Nokea
Choose which one fo the following microorganisms is an example of obligate parasitism
- A. HIV
- B. E. Coli
- C. S. Epidermidis
- D. Pseudomonas aeruginosa
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.
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Rocky Mountain spotted fever is caused by:
- A. Borrelia burgdorferi
- B. Rickettsia rickettsii
- C. Rickettsia prowazekii
- D. Ehrlichia chaffeensis
Correct Answer: B
Rationale: The correct answer is B: Rickettsia rickettsii. Rocky Mountain spotted fever is caused by Rickettsia rickettsii, a bacterium transmitted by ticks. This pathogen primarily affects the endothelial cells of blood vessels, leading to vasculitis. Borrelia burgdorferi (choice A) causes Lyme disease, not Rocky Mountain spotted fever. Rickettsia prowazekii (choice C) causes epidemic typhus, and Ehrlichia chaffeensis (choice D) causes human monocytic ehrlichiosis, both of which are distinct from Rocky Mountain spotted fever. Therefore, the correct choice is B based on the specific etiology of the disease.
The specific antidote to botulinum toxin is:
- A. sodium bicarbonate
- B. antitoxin
- C. pralidoxime
- D. naloxone
Correct Answer: B
Rationale: The correct answer is B: antitoxin. Antitoxin is the specific antidote to botulinum toxin as it helps neutralize the toxin in the body. Sodium bicarbonate (A) is used to treat acidosis, pralidoxime (C) is used for organophosphate poisoning, and naloxone (D) is used for opioid overdose. Antitoxin directly targets and counteracts the botulinum toxin, making it the appropriate antidote in cases of botulism.
What type of immune reaction is most common in identification of vibrio cholerae
- A. Agglutination
- B. Precipitation
- C. Western blot
- D. ELISA
Correct Answer: A
Rationale: The correct answer is A: Agglutination. Agglutination is commonly used to identify Vibrio cholerae by clumping together the bacteria with specific antibodies. This reaction occurs when antibodies bind to antigens on the bacterial surface, leading to visible clumps. In contrast, Precipitation, Western blot, and ELISA are not typically used to identify Vibrio cholerae. Precipitation involves antigen-antibody complexes becoming insoluble, Western blot is used for protein detection, and ELISA is commonly used for detecting specific antigens or antibodies in a sample. Agglutination is the most appropriate choice for identifying Vibrio cholerae due to its specific interaction with the bacteria's surface antigens.
A wound infection culture revealed Gram-positive cocci in clusters. The bacteria were catalase-positive and coagulase-positive. What is the most likely causative agent?
- A. Staphylococcus aureus
- B. Staphylococcus epidermidis
- C. Streptococcus pyogenes
- D. Enterococcus faecalis
Correct Answer: A
Rationale: The correct answer is A: Staphylococcus aureus. This bacterium is characterized by Gram-positive cocci in clusters, catalase-positive, and coagulase-positive. Staphylococcus aureus is a common pathogen causing wound infections due to its ability to produce toxins and enzymes. Staphylococcus epidermidis (B) is catalase-positive but coagulase-negative, making it less likely to cause infections. Streptococcus pyogenes (C) is catalase-negative and typically causes streptococcal infections, not wound infections. Enterococcus faecalis (D) is catalase-negative and typically associated with urinary tract infections. Thus, Staphylococcus aureus is the most likely causative agent based on the given characteristics.
Which of the following media is often used to grow Mycobacterium tuberculosis?
- A. Middlebrook’s medium
- B. Heart infusion broth
- C. MacConkey agar
- D. Middlebrook’s medium
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.