USMLE Step 1 Practice Questions (2026) — Free Sample Questions with Answers

14 free USMLE Step 1 questions from our 150-question bank, grouped by content domain. Each answer explains why the right option is right — and why every wrong option is wrong.

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Blood & Lymphoreticular/Immune Systems

Question 1

A 58-year-old woman with a 12-year history of rheumatoid arthritis comes to the physician because of increasing fatigue for 3 months. She takes hydroxychloroquine and low-dose prednisone. She has no gastrointestinal symptoms, and her last menstrual period was 6 years ago. Temperature is 37.2°C (99.0°F), pulse is 88/min, respirations are 14/min, and blood pressure is 124/78 mm Hg. Examination shows pallor and symmetric swelling of the metacarpophalangeal joints. A test of the stool for occult blood is negative. Laboratory studies are shown. Increased hepatic production of which of the following most likely contributes to this patient's anemia?

Laboratory studies
TestResult
Blood
Hemoglobin10.1 g/dL
Mean corpuscular volume79 µm³
Erythrocyte sedimentation rate58 mm/h
Serum
Iron38 µg/dL
Total iron-binding capacity210 µg/dL
Ferritin340 ng/mL
Reference ranges: iron 50–170 µg/dL; total iron-binding capacity 250–400 µg/dL; ferritin, female 10–120 ng/mL.
Show answer & explanation

Answer: D

Erythropoietin: Erythropoietin is made by renal peritubular interstitial cells, not the liver, and in chronic inflammation its production is inappropriately low rather than increased.

Ferroportin: Ferroportin is the iron exporter on enterocytes and macrophages; in this condition it is internalized and degraded, so its activity falls.

Haptoglobin: Haptoglobin is an acute-phase protein that binds free hemoglobin; it does not regulate iron availability for erythropoiesis.

Hepcidin: Correct. Low serum iron with low TIBC and high ferritin in a patient with active rheumatoid arthritis indicates anemia of chronic inflammation. IL-6 stimulates hepatocytes to make hepcidin, which binds ferroportin and causes its degradation; iron is trapped in macrophages and enterocytes, so iron is unavailable to erythroid precursors despite ample stores.

Transferrin: Transferrin synthesis decreases in inflammation (it is a negative acute-phase protein), which is why TIBC is low; increased transferrin (high TIBC) is the pattern of iron deficiency.

Inflammation → IL-6 → hepatic hepcidin → ferroportin degradation → iron sequestration: low iron, low TIBC, normal or high ferritin.

Behavioral Health & Nervous Systems/Special Senses

Question 2

A 46-year-old woman is brought to the emergency department 1 hour after the onset of a severe occipital headache, palpitations, and sweating that began during a dinner party at which she ate from a platter of aged cheeses and dry-cured sausage. She has major depressive disorder with atypical features (hypersomnia, increased appetite, and mood reactivity) that did not respond to a selective serotonin reuptake inhibitor or to bupropion; 2 months ago, her psychiatrist switched her to an antidepressant of a different class. She takes no other medications. Temperature is 37.4°C (99.3°F), pulse is 112/min, respirations are 20/min, and blood pressure is 218/124 mm Hg. She is diaphoretic. Pupils are 6 mm and reactive to light. Deep tendon reflexes are 2+ without clonus. Which of the following best explains this patient's hypertension?

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Answer: E

Blockade of norepinephrine reuptake: Blockade of the norepinephrine transporter (cocaine, tricyclics) raises synaptic catecholamines, but nothing suggests cocaine use, and a crisis triggered by a meal points to a diet–drug interaction.

Catecholamine secretion by a chromaffin tumor: Pheochromocytoma causes episodic headache, palpitations, and sweating, but onset during a meal of aged cheese and cured meat soon after starting a new antidepressant class makes a tyramine–MAO inhibitor interaction far more likely.

Direct agonism at α1-adrenergic receptors: Tyramine is an indirect sympathomimetic; it acts mainly by releasing stored norepinephrine, not by binding α1 receptors directly (as phenylephrine does).

Excess stimulation of serotonin receptors: Serotonin syndrome causes hyperthermia, clonus, and hyperreflexia after combining serotonergic drugs; this patient has no clonus and ingested tyramine rather than a second serotonergic agent.

Indirect release of stored norepinephrine: Correct. A hypertensive crisis during a meal of aged cheese and cured meat, in a patient with atypical depression recently switched to a new antidepressant class, indicates a monoamine oxidase inhibitor (such as phenelzine) interacting with dietary tyramine. MAO inhibitors irreversibly inhibit MAO-A and MAO-B, including intestinal and hepatic MAO-A that normally inactivates tyramine. Unmetabolized tyramine, an indirect sympathomimetic, is taken up into sympathetic nerve terminals by the norepinephrine transporter and displaces vesicular norepinephrine, whose stores are enlarged by MAO inhibition; massive norepinephrine release causes the crisis.

Patients taking MAO inhibitors must avoid tyramine-rich foods: unmetabolized tyramine (an indirect sympathomimetic) releases large stores of norepinephrine and causes hypertensive crisis.

Question 3

A 67-year-old man is brought to the emergency department 2 hours after the sudden onset of weakness of his right arm and leg. He has hypertension and type 2 diabetes mellitus. Temperature is 37.0°C (98.6°F), pulse is 84/min, respirations are 16/min, and blood pressure is 172/96 mm Hg. Examination shows 2/5 strength in the right upper and lower extremities and a right extensor plantar response. Vibratory and position sense are decreased over the right side of the body; pain and temperature sensation are intact throughout. Facial sensation and strength are normal. MRI of the brain shows an acute infarct in the shaded area of the axial section of the medulla shown. Which of the following additional findings is most likely?

Axial section of the medulla (schematic)
PosteriorAnteriorPatient's leftPatient's right
Shaded area = infarct. Dashed line = midline. Figure not drawn to scale.
Show answer & explanation

Answer: A

Deviation of the tongue to the left: Correct. Contralateral (right) hemiparesis and loss of vibration/position sense with spared pain and temperature sensation localize to the left medial medulla (medial medullary syndrome, anterior spinal artery or vertebral paramedian branches). The infarct involves the left pyramid (corticospinal tract, decussates below), the left medial lemniscus (already crossed), and the exiting left hypoglossal nerve fibers/nucleus. Paralysis of the left genioglossus lets the intact right genioglossus push the tongue toward the side of the lesion.

Deviation of the tongue to the right: Tongue deviation to the right follows a right hypoglossal (lower motor neuron) lesion or a left cortical lesion of corticobulbar fibers; with a left medullary lesion of CN XII the tongue deviates left.

Hoarseness and dysphagia: Hoarseness and dysphagia result from involvement of the nucleus ambiguus in the lateral medulla (posterior inferior cerebellar artery territory), which lies outside the shaded medial area.

Left-sided ptosis and miosis: Horner syndrome results from interruption of descending hypothalamospinal sympathetic fibers in the lateral medulla (lateral medullary syndrome), not the medial medulla.

Loss of pain sensation over the left face: Ipsilateral facial pain and temperature loss reflects the spinal trigeminal nucleus and tract in the lateral medulla; this patient's facial sensation is normal and pain sensation is spared.

Medial medullary syndrome = contralateral hemiparesis (pyramid) + contralateral loss of vibration/proprioception (medial lemniscus) + ipsilateral tongue deviation (CN XII); lateral medullary findings (ambiguus, spinal trigeminal, spinothalamic, sympathetic) are absent.

Musculoskeletal, Skin & Subcutaneous Tissue

Question 4

A 58-year-old man comes to the physician because of a 3-month history of a red, scaly, burning rash on the backs of his hands, forearms, and neck that worsens after sun exposure. For the past year, he has had episodes of facial flushing with wheezing and watery diarrhea. His wife says he has also become forgetful. He does not drink alcohol and eats a varied diet. Temperature is 37.0°C (98.6°F), pulse is 88/min, respirations are 16/min, and blood pressure is 122/78 mm Hg. Examination shows sharply demarcated, hyperpigmented, scaly plaques on sun-exposed skin, including a band around the neck. The liver edge is firm and nodular, 4 cm below the right costal margin. A grade 3/6 holosystolic murmur is heard at the left lower sternal border and increases with inspiration. Urinary 5-hydroxyindoleacetic acid excretion is increased. Deficiency of which of the following is the most likely cause of this patient's skin findings?

Show answer & explanation

Answer: A

Niacin: Correct. Flushing, bronchospasm, diarrhea, right-sided valve disease (tricuspid regurgitation), hepatic metastases, and increased urinary 5-HIAA indicate carcinoid syndrome. The tumor diverts up to half of dietary tryptophan to serotonin synthesis instead of the kynurenine pathway that normally supplies part of the body's niacin (as NAD⁺/NADP⁺). The resulting niacin deficiency causes pellagra: photosensitive dermatitis (including the neck band, Casal necklace), diarrhea, and dementia.

Riboflavin: Riboflavin (FAD/FMN) deficiency causes angular cheilitis, glossitis, and seborrheic dermatitis, not a photosensitive, sharply demarcated rash.

Thiamine: Thiamine deficiency causes beriberi (neuropathy, high-output heart failure) and Wernicke encephalopathy, without dermatitis.

Vitamin A: Vitamin A deficiency causes night blindness, xerophthalmia, and follicular hyperkeratosis (phrynoderma), not a photosensitive dermatitis linked to carcinoid tumor.

Zinc: Zinc deficiency (acrodermatitis enteropathica, or acquired) causes periorificial and acral eczematous dermatitis, alopecia, and diarrhea, but not in a photosensitive distribution, and is not caused by tryptophan diversion.

Carcinoid tumors divert tryptophan to serotonin and can cause niacin deficiency (pellagra: dermatitis, diarrhea, dementia); isoniazid and Hartnup disease are other causes.

Cardiovascular System

Question 5

A 24-year-old woman comes to the office for a preparticipation sports examination. She has occasional palpitations but no chest pain or syncope. She is 180 cm (5 ft 11 in) tall and weighs 59 kg (130 lb); BMI is 18 kg/m². Her arm span exceeds her height, and the thumb protrudes beyond the ulnar border of the clenched fist. With the patient supine, auscultation shows a midsystolic click followed by a late systolic murmur heard best at the apex. Which of the following maneuvers is most likely to cause the click to occur earlier in systole?

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Answer: E

Isometric handgrip: Handgrip raises afterload and left ventricular volume, so the leaflets prolapse later and the click moves toward S2.

Passive leg elevation: Raising the legs increases venous return and left ventricular end-diastolic volume, delaying prolapse and the click.

Phenylephrine infusion: Phenylephrine increases afterload through α1 vasoconstriction, enlarging the ventricle and delaying the click.

Squatting: Squatting increases both venous return and afterload, so the ventricle is larger and the click occurs later.

Standing abruptly: Correct. Marfan-like features with a midsystolic click and late systolic murmur indicate mitral valve prolapse (myxomatous degeneration). Standing abruptly reduces venous return and left ventricular volume; the redundant leaflets reach the point of prolapse earlier in systole, so the click and murmur move toward S1.

In mitral valve prolapse, anything that makes the left ventricle smaller (standing, Valsalva strain) moves the click earlier; anything that enlarges it (squatting, leg raise, handgrip) moves it later.

Respiratory & Renal/Urinary Systems

Question 6

A 34-year-old woman comes to the emergency department 2 hours after the sudden onset of shortness of breath and sharp right-sided chest pain that worsens with inspiration. Yesterday she returned from a 14-hour flight. She takes an oral contraceptive. Pulse is 116/min, respirations are 28/min, and blood pressure is 122/76 mm Hg. Her left calf is swollen and tender. The lungs are clear to auscultation. While she breathes room air, arterial blood gas analysis shows pH 7.49, Pco₂ 30 mm Hg, and Po₂ 62 mm Hg. With supplemental oxygen by face mask, her Po₂ increases to 145 mm Hg. Which of the following is the most likely mechanism of this patient's hypoxemia?

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Answer: E

Alveolar hypoventilation: Hypoventilation raises Pco₂ and leaves the A–a gradient normal; she is hyperventilating (Pco₂ 30 mm Hg) and her A–a gradient is about 50 mm Hg.

Decreased inspired oxygen tension: Low inspired O₂ (high altitude) causes hypoxemia with a normal A–a gradient; she is at sea level breathing room air.

Diffusion impairment: Diffusion limitation occurs with thickened alveolar membranes (interstitial fibrosis) and is usually revealed by exercise; it does not explain sudden hypoxemia after a long flight.

Right-to-left shunt: Hypoxemia from a true shunt improves little with supplemental oxygen; her Po₂ rose substantially.

Ventilation–perfusion mismatch: Correct. Sudden pleuritic chest pain and dyspnea after a long flight in a woman taking an oral contraceptive, with a swollen calf, indicate pulmonary embolism. The A–a gradient is elevated (150 − 30/0.8 = 112.5 mm Hg alveolar; 112.5 − 62 ≈ 50 mm Hg). Emboli redirect blood flow to the remaining vessels, overperfusing those lung units relative to their ventilation (low V/Q areas); because those units still receive some ventilation, the hypoxemia corrects with supplemental oxygen.

Hypoxemia with an increased A–a gradient that corrects with supplemental oxygen indicates ventilation–perfusion mismatch, the main mechanism in pulmonary embolism; shunt is refractory to oxygen.

Question 7

A 71-year-old man with heart failure is admitted to the hospital because of worsening shortness of breath and leg edema. He is treated with an intravenous diuretic and loses 4 kg (9 lb) over 3 days. On the third day, he reports ringing in his ears. Serum studies show K⁺ 3.1 mEq/L, HCO₃⁻ 33 mEq/L, Mg²⁺ 1.3 mg/dL, and Ca²⁺ 8.3 mg/dL; urinary calcium excretion is increased. The diuretic most likely acts at which of the labeled sites in the schematic?

Nephron and collecting duct
CortexMedullaABCDE
Schematic of a juxtamedullary nephron and its collecting duct; letters mark tubular segments. Figure not drawn to scale.
Show answer & explanation

Answer: D

A: Site A is the proximal convoluted tubule, where carbonic anhydrase inhibitors act; they cause a metabolic acidosis, not alkalosis.

B: Site B is the thin descending limb, which is permeable to water and has no diuretic target.

C: Site C is the distal convoluted tubule, where thiazides inhibit the Na⁺-Cl⁻ cotransporter; thiazides decrease urinary calcium and are not ototoxic.

D: Correct. A potent intravenous diuretic causing hypokalemia, metabolic alkalosis, hypomagnesemia, hypercalciuria, and tinnitus is a loop diuretic. Loop diuretics inhibit the Na⁺-K⁺-2Cl⁻ cotransporter in the thick ascending limb (site D), which abolishes the lumen-positive potential that drives paracellular Ca²⁺ and Mg²⁺ reabsorption; ototoxicity reflects inhibition of the same transporter in the inner ear.

E: Site E is the collecting duct, where ENaC blockers and aldosterone antagonists act; these spare or raise potassium.

Loop diuretics block the Na⁺-K⁺-2Cl⁻ cotransporter in the thick ascending limb, causing hypokalemic alkalosis, hypercalciuria, hypomagnesemia, and dose-related ototoxicity.

Gastrointestinal System

Question 8

A 34-year-old man comes to the emergency department because of severe, colicky left flank pain radiating to the groin for 6 hours. He has Crohn disease and underwent resection of 100 cm of the terminal ileum 2 years ago. Since the operation, he has had four to five loose, greasy, foul-smelling stools daily. He takes no medications. Urinalysis shows blood and envelope-shaped crystals. CT of the abdomen shows a 6-mm radiopaque stone in the left ureter. Which of the following best explains the formation of this patient's stone?

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Answer: C

Increased hepatic oxalate synthesis: Hepatic overproduction of oxalate occurs in the inherited primary hyperoxalurias, which present in childhood; this patient has an acquired, intestinal cause.

Increased intestinal calcium absorption: Absorptive hypercalciuria is the opposite of what happens here: unabsorbed fatty acids bind luminal calcium and decrease its absorption.

Oxalate hyperabsorption in the colon: Correct. Envelope-shaped crystals and a radiopaque stone indicate calcium oxalate nephrolithiasis. Ileal resection depletes the bile acid pool, causing fat malabsorption (steatorrhea). Unabsorbed fatty acids bind dietary calcium, leaving oxalate free in the lumen, and bile acids and fatty acids reaching the colon increase its permeability, so the colon absorbs excess oxalate, which the kidney excretes (enteric hyperoxaluria).

Urease-producing bacterial infection: Urease-producing organisms (e.g., Proteus) form struvite stones with coffin-lid crystals in alkaline urine.

Uric acid supersaturation in acidic urine: Uric acid stones also occur in patients with chronic diarrhea, but they are radiolucent and form rhomboid crystals.

Fat malabsorption leaves oxalate unbound by calcium, so the colon absorbs it and calcium oxalate stones form; this enteric hyperoxaluria requires an intact colon.

Reproductive & Endocrine Systems

Question 9

A 46-year-old woman comes to the physician because of a 6-month history of diffuse bone pain and difficulty climbing stairs. Five years ago she underwent Roux-en-Y gastric bypass surgery and has since lost 50 kg (110 lb). She stopped taking her prescribed vitamin and mineral supplements 3 years ago. Examination shows tenderness to palpation over the ribs and tibias, a waddling gait, and weakness of the hip flexors. X-rays show bilateral, symmetric transverse lucencies at the medial femoral necks. Serum alkaline phosphatase activity is increased. Which of the following sets of serum findings is most likely in this patient? Each option lists, in order: calcium / phosphate / parathyroid hormone (PTH).

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Answer: A

↓ / ↓ / ↑: Correct. Bone pain, proximal myopathy, pseudofractures (Looser zones), and high alkaline phosphatase after gastric bypass with no supplements indicate osteomalacia from vitamin D (and calcium) malabsorption. Low 1,25-dihydroxyvitamin D reduces intestinal calcium and phosphate absorption; the low calcium drives secondary hyperparathyroidism, and PTH further lowers phosphate by inhibiting proximal tubular reabsorption.

↑ / ↑ / ↓: The calcium and phosphate entries are wrong: high calcium and phosphate with suppressed PTH is vitamin D toxicity, the opposite of deficiency.

↓ / ↑ / ↓: The phosphate and PTH entries are wrong: low calcium with high phosphate and low PTH is hypoparathyroidism.

↓ / ↑ / ↑: The phosphate entry is wrong: low calcium and high phosphate with high PTH is the pattern of chronic kidney disease, in which phosphate cannot be excreted.

↑ / ↓ / ↑: The calcium entry is wrong: high calcium with high PTH and low phosphate is primary hyperparathyroidism.

Vitamin D deficiency gives low (or low-normal) calcium, low phosphate, and high PTH with high alkaline phosphatase; bypassing the duodenum and proximal jejunum predisposes to it.

Question 10

A 34-year-old woman comes to the physician because of a 3-month history of palpitations, heat intolerance, and a 5-kg (11-lb) weight loss despite an increased appetite. Temperature is 37.4°C (99.3°F), pulse is 112/min, respirations are 16/min, and blood pressure is 138/62 mm Hg. Examination shows bilateral proptosis with lid retraction, a diffusely enlarged, nontender thyroid gland with a bruit, and nonpitting, waxy thickening of the skin over both shins. Laboratory studies are shown. Which of the following is the most likely mechanism of this patient's hyperthyroidism?

Laboratory studies
TestResult
Serum
Thyroid-stimulating hormone (TSH)<0.01 μU/mL
Free thyroxine (T4)4.2 ng/dL
Thyroid scan
Radioactive iodine uptake (24 hours)68%, diffuse and homogeneous
Reference ranges: TSH 0.4–4.0 μU/mL; free T4 0.9–1.7 ng/dL; radioactive iodine uptake at 24 hours 8%–30%.
Show answer & explanation

Answer: A

Antibodies that stimulate the TSH receptor: Correct. Graves disease: diffuse goiter with a bruit, ophthalmopathy, pretibial myxedema, suppressed TSH, and diffusely increased iodine uptake. IgG antibodies (thyroid-stimulating immunoglobulins) bind and activate the TSH receptor, driving follicular hyperplasia and hormone synthesis; the same autoimmunity activates orbital and pretibial fibroblasts to deposit glycosaminoglycans.

Autonomous hyperfunctioning nodule: A toxic adenoma shows a single hot nodule with suppressed uptake in the rest of the gland and does not cause ophthalmopathy or dermopathy.

Excess pituitary TSH secretion: A TSH-secreting pituitary adenoma would give an inappropriately normal or high TSH, not a suppressed one.

Exogenous thyroid hormone ingestion: Factitious thyrotoxicosis suppresses iodine uptake and causes a small, not enlarged, gland.

Release of preformed hormone from damaged follicles: Thyroiditis (subacute or painless) leaks stored hormone, so radioactive iodine uptake is low, not increased.

Thyrotoxicosis with a high, diffuse iodine uptake plus ophthalmopathy and pretibial myxedema means Graves disease, caused by stimulating antibodies to the TSH receptor (type II hypersensitivity).

Multisystem Processes & Disorders

Question 11

A 38-year-old man comes to the physician because of a 2-month history of a painful rash, hair loss, and a diminished sense of taste. He has Crohn disease, and 80 cm of distal jejunum and ileum were resected 6 months ago; since then, he has had six to eight watery stools daily. Temperature is 37.0°C (98.6°F), pulse is 88/min, respirations are 14/min, and blood pressure is 114/72 mm Hg. Examination shows sharply demarcated, erythematous, crusted, scaly plaques around the mouth, nostrils, and anus and on the fingers and soles. Sun-exposed skin of the neck and forearms is normal. There is patchy alopecia of the scalp, and a 2-cm segment of the surgical incision has not healed. Serum alkaline phosphatase activity is decreased. Deficiency of which of the following is the most likely cause of these findings?

Show answer & explanation

Answer: E

Biotin: Biotin deficiency (raw egg white avidin, prolonged biotin-free parenteral nutrition) can cause periorificial dermatitis and alopecia, but not loss of taste, poor wound healing, or a low alkaline phosphatase.

Copper: Copper deficiency (after bariatric surgery or excess zinc intake) causes anemia, neutropenia, and a myeloneuropathy resembling B12 deficiency, not acral and periorificial dermatitis.

Niacin: Pellagra causes dermatitis of sun-exposed skin (such as a necklace distribution), diarrhea, and dementia; this patient's sun-exposed skin is spared.

Selenium: Selenium deficiency (prolonged unsupplemented parenteral nutrition) causes cardiomyopathy, myopathy, and whitened nail beds from loss of glutathione peroxidase activity, not periorificial dermatitis or loss of taste.

Zinc: Correct. Chronic diarrhea after small-bowel resection causes zinc malabsorption and loss. Acquired zinc deficiency mimics acrodermatitis enteropathica: crusted periorificial and acral plaques, alopecia, diarrhea, impaired taste, and poor wound healing. Zinc is a catalytic or structural cofactor for hundreds of enzymes and for zinc-finger transcription factors, so rapidly renewing epithelium and healing wounds are affected first; low activity of alkaline phosphatase, a zinc metalloenzyme, supports the diagnosis.

Zinc deficiency (chronic diarrhea, malabsorption, alcohol use disorder, unsupplemented parenteral nutrition) causes periorificial and acral dermatitis, alopecia, dysgeusia, poor wound healing, and a low alkaline phosphatase.

Question 12

A 3-year-old girl is brought to the physician because of irritability, abdominal pain, and constipation for 2 months. Her family lives in a house built in 1938 that is being renovated, and she has been seen chewing on painted windowsills. Vital signs are within normal limits. Examination shows pallor. Her hemoglobin concentration is 9.6 g/dL, and mean corpuscular volume is 70 µm³. A peripheral blood smear shows erythrocytes with coarse basophilic stippling. Her erythrocyte zinc protoporphyrin concentration is markedly increased. The diagram shows the pathway of heme synthesis. Inhibition of the step labeled by which of the following letters best explains the increased erythrocyte protoporphyrin concentration in this patient?

Heme synthesis pathway
MitochondrionCytosolGlycine + succinyl-CoAδ-Aminolevulinic acidPorphobilinogenUroporphyrinogen IIICoproporphyrinogen IIIProtoporphyrin IXHemeDABEC
Simplified pathway. The dashed arrow represents two enzymatic steps (via hydroxymethylbilane), and step E also combines two enzymatic steps (via protoporphyrinogen IX). Steps A, B, C, and D are single enzymatic steps. Figure not drawn to scale.
Show answer & explanation

Answer: C

A: Step A is δ-aminolevulinic acid dehydratase. Lead does inhibit it, but that block raises δ-ALA (contributing to neurotoxicity), not protoporphyrin; acute intermittent porphyria involves the next (dashed) step.

B: Step B is uroporphyrinogen decarboxylase, deficient in porphyria cutanea tarda, which causes photosensitive blistering and uroporphyrin accumulation.

C: Correct. Pica in an old house with peeling paint, abdominal pain, constipation, microcytic anemia and coarse basophilic stippling indicate lead poisoning. Lead inhibits ferrochelatase (step C), which inserts Fe²⁺ into protoporphyrin IX in the mitochondrion; protoporphyrin accumulates and zinc is incorporated instead, so erythrocyte zinc protoporphyrin rises. Lead also inhibits δ-ALA dehydratase and erythrocyte pyrimidine 5′-nucleotidase (causing stippling).

D: Step D is δ-aminolevulinic acid synthase, the rate-limiting, vitamin B6–dependent step; its deficiency causes X-linked sideroblastic anemia with low protoporphyrin.

E: Step E (coproporphyrinogen oxidase and protoporphyrinogen oxidase) is defective in hereditary coproporphyria and variegate porphyria, not lead poisoning, and would lower rather than raise protoporphyrin formation.

Lead inhibits ferrochelatase and δ-ALA dehydratase: ↑ δ-ALA, ↑ (zinc) protoporphyrin, microcytic anemia with basophilic stippling.

Biostatistics & Epidemiology/Population Health

Question 13

Investigators evaluate a new serum assay for detecting a chronic liver disease. They test 1000 adults referred to a hepatology clinic, and every participant also undergoes liver biopsy, the reference standard. The results are shown. A 52-year-old woman from this same referral population has a positive result on the assay. Based on these data, which of the following is the probability that she has the disease?

Assay results compared with liver biopsy
Assay resultDisease present on biopsyDisease absent on biopsy
Positive9060
Negative10840
Show answer & explanation

Answer: B

  1. Set up the 2 × 2 table: a (true positive) = 90, b (false positive) = 60, c (false negative) = 10, d (true negative) = 840.
  2. PPV = a/(a + b): the proportion of positive results that are true positives.
  3. PPV = 90/(90 + 60) = 90/150.
  4. 90/150 = 3/5 = 60%.

10%: This is the prevalence of disease in the sample (100/1000), the pretest probability, not the probability after a positive result.

60%: Correct. The question asks for the positive predictive value: true positives divided by all positive results, 90/(90 + 60) = 90/150 = 60%. Because the woman comes from the same referral population, the PPV calculated in this sample applies to her.

90%: This is the sensitivity, 90/(90 + 10): the probability of a positive test in a person who has the disease, not the reverse.

93%: This is the specificity, 840/(840 + 60): the probability of a negative test in a person without the disease.

99%: This is the negative predictive value, 840/(840 + 10), which applies to a patient with a negative result.

The probability of disease after a positive test is the PPV, a/(a + b). It is read across the positive-test row, whereas sensitivity and specificity are read down the disease columns.

Social Sciences: Communication and Interpersonal Skills

Question 14

A 64-year-old man comes to the office with his wife to receive the results of a CT-guided biopsy of a 3-cm mass in his right lung that was found on a chest x-ray obtained because of a persistent cough. The biopsy specimen shows adenocarcinoma. The patient has not been told anything about the results. The physician has arranged a private room, both are seated, and the patient says he would like his wife to stay for the discussion. Which of the following is the most appropriate next statement by the physician?

Show answer & explanation

Answer: E

"Your biopsy showed lung cancer, but there are many good treatments.": Stating the diagnosis immediately, without first learning what he understands or expects, can overwhelm him; the added reassurance may also be premature.

"Don't worry, we found this at a stage where it can be treated.": False or premature reassurance is inappropriate before staging is complete and before the diagnosis has been shared.

"Before we start, have you thought about whether you would want chemotherapy?": Raising treatment choices before he has heard the diagnosis presumes the result, skips assessing his perception, and leaves him to guess the news.

"The pathologist found adenocarcinoma cells in the tissue sample.": Technical jargon given abruptly is hard for patients to understand, and it skips assessing his perception.

"What is your understanding of why we did the biopsy?": Correct. The setting has been arranged, so the next step in delivering bad news is to assess the patient's perception: what he knows and expects. His answer lets the physician start from his understanding, correct misconceptions and judge how much information he wants before giving the diagnosis.

When breaking bad news (SPIKES), arrange the setting, then assess the patient's perception and invitation, before giving the information with a warning shot in plain language and responding to emotions.

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