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

USMLE USMLE-STEP-1 Online Questions & Answers

• Question 11:

  A 70-year-old man presents with Parkinson's disease in an early stage. Which of the following drugs acts directly on dopamine receptors to reduce the signs and symptoms of Parkinson's disease?

  A. amantadine
  B. carbidopa
  C. entacapone
  D. orphenadrine
  E. pramipexole
  E. pramipexole

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  Explanation

  Section: Pharmacology Pramipexole is a dopamine receptor agonist unrelated to levodopa or ergot derivatives such as bromocriptine and pergolide (which are also useful dopamine agonists). Amantadine (choice A) acts to facilitate the release or slow the reuptake of dopamine, but does not act on the dopamine receptor. Carbidopa (choice B) blocks the peripheral conversion of levodopa to dopamine, it does not act on dopamine receptors. Entacapone (choice C) blocks the peripheral metabolism of levodopa by catechol- O- methyltransferase. Orphenadrine (choice D) is a centrally acting anticholinergic agent that is sometimes used in Parkinson's disease.

• Question 12:

  A biopsy from a 55-year-old woman with breast cancer was submitted to an estrogen receptor assay. The binding of estrogen to the cancer cells below figure, solid line was plotted in the form of a Scatchard plot and compared to control cells below figure, dotted line). Which of the following would you conclude for the patient's estrogen receptors?

  

  A. they are equal in number and have a higher affinity
  B. they are equal in number and have a lower affinity
  C. they are fewer in number and have a higher affinity
  D. they are fewer in number and have a lower affinity
  E. they are higher in number and have an equal affinity
  B. they are equal in number and have a lower affinity

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  Explanation

  Section: Physiology The slope of the patient's line is decreased compared to normal. This indicates a low affinity of estrogen for the estrogen receptor, which makes choice B correct and choice Aincorrect. The x- intercept is called Bmax and represents the total number of estrogen receptors. This is unchanged in the patient compared to normal so that choices C, D, and E are excluded. The information about the numbers and status of estrogen and progesterone receptors on cancer cells is of high clinical value to determine the efficiency of consequent hormonal therapy.

• Question 13:

  The prevalence for disease X is 1%. Anew laboratory test is positive in 99% of the patients with disease X, but half of the total of positive test results obtained correspond to patients without disease X. Based on this information, the test would be classified as which of the following?

  A. sensitive and nonspecific
  B. sensitive and specific
  C. nonsensitive and nonspecific
  D. nonsensitive and specific
  B. sensitive and specific

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  Explanation

  Section: Pathology and Path physiology Let us take a population of 100,000 individuals. Since the prevalence of disease X is 1%, 1000 individuals have disease X and 99,000 are disease-free. The test is positive in 99% of individuals with the disease (i.e., of the 1000 with disease, there are 990 true positives and 10 false negatives), thus the test is clearly sensitive. However, half of all positive test results are occurring in disease-free individuals. If 990 individuals with disease had a positive test result (true positives), then there must also be 990 disease-free individuals who had a positive test result. It was stated above that 99,000 are disease-free, so of these 990 have a positive test result (false positives) and the remainder (99,000 990 = 98,010) must have a negative test result (true negatives). Specificity may be defined as the percentage of individuals in a disease-free population who have a negative test result. This is mathematically equivalent to (true negatives)/(true negatives + false positives) = (98,010)/(98,010 + 990) = 98,010/99,000 = 99% . Thus the test is both sensitive and specific, and the other answers (choices A, C, D, and E) are incorrect.

• Question 14:

  A 53-year-old man is being treated for hypertension and diabetes. His medications include insulin and propranolol. He presents at his physician's office complaining of muscle weakness. Blood tests reveal hyperkalemia (elevated serum potassium) as well as elevated BUN (blood urea nitrogen). Propranolol is gradually eliminated and his insulin dosage is adjusted. His serum potassium normalizes and his muscle weakness is alleviated. What probably caused his muscle weakness?

  A. high potassium-mediated block of acetylcholine receptors
  B. high potassium-mediated block of skeletal muscle calcium channels
  C. motor neuron hyperpolarization
  D. skeletal muscle depolarization with resultant Na-channel inactivation
  E. skeletal muscle hyperpolarization with resultant Na-channel blockade
  D. skeletal muscle depolarization with resultant Na-channel inactivation

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  Explanation

  Section: Physiology Elevated serum potassium levels cause membrane depolarization with a resulting Na-channel inactivation. Fibers are thus less able to fire action potentials, leading to impaired excitation contraction coupling, with muscle weakness (choice D is correct). Though hyperpolarization would also impede action potential generation, by moving the Na-channel away from its activation threshold, choices C and E are incorrect since high potassium causes membrane depolarization. Calcium required for skeletal muscle contraction is derived from internal stores (the sarcoplasmic reticulum) and is not dependent on calcium influx through surface membrane channels (choice B). There is no evidence that potassium interferes with the acetylcholine receptor (choice A). Hyperkalemia in this patient is probably due to multiple factors. Since insulin promotes potassium uptake into cells, too low an insulin dosage in the diabetic can lead to hyperkalemia. In addition, propranolol can cause a shift of potassium from cell to blood.

  Finally, the elevated BUN indicates some renal failure, and failing kidneys cannot efficiently excrete potassium into the urine.

• Question 15:

  An outbreak of severe, atypical pneumonia occurred in a Southeast Asian community. Symptoms included high fever, chills, headache, and a cough or breathing difficulty. Avirus with the structure shown in below figure was implicated as the cause of the infections. Which of the viruses listed below caused this outbreak?

  

  A. arenavirus
  B. filovirus
  C. poxvirus
  D. rabies virus
  E. SARS virus
  E. SARS virus

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  Explanation

  Section: Microbiology/Immunology It has recently been shown that severe acute respiratory syndrome or SARS is caused by a corona- associated virus. This virus is a positive, single-stranded RNA, enveloped virion which has petal-like projections called corona, and is shown in below figure. The structure of rabies virus is bullet-shaped (choice D). Poxvirus is a very large, brick-shaped, very complex virus (choice C). Filovirus, such as the Ebola, or Marburg virus is very long and filamentous (choice B). Arenaviruses, like the lymphocytic choriomeningitis, or Lassa fever virus are circular, and have granules (arena) of nonfunctional ribosomes on their surfaces (choice A).

• Question 16:

  A Gram stain of pus from the oral abscess of a 62-year-old man with periodontitis shows that the patient's infection is polymicrobic. Which of the following is the most likely source of this man's infection?

  A. israelii following tooth extraction
  B. autoinfection with oral anaerobes
  C. autoinfection with oral Staphylococcus spp.
  D. periodontitis is not an infectious process
  E. P. aeruginosa following primary viral infection
  B. autoinfection with oral anaerobes

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  Explanation

  Section: Microbiology/Immunology Medically important infections due to anerobic bacteria are common. The infections are often polymicrobial, anerobes being mixed with other anerobes, facultative anerobes, and aerobes. Anerobic bacteria are found throughout the body's normal or usual flora. They comprise part of the bacteria on the skin, on mucosal surfaces, in the mouth, and in the intestinal tract. Infection occurs when anerobes contaminate body sites that are normally sterile. Such conditions may also arise with the use of broad spectrum antimicrobials. In the oral cavity, anerobes are present in small numbers. Bacteroides, Fusobacterium, Prevotella, and Porphyromonas make up the most common genera. The main etiological agents of periodontitis (soft oral tissue infections) and endodontic (root canal nfections) belong to those four genera. Therefore, choice B represents a situation where low numbers of endogenous, normal anerobic bacteria were introduced into a lesion area and made the infection orse. Such oral infections will respond to dental manipulations and antibiotics. Periodontitis (choice D) is definitely an infectious process. Another important factor is the polymicrobial etiology from endogenous sources can be found in the lesions. Choices A (Actinomyces), C (Staphylococcus), and E (Pseudomonas) ould be incorrect since they present a single genus in each case for etiology.

• Question 17:

  A football player was simultaneously tackled from the front above the knee and from the back below the knee. In the emergency room, he displays a positive anterior drawer test which becomes negative if the knee is internally or externally rotated. Which of the following has ruptured?

  A. anterior cruciate ligament
  B. fibula
  C. fibular (lateral) collateral ligament
  D. posterior cruciate ligament
  E. tibial (medial) collateral ligament
  A. anterior cruciate ligament

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  Explanation

  Section: Anatomy In the anterior drawer test, the patient's tibia is pulled forward and anterior tibial displacement reveals rupture of the anterior cruciate ligament. The drawer test becomes negative when the knee is internally rotated because of the tightening of the posterior cruciate ligament (choice D); in this case, the result reveals that the posterior cruciate ligament is not affected. External rotation is limited by the tibial (medial) collateral ligament (choice E) and since the drawer sign is negative with the knee in this position, this ligament is also intact. The fibula (choice B) and fibular colateral ligament (choice C) do not participate in the knee joint and have no effect on the drawer test.

• Question 18:

  Exhbit:

  

  Please refer to the exhibit.

  Hypersensitive individuals have IgE to specific antigens (e.g., pollen, bee venom) on the surface of their leukocytes (monocytes, macrophages, basophils, eosinophils). When these individuals are challenged with antigen, the antigen-IgE complexes induce synthesis and release of which of the following physiologically potent lipids?

  A. A
  B. B
  C. C
  D. D
  E. E
  C. C

• Question 19:

  Which of the following correctly describes the mechanism of action of streptokinase?

  A. combines with plasminogen to form an enzymatically active complex
  B. competitively blocks binding of plasminogen to fibrin
  C. converts plasmin to plasminogen
  D. inhibits platelet cyclooxygenase (COX) activity
  E. provides a template for combination of thrombin and antithrombin III
  A. combines with plasminogen to form an enzymatically active complex

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  Explanation

  Section: Pharmacology Streptokinase has no intrinsic enzymatic activity, but instead forms a stable complex with the patient's plasminogen, making it enzymatically active in cleaving free plasminogen to plasmin. The streptokinase- plasminogen complex is not inhibited by antiplasmin. The other thrombolytic agents-- t-PA, reteplase, tenecteplase, and urokinase-- activate plasminogen directly. Competitive blocking of binding of plasminogen to fibrin (choice B) is a property of aminocaproic acid (AMICAR), a lysine analog used to inhibit fibrinolysis. Conversion of plasmin to plasminogen (choice C) is incorrect; it is plasminogen that is converted to plasmin in the thrombolytic action. Inhibition of cyclooxygenase activity (choice D) is a property of the NSAIDs that contributes to prevention of platelet aggregation and thrombosis. Heparin exerts its anticoagulant actions by providing a template for the combination of thrombin and antithrombin III (choice E).

• Question 20:

  Which of the following drugs reduces LDL cholesterol by inhibiting an intestinal transport protein?

  A. atorvastatin
  B. cholestyramine
  C. ezetimibe
  D. gemfibrozil
  E. niacin
  C. ezetimibe

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  Explanation

  Section: Pharmacology Ezetimibe is an important newer drug that inhibits cholesterol and bile acid absorption from the gut by inhibiting the epithelial transport protein NPC1L1. Atorvastatin (choice A) reduces both LDL and triglycerides, but does not inhibit VLDL secretion. Cholestyramine (choice B) inhibits the reabsorption of bile acids and cholesterol from the gut and actually increases triglycerides in some patients. Gemfibrozil (choice D) activates lipoprotein lipase in peripheral tissues and reduces VLDL levels and triglycerides. It does not reduce hepatic secretion of VLDL. Niacin (choice E) is the most effective drug available for reducing triglycerides and also reduces LDL cholesterol by reducing the hepatic secretion of its precursor, VLDL particles.