Nokea
A sample stained using Romanovsky-Giemsa method revealed crescent-shaped protozoa with pointed ends and blue cytoplasm. What is the causative agent?
- A. Toxoplasma
- B. Plasmodium
- C. Leishmania
- D. Trypanosoma
Correct Answer: A
Rationale: The correct answer is A: Toxoplasma. The crescent-shaped protozoa with pointed ends and blue cytoplasm described in the question are characteristic of Toxoplasma gondii, the causative agent of toxoplasmosis. Toxoplasma staining with Romanovsky-Giemsa method typically shows these specific morphological features. Plasmodium causes malaria and does not exhibit the described morphology. Leishmania and Trypanosoma are both kinetoplastid parasites, but they do not typically present as crescent-shaped protozoa with blue cytoplasm. Therefore, based on the specific characteristics described in the question, the correct answer is Toxoplasma.
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The sepsis is:
- A. Presence of bacteria in the blood
- B. Bacteremia, accompanied with chemical and laboratory data for systemic information
- C. Massive proliferation of toxins in the blood with organ failure
- D. All of the above
Correct Answer: D
Rationale: The correct answer is D. Sepsis is a systemic response to infection resulting in organ dysfunction.
A: Sepsis can be caused by bacterial infection, but it is not solely defined by the presence of bacteria in the blood.
B: Bacteremia is the presence of bacteria in the blood, which can be a component of sepsis, but sepsis involves a broader systemic response beyond just bacteremia.
C: Sepsis can lead to the release of toxins and organ failure, but sepsis itself is not defined solely by the massive proliferation of toxins.
Therefore, the correct answer is D, as sepsis can involve the presence of bacteria, bacteremia, systemic manifestations, and organ failure.
The incubation period for hepatitis B is:
- A. One week - 10 days
- B. One month - 45 days
- C. Up to 3 months
- D. Up to 6 months
Correct Answer: C
Rationale: The correct answer is C: Up to 3 months. Hepatitis B has an average incubation period of 60-90 days, which aligns with the timeframe of up to 3 months. This period refers to the time between exposure to the virus and the onset of symptoms. One week to 10 days (choice A) is too short for hepatitis B, as it typically takes longer to manifest. One month to 45 days (choice B) is also shorter than the average incubation period for hepatitis B. Up to 6 months (choice D) is too long for hepatitis B, as the virus typically shows symptoms within 3 months.
The lipophilization is:
- A. Dehydration of the cell under deep vacuum in frozen state (-20.-30 degrees)
- B. Hydration of frozen cells
- C. Enrichment of the cells with proteins under deep vacuum
- D. Removing the protein content of the cell under deep vacuum
Correct Answer: A
Rationale: The correct answer is A because lipophilization is the process of removing water from cells under deep vacuum in a frozen state. Dehydration of the cell at a low temperature helps to preserve the cellular structure and components. Choice B is incorrect as hydration does the opposite of what lipophilization aims to achieve. Choice C is incorrect as lipophilization does not involve enriching cells with proteins. Choice D is incorrect as lipophilization does not involve removing protein content from cells.
The most common test for the diagnosis of Lyme borreliosis is:
- A. Widal agglutination test
- B. Cultivation on blood and chocolate agar
- C. ELISA, Western blot
- D. Ascoli thermoprecipitation test
Correct Answer: C
Rationale: The correct answer is C: ELISA, Western blot. These tests are commonly used for diagnosing Lyme borreliosis due to their high sensitivity and specificity in detecting antibodies against Borrelia burgdorferi, the causative agent. ELISA is the initial screening test, while Western blot is used for confirmation. Widal agglutination test (A) is used for diagnosing typhoid fever, not Lyme disease. Cultivation on blood and chocolate agar (B) is not commonly used for diagnosing Lyme borreliosis as Borrelia species are difficult to culture. Ascoli thermoprecipitation test (D) is used for diagnosing pneumonia caused by Streptococcus pneumoniae, not Lyme disease.
Which of the following bacteria is responsible for causing the disease known as syphilis?
- A. Treponema pallidum
- B. Neisseria gonorrhoeae
- C. Chlamydia trachomatis
- D. Escherichia coli
Correct Answer: A
Rationale: The correct answer is A: Treponema pallidum. This bacterium is responsible for causing syphilis due to its unique spiral shape and ability to penetrate mucous membranes. It is transmitted through sexual contact or from mother to child during childbirth. Neisseria gonorrhoeae causes gonorrhea, Chlamydia trachomatis causes chlamydia, and Escherichia coli is a common bacterium found in the gut. Therefore, they are not responsible for syphilis.