dichotomus key gram postive bacteria coccus

| April 27, 2015

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Dichootomus key graph diagram Microbiology lab
Gram Positive Bacteria
Identify if it is Stapylococcus aureus gram + coccus
versus
Staphylococcus epidermis,
versus
Micrococcus luteus
versus
Micrococcus Rosea

Test results Listed below:
Simmons Sodium Citrate: sucrose negative, lactose Acid no gas, dextrose acid no gas, maltose acid no gas, manitol negative, citrase negative, urease negative
SIM: Sulfur Negative, Indole Negative, Motility Negative, Maltose Negative, Glucose Positive, lactose Positive, H2S negative, Gas Negative

Gram stain positive: for Gram positive Coccus
Acid Fast Stain: Negative
Capsule: Negative
Endospores: Negative
Motility: Negative
H2S Production: Negative
Indole: Negative
Kliger Iron Agar: Glucose= + positive, Lactose + positive, H2S negative,
Oxygen reqirements: Obligate Anerobe
Manitol Salt Agar : + positive
Eosin Methylene Blue Agar : Negative
MacConkey Agar : Negative
Blood Agar : Positive, Gamma
Phenol Red- Dextrose : Acid , no gas
Phenol Red Lactose: Acid, no gas
Phenol Red Sucrose: negative
Phenol Red Mannitol: Negative
Phenol Red Maltose: Acid, No gas
Catalase: Positive
Citrase: Negative
Urease: Negative
Oxidase: Negative

Simple Stain
Gram Stain
Negative Stain
Nutrient Agar
Nutrient Broth
TSI agar
Phenol Red broth
Citrate Agar
Urea Broth
Blood Agar (TSA 5% sheep blood)
Mannitol Salt Agar
MacConkey Agar
Motility Media

The following is a list of possible unknowns:

Micrococcus luteus
Micrococcus rosea
Staphyloccocus epidermis
Staphylococcus epidermidis

Instructions:
Create Dichotomus key for listed test and results for listed gram positive Coccus above. Give Rationale behind it, Show what the bacteriais going to do. List the characteristics to determine what organism you have and show leading to test results identifying the unknown.

With the organisms given, you have to be aware of their reactions with the tests and
media. Your final key will be simpler if you can find as many tests as possible, that
split as many organisms as possible into 2 relatively equal groups. Theoretically, with
11 organisms, one test might split them into 5 organisms (group 1) and 6 organisms
(group 2). Group 2 could then be split by another test into two groups of 3. Then either
of those groups could be fully differentiated by 2 more tests. So, at the least, you
would need a total of 4 tests. Usually, more tests than this minimum are required.
It’s mainly a matter of writng down each bacterium’s test result, then finding the
minimum number (with reservations) of tests that will differentiate all the organisms.
Then you can draw your key.

Sometimes though, it’s better to eliminate an unusual organism first, such as example for
M. smegmatis, which is quite different from the rest, in that one test, the Negative
Stain, may remove it onto one branch of the tree, with the others on another branch.
In this way, it will not rear it’s ugly head again if it has many variable reactions, or
unknown reactions, with some of the other tests.

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