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BIOD 171 Module 5 Exam (2 Versions, Latest-2023)/ BIOD171 Module 5 Exam / BIOD 171 Microbiology Module 5 Exam: Essential Microbiology W/ Lab: Portage Learning |100% Verified and Correct Q & A| $19.49   Add to cart

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BIOD 171 Module 5 Exam (2 Versions, Latest-2023)/ BIOD171 Module 5 Exam / BIOD 171 Microbiology Module 5 Exam: Essential Microbiology W/ Lab: Portage Learning |100% Verified and Correct Q & A|

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BIOD 171 Module 5 Exam (2 Versions, Latest-2023)/ BIOD171 Module 5 Exam / BIOD 171 Microbiology Module 5 Exam: Essential Microbiology W/ Lab: Portage Learning |100% Verified and Correct Q & A|

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  • June 30, 2023
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BIOD 171 Module 5 Exam

, 4. The presence of any chemical reactions
5. Changes in color localized to the organism or the surrounding
media
6. Capture (or draw) images of any of the characteristics described
above


Question 3
pts
While observing an unknown sample of limited amounts, a researcher must determine
the following observations: (1) the presence of any motility and (2) its Gram status using
the same sample—the liquid sample cannot be divided. Which would you determine first
and why?
Your Answer:
You would need to determine the motility before gram status. You can observe motility
on a wet mount and then heat fix your sample to gram stain your sample. If you were to
begin with using gram staining, you would kill your sample with heat fixing, you would
not be able to observe any motility.
You must determine motility before determining the Gram status. Motility requires
a wet mount, while Gram staining requires heat fixing the sample. If one were to
begin with the Gram stain the heat fixation process would kill the organism,
making any observations regarding motility impossible. The correct approach
would be to place the liquid culture on a glass slide and determine its motility
status. Next, the same liquid culture can be heat fixed and Gram stained.


Question 4
pts
A facultative anaerobe is a microorganism capable of growth under what conditions?
Your Answer:
aerobic and anaerobic conditions
A facultative anaerobe is capable of growth under aerobic (with oxygen) and
anaerobic (without oxygen) conditions.




Question 5
pts
As Streptococcus is catalase negative would it thrive or die in the presence of
peroxides? Why?

,Your Answer:
Streptococcus would thrive in the presence of peroxides because it does not bubble
when it comes in contact. Streptococcus is gram negative, anaerobic.
Catalase negative=no bubble formation
catalase positive =bubble formation
Streptococcus would not survive in the presence of peroxides—it is unable to
breakdown peroxides (catalase negative). Left unchecked, peroxides would
damage the cellular integrity of Strep causing lysis/cell death.


Question 6
pts
Streptococcus is most often streaked onto:


Chocolate agar



EMB agar

Correct!

Blood agar

Strep is often cultured on Blood agar plates to determine its hemolytic properties, which aids in
the classification (and differentiation) process.


Spirit Blue agar



Question 7
pts
True or False. The Lancefield groups are used to subdivide antigenic groups of gamma-
hemolytic Streptococcus.


True

Correct!

, False

The Lancefield groupings are used to subdivide beta-hemolytic Strep.


Question 8
pts
The distinctions for Lancefield subgroupings lie in its: (select all that apply)


Hemolytic activity



Catalase activity

Correct!

Carbohydrate composition of antigens

Carbohydrate composition of antigens present in the cell wall determines the Lancefield
groupings (A, B, C, etc). Note, ALL strep under Lancefield groupings are (by definition) catalase
negative and beta-hemolytic. Thus, answers A and C cannot be used to subdivide streptococcal
species.


All of the above



Question 9
pts
Left untreated, strep throat can progress to ______ , which displays _____ hemolytic
activity.


Strep. Septicemia; gamma



Strep. Pharyngitis; alpha

Correct!

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