Detection of Biological Molecules Essay
Introduction: Without carbon, nitrogen, hydrogen, sulfur, oxygen and phosphorus, life wouldn’t exist. These are the most abundant elements in living organisms. These elements are held together by covalent bonds, ionic bonds, hydrogen bonds, and disulfide bonds. Covalent bonds are especially strong, thus, are present in monomers, the building blocks of life. These monomers combine to make polymers, which is a long chain of monomers strung together. Biological molecules can be distinguished by their functional groups. For example, an amino group is present in amino acids, and a carboxyl group can always be found in fatty acids.
The groups can be separated into two more categories, the polar, hydrophilic, and the nonpolar, hydrophobic. A fatty acid is nonpolar, hence it doesn’t mix with water. Molecules of a certain class have similar chemical properties because they have the same functional groups. A chemical test that is sensitive to these groups can be used to identify molecules that are in that class. This lab is broken down into four different sections, the Benedict’s test for reducing sugars, the iodine test for the presence of starch, the Sudan III test for fatty acids, and the Biuret test for amino groups present in
Introduction: Monosaccharides and disaccharides can be detected because of their free aldehyde groups, thus, testing positive for the Benedict’s test. Such sugars act as a reducing agent, and is called a reducing sugar. By mixing the sugar solution with the Benedict’s solution and adding heat, an oxidation- reduction reaction will occur. The sugar will oxidize, gaining an oxygen, and the Benedict’s reagent will reduce, loosing an oxygen. If the resulting solution is red orange, it tests positive, a change to green indicates a smaller amount of reducing sugar, and if it remains blue, it tests negative. Materials: onion juice5 test tubes1 beaker potato juice rulerhot plate deionized waterpermanent marker5 tongs glucose solutionlabels starch solution6 barrel pipettes Benedict’s reagent5 toothpicks Procedure: 1.Marked 5 test tubes at 1 cm and 3 cm from the bottom. Label test tubes #1-#5. 2.Used 5 different barrel pipettes, added onion juice up to the 1 cm mark of the first test tube, potato juice to the 1 cm mark of the second, deionized water up to the 1 cm mark of the third, glucose solution to the 1 cm mark of the fourth, and the starch solution to the 1 cm mark of the fifth test tube. 3.Used the last barrel pipette, added Benedict’s Reagent to the 3 cm mark of all 5 test tubes and mix with a toothpick. 4.Heated all 5 tubes for 3 minutes in a boiling water bath, using a beaker, water, and a hot plate. 5.Removed the tubes using tongs.
Recorded colors on the following table. 6.Cleaned out the 5 test tubes with deionized water. Data: Benedict’s Test Results Discussion: From the results, the Benedict’s test was successful. Onion juice contains glucose, and of course, glucose would test positive. Starch doesn’t have a free aldehyde group, and neither does potato juice, which contains starch. Water doesn’t have glucose monomers in it, and was tested to make sure the end result would be negative, a blue color. IODINE TEST Introduction:The iodine test is used to distinguish starch from monosaccharides, disaccharides, and other polysaccharides. Because of it’s unique coiled geometric configuration, it reacts with iodine to produce a blue- black color and tests positive. A yellowish brown color indicates that the test is negative. Materials: 6 barrel pipettespotato juicestarch solution 5 test tubeswateriodine solution onion juice glucose solution5 toothpicks
Procedure: 1.Used 5 barrel pipettes, filled test tube #1 with onion juice, second with potato juice, third with water, fourth with glucose solution, and fifth with starch solution. 2.Added 3 drops of iodine solution with a barrel pipette, to each test tube. Mixed with 5 different toothpicks. 3.Observed reactions and recorded in the table below. Cleaned out the 5 test tubes. Data: Iodine Test Results Discussion:The iodine test was successful. Potato juice and starch were the only two substances containing starch. Again, glucose and onion juice contains glucose, while water doesn’t contain starch or glucose and was just tested to make sure the test was done properly. SUDAN III TEST Introduction: Sudan III test detects the hydrocarbon groups that are remaining in the molecule. Due to the fact that the hydrocarbon groups are nonpolar, and stick tightly together with their polar surroundings, it is called a hydrophobic interaction and this is the basis for the Sudan III test. If the end result is a visible orange, it tests positive. Material: scissorsdeionized watermargarineSudan III solution petri dishstarchethyl alcohol forceps lead pencilcream5 barrel pipettes filter paper cooking oilblow dryer
Procedure: 1.Cut a piece of filter paper so it would fit into a petri dish. 2. Used a lead pencil, and marked W for water, S for starch, K for cream, C for cooking oil and M for margarine. Draw a small circle next to each letter for the solution to be placed. 3.Dissolve starch, cream, cooking oil and margarine in ethyl alcohol. 4.Used a barrel pipette for each solution, added a small drop from each solution to the appropriate circled spot on the filter paper. 5.Allowed the filter paper to dry completely using a blow dryer. 6.Soaked the paper in the Sudan III solution for 3 minutes. 7.Used forceps to remove the paper from the stain. 8.Marinated the paper in a water bath in the petri dish, changed water frequently. 9.Examined the intensity of orange stains of the 5 spots. Record in the table below. 10. Completely dried the filter paper, and washed the petri dish. Data: Sudan III Test Results Filter paper: Discussion: The results indicate that the Sudan III test was sucessful. Water and starch definitely doesn’t contain any fatty substances. Cream and cooking oil no doubtedly does contain lipids. It was surprising to find that margarine doesn’t contain any fat.
Introduction: In a peptide bond of a protein, the bond amino group is sufficiently reactive to change the Biuret reagent from blue to purple. This test is based on the interaction between the copper ions in the Biuret reagent and the amino groups in the peptide bonds. Materials: 6 test tubesegg white solutionstarch solution6 toothpicks rulerchicken soup solutiongelatin6 parafilm sheets permanent markerdeionized watersodium hydroxide labels glucose solutioncopper sulfate Procedures: 1.Used 6 test tubes, and labeled them at 3cm and 5cm from the bottom. Labeled each #1 to #6. 2.Added egg white solution to the 3cm mark of the first tube, chicken soup solution to the 3-cm mark of the second tube, water to the 3 cm mark of the third test tube, glucose solution to the fourth, starch to the fifth, and gelatin to the sixth, all at the 3 cm mark. 3.Added sodium hydroxide to the 5 cm mark of each tube and mix with 6 different toothpicks. 4.Added 5 drops of Biuret test reagent, 1% copper sulfate, to each tube and mix by placing a parafilm sheet over the test tube opening, and shake vigorously. 5.Held the test tubes against a white piece of paper, and recorded the colors and results. Discarded the chemicals, and washed the test tubes.
Data: Biuret Test
Results Discussion: The Biuret test seemed to have been successful. Glucose and starch are both carbohydrates, while water has no proteins. Egg white definitely has proteins, and so does gelatin. Chicken soup had a hint of protein content. Unknown Chemical # 143 Introduction: By performing the Benedict’s Test, the Iodine Test, the Sudan III Test, and the Biuret Test, chemical #143 should be identified. Materials: materials from the Benedict’s Testmaterials from the Sudan III Test Materials from the Iodine Testmaterials from the Biuret Test Procedures: 1.Performed the Benedict’s Test, and recorded results. 2. Performed the Iodine Test, and recorded results. 3.Performed the Sudan III Test, and recorded results. 4.Performed the Biuret Test, and recorded results. Data: Properties of Chemical #143 chemical #143 was a white powderish substance.
Conclusion: After ruling out the obvious wrong substances from the list like ground coffee, egg white and yolk, table sugar and salt, syrup and honey, the small amount of proteins was taken into factor. That also eliminated powdered skim milk, and soy flour. The low, or none fat content ruled out some more choices like enriched flour. The only choices left was corn starch, glucose, and potato starch. Because of the low reducing sugar, glucose can be ruled out also. The starch content of substance #143 was very high. The protein content was around the 10% range, so potato starch would be a better guess then corn starch. But corn starch contained only a trace of fat when potato starch contained 0.8%. But 0.8% is very insignificant. The most educated guess to what chemical #143 is potato starch.