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A patient with a deep tissue infection had a wound culture revealing Gram-positive cocci in clusters. The bacteria were catalase-positive and coagulase-negative. What is the most likely causative agent?
- A. Staphylococcus epidermidis
- B. Staphylococcus aureus
- C. Streptococcus pyogenes
- D. Enterococcus faecalis
Correct Answer: A
Rationale: The correct answer is A: Staphylococcus epidermidis. The presence of Gram-positive cocci in clusters that are catalase-positive and coagulase-negative points towards coagulase-negative staphylococci like Staphylococcus epidermidis. Staphylococcus aureus is catalase-positive and coagulase-positive. Streptococcus pyogenes is catalase-negative. Enterococcus faecalis is catalase-negative and not typically seen in clusters. Therefore, the most likely causative agent in this scenario is Staphylococcus epidermidis.
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On the 5th day after a surgery for colon injury a patient complains of bursting pain in the postoperative wound, weakness, drowsiness, headache, fever up to 40oC. Objectively: the skin around the wound is swollen, there is gas crepitation. The wound discharges are scarce foul-smelling, of dark-gray color. What is the most likely diagnosis?
- A. Anaerobic clostridial wound infection
- B. Abscess
- C. Postoperative wound infection
- D. Erysipelas
Correct Answer: A
Rationale: The correct answer is A: Anaerobic clostridial wound infection. The symptoms and signs described (bursting pain, weakness, drowsiness, fever, swollen skin, gas crepitation, foul-smelling dark-gray discharge) are classic for gas gangrene, caused by Clostridium perfringens. Gas gangrene is an anaerobic infection that thrives in devitalized tissue, such as postoperative wounds. The gas crepitation is due to gas production by the bacteria. Abscess (B) typically presents with localized swelling and fluctuance, not gas crepitation. Postoperative wound infection (C) may present with redness, warmth, and increased wound discharge but not the specific features mentioned. Erysipelas (D) is a superficial bacterial skin infection, typically caused by Streptococcus pyogenes, presenting with well-defined raised borders and redness.
Production of auto-antibodies may be due to:
- A. Emergence of mutant clones of B cells
- B. Production of antibodies against hidden tissues
- C. Genetic factors
- D. All of these
Correct Answer: D
Rationale: Production of auto-antibodies can be due to various factors. Mutant B cell clones can lead to the production of self-reactive antibodies. Antibodies against hidden tissues can also trigger autoimmune responses. Genetic factors play a crucial role in predisposing individuals to autoimmunity. Therefore, the correct answer is D, as all these factors can contribute to the production of auto-antibodies. Choice A is incorrect as it only addresses one possible cause. Choice B is incorrect as it focuses solely on antibodies against hidden tissues. Choice C is incorrect as it overlooks the other potential reasons for auto-antibody production.
Which bacterial structure allows for attachment to host cells?
- A. Flagella
- B. Fimbriae
- C. Capsule
- D. Endospore
Correct Answer: B
Rationale: The correct answer is B: Fimbriae. Fimbriae are hair-like appendages found on the surface of bacteria that enable attachment to host cells. This attachment is crucial for establishing infection. Flagella (A) are used for movement, not attachment. Capsules (C) provide protection and help evade the host immune system, but do not directly facilitate attachment. Endospores (D) are dormant structures for survival, not for attachment to host cells. Fimbriae specifically aid in adherence to host cells, making them the correct choice in this context.
A bacteriological laboratory received sputum sample of a patient suffering from tuberculosis. Bacterioscopic examination of smears and detection of tuberculosis bacillus can be realized by one of enrichment methods that involves processing of sputum only with solution of caustic soda. What is this method called?
- A. Homogenization
- B. Inactivation
- C. Flotation
- D. Filtration
Correct Answer: B
Rationale: The correct answer is B: Inactivation. In this method, the sputum sample is processed with a solution of caustic soda, which helps to kill any contaminants present and preserve the tuberculosis bacillus for microscopic examination. This step is crucial in ensuring accurate detection of the bacillus.
A: Homogenization involves breaking down the sample to make it uniform, but it does not necessarily involve using caustic soda for inactivation.
C: Flotation is a method used to separate different components based on their density and is not specific to inactivating tuberculosis bacillus.
D: Filtration involves passing a sample through a filter to separate particles based on size and is not related to inactivating the tuberculosis bacillus.
Chlorine belongs to which of the following chemical groups?
- A. Halogens
- B. Heavy metals
- C. Phenols
- D. Alcohols
Correct Answer: A
Rationale: The correct answer is A: Halogens. Chlorine belongs to the halogens group because it is a nonmetal element in group 17 of the periodic table. Halogens are highly reactive elements that have 7 electrons in their outer shell, making them eager to gain an electron to achieve stability. Chlorine specifically is a diatomic molecule with the chemical symbol Cl and atomic number 17.
Option B: Heavy metals are elements with high atomic weights and densities, such as lead and mercury, which are not related to chlorine.
Option C: Phenols are a class of organic compounds containing a hydroxyl group bonded to an aromatic ring, not related to chlorine.
Option D: Alcohols are organic compounds characterized by the presence of a hydroxyl group (-OH), also not related to chlorine.