Identifying Macromolecules in Common Foods

And finally into test tube 3, I pipetted 1.0 ml turnip extract and 4.0 ml of water. The contents of test tube 1 was poured into a spectrometer tube and labeled it “B” for blank. “B” tube was now inserted it into the spectrometer. An adjustment to the control knob was made to zero the absorbance reading on the spectrometer since one cannot continue the experiment if the spectrometer is not zeroed. A combination of two people and a stop watch was now needed to not only record the time of the reaction, but to mix the reagents in a precise and accurate manner. As my partner recorded the time, I quickly poured tube 3 into tube 2. I then poured tube 2 into the experiment spectrometer tube labeled “E” and inserted it into the spectrometer. A partner then recorded the absorbance reading for every 20 seconds for a total of 120 seconds. After the experiment, a brown color in the tube should be observed to indicate the reaction was carried out. Using sterile techniques, any excess liquid left was disposed

Step 4: Place the test tubes into separate coffee cups to maintain the upward position. Add 2 mls of the catalase solution to each of the test tubes and then place tubes1, 2, and 3 in the conditions described above. For test tube 4, fill the coffee mug half full of boiling

The purpose of this lab was to test different substances using various procedures to see what biomolecules were present and ultimately find out what restaurant Anna Lyza had eaten at before she died. For the first control test, we used vegetable oil to test for lipids. So, if the solution does not contain lipids, it does not become translucent when placed onto a paper bag square and held up to a light. So, it is a negative result. However, in the presence of lipids, the solution will become translucent when placed onto a paper bag square and held up to a light. Therefore in this case, the result is positive. On the other hand, we used albumin egg to test for proteins in another control test. If the solution does not contain proteins, it will not experience any color change and so it is a negative result. When there are proteins existing in the solution, it will turn bluish/purplish and for this reason it is a positive result. Furthermore in the third control test, we used dextrose to test for simple carbohydrates such as glucose. If the solution does not contain simple carbohydrates, it will not undergo any color change and will remain a blue color. So, it is a negative result in this circumstance. If there are simple carbohydrates present in the solution, the solution will turn reddish and so the result is positive. For the last control test, we used starch solution to test

The null hypothesis will be that the test tubes with an increase in temperature, pH values, enzyme concentrations, and substrate concentration will have a very small color change or no color change at all. The alternate hypothesis is that the test tubes containing an increase in temperature, pH values, enzyme concentrations, and substrate concentration will all have an intense color change; the more the change, the more intense the color change will be.

12.Stir then pour 2.5ml of the enzyme mixture into one of the test tubes not allowing any water from the saucepan into the test tube.

The purpose of this experiment was to find out the nutrient content of unidentified food samples and to use chemical reagents to see if the unidentified foods for sugar, starch, protein and fat (lipid) content. After comparing the color changes in each unknown substance then you will be able to see the amount of sugar, starch, protein or fat there is in the unknown substance.

In Part C of the experiment, the process completed in Part B was performed again on the six Kool-Aid drink samples: black cherry, cherry limeade, grape, mixed berry, strawberry, and tropical punch. The cherry limeade, grape, mixed berry, and tropical punch samples were undiluted stock solutions that were used in the SpectroVis. The black cherry solution was diluted with a 2-fold dilution. The 3 mL of black cherry solution was diluted with 3 mL of deionized water. This was done because the Red 40 dye was too concentrated. The strawberry solution was first diluted by a 2-fold but the absorbances values were too high and fell outside the acceptable range of 0-1. Therefore, 8 mL of strawberry solution was diluted with 12 mL of deionized water. These values for all of the solutions were recorded in Table 5. The solutions were emptied down the sink. All the glassware and cuvettes were rinsed with deionized water and dried. The volumetric flasks and caps were returned to the appropriate containers.

7)a) You would do the Benedict’s reagent test for simple sugars and the Lugol’s solution or Iodine test for polysaccharides and starch. To test the lipids you would use the Sudan iV test and Biurets reagent test for protein.

2.List whether each of the following substances was positive or negative for reducing sugar, as indicated by the Benedict's test. (6 points)

In part II of the lab six small glass tubes were obtained in a test tube rack. Ten drops of distilled water were then added to test tube 1, five drops to tubes 2-4, and no drops in tubes 5 and 6. Five drops of 0.1M HCl were added to test tube 5 and five drops of 0.1M NaOH to test tube 6. Five drops of enzyme were then added to all tubes except tube 1. Tube 3 was then placed in the ice bucket and tube 4 was placed in the hot bucket at 80-900C for five minutes, the remaining tubes were left in the test tube rack. After the five minutes five drops of 1% starch was added to every tube and left to sit for ten minutes. After ten minutes five drops of DNSA were then added to all the tubes. All the tubes were then taken and placed in the

Usually, the higher the concentration of the solute for a specific taste in the sample solution, the easier the panellist can identify the taste attribute. In this experiment, the concentrations of chemical compounds for each taste attribute in the sample solutions were low. For instance, the concentration of sucrose in the sucrose sample solution was 1%, concentration of sodium chloride in the sodium chloride solution was 0.2%, concentration of citric acid in the sample solution was 0.04% and concentration of MSG in the sample solution was 0.08%. The low concentration of the sucrose may cause some difficulties in differentiating the sweet and neutral taste of the plain water. Panellists may also find difficulty in differentiating the salt

There are exceptions to conducting this test. For example, non-reducing sugars such as sucrose cannot be detected using the Benedict’s test (Mrothery, 2004). In order for non-reducing sugars to be tested with a Benedict reagent, hydrochloric acid (HCl) and sodium hydroxide (NaOH) must be placed into the solution before adding a Benedict reagent (Mrothery, 2004). As a result, sucrose (C12H22O11), made up of two monosaccharides or two carbonyl carbons, is hydrolysed and broken down its glyosidic bond to form glucose and fructose or one carbonyl carbon when heated (Ausetute, 2014 and Chem-Guide, 2010) (see Appendix 4). The sodium hydroxide then neutralises the solution to ensure a positive Benedict Test (Chem-Guide, 2010).

1. Number four clean test tubes 1-4. 2. Place 1 ml (or 20 drops) of the following solutions into each tube: water, egg albumin, starch, and chicken broth. 3. Add 3 drops of Buiret reagent to each tube and gently shake

Solutions and color reaction for Benedict’s test for reducing sugars and 2 iodine test for starch

Niacin is found in its natural form in fish, peanuts, yeast, and grains. In the milling process of grains, it is usually removed. The government’s action of trying to reduce niacin deficiency has led to the fortification of niacin in food. Fortification is an attempt to replenish the niacin lost in food processing. Niacin helps the energy from food reach the cells in the body. It’s important to the health of the live and production of hormones.