USMLE USMLE-STEP-1 Online Practice Questions and Exam Preparation

USMLE USMLE-STEP-1 Online Questions & Answers

• Question 171:

  A 56-year-old woman returns to clinic for a follow-up after undergoing a lung biopsy of a 1.4 ?2.1 cm, left upper lobe lesion originally discovered on chest x-ray. Her physician checks her record for the pathology report and reads that histology of the biopsy demonstrated "scattered spindle cells with a minimal mononuclear inflammatory infiltrate and few blood vessels in a background of abundant pale connective tissue. Small areas of calcification are observed and there is no evidence of normal lung parenchyma." Which of the following is most likely responsible for these findings?

  A. deposition of amyloid protein
  B. hamartoma formation
  C. healing of an abscess
  D. malignant transformation of type 1 pneumocytes
  E. unresolved infection with Mycobacterium tuberculosis
  C. healing of an abscess

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  Explanation

  Section: Pathology and Path physiology The first sentence of the pathology report describes the histological morphology of granulation tissue, representing evidence of healing within the biopsied lesion. The absence of normal lung parenchyma indicates that injury to the lung was extensive enough to prohibit regeneration, and this would be expected due to the liquefactive necrosis occurring with abscess formation. Areas of calcification can develop in any necrotic tissue and represent a nonspecific finding. Deposition of amyloid protein (choice A), or amyloidosis, is associated with particular chronic diseases (AA protein) or certain immunoglobulin- secreting neoplasms (AL protein). It appears histologically as extracellular, diffusely distributed, acellular material and is best demonstrated using Congo-red stain. The pathology report was not consistent with the morphology of a hamartoma (choice B), which refers to a disorganized mass of mature tissue indigenous to the organ in which it arises. There was also no cytological description suggestive of malignancy (choice D, malignant transformation of type 1 pneumocytes), such as pleomorphism or hyperchromatism. Granulation tissue could potentially be seen in an unresolved infection with M. tuberculosis (choice E); however, evidence of granuloma formation is necessary to establish the diagnosis.

• Question 172:

  Below figure shows an intracellularly recorded action potential from a mammalian neuron. Which of the following statements is correct concerning events taking place at the point in time indicated by the arrow?

  

  A. calcium conductance is higher than its normal resting level
  B. chloride conductance is higher than its normal resting level
  C. potassium conductance is higher than its normal resting level
  D. sodium conductance is higher than its normal resting level
  E. sodium-potassium ATPase activity is depressed
  C. potassium conductance is higher than its normal resting level

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  Explanation

  Section: Physiology The neuron, at the arrow, is in the refractory period following the action potential. The membrane potential is more negative (hyperpolarized) than its normal resting level because potassium conductance is elevated. The after hyperpolarization is also due to the fact that sodium channels are inactivated causing lower than normal sodium conductance (choice D). In skeletal muscle neither calcium (choice A) nor chloride (choice B) conductances should be elevated. After the action potential, intracellular sodium concentration could be slightly elevated due to the sodium influx that occurred during the action potential. This could stimulate the Na, K-ATPase (choice E).

• Question 173:

  Askydiver landing forcefully on his right lower limb suffered a central fracture of the acetabulum with dislocation of the femoral head into the pelvis. The acetabulum is formed by the joining of the ilium, ischium, and pubis. These three bones are completely fused by which of the following periods?

  A. birth
  B. 6 years
  C. puberty
  D. 16 years
  E. 23 years
  E. 23 years

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  Explanation

  Section: Anatomy Fusion of the ilium, ischium, and pubis at the acetabulum is usually complete by age 23. From birth to the early 20s (choices AD), the three bones are held together by a Y-shaped cartilage.

• Question 174:

  Exogenous peripherally injected insulin differs from endogenously secreted insulin in a number of aspects, including which of the following?

  A. achieves a higher concentration in the periphery than in the liver, contrary to endogenous insulin
  B. contains C-peptide, which is missing from secreted endogenous insulin
  C. is able to bypass insulin resistance observed with endogenous insulin
  D. is always extracted from animal sources and, therefore, is less effective due to sequence differences and anti-insulin antibodies
  E. is in the form of proinsulin, whereas endogenous insulin has had C-peptide removed
  A. achieves a higher concentration in the periphery than in the liver, contrary to endogenous insulin

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  Explanation

  Section: Physiology The concentration of exogenous insulin is higher at the site of injection in the periphery, compared to its concentration in the liver. On the other hand, endogenous insulin is higher concentrated in the liver than in the periphery. Insulin is normally secreted by the endocrine pancreas into the portal venous drainage. Thus, it passes through the liver before being seen by the periphery. Acertain fraction of insulin is extracted by the liver; so that the concentration of insulin seen by the liver is normally higher than that seen by the periphery. This discrepancy between exogenous and endogenous insulin might contribute to the problems experienced by diabetics such as hypertension and cardiovascular disease. Injected insulin is missing C- peptide (choice B), while endogenous insulin is cosecreted with C- peptide, the physiologic functions of which remain unknown. Injected insulin is biologically active and not in the proinsulin form (choice E). Exogenous insulin generally used today in the United States is recombinant human insulin, not from animal sources (choice D). The problem of developing anti- insulin antibodies is now rarer since the injected insulin is not from animals. Insulin resistance (choice C) is generally due to receptor downregulation or desensitization, or anti-insulin antibodies. Tissues become equally resistant to both the body's own (endogenous) and administered (exogenous) insulin.

• Question 175:

  Methadone has been used in opioid addiction maintenance clinics with partial success. Its ability to reduce the use of heroin and morphine by recovering addicts can be ascribed to which of the following mechanisms?

  A. agonist effects at central dopamine receptors
  B. antidepressant effects at serotonin (5-HT) receptors
  C. antagonist actions at the mu opioid receptor
  D. blockade of the metabolic activation of heroin and morphine to their active forms
  E. long-lasting saturation of the mu opioid receptor and development of tolerance
  E. long-lasting saturation of the mu opioid receptor and development of tolerance

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  Explanation

  Section: Pharmacology Methadone is a typical strong agonist at opioid mu receptors and differs from morphine chiefly in its long duration of binding to the receptor and long elimination half-life. It therefore produces a prolonged opioid effect that causes tolerance to the euphoric action of short-acting heroin and morphine. Methadone has no opioid antagonist action (choice C), nor does it act on metabolic pathways (choice D), serotonin receptors (choice B), or dopamine receptors (choice E).

• Question 176:

  A Boy Scout troop camped for 1week in the woods in the New England area. After arriving home, two experienced an insect bite lesion that was characterized by a clear center surrounded by a flat reddened area that expanded. They also had fever, chills, fatigue, and headache. Which disease listed below was diagnosed?

  A. epidemic typhus
  B. lyme disease
  C. Q fever
  D. rickettsial pox
  E. trench feve
  B. lyme disease

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  Explanation

  Section: Microbiology/Immunology This patient's symptoms point to a diagnosis of Lyme disease, caused by B. burgdorferi, and transmitted to humans by ticks that harbor this organism. The initial symptoms of Lyme disease are fever, chills, fatigue, and headache, but the pathognomonic feature is a spreading, circular rash with a clear center. The rash begins 318 days after the tick bite and is called erythema chronicum migrans. Q fever (choice C), epidemic typhus (choice A), rickettsial pox (choice D), and trench fever (choice E) do not produce erythema chronicum migrans.

• Question 177:

  Arrow 4 in following figure, is pointing to which of the following structures?

  

  A. abdominal aorta
  B. colon
  C. liver
  D. spleen
  E. stomach
  D. spleen

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  Explanation

  Section: Anatomy The spleen (arrow 4) lies to the left of the abdominal cavity. It is in contact with the left side of the stomach (arrow 2) and lodges against the left paravertebral gutter. The abdominal aorta (choice A, arrow 5) is seen as the circular structure immediately anterior to the vertebra. The colon (choice B, arrow 3) is the convoluted structure to the left anterior aspect of the abdominal cavity. The large liver (choice C, arrow 1) occupies most of the right side of the abdominal cavity. The stomach (choice E, arrow 2) is located between the colon and the liver, and in this case, contains liquid contrast material.

• Question 178:

  Heparin is a rapidly acting, potent anticoagulant that has many important clinical uses. Which of the following is an action of heparin?

  A. activates prothrombin
  B. acts with antithrombin to inhibit thrombin activity
  C. decreases prothrombin time
  D. inhibits calcium action
  E. promotes vitamin K activity
  B. acts with antithrombin to inhibit thrombin activity

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  Explanation

  Section: Physiology Thrombin is a critical enzyme in the coagulation cascade. It not only can activate factors VIII and V, it also acts on fibrinogen to form fibrin. Thrombin is essential for clot formation. Antithrombin III modulates the coagulation cascade, serving to inhibit thrombin activity. Heparin acts as an anticoagulant because it accelerates the action of antithrombin III. Heparin does not stimulate vitamin K activity (choice E) and, in any case, anything that promoted vitamin K activity would increase coaguability, since vitamin K is necessary for synthesis of multiple coagulation factors. Activation of prothrombin (choice A) promotes coagulation. Though calcium is an essential cofactor for coagulation, heparin does not act via calcium inhibition (choice D). Because heparin inhibits thrombin action, it would prolong prothrombin time (choice C).

• Question 179:

  Reticulocytes control the rate of globin synthesis as a consequence of the level of heme in the cell. This prevents globin protein from being made when there are insufficient amounts of heme. Which of the following best explains the effects of heme on protein synthesis in these cells?

  A. A heme-controlled phosphatase dephosphorylates cap-binding factor, which prevents recognition of globin mRNA by the ribosomes.
  B. A tRNA-degrading enzyme is active in the absence of heme.
  C. Heme normally activates peptidyltransfersase in reticulocytes.
  D. RNA polymerase activity is decreased in reticulocytes by low heme.
  E. The initiation factor eIF-2 becomes phosphorylated, reducing its level of activity.
  E. The initiation factor eIF-2 becomes phosphorylated, reducing its level of activity.

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  Explanation

  Section: Biochemistry One mechanism by which initiation of translation in eukaryotes is effected is by phosphorylation of a ser(S) residue in the alphasubunit of eIF-2. The factor eIF-2 requires activation by interaction with GTP. The energy of GTP hydrolysis is used during translational initiation, thereby allowing eIF-2 to have GDP bound instead of GTP. In order to reactivate eIF-2, the GDP must be exchanged for GTP. This requires an additional protein of the GEF family known as eIF-2B. The phosphorylated form of eIF-2, in the absence of the eIF-2B, is just as active an initiator of translation as the nonphosphorylated form. However, when eIF-2 is phosphorylated, the GDP-bound complex is stabilized and exchange for GTP is inhibited. When eIF-2 is phosphorylated, it binds eIF-2B more tightly thus slowing the rate of exchange. It is this inhibited exchange that affects the rate of initiation. Within reticulocytes the phosphorylation of eIF-2 is the result of an activity called heme-controlled inhibitor, HCI (see below figure).

  

  The presence of HCI was first seen in an in vitro translation system derived from lysates of reticulocytes. When heme is limiting it would be a waste of energy for reticulocytes to make globin protein, since active hemoglobin could not be generated. Therefore, when the level of heme falls, HCI becomes activated, leading to the phosphorylation of eIF-2 and reduced globin synthesis. Removal of phosphate is catalyzed by a specific eIF-2 phosphatase which is unaffected by heme. There is no heme- controlled phosphatase activity (choice A) in any cell. No tRNA-specific degrading enzymes (choice B) are present in cells. There is no effect of heme levels on peptidyltransferase activity (choice C) or RNA polymerase activity (choice D).

• Question 180:

  Which of the following drugs is considered a first-line agent in the chronic treatment of rheumatoid arthritis and in the treatment of choriocarcinoma?

  A. cyclosporine
  B. hydroxychloroquine
  C. methotrexate
  D. sulfasalazine
  E. thalidomide
  C. methotrexate

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  Explanation

  Section: Pharmacology Methotrexate in very low dosage has proven effective in slowing the progression of rheumatoid arthritis and has low toxicity. It is also one of the most effective (in much higher dosage) antimetabolite cancer chemotherapeutic drugs used in choriocarcinoma. Cyclosporine (choice A) is not used in either condition but is a first-line drug in transplant immunopharmacology. Hydroxychloroquine (choice B) is a disease- modifying antirheumatoid arthritis drug, but has no efficacy in the treatment of cancer. It is more toxic and less effective in rheumatoid arthritis than methotrexate. Sulfasalazine (choice D) is also effective in slowing the progression of joint damage in rheumatoid arthritis but has no efficacy in cancer. Thalidomide (choice E) is effective in erythema nodosum leprosum and multiple myeloma but has not been demonstrated to be effective in rheumatoid arthritis or choriocarcinoma.