synthesize melanin from tyrosine. Other derivatives of tyrosine are thyroxine,
dopamine, norepinephrine, and epinephrine.
Choice A - derivatives of tryptophan are niacin, serotonin, and melatonin.
Choice C - derivatives of arginine are creatine and urea.
Choice D - porphyrin is derived from glycine and eventually from heme.

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10
Q
10. A chronic alcoholic develops severe memory loss with marked confabulation. Deficiency of
which of the following vitamins would be most likely to contribute to the neurologic damage
underlying these symptoms?
A. Folic acid
B. Niacin
C. Riboflavin
D. Thiamine
E. Vitamin B12
A

The correct answer is D.Wernicke-Korsakoff syndrome refers to the constellation of neurologic
symptoms caused by thiamine deficiency. Among these, a severe memory deficit, which the patient
may attempt to cover by making up bizarre explanations (confabulation), is prominent. Wernicke’s
encephalopathy consists of the of confusion, ataxia and ophthalmoplegia. Anatomical damage to the
mamillary bodies and periventricular structures has been postulated as the cause. In the U.S., severe
thiamine deficiency is seen most commonly in chronic alcoholics. Thiamine deficiency can also
damage peripheral nerves (“dry” beriberi) and the heart (“wet” beriberi).
Folic acid deficiency (choice A) produces megaloblastic anemia without neurologic symptoms.
Niacin deficiency (choice B) produces pellagra, characterized by depigmenting dermatitis, chronic
diarrhea, and anemia.
Riboflavin deficiency (choice C) produces ariboflavinosis, characterized by glossitis, corneal
opacities, dermatitis, and erythroid hyperplasia.
4
Vitamin B12 deficiency (choice E) produces megaloblastic anemia accompanied by degeneration of
the posterolateral spinal cord. This megaloblastic anemia is commonly known as pernicious anemia.

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11
Q
  1. One of the enzymes of the citric acid cycle, which is on the inner mitochondria membrane,
    also functions as complex II of the electron transport chain. This enzyme is also responsible
    for converting succinate to fumarate, producing FADH2. The enzyme is
    A. isocitrate dehydrogenase
    B. a ketoglutarate dehydrogenase
    C. succinate dehydrogenase
    D. succinyl CoA synthase
A

The correct answer is choice C. All Krebs cycle components are in the matrix of the mitochondria
except succinate dehydrogenase, which is in the inner mitochondrial membrane. Succinate
dehydrogenase reoxidizes FADH2 and passes electrons directly to Coenzyme Q.
Choice A - Isocitrate dehydrogenase, the rate limiting enzyme of citric acid cycle, produces NADH. It
is activated by ADP and inhibited by NADH.
Choice B - a ketoglutarate dehydrogenase, like pyruvate dehydrogenase, requires thiamine, lipoic
acid, CoA, FAD, NAD, and produces NADH.
Choice D - Succinyl CoA synthase catalyses a substrate level phosphorylation of GDP to GTP.

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12
Q
12. The extremely potent vasodilator nitric oxide (NO) is produced naturally by the body from
which of the following amino acids?
A. Arginine
B. Aspartate
C. Glutamate
D. Methionine
E. Proline
A

The correct answer is A. In the body, the very short acting (5-second half-life) signaling molecule
nitric oxide (NO) is synthesized by nitric oxide synthase (NOS). This enzyme is found in many tissues
and converts arginine to citrulline and NO. The vasodilation produced by NO normally occurs in
response to appropriate local biologic triggers, such as infection. However, this process may also
produce pathology if uncontrolled, since it is thought that the endotoxic shock that can complicate
bacterial septicemia may be related to continued activity of NOS. NO also plays a role in macrophage
activity, where its metabolites are toxic to ingested microorganisms.
Aspartate (choice B) is associated with donation of one of the nitrogens in urea, with the liver enzyme
aspartate aminotransferase (AST), and with nucleotide synthesis.
Glutamate (choice C) is associated with amino acid transport into the cell via the gamma-glutamyl
cycle, and with the liver enzymes aspartate aminotransferase (AST) and alanine aminotransferase
(ALT).
Methionine (choice D) is associated with 1-carbon transfer reactions.
Proline (choice E) is associated with the collagen helix.

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13
Q
13. Which of the following amino acids would most likely be found on the surface of a protein
molecule?
A. Alanine
B. Arginine
C. Isoleucine
D. Leucine
E. Phenylalanine
A

The correct answer is choice B. This question requires two logical steps. First, you need to
appreciate that hydrophilic amino acids are more likely to appear on the surface of a protein molecule,
whereas hydrophobic amino acids are most likely be found in the interior. Next, you need to figure out
which of the amino acids listed is hydrophilic. If you recall that arginine is a basic amino acid that is
positively charged at physiologic pH, you should be able to answer this question right away.
All the other choices have neutral side chains and are uncharged at physiologic pH. They would most
likely be found in the hydrophobic core of the protein structure. Alanine (choice A), isoleucine (choice
C), and leucine (choice D) all have aliphatic side chains; phenylalanine (choice E) and tryptophan
have aromatic side chains.

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14
Q
14. Dietary intake of which of the following amino acids can substitute for a portion of the
daily requirement of niacin?
A. Alanine
B. Asparagine
C. Methionine
D. Proline
E. Tryptophan
A

The correct answer is E. Tryptophan is an aromatic amino acid that contains an indole group. By a
complex series of minor enzymatic reactions, a small amount (~2%) of the tryptophan can be
converted to quinolinate, which can then be used in place of niacin (nicotinic acid) in NAD
(nicotinamide adenine dinucleotide) synthesis. Very high tryptophan levels can replace a portion of
the dietary requirements for niacin. The nutritional disease pellegra (characterized by swollen tongue,
dermatitis, neurologic dysfunction, and gastrointestinal dysfunction) usually occurs in the setting of
combined tryptophan and niacin deficiency.
Alanine (choice A), the amino acid with a methyl R group, is a substrate of the liver enzyme alanine
amino transferase (ALT, formerly called SGPT). Asparagine (choice B) is one of the sources of
ammonia for the urea cycle. Methionine (choice C) is a sulfur-containing amino acid that is
associated with methyl group transfer. Proline (choice D) is technically an imino acid, rather than an
amino acid, with a ring structure. You should remember that collagen has a high proline
concentration.

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15
Q
  1. A child is noted to be severely retarded. Physical examination reveals a pot-bellied, pale
    child with a puffy face. The child’s tongue is enlarged. Dietary deficiency of which of the
    following substances can produce this pattern?
    A. Calcium
    B. Iodine
    C. Iron
    D. Magnesium
    E. Selenium
A
The correct answer is choice B. The disease is cretinism, characterized by a profound lack of
thyroid hormone in a developing child, leading to mental retardation and the physical findings
described in the question stem. Cretinism can be due to dietary deficiency of iodine (now rare in this
country because of iodized salt), to developmental failure of thyroid formation, or to a defect in
thyroxine synthesis. Iodine in the diet is absorbed at the digestive tract as iodide (I-). The follicle cells
in the thyroid gland absorb 120-150 μg of iodide ions per day. Iodide ions are actively transported into
the thyroid follicle cells. The active transport mechanism for iodide is stimulated by TSH.
Calcium deficiency (choice A) in children can cause osteoporosis or osteopenia.
Iron deficiency (choice C) can cause a hypochromic, microcytic anemia.
Magnesium deficiency (choice D) is uncommon, but can cause decreased reflexes, and blunts the
parathyroid response to hypocalcemia.
Selenium deficiency (choice E) is rare, but may cause a reversible form of cardiomyopathy.
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16
Q
  1. A chronically malnourished patient notices her “hair falling out.” She is on a strict fat-free
    diet. She probably has a deficiency in which vitamins?
    A. Vitamin A
    B. Vitamin C
    C. Vitamin D
    D. Vitamin E
    E. Vitamin K
A
The correct answer is A. While it is hard to develop a deficiency in oil-soluble vitamins (A, D, E, K)
because the liver stores these substances, deficiency states can be seen in chronic malnutrition
(specifically chronic fat deprivation) and chronic malabsorption. Vitamin A deficiency is one of the
most common deficiencies in developing countries. Vitamin A is necessary for formation of retinal
pigments (deficiency can cause night blindness) and for appropriate differentiation of epithelial tissues
(including hair follicles, mucous membranes, skin, bone, and adrenal cortex).
Vitamin C (choice B), which is water soluble rather than oil soluble, is necessary for collagen
synthesis. Deficiency is associated with development of scurvy.
Vitamin D (choice C) is important in calcium absorption and metabolism. Calcium and phosphate
irregularities can occur in deficiencies.
Vitamin E (choice D) is a lipid antioxidant that is important in the stabilization of cell membranes.
Vitamin K (choice E) is necessary for normal blood coagulation. When vitamin K is deficient, clotting
factors 2, 7, 9, and 10 will be decreased leading to an increased pT/INR.
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17
Q
17. Which of the following amino acids is most responsible for the buffering capacity of
hemoglobin and other proteins?
A. Arginine
B. Aspartic acid
C. Glutamic acid
D. Histidine
E. Lysine
A

The correct answer is D.Histidine, lysine and arginine are amino acids with basic side chains.
Remember that a buffer is most effective when its pKa is within the pH range of the surrounding
7
medium. Histidine is the only amino acid with good buffering capacity at physiologic pH. The
imidazole side chain of histidine has a pKa around 6.5 and can reversibly donate and accept protons
at physiologic pH.
Arginine (choice A) and lysine (choice E) are basic amino acids with pKa’s of 12.5 and 10.5,
respectively; at physiologic pH both will behave as bases and accept protons.
Aspartic acid (choice B) and glutamic acid (choice C) are acidic amino acids with pKa’s of
approximately 4; at physiologic pH they will behave as acids and donate protons. Their side chains
are almost always negatively charged.

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18
Q
18. Which of the following cofactors is required for decarboxylation of alpha-ketoacids?
A. Vitamin B1
B. Vitamin B2
C. Vitamin B3
D. Vitamin B5
E. Vitamin B6
A

The correct answer is A. Vitamin B1, or thiamine, is the coenzyme required (as the pyrophosphate)
for the decarboxylation of alpha-ketoacids. An example of this reaction is pyruvate decarboxylase
reaction in alcoholic fermentation. Other reactions such as that catalyzed by pyruvate dehydrogenase
also rely on thiamine pyrophosphate for decarboxylation, but require other cofactors as well. Thiamine
is also required for the generation of pentose phosphates for nucleotide synthesis in the pentose
phosphate pathway (hexose monophosphate shunt), serving as a cofactor for transketolase. The
primary cause of thiamine deficiency in the US is due to alcoholism. Thiamine deficiency is known as
beri beri. As this condition progresses, Wernicke-Korsakoff syndrome, which is seen in chronic
alcoholics, can develop.
Vitamin B2 (choice B), or riboflavin, is a constituent of FMN (flavin mononucleotide) and FAD (flavin
adenine dinucleotide). It functions in hydrogen and electron transport. Deficiency is associated with
the development of cheilosis, angular stomatitis and glossitis.
Vitamin B3 (choice C), or niacin (nicotinic acid), is a coenzyme that is also involved in hydrogen and
electron transport. Nicotinic acid functions in the form of NAD and NADP. Niacin is used clinically for
the treatment of hypercholesteremia and hypertriglyceredemia.
Vitamin B5 (choice D), or pantothenic acid, is conjugated with coenzyme A to act as a carboxylic acid
carrier.
Vitamin B6 (choice E), or pyridoxine, is required as a cofactor for pyridoxal phosphate and
pyridoxamine phosphate. Both of these cofactors are essential to protein metabolism and energy
production. Deficiencies can lead to the development of mouth soreness, glossitis, cheilosis,
weakness and irritability. Severe deficiency can cause peripheral neuropathy and anemia.

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19
Q
19. A stretch of 25 hydrophilic amino acids in a protein could be found in a
A. signal sequence
B. start transfer sequence
C. stop transfer sequence
D. transmembrane domain
E. triple helix
A

The correct answer is E. A triple helix, such as the one found in collagen, is composed of three
polypeptide chains wound together to form one structure. The chains are highly enriched with the
hydrophilic amino acids glycine and proline.
8
A signal sequence (choice A) contains a stretch of 30 hydrophobic amino acids.
The start transfer sequence (choice B) and stop transfer sequence (choice C) are each composed of
25-30 hydrophobic amino acids.
Transmembrane domains (choice D) contain 25-30 hydrophobic amino acids. This is enough to form
an alpha helix that will span a lipid bilayer.

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20
Q
20. The process involved in formation of protein from RNA is
A. Replication
B. Translation
C. Transcription
D. Transformation
A

The correct answer is choice B.In translation, protein synthesis occurs by peptide bond formation
between successive amino acids, whose order is specified by a gene and thus by an mRNA
Choice A - Replicationis a process in which two daughter DNA molecule are produced, that are each
identical to the parental DNA molecule
Choice C - Transcription is a process in which mRNA is synthesized using DNA as template.
Choice D - Transformation is not involved in protein synthesis either from DNA or RNA
Putting the processes together:

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21
Q
  1. Folate plays a role in single-carbon unit transfer in the synthesis of nucleotides. Which of
    the following nucleotides require folate for synthesis?
    A. Adenosine, cytosine, and uracil
    B. Adenosine, guanine, and thymidine
    C. Adenosine, guanine, and uracil
    D. Cytosine, thymidine, and uracil
    E. Guanine, thymidine, and uracil
A

The correct answer is choice B. In summary, a nucleoside consists of a purine or pyrimidine base
linked to a pentose, and a nucleotide is a phosphate ester of a nucleoside. For example, adenine is
the base of the ribonucleoside adenosine and the ribonucleotide adenylate (AMP). Folate is involved
in the transfer of carbons 2 and 8 of the purine nucleus (affecting adenosine and guanosine) and the
5-methyl group of thymidine. This means that folate is required for synthesis of 3 of the 4 nucleic acid
bases of DNA and 2 of the 4 nucleic acid bases in RNA. It is thus no wonder that folate deficiency has
effects on so many tissues with high mitotic rate. Megaloblastic changes analogous to those seen in
erythrocytes and their precursors can also be seen in other cells produced by bone marrow
(neutrophil, eosinophil, basophil, and macrophage as well as megakaryocyte lines) and in epithelia
throughout the body, including skin, mucous membranes such as the mouth and vagina (where the
changes can be seen on Pap smear), stomach and intestinal linings, and cells from lung or liver.
Similar megaloblastic changes are observed throughout the body when cobalamin (vitamin B12)
deficiency is present, since cobalamin plays a role in methionine synthesis, which is the source of the
one-carbon unit “active-formate.” Cobalamin is also involved in the conversion of methylmalonic acid
to succinic acid and is required to maintain the integrity of nerve cells via an unknown biochemical
pathway.

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22
Q
  1. A 45-year-old mother has a 5-year-old daughter with a history of repeated respiratory
    infections. The child has mental retardation, short stature, hypotonia with depressed nasal
    bridge, upslanting palpebral fissure, and epicanthal fold. She is also at an increased risk for
    leukemia. What is the probable chromosome abnormality?
    A. Trisomy 13
    B. Trisomy 18
    C. Trisomy 21
    D. Turner syndrome (45,X)
A
The correct answer is C. Trisomy 21 is the most common autosomal trisomy, and it causes Down
syndrome. The increased risk of Down syndrome with increased maternal age is well documented.
Trisomy 13 (choice A) features include oral facial clefts, microophthalmia, polydactyly, and renal
defects. They usually do not survive to 1 year.
Trisomy 18 (choice B) features include small mouth and ears, congenital heart defects, and
overlapping fingers. Ninety percent die during the first year of life.
Turner syndrome (choice D) features include reduced stature, webbed neck, and lymphedema of the
wrists and ankles.
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23
Q
  1. The biochemical structure of all the hormones secreted by the anterior pituitary, posterior
    pituitary, and pancreas can best be described as which of the following?
    A. Amino acid derivatives
    B. Catecholamines
    C. Glucocorticoids
    D. Peptides
    E. Steroid hormones
A

The correct answer is D.In the human body there are three basic types of hormones: peptides
(protein derivatives), amino acid derivatives, and steroid hormones. Some examples of peptide
hormones are those produced in the anterior pituitary (growth hormone, adrenocorticotropin, thyroid
stimulating hormone, luteinizing hormone, and prolactin), the posterior pituitary (vasopressin and
oxytocin), and the pancreas (insulin and glucagon).
Amino acid derivatives (choice A) include tyrosine and triiodothyronine, as well as the
catecholamines (choice B) dopamine, epinephrine, and norepinephrine.
Glucocorticoid hormones (choice C; e.g., cortisol) are a specific type of steroid hormone produced by
the adrenal cortex.
Steroid hormones (choice E) include the hormones produced in the adrenal cortex (cortisol and
aldosterone), the ovaries (estrogen and progesterone), the testes (testosterone), and the placenta
(estrogen and progesterone).

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24
Q
24. Which of the following enzymes is stimulated by glucagon?
A. Acetyl-CoA carboxylase
B. Glycogen phosphorylase
C. Glycogen synthase
D. HMG-CoA reductase
E. Pyruvate kinase
A

The correct answer is choice B. Before you started analyzing all of the answer choices you should
have reminded yourself that glucagon increases serum glucose. So an enzyme stimulated by
glucagon might be involved in either the breakdown of glycogen to glucose (glycogenolysis) or in the
creation of glucose from noncarbohydrate precursors (gluconeogenesis). Glycogen phosphorylase
catalyzes the first step in glycogenolysis; it makes sense that it would be stimulated by glucagon.
Clinical correlate: patients with Type I diabetes lose their response to hypoglycemia (but not to amino
acids in protein containing meals) within a year or so after developing diabetes. Type I diabetics rely
on the sympathetic nervous system to counter regulate hypoglycemia.
Acetyl-CoA carboxylase (choice A) catalyzes the first step in fatty acid synthesis, an anabolic process
that would be stimulated by insulin, not glucagon.
As its name implies, glycogen synthase (choice C) is involved in the synthesis of glycogen. Glucagon
(and epinephrine) stimulates the phosphorylation and inactivation of glycogen synthase.
HMG-CoA reductase (choice D) is the key enzyme involved in the synthesis of cholesterol. Since this
is an anabolic process that occurs in the well-fed state, you would expect it to be stimulated by insulin
and inhibited by glucagon (which it is).
Pyruvate kinase (choice E) catalyzes the last reaction of glycolysis. You would expect it to be
inhibited by glucagon (thus decreasing the amount of glucose consumption). Glucagon promotes the
phosphorylation of pyruvate kinase, which renders it inactive.

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25
Q
25. The majority of ATP generated during a 100 meter race is derived from which of the
following?
A. ATP stores
B. Creatine phosphate
C. Gluconeogenesis
D. Glycolysis
E. Lipolysis
A

The correct answer is D.The key to this question is understanding how and when the body utilizes
fuel stores. The stores of ATP (choice A) will be used up in less than 1 second once the race has
started. Creatine phosphate (choice B) will be the primary source of energy for the next 3 or 4
seconds. After the creatine phosphate stores are depleted, the majority of ATP needed to complete
the race will be derived from glycolysis (anaerobic respiration). If the race were to last for an extended
period of time, then the processes of gluconeogenesis (choice C) and lipolysis (choice E) might be
utilized. Gluconeogenesis is the process of synthesizing glucose in the liver from non-carbohydrate
sources, such as amino and fatty acids. Lipolysis is the splitting up or decomposition of fat in the
body.

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26
Q
  1. As cells in the erythrocytic lineage mature and lose their nuclei, mitochondria, and
    ribosomes, which of the following pathways can still be used to produce ATP?
    A. Citric acid cycle
    B. Electron transport chain
    C. Glycolysis
    D. Malate shuttle
    E. Urea cycle
A

The correct answer is C. Circulating erythrocytes have a life span of about 60 days and are
dependent on a functioning Na+/K+ ATPase in the plasma membrane. This pump provides the
electrochemical force across the plasma membrane that helps to maintain the volume of the red cell
11
at a constant level by regulating salt, and consequently water, flow into and out of the cell. When the
pump stops, the erythrocytes tend to lyse. It is therefore of extreme importance to the erythrocyte to
supply the ATP necessary to keep the pump running. This is accomplished through the use of
glycolysis, which converts glucose to pyruvate, with a net production of 2 ATP per glucose molecule.
The Krebs, or citric acid, cycle (choice A) is the central metabolic degradative pathway of aerobic
cells. However, it is located in mitochondria and requires the electron transport chain, making it
unavailable for mature red cells to use.
The electron transport chain (choice B) is located on the mitochondrial cristae and helps to convert
the energy of NADH and FADH2 to ATP.
The malate shuttle (choice D) is used to transport cytoplasmic NADH electrons into mitochondria and
is not available to erythrocytes.
The urea cycle (choice E) is used by liver cells for processing nitrogenous wastes, not generating
energy.

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27
Q
27. Megaloblastic anemia with folate deficiency is linked to an inability to perform which type
of enzymatic process?
A. Acyl transfer
B. Carboxylation
C. Decarboxylation
D. Hydroxylation
E. Methylation
A

The correct answer is E. Folic acid is a pteridine vitamin that exists as tetrahydrofolate (TH4) in its
most reduced form. TH4 can accept methyl, methylene, or formyl carbons and transfer them as
methyl groups. This function is vital in nucleotide and amino acid synthesis.
Pantothenic acid is a key vitamin in acyl transfer reactions (choice A). It forms part of coenzyme A,
which transfers acyl groups in thiol esters as acetyl-CoA, succinyl-CoA, and other acyl-CoA forms.
Important vitamins in carboxylation reactions (choice B) include biotin and vitamin K. Biotin carries
the carboxyl group in the pyruvate carboxylase and acetyl-CoA carboxylase reactions. Vitamin K is
used in post-translational carboxylation of amino acid residues in blood clotting factors.
Oxidative decarboxylation reactions (choice C) require thiamine (vitamin B1). Examples include the
pyruvate dehydrogenase and alpha-ketoglutarate dehydrogenase complexes.
Ascorbic acid (vitamin C) is a coenzyme in the hydroxylation (choice D) of lysyl and prolyl residues
for collagen synthesis.

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28
Q
28. Long term broad spectrum antibiotics with reduced clotting time in localized areas is an
indication of a deficiency in what?
A. Vitamin C
B. Vitamin K
C. Cyanocobalamine (B12)
D. Thiamine (B1)
A

The correct answer is choice B. Prolonged treatment with broad-spectrum antibiotics eliminates
intestinal bacteria that supply vitamin K. Vitamin K catalyses gamma carboxylation of glutamic acid
residues on various proteins concerned with blood clotting. Deficiency is characterized by prolonged
PT but normal bleeding time.
12
Choice A - Vitamin C deficiency leads to scurvy with increased bleeding time but normal PT. A diet
deficient in vitamin C results in scurvy.
Choice C - Vitamin B12 deficiency is usually caused by malabsorption and results in megaloblastic
anemia and progressive peripheral neuropathy.
Choice D - The most common cause of thiamine deficiency is alcoholism. Because alcohol interferes
with absorption, deficiency results in dry and wet beri-beri.

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29
Q
  1. The finding that almost all of the melanoma cells have very large, visible nucleoli suggests
    that these cells are making large amounts of which of the following?
    A. Cell surface markers
    B. Golgi apparatus
    C. Immunoglobulins
    D. New DNA
    E. Ribosomes
A

The correct answer is E. The nucleolus is the site of manufacture of ribosomal RNA with its
subsequent packaging into ribosomes. Consequently, very large nucleoli indicate an increased rate of
ribsome production, which, in turn, suggests an increased rate of production of proteins. Large
nucleoli are seen in many active cancers but can also be prominent in benign conditions
characterized by high metabolic rate, such as tissue repair after trauma.
Depending on the cells involved, the increased protein production associated with more ribosomes
might produce more cell surface markers (choice A), but it is impossible to predict this simply from
the presence of a large nucleolus.
You should associate a large Golgi apparatus (choice B) with transport of material to the cell surface.
Plasma cells that produce large amounts of immunoglobulins (choice C) may have large nucleoli, but
more specifically have a very large endoplasmic reticulum and Golgi apparatus.
The nucleolus is not involved with new DNA (choice D) synthesis; this is done elsewhere in the
nucleus.

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30
Q
30. Which type of enzyme reaction is effected by a folic acid deficiency?
A. Acyl transfer
B. Carboxylation
C. Decarboxylation
D. Hydroxylation
E. Methylation
A

The correct answer is E. Folic acid deficiency results in the development of macrocytic anemia that
yields macro-ovalocytes and hypersegmented neutrophils on the peripheral blood smear. Folic acid is
a pteridine vitamin that exists as tetrahydrofolate (TH4) in its most reduced form. TH4 can accept
methyl, methylene, or formyl carbons and transfer them as methyl groups. This function is vital in
nucleotide and amino acid synthesis. By far the most common cause of folate deficiency is
inadequate dietary intake. Alcoholics and those with poor diets are at the highest risk for developing
this type of anemia.
Pantothenic acid is a key vitamin in acyl transfer reactions (choice A). It forms part of coenzyme A,
which transfers acyl groups in thiol esters as acetyl CoA, succinyl CoA, and other acyl CoA forms.
13
Important vitamins in carboxylation reactions (choice B) include biotin and vitamin K. Biotin carries
the carboxyl group in the pyruvate carboxylase and acetyl CoA carboxylase reactions, and vitamin K
is utilized in post-translational carboxylation of amino acid residues in blood clotting factors.
Oxidative decarboxylation reactions (choice C) require thiamine (vitamin B1). Examples include the
pyruvate dehydrogenase and alpha-ketoglutarate dehydrogenase complexes.
Ascorbic acid (vitamin C) is a coenzyme in the hydroxylation (choice D) of lysyl and prolyl residues of
collagen.

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31
Q
31. Which of the following inhibits the activity of acetyl-CoA carboxylase?
A. Citrate
B. Glucagon
C. High-carbohydrate, low-fat diet
D. Insulin
A

The correct answer is choice B. The key thing to remember here is that acetyl-CoA carboxylase
catalyzes the first and rate-limiting step of fatty acid synthesis. If you got that far, you could have
figured out which of the choices would inhibit the synthesis of fatty acids. Certainly glucagon, a
catabolic hormone released in response to low blood glucose, would be a likely candidate to inhibit
the synthesis of fatty acids. In fact, glucagon inhibits fatty acid synthesis by a cAMP-dependent
phosphorylation of acetyl-CoA carboxylase. Conversely, glucagon stimulates fatty acid oxidation.
Clinical correlate: Glycogen metabolism is profoundly affected by specific hormones. Insulin, a
polypeptide hormone, increases the capacity of the liver to synthesize glycogen. When insulin levels
are high, the production of glycogen is high. The action of insulin is opposed by both glucagon and
epinephrine which will increase blood glucose levels.
Citrate (choice A) is a key player in fatty acid synthesis (citrate shuttle). Therefore, the presence of
citrate would stimulate, not inhibit, acetyl-CoA carboxylase.
A high-carbohydrate, low-fat diet (choice C) would stimulate, not inhibit, the synthesis of fatty acids.
In contrast to glucagon, insulin (choice D) is an anabolic hormone that promotes fatty acid synthesis
and therefore would stimulate acetyl-CoA carboxylase. It does so by dephosphorylating the enzyme.

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32
Q
  1. From which intermediate in the glycolytic pathway does the pentose phosphate pathway
    (also known as the hexose monophosphate, or pentose, shunt) “shunt”?
    A. Fructose-1,6-bisphosphate
    B. Fructose-6-phosphate
    C. Glucose-6-phosphate
    D. Phosphoenolpyruvate
    E. Pyruvate
A

The correct answer is C. The hexose monophosphate shunt is an alternative route for the oxidation
of glucose; it supplies the cell with NADPH and pentose sugars. The NADPH is used in many
biosynthetic processes (e.g., fatty acid and cholesterol synthesis), whereas the pentoses are involved
in the synthesis of nucleotides and some coenzymes. The pathway “shunts” from glucose-6-
phosphate, which is oxidized in a series of NADPH-generating reactions, to ribulose-5-phosphate.
The nonoxidative phase, which involves the transfer of C2 and C3 units from one sugar to another,
follows. One resulting intermediate is fructose-6-phosphate (choice B), which can serve as a re-entry
point to glycolysis, thereby closing the”shunt” loop.
14
Although all the other choices are glycolytic intermediates, they are not involved in the hexose
monophosphate shunt.

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33
Q
  1. Loss of which of the following classes of molecules on the surface of a tumor cell target
    would result in loss of susceptibility to killing by host immune cells?
    A. CD3
    B. CD4
    C. CD8
    D. MHC class I
    E. MHC class II
A

The correct answer is D.Major histocompatibility complex (MHC) class I proteins are found in the
membranes of all nucleotide cells. If the cell is healthy and the peptides are normal, T cells will ignore
them. If the cytoplasm contains abnormal peptides, they will appear in the cell membrane and the T
cells will be activated which will lead to the destruction of the cell. After the MHC class I molecule has
moved to the surface of the tumor cell, peptide fragments from the tumor are presented in a groove of
the class I molecule. The peptide fragments are presented to cytotoxic CD8 T cells, which recognize
the MHC class I molecules on the cell surface and kill the tumor cell. Loss of this molecule would
therefore prevent the tumor cell from being killed.
The CD3 molecule (choice A) is a marker on all T cells. It is involved in signal transduction, but not
antigen recognition. This molecule would not be on the surface of tumor cells.
The CD4 molecule (choice B) is not on the surface of a tumor cell, but it is on the surface of a CD4+
T helper lymphocyte.
The CD8 molecule (choice C) is not on the surface of a tumor cell, but it is on the surface of a CD8+
cytotoxic T lymphocyte.
MHC class II antigens (choice E) are not involved in killing of tumor cell targets. They present peptide
fragments (derived from intracellular killing of extracellular organisms by macrophages) to CD4 T
lymphocytes.

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34
Q
34. Which of the following citric acid cycle intermediates is required for heme synthesis?
A. a±-Ketoglutarate
B. Fumarate
C. Isocitrate
D. Oxaloacetate
E. Succinyl-CoA
A

The correct answer is E. The porphyrin ring of heme is derived from the citric acid cycle intermediate
succinyl-CoA and the amino acid glycine. The initial synthetic step, which is rate-limiting, is catalyzed
by aminolevulinic acid synthase (ALA synthase).

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35
Q
35. Which of the following metabolic processes occurs exclusively in the mitochondria?
A. Cholesterol synthesis
B. Fatty acid synthesis
C. Gluconeogenesis
D. Ketone body synthesis
E. Urea cycle
A

The correct answer is D. Of the processes listed, only ketone body synthesis occurs exclusively in
the mitochondria. Other mitochondrial processes include the production of acetyl-CoA, the TCA cycle,
the electron transport chain, and fatty acid oxidation. The most common cause of ketone body
formation is ketoacidosis secondary to diabetes. The essential diagnostic characteristics include:
hyperglycemia, acidosis (with blood pH < 7.3), serum bicarbonate < 15 and serum positive for
ketones.
Processes that occur exclusively in the cytoplasm include cholesterol synthesis (choice A; in cytosol
or in ER) and fatty acid synthesis (choice B).
Note that gluconeogenesis (choice C) and the urea cycle (choice E) occur in both the mitochondria
and the cytoplasm.

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36
Q
  1. An infant diagnosed with phenylketonuria would be expected to be deficient in which of the
    following nonessential amino acids, assuming that it is not obtained from dietary sources?
    A. Asparagine
    B. Cysteine
    C. Glutamine
    D. Proline
    E. Tyrosine
A

The correct answer is E. The human body is able to synthesize roughly half the amino acids
necessary to build protein. These amino acids, termed nonessential, include alanine, arginine,
asparagine, aspartate, cysteine, glutamate, glutamine, glycine, proline, serine, and tyrosine. The
amino acids that must be supplied in the diet are termed essential; these include histidine, isoleucine,
leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. Phenylalanine
undergoes hydroxylation to tyrosine, catalyzed by the enzyme phenylalanine hydroxylase. It is
noteworthy that tyrosine becomes an essential amino acid in individuals lacking this enzyme. The
hyperphenylalaninemias, which include phenylketonuria, result from impaired conversion of
phenylalanine to tyrosine, and are also asociated with mental retardation. This condition is associated
with increased phenylalanine in blood and increased phenylalanine and its by-products (e.g.,
phenylpyruvate, phenylacetate, and phenyllactate) in urine.
Asparagine (choice A) is a member of the “oxaloacetate family”; its immediate precursor is aspartate.
The immediate precursor of cysteine (choice B) is serine. Serine is also the precursor of the
nonessential amino acid glycine.
Glutamine (choice C), proline (choice D), and arginine are produced from glutamate. The synthesis
of glutamate occurs by the reductive amination of alpha-ketoglutarate.

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37
Q
37. Which of the following cofactors is required for decarboxylation of alpha-ketoacids?
A. Vitamin B1
B. Vitamin B2
C. Vitamin B3
D. Vitamin B5
E. Vitamin B6
A

The correct answer is A. Vitamin B1, or thiamine, is the coenzyme required for the decarboxylation
of alpha-ketoacids. An example of this reaction is pyruvate → acetyl CoA, catalyzed by the pyruvate
dehydrogenase complex (although this particular reaction requires other cofactors, thiamine is
required in all alpha-ketoacid decarboxylations). Thiamine is also required for the generation of
pentose phosphates for nucleotide synthesis in the pentose phosphate pathway (hexose
monophosphate shunt), serving as a cofactor for transketolase.
16
Vitamin B2 (choice B), or riboflavin, is a constituent of FMN (flavin mononucleotide) and FAD (flavin
adenine dinucleotide). It functions in hydrogen and electron transport.
Vitamin B3 (choice C), or niacin (nicotinic acid), is a coenzyme that is also involved in hydrogen and
electron transport. Nicotinic acid functions in the form of NAD and NADP.
Vitamin B5 (choice D), or pantothenic acid, is conjugated with coenzyme A to act as a carboxylic acid
carrier.
Vitamin B6 (choice E), or pyridoxine, is used as pyridoxal phosphate and pyridoxamine phosphate.
Both of these cofactors are essential to protein metabolism and energy production.

38
Q
38. Albinism is a disorder involving what substance?
A. Aromatic amino acids
B. Branched chain amino acids
C. Glycolipids
D. Glycoproteins
E. Sulfur-containing amino acids
A

The correct answer is A. The disease is albinism. The most common form of albinism is caused by a
deficiency of copper-dependent tyrosinase (tyrosine hydroxylase), blocking the production of melanin
from the aromatic amino acid tyrosine. Affected individuals lack melanin pigment in skin, hair, and
eyes, and are prone to develop sun-induced skin cancers, including both squamous cell carcinomas
and melanomas.
Maple syrup urine disease is an example of a disorder of branched chain amino acids (choice B)
causing motor abnormalities and seizures.
Tay-Sachs disease is an example of a disorder of glycolipids (choice C). In this disorder, a deficiency
of hexosaminidase A leads to accumulation of ganglioside GM2.
Hunter’s disease is an example of a disorder of glycoproteins (choice D). This
mucopolysaccharidosis is inherited as an X-linked recessive trait.
Homocystinuria disease is an example of a disorder of sulfur-containing amino acids (choice E).

39
Q
39. A woman vegetarian is found to have a severe riboflavin deficiency. The function of which
of the following enzymes in the citric acid cycle would be most directly affected by the
riboflavin deficiency?
A. Aconitase
B. Citrate synthase
C. Isocitrate dehydrogenase
D. Malate dehydrogenase
E. Succinate dehydrogenase
A

The correct answer is E. To answer this question you need two separate pieces of information. First,
riboflavin (one of the B vitamins) is used to make the flavin part of FAD (flavin adenine dinucleotide).
Second, of the citric acid cycle enzymes listed, only succinate dehydrogenase (which catalyzes the
conversion of succinate to fumerate) used FAD (which is converted to FADH2) as a cofactor.
Aconitase (choice A) converts citrate to isocitrate and does not require a cofactor.
17
Citrate synthetase (choice B) combines acetyl-CoA and oxaloacetate to make citrate with release of
coenzyme A (which requires pantothenic acid for synthesis).
Isocitrate dehydrogenase (choice C) converts isocitrate to alpha-ketoglutarate and uses NAD+ (which
is converted to NADH + H+). The NAD+ requires the vitamin niacin for synthesis.
Malate dehydrogenase (choice D) converts malate to oxaloacetate and uses NAD+ (which converted
to NADH + H+). The NAD+ requires the vitamin niacin for synthesis.

40
Q
  1. Which of the following pairs of enzymes is required for the process of gluconeogenesis?
    A. Fructose-1,6-bisphosphatase and pyruvate carboxylase
    B. Glucose-6-phosphatase and phosphofructokinase-1
    C. Glucose-6-phosphatase and pyruvate dehydrogenase
    D. Phosphoenolpyruvate carboxykinase and glucokinase
    E. Pyruvate kinase and pyruvate carboxylase
A

The correct answer is A. Gluconeogenesis is the process that results in the synthesis of glucose
from non-carbohydrate precursors. This pathway is very important because the brain is highly
dependent on glucose as the primary source of fuel. The three irreversible steps of glycolysis are
catalyzed by hexokinase, phosphofructokinase-1 (choice B), and pyruvate kinase. In
gluconeogenesis, other enzymes are needed to bypass these key steps. Pyruvate cannot be directly
converted to phosphoenolpyruvate in gluconeogenesis. Therefore, pyruvate carboxylase (a
mitochondrial enzyme; choice A) converts pyruvate to oxaloacetate, which can be converted to
phosphoenolpyruvate by phosphoenolpyruvate carboxykinase (choice D), using two ATP equivalents
per molecule of phosphoenolpyruvate. Fructose-1,6-bisphosphatase (choice A) is the enzyme that
splits fructose-1,6-bisphosphate into fructose-6-phosphate and inorganic phosphate. It is also required
for gluconeogenesis.
Glucose-6-phosphatase (choices B and C) is a liver enzyme that hydrolyzes glucose-6-phosphate to
glucose. A deficiency of this enzyme leads to von Gierke disease, also known as glycogen storage
disease type I.
Pyruvate dehydrogenase (choice C) is a mitochondrial enzyme that converts pyruvate to acetyl CoA.
This enzyme requires thiamine pyrophosphate, lipoamide, and FAD as cofactors.
Glucokinase (choice D) is a liver enzyme that converts glucose to glucose-6-phosphate. Unlike
hexokinase, it is specific for glucose and is unresponsive to the level of glucose-6-phosphate. Its
function is to store excess glucose, so it has a very high Km (ie, a low affinity) for glucose, becoming
active only when the concentration of glucose is very high.
Pyruvate kinase (choice E) catalyzes the conversion of phosphoenolpyruvate to pyruvate in the
glycolytic pathway. It is activated by fructose-1,6-bisphosphate, the product of the committed step of
glycolysis, and is allosterically inhibited by ATP, alanine, and acetyl CoA.

41
Q
  1. Cyanide poisoning will directly affect which anabolic process?
    A. Breaking of covalent bonds in glucose molecules
    B. Formation of carbon dioxide
    C. Formation of GTP from GDP
    D. Movement of hydrogen ions through channels in the respiratory enzymes
    E. Splitting of water molecules into hydrogen and oxygen atoms
A

The correct answer is D.Cyanide binds to the heme Fe3+ of cytochrome a3 in the electron transport
system, blocking the transfer of electrons to oxygen and consequently the synthesis of ATP. In other
18
words, cyanide blocks the process of oxidative phosphorylation, which produces greater than 90% of
the ATP used by the cells in our body. The major steps involved in this process occur within the
“electron transport system (ETS),” or “respiratory chain,” of the mitochondria. The steps at the end of
the elctron transport system, where ATP is generated, are as follows: along the ETS, the repiratory
enzymes continually pump hydrogen ions from the matrix of the mitochondria to the intermembrane
space, which creates a large concentration gradient. At the end of the ETS, hydrogen ions pass
through channels in the respiratory enzymes along the concentration gradient. As the hydrogen ions
pass through these enzymes, the energy created is used to phophorylate ADP to ATP. Inhibition of
mitochondrial respiration can lead to coma and death unless diagnosed and treated early.
The breaking of covalent bonds in glucose molecules (choice A), the formation of carbon dioxide in
the TCA cycle (choice B), and the splitting of water molecules into hydrogen and oxygen atoms
(choice E) all require energy; these processes do not create energy.
The formation of GTP from GDP (choice C) is a process that creates high energy phosphates in the
tricarboxylic acid cycle.

42
Q
42. An individual lacking the enzyme tyrosinase would be particularly predisposed to develop
which of the following?
A. Glioblastoma multiforme
B. Hemangioblastoma
C. Hepatoma
D. Melanoma
E. Renal cell carcinoma
A

The correct answer is D.This question is simple if you know that tyrosinase is an enzyme in the
biosynthetic pathway for melanin formation from tyrosine. A lack of tyrosinase causes one form of
albinism; a second form is caused by defective tyrosine uptake. Patients with albinism are vulnerable
to developing cancers of the skin of all types, including basal cell carcinoma, squamous cell
carcinoma, and melanoma. The melanomas are unusual in that they are non-pigmented (amelanotic)
rather than black, since the patients cannot form melanin. Malignant melanoma is the leading cause
of death from skin disease. Tumor thickness (Breslow’s Classification) is the single most important
prognostic factor. These lesions are often flat or raised. The borders are typically irregular. Remember
many cases of melanoma occur with lesions on the head and neck.

43
Q
43. Thiamine is used by which coenzyme in order to prevent a lactic acid acidosis?
A. Lactate dehydrogenase
B. Pyruvate carboxylase
C. Pyruvate dehydrogenase
D. Pyruvate kinase
E. Transketolase
A

The correct answer is C. Thiamine is a water-soluble vitamin that is converted to the coenzyme
thiamine pyrophosphate. This coenzyme is used by pyruvate dehydrogenase to convert pyruvate to
acetyl coenzyme A. In the absence of thiamine, pyruvate accumulates and can be converted by
lactate dehydrogenase to lactate, which is spilled in the blood causing lactic acidosis.
As a side note, thiamine deficiency is associated with the development of beri-beri - a neurological
and cardiovascular disorder. The most common cause of beri-beri in the US today is seen in
alcoholics. Damage to the nerves is expressed in terms of pain in the limbs and weakness of
musculature. The heart can become enlarged and cardiac output can be decreased.
Lactate dehydrogenase (choice A) produces lactate from pyruvate but does not use thiamine
pyrophosphate.
19
Some lactic acidosis might be produced by decreased pyruvate carboxylase activity (choice B), but
the enzyme requires biotin rather than thiamine pyrophosphate.
Pyruvate kinase (choice D) makes pyruvate from phosphoenolpyruvate, but does not use thiamine
pyrophosphate.
Transketolase (choice E) requires thiamine pyrophosphate, but operates in another pathway
(pentose phosphate pathway). Decreased transketolase activity is not associated with the
development of lactic acidosis.

44
Q
  1. A patient’s lab studies demonstrate a deficiency of cytochrome C oxidase activity. A defect
    in which of the following subcellular organelles would cause this deficiency?
    A. Golgi apparatus
    B. Lysosomes
    C. Mitochondria
    D. Ribosomes
    E. Smooth endoplasmic reticulum
A

The correct answer is C. Cytochrome C oxidase is an electron transport chain component, in the
mitochondria of muscle and brain. Cytochromes and the electron transport chain are involved in
oxidative respiration. Defects can produce mental retardation, weakness, ataxia or seizures.
The Golgi apparatus (choice A) is involved in packaging materials for secretion outside the cell. You
should associate mucolipidosis I-cell disease with Golgi apparatus problems.
Lysosomes (choice B) are the organelles that degrade many cellular products. Defects can produce
a wide variety of lysosomal storage diseases, including Hurler and Hunter syndromes.
Ribosomes (choice D) are the organelles that traslate mRNA into proteins. There are no known
diseases of ribosomes, probably because their function is so crucial that any problems produce death
in utero.
Hypertrophy of the smooth endoplasmic reticulum (choice E) in the liver is associated with conditions
that stimulate the cytochrome p450 detoxification systems (e.g., barbiturate use and alcoholism).

45
Q
  1. A genetic mutation that results in abnormal stimulatory G protein (Gs) structure would
    adversely affect which of the following mechanisms?
    A. Active transport
    B. Facilitated diffusion
    C. Pinocytosis
    D. Receptor-mediated endocytosis
    E. Signal transduction
A

The correct answer is E. G proteins are involved in signal transduction, which allows
macromolecules to affect a cell’s biological functions without crossing the plasma membrane. The
binding of an agonist, such as a hormone, to a receptor on the plasma membrane causes a
conformational change in the receptor, which then interacts with a stimulatory G protein (Gs),
accompanied by the binding of GTP to the alpha subunit of the G protein. The alpha subunit of the G
protein (with attached GTP) dissociates from the beta and gamma subunits, and interacts with
intracellular effector molecules such as adenylate cyclase, which it activates.
Active transport (choice A) is the movement of molecules across a membrane from a region of low
concentration to a region of high concentration; it requires both a carrier protein and the expenditure
of energy (usually supplied by the hydrolysis of ATP by an ATPase).
20
Facilitated diffusion (choice B) is the movement of a molecule down its concentration gradient by
means of a protein carrier. This type of diffusion is suited for molecules with low permeability resulting
from inappropriate size or polarity.
Pinocytosis (choice C) is also known as fluid-phase endocytosis and refers to the uptake of
molecules that are in solution. Receptors are not involved.
Receptor-mediated endocytosis (choice D) refers to the engulfment of a molecule that binds to a
receptor on the surface of the cell, occurring commonly at the clathrin-coated pits on the plasma
membrane.

46
Q
  1. Which of the following is the action of dietary fiber?
    A. Increases the blood cholesterol
    B. Decreases the blood cholesterol
    C. Decreases bowel motility
    D. Increases the exposure of gut to carcinogens
A

The correct ansewer is choice B. Dietary fiber lowers blood cholesterol. Dietary fiber consists of
nondigestable carbohydrate, including cellulose, lignin, and pectin. Other actions include the
increasing of bowel motility, reduced exposure of the gut to carcinogens, softer stools, and
interference with mineral absorption.

47
Q
47. Which of the following inhibits the activity of acetyl-CoA carboxylase?
A. Citrate
B. Glucagon
C. High-carbohydrate, low-fat diet
D. Insulin
A

The correct answer is choice B. The key thing to remember here is that acetyl-CoA carboxylase
catalyzes the first and rate-limiting step of fatty acid synthesis. If you got that far, you could have
figured out which of the choices would inhibit the synthesis of fatty acids. Certainly glucagon, a
catabolic hormone released in response to low blood glucose, would be a likely candidate to inhibit
the synthesis of fatty acids. In fact, glucagon inhibits fatty acid synthesis by a cAMP-dependent
phosphorylation of acetyl-CoA carboxylase. Conversely, glucagon stimulates fatty acid oxidation.
Citrate (choice A) is a key player in fatty acid synthesis (citrate shuttle). Therefore, the presence of
citrate would stimulate, not inhibit, acetyl-CoA carboxylase.
A high-carbohydrate, low-fat diet (choice C) would stimulate, not inhibit, the synthesis of fatty acids.
In contrast to glucagon, insulin (choice D) is an anabolic hormone that promotes fatty acid synthesis
and therefore would stimulate acetyl-CoA carboxylase. It does so by dephosphorylating the enzyme.

48
Q
  1. Addition of which of the following exhaustively 14C labeled substrates would lead to
    evolution of 14CO2 from a cell-free suspension containing all the enzymes and substrates
    required for the synthesis of uridylic acid?
    A. Aspartate
    B. Carbamoyl phosphate
    C. Glutamine
    D. Glycine
    E. N10-Formyltetrahydrofolate
A

The correct answer is A. In the first step of pyrimidine synthesis, carbamoyl phosphate condenses
with aspartate to form carbamoyl aspartate, in a reaction catalyzed by aspartate transcarbamoylase.
In subsequent steps, ring closure occurs with the loss of water, followed by oxidation to yield orotic
acid. Addition of ribose-5-phosphate produces orotidylic acid, which is decarboxylated by orotidylate
decarboxylase to yield uridylic acid. The carbon dioxide that is evolved is derived from the alpha
carboxyl group of aspartate.
Carbamoyl phosphate (choice B) condenses with aspartate with the loss of inorganic phosphate to
produce carbamoyl aspartate. The carbamoyl moiety of carbamoyl phosphate is retained.
Glutamine (choice C), glycine (choice D) and N10-formyltetrahydrofolate (choice E) are all used in
purine synthesis. Glutamine also donates an amino group to UTP to form CTP, but this step occurs
after the synthesis of uridylic acid is complete.

49
Q
  1. The receptors for the hormone that causes blood glucose level to quickly drop are located
    on which cellular component in target cells?
    A. Mitochondrial matrix
    B. Nuclear matrix
    C. Nuclear membrane
    D. Plasma membrane
    E. Smooth endoplasmic reticulum
A

The correct answer is E. The hormone in question is insulin, which helps the body utilize glucose in
the bloodstream. Circulating insulin binds to a receptor on the plasma membrane of target cells. The
insulin receptor is a transmembrane protein that stimulates cytosolic tyrosine kinase activity when
activated. One of the important proteins activated by the tyrosine kinase is an insulin receptor
substrate, which in turn activates the wide variety of enzymes and transport molecules that produce
the actual metabolic changes in sugar, protein, and fat metabolism seen in the presence of insulin.
The mitochondrial matrix (choice A) is not a significant receptor site for hormones.
The nuclear matrix (choice B), sometimes together with cytoplasm, contains receptors for steroid
hormones such as cortisol.
The nuclear membrane (choice C) is not a significant receptor site for hormones.

50
Q
50. Megaloblastic anemia, due to folate deficiency folate deficiency from dietary causes, is
linked to which dietary problem?
A. Lack of leafy green vegetables
B. Lack of milk products
C. Lack of red meat
D. Lack of yellow vegetables
E. Overcooked food
A

The correct answer is E. Folates (pteroylglutamic acid and related compounds) are widely
distributed in foodstuffs. Folic acid serves as an important mediator of one-carbon transfers Dietary
deficiency is usually due to overcooked (folates are very labile) and old (folates rapidly decay with
time) food. Folic acid deficiency is a macrocytic anemia that reveals macro-ovalocytes and
hypersegmented neutrophils on the peripheral blood smear. Reduced folate levels can be seen in red
blood cells or serum.

51
Q
51. Absence of which of the following enzymes would impair the rate-limiting step of
glycogenolysis?
A. α ±-1,4-Glucan transferase
B. Glycogen phosphorylase
C. Glycogen synthase
D. Phosphoglucomutase
E. UDP-glucose pyrophosphorylase
A

The correct answer is choice B. A key to excelling in biochemistry on the NBDE part 1 is to master
the most clinically important elements of metabolic pathways, e.g., rate-limiting steps, irreversible
steps, and steps involving enzymes affected by genetic diseases. In this case, glycogen
phosphorylase is the enzyme involved in the rate-limiting step of glycogenolysis:
Glycogen phosphorylase: (Glucose)n + Pi→ (Glucose)n–1 + Glucose-1-P
Note that this enzyme is activated in response to the binding of glucagon to liver cell receptors or the
binding of epinephrine to muscle receptors, via signal transduction.
α±-1,4-Glucan transferase(choice A) is a debranching enzyme that removes three of four glucose
units from a branch point and transfers them to the end of another chain. In this reaction, one α±-1,4
bond is cleaved and another is formed. The elongated chain then becomes a substrate for glycogen
phosphorylase.
Glycogen synthase (choice C) is the enzyme involved in the rate-limiting step of glycogen synthesis:
Glycogen synthase: (Glucose)n + UDP-glucose → (Glucose) n+1 + UDP
Note that this enzyme is inactivated in response to the binding of glucagon to liver cell receptors or
the binding of epinephrine to muscle receptors.
Phosphoglucomutase (choice D) is a key enzyme in glycogen synthesis that reversibly converts
glucose-6-phosphate to glucose-1-phosphate.
UDP-glucose pyrophophorylase (choice E) is another key enzyme in glycogen synthesis:
UDP-glucose pyrophosphorylase: Glu-1-P + Uridine-P-P-P → UDP-Glu + PPi
Note that the linkage formed between UDP and glucose is a high-energy bond that can provide
energy to many biosynthetic reactions.

52
Q
52. Which of the following will be unchanged in a Lineweaver-Burk plot of an enzyme with and
without a competitive inhibitor?
A. Km
B. Slope
C. x-Intercept
D. y-Intercept
A

The correct answer is D.It is worth taking the time to learn how to read a Lineweaver-Burk plot.
Lineweaver-Burk plots are used to determine the Vmax and Km of an enzyme; they are also used to
differentiate between competitive and noncompetitive inhibition.
Note that in a Lineweaver-Burk plot, the slope is Km/Vmax, the x-intercept is -1Km, and the yintercept
is 1/Vmax. In the presence of a competitive inhibitor, the Km (choice A), and therefore the
23
slope (choice B), is increased. Similarly, if Km is increased, -1/Km will become less negative and the
x-intercept (choice C) will shift to the right. Intuitively, this makes sense since a competitive inhibitor
will increase the amount of substrate needed to reach half-maximal velocity (definition of Km). In
contrast, the Vmax, and hence the y-intercept, is unchanged.

53
Q
53. Which of the following structures is common to all sphingolipids?
A. Carnitine
B. Ceramide
C. Diacylglycerol
D. Sphingomyelin
E. Squalene
A

The correct answer is choice B. Sphingolipids are a class of lipids that are structural components of
membranes. Specifically, sphingolipids are any lipid containing a long chain base like that of
sphingosine (e.g., ceramides, cerebrosides, gangliosides). They are a common constituent of nerve
tissue. Ceramide is a component of sphingolipids. Ceramide is composed of sphingosine, a longchain
amino alcohol with a saturated fatty acid linked to the amino group. Sphingolipids can be
differentiated on the basis of the “X” group that is esterified to the terminal hydroxyl group of
ceramide.
Carnitine (choice A) is involved in the oxidation of fatty acids. Carnitine is important in transferring
fatty acids from the cytoplasm into the mitochondria (the carnitine shuttle).
Diacylglycerol (choice C) is the alcohol common to all phospholipids. The second alcohol (e.g.,
choline, ethanolamine, serine) contributes the polar head that distinguishes the different classes of
phospholipids. Like sphingolipids, phospholipids are found in membranes.
Sphingomyelin (choice D) is a sphingolipid with phosphocholine as its “X” group. It is a component of
the myelin sheath.
Squalene (choice E) is a 30-carbon intermediate in the synthesis of cholesterol.

54
Q
54. Which of the following structures is common to all sphingolipids?
A. Carnitine
B. Ceramide
C. Diacylglycerol
D. Sphingomyelin
E. Squalene
A

The correct answer is choice B. Sphingolipids are a class of lipids that are structural components of
membranes. Ceramide is a component of sphingolipids. Ceramide is composed of sphingosine, a
long-chain amino alcohol with a saturated fatty acid linked to the amino group. Sphingolipids can be
differentiated on the basis of the “X” group that is esterified to the terminal hydroxyl group of
ceramide.
Carnitine (choice A) is involved in the oxidation of fatty acids. Carnitine is important in transferring
fatty acids from the cytoplasm into the mitochondria (the carnitine shuttle).
Diacylglycerol (choice C) is the alcohol common to all phospholipids. The second alcohol (e.g.,
choline, ethanolamine, serine) contributes the polar head that distinguishes the different classes of
phospholipids. Like sphingolipids, phospholipids are found in membranes.
24
Sphingomyelin (choice D) is a sphingolipid with phosphocholine as its “X” group. It is a component of
the myelin sheath.
Squalene (choice E) is a 30-carbon intermediate in the synthesis of cholesterol.

55
Q
  1. A young child is well for the first 2 years of his life, and receives all his immunizations
    without any complications. Starting around his 2nd birthday he contracts frequent upper
    respiratory tract infections, and is hospitalized three times for pneumonia. Laboratory testing
    would most likely reveal a deficiency of which of the following immunoglobulins in this child?
    A. IgA
    B. IgD
    C. IgG
    D. IgM
A

The correct answer is A. Selective IgA deficiency (<5 mg/dL) is the most common of all the primary
immunodeficiency diseases. The incidence reported in the US has ranged from 1:250 to 1:1000. IgA
has two subclasses, IgA1 and IgA2. IgA1 predominates in the serum, while IgA2 predominates in
mucosal secretions as a dimer bound together by a J chain with a secretory piece attached. Recurrent
bacterial and viral infections of the respiratory tract can be attributed to a lack of secretory IgA (sIgA),
the predominant immunoglobulin of the mucosal immune system. The most common types of
infection include sinusitis, otitis and bronchitis. Atopic disease and autoimmune disorders can be
associated with IgA deficiency.
IgD (choice B) has not been given any particular function other than to act as a receptor on the B cell.
It can be found in very low levels in serum.
IgG (choice C) is the major immunoglobulin found in the humoral immune response. A patient with a
low IgG will experience pyogenic infections.
IgM (choice D) is found in the early response to an antigen. If the patient was deficient in IgM he
would have also been characteristically low in IgG and would have experienced recurrent pyogenic
infection, usually commencing by the age of 5-6 months.

56
Q
  1. In the citric acid cycle, succinate thiokinase (succinyl-CoA synthetase) catalyzes the
    cleavage of the succinyl-CoA thioester bond with formation of a high-energy compound. This
    compound can then be used by the body in which of the following biochemical pathways?
    A. Cysteine degradation
    B. Elongation of the polypeptide chain
    C. Epinephrine synthesis from tyrosine
    D. Isopentyl pyrophosphate synthesis
    E. Oxidative phosphorylation
A

The correct answer is choice B. In this question, you need to know that GTP is synthesized when
CoASH is cleaved from succinyl-CoA to form succinate in the citric acid cycle. You also need to know
that GTP, rather than ATP, is used as the energy source in protein synthesis, specifically in the
formation of the activated elongation factor to which tRNA binds, and in the transfer of the elongating
chain from the P to the A site in the ribosome.
Cysteine degradation (choice A) requires O2 and produces sulfite and then sulfate, which is secreted
in urine.
Epinephrine synthesis from tyrosine (choice C) requires tetrahydrobiopterin, pyridoxal phosphate,
oxygen, copper, and S-adenosylmethionine.
25
Isopentyl pyrophosphate synthesis from HMG CoA (choice D) requires NADPH and ATP, and
releases CO2.
Oxidative phosphorylation (choice E) used NADH, FADH2, coenzyme Q, oxygen, and a variety of
cytochromes to produce ATP.

57
Q
  1. A 25-year-old woman with sickle cell anemia is diagnosed with gallstones. Of which of the
    following compounds are these stones most likely composed?
    A. Calcium bilirubinate
    B. Calcium oxalate
    C. Cholesterol
    D. Cholesterol and calcium bilirubinate
    E. Cystine
A

The correct answer is A. Bilirubin is a degradative product of hemoglobin metabolism. Bilirubin
(pigment) stones are specifically associated with excessive bilirubin production in hemolytic anemias,
including sickle cell anemia. Bilirubin stones can also be seen in hepatic cirrhosis and liver fluke
infestation.
Calcium oxalate stones (choice B) and cystine stones (choice E) are found in the kidney, rather than
the gallbladder.
Pure cholesterol stones (choice C) are less common than mixed gallstones, but have the same risk
factors, including obesity and multiple pregnancies.
Mixed stones (choice D) are the common “garden variety” gallstones, found especially in obese,
middle aged patients, with a female predominance.

58
Q
  1. A newborn with severe acidosis and vomiting has elevated serum levels of lactate and
    alanine. This suggests a deficiency in which enzyme?
    A. Alanine aminotransferase
    B. Glutamate dehydrogenase
    C. Lactate dehydrogenase
    D. Pyruvate carboxylase
    E. Pyruvate dehydrogenase
A

The correct answer is E. Pyruvate dehydrogenase (PDH) catalyzes the irreversible conversion of
pyruvate to acetyl-CoA. Thiamine pyrophosphate (TPP), lipomide and FAD serve as catalytic
cofactors in addition to CoA and NAD+, the stoichiometric cofactors in this reaction. If PDH is absent,
pyruvate will be used in other pathways instead. Pyruvate will be converted to alanine via alanine
aminotransferase (choice A) and to lactate via lactate dehydrogenase (choice C).
Glutamate dehydrogenase (choice B) is involved in oxidative deamination, releasing ammonium ion
for urea synthesis. Deficiency of this enzyme would not cause the symptoms described.
Pyruvate carboxylase (choice D) is a gluconeogenic enzyme that catalyzes the conversion of
pyruvate to oxaloacetate. Deficiency of this enzyme would not cause the symptoms described.

59
Q
  1. Observation of a hematoxylin and eosin-stained microscope slide reveals that the nuclei
    are blue. What is the basis for this observation?
    A. Eosin binds to carbohydrates
    B. Eosin binds to lipids
    C. Eosin binds to nucleic acids
    D. Hematoxylin binds to lipids
    E. Hematoxylin binds to nucleic acids
A

The correct answer is E. Blue hematoxylin binds to polyanions such as RNA and DNA. Nuclei
contain large amounts of DNA and RNA, and they are consequently almost always blue. The nuclei of
dysplastic and cancerous cells are often enlarged and hyperchromatic (e.g., darker blue) compared to
normal cells of similar cell lines, because these altered cells often have extra DNA (are aneuploid)
and/or RNA (are metabolically active).
Pink eosin binds relatively nonselectively to cellular components, particularly proteins. Cytoplasm of
different cell lines can be pink, purple, or blue, depending principally on the number of ribosomes in
the cytoplasm. Consequently, blue-tinged cytoplasm tends to suggest high synthetic activity (e.g.,
abundant ribosomes).

60
Q
  1. In achondroplasia, mutation results in overexpression of FGFR3 and excess inhibition of
    bone growth, particularly in long bones. Sons and daughters are equally affected. What is the
    mode of inheritance in achondroplasia?
    A. Autosomal dominant
    B. Autosomal recessive
    C. X-linked dominant
    D. X-linked recessive
A

The correct answer is A. Achondroplasia is an autosomal-dominant disease, typically seen in
multiple generations of a pedigree. Men and women are affected in roughly equal frequencies.
Autosomal recessive diseases (choice B) are clinically seen only in the homozygous individual.
Skipped generations are typically seen. Both men and women are affected.
An x-linked dominant disease phenotype (choice C) is seen in multiple generations. Both men and
women are affected.
X-linked recessive diseases (choice D) are seen much more commonly in men than women. Skipped
generations are commonly seen. Male-to-male transmission is not seen in X-linked inheritance.

61
Q
  1. Which of the following vitamins has antioxidant properties, reduces the incidence of heart
    attacks and deficiency of which can lead to hemolysis and neurological problems ?
    A. Vitamin A
    B. Vitamin D
    C. Vitamin E
    D. Vitamin K
A

The correct answer is choice C. Vitamin E acts as antioxidant, and may prevent oxidation of LDL
(low density lipoproteins). It protects membrane lipids from peroxidation. Oxidized LDL is thought to
promote heart disease.
27
Choice A - Vitamin A is antioxidant, maintains vision and growth regulation. Consumption of betacarotene
in the diet decreases the incidence of lung and skin cancer. Vitamin A deficiency results in
night blindness and follicular hyperkeratosis.
Choice B - In response to hypocalcemia, vitamin D helps normalize serum calcium level. Deficiency
results in Rickets and osteomalacia.
Choice D - Vitamin K exposes calcium binding sites on several calcium dependent proteins.
Deficiency leads to prolonged bleeding and easy bruising with normal bleeding time.

62
Q
  1. Aerobic glycolysis is utilized for energy during intense aerobic exercise. In which of the
    following forms will the carbons derived from glucose enter the citric acid cycle?
    A. Acetyl-CoA
    B. Citrate
    C. Oxaloacetate
    D. Pyruvate
    E. Succinate
A

The correct answer is A. In aerobic glycolysis, glucose is degraded to pyruvate, which is then
converted to acetyl-CoA, the form in which it actually enters the citric acid cycle. Two acetyl-CoAs,
each containing two of the glucose’s carbons, are produced from each glucose molecule. In addition,
a total of two carbons from glucose are released as CO2 when each of the two pyruvates is converted
to acetyl-CoA.
Citrate (choice B) is the product formed in the citric acid cycle when acetyl-CoA condenses with
oxaloacetate.
Oxaloacetate (choice C) is the citric acid cycle intermediate that condenses with acetyl-CoA to form
citrate.
Two pyruvates (choice D) are produced from degradation of glucose. These are then converted to
the two acetyl-CoAs that enter the citric acid cycle.
Succinate (choice E) is another citric acid cycle intermediate that forms when coenzyme A is
removed from succinyl-CoA.

63
Q
63. A patient has an inherited disorder in which the administration of aspartame could be
detrimental to her health. This patient most likely has which of the following genetic
disorders?
A. Hyperornithinemia
B. Hyperuricemia
C. Hypervalinemia
D. Phenylketonuria
E. Wilson disease
A

The correct answer is D.The administration of any product that contains phenylalanine, such as
aspartame (an artificial sweetener), to an individual with any of the hyperphenylalaninemias could be
detrimental to his or her general health. The hyperphenylalaninemias result from an impaired
conversion of phenylalanine to tyrosine. The most common and clinically important of these is
phenylketonuria, which is characterized by an increased concentration of phenylalanine in blood,
increased concentration of phenylalanine and its by-products (such as phenylpyruvate, phenylacetate,
and phenyllactate) in urine, and mental retardation. Phenylketonuria is caused by a deficiency of
phenylalanine hydrolase.
28
Hyperornithinemia (choice A) is an inherited disorder of amino acid metabolism that results from a
defect of the enzyme ornithine decarboxylase. This condition is associated with mental retardation,
neuropsychiatric dysfunction, and protein intolerance.
Hyperuricemia (choice B) is a condition associated with higher than normal blood levels of uric acid.
Gout may be produced if the hyperuricemia persists.
Hypervalinemia (choice C) is a genetic disorder of amino acid metabolism that results from a defect
of the enzyme valine aminotransferase. This condition is associated with mental retardation,
neuropsychiatric dysfunction, and protein intolerance.
Wilson disease (choice E) is an autosomal recessive disorder associated with an abnormality of the
hepatic excretion of copper, resulting in toxic accumulations of the metal in the brain, liver, and other
organs.

64
Q
64. What hormone is responsible for protein breakdown and increased gluconeogenesis?
A. Aldosterone
B. Cortisol
C. Estrogen
D. Testosterone
A

The Correct Answer is choice B. Cortisol increases gluconeogenesis and is responsible for protein
breakdown. It also has anti-inflammatory action.
Aldosterone (choice A) stimulates renal reabsorption of sodium and excretion of potassium.
Estrogen (choice C) controls the menstrual cycle and promotes the development of female secondary
sex characteristics.
Testosterone (choice D) stimulates spermatogenesis and promotes the development of male
secondary sex characteristics.

65
Q
  1. A complete blood count with differential for a pregnant woman reveals a hematocrit of
    30%, with hypersegmented neutrophils and large, hypochromic red cells. Deficiency of which
    of the following would be most likely to produce these findings?
    A. Ascorbic acid
    B. Calcium
    C. Copper
    D. Folate
    E. Iron
A

The correct answer is D.The patient has a megaloblastic anemia, which can be due to deficiency of
folate or B12. Pregnancy increases the need for folate and other nutrients used by both baby and
mother, and may “unmask” a borderline dietary deficiency. For this reason, most obstetricians
recommend vitamin supplements for pregnant women. Folic acid deficiency is a macrocyte anemia.
Macro-ovalocytes and hypersegmented neutrophils are revealed on peripheral blood smear. Folic
acid is the common term for pteroylmonoglutamic acid.
Ascorbic acid (choice A) is vitamin C, and its deficiency predisposes for capillary fragility and oral
lesions.
Calcium deficiency (choice B) predisposes for osteoporosis/osteopenia.
Copper deficiency (choice C) is rare; when it occurs, it may cause a hypochromic anemia,
neutropenia, osteoporosis, or hypotonia.
29
Iron deficiency (choice E) causes a microcytic, hypochromic anemia, with reduced mental and
physical performance.

66
Q
66. At which of the following sites is the characteristic triple helical structure of the collagen
initially formed?
A. Extracellular space
B. Golgi body
C. Nucleus
D. Rough endoplasmic reticulum
E. Smooth endoplasmic reticulum
A

The Correct Answer is choice B. Collagen formation begins with transcription of mRNA from
appropriate DNA genes in the nucleus. While still within the nucleus, the mRNA is spliced. It is then
transported through the cytoplasm to the ribosomes on the rough endoplasmic reticulum. Individual
chains are translated on the ribosomes, with the ends feeding into the endoplasmic reticulum lumen.
Within the lumen, glycosylation of the individual chains occurs. The material then moves toward the
Golgi bodies (whose lumens are connected to the endoplasmic reticulum) where the triple helices of
procollagen form. The procollagen is then secreted into the extracellular space, where cleavage of
pro-peptides and cross- linking of different triple helices occurs, maturing the collagen.
The extracellular space (choice A) is the site of procollagen cleavage and cross- linking.
The nucleus (choice C) is the site of mRNA transcription and splicing.
The rough endoplasmic reticulum (choice D) is the site of chain translation and glycosylation.
The smooth endoplasmic reticulum (choice E) does not participate in collagen synthes

67
Q
67. The biochemical basis of a genetic disease is an inability to add the recognition marker
mannose phosphate to enzymes. In which of the following organelles does this step usually
occur?
A. Endoplasmic reticulum
B. Golgi apparatus
C. Lysosome
D. Mitochondria
E. Ribosome
A

The Correct answer is choice B. The disease is I cell disease, which is a rare genetic disease that
may receive disproportionate attention both because it is a mucolipidosis (a form of generalized
lysosomal disorder) with accumulation of abnormal chemical material in lysosomes, and because the
biochemical basis illustrates an interesting mechanism. Specifically, the Golgi apparatus in cells of
these patients has an abnormal N-acetyl-glucosaminotransferase (N- acetylglucosamine-1-
phosphotransferase), and is not able to add the necessary recognition marker mannose phosphate to
enzymes usually destined to enter lysosomes. Physiologically, the Golgi apparatus is a cellular
organelle consisting of a series of membranous plates that give rise to lysosomes and secretory
vesicles. A complete deficiency of this enzyme (type I form of I cell disease) causes death early in life.
Endoplasmic reticulum (choice A) receives the growing peptide chain of the enzymes, but is not the
site of addition of recognition markers.
Lysosomes (choice C) are the normal destination for the enzymes with the mannose phosphate
marker, but do not receive them if the marker is not present.
30
Mitochondria (choice D) supply the ATP to drive these chemical reactions.
Ribosomes (choice E) are the site of synthesis of the amino acid chains of proteins, but not the site of
addition of the destination marker.

68
Q
  1. A patient has been on a self-imposed “starvation diet” for four months, and has lost 60
    pounds while consuming only water and vitamin pills. If extensive blood studies were
    performed, which of the following would be expected to be elevated?
    A. Acetoacetic acid
    B. Alanine
    C. Bicarbonate
    D. Chylomicrons
    E. Glucose
A

The correct answer is A. Long-term starvation induces many biochemical changes. Much of the
body’s energy requirements are normally supplied by serum glucose, but in starvation are supplied by
both glucose and lipid-derived ketone bodies, including acetoacetic acid and beta-hydroxybutyric acid.
Glucose cannot be synthesized from lipids, and is instead made from amino acids such as alanine in
the process of gluconeogenesis. Ketosis as seen in diabetic patients would also be expected.
Serum alanine (choice B) drops dramatically in starvation, due to its conversion to glucose.
Bicarbonate (choice C) levels drop as the bicarbonate buffers the hydrogen ions produced by the
ketone bodies.
Chylomicrons (choice D) are the lipid form seen after absorption of dietary fat, and would drop
because the person is not feeding.
Glucose (choice E) is maintained in the blood at a much lower than normal level during starvation.

69
Q
  1. A recent immigrant complains of fatigue, weight gain, constipation, and cold intolerance.
    Physical examination demonstrates a diffuse mass in the anterior aspect of the neck. Dietary
    deficiency of which of the following nutrients is most likely to have contributed to the patient’s
    problem?
    A. Copper
    B. Iodine
    C. Iron
    D. Selenium
    E. Zinc
A

The correct answer is choice B. Endemic goiter, such as in this patient, is due to dietary iodine
deficiency. This disorder is common world-wide in mountainous areas (where fish are not available),
although the use of iodized salt in the United States has limited its prevalence here. Frank symptoms
of hypothyroidism may or may not be present, possibly because of the increased synthesis of the
more potent triiodothyronine (T3) at the expense of thyroxine (T4). Goiters may become multinodular
and grow to great size. Impaired cognition and hearing may be severe in congenital hypothyroidism.
Copper deficiency (choice A) can cause anemia, neutropenia, hypotonia, psychomotor retardation,
osteoporosis, depigmentation of hair, and glucose intolerance.
Iron deficiency (choice C) can cause anemia, cognitive dysfunction, impaired immunity, impaired
thermoregulation, and reduced levels of physical activity. Iron deficiency most commonly occurs
during pathological bleeding, such as an ulcer.
31
Selenium deficiency (choice D) can cause congestive cardiomyopathy and skeletal muscle
degeneration.
Zinc deficiency (choice E) causes rash, growth retardation, and impairments of immunity, wound
healing, mentation, sexual function, and night vision. Zinc deficiency most commonly occurs in the
elderly.

70
Q
70. Histones can bind DNA tightly because they have a high isoelectric point. This is due to an
enrichment in the amino acid
A. aspartate
B. glycine
C. lysine
D. proline
E. tyrosine
A

The correct answer is C. Histones are enriched in the amino acids lysine and arginine. They are
positively charged and bind tightly to DNA, which is negatively charged from phosphate groups.
Proline (choice D) and glycine (choice B) are neutral, aliphatic amino acids. Aspartate (choice A) is
negatively charged. Tyrosine (choice E) is a neutral, aromatic amino acid.

71
Q
71. Which of the following amino acids is post-translationally hydroxylated in the cytoplasm of
fibroblasts?
A. Cysteine
B. Glycine
C. Proline
D. Serine
E. Tyrosine
A

The correct answer is C. Hydroxylation of proline in fibroblasts generates the modified amino acid
hydroxyproline. This is an example of post-translational modification. Hydroxyproline is involved in
stabilizing the three-dimensional triple helix structure of collagen. Proline also has an alphatic side
chain but differs from other amino acids in that its side chain is bonded to both the nitrogen and
carbon atoms. Note: Proline contains a secondary rather than a primary amino group, which
specifically makes it an imino acid.
Cysteine (choice A) is unique in its ability to form a covalent disulfide bond with another cysteine
residue elsewhere in the protein molecule, thereby forming a cysteine bond. Such strong disulfide
bonds stabilize the three-dimensional structure of the protein.
Glycine (choice B) is abundant in fibroblasts since it constitutes every third amino acid in the primary
sequence of collagen. However, glycine is not hydroxylated.
Serine (choice D), tyrosine (choice E), and threonine can all be phosphorylated post-translationally
to form phosphoserine, phosphotyrosine, and phosphothreonine, respectively. These phosphorylated
amino acids are believed to play a role in signal transduction.

72
Q
72. Which of the following will be unchanged in a Lineweaver-Burk plot of an enzyme with and
without a competitive inhibitor?
A. Km
B. Slope
C. x-intercept
D. y-intercept
A

The correct answer is D.It is worth taking the time to learn how to read a Lineweaver-Burk plot.
Lineweaver-Burk plots are used to determine the Vmax and Km of an enzyme; they are also used to
differentiate between competitive and noncompetitive inhibition.
Note that in a Lineweaver-Burk plot, the slope is Km/Vmax, the x-intercept is -1/Km, and the y-intercept
is 1/Vmax. In the presence of a competitive inhibitor, the Km (choice A) and therefore the slope (choice
B) are both increased. Similarly, if Km is increased, -1/Km will become less negative and the xintercept
will shift to the right. Intuitively, this makes sense since a competitive inhibitor will increase
the amount of substrate needed to reach half-maximal velocity (definition of Km). In contrast, the Vmax,
and hence the y-intercept, is unchanged (choice D). It is important to note that it is convenient to
transform the Michaelis-Menton equation into one that gives a straight line plot, which can be done by
taking the reciprocal of both sides to
1/V = 1/Vmax + Km/Vmax + 1/[s]
A plot of 1/V versus 1/[s] is called the Lineweaver-Burk plot.

73
Q
  1. A patient has Lesch-Nyhan syndrome caused by an absence of hypoxanthine-guanine
    phosphoribosyl transferase. The patient’s condition is caused by an enzyme deficiency in
    which of the following biochemical pathways?
    A. Ganglioside metabolism
    B. Monosaccharide
    C. Purine metabolism
    D. Pyrimidine metabolism
    E. Tyrosine metabolism
A

The correct answer is C. The patient has a classic case of Lesch-Nyhan syndrome, an X-linked
recessive disorder caused by severe deficiency of the purine salvage enzyme hypoxanthine-guanine
phosphoribosyl transferase (HPRT). This defect is associated with excessive de novo purine
synthesis, hyperuricemia. The clinical and symptoms include spasticity, writhing movements,
compulsive biting of fingers and lips, and head-banging. At puberty, the child often develops arthritis
and death from renal failure occurs at age 25 years. The biochemical basis of the often striking selfmutilatory
behavior (which may require restraints and even tooth extraction) has never been
established. Treatment with allopurinol inhibits xanthine oxidase and reduces gouty arthritis, urate
stone formation, and urate nephropathy. It does not, however, modify the neurologic/psychiatric
presentation. Mental deficiency, spasticity, and self-mutilation are the most common characteristics of
this condition.

74
Q
  1. Patients with gout will have localized concentrations of needle shaped, negatively
    birefringent crystals of which substance?
    A. Bile pigments
    B. Calcium pyrophosphate
    C. Cystine
    D. Monosodium urate
    E. Struvite
A
The correct answer is D.The patient has gout, which is due to precipitation of monosodium urate
crystals in joint spaces (notably the great toe) and soft tissues (causing tophi, which are often found
on the external ears). Gout is often an acute condition that is typically nocturnal and usually
monarticular, often involving the first metatarsophalangeal joint. Hyperuricemia occurs in most cases.
Identification of urate crystals in joint fluid or tophi is diagnostic.
Bile pigments (choice A) are found in some gallstones.
Calcium pyrophosphate (choice B) crystals are deposited in pseudogout, which classically affects the
knee or other large joints.
Cystine (choice C) and struvite (choice E) can form kidney stones
75
Q
  1. A young woman with anorexia nervosa consumes about 125g carbohydrate, 15g protein
    and 10 g fat daily. Her daily caloric intake is roughly equal to which of the following values?
    A. 450 kcal/day
    B. 650 kcal/day
    C. 850 kcal/day
    D. 1050 kcal/day
    E. 1250 kcal/day
A

The correct answer is choice B. You may be asked to make this type of calculation on the NBDE
part 1. You should know that 1 g of either protein or carbohyrate produces about 4 kcal = 4 Calories
(kcal) of energy, and 1 g of fat produces 9 kcal = 9 Calories of energy. The calculation is then
straightforward. The Calories from carbohydrates are 125 X 4 = 500; from protein are 15 X 4 = 60;
and from fat are 10 X 9 = 90. The total is 500 + 60 + 90 = 650 kcal/day.

76
Q
76. The activity of which of the following enzymes is directly affected by citrate?
A. Fructose-2,6-bisphosphatase
B. Isocitrate dehydrogenase
C. Phosphofructokinase I
D. Pyruvate carboxylase
E. 6-phosphogluconate dehydrogenase
A

The correct answer is C. Citrate is produced by citrate synthase from acetyl CoA and oxaloacetate.
Note that this is the first step in the citric acid cycle. This reaction takes place in the mitochondria, but
citrate can move freely from the mitochondria into the cytosol. When the citric acid cycle slows down,
citrate accumulates. In the cytosol, it acts as a negative allosteric regulator of phosphofructokinase I,
the enzyme that catalyzes the committed step of glycolysis.
Fructose-2,6-bisphosphatase (choice A) breaks down fructose-2,6-bisphosphate, a potent allosteric
activator of phosphofructokinase I. Fructose-2,6-bisphosphatase is activated by cyclic AMPdependent
protein kinase.
Isocitrate dehydrogenase (choice B) converts isocitrate to alpha-ketoglutarate in the citric acid cycle.
It is allosterically stimulated by ADP and inhibited by ATP and NADH. This reaction produces NADH
and CO2.
Pyruvate carboxylase (choice D) is a mitochondrial enzyme that converts pyruvate to oxaloacetate. It
is important in gluconeogenesis and replenishes the oxaloacetate in the citric acid cycle.
6-phosphogluconate dehydrogenase (choice E) converts 6-phosphogluconate to ribulose 5-
phosphate in the pentose phosphate shunt pathway.

77
Q
  1. Which of the following enzymes is responsible for maintaining blood glucose levels by
    releasing glucose from its storage form in the liver?
    A. Acetyl-CoA carboxylase
    B. Glucose 6-phosphate dehydrogenase
    C. Glycogen phosphorylase
    D. Glycogen synthase
    E. Thiolase
A

The correct answer is C. Glycogen is the storage form of glucose and a readily mobilizable fuel
store. When individuals do not consume adequate quantities of carbohydrates, the body responds by
breaking down glycogen stores in the liver to maintain normal blood glucose levels. The two enzymes
involved in the degradation and synthesis of glycogen are glycogen phosphorylase (choice C) and
glycogen synthase (choice D), respectively. The processes of glycogen synthesis and degradation
are coordinated by a hormone-triggered cascade that ensures that when one enzyme is active, the
other enzyme is inactive.
Acetyl-CoA carboxylase (choice A) is the key enzyme involved in fatty acid synthesis.
Glucose 6-phosphate dehydrogenase (choice B) is involved in the pentose phosphate pathway.
Thiolase (choice E) converts acetoacetyl-CoA into acetyl-CoA.

78
Q
  1. A cell biologist wants to examine the microstruture of an integral membrane protein. She
    solubilizes the protein by destabilizing its association with the membrane lipid bilayer. Which
    of the following techniques did she most likely use?
    A. Alterations in pH
    B. Detachment of protein phenyl groups
    C. Dissociation of phospholipid polar head groups
    D. Increase in ionic strength
    E. Interruption of hydrophobic interactions
A

The correct answer is E. This question is asking you what forces are responsible for retaining
integral membrane proteins within the lipid bilayer, i.e., hydrophobic interactions. Amphipathic agents,
such as detergents, are used to solubilize integral membrane proteins; they do so by disrupting
hydrophobic interactions between integral membrane proteins and other membrane constituents,
such as phospholipids.
Choices A, B, C, and D would destabilize the association of peripheral membrane proteins with the
membrane lipid bilayer.

79
Q
  1. Supplementation of calcium with which vitamin would reduce the risk for fracture caused
    by bone loss, particularly in elderly patients?
    A. Calcitonin
    B. Cholecalciferol
    C. Carotene
    D. Alpha tocopherol
A

The correct answer is choice B. Most of the calcium in the body is found as calcium phosphate
crystals in the bones and teeth that contribute to the physical strength of these structures. Vitamin D
(cholecalciferol) increases intestinal absorption of calcium and phosphate.
35
Choice A - Calcitonin is secreted by specialized parafollicular cells of the thyroid gland. It decreases
the resorption of bone, reduces serum calcium, and improves bone architecture.
Choice C - Vitamin A (carotene) is converted to several active forms in the body and is associated
with important functions, including maintenance of healthy epithelium and vision.
Choice D - Vitamin E (alpha tocopherol) prevents peroxidation of fatty acids in cell membranes,
helping to maintain their normal fluidity.

80
Q
80. Children with dry and wet beriberi typically have a deficiency of which vitamin?
A. Ascorbic acid
B. Retinol
C. Riboflavin
D. Thiamine
E. Vitamin K
A

The correct answer is D.Thiamine deficiency is most frequently encountered in alcoholics and in
developing countries. Deficiency of this vitamin can take several forms: dilated cardiomyopathy (wet
beriberi), polyneuropathy (dry beriberi), and mamillary body degeneration (Wernicke-Korsakoff
syndrome).
Ascorbic acid (choice A, Vitamin C) deficiency causes scurvy, associated with capillary fragility, bony
abnormalities, and poor wound healing.
Retinol (choice B, Vitamin A) deficiency causes blindness and impaired immune responses.
Riboflavin (choice C) deficiency causes cheilosis, glossitis, and dermatitis.
Vitamin K (choice E) deficiency causes impaired blood clotting because of decreased production of
factors II, VII, IX, and X.

81
Q
  1. Transcription of a prokaryotic gene by RNA polymerase yields an mRNA corresponding to
    a single polypeptide chain containing 64 amino acids. The mRNA has a 5’ untranslated region
    of 120 nucleotides and a 3’ untranslated region of 240 nucleotides. Approximately how many
    nucleotides are in the coding region of the mRNA?
    A. 64
    B. 128
    C. 192
    D. 424
    E. 552
A

The correct answer is C. Regardless of how long the untranslated regions are, the number of
nucleotides in the coding region of an mRNA is three times the number of amino acids, since three
nucleotides are required to code for each amino acid, and 3 X 64 = 192. In reality, three nucleotides
code for the first amino acid (formyl-methionine in prokaryotes, methionine in eukaryotes), which may
be removed in posttranslational steps, and three nucleotides at the 3’ end (are needed to terminate
the process (i.e., a STOP codon), so the actual number would likely be slightly higher.

82
Q
  1. Which of the following enzymes does the neutrophil use to initiate the production of toxic oxygen
    compounds that kill bacteria abscesses and in other bacterial infections?
    A. Hydrogen peroxide
    B. Myeloperoxidase
    C. NADPH oxidase
    D. Peroxidase
    E. Superoxide dismutase
A

The correct answer is C. The first step in killing bacteria is the production of superoxide ion, O2- , by
the action of NADPH oxidase on NADPH and O2. The superoxide is then converted to hydrogen
peroxide, either spontaneously or through the action of superoxide dismutase. The hydrogen peroxide
can also be converted to the toxic HOCl. radical by the action of myeloperoxidase.
Hydrogen peroxide(choice A) is not an enzyme but is used as an antiseptic for dental procedures and
for the treatment of apthous ulcers in the mouth.
Myeloperoxidase (choice B) is part of the killing pathway, but is active later in the sequence, when
H2O2
Peroxidase (choice D) is also a heme enzyme. It catalyzes the reaction in which hydrogen peroxide is
reduced to water.
Superoxide dismutase (choice E) is part of the killing pathway, but is active later in the sequence,
when superoxide is converted to hydrogen peroxide. This enzyme is present in all aerobic organisms.

83
Q
83. Which of the following amino acids is post-translationally hydroxylated in the cytoplasm of
fibroblasts?
A. Cysteine
B. Glycine
C. Proline
D. Serine
E. Tyrosine
A

The correct answer is C. Hydroxylation of proline in fibroblasts generates the modified amino acid
hydroxyproline. This is an example of post-translational modification. Hydroxyproline is involved in
stabilizing the three-dimensional triple helix structure of collagen.
Cysteine (choice A) is unique in its ability to form a covalent disulfide bond with another cysteine
residue elsewhere in the protein molecule, thereby forming a cystine residue. Such strong disulfide
bonds stabilize the three-dimensional structure of the protein.
Glycine (choice B) is abundant in fibroblasts since it constitutes every third amino acid in the primary
sequence of collagen. However, glycine is not hydroxylated.
Serine (choice D), tyrosine (choice E), and threonine can all be phosphorylated post-translationally
to form phosphoserine, phosphotyrosine, and phosphothreonine, respectively. These phosphorylated
amino acids are believed to play a role in signal transduction.

84
Q
  1. During DNA replication, which of the following enzymes produces a short strand of RNA
    complementary to the template DNA with a free 3’-OH end?
    A. DNA ligase
    B. Polymerase I
    C. Polymerase III
    D. Primase
    E. Topoisomerase
A

The correct answer is D.The primer molecule required by DNA polymerase is a short strand of RNA
(4-10 bases) complementary to the template strand of the DNA molecule. The primer is synthesized
by a specific RNA polymerase known as primase. The growing end of the RNA primer is a free 3’-OH
group. The primer RNA is excised at a later stage of replication. The primase does not itself require a
primer for initiation of nucleotide synthesis.
DNA ligase (choice A) catalyzes the formation of a phosphodiester bond between the 3’-OH of one
fragment of DNA and the 5’-monophosphate group of an adjacent DNA fragment.
Polymerase I (choice B) catalyzes the polymerization of nucleotides and also functions in processing
and repair mechanisms.
Polymerase III (choice C) is a part of a multiprotein complex and is the major replicating enzyme in E.
coli.
Topoisomerases (choice E) produce swivel points in the DNA molecule that relieve the strain induced
by the replication fork. These enzymes cut and reseal the DNA.

85
Q
85. Histones can bind DNA tightly because they have a high isoelectric point. This is due to an
enrichment in the amino acid
A. aspartate
B. glycine
C. lysine
D. proline
E. tyrosine
A

The correct answer is C. Histones are enriched with the amino acids lysine and arginine. They are
positively charged and bind tightly to DNA, which is negatively charged because of its phosphate
groups.
Proline (choice D) and glycine (choice B) are neutral, aliphatic amino acids. Aspartate (choice A) is
negatively charged. Tyrosine (choice E) is a neutral, aromatic amino acid.

86
Q
86. Which of the following amino acids would most likely be found on the surface of a protein
molecule?
A. Alanine
B. Arginine
C. Isoleucine
D. Leucine
E. Phenylalanine
A

The correct answer is choice B. This question requires two logical steps: first, you need to
appreciate that the hydrophilic amino acids are more likely to appear on the surface of a protein
molecule, while hydrophobic amino acids are most likely be found in its interior. Next, you need to
figure out which of the amino acids listed is hydrophilic. If you recall that arginine is a basic amino acid
that is positively charged at physiologic pH, you should be able to answer this question right away. In
addition to arginine, lysine and histidine also have basic side chains. Structurally the planar outer part
of arginine’s side chain consists of 3 nitrogens bonded to a carbon atom, which is called a guanidium
group.
All of the other choices have neutral side chains and are uncharged at physiologic pH. They would
most likely be found in the hydrophobic core of the protein structure. Alanine (choice A), isoleucine
(choice C), and leucine (choice D) all have aliphatic side chains; phenylalanine (choice E) has
aromatic side chains.

87
Q
  1. A 25-year-old man with known history of drug abuse walks into the clinic. A routine
    screening test for HIV exposure is positive. To confirm HIV exposure, which one of these tests
    is ideal?
    A. ELISA
    B. Southern blot
    C. Northern blot
    D. Western blot
A

The correct answer is D.Western blot is used as the confirmatory test for HIV exposure and is highly
specific. The combination of ELISA and Western blot has a positive predictive value greater than
99%. The Western blot primarily eliminates a small percentage of false positives found by the ELISA
test alone.
Choice A - ELISA is used as the primary screening assay because it is very sensitive.
Choice B - Southern blotting is a technique that can be used to detect mutations in DNA.
Choice C - Northern blots analyze RNA extracted from a tissue and are typically used to determine
which genes are being expressed.

88
Q
  1. A 30-year-old man has been fasting for several days. His blood glucose level is now about
    60% of its normal value, but he does not feel lightheaded because his brain has reduced its
    need for serum glucose by using which of the following substances as an alternate energy
    source?
    A. Apoprotein B
    B. Beta-carotene
    C. Beta-hydroxybutyrate
    D. C-reactive protein
    E. Coenzyme A
A

The correct answer is C. Ketone bodies, which include acetoacetate, beta-hydroxybutyrate, and
acetone, are produced by the liver in the fasting state by beta-oxidation of fatty acids. They are then
released into the blood stream, where they can be used as alternative energy sources for other
organs, such as muscle, kidney, and brain. The brain specifically still requires a small amount of
circulating glucose to function, but the amount required is reduced when ketone bodies are available.
Apoprotein B (choice A) is one of the proteins that hold lipoproteins together.
Beta-carotene (choice B) is a vitamin with antioxidant properties.
C-reactive protein (choice D) is a serum protein produced by the liver that rises during infections and
in inflammatory states.
39
Coenzyme A (choice E) is found in mitochondria and carries acetyl groups into the citric acid, or
tricarboxylic acid, cycle.

89
Q
  1. When alcohol is consumed, it is metabolized by the liver, with almost half the alcohol
    being oxidized to acetaldehyde. In which of the following sites does the reaction occur?
    A. Golgi bodies
    B. Lysosomes
    C. Mitochondrial matrix
    D. Peroxisomes
    E. Ribosomes
A

The correct answer is D.Peroxisomes are cell organelles that are present in only small numbers in
most mammalian cells. In the liver, however, these single membrane-bound organelles are present in
large numbers and are import in detoxification and long chain fatty acid metabolism. The clinically
important degradation of ethanol to (potentially toxic) acetaldehyde occurs in humans in both
peroxisomes and the smooth endoplasmic reticulum (P450 system). Acetaldehyde is then oxidized to
(non-toxic) acetate. This reaction, catalyzed by aldehyde dehydrogenase, occurs in the mitochondria.
Golgi bodies (choice A) are involved with packaging substances that are transported out of the cell.
Lysosomes (choice B) are sites of degradation of intracellular waste products.
The mitochondria (choice C) are involved with ATP production and contain both the electron transport
chain and the citric acid (Krebs) cycle.
Ribosomes (choice E) are the sites of protein synthesis.

90
Q
  1. In the CNS, the drug levodopa is metabolized to dopamine, which can be subsequently
    converted by some neurons to which of the following substances?
    A. Choline
    B. Dihydroxyphenylalanine (DOPA)
    C. Epinephrine
    D. Norepinephrine (NE)
    E. Tyrosine
A

The correct answer is D.Levodopa is used in the treatment of Parkinson disease, a condition caused
by dopamine deficiency in the CNS. When levodopa enters the CNS, it is metabolized to dopamine,
some of which is subsequently metabolized to norepinephrine (NE) in noradrenergic neurons. In other
words, dopamine is the immediate precursor to NE. The synthesis of NE begins in the axoplasm of
the terminal nerve endings of adrenergic fibers; however, its synthesis is completed inside the
vesicles of these fibers.
With respect to the endogenous syntheses of NE, the basic steps are as follows: tyrosine (choice E)
is converted to dihydroxyphenylalanine (DOPA) through the process of hydroxylation. Then, DOPA
(choice B) undergoes decarboxylation to the neurotransmitter dopamine. Dopamine is then
transported into the vesicles of the adrenergic fibers, where it undergoes hydroxylation to NE (choice
D). In the adrenal medulla, NE is transofrmed into epinephrine (choice C) through the process of
methylation.
Choline (choice A) is combined with acetyl-CoA to become acetylcholine.

91
Q
91. The glycolytic degradation of ingested glucose commences with the action of which of the
following enzymes?
A. Aldolase
B. Hexokinase
C. Phosphofructokinase
D. Phosphoglucose isomerase
E. Pyruvate kinase
A

The correct answer is choice B. The process of glycolysis is defined as the sequence of reactions
that converts glucose into pyruvate with the concomitant production of ATP.
Glycolysis begins when glucose is converted by hexokinase to glucose-6-phosphate. When this
compound interacts with the enzyme phosphoglucose isomerase (choice D), fructose-6-phosphate is
formed. Fructose-6-phosphate is then converted by phosphofructokinase (choice C) to form fructose
1,6-biphosphate, which is subsequently converted to glyceraldehyde 3-phosphate by aldolase
(choice A). After a number of enzymatic reactions, phosphoenolpyruvate is formed.
Phosphoenolpyruvate is converted to pyruvate by pyruvate kinase (choice E), and the glycolytic
pathway is then completed.