USMLE-STEP-1 Online Practice Questions and Answers

Correct Answer: E

Section: Anatomy In TGA, the aorta arises from the right ventricle and the pulmonary trunk arises from the left ventricle (choice C). This is the reverse of the normal situation and gives rise to the cyanotic condition in the newborn. The large arteries arise from the truncus arteriosus in the developing heart and thus could not develop from the atria (choices B and D), which are formed from the sinus venosus. The truncus arteriosus and the sinus venosus are at opposite ends of the heart. The ductus arteriosus (choice A) is the vessel that shunts oxygenated blood from the pulmonary trunk to the arch of the aorta in the fetus. It does not give rise to the aorta.

Correct Answer: B

Section: Anatomy Arrow 2 points to the base of the gastric glands where chief or peptic (zymogenic) cells tend to be clustered. Arrow 1 points to the luminal surface of the stomach where mucus-secreting cells are found. Arrow 3 points to the muscularis mucosae. Arrow 4 points to the middle of the gastric glands where parietal or oxyntic cells tend to be most numerous. Arrow 5 point to the side of a gastric pit where mucus-secreting cells are also found.

Correct Answer: D

Section: Physiology The primary change in the cochlea due to an increase in the frequency of a sound wave is a change in the position of maximal displacement of the basilar membrane. A low-pitched sound produces the greatest displacement toward the apex of the cochlea, which results in activation of hair cells at that location. As pitch increases, the position of greatest displacement moves closer to the base of the cochlea. Increases in the number of hair cells that are activated (choice A), increased amplitude of basilar membrane displacement (choice B), increases in the frequency of discharge of units in the auditory nerve fibers (choice C), and an increase in range of frequencies to which such units respond (choice E), are all more likely to be observed in response to increases in the intensity of the sound stimulus (sound pressure) rather than to increases in pitch.

Correct Answer: C

Section: Physiology In clinical trials, a class of drugs, the aminopyridines, that blocks certain potassium channels has shown some promise for symptomatic relief of the symptoms of multiple sclerosis. The neuronal action potential is terminated by sodium channel inactivation and potassium channel activation. A drug that blocks potassium channels would thus prolong the action potential. Activating potassium channels (choice A) or increasing sodium channel inactivation (choice E) would shorten the action potential. Blocking sodium channels (choice D) would either shorten the action potential or block it altogether. Since calcium channels are not involved in the neuronal action potential, calcium channel blockade would not be expected to have much effect (choice B).

Correct Answer: C

Section: Microbiology/Immunology Point C represents the zone of equivalence, that is, that point of the antigen-antibody reaction when optimal concentrations of antigen and antibody combine. Thus, there is little, if any, free antigen and antibody, and as a consequence maximal amounts of antigen and antibody will be found in the precipitate resulting from the antigen-antibody reaction. Points A and B represent areas in the antigen- antibody precipitin curve where there is an excess of antibody (choices A and B). Points D and E show areas in the precipitin curve where there is an antigen excess in the antigen-antibody reaction mixture (choices D and E).

Correct Answer: B

Section: Microbiology/Immunology

A. israelii is an opportunistic microorganism that does not produce any known virulence factor. It seems that its pathogenesis is related to the ability of this organism to survive the host's inflammatory responses. Devitalized tissues, presence of foreign bodies, broken mucous membranes, and immunosuppression promote development of actinomycosis. A. israelii is an anaerobic organism. Thus high levels of oxygen will kill this microbe (choice A). This bacterium does not produce leukocidin (choice D), exotoxin (choice C), or pili (choice E).

Correct Answer: E

Section: Microbiology/Immunology Streptococcal infections in chronic granulomatous disease are rare, because the defective myeloperoxidase-hydrogen peroxide-halide system can use hydrogen peroxide accumulated by phagocytized streptococci, which are catalase negative and cannot convert hydrogen peroxide to water and oxygen. Administration of pooled gamma globulin (choice C) or antistreptolysin (choice A) is likely to provide only brief, but not chronic, protection against streptococci. C-reactive protein (CRP) is a beta serum globulin which reacts with the C carbohydrate of the cell wall of S. pneumoniae. It is not an antibody and thus will not provide even brief protection against streptococci (choice B). Phagocytosis plays a major, not a minor, role in the killing of streptococci (choice D).

Correct Answer: A

Section: Pathology and Path physiology Pemphigus vulgaris is an autoimmune disorder. The autoantibodies are directed against keratinocyte antigens with subsequent dyshesion and fluid-filled blister formation. Bacterial infections (choice B), dietary deficiencies (choice C), chemical toxins (choice D), and local ischemia (choice E) are not thought to be the causative agent of pemphigus vulgaris.

Correct Answer: C

Correct Answer: B

Section: Behavioral Science and Biostatics Although all the other items (choices A, C, D, and E) may be significant, schizophrenia cannot be diagnosed if the patient has never had an episode of psychotic symptoms lasting for at least 6 months.