LAB Final

Question 1

What are the the Four Key Considerations in Sampling?

Answer 1

Statistical Selection: using proper statistical procedures when selecting samples from a food lot to ensure representativeness and reliability.
Aseptic Collection: used in sample collection to avoid contamination.
Transport and Handling: including temperature control and minimizing delays.
Sample Preparation: ensuring the sample integrity is maintained.

Question 2

What are the Procedures for Obtaining and Processing Samples?

Answer 2

Aseptic Collection and Packaging: aseptically obtain and package samples, including the correct use of original containers for pre-packaged samples.
Temperature and Timing: samples must be delivered at the optimal temperature and analyzed as soon as possible, and not opening samples until ready for analysis.

Question 3

What are the Drawbacks of Blenders?

Answer 3

 Recognize that blender jars are heavy, breakable, and expensive.
 Understand issues such as potential leakage, noise, and the risk of heating the food (which may damage sensitive microorganisms).

Question 4

What are the Advantages and Disadvantages of Stomachers?

Answer 4

 Explain how a stomacher prevents heating of the sample and uses inexpensive, disposable presterilized bags that save storage space.
 Identify limitations such as the high initial cost, inability to process samples with sharp objects, non-biodegradable bag waste, and uneven distribution of microbes in fatty samples.

Question 5

Differentiate Microbial Colony Morphologies of Yeast vs. Molds on Petri Films:

Answer 5

Yeasts are typically identified by small, blue-green colonies with defined edges.
Molds usually form large, variably colored colonies with diffuse edges and central foci.

Question 6

Interpret Results from Compact Dry Plates for E. coli and Coliform Counts (2)

Answer 6

For total coliform counts, both purple and blue colonies are counted, with blue colonies specifically indicating E. coli (fecal coliform).
The Compact Dry Plates method is applied in routine analysis.

Question 7

Understand Regulatory Standards and Guidelines in Microbiological Sampling (2)

Answer 7

The role of the Health Products and Food Branch of the Public Health Agency of Canada (PHAC) and the importance of the Compendium of Analytical Methods, which serves as a foundational guide for industry sampling practices.

Question 8

What are the contamination factors and regulatory responses of the Romaine Lettuce Outbreaks?

Answer 8

Contamination Factors: water runoff, wind-blown material, animal intrusion
Regulatory Responses: temporary import requirements and sampling protocols (both finished-product and pre-harvest options) imposed to enhance food safety.

Question 9

Explain the process of Case 1: Only One Dilution Yields Acceptable Counts (25–250 Colonies)

Answer 9

That single dilution is considered statistically reliable.
o Use only the data from the dilution that meets the criteria. For example, if the 10⁻² dilution yields 88 and 99 colonies on duplicate plates, calculate the average (i.e., 93.5) and then multiply by the dilution factor (10²) to obtain the MPN value per gram.
 Average count = (88 + 99) / 2 = 93.5
 Multiply by the dilution factor (10² = 100)
 Result: 93.5 × 100 = 9,350 CFU/g

Question 10

Explain the process of Case 2: Both Dilutions Yield Acceptable Counts

Answer 10

You have two sets of data for potential averaging.
o If the Count Ratio is < 2: Average the duplicate counts from both dilutions to obtain a single, combined average. Multiply this average by the appropriate dilution factor (often that of the lower dilution) to calculate the MPN
o If the Count Ratio is > 2: use the count from the dilution with the lower factor (which is expected to be more reliable) for the final MPN calculation.

Question 11

Go through an example of Case 2, when the Count Ratio > 2.
Dilution 10⁻² (duplicate plates): 150 and 155 colonies
Dilution 10⁻³ (duplicate plates): 30 and 35 colonies

Answer 11

 Dilution 10⁻² (duplicate plates): 150 and 155 colonies → Average = 152.5
 Dilution 10⁻³ (duplicate plates): 30 and 35 colonies → Average = 32.5
 Count ratio = 152.5 / 32.5 ≈ 4.7 (which is >2)
 Approach: Use the lower dilution count (i.e., the more diluted sample with the lower colony count)
 CFU/g = 32.5 × 10³ = 32,500 CFU/g

Question 12

Go through an example of Case 2, when the Count Ratio < 2.
Dilution 10⁻² (duplicate plates): 80 and 85 colonies
Dilution 10⁻³ (duplicate plates): 45 and 50 colonies

Answer 12

 Dilution 10⁻² (duplicate plates): 80 and 85 colonies → Average = 82.5
 Dilution 10⁻³ (duplicate plates): 45 and 50 colonies → Average = 47.5
 Count ratio = 82.5 / 47.5 ≈ 1.74 (which is <2)
 Approach: Average the counts from both dilutions after adjusting for their dilution factors. One method is to compute the CFU/g for each dilution and then average those values:
 From 10⁻²: 82.5 × 100 = 8,250 CFU/g
 From 10⁻³: 47.5 × 1,000 = 47,500 CFU/g
 Average CFU/g = (8,250 + 47,500) / 2 ≈ 27,875 CFU/g

Question 13

Explain the process of Case 3: Neither Dilution Yields Acceptable Counts

Answer 13

o When both dilutions fall outside the 25–250 range (either too few or too many colonies)
o Choose the dilution that provides a count closest to the upper threshold of 250 colonies.
o A count near 250 is preferable because higher counts typically provide a better approximation of the true microbial load, assuming overcrowding is not an issue.
o Use the average from this dilution along with its dilution factor to estimate the MPN
o Example
 Dilution 10⁻²: 260 colonies (slightly above 250)
 Dilution 10⁻³: 20 colonies (below 25)
 The 10⁻² dilution is chosen because 260 is closer to the acceptable upper limit of 250.
 CFU/g = 260 × 100 = 26,000 CFU/g

Question 14

Explain the process of Case 4: Both Dilutions Yield Low Colony Counts <25 Colonies)

Answer 14

o Counts under 25 colonies in both dilutions indicate a very low microbial concentration
o Use the mean of the duplicate counts from the lowest dilution. Multiply the average by the corresponding dilution factor to derive the MPN
o Example:
 Dilution 10⁻² (duplicate plates): 10 and 12 colonies → Average = 11
 Dilution 10⁻³ (duplicate plates): 4 and 5 colonies → Average = 4.5
 Use the mean from the dilution with higher counts (10⁻²) because it is less affected by low-count variability.
 CFU/g = 11 × 100 = 1,100 CFU/g

Question 15

Explain the process of Case 5: Both Dilutions Yield No Colonies

Answer 15

o The absence of any colonies in both dilutions indicates that the microbial concentration is below the detectable limit
o Estimate the MPN as being less than one times the lowest dilution factor
o Example:
 Dilutions 10⁻² and 10⁻³: Both yield 0 colonies on duplicate plates.
 The MPN is reported as less than one times the lowest dilution factor.
 For the 10⁻² dilution: CFU/g is reported as < 1 × 100 = < 100 CFU/g ESPC

Question 16

Explain the process of Case 6: Only One Plate in One Dilution Has Acceptable Counts

Answer 16

Typically, you would use the acceptable plate’s count to compute the MPN. The calculation is then performed using the corresponding dilution factor.
o Example:
 Dilution 10⁻² (duplicate plates): One plate shows 90 colonies (acceptable range) while the second shows 20 colonies (below 25, not acceptable).
 Although one plate is below the acceptable range, the method directs that both plates be used if possible.
 Average count = (90 + 20) / 2 = 55 colonies
 CFU/g = 55 × 100 = 5,500 CFU/g

Question 17

Understand the Purpose and Applications of LST Broth

Answer 17

LST broth is specifically formulated to detect coliform organisms. It is the medium of choice during the presumptive phase of the Standard Total Coliform MPN Test, widely used for microbiological examinations of water and beverages. Its effectiveness is supported by its inclusion in the Official Methods of Analysis of AOAC International.

Question 18

Explain the Role of Lactose and Sodium Lauryl Sulfate in LST Broth

Answer 18

In LST broth, lactose serves as a fermentable carbohydrate that coliform bacteria can metabolize, producing gas as a by-product. This gas formation acts as a presumptive indicator of coliform presence. Sodium lauryl sulfate enhances the selectivity of the medium by inhibiting the growth of non-coliform organisms, ensuring that the observed gas production is primarily due to coliform activity.

Question 19

The Millipore Filtration technique uses different filter pore sizes depending on the ?

Answer 19

target organisms

Question 20

In Millipore Filtration - 0.22 µm: Used for

Answer 20

bacteria (e.g., Brevundimonas diminuta, Pseudomonas aeruginosa) and bacteriophage in air filtration.

Question 21

In Millipore Filtration - 0.45 µm: Suitable for bacteria like ?

Answer 21

Escherichia coli, Leuconostoc oenos, Pediococcus damnosus, Lactobacillus hilgardii, and Oenococcus oeni.

Question 22

In Millipore Filtration - 0.55 µm: Captures both ?

Answer 22

bacteria and yeast, such as Saccharomyces cerevisiae, Pediococcus damnosus, and Lactobacillus brevis.

Question 23

In Millipore Filtration - 0.65 µm: Primarily used for _____, for example, Saccharomyces cerevisiae.

Answer 23

yeast

Question 24

In Millipore Filtration - 1.0 µm: Effective for ?

Answer 24

protozoa like Giardia and Cryptosporidium.

Question 25

Evaluate the Advantages of the Membrane Filter Technique (7)

Answer 25

Large Sample Volumes Reduced Labor Intensity Less Glassware Required Flexibility Discrete Colony Isolation Field Usability Rapid Results

Question 26

Describe the Uses and Calculation Methods in Millipore Filtration

Answer 26

o Millipore Filtration is commonly used to test wastewater, drinking water, and beverages (provided they do not have high fiber content) as well as to detect specific pathogens. For beverage samples, the level of indicator organisms is calculated using the formula:  Indicator organisms per 100 mL= [(Average # of colonies per plate) / (mL of sample)] ×100

Question 27

Understand the Function and Application of Endo Agar Base

Answer 27

Endo Agar Base is a specialized culture medium that differentiates coliforms from other microorganisms. It is moderately selective, favoring the growth of enteric bacteria while suppressing non-target organisms. This medium is commonly used for testing water, milk, dairy products, beverages, and other food items for the presence of coliforms and related enteric bacteria.

Question 28

Differentiate Between Lactose-Fermenting and Non-Lactose-Fermenting Organisms on Endo Agar

Answer 28

On Endo Agar plates, lactose-fermenting organisms such as Escherichia coli produce a distinctive green metallic sheen due to acid production from lactose fermentation. In contrast, non-lactose-fermenting organisms form clear, colorless colonies because they do not metabolize lactose. This visual differentiation is crucial for preliminary identification of coliforms in a sample.

Question 29

Identify Typical Colonial Morphologies of Escherichia coli, Enterobacter, and Klebsiella on Endo Agar

Answer 29

o Escherichia coli: Exhibits pink to rose-red colonies with a green metallic sheen, indicative of strong lactose fermentation. o Enterobacter: Forms large, mucoid, pink colonies. o Klebsiella: Also appears as large, mucoid, pink colonies.

Question 30

Define Staphylococcal aureus Food Poisoning and Its Etiology

Answer 30

Staphylococcal aureus food poisoning occurs when foods become contaminated with toxins produced by S. aureus. These toxins are preformed in the food before consumption. Even though the bacteria may be destroyed during cooking, the toxins remain stable and can cause illness. The condition is characterized by a rapid onset of gastrointestinal symptoms.

Question 31

Describe the Role of Carriers and Food Handlers in Contamination

Answer 31

A significant portion of the population naturally harbors S. aureus on their skin or in their nasal passages. Food handlers, in particular, pose a major risk if they do not adhere to proper hand-washing protocols. When food is handled without adequate hygiene, enterotoxin-producing S. aureus from the carrier’s skin or nose can contaminate the food, leading to toxin production when conditions favor bacterial multiplication.

Question 32

Outline the Symptomatology and Time Frame of the Staph Illness

Answer 32

Symptoms of S. aureus food poisoning typically include a sudden onset of nausea, vomiting, and stomach cramps, with some individuals also experiencing diarrhea. Symptoms usually develop within 30 minutes to 8 hours after ingestion of contaminated food and generally last no longer than one day. Severe illness is uncommon, and the condition is not transmitted from person to person.

Question 33

Explain the Impact of Cooking and Toxin Stability

Answer 33

Although cooking can effectively kill S. aureus bacteria, the enterotoxins produced by these bacteria are heat stable. This means that once the toxin is present in the food, it is not destroyed by normal cooking temperatures, and consuming the toxin-contaminated food can lead to illness.

Question 34

Identify Optimal Conditions for Bacterial Replication and Toxin Production

Answer 34

S. aureus replicates most efficiently at temperatures between 35°C and 41°C, while the optimal range for enterotoxin production is slightly lower, between 34°C and 40°C. These temperature ranges underscore why foods held in the “danger zone” (between 4°C and 60°C) are particularly at risk for bacterial growth and toxin accumulation.

Question 35

Recommend Food Storage and Handling Practices to Prevent Toxin Formation

Answer 35

To prevent the proliferation of S. aureus and the subsequent production of toxins, food should be rapidly cooled after preparation. This can be achieved by storing food in small, shallow containers that are loosely covered to allow adequate airflow and faster heat transfer. Importantly, food should not be held at temperatures between 4°C and 60°C for more than 2 hours, as prolonged exposure to this temperature range increases the risk of bacterial multiplication and toxin production.

Question 36

Define the Parameters of the Three Class Plan (n, m, M, c)

Answer 36

o n: The number of sample units chosen independently and randomly from the lot. o m: The microbiological limit that distinguishes good quality from marginally acceptable products. o M: The microbiological limit above which the product is considered unacceptable or defective. o c: The maximum allowable number of sample units that can yield results between m and M (i.e., marginally acceptable). Typically, no sample units are allowed to exceed the limit M.

Question 37

Explain How the Three Class Plan is Used to Classify Food Samples

Answer 37

o In the Three Class Plan, a predetermined number (n) of sample units are randomly selected from a lot. Each unit is tested, and its microbial count is compared against the established limits: o Units with counts below m are considered acceptable. o Units with counts between m and M are deemed marginally acceptable. o Units with counts above M are unacceptable. The plan allows for a certain number (c) of marginal results but typically permits no units to exceed M.

Question 38

Describe the Characteristics and Significance of Escherichia coli

Answer 38

Escherichia coli are facultatively anaerobic rods naturally present in the intestines of warm-blooded animals. Because they are commonly associated with fecal contamination, their presence in food or water is used as an indicator of sanitation levels. While most strains are harmless, some—such as E. coli O157:H7—can cause severe illness. Food poisoning from E. coli is typically characterized by symptoms like stomach pains, cramps, diarrhea (which may range from watery to bloody), fatigue, loss of appetite, nausea, and vomiting. Additionally, E. coli is a leading cause of traveler’s diarrhea

Question 39

Describe the Characteristics and Pathogenicity of Salmonella

Answer 39

Salmonella are facultatively anaerobic rods known for being inherently pathogenic. They are transmitted primarily through the ingestion of contaminated food and water, typically via fecal matter. Infection with Salmonella can lead to a range of symptoms including fever, diarrhea, abdominal cramps, headache, and vomiting. In cases where Salmonella invades the bloodstream, the infection can become severe or life-threatening. The common treatment for Salmonella food poisoning is antibiotics.

Question 40

Summarize the Key Features of Shigella and Campylobacter

Answer 40

Shigella are facultatively anaerobic rods that are universally pathogenic and known for causing severe gastrointestinal illness, often manifesting as dysentery. Campylobacter, which appear as spirally curved rods, are commonly found in the intestinal and reproductive tracts of both humans and animals. Among the various species, Campylobacter jejuni is a leading cause of bacterial gastroenteritis. Both pathogens are important considerations in foodborne illness outbreaks.

Question 41

Explain the Components and Function of Neogen EC Petri Film (5)

Answer 41

Selective, and differential system designed for the detection of E. coli. It contains: o Modified Violet Red Bile (VRB) Nutrients: Promote the growth of coliform organisms. o Cold-Water-Soluble Gelling Agent: Provides a stable matrix for colony formation. o 5-Bromo-4-chloro-3-indolyl-D-glucuronide (BCIG): Serves as an indicator for glucuronidase activity—a common trait among E. coli strains. o Tetrazolium Indicator: Helps in the visualization and enumeration of colonies by producing color changes.

Question 42

Discuss the Modifications in MacConkey-Sorbitol Media for Detecting E. coli O157:H7

Answer 42

MacConkey-Sorbitol media is a modified version of the standard MacConkey agar, where lactose is replaced with sorbitol. Most E. coli strains ferment sorbitol, but E. coli O157:H7 lacks this ability. Instead, it uses peptone as a nutrient, which results in a pH increase in the medium. This change in pH is detected by a pH indicator within the medium, differentiating non-sorbitol fermenters like E. coli O157:H7 from other E. coli strains that ferment sorbitol and produce a distinct color change. This selective differentiation is crucial for identifying the pathogenic E. coli O157:H7 in contaminated samples.

Question 43

Identify the Colony Characteristics of Bacteria on SS Media (6)

Answer 43

o SS (Salmonella-Shigella) Media is formulated to differentiate enteric bacteria based on their growth characteristics: o Escherichia coli: Typically shows slight growth with pink or red colonies, reflecting its lactose fermentation. o Enterobacter/Klebsiella: Also produces slight growth with a pink hue; these genera ferment lactose to varying degrees. o Salmonella: Colonies are usually colorless and may exhibit a black center due to hydrogen sulfide production. o Shigella: Forms colorless colonies because it does not ferment lactose. o Gram-positive bacteria: Do not grow on SS media due to the selective agents included in the formulation.

Question 44

Describe the API 20 system as a standardized biochemical test panel used for identifying bacteria.

Answer 44

A standardized identification tool commonly used in clinical and food microbiology. It comprises 20 miniaturized biochemical tests arranged in a strip. Each test uses a specific substrate to evaluate particular enzymatic activities or fermentation capabilities. The pattern of positive and negative reactions generates a numerical profile that can be compared with an identification manual to determine the bacterial species.

Question 45

______: Detects beta-galactosidase activity. A negative result is colorless, while a positive result turns yellow.

Answer 45

ONPG

Question 46

For these tests, a yellow color typically indicates a negative reaction, whereas a change to red/orange is positive. [5 tests total]

Answer 46

A (ADH) Lovely (LDC) Owl (ODC) Uses (URE) Instruments (IND)

Question 47

CIT: Evaluates citrate utilization. A pale ______ result is negative, while ______ signifies a positive reaction.

Answer 47

green/yellow, blue-green/blue

Question 48

H2S: Measures hydrogen sulfide production from sodium thiosulfate. _______ is negative; _____ indicates a positive result.

Answer 48

No black deposit (colorless/gray), black deposit

Question 49

TDA: Assesses tryptophan deaminase activity, turning from _____ (negative) to _______ (positive).

Answer 49

yellow, brown-red

Question 50

GEL: Checks for gelatinase activity using charcoal gelatin. ______ indicates a negative result; ______ signifies a positive reaction.

Answer 50

No diffusion of black color, diffuse black

Question 51

These tests check for fermentation/oxidation of various carbohydrates. A blue/blue-green color typically indicates a negative reaction, whereas a change to yellow is positive. [9 tests total]

Answer 51

Great (GLU) Minds (MAN) In (INO) Science (SOR) Rarely (RHA) Succumb (SAC) to Mediocrity (MEL) And (AMY) Apathy (ARA)

Question 52

OX: Tests for oxidase activity. A negative result appears as ______, while a _____ color denotes a positive result.

Answer 52

colorless/yellow, violet

Question 53

Learn How to Generate an API 20 Profile Number

Answer 53

After completing the 20 tests, the results are interpreted by grouping them into sets of three tests. Each test within a group is assigned a specific numerical value (for example, the first test might be assigned 1, the second 2, and the third 4). For each group, you add the values of the tests that are positive. This yields a number for each group. When the group values are combined sequentially, they form a unique profile number (e.g., 5-2-0-... etc.). This profile number is then cross-referenced with the API 20 identification manual, where each profile number correlates with a specific bacterial species.

Question 54

When performing MPN analysis, If Only 3 Dilutions Are Made...?

Answer 54

When you have only 3 dilutions, you simply use the results from these three tubes. There is no additional selection needed

Question 55

When performing MPN analysis, If More Than 3 Dilutions Are Made...?

Answer 55

Choose results from only three consecutive dilutions: Select the highest dilution in which all three tubes test positive. Then, include the two subsequent higher dilutions

Question 56

When performing MPN analysis, If More Than 3 Dilutions Are Made but None Have All Positive Tubes..?

Answer 56

Use the results from the first (lowest) dilution made

Question 57

When performing MPN analysis, If a Positive Tube Occurs in a Dilution Higher than the Three Chosen..?

Answer 57

At that higher dilution, add the number of positive tubes to those of the next lower dilution

Question 58

When performing MPN analysis, If All Sets of Dilutions Are 3/3 Positive..?

Answer 58

Choose the three highest dilutions of the series. Indicate this selection with a “greater than” symbol

Question 59

When performing MPN analysis, If All Sets of Dilutions Are Negative..?

Answer 59

Choose the three lowest dilutions of the series. Indicate this selection with a “less than” symbol

Question 60

Example: Suppose you prepare only three dilutions, how would the MPN be calculated? Results: 10⁻¹: 3/3 positive 10⁻²: 2/3 positive 10⁻³: 0/3 positive

Answer 60

1. use these tube results (3, 2, 0) 2. Value on MPN table: 0.93 3. Middle dilution is 10² = 93 coliform 4. x100ml = 9300 = 9.3x10^3 coliform/100 ml

Question 61

Example: 10⁻¹, 10⁻², 10⁻³, 10⁻⁴, and 10⁻⁵. 10⁻¹: 3/3 positive 10⁻²: 3/3 positive 10⁻³: 3/3 positive 10⁻⁴: 1/3 positive 10⁻⁵: 0/3 positive How would the MPN be calculated?

Answer 61

1. select the highest dilution with 3/3 positive, then the 2 consecutive ones (310) 2. Value on MPN table: 0.43 3. Middle dilution is 10⁴ = 4300 coliform 4. x100ml = 430000 = 4.3x10^5 coliform/100 ml

Question 62

Example: 10⁻¹, 10⁻², 10⁻³, 10⁻⁴. 10⁻¹: 2/3 positive 10⁻²: 1/3 positive 10⁻³: 0/3 positive 10⁻⁴: 0/3 positive How would the MPN be calculated?

Answer 62

1. use lowest dilution, then the 2 consecutive (210) 2. Value on MPN table: 0.15 3. Middle dilution is 10² = 15 coliform 4. x100ml = 1500 = 1.5x10^3 coliform/100 ml

Question 63

Example: 10⁻², 10⁻³, 10⁻⁴, and 10⁻⁵. 10⁻²: 3/3 positive 10⁻³: 2/3 positive 10⁻⁴: 0/3 positive 10⁻⁵: 1/3 positive How would the MPN be calculated?

Answer 63

1. initially use the ones starting at the 3/3, but add the value of the positive onto last number (320 -> 321) 2. Value on MPN table: 1.50 3. Middle dilution is 10³ = 1500 coliform 4. x100ml = 150000 = 1.5x10^5 coliform/100 ml

Question 64

Example: 10⁻¹, 10⁻², 10⁻³, 10⁻⁴, and 10⁻⁵. 10⁻¹: 3/3 positive 10⁻²: 3/3 positive 10⁻³: 3/3 positive 10⁻⁴: 3/3 positive 10⁻⁵: 3/3 positive How would the MPN be calculated?

Answer 64

1. choose 3 highest dilutions (333) 2. Value on MPN table: 24 3. Middle dilution is 10⁴ = 240000 coliform 4. x100ml = 24000000 > 2.4x10^7 coliform/100 ml

Question 65

Example: 10⁻¹, 10⁻², and 10⁻³. 10⁻¹: 0/3 positive 10⁻²: 0/3 positive 10⁻³: 0/3 positive How would the MPN be calculated?

Answer 65

1. choose 3 lowest dilutions (000) 2. Value on MPN table: 0.03 3. Middle dilution is 10² = 3 coliform 4. x100ml = 300 < 3.0x10^2 coliform/100 ml

Question 66

What does IMViC stand for?

Answer 66

Indole, Methyl Red, VP, and Citrate

Question 67

Within the MPN system, the viability of cells is based on? (4)

Answer 67

Turbidity, pH colour change, gas production, coloured pigments

Question 68

What are the four assumptions with MPNs?

Answer 68

1. Random distribution of bacteria 2. Organisms are not in clusters 3. If present, it will grow 4. Proper conditions for growth exist

Question 69

What are the advantages for MPN? (4)

Answer 69

Detects low densities of indicator organisms Suitable for foods Preferred if growth kinetics are variable Preferred if agar is affecting reliability

Question 70

Describe the colonial morphology of S. aureus on BP agar and why is this?

Answer 70

Looks black, shiny, and convex surrounded by a clear zone This is due to the reduction of telluride and proteolysis

Question 71

What are the ingredients in BP agar that make it selective and differential?

Answer 71

Pyruvate, egg yolk, and telluride

Question 72

What colour is S. aureus on STX plate?

Answer 72

Looks red-violet

Question 73

What modified media is used in STX?

Answer 73

Modified BP media, it’s chromogenic, meaning it contains a dye that reacts with bacterial enzymes. It also contains DNA, which for microbes that have DNAse, break down to form pink zones

Question 74

The presence of S. aureus can be determined for the presence of what enzyme? And what test is this called?

Answer 74

Thermostable DNAse, it’s called a thermonuclease test

Question 75

What are the selective and differential agents in MSA?

Answer 75

Selective: high salt concentration Differential: mannitol and phenol red, if fermented, causes a lower pH, changes from red to yellow

Question 76

why is mannitol salt not a preferred median for isolation of S. aureus from foods?

Answer 76

isn't a universally preferred medium for food samples due to its high salt concentration, which can inhibit the growth of other foodborne pathogens, and because some Staphylococcus species other than S. aureus can ferment mannitol