carbohydrates and it's properties, Exercises of Biochemistry

Download carbohydrates and it's properties and more Exercises Biochemistry in PDF only on Docsity! Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT Republic of the Philippines University of Northern Philippines Tamag, Vigan City 2700, Ilocos Sur CARBOHYDRATES I. Objective: To characterize carbohydrates present in an unknown solution on the basis of various chemical assays. II. Theory: Carbohydrates are polyhydroxy aldehydes and ketones or substances that hydrolyze to yield polyhydroxy aldehydes and ketones. Aldehydes (–CHO) and ketones (= CO) constitute the major groups in carbohydrates. Carbohydrates are mainly divided into monosaccharides, disaccharides and polysaccharides. The commonly occurring monosaccharides includes glucose, fructose, galactose, ribose, etc. The two monosaccharides combine together to form disaccharides which include sucrose, lactose and maltose. Starch and cellulose fall into the category of polysaccharides, which consist of many monosaccharide residues. Molisch’s Test:  https://www.youtube.com/watch?v=zHcmxBi_wG0 This is a common test for all carbohydrates larger than tetroses. The test is on the basis that pentoses and hexoses are dehydrated by conc. Sulphuric acid to form furfural or hydroxymethylfurfural, respectively. These products condense with α-naphthol to form purple condensation product. Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT Furfural α-Naphthol Fehling’s Test: This forms the reduction test of carbohydrates. Fehling’s solution contains blue alkaline cupric hydroxide solution, heated with reducing sugars gets reduced to yellow or red cuprous oxide and is precipitated. Hence, formation of the yellow or brownish-red colored precipitate helps in the detection of reducing sugars in the test solution. Benedict’s test:  https://www.youtube.com/watch?v=uY98fzHu4M4 As in Fehling’s test, free aldehyde or keto group in the reducing sugars reduce cupric hydroxide in alkaline medium to red colored cuprous oxide. Depending on the concentration of sugars, yellow to green color is developed. All monosaccharides are reducing sugars as they all have a free reactive carbonyl group. Some disaccharides, like maltose, have exposed carbonyl groups and are also reducing sugars, but less reactive than monosaccharides Barfoed’s Test:  https://www.youtube.com/watch?v=yQfMqvOxPrc Barfoed's test is used to detect the presence of monosaccharide (reducing) sugars in solution. Barfoed's reagent, a mixture of ethanoic (acetic) acid and copper (II) acetate, is combined with the test solution and boiled. A red copper (II) oxide precipitate is formed will indicates the presence of reducing sugar. The reaction will be negative in the presence of disaccharide sugars because they are weaker reducing agents. This test is specific for monosaccharides. Due to the weakly acidic nature of Barfoed's reagent, it is reduced only by monosaccharides. Seliwanoff’s Test: https://www.youtube.com/watch?v=KcODIk9EKQo Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT 3) Benedict’s Test: In the test tube with 2 ml of Benedict's reagent, add 5-6 drops of the test carbohydrate solution and mix well. Place the test tube in a boiling water bath for 5 minutes and observe any change in color or precipitate formation. Cool the solution. Observe the colour change from blue to green, yellow, orange or red depending upon the amount of reducing sugar present in the test sample. 4) Barfoed’s Test: To 2 mL of the test solution add about 2-3 mL of Barfoed’s reagent. Mix it well and boil it for one minute in the water bath and allow to stand for a few minutes. Formation of a red precipitate of cuprous oxide in the bottom and along the sides of the test tube immediately, only monosaccharides answer this test. Since Barfoed’s reagent is slightly acidic, this test is specific for monosaccharides. 5) Seliwanoff’s Test: To 2 mL of Seliwanoff’s reagent, add two drops of test solution. The mixture is heated to just boiling. A cherry red condensation product will be observed indicating the presence of ketoses in the test sample. There will be no significant change in colour produced for aldose sugar. 6) Bial’s Test: To 5 mL of Bial’s reagent add 2–3 mL of test solution and warm gently in a hot water bath for 2minutes. The formation of a bluish green product is indicative of pentoses. Hexoses generally react to form muddy brown products. 7) Iodine Test: Add 2 drops of iodine solution to about 2 mL of the carbohydrate containing test solution. A blue- black colour is observed which is indicative of presence of polysaccharides. 8) Osazone Test: To 0.5 g of phenylhydrazine hydrochloride add 0.1 gram of sodium acetate and ten drops of glacial acetic acid. Add 5 mL of test solution to this mixture and heat under boiling water bath for about half an hour. Cool the solution slowly and examine the crystals under a microscope. Needle-shaped yellow osazone crystals will be observed for glucose and fructose, whereas lactosazone shows mushroom shaped and maltose produces flower-shaped crystals. No. Test Observation Inference Reaction Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT 1 Molisch’s Test 2-3 drops of beta- naphthol soluti on are added to 2ml of the test solution. Very gently add 1ml of Conc. H2SO4 along the side of the test tube. A deep violet coloration is produced at the junction of two layers. Presence of carbohydrates. This is due to the formation of an unstable condensation product of beta- naphthol with furfural (produced by the dehydration of the carbohydrate). 2 Iodine test 4-5 drops of iodine solution are added to 1ml of the test solution and contents are mixed gently. Blue colour is observed. Presence of polysaccharide. Iodine forms coloured adsorption complexes with polysaccharides. 3 Fehling's test About 2 ml of sugar solution is added to about 2 ml of Fehling’s solution taken in a test-tube. It is then boiled for 10 min A red precipitate is formed Presence of reducing sugar This is due to the formation of cuprous oxide by the reducing action of the sugar. 4 Benedict’s test To 5 ml of Benedict's Formation of a green, red, or yellow precipitate Presence of reducing sugars If the saccharide is a reducing sugar it will reduce Copper Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT solution, add 1ml of the test solution and shake each tube. Place the tube in a boiling water bath and heat for 3 minutes. Remo ve the tubes from the heat and allow them to cool. [Cu] (11) ions to Cu(1) oxide, a red precipitate 5 Barfoed’s test To 2 ml of the solution to be tested added 2 ml of freshly prepared Barfoed's reagent. Place test tubes into a boiling water bath and heat for 3 minutes. Allow to cool. A deep blue colour is formed with a red ppt. settling down at the bottom or sides of the test tube. Presence of reducing sugars. Appearance of a red ppt as a thin film at the bottom of the test tube within 3-5 min. is indicative of reducing mono- saccharide. If the ppt formation takes more time, then it is a reducing disaccharide. If the saccharide is a reducing sugar it will reduce Cu (11) ions to Cu(1) oxide 6 Seliwanoff test To 3ml of of Seliwanoff’s reagent, add 1ml of the test solution. Boil in water bath for 2 minutes. A cherry red colored precipitate within 5 minutes is obtained. Presence of ketoses [Sucrose gives a positive ketohexose test ] Presence of When reacted with Seliwanoff reagent, ketoses react within 2 minutes forming a cherry red condensation product Aldopentoses Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT 6) Out of the following the colour reaction specific for ketoses is: Bial’s test Seliwanoff’s test Osazone test Barfoed’s test 7) Flower shaped osazone crystals are ________________. Lactose Glucose Maltose Mannose 8) The Seliwanoff's test is answered by Sucrose. True False 9. Will disaccharides and polysaccharides give a positive result for Molish’s test? Yes, because all carbohydrates such as monosaccharides, disaccharides and polysaccharides gives a positive result for Molish’s test wherein a reddish violet ring is found at the junction of two liquids. 10. Which of the different carbohydrate tests would give a positive result for maltose? Describe the color change. Molish’s Test - A deep violet coloration is produced at the junction of two layers (Formation of violet ring). There is a presence of carbohydrate and maltose is a carbohydrate. This is due to the formation of an unstable condensation product of beta-naphthol with furfural (produced by the dehydration of the carbohydrate). Fehling’s Test - A red precipitate is formed. Maltose contain reducing sugar. This is due to the formation of cuprous oxide by the reducing action of the sugar. Benedict's Test - Formation of a green, red, or yellow precipitate. There is a presence of reducing sugar. If the saccharide is a reducing sugar it will reduce Copper [Cu] (11) ions to Cu (1) oxide, a red precipitate. Biochemistry Laboratory April Ellen B. Baccay, BSChem Eng, BSChT Osazone Test - Formation of beautiful yellow crystals of osazone, needle shaped crystals, hedgehog crystals and sunflower shaped crystals wherein there is a presence of maltose and (glucose/fructose). Reducing sugars forms osazone on treating with phenylhydrazine. 11. Can Seliwanoff’s test tube be used to distinguish sucrose from fructose? Explain your answer. Yes, because sucrose is a disaccharide consisting of two monosaccharides - a glucose and a fructose, linked together by a glycosidic linkage covalent bond. Therefore, when the Seliwanoff's test is performed on a sucrose, the said disaccharide is hydrolyzed, which leads to having a fructose and a glucose. The fructose, having a ketone group, is what causes the positive result. 12. Inulin is a polysaccharide composed entirely of fructose unit. Which test should be used to best identify the presence of fructose? The best way to identify the presence of fructose is the Osazone Test. The Osazone Test determines the presence of fructose or glucose by reducing the sugars present that forms osazone. This test can also identify the presence of lactose or maltose. As I search on the internet, inulin is a complex of sugar present in the roots of various plants and used medically to test kidney function and it is a polysaccharide based on fructose. Thus, the formation of the beautiful yellow crystals of osazone, needleshaped crystals, hedgehog crystals, or Sunflower shaped crystals determines if the variable inhibits the presence of the aforementioned sugars.