Chapter 4 - Cell Biology, Signal Transduction, and the Molecular Biology of Cancer

Question 1

Bacteria grown at 15°C contain a different fatty acid composition in their membranes as compared to bacteria grown at 37°C. Which one of the following would best represent the composition of the fatty acids at these two different temperatures?
(A) Bacteria at the lower temperature would contain a higher percentage of saturated fatty acids than bacteria grown at the higher temperature.
(B) Bacteria grown at the lower temperature would have a higher percentage of long-chain fatty acids than bacteria grown at the higher temperature.
(C) Bacteria grown at the lower temperature would have a higher percentage of unsaturated fatty acids than bacteria grown at the higher temperature.
(D) Bacteria grown at the lower temperature would have an increased level of cholesterol as compared to bacteria grown at the higher temperature.
(E) Bacteria grown at the lower temperature would have a decreased level of cholesterol as compared to bacteria grown at the higher temperature.

Answer 1

The answer is C. Membrane fluidity needs to be constant at the two different temperatures for the bacteria to grow. At the lower temperature, a higher percentage of unsaturated fatty acids and short-chain saturated fatty acids would be present as those fatty acids have a lower melting point than long-chain saturated fatty acids. Cholesterol is not found in bacterial membranes.

Question 2

A 57-year-old pathologist, who had often cut himself while performing autopsies, develops blurred vision, dementia, personality changes, and muscle twitching in a very short period of time. The protein that is leading to these behavioral changes is best described as which one of the following?
(A) A soluble, cytoplasmic protein
(B) A peripheral membrane protein
(C) An embedded membrane protein
(D) A GPI-anchored membrane protein
(E) A secreted protein

Answer 2

The answer is D. The pathologist has obtained a prion disease, which results from the prion protein adopting an alternative conformation and precipitating in neural tissue. The normal prion protein is a GPI-anchored protein.

Question 3

Many growth factors, upon binding to their receptor, exhibit downregulation, in which the receptor number on the cell surface decreases. This occurs due to which one of the following processes?
(A) Endocytosis
(B) Exocytosis
(C) Pinocytosis
(D) Potocytosis
(E) Phagocytosis

Answer 3

The answer is A. Receptor-mediated endocytosis refers to the clustering of receptors over clathrin-coated pits in the inner membrane, and then invagination of the membrane to form an intracellular vesicle that contains the receptor–growth factor complex. Exocytosis is the opposite effect–an intracellular vesicle fuses with the plasma membrane to release its contents into the extracellular space. Pinocytosis refers to endocytosis without the receptors–small particles can be taken into the cell through vesicle formation on the cell surface. Potocytosis refers to receptor-mediated entry into a cell through caveolae, and not through clathrin-coated pits. Phagocytosis refers to the forming of a membrane around a particle, and then the endocytosis of that membrane containing the particle (or bacteria).

Question 4

Proton gradients across membranes are essential for the functions of which of the following organelles? Choose the one best answer.
(A) Lysosomes
(B) Mitochondria
(C) Nucleus
(D) Lysosomes and mitochondria
(E) Lysosomes and nucleus
(F) Nucleus and mitochondria

Answer 4

The answer is D. Lysosomes depend on a proton gradient to acidify their intracellular milieu, such that the lysosomal hydrolases will be at their pH optima (around 5.5). Mitochondria re- quire a proton gradient across their inner membrane in order to synthesize ATP via oxidative phosphorylation. The nucleus does not concentrate protons; the intranuclear space has the same pH as the cytoplasm.

Question 5

Assume there is a microRNA that participates in regulating the expression
of a particular cyclin kinase inhibitor. How might an alteration in this microRNA lead to uncontrolled cell proliferation?
(A) Overexpression of the microRNA, so it acts as an oncogene.
(B) Reduced expression of the microRNA, so it acts as a tumor suppressor.
(C) A total loss of activity of the microRNA, so it cannot bind to its target mRNA.
(D) A loss of specificity of the microRNA for its target, so different mRNAs are not targeted.
(E) No change in microRNA activity can lead to uncontrolled cell proliferation.

Answer 5

The answer is A. Cyclin kinase inhibitors act as brakes on the cell cycle. If the cyclin kinase inhibitor can be removed from the cell, then the cell cycle could proceed in an uncontrolled fashion. MicroRNAs reduce the amount of protein product formed from the target mRNA. In order to eliminate the production of the cyclin kinase inhibitor, the microRNA would need to be overexpressed, such that all target mRNAs are bound, and translation of the gene product is halted. Reducing the expression of the microRNA would lead to overexpression of the cyclin kinase inhibitor, and more control of the cell cycle. This is also the case if the microRNA lost all activity (overproduction of its target), or lost its specificity (again, overproduction of the target).

Question 6

The regulation of heterotrimeric G-proteins is similar to the regulation of which one of the following processes?
(A) DNA synthesis
(B) RNA synthesis
(C) Protein synthesis
(D) Active transport
(E) Facilitated transport

Answer 6

The answer is C. The heterotrimeric G-proteins bind GTP on the α subunit, which activates the subunit. The activation is self-controlled by a built-in GTPase that is present within the α subunit. This activity slowly hydrolyzes the bound GTP to GDP, thereby inactivating the sub- unit, and allowing the heterotrimer to re-form. Similar events occur in protein synthesis with both initiation factors and elongation factors. These factors are active when GTP is bound to them, and a built-in GTPase activity within these factors limits the length of time they are active. Such control systems are not observed in DNA or RNA synthesis, or in carrier-mediated transport across a cellular membrane.

Question 7

A message transmitted by which example of a chemical messenger would most likely be negatively affected by a mutation that greatly reduced the fluidity of the plasma membrane?
(A) Cytokine
(B) Steroid hormone
(C) A transforming growth factor
(D) Insulin
(E) Glucagon

Answer 7

The answer is B. Steroid hormones must enter the cell by passive diffusion, and if the membrane is less fluid, it will be more difficult for the hormone to enter the cell to bind to its receptor. Cytokines, transforming growth factors (TGFs), insulin, and glucagon all bind to transmembrane receptors, which transmit their signal to the cytoplasmic portion of the receptor. It is less likely that those conformational signals will be affected by the fluidity of the membrane than the passage of the steroid hormone through the membrane. Decreased membrane fluidity may impair dimerization of the receptors, but the initial binding events should still occur normally.

Question 8

Li–Fraumeni syndrome results from which one of the following? Choose the one best answer.
(A) Inability to recognize DNA damage
(B) Inability to regulate CDKs
(C) Inability to regulate a tyrosine kinase
(D) Inability to regulate gene transcription
(E) Inability to activate a heterotrimeric
G-protein

Answer 8

The answer is A. Li–Fraumeni syndrome results from an inherited mutation in p53, the guardian of the genome. This protein monitors the DNA for damage, and if damage is found, acts as a transcription factor to arrest the cell cycle, allow the DNA damage to be repaired, and then to allow the cycle to proceed. If the DNA damage cannot be repaired, then apoptosis is induced so that the cell will not replicate the damaged DNA. P53 does not regulate the CDKs, tyrosine kinases, or G-proteins. Loss of p53 expression will alter gene transcription (repair enzymes will not be induced, nor will apoptosis be induced if the damage cannot be repaired), but does not hinder the normal regulation of gene transcription.

Question 9

Chromosomal translocations can lead to uncontrolled cell growth due to which one of the following?
(A) Interference with mitosis
(B) Interference with DNA synthesis
(C) Unequal crossing over during mitosis
(D) Inappropriate expression of translocated
genes
(E) Loss of gene expression

Answer 9

The answer is D. For most translocations that lead to uncontrolled cell growth, a gene is inappropriately expressed because it has been moved adjacent to a constitutive promoter (such as the myc gene next to the immunoglobulin promoter in Burkitt lymphoma). The dysregula- tion of cell proliferation does not occur owing to problems with mitosis or DNA replication, nor with crossing over. In most cases, the problem is an increased or inappropriate expression of the translocated gene, and not a loss of gene expression.

Question 10

Interference in generalized cytokine signaling can lead to which one of the following disorders?
(A) Adrenoleukodystrophy
(B) SCID
(C) Influenza
(D) Myasthenia gravis
(E) Type 1 diabetes

Answer 10

The answer is B. X-linked SCID is due to the lack of a common cytokine receptor subunit (the γ subunit), which affects the ability of a variety of cytokines to transmit signals to hematopoietic cells. Adrenoleukodystrophy is due to the buildup of very long-chain fatty acids, for a variety
of reasons. Influenza is due to a virus. Myasthenia gravis is due to the production of autoantibodies directed against the acetylcholine receptor. Type 1 diabetes results from an inability to produce insulin.

Question 11

A bodybuilder has gained 50 pounds of muscle over the last 6 months, facilitated by both increased weight lifting and black-market pharmaceutical injection of one substance. He never experienced hypoglycemia during this time frame.

The mechanism whereby the illegal substance is entering the muscle cells of the bodybuilder is most likely which one of the following?
(A) Simple diffusion
(B) Active transport
(C) Endocytosis
(D) Facilitative diffusion (E) Pinocytosis

Answer 11

The answer is A. The bodybuilder is injecting (most probably) testosterone, a steroid hormone, which aids in building muscle mass. Steroid hormones are lipid-soluble substances, and cross membranes by simple diffusion. The receptor for steroid hormones is present inside the cell (either the cytoplasm or nucleus), and once the steroid hormone enters the cell, it will bind to the receptor in a saturable manner. Once the concentration of the hormone inside and outside the cells is equal, transport will stop. Active transport refers to using energy to concentrate a solute against its concentration gradient, which is not the case for steroid hormone transport across the membrane. Facilitative diffusion requires a membrane-bound carrier (no energy), but as indicated previously, the carrier (receptor) for these hormones is intracellular. Steroid hormones do not enter cells through either endocytosis or pinocytosis. The fact that the bodybuilder never became hypoglycemic after taking the drug suggests that it was not insulin being injected.

Question 12

A bodybuilder has gained 50 pounds of muscle over the last 6 months, facilitated by both increased weight lifting and black-market pharmaceutical injection of one substance. He never experienced hypoglycemia during this time frame.

How much energy is needed for the transport of the majority of the illegal substance that the bodybuilder is using for the drug to enter cells?
(A) No energy is required.
(B) One ATP molecule is used for each
molecule of the substance transported. (C) Only a few ATP molecules are being used to open and close the channel through
which many substance molecules diffuse.
(D) This is an example of cotransport, in which
the energy generates a sodium gradient across the membrane, and it is difficult to calculate an exact energy amount.
(E) The transporter has to be phosphorylated once to allow transport to occur for many solutes.

Answer 12

The answer is A. No energy is needed for simple diffusion, which is the case if this is a steroid hormone. The other answer choices require ATP, which would be necessary for an active trans- port process, whether it be activation of a channel by phosphorylation, or generation of a gradient across the membrane for cotransport. Simple and facilitated diffusion do not require any energy sources for transport.

Question 13

A 12-year-old boy is admitted to the hospital in ketoacidosis with a blood glucose level of 700 mg/dL (normal fasting levels are between 80 and 100 mg/dL). The boy is shown to have no detectable C-peptide upon further testing.

A potential reason for the elevated blood glucose is which one of the following?
(A) A lack of a sodium gradient across cellular membranes
(B) A lack of a calcium gradient across cellular membranes
(C) A lack of a chloride gradient across cellular membranes
(D) A reduced number of glucose transport molecules in the brain membrane
(E) A reduced number of glucose transport
molecules in the muscle membrane
(F) A reduce number of glucose transport
molecules in the liver membrane

Answer 13

The answer is E. The boy has Type 1 diabetes, and is producing no insulin. One of insulin’s effects is to stimulate the translocation of GLUT4 transporters from internal vesicles to the plasma membrane of muscle and fat cells. The increase in glucose transport molecules on the cell surface is important for rapidly reducing blood glucose levels. The GLUT4 transporter is for facilitative diffusion, and is not dependent on an ion gradient across the membrane for effective transport, as are the glucose transporters in the small intestine.

Question 14

A 12-year-old boy is admitted to the hospital in ketoacidosis with a blood glucose level of 700 mg/dL (normal fasting levels are between 80 and 100 mg/dL). The boy is shown to have no detectable C-peptide upon further testing.

The child is treated appropriately such that the glucose levels have been reduced, and he does not become dehydrated. Once glucose is transported into his cells, which organelle is responsible for generating energy from the oxidation of glucose to carbon dioxide and water?
(A) Lysosome
(B) Golgi complex
(C) Mitochondria
(D) Nucleus
(E) Peroxisome

Answer 14

The answer is C. The mitochondria are organelles of fuel oxidation and ATP generation. Lysosomes contain hydrolytic enzymes that degrade proteins and other large molecules. The Golgi form vesicles for transport of molecules to the plasma membrane and for secretion. The nucleus carries out gene replication and transcription of DNA.

Question 15

A person is diagnosed with group A streptococcal bacteremia. One of the body’s major defenses in this type of disease is for eosinophils to phagocytize the bacteria. Once internalized, the bacteria are destroyed by fusing the phagosome with a particular intracellular organelle. Which one of the following would destroy the activity of that organelle such that the bacteria would not be incapacitated?
(A) Inhibiting sodium–potassium ATPase activity
(B) Interfering with mitochondrial protein synthesis
(C) Blocking transport through nuclear pores
(D) Inhibiting a proton-translocating ATPase
(E) Inhibiting a calcium-activated ATPase

Answer 15

The answer is D. The phagosomes fuse with lysosomes, where the acidity and digestive enzymes within the lysosomes destroy the contents of the phagosome (in this case, the bacteria within the phagosome). The digestive enzymes have a pH optimum of 5.5, which is maintained within the lysosome through the actions of a proton pump (a proton-translocating ATPase activity). The nucleus and mitochondria are not involved in the lysosomal digestion of phagosome contents. Blocking either the sodium ATPase, potassium ATPase, or a calcium-activated ATPase will greatly affect other organs, but will not affect the ability of the lysosome to degrade its contents.

Question 16

Lysosomal hydrolases are targeted to the lysosome by the addition of a carbohydrate residue to the protein. An inability to add this carbohydrate leads to a disease in which the lysosomal hydrolases are treated as secreted proteins, and are exported from the cell, rather than taken to the lysosomes. The secreted pro- teins will have which one of the following effects on the cells and proteins in the circulation?
(A) The blood cells will have their membrane proteins digested.
(B) The blood cells will have their carbohydrates on the cell surface removed.
(C) The blood cell membranes will become
leaky, leading to the death of the blood
cells.
(D) Circulating proteins will be degraded,
whereas the blood cells will be protected
against the enzymes.
(E) Circulating proteins will be targeted to the
spleen for removal.
(F) There will be no effect on the proteins and
cells in the circulation.

Answer 16

The answer is F. Most lysosomal hydrolases have their highest activity near an acidic pH of 5.5 (pH optimum) and little activity in a neutral or basic environment. The intralysosomal pH is maintained near 5.5 by vesicular ATPases, which actively pump protons into the lysosome. The cytosol and other cellular components have a pH near 7.2, and are therefore protected from escaped hydrolases. The pH of the blood is maintained between 7.2 and 7.4, so the escaped lysosomal enzymes will have no activity at that pH, and will not affect the proteins and cells
in the circulation. I-cell disease results from the inability to appropriately target lysosomal proteins, and it is a lysosomal storage disease.

Question 17

A 25-year-old female presents with intense fear whenever she has to drive her car through a tunnel. She feels faint, sweats profusely, has palpitations, and hyperventilates. She is prescribed diazepam to reduce her symptoms. The type of chemical messenger enhanced by this treatment is best described as which one of the following?
(A) Neuropeptide
(B) Biogenic amine
(C) Large-molecule neurotransmitter
(D) Cytokine
(E) G-protein

Answer 17

The answer is B. The patient is experiencing an anxiety disorder and panic attacks. These symptoms are often treated with psychotherapy and benzodiazepams. Patients with anxiety disorders have low gamma-aminobutyric acid (GABA). Benzodiazepams, such as diazepam, increase the efficiency of the synaptic transmission of GABA, helping to make any existing GABA more efficacious. This is through the drug binding to GABA receptors such that when GABA binds to the receptor, the response to GABA is greater than in the absence of the drug (one effect is to leave chloride channels open for greater periods of time in response to GABA, thereby depolarizing the membrane and sending an inhibitory signal). GABA is the chief inhib- itory neurotransmitter. GABA is a biogenic amine or “small-molecule” neurotransmitter, and is derived from the decarboxylation of glutamate. Neuropeptides are the other type of chemical messenger secreted by the nervous system, and are usually small peptides. Cytokines are small protein messengers of the immune system. G-proteins aid in transmitting the signals induced by proteins that bind to heptahelical receptors (such as the epinephrine or glucagon receptors). GABA does not transmit its signal through a G-protein.

Question 18

A 15-month-old girl has been given an MMR immunization. Which of the following chemical messengers is responsible for the body’s ability to mount an immune response to this vaccination?
(A) Neuropeptides
(B) Biogenic amines
(C) Steroid hormones
(D) Cytokines
(E) Amino acids

Answer 18

The answer is D. Cytokines are the messengers of the immune system. Neuropeptides and biogenic amines (small-molecule neurotransmitters) are messengers of the nervous system. Steroid hormones are messengers of the endocrine system. Amino acids (such as glycine and glutamate) can act as mediators within the nervous system. Once the shot is given, immune cells secrete cytokines to induce the synthesis of antibodies against the antigens injected into the girl.

Question 19

A 30-year-old female presents with a 15-pound weight loss over 1 month, heat intolerance, tachycardia, tremor, bilateral exophthalmos, and a mass in the anterior neck. The hormone overproduced in this condition requires which one of the following?
(A) Arachidonic acid
(B) Cholesterol
(C) Tyrosine
(D) Tryptophan
(E) Glutamate

Answer 19

The answer is C. The patient has hyperthyroidism, or Grave disease, an overproduction of thyroid hormone. Thyroid hormone is derived, in part, from tyrosine, which is iodinated to produce the active forms of thyroid hormone, T3 and T4. Cholesterol is a precursor to ste- roid hormones. Arachidonic acid is a precursor to eicosanoids (prostaglandins). Tryptophan is a necessity in the production of serotonin, and glutamate is needed to produce GABA. The symptoms described do not occur if there is an overproduction of steroid hormones, eicosanoids, serontonin, or GABA.

Question 20

A 3-year-old boy presents with 3 days of a low-grade fever, joint pain, and a “lacy-”appearing rash on his arms and legs. His rash began on his face and he appeared to have “slapped cheeks.” The chemical messenger that caused the symptoms (vasodilatation presenting clinically as a “rash”) can be classified as which one of the following?
(A) Cytokine
(B) Neuropeptide
(C) Eicosanoid
(D) Steroid hormone
(E) Amino acid

Answer 20

The answer is C. The patient has Fifth disease, a viral illness caused by parvovirus B19. The “slapped cheek” appearance of this rash is very distinctive. The eicosanoids control cellular function in response to injury (in this case, a viral infection). In response to the viral infection, vascular endothelial cells will secrete prostaglandins that act on smooth muscle cells to cause vasodilation, which leads to the reddish appearance of the infected individual. The release
of eicosanoids may also be responsible for the fever that sometimes accompanies Fifth dis- ease. Neuropeptides, cytokines, steroid hormones, or amino acids are not responsible for the vasodilation that occurs in this disease.

Question 21

A 62-year-old male has a reddish, rough patch with white scales on the top of his ear. He does not get this treated, and 3 years later it has become an enlarged, raised lesion with a central ulcerated area that will not heal. Of the following, which is the most likely causative factor for this malignancy?
(A) Creation of pyrimidine dimers
(B) Creation of hydroxyl radicals
(C) Oncogenic RNA virus
(D) TNF receptor mutation
(E) Double-strand breaks in the DNA

Answer 21

The answer is A. This man originally displayed an actinic keratosis that has, over the intervening 3-year period, become a squamous cell carcinoma. Actinic keratosis develops in areas of the skin that are frequently exposed to sunlight, such as the top of the ear. The frequent exposure to UV light led to the creation of pyrimidine dimers in the DNA. If the cells cannot repair the DNA damage rapidly enough, cancerous changes do occur over time. The presentation of actinic keratosis may represent a precancerous state toward squamous cell carcinoma. Removal of the actinic keratosis would have prevented the development of the tumor. Hydroxyl radicals are created by ionizing radiation such as X-rays, not by sunlight. Oncogenic RNA viruses such as HTLV-1 could cause lymphomas or leukemias, but have not been implicated in squamous cell carcinoma. TNF receptor mutations can occur in immune system cells, leading to apoptosis, which would lead to an immune defect, but not the cancer observed. UV light does not lead to the creation of double-strand breaks in DNA.

Question 22

A 32-year-old female has developed breast cancer. Her mother and one maternal aunt had breast cancer and her maternal grandmother had ovarian cancer. Which of the following best describes the mechanism behind this inherited problem?
(A) A tumor suppressor leading to loss of apoptosis
(B) A tumor suppressor leading to an inability to repair DNA
(C) A tumor suppressor leading to a constitutively active MAP kinase pathway
(D) An oncogene leading to loss of apoptosis
(E) An oncogene leading to an inability to re-
pair DNA
(F) An oncogene leading to a constitutively
active MAP kinase pathway

Answer 22

The answer is B. Hereditary breast cancer is due to inherited mutations in either of the tumor suppressor genes BRCA1 or BRCA2. These genes encode proteins that play important roles in DNA repair (primarily single- and double-strand breaks), and it is the loss of this function that predisposes the patient to breast and ovarian cancers. The inability to repair the breaks in the backbone leads to errors during replication, and mutations will develop that eventually lead to a loss of growth control. This is a loss-of-function disorder, which characterizes the genes involved as tumor suppressors. Inheriting one mutated copy of BRCA1 means that the other, normal copy of BRCA1 must be lost in a particular cell in order to initiate the disease (loss of heterozygosity). For breast cancer, this occurs 85% of the time (penetrance upon inheriting a BRCA1 or BRCA2 mutation). An oncogene is a dominant gene, so only one mutated copy can bring about the disease. BRCA1 or BRCA2 mutations do not directly lead to a loss of apoptosis, or to a constitutively active MAP kinase pathway.

Question 23

A 4-year-old boy went to the beach with his parents, and they found some clams, which they later ate for dinner. A few hours later, the boy developed a fever, started vomiting, and had profuse watery diarrhea. After being taken to the emergency room (ER), the boy was treated for potential dehydration, and recov- ered uneventfully. The root molecular cause of his symptoms was which one of the following?
(A) ADP-ribosylation of a Gαs protein
(B) Phosphorylation of a Gαs protein
(C) Acetylation of a Gαs protein
(D) ADP-ribosylation of a Gαi protein
(E) Phosphorylation of a Gαi protein
(F) Acetylation of a Gαi protein

Answer 23

The answer is A. Cholera is caused by Vibrio cholerae, found in fecally contaminated food or water and in shellfish. Cholera toxin, which is composed of multiple subunits, utilizes some subunits to allow one particular subunit with enzymatic activity to enter the intestinal epithelial cell. This toxin catalyzes the ADP-ribosylation of Gαs, inhibiting the GTPase activity
of the α subunit of the G-protein, leading to constitutive activation of adenylate cyclase, and high cAMP levels. This leads to the activation of ion channels, having potassium, sodium, and chloride ions leave the intestinal epithelial cells into the lumen, along with water, leading to the watery diarrhea. The treatment consists of rehydration with electrolytes. Owing to the volume of water lost, the disease is usually self-limiting, as the bacteria causing the disorder are washed out of the intestine.

Question 24

A 4-year-old boy has had multiple episodes of pneumonia, steatorrhea, and has fallen off his normal growth curve. A sweat test was positive for chloride ions. The reason this boy is at risk for repeat episodes of pneumonia is which one of the following?
(A) Elastase destruction of lung cells
(B) Defective α1-antitrypsin activity
(C) Excessive water in the lungs
(D) Dried mucus accumulation in the lungs
(E) Loss of lung cells due to a defect in DNA
repair

Answer 24

The answer is D. The boy is exhibiting the symptoms of cystic fibrosis, which is due to a mutation in the CFTR. The CFTR is required for chloride transport across the membrane, is activated by phosphorylation by the cAMP-activated protein kinase, and when activated allows chloride to flow down its electrochemical gradient. A defective CFTR also alters the ion composition of mucus, reducing its ability to absorb water through osmosis, leading to the drying of mucus in various ducts and tissues, including the lung cells. The lung cells normally secrete a thin, watery mucus designed to trap small particles, which are moved through the lung so they can be swallowed or removed by coughing. When water cannot leave the lung cells, the mucus dries out, leading to pulmonary dysfunction due to clogged bronchi.

Question 25

An experimental drug has been added to a eukaryotic cell, and while the drug was designed to interfere with a membrane transport process, the investigators found that in cells treated with the drug the lysosomes quickly turn into inclusion bodies. None of the mate- rial directed to the lysosomes for removal was being digested in the lysosome, and remained intact inside the organelle. An analysis of lysosomal contents in drug-treated cells indicated that the full complement of lysosomal enzymes were present in the organelle. Assuming that the drug is targeting just one protein, which one of the following proteins is most likely the target?
(A) An outer membrane protein that allows the lysosomal membrane to become permeable to small molecules
(B) A proton-translocating ATPase in the lysosomal membrane
(C) A chloride pump in the lysosomal membrane
(D) The enzyme that adds mannose- 6-phosphate to lysosomal enzymes
(E) The mannose-6-phosphate receptor

Answer 25

The answer is B. Lysosomes contain a single membrane (so there is no outer membrane, such as in mitochondria) that contains a proton-translocating ATPase. The ATPase will concentrate protons inside of the lysosome, at the expense of ATP hydrolysis, to acidify the intraorganelle pH such that the lysosomal enzymes will be active. If the intravesicular pH cannot be lowered, the digestive enzymes will be inactive, and no digestion will take place. Targeting a chloride pump in the lysosomal membrane will not affect the activity of the lysosomal enzymes. If the lysosomal enzymes were not marked with a mannose-6-phosphate residue in the Golgi apparatus, they would not be able to bind to the mannose-6-phosphate receptor to be targeted to the lysosomes. If the drug altered either of those proteins (the enzyme responsible for adding the mannose-6-phosphate or the mannose-6-phosphate receptor), then the lysosomal enzymes would not be in the lysosomes, which is not the case. Such a drug would bring about the symptoms of I-cell disease.

Question 26

A 5-year-old boy begins to regress in terms of developmental milestones, particularly neurologically. Shortly thereafter, the child enters a coma, and dies 2 years into the coma. Upon autopsy, the myelin sheath in the brain was found to be abnormal, as it contained a large quantity of very long-chain fatty acids in its phospholipids. The adrenal glands were also abnormal in appearance. The child, at the molecular level, had inherited a mutation that led to an inability to catalyze reactions that occur in which one of the following intracellular organelles?
(A) Lysosomes
(B) Nucleus
(C) Mitochondria
(D) Peroxisomes
(E) Golgi apparatus
(F) Nucleolus

Answer 26

The answer is D. The child has the symptoms of X-linked adrenoleukodystrophy, which is an X-linked disorder with a mutation in the ABCD1 gene. The ABCD1 gene is required for the transport of very long-chain fatty acids into the peroxisome for catabolism. In the absence of this activity, the very long-chain fatty acids accumulate, become incorporated into phospholipids, and alter the structure of myelin, leading to the neurological disorders observed. The lysosomes, nucleus, and Golgi apparatus are not involved in very long-chain fatty acid oxidation. The nucleolus is found in the nucleus and is the site of ribosome formation. Mitochondria oxidize fatty acids, but not when they are very long-chain fatty acids (greater than 20 carbons). In those cases, the initial steps of oxidation occur in the peroxisome, and when the chain length has been reduced, the partially oxidized fatty acid is transferred to the mitochondria to finish the oxidation of the compound.

Question 27

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses.

A drug that may help to stabilize this condi- tion would do which one of the following?
(A) Stimulate the production of immune cells
(B) Stimulate the production of epinephrine
(C) Inhibit the production of acetylcholine
(D) Inhibit acetylcholinesterase
(E) Stimulate catechol-O-transferase

Answer 27

The answer is D. The woman has myasthenia gravis, which is due to autoantibodies directed against the acetylcholine receptor. As such, acetylcholine stimulation of muscle cells is de- creased, owing to a reduced number of functional acetylcholine receptors at the neuromuscu- lar junction. One way to treat this condition is to inhibit acetylcholinesterase, the enzyme that degrades acetylcholine at the neuromuscular junction. By keeping the levels of acetylcholine high at the junction, there is a greater probability that the receptors that are active are occupied, and the signal from the neuron can be transmitted. Inhibiting the production of acetylcholine would exacerbate the problem, as would stimulating the production of immune cells (more autoantibodies would potentially be generated). Epinephrine is not involved at the neuromuscular junction, and stimulation of catechol-O-transferase is a mechanism to inhibit the action of catecholamines in nonneuronal tissues, and does not contribute to the progression of myasthenia gravis.

Question 28

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses.

In addition to the answer to the previous question, a drug that may help to stabilize this condition would do which one of the following?
(A) Stimulate apoptosis
(B) Inhibit apoptosis
(C) Stimulate cell growth
(D) Inhibit cell growth
(E) Induce muscle growth
(F) Inhibit muscle growth

Answer 28

The answer is A. The woman has myasthenia gravis, which is due to an autoimmune disorder in which antibodies directed against the acetylcholine receptor block the ability of acetylcholine to stimulate the muscle cells at the neuromuscular junction. Immunosuppressants can be taken to reduce the autoantibody production. Such drugs work, in part, through the activation of the tumor necrosis factor receptor, which activates apoptosis in the cells, leading to their destruction. Inhibiting apoptosis would exacerbate the problem, as the antibody-producing cells would survive longer and continue to produce the antibodies directed against the acetylcholine receptor. Drugs affecting the muscle would not help with this disorder, as it is a problem unique to the acetylcholine receptor expressed on the muscle surface. Stimulation or inhibition of cell growth does not stop the antibody-producing cells from continuing to make antibodies, and would not be an effective drug target for this disease.

Question 29

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses.

In order to diagnose the disease the patient is experiencing, a Western blot was run using the patient’s sera as the source of antibodies. The protein run on the gel would need to be which one of the following?
(A) Acetylcholinesterase
(B) Acetylcholine receptor
(C) Epinephrine receptor
(D) Catechol-O-methyltransferase
(E) Glucocorticoid receptor
(F) HMG-CoA reductase

Answer 29

The answer is B. The woman has myasthenia gravis, which is due to the presence of autoim- mune antibodies directed against the acetylcholine receptor in the neuromuscular junction. To confirm the diagnosis by Western blot, a sample of acetylcholine receptor would be run through a polyacrylamide gel, the protein transferred to filter paper, and the filter paper incubated with the patient’s sera. If the sera contain antibodies that bind to the acetylcholine receptor, the antibodies will be bound to the filter paper, and then visualized using a secondary antibody linked to a reporter enzyme. Controls would be done to indicate that sera from an individual who did not have myasthenia gravis did not allow for the formation of a band on the Western blot. Running acetylcholinesterase, the epinephrine receptor, catechol-O-methyltransferase, the glucocorticoid receptor, or HMG-CoA reductase on the gel would not allow detection of antibodies against the acetylcholine receptor in the patient’s blood sample.

Question 30

A 12-year-old boy is displaying tiredness and lethargy, and is found to have a hypochro- mic, microcytic anemia. Microscopic examina- tion of the boy’s red blood cells demonstrated a spherical shape, rather than concave. The mu- tation in this child is most often found in a pro- tein located in which part of the red blood cell?
(A) The cytoskeleton
(B) The nucleus
(C) The mitochondria
(D) The endoplasmic reticulum
(E) The plasma membrane

Answer 30

The answer is A. The boy has hereditary spherocytosis, which is due to a mutation in a red blood cell cytoskeletal protein. The most common mutation is in spectrin, although mutations in ankyrin, band 3, and protein 4.2 can also lead to this phenotype. Owing to the mutation in the cytoskeletal protein, the membrane shape becomes spherical instead of concave. This leads to the removal of the spherical cells by the spleen, leading to both anemia and splenomegaly. Mutations in the proteins in the plasma membrane or the endoplasmic reticulum will not lead to this disorder. Red blood cells do not have a nucleus or mitochondria.

Question 31

A 39-year-old man is brought to the ER for a suspected suicide attempt. He has blurred vi- sion; very dry, hot, red skin; dry mouth; urinary retention; confusion; hallucinations; loss of balance; and tachycardia. Emergency medical technicians (EMTs) found an empty bottle of amitriptyline in his apartment. The date on the bottle was just 1 week ago, yet all the pills were missing. The effects the man is experiencing is due to an inhibition of which of the following processes?
(A) Muscarinic acetylcholine receptor signaling
(B) Nicotinic acetylcholine receptor signaling
(C) GABA signaling
(D) Serotonin signaling
(E) Catecholamine signaling

Answer 31

The answer is A. Many classes of drugs, including some antihistamines (e.g., Benadryl), some antipsychotics (e.g., olanzapine), tricyclic antidepressants (e.g., amitriptyline), and atropine- like drugs (e.g., atropine, scopolamine) have anticholinergic effects or side effects, and function as antagonists to the acetylcholine receptor. Muscarinic acetylcholine receptors act through G-protein activation, whereas nicotinic acetylcholine receptors act as an ion channel, allowing sodium to flow through the receptor once it has been activated. The drug overdose
in this case is inhibiting the muscarinic receptors, which occur in the autonomic and central nervous symptoms. This is a very typical case of anticholinergic overdose, with the “classic” symptoms classified as “blind as a bat,” “dry as a bone,” “red as a beet,” “mad as a hatter,” and “hot as a hare.” The overdose is not affecting the nicotinic acetylcholine receptors, or receptors for GABA, serotonin, or catecholamines.

Question 32

A 21-year-old patient is being evaluated for a major depressive disorder. During the interview, he admits to having several episodes in the past
of feeling “on top of the world,” able to function very well with only 4 hours of sleep per night, maxing out his credit cards (a very unusual char- acteristic for him), and “indiscriminate” sexual encounters with multiple partners. The elemen- tal medication most commonly used to treat this patient’s disorder will lead to the accumulation of which one of the following compounds?
(A) Inositol
(B) Phosphatidylinositol
(C) Inositol phosphate
(D) Inositol bisphosphate
(E) Inositol trisphosphate
(F) Diacylglycerol

Answer 32

The answer is C. The patient has presented with classic symptoms of bipolar disorder. Lithium is a first-line treatment of bipolar disorder whose mechanism of action is to interrupt the PI cycle by blocking the action of inositol monophosphatases, the enzyme that converts inositiol phosphate to free inositol, such that phosphatidylinositol can be resynthesized from CDP-diacylglycerol and inositol. Through an interruption in the cycle, the key PI-cycle second messengers cannot be continually generated, leading to a reduction in signaling capabilities.

Question 33

On a routine newborn exam, it is noted that the red reflex is absent in one eye. An MRI shows a tumor blocking the retina. Regulation at which phase of the cell cycle would be af- fected by the mutation that leads to this tumor?
(A) G0 to G1
(B) G1 to S
(C) S to G2
(D) G2 to M
(E) M to G1
(F) G1 to G0

Answer 33

The answer is B. The child has hereditary retinoblastoma, which is due to an inherited mutation in the rb gene. As the rb gene is a tumor suppressor gene, once loss of heterozygosity occurs, the function of rb in the cell cycle is lost. Rb helps to regulate the E2F family of transcription factors. Once cyclin D is synthesized, and activates a pair of CDKs, rb protein is phosphorylated, which causes it to leave a complex with the E2F factors. The removal of rb from the protein complex activates the E2F proteins, which initiate new gene transcription to allow the cell to transition to the S phase of the cell cycle. In the absence of any functional rb gene product, the transition to S phase is unregulated, and occurs continuously, leading to tumor growth. The rb gene product is not required for any other checkpoints in the cell cycle.

Question 34

A 45-year-old man presents with blood in his stool. Workup reveals a stage 3 (Dukes 3) colon carcinoma, with multiple polyps within the colon. A family history reveals that his father and grandfather both had colon cancer in their fifth decade of life. A potential initiating activating event in the development of this tumor is which one of the following?
(A) Loss of β-catenin activity
(B) Activation of β-catenin activity
(C) Loss of transcription factor myc activity
(D) Activation of Bcl-2 activity
(E) Loss of cyclin expression
(F) Gain-of-cyclin expression

Answer 34

The answer is B. The patient has hereditary colon cancer, specifically adenomatous polyposis coli, which presents in the fourth or fifth decade of life, with multiple polyps lining the lumen of the colon. The defective protein is APC, which regulates β-catenin activity. The loss of APC function leads to inappropriately activated β-catenin, which is a transcription factor and can stimulate the expression of myc and cyclin D1, to promote cell growth. The inappropriate ex- pression of myc and cyclin D1 due to the loss of APC is an initiating event in tumorigenesis. The APC mutation does not affect BCl-2 activity, which is an antiapoptotic activity. While the loss of APC activity will lead to a gain-of-cyclin expression, that gain is due to the activation of β-catenin, which would be the initiating event for tumor formation.

Question 35

A 29-year-old female presents with chronic fatigue for the past 9 months. She did have mononucleosis in the past, and blood work reveals a chronic viral infection. An analysis
of a liver biopsy indicated that when placed under conditions in which apoptosis should be initiated, the cells continued to grow. The viral infection was most likely caused by which one of the following?
(A) Epstein–Barr virus
(B) Influenza virus
(C) Simian sarcoma virus
(D) Polio virus
(E) Herpes simplex virus

Answer 35

The answer is A. Infection by the Epstein–Barr virus will lead to the synthesis of a Bcl-2-like factor that antagonizes apoptosis, and allows the virus-infected cells to survive and continue producing more viruses. This factor is not present in the other viruses listed as potential answers.