Nokea
The basic method for routine mumps diagnosis is;
- A. PCR
- B. Isolation on cell cultures
- C. ELISA
- D. Immunofluorescence method
Correct Answer: B
Rationale: The correct answer is B: Isolation on cell cultures. This method is the gold standard for routine mumps diagnosis as it involves isolating the virus in cell cultures to confirm the presence of the mumps virus. PCR (choice A) is used for detecting viral RNA, not for routine diagnosis. ELISA (choice C) detects antibodies, not the virus itself. Immunofluorescence method (choice D) is used for visualizing viral antigens but is not as reliable as isolating the virus in cell cultures for diagnosis.
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Botulism agent causes severe food toxicoinfection. Point out the most characteristic morphologic feature of botulism agent.
- A. Gram-positive spore-forming bacilli with subterminal spore
- B. Thick gram-positive non-sporeforming bacilli
- C. Gram-positive spore-forming bacilli with terminal spore
- D. Thin mobile spore-forming bacilli with central spore
Correct Answer: B
Rationale: The correct answer is B: Thick gram-positive non-sporeforming bacilli. In botulism, the causative agent, Clostridium botulinum, is a thick gram-positive bacilli that does not form spores. This is a key characteristic morphologic feature.
A: Gram-positive spore-forming bacilli with subterminal spore - This is incorrect as Clostridium botulinum does not have subterminal spores.
C: Gram-positive spore-forming bacilli with terminal spore - Incorrect because Clostridium botulinum does not form spores.
D: Thin mobile spore-forming bacilli with central spore - Incorrect because Clostridium botulinum is not thin, mobile, or spore-forming.
In summary, the most characteristic morphologic feature of the botulism agent is being thick, gram-positive, and non-sporeforming, making choice B the correct answer.
Bacteria that use organic compounds as the source of both carbon and energy are referred to as:
- A. photoautotrophs
- B. chemoautotrophs
- C. photoheterotrophs
- D. chemoheterotrophs
Correct Answer: D
Rationale: The correct answer is D: chemoheterotrophs. Chemoheterotrophs use organic compounds as both their carbon source and energy source. They obtain energy by breaking down organic molecules through cellular respiration. Photoautotrophs (A) use light energy to produce organic compounds, chemoautotrophs (B) use inorganic compounds for energy, and photoheterotrophs (C) use light energy but require organic compounds as a carbon source. Therefore, chemoheterotrophs best describe bacteria that obtain both carbon and energy from organic compounds.
Which of the following is likely to be the most frequently used portal of entry for pathogens?
- A. Skin
- B. Gastrointestinal tract
- C. Respiratory tract
- D. Conjunctiva
Correct Answer: C
Rationale: The correct answer is C: Respiratory tract. The respiratory tract is likely the most frequently used portal of entry for pathogens because it is constantly exposed to the external environment and has a large surface area lined with mucous membranes that can trap and facilitate the entry of pathogens. Additionally, the respiratory tract serves as a direct pathway for pathogens to reach the bloodstream and other organs. Skin (choice A) is a physical barrier that can prevent pathogen entry, while the gastrointestinal tract (choice B) has acidic pH and enzymes that can kill many pathogens. Conjunctiva (choice D) is a less common portal of entry compared to the respiratory tract due to its smaller surface area and protective mechanisms such as tears.
What differentiates facultative anaerobes from obligate anaerobes?
- A. Facultative anaerobes require oxygen for survival
- B. Facultative anaerobes can grow with or without oxygen
- C. Obligate anaerobes can grow with oxygen
- D. Obligate anaerobes can tolerate small amounts of oxygen
Correct Answer: B
Rationale: Facultative anaerobes can switch between aerobic and anaerobic metabolism based on oxygen availability. This flexibility enables them to grow in both oxygen-rich and oxygen-poor environments, making choice B correct. Choice A is incorrect because facultative anaerobes do not require oxygen for survival. Choice C is incorrect as obligate anaerobes cannot grow in the presence of oxygen. Choice D is incorrect because obligate anaerobes cannot tolerate any amount of oxygen.
In a bacteriological laboratory some bacterial smears had to be stained by Gram's method. For this purpose the following reagents were prepared: gentian violet, Lugol's solution, aqueous fuchsin solution. What other reagent is required?
- A. 96% ethanol
- B. 5% sulfuric acid
- C. Methylene blue solution
- D. Carbolic fuchsin
Correct Answer: A
Rationale: The correct answer is A: 96% ethanol. In Gram's staining method, after applying gentian violet, Lugol's solution, and aqueous fuchsin, the next step is to use a decolorizing agent like 96% ethanol to wash away the excess stain from the Gram-negative bacteria. This step is crucial as it helps differentiate between Gram-positive and Gram-negative bacteria based on their cell wall properties. The other choices are incorrect because sulfuric acid is not used in Gram's staining method, methylene blue is typically used in other staining techniques like the simple stain, and carbolic fuchsin is not a standard reagent in the Gram's staining process.