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Lab Practical 2 Review
Exercise 14 (Blood Agar)
1. Be prepared to identify the different hemolytic reactions of blood:
A). Beta hemolysis (β) – Complete lysis of red blood cells and breakdown of hemoglobin. Clearing zone around bacteria. (Streptococcus pyogenes)
B). Alpha hemolysis (α) – Small amount of lysis. Small greenish halo around bacteria.
C). Gamma hemolysis (γ) – No lysis. Organism will grow, but will not lyse
2. Know why blood agar is both enriched and differential:
-‐ It is enriched with blood (including iron and other nutrients in blood) so almost all organisms will grow on this agar. It is beneficial to fastidious microorganisms, like streptococci.
-‐ It differentiates between organisms that will hemolyze red blood cells and those that will not.
3. Be able to name the genre of organisms that is usually capable of beta lysis:
A). Streptococcus (and Staphylococcus)
4. Know the substrate, enzyme, and end product for beta and alpha hemolysis:
-‐ substrate: red blood cell
-‐ enzyme: hemolysin
-‐ end product: lysis of RBCs (alpha hemolysis) and digestion of hemoglobin (beta hemolysis)
5. The bacitracin test identifies Group A Streptococcus pyogenes from other Streptococcus species because Group A is especially sensitive to the antibiotic bacitracin. There will be diminished growth of the bacteria in the area surrounding the bacitracin. Left quadrant is Streptococcus pyogenes.
Exercise 15 (Mannitol Salt Agar)
1. Be able to define:
a). Selective media: Selective media allow certain types of organisms to grow, and inhibit
the growth of other organisms.
b). Differential media: Differential media are used to differentiate closely related organisms or
groups of organisms.
c). Chemically defined media: a medium in which all of the chemical components are known.
d). Complex media: Complex media are rich in nutrients, they contain water soluble extracts of plant or animal tissue (e.g., enzymatically digested animal proteins such as peptone and tryptone). Usually a sugar, often glucose is added to serve as the main carbon and energy source. The combination of extracts and sugar creates a medium which is rich in minerals and organic nutrients, but since the exact composition is unknown, the medium is called complex.
e). Substrate: a molecule upon which an enzyme acts.
f). Inhibitor: a molecule, which binds to enzymes and decreases their activity.
g). Reagent/Indicator: a "substance or compound that is added to a system in order to bring about a chemical reaction, or added to see if a reaction occurs.
h). Exoenzyme: an enzyme that is made inside the cell then secreted and functions outside of that cell.
i). Endoenzyme: an enzyme that functions within the cell in which it was produced.
2. Know why mannitol sugar is both selective and differential:
-‐ This medium contains a high concentration of salt, inhibiting growth of most bacteria, but allowing salt-‐tolerant staphylococci (halophiles) to grow, which makes it selective.
-‐ This medium is differential whereby it differentiates between pathogenic and non-‐pathogenic strains of staphylococci. It contains the sugar mannitol and the pH indicator phenol red. Disease-‐causing staphylococci ferment the mannitol, which in turn releases acid, turning the phenol red to yellow. Non-‐pathenogenic staphylococci will grow, but not change the color of the phenol red.
3. Be prepared to differentiate between growth with and without mannitol fermentation:
S. aureus and S. epidermidis(although not evident here) are the only organisms that grew since they are halophiles (salt-‐tolerant). The yellow produced in the agar is due to the production of acid from the fermentation of the mannitol.
4. Be able to name the genera of organsims that are selected for by this agar:
A). Staphylococcus
5. Know the genus and species of the organism capable of growth and mannitol fermentation:
A). Staphylococcus aureus (pathogenic)
Exercise 16 (Eosin Methylene Blue Agar EMB – identifies gram-‐negative enteric organisms)
1. Be able to define:
a). Coliform: gram-‐negative, rod-‐shaped bacteria (e.g., E. coli).
b). Enteric: relating to/residing in the intestines.
2. Why is EMB media both selective and differential?
A). Eosin and Methylene blue together make the medium selective by inhibiting gram-‐positive growth but allowing gram-‐negative (including coliforms).
It is differential because it differentiates between lactose fermenters and non-‐lactose fermenters.
3. Differentiate between:
-‐ Non-‐lactose fermenters – white colonies
-‐ Lactose fermenters – pink or purple colonies
-‐ Strong lactose fermenters – metallic green
4. Organisms that causes the metallic green sheen is E. coli.
Exercise 17 (MacConkey Agar – identifies gram-‐negative enteric organsims)
1. MacConkey is both selective and differential:
-‐ Inhibits (or causes poor growth) the growth of gram-‐positive bacteria. Allows the growth of gram-‐negative (including coliforms). This is what makes it selective.
-‐ This media also contains a sugar lactose and the pH indicator neutral red, which differentiates between non-‐lactose fermenters (Pseudomonas) and lactose fermenters (Klebsiella and Escherichia). This is what makes it differential.
2. Identify growth with lactose fermentation:
-‐ E. coli is gram-‐negative and a strong lactose fermenter and produces pink growth and causes pink dye to be deposited in the agar surrounding the bacteria. Non-‐lactose fermenters produce colorless colonies.
3. Know the substrate, enzyme, and end product of lactose fermentation:
-‐ substrate: lactose
-‐ enzyme: lactase
-‐ end result: The production of acid. Bacteria will turn pink if positive for lactose fermentation.
Exercise 22 (Catalase)
1. Be prepared to identify a positive catalase test:
When hydrogen peroxide is added, if the organism is positive for catalase, there will be bubbles.
2. Know the substrate, enzyme, and end product for the reaction:
- substrate: hydrogen peroxide (H2O2)
- enzyme: catalase
- end product: You will see “fizzing” or bubbles if positive for catalase which means that
catalase converted H2O2 ! H2O + O2. The bubbles are oxygen.
3. The ability to produce enzymes which neutralize toxic oxygen, like catalase, is a capability shared by many aerobic and facultative anaerobic organisms, a capability most strict anaerobes lack.
Exercise 23 (Lipid Hydrolysis – Spirit Blue)
1. Know the substrate, enzyme, and end product of the reaction:
-‐ substrate: triglyceride
-‐ enzyme: lipase
-‐ end product: triglycerides are hydrolyzed into glycerol and fatty acids.
2. Be prepared to identify a positive and negative test:
-‐ The plates start out blue. Look for a clearing zone around the bacteria. P. vulgaris was positive for lipid hydrolysis in our lab exercise. Below, S. epidermidis is negative and E. coli is positive for lipid hydrolysis.
Exercise 36 (DNase Test)
1. Be able to define:
a). Extracellular enzyme: an enzyme produced within the cell and secreted outside of the cell wall.
b). Extracellular digestion: A form of digestion wherein the breaking down of materials into smaller, absorbable components takes place outside the cell (Saprobes, i.e. fungi)
2. Be prepared to identify a positive test:
3. Know the substrate, enzyme, and end product for DNase reaction:
-‐ substrate: DNA
-‐ enzyme: DNase
-‐ end product: DNase breaks down DNA into nucleotides. This causes a clearing zone around the bacteria (as seen on the right in the figure above). In lab we used agar plates containing a blue substrate.
*Most commonly used organisms will grow on this plate.*
4. Know the clinical importance of organisms capable of secreting DNase:
A). This gives an organism the ability to degrade biofilms, which enhances the organisms ability to invade the host or spread from the site of origin.
5. Staphlycoccus aureus is detected using this test.
Exercise 37 (Gelatin Hydrolysis Test)
1. Know the substrate, enzyme, and end result of the reaction:
-‐ substrate: gelatin (derived from collagen)
-‐ enzyme: gelatinase
-‐ end result: if positive, the gelatinase hydrolyzes the gelatin to produce short peptides and amino acids and the medium is now a liquid. These smaller components diffuse more easily and can be transported into the cell.
2. Be prepared to identify a positive and negative reaction:
A). In a positive reaction, the gelatin will be liquid. In a negative one, the gelatin will still be solid (gelatinous, like jello), indicating it was not hydrolyzed.
3. A positive test differentiates between Staphylococcus aureus (gelatinase-‐positive) and Staphylococcus epidermidis (gelatinase-‐negative).
4. Know why the tubes must be chilled before interpreting:
A). During incubation, both tubes will liquefy due to the temperature, so both tubes must then be chilled again to see which is actually liquid from being hydrolyzed and which isn’t.
Exercise 38 (Starch Hydrolysis)
1. Be able to define:
a). Polysaccharide: Polymeric carbohydrate structures, formed of repeating units (either mono-‐ or di-‐saccharides) joined together by glycosidic bonds.
b). Monosaccharide: the most basic units of carbohydrates. They are the simplest form of sugar and are usually colorless, water-‐soluble, crystalline solids.
c). Disaccharide: macromolecules consisting of two monosaccharides connected by a glycosidic bond.
d). Glycosidic bond: A covalent bond that holds a carbohydrate (sugar) to another group that can or cannot be another sugar.
e). Hydrolysis: A chemical reaction in which the interaction of a compound with water results in the decomposition of that compound.
2. Know the substrate, enzyme, and end product of the reaction:
-‐ substrate: starch
-‐ enzyme: amylase
-‐ end product: amylase breaks the glycosidic bonds to produce maltose (disaccharide), glucose,and short chain polysaccharides. These smaller subunits can easily diffuse through the medium and be transported into the cell.
3. Know the reagent used in this test:
A). Iodine is used to identify intact starch (turns black)
4. Be prepared to identify a positive and negative test:
A clearing zone (on the left on both plates) around the bacteria indicates a positive test.
Exercise 39 (Casein Hydrolysis Test – skim milk plate)
Casein is a good source of protein but is too large to be transported across the membrane intact.
1. Know the substrate, enzyme, and end product of the reaction:
-‐ substrate: casein (large, globular protein that gives milk its white, opaque color)
-‐ enzyme: casease (caseinase) (exoenzyme, secreted outside of the cell)
-‐ end result: casein is hydrolyzed into amino acids and peptides.
2. Be prepared to identify a positive and negative test:
-‐ If casease is present and breaks down the casein, the protein is no longer structurally intact and the white color will disappear (clearing zone, seen on the right of both plates below). The agar surrounding a colony in which the organism does not produce casease will appear white.
Exercise 40 (Phenol Red Broth)
1. Define:
Reversion: This happens when bacteria break down the peptone in the medium as a result of exhausting the available carbohydrate. The breakdown of protein releases amines (ammonia) into the medium which increases the pH and the phenol red will revert from yellow back to red. This happens when tubes are incubated for an extended period of time.
2. Know the end product of the reaction:
-‐ end product: Positive sugar fermentation tubes will be yellow. This indicates acid production and indirectly indicates fermentation has occurred. Phenol red is a pH indicator that will stay red at neutral pH values and turn yellow when pH levels decrease and become more acidic. Sometimes gas is a byproduct of this and will be present in the Durham tube. If the broth turns a bright fuschia, this indicates protein metabolism, not carb metabolism.
3. Know how to tell if gases were produced:
A). Gas bubble in Durham tube
4. Know the indicator dye used in these tubes:
A). Phenol red
5. Be prepared to identify a positive and negative fermentation test:
6. Know why tubes must be interpreted within 24 hours at 37 degrees C or 48 hours at 25 degrees C and how to identify a tube that may have undergone reversion:
A). Tubes must be checked within that time frame because bacteria generally like to metabolize carbohydrates. If carb supplies run out, the organism will catabolize peptone (proteins). This will release ammonia and increase the pH of the broth and change the phenol red indicator and yield inconclusive results. A tube that is yellow at 24 hours can revert back to red or appear fuschia or yellow with red on top.
Exercise 41 (Triple Sugar Iron Agar)
1. Be prepared to identify acid or alkaline products of the slant or butt of the tube (ex: A/A, or A/K or K/K (second pic, tube all the way on the right (all red), etc):
a.) red/yellow = K/A. Glucose fermentation + or -‐ peptone catabolism.
b). red/yellow plus bubbles = K/A, Gas. Glucose fermentation; gas produced.
c). yellow/yellow = A/A. Glucose + lactose or sucrose fermentation.
d). yellow/yellow plus gas bubbles = A/A, Gas. Glucose + lactose fermentation; gas produced.
e). yellow/yellow plus black precipitate = A/A, H2S. Glucose + lactose fermented hydrogen
sulfide produced.
f). red/yellow plus black precipitate = Glucose fermentation + or – peptone catabolism,
hydrogen sulfide produced.
g). red/red = K/K. Protein catabolism
2. Be prepared to identify H2S production (know substrate and product H2S + Fe = FeS):
A). The contents of the tube will turn black. Pictured in the 3rd tube (both pics). When hydrogen
sulfide interacts with iron, it produces an insoluble black iron sulfide.
3. Be prepared to identify CO2 production:
A). A tube containing an air bubble but no black precipitate, indicated on the second pic in the
first tube.
4. Know what reversion is and why incubation times are important with this test:
A). Reversion is what occurs when sugar (carb) supplies have been exhausted and the organism
begins to hydrolyze protein (peptides) instead. This happens if incubation occurs for a longer period of
time than recommended. It is also possible to have a result that appears to be reversion when in
fact, the organism just doesn’t use any of the sugars and instead catabolizes protein (peptides).
5. Know the general reactions that result in acid or alkaline products:
-‐ acid product = fermentation of sugars (carbohydrates, yellow).
-‐ alkaline product = organism catabolized peptones instead of sugars. Deamination causes the
production of ammonia (basic) and causes a magenta color.
Exercise 43 (SIM Media/Indole Test IMVIC)
1. Know the substrate, enzyme, and end product of the Indole reaction:
-‐ substrate: tryptophan
-‐ enzyme: tryptonphanase
-‐ end product: tryptophanse hydrolyzes tryptophan to produce indole, pyruvic acid, and
ammonia (through deamination). SIM (sulfide indole motility) Once Kovac is added, if the
reagent turns cherry red, indole was produced. If it stays the same, it is indole-‐negative.
2. Know the reaction that results in FeS accumulation:
A). When thiosulfate is reduced, the hydrogen sulfide (H2S) gas reacts with the iron salt in the medium to form a black iron sulfide precipitate).
3. Be prepared to identify motility, H2S production, and Indole production:
-‐ If the organism is motile, the stab line will be spread out (D). If it is non-‐motile, the stab line will be thin and still intact (B).
-‐ if H2S was present, a black precipitate is produced (C).
-‐ if indole is present (indole-‐positive), Kovac’s reagent will be cherry-‐red (A).
4. Know what reagent is used in the Indole test:
A). Kovac’s reagent
5. Know how to perform a stab inoculation:
A). All the way down through the medium to the bottom, and back up through the same pathway.
Exercise 43 (Methyl Red and Vogues-‐Proskaur Test MRVP)
1. Know the reagent and/or indicator dye used in each test:
-‐ MR test = Methyl Red
-‐ VP test = Barritts A and Barritts B
2. Be prepared to identify a positive and negative result for each:
-‐ MR test -‐ after adding methyl red (left image), if the broth turns red, it is positive for mixed acid fermentation. Yellow = negative.
-‐ VP test -‐ after adding both Barritts A and B (right image), if it turns red, that is a positive result for the production of the neutral end products butanediol and/or acetoin and/or acetyl methyl carbinol. Yellow = negative.
Glucose -‐> glucose metabolism -‐> pyruvic acid -‐>Acetoin.
3. The Methyl Red test is used to identify mixed acid fermenters (Escherichia coli and Proteus vulgaris) when supplied with glucose.
4. The VP test is used to identify organisms that are capable of producing butanediol or acetoin. The above organisms use a butylene glycol pathway to convert pyruvate to these neutral end products.
Exercise 43 (Citrate Test)
1. Be prepared to identify a positive and negative test as well as state the pH of the media:
-‐ the initial pH of the citrate agar is 6.9
-‐ growth on the media indicates a positive test for citrate utilization
-‐ growth usually results in release of ammonia, increased pH and a color change to blue
-‐ occasionally, acid products are formed and the color may change to orange or yellow
-‐ little or no growth is a negative result (remains green)
2. Know the carbon source of this media and the enzyme required to utilize citrate:
A). Citrate is the sole carbon source of this media. Citrate permease is required for bacteria to import the citrate into the cell.
Exercise 44 (Urea Hydrolysis)
1. Know the substrate, enzyme, and end product of the reaction:
-‐ substrate: urea
-‐ enzyme: urease
-‐ end product: ammonia, which makes the urine more basic.
2. Be prepared to identify a positive and negative test:
-‐ pink tube is positive for urease.
-‐ straw colored one is negative.
3. Know the genus of the organism that yields a strong, positive result:
A). Proteus vulgaris
Exercise 45 (Nitrate Reduction)
1. Explain the chemical process of nitrate reduction to nitrite and nitrite reduction to N2O and N2 and be able to name the enzymes in both processes:
Nitrogen exists in several forms and bacteria play a key role in cycling nitrogen through its various forms in the biosphere.
-‐ Nitrate reductase catalyzes the reduction of nitrate (NO3) to nitrite (NO2).
-‐ Nitrite reductase catalyzes the reduction of nitrite (NO2) to nitrogen gas (N2) or nitrous oxide (N2O).
2. Describe the purpose and function of nitrate broth and be able to identify positive tests for nitrate, nitrite, and N2 or N2O.
A) Nitrate broth is a minimal medium containing nitrate and a protein digest with no carbs.
B) See above flow chart for possible results. Briefly: After adding reagents A/B – Red is positive for nitrate reductase only After adding reagents A/B and zinc – Clear is positive for both nitrate and nitrite reductases After adding reagents A/B and zinc – Red is negative for both nitrate and nitrite reductases
3. Know the reagents used for the identification of nitrate, nitrite, and N2: -‐ Reagent A = alpha-‐napthylamine -‐ Reagent B = sulfonic acid -‐ zinc powder 4. Know the medical significance of anaerobic and facultative anaerobic organisms capable of denitrification: -‐ Clostridium (anaerobe) and Pseudomonas (facultative anaerobe) are 2 organisms capable of denitrification. Denitrification decreases the amount of nitrogen available to plants and decreases plant productivity. This leaves farmers adding fertilizers to the soil to replenish lost nitrogen and this, in turn, increases the cost of plant products. Overuse of fertilizers due to denitrification also has detrimental effects. Nitrate reduction is characteristic of the Enterobacteriaceae.
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