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Free STEP1 Exam Braindumps (page: 78)

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PDF with 845 Questions

During the process of protein synthesis, the factor eEF-2 induces the hydrolysis of GTP. The energy of this hydrolysis is coupled to which of the following?

  1. amino acid activation by attachment to a tRNA
  2. correct alignment of the mRNA on the 40S ribosome
  3. formation of the 80S initiation complex
  4. formation of the peptide bond
  5. translocation of the ribosome

Answer(s): E

Explanation:

The translation factor, eEF-2, is involved in the process of peptide elongation. Specifically, eEF-2 catalyzes the GTP-dependent translocation of the ribosomes along the mRNA to the next codon. Attachment of an amino acid to a tRNA (choice A) is catalyzed by the family of aminoacyl-tRNA synthetases. Alignment of the mRNA on the 40S ribosome (choice B) is facilitated by the initiation factor eIF-1. Formation of the 80S initiation complex (choice C) occurs after complete assembly of the 40S preinitiation complex and requires the activity of eIF-5. Formation of the peptide bond (choice D) occurs through the action of peptidyltransferase.



Patients with poorly controlled diabetes mellitus have elevated levels of blood glucose. One severe consequence of the hyperglycemia is an increase in glucose attachment to serum proteins. Which of the following proteins, when glycosylated, is an excellent measure of the length of time someone has suffered from an episode of hyperglycemia?

  1. albumin
  2. cholesterol
  3. fatty acids
  4. hemoglobin
  5. transferrin

Answer(s): D

Explanation:

The formation of glycosylated hemoglobin occurs spontaneously (i.e., nonenzymatically through a reaction known as the Amadori rearrangement) in red blood cells. The amino terminal groups of the beta-chains of hemoglobin complex with the aldehyde groups of glucose to form an amino ketone linkage. This form of hemoglobin is known as Hb . Measurement of the circulating level of glycosylated hemoglobin is a diagnostic tool used to determine the relative length of hyperglycemia and can be used as a measure of treatment effectiveness. Glucose does not form covalent bonds with any of the other choices (A, B, C, and E).



The ability of rod cells in the eye to respond to light and transmit that response to the optic nerve requires that the 11-cis form of vitamin A be attached to which of the following proteins?

  1. cGMP phosphodiesterase
  2. Na+ channel
  3. rhodopsin
  4. scotopsin
  5. transducin

Answer(s): D

Explanation:

Both rod and cone cells contain a photoreceptor pigment in their membranes. The photosensitive compound (photoreceptor) of most mammalian eyes is a complex of protein and an aldehyde form of vitamin A. The protein component is a member of the opsin family called scotopsin. The photoreceptor of rod cells is specifically called rhodopsin (choice C) or visual purple, which is a complex between the protein scotopsin and the 11-cis-retinal form of vitamin A. Intracellularly, rhodopsin is coupled to a specific G- protein called transducin (choice E). When the rhodopsin is exposed to light it is bleached releasing the 11- cis-retinal from opsin. Absorption of photons by 11-cis-retinal triggers a series of conformational changes on the way to conversion all-trans-retinal. The release of opsin results in a conformational change in the photoreceptor. This conformational change activates transducin, leading to an increased GTP-binding by the alpha-subunit of transducin. Binding of GTP releases the alpha- subunit from the inhibitory beta- and gamma-subunits. The GTPactivated alpha-subunit in turn activates an associated phosphodiesterase (choice A), an enzyme that hydrolyzes cyclic-GMP (cGMP) to GMP. Cyclic GMP is required to maintain the Na+ channels (choice B) of the rod cell in the open conformation. The drop in cGMP concentration results in complete closure of the Na+ channels.



In an assay for the presence of a specific disease gene alelle in several individuals, you isolate genomic DNA from each and perfom polymerase chain reaction (PCR) using gene-specific primers. The PCR product is expected to have a recognition site for the restriction endonuclease BamHI. Following PCR and BamHI digestion, the products are separated by gel electrophoresis and the results are shown in below figure. Which lane corresponds to the individual demonstrating heterozygosity for the BamHI site?

  1. 1
  2. 2
  3. 3
  4. 4
  5. 5

Answer(s): B

Explanation:

Someone who exhibits heterozygosity would harbor two distinct alelles. In the case of this analysis, they would be identified as containing a copy of the gene that does not harbor the BamHI site and a copy that does harbor the site. The presence of the BamHI site in the PCR product in this example would result in the generation of 300 and 100 bp fragments. The lack of the site would yield a product of 400 bp. Thus, following PCR amplification and BamHI digestion of DNA from a heterozygote, one would be able to observe three bands of equal intensity. A person who was homozygous for having the BamHI-containing sequence would result in two bands (choice A) whose intensities would be greater than the three of the heterozygote. A person homozygous for the lack of the BamHI site would yield a single 400 bp band (choice D). The banding patterns observed in lanes 3 (choice C) and 5 (choice E) would not be possible from this analysis.






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