Free USMLE Step 1 practice test — 280 questions, 14 blocks 🇺🇸

8 original practice questions written by Myndaq in the USMLE Step 1 style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.

1. Question 1Pharmacology - CardiovascularOne correct answer

   A 67-year-old man with heart failure with reduced ejection fraction (left ventricular ejection fraction 30%) still becomes short of breath after climbing one flight of stairs despite taking lisinopril, metoprolol succinate, spironolactone and furosemide. His blood pressure is 118/72 mm Hg, and serum potassium is 4.6 mEq/L (4.6 mmol/L). His cardiologist decides to replace lisinopril with a fixed-dose combination of sacubitril and valsartan. He is told to take his last lisinopril dose tonight and to wait at least 36 hours before swallowing the first tablet of the new combination. This waiting period mainly protects him against a serious adverse effect caused by accumulation of which substance?

   1. A. Angiotensin II
   2. B. Bradykinin
   3. C. Aldosterone
   4. D. Atrial natriuretic peptide
   5. E. Norepinephrine
   Show answer and explanation

   Answer: B

   Sacubitril is a prodrug whose active metabolite inhibits neprilysin, the neutral endopeptidase that degrades natriuretic peptides and also bradykinin. Angiotensin-converting enzyme (also called kininase II) is the other main enzyme that breaks down bradykinin. If an ACE inhibitor and a neprilysin inhibitor act at the same time, bradykinin builds up and the risk of angioedema rises sharply. Letting lisinopril wash out for about 36 hours before the first dose of sacubitril-valsartan, and never giving the two together, prevents this.

   Angiotensin II is not the concern: lisinopril reduces its formation and valsartan blocks its AT1 receptor. Aldosterone falls rather than rises with these drugs. Natriuretic peptide levels do rise with sacubitril, but that is the intended therapeutic effect, and ACE inhibitors do not add to it. Norepinephrine is not a substrate of either enzyme.

2. Question 2MicrobiologyOne correct answer

   A 23-year-old man comes to the clinic 2 days after returning from a 3-week volunteer construction project in a rural coastal town in Peru. He has had 6 to 8 large, watery, nonbloody stools each day with mild abdominal cramping. His temperature is 37.4°C (99.3°F), and he shows no signs of dehydration. Stool microscopy shows no leukocytes. Culture grows a lactose-fermenting, gram-negative rod, and molecular testing of the isolate detects the gene for a heat-stable enterotoxin but no other toxin or invasion genes. How does the toxin made by this isolate most likely produce his diarrhea?

   1. A. ADP-ribosylation of the Gs alpha subunit, raising enterocyte cAMP
   2. B. Stimulation of guanylyl cyclase C, raising enterocyte cGMP
   3. C. Removal of an adenine base from 28S ribosomal RNA, halting protein synthesis
   4. D. ADP-ribosylation of elongation factor 2
   5. E. Invasion of colonic mucosa with actin-based spread from cell to cell
   Show answer and explanation

   Answer: B

   This is traveler's diarrhea caused by enterotoxigenic Escherichia coli (ETEC): watery, noninflammatory diarrhea after travel, with a lactose-fermenting gram-negative rod. ETEC can make two toxins. The heat-stable toxin binds guanylyl cyclase C on the apical membrane of enterocytes and raises intracellular cGMP. This inhibits NaCl absorption and promotes chloride secretion through CFTR, and water follows into the lumen.

   ADP-ribosylation of Gs with a rise in cAMP is the action of the heat-labile toxin of ETEC and of cholera toxin, which this isolate lacks. Removing an adenine from 28S rRNA is how Shiga toxin and Shiga-like toxin stop protein synthesis, typically causing bloody diarrhea. ADP-ribosylation of elongation factor 2 is the action of diphtheria toxin and Pseudomonas exotoxin A. Actin-based spread through colonic cells describes Shigella, which causes inflammatory, bloody stools with leukocytes.

3. Question 3BiochemistryOne correct answer

   A 7-month-old girl is brought to the emergency department 40 minutes after she became pale, sweaty and limp and vomited twice, shortly after a serving of fruit puree. Until 10 days ago she had been fed only breast milk and iron-fortified rice cereal; since then her grandmother has given her sweetened pear and apple puree most days, and she has vomited after several of these feeds. She is lethargic, and the liver edge is palpable 4 cm below the right costal margin. Serum glucose is 36 mg/dL (2.0 mmol/L). A urine test for reducing substances is positive, while a glucose-specific urine dipstick is negative. What most directly accounts for her low blood glucose?

   1. A. Absent fructokinase activity, allowing unmetabolized fructose to spill into the urine
   2. B. Trapping of inorganic phosphate in a phosphorylated sugar, limiting hepatic glycogen breakdown
   3. C. Deficient glucose-6-phosphatase, blocking the final step of hepatic glucose release
   4. D. Accumulation of galactose-1-phosphate derived from dietary lactose
   5. E. Fructose-stimulated insulin secretion from pancreatic beta cells
   Show answer and explanation

   Answer: B

   Vomiting and hypoglycemia after meals since fruit and sucrose entered the diet, with hepatomegaly and a non-glucose reducing sugar in the urine, point to hereditary fructose intolerance, an autosomal recessive deficiency of aldolase B. Fructokinase still converts fructose to fructose-1-phosphate, but aldolase B cannot cleave it, so fructose-1-phosphate accumulates in the liver. This ties up inorganic phosphate and depletes ATP. Glycogen phosphorylase needs inorganic phosphate to release glucose-1-phosphate from glycogen, and gluconeogenesis is also impaired, so blood glucose falls quickly after fructose intake. Treatment is strict avoidance of fructose, sucrose and sorbitol.

   Fructokinase deficiency (essential fructosuria) also gives a positive reducing-substance test but is harmless and does not cause hypoglycemia. Glucose-6-phosphatase deficiency and galactosemia would have caused trouble earlier, on breast milk or between feeds, not only after fruit. Fructose is a weak insulin secretagogue, and hyperinsulinism would not explain the hepatomegaly or the sugar in the urine.

4. Question 4Physiology - RenalOne correct answer

   During a phase 1 study, a healthy 28-year-old volunteer receives constant intravenous infusions of inulin and of an investigational compound Z. Compound Z is not bound to plasma proteins and is freely filtered at the glomerulus.

   Steady-state measurements: plasma inulin 0.30 mg/mL; urine inulin 45 mg/mL; urine flow rate 0.8 mL/min; plasma compound Z 0.50 mg/mL; urine compound Z 7.5 mg/mL.

   Which statement best describes how the kidneys handle compound Z in this volunteer?

   1. A. Net tubular secretion of 6 mg/min
   2. B. Net tubular reabsorption of 54 mg/min
   3. C. Net tubular reabsorption of 6 mg/min
   4. D. Net tubular secretion of 54 mg/min
   5. E. Filtration only, with no net tubular transport
   Show answer and explanation

   Answer: B

   Inulin is freely filtered and is neither reabsorbed nor secreted, so its clearance equals the GFR: GFR = (U × V) / P = (45 mg/mL × 0.8 mL/min) / 0.30 mg/mL = 120 mL/min.

   Filtered load of Z = GFR × plasma Z = 120 mL/min × 0.50 mg/mL = 60 mg/min. Excretion rate of Z = urine Z × urine flow = 7.5 mg/mL × 0.8 mL/min = 6 mg/min. Less is excreted than filtered, so the tubules reabsorb the difference: 60 - 6 = 54 mg/min. As a check, the clearance of Z is 6 / 0.50 = 12 mL/min, well below the GFR of 120 mL/min (fractional excretion 10%), which confirms net reabsorption.

   6 mg/min is the excretion rate, not a transport rate. Net secretion would require excretion greater than the 60 mg/min filtered load, and filtration alone would make the clearance of Z equal to the GFR.

5. Question 5Pathology - HematologyOne correct answer

   A 36-year-old woman comes to the emergency department with 2 days of right upper quadrant pain and increasing abdominal girth. For the past month she has felt tired and has noticed dark brown urine, most obvious when she first voids in the morning. She takes no medications. She has scleral icterus, tender hepatomegaly and shifting dullness. Laboratory studies show hemoglobin 8.4 g/dL (84 g/L), reticulocytes 7%, leukocyte count 3,100/mm³, platelet count 91,000/mm³, markedly elevated lactate dehydrogenase, undetectable haptoglobin and raised indirect bilirubin. The direct antiglobulin test is negative. The peripheral smear shows neither spherocytes nor schistocytes, and her urine contains hemosiderin. Doppler ultrasonography shows occlusive thrombi in the hepatic veins. The underlying defect in this patient most likely involves which of the following?

   1. A. IgG autoantibodies bound to erythrocyte surface antigens
   2. B. An inherited defect in an erythrocyte cytoskeletal protein such as spectrin or ankyrin
   3. C. An acquired stem cell mutation that prevents synthesis of glycosylphosphatidylinositol (GPI) anchors
   4. D. An activating mutation of the JAK2 tyrosine kinase in hematopoietic stem cells
   5. E. Severe deficiency of ADAMTS13, the von Willebrand factor-cleaving protease
   Show answer and explanation

   Answer: C

   Coombs-negative intravascular hemolysis (high LDH, absent haptoglobin, hemosiderinuria, dark morning urine), low counts in all three cell lines and thrombosis at an unusual site (hepatic vein thrombosis, Budd-Chiari syndrome) together indicate paroxysmal nocturnal hemoglobinuria. A somatic mutation of the X-linked PIGA gene in a hematopoietic stem cell blocks GPI anchor synthesis, so the cell's progeny lack the GPI-anchored complement regulators CD55 (decay-accelerating factor) and CD59. Their erythrocytes are lysed by the membrane attack complex, and complement-activated platelets and released hemoglobin promote thrombosis. Flow cytometry showing loss of CD55 and CD59 confirms the diagnosis, and C5 complement inhibitors are used in treatment.

   Warm autoimmune hemolysis would give a positive direct antiglobulin test and spherocytes. Hereditary spherocytosis shows spherocytes and extravascular hemolysis without leukopenia or thrombosis. JAK2-positive polycythemia vera can cause hepatic vein thrombosis but raises the red cell mass rather than causing hemolytic anemia. ADAMTS13 deficiency (thrombotic thrombocytopenic purpura) produces schistocytes and does not cause leukopenia.

6. Question 6Biostatistics & EpidemiologyOne correct answer

   Investigators assess a new serum biomarker assay in 2,000 adults referred for evaluation of a pancreatic mass. Every participant also undergoes tissue sampling, which serves as the reference standard. Adenocarcinoma is confirmed in 400 participants, and the assay is positive in 340 of them. Among the 1,600 participants without adenocarcinoma, the assay is negative in 1,520. In this population, what is the probability that a participant with a positive assay result has pancreatic adenocarcinoma?

   1. A. 19%
   2. B. 81%
   3. C. 85%
   4. D. 95%
   5. E. 96%
   Show answer and explanation

   Answer: B

   Build the 2 × 2 table. True positives = 340 and false negatives = 400 - 340 = 60. True negatives = 1,520 and false positives = 1,600 - 1,520 = 80.

   The question asks for the positive predictive value: TP / (TP + FP) = 340 / (340 + 80) = 340 / 420 ≈ 0.81, or 81%.

   85% is the sensitivity (340/400) and 95% is the specificity (1,520/1,600). 96% is the negative predictive value (1,520/1,580). 19% is the share of positive results that are false (80/420). Predictive values depend on prevalence, which is 20% in this referred group, so the same assay would have a much lower positive predictive value in an unselected screening population.

7. Question 7ImmunologyOne correct answer

   A 29-year-old woman receives a unit of packed red blood cells for anemia after a postpartum hemorrhage. Within 5 minutes of the start of the transfusion, she develops generalized hives, swelling of the lips, wheezing and a blood pressure of 78/40 mm Hg. Her temperature is unchanged at 36.9°C (98.4°F). She has had frequent sinus infections since childhood and was treated twice for giardiasis. She has never received blood products before, and the pretransfusion crossmatch was compatible. Which mechanism best accounts for her reaction?

   1. A. Recipient IgM antibodies against ABO antigens on donor erythrocytes
   2. B. Recipient antibodies against IgA present in the donor plasma
   3. C. Donor anti-HLA antibodies activating neutrophils in the recipient's pulmonary capillaries
   4. D. Cytokines released from donor leukocytes during blood storage
   5. E. Circulatory overload from the transfused volume
   Show answer and explanation

   Answer: B

   Anaphylaxis (urticaria, angioedema, bronchospasm and hypotension) within minutes of starting a transfusion, without fever, in a patient with recurrent sinopulmonary infections and giardiasis suggests selective IgA deficiency. Many such patients have preformed anti-IgA antibodies, which can be present even without earlier transfusion. The small amount of IgA in the residual plasma of a red cell unit is enough to trigger mast cell activation. Future transfusions should use washed cellular products or components from IgA-deficient donors.

   An ABO-incompatible (acute hemolytic) reaction causes fever, flank pain, hemoglobinuria and sometimes disseminated intravascular coagulation, not isolated urticaria and wheeze. Transfusion-related acute lung injury from donor anti-HLA or antineutrophil antibodies causes hypoxemic noncardiogenic pulmonary edema, often with fever, rather than hives and lip swelling. Cytokines from stored leukocytes cause febrile nonhemolytic reactions. Circulatory overload produces hypertension and pulmonary edema, not hypotension with urticaria.

8. Question 8Behavioral ScienceOne correct answer

   A 52-year-old man who has smoked 1 pack of cigarettes daily for 30 years comes for a routine visit. When asked about smoking, he says, 'My cough keeps getting worse, and I know the cigarettes are behind it. I really think I should quit sometime in the next few months, but I haven't set a date or done anything about it yet.' Which stage of change best describes this patient at present?

   1. A. Precontemplation
   2. B. Contemplation
   3. C. Preparation
   4. D. Action
   5. E. Maintenance
   Show answer and explanation

   Answer: B

   In the transtheoretical (stages of change) model, contemplation means the person recognizes the problem and intends to change within roughly the next 6 months but has not yet committed to a plan. This patient links his cough to smoking and wants to quit in the coming months, yet he has set no date and taken no steps.

   Precontemplation would mean he sees no problem or has no intention to change. Preparation requires an intention to act within about the next month, usually with concrete steps such as choosing a quit date or obtaining nicotine replacement. Action is active change that has lasted less than 6 months, and maintenance is change sustained beyond 6 months. The fitting approach in contemplation is motivational interviewing to explore his ambivalence and move him toward setting a quit date.