microbiology an introduction 12th edition test bank

Correct Answer: B

Rationale: The correct answer is B: Klebsiella pneumoniae. This bacterium is gram-negative, capsulated, and forms lactose-positive colonies. It was first isolated in pneumonia but is more commonly associated with urinary infections, otitis, meningitides, sepsis, and nosocomial infections. Klebsiella pneumoniae fits all the characteristics mentioned in the question, making it the correct choice. Incorrect choices: A: Staphylococcus aureus is gram-positive and does not have a capsule. C: Streptococcus pneumoniae is also gram-positive and does not fit the description given. D: Mycobacterium tuberculosis is an acid-fast bacterium associated with tuberculosis, but it is neither gram-negative nor lactose-positive.

Correct Answer: B

Rationale: Facultative anaerobes can switch between aerobic and anaerobic metabolism based on oxygen availability. This flexibility enables them to grow in both oxygen-rich and oxygen-poor environments, making choice B correct. Choice A is incorrect because facultative anaerobes do not require oxygen for survival. Choice C is incorrect as obligate anaerobes cannot grow in the presence of oxygen. Choice D is incorrect because obligate anaerobes cannot tolerate any amount of oxygen.

Correct Answer: A

Rationale: The correct answer is A: 20 minutes. Escherichia coli has a short generation time, typically around 20 minutes. This means that under ideal conditions, one bacterium can divide and produce two daughter cells every 20 minutes. This rapid growth rate is a characteristic of E. coli and is important in various fields such as microbiology and biotechnology. Choice B (60 minutes) and Choice C (20 hours) are incorrect as they do not align with the typical generation time of E. coli. Choice D (24 hours) is also incorrect as it is too long for the generation time of E. coli.

Correct Answer: A

Rationale: The correct answer is A: 96% ethanol. In Gram's staining method, after applying gentian violet, Lugol's solution, and aqueous fuchsin, the next step is to use a decolorizing agent like 96% ethanol to wash away the excess stain from the Gram-negative bacteria. This step is crucial as it helps differentiate between Gram-positive and Gram-negative bacteria based on their cell wall properties. The other choices are incorrect because sulfuric acid is not used in Gram's staining method, methylene blue is typically used in other staining techniques like the simple stain, and carbolic fuchsin is not a standard reagent in the Gram's staining process.

Correct Answer: C

Rationale: The correct answer is C (Both A and B) because both Bacillus anthracis and Clostridium botulinum are capable of forming spores to survive in harsh environments. Bacillus anthracis causes anthrax and forms spores that can survive extreme conditions. Clostridium botulinum causes botulism and also forms spores for survival. Option A (Bacillus anthracis) and Option B (Clostridium botulinum) are incorrect on their own because they only represent one of the bacteria capable of forming spores. Option D (None of the above) is incorrect because it disregards the fact that both bacteria can form spores.