Chem 5006 Expt 4-6 Carbohydrates & Lipids PDF

Expt. 4 Carbohydrates Carbohydrates- substances that are polyhydroxyaldehyde or polyhydroxyketones that yield them upon hydrolysis. Monosaccharides- are the simplest sugar that can’t be hydrolyzed further. D-glucose – known as dextrose or grape sugar. Component of maltose, starch, dextrin,...

Expt. 4 Carbohydrates Carbohydrates- substances that are polyhydroxyaldehyde or polyhydroxyketones that yield them upon hydrolysis. Monosaccharides- are the simplest sugar that can’t be hydrolyzed further. D-glucose – known as dextrose or grape sugar. Component of maltose, starch, dextrin, glycogen and cellulose -Blood sugar D-fructose – known as levulose and fruit sugar. Occurs in fruits and found in 1:1 ratio with glucose in honey -Component of table sugar (sucrose) and inulin. Disaccharides- composed of 2-10 monosaccharides linked by acetal or ketal linkages and produces 2 mono-saccharides upon hydrolysis. Maltose (malt sugar) – intermediate in the hydrolysis of starch to glucose Sucrose (cane or beet sugar) – occurs free in the plant kingdom. Ex. sugar cane, sugar maple, sugar beets Lactose (milk sugar) – found in milk, not very sweet, not fermented by yeast. Polysaccharide- made of monosaccharides, may also contain sugar derivatives e.g. sugar acids or amino sugars -structural function cellulose, agar – agar -nutrient function starch, dextrin, glycogen, insulin General Test for Carbohydrates 1. Molisch test- In the presence of conc. H2SO4 glycosidic bonds are hydrolyzed giving monosaccharides which are then dehydrated to furfural, hydroxymethylfurfural, etc. These then react with α-napthol forming purple-colored condensation products All sugars are positive for Molisch test. 2. Fehling’s test- test for the presence of aldehyde functional group but not the ketones. A test for reducing action of carbohydrates. Fehling's A – a blue solution which consists of CuSO4 in distilled H2O + dil.H2SO4 Fehling's B – an alkaline colorless solution containing Potassium Sodium Tartrate (KNaC4H4O6 4H2O) + NaOH in dist.H2O. Aldehydes are reducing agents, they reduce the blue copper (II) ions to brick-red copper (I) oxide, Cu2O, which forms a brick-red precipitate. Ketones are not reducing agents; no color change is observed. 3. Benedicts test- uses reducing property of some sugars. Cupric ion is reduced and cuprous oxide precipitates out of the alkaline solution as brick red precipitate. Glucose, fructose, maltose, and lactose are reducing sugars while sucrose, starch and cellulose are non-reducing sugars. Iodine test for Polysaccharides- differentiates helical from non-helical polysaccharides Starch produces a blue to black solution. Agar-agar produces a yellowish-brown solution and gum Arabic produces a yellow solution. Only starch gave a positive reaction to iodine test because they have helical structures. Hydrolysis of sucrose- upon hydrolysis, sucrose would yield glucose and fructose. When subjected to Benedict’s test, it would form a brick-red precipitate indicating that glucose and fructose were separated from sucrose upon hydrolysis. Unhydrolyzed sucrose would give negative results for benedicts test. Hydrolysis of Polysaccharide- hydrolysis of starch: starch erythrodextrin maltose glucose Hydrolyzed starch will be converted to glucose units win which it would yield a positive brick red precipitate on Benedict’s test. Why does fructose, a ketose reduce Benedict’s reagent? Fructose, based on its structure has an aldehyde available for reduction reaction that is why it was able to reduce Benedict’s reagent. Why sucrose was negative to reduction test? In sucrose (1,2 glycosidic structure), the glucose part had the aldehyde at C-1 and the fructose part had the ketone group at C-2. Since the linkage is 1,2- neither is free thus sucrose is not a reducing sugar. Why sucrose and starch become positive for Benedict’s after hydrolysis? Hydrolyzed sucrose become positive for Benedict’s test because the product; glucose and fructose are reducing sugars. Also, hydrolyzed starch become positive too to Benedict’s test because the end product of complete starch hydrolysis is glucose. Expt. 5 Lipids Lipids- are the second group of organic compounds that serve as food for the body. - a collection of organic molecules united by solubility in nonpolar solvents. -Uses of fats in our body: - 1. As fuel Fat: 9kcal/g Carbohydrate: 4kcal/g 2. Reserve supply of food- Stored in the adipose tissue 3. Protector of vital organ as shock absorber- Surround the vital organ to keep them in place 4. As heat and electrical insulator- Keep our body warm and allow propagation of nerve impulses  A. Physical properties: 1. Physical examination Properties Beef/pork fat butter Cottonseed oil Odor unpleasant pleasant Unpleasant Appearance Dirty white yellow Yellow Physical state Soft solid Soft solid Liquid Texture greasy greasy Greasy 2. Solubility of coconut oil: Water- immiscible cold ethanol- immiscible chloroform – miscible ether- miscible 3. Formation of translucent spot: Ether- evaporated first and did not produce a translucent spot Ethanol- 2nd to evaporate and did not form a translucent spot Coconut oil- did not evaporate and formed a translucent spot (the oil did not evaporate and light can pass through both side of the paper). B. Emulsification- a process in which a liquid like soap mixes two insoluble liquids like water and oil to produce one colloidal suspension. - Water + cottonseed oil → formation of two layers → adding soap soln → a white turbid mixture in which oil and water mixed. - Water + cottonseed oil → formation of two layers → adding bile soln → a yellow turbid mixture in which oil and water mixed. - Soap contains a polar and a non-polar molecule at it ends so it can dissolve both polar and non-polar molecules. - Bile contains bile acids which emulsify the lipids within the water-insoluble droplets during the digestion. - Bile acids solubilize lipids by forming micelles which act as emulsifying agents C. Acrolein test- When a fat is heated strongly in the presence of KHSO4 (dehydrating agent) the glycerol portion of the molecule is dehydrated to form the unsaturated aldehyde called “ACROLEIN” distinguished by a “BURNT OIL” odor. Coconut oil- (+) result glycerol- (+) result oleic acid- (-) result (no glycerol portion in its molecule) D. Saponification of lard (soap formation)- heating of a fat with strong alkali or base will form a soap. E. Formation of insoluble soap: (CH3(CH2)10COO-K+) + CaCl2 → (CH3(CH2)10COO-)2Ca2+ + 2KCl calcium soap, white ppt. (CH3(CH2)10COO-K+) + MgSO4 → (CH3(CH2)10COO-)2Mg2+ + 2K2SO4 magnesium soap, white ppt. Calcium and magnesium soaps form insoluble soaps. These soaps are hard to dissolve in water. F. Preparation of cholesterol from oily foods. Description of extract: yellowish liquid Chemical tests for sterols: 1. Lieberman-Burchard reaction (acetic anhydride test) - blue to blue green color 2. Salkowski test - reddish brown ring at the junction of the two liquids Expt. 7 Nucleic acids Nucleic acids- Are informational molecules made up of polymers of nucleotides linked together by phosphodiester bonds. They act as repositories and transmitters of genetic information for every cell, tissue and organism. Nucleotide differences between DNA and RNA 1. Why is thymus or spleen best recommended for DNA isolation? Thymus and spleen are lymphoid tissues. -contain large cells and therefore large amount of DNA. 2. What is the purpose of adding a mixture of 0.15M NaCl and 0.01 M Na citrate? 0.15M NaCl ≈ 0.9% NaCl: saline solution, physiological isotonic salt solution. Isotonic meaning same solute conc. with body fluid. 0.01M Na citrate - reduces the activity of the enzymes by binding with Ca+2 or Mg+2 (enzyme activators). 3. Why is ice cold condition necessary for the isolation of DNA? To reduce the residual enzyme action of DNA. Alcohol precipitate (fibrous precipitate of nucleoprotein) under the microscope: Rod-shaped aggregates RNA isolation: yeast contains about 4% RNA warm NaOH – to dissolve the yeast glacial acetic acid – to precipitate out proteins 95% ethanol – precipitating reagent for RNA washing with ethanol and ether- to wash out lipid materials Isolation of RNA from yeast under the microscope: Cubic-shaped aggregates Hydrolysis products of DNA: DNA + HCl + heat → N-bases (A,G,C,T) + deoxyribose + phosphate Hydrolysis products of RNA: RNA + H2SO4 + heat → N-bases (A,G,C,U) + ribose + phosphate Qualitative Chemical tests on DNA and RNA: Tests DNA RNA Benedict’s test for reducing sugars Brick-red ppt (+) Brick-red ppt (+) Orcinol test for pentoses faint green coloration (+) Green coloration (+) Purine bases White ppt. (+) White ppt. (+) Inorganic phosphates Yellow ppt. (+) Yellow ppt. (+) Orcinol will react with the deoxyribose of DNA in addition to the ribose of RNA, although the color intensity of this reaction is only about 10% as that of the reaction with ribose of RNA. The precipitation mechanism of purine bases is by a reaction of an alkaline AgNO3 with the nitrogen of purines.