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
One bacterium was gram negative. It underwent four different tests. These tests were the EMB test (Eosin Mehylene Blue), the Sulfur Indole Motility (SIM) test, the Urease test, and the Simmon’s Citrate Utilization test. The EMB test checks for a bacteria’s ability to ferment lactose. This test is accomplished by placing the bacteria on Eosin Methylene Blue agar. The agar is selective for gram negative bacteria and those bacteria that can ferment lactose will have colored growth, usually a metallic green sheen.
The results of the gram stain test were cocci and purple. This indicated that the unknown bacteria were gram positive. The gram stain test eliminated Escherichia coli, Klebsiella pneumonia, Salmonella enterica, and Yersinia enterocolitica as choices because these bacteria are gram negative. Next a Blood Agar plate was used because in order to do a MSA or a Catalase test there needs to be a colony of the bacteria. The result of the Blood Agar plate was nonhemolytic.
While the tube for specimen Cb turned a tannish white in the lower half of the tube while the top stayed the lavender inoculated tube color. Do to this evidence I determined that both specimens Ca and Cb cannot use the process Casein hydrolysis or Casein coagulation due to lack of soft or firm curds in both tubes. Since there was no casein curds formed, I concluded that specimens Ca and Cb also cannot perform the process of proteolysis. My conclusion is supported by the fact that there was no clearing of the medium. I have also determine that neither of my organisms can make the enzymes rennin, proteolytic or even proteases. I know my specimens cannot produce proteases due to the fact that there was no blue coloring in the tubes which means that the byproduct Ammonia was not produced to increase the pH. Since neither of my specimens can make these enzymes, I concluded that my specimens cannot break down lactose or casein. Although I did learn that specimen Cb can reduce litmus due to the evidence that the lower part of the tube turned a tannish white color with a purple ring at the top. This color change from a purple to a white means that the litmus was reduced turning it clear and leaving the white of the milk to show. Finally I know that specimen Ca cannot reduce litmus due to the fact that the tube had no change in
The purpose of this experiment was to discover the specificity of the enzyme lactase to a spec...
Lactose alone cannot induce the lac operon. E. Coli prefers to utilize glucose rather than lactose because glucose is simpler and more abundant.
The first step to the unknown is selecting an actual organism. The best way to select a culture is based on a high-quality distribution. Equally important, shaking up the broth tube facilitates in the distribution. Upon selection, a gram check for purity is performed. Step by step instructions for this procedure can be found in Benson’s, Microbiological Applications p. 99. Furthermore, an aseptic technique must be performed for this test and the entire tests following the unknown. The purpose of this test is to differentiate between gram positive and gram-negative bacteria. The key indicator of gram-positive bacteria is a purple stain and a pink stain for gram-negative bacteria. A slide is viewed with a microscope under oil immersion. Equally
This is because the cells are normally stained with Gram's iodine solution which forms a complex with crystal violet stain that is insoluble in water. Addition of decolourizer dehydrates the peptidoglycan layer; tightening it and shrinking peptidoglycan layer of the gram positives, making the large complex not penetrate the layer. When a counter stain is added, it does not disrupt the complex i.e. purple colouration of the Gram positive cells. Therefore, in absence of Gram's iodine solution, the Gram positive cells would stain brownish red as gram negative cells, making it difficult for the identification of the unknown bacteria (Leboffe and Pierce, 2010). 13. (2 points) Control slides are missing in The Gram Stain Investigation. What are they and what is their importance in the straining
Generally, lactic acid bacteria (LAB) can be defined as Gram positive, non-spore forming, catalase negative, devoid of cytochromes, acid tolerant, and facultative anaerobe group that produce lactic acid as the major end-product during fermentation of carbohydrates. According to carbohydrate metabolism, they can be divided into two main groups:
I identified the genus and species of an unknown bacterial culture, #16, and I applied the following knowledge of morphologic, cultural and metabolic characteristics of the unknown microorganism according to the laboratory manual as well as my class notes and power point print outs. I was given an incubated agar slant labeled #16 and a rack of different tests to either examine or perform myself; the tests are as follows: Gram Stain; Nutrient Gelatin Test; Carbohydrate Fermentation; Dextrose, Lactose and Sucrose; IMVIC tests; Citrate, Indole, Mythel-Red and Vogues Proskauer test; as well as a Urease and TSI Test.
The sugar molecule lactose is found in dairy products and people with Lactose Intolerance are unable to digest lactose because the enzyme lactase is unable to split the sugar molecule into glucose. Lactase is an enzyme that breaks lactose down into galactose and glucose. Lactase functions best between 21 and 48 degrees Celsius (or 70 and 120 degrees Fahrenheit). Cooler temperatures will slow down lactase’s function, whereas high temperatures can denature it or lactase will lose its shape. If lactase is rendered nonfunctional because of temperature or pH extremes, the breakdown of lactose stops. Lactose intolerance occurs when this breakdown fails due to insufficient or ineffective lactase. The experiment will stimulate the process
... result of the Gelatin test. The only test that did not match the E. aerogenes identification was the VP test. Possible errors made to not achieve an accurate test could be not waiting a full hour to view the color change or even not using the correct Reagent.
= Before conducting the experiment I would conduct a simple test for the protein by placing a sample of the albumen into a test tube and add biurett reagent. This contains copper (II) sulphate and sodium hydroxide.
Bacteria are grouped into two categories as Gram positive and Gram negative. The bacteria that retain the color of the primary stain are gram positive and the bacteria that lose the color of primary stain are called gram negative bacteria. Gram positive cells have a thicker peptidoglycan layer and doesn 't contain the outer membrane. Gram positive cells will retain the primary stain of crystal violet because they have a thick outer layer of peptidoglycan that traps the dye among its high degree of teichoic acid crosslinks. On the other hand, Gram (-) cells have a thin peptidoglycan layer and an outer membrane, gram negatives have an outer membrane made up of Lipopolysaccharide, proteins and prions and have thin peptidoglycan layer with higher lipid content that are targeted by the alcohol/ acetone decolorizer and makes the cell’s outer layers more porous, thus they are unable to retain the primary stain.
While Pasteur was at Lille, a local distiller sought help controlling the fermentation of beet sugar. Pasteur realized that fermentation was not a simple chemical process but that living organisms were involved. This led him to discover that fermentation, infection, and spoiling were the result of microbes. The first paper he published discussed lactic acid and it’s role in souring milk. He spent numerous years studying microbes and proving that they do not originate from within matter, but that they come in from the outside. He eventual...
Biology Experiment Hypothesis I predict that the potatoes in the solution with high sugar concentration will increase in mass the most, the potatoes in the water with no sugar will lose the most mass. Osmosis is the movement of water from a high concentration to a low concentration across a particularly permeable membrane. The water can move between the potato and the solution but the sugar can’t because the molecules are too big to pass through the membrane. The water will move between the solution and the potato. If the net movement of water is from the potato then the potato will decrease in mass, it will also become flaccid.