Vinegar, Antacid, and Soap all became positive solutions after I put in the enzyme. A change in pH disrupts an enzyme's shape and structure. pH measure acidity--water is neutral and has a pH of 7. When the pH changes an enzyme's structure, the enzyme can't do its job. Changes in pH break the delicate bonds that maintain an enzyme's shape.
Test Result Oxidase This laboratory test is based on detecting the production of the enzyme cytochrome oxidase by Gram-negative bacteria if no color change to purple or blue is evident at 30 seconds-1 minute then the result is positive. It is important that the test is read by one minute to ensure accurate results (avoid false negatives and false positives). Unknown 12 resulted in a negative test with no color change. These were similar results seen in Eschericia coli, Proteus vulgaris,
Calculate the Normality of the vinegar using the previously given equation. Na = (Nb)(Volumeb) (Volumea) C. Calculate the mass of the acetic acid in grams using the previously given equation. Massa = (Na)(GMWa) D. Calculate the percentage of acetic acid using the previously given equation. % Acid = Massa(g/L) x 100 1000g/L Discussion and Conclusion: Questions: LabPaq question guidelines: Answer questions A and G in the lab manual. Skip questions B, C, D, E and F in the lab manual, and answer these instead: A.
Online General Chemistry Titration of Acetic Acid in Vinegar Lab Manual pp158-167 When performing step 9 you want to remove all the NaOH from the beaker before adding acid (vinegar) to the beaker. If you do not remove the NaOH, it will react with the acid in the vinegar prior to starting the titration, making the vinegar appear less acidic than it really is. The next lab lists white vinegar as provided by the student. This one asks for brand and label information, yet implies the vinegar is provided by LabPaq. Results Sheet, Experiment 13 Brand of Vinegar used:________________ Acetic Acid % from bottle label_____________ | Initial NaOH Reading(estimate to 0.1 mL) | Final NaOH Reading(estimate to 0.1 mL) | Volume NaOH used(Final – Initial) | Trial 1 | 9.5 | 1.3 | 8.2 | Trial 2 | 9.6 | 1.4 | 8.2 | Trial 3 | 9.7 | 1.4 | 8.3 | Average Volume of NaOH used | 8.2 | Calculations: Normality of the acetic acid: .82 Mass of acetic acid: 49.2 % acetic acid (divide the grams, above, by the volume you used to get g/L): 4.92 Answer questions A-G in the lab manual: A.
Eosin methylene blue and Mannitol agar plates will also be used to determine pH values. Test Three: Gram Stain – A basic technique which will help to distinguish between a Gram-positive and Gram-negative bacterium. Test Four: Catalase assay – Is to be used to determine whether the unknown bacteria will liberate free O₂ gas. Test Five: Oxidase assay – The cytochrome Oxidase will aid in determining whether the unknown bacteria has an absence of Oxidase or not. Test Six: Glucose assay – Whether an organism can respire or ferment glucose can be tested with this Glucose O/F medium assay.
The purpose of these tests were to determine how the bacterium reacted to glucose with and without oxygen. My findings reported a gram stain negative along with positive tests for glucose oxidation and fermentation. Next I performed an Indole Production Test. This test determines whether the microbe produces indole from the amino acid tryptophan. The results from this test were negative.
One mixture will contain vinegar, and the other will contain distilled water. What I expect to find it that the mixtures containing the vinegar’s pH will remain relatively constant, while the distilled water mixture will change drastically. Results and Observations: 1) The colors of the mixtures varied from drop to drop. One mixture turned blue, while the other turned a gold-ish color. Data Table 1: Add 0.1 M HCl| Drops|pH|Color| 0|6.0|Yellow| 1|6.0 |Yellow| 2|6.0|Yellow| 3|6.0|Yellow| 4|6.0|Yellow| 5|6.0|Yellow| 6|6.0 |Yellow| 7|6.0 |Yellow| 8|6.0 |Yellow| 9|6.0 |Yellow| 10|6.0 |Yellow| 11|6.0 |Yellow| 12|6.0 |Yellow| 13|6.0 |Yellow| 14|6.0 |Yellow| 15|6.0 |Yellow| Data Table 2: Add 0.1 M NaOH| Drops|pH|Color| 0|6.0|Yellow| 1|6.0|Yellow| 2|6.0|Yellow| 3|6.0|Yellow| 4|6.0|Yellow| 5|6.0|Yellow| 6|6.0|Yellow| 7|6.0|Yellow| 8|6.0|Yellow| 9|6.0|Yellow| 10|6.0|Yellow| 11|6.0|Yellow| 12|6.0|Yellow| 13|6.0|Yellow| 14|6.0|Yellow| 15|6.0|Yellow| Data Table 3: Add 6M HCl| Drops|pH|Color| 0|6.0|Yellow| 1|5.8|Yellow Orange| 2|5.6|Yellow Orange| 3|5.4|Light Orange| 4|5.0|Orange| 5|4.5|Orange| Data Table 4: Add 6M NaOH| Drops|pH|Color| 0|6.0 |Yellow| 1|6.2|Yellow Green| 2|6.4|Yellow Green|
Experiment-Specific Questions Experiment 1: Monosaccharide Test Fill in the table below with the results from the monosaccharide test experiment, and your conclusions based on those results. Results Monosaccharide Test Solution Initial Color Color with Benedict's Solution Color After Heating Monosaccharide? glucose solution water sucrose solution fructose solution Benedict’s solution is added to white grape juice and heated. The color changes from blue to orange. Based on this result, what biological molecules are present in the white grape juice?
Protects the bacteria from phagocytosis allowing the bacteria to stay in the body 6. pure culture 7. It is differential based on hemolysis of the agar. Hemolysis can be wide-narrow band beta, alpha, gamma, or none. 8. candle jar in microbiology is used for anaerobiosis in which a lit candle is placed in an air tight jar and if it went out, it would be because it used up all the available oxygen. 9. any streptococcus capable of hemolyzing erythrocytes, classified as α-hemolytic type, producing a zone of greenish discoloration much smaller than the clear zone produced by
The agar plate with the –pGLO LB / amp add no bacteria living in it because it wasn’t ampicillin resistant. But on the other hand the +pGLO LB/amp had a colonies of bacteria because when there is the pGLO plasmid it makes the e-coli bacteria ampicillin resistant so that is the reason why there are colonies of bacteria on the agar plate. For the last agar plate left the +pGLO LB/amp/ara this agar plate is the one that genetically transformed this is the one that could glow under the UV light <Fig 3 bottom right>. The reason that it glowed is because of the arabinose gene on the agar plate, that arabinose gene is the gene that triggers the GFP gene to express