Report 5 Biochemical Unknown 2 2 PDF

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Name:

Mahi Solomon

Unknown Number

63

REPORT 5 (50 pts): Biochemical identification of Bacteria 1. 2. 3. 4.

Fill out responses on THIS document and print it out. Do not reorganize/change this report . You can format the table/spacing to fit in your responses Handwritten report is not acceptable Email submission is not acceptable

_________(12.0 pts) Unknowns Purpose (4.0pts.)

Observation (4.0pts.) May include colored images for each.

Organism A Enterobacter aerogenes

Gram stain

TSA

PEA

MAC

To be able to differentiate between bacteria by cell wall composition.

Allows a wide variety of bacteria to be grown, not selective. Although fastidious bacteria may not grow well do to environment requirements.

Allows (selects) for the growth of gram-positive bacteria.

Allows the growth (selects) of gramnegative bacteria and differentiates between lactose fermenter and nonfermenator.

Organism A is gram negative rod due to the pink/slightly red color present caused by the counter stain safranin.

Organism A did produce plenty of bacteria but not isolated colonies on the TSA plate, creating a lawn growth. Having a mass of bacteria growth present can be caused by the environment requirement for organism A. Although the streak plate is poor it is a pure culture by having the same morphological characteristics and no inconsistency.

Organism A was not grown on PEA plate since the PEA plate is selective for gram positive while organism A is gram negative.

Organism A bacteria growth has a pink appearance suggesting that it is a lactose fermenter and also supports that it is a gram-negative bacterium.

*poor representation of streak plate

Organism B Streptococcus bovis

Organism B is gram positive cocci due to the purple appearance by the primary stain.

Organism B did produce isolated colonies on the TSA plate, but not many colonies were formed. Having a few colonies formed can be caused by the environment requirement for Organism B. By colonies forming we can conclude it is a pure culture.

Organism B was grown on the PEA plate. Not many colonies were formed on the plate, again most likely due to the growth requirement.

Organism B was not present on the plate since MAC selects for gram- negative bacteria. This also supports our finding with the gram stain, grampositive.

*colonies are present/hard to see

Conclusion* (4.0pts.) Do not make conclusion about lactose fermentation if you observe ‘no growth’

Organism A Enterobacter aerogenes

Organism B Streptococcu s bovis

Organism A is gram-negative rod because it was had purple appearance.

Organism A grown well in this environment considering how much bacteria formed.

Organism A did not grow because it is a gramnegative bacteria.

Since organism A is gramnegative it is able to grow, also it’s positive for a lactose fermenter.

Organism B is gram-positive cocci because it was pink/reddish colored.

Organism B did not grow so well, but growth is apparent possibly due to special environmental requirements.

Organism B did grow because it is a gram-positive bacteria.

Since organism B is grampositive it will not be able to grow therefore can’t differentiate if it’s a LF or NLF.

________(2.0 pts) Does growth on selective &/or differential plates support your Gram staining results? Explain your response.  Yes, organism A was determined to be gram-negative rod through gram staining. The MAC plate selects for a gram-negative bacterium, on this plate organism A was present. On the PEA plate

organism, A is absent because it only selects for gram positive. Furthermore, these analyses conclude that organism A is gram-negative.  Yes, organism B is gram-positive determined through gram staining. The PEA plate selects the growth of gram-positive organisms, on this plate organism B was present. On the MAC plate it selects for gram-negative organisms, but on this plate organism B was not present since it’s a gram-positive organism. Therefore, these separate analyses conclude that it organism B is grampositive. ________(16.0 pts) Briefly summarize each test (in your flow chart) that allowed you to identify your unknown. Test: name of the test Purpose: objective of the test, include names of enzymes, substrates & products (if applicable) and expected result for positive test. Media & Reagents: name of the media, pH indicator (if any) and reagent(s) if added, temperature Observations & conclusion: describe your observation for your unknown and positive or negative for the test. Gram negative bacteria: Enterobacter aerogenes (Genus 5pts) (species 5 pts) Pts. (2.0) will be subtracted for incorrect formatting of the scientific name. Tests

Lactose Fermentation Indole

Purpose To be able to determine if the organism is able to ferment lactose. The substrate associated is lactose (galactose and glucose). There is no enzyme associated with this reaction. The products are pyruvic acid, possibly others. The expected positive result is a yellow colored appearance with or without gas indicating an acidic pH. To be able to determine which bacteria can decompose tryptophan into indole. he substrate associated is tryptophan. The enzyme associated in

Media & Reagents

Observations & conclusion

The broth used is a red lactose broth. The pH indicator is phenol red. Since temperature was not specified, room temperature.

After incubation the broth turned from red into a yellow color. The yellow color is an indication that fermentation has taken place, indicating a positive result.

The broth used is a tryptone broth. The reagent used is kovac reagent. The pH indicator is phenol red. Since

After the addition of kovac reagent nothing formed at the top of the solution, leading to a negative result. For that reason, it does not have the enzyme tryptophanase to breakdown tryptophan into indole.

Methyl Red (MR)

Voges Proskauer (VP)

Citrate (+) Urease (-)

this reaction is tryptophanase. The products would be indole. The expected result is a fuchsia color ring at the top. To determine if acid was produced. The substrates involved in this reaction is glucose. There is a variable of enzymes involved in this reaction. The products from this reaction would be mixed acids. The expected results from this test is turning red after the addition of the reagent methyl red. To determine if acetoin was produced. The substrates associated is glucose. There is no enzyme associated with the test. The products of the test are acetyl methyl carbonyl or acetoin. The expected positive results are a brown or dark red color after the addition of the reagents A and B. Tests the ability of an organism to use citrate as a carbon source and ammonium phosphate as the only nitrogen source. The substrate associated is citrate. The enzyme involved is citrate permease. The products are Ammonia or Ammonium hydroxide. The expected positive results are blue agar slant after incubation. To determine if the Helicobacter pylori is present causing a secretion of urease and hydrolyzes to create ammonia and carbon dioxide. The substrate involved is urea.

temperature was not specified, room temperature.

The broth used is MR-VP broth (contains glucose, peptone, phos. buffer). The reagent is methyl red. The pH indicator present is methyl red. Since temperature was not specified, room temperature.

The broth used is MR-VP broth (contains glucose, peptone, phos. buffer). The reagents are Alphanaphthol (A) and Potassium hydroxide (B). Since temperature was not specified, room temperature.

The media used is simmons agar. The pH indicator is bromothymol blue. No reagent was needed. Since temperature was not specified, room temperature. The broth used is Christensen urea broth. The pH indicator used is Phenol red. No reagent was needed. temperature was not specified, room temperature.

After the addition of the reagent methyl red the solution did not change. Since there was no apparent change it can be concluded that there was no acid produced and indicating a negative result.

After the addition of reagent, A and B it took approximately 20 minutes to form a brown colored ring at the top. After the color appeared, it can be confirmed that acetoin was produced, therefore a positive result.

Originally the agar slant was green after incubation the color of the slant changed drastically into a deep blue color. Meaning that the organism is able to use citrate as a carbon source and ammonium phosphate as a nitrogen source, indicating a positive result. After incubation the color turned into a yellow/orange color indicating urease was not secreted to hydrolyze creating carbon dioxide and ammonia, therefore a negative result.

H2S (-)

The enzymes associated is urease. The products are Ammonia and carbon dioxide. The expected positive results are the agar turns a fuchsia color. To determine if the organism is able to reduce a sulfur compound into H2S. The substrate involved is cysteine or inorganic sulfur compounds and FeSO4. Th enzyme involved is cysteine desulfurase or thiosulfate reductase. The products would be H2S gas then reacts with FeSO4 —> FeS. The expected positive results are a black precipitate of FeS forms in the agar.

The media used is SIM media. No reagent was needed. Since temperature was not specified, room temperature.

After incubation no black precipitate formed, indicating that the organism was not able to reduce H2S, therefore a negative result.

Gram positive bacteria: Streptococcus bovis (Genus 5pts) (species 5 pts) Pts. (2.0) will be subtracted for incorrect formatting of the scientific name. Tests

Catalase Test Growth @ 10 ° C

Purpose To determine if the organism has the enzyme catalase. The substrate associated is hydrogen peroxide. The enzyme involved is catalase. The products would be water and oxygen. The expected positive result would be bubbles (oxygen production) appearing after the addition of hydrogen peroxide. To determine if the organism is able to grow in 10 ° C environment. There are no enzymes and substrates involved in this test. The products would just be growth (turbid) of the organism. The expected positive result would be

Media & Reagents

TSA plate for growth of organism used in the catalase test. The reagent was hydrogen peroxide. Since temperature was not specified, room temperature. The broth used to inoculate is TSA broth. No reagent added. The temperature that is set is at 10 ° Celsius.

Observations & conclusion

There was no bubbles (oxygen) produced when hydrogen peroxide was added. Therefore, the organism does not have the enzyme catalase indicating a negative result. Growth was not apparent at 10 ° Celsius. This can be due to the failure to reach a specific requirement for the organism therefore no organisms were able to grow indicating a negative result.

Growth @ 45 ° C

Growth in 6.5 % NaCl

Starch Hydrolysis

growth of the organism To determine if the organism is able to grow in 45 ° C environment. There are no enzymes and substrates involved in this test. The products would just be growth (turbid) of the organism. The expected positive result would be growth of the organism. To determine if the organism is able to grow in salt like conditions (6.5% NaCl). There are no enzymes and substrates involved in this test. The product would be growth (turbid) of the organism. The expected positive result would be growth of the organism. To determine if the organism produces amylase to hydrolyze starch. The enzyme involved is amylase. The products are glucose and maltose. The expected positive result is a clear hydrolyzed zone on the plate.

The broth used to inoculate is TSA broth. No reagent added. The temperature that is set is at 45 ° Celsius.

Growth at 45 ° Celsius was very noticeable, it formed a very turbid solution. This organism can be considered a mesophilic due to the mass or organisms produced therefore, can be considered a positive result.

The broth used is 6.5% NaCl broth. No reagent added. Since temperature was not specified, room temperature.

There was no growth in apparent in the 6.5% NaCl broth. We can infer that the organism does not grow well in halophilic (salt) conditions therefore, a negative result.

The media used is starch agar. The reagent used is iodine. Since temperature was not specified, room temperature.

There was a clear zone of where hydrolysis took place. Therefore, the organism does have the enzyme amylase to breakdown starch, indicating a positive result....