ATI RN
foundations in microbiology test bank Questions
Question 1 of 5
Gram-negative rods producing blue-green pigment and a fruity odor were isolated from a wound infection. What is the causative agent?
Correct Answer: A
Rationale: The correct answer is A: Pseudomonas aeruginosa. P. aeruginosa is known for producing a blue-green pigment called pyocyanin, which gives a characteristic color to the colonies. The fruity odor is due to the production of a compound called 2-aminoacetophenone by P. aeruginosa. These features are classic for P. aeruginosa and help differentiate it from other Gram-negative rods. Proteus mirabilis (B) typically does not produce blue-green pigment or fruity odor. Klebsiella pneumoniae (C) and Escherichia coli (D) also do not exhibit these specific characteristics associated with P. aeruginosa.
Question 2 of 5
Microscopic examination of a smear from a chancre revealed spiral microorganisms. The bacteria were thin and mobile with 8-12 regular coils. What is the likely diagnosis?
Correct Answer: A
Rationale: The correct diagnosis is A: Syphilis. The spiral microorganisms described are characteristic of Treponema pallidum, the causative agent of syphilis. The thin and mobile bacteria with 8-12 regular coils match the morphology of Treponema pallidum. Leptospirosis is caused by Leptospira interrogans, which are thicker and have hook-like ends, different from the described bacteria. Lyme disease is caused by Borrelia burgdorferi, which has a different appearance than the thin, coiled bacteria in the smear. Relapsing fever is caused by Borrelia recurrentis, which typically have irregular coils, unlike the regular coils seen in the smear.
Question 3 of 5
A 32 year old patient complains about heartburn and dull pain in the epigastrium that appear 2-3 hours after meal. Exacerbations happen in spring and in autumn. The patient has food intolerance of eggs and fish. Objectively: stomach palpation reveals painfulness in the gastroduodenal area. Electrophasoduodenoscopy revealed a 5 mm ulcer on the anterior wall of duodenum. Urease test is positive. What is the most probable leading mechanism of disease development?
Correct Answer: A
Rationale: The correct answer is A: Helicobacterial infection. This is the most probable leading mechanism of disease development in this case because the patient's symptoms, such as heartburn, epigastric pain, food intolerance, and presence of a duodenal ulcer with a positive urease test, are classic signs of Helicobacter pylori infection. This bacterium colonizes the stomach lining, leading to inflammation and ulcer formation. The exacerbations in spring and autumn could be due to seasonal variations in immune response. Choice B: Dietary allergy is incorrect because the symptoms and findings in the patient are more indicative of an infectious etiology rather than an allergic reaction to specific foods. Choice C: Autoantibody production is unlikely as there are no indications of an autoimmune process in this case. Choice D: Reduced prostaglandin synthesis is not the leading mechanism as the symptoms and findings are more consistent with an infectious cause rather than a deficiency in prostaglandin synthesis.
Question 4 of 5
How is viral multiplication detected in cell culture?
Correct Answer: D
Rationale: The correct answer is D because viral multiplication in cell culture can be detected by observing a specific cytopathic effect (A) caused by viral infection in cells and by using an immunofluorescence detection method (B) to visualize viral antigens within the cells. Choice C, the characteristic type of viral colonies, is incorrect as viral multiplication is not typically detected by observing viral colonies in cell culture. Option D is the correct choice as both A and B methods are commonly used to detect viral multiplication in cell culture.
Question 5 of 5
A 60-year-old patient was hospitalized to the surgical department because of infection caused by blue pus bacillus (Pseudomonas aeruginosa) which is sensative to penicillin antibiotics. Indicate which of the given penicillins has marked activity to the Pseudomonas aeruginosa?
Correct Answer: A
Rationale: The correct answer is A: Carbenicillin disodium. Carbenicillin is a semisynthetic penicillin that has marked activity against Pseudomonas aeruginosa. This is due to its extended spectrum of activity and increased stability against beta-lactamases produced by Pseudomonas aeruginosa. Carbenicillin is commonly used to treat infections caused by this bacterium. Choice B: Benzylpenicillin (Penicillin G) has limited activity against Pseudomonas aeruginosa. Choice C: Methicillin is not effective against Pseudomonas aeruginosa. Choice D: Oxacillin also has limited activity against Pseudomonas aeruginosa. In summary, Carbenicillin is the correct choice due to its marked activity against Pseudomonas aeruginosa compared to the other penicillins listed.
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