Detection of Biological MoleculesIntroduction: 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 inmonomers, the building blocks of life. These monomers combine to make polymers,which is a long chain of monomers strung together. Biological molecules can bedistinguished by their functional groups. For example, an amino group ispresent in amino acids, and a carboxyl group can always be found in fatty acids.Order now
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 mixwith water. Molecules of a certain class have similar chemical propertiesbecause they have the same functional groups. A chemical test that is sensitiveto these groups can be used to identify molecules that are in that class. Thislab is broken down into four different sections, the Benedict’s test forreducing sugars, the iodine test for the presence of starch, the Sudan III testfor fatty acids, and the Biuret test for amino groups present in proteins.
Thelast part of this lab takes an unknown substance and by the four tests,determine what the substance is. BENEDICT’S TESTIntroduction: Monosaccharides and disaccharides can be detected because oftheir 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 mixingthe sugar solution with the Benedict’s solution and adding heat, an oxidation-reduction reaction will occur. The sugar will oxidize, gaining an oxygen, andthe Benedict’s reagent will reduce, loosing an oxygen.
If the resulting solutionis red orange, it tests positive, a change to green indicates a smaller amountof reducing sugar, and if it remains blue, it tests negative. Materials: onion juice5 test tubes1 beaker potato juicerulerhot plate deionized waterpermanentmarker5 tongs glucose solutionlabels starch solution6 barrelpipettes Benedict’s reagent5 toothpicksProcedure: 1. Marked 5 test tubes at 1 cm and 3 cm from the bottom. Label testtubes #1-#5. 2.
Used 5 different barrel pipettes, added onion juice upto the 1 cm mark of the firsttest tube, potato juice to the 1 cm mark of the second, deionized waterup to the 1cm mark of the third, glucose solution to the 1 cm mark of the fourth,and thestarch solution to the 1 cm mark of the fifth test tube. 3. Used thelast barrel pipette, added Benedict’s Reagent to the 3 cm mark of all 5test tubes and mix with a toothpick. 4. Heated all 5 tubes for 3minutes in a boiling water bath, using a beaker, water, anda hot plate. 5.
Removed the tubes using tongs. Recorded colorson the following table. 6. Cleaned out the 5 test tubes with deionizedwater. Data:Benedict’s Test ResultsDiscussion: From the results, the Benedict’s test was successful. Onion juicecontains glucose, and of course, glucose would test positive.
Starch doesn’thave 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 endresult would be negative, a blue color. IODINE TESTIntroduction:The iodine test is used to distinguish starch frommonosaccharides, disaccharides, and other polysaccharides. Because of it’sunique coiled geometric configuration, it reacts with iodine to produce a blue-black color and tests positive. A yellowish brown color indicates that the testis negative.
Materials: 6 barrel pipettespotato juicestarch solution 5 testtubeswateriodine solution onion juiceglucose solution5 toothpicksProcedure: 1. Used 5 barrel pipettes, filled test tube #1 with onion juice,second with potatojuice, third with water, fourth with glucose solution, and fifth withstarch solution. 2. Added 3 drops of iodine solution with a barrel pipette,to each test tube. Mixedwith 5 different toothpicks.
3. Observed reactions and recordedin the table below. Cleaned out the 5 test tubes. Data:Iodine Test ResultsDiscussion:The iodine test was successful.
Potato juice and starch werethe only two substances containing starch. Again, glucose and onion juicecontains glucose, while water doesn’t contain starch or glucose and was justtested to make sure the test was done properly. SUDAN III TESTIntroduction: Sudan III test detects the hydrocarbon groups that are remainingin the molecule. Due to the fact that the hydrocarbon groups are nonpolar, andstick tightly together with their polar surroundings, it is called a hydrophobicinteraction and this is the basis for the Sudan III test. If the end result isa visible orange, it tests positive.
Material: scissorsdeionized watermargarineSudanIII solution petri dishstarchethyl alcoholforceps lead pencilcream5 barrel pipettes filter papercooking oilblow dryerProcedure: 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, Cforcooking oil and M for margarine. Draw a small circle next to eachletter for thesolution to be placed. 3.
Dissolve starch, cream, cooking oil andmargarine in ethyl alcohol. 4. Used a barrel pipette for each solution, added asmall drop from each solution tothe appropriate circled spot on the filter paper. 5.
Allowed thefilter paper to dry completely using a blow dryer. 6. Soaked the paper in theSudan III solution for 3 minutes. 7. Used forceps to remove the paper fromthe stain. 8.
Marinated the paper in a water bath in the petri dish, changedwater frequently. 9. Examined the intensity of orange stains of the 5 spots. Record in the table below.
10. Completely dried the filter paper, andwashed the petri dish. Data: Sudan III Test ResultsFilter paper:Discussion: The results indicate that the Sudan III test was sucessful. Waterand starch definitely doesn’t contain any fatty substances. Cream and cookingoil no doubtedly does contain lipids.
It was surprising to find that margarinedoesn’t contain any fat. BIURET TESTIntroduction: In a peptide bond of a protein, the bond amino group issufficiently reactive to change the Biuret reagent from blue to purple. Thistest is based on the interaction between the copper ions in the Biuret reagentand the amino groups in the peptide bonds. Materials: 6 test tubesegg white solutionstarchsolution6 toothpicks rulerchicken soupsolutiongelatin6 parafilm sheets permanentmarkerdeionized watersodium hydroxide labelsglucose solutioncopper sulfateProcedures: 1. Used 6 test tubes, and labeled them at 3cm and 5cm from thebottom.
Labeledeach #1 to #6. 2. Added egg white solution to the 3cm mark of thefirst tube, chicken soup solutionto the 3-cm mark of the second tube, water to the 3 cm mark of the thirdtest tube,glucose solution to the fourth, starch to the fifth, and gelatin to thesixth, all at the3 cm mark. 3.
Added sodium hydroxide to the 5 cm mark of each tube andmix with 6 differenttoothpicks. 4. Added 5 drops of Biuret test reagent, 1% copper sulfate,to each tube and mixby placing a parafilm sheet over the test tube opening, and shakevigorously. 5. Held the test tubes against a white piece of paper, and recordedthe colors andresults.
Discarded the chemicals, and washed the test tubes. Data:Biuret Test ResultsDiscussion: The Biuret test seemed to have been successful. Glucose and starchare both carbohydrates, while water has no proteins. Egg white definitely hasproteins, and so does gelatin. Chicken soup had a hint of protein content.
Unknown Chemical # 143Introduction: By performing the Benedict’s Test, the Iodine Test, the Sudan IIITest, and the Biuret Test, chemical #143 should be identified. Materials: materials from the Benedict’s Testmaterials from the SudanIII Test Materials from the Iodine Testmaterials from theBiuret TestProcedures: 1. Performed the Benedict’s Test, and recorded results. 2.
Performed the Iodine Test, and recorded results. 3. Performed the Sudan IIITest, and recorded results. 4. Performed the Biuret Test, and recorded results.
Data: Properties of Chemical #143chemical #143 was a white powderish substance. Conclusion: After ruling out the obvious wrong substances from the list likeground coffee, egg white and yolk, table sugar and salt, syrup and honey, thesmall amount of proteins was taken into factor. That also eliminated powderedskim milk, and soy flour. The low, or none fat content ruled out some morechoices like enriched flour.
The only choices left was corn starch, glucose,and potato starch. Because of the low reducing sugar, glucose can be ruled outalso. The starch content of substance #143 was very high. The protein content wasaround 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 #143is potato starch.