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Exam Bank: Portage Learning BIOD 171 Essential Microbiology Module 5 Exam | 2025 Updated Edition with 100% Verified Answers

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BIOD 171 Module 5 Exam Bank (Latest-2025) / BIOD171 Essential Microbiology with Lab: Portage Learning - This Exam Bank features multiple complete sets of the most current exam questions, reviewed and updated to reflect the 2025 version of BIOD171 as offered through Portage Learning. With over two full exam sets, this resource offers a broader and more comprehensive preparation tool. Students can study with confidence, knowing they’re reviewing the latest question formats and lab-related content likely to appear in their module assessments.

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M5: Exam - Requires Respondus LockDown
Browser + Webcam
Due No due date Points 100 Questions 29
Time Limit 60 Minutes Requires Respondus LockDown Browser




Questions and Answers:
# Question Points Answer
1 a. Define the concept of universal 3 a. Universal precautions means any and all
precautions. samples, whether known or unknown, are to
be treated as potentially hazardous (or
b. The concept of treating all samples, pathogenic) materials.
whether known or unknown, as
potentially hazardous (or pathogenic) b. Universal precautions
materials is known as __________.

2 a. List at least 3 observations a researcher 3 a. A lab researcher would be certain to note:




ta
would be sure to note while assessing an 1. Size and shape
unknown microbial sample. 2. Any observable motility
3. Gram status (positive or negative)
4. The presence of any chemical reactions
da
5. Changes in color localized to the organism or
the
surrounding media
6. Capture (or draw) images of any of the
characteristics
on

described above

3 a. While observing an unknown sample of 10 a. You must determine motility before
limited amounts, a researcher must determining the Gram status. Motility requires
determine the following observations: (1) a wet mount, while Gram staining requires
the presence of any motility and (2) its heat fixing the sample. If one were to begin
ta


Gram status using the same sample—the with the Gram stain the heat fixation process
liquid sample cannot be divided. Which would kill the organism, making any
would you determine first and why? observations regarding motility impossible. The
correct approach would be to place the liquid
da




culture on a glass slide and determine its
motility status. Next, the same liquid culture
can be heat fixed and Gram stained.

4 a. A facultative anaerobe is a 3 a. A facultative anaerobe is capable of growth
microorganism capable of growth under under aerobic (with oxygen) and anaerobic
what conditions? (without oxygen) conditions.
5 a. As Streptococcus is catalase negative 10 a. Streptococcus would not survive in the
would it thrive or die in the presence of presence of peroxides—it is unable to
peroxides? Why? breakdown peroxides (catalase negative). Left
unchecked, peroxides would damage the
cellular integrity of Strep causing lysis/cell
death.

6 a. Streptococcus is most often streaked 2 a. Blood agar.
onto:

, M5: Exam - Requires Respondus LockDown
Browser + Webcam
Due No due date Points 100 Questions 29
Time Limit 60 Minutes Requires Respondus LockDown Browser




# Question Points Answer
- Blood agar (Strep is often cultured on Blood agar plates to
- EMB agar determine its hemolytic properties, which aids
- Chocolate agar in the classification (and differentiation)
- Spirit Blue agar process).

7 a. True or False. The Lancefield groups are 2 a. False.
used to subdivide antigenic groups of (The Lancefield groupings are used to subdivide
alpha-hemolytic Streptococcus. beta-hemolytic Strep).

b. True or False. The Lancefield groups are b. False.
used to subdivide antigenic groups of




ta
gamma-hemolytic Streptococcus.

8 a. Greater than 90% of all human 2 a. Group A.
streptococcal infections belong to: (Group A is the causative agent behind > 90%
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- Group D of all Strep infections).
- Group C
- Group B b. Carbohydrate composition of antigens
- Group A (Carbohydrate composition of antigens present
in the cell wall determines the Lancefield
on

b. The distinctions for Lancefield groupings (A, B, C, etc). Note, ALL strep under
subgroupings lie in its: (select all that Lancefield groupings are (by definition)
apply) catalase negative and beta-hemolytic. Thus,
- Carbohydrate composition of antigens answers A and C cannot be used to subdivide
- Catalase activity streptococcal species).
- All of the above
ta


- Hemolytic activity

9 a. Left untreated, strep throat can progress 2 a. Rheumatic fever; beta.
to ______ , which displays _____ Strep throat (also known as strep. Pharyngitis)
da




hemolytic activity. can lead to Rheumatic fever if left untreated.
- Rheumatic fever; beta Strep. Pharyngitis displays beta- hemolytic
- Strep. Pharyngitis; alpha activity.
- Strep. Pharyngitis; beta
- Strep. Septicemia; gamma b. Beta; Strep. Pharyngitis

b. Rheumatic fever displays ______
hemolytic activity and occurs when
_______ is left untreated.
- Beta; Rheumatic fever
- Alpha; Strep. Pharyngitis
- Beta; Strep. Pharyngitis
- Gamma; Strep. Septicemia

10 a. True or False. Under most 2 a. False.
circumstances, staphylococcus can be

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