USMLE Full Length Exam 1
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Questions and Answers
- 1.
A new patient presents to a clinic in Denver, Colorado with complaints of dyspnea and fatigue on exertion. On history, the patient informs the doctor that she just moved from the California coastline less than 1 week ago, and has been attempting to do her 3-mile run every morning. The patient is told to limit her running until her body adjusts to the new altitude. Which of the following will occur in this woman’sbody in response to the high altitude?
- A.
An increased partial oxygen pressure at 50% saturation on the hemoglobin-oxygen dissociation curve
- B.
Decreased pulmonary vascular resistance
- C.
Decreased RBC 2,3-diphosphoglycerate concentration
- D.
Increased arterial partial oxygen pressure
- E.
Right ventricular atrophy
Correct Answer
A. An increased partial oxygen pressure at 50% saturation on the hemoglobin-oxygen dissociation curveExplanation
The correct answer is A. In response to high altitude, the hemoglobin-oxygen curve shifts right (to release oxygen more easily at the tissues). The P50 (the partial oxygen pressure) at 50% saturation of hemoglobin) will increase.
Answer B is incorrect. Pulmonary vasoconstriction (hypoxic vasoconstriction) is a result of hypoxemia(decreased arterial partial oxygen pressure), thus increasing pulmonary arterial pressure (i.e., pulmonary vascular resistance).
Answer C is incorrect. 2,3-Diphosphoglycerate (2,3-DPG) binds to the β chains of deoxyhemoglobin, decreasing the affi nity of hemoglobin for oxygen. An increase in 2,3-DPG shifts the hemoglobin-oxygen dissociation curve to the right, which is what happens in response to high altitudes.
Answer D is incorrect. At increased altitudes, barometric pressure decreases, which decreases alveolar partial oxygen pressure. As a result, arterial partial oxygen pressure is decreased hypoxemia).
Answer E is incorrect. As a result of increased pulmonary vascular resistance, there is increased work on the right side of the heart, resulting in hypertrophy of the right ventricle to counteract the increased afterload.Rate this question:
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- 2.
A 32-year-old woman with pheochromocytoma is being treated with phenoxybenzamine. After surgical excision of the tumor, the patient has an episode of hypotension requiring 30 seconds of cardiopulmonary resuscitation and subsequent treatment in the intensive care unit. The attending physician asks his intern what physiologic responses he would expect to see if the patient had been given epinephrine during resuscitation. What would have been observed following administration of epinephrine?
- A.
Decrease in blood pressure
- B.
Decrease in heart rate
- C.
Increase in blood pressure
- D.
Increase in respiratory rate
- E.
No changes in vital signs
Correct Answer
A. Decrease in blood pressureExplanation
The correct answer is A. Phenoxybenzamine is a nonselective α-antagonist that will block both α1- and α2-receptors. In this patient, the administration of high-dose epinephrine (which is both an α- and a β-agonist) would result in unopposed β1- (increased heart rate, increased contractility) and β2-(vasodilation, bronchodilation) agonist effects because the α-effects of epinephrine are blocked by prior phenoxybenzamine administration. The net effect will be β-agonist effects, including an increase in heart rate and a decrease in blood pressure.
Answer B is incorrect. Unopposed β-agonist effects will cause an increase in heart rate because β1-agonists result in increased heart rate.
Answer C is incorrect. Unopposed β-agonist effects will cause a decrease in blood pressure because β2-agonists result in vasodilation.
Answer D is incorrect. Unopposed β-agonist effects do not have a signifi cant effect on respiratory rate.
Answer E is incorrect. Unopposed β-agonist effects will cause changes in both blood pressure and heart rate.Rate this question:
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- 3.
A 35-year-old man with no signifi cant past medical history presents to his primary care physician complaining of shortness of breath on exertion for the past several months. The patient acknowledges recent heart palpitations, but denies chest pain, cough, lower extremity edema, paroxysmal nocturnal dyspnea, or weakness. He further denies any recent illness, and he states that he does not smoke. Cardiac examination shows an irregularly irregular rhythm, a widely split fixed S2, and a midsystolic ejection murmur over the left upper sternal border. ECG reveals that the patient is in atrial fibrillation. Based on these findings, the physician concludes that these symptoms are due to pathology of which of the following fetal structures?
- A.
Aorticopulmonary septum
- B.
Ductus arteriosus
- C.
Ductus venosus
- D.
Foramen ovale
- E.
Interventricular septum
Correct Answer
D. Foramen ovaleExplanation
The correct answer is D. The clinical picture is consistent with a patent foramen ovale (PFO), in which the foramen ovale fails to fully close at birth. A small PFO may go undetected well into adulthood, and it is the most commonly found congenital heart disease in adults. Atrial fi brillation, characterized by an irregularly irregular pulse and the sensation of palpitations, commonly accompanies PFO. A stroke in a young patient is another possible manifestation of PFO, as venous thrombi are able to bypass the pulmonary circulation via the PFO and lodge in the cerebral vasculature as “paradoxical emboli.”
Answer A is incorrect. The aorticopulmonary septum forms the division between the pulmonary trunk and the aorta, which are both derived from the truncus arteriosus. Pathology of this septum is associated with transposition of the great vessels and tetralogy of Fallot. Both of these conditions cause early cyanosis and are present in infancy.
Answer B is incorrect. When the ductus arteriosus fails to close at birth, this produces a patent ductus arteriosus (PDA). While this can cause exertional dyspnea, it is typically discovered much earlier in life and produces a continuous “machine-like” murmur, not a purely systolic murmur. PDA is associated with maternal rubella infection during pregnancy and with premature birth. There is no association with atrial fi brillation.
Answer C is incorrect. The ductus venosus shunts blood from the portal vein to the inferior vena cava and thus is not associated with cardiac pathology.
Answer E is incorrect. The interventricular septum divides the fetal primitive ventricle into right and left ventricles. Pathology of this structure produces a membranous ventricular septal defect (VSD). A VSD produces exertional dyspnea and is usually discovered during infancy. VSD is characterized by a harsh pansystolic murmur heard best over the left lower sternal border.Rate this question:
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- 4.
A 10-year-old boy is referred to the neurologist with intellectual deterioration, personality changes, generalized seizures, and visual disturbances that have worsened over the last few months. The patient’s cerebrospinal fluid culture shows no bacterial growth. Further analysis shows normal glucose levels and normal protein. The patient is afebrile and reports no headache. The child’s parents say that he has not received any vaccinations since arriving in the United States last year. They also say that he has had only one major illness prior to this. The child was approximately 2 years old when he developed a high fever, cough, and runny nose. Soon after the onset of these symptoms, he developed a red maculopapular rash that spread downward from his head. Antibodies against which of the following are likely to be found in this patient’s cerebrospinal fluid?
- A.
Herpes simplex virus type 2
- B.
Measles virus
- C.
Mumps virus
- D.
Neisseria meningitidis
- E.
Rubella virus
- F.
Treponema pallidum
Correct Answer
B. Measles virusExplanation
The correct answer is B. This patient is most likely suffering from subacute sclerosing panencephalitis. This is a rare progressive demyelinating disease associated with chronic central nervous system infection with measles virus. There is often a history of primary measles infection at an early age (approximately 2 years) followed by a latent interval of 6–8 years. Initial manifestations include poor school performance and mood and personality changes. Fever and headache do not occur. As the disease progresses, patients develop progressive intellectual deterioration, focal and/or generalized seizures, myoclonus, ataxia, and visual disturbances. The cerebrospinal fluid (CSF) is acellular with normal or mildly elevated protein and markedly elevated gamma globulin (>20% of total CSF protein). CSF anti-measles antibodies are elevated. CT and MRI show evidence of multifocal white matter lesions, cortical atrophy, and ventricular enlargement.
Answer A is incorrect. Herpes simplex virus 2 (HSV-2) can cause a recurrent meningitis. As with the other examples, one would expect to see signs of meningeal irritation as well as an increase in lymphocytes. Infection with HSV-2 is often associated with genital lesions.
Answer C is incorrect. The patient has no evidence of current mumps infection. Mumps virus can cause acute viral meningitis, but one would expect to see classic signs of meningitis as well as an increase in lymphocytes in the cerebrospinal fluid.
Answer D is incorrect. Neisseria meningitidis can cause bacterial meningitis. These organisms would likely be discovered on culture of the CSF. Bacterial meningitis would also manifest in a high fever with meningeal signs (headache, nuchal rigidity) as well as decreased CSF glucose, increased CSF protein, and mononuclear and/or polymorphonuclear cells.
Answer E is incorrect. Rubella virus causes German measles, which is generally characterized by fever and upper respiratory symptoms that resolve with subsequent rash. The maculopapular rash usually starts with the face and descends to the extremities, lasting only several days.
Answer F is incorrect. Infection with Treponema pallidum can eventually lead to neurosyphilis, which can include some of the symptoms described in this case. The patient has neither elevated CSF protein nor the presence of mononuclear cells. There is also little evidence in the history of prior or current infection with T. pallidum.Rate this question:
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- 5.
A 34-year-old man comes to the clinic because his gums have become swollen and have exhibited a tendency to bleed. He states that he has been brushing his teeth at least twice a day. On examination, he is found to have several bruises on his legs in different stages of healing. The patient adds that his bruises have been taking longer than usual to heal. His complete blood cell count and coagulation panel are within normal limits. What is the likely cause of this patient’s symptoms?
- A.
Decreased platelet count
- B.
Decreased von Willebrand factor
- C.
Defect in hydroxylation of collagen residues
- D.
Mutation of dystrophin gene
- E.
Mutation of spectrin
Correct Answer
C. Defect in hydroxylation of collagen residuesExplanation
The correct answer is C. A defect in the hydroxylation of proline and lysine residues of collagen is a result of vitamin C deficiency and is associated with scurvy. Patients with scurvy can present with swollen gums and poor wound healing.
Answer A is incorrect. Thrombocytopenia, or decreased platelet counts, may account for easy bleeding. Thrombocytopenia may be a result of many factors, including decreased platelet production (secondary to viral infections or to chemotherapy or radiation), increased platelet destruction (which may be idiopathic or secondary to disseminated intravascular coagulation, thrombotic thrombocytopenic purpura, or hemolytic uremic syndrome), or distribution (splenomegaly). The characteristic symptom of thrombocytopenia is mucosal or cutaneous bleeding. However, this diagnosis is ruled out by a normal platelet count.
Answer B is incorrect. Von Willebrand’s factor (vWF) plays an important role in primary hemostasis by binding to both platelets and endothelial components, forming an adhesive bridge between platelets and vascular subendothelial structures as well as between adjacent platelets at sites of endothelial injury. It also decontributes to fibrin clot formation by acting as a carrier protein for factor VIII, which has a greatly shortened half-life and abnormally low concentration unless it is bound to vWF. Von Willebrand’s disease is characterized by mutations that lead to impairment in the synthesis or function of vWF. Patients with von Willebrand’s disease have a tendency to bleed, as the disease is associated with an increased partial thromboplastin time and bleeding time, both of which are normal in this patient.
Answer D is incorrect. Dystrophin is a protein that is located on the cytoplasmic face of the plasma membrane of muscle fibers. It functions as a component of a large, tightly associated glycoprotein complex and shields the complex from degradation. Mutations in the dystrophin gene lead to digestion of the glycoprotein complex by proteases. Loss of these membrane proteins may initiate the degradation of muscle fi bers, resulting in muscle weakness characteristic of Duchenne’s muscular dystrophy.
Answer E is incorrect. Spectrin is a protein that ties the skeleton of an RBC to its outer lipid bilayer. Mutations of spectrin can lead to a disease called hereditary spherocytosis. Patients with this RBC membrane defect typically present with hemolytic anemia, jaundice, and splenomegaly.Rate this question:
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- 6.
A 21-year-old woman with no family or personal history of breast cancer presents with a small, firm mass in the lower inner quadrant of her right breast that seems mobile when palpated. It is nontender. There are no overlying skin changes or nipple discharge. Which of the following would most likely be found on biopsy of this mass?
- A.
Blue dome cysts and some atypical epithelial hyperplasia
- B.
Cells in a single fi le formation
- C.
Fibrosing stroma around normal-looking glands
- D.
Large cells with clear “halos”
- E.
Multicentric lobes with lymphocytic infi ltrate
Correct Answer
C. Fibrosing stroma around normal-looking glandsExplanation
The correct answer is C. Fibroadenomas are the most common tumor in young women, presenting as small, fi rm, mobile masses. They are not associated with malignancy. On histology, fibrosing stroma is seen around normal duct and gland structures.
Answer A is incorrect. Blue dome cysts are seen in fibrocystic diseases of the breast. These lesions have associated risks of carcinoma with the presence of atypia.
Answer B is incorrect. Infi ltrating lobular carcinomas are often multilocular and bilateral. These cells are found in clusters or in a linear formation.
Answer D is incorrect. Paget’s disease presents with eczematous skin fi ndings with underlying ductal carcinomas. Paget’s cells are large cells with halo-like clearings.
Answer E is incorrect. Any lymphocytic infi ltrate suggests infl ammatory carcinoma with a poor prognosis.Rate this question:
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- 7.
A 50-year-old woman who works as a secretary comes to the physician because of numbness and tingling in her hands. On examination, the patient is found to have decreased sensation in all of her fingers except her fifth digit. Which of the following muscles is most commonly weakened in patients with this condition?
- A.
Adductor pollicis
- B.
Dorsal interossei
- C.
Lumbricals (3 and 4)
- D.
Opponens digiti minimi
- E.
Opponens pollicis
Correct Answer
E. Opponens pollicisExplanation
The correct answer is E. This woman has carpal tunnel syndrome, which occurs in individuals whose work involves repetitive hand motions. The median nerve is compressed in the carpal tunnel, leading to decreased sensation on the first three and one-half digits and loss of strength of the thumb due to weakness of the abductor pollicis brevis and opponens pollicis. The opponens pollicis is supplied by the median nerve and functions to aid in opposition.
Answer A is incorrect. The adductor pollicis is innervated by a branch of the ulnar nerve and thus would not be affected by carpal tunnel syndrome. It functions to adduct the thumb toward the middle digits.
Answer B is incorrect. Dorsal interossei muscles are innervated by a branch of the ulnar nerve, which functions to abduct the digits.
Answer C is incorrect. The third and fourth lumbricals are innervated by a branch of the ulnar nerve.
Answer D is incorrect. The opponens digitiminimi muscle is innervated by a branch of the ulnar nerve and brings the fifth digit in opposition with the thumb.Rate this question:
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- 8.
A 19-year-old man presents to the emergency department complaining of fatigue, lethargy, and a history of a recent upper respiratory infection. His temperature is 37° C (98.6° F) and his physical examination shows jaundice and prominent splenomegaly. Blood counts show decreased hemoglobin (9 g/dL), elevated mean cell hemoglobin concentration, and increased reticulocyte count. A peripheral blood smear is shown in the image. Which of the following is the definitive choice of therapy for this condition?
- A.
Blood transfusion
- B.
Chemotherapy
- C.
Folic acid supplementation
- D.
Iron chelation therapy
- E.
Iron supplementation
- F.
Splenectomy
Correct Answer
F. SplenectomyExplanation
The correct answer is F. The most likely diagnosis in this patient is hereditary spherocytosis (HS)caused by a defect in cytoskeletal proteins in RBCs, such as spectrin and ankyrin. HS is most commonly inherited as an autosomaldominant disorder and presents with a triad of anemia, jaundice, and splenomegaly. Anemia may be intermittent and can be aggravated by infection or bone marrow suppression. Laboratory indices that are supportive of HS include evidence of hemolytic anemia (elevated reticulocyte count and low hemoglobin), elevated mean cell hemoglobin concentration due to loss of membrane stability, and increased osmotic fragility. HS is most characteristically defined on peripheral blood smear by the presence of spherocytic RBCs. Spherocytes are formed by the activity of splenic reticuloendothelial cells, which remove portions of abnormal membrane from the cytoskeletal defects found on these RBCs. Symptomatic treatment for anemic crises includes folic acid and blood transfusions. In more severe cases, the only treatment for HS anemia is splenectomy. Even following splenectomy, however, spherocytes will still be observed due to the underlying defect in the RBC membrane.
Answer A is incorrect. Both folic acid supplements and blood transfusion are important in treating the symptoms of anemia in hereditary spherocytosis, but they are not curative measures.
Answer B is incorrect. Chemotherapy is not appropriate and would induce a state of aplastic anemia.
Answer C is incorrect. Both folic acid supplements and blood transfusion are important in treating the symptoms of anemia in hereditary spherocytosis, but they are not curative measures.
Answer D is incorrect. Iron chelation therapy would be appropriate in iron-overloaded states, such as in thalassemic patients requiring many transfusions or in hemochromatosis.
Answer E is incorrect. Iron supplementation would be appropriate in iron deficiency anemiaRate this question:
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- 9.
A 22-year-old woman presents to her familyphysician because of increasing fatigue and because she looks “pale” despite spending many hours outside as a camp counselor. She also states that her urine looks “cola-colored” when she first goes to the bathroom in the morning. The patient feels well otherwise. Blood analysis shows a low platelet count, a low RBC count, and a low WBC count. The patient’s RBCs are mixed with acidifi ed normal serum and compared to normal RBCs at room temperature and at 37° C (98.6° F); both temperatures cause the patient’s, but not the normal, RBCs to lyse. Based on this clinical picture and the laboratory tests, this patient most likely has which of the following disorders?
- A.
Alkaptonuria
- B.
Cystinuria
- C.
Hemophilia A
- D.
Maple syrup urine disease
- E.
Paroxysmal nocturnal hemoglobinuria
Correct Answer
E. Paroxysmal nocturnal hemoglobinuriaExplanation
The correct answer is E. Paroxysmal nocturnal hemoglobinuria (PNH) is caused by a defect in synthesis of the cellular anchor used to hold surface proteins to the cell membranes of RBCs, WBCs, and platelets. This defect leads to the clinical manifestations of the disorder: anemia caused by intravascular hemolysis (leading to hemoglobinuria and the darkened urine), thromboses in unusual veins, and hematopoietic deficiencies leading to pancytopenia. The hemolysis occurs all day, but the concentrated urine formed overnight shows an obvious color change. The Ham test (mixing the patient’s RBCs with acidifi ed serum) is used to diagnose PNH. Lysis of the patient’s RBCs indicates PNH.
Answer A is incorrect. Alkaptonuria presents with urine that darkens after exposure to air as a result of alkapton bodies (accumulations of homogentisic acid), as well as darkening of connective tissues. Patients may also have arthralgias.
Answer B is incorrect. Cystinuria is due to a defect in the tubular amino acid transporter in the kidneys. Patients can form cystine kidney stones due to excess cystine in the urine.
Answer C is incorrect. Hemophilia A is an X-linked disease characterized (in moderate to severe deficiency) by spontaneous bleeding, easy bruising into soft tissues, and hemarthrosis into weight-bearing joints (hip, knee, and ankle).
Answer D is incorrect. Maple syrup urine disease is caused by a deficiency of α-ketoacid ehydrogenase. Patients present with lethargy, seizures, failure to thrive, mental retardation, and urine that smells like maple syrup.Rate this question:
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- 10.
A 34-year-old man presents to the emergency department complaining of a 2-day history of fatigue and double vision. Physical examination shows a right nystagmus. A detailed history reveals that he recently began treatment for recurrent tonic-clonic seizures. Laboratory studies show: Na+: 143 mEq/L K+: 4.5 mEq/L Cl-: 103 mEq/L HCO3-: 26 mEq/L Blood urea nitrogen: 45 mg/dL Creatinine: 4.3 mg/dL Which of the following agents is most likely responsible for this patient’s condition?
- A.
Clozapine
- B.
Imipramine
- C.
Lithium
- D.
Phenytoin
- E.
Sumatriptan
- F.
Valproic acid
Correct Answer
D. PhenytoinExplanation
The correct answer is D. Phenytoin toxicity, as with toxicity of many antiepileptic medications, can lead to nystagmus, diplopia, lethargy, and ataxia. It can also lead to tubulointerstitial nephritis, which can cause a sharp increase in creatinine levels.
Answer A is incorrect. Clozapine is an antipsychotic that possesses extrapyramidal and anticholinergic adverse effects.
Answer B is incorrect. Imipramine is a tricyclic antidepressant that can lead to the “3 C’s”: Convulsion, Coma, and Cardiotoxicity.
Answer C is incorrect. Lithium toxicity presents with tremor, polyuria, slurred speech, and poor coordination.
Answer E is incorrect. Sumatriptan is a 5-HT1 agonist used to treat migraines. Sumatriptan toxicity can present with chest discomfort and tingling.
Answer F is incorrect. Valproate, an antiepileptic that works primarily by increasing GABA in the central nervous system, can cause nystagmus. Although it does not cause nephritis, valproate has been known to disrupt the urea cycle, leading to hyperammonemia and cerebral edema. Other adverse effects include rare but severe hepatotoxicity, thrombocytopenia, and alopecia. Valproate can increase phenytoin levels by inhibiting CYP450 metabolism.Rate this question:
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- 11.
A 4-year-old boy has a sublingual mass. A scan using 99mTc pertechnetate, which behaves as iodine and approximates iodine uptake, shows significant uptake in this region with little activity lower in the neck. Which of the following is the embryologic explanation for thismass?
- A.
The third and fourth branchial (pharyngeal) arches have hypertrophied
- B.
The thymus has developed ectopically
- C.
The thymus has hypertrophied
- D.
The thyroid has failed to migrate caudally
- E.
The thyroid has migrated too far rostrally
Correct Answer
D. The thyroid has failed to migrate caudallyExplanation
The correct answer is D. The uptake of 99mTc
pertechnetate (which is captured by thyroid tissue
just as iodine is) in this mass and its sublingual position strongly suggest that it is composed of ectopic thyroid tissue. Normally, the thyroid diverticulum develops from the floor of the primitive pharynx and descends from there into the neck. Therefore, fi nding thyroid tissue still attached to the tongue implies that it has failed to migrate caudally. The tongue is the most common site of ectopic thyroid tissue for this reason.
Answer A is incorrect. The third and fourth branchial (pharyngeal) arches form the posterior third of the tongue. However, the 99mTc pertechnetate uptake in this mass indicates that it is composed of thyroid, and not lingual, tissue.
Answer B is incorrect. The thymus is located in the anterior mediastinum, deep to the sternum. There would be no embryologic explanation for finding thymic tissue in the upper neck. Furthermore, the uptake of 99mTc pertechnetate implies that this mass is composed of thyroid, and not thymus, tissue.
Answer C is incorrect. The thymus is not normally found in the neck; it is instead located in the anterior mediastinum. Thymic hypertrophy would not explain this location. Furthermore, the uptake of 99mTc pertechnetate indicates that this mass is composed of thyroid, and not thymus, tissue.
Answer E is incorrect. The thyroid does not migrate rostrally during development. Instead, it develops near the tongue and migrates caudally (descends) to its normal position in the lower neck.Rate this question:
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- 12.
A 27-year-old healthy man presents because heand his wife have been repeatedly unsuccessful in conceiving a child. His wife has been tested and determined to be fertile. Upon questioning, the patient denies coronary or lipid abnormalities but admits to having multiple sinus infections and a chronic productive cough. Further analysis of his semen shows a normal number of sperm. Which of the following is the most likely etiology for the patient’s infertility?
- A.
Age-related increase in estradiol with possible prostate dihydrotestosterone sensitization
- B.
Autosomal recessive dysfunction of a chloride ion channel
- C.
Failure of testicles to descend into the scrotum
- D.
Familial disease causing early atherosclerosis leading to erectile dysfunction
- E.
Lack of dynein ATPase arms in microtubules of cilia
Correct Answer
E. Lack of dynein ATPase arms in microtubules of ciliaExplanation
The correct answer is E. This patient has Kartagener’s syndrome, which is caused by a lack of dynein arms in microtubules in cilia, rendering them immotile. It results in infertility due to immotile sperm, as well as recurrent sinusitis due to deficient removal of bacteria and other infectious particles. It is also associated with situs invertus, in which the major organs are reversed or mirrored from their original locations.
Answer A is incorrect. Benign prostatic hypertrophy could cause impairment of ejaculation by not allowing semen to be expelled from the body. Because the patient is without an enlarged prostate and is only 27 years old, this diagnosis is highly unlikely.
Answer B is incorrect. Cystic fibrosis does cause infertility, but usually because of bilateral absence of the vas deferens, which would lead to lack of sperm in semen.
Answer C is incorrect. Undescended testicles are associated with infertility and an increased risk of testicular cancer. It is usually found at a very young age and resolves by itself or is surgically corrected before serious complications occur.
Answer D is incorrect. Familial hypercholesterolemia can cause atherosclerosis of the vessels of the male genitalia, causing erectile dysfunction. Without a history of erectile dysfunction or elevated lipid levels, this diagnosis is highly unlikely.Rate this question:
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- 13.
An anxious young woman presents to the emergency department with an acute onset of severe abdominal pain. She states that she “partied a little bit last night” and consumed approximately 8 or 9 alcoholic drinks. She also admits to using diuretics to “lose water weight.” Her stool is guaiac-negative, but she has periumbilical tenderness to palpation. An arterial blood gas study shows that her pH is 7.55 and her bicarbonate level is 21 mEq/L, with a partial pressure of carbon dioxide of 25 mm Hg. Her serum shows normal sodium chloride levels. Which of the following is the origin of her acid-base disturbance?
- A.
A build-up of unmeasured anions due to hepatic metabolism of alcohol
- B.
Electrolyte imbalance due to diuretic use
- C.
Hyperventilation secondary to pain and anxiety
- D.
Hypoventilation due to the respiratory depression caused by alcohol ingestion
- E.
Vomiting due to alcohol toxicity
Correct Answer
C. Hyperventilation secondary to pain and anxietyExplanation
The correct answer is C. According to her lab data, this young woman has an acute respiratory alkalosis. Respiratory alkalosis is caused by a loss of CO2, which is balanced by an increased excretion of HCO3 -. Hence, a low CO2 and low HCO3- level indicate respiratory alkalosis. The key to this question is to recognize that respiratory alkalosis can be caused only by an increase in ntilation, which can be caused by low oxygen (in high altitudes) or by sympathetic stimulation such as anxiety, panic attack, or pain. In this case, the patient is described as anxious and presents with severe glucuronoabdominal pain, which is most likely the result of acute alcohol-induced pancreatitis. Both the anxiety and the pain could be causing her to hyperventilate.
Answer A is incorrect. An increase in anions would be consistent with anion-gap metabolic acidosis. Metabolic acidosis is indicated by the presence of a low pH with a low plasma HCO3- and a low CO2, and an increased anion gap, measured by ([Na+] - [Cl-] - [HCO3-]), which is normally between 10 and 16 mEq/L.
Answer B is incorrect. Diuretic use can cause metabolic alkalosis by volume contraction. This causes the kidney to compensate by reabsorbing sodium and excreting hydrogen ions. A metabolic alkalosis would present with elevated pH, elevated CO2, and elevated HCO3-.
Answer D is incorrect. Hypoventilation causes a reduction in pH due to CO2 retention. This will lead to a respiratory acidosis with a low pH, a high CO2, and a high HCO3-. The compensatory mechanism for respiratory acidosis is an increase in HCO3- retention by the kidneys to normalize the pH.
Answer E is incorrect. Vomiting causes a metabolic alkalosis secondary to the loss of acid and chloride from the stomach. If this were the cause, this patient’s lab results would show a high pH, a high HCO3-, and (with respiratory compensation) a high CO2. The causes of metabolic alkalosis include vomiting, diuretic therapy, and chloride restriction. The compensation for metabolic alkalosis is hypoventilation.Rate this question:
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- 14.
A 38-year-old white woman presents to the physician with a 2-week history of aching painin her left calf that is made worse by dorsiflexion of her foot. On physical examination, her left calf is found to be erythematous, warm, and swollen. Which of the following measures should she take to decrease similar problems in the future?
- A.
Begin taking a bile acid resin
- B.
Begin taking a statin
- C.
Begin taking low-dose oral contraceptives
- D.
Exercise 30 minutes three times per week
- E.
Quit smoking
- F.
Reduce alcohol consumption to one or two glasses of red wine per week
Correct Answer
E. Quit smokingExplanation
The correct answer is E. This patient presents with deep venous thrombosis (DVT). Erythematous,
warm, and tender unilateral calf swelling is classic for DVT. Risk factors for DVT and subsequent pulmonary thromboembolism include Virchow’s triad, which consists of stasis (eg., immobility, obesity, congestive heart failure), endothelial injury (e.g., trauma, surgery, previous DVT), and hypercoagulable state (e.g., pregnancy, oral contraceptive use, coagulation disorders, malignancies, smoking). This patient also has a positive Homans’ sign (calf pain on forced dorsifl exion), which further supports the diagnosis. Not only should this patient be anticoagulated with heparin or warfarin upon presentation, but she should quit smoking to decrease her clotting tendencies.
Answer A is incorrect. Bile acid resins such as cholestyramine and colestipol decrease serum triglycerides and cholesterol, which may indirectly, although not directly, improve vascular health.
Answer B is incorrect. Statins decrease LDL cholesterol but do not affect the rate of DVT formation.
Answer C is incorrect. Oral contraceptives are associated with hypercoagulable state, so they would make DVT more likely.
Answer D is incorrect. Moderate exercise has been linked to improved cardiovascular health and a decreased incidence of acute coronary syndromes, although it is not specifi cally linked to DVT. stasis, however, can make DVT thrombosis more likely.
Answer F is incorrect. Modest alcohol consumption has been associated with improved cardiovascular health, although no specific linkto DVT has been proven.Rate this question:
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- 15.
An 18-year-old woman presents to her physician’s office with symptoms of nausea and vomiting as well as jaundice. She is taking oral contraceptives, but also reports consistent condom use. She denies illicit drug use. On physical examination the patient shows no hepatomegaly or right upper quadrant tenderness to deep palpation. Laboratory studies show a total bilirubin level of 4 mg/dL, direct bilirubin level of 1mg/dL, and indirect bilirubin level of 3 mg/dL. Liver enzyme levels are within normal limits. A direct Coombs’ test is negative. Which of the following is the most likely diagnosis?
- A.
Autoimmune hemolytic anemia
- B.
Crigler-Najjar syndrome type I
- C.
Gilbert’s disease
- D.
Hepatitis C
- E.
Oral contraceptive-associated cholestasis
Correct Answer
E. Oral contraceptive-associated cholestasisExplanation
The correct answer is E. This patient has an increase in both direct and indirect bilirubin levels. Oral contraceptives can cause jaundice, which is very similar to intrahepatic cholestasis. This condition could lead to increased direct and indirect bilirubin levels. This condition is reversible when the medication is discontinued.
Answer A is incorrect. Autoimmune hemolytic anemia is a type of hemolytic anemia in which the body’s immune system attacks its own RBCs. The antibodies involved in this reaction can be detected with the direct Coombs’ test. Patients may present with a conjugated or unconjugated hyperbilirubinemia and right upper quadrant tenderness.
Answer B is incorrect. Crigler-Najjar syndrome type I is an indirect bilirubinemia that is caused by a complete absence of glucuronosyltransferase. It presents in infancy with high unconjugated bilirubin.
Answer C is incorrect. Gilbert’s disease is an indirect bilirubinemia that typically presents in the second decade of life in response to a stressor (infection, surgery, excessive alcohol consumption, exertion, or fasting). These patients typically present with normal liver function tests, high serum bilirubin, and low/normal direct bilirubin levels.
Answer D is incorrect. Patients with hepatitis typically present with increased levels of both indirect and direct bilirubin. This patient also denies two of the common risk factors for hepatitis, unprotected intercourse and drug use.Rate this question:
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- 16.
A 74-year-old man is brought to the emergency department after he became combative and was screaming that “the little people” were after him. The staff is unable to obtain a history or physical examination because of his agitation, although the triage nurse is able to obtain his vital signs, which are signifi cant for a temperature of 40o C (104o F) and a blood pressure of 90/50 mm Hg. His family reports that he was fi ne earlier in the day except for an occasional cough, and that he has never had any psychiatric issues before. By the time the psychiatrist arrives, the patient is somnolent and somewhat confused. Which of the following will confi rm this patient’s most likely diagnosis?
- A.
Blood alcohol level
- B.
Blood, urine, and sputum cultures
- C.
CT of the head
- D.
Electroencephalography
- E.
Urine toxicology screen
Correct Answer
B. Blood, urine, and sputum culturesExplanation
The correct answer is B. This patient is experiencing
delirium, which is characterized by a
decreased attention span and level of arousal,
disorganized thinking, hallucinations, illusions,
misperceptions, disturbance in sleep-wake cycle,
fl uctuating levels of consciousness, and
cognitive dysfunction. Delirium is the most
common psychiatric diagnosis on medical and
surgical fl oors, particularly among elderly patients.
Delirium may be caused by infection,
metabolic disturbances, electrolyte abnormalities,
hypoperfusion, drug intoxication, alcohol
intoxication or withdrawal, and adverse effects
due to medication. In this patient, the combination
of a high fever and hypotension suggests
an infectious etiology. The causative agent is
likely to be identifi ed through blood, urine,
and sputum cultures.
Answer A is incorrect. A blood alcohol level
will give the physician insight into the presence
of alcohol intoxication or withdrawal,
both of which can cause delirium. In this patient,
however, infection is the more likely
cause given his fever.
Answer C is incorrect. CT of the head can detect
the presence of structural brain abnormalities,
hematomas, hemorrhages, infarcts, and
masses. Although these processes can certainly
lead to changes in mental status, this patient is
presenting with clear signs of delirium most
likely related to an infection. CT of the head is
unlikely to be useful in this setting.
Answer D is incorrect. An electroencephalogram
is useful in documenting a seizure or
the postictal state following a seizure. While
complex partial seizures can be associated
with hostile, aggressive behavior, they are also
typically associated with automatisms such as
lip-smacking and grimacing, which this patient
is not experiencing.
Answer E is incorrect. A urine toxicology
screen will detect the presence of metabolites
of recently ingested drugs. While drug intoxication
is a well-established cause of delirium,
infection is the more likely cause in this patient
given his high fever. However, in delirious
patients in whom the cause is not obvious,
it is crucial to perform toxicology screens of the
urine and blood.Rate this question:
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- 17.
A 34-year-old woman who is at 26 weeks of gestation and who has a history of multiple spontaneous abortions presents with severe abdominal pain, jaundice, ascites, and mental status change. Ultrasonography reveals an obscure hepatic venous connection to the inferior vena cava and absence of any waveform in the hepatic veins. She has a positive serum antiphospholipid antibody titer. Which of the following is the most likely diagnosis?
- A.
Budd-Chiari syndrome
- B.
Congestive heart failure
- C.
Polymyalgia rheumatica
- D.
Portal vein thrombosis
- E.
Veno-occlusive disease
Correct Answer
A. Budd-Chiari syndromeExplanation
The correct answer is A. Budd–Chiari syndrome
(BCS) is a nearly complete obstruction
to blood fl ow by an acute clot in the hepatic
veins or in the inferior vena cava. This sudden
event is followed by the onset of hepatomegaly,
pain, ascites, and jaundice. The patient has
classical hypercoagulable risk factors for developing
BCS, including pregnancy and antiphospholipid
antibody syndrome (positive antiphospholipid
antibody titer, seizures, and multiple
abortions). In addition to the clinical presentation
and risk factors, imaging study further supports
the diagnosis by suggesting hepatic venous
occlusion.
Answer B is incorrect. Although pregnancy increases
the patient’s risk for congestive heart
failure (CHF), this patient’s clinical presentation,
combined with the ultrasound fi ndings, is
unlikely to result from CHF.
Answer C is incorrect. Polymyalgia rheumatica
is a rheumatologic disorder characterized
by pain in several muscle groups with an increased
erythrocyte sedimentation rate. It does
not involve the liver or related vasculature.
Answer D is incorrect. Unlike BCS, the presenting
symptom of portal vein thrombosis is
almost always variceal hemorrhage with melena.
In contrast to BCS, ultrasound testing reveals
an echogenic thrombus in the portal
vein.
Answer E is incorrect. Veno-occlusive disease
is characterized by occlusion of terminal hepatic
venules and hepatic sinusoids. It can clinically
resemble BCS; however, the risk factors
for developing this disorder are different. They
include bone marrow transplantation, chemotherapy,
hepatic irradiation, and Jamaican
bush tea. Hypercoagulable states such as pregnancy
and antiphospholipid syndrome are not
risk factors for the disorder. Hence, the most
likely diagnosis is not veno-occlusive disease.Rate this question:
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- 18.
A 5-year-old girl is brought to the pediatrician by her mother for evaluation of readiness to enter school. The mother is worried because the child had to be withdrawn from preschool last year because of an inability to cope with the other children. She reports that the child becomes very upset if her daily routine is interrupted. The child’s birth history and past medical history are unremarkable. She reached all of her neurodevelopmental milestones, including speech development, on schedule. Which of the following is the most likely diagnosis?
- A.
Asperger’s syndrome
- B.
Autistic disorder
- C.
Childhood disintegrative disorder
- D.
Expressive language disorder
- E.
Rett’s disorder
Correct Answer
A. Asperger’s syndromeExplanation
The correct answer is A. Asperger’s syndrome
is now considered to be an autistic spectrum
disorder. These children have normal intelligence
but are unable to show emotion or attachment
to other people. They exhibit some
characteristics of autism, including repetitive
behaviors and relationship problems, but what
distinguishes Asperger’s from autism is the lack
of delay in language acquisition. This condition
is associated with poor visuospatial skills,
diffi culty with subtle aspects of social interactions,
and clumsiness. Many of the diffi culties
evident in patients with Asperger’s syndrome
are closely associated with right hemisphere
dysfunction.
Answer B is incorrect. Children with autistic
disorder exhibit pervasive cognitive and behavioral
defi cits. They display impaired social interaction
and formation of peer relationships,
delay in speech, and stereotyped patterns.
Their intelligence is generally below normal.
This child does not exhibit such a severe disability.
Answer C is incorrect. Children with childhood
disintegrative disorder develop normally
and reach developmental milestones until
about 2 years of age, after which they show loss
of abilities. Areas of loss include language ability,
social skills, bowel/bladder control, play, or
motor skills. Eventually these children become
severely mentally retarded.
Answer D is incorrect. Expressive language
disorder is a disorder of communication. These
children are not mentally retarded and exhibit
no diffi culties other than with speech.
Answer E is incorrect. Rett’s disorder is an Xlinked
condition that is seen only in girls (affected
boys die at birth). Most of the classic
signs and symptoms, including neurodevelopmental
regression and mental retardation, are
fi rst noticed after the age of 4 years. Specifi -
cally, these children will display stereotypic
hand movements, seizures, ataxia, and dementia.Rate this question:
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- 19.
A 4-year-old boy with a history of mental retardation and seizures is brought to the physician with a 3-month history of worsening shortness of breath. During physical examination, the physician notices numerous acnelike papules on the patient’s face. Echocardiography shows signifi cant left ventricular outfl ow obstruction. Which of the following is the most likely diagnosis for this patient’s heart condition?
- A.
Coronary artery disease
- B.
Dilated cardiomyopathy
- C.
Myxoma
- D.
Rhabdomyoma
- E.
Transposition of the great vessels
Correct Answer
D. RhabdomyomaExplanation
The correct answer is D. Tuberous sclerosis is
a genetic condition (autosomal dominant)
characterized by nodular proliferation of multinucleated
atypical astrocytes. These form tubers,
which are found throughout the cerebral
cortex and periventricular areas. The classic
triad, which is manifest in only the most severe
of cases, consists of seizures, mental retardation,
and facial angiofi bromas (also known as
adenoma sebaceum). Half of patients with tuberous
sclerosis develop rhabdomyomas, primary
tumors of cardiac muscle that, although
benign, may compromise cardiac function, especially
of the atrioventricular valves. Tuberous
sclerosis is also notable for a link to angiomyolipomas
of the kidney.
Answer A is incorrect. Coronary artery disease
(CAD) can cause progressive shortness of
breath, but in conjunction with the facial lesions,
the outfl ow obstruction on echo, and the
patient’s age, CAD is an unlikely diagnosis.
Answer B is incorrect. Dilated cardiomyopathy
is often idiopathic. It involves four-chamber
hypertrophy and dilation, and eventually heart
failure. This condition is not associated with
tuberous sclerosis. Note that hypertrophic cardiomyopathy
also causes ventricular outfl ow
obstruction and is often responsible for sudden
death in young athletes.
Answer C is incorrect. Myxomas, like rhabdomyomas,
are capable of obstruction. However,
these are seen in adults and are often located
in the atria.
Answer E is incorrect. Transposition of the
great vessels is a situation in which the pulmonary
trunk arises from the left ventricle and the
aorta arises from the right ventricle. This arrangement
is incompatible with life, and a histopathocompensatory
anomaly such as a patent ductus
arteriosus is necessary.Rate this question:
-
- 20.
A 28-year-old male physician is seeing a new patient, who is a 25-year-old woman. At the beginning of the encounter he recognizes her as someone he met at a bar several weeks ago. He was highly attracted to her at that time, but she left the bar before he could invite her on a date. What is the physician’s best course of action during this visit?
- A.
Ask her on a date
- B.
Continue the visit as normal
- C.
Refer her to another doctor
- D.
Refer her to another doctor, then ask her on a date
- E.
Return to the room with a chaperon and conduct the visit as normal
Correct Answer
B. Continue the visit as normalExplanation
The correct answer is B. Personal (especially
sexual) relationships with patients are inappropriate
and should not be actively sought by either
party of an existing physician-patient relationship.
The doctor’s best course of action
here is to continue the interview and examination,
while making no mention of his feelings
for the woman, who is now his patient. If the
physician had a prior or previously existing relationship
with the patient, he could refuse to
see her on that basis because the personal relationship
preceded any would-be physicianpatient
relationship.
Answer A is incorrect. The physician should
not actively seek romantic (or other personal)
liaisons with patients.
Answer C is incorrect. Generally, referring the
patient to another physician will be an incorrect
answer to physician-patient encounter
questions, unless the referring physician seeks
the particular medical training, expertise, or
knowledge of another doctor that is pertinent
to the patient. Furthermore, differences in religious
or moral beliefs about health care
choices do not alone constitute a reason to refer
because the patient should always be counseled
about alternatives and provide informed
consent. In these cases, referral should occur
only if the patient’s health care choice lends itself
to better (i.e., more expert) care from another
provider.
Answer D is incorrect. This physician should
not refer the patient in order to escape from
the physician-patient relationship. The woman
is now this physician’s patient unless a proper
hand-off is made.
Answer E is incorrect. Chaperons are typically
helpful in cases in which a patient makes inappropriate
seductive gestures or remarks directed
at the health care provider. The chaperon
thus neutralizes the environment during
an interview and physical examination, allowing
the physician to comfortably serve the patient.
Here, the female patient has not indicated
any interest in the male physician, so he should not behave any differently than normal
during this visit.Rate this question:
-
- 21.
A 45-year-old man visited his primary care physician 1 month ago because of chest pain that he had experienced four times in the past 4 months. The onset of the pain is sudden and radiates to his left jaw. He usually feels the pain when he is watching television but has never felt it during exertion. During last month’s visit, the physician prescribed sublingual nitroglycerin, and the patient reports that this has shortened the duration of his episodes. Last month’s ECG is shown in the image. Which of the following is the most likely cause of this patient’s chest pain?
- A.
Myocardial infarction
- B.
Pericarditis
- C.
Prinzmetal’s angina
- D.
Stable angina
- E.
Unstable angina
Correct Answer
C. Prinzmetal’s anginaExplanation
The correct answer is C. This patient has classic
symptoms of cardiac ischemia: chest pain
with sudden onset that radiates to his left shoulder
or jaw and is relieved by sublingual nitroglycerin.
However, the patient is young, and
the pain is not prompted by activity but occurs
at rest. Additionally, his ECG is normal, showing
no evidence of infarct or ischemia. As a result,
he probably suffers from coronary vasospasm,
also known as Prinzmetal’s angina
Answer A is incorrect. If myocardial infarctions
were the etiology of the four episodes of
chest pain in the past 4 months, his ECG
would show evidence of infarct (T-wave inversion,
pathologic Q waves, etc.).
Answer B is incorrect. Pericarditis can cause
sudden onset of chest pain without exertion,
but the pain would be not relieved with nitroglycerin.
Typically, an ECG would also show
diffuse ST-segment elevations.
Answer D is incorrect. Although the patient’s
clinical symptoms are of cardiac ischemia, they
are not induced by a specifi c amount of exercise,
which is the classic defi nition of stable angina.
Answer E is incorrect. Because the patient’s
symptoms are not prompted by a light amount
of exercise or strain, it is unlikely that they are
due to unstable angina.Rate this question:
-
- 22.
A clinical study is performed on young male subjects who have deafness, ocular abnormalities, and a nephritic syndrome. Kidney biopsies of these subjects reveal no pathology under immunofl uorescence or light microscopy. Which of the following is the most common glomerular pathological characteristic likely to be seen under electron microscopy?
- A.
Diffuse epithelial foot process fusion
- B.
Immune complex deposits
- C.
Split basement membrane
- D.
Wire-loop appearance
Correct Answer
C. Split basement membraneExplanation
The correct answer is C. Alport’s syndrome is
a heterogeneous (although most commonly Xlinked)
genetic disorder with absent or mutated
type IV collagen. It is characterized by renal
disease, nerve disorders (deafness), and ocular
disorders. There is no evidence of disease under
low-power light or immunofl uorescence
microscopy, as it is not an immune-mediated
disease. However, under electron microscopy,
there is evidence of a split basement membrane
due to the collagen IV mutation.
Answer A is incorrect. Minimal change disease,
the most common cause of nephrotic
syndrome in children, includes a histopathologic fi nding of diffuse epithelial foot process
fusion on electron microscopy.
Answer B is incorrect. Alport’s syndrome is genetic,
not immune-mediated. Therefore, there
will be no immune complex deposits visible
under electron microscopy.
Answer D is incorrect. A wire-loop appearance
under electron microscopy is peculiar to systemic
lupus erythematosus (SLE), which is accompanied
by subendothelial basement membrane
deposits. SLE is a chronic autoimmune
disease that affects multiple organ systems, including
the kidney. SLE affects young adults
(women more than men) and usually presents
with a combined nephritic and nephrotic picture.Rate this question:
-
- 23.
A researcher is designing an in vitro experimental system to study the kinetics of GLUT4- mediated glucose transport into mammalian cells. The system will measure radiolabeled glucose concentrations in cell culture media both before and at intervals following the addition of insulin. Which of the following cell types is the best choice for use in this experimental system?
- A.
Adipocytes
- B.
Cortical neurons
- C.
Erythrocytes
- D.
Hepatocytes
- E.
Pancreatic β cells
Correct Answer
A. AdipocytesExplanation
The correct answer is A. Adipocytes are the cells
that comprise adipose tissue. GLUT4-mediated
glucose transport occurs in only two tissue types:
adipose (fat) and skeletal muscle. This is the only
choice among those listed that could be used in
the hypothetical experimental system described.
Answer B is incorrect. Cortical neurons are
derived from the brain, where glucose transport
occurs independent of insulin stimulation.
Thus, these cells could not be used in this hypothetical
system.
Answer C is incorrect. Insulin has no effect on
glucose uptake in erythrocytes, so this cell type
could not be used in this hypothetical system.
Answer D is incorrect. Insulin has no effect on
glucose uptake in hepatocytes, so this cell type
could not be used in this hypothetical system.
Answer E is incorrect. Pancreatic β cells express
GLUT2 transporters, which serve as glucose
sensors. These cells do not express
GLUT4 transporters and would not be appropriate
for use in this hypothetical system.Rate this question:
-
- 24.
Myasthenia gravis is an autoimmune disorder that affects approximately 3 in 100,000 people. Individuals with mysasthenia gravis classically present with complaints of muscle weakness and fatigue secondary to the formation of autoantibodies directed against the acetylcholine receptors at neuromuscular junctions. The most accurate method of diagnosis involves the detection of these autoantibodies. On average, this test is approximately 80% sensitive and 90% specifi c. If an individual has a positive test for autoantibodies against the acetylcholine receptor, what is the approximate post-test probability of having this disease, assuming a pre-test probability of 50%?
- A.
80%
- B.
85%
- C.
89%
- D.
90%
- E.
95%
- F.
99%
Correct Answer
C. 89%Explanation
The correct answer is C. The positive predictive
value (PPV) of the test can be calculated
with the following formula, where TP is truepositive
results and FP is false-positive results:
TP / (TP + FP). Given the pretest probability
of 50%, we need to set up a hypothetical 2 × 2 table in which the number of subjects with the
disease is equal to the number not having the
disease (or to be said differently, the pretest
probability becomes the prevalence). If we set
the number of those with the disease as 10,
then TP = 8 and FP = 1, given the sensitivity
of 80% and specifi city of 90%. Therefore, the
PPV would be calculated as 8 / (8 + 1) = 89%,
or about 90%. The same answer can also be
obtained by converting the pretest probability
to an odds ratio (1:1) and multiplying it by the
test’s positive likelihood ratio (LR+), which can
be calculated using the formula LR+ = sensitivity
/ (1 - specifi city) = 0.80 / (1 - 0.90) = 8.
Therefore, the posttest odds of having the disease
is 8:1 or 8/9 = 89% once the fi gure is converted
back into a probability.
Answer A is incorrect. This value is too low to
be the correct answer.
Answer B is incorrect. This value is too low to
be the correct answer.
Answer D is incorrect. This value is too high
to be the correct answer.
Answer E is incorrect. This value is too high
to be the correct answer.
Answer F is incorrect. This value is too high
to be the correct answer.Rate this question:
-
- 25.
An 85-year-old man is rushed to the emergency department from his primary care physician’s offi ce after his physician palpates a pulsating mass in his abdomen. The patient is diagnosed with an abdominal aortic aneurysm. Instead of repairing the aneurysm by surgically opening the abdomen, the surgeon decides to perform endovascular stenting and grafting. The stent is inserted into the femoral artery and threaded up toward the aortic defect. To access the femoral artery the surgeon must open the femoral sheath and expose its contents. Which of the following structures is enclosed inside the femoral sheath?
- A.
Cooper’s ligament
- B.
Femoral canal
- C.
Femoral nerve
- D.
Obturator nerve
- E.
Tunica vaginalis
Correct Answer
B. Femoral canalExplanation
The correct answer is B. The femoral canal
contains the deep inguinal lymph nodes and is
enclosed inside the femoral sheath with the
femoral artery and vein. In a femoral hernia,
this is the potential space into which abdominal
contents herniate. A mnemonic for the
contents of the femoral triangle is “N(AVEL)”
(laterally to medially) for Nerve, Artery, Vein,
Empty space, Lymphatics.
Answer A is incorrect. Cooper’s ligament (lacunar
ligament) is an extension of the inguinal
ligament and forms the medial border of the
femoral ring.
Answer C is incorrect. The femoral nerve is
found outside of the femoral sheath.
Answer D is incorrect. The obturator nerve
runs along the medial edge of the psoas muscle
and is posterior to the femoral triangle.
Answer E is incorrect. The tunica vaginalis is
the refl ection of the peritoneal membrane that
invests the testis and spermatic cord. It is
formed when the testis descends from the abdomen
into the outpouching of peritoneum
known as the processus vaginalis. The processus
vaginalis normally closes but can remain
patent in up to 20% of men. A patent processus
vaginalis can lead to hydroceles or indirect hernias,
but this structure is not found in the femoral
triangle.Rate this question:
-
- 26.
A 6-year-old girl is found to be nearsighted during a vision screening at school, and the school nurse tells the parents the child should be fi tted for corrective lenses. Her mother is upset because her daughter is already much taller than her classmates, has an awkward gait, and was recently diagnosed with scoliosis. She is afraid that the glasses will only add to her daughter’s problems at school, where her classmates frequently tease her. When the ophthalmologist observes that the patient’s right lens is dislocated, he suspects that her symptoms are in fact related to an enzyme defi ciency. As a result of this defi ciency, which of the following amino acids is essential in this patient’s diet?
- A.
Cysteine
- B.
Lysine
- C.
Methionine
- D.
Tryptophan
- E.
Tyrosine
Correct Answer
A. CysteineExplanation
The correct answer is A. Homocystinuria is an
inborn error of metabolism caused by a defect
in cystathionine synthase, the enzyme that
converts homocysteine to cystathionine. Cystathionine
is later converted to cysteine, so patients
with this enzyme defi ciency are required
to supplement their diets with exogenous
cysteine. In addition to marfanlike features and
subluxation of the lens, these patients are at increased
risk of a variety of cardiovascular derangements,
including premature vascular disease
and early death.
Answer B is incorrect. Lysine is another of the
essential amino acids (recall the mnemonic
PVT TIM HALL used to remember the 10 essential
amino acids: Phenylalanine, Valine,
Tryptophan, Threonine, Isoleucine, Methionine,
Histidine, Arginine, Lysine, and Leucine).
It is not related to homocystinuria.
Answer C is incorrect. Homocystinuria is a
disorder of methionine metabolism; this patient
would actually have an excess of methionine
as opposed to a defi ciency.
Answer D is incorrect. Tryptophan is an
amino acid often confused with tyrosine. It is
already an essential amino acid and does not
have any relationship to homocystinuria.
Answer E is incorrect. Tyrosine is the amino
acid affected in phenylketonuria (PKU), a defi -
ciency of phenylalanine hydroxylase. This enzyme
defi ciency results in an inability to convert
phenylalanine to tyrosine, making the
latter an essential amino acid in patients with
PKU. However, it has no role in homocystinuria.Rate this question:
-
- 27.
A 57-year-old man presents to his primary care physician with complaints of fatigue and nausea over the past month. On physical examination, the patient is found to have a low-grade fever, scattered lymphadenopathy including two fi rm 2-cm lymph nodes in the left axilla, and an enlarged spleen. Biopsy of the lymph nodes yields a diagnosis of diffuse large cell lymphoma. Upon consultation with an oncologist, the patient begins a multidrug regimen that includes cyclophosphamide. Although generally safe, cyclophosphamide treatment can produce severe adverse effects in some patients. Which of the following drug-symptom combinations correctly states the treatment strategy for a common adverse event associated with this medication?
- A.
Acrolein for hemorrhagic cystitis
- B.
Acrolein for myelosuppression
- C.
Acrolein for nausea and vomiting
- D.
N-acetylcysteine for hemorrhagic cystitis
- E.
N-acetylcysteine for myelosuppression
- F.
N-acetylcysteine for nausea and vomiting
Correct Answer
D. N-acetylcysteine for hemorrhagic cystitisExplanation
The correct answer is D. The adverse effects of
cyclophosphamide include nausea, vomiting,
myelosuppression, and hemorrhagic cystitis.
During its metabolism, cyclophosphamide is
converted to its active form by the hepatic cytochrome
P-450 enzymes. The fi nal step in the
production of the active form is nonenzymatic
and produces phosphoramide mustard, the desired
cytotoxic agent, and acrolein, an unwanted
cytotoxic compound that is directly responsible
for hemorrhagic cystitis. This dreaded adverse
effect is ameliorated by increasing fl uid intake
and administering N-acetylcysteine, a sulfhydryl
donor. N-acetylcysteine has little impact on the
other adverse effects. Mesna is another thiol
compound commonly used to prevent cyclophosphamide-
induced hemorrhagic cystitis.
Answer A is incorrect. Acrolein is an unwanted
cytotoxic compound and is not a treatment.
Answer B is incorrect. Acrolein is an unwanted
cytotoxic compound and is not a treatment.
Answer C is incorrect. Acrolein is an unwanted
cytotoxic compound and is not a treatment.
Answer E is incorrect. N-acetylcysteine has little
impact on myelosuppression.
Answer F is incorrect. N-acetylcysteine has little
impact on nausea and vomitingRate this question:
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- 28.
An obese 46-year-old, multiparous woman presents to the physician with nonradiating right upper quadrant pain and fever that was preceded by nausea and vomiting. Ultrasonography shows hyperechogenic structures in the right upper quadrant. Laboratory testing reveals a WBC count of 14,500/mm 3, an erythrocyte sedimentation rate of 40 mm/hr, and a serum amylase level of 70 U/L. Which of the following is the most likely diagnosis in this patient?
- A.
Acute acalculous cholecystitis
- B.
Acute calculous cholecystitis
- C.
Acute pancreatitis
- D.
Carcinoma of the pancreas
- E.
Cholesterolosis
Correct Answer
B. Acute calculous cholecystitisExplanation
The correct answer is B. Right upper quadrant
pain in an obese, middle-aged, multiparous
woman with ultrasonographic fi ndings consistent
with gallstones is a classic presentation of
acute calculous cholecystitis. Acute calculous
cholecystitis is an acute mechanical infl ammation
of the gallbladder commonly resulting from
a gallbladder stone obstructing the gallbladder
neck or cystic duct, chemical infl ammation,
and/or bacterial infl ammation. Risk factors are
the “4F’s”: Female, “Fat,” Fertile, and Forty.
Defi nitive treatment is cholecystectomy.
Answer A is incorrect. Acute acalculous cholecystitis,
in contrast to acute calculous cholecystitis,
occurs in the absence of gallstones, generally
in a severely ill patient.
Answer C is incorrect. Acute pancreatitis usually
presents with radiating epigastric pain and
increased serum amylase levels. This is in contrast
to the nonradiating right upper quadrant
abdominal pain and normal amylase levels
seen in this patient.
Answer D is incorrect. Carcinoma of the pancreas
often presents with jaundice and abdominal
pain radiating to the back. It may also present
with migratory thrombophlebitis (Trousseau’s
sign). This presentation differs signifi cantly from
the one described in the question.
Answer E is incorrect. Cholesterolosis, or
strawberry gallbladder, is characterized by yellow
cholesterol-containing fl ecks in the mucosal
surface. In contrast to this patient’s diagnosis,
it is not associated with infl ammatory
changes (normal erythrocyte sedimentation
rate and WBC count) and has no special association
with cholelithiasis.Rate this question:
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- 29.
A 60-year-old woman with a 25-year history of type 2 diabetes mellitus presents with pruritus, diffuse bone pain, and proximal muscle weakness. Laboratory studies show a serum Ca2 + level of 6.5 mg/dL, a serum phosphate level of 6.0 mg/dL, a serum creatinine level of 2.7 mg/ dL, and an intact parathyroid hormone level of 300 pg/mL. The laboratory fi ndings in this patient are most likely due to which of the following conditions?
- A.
Parathyroid adenoma
- B.
Parathyroid insuffi ciency
- C.
Renal failure
- D.
Underlying malignancy
- E.
Vitamin D intoxication
Correct Answer
C. Renal failureExplanation
The correct answer is C. This patient has secondary
hyperparathyroidism due to chronic renal
insuffi ciency or renal osteodystrophy.
There are numerous etiologies of chronic renal
failure; the most common cause in the U.S. is
diabetic nephropathy secondary to diabetes
mellitus. The central problems in secondary
hyperparathyroidism are impaired Ca2+ reabsorption
and phosphate excretion from the kidneys
due to nephron loss. The resulting hypocalcemia
stimulates increased secretion of
parathyroid hormone (secondary hyperparathyroidism),
causing increased bone turnover and
contributing to the hyperphosphatemia. Moreover,
nephron loss results in impaired conversion
of 25-OH vitamin D to 1,25-dihydroxy
vitamin D, reducing Ca2+ absorption from the
intestines and thus exacerbating hypocalcemia
in this syndrome.
Answer A is incorrect. Parathyroid adenoma
would cause increased secretion of parathyroid
hormone, resulting in hypercalcemia and hypophosphatemia
rather than hypocalcemia and
hyperphosphatemia.
Answer B is incorrect. Parathyroid insuffi -
ciency would result in hypocalcemia but cannot account for the hyperphosphatemia presented
in this case.
Answer D is incorrect. Malignancy usually results
in hypercalcemia either due to lytic metastases
to bone (with increased serum alkaline
phosphatase activity and hyperphosphatemia)
or due to production of parathyroid hormonerelated
peptide (with hypophosphatemia).
Answer E is incorrect. Vitamin D intoxication
results in hypercalcemia and hyperphosphatemia
and thus would be inconsistent with
the low calcium value presented in the vignette.
However, vitamin D intoxication may
indeed present with the clinical fi ndings stated
above, including pruritis, bone pain, weakness,
and renal dysfunction.Rate this question:
-
- 30.
A 73-year-old man who has atrial fi brillation has been treated pharmacologically for 10 years. He presents to his primary care physician complaining of generalized dyspnea. Pulmonary function tests show forced expiratory volume in 1 second (FEV 1) and forced vital capacity (FVC) are both less than 70% of the predicted value, with a ratio of FEV 1 to FVC of 81%. The fl ow-volume curve is shown in the image. Which of the following is a possible etiology of this presentation?
- A.
Adult-onset asthma
- B.
Amiodarone
- C.
Diltiazem
- D.
Sotalol
- E.
Tobacco
Correct Answer
B. AmiodaroneExplanation
The correct answer is B. This is a clinical picture
of restrictive lung disease; the FEV1:FVC
ratio is approximately normal, but both are dramatically
reduced. Amiodarone is an antiarrhythmic
that can cause pulmonary fi brosis, a
restrictive lung disease.
Answer A is incorrect. Asthma is a cause of
chronic obstructive pulmonary disease.
Answer C is incorrect. Diltiazem is an antiarrhythmic
that is sometimes used in intravenous
form to treat atrial fi brillation. It infrequently
causes hypotension or bradyarrhythmias, but is
not known to cause pulmonary fi brosis.
Answer D is incorrect. Sotalol has both β-blocking
and action potential-prolonging activity. It is
used for treatment of ventricular and supraventricular
arrhythmias in children and for lifethreatening
ventricular arrhythmias in adults. It
can sometimes cause torsades de pointes when
taken at higher doses. However, sotalol does not
cause pulmonary fi brosis.
Answer E is incorrect. Tobacco is a known risk
factor for chronic obstructive pulmonary disease
(COPD). COPD presents with an FEV1:FVC
ratio ofRate this question:
-
- 31.
A 23-year-old man comes to the emergency department complaining of bloody diarrhea and a fever. Laboratory tests of blood and stool cultures show an oxidase-negative, motile, gramnegative bacillus that grows as clear colonies on MacConkey agar. Which of the following is the most likely causative organism?
- A.
Escherichia coli
- B.
Pseudomonas aeruginosa
- C.
Salmonella species
- D.
Shigella species
- E.
Vibrio cholerae
Correct Answer
C. Salmonella speciesExplanation
The correct answer is C. Bloody diarrhea and
fever can have a number of bacterial causes,
including Escherichia coli (O157:H7), Salmonella,
and Shigella species. Other bacterial causes include Campylobacter jejuni and Yersinia
enterocolitica. Both Salmonella and Shigella
species are gram-negative rods that do not
ferment lactose and are oxidase-negative. Salmonella,
however, is motile, while Shigella is
not. When a bacterium that ferments lactose is
plated on MacConkey agar, the colonies are
pink/red. If the plated bacteria do not ferment
lactose, the colonies are clear.
Answer A is incorrect. Certain strains of E.
coli (O157:H7) can cause bloody diarrhea.
However, E. coli does ferment lactose, which
would result in the growth of pink colonies on
MacConkey agar.
Answer B is incorrect. Pseudomonas aeruginosa
is a frequent cause of nosocomial pulmonary infection
in intubated patients and those with cystic
fi brosis. It does not cause bloody diarrhea,
and although it is a non-lactose-fermenting,
gram-negative bacillus, it is oxidase-positive.
Answer D is incorrect. Enteric shigellosis can
present similarly to Salmonella species infection.
They also are oxidase-negative, non-lactose-
fermenting, gram-negative rods. However,
Shigella species are nonmotile, while
Salmonella species are motile.
Answer E is incorrect. Vibrio species are
curved, motile, gram-negative rods. However,
Vibrio cholerae causes rice-water stools, not
bloody diarrhea.Rate this question:
-
- 32.
A 35-year-old man is brought to the emergency department by ambulance after having a tonicclonic seizure at work. The patient reports that he has always been healthy and has never had a seizure before. On further questioning, the patient reports that he has been having intermittent bloody stools for the past 4 months. CT of the head reveals an irregular 3-cm × 4-cm mass extending from the right to the left hemisphere. CT of the abdomen shows multiple polypoid masses in the sigmoid colon. Which of the following is the most likely diagnosis?
- A.
Familial adenomatous polyposis
- B.
Gardner syndrome
- C.
Hereditary nonpolyposis colorectal carcinoma
- D.
Tuberous sclerosis
- E.
Turcot syndrome
Correct Answer
E. Turcot syndromeExplanation
The correct answer is E. This patient has Turcot
syndrome, an autosomal dominant disease.
All familial polyposis syndromes, with the exception
of Peutz-Jeghers syndrome, predispose
to colorectal cancer. Turcot syndrome is associated
with two separate dominant mutations.
The fi rst is a mutation of the APC gene leading
to polyposis and medulloblastoma, and the second
is associated with the hMLH1 DNA mismatch
repair gene leading to polyposis and
glioblastoma multiforme.
Answer A is incorrect. Familial adenomatous
polyposis is associated with hundreds of colorectal
polyps, and nearly all affected patients
will develop colorectal cancer.
Answer B is incorrect. Gardner syndrome is
characterized by colorectal polyposis and osteomas
or other bone and soft tissue tumors.
Answer C is incorrect. Hereditary nonpolyposis
colorectal carcinoma is associated with dozens
of colorectal polyps, and a majority of affected
patients will develop colorectal cancer.
Answer D is incorrect. Tuberous sclerosis is an
autosomal dominant condition characterized
by mental retardation, seizures, tuberous central
nervous system tumors, angiomyolipomas
of the kidneys, leptomeningeal tumors, and
skin lesions such as ash-leaf spots and shagreen
patches.Rate this question:
-
- 33.
A patient’s serum is placed on a plate that is precoated with a specifi c antigen. The plate is then washed free of non-antigen-binding antibodies. Anti-immunoglobulin antibodies coupled to an enzyme are then added to the mixture. Excess anti-immunoglobulin antibodies are washed free, and a substrate that changes color when cleaved by the enzyme is added to the plate. Which of the following laboratory techniques does this describe?
- A.
Allele-specifi c oligonucleotide probe
- B.
Enzyme-linked immunosorbent assay
- C.
Northern blot
- D.
Polymerase chain reaction
- E.
Sequencing
- F.
Southern blot
- G.
Western blot
Correct Answer
B. Enzyme-linked immunosorbent assayExplanation
The correct answer is B. This question describes
enzyme-linked immunosorbent assay
(ELISA). ELISA is an immunologic technique
used in laboratories to determine whether a
particular antibody is present in a patient’s
blood. Labeled antibodies are used to detect
whether the serum contains antibodies against
a specifi c antigen precoated on an ELISA
plate. The patient’s serum can also be challenged
with a specifi c antibody to determine
whether the corresponding antigen is present
in the patient’s blood.
Answer A is incorrect. Allele-specifi c oligonucleotide
probes are short labeled DNA sequences
complementary to an allele of interest.
These probes can be used to detect the
presence of disease-causing mutations.
Answer C is incorrect. Northern blots are similar
to Southern blots except that in Northern
blotting, mRNA is separated by electrophoresis
instead of DNA. This is not the technique described
above.
Answer D is incorrect. Polymerase chain reaction
is a laboratory technique used to produce
many copies of a segment of DNA. In the procedure,
DNA is mixed with two specifi c primers,
deoxynucleotides and a heat-stable polymerase.
The solution is heated to denature the
DNA and is then cooled to allow synthesis.
Twenty cycles of heating and cooling amplify
the DNA over a million times. This is not the
procedure described above.
Answer E is incorrect. Sequencing is a laboratory
technique that utilizes dideoxynucleotides
to randomly terminate growing strands of
DNA. Gel electrophoresis is used to separate
the varying lengths of DNA. The DNA sequence
can then be read based on the position
of the bands on the gel. This is not the technique
described above.
Answer F is incorrect. In a Southern blot procedure,
DNA is separated with electrophoresis,
denatured, transferred to a fi lter, and hybridized
with a labeled DNA probe. Regions on
the fi lter that base-pair with the labeled DNA
probes can be identifi ed when the fi lter is exposed
to fi lm that is sensitive to the radiolabeled
probe. This is not the technique described
above.
Answer G is incorrect. In a Western blot procedure,
protein is separated by electrophoresis
and labeled antibodies are used as a probe.
This technique can be used to detect the existence
of an antibody to a particular protein.Rate this question:
-
- 34.
A 3-year-old boy comes to the physician because of fever and erythema in his conjunctivae, oral mucosa, palms, and soles for the past week. Physical examination is signifi cant for fever, enlarged cervical lymph nodes, and edema of the hands and feet. Although the precise cause of the patient’s disease is unknown, it is speculated that autoantibodies may play a role. Based on the known structure that is primarily affected in the patient’s disease, what autoantibodies are suspected to be associated with this condition?
- A.
Anticentromere antibodies
- B.
Anti-endothelial cell antibodies
- C.
Antihistone antibodies
- D.
Anti-IgG antibodies
- E.
Antinuclear antibodies
Correct Answer
B. Anti-endothelial cell antibodiesExplanation
The correct answer is B. To answer this question,
one must know that Kawasaki’s syndrome
(also referred to as mucocutaneous lymph node
syndrome) is an arteritis that primarily affects
medium- and small-sized arteries. Hence, it
makes sense that there is evidence suggesting
the formation of anti-endothelial cell (and anti–
smooth muscle cell) autoantibodies in patients
with this disease. The clinical manifestations of
this disease include fever for more than 5 days,
cervical lymphadenopathy, a skin rash (which
often has desquamation, or shedding of the
skin), and erythema of the conjunctivae, oral
mucosa, palms, and soles. Eighty percent of patients
are under the age of 4 years. Twenty percent
of patients develop cardiovascular disease,
including coronary artery vasculitis and coronary
artery aneurysm.
Answer A is incorrect. Anticentromere antibodies,
which are found in 90% of patients
with the CREST variant of scleroderma, are
not particularly associated with Kawasaki’s syndrome.
Answer C is incorrect. Antihistone antibodies,
which are found in over 95% of patients with drug-induced lupus erythematosus, are not particularly
associated with Kawasaki’s syndrome.
Answer D is incorrect. Anti-IgG (rheumatoid
factor) is not particularly associated with Kawasaki’s
syndrome. Elevated levels of serum rheumatoid
factor are present in 80% of patients
with rheumatoid arthritis.
Answer E is incorrect. Antinuclear antibodies,
which are present in over 95% of patients with
systemic lupus erythematosus, are not particularly
associated with Kawasaki’s syndrome.Rate this question:
-
- 35.
A 6-year-old boy arrives at the emergency department breathing rapidly and complaining of tinnitus and nausea. His parents explain that he swallowed half a bottle of aspirin that they had accidentally left out. The physician decides to administer a medication that alters the pH of the boy’s urine in order to improve excretion of the drug. How does altering the pH of the urine improve the excretion of aspirin?
- A.
Acidifi cation of the urine traps ionized molecules in the tubule
- B.
Acidifi cation of the urine traps nonionized molecules in the tubule
- C.
Acidifi cation of urine will increase the glomerular fi ltration rate
- D.
Alkalinization of the urine traps ionized molecules in the tubule
- E.
Alkalinization of the urine traps nonionized molecules in the tubule
- F.
Alkalinization of urine will increase the glomerular fi ltration rate
Correct Answer
D. Alkalinization of the urine traps ionized molecules in the tubuleExplanation
The correct answer is D. Because aspirin is a
weak acid with a pKa near 3.5, it can interconvert
between neutral and negatively charged
forms depending on the pH. Increasing the pH
of tubular fl uid shifts the equilibrium toward
the deprotonated charged state of the molecule.
Thus, neutral molecules diffusing into
the tubule will become ionized. Once in the
charged state, molecules cannot diffuse back
across tubular epithelial membranes to the
bloodstream. Thus the clearance of aspirin is
greatly increased when urine pH is alkalinized.
Answer A is incorrect. Acidifi cation of the
urine would lower the pH and shift the equilibrium
toward the protonated neutral form of
aspirin. These nonionized molecules could
then move back into the bloodstream and
clearance of aspirin would be decreased.
Answer B is incorrect. Acidifi cation of the
urine would lower the pH and shift the equilibrium
toward the protonated neutral form of
aspirin, but these molecules can diffuse across
cell membranes back into the bloodstream and
would not be excreted.
Answer C is incorrect. Acidifi cation of the
urine has no effect on the glomerular fi ltration
rate (GFR). GFR is affected by the difference
in pressures across the glomerulus and glomerular
permeability.
Answer E is incorrect. Alkalinization of urine
promotes ionization of aspirin in the urine; the
concentration of nonionized molecules of aspirin
in the tubule would decrease as the urine is
alkalinized.
Answer F is incorrect. Alkalinization of the
urine has no effect on the GFR. GFR is affected
by the difference in pressures across the
glomerulus and glomerular permeability.Rate this question:
-
- 36.
A 61-year-old man with a past medical history of cancer presents with a 2-week history of constant and severe headaches. He also notes changes in his vision associated with the headaches. Physical examination shows a healing ecchymotic lesion on the right forearm, papilledema in the left eye, a right-sided pronator drift, and weakness of the right arm. The diagnosis of an intracranial hemorrhage is confi rmed with CT of the head. Which of the following cancers is most likely to have resulted in this patient’s presentation?
- A.
Angiosarcoma
- B.
Basal cell carcinoma
- C.
Colorectal carcinoma
- D.
Melanoma
- E.
Prostate cancer
Correct Answer
D. MelanomaExplanation
The correct answer is D. Intracranial metastases
represent nearly half of all brain tumors, yet
only 15% of tumors metastasize to the brain.
Intracranial hemorrhages are a recognized but
relatively uncommon complication of brain tumors
and can result in intraparenchymal, subarachnoid,
subdural, and epidural hematomas.
Focal neurologic signs are frequently evident
and are due to pressure exerted on the brain
parenchyma. Renal cell carcinomas, choriocarcinomas,
melanomas, retinoblastomas, and
lung and breast cancers can result in hemorrhagic
brain metastases. Since melanoma is a
relatively frequent source of metastatic lesions
to the brain (although less common than breast
or lung carcinoma) and demonstrates a tendency
to hemorrhage, melanoma is the correct
answer in this case.
Answer A is incorrect. Angiosarcomas are malignant
endothelial neoplasms that resemble
hemangiomas. Although these tumors may
bleed, angiosarcomas rarely metastasize, and
only a few case reports exist of hemorrhage of
cerebral metastasis from angiosarcoma.
Answer B is incorrect. Some cancers rarely
metastasize to the brain; these include carcinomas
of the oropharynx, esophagus, and prostate,
as well as nonmelanoma skin cancers.
Answer C is incorrect. Colorectal carcinoma
does metastasize to the brain (though less frequently
than melanoma) but does not typically
result in intracranial hemorrhage. Since colorectal
carcinoma is less likely than melanoma
to result in brain metastases and is not as likely
to hemorrhage, melanoma is a better answer.
Answer E is incorrect. Carcinoma of the prostate
almost never results in metastatic brain
disease and therefore represents an extremely
unlikely etiology for this patient’s disease.Rate this question:
-
- 37.
A 7-year-old boy presents to the physician with acute-onset edema and facial swelling. Dipstick urinalysis reveals 4+ proteinuria. Renal biopsy shows no appreciable changes under light and fl uorescence microscopy, but electron microscopy demonstrates glomerular epithelial cell foot process effacement. A diagnosis of minimal change disease is made. How does this disease affect the pressures governing the fl ow of fl uid across the glomeruli?
- A.
Bowman’s space hydrostatic pressure will be decreased
- B.
Bowman’s space hydrostatic pressure will be increased
- C.
Bowman’s space oncotic pressure will be decreased
- D.
Glomerular capillary hydrostatic pressure will be increased
- E.
Glomerular capillary oncotic pressure will be decreased
Correct Answer
E. Glomerular capillary oncotic pressure will be decreasedExplanation
The correct answer is E. Minimal change disease
results in nephrotic syndrome, which is
primarily manifested in the loss of signifi cant
protein in the urine. As a result of this protein
loss, the plasma protein concentration will go
down, thus decreasing the oncotic pressure in
the glomerular capillary. According to the Starling
equation (GFR = Kf [(PGC - PBS) - (pGC -
pBS)]), this change will lead to a higher GFR
by decreasing the oncotic forces that normally
oppose ultrafi ltration.
Answer A is incorrect. Tubular hydrostatic
pressures are not affected by nephrotic syndrome.
The Bowman’s space hydrostatic pressure
generally does not decrease.
Answer B is incorrect. Tubular hydrostatic
pressures are not affected by nephrotic syndrome.
The Bowman’s space hydrostatic pressure
could be increased in a patient with an
obstruction to urine fl ow.
Answer C is incorrect. Bowman’s space oncotic
pressure will increase, not decrease, as
protein is fi ltered into Bowman’s space and
thus increases the protein concentration there.
Answer D is incorrect. Hydrostatic pressures
are not affected in minimal change disease.
The glomerular capillary hydrostatic pressure
could be increased with constriction of the efferent
arteriole, for example.Rate this question:
-
- 38.
A 15-year-old boy is riding his skateboard down a rail when the board slips and he falls, straddling the rail. He comes to the emergency department in extreme pain. On physical examination, he is found to be febrile and tachycardic. His genital examination is notable for ecchymosis and swelling of the scrotum and perineal region due to urinary leakage. Which of the following is the source of this urinary leakage?
- A.
Anterior bladder wall rupture
- B.
Penile urethra rupture
- C.
Superior bladder wall rupture
- D.
Urethral rupture above the urogenital diaphragm
- E.
Urethral rupture below the urogenital diaphragm
Correct Answer
E. Urethral rupture below the urogenital diaphragmExplanation
The correct answer is E. The male urethra is
made up of three parts. The prostatic urethra
runs through the prostate. The membranous
urethra runs through the urogenital diaphragm,
and the penile urethra runs through
the penis. Rupture of the urethra below the
urogenital diaphragm (at the junction between
the membranous and the penile urethra) from
a so-called “straddle injury” causes urine to
fl ow into the scrotum and the perineal region.
Answer A is incorrect. Anterior bladder wall
rupture is caused by a fractured pelvis. In this
kind of injury, urine will fl ow into the retropubic
space.
Answer B is incorrect. Penile urethra rupture
occurs following a crush injury. Urine will fl ow
into the deep fascia of Buck within the penis.
Answer C is incorrect. Superior bladder wall
rupture, also called dome rupture, is caused by forceful compression of a full bladder. This
form of bladder rupture causes urine to fl ow
into the peritoneal cavity.
Answer D is incorrect. Urethral rupture above
the urogenital diaphragm (at the junction of
the prostatic and membranous urethra) due to
a fractured pelvis or improper catheter insertion
causes urine to fl ow into the retropubic
space.Rate this question:
-
- 39.
A gram-positive organism is isolated and cultured. Analyses show the organism is catalasenegative with no hemolysis, and is resistant to optochin and penicillin. The organism is able to grow in 40% bile salts and in 6.5% sodium chloride solution. Which of the following organisms has been isolated?
- A.
Enterococcus faecalis
- B.
Staphylococcus epidermidis
- C.
Streptococcus agalactiae
- D.
Streptococcus pneumoniae
- E.
Streptococcus sanguis
Correct Answer
A. Enterococcus faecalisExplanation
The correct answer is A. Enterococcus faecalis
and E. faecium (Lancefi eld group D streptococci)
are normal fl ora of the intestine and can
cause both urinary tract infection and infectious
endocarditis. They are very hardy organisms that
are able to grow in salt and bile solutions. They
can be γ- or α-hemolytic and are optochin- and
penicillin-resistant. Recently, strains of enterococcus
have become resistant to vancomycin.
Vancomycin-resistant enterococci can cause a
life-threatening nosocomial infection.
Answer B is incorrect. Staphylococcus epidermidis
is a catalase-positive, coagulase-negative,
novobiocin-sensitive organism. It is the cause of
infection in many patients with artifi cial prosthetic
devices such as catheters, heart valves,
and vascular shunts.
Answer C is incorrect. Streptococcus agalactiae
is a bacitracin-resistant, β-hemolytic group
B streptococcus. It is the leading cause of neonatal
meningitis but does not usually cause
symptomatic infection in adults. S. agalactiae
is usually penicillin-sensitive.
Answer D is incorrect. Streptococcus pneumoniae
is a signifi cant cause of many different types
of infections, including pneumonia. S. pneumoniae
is an α-hemolytic, optochin-sensitive organism.
The majority of S. pneumoniae strains are
still sensitive to penicillin and do not grow in
bile.
Answer E is incorrect. Streptococcus sanguis is
similar to E. faecalis in that it is optochinresistant
and bile-tolerant. Although resistance
to penicillin among the viridans group of streptococci
(which includes S. sanguis) is increasing,
most strains are still susceptible. The viridans streptococci are α-hemolytic, while the
enterococci can be either α- or γ-hemolytic.Rate this question:
-
- 40.
The mother of a 16-month-old girl is concerned because she noticed that her daughter became lethargic and irritable 2 days after an ear infection. The mother recalls periodically noticing sweet-smelling urine in her child’s diaper but that otherwise her development has been normal. An inability to metabolize which of the following would explain these symptoms?
- A.
Glucosylceramide
- B.
Histidine
- C.
Phenylalanine
- D.
Sphingomyelin
- E.
Tyrosine
- F.
Valine
Correct Answer
F. ValineExplanation
The correct answer is F. This child has intermittent
maple syrup urine disease (MSUD).
MSUD is caused by an inability to degrade
the carbon skeleton of the three branched
chain amino acids valine, leucine, and isoleucine.
MSUD has many subtypes; the two most
common are classical and intermittent. Classical
MSUD presents with ketonuria 48 hours
to 1 week after birth. The intermittent form
presents during times of catabolism such as
after infections. If left untreated, MSUD can
lead to seizure, coma, and death. Dietary restriction
of branched-chained amino acids is
the mainstay of treatment.
Answer A is incorrect. Gaucher’s disease results
from an accumulation of glucosylceramide
in macrophages of bone marrow, spleen,
and liver due to defi ciency of glucocerebrosidase.
Patients may present with bone pain and
fractures, thrombocytopenia, and hepatosplenomegaly.
Answer B is incorrect. Histidine breakdown is
impaired in histidase defi ciency, which can
lead to histidinemia. Mental retardation is
common but not always present in this disease.
It is 20 times more common than MSUD.
Answer C is incorrect. Phenylalanine degradation
defi ciencies are characteristic of PKU.
This is screened for at birth and the main
symptom is mental retardation.
Answer D is incorrect. Defi ciency of sphingomyelinase
can lead to Niemann–Pick disease,
an autosomal recessive disorder characterized
by hepatosplenomegaly, macular cherry-red
spot, and loss of motor skills and neurologic
function.
Answer E is incorrect. Phenylalanine is converted
to tyrosine, which is degraded by homogentisate
oxidase. Defi ciency of this enzyme
leads to a buildup of phenylalanine and tyrosine.
This disorder is known as alkaptonuria,
which results in black-appearing urine when
left untreated.Rate this question:
-
- 41.
A 35-year-old man presents to the physician with a 2-month history of non-bloody, non-mucoid, non-oily watery diarrhea. He has a diastolic murmur that gets louder with inspiration and is best heard over the left lower sternal border. His face is warm and appears to be engorged with blood for several minutes during the examination. Laboratory studies show: Vanillylmandelic acid: 5 mg/day (normal 0–7mg/day) Metanephrine, urine: 250 μg/g of creatinine (normal 0–00 μg/g) Homovanillic acid, urine: 14 mg/day (normal 0–5 mg/day) 5-HIAA: 28 mg/day (normal 0–9 mg/day) Gastrointestinal endoscopy is most likely to show a lesion located near which of the following?
- A.
Gastroesophageal junction
- B.
Ligament of Treitz
- C.
Pancreaticoduodenal junction
- D.
Rectosigmoid junction
- E.
Splenic flexure
Correct Answer
B. Ligament of TreitzExplanation
The correct answer is B. This patient presents with chronic diarrhea, intermittent facial fl ushing, and a murmur consistent with tricuspid stenosis, a triad of fi ndings classic for carcinoid syndrome. One-third of carcinoid tumors of the gastrointestinal tract occur in the midgutderived small bowel, which begins at the ligament of Treitz and ends at the mid transverse colon. While adenocarcinoma is the most common type of small bowel tumor, carcinoid tumors are most likely to occur in the small bowel. Carcinoid tumors of the small intestine secrete serotonin, which is usually metabolized by the liver and doesn’t cause the symptoms of the carcinoid syndrome. However, when metastases to the liver are present, the bioactive amines can no longer be metabolized and enter the systemic circulation causing diarrhea, abdominal cramps, gastrointestinal bleeding, malabsorption, fl ushing, bronchospasm, and right heart valvular disease from serotoninmediated fibroelastosis. Electron microscopy reveals “salt and pepper” granulation of cells, consistent with their neuroendocrine origin.
An elevated urinary 5-HIAA level is diagnostic of carcinoid syndrome.
Answer A is incorrect. The gastroesophageal junction is affected by gastroesophageal refl ux
disease, not carcinoid tumors.
Answer C is incorrect. The pancreaticoduodenal junction is the site where pancreatic endocrine and exocrine secretions empty into the small bowel to aid in digestion. It is part of the
foregut-derived intestine, and it is a rare site for carcinoid tumors.
Answer D is incorrect. The rectosigmoid junction is not a common location for carcinoid tumors.
Answer E is incorrect. The splenic fl exure is a watershed area that is susceptible to ischemic
damage if cardiac output becomes low. It is not, however, a common site for carcinoid tumorsRate this question:
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- 42.
A 57-year-old woman presents with violet discoloration of her upper eyelids, periorbital edema, and erythematous patches over her knuckles, elbows, and knees for the past several months. She also complains of bilateral muscle weakness that causes diffi culty swallowing and trouble getting up from a chair. Which additional disease process is this patient most likely to have?
- A.
Cancer of a visceral organ
- B.
Osteoarthritis
- C.
Psoriasis
- D.
Secondary syphilis
- E.
Zenker’s diverticulum
Correct Answer
A. Cancer of a visceral organExplanation
The correct answer is A. The clinical vignette gives a description of dermatomyositis, an immune-mediated disorder that involves the skin and skeletal muscles. The distinctive rash of this disease is characterized by a violet discoloration of the upper eyelids together with periorbital edema. Patients often develop a gradual symmetric muscle weakness of the proximal muscles, which may manifest as diffi culty getting up from chairs; one-third of patients develop muscle weakness that causes dysphagia. The disease often also presents with Gottron’s lesions, which are erythematous patches over the knuckles, elbows, and knees. Between 6% and 45% of patients with dermatomyositis have
an underlying visceral cancer.
Answer B is incorrect. Osteoarthritis is a degenerative joint disease characterized by joint inflammation and destruction secondary to wear and tear. It is not notably associated with dermatomyositis.
Answer C is incorrect. The diagnosis of psoriasis is characterized by nonpruritic scaly/silvery
erythematous plaques with well-defi ned borders. Nevertheless, psoriasis is not notably associated
with dermatomyositis.
Answer D is incorrect. Secondary syphilis often develops 6 weeks after an untreated primary chancre of syphilis has healed. It is characterized by fever, lymphadenopathy, skin rashes (widespread small, fl at lesions that particularly involve the palms, soles, and oral mucosa), and condylomata lata (painless wartlike lesions that present in the vulva, the scrotum, or other warm, moist areas of the body). This disease is not notably associated with dermatomyositis.
Answer E is incorrect. Zenker’s diverticulum, also known as pharyngoesophageal diverticulum, is an outpouching of the esophageal wall above the level of the upper esophageal sphincter that results from herniation of mucosa through a defective muscular layer. While this disease may cause dysphagia, it is not notably associated with dermatomyositis.Rate this question:
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- 43.
A 20-year-old man presents to the emergency department with a 3-day history of worsening fever and swelling and redness over his left leg. On physical examination, the patient has erythema and edema of his left leg that is exquisitely tender. The patient is admitted to the hospital and given intravenous antibiotics. The erythema and swelling decrease over the next 3 weeks. During the fourth week, the patient develops new onset of weakness, fatigue, fever, and a maculopapular rash. Significant laboratory findings include a blood urea nitrogen level of 45 mg/dL and creatinine level of 2.8 mg/dL. Blood studies show increased WBCs and eosinophils. Urinalysis shows hematuria, mild proteinuria, and increased WBCs, with a high number of eosinophils. Which of the following is the most likely cause of this patient’s new onset of symptoms?
- A.
IgA nephropathy
- B.
Interstitial nephritis due to medications
- C.
Poststreptococcal glomerulonephritis
- D.
Rapidly progressive glomerulonephritis
- E.
Systemic lupus erythematosus
Correct Answer
B. Interstitial nephritis due to medicationsExplanation
The correct answer is B. Allergic interstitial nephritis is an intrarenal cause of acute renal failure. The cause of interstitial nephritis is usually medications, but infections and immunologic disorders occasionally precipitate the disorder. Medications that cause interstitial nephritis include penicillins (particularly methicillin and nafcillin), cephalosporins, sulfonamides, and nonsteroidal anti-inflammatory drugs. Clinical findings include fever, rash, and peripheral eosinophilia. Eosinophilia is also seen on urinalysis, along with RBCs and proteinuria.
Answer A is incorrect. IgA nephropathy (Berger’s disease) is a primary renal disease with deposition
of IgA in the glomerulus. Patients usually present with gross hematuria of little clinical signifi cance following an infection. Unlike postinfectious glomerulonephritis, no latent period exists between infection and renal symptoms. Eosinophilia is not present.
Answer C is incorrect. Poststreptococcal glomerulonephritis is a type of immune complex
glomerulonephritis that causes enlargement of glomeruli with infi ltration of neutrophils and mesangial cell proliferation. This condition usually presents in children about 2 weeks after skin or throat infection with Streptococcus pyogenes. Most patients have antistreptolysin O, which can be helpful in making the diagnosis. The classic presentation is oliguric renal failure with nephritic syndrome. It is not associated with eosinophilia.
Answer D is incorrect. In rapidly progressive glomerulonephritis (RPGN), patients develop renal failure over a period of weeks to months. They present with renal failure and nephritic syndrome. Pathologically, RPGN is characterized by crescent formation involving most glomeruli. There are many etiologies for nephritic syndrome/RPGN. One would not expect to see eosinophils
Answer E is incorrect. SLE is an infl ammatory autoimmune disorder that affects multiple organ
systems. SLE frequently causes renal damage that can present as either nephrotic or
nephritic syndrome or be asymptomatic. Urinalysis typically shows microscopic hematuria
and proteinuria but not eosinophilia. End-stage renal disease is a common cause of death in SLE.Rate this question:
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- 44.
A 10-year-old boy is brought to the emergency department after becoming less responsive following several bouts of nausea and vomiting. The patient is tachycardic and is breathing deeply and slowly. Laboratory studies are remarkable for a serum pH of 7.21, a serum glucose level of 700 mg/dL, a serum bicarbonate level of 16 mEq/L, and a serum anion gap of 22 (normal 7–16). Intravenous fluids and insulin are administered. Measurement and management of which of the following electrolytes are most critical in this patient?
- A.
Bicarbonate
- B.
Calcium
- C.
Chloride
- D.
Potassium
- E.
Sodium
Correct Answer
D. PotassiumExplanation
The correct answer is D. This patient is in diabetic ketoacidosis (DKA), which is often the presenting syndrome in type 1 diabetes mellitus. The initial management of this condition requires aggressive fl uid resuscitation and correction of hyperglycemia with insulin. Insulin stimulates the shift of potassium from the extracellular compartment to the intracellular compartment, causing a decrease in serum potassium levels and possible cardiac conduction abnormalities. In addition, the rise in serum pH (as a result of correcting the ketoacidosis with insulin) will cause hydrogen ions to come out of the cells, and their positive charge will be replaced by potassium ions moving intracellularly, leading to further hypokalemia. Thus, patients with a low serum potassium level before administering insulin should be considered to have a potentially life-threatening low total body potassium level. Hence, after the administration of insulin, judicious administration of potassium is the most important step in the treatment of DKA. In addition, calcium gluconate should be administered to protect cardiac myocytes and prevent arrhythmias.
Answer A is incorrect. Bicarbonate does not undergo insulin-mediated transcellular shifts as is the case with potassium. Bicarbonate levels often normalize with the correction of hyperglycemia and fluid administration with diuresis of serum ketoacids. Bicarbonate should be administered only when serum levels areRate this question:
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- 45.
A 51-year-old man presents to the emergency department 30 minutes after his wife noticed drooping of the left side of his face and difficulty speaking. The patient is alert and oriented to person and place. His wife states that he has a history of benign prostatic hypertrophy, peptic ulcer disease, and high blood pressure. According to the wife, the patient has never experienced symptoms like this before and has never had surgery of any type. The physicians determine that the patient is hemodynamically stable and they obtain the CT scan shown in the image. Which of the following is the best next step in management?
- A.
Echocardiogram
- B.
Hemicraniectomy
- C.
Heparin
- D.
Insulin
- E.
Tissue plasminogen activator
Correct Answer
E. Tissue plasminogen activatorExplanation
The correct answer is E. The history and physical examination suggest a possible cerebrovascular accident. Before any therapeutic intervention is done, an emergent CT scan must be performed to rule out hemorrhage. In the emergent setting, CT scans are favored over MRI. The CT scan shown does not demonstrate any signs of intracranial hemorrhage. Given the high clinical suspicion for stroke, your attention should focus on the likely possibility of an ischemic etiology. It is important to
remember, however, that an ischemic infarct will often not be visible on the initial scan, especially
if the scan is done within a few hours of symptom onset. Tissue plasminogen activator (t-PA), a thrombolytic agent, is the best next step in management given that the patient does not have any obvious contraindications to thrombolytic therapy. Treatment with t-PA has been shown to be very effective in the management of acute ischemic stroke, especially if administered within 3 hours of symptom onset. This form of treatment does, however, carry a risk of hemorrhage.
Answer A is incorrect. Once a hemorrhagic stroke has been ruled out by CT scan, the possibility
of a cardioembolic source should be investigated with an echocardiogram. Patients with a history of atrial fi brillation are at significantly increased risk of thromboembolism. Thrombolysis/anticoagulation, however, helps restore perfusion to the brain and therefore should be done first.
Answer B is incorrect. Some strokes, particularly those involving the middle cerebral artery, are associated with signifi cant parenchymal edema and subsequent mass effect, with possible sequelae including herniation and death. Hemicraniectomy is a rather novel therapeutic intervention that involves temporarily removing half of the skull over the edematous area with the goal of relieving pressure and reducing the chance of herniation. This intervention is not widely used and is not considered the standard of care in the management of acute stroke.
Answer C is incorrect. Heparin therapy is the next step in management for patients who have
contraindications to thrombolytic therapy. Such contraindications include past history of hemorrhagic
stroke, active internal bleed, history of surgery within the past 3 weeks, and any form of coagulopathy. As with tissue plasminogen activator, it is imperative to rule out the presence of
intracranial hemorrhage with a CT scan before initiating therapy.
Answer D is incorrect. Hyperglycemia worsens functional outcomes in cases of ischemic stroke. It has been hypothesized that hyperglycemia may increase local tissue acidosis and blood-brain barrier permeability. While glucose control with insulin would help minimize the harmful effects of hyperglycemia, it is not the next step in management.Rate this question:
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- 46.
3-year-old girl is brought to the emergency department because she is feeling sick and has had a temperature of 38.9° C (102° F) for 3 days. The intern notices a shallow, healing laceration on the girl’s right calf with an erythematous papule in the same area. On questioning, her brother states that a cat may have scratched the toddler because he “saw her playing with a stray.” Which of the following organisms is the most likely cause of this illness?
- A.
Bartonella henselae
- B.
Borrelia burgdorferi
- C.
Eikenella corrodens
- D.
Francisella tularensis
- E.
Pasteurella multocida
Correct Answer
A. Bartonella henselaeExplanation
The correct answer is A. Bartonella henselae is a gram-negative bacillus that is the cause of cat-scratch disease. Typically (in 60% of cases), a child is scratched by a bacteremic young cat, and a papule or pustule develops in the area 3–5 days later. Tender regional lymphadenopathy develops in 1–2 weeks. Most patients present with systemic symptoms such as anorexia, fever, and malaise.
Answer B is incorrect. The spirochete Borrelia burgdorferi is the cause of Lyme disease. The
spirochete is carried by the Ixodes tick, which is most common in the northeastern United States. It initially presents with an expanding ring-shaped lesion known as erythema migrans, which begins at the site of the tick bite.
Answer C is incorrect. Eikenella corrodens is a gram-negative organism that is part of the normal fl ora of the mouth and nasopharynx. It is associated with infections resulting from human bites.
Answer D is incorrect. Francisella tularensis is the cause of tularemia. This disease is carried by wild rabbits and ticks in the southeastern United States. It often presents with lymphadenopathy and an ulcer at the site of entry as well as with fever.
Answer E is incorrect. Pasteurella multocida is caused by cat bites and dog bites. This infection causes a rapid infl ammation (often within hours) and is accompanied by purulent drainage.Rate this question:
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- 47.
A 35-year-old man presents comes to his primary care physician with a chief complaint of palpitations and occasional chest pain. Further questioning reveals a recent history of weight loss, diarrhea, and heat intolerance. Laboratory evaluation shows anti-thyroid-stimulating hormone (TSH) receptor antibodies in the patient’s serum. Which of the following best describes this patient’s TSH and thyroid hormone levels relative to normal baseline values?
- A.
A
- B.
B
- C.
C
- D.
D
- E.
E
Correct Answer
D. DExplanation
The correct answer is D. The vignette describes a classic history of an autoimmune hyperthyroidism, Graves’ disease. In this disorder, thyroid follicular cells are stimulated to synthesized and secrete thyroid hormone by anti- TSH receptor antibodies, leading to increased levels of thyroxine (T4) and triiodothyronine (T3 ) in the blood, which results in negative feedback on the anterior pituitary and suppression of TSH secretion. Thus, both free T4 and total T4, which includes free T4 and T4 bound to proteins in the blood (e.g., albumin and thyroxine- binding globulin) will be increased, while blood TSH levels will be low relative to the normal baseline.
Answer A is incorrect. An elevated TSH level is not characteristic of Graves’ disease, and elevated
T4 levels should result in a lower TSH level due to negative feedback on the anterior pituitary.
Answer B is incorrect. An elevated TSH level is not characteristic of Graves’ disease, and an elevated T4 level should result in a lower TSH level due to negative feedback on the anterior
pituitary. Furthermore, thyroid hormone binding to proteins in the blood should not be decreased
but instead should be increased in the setting of increased free T4. Therefore, the total T4 level should be elevated rather than low.
Answer C is incorrect. Graves’ disease is characterized by a low TSH due to the circulating thyroid-stimulating immunoglobulins which elevate the T3 and T4 levels and, via negative feedback, downregulate the level of TSH. In this answer choice, the level of TSH is elevated, which would lead to elevated, not diminished, levels of T3 and T4.
Answer E is incorrect. Total and free T4 levels are expected to be low in the setting of low TSH levels. However, in Graves’ disease, stimulation of TSH receptors on the thyroid follicular cells by anti-TSH receptor antibodies stimulates the secretion of thyroid hormones and results in increased total and free T4 levels in the setting of normal or even low TSH levels. The resulting negative feedback loop to the anterior pituitary leads to reduced TSH levels.Rate this question:
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- 48.
A 35-year-old woman presents to her primary care physician with a fever of 38.C (101 F), night sweats, and fatigue. The patient says that she has lost about 6.7 kg (15 lb) over the past year. A CT scan demonstrates mediastinal lymphadenopathy. Biopsy of the nodes shows a small number of large cells with “owl-eye” nucleoli, multiple nuclei, and an abundance of pale cytoplasm on a background of many reactive lymphocytes, macrophages, and granulocytes. Which of the following drugs could be used to treat this condition?
- A.
Azathioprine
- B.
Cisplatin
- C.
Doxorubicin
- D.
Paclitaxel
- E.
β-Interferon
Correct Answer
C. DoxorubicinExplanation
The correct answer is C. The classic symptoms of Hodgkin’s disease include nonspecifi c constitutional symptoms such as night sweats, fatigue, fever, and weight loss. Additionally, mediastinal lymphadenopathy is common, and biopsy of affected nodes will show Reed-Sternberg cells on a background of reactive infl ammatory cells, just as described in the question stem. A variety
of chemotherapeutic agents can be used for treatment of Hodgkin’s disease, including doxorubicin.
Answer A is incorrect. Azathioprine is used as an immunosuppressant in kidney transplant patients and those with autoimmune disorders.
Answer B is incorrect. Cisplatin (as well as other platinum-based chemotherapeutics) is used in the treatment of testicular, ovarian, and lung cancers.
Answer D is incorrect. Paclitaxel (Taxol) is used for treatment of ovarian and breast cancers.
Answer E is incorrect. β-Interferon is used in treatment of multiple sclerosis.Rate this question:
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.JPG "A 19-year-old man presents to the emergency department complaining of fatigue, lethargy, and a history of a recent upper respiratory infection. His temperature is 37° C (98.6° F) and his physical examination shows jaundice and prominent splenomegaly. Blood counts show decreased hemoglobin (9 g/dL), elevated mean cell hemoglobin concentration, and increased reticulocyte count. A peripheral blood smear is shown in the image. Which of the following is the definitive choice of therapy for this condition?")
.JPG "A 45-year-old man visited his primary care physician
1 month ago because of chest pain that
he had experienced four times in the past 4
months. The onset of the pain is sudden and
radiates to his left jaw. He usually feels the pain
when he is watching television but has never
felt it during exertion. During last month’s
visit, the physician prescribed sublingual nitroglycerin,
and the patient reports that this has
shortened the duration of his episodes. Last
month’s ECG is shown in the image. Which of
the following is the most likely cause of this patient’s
chest pain?")
.JPG "A 73-year-old man who has atrial fi brillation
has been treated pharmacologically for 10
years. He presents to his primary care physician
complaining of generalized dyspnea. Pulmonary
function tests show forced expiratory volume
in 1 second (FEV
1) and forced vital capacity
(FVC) are both less than 70% of the
predicted value, with a ratio of FEV
1 to FVC
of 81%. The fl ow-volume curve is shown in the
image. Which of the following is a possible etiology
of this presentation?")
.JPG "A 51-year-old man presents to the emergency department 30 minutes after his wife noticed drooping of the left side of his face and difficulty speaking. The patient is alert and oriented to person and place. His wife states that he has a history of benign prostatic hypertrophy, peptic ulcer disease, and high blood pressure.
According to the wife, the patient has never experienced symptoms like this before and has never had surgery of any type. The physicians determine that the patient is hemodynamically stable and they obtain the CT scan shown in the image. Which of the following is the best next step in management?")
.JPG "A 35-year-old man presents comes to his primary
care physician with a chief complaint of palpitations and occasional chest pain. Further questioning reveals a recent history of weight loss, diarrhea, and heat intolerance. Laboratory evaluation shows anti-thyroid-stimulating hormone (TSH) receptor antibodies in the patient’s serum. Which of the following best describes this patient’s TSH and thyroid hormone levels relative to normal baseline values?")
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Mar 20, 2023Quiz Edited by
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