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Free USMLE Step 1 Gastrointestinal practice questions

10 original exam-style questions from the StepStash Step 1 qbank, each with a full explanation, an educational objective and a link to the matching study note. Try each one before you open the answer. Reviewed by Veli Bakalov, MD · @HemeOncBuddy.

Open the full 2,500-question Step 1 qbank

Question 1 · Anal Canal Anatomy · easy

A 52-year-old man with HIV infection comes to the physician because of anal pain, itching and spots of blood on the toilet paper for 3 months. Anoscopy shows a 2-cm indurated, ulcerated lesion arising from the anoderm below the pectinate line. Biopsy shows invasive keratinizing squamous cell carcinoma. In situ hybridization of tumor cells is positive for HPV-16.

Which of the following lymph node groups is the most likely first site of nodal metastasis from this tumor?

  1. Internal iliac nodes
  2. Superficial inguinal nodes
  3. Inferior mesenteric nodes
  4. Para-aortic nodes
  5. Pararectal nodes along the superior rectal artery
Show answer and explanation

Correct answer: B. Superficial inguinal nodes

The pectinate (dentate) line marks the junction of hindgut endoderm above and proctodeum ectoderm below, and nearly every feature switches there. Below the line the canal is lined by stratified squamous epithelium, is innervated by the somatic inferior rectal nerve (so lesions hurt), is supplied by inferior rectal vessels draining to the internal iliac system and IVC, and drains lymph to the superficial inguinal nodes. This man has an HPV-16-driven squamous cell carcinoma below the line, a tumor associated with HIV infection and immunosuppression, so his first nodal spread is expected in the groin. Anal squamous cell carcinoma is treated first with chemoradiation using 5-FU and mitomycin, which preserves the sphincter.

A. Internal iliac nodes drain the canal above the pectinate line, where adenocarcinoma of columnar epithelium arises, and the middle rectal territory. C. Inferior mesenteric nodes drain the rectum and sigmoid along the superior rectal and sigmoid vessels, a hindgut pathway above the line. D. Para-aortic nodes are reached secondarily from iliac and mesenteric chains, not as a first station for a lesion below the line. E. Pararectal nodes along the superior rectal artery drain upper rectal mucosa and would be the first station for a rectal adenocarcinoma, not for a squamous lesion of the anoderm.

Educational objective: The key takeaway is that lesions below the pectinate line are squamous, painful and drain to the superficial inguinal nodes, whereas lesions above it drain to internal iliac and mesenteric nodes.
Schematic of the opened anal canal divided by the scalloped pectinate line, with rules on the left (above: painless, superior rectal vein to IMV, splenic and portal veins, internal iliac nodes; below: painful, inferior rectal vein to internal pudendal and internal iliac veins and the IVC, superficial inguinal nodes) and lesions pinned on the right (adenocarcinoma and internal hemorrhoid above, anal crypt abscess at the line, fissure, anal SCC and external hemorrhoid below).
Lesions above the pectinate line are painless and drain to the portal system and internal iliac nodes; lesions below it are painful and drain to the IVC and superficial inguinal nodes.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 2 · Groin hernias · easy

A 3-month-old boy, born at 34 weeks' gestation, is brought to the physician because of an intermittent swelling in his right groin that enlarges when he cries and disappears while he sleeps. On examination during crying, a soft, reducible bulge extends from the right groin into the right hemiscrotum. Both testes are palpable in the scrotum. The bulge cannot be transilluminated. At operation, the hernia sac is found entering the inguinal canal lateral to the inferior epigastric vessels and running within the spermatic cord.

Which of the following is the most likely cause of this patient's condition?

  1. Patent processus vaginalis
  2. Weakness of the transversalis fascia within Hesselbach triangle
  3. Enlarged femoral ring beneath the inguinal ligament
  4. Incomplete descent of the testis along the gubernaculum
  5. Persistent communication of the allantois with the bladder
Show answer and explanation

Correct answer: A. Patent processus vaginalis

A reducible groin bulge extending into the scrotum in a premature infant boy, with a sac that enters the deep inguinal ring lateral to the inferior epigastric vessels and runs within the spermatic cord, is an indirect inguinal hernia, the most common hernia overall. As the testis descends, it is preceded by an outpouching of peritoneum, the processus vaginalis, which normally obliterates around birth except for the tunica vaginalis surrounding the testis. If it stays patent, abdominal contents can follow it through the deep ring (an opening in the transversalis fascia), along the inguinal canal and out the superficial ring into the scrotum, picking up all three spermatic fascial coverings. Prematurity and male sex raise the risk, and infant indirect hernias can incarcerate, so they are repaired promptly. A narrower patency that admits only peritoneal fluid causes a communicating hydrocele, which transilluminates.

Remember: MDs don't LIe, Medial = Direct, Lateral = Indirect.

B. Weak transversalis fascia in Hesselbach triangle causes a direct inguinal hernia, which emerges medial to the inferior epigastric vessels in older men. C. A femoral hernia passes below the inguinal ligament, medial to the femoral vein, and is more common in women. D. Incomplete testicular descent causes cryptorchidism, but both testes are in the scrotum. E. A patent urachus, the allantois remnant, leaks urine from the umbilicus.

Educational objective: The key takeaway is that an indirect inguinal hernia passes through the deep ring lateral to the inferior epigastric vessels into a patent processus vaginalis, the most common hernia in infant boys.
Front-view schematic of the right groin: the deep ring lies lateral to the inferior epigastric vessels and an indirect hernia runs from it through the canal and superficial ring into the scrotum; Hesselbach triangle, bounded by the inferior epigastric vessels, rectus abdominis and inguinal ligament, is where a direct hernia pushes through; below the inguinal ligament the femoral nerve, artery, vein, canal and lymphatics run lateral to medial (NAVEL), and a femoral hernia enters the canal medial to the vein.
The inferior epigastric vessels sort inguinal hernias (lateral = indirect through the deep ring, medial = direct through Hesselbach triangle); femoral hernias pass below the inguinal ligament, medial to the femoral vein.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 3 · Cellular Adaptations · easy

A 58-year-old man comes to the physician because of heartburn after meals for 10 years that has been partly relieved by over-the-counter antacids. He is obese and smokes one pack of cigarettes daily. Upper endoscopy shows salmon-colored mucosa extending 4 cm above the gastroesophageal junction. Biopsy specimens from this area show columnar epithelium with goblet cells; the nuclei are small, basally located and uniform, and there is no architectural disarray.

The biopsy findings are best described as which of the following?

  1. Hyperplasia
  2. Low-grade dysplasia
  3. Metaplasia
  4. Carcinoma in situ
  5. Hypertrophy
Show answer and explanation

Correct answer: C. Metaplasia

Chronic acid and bile reflux injures the stratified squamous epithelium of the distal esophagus. Stem cells adapt by producing a cell type better able to withstand the injury: intestinal-type columnar epithelium with goblet cells. Replacement of one adult cell type by another is metaplasia, and in the esophagus it is called Barrett esophagus. Metaplasia is an adaptive change and can regress if the stimulus is removed, but persistent reflux can push it to dysplasia and then adenocarcinoma, so patients undergo surveillance endoscopy. In this biopsy the nuclei are small, basal and uniform without architectural disarray, which means there is no dysplasia yet.

A. Hyperplasia is an increase in the number of cells of the same type, such as endometrial hyperplasia from unopposed estrogen; it does not change the cell type. B. Low-grade dysplasia would show nuclear enlargement, hyperchromasia, stratification and loss of maturation toward the surface, which are absent here. D. Carcinoma in situ shows full-thickness atypia with an intact basement membrane; the cells here are cytologically bland. E. Hypertrophy is an increase in cell size without new cells, the only growth option for permanent tissues such as cardiac and skeletal muscle.

Educational objective: The key takeaway is that Barrett esophagus is metaplasia, a reversible replacement of squamous by intestinal-type columnar epithelium with goblet cells, which can progress to dysplasia and adenocarcinoma.
Four panels with before and after cell sketches contrast hypertrophy (bigger cells), hyperplasia (more cells), atrophy (smaller, fewer cells) and metaplasia (squamous to columnar with a goblet cell), above a strip showing metaplasia progressing to dysplasia, carcinoma in situ and invasive carcinoma, with the first two stages marked reversible.
Hypertrophy, hyperplasia, atrophy and metaplasia change cell size, number or type; a persistent stimulus can push metaplasia to dysplasia and carcinoma.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 4 · Antimetabolites · medium

A 58-year-old man undergoes right hemicolectomy for adenocarcinoma of the ascending colon with 3 of 18 regional lymph nodes involved. He begins adjuvant chemotherapy every 2 weeks with oxaliplatin, an intravenous bolus and 46-hour infusion of 5-fluorouracil, and folinic acid (leucovorin) infused just before the 5-fluorouracil on the same day. Pretreatment DPYD genotyping is normal. A medical student asks why the folinic acid is given together with the 5-fluorouracil rather than afterward.

Which of the following is the primary reason leucovorin is included in this regimen?

  1. Bypassing dihydrofolate reductase to rescue normal marrow cells
  2. Inhibiting dihydropyrimidine dehydrogenase to slow 5-FU breakdown
  3. Stabilizing the complex of 5-FdUMP with thymidylate synthase
  4. Increasing activation of 5-FU by thymidine phosphorylase
  5. Chelating platinum to prevent oxaliplatin neuropathy
Show answer and explanation

Correct answer: C. Stabilizing the complex of 5-FdUMP with thymidylate synthase

5-Fluorouracil is converted to 5-fluorodeoxyuridine monophosphate (5-FdUMP), which binds thymidylate synthase together with the folate cofactor 5,10-methylene-THF. In this covalent ternary complex the enzyme cannot methylate dUMP to dTMP, so DNA synthesis stops and cells die in S phase. The complex is stable only when methylene-THF is plentiful.

Leucovorin (5-formyl-THF) is an already reduced folate that enters the folate pool and expands the methylene-THF supply. This stabilizes and prolongs thymidylate synthase inhibition, so with 5-FU leucovorin potentiates cytotoxicity rather than rescuing cells. That is why it is given with or just before the 5-FU in FOLFOX and FOLFIRI, and why it also increases 5-FU toxicity such as mucositis and diarrhea.

A. Bypassing the DHFR block to rescue normal cells is leucovorin's role after high-dose methotrexate, not with 5-FU, which does not inhibit DHFR. B. Leucovorin does not inhibit dihydropyrimidine dehydrogenase; DPD deficiency is an inherited cause of severe 5-FU toxicity, and his DPYD genotype is normal. D. Thymidine phosphorylase performs the final step converting oral capecitabine to 5-FU in tissues; leucovorin does not induce it. E. Leucovorin has no chelating activity and does not prevent the cold-induced and cumulative sensory neuropathy of oxaliplatin.

Educational objective: The key takeaway is that leucovorin expands the 5,10-methylene-THF pool and stabilizes the 5-FdUMP and thymidylate synthase complex, potentiating 5-FU, whereas after methotrexate it rescues normal cells by bypassing DHFR.
Pathway diagram of the thymidylate cycle: DHFR reduces dihydrofolate to THF, THF becomes 5,10-methylene-THF and also feeds purine synthesis, and thymidylate synthase uses methylene-THF to turn dUMP into dTMP for DNA, returning dihydrofolate. Methotrexate blocks DHFR, 5-FU as 5-FdUMP blocks thymidylate synthase, hydroxyurea and gemcitabine block ribonucleotide reductase, leucovorin enters past the DHFR block, and folic acid cannot.
Methotrexate blocks DHFR and 5-FdUMP blocks thymidylate synthase; already-reduced leucovorin bypasses DHFR to rescue methotrexate toxicity but enlarges the methylene-THF pool to potentiate 5-FU.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 5 · Immunohistochemistry · medium

A 67-year-old man comes to the physician because of fatigue, early satiety and an 8-kg weight loss over 3 months. He has smoked one pack of cigarettes daily for 40 years. CT shows multiple round masses throughout the liver; the lungs are clear. A core biopsy of a liver lesion shows a gland-forming adenocarcinoma with tall columnar cells and central necrotic debris. Immunohistochemistry shows the tumor cells are cytokeratin 20-positive, cytokeratin 7-negative and positive for nuclear CDX2; TTF-1 is negative.

Which of the following is the most likely primary site of this tumor?

  1. Lung
  2. Breast
  3. Pancreas
  4. Urinary bladder
  5. Colon
Show answer and explanation

Correct answer: E. Colon

Immunohistochemistry identifies the lineage and site of origin of a metastasis when the primary is unknown. Cytokeratin positivity (here CK20) establishes an epithelial tumor, and the CK7 and CK20 profile narrows the organ. A CK7-negative, CK20-positive adenocarcinoma is colorectal until proven otherwise, and nuclear CDX2, an intestinal transcription factor, strongly supports intestinal origin. The morphology fits as well: tall columnar cells forming glands with central "dirty" necrosis are typical of colorectal adenocarcinoma. The liver is the most common site of colorectal metastasis because the colon drains through the portal vein, so the liver is the first capillary bed tumor cells meet.

A. Lung adenocarcinoma is usually CK7-positive, CK20-negative and TTF-1-positive; his smoking history is a distractor, and TTF-1 is negative. B. Breast carcinoma is CK7-positive, CK20-negative and typically GATA3-positive. C. Pancreatic adenocarcinoma is usually CK7-positive, often with variable CK20, and is not typically strongly CDX2-positive with a CK7-negative profile. D. Urothelial carcinoma of the bladder is characteristically positive for both CK7 and CK20 and expresses GATA3, and it does not form tall columnar glands.

Educational objective: The key takeaway is that a CK7-negative, CK20-positive, CDX2-positive adenocarcinoma in the liver indicates a colorectal primary.
Decision diagram: an undifferentiated tumor is screened with cytokeratin, CD45, S-100 and vimentin to assign carcinoma, lymphoma, melanoma or sarcoma, with confirming stains for each lineage and, for carcinoma, the CK7/CK20 pattern and site-specific markers.
Work up an unknown primary in two steps: four screening stains for lineage, then CK7/CK20 and a site marker for the organ.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 6 · Esophageal Diverticula · medium

A 76-year-old man comes to the physician because of difficulty initiating swallowing for 1 year. He often regurgitates undigested food that he ate the previous day, has a gurgling sound in his neck when he drinks, and his wife complains of his bad breath. He was treated for aspiration pneumonia 2 months ago. Examination shows a soft, compressible fullness in the left side of the neck. A barium swallow shows a contrast-filled pouch projecting posteriorly from the junction of the pharynx and esophagus.

Which of the following best describes the site at which this patient's lesion formed?

  1. Mid esophageal wall pulled outward by scarred mediastinal lymph nodes
  2. Distal esophageal wall just above the lower esophageal sphincter
  3. Area between the thyropharyngeal and cricopharyngeal fibers of the inferior constrictor
  4. Esophageal hiatus of the diaphragm
  5. Thyrohyoid membrane at the entry of the internal laryngeal nerve
Show answer and explanation

Correct answer: C. Area between the thyropharyngeal and cricopharyngeal fibers of the inferior constrictor

An older man with oropharyngeal dysphagia, regurgitation of undigested food eaten hours to days earlier, gurgling, halitosis, aspiration pneumonia, and a posterior pouch at the pharyngoesophageal junction has a Zenker diverticulum. It is a pulsion, or false, diverticulum, containing only mucosa and submucosa. The driving problem is cricopharyngeal (upper esophageal sphincter) dysfunction: the sphincter relaxes poorly, so each swallow raises hypopharyngeal pressure. The mucosa herniates posteriorly through Killian triangle, a weak area between the oblique thyropharyngeal fibers and the transverse cricopharyngeal fibers of the inferior pharyngeal constrictor. The pouch traps food, which ferments (halitosis) and is later regurgitated or aspirated. It usually extends to the left. A barium swallow is the diagnostic test, because blind endoscopy or nasogastric tube placement can perforate the pouch; treatment is cricopharyngeal myotomy with diverticulectomy or endoscopic septotomy.

A. A mid esophageal outpouching pulled by scarred nodes (after tuberculosis or histoplasmosis) is a traction diverticulum, a true diverticulum that is usually asymptomatic.

B. A pouch just above the lower esophageal sphincter is an epiphrenic diverticulum, a false pulsion pouch associated with achalasia or esophageal spasm.

D. Herniation through the esophageal hiatus is a hiatal hernia of the stomach, not a pharyngeal pouch.

E. Herniation through the thyrohyoid membrane produces a lateral pharyngeal pouch or laryngocele, which is anterolateral and much rarer.

Educational objective: The key takeaway is that Zenker diverticulum is a false pulsion diverticulum through Killian triangle caused by cricopharyngeal dysfunction and is diagnosed by barium swallow.
Decision tree splitting dysphagia into oropharyngeal and esophageal types, then esophageal dysphagia into solids only (mechanical causes) and solids plus liquids from the start (motility disorders), with the classic causes listed under each branch.
Trouble starting the swallow is oropharyngeal; food sticking behind the sternum is esophageal, where solids only means a ring, EoE, stricture, cancer or web, and solids plus liquids from the start means achalasia, spasm or scleroderma.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 7 · Ethanol metabolism · medium

A 43-year-old man comes to the physician for a routine examination. He drinks 8 to 10 beers daily. He has no abdominal pain or jaundice. Examination shows a smooth, nontender, enlarged liver. Serum AST is 84 U/L and ALT is 38 U/L. Ultrasonography shows a diffusely echogenic liver. A liver biopsy shows that most hepatocytes contain a single large cytoplasmic vacuole that displaces the nucleus to the periphery, with no fibrosis.

Which of the following biochemical changes in hepatocytes contributes most directly to this lesion?

  1. Decreased NADH, increasing beta-oxidation of fatty acids
  2. Increased NADH, converting pyruvate to alanine
  3. Decreased NADH, activating hormone-sensitive lipase in hepatocytes
  4. Increased NADH, converting dihydroxyacetone phosphate to glycerol-3-phosphate
  5. Increased NAD+, accelerating the TCA cycle
Show answer and explanation

Correct answer: D. Increased NADH, converting dihydroxyacetone phosphate to glycerol-3-phosphate

Heavy daily drinking, hepatomegaly, an AST to ALT ratio above 2, and hepatocytes filled with a single large lipid vacuole that pushes the nucleus aside indicate macrovesicular hepatic steatosis, the earliest and reversible alcoholic liver lesion. Both steps of ethanol oxidation, alcohol dehydrogenase in the cytosol and aldehyde dehydrogenase in mitochondria, reduce NAD+ to NADH, so the hepatic NADH/NAD+ ratio rises sharply. Excess NADH drives the glycerol-3-phosphate dehydrogenase reaction toward reduction of dihydroxyacetone phosphate (DHAP) to glycerol-3-phosphate, which supplies the backbone for esterifying fatty acids into triglycerides. At the same time, high NADH inhibits the NAD+-dependent steps of beta-oxidation and the TCA cycle, so fatty acids are esterified rather than burned. Triglyceride synthesis outpaces export as VLDL, and fat accumulates.

A and C are wrong because ethanol increases, not decreases, NADH; less beta-oxidation, not more, promotes fat accumulation, and hormone-sensitive lipase acts in adipose tissue rather than hepatocytes. B is wrong because high NADH pushes pyruvate toward lactate through lactate dehydrogenase, causing lactic acidosis and hyperuricemia, not toward alanine. E is wrong because ethanol lowers NAD+ availability and inhibits the TCA cycle, diverting acetyl-CoA to ketone bodies and fatty acid synthesis.

Educational objective: The key takeaway is that ethanol raises the hepatic NADH/NAD+ ratio, which drives DHAP to glycerol-3-phosphate and suppresses fatty acid oxidation, causing macrovesicular steatosis.
Pathway diagram: ethanol becomes acetaldehyde (ADH, cytosol, blocked by fomepizole), then acetate (ALDH, mitochondria, blocked by disulfiram), then acetyl-CoA, each oxidation step making NADH; the raised NADH/NAD+ ratio pushes pyruvate to lactate, oxaloacetate to malate, DHAP to glycerol-3-P and inhibits the TCA cycle, giving lactic acidosis and gout, fasting hypoglycemia, fatty liver and ketoacidosis.
Both ethanol oxidation steps make NADH, and the high NADH/NAD+ ratio explains the four metabolic hallmarks of heavy drinking; fomepizole and disulfiram act at the two steps.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 8 · Head and Neck Cancer · medium

A 63-year-old man comes to the physician because of difficulty swallowing solids for 2 months. Two years ago he was treated with surgery and radiation for a keratinizing squamous cell carcinoma of the floor of the mouth; the tumor was p16-negative. He has smoked 1.5 packs of cigarettes daily for 45 years and drinks 6 beers daily. Endoscopy shows an ulcerated mass in the midesophagus, and biopsy shows invasive squamous cell carcinoma surrounded by areas of high-grade squamous dysplasia in the adjacent mucosa. Sequencing shows a TP53 mutation that differs from the one found in his oral cancer.

Which of the following best explains the development of the esophageal tumor?

  1. Hematogenous metastasis from the original oral carcinoma
  2. Germline mutation in the TP53 tumor suppressor gene
  3. Integration of high-risk HPV DNA into esophageal epithelium
  4. Radiation-induced malignancy within the treatment field
  5. Carcinogen exposure of the entire upper aerodigestive mucosa
Show answer and explanation

Correct answer: E. Carcinogen exposure of the entire upper aerodigestive mucosa

This patient has a second primary squamous cell carcinoma, not a recurrence. The clues are surrounding dysplasia in the adjacent esophageal mucosa (the tumor arose in place from a precursor lesion) and a TP53 mutation different from that of the first tumor (an independent clone). The explanation is field cancerization: tobacco smoke and alcohol bathe the whole upper aerodigestive tract, including the oral cavity, pharynx, larynx, esophagus and lung, and seed many independent mutated clones throughout that mucosa. A patient with one tobacco-related head and neck SCC therefore has a high risk of synchronous or later second primaries in the head and neck, esophagus or lung, which is why smoking cessation and lifelong surveillance are essential.

Tobacco-related head and neck SCCs occur in older adults, involve the oral cavity, larynx and hypopharynx, are p16-negative and characteristically carry TP53 mutations, as in this patient.

A. A metastasis would share the founding mutations of the primary tumor and would not arise from adjacent dysplastic mucosa. B. A germline TP53 mutation (Li-Fraumeni syndrome) would be present in every cell, so all tumors would share the same mutation, and it causes sarcomas, breast cancer, brain tumors and adrenocortical carcinoma at young ages. C. High-risk HPV drives oropharyngeal SCC in younger nonsmokers and produces p16-positive tumors; his first tumor was p16-negative and tobacco-related. D. Radiation for a floor-of-mouth tumor does not include the midesophagus, and radiation-induced cancers usually appear many years later, often as sarcomas.

Educational objective: The key takeaway is that tobacco and alcohol produce field cancerization of the upper aerodigestive mucosa, so patients with one head and neck SCC frequently develop independent second primaries in the head and neck, esophagus or lung.
Schematic map of the upper airway: the nasal cavity, oral cavity and larynx in front each open into their level of the pharynx behind (nasopharynx, oropharynx, hypopharynx), and each region lists its classic lesions, colored by cause: EBV, HPV, or tobacco and alcohol.
Each upper airway region has its own classic lesions: EBV carcinoma in the nasopharynx, HPV-16 SCC in the oropharynx, and tobacco and alcohol SCC in the oral cavity, larynx and hypopharynx.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 9 · Nutrient Absorption · hard

A 7-year-old boy is evaluated for recurrent episodes of a red, scaly rash on sun-exposed skin of the face and forearms, accompanied by unsteady gait. Growth is at the 15th percentile. Urinalysis shows large amounts of neutral amino acids, including tryptophan, with normal plasma levels of basic amino acids. A defect in a neutral amino acid transporter of the intestine and proximal tubule is confirmed. A dietitian recommends a high-protein diet in addition to nicotinamide, and his symptoms and growth improve.

Which of the following best explains why a high-protein diet supplies tryptophan to this patient despite his transporter defect?

  1. Increased paracellular diffusion of free amino acids across the jejunum
  2. Uptake of free tryptophan through the basic amino acid transporter
  3. Uptake of dipeptides and tripeptides by an H+-coupled apical transporter
  4. Endocytosis of intact dietary proteins by duodenal enterocytes
  5. Colonic bacterial synthesis of tryptophan that is absorbed in the colon
Show answer and explanation

Correct answer: C. Uptake of dipeptides and tripeptides by an H+-coupled apical transporter

This boy has Hartnup disease: a defective sodium-dependent neutral amino acid transporter in the intestine and proximal tubule causes neutral aminoaciduria and poor absorption of free tryptophan. Because tryptophan is a precursor for niacin synthesis, deficiency produces pellagra-like photosensitive dermatitis and cerebellar ataxia.

The defect affects only the uptake of free amino acids. Dietary protein is digested by pancreatic proteases and brush-border peptidases into a mixture of free amino acids and small peptides. Di- and tripeptides enter enterocytes through PepT1, an apical transporter driven by an inward H+ gradient rather than by Na+. Peptide uptake is actually faster than uptake of free amino acids, and once inside the cell the peptides are split by cytosolic peptidases, releasing tryptophan and other neutral amino acids, which then leave across the basolateral membrane into portal blood. A high-protein diet therefore bypasses the defective carrier, and nicotinamide treats the niacin deficit directly.

A is wrong because paracellular diffusion of amino acids is trivial and does not compensate for a transporter defect. B is wrong because basic amino acids (lysine, arginine, ornithine, cystine) use a separate carrier that does not move tryptophan; that carrier is defective in cystinuria. D is wrong because uptake of intact proteins is negligible in children and adults (it matters only for antibody transfer in some neonatal mammals). E is wrong because colonic bacteria do not supply meaningful absorbable tryptophan.

Educational objective: The key takeaway is that di- and tripeptides enter enterocytes through the H+-coupled PepT1 transporter, so a high-protein diet bypasses the free neutral amino acid transporter defect of Hartnup disease.
Schematic enterocyte showing glucose and galactose entering through SGLT1 with sodium, fructose through GLUT5, peptides through PepT1 and amino acids through sodium-coupled carriers, with all sugars leaving through GLUT2 and a basolateral Na+/K+-ATPase, plus boxes on cholera and oral rehydration, SGLT1 mutation and Hartnup disease.
Glucose and galactose ride in with Na+ on SGLT1, fructose uses GLUT5, and all three leave on GLUT2; cholera spares SGLT1 (why ORS works), and PepT1 lets peptides bypass the Hartnup defect.

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Reviewed by Veli Bakalov, MD · @HemeOncBuddy

Question 10 · Intestinal Protozoa · hard

A 37-year-old man returns from 6 weeks in rural India with 3 weeks of intermittent fever and right upper quadrant pain. He had bloody diarrhea with tenesmus during the trip, which resolved. Temperature is 38.6 C (101.5 F). The liver is tender and enlarged. Leukocyte count is 15,200/mm3 with no eosinophilia; alkaline phosphatase is elevated. Ultrasound shows a single 7-cm hypoechoic lesion in the right hepatic lobe. Serology for Entamoeba histolytica is positive, and he improves promptly on metronidazole without drainage.

Addition of paromomycin after the metronidazole course is necessary for which of the following reasons?

  1. Metronidazole does not penetrate hepatic abscess cavities
  2. Paromomycin eradicates luminal cysts that metronidazole does not kill
  3. Paromomycin prevents the disulfiram-like reaction caused by metronidazole
  4. Metronidazole lacks activity against coexisting anaerobic bacteria
  5. Paromomycin is required to clear tissue trophozoites resistant to nitroimidazoles
Show answer and explanation

Correct answer: B. Paromomycin eradicates luminal cysts that metronidazole does not kill

A traveler with resolved bloody dysentery followed by fever, right upper quadrant pain, and a single right-lobe liver lesion with positive serology has an amebic liver abscess. Trophozoites of Entamoeba histolytica invade the colonic mucosa, creating flask-shaped ulcers, and reach the liver through the portal vein, so the right lobe, which drains most of the right colon, is usually affected. Metronidazole is a prodrug reduced to DNA-damaging nitro radicals only by the low-redox ferredoxin-type carriers of anaerobes and protozoa; it is well absorbed and reaches tissue, so it kills invasive trophozoites and cures most abscesses without drainage. However, because it is absorbed high in the gut, it achieves poor concentrations in the colonic lumen and does not eliminate the cysts that remain there. Those cysts cause relapse and continued transmission, so every tissue-active course is followed by a luminal agent, paromomycin (a nonabsorbed aminoglycoside) or iodoquinol. Asymptomatic cyst passers receive the luminal agent alone.

A. Metronidazole penetrates tissue and abscesses well, which is precisely why most amebic abscesses resolve without drainage.

C. The disulfiram-like reaction comes from aldehyde dehydrogenase inhibition and is avoided by abstaining from alcohol; paromomycin does not prevent it.

D. Metronidazole has excellent activity against anaerobes such as Bacteroides fragilis, so this is the opposite of its gap.

E. Nitroimidazole resistance in tissue trophozoites is not the issue, and paromomycin is not absorbed, so it cannot treat tissue disease at all.

Educational objective: The key takeaway is that metronidazole kills invasive trophozoites but not luminal cysts, so amebiasis treatment is always followed by a nonabsorbed luminal agent such as paromomycin or iodoquinol.
Side-by-side panels show Giardia trophozoites clinging to duodenal villi without invading and E. histolytica trophozoites undermining the colonic mucosa in a flask-shaped ulcer, above a sketch of trophozoites traveling from the right colon through the portal vein to a right-lobe liver abscess and a two-step plan of a tissue drug followed by a luminal agent.
Giardia stays on the surface and causes malabsorption; E. histolytica invades, causes dysentery, seeds the right hepatic lobe through the portal vein, and needs a tissue drug followed by a luminal agent.

Read the study note on this topic

Reviewed by Veli Bakalov, MD · @HemeOncBuddy

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