2003 official step 1 explanations

7/31/2019 2003 Official Step 1 Explanations

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USMLE Step I

Explanations to Block I of the2000 USMLE Step I Released Items

1. The correct answer is B. Finasteride is an inhibitor of 5-!-reductase, theenzyme that converts testosterone (choice E) to dihydrotestosterone(dihydrotestosterone causes prostate growth). Therefore, use of this agent wouldinhibit dihydrotestosterone synthesis.

Androstenedione (choice A) is the precursor to testosterone. Androstenedionesynthesis would be unaffected by finasteride administration.

Testosterone is also a precursor to estradiol (choice C). Testosterone synthesiswould not be decreased by finasteride administration.

Estradiol is the precursor to estrone (choice D). Estradiol synthesis would not be

decreased by finasteride administration.

2. The correct answer is E. Stranger anxiety develops in most infants betweenseven and nine months of age. The infant will become very apprehensive in thepresence of a stranger, and will respond by turning away, clinging to the primarycaregiver, or crying. Stranger anxiety is thought to signify a major step inpsychological development because it indicates that the infant is capable ofdistinguishing one person from another, and of recognizing the relativeimportance of different people.

Separation anxiety is also a normal developmental phase. It occurs later thanstranger anxiety (at 10 months to 18 months), and is precipitated by separationfrom a person to whom the infant is attached. Separation anxiety disorder(choice D) is characterized by extreme distress due to separation or anticipatedseparation from caretakers, the home, or familiar surroundings. It is the mostcommon anxiety disorder of childhood, but is nonetheless rare before age six.

The other choices are not supported by the patient's history.

3. The correct answer is B. The symptoms described (fever, malar rash,pericardial rub, arthritis) are suggestive of a lupus-like syndrome. Hydralazinecan cause a lupus-like syndrome in a significant number of patients, especially ifthe dosage is high, and if therapy continues for more than six months. Femalesare more likely to suffer from this complication than are males, and so-calledslow acetylators (acetylation is important in the metabolism of hydralazine) aremore likely to develop this complication.

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Captopril (choice A) is an ACE inhibitor, and can cause hyperkalemia, cough,and acute renal failure in patients with preexisting renal disease.

Minoxidil (choice C) is a vasodilator, and its toxic effects include salt and water

retention, tachycardia, pericardial effusion and hirsutism.

Nitroprusside (choice D) is a vasodilator. Toxic effects include hypotension,tachycardia, and accumulation of cyanide or thiocyanate in the blood.

Propranolol (choice E) is a nonselective beta antagonist, and its toxic effectsinclude bradycardia, atrioventricular blockade, congestive heart failure, andasthmatic attacks in patients prone to airway disease.

4. The correct answer is E. The interosseous membrane holds the radius andulna together; although it does permit the ulna and radius to move in pronationand supination, this membrane prevents the ulnar and radius from movinglongitudinally apart. By preventing longitudinal movement, the interosseousmembrane helps transmit force from the radius to the ulna.

The annular ligament (choice A) binds the proximal (elbow) end of the radius tothe ulna, but is much smaller than the interosseous ligament, and is not asimportant in transmitting longitudinal force.

The bicipital aponeurosis (choice B) arises from the distal tendon of the bicepsand inserts into the ulna.

The flexor retinaculum (choice C) covers the carpal tunnel at the wrist.

The intermuscular septa (choice D) extend from the deep fascia to the radius andulna.

5. The correct answer is D. The lesion described is a pleomorphic adenoma(mixed tumor), which typically involves the parotid gland. This benign lesion

tends to have poorly defined “fingers” extending into the parotid gland that maycause the tumor to recur if they are not completely excised (complete excisioncan be difficult as the surgeon must try to spare all of the branches of the facialnerve, which runs in the substance of the parotid gland).

Contralateral immune-mediated parotitis (choice A) would be seen in Sjögren’ssyndrome.

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USMLE Step I

Hematogenous metastases (choice B) would not be seen with pleomorphicadenoma because this is a benign tumor.

Pleomorphic adenomas do not usually involve the submaxillary gland (choice

C).

Regional lymph node metastases (choice E) would not be seen because this is abenign tumor.

6. The correct answer is C. In bacteria, the cell envelope consists of all the layersthat encircle and retain the cytoplasmic contents. The cell envelope of a gram-positive bacterium is composed of a thick layer of peptidoglycan and thecytoplasmic membrane. A capsule may be external to the peptidoglycan.

Penicillin G can access its targets, the penicillin-binding proteins (PBP), because itcan easily penetrate into the peptidoglycan and the cytoplasmic membrane. Ingram-negative bacteria, the cell envelope is composed of the outer membrane,the periplasmic space (which contains the peptidoglycan), and the cytoplasmicmembrane. A capsule may be present. The outer membrane acts as a sieve torestrict the entry of molecules into the gram-negative cell. Porins, which formaqueous channels through the outer membrane, can permit the entry of small,hydrophilic antibiotics; however, the diffusion of penicillin G into most gram-negative bacteria is very slow.

The cytoplasmic membrane (choice A) does not effectively exclude antibiotics. Itfunctions as an osmotic barrier, actively transports nutrients, and exportsproteins since the Golgi apparatus is absent in bacteria. If the bacteria arecapable of oxidative phosphorylation, the electron transport chain will be presentin the cytoplasmic membrane.

Lipoproteins (choice B) covalently anchor the outer membrane onto thepeptidoglycan.

Peptidoglycan (choice D) comprises the rigid cell wall that functions to maintainthe shape of the bacterial cell, and helps to prevent lysis.

Teichoic acids (choice E) consist of a substituted ribitol or glycerol residue withphosphodiester links attached to muramic acid residues in some gram-positivebacteria.

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7. The correct answer is B. The disease in question is irreversible and fatal, soclearly you don’t want to miss a single case. Setting the cutoff at point Baccomplishes this and keeps the number of false positives (healthy people whotest positive on the screening test) minimal. In essence, the sensitivity (how wellthe test identifies people who have the disease) of the screening procedure is

100% at point B. The specificity of the test (how well it identifies people who donot have the disease) is acceptable at this point; the majority of healthyindividuals will test negative.

The false-positive rate associated with a cutoff of point A is too high, as fullyone-half of the healthy individuals will be told they have a fatal disease.

If points C, D, or E are used as cutoff points, too many individuals with diseasewill escape detection and therefore cannot be treated (it should be assumedtreatment does not involve great risk to the patient as it would be otherwise be

unacceptable to treat the false positives).

8. The correct answer is D. A peak of 17-"-estradiol (an estrogen), in theabsence of progesterone, will act through positive feedback to cause an LH surge,which will peak about 24 hours after the estradiol peak. Ovulation will followthe LH peak by about 12 hours.

Basal temperature rises slightly immediately after ovulation (compare to choice

B). After ovulation, the corpus luteum will secrete both estradiol and

progesterone; in this combination, the ability of estrogen to cause endometrialproliferation is inhibited, and progesterone causes differentiation of thepreviously thickened endometrium into its secretory form, ready forimplantation. Continued low plasma LH levels after the peak initiate luteolysis(death and regression of the corpus luteum) on about day 25 (choice E). Thiscauses a drop in estrogen and progesterone levels; the drop in progesteronedestabilizes the lysosomes of the endometrial cells, causing inflammation, andmenstruation begins about three days later (day 1 of the next menstrual cycle;choice C). Menstruation lasts for about four to seven days in most women(choice A).

9. The correct answer is C. Loss of a plasmid carrying a gene for ampicillinresistance accounts for the relatively abrupt loss of resistance seen here in strainY.

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USMLE Step I

Down-regulation (choice A) would be expected to be reversed when antibioticsare reintroduced, so a delay in growth in the presence of the antibiotic might beseen rather than an outright loss of resistance.

Insertion of a transposon (choice B), point mutations (choice D), and

recombination (choice E) would be expected to affect only a small number of thebacteria, with the rest retaining the antibiotic resistance.

10. The correct answer is A. Peptide hormones that regulate growth,differentiation, and development often act at membrane-bound protein kinasereceptors. These receptors generally phosphorylate tyrosine residues, but somephosphorylate threonine or serine residues. Receptors with tyrosine kinaseactivity include receptors for platelet-derived growth factor, insulin, epidermalgrowth factor, and some lymphokines.

G protein–coupled receptors (e.g., adrenergic receptors, muscarinic cholinergicreceptors) mediate their effects via a group of GTP-binding proteins called Gproteins. Stimulation of these receptors facilitates the binding of GTP to Gproteins (choice B), thereby activating the G proteins, which will subsequentlychange the activity of membrane-bound enzymes (e.g., adenylate cyclase,phospholipase C) or open ion channels.

Steroid receptors are intracellular receptors that, when activated, bind to DNA(choice C) and alter transcription.

Activation of phospholipase C via a G protein produces inositol triphosphate,which releases calcium stores from the endoplasmic reticulum, thus increasingintracellular calcium concentrations (choice D).

Ligand-gated ion channel receptors are generally composed of several peptidesubunits. When activated by an agonist, the subunits change conformation,allowing ions to flow through the channel (e.g., nicotinic cholinergic receptor;choice E).

11. The correct answer is E. The question concerns the acute allergic reaction toa bee sting. The swelling and redness observed with bee, wasp, or ant stings aredue to vasodilation with leakage of fluid into the interstitium (part of the acuteinflammatory process).

The foreign body reaction (choice A) is a granulomatous reaction to foreignmaterial that develops over a period of weeks to months.

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The redness of the bee sting is not due to hemorrhage (choice B); the red cellsusually remain within the capillaries and other vessels while fluid and whiteblood cells enter the tissues.

Infiltration by lymphocytes (choice C) and neutrophils (choice D) can occur ininflammation in response to chemotactic factors released during theinflammatory process, but would not be predominant in the first five minutes ofthe process.

12. The correct answer is A. Presbycusis is the term for the progressive loss ofsensitivity to high-frequency sounds that occurs with age. The cause may bedegeneration or loss of elasticity of the basilar membrane in the first fewmillimeters of the basal end of the cochlea (the part that subserves high

frequencies).

A loss of low-frequency tones (choice B) could occur with sensory–neuraldeafness (e.g., damage to the hair cell of the cochlea and/or vestibulocochlearnerve); however, a selective loss of sensitivity to higher frequency sounds ismore typical.

Conduction deafness (e.g., wax in the external ear or impaired ossicularmovement) typically produces a hearing loss across all frequencies (choice C).

Presbycusis would affect both air and bone conduction (compare with choices Dand E).

13. The correct answer is A. The patient is described as having both uppermotor neuron signs (hyperreflexia of lower limbs, extensor plantar responses,weakness) and lower motor neuron signs (atrophy of the forearm, fasciculationsof the muscles of the chest, flaccid paralysis) with no impairment of sensation.Amyotrophic lateral sclerosis (ALS) is characterized by degeneration of bothupper and motor neurons, while sparing sensation.

Dementia of the Alzheimer type (choice B) would not cause any motor problems.

Guillain-Barré syndrome (choice C) is a demyelinating disorder of peripheralnerves, and therefore would produce only lower motor neuron signs (ascendingweakness/paralysis). Mild sensory loss can also occur.

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USMLE Step I

Multiple cerebral infarcts (choice D) could produce upper motor neuron signs,but would not lead to lower motor neuron symptoms.

Multiple sclerosis (choice E) is a demyelinating disorder of the central nervoussystem, and could therefore produce upper but not lower motor neuron

symptoms.

14. The correct answer is A. Oral ingestion of 1 L of water will decrease plasmaosmolarity (Posm), which will decrease antidiuretic hormone (ADH), leading toa decrease in urine osmolarity (Uosm) and an increase in urine flow rate (V).This explains the change from X to Y, which indicates an increase in flow rate (V)and a decrease in the ratio of urine osmolarity (Uosm) to plasma osmolarity(Posm), or (U/P)osm. Uosm normally ranges from ~ 1,200 mOsm/L to ~ 50mOsm/L, while the normal Posm range is 275–295 mOsm/L, so the decrease in

(U/P)osm is primarily due to a decrease in Uosm. ADH is the primary regulatorof Uosm and V, and acts by regulating the water permeability of the collectingduct (distal nephron).m. ADH is the primary regulator of Uosm and V, and actsby regulating the water permeability of the collecting duct (distal nephron).

Ingestion of 200 ml of isotonic fluid (choice B) will not change Posm, and is toosmall a change in plasma volume to affect ADH.

ADH secretion is primarily regulated by Posm: increased Posm (choices C and

D) will increase ADH, thus increasing Uosm and decreasing V.

Increased ADH (choice E) will increase water permeability and waterreabsorption, causing an increase in Uosm and a decrease in V.

15. The correct answer is C. Hereditary spherocytosis is an autosomal recessivecondition associated with hemolysis of varying degrees, spherocytosis, andincreased osmotic fragility of red blood cells. An abnormality in spectrin oranother proteins in the membrane skeleton results in spherocytes, which gettrapped in the spleen because they are less deformable. Typically, hereditaryspherocytosis presents with mild anemia, splenomegaly, and mild jaundice.

Treatment is splenectomy. Labs will show a mildly elevated indirect bilirubintest, and a negative Coombs test, but the most salient finding is increasedosmotic fragility of the red blood cells. For the purposes of the USMLE,increased osmotic fragility means hereditary spherocytosis (although it alsooccurs in some other rare hematologic disorders).

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Ovalocytes (choice A) are seen in megaloblastic anemias, and in myelofibrosiswith extramedullary myeloid metaplasia.

Schistocytes (choice B) are irregularly contracted, fragmented red blood cells,often exhibiting marked poikilocytosis. They are produced in disorders in which

fibrin is deposited into small blood vessels.

Target cells (choice D) are seen in iron deficiency, liver disease, thalassemicsyndromes, and hemoglobinopathies.

Teardrop cells (choice E) reflect disordered erythropoiesis. They are found insevere iron deficiency, and in myelofibrosis with extramedullary myeloidmetaplasia.

16. The correct answer is B. The question stem describes an intention tremor,which is an uncontrolled shaking of the affected extremity that is present onlywith purposeful movement, and is worse at the end of the movement. A lesionof a cerebellar hemisphere will produce motor disorders, including intentiontremor in the ipsilateral extremities. Other movement disorders that would bepredicted in this patient include dysmetria (inability to stop movements at thedesired point), adiadochokinesia (inability to perform rapid alternatingmovements smoothly and regularly), and decomposition of movements (inabilityto coordinate movements that require several joints).

Lesions of the basal ganglia (choice A) do not produce intention tremor. Basalganglia lesions can produce (depending on where the lesion is) resting tremor(which improves with purposeful movement), chorea, athetosis, and dystonia.

A lesion of the cerebellar vermis (choice C) can produce truncal ataxia anddysarthria.

A lesion of the frontal eye field (choice D) results in deviation of the eyesipsilateral to the lesion.

The motor nucleus of the thalamus (choice E) is an imprecise term, but probably

refers to the ventral anterior and ventral lateral nuclei, which receive inputs fromthe basal ganglia and cerebellum. Lesion of this area could produce motordeficits, but would not cause an intention tremor.

17. The correct answer is E. Elderly women are particularly prone to developvarying degrees of prolapse of the bladder and uterus. In the early stages, this

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USMLE Step I

problem can present with incontinence in situations that raise intra-abdominalpressure, including coughing, sneezing, and laughing (stress incontinence).Exercises that strengthen the urogenital diaphragm can sometimes ameliorate theproblem.

The detrusor muscle (choice A) is the intrinsic muscle of the bladder thatcontracts during urination, expelling the urine into the urethra.

The obturator internus (choice B) and the piriformis muscles (choice C) stabilizethe hip.

The rectus abdominis (choice D) is a major muscle of the abdominal wall.

18. The correct answer is B. Growth hormone stimulates the growth of long

bones prior to epiphyseal plate closure. The increased osteoblastic activity isassociated with measurable increases in alkaline phosphatase activity.

Insulin-like growth factor I (IGF1) secretion is increased, not decreased, bygrowth hormone (choice A).

Whereas growth hormone stimulates gastrointestinal calcium absorption, serumcalcium levels are tightly regulated, largely by parathyroid hormone and vitaminD. Serum calcium levels would therefore likely be normal, rather than increased(choice C), after three months of growth hormone therapy.

Growth hormone has an anabolic effect, increasing protein and lean body mass,and decreasing the concentration of free amino acids and urea nitrogen in theblood (compare with choice D).

Growth hormone produces an acute increase in plasma phosphorous levels, butafter three months of therapy, phosphorous levels would probably be normal(compare with choice E). Phosphorous levels are regulated by parathyroidhormone and vitamin D levels.

19. The correct answer is D. The arterial PO2 of the mother is approximately100 mm Hg and the PO2 of the umbilical artery is approximately 20-23 mm Hg.

The mean P02 of fetal blood after it is oxygenated in the placenta is only about 30

mm Hg. If you recall that the umbilical arteries carry deoxygenated blood backto the placenta to be oxygenated, then the PO2 of the umbilical artery blood must

be much lower than that of the maternal arterial blood.

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The normal hematocrit (choice A) for a pregnant woman is 30-35%. At term, thefetal hematocrit is approximately 44%.

The O2 affinity (choice B) of Hb is greater in the fetus than the mother. This is

because the Hb of the fetus is mainly fetal Hb, which has a left-shifted oxygen-

hemoglobin dissociation curve compared to adult Hb. Because fetal Hb has ahigher affinity for oxygen, fetal Hb can carry 20-50% more oxygen than maternalHb at a low PO2.

The O2 capacity (choice C) is the maximum amount of O2 that can be combined

with hemoglobin (Hb). The normal adult oxygen capacity is about 21 ml O2/100

mL blood, assuming a hemoglobin concentration of 15 gm/100 mL blood. In thefetus, the hemoglobin concentration is 50% greater than that of the mother, andtherefore fetal O2 capacity is greater.

Hematocrit is the most important determinant of blood viscosity (choice D), andbecause the fetal hematocrit is higher than the maternal hematocrit, the viscosityis also higher.

20. The correct answer is B. There are four conditions that can lead to a right-shift of the oxygen dissociation curve, facilitating oxygen unloading. These areincreased PCO2, increased 2,3-bisphosphoglycerate, increased temperature, and

decreased pH. Choice B is the only curve that is shifted to the right of theoriginal curve.

21. The correct answer is H. Ulcerative colitis (UC) is an inflammatory boweldisease characterized histologically by a hyperemic, edematous mucosa withsmall hemorrhages that develop into crypt abscesses (small, focal collections ofinflammatory cells and necrosis in the colonic crypts). Crypt abscesses mayprogress to small ulcers, which coalesce into larger ones. Ulcerative colitistypically only affects the mucosa, unlike Crohn’s disease (choice C), whichaffects the entire thickness of the bowel wall (transmural involvement). Both UCand Crohn’s can produce bloody diarrhea, but blood is usually more prominent

in UC. UC characteristically begins at the rectum and works backwards; Crohn’slesions can “skip” areas of bowel, and tend to occur in the right colon. Crohn’s isalso characterized by granuloma formation and the development of strictures orfissures; UC is much less likely to show these.

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USMLE Step I

In amebiasis (choice B), which is caused by Entamoeba histolytica, characteristicflask-shaped ulcers are produced, but these extend through the mucosa into thesubmucosa.

None of the other diseases listed in the question produce a histological/clinical

picture compatible with the stated case; most produce a nonspecific enterocolitis:

AIDS-associated gastroenteritis (choice A), or HIV enteropathy, is associatedwith diarrhea in some cases; however, the clinical picture is strongly suggestiveof UC, and there is no suggestion that the patient is infected with HIV.

Clostridium difficile (choice D) is notable for producing pseudomembranouscolitis, usually in response to prior or concurrent antibiotic therapy.

Escherichia coli-associated colitis (choice E) is unlikely, given the biopsy results

and the pattern of involvement of the bowel.

Ischemic colitis (choice F) is often transmural, and could conceivably be limitedto the mucosa. The presence of acute and chronic inflammation in addition tothe areas of the bowel involved, however, suggest inflammatory bowel diseaseinstead.

Salmonella gastroenteritis (choice G) typically involves the ileum (includingPeyer’s patches) and the colon.

22. The correct answer is D. Parathion, an organophosphorusacetylcholinesterase inhibitor, increases synaptic acetylcholine (ACh)concentrations at parasympathetic effector sites, sympathetic cholinergic effectorsites, neuromuscular junctions, and in the central nervous system. Increasedparasympathetic cholinergic tone to the lungs can cause wheezing because ofbronchial hypersecretion and bronchospasm. Increased parasympathetic tone tothe gastrointestinal tract can cause diarrhea; nausea and vomiting also occur.Increased parasympathetic tone to salivary glands leads to excessive salivation.Sweating results from increased ACh at the sympathetic cholinergic synapses atsweat glands. CNS effects (depending on the degree of toxicity) include

confusion, ataxia, slurred speech, loss of reflexes, convulsions, coma, and centralrespiratory paralysis. Generalized muscle weakness occurs because increasedACh at the neuromuscular junction produces a depolarizing blockade.

Glutethimide (choice A) is a sedative–hypnotic that can cause a variety ofsymptoms depending on the degree of toxicity. Symptoms include disinhibition,lethargy, stupor, coma, and nystagmus.

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USMLE Step I

By itself the patient's behavior does not suggest a psychotic (choice B) orpersonality (choice C) disorder.

If the patient was oversedated (choice D), he would not appear alert andoriented.

25. The correct answer is E. The urachus is a thick fibrous cord that remainsupon obliteration of the allantois, the connection between the yolk sac and thepresumptive bladder. The intraembryonic portion of the urachus extends fromthe apex of the bladder to the umbilicus. If the lumen of the allantois persistsover a small region (i.e., is not obliterated during normal formation of theurachus), a cystic dilatation will form, which may become infected and enlarge.This is apparently the case in the patient described.

A hydrocele (choice A) results from a small opening at the abdominal end of theprocessus vaginalis that is too small to permit intestinal herniation, but fills withperitoneal fluid. The processus vaginalis is the evagination of the peritoneumthat forms the inguinal canal. Normally it obliterates, leaving an isolated tunicavaginalis (a peritoneal sac) related to the testis, but in hydrocele the closure isincomplete.

A Meckel's cyst (choice B) is a variant of a Meckel's diverticulum in which theproximal portion of the vitelline duct (yolk stalk), which in the early embryo isconnected to the yolk sac, remains patent somewhere along its course. The yolk

sac functions in transferring nutrients to the embryo before the uteroplacentalcirculation is established, so it must be connected to the developing midgut (bythe vitelline duct). In this condition, both ends of the vitelline duct aretransformed into fibrous cords, but somewhere in the middle portion a lumenpersists, forming a cyst hanging from the antimesenteric border of the ileum.

A Meckel’s diverticulum (choice C) is a small outpocketing of the antimesentericborder of the small intestine located 40–60 cm from the ileocecal junction. Itrepresents the remnant of the proximal portion of the vitelline duct (yolk stalk),which in the early embryo is connected to the yolk sac. In normal development,the stalk is obliterated, forming a fibrous cord, but in approximately 2% of

individuals the proximal portion (two inches or so) persists.

An omphalocele (choice D) is a herniation of the abdominal viscera through anenlarged umbilical ring resulting from failure of the bowel to return to theperitoneal cavity after normal, physiological herniation at 6–10 weeks ofdevelopment.

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26. The correct answer is D. The woman described has systemic lupuserythematosus, as evidenced by fever, rash, joint involvement, and renal lesions.The positive antinuclear antibody (ANA), and especially the anti-native DNA(anti-double stranded DNA) tests, strongly suggest this diagnosis. In lupus

nephritis, deposits of immunoglobulins and complement are seen in theglomerular mesangium, or in subepithelial or subendothelial locations.Subendothelial immune complex deposition is strongly suggestive of lupusnephritis.

27. The correct answer is D. The pulmonary artery carries mixed venous blood,which has an oxygen saturation typically ! 75% (possibly as low as 60%); theupper limit of normal for pulmonary artery pressure is 30/15, so 26/14 isnormal.

The ductus arteriosus (choice A) is closed in a healthy person, but if open in anadult, would carry highly oxygenated blood (probably > 80% saturation) fromthe aorta (upper limit of normal pressure is 140/90) to the pulmonary artery, andso would have a pressure somewhere between these values.

The foramen ovale (choice B) is also closed in a healthy adult, but if open, wouldcarry highly oxygenated blood (probably > 80% saturation) from the left atrium(choice C) to the right atrium at atrial pressures; left atrial pressure ranges from afew mm Hg to the v wave peak of " 20 mm Hg.

Right atrial blood (choice E) has low oxygen saturation (possibly 60%), but alsolow pressure (about 0–15 mm Hg).

28. The correct answer is D. The unfortunate girl described succumbed tocongestive heart failure secondary to rheumatic heart disease. Rheumatic fevertypically occurs one to five weeks after a bout of acute streptococcal pharyngitis.Migratory polyarthritis occurs commonly with rheumatic fever, typicallyaffecting the large joints. An expanding erythematous rash with clearing in thecenter (erythema marginatum) is often seen, especially in children, and tends to

occur in a "bathing suit distribution", but may also affect the legs or face.Rheumatic fever is associated with pancarditis, an inflammation of all the layersof the heart. Inflammation of the endocardium may lead to valvular dysfunction,a serious consequence of the disease. The mitral valve is most commonlyaffected, followed by involvement of the mitral and aortic valves together;tricuspid and pulmonic involvement are relatively rare. Mitral stenosis, aorticregurgitation and mitral regurgitation are frequent sequelae of rheumatic

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USMLE Step I

carditis. Myocarditis (i.e., inflammation of the myocardial layer) is the mostserious manifestation of acute rheumatic fever, and may lead to ventriculardilatation and acute decompensation.

Recent histories of cyanosis (choice A), jaundice (choice B), meningitis (choice

C), or skin infection (choice E) are likely unrelated to the girl's clinical course.

29. The correct answer is D. This question asks which drug would dilate thepupil (mydriasis) without producing cycloplegia. Cycloplegia refers to paralysisof the ciliary muscle (which controls the shape of the lens) so that the lens cannotchange focus for different distances. To answer this question quickly, note thatonly phenylephrine and atropine would produce mydriasis. Phenylephrine, an

!1 agonist, produces mydriasis by stimulating ! receptors on the radial dilator

muscle, which constricts it. Cycloplegia would not occur because the ciliary

muscle, which has muscarinic receptors but lacks ! receptors, is not affected byphenylephrine.

Atropine (choice A) produces mydriasis by blocking parasympathetic tone to thepupillary constrictor muscles, but would also cause cycloplegia by blockingparasympathetic tone to the ciliary muscles.

Neostigmine (choice B) is an intermediate-acting carbamylating inhibitor ofacetylcholinesterase; it would increase cholinergic tone to the pupillaryconstrictor muscle, producing miosis, and would cause ciliary muscle

contraction.

Phentolamine (choice C) is a nonspecific !-adrenergic antagonist. It would block

! receptors on the radial dilator muscle, producing at best a small degree ofmiosis, since the predominant tone of the eye is parasympathetic. Phentolamineshould not affect the ciliary muscles.

Pilocarpine (choice E) is a nonspecific muscarinic agonist, and would thereforeproduce miosis and focusing for near vision.

30. The correct answer is A. Amiodarone is one of the most effectiveantiarrhythmic medications; however, its use is limited by severe adversereactions, including pneumonitis and pulmonary fibrosis. Pneumonitis andfibrosis present with progressive cough, pleuritic chest pain, dyspnea, andmalaise. Other adverse effects of amiodarone include agranulocytosis, hepatitis,skin changes, effects on thyroid function, GI effects, and arrhythmias.

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USMLE Step I

Primary atrophy of the thyroid gland (choice C) would produce a small thyroidrather than a goiter.

Riedel’s struma (choice D) produces a hard thyroid attached to the adjacentmuscle.

Subacute thyroiditis (choice E) characteristically produces an exquisitely tenderthyroid.

32. The correct answer is B. The photograph shows a diverticulum in which amucosal outpouching penetrates through a defect in the muscular wall of thesmall intestine. Diverticular disease can cause abdominal pain, constipation, andbleeding.

Adenocarcinoma (choice A) forms an irregular mass grossly resembling thecovering mucosa.

A polypoid adenoma (choice C) or a villous adenoma (choice D) would protrudeinto the lumen above the level of the mucosa.

Grossly, a volvulus (choice E) would appear as a twisted loop of bowel.

33. The correct answer is J. Although several causes of exudative pharyngitis

are among the choices, the diagnosis can be made from the culture results.Streptococcus pyogenes, or Group A "-hemolytic streptococci, can be differentiated

from the other "-hemolytic streptococci by its exquisite sensitivity to bacitracin.

Neither adenovirus (choice A), Coxsackie virus (choice D), Epstein-Barr virus(choice E), nor rhinovirus (choice I) can be cultured on ordinary laboratorymedia.

Candida albicans (choice B) is a cause of oral thrush, but does not commonly

produce pharyngitis, or "-hemolytic colonies on blood agar.

Corynebacterium diphtheriae (choice C) is the cause of diphtheria. The organismcan be identified by its ability to reduce tellurium salts to elemental tellurium,which is black.

Haemophilus influenzae (choice F) causes meningitis, epiglottitis, sinusitis, andotitis in children, and may cause pneumonia in adults. It is cultured on chocolateagar supplemented with factors X and V, and is not beta hemolytic.

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Infection with Mycobacterium tuberculosis (choice G) does not produce the clinicalpicture described, nor will the organism grow on blood agar.

Mycoplasma pneumoniae (choice H) is a cause of community-acquired atypical

pneumonia, and is difficult to culture without special media.

34. The correct answer is C. Halving the amount of enzyme will halve thevelocity of the reaction at each higher substrate concentration. Therefore, curve 3is correct. At very low substrate concentrations, the amount of enzyme is notlimiting, and the curve resembles that of the control. This feature is also seen incurve 3.

Curve 1 (choice A) has the same maximum velocity as the control curve, but has

a faster velocity at lower substrate concentrations. This pattern can be seen withenzymes with a lower Michaelis-Menten constant (KM).

Curve 2 (choice B) has the same maximal velocity, but has a slower velocity atlower substrate concentrations. This pattern can be seen with competitiveinhibition or with enzymes with a higher KM.

If the amount of enzyme is halved, the maximal velocity should be only one-halfthe maximal velocity indicated by the control curve (choice D).

35. The correct answer is A. The history and the presence of fever and coughproductive of purulent sputum are highly suggestive of pulmonary abscess; thex-ray findings are characteristic. Oral surgeries, dental caries, sinus infections,and bronchial infections/bronchiectasis all predispose to pulmonary abscess.Aspiration of infectious material is the most important pathogenetic mechanism;alcoholism makes aspiration more likely.

In bronchiectasis (choice B), chronic infection of the bronchi and bronchiolesleads to dilatation of airways (usually in the lower lobes), and may predispose tolung abscess.

A pulmonary infarct (choice C) is likely to appear as a wedge-shaped infiltrate inthe chest x-ray rather than a fluid-filled cavity; an infarct is typically sterileunless it arises from septic emboli.

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USMLE Step I

Squamous cell carcinoma (choice D) might produce bronchiectasis or lungabscess secondary to obstruction; such a tumor would likely be evident on thechest x-ray.

Tuberculosis (choice E) can certainly produce cavitating lesions in the upper

lobes, but the history, the foul, purulent sputum, and the appearance of thelesion on x-ray all suggest an abscess.

36. The correct answer is C. Cytomegalovirus infection is a feared andunfortunately fairly frequent complication of profound immunosuppressionsuch as that seen in transplant patients and AIDS patients. Many individualsharbor quiescent CMV virus in their bodies that may become active, involvingmultiple organs in the immunosuppressed. The clinical scenario illustrated istypical.

37. The correct answer is A. The ABO blood group has three alleles: A, B, andO. The A and B alleles code for the A and B antigens, respectively, whereas theO allele is the absence of antigen. A and B are codominant to each other. If anindividual is type AB, he will have both A and B antigens on the surface of hisred cells. A and B are dominant to O. If an individual is AO or BO, only the A orB allele will produce surface antigen. The Rhesus or Rh blood group has two

alleles: Rh+ and Rh-. Rh+ indicates the presence of the rhesus antigen on the

surface of the red cell; Rh- indicates the absence of antigen. Rh+ is dominant to

Rh-.

Saline plus child’s RBC - ControlConclusion: The negative control is working.

Anti A plus child’s RBC - Antibody against blood type A does not leadto hemagglutination of the child’s bloodConclusion: The child does not have the Aantigen on the surface of the red cells.

Anti B plus child’s RBC + Antibody against blood type B leads tohemagglutination of the child’s blood

Conclusion: The child does have the Bantigen on the surface of the red cells.

Anti D plus child’s RBC - Antibody against the Rhesus factor does notlead to hemagglutination of the child’s bloodConclusion: The child does not have the Rhantigen on the surface of the red cells.

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Child’s serum plus type ARBC

+ Antibody in the child’s serum leads tohemagglutination of type A bloodConclusion: The child has antibody againsttype A blood.

Child’s serum plus type B RBC - Antibody in the child’s serum does not lead

to hemagglutination of type B bloodConclusion: The child does not haveantibody against type B blood.

Child’s serum plus type DRBC

- Antibody in the child’s serum does not leadto hemagglutination of Rh+ bloodConclusion: The child does not haveantibody against Rh+ blood.

Overall conclusion: The child’s blood type is B negative.

The phenotype and genotype of the father is type AB negative. Since thephenotype of the mother is type B negative, she could either be BB or BO

negative. The Rh- allele is recessive, so two phenotypic Rh- individuals can only

produce an Rh- child. The two types of matings (AB x BB or AB x BO) couldproduce type AB, type B, or type A children. Since the child is B negative, thereare no inconsistencies with the conclusion that the child could be the offspring ofthis couple.

38. The correct answer is C. The woman in the question is experiencing a

normal grief reaction one month after the traumatic loss of her daughter.Uncomplicated bereavement may last for up to three months (longer for closerelations, and in some cultures), and is typically characterized by the type ofsymptoms described in the question stem.

Dysthymic disorder (choice A) is characterized by a depressed mood lastingmost of the day, for the majority of time over a two year period, with theadditional qualification that no major depressive episodes have occurred duringthat period.

Major depressive disorder (choice B) is characterized by the occurrence of atleast one major depressive episode (depressed mood, loss of interest in everydayactivities, anhedonia, feelings of worthlessness, etc.), and no manic or hypomanicepisodes. Because the woman is experiencing depression secondary to the loss ofa loved one, the diagnosis of major depressive episode does not apply, and majordepressive disorder is ruled out.

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USMLE Step I

In obsessive-compulsive disorder (choice D), patients experience recurrent,intrusive thoughts (obsessions) that are relieved, to some extent, by engaging inrepetitive, ritualistic behaviors (compulsions).

Schizoaffective disorder (choice E) is a psychotic disorder with prominent

affective symptoms. There is no evidence to support a diagnosis of psychosis inthis woman.

39. The correct answer is A. Neostigmine is a carbamylatingacetylcholinesterase inhibitor. In this patient, neostigmine is causing excessiveparasympathetic stimulation secondary to increased acetylcholine atparasympathetic neuroeffector sites. Atropine, a muscarinic cholinergicantagonist, would block the excessive stimulation at these sites.

Carbachol (choice B) is a muscarinic and nicotinic cholinergic agonist.Administration of this drug would exacerbate this patient’s symptoms.

Edrophonium (choice C) is a short-acting acetylcholinesterase inhibitor. Thisdrug would exacerbate this patient’s symptoms by causing even greaterinhibition of acetylcholinesterase.

Epinephrine (choice D) is an alpha and beta adrenergic agonist. Whereas itcould counteract some of the symptoms resulting from excessiveparasympathetic stimulation, atropine would be more effective.

Pralidoxime (choice E) is an acetylcholinesterase reactivating agent. This drugregenerates the acetylcholinesterase enzyme following administration of aorganophosphorus acetylcholinesterase inhibitor. It would have no effect in thiscase because neostigmine is a carbamylating acetylcholinesterase inhibitor.

40. The correct answer is B. Increased amniotic fluid (polyhydramnios) occurswith duodenal atresia. The normal fetus will swallow amniotic fluid, however,this is prevented by the intestinal obstruction of duodenal atresia, and theamniotic fluid accumulates. Duodenal atresia presents in an infant with

evidence of upper GI obstruction, such as early vomiting and a "double bubble"sign on x-ray from the stomach and the dilated proximal duodenum.

Congenital heart disease (choice A) may present after birth with various featuressuch as dyspnea, murmurs, tachycardia, failure to thrive, and cyanosis, amongothers.

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Erythroblastosis fetalis (choice C) is a hemolytic disorder of the newborn. Itmost commonly occurs when an Rh-negative mother is sensitized to Rh factorfrom fetal blood (typically during delivery of a previous Rh-positive child). Themother will produce anti-Rh IgG, which crosses the placenta to cause fetalhemolysis, often causing anemia, hydrops (generalized edema), and

hepatosplenomegaly. The disease can also occur with sensitization to non-Rhblood factors.

Horseshoe kidney (choice D) is quite common, and is generally withoutconsequence. The upper or lower poles of the kidneys are fused, essentiallymaking one large kidney.

Neural tube defects (choice E) are associated with improper development andclosure of the neural tube. They include anencephaly, spina bifida withmeningomyelocele or meningocele, and sacral agenesis. Open defects may result

in elevated levels of alpha-fetoprotein (AFP) in the maternal serum (usuallyscreened at 16–18 weeks of gestation).

41. The correct answer is E. Wernicke-Korsakoff syndrome, seen most often inthis country in chronic alcoholics, is due to profound thiamine deficiency.Initially, confusion, ataxia, and ophthalmoplegia dominate the clinical picture,but if the thiamine deficiency is uncorrected, an amnestic syndrome calledKorsakoff’s syndrome supervenes. Korsakoff’s syndrome is characterized by astriking loss of drives, and by dense anterograde amnesia with some retrograde

amnesia with regard to recent events. Characteristic hemorrhagic lesions occurin the mammillary bodies, dorsomedial nucleus of the thalamus, and in theperiventricular regions from the third to the fourth ventricle.

The amygdala (choice A) is an almond-shaped collection of nuclei deep to theuncus of the parahippocampal gyrus. It is not damaged in Wernicke-Korsakoffsyndrome.

The caudate nucleus (choice B) is a major component of the neostriatum thatfunctions in the initiation of voluntary movements. It is not damaged inWernicke-Korsakoff syndrome.

The hippocampus (choice C) is involved in processing new memories. Althoughhippocampal lesions produce amnesia, this structure is not damaged inWernicke-Korsakoff syndrome.

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The jugulodigastric lymph node (choice A), which drains the anterior tongue, isnot a derivative of the third pharyngeal pouch.

The lingual tonsils (choice B) are not derivatives of the third pharyngeal pouch.

The submandibular glands (choice D) are derived from endoderm of thedeveloping oral cavity, and are not derivatives of the third pharyngeal pouch.

The thyroid gland (choice E) is derived from a diverticulum arising from theendodermal lining of the foregut.

45. The correct answer is C. Carbohydrates and proteins both containapproximately 4 kcal of energy per gram. Fats are the more energy-efficientstorage form, containing 9 kcal of energy per gram. 100 grams of carbohydrate

would yield 400 kcal of energy; 25 grams of fat would yield 225 kcal of energy,and 20 grams of protein would yield 80 kcal of energy, for a total ofapproximately 700 kcal of energy. Note that each gram of glycogen, the storageform of glucose, is hydrated with 2 grams of water. Taking that intoconsideration, energy storage in the form of fat is six times more efficient thanglycogen because 1 gram of fat takes up 1 gram worth of space, while one gramof glucose take up 3 grams worth of space.

46. The correct answer is A. Apoptosis is a specialized form of programmed cell

death characterized by synthesis of specific RNAs and proteins, chromatincondensation, and DNA breakdown in the absence of a substantial inflammatoryresponse. The changes in morphology that occur during the formation of digitsare a good example of apoptosis. Apoptosis also occurs in association withhormonal changes (during the menstrual cycle, in the breast after weaning), inthe intestinal crypts, in the immune system, in pathologic atrophy of organs afterduct atrophy (e.g., pancreas), and following damage to cells by a variety ofagents (viruses, cytotoxic cells, radiation).

Differentiation (choice B) is the process by which pluripotent cells becomesuccessively more specialized, rather than undergoing programmed cell death.

Fusion of two or more structures to form a single structure is a common theme indevelopment; failure of such fusion (choice C) is not thought to be responsiblefor syndactyly.

Abnormalities in cellular migration (choice D) during development can result ina wide range of abnormalities, but syndactyly is not among them.

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USMLE Step I

Failure of cellular proliferation (choice E) can result in underdevelopment ofmany structures during embryonic development, but is not thought to beinvolved in the production of syndactyly.

47. The correct answer is B. Bradykinin is a nonapeptide that producesvasodilation via production of nitric oxide, which relaxes vascular smoothmuscle, lowering blood pressure. Because it is a peptide, it can be inactivated bypeptidase treatment. Two different peptidases, called kininase I and II, inactivatebradykinin in vivo. Note that kininase II is the same enzyme as angiotensin-converting enzyme.

Angiotensin II (choice A) is produced from angiotensin I by the actions ofangiotensin-converting enzyme. Angiotensin is a potent vasoconstrictor, and

would significantly increase mean arterial pressure.

Histamine (choice C) is a vasodilator that would produce a decrease in bloodpressure. Histamine would not be inactivated by a peptidase since it is not apeptide; it is an amino acid derivative formed by decarboxylation of histidine.

Neuropeptide Y (choice D) is colocalized with norepinephrine in autonomicganglia, and acts to potentiate the effects of alpha-adrenergic agonists on bloodpressure.

Serotonin (choice E) is produced by hydroxylation and decarboxylation oftryptophan. It would not be susceptible to peptidase treatment.

48. The correct answer is B. The most likely cause of hereditaryhyperammonemia is a deficiency of the enzyme carbamoyl phosphate synthetaseI. This mitochondrial form of the enzyme is distinct from the cytosolic form,which functions in pyrimidine biosynthesis. In an ATP-requiring reaction,carbamoyl phosphate synthetase I converts the ammonia generated in livermitochondria and the bicarbonate produced during respiration into carbamoylphosphate. Carbamoyl phosphate then enters the urea cycle.

The other choices are enzymes associated with amino acid metabolism:

Asparagine synthetase (choice A) amidates aspartate to form asparagine.

Fumarate, produced in the argininosuccinate lyase reaction of the urea cycle,enters the TCA cycle and is converted to oxaloacetate by fumarase (choice C).

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Glutamate-oxaloacetate aminotransferase (choice D) is an enzyme involved inthe degradation of hydroxyproline.

Extrahepatic tissues must convert ammonia produced by their metabolism to

glutamine for transport to the liver. Ammonia is combined with glutamate byglutamine synthetase. Glutamine is a nontoxic compound that can easily passthrough cell membranes and into the bloodstream. In liver mitochondria,

glutaminase (choice E) converts glutamine to glutamate and NH4+, which can

then enter the urea cycle.

49. The correct answer is B. Normally, for each molecule of ATP hydrolyzed,

the cellular Na+/K+ ATPase moves three sodium out of the cell and twopotassium into the cell. When the ATPase is inhibited, sodium begins

accumulating in the cell, and accumulated intracellular potassium begins leakingout. Digitalis inhibits the Na+/K+ ATPase, leading to increased intracellularsodium levels and decreased intracellular potassium levels.

Intracellular Cl- would tend to increase (compare with choice A), approaching

extracellular (serum) values if the Na+/K+ ATPase were inhibited by digitalis.

Digitalis would increase, rather than decrease (choice C), intracellular sodium.

The amounts of sodium and potassium exchanged by the Na+/K+ ATPase are

very small, so blocking the Na+/K+ ATPase would not produce an immediateblock in propagation of action potentials (choice D), but over time, it wouldeventually lead to a conduction block.

Inhibition of the membrane Na+/K+ ATPase would produce depolarization, not

hyperpolarization (choice E), since the Na+/K+ ATPase is an electrogenic pump,removing more positive charge from the cell than it lets in.

50. The correct answer is D. The subscapular artery is the largest and most

distal branch of the axillary artery, and forms an anastomosis with branches ofthe subclavian artery. These anastomoses can permit blood to flow from thesubclavian to the axillary systems, or vice versa.

The vessels in the other answer choices do not anastomose with the distalaxillary system.

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USMLE Step I

Explanations to Block II of the2000 USMLE Step I Released Items

1. The correct answer is C. "-hemolytic streptococci are subdivided into groupsA–D, F, and G using antibodies against the heat- and acid-stable carbohydrateantigens in their cell walls. Streptococcus agalactiae, a group B "-hemolyticstreptococcus, colonizes the vagina in some women. It is primarily a pathogenof newborns (acquired during delivery or from contact with the mother),pregnant women, and diabetics. Its main virulence factor is a capsule thatprevents phagocytosis. Meningitis due to Streptococcus agalactiae is primarilycaused by organisms with capsule type III, and occurs mostly in infants betweenthe ages of approximately 7 and 30 days, although later onset infection can occurup to about three months of age.

2. The correct answer is B. Adult respiratory distress syndrome (ARDS) is theend result of diffuse injury to alveolar capillaries. ARDS may occur inassociation with a number of conditions, including sepsis, direct chemical injury,disseminated intravascular coagulation, radiation injury, diffuse pulmonaryinfections, severe burns, embolic phenomena, dialysis, cardiopulmonary bypass,metabolic disorders, and acute pancreatitis. The injured capillary wall allowsleakage of fluid and proteins into the interstitium or alveoli, producing severepulmonary edema. Pulmonary edema makes the lung less compliant, or stiffer,while thick hyaline membranes formed from edema fluid and cellular debris line

the alveoli, and act as a barrier to gas diffusion, producing hypoxemia.

The alveolar–arterial PO2 difference is increased due to hypoxemia, rather than

decreased (choice A) in ARDS.

The oncotic pressure of alveolar fluid is increased (compare with choice C) asproteins and other macromolecules leak into the alveoli.

The surface tension of alveolar fluid is increased (compare with choice D), as theedema fluid and associated proteins dilute the surfactant, making it less effective.

The work of breathing is increased (compare with choice E) in ARDS, primarilybecause the lungs are less compliant.

3. The correct answer is D. The organism in question is Clostridium difficile, thecause of pseudomembranous colitis (antibiotic-associated colitis). C. difficile is ananaerobic gram-positive rod that forms spores, necessitating sterilization with

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wet heat at 121°C for at least 15 minutes. None of the other procedures listedwould effectively eliminate spores from the bedpan.

4. The correct answer is C. After initial immunization, a booster injection is

essentially a reexposure to the immunizing antigen (tetanus toxoid, in this case).The reaction described is sometimes called the Arthus reaction, a localized formof Type III hypersensitivity, induced by fixation of complement by preformedcirculating antibodies. In severe cases, the degree of complement fixation can beso substantial that it induces local tissue necrosis.

This reaction is mediated predominately by vascular changes induced by thecomplement reaction rather than by accumulation of mononuclear cells (choice

A).

This reaction consumes preformed (usually IgG) antibodies, and does not requireantigen capture by Langerhans’ cells in the epidermis (choice B).

This reaction does not involve the IgE/histamine pathway (choice D).

This reaction utilizes preformed (usually IgG) antibodies rather than requiringIgM synthesis (choice E).

5. The correct answer is D. The suprachiasmatic nucleus is a hypothalamic

nucleus involved, along with the pineal gland, with generation of circadianrhythms. A lesion of this nucleus would disrupt the regulation of circadianrhythms, producing the symptoms described in the question stem.

The accessory optic nucleus (choice A) is involved in eye movements rather thancircadian rhythms.

The lateral preoptic nucleus (choice B) is an interstitial nucleus of the medialforebrain bundle rostral to the lateral hypothalamic area. This nucleus is notinvolved in the generation or maintenance of circadian rhythms.

The pretectal nucleus (choice C) is thought to be involved in the pupillary lightreflex and some types of eye movements, rather than circadian rhythms.

The supraoptic nucleus (choice E) is one of the magnocellular hypothalamicnuclei involved in oxytocin and vasopressin secretion.

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USMLE Step I

6. The correct answer is C. Lymphatic fluid from the scrotal skin drains first tothe superficial inguinal nodes. These nodes receive drainage from the superficialstructures of the lower part of the anterior abdominal wall; from the penis,scrotum, perineum, and buttocks; terminal portions of the urethra, vagina, andanal canal; and superficial structures of the lower limb.

The external iliac nodes (choice A) receive lymphatic drainage from part of theuterus and cervix.

The internal iliac nodes (choice B) receive lymphatic drainage from the bladder,the male internal genital organs, and part of the uterus.

The deep inguinal nodes (choice D) receive lymphatic drainage from thesuperficial nodes, and also directly receive lymphatics from the deeper legstructures.

The lumbar (para-aortic) nodes (choice E) receive lymphatic drainage from theovary and testes.

7. The correct answer is B. The dashed curve is shifted to the right along the x-axis, indicating an increase in volume of the cells. The volume of the cells isgreater in solution Y because of cellular swelling as water flows into therelatively hypertonic cells from the relatively hypotonic solution (osmosis).

If solution Y were relatively hypertonic (choice A), the cells would shrink aswater flowed out of the cells into the solution, and curve Y would be left-shiftedwith respect to curve X.

If solution Y were isoosmotic (choice C) with solution X, there would be novolume change, and the two distributions would be identical. Another way ofsaying this is that the two solutions are isotonic (choice D) with respect to eachother.

8. The correct answer is E. Ecchymosis, perifollicular petechiae, and swelling of

the gums all point to a diagnosis of scurvy, caused by a deficiency of vitamin C.Vitamin C (ascorbic acid) is a water-soluble antioxidant. It is necessary forhydroxylation of proline in collagen synthesis, degradation of tyrosine,epinephrine synthesis, and bile acid formation. Defective collagen in vessel wallsleads to bleeding, producing ecchymoses and petechiae. The man's diet ofprocessed foods lacks a significant source of vitamin C (e.g., citrus fruits,strawberries, green vegetables, potato skins, and tomatoes).

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The incorrect answers are conditions caused by other dietary deficiencies:

Beri-beri (choice A) is caused by a deficiency of vitamin B1 (thiamine). It ischaracterized by peripheral neuropathy and exhaustion. Beri-beri can progress

to cardiovascular, neurological, and muscular degeneration.

Kwashiorkor (choice B) is a severe protein deficiency characterized by weightloss and edema.

Pellagra (choice C) is caused by a niacin deficiency, and is characterized byweight loss, dermatitis, digestive problems, depression, and dementia.

Rickets (choice D) is a disease in young children caused by a deficiency ofvitamin D, leading to defective mineralization of bone matrix. In adults who are

not exposed to sunlight and are deficient in vitamin D, it presents asosteomalacia, a softening of the bones.

9. The correct answer is E. Vomiting causes loss of hydrochloric acid from thestomach, producing an acute metabolic alkalosis. Therefore, the answer choicescan immediately be narrowed to D or E. The respiratory response ishypoventilation, leading to CO2 accumulation, partly normalizing the blood pH.

Normal PCO2 is 40 mm Hg; the correct answer must reflect increased PCO2, so the

answer must be E. If the vomiting is protracted, loss of blood volume stimulates

synthesis of angiotensin II and secretion of aldosterone, increasing acid secretionby the kidney and making the alkalosis worse. Therefore, chronic vomiting in anotherwise healthy person produces alkalosis, mild hypercapnia, and significantlyincreased serum bicarbonate.

The pattern of values in choice A probably reflects metabolic acidosis withrespiratory compensation (hyperventilation).

The pattern of values in choice B probably reflects respiratory acidosis.

The pattern of values inchoice C

are those of a normal person.

The pattern of values in choice D probably reflects respiratory alkalosis.

10. The correct answer is A. Clumping of latex beads coated with antibody(usually IgG) and fibrinogen can be used to identify certain bacterial species,including gram-positive cocci. These tests depend on the availability of very

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USMLE Step I

specific antibodies directed against bacterial antigens. Coating the beads withfibrinogen, as well as antibody, allows detection of fibrinogen receptors on somebacteria (e.g. staphylococcal clumping factor). Staphylococci also express proteinA, which binds IgG.

In choice B, interleukin-1 is a chemotactic factor, and factor VIII is a bloodclotting factor.

In choice C, properdin is a factor that helps to activate complement, and plateletfactor 3 helps to aggregate platelets.

In choice D, prothrombin is the precursor for thrombin, and C3b is acomplement factor.

In choice E, transferrin is an iron transport protein, and plasminogen is the

precursor for the clot-dissolving enzyme plasmin.

11. The correct answer is D. The symptoms described are those of amyotrophiclateral sclerosis, a degenerative neurological disease in which both upper andlower motor neurons are lost. The disease generally first presents in the fortiesor fifties, and is typically initially characterized by asymmetric weakness of alimb or the face or tongue. The disease progresses to symmetrically involve allmuscles of the body. The clinical picture may be one of predominantly lowermotor neuron involvement (flaccid paralysis, fasciculations) or upper motor

neuron involvement (spastic paralysis, hyperreflexia), but eventually both areseen. Cognition is generally preserved, even in the late stages of the disease.Death usually occurs from paralysis of respiratory muscles. Pathologically,atrophy of the corticospinal tracts (and occasionally the precentral gyrus) is seen,along with loss of neurons in the anterior horns of the spinal cord and in themotor nuclei of the brainstem (e.g., the hypoglossal nucleus). Skeletal musclesexhibit neurogenic atrophy.

Decreased pigmentation in the substantia nigra and locus ceruleus (choice A) ischaracteristic of Parkinson disease.

Demyelination of the posterior columns, corticospinal tracts, and spinocerebellartracts (choice B) is typical of Friedrich’s ataxia, although similar findings may beseen in subacute combined degeneration of the spinal cord, caused by vitaminB12 deficiency.

Granulovacuolar degeneration with neuritic plaques and neurofibrillary tangles(choice C) is diagnostic of Alzheimer disease.

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Symmetrical atrophy of the caudate nuclei and frontal cortex (choice E) ischaracteristic of Huntington disease.

12. The correct answer is D. In order for small amounts of virions to beproduced from the injected nucleic acid, the needed proteins must besynthesized. This means that single-stranded positive-sense RNA, which is theonly one from which proteins can be transcribed, must be present. It would notmatter if the RNA were segmented or not.

Double-stranded DNA (choice A), negative-sense RNA (choices B and E), andsingle-stranded DNA (choice C) cannot be directly transcribed.

13. The correct answer is A. These data indicate a left-to-right shunt ofoxygenated blood into the right atrium, as evidenced by the elevated oxygensaturation of the right atrium, right ventricle, and pulmonary artery.Additionally, the mildly elevated pressures in the right atrium, right ventricle,and pulmonary artery can be explained by increased flow in all of these areas.Atrial septal defect would shunt blood from the left atrium to the right atrium,and the increased pulmonary flow compared to left ventricle cardiac outputwould explain all of the elevated right heart pressures. The most common typeof atrial septal defect is an ostium secundum type, which occurs near theforamen ovale (not to be confused with the hemodynamically insignificant

patent ductus arteriosus, a slitlike remnant of the foramen ovale).

Mitral stenosis (choice B) would elevate left atrial pressure, but would notprovide a shunt. Therefore, oxygen saturation in the various chambers wouldnot be affected.

Patent ductus arteriosus (choice C) would provide a left-to-right shunt from theaorta into the pulmonary artery, and in time could lead to pulmonaryhypertension, but would not be expected to elevate oxygen saturation in the rightatrium.

Pulmonic stenosis (choice D) would not elevate pulmonary artery pressure sincethe resistance is between the pulmonary artery and the right ventricle. Thisdisorder would not cause a change in oxygen saturation in the right heartbecause there is no shunt.

Tricuspid insufficiency (choice E) would not provide a shunt to change oxygenlevels, and would not, by itself, increase pulmonary artery pressure, since the

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USMLE Step I

"extra" blood pumped by the ventricle is regurgitated back into the atrium, notinto the artery.

14. The correct answer is D. Warfarin is a coumarin anticoagulant that works

by blocking the vitamin-K–dependent gamma-carboxylation of prothrombin,and factors VII, IX, and X, as well as protein C. It is used in the prophylaxis andtreatment of deep vein thrombosis and pulmonary embolism, and for thetreatment of atrial fibrillation with embolism. It is also used as an adjunct in thetreatment of coronary artery occlusion, cerebral transient ischemic attacks, and asa prophylactic in patients with prosthetic cardiac valves. Since the patient isgoing to receive trimethoprim-sulfamethoxazole (TMP-SMX) therapy forrecurring urinary tract infections, the dosage of the warfarin will need to bedecreased because the interaction between these two medications will increasethe patient's prothrombin (PT) time. The mechanism by which the PT time will

be elevated is probably related to one or more of the following factors: 1) TMP-SMX displaces warfarin from circulating plasma proteins, thus increasingwarfarin levels; 2) since SMX-TMP and warfarin are both metabolized by theliver in a similar manner, the blood levels of both medications will increase dueto a decreased rate of metabolism; 3) since intestinal bacteria produce nearlyone-half of our daily vitamin K requirements, SMX-TMP can also decreaseproduction of this vitamin by eliminating these bacteria in the intestinal tract.

Institution of vitamin K therapy (choice A) would not be recommended, sincethis would antagonize the effects of the warfarin.

Increasing the dose of the warfarin (choice B), would increase the patient'sprothrombin time.

Stopping the warfarin and beginning aspirin therapy (choice E) would not berecommended, since aspirin therapy alone would not provide the level ofanticoagulation therapy needed for a patient with a recent prosthetic heart valvereplacement.

15. The correct answer is B. The uterine tubes develop from the cranial portions

of the müllerian (paramesonephric) system. Regression of the müllerian ducts inthe male normally occurs in response to secretion of müllerian inhibitorysubstance (MIS) by the Sertoli cells of the embryonic testes; failure to secrete MIS(or an abnormality in the receptor for MIS) can result in a genetic male withtestes and normal male phenotypic development, but additionally, uterine tubes,a uterus, and an upper vagina.

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The embryonic testes secrete testosterone rather than estrogen (choice A).

Production of testosterone by the embryonic testes (choice C) is necessary for thedevelopment of the wolffian duct system derivatives (e.g., vas deferens) in themale.

An abnormal response to estrogen (choice D) is not likely to cause persistence ofthe müllerian system. The predominant sex steroid in this genetic male would betestosterone.

Testosterone (choice E) does not cause regression of the müllerian system,although it is essential for the development of wolffian (mesonephric) structuresin the male.

16. The correct answer is B. Delirium is an acute, usually reversible disorderaffecting orientation, perception, cognition, and reality testing. Clouding ofconsciousness is a prominent attribute, as are disorientation to place and time,agitation, and confusion. The symptoms are rarely consistent, and often varyover time. Individuals often cannot maintain a coherent train of thought, andhave difficulty sustaining or shifting attention. Psychomotor functioning and thesleep–wake cycle are often disturbed. Delirium is associated with many diverseetiologies: drug withdrawal (e.g., alcohol, barbiturates), bacterial and viralinfections, intoxications, metabolic derangements, and certain neurologicaldisorders. It is more common in the elderly.

Brief reactive psychosis (choice A) is diagnosed in patients who developpsychotic symptoms and emotional turmoil in response to a severe emotionalstressor. Orientation would be preserved in this condition.

Mania (choice C) is an elevated, expansive, or irritable mood that is characterizedby hyperactivity, pressured speech, grandiosity, decreased need for sleep,increased sex drive, and distractibility. Orientation would be preserved in thiscondition.

Psychotic depression (choice D) is major depression with delusions and/or

hallucinations, but with preservation of orientation. The delusions can be mood-congruent (i.e., consistent with the typically depressive themes of worthlessness,guilt, punishment and death) or mood-incongruent (e.g., persecution, thoughtinsertion).

Schizophrenia (choice E) is a group of disorders characterized by incoherence ofspeech and thought, hallucinations, delusions, inappropriate affect, and

Kaplan/NMSR



USMLE Step I

deterioration in social and occupational functioning. Lack of orientation in apatient diagnosed with schizophrenia should alert the clinician to the possibilityof delirium, rather than schizophrenia.

17. The correct answer is E. The cricoarytenoid muscles are innervated by therecurrent laryngeal nerve (a branch of the vagus), which is given off in the thoraxas the vagus passes anteriorly to the aortic arch. The left recurrent laryngealnerve winds back under the arch of the aorta to ascend in the neck, and couldtherefore be compressed by an aortic aneurysm. Compression of this nervewould produce paralysis of the cricoarytenoids, and therefore hoarseness. Theother muscles listed do not directly contribute to vocalization, with the exceptionof the cricothyroid.

The digastric muscle, which is one of the group of suprahyoid muscles that

depress the mandible, raises the hyoid and steadies it during chewing andswallowing. The anterior belly of the digastric (choice A) is innervated by themandibular branch of the trigeminal nerve (from the inferior alveolar), whichwould probably not be affected by an aortic arch aneurysm.

While the cricothyroid (choice B) is a laryngeal muscle, its innervation is fromthe superior laryngeal nerve (off the vagus). In fact, it is the only laryngealmuscle innervated by this nerve, and not by the recurrent laryngeal nerve. Sincethe superior laryngeal nerve approaches the larynx superiorly, it is unlikely thatit would be affected by an aortic arch aneurysm.

The omohyoid (choice C), which is one of the group of infrahyoid muscles,depresses, retracts, and steadies the hyoid during chewing and swallowing. It isinnervated by a branch of the ansa cervicalis off the cervical plexus, which isfound superiorly in the neck, and would be unlikely to be affected by an aorticarch aneurysm.

The posterior belly of the digastric (choice D) is the other half of the digastricmuscle, and is one of the group of suprahyoid muscles. This muscle depressesthe mandible, and raises and steadies the hyoid bone during chewing andswallowing. The muscle is innervated by the facial nerve, which would not

likely be affected by an aortic arch aneurysm.

18. The correct answer is E. The artificial sweetener aspartame is the dipeptideof the methyl ester of phenylalanine and aspartate. In phenylketonuria (PKU; anautosomal recessive disorder), individuals lack the enzyme phenylalaninehydroxylase, which catalyzes the hydroxylation of phenylalanine to tyrosine (the

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first reaction in the catabolism of phenylalanine). In the absence ofphenylalanine hydroxylase, phenylalanine is transaminated to yieldphenylpyruvate, which is then decarboxylated to produce phenylacetate, orreduced to form phenyllactate (which causes individuals with PKU to have acharacteristic mousy odor). Accumulation of phenylalanine or its metabolites in

early life impairs normal development of the brain, and leads to severe mentalretardation. If phenylketonuria is recognized in the newborn, most of themanifestations of the condition can be prevented by strict adherence to a specialdiet that is low in phenylalanine and tyrosine. In adolescence, the dietaryrestrictions can be eased because the nervous system has developed; however,female phenylketonurics need to adhere to the diet during pregnancy to protectthe developing fetus.

Restriction of phenylalanine intake is not necessary for any of the otherconditions listed in the question.

Diabetes mellitus (choice A) is caused by lack of insulin; long-term complicationsof the condition involve the eyes, kidneys, nerves, and blood vessels.

Hereditary fructose intolerance (choice B) is caused by a deficiency of fructose-1-phosphate aldolase. Acute ingestion of fructose leads to nausea, vomiting,sweating, and sometimes convulsions. Chronic exposure can lead to jaundice,hepatomegaly, diarrhea, and failure to thrive.

Lactose intolerance (choice C) is a common condition characterized by

abdominal cramps and diarrhea. It is caused by a lack of the enzyme lactase,which breaks lactose into glucose and galactose.

Maple syrup urine disease (choice D) is caused by a defect in the branched-chain

-keto acid dehydrogenase complex, affecting the degradation of isoleucine,leucine, and valine. It is characterized by vomiting, convulsions, mentalretardation, and early death.

19. The correct answer is E. The median nerve lies between flexor carpi radialisand flexor digitorum superficiale in the wrist. The recurrent branch of themedian nerve supplies three of four thenar muscles: the abductor pollicis brevis,the flexor pollicis brevis, and the opponens pollicis, all of which participate inopposition of the thumb. Thus, damage to the median nerve would paralyzethese muscles, preventing opposition of the thumb. The muscles that adduct and abduct the fingers (choice A; palmar and dorsalinterossei, respectively) are supplied by the ulnar nerve (deep branch), which

Kaplan/NMSR



USMLE Step I

runs more medially in the wrist through a passage formed by the pisiform bone,the hook of the hamate bone, and the flexor retinaculum.

The muscles that extend the index finger (choice B; extensor digitorum andextensor indicis) are supplied by the radial nerve (posterior interosseus branch),

which runs deep in the posterior compartment of the forearm, and would notlikely be damaged by cutting the wrist.

Although damage to the median nerve might be expected to affect finger flexiondue to loss of innervation of flexor digitorum superficiale, the wrist injury islikely to be distal to the branches that supply this muscle. In addition, two othermuscles specifically involved in fourth (ring) and fifth (little) finger flexion(choice C)—the medial half of flexor digitorum profundus and the medial twolumbricals—are supplied by the ulnar nerve.

Sensation to skin at the base of the little finger (choice D) is supplied by the ulnarnerve (dorsal or cutaneous branch), which runs more medially between the ulnaand flexor carpi ulnaris.

20. The correct answer is E. The symptoms (anorexia, nausea, and brown urine)are typical of hepatitis. The presence of HBV surface antigen and anti-HBV coreantigen suggest that this patient has an active (and recently acquired, because ofthe absence of antibody to HBV surface antigen) hepatitis B infection. The bodyfights viral infections by CD8+ T cell-mediated lysis of infected cells that present

viral antigens on their surface.

The viral surface antigen does not directly injure hepatocyte membranes (choice

A).

IgM directed against core antigen does not cause the hepatocytes to lyse (choice

B).

Inhibition of hepatocyte DNA replication (choice C) does not cause hepaticinjury in hepatitis B.

Inhibition of mRNA translation (choice D) is not responsible for hepatocellularinjury in hepatitis B.

21. The correct answer is C. An abnormally increased amount of hepaticglycogen of normal structure indicates that glycogen is being synthesizednormally. However, once made, it cannot be broken down and exported from

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the liver as glucose. After administration of oral fructose, there is not an increasein serum glucose, indicating that fructose has not been converted into glucoseand exported. The defect that could tie both of these clinical observationstogether is a deficiency of the enzyme glucose-6-phosphatase. Glucose-6-phosphatase is the liver enzyme that removes the phosphate group from glucose,

allowing it to leave the cell to supply the needs of the periphery. Both glycogenand fructose must first be transformed by the liver into glucose 6-phosphate, andthen to glucose in order to leave the cell.

A deficiency in fructokinase (choice A) would result in the inability of any cell totrap fructose by means of phosphorylation. After oral administration, blood andurine levels of fructose would rise.

A deficiency in glucokinase (choice B) would not allow the liver to removeexcess glucose from the blood for storage.

A deficiency in phosphoglucomutase (choice D) would not allow theinterconversion of fructose and glucose.

UDPG-glycogen transglucosylase (choice E) is another name for glycogensynthase.

22. The correct answer is E. A normal loading dose and increased maintenancedose would be the most rational drug regimen for fast metabolizers. A loading

dose is often given because a drug is heavily bound to plasma proteins; thisquickly saturates plasma proteins so that therapeutic levels of free drug will beavailable rapidly. The amount of plasma protein binding is not affected by therate of drug metabolism, so increasing the loading dose is unnecessary. Themaintenance dose is given to maintain adequate levels of free drug in the plasma.Drug clearance is accelerated in rapid metabolizers, and free drug levels declinequickly. Increasing the maintenance dose can help prevent this decline.

23. The correct answer is B. The patient described suffers from apheochromocytoma. In this disorder, a neoplasm of the adrenal medulla cells

produces and secretes catecholamines (epinephrine and norepinephrine), givingrise to hypertension, tremor, anxiety, sweating, palpitations, and increasedurinary excretion of catecholamines and their metabolites. Most occursporadically, but a small percentage occur in association with certain familialsyndromes (e.g., multiple endocrine neoplasia types IIa and IIb, von Hippel-Lindau syndrome). Most pheochromocytomas are benign, although some aremalignant (choice D), and may metastasize.

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USMLE Step I

Adrenal cortical masses (choices A and C) are unlikely to be associated withcatecholamine production; instead they may produce excess steroid hormones.

Diffuse hyperplasia of the adrenal cortex (choice E) is associated with

adrenocortical hyperfunction rather than medullary hyperfunction.

Diffuse hypoplasia (choice F) of the adrenal medulla would lead to decreasedcatecholamine production and excretion.

24. The correct answer is C. Isoniazid is indicated for prevention ofactive tuberculosis in a person known to be infected with the tuberclebacillus. Isoniazid causes drug-induced hepatitis in about 1% of patientsreceiving the drug. Once hepatitis occurs, the drug must be discontinued

to prevent an increase in the severity of the damage. Age is the mostimportant factor in determining the risk of isoniazid-induced hepatitis. Inpatients under the age of 20, the risk of hepatitis is less than 0.3%, but in apatient older than 50 years, the risk is approximately 2.3%.

Hepatic tuberculosis (choice A) is a rare condition that presents with feverand abnormalities of liver function tests. It can produce extrahepaticobstruction with ascending cholangitis.

Infection with hepatitis B virus (choice B) could produce the symptoms in

the problem; however, there is no evidence in the history that thisindividual has an acute hepatitis B infection, and chronic carriers ofhepatitis B virus tolerate isoniazid very well.

Pyridoxine (choice D) is given with isoniazid to minimize the incidence ofperipheral neuritis. Such administration has not been associated withcholecystitis.

Tubercular pancreatitis (choice E) is very rare, and resembles otherinfections or neoplastic changes of the pancreas.

25. The correct answer is C. Lovastatin is an HMG-CoA reductase inhibitorused to decrease cholesterol synthesis in patients with hypercholesterolemia.About 10% of patients taking reductase inhibitors intermittently show increasesin creatine kinase (CK) activity, often in association with heavy physical exercise.Patients may rarely present with striking elevations in creatine kinase activityaccompanied by generalized pain in skeletal muscles. In such cases, if the drug

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is not discontinued, the ensuing rhabdomyolysis can cause myoglobinuria,which can produce renal failure.

Captopril (choice A) is an ACE inhibitor; toxic effects include hyperkalemia andcough.

Hydrochlorothiazide (choice B) is a thiazide diuretic; toxic effects includeincreased serum lipids, uric acid, and glucose, and potassium wasting.

Nicotinic acid (choice D), or niacin, is used to treat some hyperlipidemias. Toxiceffects include flushing, pruritus, rashes, dry skin, acanthosis nigricans,hyperuricemia, and hypotension if the patient is receiving antihypertensivetherapy.

Propranolol (choice E) is a nonselective beta adrenergic antagonist. Toxic effects

include bradycardia, atrioventricular blockade, congestive heart failure, andasthma attacks.

26. The correct answer is D. Cocaine inhibits the process of monoamine(dopamine, norepinephrine, and serotonin) reuptake. Its effects on the heart areprimarily due to elevation of synaptic norepinephrine levels. Cocaine indirectlystimulates alpha receptors on the coronary arteries via increased norepinephrine,producing vasoconstriction and myocardial ischemia.

Cocaine is not a beta blocker (choice A).

Cocaine is not a direct adenosine receptor agonist (choice B).

Cocaine is not a direct beta receptor agonist (choice C).

Cocaine does not indirectly inhibit alpha adrenergic receptors (choice E).

27. The correct answer is D. Eukaryotic DNA is complexed with proteins calledhistones to form chromatin. Nucleosomes, the basic structural unit of chromatin,

are formed from an octamer of histone proteins (two copies of H2A, two copiesof H2B, two copies of H3, and two copies of H4) about which the DNA iswrapped. Treatment with micrococcal nuclease will digest free DNA, leavingDNA that was wrapped into a nucleosomal arrangement intact. Digestion of theintervening DNA would leave double-stranded DNA bound to these eightproteins.

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USMLE Step I

Histone H1 (choice A) is important in higher-order packing of nucleosomes into30 nm fibers, but it does not protect the DNA in the same manner as the basichistone octamer.

The nuclear membrane (choice B) would not substantially protect the DNA from

digestion.

The nucleolus (choice C) is the site of ribosomal assembly, and would not serveto protect the majority of nuclear DNA from nuclease digestion.

28. The correct answer is A. Disseminated intravascular coagulation (DIC) is acoagulopathy characterized by bleeding from mucosal surfaces,thrombocytopenia, prolongation of the prothrombin time and the activatedpartial thromboplastin time, decreased fibrinogen level, elevated fibrin split

products, and fragmented red blood cells (schistocytes) on peripheral bloodsmear. In this disorder, both the coagulation and fibrinolytic systems areactivated, leading to widespread clotting (and bleeding) with consumption ofplatelets and clotting factors. DIC is thought to be etiologically related either toendothelial injury or the release of thromboplastic substances. It is a fearedcomplication of gram-negative sepsis or other overwhelming infections, and mayoccur in association with obstetrical complications, severe burns, trauma, and avariety of other conditions. Therapy in DIC is always problematic, and may beunsuccessful since the patient can either die of hemorrhage secondary tobleeding, or of widespread infarction and ischemia secondary to clot formation.

The most specific laboratory test for DIC is theD

-dimer assay for fibrin splitproducts, which measures a degradative product produced during fibrinolysis.

The activated partial thromboplastin time (choice B) is increased due toconsumption of clotting factors, but this is not the most specific test for DIC.

Plasma fibrinogen concentration (choice C) can be decreased due toconsumption, but this is not the most specific test for DIC.

Platelet count (choice D) can be decreased due to consumption, but this is not themost specific test for DIC.

Prothrombin time (choice E) is increased due to consumption of clotting factors,but this is not the most specific test for DIC.

29. The correct answer is C. After the leg has been restrained by a cast for sixweeks, one would expect atrophy of the gastrocnemius muscle. Atrophy is

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characterized by a loss of muscle cell mass due to a decrease in the number ofmyofibrils per cell, as in choice C, but not in the number of muscle cells (choice

B).

Conversion of the muscle to fast fibers (choice A) would be expected to occur

after denervation and reinnervation by a different nerve branch, because themuscle fiber type (fast or slow fiber) is determined by the type of innervation.No evidence for denervation injury is given in the question.

Considering that exercise, which requires aerobic metabolism, increases themitochondrial content of muscle (choice D), immobility would be expected todecrease the mitochondrial content.

Satellite cells are normally quiescent cells in skeletal muscle that become active inregeneration and hypertrophy, and so would not be expected to increase in

number with immobility (choice E).

30. The correct answer is E. There are two muscles that separately control thediameter of the pupil. The radial dilator muscle, which contracts (producing

mydriasis) with -adrenergic receptor stimulation, is under the control of thesympathetic nervous system. The pupillary sphincter muscle (sometimes calledthe pupillary constrictor muscle) produces pupillary constriction with muscariniccholinergic receptor stimulation, and is controlled by the parasympatheticnervous system. Without treatment, the left eye is slightly smaller than the right,

which by itself could indicate either decreased sympathetic tone or increasedparasympathetic tone in the left eye.

Tyramine is an indirect-acting sympathomimetic (it is taken up by sympatheticnerve terminals and displaces the norepinephrine [NE] into the synapse). Fortyramine to work, sympathetic nerves must be intact. Tyramine had an effect onthe right eye, but failed to affect the left, which is consistent with sympathetic

denervation (choice E). However, choice A cannot be ruled out yet because !-adrenergic receptor blockade could have prevented NE's actions on the radialdilator muscle. The treatment with EPI can distinguish between choices A and

E. If the answer was A, then EPI should have no effect. However, EPI producedan unusually large effect in the left eye, which is consistent with sympatheticdenervation supersensitivity.

Blockade of "-adrenergic receptors (choice B) can be quickly eliminated because

" receptors do not play a role in controlling the diameter of the eye.

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USMLE Step I

Blockade of muscarinic receptors (choice C) can be quickly eliminated becausewithout treatment, the left eye would be dilated.

If the left eye had been previously treated with cholinesterase inhibitors (choice

D), tyramine still could have produced some dilation.

31. The correct answer is E. The first tip-off to the diagnosis is the “worm-like”mass, which suggests either ectasia of the spermatic duct or varicocele (dilated,tortuous veins). These two conditions can be distinguished by the effect ofmoving to a supine position, only the varicocele will “disappear” (or at leastsignificantly decrease in size).

Cystocele (choice A) generally occurs in women with uterine prolapse,producing a round, rather than a worm-like, lesion.

Ectasia of the spermatic cord (choice B) will produce a worm-like mass, but itwill not disappear when the patient lies down.

Indirect inguinal hernia (choice C) will produce a thicker mass that may or maynot disappear when the patient lies down.

Spermatocele (choice D) will produce a round, rather than worm-like, lesion.

32. The correct answer is A. Sexual desire often decreases in the first trimesterbecause of fatigue and nausea. This desire may rebound in the second trimester,for a variety of reasons. Energy levels typically increase and nausea usuallywanes in the second trimester. Also, by this point, the woman is betterpsychologically and physically adjusted to the pregnancy, and at this time, hersize is not as much of an issue as it will be later in her pregnancy.

During the third trimester (choice B), interest often wanes again for obviousreasons. The size of the woman and the aches and discomforts that accompanyadvancing pregnancy can put a damper on sexual desire. Also, psychologically,the woman may be focusing more on the impending birth.

The woman will most likely have little sexual interest one week postpartum(choice C). First of all, sexual intercourse is contraindicated for medical reasonsone week after delivery. Also, the radically changing hormone levels, physicallyrecovering from the trauma of childbirth, and exhaustion because of childbirthand the subsequent care of the newborn can often diminish sexual desire.

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Sexual desire may still be dampened one month postpartum (choice D). It oftentakes at least six weeks to recover from childbirth. Also, the new mother is stilladjusting to not being pregnant, and is still fatigued by care of the newborn. Shemay also be afraid of pain or of becoming pregnant again.

33. The correct answer is E. Insertion of a uracil after the end of codon 135changes the reading frame, and produces the following codons: UUU, AUG,UGU, and UAA (a stop codon). Thus, three amino acids are added after codon135, before the stop codon UAA, and the total number of amino acids is 138.

34. The correct answer is B. All nucleated cells in the body display MHC I classantigens. When the abnormal proteins from the bacteria or viruses are degradedin the cytoplasm of a macrophage (or other nucleated cells), parts of them are

displayed bound to MHC I sites on the plasmalemma. CD8+ cytotoxic cellsrecognize these antigens as “nonself”, and induce lysis of the macrophage orother cells, helping to clear the infection. Macrophages display MHC II antigensas well, and present antigens resulting from engulfment and phagocytosis ontheir surface in association with MHC II to CD4+ cells. In this case, however, theantigenic material escaped from the phagosome and entered the cytoplasm. Themacrophage is therefore treated like any other infected cell, and is killed byCD8+ cells.

35. The correct answer is E. This patient has symptoms of nonproductive coughwith dyspnea on exertion, and signs of consolidation in the right lower lobe.Sounds in the lungs are always transmitted better in areas of consolidation thanin areas where infiltrates are primarily interstitial and spare the alveoli (e.g.,atypical pneumonia due to Mycoplasma pneumoniae). Lung consolidation signsdetected by the clinician include: (1) decreased percussion, (2) increased tactilefremitus (increased vibrations over the normal that are transmitted to the chestwall when the patient speaks), (3) bronchial breath sounds (short inspiratoryphase and long expiratory phase from compression of the bronchi by aninfiltrate), and (4) wet crackles (sounds created by airflow through fluid in smallairways and alveoli). The clinical findings in this patient are most consistent

with a right lower lobe consolidation secondary to a bacterial lobar pneumonia,most likely due to Streptococcus pneumoniae, the most common community-acquired typical pneumonia.

Asthmatic bronchitis (choice A) presents with cough and generalized wheezing,the latter due to inflammation of the terminal bronchioles with narrowing of theairways. Consolidation signs are absent.

Kaplan/NMSR



USMLE Step I

In bullous emphysema (choice B) there is destruction of alveoli with formation oflarge air-filled bullae at the periphery of the respiratory unit (e.g., respiratorybronchiole, alveolar duct, and alveoli). Hyperresonance to percussion and drycrackles are commonly present rather than signs of consolidation.

Chronic bronchitis (choice C), like bullous emphysema, is most often secondaryto smoking. The primary sites of inflammation are: (1) the segmental bronchi,inflammation of which leads to increased mucous production (productive cough)and sibilant rhonchi (whistling sounds that clear with coughing), and (2) theterminal bronchioles, in which inflammatory obstruction of airways leads towheezing and trapping of air. Consolidation signs are absent.

Signs of consolidation may occur in congestive heart failure (choice D); however,unlike lobar pneumonia, the consolidation signs are bilateral, involving both

lung bases.

A pleural effusion (choice F) refers to a collection of fluid in the pleural cavity.On physical examination, there is dullness to percussion. However, unlike lobarpneumonia, fluid in the pleural cavity blocks transmission of sounds from thelung parenchyma. Therefore, tactile fremitus and crackles are absent. Signs ofconsolidation are absent.

Pleuritis (choice G) refers to inflammation of the pleura. The classic finding ofpleuritis is a pleural friction rub. Patients experience severe, knife-like pain with

inspiration from stretching of the inflamed pleura. Signs of consolidation areabsent.

A pneumothorax (choice H) refers to collapse of the lung due to an opencommunication between the pleural cavity and the airways. It is usuallysecondary to rupture of a bleb (air filled cavity) in or directly beneath the pleura.There is hyperresonance to percussion; breath sounds and tactile fremitus areabsent. There are no consolidation signs.

A pulmonary embolism (choice I) produces sudden onset of tachypnea anddyspnea. Small emboli usually lodge in peripheral pulmonary arterial branches,

and produce a wedge-shaped area of hemorrhagic infarction extending to thepleural surface. Inflammation of the pleura produces pleuritic chest pain, apleural friction rub, and a hemorrhagic pleural effusion.

36. The correct answer is D. The physical findings are consistent with volumedepletion (e.g., dizziness and hypotension on standing, sinus tachycardia).

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Diverticulosis does not predispose for colon cancer (choice B) or Crohn’s disease(also known as granulomatous colitis; choice D).

Diverticulosis does not predispose for malabsorption (choice E), as nutrient

absorption principally occurs in the small intestine, whereas diverticula occurmainly in the distal colon.

39. The correct answer is D. The relationship between flow and pressure is

described by the equation: pressure gradient #P = flow (Q) $ resistance (R). Ifresistance is constant, as in a rigid tube, flow increases linearly with upstreamperfusion pressure (P1, or arterial pressure in a vascular bed) as shown in choice

A.

Autoregulation refers to the ability of the arterioles in a vascular bed to regulatetheir resistance in order to maintain a relatively constant flow in the face of achanging perfusion pressure gradient, for example, when mean arterial pressure(MAP) changes due to increased cardiac output, as shown in choice D. At verylow MAP, the arterioles are maximally dilated, but as the MAP reaches theautoregulation range, the arterioles begin to constrict to compensate for theincreased pressure (myogenic response). At very high MAP, arterioles reachmaximum constriction, and further MAP increases will again raise flow (as seenon the far right-hand side of the curve). Many vascular beds show some degreeof autoregulation, with the cerebral, renal, and coronary circulations exhibiting

the strongest effects.

40. The correct answer is A. Breast abscess is caused by bacterial infection(usually S. aureus) of the breast, which generally occurs if the normal barrier toinfection provided by the skin of the nipple is compromised (usually by traumadue to nursing). The condition usually will usually respond to antibiotic therapy,but the resulting scar may mar the shape of the breast.

A fibroadenoma (choice B) usually causes a round, firm breast mass that wouldnot resolve with antibiotic administration.

Fibrocystic change (choice C) can cause round, fluid-filled masses (cysts) or areasof thickened breast tissue (fibrosis) that would not resolve with antibiotics.

Infiltrating mammary carcinoma (choice D) can produce skin retraction, butwould be very unusual in a 24-year-old, and would not have responded toantibiotics.

Kaplan/NMSR



USMLE Step I

Traumatic fat necrosis (choice E) follows trauma to the breast, and would not beexpected to respond to antibiotics.

41. The correct answer is B. Coccidioidomycosis is caused by the dimorphic,primary pathogenic fungus Coccidioides immitis, native to the deserts of theSouthwestern United States. Arthroconidia, the infectious mycelial form of thefungus, are inhaled, and develop into endospore-containing spherules thateventually burst, releasing endospores into the surrounding tissue. Eachendospore then develops into a new spherule. Most infections are subclinical,but the disease can present as an acute pneumonia that can be complicated bydissemination of the organism to other sites via the bloodstream. The clinicalspectrum associated with disseminated disease is very wide, and includesosteomyelitis, arthritis, subcutaneous nodules, and an extremely virulent form of

meningitis. Dissemination is most often seen in those with defects in cell-mediated immunity, although Blacks and Filipinos have a tenfold greaterincidence of dissemination.

Blastomycosis (choice A) is caused by the dimorphic fungus Blastomycesdermatitidis, which is endemic to the Central United States from the Great Lakesto Arkansas. It usually affects the lungs, but may disseminate.

Histoplasmosis (choice C) is caused by the dimorphic fungus Histoplasmacapsulatum, which is most common in soil contaminated with bird or bat

droppings in the Mississippi and the Ohio River Valleys. It usually affects thelungs, but may disseminate.

Mycobacterium marinum (choice D) is an atypical mycobacterium found inswimming pools and fish tanks. It is the cause of “fish fanciers finger”, a chroniclocalized granulomatous infection found among tropical fish keepers.

Mycoplasma pneumoniae (choice E) is a common cause of atypical pneumonia.

42. The correct answer is D. In “starvation”(after glycogen stores are depleted),

the body eventually shifts to using fatty acids and other fat-derived substances.Muscle specifically uses serum fatty acids for nutrition in this setting.

Creatine phosphate (choice A) is a temporary energy storage form for muscle.

Muscle glycogen (choice B) would be exhausted after two days.

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Triglycerides are not stored in muscle (choice C) for energy, although thoughthey are present in muscle membranes.

Serum glucose (choice E) will be relatively low in this setting, and is “reserved”primarily for the central nervous system.

43. The correct answer is E. Regression is reversion to an early or infantilepattern of behavior. Regression may be expressed as helplessness anddependency (seeking frequent reassurance and advice) in a patient as a responseto serious illness.

Apathy and listlessness (choice A) might occur as an outward manifestation ofdepression.

Suspiciousness about a diagnosis of myocardial infarct (choice B) could berelated to denial of the patient’s condition.

Anger because the dietician delivers the wrong food (choice C) might be due todisplacement of the patient’s anger about his illness onto a safe target.

A patient in the cardiac care unit who continues to make phone calls to his office(choice D) may be exhibiting elements of denial and rationalization.

44. The correct answer is B. Ignoring the tantrums as much as possible wouldlead to extinction of the behavior. Extinction involves removing whateverreinforcement is maintaining the response, gradually causing it to die out. Thischild is attempting to get attention through acting out; ignoring him will causehim to try a different tactic for receiving attention.

Distracting him (choice A) with a treat positively reinforces the behavior, andmay increase the frequency with which it occurs.

Spanking him (choice C) is an attempt to use punishment to suppress thetantrums; in this case, however, the effect might be canceled out by the

conflicting reinforcer of giving him attention.

Children should never be threatened with desertion (choice D) when theymisbehave. In addition, parents lose credibility by not following through on astatement.

Kaplan/NMSR



USMLE Step I

45. The correct answer is D. Since the patient has a fixed intake of NaCl of 200mmol/day, after four days the total intake of NaCl will be 800 mmol. The totalamount of NaCl excreted in the urine over the four days is 500 mmol; therefore,the patient has retained 300 mmol of NaCl. The patient will retain water tomaintain a normal serum sodium concentration. If 1 liter of normal saline

containing 150 mmol of NaCl weighs 1 kg, then the 300 mmol of NaCl in thepatient will increase the patient's weight by 2 kg. Hence, instead of 70 kg, thepatient will weigh 72 kg after four days.

The most common cause of increasing weight in a hospitalized patient is sodiumretention (any time a patient retains sodium, the body weight increases).Therefore, choice A (66 kg), choice B (68 kg), and choice C (70 kg) are incorrectsince they do not demonstrate weight gain. Choice E (74 kg) is also incorrect,since the patient did not retain 600 mmol of NaCl after four days.

46. The correct answer is E. People with social phobia have a persistent (andadmittedly irrational) fear of being embarrassed or humiliated while in socialsituations. Many men report some degree of apprehension about urinating inpublic restrooms because the process is so visible. Most men are able to copewith this form of “performance anxiety”; however, the degree of apprehensionabout public scrutiny leads to the diagnosis of social phobia in this individual.

Agoraphobia is characterized by fears of being trapped in situations from whichthere is no easy escape if anxiety or panic develops. Examples include formal

dinners, attending events in auditoriums, waiting in long lines, and the like.Agoraphobia can occur with or without panic disorder (choice A), but thescenario described is typical for social phobia rather than agoraphobia.

In generalized anxiety disorder (choice B), the patient experiences excessiveanxiety or worry about events such as work or school. Apprehension aboutbeing embarrassed in public (social phobia) is specifically excluded from thisdiagnostic category.

In obsessive-compulsive disorder (choice C), patients experience recurrent,intrusive thoughts (obsessions) that are relieved, to some extent, by engaging in

repetitive, ritualistic behaviors (compulsions).

In the case of prostatism (choice D), urination is physically impaired due tobladder neck obstruction. This condition would not be expected to improve inthe absence of an audience.

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47. The correct answer is E. The insulin receptor has an extracellular hormone-binding domain and an intracellular cytoplasmic enzyme domain which acts as atyrosine kinase.

Adenylyl cyclase (choice A) is an enzyme that is regulated by receptors linked to

the G proteins, Gs (stimulates the enzyme) and Gi (inhibits the enzyme).Adenylyl cyclase converts ATP to cAMP.

Phosphodiesterase (choice B) is the enzyme that metabolizes cAMP.

Phospholipase C (choice C) is an enzyme that is regulated by receptors linked tothe Gq G protein. Phospholipase C converts phosphatidylinositol-4,5-bisphosphate (PIP2) to inositol-1,4,5-trisphosphate (IP3) and diacylglycerol

(DAG).

Phosphoprotein phosphatase (choice D), also called protein phosphatase, is ahydrolase that cleaves the phosphoryl groups from phosphoproteins. Theseenzymes are involved in digestion, and in the regulation of several enzymes thatundergo a phosphorylation-dephosphorylation cycle.

48. The correct answer is A. Amphetamines are indirect-actingsympathomimetics. These drugs enter the nerve terminals of postganglionicsympathetic neurons and displace norepinephrine, which exits the terminal via

the uptake carrier. Norepinephrine then stimulates !1-adrenergic receptors onblood vessels, leading to vasoconstriction and elevated blood pressures.

Amphetamines do inhibit monoamine oxidase (MAO) (choice B) to some extent,but releasing endogenous catecholamines is the primary mechanism by whichblood pressure is increased.

Amphetamines are not converted to false neurotransmitters (choice C).

Amphetamines are not direct agonists at the !1-adrenergic (choice D) or "1-

adrenergic (choice E) receptors.

49. The correct answer is B. The blood vessels that supply the skin are found inthe deep and superficial layers of the dermis. The vascular plexus in the deepdermis is horizontally oriented, and communicates vertically with the plexus inthe superficial layer of the dermis.

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USMLE Step I

The basal layer (choice A), the stratum corneum (choice C), and the stratumlucidum (choice D) are located in the epidermis, which does not have vessels.

The vessels of the subcutaneous fat (choice E) supply the fat, not the skin.

50. The correct answer is A. Markedly distended loops of small bowel andcolon in a neonate strongly suggests Hirschsprung’s disease (aganglionosis of therectum). Hirschsprung’s disease is a relatively common disorder characterizedby absence of ganglionic neurons of Meissner’s and Auerbach’s plexuses,classically involving the rectum. The muscle in this area is unable to relax, andpassage of bowel contents is impaired. The proximal bowel dilates, and there isabdominal distention, constipation, and vomiting. Microscopic examinationshows an absence of ganglion cells in the submucosa and muscularis.

Atrophy of colonic mucosa (choice B) would not cause bowel distension.

Hypertrophic pyloric stenosis (choice C) presents at a slightly older age (two tothree weeks) with projectile vomiting. Physical examination reveals a palpableovoid mass in the epigastrium; surgical muscle splitting is curative.

A Meckel’s diverticulum (choice D) is a small, usually asymptomaticoutpouching in the distal ileum near the ileocecal valve, representing persistenceof a portion of the vitelline duct. Meckel’s diverticula may contain ectopicpancreatic tissue or gastric mucosa (acid secretion from the latter may produce a

small peptic ulcer in adjacent intestinal mucosa).

Colonic polyps (choice E) are unusual in the neonate.

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Explanations to Block III of the2000 USMLE Step I Released Items

1. The correct answer is A. While the measles virus contains a variety ofproteins against which antibodies can be formed, the hemagglutinin proteins,

which are located on the viral surface, are most likely to be clinically protective.Measles, mumps, and parainfluenza viruses all have hemagglutinin spikes ontheir outer lipoprotein envelope.

Matrix antigens (choice B), nonstructural antigens (choice C), nucleocapsidantigens (choice D), and polymerase antigens (choice E) are not exposed, and theantibody would not be as effective.

2. The correct answer is A. Sickle cell anemia is a chronic hemolytic disease

caused by the substitution of valine for glutamic acid in the beta-chain ofhemoglobin to produce hemoglobin S. This substitution creates a “sticky” site onboth the oxygenated and deoxygenated forms of hemoglobin S that can thenbind a hydrophobic site on deoxyhemoglobin S. Sickling results frompolymerization of the deoxyhemoglobin S under conditions of low oxygentension into rigid fibers that can extend the length of the cell. Newbornshomozygous for sickle cell trait still have high levels of fetal hemoglobin (whichcontains a gamma-chain rather than a beta-chain, and hence does not sickle) thatare protective for the first 8–10 weeks of life.

Maternal IgG (choice B) can be transferred transplacentally prior to birth, butthese antibodies do not afford any protection against sickling.

Maternal erythrocytes (choice C) never enter the fetal circulation, and hence arenot present in the neonate.

Neonatal erythrocytes (choice D) are not exposed to a higher oxygenconcentration than adults.

The spleen of the neonate can filter out sickled cells (choice E), but this processwill lead in time to autoinfarction of the spleen.

3. The correct answer is B. The Fick principle allows calculation of cardiacoutput from the oxygen consumption divided by the arterial–venous O2

difference. In order to measure the arterial–venous O2 difference accurately,

mixed venous blood reflecting an average of the oxygen extraction from the

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USMLE Step I

various organs is needed. Ideally, this can be achieved by taking blood from thepulmonary artery just prior to its oxygenation in the lungs.

Jugular vein (choice A) blood reflects oxygen extraction by the brain, which is avery metabolically active organ.

Pulmonary venous (choice B) blood is highly oxygenated.

Saphenous vein (choice D) blood reflects oxygen extraction by the lowerextremities.

Superior vena cava (choice E) blood largely reflects oxygen extraction from theupper body.

4. The correct answer is C. Thiamine functions in oxidative decarboxylation ofalpha-ketoacids, is a cofactor for transketolase, and maintains the integrity ofnerve cells. The patient described suffers from Wernicke’s encephalopathy,which is characterized by the clinical triad of ataxia, eye movement problems(ophthalmoplegia and nystagmus), and severe cognitive deficits, and is causedby protracted thiamine deficiency. In this patient’s case, ingestion of a meal thatpresumably contained carbohydrate exhausted his limited stores of thiamine,and actually worsened his condition. If untreated, Korsakoff’s psychosis,characterized by severe anterograde and some retrograde amnesia, can result.

Folate deficiency (choice A) is characterized by megaloblastic anemia.

Vitamin A deficiency (choice B) causes epithelial metaplasia, and especiallyaffects the eyes, causing dryness of the conjunctiva, eventual keratomalacia, orerosion of the cornea, which may lead to blindness.

Pyridoxine (choice D) deficiency causes seborrheic dermatitis, cheilosis, glossitis,peripheral neuropathy, or even seizures.

Deficiency of cyanocobalamin (vitamin B12; choice E) causes megaloblasticanemia, peripheral neuropathy, spinal cord lesions, and decreased fertility.

5. The correct answer C. Gap junctions form hydrophilic channels between twocells, permitting the interchange of molecules with a molecular weight less than1,500. Substances like cAMP, cGMP, ions, and some hormones can pass betweencells and spread information. This allows cells in a tissue to integrate theirmetabolic activities by sharing signals and metabolites.

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The basal lamina (choice A) is the thin matrix of collagen and laminin uponwhich cells sit, tightly bound by specific receptor proteins.

Desmosomes (choice B) bind cells together to give tissues strength. There are

three types of desmosomes: belt desmosomes, spot desmosomes, andhemidesmosomes. Belt desmosomes form a belt of cell-to-cell adhesion justunder the tight junction. Spot desmosomes are button-like points of contactbetween cells. Hemidesmosomes anchor the plasma membrane to theextracellular matrix.

Glycosaminoglycans (choice D) are long repeating linear polymers of specificdisaccharides. The backbone of peptidoglycan is an example.

Tight junctions (choice E) seal adjacent cells to prevent the passage of fluids

between cell layers.

6. The correct answer is D. In an immediate hypersensitivity reaction (Type Ihypersensitivity reaction), exposure to an antigen causes elaboration of IgEmolecules, which bind to mast cells and basophils via interaction with Fcreceptors on the basophil and mast cell surfaces. Reexposure to the antigen, andbinding of the antigen to the cell-surface IgE molecules, triggers the release ofprimary (preformed) and secondary (synthesized) mediators. Primary mediatorsinclude histamine, eosinophil chemotactic factors, neutrophil chemotactic factors,

heparin, and neutral proteases. Secondary mediators include leukotrienes (e.g.,LTC4, LTD4, and LTE4 = slow-reacting substance of anaphylaxis), prostaglandin

D2, and platelet-activating factor. Generally, these substances increase vascular

permeability and function as chemotactic agents; a number also producebronchoconstriction. Immediate hypersensitivity reactions underlie allergicrhinitis, hay fever, hives, allergic gastroenteritis, or, especially if the antigen isinjected intravenously, systemic anaphylaxis.

Activated T-lymphocytes on smooth muscle cells (choice A) do not participate inthe immediate hypersensitivity response.

IgA (choices B and C) is not important in the immediate hypersensitivityreaction.

Eosinophils (choice E) contain enzymes to degrade histamine and heparin, thusmoderating an existing Type I reaction rather than initiating a new one.

Kaplan/NMSR



USMLE Step I

7. The correct answer is B. This is actually a straightforward question and it isnot even necessary to “work through” all of the drug choices to arrive at theanswer. Nadolol (choice C), pindolol (choice D), and propranolol (choice E) areall nonspecific beta blockers. These drugs can be eliminated immediatelybecause they are fairly indistinguishable from one another. Isoproterenol (choice

A) is a nonselective beta agonist, and would clearly cause bronchiolar dilation(decrease bronchiolar resistance) when administered alone. Because Drug X didnot decrease bronchiolar resistance when administered alone, isoproterenol canbe eliminated, leaving only metoprolol (choice B).

A more rigorous analysis of the results yields the same conclusion. Drug X alonedid not increase heart rate or decrease bronchiolar resistance. Metoprolol (abeta1 antagonist) would decrease heart rate by blocking beta1 receptors on the

heart, and would likely not decrease bronchiolar resistance because beta2, not

beta1, receptors are present in the airways. Epinephrine (nonselective alpha and

beta agonist) alone caused an increase in heart rate (beta1 receptors) and adecrease in bronchiolar resistance (beta2 receptors). Only a beta1 antagonist

could prevent epinephrine-induced tachycardia without greatly affecting thedecrease in bronchiolar resistance. Metoprolol did diminish epinephrine’s effecton bronchiolar resistance to some extent, but this is probably because metoprololis not entirely specific, and also weakly blocks beta2 receptors.

If drug X was isoproterenol (choice A), both a decrease in bronchiolar resistanceand a reflex increase in heart rate (isoproterenol lowers blood pressure) would bepredicted. Furthermore, the combination of epinephrine and isoproterenol

should have additive effects on both heart rate and bronchiolar resistance viabeta receptor stimulation.

If drug X was a nonspecific beta blocker (choices C, D, and E) it would preventthe epinephrine-induced bronchiolar dilatation by blocking beta-2 receptors.

8. The correct answer is E. A systolic murmur with a systolic thrill in anacyanotic infant is consistent with a ventricular septal defect (VSD). VSD is thesingle most common congenital heart malformation, accounting for

approximately 20% of all congenital heart malformations. Cyanosis occurs whenthere is a shunt of deoxygenated blood from the right heart to the left heart, butVSD initially produces a left-to-right shunt. VSD, atrial septal defect (ASD), anda patent ductus arteriosus (PDA) all are characterized by (acyanotic) left-to-rightshunts. However, all of these lesions may develop into cyanotic right-to left-shunts with the development of pulmonary hypertension (Eisenmengersyndrome).

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An ASD (choice A) presents with a widely split S2, a midsystolic pulmonary

ejection murmur, and a middiastolic rumbling murmur from increased flowacross the tricuspid valve. An apical thrill and a holosystolic murmur may alsobe present if there is associated mitral or tricuspid incompetence. ASD shouldnot be confused with a patent foramen ovale, which is of little or no

hemodynamic significance.

Coarctation (choice B), or narrowing of the aorta, causes a lag in or weakening ofthe femoral pulses, along with upper extremity hypertension. There may be amidsystolic or a continuous murmur. It is an acyanotic lesion without a shunt.

A patent ductus arteriosus (PDA; choice C) causes a continuous, rough"machinery" murmur, and has a characteristic thrill.

Tetralogy of Fallot (choice D) includes a VSD and right ventricular outflow

obstruction. The amount of cyanosis is directly dependent on the extent of rightventricular outflow obstruction. Most patients develop cyanosis by four monthsof age, and the cyanosis is usually progressive.

9. The correct answer is D. 5’TGATCA3’ base-pairs with 5’ACTAGT3’, which issimply the first sequence reversed. This sequence is a palindrome, because it“reads” the same way in both directions. From the diagram, one can appreciatean inverse, or dyadic symmetry of these two nucleotide sequences as they base-pair. For example, the trinucleotide 5’AGT3’ at the lower right of the figure is

identical with the mirror-image of the 3’TGA5’ trinucleotide above it and to theleft of it. Restriction endonucleases typically cut DNA at sites such as this,leaving either “blunt ends” or “sticky ends” (if the cut is staggered).

10. The correct answer is B. Estrogens serve to maintain the femalereproductive system and secondary sex characteristics as well as promote thegrowth and development of the vagina, uterus, and breasts. They also contributeto shaping the skeleton by conserving calcium and phosphorus in addition topromoting bone formation. Estrogens are indicated for treatment and preventionof osteoporosis in postmenopausal women, or in those withhysterectomy/oophorectomy.

Although oral calcium (choice A) is an essential part of a women’s diet, the useof estrogen would be a more appropriate agent to prevent bone loss in thispatient with a recent hysterectomy/oophorectomy.

Kaplan/NMSR



USMLE Step I

Prednisone (choice C) is a glucocorticoid indicated for the treatment of varioushormone deficiency states and a number of inflammatory conditions. This agentshould not be used in this patient since it promotes bone degradation andcalcium excretion.

Progesterone (choice D) is used for treating amenorrhea, abnormal uterinebleeding, and endometriosis as well as various other conditions. Progesterone isnot indicated for the treatment or prevention of osteoporosis, but is often givenalong with estrogen replacement to women with intact uteruses to help preventendometrial cancer from unopposed estrogen.

Vitamin D (choice E) is often combined with oral calcium preparations because itfacilitates calcium absorption from the intestinal tract. It would not be indicatedas the sole treatment modality in this patient.

11. The correct answer is A. Cessation of smoking is unquestionably the bestway to reduce this woman’s risk of dying from cancer. She has roughly threetimes the chance of developing breast cancer than lung cancer, yet because of itslethality, she is more likely to die from lung cancer than she is from any othertype of cancer.

At age 21, she is not yet old enough to require yearly mammography (choice B) to screen for breast cancer.

At age 21, she is not yet old enough to require annual tests for blood in stool(choice C) to screen for colon cancer.

Annual pap tests (choice D) are important, but the risk of dying from cervicalcarcinoma is lower than that of dying from lung cancer.

The nuclear power plant (choice E) will probably not affect her cancer risk at all.

Sunscreen (choice F) is a good idea to prevent melanoma and other types of skincancer, but melanoma, which affects about three percent of women, kills far lessoften than lung cancer.

12. The correct answer is C. Deferoxamine is the drug of choice for ironpoisoning. It effectively chelates ingested iron, but does not compete for iron inhemoglobin and cytochromes.

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Acetylcysteine (choice A) is the drug of choice for acetaminophen (Tylenol)poisoning. Because acetylcysteine provides reducing equivalents in the form ofsulfydryl compounds, it probably acts in part by replenishing the hepaticglutathione stores that are depleted by acetaminophen overdose.

Calcium disodium edetate (EDTA) (choice B) is a chelating agent used to treatlead poisoning.

Dimercaprol (choice D), also known as BAL, is a chelator used in arsenic, lead,and mercury poisonings.

Penicillamine (choice E) is used as a chelating agent in copper, mercury, and leadpoisoning.

13. The correct answer is A. Erythropoietin is a natural human hormoneproduced for therapeutic use using recombinant DNA technology. Its maineffect is stimulation of red blood cell production by the bone marrow.Erythropoietin is produced in the kidneys; chronic renal failure is oftenaccompanied by normocytic normochromic anemia, which can be treated witherythropoietin.

Ferrous sulfate (choice B) provides iron, an essential component of hemoglobin,myoglobin, and a number of enzymes in the body. This agent is indicated in thetreatment of iron deficiency anemia, a microcytic, hypochromic anemia.

Folic acid (choice C) is required for nucleoprotein synthesis and maintenance ofnormal erythropoiesis. Deficiency results in a macrocytic, rather than anormocytic, anemia.

Vitamin B6 (pyridoxine; choice D) acts as a coenzyme in the metabolism ofprotein, carbohydrates, and fatty acids. Deficiency of this vitamin is associatedwith the development of anemia in several hereditary disorders, but it is notinvolved in the anemia of chronic renal failure.

Vitamin B12 (cyanocobalamin) (choice E) is essential for cell growth, cell

reproduction, and hematopoiesis. Deficiency produces megaloblastic anemia, atype of macrocytic anemia.

14. The correct answer is B. Hydrochlorothiazide is a thiazide diuretic used forthe treatment of hypertension and edema associated with congestive heartfailure, and other conditions causing edema. One side effect of this agent is

Kaplan/NMSR



USMLE Step I

development of electrolyte abnormalities such as hypokalemia, hyponatremia,hypochloremia, hypomagnesemia, hypercalcemia, and hyperuricemia. Since thepatient in this question is hypokalemic, the most appropriate next step would beto add a “potassium-sparing” diuretic to this current regimen. Amiloride is anagent that inhibits sodium reabsorption induced by aldosterone in the distal

tubule and collecting duct, and inhibits the active excretion of potassium into theurine, thereby increasing potassium levels. Such agents are indicated for theadjunctive therapy of congestive heart failure and hypertension with thiazide orloop diuretics.

Acetazolamide (choice A) is a carbonic anhydrase inhibitor indicated for thetreatment of open-angle glaucoma, and as an adjunctive agent in the treatment ofedema; however, this agent does not increase potassium levels.

Furosemide (choice C) is a “loop” diuretic that is often used to control edema in

congestive heart failure, but since this agent also causes hypokalemia, it wouldnot be recommended in this patient.

Mannitol (choice D) is an osmotic diuretic indicated for treatment of the oliguricphase of acute renal failure and reduction of intracranial pressure and cerebraledema. It would not be recommended in this patient.

Metolazone (choice E) is thiazide diuretic with a pharmacological profile similarto that of hydrochlorothiazide; therefore, this agent should not be used in thispatient.

15. The correct answer is D. Alcohol increases the concentration of high-densitylipoprotein 1 (HDL1) by increasing the synthesis of apolipoprotein AI. Alcoholalso decreases the synthesis of LCAT in the liver, which reduces the conversionof HDL1 into HDL3, thereby increasing HDL1. In the blood, HDL1 picks upcholesterol and triglycerides from other lipoproteins, attaches apolipoproteins Eand CII to chylomicrons and VLDL, and extracts free cholesterol from foam cells(reverse cholesterol transport), essentially "defoaming the foam cells."

Alcohol does not have a major effect on LDL (choice A), which is the primary

vehicle for carrying cholesterol.

Alcohol increases, rather than decreases, the concentration of triglycerides(choice B) in the blood by increasing VLDL synthesis. The effect is indirect;increased formation of reduced nicotinamide adenine dinucleotide (NADH)favors the accumulation of glycerol-3-phosphate levels, which is used tosynthesize VLDL.

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Alcohol has no effect on the contractile proteins (choice C) in coronary arteries.

Thromboxane B2 (TXB2, choice E) is the inactive metabolic end-product ofthromboxane A2 (TXA2) in platelets.

16. The correct answer is C. Area C represents the location of the spinothalamictract (spinal lemniscus) which carries pain and temperature information from thecontralateral body. Pain and temperature information is initially carried byprimary sensory neurons whose cell bodies reside in the dorsal root ganglia.Fibers of these neurons enter the dorsal spinal cord and synapse in the dorsalhorn. The secondary neurons cross in the anterior white commissure, ascend asthe spinothalamic tract in the ventrolateral cord, ascend in the brainstem as thespinal lemniscus, and synapse in the VPL nucleus of the thalamus.

Choice A is the hypoglossal nucleus, from which the hypoglossal nerveemanates. The hypoglossal nerve provides motor innervation to the intrinsic andextrinsic muscles of the tongue; a lesion of the hypoglossal nucleus wouldproduce hemiparalysis of the ipsilateral tongue.

Choice B primarily depicts the descending tract of V (trigeminal nerve). Thistract is composed of axons of primary sensory neurons that carry pain andtemperature information from the ipsilateral face to synapse in the spinal nucleusof V (the structure just medial to B). A lesion in this area would cause ipsilateralloss of pain and temperature sensation of the face.

Choice D is a large lesion including the medullary pyramid, part of the mediallemniscus, and part of the inferior olivary nucleus. This lesion would produceleft-sided paralysis of the body (pyramid), some left-sided loss of proprioceptionand discriminative touch (medial lemniscus), and possibly cerebellar signs(inferior olivary nucleus).

Choice E is a lesion of the medial lemniscus. This would produce a left-sidedloss of proprioception and discriminative touch over the body.

17. The correct answer is C. In the polyadenylation process, an AAUAAAsequence near the 3’ end is recognized, the RNA is cleaved by an endonuclease,and then a poly-A polymerase adds 100 to 200 adenylate residues to the RNA.Failure to recognize this sequence (as well as a sequence downstream from it)would result in failure of polyadenylation.

Kaplan/NMSR



USMLE Step I

Capping (choice A) occurs almost immediately after synthesis of the first 30nucleotides or so. The triphosphate of GTP condenses with the available 5’diphosphate on the growing RNA chain to form a “cap”, recognized duringprotein synthesis, that also protects the RNA from degradation.

Hybridization (choice B) is the process by which two molecules of nucleic acidanneal to each other based on nucleotide base-pairing.

Splicing (choice D) is required in the excision of intervening sequences (introns)to produce the mRNA molecule from heterogeneous nuclear RNA (hnRNA).

Transport (choice E) of mRNA molecules to the cytoplasm occurs after splicing iscomplete, and involves recognition of the mRNA molecule by receptor proteinsin the nuclear pore complex.

18. The correct answer is C. Anaerobic organisms such as clostridia can causepelvic infection after a self-induced abortion. These microorganisms are bestdetected by taking samples of the discharge and directly inoculating intoanaerobic transport medium.

Cultured mammalian epithelial cells (choice A) are used for viral detection.

Swabs in transport medium at room temperature (choice B) are used fordetecting conventional aerobic bacteria.

Use of a calibrated inoculating loop (choice D) can help enumerate bacteria, butthis technique is usually used for urine culture, and is not specifically used foranaerobes.

19. The correct answer is B. Alveolar ventilation (V%

A) reflects the ability of the

lungs to blow off CO2, and depends on three variables: tidal volume (VT), dead

space (VDS), and the frequency of breathing (f). VT is the amount of air that fills

the alveoli and airways during quiet breathing (normally 500 mL). VDS is the

volume of the airways and lungs that does not participate in gas exchange.Anatomic dead space is the volume of the conducting airways from the nose ormouth all the way to the end of the terminal (nonrespiratory) bronchioles(normal anatomic VDS is 150 mL). Physiologic dead space is the volume of the

lungs that does not participate in gas exchange, which includes the anatomicdead space plus any alveoli that are not participating in gas exchange. The

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relationship between these factors is expressed as follows: V

%

A = (VT - VDS) x f.

Placing this formula into the alveolar ventilation equation, PACO2 = V CO2 /V

% %

A,

the relationship is as follows: PACO2 = V

%

CO2/ (VT - VDS) x f. Since PACO2

(alveolar PCO2) is equal to PaCO2 (arterial PCO2), whatever value is found for thePaCO2 can be substituted in the equation for the PACO2.

In this question, both patients X and Y have the same respiratory rate (f) and the

same VT, yet they have different PaCO2 values. Assuming that V%

CO2 (CO2

production) is a constant, the only variable that could change the PaCO2 in

patients X and Y is VDS. Patient X must have less VDS than patient Y.

Regarding the other choices, cardiac output (choice A), forced expiratory volume

in 1 sec (choice C), functional residual capacity (choice D, amount of airremaining in the lungs after a normal expiration), and lung compliance (choice E,distensibility of the lungs) are not directly related to any of the componentsinvolved in normal ventilation.

20. The correct answer is E. The patient is hypokalemic and alkalotic, yet is stillspilling abnormal amounts of potassium into the urine. A plausible explanationis that she is taking a diuretic drug. Athletes often use diuretics to reduceweight, and diuretic use can trigger a metabolic alkalosis with increased

bicarbonate levels, and significant potassium loss.

Aldosterone deficiency (choice A) would produce hyponatremia andhyperkalemia.

Hyperventilation (choice B) would produce a respiratory alkalosis withdecreased bicarbonate.

Diabetes mellitus with ketoacidosis (choice C) would cause a metabolic acidosis,characterized by a decreased serum bicarbonate.

Anabolic steroid use (choice D) would probably not produce any markedchanges in electrolytes.

21. The correct answer is C. Very often, the doctor–patient relationship can bestrengthened by being carefully attentive to the patient’s needs, fears, and beliefs.

Kaplan/NMSR



USMLE Step I

Offering the patient analgesia addresses her stated reason for not having themammograms.

Exaggerating the risk of breast cancer (choice A) is unethical and likely to beineffective.

There is no indication that the patient has unresolved grief issues regarding themother and grandmother. Insisting that she obtain counseling (choice B) wouldtherefore be unproductive, or counterproductive.

Showing the woman photographs of untreated breast cancer (choice D) willprobably shock her, but will leave a lasting negative impression regarding herphysician, and may damage the therapeutic relationship.

Telling the patient that the therapeutic relationship will be terminated unless she

complies with medical advice (choice E) is an extreme move that should not beundertaken as long as open lines of communication exist between the doctor andpatient.

22. The correct answer is E. The woman described likely suffers from primary(idiopathic) cardiomyopathy. In this disorder, the heart muscle growsprogressively weaker, dilates, and failure ensues. The patient develops fatigue,dyspnea on exertion, paroxysmal nocturnal dyspnea, and orthopnea with leftventricular failure. Pulmonary edema occurs due to failure of the left heart;

peripheral edema may occur if the right heart fails as well. Typically, thecoronary arteries are normal. Various arrhythmias may occur, and contribute tothe high death rate in this disorder.

The history was not significant for recent streptococcal infection, and nomurmurs are present, making acute rheumatic fever (choice A) unlikely.

Congenital fibroelastosis (choice B) is most common in infancy and earlychildhood (but is nonetheless rare).

Constrictive pericarditis (choice C) is not classically associated with pulmonary

edema, and the heart is frequently normal-sized, or slightly enlarged. Apericardial knock or pulsus paradoxus may be appreciated.

The diagnosis of myocardial infarction (choice D) is unlikely, based on thehistory and the normal coronary angiogram.

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23. The correct answer is C. The unfortunate individual described probablysuffered from a fat embolism. Fracture of a long bone, or rarely, severe softtissue injury, can release fat to the venous circulation. The fat will reach thelungs, where most of it will be trapped. However, some fat globules may be ableto squeeze into the systemic circulation, and can end up in the brain. After

several days, microscopic hemorrhages appear in the highly vascular whitematter. Respiratory difficulty, thrombocytopenia, petechiae, and obtundation orcoma are characteristic of the fat embolism syndrome (which occurs in only asmall minority of patients with demonstrable fat globules in the circulation aftera large fracture).

Adult respiratory distress syndrome (ARDS; choice A) refers to a constellation ofclinical symptoms and pathologic changes occurring in the lungs as a result ofacute alveolar injury due to any of a number of causes. ARDS can occur in thecontext of fat embolism (so this is not the world’s best question).

A contrecoup injury (choice B) refers to an injury to the side of the brain oppositethe side that was struck.

In septicemia (choice D), small cerebral vessels or capillaries are likely to beinfected, leading to development of cerebral abscesses.

A subdural hematoma (choice E) is a collection of blood below the dura mater. Itis typically associated with head trauma, and has been postulated to be the resultof tearing of bridging veins between the dural sinuses and large cerebral veins.

24. The correct answer is A. Cimetidine is the most likely cause of thetheophylline toxicity because it inhibits the cytochrome P450 oxidative drugmetabolizing system. Theophylline’s metabolism would be impaired, leading totoxic plasma levels.

Hydrochlorothiazide (choice B) and prazosin (choice C) should not alter levels oftheophylline.

Rifampin (choice D) would actually lower plasma levels of theophylline by

inducing the cytochrome P450, thus hastening theophylline’s metabolism.

25. The correct answer is D. The patient has hypertension and hypokalemia.This could be due either to primary hyperaldosteronism (Conn's syndrome) orsecondary hyperaldosteronism (e.g., diuretic use in a volume-depletedhypertensive patient, renovascular hypertension).

Kaplan/NMSR



USMLE Step I

Conn's syndrome is most often caused by an aldosterone-secreting adenoma inthe zona glomerulosa of the adrenal cortex. Excess aldosterone increases the rateof exchange of sodium for potassium in the distal and collecting tubules, leadingto progressive sodium retention (causing volume expansion) and significant

potassium loss in the urine, resulting in hypokalemia. The increase in plasmavolume increases the cardiac output, which in turn, increases renal blood flow.An increase in renal blood flow inhibits the release of renin from the juxtaglomerular apparatus.

Secondary aldosteronism can be caused by primary renin overproduction or byrenin overproduction due to decreased renal blood flow. Therefore, of all thetests listed in the question, the measurement of plasma renin activity (PRA;choice D) would best separate primary aldosteronism excess (PRA would besuppressed) from secondary aldosteronism (PRA would be increased).

Plasma ACTH (choice A) is a useful test for the workup of Cushing's syndrome(which could cause hypertension due to oversecretion of weakmineralocorticoids).

Plasma cortisol (choice B), though useful in the workup of Cushing's syndrome,is not useful for differentiating primary from secondary hyperaldosteronism.

Plasma prolactin (choice C) has no role whatsoever in the production ofhypertension and hypokalemia.

Urinary sodium (choice E) is rarely used as an initial screen for any cause ofhypertension, and is most often used to distinguish renal from nonrenal causesof hypo- or hypernatremia.

26. The correct answer is C. This girl has not had a tetanus booster for over tenyears. Therefore, she needs not only a toxoid booster shot, but also antitoxin toblock any toxin produced by the bacteria. Had she received immunization(toxoid) during the past ten years, a simple booster shot of toxoid would havebeen sufficient.

27. The correct answer is C. Spectrin is a red cell structural protein that helps toform a lattice on the cytoplasmic side of the cell membrane, contributing to thepliability and deformability of the red cell. In hereditary spherocytosis, anautosomal dominant defect in the spectrin molecule results in spherical red cellsthat are vulnerable to destruction by the spleen (extravascular hemolysis)

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because the membrane is less pliable. The peripheral blood smear showsspherical or ovoid red cells lacking central pallor with superimposedreticulocytosis. The spleen may be moderately enlarged.

Integrins (choice A) are a large class of receptor proteins that bind the cell to the

extracellular matrix and mediate transmembrane signaling.

Myosin (choice B) is the main constituent of thick fibers in muscle, mediatingcontraction via its ATPase activity.

Tubulin (choice D) is a constituent protein of microtubules.

28. The correct answer is A. Anemia and granulocytopenia are the mostsignificant adverse effects encountered with the nucleoside polymerase

inhibitors, such as zidovudine (AZT). Anemia is also caused by the proteaseinhibitors. The associated anemia typically occurs four to six weeks after theonset of therapy, and is caused by a decreased formation of erythrocytes by themarrow. Erythropoietin is the agent of choice to treat anemia in patients takingnucleoside polymerase inhibitors.

This type of anemia is not related to increased formation of erythrocyteantibodies (choice C) or increased fragility of erythrocytes (choice D).

Since there is no indication that the anemia is macrocytic or microcytic in nature,

it is unlikely that the anemia is caused by folate (choice B) or iron (choice E)deficiency respectively.

29. The correct answer is D. Chronic ethanol use produces increased NADH +

H+, with resulting proportional loss of NAD+. This causes a shift in the

oxidation–reduction potential of the cytoplasm. Less NAD+ is available toabsorb hydrogens from lactate, so less pyruvate is formed and available forgluconeogenesis.

The steps described in choices A, B, C, and E do not require NAD+, and are notlimited by alcohol use.

30. The correct answer is B. Hydatidiform mole is a form of gestationaltrophoblastic disease (GTD). In GTD, there is proliferation of trophoblastictissue; the proliferation may be benign, invasive, or frankly malignant.Hydatidiform mole is a benign proliferation of trophoblastic tissue with cystic

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USMLE Step I

(hydropic) swelling of chorionic villi. Therapy consists of removing the mole bycurettage or hysterectomy. hCG levels should be monitored in this disorder;persistent or rising levels of beta-hCG indicate molar remnants, or thedevelopment of an invasive mole or choriocarcinoma.

An adrenal adenoma (choice A) would not be expected to produce hCG.

Hydatidiform mole is essentially a failed pregnancy; a second, ectopic pregnancy(choice C) would be unlikely.

Pituitary insufficiency (choice D) would not lead to hCG production becausehCG is made by trophoblast cells.

A second, noninvasive mole (choice E), would denote a second molar pregnancy,which would be unlikely compared to choriocarcinoma, which develops in about

4% of molar pregnancies. An invasive mole, or remnants of the first mole, arepossible causes of the rising hCG levels, but they are not included in thealternatives.

31. The correct answer is D. The patient described cannot adduct his right eyeduring conjugate gaze, but is able to adduct it during visual convergence. Thispatient has internuclear ophthalmoplegia (INO), which is caused by a lesion ofthe medial longitudinal fasciculus (MLF). The MLF connects the three brain stemnuclei that supply the extraocular muscles (oculomotor, trochlear, and abducens

nuclei), enabling the nuclei to coordinate together for conjugate eye movements.Therefore, lesions of the MLF disrupt conjugate eye movements. The only otherplausible answer might have been a lesion of the medial rectus muscle (choice

B), the muscle that adducts the eye. However, this clearly cannot be the answerbecause the patient was able to adduct his eye during visual convergence. Infact, INO can look very similar to a medial rectus lesion or to an oculomotornerve or nucleus lesion, except for the fact that patients can converge.

The lateral rectus muscle, which is supplied by CN VI, abducts the eye.

A lesion of the medial rectus muscle (choice B) would result in the inability of

the patient to adduct his eye during conjugate gaze and visual convergence.

The medial forebrain bundle (choice C) is a tract that originates in the olfactoryregion and septal nuclei, travels through the lateral hypothalamus, and continuescaudally into the midbrain tegmentum. It is not involved in eye movementcontrols.

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The oculomotor nucleus (choice E) supplies the medial rectus muscle. A lesionin this nucleus would result in the patient's inability to adduct his eye duringconjugate gaze and visual convergence.

32. The correct answer is D. Pyogenic (pus-forming) infections are almostalways caused by bacteria, (especially Staphylococcus or Haemophilus);recurrent bacterial infections with normal antibody production suggests aneutrophil defect. The history given is typical for children with congenitaldefects in neutrophil function. Normal cellular immunity (e.g., T cells,macrophages) is evidenced by successful immunizations and uneventfulrecovery from viral infections.

If the patient’s B lymphocytes (choice A) were defective, the child would nothave normal antibody responses to immunizations.

Defective eosinophil production (choice B) does not seem to produce muchclinical disease.

Defective macrophage function (choice C) is a possibility, but the specificvulnerability to pyogenic infections more strongly suggests neutrophildysfunction.

Defective T lymphocytes (choice E) predisposes for problems with diseases thatnormally cause granuloma formation.

33. The correct answer is D. The normal human karyotype consists of 46chromosomes. A person who has Down syndrome, or trisomy 21, and only 46chromosomes, has an extra chromosome worth of material physicallytranslocated onto another chromosome. Most Down cases are due tonondisjunction events during oogenesis in females over 40 years old, however,4% of Down cases are due to a Robertsonian translocation between theacrocentric 21 chromosome and another acrocentric chromosome (usually the14th). During meiosis, the normal 14 and 21 both pair with the translocatedchromosome, and the translocation is randomly passed into a daughter cell at the

end of meiosis I. This leads to three possible segregation events, which give riseto three surviving gametes: a gamete with the normal 21 and 14, a gamete withthe translocated chromosome, and a gamete with the translocated chromosomeand the normal 21. Phenotypically normal individuals who inherit thetranslocated chromosome will only have 45 chromosomes. A gamete with thetranslocated chromosome and the normal 21 gives rise to a Down child when itcombines with the other parent’s gamete. In reality, such an individual has three

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USMLE Step I

copies of each of the genes on the 21 chromosome, but because of the methodused to count chromosomes (counting centromeres in a metaphase spread), it isreported as a normal chromosome number. Evidence of a Robertsoniantranslocation can often be detected through pedigree analysis. Spontaneousabortions and a high frequency of Down progeny relative to maternal age are

hallmarks.

A deletion (choice A) could not account for the observed (partial) trisomy.

Mosaicism (choice B) is present in about 1% of Down patients. It is generally theresult of mitotic nondisjunction during early embryogenesis, and would yield amixture of cells with 46 and 47 chromosomes.

A somatic mutation (choice C), if it occurred early enough in development, couldproduce a mosaic, but not a Down’s child with 46 chromosomes.

Undetected trisomy (choice E) is unlikely given the accuracy of the karyotypingprocedure.

34. The correct answer is A. The dermis is attached to the underlying tissues viathe suspensory ligaments (retinacula cutis). When the dermis swellsconsiderably in association with lymphedema caused by blockage of thelymphatics, the sites of attachment become visibly dimpled. In the entity knownas inflammatory breast carcinoma, widespread lymphatic involvement with

corresponding lymphedema is especially likely to produce the dermatologicchanges described in the question stem.

Scarring of subcutaneous tissue (choice B) causes a broad retraction of the skin.

Focal invasion of the dermis by neoplastic cells (choice C) contributes tolymphatic blocking with consequent swelling, but does not directly cause the“orange-peel” appearance.

The dimples are not openings of sebaceous (choice D) or sweat glands (choice

E).

35. The correct answer is D. This woman is dehydrated from vomiting anddiarrhea, and should be trying to retain sodium and water, and excrete urine.The fact that her serum sodium is so low suggests that the drive to retain water isgreater than the drive to retain salt. Of all the answer choices, only increasedserum ADH is plausible under these circumstances. ADH conserves water by

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increasing the permeability of the collecting ducts to water, resulting in waterreabsorption.

The patient will be trying to save sodium, so aldosterone will be increased, notdecreased (choice A).

The patient will be trying to save sodium, so atrial natriuretic peptide willdecrease, not increase (choice B).

The circulating volume is decreased, not increased (choice C).

Both ADH and aldosterone should be high, but the drive to retain water isstronger than the drive to retain sodium, so urine osmolality is greater than, notless than, serum osmolality (choice E).

36. The correct answer is C. Stage is the most important determinant of poorprognosis in adenocarcinoma of the colon. Specifically, extension to the serosa(the outermost layer) is a poor prognostic sign, as the likelihood of metastasisincreases dramatically.

Circumferential growth (choice A) is not necessarily a poor prognostic sign.

A tumor that extends only to the muscularis mucosa (choice B) is less likely tometastasize.

Polypoid growth (choice D) and surface ulceration (choice E) are less significantthan penetration to the serosa.

37. The correct answer is A. In periods of severe stress, such as this case ofburns, cortisol is secreted in large amounts. Cortisol has many metabolic actions,including increasing protein turnover (for synthesizing new proteins). A sideeffect of this is an increase in nitrogen loss.

Erythropoietin (choice B) stimulates erythrocyte production from the marrow.

Insulin (choice C) acts principally on glucose and fat metabolism.

Parathyroid hormone (choice D) regulates calcium and phosphate metabolism.

Thyroxine (T4) (choice E) increases metabolic rate, but would not specifically be

expected to increase nitrogen loss.

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USMLE Step I

38. The correct answer is B. The photograph shows a segment of bowel“telescoped” into the adjacent bowel. This is intussusception, which can causeabdominal pain and vomiting. The involved bowel may become infarcted and

necrotic (a surgical emergency).

In appendicitis (choice A), the appendix would be inflamed, and probablycovered with pus. Involvement of the surrounding bowel would be unusual.Anorexia is common, and tenderness is invariably present eventually.

Meckel’s diverticulum (choice C) causes a fat outpouching of bowel coming offthe small intestine.

Necrotizing enterocolitis (choice D) is usually a disease of neonates.

In strangulated hernia (choice E), the appearance would be similar to that in thephotograph, but you would not see one loop of bowel telescoped into another.

39. The correct answer is C. This woman is exhibiting regression, an automaticretreat to a less mature level of behavior in times of stress. Regression is an egodefense mechanism in which there is a return to an earlier (often infantile) stageof development. It occurs in many mental illnesses and in normal individualsexperiencing tragic or extremely stressful events.

Denial (choice A) is an extremely common defense mechanism, especially forindividuals who receive devastating medical news. Elizabeth Kubler-Rossconsidered it the first step toward eventual acceptance of one’s own mortality.

Displacement (choice B) involves transferring feelings to an inappropriateperson, situation, or object (e.g., a man who has been yelled at by his boss takesout his anger on his wife).

Repression (choice D) occurs when conflict-provoking thoughts or feelings arehidden from the person’s awareness. Forgetting an emotionally charged event is

an example of repression.

Sublimation (choice E) is a mature defense mechanism that involves consciouslyturning socially unacceptable impulses into acceptable or more benign forms inorder to allow their expression. For example, a young college girl immersesherself in athletics rather than engage in premarital sex.

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40. The correct answer is C. Separation anxiety is a normal feature of childhoodemotional development, occurring between the ages of 8 and 18 months, whenthe infant is separated from the mother. Between the ages of 2 months and 2years, children might show preference for a comforting “transitional object” (e.g.,

teddy bear), which is usually discarded by age 4, when the transition fromdependence on the mother to independence is more complete. Separationanxiety disorder is characterized by excess anxiety for at least two weeksfollowing separation from persons to whom the child is attached. At separation,anxiety may be experienced to the point of panic. Physical complaints such asstomach aches and headaches are common. These children often have problemsfalling asleep, and frequently experience nightmares. Often, they will come intothe parents’ room and try to sleep with them.

In childhood schizophrenia (choice A), schizophrenic symptoms occur before

puberty. Schizophrenia may present with grossly immature behavior, failure todevelop a separate identity from the mother, withdrawal, and may includeautistic behaviors.

According to Freud, the latency phase (choice B) occurs between 5 and 12 yearsof age, and is characterized by a quiescent sexual drive because of the resolutionof the Oedipal complex. Sexual energy is sublimated into socially acceptableoptions such as schoolwork and sports. During this phase, the superego,responsible for moral and ethical development, is formed.

In socialized conduct disorder (choice D), conduct problems manifest asantisocial activity (stealing, fighting, cruelty to animals, etc.) with peers. Thesechildren are very loyal to their group, and have an ethical code, although it maydiffer from the rest of society. They are likely to be diagnosed with antisocialpersonality disorder in adulthood.

Symbiotic psychosis (choice E) is a developmental disorder of early childhood. Itis characterized by profound reaction to separation, but also by severedevelopmental and social retardation.

41. The correct answer is E. Gestational onset of hypertension (usually afterweek 20), proteinuria, and edema are classic symptoms of preeclampsia, amultisystem disorder. Preeclampsia is associated with arteriolar spasm, possiblydue to an unusually high ratio of thromboxane to prostacyclin. Renal vasospasmresults in decreased glomerular filtration with increased blood urea nitrogen(BUN), increased creatinine, and proteinuria. Uterine vasospasm can causeplacental ischemia and poor fetal growth; cerebral vasospasm can cause

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USMLE Step I

headaches. High levels of thromboxane can also cause aggregation of platelets,contributing to thrombosis and thrombocytopenia. Evidence of intravascularhemolysis and liver damage indicate that the condition is severe.

Acute glomerulonephritis (choice A) is unlikely since there was no hematuria,

and red cell casts were absent.

The peripheral edema observed would require severe right-sided as well as left-sided heart failure, which is unlikely. Congestive heart failure (choice B) in thecontext of peripartum cardiomyopathy is typically left-sided, and does notinvolve hypertension.

Seizures indicate eclampsia (choice C), a severe complication of preeclampsia.

Nephrotic syndrome (choice D) is characterized by proteinuria,

hypoalbuminemia, and edema, but is not associated with hypertension.

42. The correct answer A. Negatively birefringent crystals strongly suggest thatthe patient has gout. Gouty arthritis is characterized by deposition of sodiumurate (the sodium salt of uric acid) crystals in joints. The crystals activateHageman factor and lead to the production of chemoattractants via thecomplement pathway. The neutrophils respond, and phagocytose the crystals,releasing lysosomal enzymes and inflammatory mediators that produce an acutesynovitis. The knee is the second most common joint (after the big toe) to be

affected by gout.

In osteoarthritis (choice B), there would not be crystals in neutrophils in the jointfluid.

Pseudogout (choice C) would be characterized by positively birefringent crystalsof calcium pyrophosphate.

Rheumatoid arthritis (choice D) would not be characterized by crystals, butrather by a sterile, chronic synovitis.

In septic arthritis (choice E) there would be bacteria, but no crystals in the jointfluid.

43. The correct answer is E. Sensitivity is defined as the ability of a test to detectthe presence of a disease in those who truly have the disease. It is calculated as

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the number of people with a disease who test positive (true positive) divided bythe total number of people who have the disease (true positive + false negative).

Disease:Test Result: Present Absent Total Patients

Positive 60 (true +) 40 (false +) 100 (those with+)Negative 20 (false -) 80 (true -) 100 (those with -)

Sensitivity =60(true +)

60(true +) + 20(false -) =6080 = 0.75

The positive predictive value is the probability that someone with a positive test

actually has the illness. Positive predictive value =60 (true +)

60 (true +) + 40(false +) =

60

100 = 0. 60 (choice C)

The specificity of a test is how well it identifies people that are truly well.

Specificity =80 (true -)

80 (true -) + 40(false +) =80120 = 0. 67 (choice D).

The negative predictive value is the probability that someone with a negative test

is actually well. Negative predictive value =80 (true -)

80 (true -) + 20(false -) =80100 =

0. 80

44. The correct answer is C. The phrenic nerve arises from C3, C4, and C5, thespinal segments that also supply sensation to the shoulder region. Thus, paindue to irritation of diaphragmatic pleura (e.g., from a biopsy of the liver below)or peritoneum (central regions) is referred to the shoulder region. Sensationfrom peripheral regions of the diaphragm are conveyed via inferior intercostalnerves.

The axillary nerve (choice A) derives from C5, C6 (the shoulder region is

supplied by C3, C4), and supplies the upper, lateral part of arm.

The intercostobrachial nerve (choice B) arises from TI, T2, and supplies skin ofthe floor of the axilla and adjacent areas of the arm.

The sympathetic chain (choice D) carries pre- and postganglionic sympatheticnerve fibers that are visceral motor, and not sensory.

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USMLE Step I

The vagus nerve (choice E) receives visceral afferent input from thoracic andabdominal viscera which may be referred to overlying somatic sites. Since thevagus originates from the brainstem, this cannot cause pain to be referred toareas that derive their sensory innervation from the spinal cord.

45. The correct answer is E. Absence of gastric acid secretion when chewingand of receptive relaxation of the proximal stomach when swallowing are bothrelated to an interruption of vagal innervation of the stomach. In the stomach,the vagus nerve directly stimulates: 1) parietal cells in the body and fundus tosecrete hydrogen ions, 2) G cells in the antrum to secrete gastrin, which entersthe blood and stimulates parietal cells to secrete hydrogen ions, 3) pepsinogen-producing chief cells, and 4) mucous-secreting cells, which are involved in themucous barrier that protects the gastric mucosa. The vagus nerve is also

involved in receptive relaxation, whereby distention of the lower esophagus byfood produces relaxation of the lower esophageal sphincter (LES) and,simultaneously, relaxation of the proximal portion of the stomach. Receptiverelaxation is a vagovagal reflex involving both the afferent and efferent limbs ofthe vagus nerve; interruption of the vagus nerve abolishes this reflex.

Absence of G cells (choice A) would only affect gastric secretion of hydrogenions, and not receptive relaxation in the proximal stomach.

Absence of chief cells (choice B) reduces production of pepsinogen, which is

activated to pepsin, an enzyme that begins the digestion of protein in thestomach. Pepsin has no effect on gastric secretion of acid or motility.

Secretin (choice C) is a hormone secreted by S cells in the duodenum. In thestomach, its main function is to decrease gastric hydrogen ion secretion byinhibiting the effect of gastrin on the parietal cells. Secretin does not alter gastricmotility.

Somatostatin (choice D) is secreted by endocrine cells of the GI mucosa and deltacells of the endocrine pancreas. Its function is to inhibit secretion of other GIhormones (e.g., gastrin, cholecystokinin, gastric inhibitory peptide, secretin). By

inhibiting gastrin, somatostatin reduces gastric acidity, but it has no direct effecton gastric motility or receptive relaxation of the proximal stomach.

46. The correct answer is B. First notice that point Y represents a patient with ahigher-than-normal PCO2, but a nearly normal pH. These data indicate that the

situation is chronic, because the pH has almost completely compensated. The

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only two conceivable ways to arrive at these values are: respiratory acidosis withrenal compensation, or metabolic alkalosis with respiratory compensation.Because the pH is slightly below 7.4, this points to the former option, becausecompensation brings the pH toward normal, but does not shoot past it. Inchronic obstructive pulmonary disease, respiratory acidosis can result from an

inability to blow off enough CO2; if this persists, the kidneys will compensate by

conserving HCO3- and secreting H+, thus correcting the body’s pH.

Because of the low oxygen availability at high altitudes (choice A), a personwould hyperventilate to try to inspire adequate oxygen. This would lower thebody’s PCO2, resulting in respiratory alkalosis. Renal compensation could occur;

however, the PCO2 would remain below 40 mm Hg. Also, respiratory alkalosis

with renal compensation would result in a low serum bicarbonate.

Both diarrhea (choice C) and ingestion of a strong acid (choice D) could producea metabolic acidosis, but in order to compensate for this, one wouldhyperventilate, which would lower the PCO2 to below 40 mm Hg. In addition,

metabolic acidosis with respiratory compensation results in a low serumbicarbonate.

Severe and prolonged vomiting (choice E) would result in a metabolic alkalosisbecause of the loss of acids. In theory, the patient could compensate byhypoventilating, thus raising PCO2; however, in reality, respiratory compensation

of metabolic alkalosis is often small or absent.

47. The correct answer is A. The gluteus maximus is the principal extensor ofthe thigh, and acts to steady it. When the thigh is fixed, it is also a strongextensor of the pelvis, used, for example, when rising from a sitting position.The muscle is relaxed when standing still, and is used little during ordinarywalking.

Gluteus minimus (choice B) is a powerful abductor of the hip joint, and is largelyresponsible for the tilt of the pelvis during walking. Thus, paralysis would havea profound effect on walking.

Although the hamstrings (choice C) cross the hip joint, and can act in trunkextension, their main action is on the leg. The muscles cross the knee joint andact as a strong flexor of the leg. Thus, if this muscle were damaged, leg flexionwould be affected.

Kaplan/NMSR



USMLE Step I

Iliopsoas (choice D) is a strong flexor of the trunk, so rising from a sittingposition, which requires trunk extension, would not be affected. In addition,iliopsoas is a muscle of the front of the thigh innervated by the Ll and L2 ventralrami, and would be unlikely to be affected by injury to the gluteal region.

Obturator internus (choice E) laterally rotates the extended thigh and abducts theflexed thigh. In addition, it stabilizes the femoral head in the acetabulum, actionsthat play a role in balance during standing and walking.

48. The correct answer is C. This child has Gaucher’s disease, caused by adefect in the body’s ability to degrade glucocerebrosides within lysosomes.Hepatomegaly and mental retardation are characteristic of type II, or the acuteneuronopathic form of Gaucher’s disease. Lysosomal hydrolase is the bestanswer of those listed, and is a generic term for a lysosomal enzyme that

hydrolyzes a chemical. The specific enzyme deficient in Gaucher’s disease isglucosylceramide beta-D-glucosidase.

Lipases (choices A, B, and E) are not involved in the critical parts of thedegradation of glucocerebrosides.

This problem is not a disorder of synthesis, so sphingolipid synthase (choice D) is wrong.

49. The correct answer is E. Of the features listed, the strongest criterion is thefinding of tumor cells in the serosal layer. This finding indicates that either thetumor has invaded through the wall to reach the serosa, or it has metastasized tothat location.

Foci of necrosis and acute inflammation (choice A) occasionally occur in benignlesions, and are sometimes related to local infarction.

Formation of new vascular channels and hemorrhage (choice B) can occur ininflammatory conditions.

Hemosiderin-filled macrophages (choice C) simply indicate prior hemorrhage,and are not specific for malignancy.

Mucin (choice D) can be produced by benign lesions as well as malignant ones.

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50. The correct answer is A. During pregnancy, estrogen increases prolactinsecretion, helping to stimulate breast development, but high levels ofprogesterone prevent prolactin from stimulating actual milk production byalveolar cells. The stimulus for milk production is the high prolactin, coupledwith rapidly reduced estrogen and progesterone immediately after delivery.

These reflexes also stimulate the secretion of oxytocin, which is required for"letdown" or filling of the nipple with milk, but not milk production.

Growth hormone (choice B) secretion is inhibited during pregnancy, but humanchorionic somatomammotropin becomes the “growth hormone” of pregnancy(see also explanation to choice C).

Human chorionic somatomammotropin (choice C), also called human placentallactogen (hPL), rises steadily during pregnancy, and acts in concert withprolactin on the breasts, but also causes insulin resistance in the mother (similar

to growth hormone), thereby providing glucose to the fetus.

Insulin, cortisol, thyroxine, and growth hormone act with estrogen, progesterone,and prolactin to cause breast development at puberty (choices D and E).Although the driving force at that time is estrogen, mild deficiencies in the otherhormones can be compensated for by an excess of prolactin.

Prolactin is cleared more slowly than the steroids, and its level is maintained byneural reflexes stimulated by suckling. Suckling can reduce dopamine, aninhibitor of prolactin secretion (choice E), thereby increasing prolactin levels.

2003 official step 1 explanations

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