Study aid only. Verify against current guidelines before clinical use.
Free USMLE Step 1 Multisystem 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.
A 68-year-old man is evaluated for 4 months of fatigue, leg swelling, and a 12-kg weight loss. He reports that his tongue feels large and that bruises appear around his eyes after he coughs. Blood pressure is 104/62 mm Hg. Examination shows an enlarged tongue with lateral indentations, periorbital purpura, waxy skin, hepatomegaly, and 3+ pitting edema to the thighs. Serum creatinine is 1.6 mg/dL and urine protein excretion is 7.2 g/24 h. Serum immunofixation shows a monoclonal free lambda light chain, and a bone marrow biopsy contains 16% clonal plasma cells. Abdominal fat pad aspiration shows amorphous eosinophilic extracellular material that stains with Congo red and shows apple-green birefringence under polarized light.
The fibrils deposited in this patient's tissues are most likely derived from which of the following precursor proteins?
Immunoglobulin light chain
Serum amyloid A
Transthyretin
Beta2-microglobulin
Islet amyloid polypeptide
Show answer and explanation
Correct answer: A. Immunoglobulin light chain
The combination of a clonal plasma cell population, a monoclonal free lambda light chain, nephrotic-range proteinuria, hepatomegaly, macroglossia, and periorbital purpura is AL (primary) amyloidosis. In AL amyloidosis a plasma cell dyscrasia (myeloma or MGUS) oversecretes free immunoglobulin light chains, most often lambda. These light chains misfold into insoluble beta-pleated sheet fibrils that resist proteolysis and accumulate extracellularly in heart, kidney, tongue, nerve, liver, and skin, which is why the illness crosses organ systems. Congo red staining with apple-green birefringence confirms amyloid of any type, so the precursor must be identified separately by free light chain studies, immunofixation, and tissue mass spectrometry before treatment, which here is directed at the plasma cell clone. Macroglossia and periorbital purpura (from fragile amyloid-laden vessels) are close to specific for AL.
B. Serum amyloid A is the precursor in AA (secondary) amyloidosis, driven by chronic inflammation such as rheumatoid arthritis, inflammatory bowel disease, tuberculosis, or familial Mediterranean fever; this patient has no chronic inflammatory disease. C. Transthyretin deposits in hereditary and wild-type ATTR amyloidosis, which cause cardiomyopathy and neuropathy but not macroglossia, purpura, or a light chain clone. D. Beta2-microglobulin accumulates only after years of hemodialysis and deposits in joints and tendons, causing carpal tunnel syndrome. E. Islet amyloid polypeptide (amylin) is a localized amyloid confined to pancreatic islets in type 2 diabetes and causes no systemic deposits.
Educational objective: The key takeaway is that AL amyloidosis arises from misfolded monoclonal immunoglobulin light chains produced by a plasma cell dyscrasia, with macroglossia, periorbital purpura, and nephrotic proteinuria as classic clues.
Start from the organ clue: purpura or a big tongue means AL, dialysis plus carpal tunnel means beta2-microglobulin, and brain, thyroid and islet deposits are made locally.
A 6-year-old girl from rural North Carolina is brought to the emergency department in June because of fever, severe headache, and a rash. Five days ago her mother removed a tick from her scalp. Fever and headache began 2 days later, and yesterday a rash appeared on her wrists and ankles and has since spread to her palms, soles, arms, legs, and trunk. Temperature is 39.8 C (103.6 F), pulse is 140/min, and blood pressure is 82/48 mm Hg. Examination shows a blanching maculopapular rash that is becoming petechial, along with edema of the hands and feet. Serum sodium is 129 mEq/L, platelet count is 61,000/mm3, and serum albumin is 2.6 g/dL.
Which of the following best explains the rash, edema, and hypotension in this patient?
Deposition of antigen-antibody complexes in dermal capillaries
Infection and injury of small-vessel endothelial cells
Superantigen cross-linking of T-cell receptors and MHC class II
Exotoxin cleavage of desmoglein in the epidermis
Septic emboli lodging in the distal digital arteries
Show answer and explanation
Correct answer: B. Infection and injury of small-vessel endothelial cells
Fever and headache after a tick bite in the southeastern United States in summer, followed by a rash that begins on the wrists and ankles, involves the palms and soles, and spreads centrally, is Rocky Mountain spotted fever. Rickettsia rickettsii, carried by Dermacentor ticks, is an obligate intracellular bacterium that preferentially infects vascular endothelial cells (and underlying smooth muscle) of small vessels throughout the body.
The organism enters endothelial cells, escapes the phagosome, and spreads from cell to cell by actin-based motility, injuring each cell it infects. The result is a widespread small-vessel vasculitis with increased vascular permeability: fluid and albumin leak into tissues, causing edema, hypoalbuminemia, hypovolemia, and hypotension; local thrombosis and bleeding make the rash petechial; and platelet consumption causes thrombocytopenia. Damage in the brain, lungs, heart, and kidneys explains encephalitis, pneumonitis, myocarditis, and renal failure, and hyponatremia is common. Doxycycline is given at any age, because delayed treatment carries high mortality.
A. Immune complex vasculitis (type III) occurs in serum sickness and IgA vasculitis; RMSF damage results from direct infection of endothelium. C. Superantigens cause toxic shock syndrome, with a diffuse erythroderma that later desquamates. D. Exfoliative toxins that cleave desmoglein-1 cause staphylococcal scalded skin syndrome. E. Septic emboli cause the splinter hemorrhages and Janeway lesions of infective endocarditis rather than a centripetally spreading rash.
Educational objective: The key takeaway is that Rickettsia rickettsii infects vascular endothelium, causing a small-vessel vasculitis with leakage that explains the petechial rash, edema, hypotension, and multiorgan damage of RMSF.
Read the smear for which cell holds the organism and the rash for which way it travels; card colors mark the vector.
A 29-year-old man returns from a 3-week trip to a rural region with ongoing transmission of a mosquito-borne flavivirus and presents with 4 days of fever, severe myalgia, headache, and a diffuse macular rash. Temperature is 39.4 C and blood pressure is 96/60 mm Hg. Platelet count is 72,000/mm3, hematocrit is rising, and aminotransferases are mildly elevated. A plasma nucleic acid amplification test is ordered. Before the thermal cycling steps of denaturation at 95 C, annealing at 55 C, and extension at 72 C can be performed, the laboratory must add an additional enzyme to the reaction because of the nature of the pathogen's genome.
The additional enzyme required catalyzes which of the following reactions?
Synthesis of RNA from a DNA template
Joining of adjacent DNA fragments through phosphodiester bonds
Relaxation of supercoiled DNA by transient strand cleavage
Synthesis of complementary DNA from an RNA template
Addition of a methyl group to cytosine residues in CpG islands
Show answer and explanation
Correct answer: D. Synthesis of complementary DNA from an RNA template
The polymerase chain reaction amplifies DNA. It requires a DNA template, two primers that flank the target, deoxynucleotides, and a heat-stable DNA polymerase (Taq, from the hot spring organism Thermus aquaticus), which survives the repeated 95 C denaturation steps. Flaviviruses such as dengue have a single-stranded RNA genome, so there is no DNA for Taq to copy. Reverse transcriptase must first be added to synthesize complementary DNA (cDNA) from the viral RNA template; the reaction is then called reverse transcription PCR, and it is how RNA viruses including dengue, SARS-CoV-2, HIV, and hepatitis C virus are detected. If fluorescence is read during each cycle, the assay also quantifies starting template, as in viral load testing. Note the distinct abbreviations: reverse transcription PCR makes RNA amplifiable, whereas real-time (quantitative) PCR measures how much target was present; the two are often combined.
A. RNA synthesis from a DNA template is transcription, carried out by RNA polymerase; that is the wrong direction for making an amplifiable DNA copy of a viral RNA genome. B. Joining DNA fragments through phosphodiester bonds is the function of DNA ligase, used in cloning and in Okazaki fragment maturation, not in PCR setup. C. Relaxing supercoils by transient strand cleavage is the action of topoisomerase; heat, not an enzyme, separates the strands in PCR. E. Methylation of cytosine in CpG islands regulates gene expression and is exploited in methylation-sensitive assays, but it plays no role in amplifying an RNA genome.
Educational objective: The key takeaway is that amplifying an RNA genome requires reverse transcriptase to generate complementary DNA before Taq polymerase can carry out the PCR cycles.
Each PCR cycle (denature, anneal, extend) doubles the target; RT-PCR starts from RNA, and qPCR reads fluorescence every cycle to measure how much template was present.
A 3-year-old girl is evaluated for obesity. She was profoundly hypotonic at birth and required nasogastric feeding for 2 months because of a weak suck. Over the past year she has developed an insatiable appetite and takes food from the trash. Her weight is above the 99th percentile. She has almond-shaped eyes, small hands and feet, and global developmental delay. Her mother was 42 years old at delivery. Fluorescence in situ hybridization shows two copies of the 15q11-q13 region, and methylation analysis of that region shows only the maternal pattern.
Which of the following best explains this child's findings?
Maternal uniparental disomy of chromosome 15
Paternal uniparental disomy of chromosome 15
Deletion of the maternal 15q11-q13 region
Point mutation in the maternal UBE3A allele
Repeat expansion silencing FMR1
Show answer and explanation
Correct answer: A. Maternal uniparental disomy of chromosome 15
Neonatal hypotonia with poor feeding followed by hyperphagia, obesity, small hands and feet, and developmental delay is Prader-Willi syndrome. Genes in 15q11-q13 such as SNRPN are imprinted so that the maternal copies are methylated and silent and only the paternal copies are expressed; disease occurs when no active paternal copy is present.
The usual cause is a deletion of the paternal 15q11-q13 region, but here FISH shows two copies, which excludes a deletion. A purely maternal methylation pattern means both chromosome 15s came from the mother, that is maternal uniparental disomy: two silenced maternal copies leave the child functionally without paternal gene expression. Uniparental disomy usually arises by trisomy rescue after maternal meiotic nondisjunction, which is why it is linked to advanced maternal age. Hypothalamic dysfunction explains the hyperphagia, hypogonadism, and short stature.
B. Paternal uniparental disomy leaves two silenced paternal UBE3A copies and causes Angelman syndrome, with a purely paternal methylation pattern. C. A maternal deletion also causes Angelman syndrome and would show only one copy on FISH. D. A maternal UBE3A point mutation causes Angelman syndrome with normal methylation. E. FMR1 silencing causes fragile X syndrome, which does not produce neonatal hypotonia with later hyperphagic obesity.
Educational objective: The key takeaway is that Prader-Willi syndrome requires loss of paternal 15q11-q13 expression, and when no deletion is present but methylation is entirely maternal the cause is maternal uniparental disomy.
Chromosome 15q imprinting: deletion or uniparental disomy removes the only active copy in Prader-Willi and Angelman syndromes.
A 38-year-old man from Alaska develops abdominal cramps and diarrhea, followed a week later by fever, severe muscle pain in the calves, jaw, and arms, and swelling around both eyes. Three weeks ago he and two friends shared undercooked black bear meat from a hunting trip, and both friends have similar symptoms. Temperature is 38.6 C (101.5 F). He has bilateral periorbital edema, subconjunctival hemorrhages, and splinter hemorrhages of the nails. Leukocyte count is 13,500/uL with 32% eosinophils, and creatine kinase is 4,200 U/L.
A biopsy of the deltoid muscle is most likely to show which of the following?
Cysticerci with an invaginated scolex in the interstitium
Macrophages packed with amastigotes
Perifascicular atrophy with perivascular inflammation
Tissue cysts containing bradyzoites without inflammation
Coiled larvae encysted within skeletal muscle fibers
Show answer and explanation
Correct answer: E. Coiled larvae encysted within skeletal muscle fibers
Gastrointestinal symptoms followed by fever, myositis with a high creatine kinase, periorbital edema, subconjunctival and splinter hemorrhages, and marked eosinophilia in a cluster of people who ate undercooked bear meat is trichinellosis. Trichinella larvae encysted in muscle of pork or wild game (bear, walrus, wild boar) are released by gastric digestion, mature into adult worms in the small intestine (causing the early enteric phase), and the females release new larvae. These larvae travel through lymphatics and blood to striated muscle, where each enters a muscle fiber, transforms it into a nurse cell, and coils and encysts, later calcifying. Tissue invasion by a helminth explains the striking eosinophilia, and vasculitis-like injury produces the hemorrhages and periorbital edema. Treatment is albendazole or mebendazole, often with corticosteroids.
A. Cysticerci with a scolex are larvae of Taenia solium acquired by ingesting eggs; they cause neurocysticercosis with seizures rather than acute febrile myositis. B. Macrophages packed with amastigotes indicate Leishmania, which involves the reticuloendothelial system and skin. C. Perifascicular atrophy with perivascular inflammation is the pattern of dermatomyositis, an autoimmune myopathy without eosinophilia or a food exposure. D. Bradyzoite tissue cysts without inflammation indicate latent Toxoplasma, which does not cause this acute eosinophilic illness.
Educational objective: The key takeaway is that trichinellosis after eating undercooked pork or bear meat causes fever, myositis, periorbital edema and eosinophilia, with coiled larvae encysted in skeletal muscle fibers.
How each high-yield worm gets in, its signature clue and its drug; Taenia solium eggs cause cysticercosis, while pork cysticerci cause only an intestinal tapeworm.
A 23-year-old man is brought to the emergency department because of profuse sweating, agitation and a feeling of burning up for 6 hours. He has been taking a yellow powder bought online as a fat-burning supplement before a bodybuilding competition and doubled the dose yesterday. Temperature is 41.2 C (106.2 F), pulse is 148/min, respirations are 34/min and blood pressure is 104/62 mm Hg. He is diaphoretic and tremulous, and skeletal muscles are rigid. Arterial blood gas shows a metabolic acidosis with a lactate of 9 mmol/L. Whole-body oxygen consumption, measured by indirect calorimetry, is markedly increased.
Which of the following is the most likely mechanism of action of the substance this patient ingested?
Inhibition of NADH dehydrogenase (complex I)
Inhibition of cytochrome c oxidase (complex IV)
Blockade of the proton channel of ATP synthase
Inhibition of the pyruvate dehydrogenase complex
Return of protons to the matrix bypassing ATP synthase
Show answer and explanation
Correct answer: E. Return of protons to the matrix bypassing ATP synthase
The yellow fat-burning powder is 2,4-dinitrophenol (DNP), an illicit weight-loss agent and a classic uncoupler of oxidative phosphorylation. DNP is a lipid-soluble weak acid that picks up protons in the intermembrane space and releases them in the matrix, carrying H+ back across the inner mitochondrial membrane without passing through ATP synthase.
The proton gradient therefore collapses while the electron transport chain is no longer held back by it. Electron flow and oxygen consumption rise, NADH and FADH2 are oxidized faster, and fat stores are burned, but ATP synthesis falls and the energy of the gradient is released as heat. This explains the severe hyperthermia, tachycardia, tachypnea and diaphoresis. As ATP falls, cells increase glycolysis, producing lactic acidosis, and muscle rigidity reflects ATP depletion. Salicylate overdose and thermogenin (UCP1) in brown fat work by the same principle.
A. Rotenone inhibits complex I, which stops electron flow; oxygen consumption falls rather than rises.
B. Cyanide, carbon monoxide and azide inhibit complex IV; they also lower oxygen consumption, producing a high venous oxygen saturation, not hypermetabolism.
C. Oligomycin blocks ATP synthase; the gradient builds up until electron transport stalls, so oxygen consumption decreases.
D. Arsenic inhibits pyruvate dehydrogenase by binding lipoic acid, causing lactic acidosis with vomiting and rice-water diarrhea, but it does not increase oxygen use or generate this degree of heat.
Educational objective: The key takeaway is that uncouplers such as 2,4-dinitrophenol return protons to the mitochondrial matrix while bypassing ATP synthase, so oxygen consumption rises while ATP synthesis falls and energy is released as heat.
Blockers stop electron or H+ flow, so O2 use and ATP both fall; uncouplers let H+ leak back, so O2 use rises while ATP falls and the energy is released as heat.
A 17-year-old girl is brought to the emergency department with 1 day of high fever, vomiting, watery diarrhea, and diffuse muscle aches. She is on the fourth day of her menstrual period and has been using high-absorbency tampons. Temperature is 40.1°C (104.2°F), pulse is 132/min, and blood pressure is 78/42 mm Hg. She has diffuse blanching, sunburn-like erythema and hyperemic conjunctivae. A tampon is removed from the vagina. Laboratory studies show a platelet count of 82,000/mm3, serum creatinine 2.3 mg/dL, and creatine kinase 4,800 U/L. Blood cultures are negative; culture of the tampon grows gram-positive cocci in clusters that are coagulase-positive.
The toxin responsible for this patient's condition most likely causes disease by which of the following mechanisms?
ADP-ribosylation of elongation factor 2 in host cells
Activation of complement through binding the Fc region of IgG
Presentation of a processed peptide in the MHC class II groove
Release of lipopolysaccharide triggering Toll-like receptor 4
Cross-linking of MHC class II to T-cell receptor Vβ chains
Show answer and explanation
Correct answer: E. Cross-linking of MHC class II to T-cell receptor Vβ chains
High fever, hypotension, diffuse sunburn-like erythroderma, mucosal hyperemia, and involvement of several organ systems (gastrointestinal, muscle with high CK, renal, platelets) in a menstruating girl using tampons, with negative blood cultures and coagulase-positive staphylococci on the tampon, is staphylococcal toxic shock syndrome, caused by TSST-1. Desquamation of the palms and soles would follow in 1 to 2 weeks.
TSST-1 is a superantigen. A normal protein antigen is processed, and a peptide sits in the groove of an MHC class II molecule, where it is recognized by the rare T cells with a matching receptor. A superantigen is not processed; it clamps the outside of the MHC class II molecule to the Vβ region of the T-cell receptor, so every T cell bearing certain Vβ families is activated, up to a large fraction of all T cells. This polyclonal activation releases huge amounts of IL-1, IL-2, TNF-α, and IFN-γ, which cause fever, capillary leak, shock, and multiorgan injury. Blood cultures are usually negative because toxin is absorbed from a local site. Treatment is source removal plus an antistaphylococcal agent with clindamycin to shut off toxin synthesis.
A. ADP-ribosylation of EF-2 is the action of diphtheria toxin and Pseudomonas exotoxin A.
B. Protein A binds the Fc region of IgG, but it prevents opsonization and complement activation rather than causing shock.
C. Groove presentation of processed peptide is conventional antigen presentation, which activates only a few clones.
D. LPS is endotoxin of gram-negative bacteria; S. aureus is gram-positive and lacks it.
Educational objective: The key takeaway is that TSST-1 is a superantigen that bridges MHC class II and TCR Vβ outside the peptide groove, activating many T cells at once and causing cytokine-driven toxic shock with negative blood cultures.
Superantigens skip the peptide groove and bridge MHC II to TCR Vβ, so a large fraction of T cells fire and the cytokine flood causes toxic shock.
A 6-year-old girl is brought to the physician because of severe obesity and constant food seeking since she was 1 year old. She was a healthy term newborn with normal feeding and no hypoglycemia. Her weight is far above the 99th percentile, and her height is at the 97th percentile, above her mid-parental target; bone age is advanced by 1.5 years. Her development is normal. She has brown hair like her parents, no dysmorphic features, normal vision and five fingers on each hand. Her father has had severe obesity since early childhood. Serum leptin is high, appropriate for her fat mass. Fasting insulin is 58 microU/mL (reference range 2 to 20). Morning cortisol and TSH are normal.
A heterozygous loss-of-function mutation in the gene encoding which of the following is the most likely cause of this patient's obesity?
Leptin
Melanocortin-4 receptor
Proopiomelanocortin
Leptin receptor
Prohormone convertase 1
Show answer and explanation
Correct answer: B. Melanocortin-4 receptor
Early-onset hyperphagia and severe obesity in a child with an affected parent, tall stature with advanced bone age, marked hyperinsulinemia, normal pigmentation, normal cortisol and a leptin level appropriate to fat mass is the picture of MC4R deficiency, the most common monogenic cause of obesity. It is usually caused by a heterozygous loss-of-function mutation with dominant (often codominant) inheritance, which fits the father's history.
In the arcuate nucleus, leptin activates POMC neurons, which cleave POMC into alpha-MSH. Alpha-MSH acts on MC4R in the paraventricular nucleus to produce satiety and increase energy expenditure, while AgRP antagonizes the same receptor. Halving MC4R signaling removes this final common satiety step, so hunger is high even though leptin, POMC and alpha-MSH are all present. Accelerated linear growth and increased lean mass are characteristic features that distinguish it from the recessive defects upstream. The MC4R agonist setmelanotide is approved for defects upstream of the receptor (POMC, PCSK1 and leptin receptor deficiency), not for MC4R deficiency itself.
A. Leptin deficiency is recessive and gives undetectable leptin with hypogonadism and T-cell dysfunction. C. POMC deficiency adds neonatal ACTH deficiency (hypoglycemia, cholestasis, low cortisol) and red hair with pale skin from lost MC1R stimulation. D. Leptin receptor deficiency is recessive, with present or high leptin, hypogonadotropic hypogonadism and no increase in linear growth. E. Prohormone convertase 1 deficiency is recessive and presents in infancy with malabsorptive diarrhea, hypoglycemia and high proinsulin.
Educational objective: The key takeaway is that heterozygous MC4R loss of function is the most common monogenic obesity, with dominant inheritance, hyperphagia, accelerated linear growth and hyperinsulinemia but normal pigmentation, cortisol and leptin regulation.
One satiety pathway, four obesity genes: leptin replacement helps only leptin deficiency, and setmelanotide activates MC4R to bypass defects upstream of it.
A 7-month-old girl has had episodes of irritability, sweating, and seizures that occur whenever feedings are delayed and resolve with glucose. On examination the abdomen is protuberant with a liver edge 7 cm below the costal margin; the spleen is not enlarged and muscle strength is normal. After a 4-hour fast, serum glucose is 32 mg/dL, lactate is 8.2 mmol/L (normal 0.5 to 2.2), triglycerides are 820 mg/dL, and uric acid is 10.4 mg/dL (normal 2.5 to 6.0). Glucagon administration does not raise her serum glucose. Liver biopsy shows hepatocytes distended with cytosolic glycogen and lipid.
The elevated serum uric acid in this patient is best explained by which of the following?
Reduced hepatic urea synthesis with retention of nitrogen
ATP depletion from trapped fructose-1-phosphate in hepatocytes
Glycogen deposition in the kidney reducing glomerular filtration
Increased purine synthesis from glucose-6-phosphate shunted into the pentose phosphate pathway
Show answer and explanation
Correct answer: E. Increased purine synthesis from glucose-6-phosphate shunted into the pentose phosphate pathway
Severe fasting hypoglycemia with massive hepatomegaly, lactic acidosis, hypertriglyceridemia, hyperuricemia, and no response to glucagon is von Gierke disease (glycogen storage disease type I), autosomal recessive deficiency of glucose-6-phosphatase. Without the final dephosphorylation step, the liver cannot release free glucose from either glycogenolysis or gluconeogenesis, so glucose-6-phosphate accumulates in hepatocytes. The trapped intermediate spills into three pathways, which explains the whole laboratory picture: into glycolysis, generating lactate; into lipogenesis, generating triglyceride; and into the pentose phosphate pathway, where ribose-5-phosphate feeds PRPP and accelerates de novo purine synthesis, raising uric acid production. The elevated lactate compounds the problem by competing with urate for proximal tubular organic anion secretion, so renal urate clearance falls. Children can develop gout and urate nephropathy. Treatment is frequent glucose feeding and nocturnal cornstarch, with avoidance of fructose and galactose, which also enter at glucose-6-phosphate.
A. Urea cycle failure causes hyperammonemia with encephalopathy and does not raise uric acid, which is a purine, not a protein nitrogen, end product. B. Hypoxanthine-guanine phosphoribosyltransferase deficiency is Lesch-Nyhan syndrome, an X-linked disorder with hyperuricemia, choreoathetosis, intellectual disability, and self-mutilation, not hepatomegaly with hypoglycemia. C. ATP depletion by trapped fructose-1-phosphate, with degradation of adenine nucleotides to urate, occurs in hereditary fructose intolerance and after fructose infusion; this child's hypoglycemia occurs with fasting rather than after fructose ingestion, and her glycogen-laden liver points to a glycogenosis. D. Glomerular filtration is initially normal in von Gierke disease; renal injury is a late consequence of urate and glomerular hyperfiltration rather than the cause of the hyperuricemia.
Educational objective: The key takeaway is that in von Gierke disease trapped glucose-6-phosphate is shunted into glycolysis, lipogenesis, and the pentose phosphate pathway, producing lactic acidosis, hypertriglyceridemia, and hyperuricemia that is worsened by lactate competing for urate secretion.
Same breakdown pathway, different tissue jobs: liver blocks cause fasting hypoglycemia, muscle blocks cause exercise intolerance, and Pompe fills lysosomes.
A 27-year-old man comes for preconception counseling. His sister has cystic fibrosis with recurrent Pseudomonas pneumonia, pancreatic insufficiency and a positive sweat chloride test; he and both of his parents are healthy. His partner has no family history of the disease. In their population, cystic fibrosis affects 1 in 2,500 newborns, and the population is assumed to be in Hardy-Weinberg equilibrium. Neither partner has had carrier testing.
Which of the following is the best estimate of the probability that their first child will have cystic fibrosis?
1 in 100
1 in 150
1 in 200
1 in 625
1 in 2,500
Show answer and explanation
Correct answer: B. 1 in 150
Three probabilities must be multiplied.
The man's carrier risk is 2/3. His sister is affected, so both parents are obligate carriers. Among their children, the outcomes are 1/4 affected, 1/2 carrier and 1/4 noncarrier. Because he is known to be unaffected, the affected outcome is excluded, leaving carrier (2 of 3 remaining parts) and noncarrier (1 of 3).
The partner's carrier risk is about 1/25. Disease frequency q² = 1/2,500, so q = 1/50. For a rare autosomal recessive disease p is close to 1, so carrier frequency 2pq is approximately 2q = 1/25.
Two carriers have a 1/4 chance of an affected child in each pregnancy.
Risk = 2/3 × 1/25 × 1/4 = 2/300 = 1/150.
The Hardy-Weinberg approach assumes random mating, no selection, mutation or migration, and a large population. Carrier testing of both partners (CFTR sequencing) would replace these estimates with actual genotypes.
A. 1 in 100 assumes the man is certainly a carrier (1 × 1/25 × 1/4), ignoring that he could be a noncarrier. C. 1 in 200 uses a 1/2 carrier risk for the man (1/2 × 1/25 × 1/4), which is the carrier probability for any child of two carriers but ignores that his being unaffected removes the affected outcome from consideration. D. 1 in 625 multiplies two population carrier frequencies (1/25 × 1/25) and omits both the family history and the 1/4 transmission factor. E. 1 in 2,500 is simply the population birth prevalence, which ignores his affected sister.
Educational objective: The key takeaway is that an unaffected sibling of a person with an autosomal recessive disease has a 2/3 carrier risk, and a rare disease's carrier frequency is about 2q, so risk equals 2/3 × 2q × 1/4.