ATI RN
microbiology chapter 10 test bank Questions
Question 1 of 5
Method for rapid diagnosis of some viral infections is
Correct Answer: D
Rationale: The correct answer is D because utilizing all three methods—immunofluorescence, viral neutralization reaction, and electron microscopy—provides a comprehensive and rapid diagnosis of various viral infections. Immunofluorescence detects viral antigens directly, viral neutralization reaction demonstrates the presence of specific antibodies, and electron microscopy visualizes viral particles. This combination ensures accurate and quick identification of different viruses. Choices A, B, and C alone may not cover all aspects of viral diagnosis, making them less effective compared to the comprehensive approach of using all three methods.
Question 2 of 5
Which of the following bacteria are capable of surviving in both aerobic and anaerobic environments?
Correct Answer: B
Rationale: The correct answer is B: Facultative anaerobes. Facultative anaerobes can survive in both aerobic and anaerobic environments. In aerobic conditions, they use oxygen for energy production, while in anaerobic conditions, they can switch to fermentation or anaerobic respiration. A: Obligate aerobes require oxygen to survive and cannot survive in anaerobic environments. C: Obligate anaerobes cannot survive in the presence of oxygen and only thrive in anaerobic conditions. D: Microaerophiles require low levels of oxygen to survive and cannot thrive in fully aerobic or anaerobic conditions.
Question 3 of 5
Reassortment of gene segments can occur in one of the following families:
Correct Answer: B
Rationale: The correct answer is B: Family Reoviridae. Reassortment of gene segments occurs in segmented viruses like Reoviruses. In Reoviridae, each segment encodes a specific protein, allowing for reassortment to generate new strains. Togaviridae (A) is a non-segmented virus, not capable of reassortment. Picornaviridae (C) and Paramyxoviruses (D) are also non-segmented viruses, limiting their ability for reassortment. Therefore, the correct choice is B due to the segmented nature of Reoviruses allowing for genetic reassortment.
Question 4 of 5
A patient with diarrhea had a stool sample that revealed Gram-negative rods that fermented lactose. What is the most likely causative agent?
Correct Answer: A
Rationale: The correct answer is A: Escherichia coli. Step-by-step rationale: 1. Gram-negative rods: E. coli is a Gram-negative rod bacterium. 2. Fermented lactose: E. coli is known to ferment lactose. 3. Common cause of diarrhea: E. coli is a common causative agent of infectious diarrhea. Summary: B: Salmonella typhi causes typhoid fever, not typically associated with lactose fermentation. C: Shigella dysenteriae causes dysentery, not known for lactose fermentation. D: Proteus vulgaris is not a common cause of infectious diarrhea and does not typically ferment lactose.
Question 5 of 5
During the incubation period S. Typhi multiplies in the
Correct Answer: D
Rationale: The correct answer is D because during the incubation period of Salmonella Typhi infection, the bacteria multiply in the macrophages of the Peyer's patches and solitary follicles. This is supported by the fact that S. Typhi is an intracellular pathogen that targets the lymphoid tissues of the intestines. Macrophages in these areas serve as the primary host cells for the bacteria, allowing for their replication and evasion of the host immune response. Choices A, B, and C are incorrect because S. Typhi does not primarily multiply in the small intestine enterocytes, the lumen of the small intestine, or hepatocytes during the initial stages of infection.
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