Researchers of a bacteriological laboratory examine tinned meat for botulinic toxin. For this purpose a group of mice was injected with an extract of the material under examination and antitoxic antibotulinic serum of A, B, E types. A control group of mice was injected with the same extract but without antibotulinic serum. What serological reaction was applied?

Correct Answer: A

Rationale: The correct answer is A: Neutralization. In this experiment, the researchers used antitoxic antibotulinic serum of different types to neutralize the botulinic toxin in tinned meat extract injected into mice. Neutralization is the process of rendering a toxin harmless by combining it with specific antibodies. The purpose of injecting the antitoxic serum was to neutralize the toxin and prevent it from causing harm to the mice. This serological reaction of neutralization directly aligns with the objective of the experiment. Summary of other choices: B: Precipitation - Precipitation involves the formation of a visible insoluble complex when an antigen reacts with its corresponding antibody. This process was not described in the scenario. C: Complement binding - Complement binding involves the activation of the complement system by antigen-antibody complexes. This was not the main focus of the experiment described. D: Double immune diffusion - This technique is used to detect the presence of specific antigens or antibodies

Correct Answer: A

Rationale: The correct answer is A: Staphylococcus epidermidis. This is supported by the characteristics of Gram-positive cocci in clusters, catalase-positive, and coagulase-negative. Staphylococcus epidermidis fits all these criteria, as it is a common skin commensal that can cause infections in immunocompromised individuals. Staphylococcus aureus (choice B) is catalase-positive and coagulase-positive. Streptococcus pyogenes (choice C) is catalase-negative. Enterococcus faecalis (choice D) is catalase-negative and not typically seen in clusters. Therefore, based on the given information, Staphylococcus epidermidis is the most likely causative agent.

Correct Answer: C

Rationale: Step-by-step rationale: 1. Shigella is a nonmotile bacterium. 2. This lack of motility is a key characteristic used to differentiate it from other enteric bacteria. 3. Nonmotile nature of Shigella helps it to cause infections by invading the intestinal epithelial cells. 4. Therefore, choice C is correct as Shigella is indeed nonmotile. Summary: A: Incorrect - Shigella does not produce hydrogen sulfide. B: Incorrect - Shigella does not ferment lactose. D: Incorrect - Shigella is not mannitol positive.

Correct Answer: A

Rationale: The correct answer is A: Trichomonas vaginalis. Trichomonas vaginalis is a flagellated protozoan parasite commonly found in vaginal infections. The presence of flagellated bacteria in the vaginal discharge stained with Romanovsky-Giemsa method indicates a protozoan infection, ruling out bacterial or viral causes. Choices B, C, and D are not correct as Leishmania donovani causes leishmaniasis, Trypanosoma gambiense causes African sleeping sickness, and Trihomonas hominis is a non-pathogenic flagellate found in the human intestine, not typically associated with vaginal infections.

Correct Answer: D

Rationale: The correct answer is D: Middlebrook's medium. This medium is specifically designed for the growth of Mycobacterium tuberculosis due to its low concentration of malachite green, which inhibits the growth of other bacteria but not Mycobacterium species. Middlebrook's medium also contains various nutrients that support the growth of Mycobacterium tuberculosis. A: Middlebrook's medium is the correct answer, not incorrect. B: Heart infusion broth is a general-purpose medium, not specific for Mycobacterium tuberculosis. C: MacConkey agar is selective for gram-negative bacteria and would not support the growth of Mycobacterium tuberculosis. In summary, Middlebrook's medium is the ideal choice for cultivating Mycobacterium tuberculosis due to its specific formulation that promotes the growth of this pathogen while inhibiting other bacteria.

Correct Answer: D

Rationale: The correct answer is D: Moeller. Moeller staining method is specifically designed to detect spores due to its ability to differentiate between the spore and the vegetative cell. It involves applying multiple stains such as malachite green, safranin, and heat to visualize the endospore structure. This method is essential for proving the formation of spores as it highlights the resistant spore structures within the bacterial cell. A: Klett and Gins is not used for spore staining. B: Ziehl-Neelsen is used for acid-fast staining, not spore staining. C: Gram and Loeffler's are used for gram staining, not spore staining.

Correct Answer: B

Rationale: Rationale: 1. Enteropathogenic E.coli (EPEC) adheres to the intestinal wall using bundle-forming pili. 2. EPEC produces toxins that disrupt cell structure and function, leading to diarrhea. 3. Shigella spp. invade intestinal cells, Salmonella spp. invade and multiply within cells, and Vibrio cholerae produces cholera toxin causing profuse watery diarrhea. Summary: A, C, and D choices are incorrect because they do not specifically adhere to the intestine wall and produce toxins like EPEC does.

Correct Answer: B

Rationale: Binary fission is a form of asexual reproduction where a single organism divides into two equal parts. Bacteria reproduce through binary fission, splitting into two identical daughter cells. Viruses are not considered living organisms and reproduce by infecting host cells. Fungi reproduce through spores or sexual reproduction. Helminths are parasitic worms that reproduce sexually. Therefore, the correct answer is B, bacteria, as they specifically utilize binary fission for reproduction.

Correct Answer: A

Rationale: The correct answer is A: HIV. HIV is an example of obligate parasitism because it requires a host cell to replicate and survive. It cannot live independently outside of a host organism. In contrast, choices B, C, and D are not obligate parasites. E. Coli and S. Epidermidis are examples of commensal bacteria that can live in the host without causing harm. Pseudomonas aeruginosa can be both an opportunistic pathogen and a free-living organism, not strictly dependent on a host for survival.