Proteus Vulgaris Experiment

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The tubes of E. coli in the phenol red lactose broth one with and the other without mineral oil on top were reddish pink, meaning that the pH was alkaline. That would then mean E. coli can not utilize lactose. However, it is know that E. coli can utilize lactose (BioCoach Activity). The E. coli in a phenol red glucose broth with mineral oil on top produced a yellow color meaning it had a low pH and thus created an acid. There was also gas in the tube which meant that it produced a gas. The E. coli in a phenol red glucose broth without mineral oil was red, that would indicate that an acid was not produced. There was also no gas that was produced. The two lactose broths containing Bacillus subtilis were both maintained the red color of the and produced no gas. This would mean that B. subtilis can not metabolize lactose. B. …show more content…

If the test was down right, Proteus vulgaris should have be able to breaks down gelatin to make it a liquid (Herter & Ten Broeck, 1911). This means that they desecrate a proteinase enzyme that breaks down gelatin. E. coli should not have been able to break down the gelatin (Exoenzymes). In the urea broth, the Proteus vulgaris broth turned pink and the E. coli did not. This means that the Proteus vulgaris is able to secrete the enzyme urease that breaks down area and forms ammonia making the broth alkaline and pink. Since E. coli did not turn pink, it does not secrete urease. Both E. coli and Lactococcus lactis had blue rings formed during the oxidize test meaning that cytochrome C is present in both. There was no bubbling of either E. coli and Lactococcus lactis during the catalase test. According to Kumar and Imlay (2013), E. coli has catalase to protect it self from hydrogen peroxide. These means that E. coli should have bubbled during the test. Lactococcus lactis does not have a catalase (Rochat et al., 2005). So, it should not have

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