BIOD 171 Module 5: Microbiology & Infectious Disease | Portage Learning Study Guide | Questions, Answers & Detailed Rationales

BIOD 171 x




Module 5: Microbiology & Infectious Disease | Portage Learning
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Study Guide | Questions, Answers & Detailed Rationales
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THIS STUDY RESOURCE INCLUDES x x x




• BIOD 171 Module 5 practice questions
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• Questions and answers with detailed rationales
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• Microorganisms and infectious diseases x x x




• Bacterial pathogens and associated diseases
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• Viral infections and disease concepts
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• Fungal and parasitic infections
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• Modes and mechanisms of disease transmission
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• Signs, symptoms, and disease progression
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• Virulence and pathogenicity concepts
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• Epidemiology and infection-control principles x x x




• Prevention and treatment concepts x x x




• Host–microbe interactions x




• Scenario-based practice questions x x




• Exam-focused microbiology review x x

, BIOD 171 Module 5: Microbiology & Infectious Disease |
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Portage Learning Study Guide | Questions, Answers &
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Detailed Rationales
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1. What is the concept of universal precautions?
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• A) Only samples known to be pathogenic require special handling.
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• B) Only samples from patients with symptoms are considered hazardous.
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• C) All samples, whether known or unknown, are to be treated as
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potentially hazardous materials.
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• D) Universal precautions only apply to blood and bodily fluids.
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Answer: C) All samples, whether known or unknown, are to be treated
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x as potentially hazardous materials.
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Rationale: Universal precautions dictate that all samples must be treated as
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potentially pathogenic to ensure maximum safety in the laboratory.
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2. While observing an unknown sample of limited amounts, which
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x observation would a researcher determine first, motility or Gram
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x status?

• A) Gram status, because it is a more important identifier.
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• B) Motility, because heat fixing for the Gram stain would kill the organism.
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• C) Either can be done first, as they are independent of each other.
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• D) Gram status, because motility can be determined from the stained
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slide.
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, Answer: B) Motility, because heat fixing for the Gram stain would kill
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x the organism.
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Rationale: Motility requires a wet mount of live cells. Heat fixation for the
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Gram stain kills the organism, making motility observations impossible.
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3. A facultative anaerobe is a microorganism capable of growth under
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x what conditions?
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• A) Only in the absence of oxygen.
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• B) Only in the presence of oxygen.
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• C) In both the presence and absence of oxygen.
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• D) Only in the presence of carbon dioxide.
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Answer: C) In both the presence and absence of oxygen.
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Rationale: A facultative anaerobe can grow in aerobic (with oxygen) and
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anaerobic (without oxygen) conditions.
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4. The Lancefield groups are used to subdivide which type of
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x Streptococcus?

• A) Alpha-hemolytic Streptococcus
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• B) Beta-hemolytic Streptococcus
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• C) Gamma-hemolytic Streptococcus
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• D) All hemolytic groups of Streptococcus
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Answer: B) Beta-hemolytic Streptococcus
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Rationale: The Lancefield groups are specifically used to subdivide beta-
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hemolytic streptococci based on carbohydrate antigens.
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