After long-term antibiotic treatment a patient has developed whitish spots on the oral mucosa. Gram- positive oval budding cells were detected in the smear preparations. What causative agents were detected?

Correct Answer: A

Rationale: The correct answer is A: Candida fungi. Candida fungi are known to cause oral thrush, characterized by whitish spots on the oral mucosa. The presence of gram-positive oval budding cells in the smear preparations is indicative of Candida species. Staphylococci (choice B) are gram-positive cocci, not oval budding cells. Sarcinae (choice C) are gram-positive bacteria that form cuboidal packets, not oval budding cells. Actinomycetes (choice D) are filamentous gram-positive bacteria, not oval budding cells like Candida fungi.

Correct Answer: D

Rationale: The correct answer is D: Structure of cell wall. When studying a bacterial smear stained by the Ozheshko method, the focus is on the cell wall structure. This staining method highlights the cell wall morphology, allowing for detailed examination. Flagella, spores, and capsules are not typically visualized using this staining method. Flagella require special staining techniques, spores are usually observed through specific staining methods like Schaeffer-Fulton, and capsules are not easily visualized with the Ozheshko staining method. Therefore, the structure of the cell wall is the most relevant feature to study in this context.

Correct Answer: A

Rationale: The correct answer is A: Isoniazid. Isoniazid is a first-line drug for treating active pulmonary tuberculosis due to its high efficacy and low toxicity. It works by inhibiting the synthesis of mycolic acids in the bacterial cell wall. Sulfalen, Cyclocerine, and Ethionamide are not first-line drugs for tuberculosis. Sulfalen is used for urinary tract infections, Cyclocerine for drug-resistant tuberculosis, and Ethionamide as a second-line drug for multidrug-resistant tuberculosis. Therefore, Isoniazid is the most appropriate initial choice for treating active focal pulmonary tuberculosis.

Correct Answer: C

Rationale: The correct answer is C: Negative stain. Negative staining uses acidic dyes which are repelled by the bacterial cell wall components, highlighting bacteria with a large amount of peptidoglycan. Acid-fast stain (A) is used to identify Mycobacterium species with waxy cell walls. Gram stain (B) differentiates bacteria into Gram-positive and Gram-negative based on cell wall composition. Option D is a duplicate of option C and is incorrect.

Correct Answer: D

Rationale: The correct answer is D: BSL4. Ebola viruses are highly infectious and pose a high risk to laboratory workers. BSL4 facilities are designed to handle agents that pose a severe threat to human health and have no available vaccines or treatments. These facilities have strict guidelines for handling infectious agents, including full-body suits, negative air pressure, and decontamination processes to prevent any possible exposure. BSL1, BSL2, and BSL3 are not adequate for handling Ebola viruses due to their lower levels of containment and safety measures.

Correct Answer: D

Rationale: The correct answer is D, Salmonella typhi, because it is a motile bacterium due to possessing flagella that aid in movement. Bacillus anthracis (A) is non-motile, Klebsiella pneumoniae (B) lacks flagella for motility, and Shigella flexneri (C) is non-motile as well. Salmonella typhi (D) is known for its flagella-driven motility, allowing it to move and spread within its host and environment. Flagella are essential for bacterial motility, enabling them to navigate and colonize different environments efficiently.

Correct Answer: D

Rationale: The correct answer is D: P.aeruginosa. This bacterium is commonly found in water and soil, making its presence in indoor air an indicator of poor sanitation. Step 1: P.aeruginosa is an opportunistic pathogen known to cause infections in hospitals. Step 2: Its presence in the air indicates potential contamination and poor air quality. Step 3: Regular monitoring of P.aeruginosa levels can help assess the effectiveness of sanitation measures. Summary: A, B, and C are not suitable indicators of air sanitation in an enclosed space as they are not typically associated with environmental contamination or poor air quality.

Correct Answer: A

Rationale: The correct answer is A: TRUE. The structures and organs listed are indeed part of the immune system. The lymphatic system transports lymph, lymphoid tissue contains immune cells, lymph nodes filter lymph, thymus matures T-cells, and spleen filters blood and produces immune cells. This comprehensive network is crucial for defending the body against pathogens. Choice B is incorrect as all listed structures are components of the immune system. Choice C is also incorrect as the statement is universally true, not situational. Choice D is incorrect as the correct answer is A.

Correct Answer: D

Rationale: The correct answer is D: Binary fission. Binary fission is a method of bacterial cell division, not a method of acquiring genetic material. Transformation, conjugation, and transduction are mechanisms through which bacteria can acquire genetic material from other sources. Transformation involves the uptake of naked DNA from the environment, conjugation is the transfer of genetic material through direct cell-to-cell contact, and transduction is the transfer of genetic material via bacteriophages. Therefore, binary fission is the only option that does not involve the acquisition of new genetic material.