Nokea
Humoral factors of the natural resistance are:
- A. Macrophages and macrophages
- B. Lysozyme, lactoferrin, interferon, complement
- C. B and T lymphocytes, NK cells
- D. Immunoglobulins A, G and M
Correct Answer: B
Rationale: The correct answer is B because lysozyme, lactoferrin, interferon, and complement are all humoral factors involved in natural resistance. Lysozyme and lactoferrin are antimicrobial proteins that help defend against pathogens. Interferon is a signaling protein that inhibits viral replication, while complement is a group of proteins that enhance the immune response by promoting inflammation and opsonization.
Choices A, C, and D are incorrect:
- Choice A mentions only macrophages, which are not humoral factors but rather phagocytic cells.
- Choice C lists immune cells involved in adaptive immunity (B and T lymphocytes, NK cells) rather than humoral factors of natural resistance.
- Choice D includes immunoglobulins, which are part of adaptive immunity, not natural resistance.
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Which of the following stains is generally used for the diagnosis of tuberculosis?
- A. Acid-fast stain
- B. Gram stain
- C. Negative stain
- D. Simple stain
Correct Answer: A
Rationale: The correct answer is A: Acid-fast stain. This stain is used for diagnosing tuberculosis due to the unique cell wall composition of Mycobacterium tuberculosis, which resists decolorization by acid-alcohol. This allows the bacteria to retain the primary stain (carbol fuchsin) and appear red under a microscope. Gram stain (B) is not suitable for Mycobacterium tuberculosis as it is gram-positive but does not retain the crystal violet-iodine complex. Negative stain (C) is not used for bacterial identification. Simple stain (D) does not differentiate acid-fast bacteria like Mycobacterium tuberculosis.
During fibergastroscopy a patient with ulcer disease of the stomach, the mucosal biopsy is taken from the area of an ulcer. Impression smear is prepared from biopsy material and stained by Gram method; the rest of biopsy material is tested for urease activity. Microscopy of the impression smear revealed gram- negative spiral-shaped microorganisms, urease activity test is positive. What bacteria were detected?
- A. Helicobacter pylori
- B. Campilobacter jejuni
- C. Treponema pallidum
- D. Spirilla minor
Correct Answer: A
Rationale: The correct answer is A: Helicobacter pylori.
Rationale:
1. Gram-negative spiral-shaped microorganisms match the morphology of Helicobacter pylori.
2. Positive urease activity is a characteristic feature of Helicjsonacter pylori.
3. Helicobacter pylori is specifically associated with gastric ulcers and is a common pathogen in the gastrointestinal tract.
Summary:
Choice B, Campilobacter jejuni, is a gram-negative curved bacterium but is not typically associated with gastric ulcers or urease activity. Choice C, Treponema pallidum, is a spirochete bacterium causing syphilis, which is not related to gastric ulcers. Choice D, Spirilla minor, is a generic term for spiral-shaped bacteria and is not a specific pathogen associated with gastric ulcers or urease activity.
A patient suffers from pulmonary tuberculosis. During treatment neuritis of visual nerve arose. What drug has caused this by-effect?
- A. Isoniazid
- B. Ethambutol
- C. Kanamycin
- D. Rifampicin
Correct Answer: B
Rationale: The correct answer is B: Ethambutol. Ethambutol is known to cause optic neuritis, affecting the visual nerve. It is a well-documented side effect of this drug. Isoniazid (Choice A) is associated with peripheral neuropathy, not optic neuritis. Kanamycin (Choice C) is known for its side effects on the inner ear, causing hearing loss. Rifampicin (Choice D) is not typically associated with optic neuritis but can cause other adverse effects like hepatitis. By process of elimination and knowledge of drug side effects, Ethambutol is the most likely culprit.
Conidiospores that bud off from a vase-shaped spore-bearing cell are referred to as:
- A. arthrospores
- B. porospores
- C. phialospores
- D. blastospores
Correct Answer: C
Rationale: The correct answer is C: phialospores. Phialospores are conidiospores that bud off from a vase-shaped spore-bearing cell called a phialide. This structure is characteristic of fungi in the genus Penicillium. Arthrospores (A) are formed by the fragmentation of hyphae, porospores (B) are not a recognized term in mycology, and blastospores (D) are produced by budding from yeast cells. Therefore, C is the correct choice based on the specific description provided in the question.
What is the structure of the bacterial cell wall?
- A. Peptidoglycan
- B. Lipopolysaccharides
- C. Phospholipids
- D. Chitin
Correct Answer: A
Rationale: The correct answer is A: Peptidoglycan. Bacterial cell walls are composed of peptidoglycan, a unique macromolecule consisting of sugars and amino acids. This structure provides rigidity and protection to the cell. Lipopolysaccharides are found in the outer membrane of Gram-negative bacteria, not in the cell wall. Phospholipids are the main components of cell membranes, not cell walls. Chitin is a structural polysaccharide found in the cell walls of fungi and some algae, not bacteria.