MICROBIAL IDENTIFICATION

ABSTRACT During the practice we performed a series of different tests to identify the microorganism we were handling, this identification is possible because not all microorganisms behave the same way, so we performed this tests to see the behavior of the microorganism against different growing cultures and depending on their growth and behavior we will be able to confirm the organisms we are handling. In this specific case we were able to confirm that the microorganism that we were handling was Escherichia coli. Beside the biochemical tests used, we used API test to identify different organisms. RESULTS AND ANALYSIS

Image 1. INDOL TEST For the indol test we were able to observe a pink ring on the surface of the broth indicating a positive result. If we have a positive result, we can affirm that the bacterium has the tryptophanase enzyme catalyses the following reaction: L-tryptophan + H2O indole + pyruvate + NH3

This enzyme participates in the nitrogen metabolism so this could mean that the bacterium needs nitrogen to keep its metabolism working properly. According to literature Escherichia coli is a positive bacterium for yhe indol test, an expected result for this test confirming that the microorganism we are handling is E.coli.

Image 2 CITRATE TEST The citrate test result was negative, if the citrate test was positive it would have to be blue. Citrate agar tests the ability of organisms to utilize citrate as a carbon source. Organisms which can utilize citrate as their sole carbon source use the enzyme citrase or citratepermease to transport the citrate into the cell. These organisms also convert the ammonium dihydrogen phosphate to ammonia and ammonium hydroxide, which creates an alkaline environment in the medium. We are able to see that E. coli gives a negative result for citrate test, result that confirms what theory says for E. coli citrate test.

Image 3 Methylene Red Test

This is a positive result for methylene red test. The Methyl Red test involves adding the pH indicator methyl red to an inoculated tube. If the organism uses the mixed acid fermentation pathway and produces stable acidic end-products, the acids will overcome the buffers in the medium and produce an acidic environment in the medium. When methyl red is added, if acidic end products are present, the methyl red will stay red. So this result means that E. coli uses the mixed fermentation pathway to produce acidic products at the end. This result is consistent with what literature says about the behavior for E. coli.

Image 4. VOGES-PROSKAUER TEST for the Voges-Proskauer (VP) test we has a doubtful result because half of the broth has a brownish color and the other half has a dark red color, the VP test detects organisms that utilize the butylene glycol pathway and produce acetoin. When the VP reagents are added to MR-VP broth that has been inoculated with an organism that uses the butylene glycol pathway, the acetoin end product is oxidized in the presence of potassium hydroxide (KOH) to diacetyl. Creatine is also present in the reagent as a catalyst. Diacetyl then reacts to produce a red color. Therefore, red is a positive result but according to what literature says, E. coli gives a negative result for this test so it is possible to say that this specific E. coli strain uses butylene glycol pathway to produce acetoin. Acetoin is a neutral, four carbon molecule used as an external energy store by a number of fermentive bacteria. Acetoin prevents the overacidification of the media which results from the accumulation of acid products which we already know E. COLI is able to produce.

Image 5. MOTILITY TEST. during the motility test we were able to observe different patterns all over the broth, this means a positive result for motility test. During motility test we will be able to see if the microorganism has flagella, appendix that allows the microorganisms movement and if it is able to move it will grow all over the broth and not only in the inoculated part. This result is also consistent with what theory says about E. coli result for motility test.

Image 6. KLIGER TEST This is a positive result for CO2 production and for acid production, and it is negative for sulfuric acid production. In this test we will be evaluating the bacteriums capacity to perform fermentation process, this includes the krebs cycle and the glycolysis pathway. If a bacterium is able to perform the krebs cycle and the glycolysis pathway it means that it has to able to ferment glucose. During the krebs cycle every organism produces carbon dioxide, which makes the big bubbles in the test tube and during glycolysis pyruvic acid is produced resulting in a acid media which gives the solution the yellow color. Hydrogen sulfide (H2S) is produced by bacterial anaerobic degradation of the two sulfur-containing amino acids, cysteine and methionine. Hydrogen sulfide is released as a by-product when carbon and nitrogen atoms in the amino acids are consumed as nutrients by the cells. Under anaerobic conditions the sulfhydryl (-SH) group on cysteine is reduced by cysteine desulfurase.

Image 7 UREA TEST this is a negative result for urea test. Urea is a nitrogenous waste product of animals. Some bacteria can break urea to produce carbon dioxide and ammonia. The ammonia is a nitrogen source for amino acid biosynthesis as well as for synthesis of other nitrogen-containing molecules in the cell.

E. coli doesnt have the urease enzymes that help the bacteria to break the urea molecules into carbon dioxide an ammonia.

Image 8. Phenylalanine test This is a negative result for the phenylalanine test. Phenylalanine medium tests the ability of an organism to produce the enzyme deaminase. This enzyme removes the amine group from the amino acid phenylalanine and releases the amine group as free ammonia. As a result of this reaction, phenylpyruvic acid is also produced. E. coli cant use phenylalanine as source to obtain ammonia.

Image 9. GLUCOSE, MALTOSE AND LACTOSE TEST

This test has positive results for lactose, glucose and maltose fermentation. This bacterium is able to perform fermentation; it uses glucose, maltose or lactose (all are sugar sources) as an energy source for this process.

For the catalase test we had a positive result. The bacterium has the enzyme that breaks hydrogen peroxide (H2O2) into H2O and O2. Hydrogen peroxide is often used as a topical disinfectant in wounds, and the bubbling that is seen is due to the evolution of O2 gas. H2O2 is a potent oxidizing agent that can cause severe damage in the cell; because of this, any cell that uses O2 or can live in the presence of O2 must have a way to get rid of the peroxide. One of those ways is to make catalase. An expected result for E. coli. The oxidase test identifies organisms that produce the enzyme cytochrome oxidase. Cytochrome oxidase participates in the electron transport chain by transferring electrons from a donor molecule to oxygen. For this test we used plastic strips that contained chomegenic reducting agent, which is a compound that changes color (it must turn purple) when it becomes oxidized. For this test we had a doubtful result, the strip looked brown in the outside and had a tiny blue spot in the center but comparing with what literature says for this test using E. coli we can say that negative would be a possible result for this test and the blue spot could be a contamination problem when taking the sample.
Gram stain.

CONCLUSION According to the Gram stain and with the biochemical tests we can confirm that the microorganism we were handling was Escherichia coli. Biochemical tests that we expected to be positive for E. coli were all positive and thanks to these biochemical tests we know that our bacterium ferments lactose glucose and maltose to produce acid compounds and carbon dioxide, Acid products will cause a noticeable color change in the pH indicator included in the medium.

Sugar fermentation does not produce alkaline products. Thats why in most of the tests we use pH indicators that change color at low pH values. We also know that the microorganism we used is able to break hydrogen peroxide so it is able to survive in hydrogen peroxides presence, if we have a wound infected with E. coli hydrogen peroxide wouldnt be a good disinfectant. E. coli has flagella so it is able to move and make chemotaxis, is a harmful substance reaches the bacterium it will be able to runaway from danger. The microorganism has the tryptophanase enzyme which reaction gives like a product pyruvic acid which is used during the krebs cycle to perform respiration. E. coli doesnt have the urease enzymes so we wont find E. coli using animal wasted as a source of ammonia. Looking at the API results we were able to identify four different microorganisms: lactobacillus fermentum, staphylococcus aureus, Listeria monocytogenes and listeria innocua.

We can say that the lactobacillus fermentum ferments maltose, glucose lactose D-mannose, fructose, Dribose and D xylose. If the bacterium ferments glucose lactose maltose it is very possible that the bacterium makes krebs cycle to release CO2 and glucolysis to generate pyruvic acid which is needed in the krebs cycle. BIBLIOGRAPHY

BBL Kliger Iron Agar Slants Quality control procedures http://www.bd.com/ds/technicalCenter/inserts/L007458(08)(201101).pdf PML microbiological technical data sheet Lab manual supplemental Spring 2008 http://inst.bact.wisc.edu/inst/index.php?module=Book&func=displayarticle&art_id=123 http://www.mesacc.edu/~johnson/labtools/Dbiochem/imvic.html

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