Vibrio Cholera Bacillus Subtilis & Escherichia Coli Microbiology Paper 2 please help with micro paperr 2 plese thank you due in 23 hourrs thank you please

Vibrio Cholera Bacillus Subtilis & Escherichia Coli Microbiology Paper 2 please help with micro paperr 2 plese thank you due in 23 hourrs thank you please work on it good and check for plagiaism Lab assignment: unknown ID
There are 5 case studies contained within the unknown lab assignment.
With these 5 case studies
•
Identify all 5 causative agents,
o Clearly describe and review the case studies.
o List the tests and pertinent results for each of the organisms in each case study.
▪ Make a note of any tests reviewed in the PowerPoint that may be used as
a substitute for the tests used in the exercise
o Provide a small summary of the organism, disorder caused, treatment etc. (no
more than a paragraph please)
With the remaining organisms
•
Report on all of the organisms on the dichotomous tree that were not listed as causative
agents
o List the determinative test results for each organism
o Provide a small summary of each organism
▪ Common sites of contamination/infection
▪ Symptomology
▪ Treatment
▪ (Keep these short, don’t kill yourself)
Focus on the purpose, components, visual identification and analysis of the tests for study
purposes in regards to your practical
The paper should be
•
•
•
•
Submitted as a word document to dk25@sunywcc.edu by June 17th
12-point font (Calibri, Ariel, New times roman…something legible)
Double spaced
Name and title at the top of the report.
Sources of information should be listed in APA or MLA format. (You don’t have to go far for
sources, the CDC should have more than enough information for the blurb I would like)
Microbiology Laboratory
Unknown Identification
Unknown Organism Identification
It’s important to understand several major concepts when attempting to identify an
unknown bacterium:
1. We start broad and narrow down by process of elimination.
2. Bacteria are impossible to distinguish by morphology alone.
• The bacteria’s metabolic requirements are usually used in the identification
process.
• What and how an organism eats and breathes are unique amongst species.
3. Most bacteria must process food before absorbing it (there’s no room), they must
instead secrete enzymes (exoenzymes) to break their food down and then absorb the
resulting nutrients.
• This is important, as it means bacterial digestion alters the environmental external
to itself
The Gram Stain
Start broad a narrow by process of
elimination:
In identifying an unknown organism,
the very first questions are :
• Gram reaction
• Morphology
The organisms on the right are all
Gram (-) bacilli
Respiration
Respiration: In microbiology respiration is the process by which an organism
attains its energy. There are several classifications
1. Obligate aerobe – Utilizes O2 to create energy.
2. Obligate anaerobe – Uses an alternative to O2 to create energy, O2 is
actually toxic to these organisms
3. Facultative anaerobe – Will utilize O2 if it is present. Has the ability to
perform fermentation or utilize an alternative to O2 if no O2 is present.
4. Aerotolerant anaerobes – Uses an alternative to O2 to create energy, can
tolerate the presence of oxygen
5. Microaerophile – Uses O2 to create energy, but requires a specific
amount (about 0.2 atm.)
Fermentation
• Some facultative anaerobes lack the
ability to use an alternate to oxygen,
they find alternate ways to harvest
energy from glycolysis.
• This process is referred to as
Fermentation, this process provides
significantly less energy but doesn’t
require oxygen to be present
• There are several paths available as
seen on the right, each of them
produces an acidic solution.
*Some forms of yeast can Ferment
also
EMB Agar
• Summary:
One of the first questions in the identification the organisms Gram reaction, a secondary question is
if it can digest lactose, and if so, what type of fermentation does it undergo?
The EMB agar inhibits the growth of gram (+) bacteria, but not to Gram (-) bacteria
This tests is performed as a streak on a petri dish, it is both
• Selective for Gram (-) organisms
• Differential for coliform bacteria
Ingredients:
The EMB (eosin-methylene blue) agar is composed of a standard agar plate supplemented with
1. Eosin – the pink acidic dye commonly used to dye human tissue cultures, this dye is also a color
indicator and turns black under acidic conditions
2. Methylene blue – A blue basic dye commonly used to dye bacteria or DNA, this dye is cytotoxic
to Gram positive organisms
3. Sucrose and Lactose – 2 forms of sugar common in the gut of human.
EMB Agar
Results:
Gram (+) organisms are actively inhibited by the presence of
methylene blue and cannot grow on EMB plates.
EMB Agar
Results: Recap
Depending on the organisms capability at fermenting
Lactose, there will be a visual difference in the appearance
of the bacterial colonies
1. Strong lactose fermenters precipitate the EMB salt,
which turns green in response to the pH change
2. Weaker lactose fermenters will provoke a range of
colors from pink to purple, sometimes even black
3. Non lactose fermenters will not develop a pigment, and
will appear white or grey
EMB Agar
Results:
Strong lactose fermenters
precipitate the EMB salt,
which turns green in
response to the pH change
EMB Agar
Results:
Weaker lactose fermenters will provoke a range of colors from pink to
purple, sometimes even black
Arthur Westman, MIT
EMB Agar
Results:
Non-lactose fermenters can still grow on the plate, provided they are Gram positive (+), the
growth will appear whitish or greyish
EMB agar plate inoculated with Pseudomonas aeruginosa
(a gram-negative non-coliform bacterium)
EMB Agar
Results:
The lactose portion of the
EMB is a good way of
distinguishing Coliform
Bacteria: bacteria that survive
in both soil as well as fecal
matter. They are defined as
being
• Gram negative(-)
• bacillus in morphology
• Non-spore forming
Some specific coliform organisms and their results
PEA agar
Summary:
Phenyl ethyl alcohol inhibits the growth of most Gram (-) organisms by inhibiting DNA synthesis.
Thus the test is selective for Gram (+) organisms.
*There are exceptions such as Pseudomonas aeruginosa
Ingredients:
The PEA (Phenylethyl alcohol) agar is a standard agar supplemented with (no surprise hopefully)
Phenylethyl alcohol.
Results:
Growth = Gram (+)
No growth = Gram (-)
*I haven’t provided an image of this test, it’s indistinguishable from a standard nutrient agar and the
organisms will either grow or not.
Thioglycolate Broth
Oxygen rich
region
Oxygen poor
region
ASM.org
Summary:
Thioglycolate is a potent reducing agent; it
converts oxygen to water on contact. This
removes oxygen from the solution before it can
penetrate deep into the tube and creates a
solution with layered oxygen concentrations.
Thioglycolate broth is utilized to determine an
organisms aerotolerance (its respiratory
capabilities in regards to Oxygen).
Ingredients:
• Oxygen reducers (Sodium thioglycolate,
thioglycolic acid, L – cysteine)
• Methylene blue – this tracks how deeply the
oxygen has penetrated into the solution.
• A small amount of agar is added to this test to
further slow oxygenation.
Thioglycolate Broth
Results:
A. Facultative anaerobe
Growth throughout tube
B. Strict Anaerobe
Growth only below the
oxygenation line
C. Flocculent (AKA biofilm)
Growth aggregates into a
semisolid film
D. Microaerophile
Growth only at the oxygenation
line
E. Strict aerobe
Growth only above the
oxygenation line
Thioglycolate Broth
Phenol Red + Sugar
Summary:
Fermentation of a sugar always results in the production of an acidic solution.
The Phenol red tests are broths that are differential and used to detect the fermentation of a single type of
sugar by an organism.
It also tests for the production of gas during this fermentation
If the organism cannot utilize any of the sugars it will survive by breaking down the peptides. This will result in
a basic solution which will turn the solution fuchsia/magenta
Ingredients:
The phenol red line of tests are composed of a very weak broth supplemented with:
• A single type of sugar (Usually Glucose, Lactose, or Sucrose)
• The color indicator phenol red
• A small amount of protein
• A small upside down test tube floating in the center (Durham tube)
MSA
Phenol Red:
• This is a very commonly used color indicator. Several of the tests in
this packet utilize it.
• It’s important to realize color indicators come in many types and
colors, so make sure you know which is being used in each test
Phenol Red + Sugar
Results
A. Acid with Gas
• the organisms can utilize the sugar, and
produce a gas as a byproduct (notice the
bubble in the Durham tube)
B. Acid without Gas
•
The organism can utilize the sugar, but no
gas is produced
C. Control (uninoculated)
•
No organisms were introduced to this tube
D. Peptide Digestion
•
The organism preferentially digests proteins
E. No acid, no gas
•
An organism was introduced but failed to
grow
MSA
Summary:
Staphylococcus as a genus is highly resistant to osmotic pressure and thrives in high salt
environments.
Further, within this genus there is only 1 species that is capable of fermenting mannitol
This test is
Selective for Staphylococcus
Differential for Staphylococcus aureus
Ingredients:
The MSA (Mannitol salt agar) is a standard agar supplemented with
1. Mannitol (a sugar alcohol)
2. NaCl (salt) 7.5%
3. Phenol red, a color indicator that detects changes in pH
Mannitol Salt Agar
Results:
The high Salt
concentration (7.5%) of
this agar results in the
inhibition of most
genera of bacteria other
than Staphylococcus.
Of all the species of
Staphylococcus, only
Staphylococcus
aureus is capable of
fermenting mannitol.
Fermentation lowers the pH
of the surrounding area
below 6.9 and the phenol in
the agar will turn yellow in
response .
Nitrate test
Summary:
Many organisms can use molecules other than oxygen as their final electron acceptor.
This test is designed to determine if the organism can utilize nitrates as an alternate for oxygen during respiration
Ingredients:
This test is performed as a broth
The broth has been supplemented with NO3 (nitrates)
There are 3 reagents for this test
Nitrate A – (sulfanilic acid)
Nitrate B – (Dimethyl-α-napthylamine)
Zinc powder
Results:
The nitrate A and Nitrate B reagents should be added
and observed before adding zinc.
1. No color change = add zinc
2. Color change to red = (+)
After Zinc addition
3. No color change = (+)
4. Color change = (-)
Nitrate test
Reasoning:
No color change after adding nitrates A&B proves there is
no nitrous acid in the solution (nitrous acid is a byproduct
of the partial breakdown of nitrites)
However, there is a chance the organism further broke
down the nitrates to nitrogen gas or ammonia
If this happens then there should be no nitrates left in the
tube at all.
Zinc binds to nitrates and produces a red pigment.
Thus a color change after the addition of Zinc means there
are still nitrates present in the solution, meaning the
organism cannot utilize nitrates
Some specific organisms and their results
Indole test
Summary:
This is a differential test designed to determine an organisms
ability to digest the amino acid tryptophan.
In breaking down tryptophan an organism will produce indole as
a necessary byproduct.
Ingredients:
• The test uses a broth supplemented with Tryptophan rich
peptides
• Kovac’s reagent is added to this test.
The presence of a red ring on the
top of the broth after addition of a
few drops of Kovac’s reagent.
SIM
Summary:
The SIM test is a differential media that incorporates 3 different
tests in 1
1. The use of Sulfide as an alternate to Oxygen
2. The ability to break down the amino acid tryptophan
3. The motility of the organism
This test is performed as a stab (seen on the right)
Ingredients:
The tube contains a semi-solid agar (.5%) supplemented with:
1. A source of Sulfur (Methionine, Cysteine, Sodium Thiosulfate)
2. Proteins rich in the amino acid Tryptophan
3. A source of Fe+ (iron) ions (Ferric Ammonium Citrate)
There test uses Kovac’s reagent to detects for indole
SIM
All possible results for the SIM test
Sample
Motility
Sulfide
Indole
A.
(-)
(-)
(+)
B.
(-)
(-)
(-)
C.
(+)
(+)
(-)
D.
(+)
(-)
(-)
E.
(+)
(+)
(+)
A
B
C
D
E
SIM
Organism
Escherichia
coli
Staphylococcus
aureus
Salmonella
typhimurium
Enterobacter
aerogenes
Proteus
vulgaris
Motility
Sulfide
Indole
(-)
(-)
(+)
(-)
(-)
(-)
(+)
(+)
(-)
(+)
(-)
(-)
(+)
(+)
(+)
A
B
C
D
E
Oxidase
Summary:
Cytochrome C oxidase is the final enzyme in the electron transport chain
before ATP synthase; thus only aerobes possess this enzyme.
The oxidase test is a differential test used to detect the enzyme
Cytochrome C oxidase.
Ingredients:
This test requires a very small amount of bacteria, usually swabbed onto a
cotton applicator
There is a reagent; N,N,N,N-Tetramethyl-p-phenylenediamine which we
will just refer to as Oxidase reagent
Results: *Because atmospheric oxygen will yield a positive result in a
short amount of time this test must be read within 30 seconds.
A blue color = (+)
No color change = (-)
Phenylalanine Deaminase
Summary
Phenylalanine deaminase is an enzyme that removes an ammonia
from the amino acid tryptophan.
This is a differential test for the presence of Phenylalanine deaminase
The product of deaminating a phenylalanine are phenyl-pyruvic acid
and NH3 (ammonia). (see below image.)
Ingredients:
This test is performed as a steak on a slanted agar that’s been
enriched with Phenylalanine rich peptides.
There is a reagent; FeCl3 (Ferric Chloride) which reacts with Phenylpyruvic acid and produces a green color
Ammonia
Phenylpyruvate
Results:
This test must be read after the addition
of FeCl3
• Solution turns Green = (+)
• Solution turns Yellow = (-)
*Ferric Chloride has a natural yellow tinge
Simmons Citrate
Summary:
Most facultative organisms can utilize citrate, however, only some
can utilize an enzyme called permease to intake exogenous
citrate.
The Simmons citrate tests is a differential test, to determine if an
organism can utilize citrate as its sole source of carbon
This test is performed as a stab on a slanted agar.
Results
Ingredients:
1. Citrate – The metabolite this test is named after
2. Ammonium salts – provide a source of Ammonium ions
3. Bromothymol blue – a color indicator that is green at a pH
below 7.6, and turns blue at a Ph above 7.6
Mechanism: The process of breaking down citrate in (+) organisms produces CO2; this starts a
series of reactions that results in the production of alkaline ammonia. The presence of
ammonia raises the Ph of the tube above 7.6 and turns the tube blue.
Urease Test
Summary
Urea is a nitrogen containing substance that animals
create as a byproduct of protein hydrolysis.
Under normal circumstances urea is removed via the
urine. This is a differential test for the production of
the enzyme urease, which catalyzes the breakdown
of Urea.
Ingredients:
Urea Phenol red – color indicator
Buffers – to prevent color change from Urea’s natural
breakdown
Mechanism:
As Urea is broken down, Ammonia is generated,
raising the pH of the solution from 7.4 to 8.1; this
turns the broth a magenta color.
MRVP
Summary
The MRVP is actually 2 tests
1. Methyl Red
2. Voges Proskauer
In the fermentation of glucose, organisms typically follow a single pathway.
These 2 tests differentiate whether an organisms will undergo pure lactic acid
fermentation, or prefers the butanediol pathway
Ingredients:
No special active ingredients
3 reagents
Methyl Red – Methyl red a color indicator
Voges Proskauer – KOH, α-Napthol
Mechanism:
If the organism prefers the lactic acid pathway, the acid produced will drop the pH
of the solution below 4.4 and the Methyl red will change from yellow to red.
If the organism prefers the butanediol pathway α-Napthol will bind to the neutral
product acetoin and turn the solution red (this reaction takes about 20 minutes.
Voges Proskauer
+
–
Methyl Red
Kligler Iron Agar
Glucose has a significantly
higher density than Lactose.
Thus when both are added to
this test the glucose will
concentrate at the bottom of
the tube while the Lactose will
float to the upper area.
• Phenol Red is used as a
color indicator
• Sources of sulfur and iron
are also added to this test,
in the same capacity as a
SIM test
Kligler Iron Agar
RESULTS:
A.) Control (no inoculation)
B.) (+) glucose, (+) Lactose, (+) gas, (-) H2S
C.) (+) glucose), (-) Lactose, (-) gas, (-) H2S
D) (-) glucose, (-) Lactose, (-) gas, (-) H2S
E) (X) glucose, (X) Lactose, (-) gas, (+) H2S
The Identification Process
Some examples of flowcharts used to determine unknown bacteria are found in the following slides
Remember
1. Start Broad; most of these charts begin with gram stains
2. Be systematic; don’t just verify the results in the flowchart, follow up with other test results that
you have that weren’t in the flow charts.
3. Many pathogens share very similar symptomologies and these tests can differentiate between
possible pathogen.
These techniques are very useful in health and epidemiology.
• For example
• Vibrio cholera is only transferred via Water, not food
• Bacillus subtilis prefers rice, E.coli prefers vegetables
• Different organisms have different incubation times; Salmonella typically takes a day to
incubate, shigella typically takes about 3 days.
• It can help determine treatment; antibiotics aren’t typically given for E.coli, but are
often recommended for Salmonella which has a chance of becoming septic.

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