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

USMLE USMLE-STEP-1 Online Questions & Answers

• Question 261:

  During a routine preschool examination, a 5-year-old boy is found to have difficulty focusing on distant objects. Which of the following is true during far accommodation of the eyes?

  A. ciliary muscles are relaxed
  B. pupils are constricted
  C. the focal length of the lens is short
  D. the lens is rounded
  E. zonula fibers are relaxed
  A. ciliary muscles are relaxed

  ---

  Explanation

  Section: Physiology This patient probably suffers from myopia (nearsightedness). Myopia is either due to eyeballs that are too long or a lens that is too strong. To focus a distant object onto the retina (far accommodation), the lens has to decrease its refractive power (i.e., increase its focal length). This is accomplished through relaxation of the ciliary muscles that oppose the pull of the sclera, resulting in a tightening of the zonula fibers and a flattening of the lens. Shortening of the focal length (choice C), rounding of the lens (choice D), and relaxation of the zonular fibers (choice E), all occur during near accommodation. The pupils also constrict during near accommodation (choice B), perhaps to increase depth of field.

• Question 262:

  Aboy picked up a grounded bat and was bitten on the hand, suffering a slight wound. In addition to flushing the wound, cleaning it surgically, and giving antitetanus prophylaxis and antibiotics as indicated, the physician should immediately perform which of the following?

  A. observe the boy very carefully
  B. order a search for the attacking bat for autopsy
  C. report the incident to the state epidemiologist
  D. start rabies vaccine
  E. start rabies vaccine and give antirabies serum
  E. start rabies vaccine and give antirabies serum

  ---

  Explanation

  Section: Microbiology/Immunology If exposure to rabies virus appears definite, as in this case, treatment with human diploid cell live- derived vaccine and hyperimmune antirabies gamma globulin should be started immediately. Serum antibodies provide an immediate barrier to the growth of virus; meanwhile, antibodies are elicited by the vaccine. If the level of exposure is minimal (e.g., no skin puncture) and the animal probably is not rabid, vaccine is not recommended. Ordering a search for the attacking fox for autopsy to determine if it has rabies is like searching for a needle in a haystack, and thus not the best approach to address a possible rabies infection, which requires immediate actions (choice B). Initiation of rabies vaccine will lead to production of antibodies against the rabies virus, but it will require approximately 2 weeks to develop protective antibodies. By that time, severe damage may have already occurred (choice D). Postexposure immunization and human rabies immunoglobulin is the best approach. Observation of the patient (choice A) or reporting of the incident to the state epidemiologist (choice C) do not address the real needs of an individual running the risk of rabies.

• Question 263:

  The forensic analytical technique identified as DNA fingerprinting refers to which of the following processes?

  A. the establishment of a complete collection of cloned fragments of DNA
  B. the identification of sequences of DNA to which specific proteins bind, thereby rendering them resistant to digestion by DNA-degrading nucleases
  C. the specific association of complimentary strands of DNA to one another
  D. the synthetic oligonucleotide-directed enzymatic amplification of specific sequences of DNA
  E. the use of repeat sequences to establish a unique pattern of fragments for any given individual
  E. the use of repeat sequences to establish a unique pattern of fragments for any given individual

  ---

  Explanation

  Section: Biochemistry DNAfingerprinting refers to the process of using polymorphic repeat sequences to establish a unique pattern of DNA fragments for any given individual. The polymorphic repeats that are identifiable by the fingerprinting technique are hypervariable repeats such as variable number tandem repeats (VNTRs). The bands are detected by Southern blotting enzyme-digested chromosomal DNA and probing the blot with various different VNTR probes. The generation of a DNA library (choice A) refers to the establishment of a complete collection of cloned fragments of DNA, either from genomic sources or cDNA. DNA footprinting (choice B) refers to the identification of sequences of DNA to which specific proteins bind, thereby rendering the DNA at that site resistant to digestion by DNA degrading nucleases. Hybridization (choice C) refers to the specific association of complimentary strands of DNA to one another. The PCR technique (choice D) uses synthetic oligonucleotides to direct the enzymatic amplification of specific sequences of DNA.

• Question 264:

  A premature female infant is born about 24 weeks after fertilization and develops rapid, labored breathing shortly after birth. She is immediately transferred to intensive care where she is diagnosed with hyaline membrane disease (HMD). Which of the following is most likely deficient in the infant?

  A. alveolar ducts
  B. lung surfactant
  C. terminal saccules
  D. type I alveolar cells
  E. type II alveolar cells
  D. type I alveolar cells

  ---

  Explanation

  Section: Anatomy HMD is also known as respiratory distress syndrome, which is most often caused by the lack of lung surfactant, due to a premature birth. Lung surfactant production begins around 20 weeks after fertilization. But it is present only in small amounts until the last 2 weeks before birth when its amount increases significantly. Alveolar ducts (choice A) branch from the respiratory bronchioles during development. Type I alveolar cells (choice D) or pneumocytes are squamous epithelial cells, which participate in gas exchange. These epithelial cells line the terminal saccules (choice C). Type II alveolar cells (choice E) synthesize surfactant.

• Question 265:

  A 23-year-old woman develops a sore throat, chills, headache, coryza, and a nonproductive cough. A chest x-ray reveals an indistinct, patchy interstitial infiltrate. She has an uneventful recovery within a few days without receiving therapy. Which of the following outcomes is most likely to occur in the region of the affected lung?

  A. healing by first intention
  B. metaplasia
  C. repair by scar formation
  D. resolution
  E. subacute inflammation
  D. resolution

  ---

  Explanation

  Section: Pathology and Path physiology Resolution, or restoration of tissue to its preinjury state, occurs in tissue injury where minimal cell necrosis has occurred, such as this case of interstitial pneumonia. Removal of debris associated with the accompanying acute inflammatory response is sufficient to restore parenchymal tissue to its normal state. Regeneration of new parenchymal cells will occur to replace necrotic cells provided the involved cells have:

  (i) regenerative capacity; (ii) exist in remaining sufficient viable number, and (iii) possess an adequate connective tissuesupporting framework. Healing by first intention (choice A), also known as primary union, describes healing of skin wounds in which the edges are in close apposition, such as a surgical incision. Metaplasia (choice B), the replacement of one mature cell type by another mature cell type abnormal to that location, occurs most commonly in epithelial tissues where chronic physical or chemical irritation is present, such as squamous metaplasia of the endocervix or bronchial mucosa, and intestinal metaplasia of the esophagus. Repair by scar formation (choice C) occurs either in injured tissue comprised of permanent cells (e.g., myocardium), when resolution fails to occur in acute inflammation, or if chronic inflammation persists with ongoing necrosis. Subacute inflammation (choice E) describes repeated attacks of acute inflammation in which incomplete resolution of individual attacks leads to increasing chronic inflammation with superimposed acute inflammation.

• Question 266:

  A patient with acute glomerulonephritis has a total plasma of 2.5 mmol/L and a GFR of 160

  

  L/day. What is the estimated daily filtered load of calcium?

  A. 64 mmol/day
  B. 120 mmol/day
  C. 240 mmol/day
  D. 400 mmol/day
  E. 800 mmol/day
  C. 240 mmol/day

• Question 267:

  Control of the rate of translational initiation can be exerted at the level of the activity of the guanosine triphosphate (GTP)-binding and hydrolyzing initiation factor, eIF-2. The efficiency with which eIF-2 recycles between the active GTP-bound form and the inactive GDPbound form is controlled by the action of the initiation factor depicted as "Factor ?" in below figure. What is the identity of this factor?

  

  A. eIF-1
  B. eIF-2B
  C. eIF-4A
  D. eIF-4E
  E. eIF-4G
  D. eIF-4E

  ---

  Explanation

  Section: Biochemistry The eIF-2 cycle consists of the translation initiation factors, eIF-2A and eIF-2B (also called guanine- nucleotide exchange factor, GEF). The cycle involves the binding of GTP by eIF-2A forming a complex that then interacts with the initiator methionyl-tRNA. When the initiator methionyl-tRNA is placed into the correct position of the 40S ribosomal subunit, the GTP ishydrolyzed (generating GDP and release of Pi) to provide the energy necessary to correctly position the incoming mRNA such that the initiator AUG codon and the initiator methionyltRNA anticodon are aligned. In order to regenerate an active eIF-2A for subsequent translation initiation events, the GDP must be exchanged for GTP. The exchange reaction is catalyzed by eIF-2B. The initiation factor, eIF-1 (choice A), facilitates the correct positioning of the initiator methionyl- tRNA and the mRNA. Initiation factor eIF-4A (choice C) binds to the mRNAand is required to "melt" any secondary structure that may exist at the 5'-end of the mRNA. Initiation factor eIF-4E (choice D) physically binds the cap structure at the 5'-end of the mRNA. Initiation factor eIF-4G (choice E) acts as a scaffold for eIF-4E facilitating cap structure binding by eIF-4E. Together, the complex composed of eIF-4A, eIF-4E, and eIF-4G is sometimes referred to as eIF-4F.

• Question 268:

  A 71-year-old man presents to his primary care physician for a check-up accompanied by his wife, who reports that she had to "push him out of the door" to go to see a doctor as his last physical examination was almost 8 years ago. He states that he has been feeling well other than having some lower back pain. Physical examination reveals a firm, nodular prostate. On further questioning, the patient states that he has not noticed any change in urination. Laboratory studies reveal a mildly elevated serum prostate-specific antigen level and a serum alkaline phosphatase level almost three times normal. These findings are most consistent with which of the following?

  A. acute prostatitis
  B. benign prostatic hyperplasia
  C. chronic prostatitis
  D. local invasion of the prostate by bladder carcinoma
  E. metastatic prostate carcinoma
  E. metastatic prostate carcinoma

  ---

  Explanation

  Section: Pathology and Path physiology Serum prostate-specific antigen can be elevated in both benign prostatic hyperplasia and prostatic carcinoma (it is worth noting that both can also be within the reference range). However, this patient has a firm, nodular prostate with no urinary changes making carcinoma the more likely of the two. The additional finding of high levels of alkaline phosphatase is compatible with osteoblastic metastases. These findings taken together make metastatic prostate carcinoma the most likely of the choices given.

  Prostate-specific antigen can be elevated in acute prostatitis (choice A), benign prostatic hyperplasia (choice B), and chronic prostatitis (choice C), but none of these conditions, or local invasion of bladder carcinoma (choice D) would be expected to produce the increase in alkaline phosphatase that is seen in this case.

• Question 269:

  Apatient is admitted to the hospital following a knife wound to the abdomen which results in extensive bleeding into the abdomen. The arterial pressure at admission is 65/30 mm Hg with a rapid, weak pulse. Compared to normal, the nerve impulses in the carotid sinus nerves and in the sympathetic nerves are changed in what way?

  A. both show decreased nerve impulse rates
  B. both show increased nerve impulse rates
  C. carotid sinus nerve impulses are decreased, sympathetic are increased
  D. carotid sinus nerve impulses are increased, sympathetic are decreased
  E. the patient is unconscious so there are no changes in nerve impulse rates
  C. carotid sinus nerve impulses are decreased, sympathetic are increased

  ---

  Explanation

  Section: Physiology The carotid sinus baroreceptors signal the magnitude of arterial pressure to the brainstem which then causes reflex inhibition of sympathetic nerves to the body. Thus, a decrease in arterial pressure will cause a reduced carotid sinus nerve impulse rate, which reflexly increases sympathetic nerve impulse rates. Hence, choices A, B, and D are directionally incorrect. Choice E is incorrect because the arterial baroreflexes function at the level of the brainstem, which does not require the conscious mind for its action.

• Question 270:

  PDH is a complex multisubunit enzyme. Which of the following exerts a positive influence on the activity of PDH?

  A. acetyl-CoA
  B. ATP
  C. dephosphorylation
  D. NADH
  E. phosphorylation
  C. dephosphorylation

  ---

  Explanation

  Section: Biochemistry Regulation of the PDH complex is effected by both allosteric means and by phosphorylation (see below figure). Allosteric effectors of the enzyme complex are of both positive and negative type. The complex is phosphorylated by a specific kinase identified as PDH kinase and phosphate is removed by PDH phosphatase. When in the unphosphorylated state the complex is much more active, therefore the activity of PDH phosphatase is important for maintaining the PDH complex in the active state. Acetyl- CoA (choice A) and NADH (choice D) are both allosteric inhibitors of the nonphosphorylated form of the PDH complex and also serves to activate PDH kinase leading to phosphorylated and inhibited PDH. Like acetyl- CoA and NADH, ATP (choice B) allosterically activates the PDH kinase leading to phosphorylated and inhbited PDH. Phosphorylation (choice E) inhibits the activity of the PDH complex.