Several patients with similar complaints came to the doctor. They all present with weakness, pain in the intestines, indigestion. Feces analysis revealed the need for urgent hospitalization of the patient, who had microbial cysts with four nuclei detected in his samples. Such cysts are characteristic of the following protozoon:

Correct Answer: A

Rationale: The correct answer is A: Entamoeba histolytica. This protozoon is known to cause amoebic dysentery, presenting symptoms of weakness, intestinal pain, and indigestion. The presence of microbial cysts with four nuclei in fecal samples is characteristic of Entamoeba histolytica. These cysts are the infective form of the parasite and can cause severe invasive disease in the intestines. Incorrect choices: B: Entamoeba coli does not typically cause invasive disease or severe symptoms like those described in the question. C: Balantidium is a ciliated protozoan that causes balantidiasis, which presents differently from the symptoms described in the question. D: Trichomonad is a flagellated protozoan that typically infects the urogenital tract, not the intestine, and does not form cysts with four nuclei. In summary, the key clues in the question were the symptoms of weakness, intestinal pain,

Correct Answer: D

Rationale: The correct answer is D: Gram-negative coccobacillus. Francisella tularensis is a Gram-negative coccobacillus because it has a unique shape that is between a coccus (spherical) and a bacillus (rod-shaped) and it stains pink/red in the Gram staining process. It is important to note that it is not a coccus (choice B) or a Gram-positive bacterium (choice A). Choice C is too general and does not specify the coccobacillus shape of F. tularensis. So, the correct answer is D due to its specific characteristics as a Gram-negative coccobacillus.

Correct Answer: A

Rationale: The correct answer is A: Bacterial ribosomes synthesize proteins. Ribosomes are cellular organelles responsible for protein synthesis by translating mRNA into amino acids. They are essential for the production of proteins needed for cell function. Choices B, C, and D are incorrect because ribosomes do not synthesize DNA, degrade toxic substances, or assist in cell division. DNA synthesis is carried out by DNA polymerase, degradation of toxic substances is typically done by enzymes, and cell division is primarily regulated by the cell cycle machinery.

Correct Answer: A

Rationale: The correct answer is A: The lysozyme hydrolyzes the glycoside bonds between the N-acetylmuramic acid and the N-acetylglucosamine in the peptidoglycan of the cell wall. This is because lysozyme is an enzyme that specifically targets the peptidoglycan layer of bacterial cell walls, breaking down the structure by cleaving the glycosidic bonds between these two specific sugar molecules. This action weakens the cell wall, leading to bacterial cell lysis. The other choices are incorrect because lysozyme does not target phosphate bonds in phospholipids (B), disulfide bonds in immunoglobulins (C), or hydrogen bonds in bacterial capsules (D).

Correct Answer: A

Rationale: The correct answer is A: Icosahedral structures. Icosahedral symmetry is commonly found in viruses due to its efficient way of packing genetic material. The icosahedron has 20 equilateral triangular faces, making it ideal for viral capsids. The other choices, B: Hexahedral, C: Tetrahedral, and D: Dodecahedral structures, do not accurately describe the cubic type of symmetry seen in viruses. Hexahedral refers to a six-faced polyhedron, tetrahedral to a four-faced polyhedron, and dodecahedral to a twelve-faced polyhedron, none of which align with the structure commonly observed in viruses.

Correct Answer: D

Rationale: The correct answer is D: Pulmonary tuberculosis. The presence of red bacilli stained with Ziehl-Nielsen method indicates acid-fast bacteria, which are characteristic of Mycobacterium tuberculosis, the causative agent of tuberculosis. The size and appearance of the bacilli described match those of M. tuberculosis. Chronic pneumonia, along with the specific staining pattern, strongly suggests tuberculosis as the likely diagnosis. Explanation for other choices: A: Pneumococcal pneumonia is caused by Streptococcus pneumoniae, which would not show red bacilli on Ziehl-Nielsen staining. B: Pulmonary actinomycosis is caused by Actinomyces species, not acid-fast bacteria like M. tuberculosis. C: Pulmonary candidiasis is caused by Candida species, which are fungi and do not appear as red bacilli on staining.

Correct Answer: C

Rationale: The correct answer is C because both Mycobacterium bovis and Mycobacterium tuberculosis are bacteria that can cause tuberculosis in humans. Mycobacterium bovis primarily infects animals but can be transmitted to humans, while Mycobacterium tuberculosis is the main causative agent of tuberculosis in humans. Therefore, both choices A and B are correct. Choice D is incorrect because we know that Mycobacterium bovis and Mycobacterium tuberculosis are causative agents of tuberculosis in humans.

Correct Answer: A

Rationale: The correct answer is A: Phaeophyta. Brown algae are commonly referred to as Phaeophyta because they contain brown pigments like fucoxanthin. Rhodophyta (B) refers to red algae, Chrysophyta (C) to golden algae, and D repeats the same term as A. Therefore, A is the correct choice based on the distinctive pigments present in brown algae.

Correct Answer: A

Rationale: The correct answer is A: Wasserman test, IFT. The patient's symptoms of diffuse alopecia and Pinkus' sign are suggestive of secondary syphilis. The Wasserman test and Immunofluorescence Test (IFT) are used to detect antibodies against Treponema pallidum, the bacterium causing syphilis. This would confirm or rule out syphilis as the underlying cause of the patient's symptoms. Choice B (T.pallidum Immobilization Test) is similar to the Wasserman test but less commonly used. Choice C (Detection of the nasal mucous membrane for M.leprae) is irrelevant as leprosy does not present with the described symptoms. Choice D (Consultation of neuropathist) is premature without ruling out syphilis first. Thus, the Wasserman test, IFT should be the initial examination to investigate the likely cause of the patient's symptoms.