Biochem Lab Experiments 8-17 PDF

EXPERIMENTS 8-17 Cut thin slices of apples, guava, chico or other available fruits and expose them to the air. Observe the darkening of the exposed sliced...

EXPERIMENTS 8-17 Cut thin slices of apples, guava, chico or other available fruits and expose them to the air. Observe the darkening of the exposed sliced EXPERIMENT 8: ENZYMES sides. I. PURPOSE Results and Observations To examine and understand enzyme activity, reactions, and how it Q Explain color change catalyzes reaction Fresh fruits exposed to air/oxygen releases: II. APPARATUS Polyphenol Oxidase (PPO) Enzyme Tyrosinase Enzyme Cheesecloth, Knife, Peeler, Grater, Test Tubes, Filter Paper, Funnel, Wire Monophenol Oxidase Enzyme Gauze, Bunsen Burner, Glowing Splint Catechol Oxidizing Enzyme A III. MATERIALS These tissues turn into o-quinones / ortho-benzoquinone / 1,2-Benzoquinone (have no color) but react with amino acids and Apple oxygen to produce melanin. Hence, brown color of the exposed Guava sides of apple slices. Chico Fruit Small Potato B. CATALASE Hydrogen Peroxide: H2O2 1% Phenol: C6H5OH 1. POTATO 1% Cathecol: C6H4(OH)2 Pyrogallol: C6H3(OH)3 Preparation Potato Extract: Source of Oxidase and Peroxidase 1 g Liver 1. Pare a small potato and grate it to a fine pulp. 2. Mix pulp with 100 mL of water. IV. PROCEDURES 3. Stand for 15 mins. 4. Strain through cheesecloth. Enzymes 5. Filter the extract. ○ Complex organic compounds with definite chemical structure, secreted by living cells. Action ○ Have the property of initiation and hastening chemical reaction without themselves being 1. Divide filtrate to 4 parts and boil one part for 1 minute. affected in the process. 2. 5 mL of boiled filtrate to one tube, 5 mL of unboiled filtrate Enzymes Activity: influenced by the to another tube. ○ concentration of the enzymes 3. Add a few drops of 3% hydrogen peroxide each. ○ concentration of the substrate ○ concentration of the products of the reaction Results and Observations ○ temperature ○ pH ○ inorganic salts Q Unboiled Filtrate ○ presence of activators and inhibitors Formation of more bubbles due to more oxygen catalyzed by All these enzyme activity factors therefore should be A the enzyme catalase. considered while performing the following experiments. Q Boiled Filtrate A. OXIDASES FROM FRUITS Formation of less bubbles due to the denaturation of the Oxidases are enzymes that catalyzes redox reactions and use A enzyme catalase by heat, making it incapable of processing the oxygen as electron acceptors. hydrogen peroxide. There is a formation of coagulation. Fruits are oxidized because of the enzyme tyrosinase, or monophenol oxidase that reacts with oxygen and Q Explain the results iron-containing phenols that basically makes a layer of rust on the surface of the fruit. Boiling the potato extract reduces the catalytic effectiveness of A the enzymes because heat can destroy the hydrogen bonding that stabilizes the structural conformation of the protein enzymes. 1  D. PEROXIDASE FROM POTATO Peroxidase ○ Unlike oxidases, may require a co-factor, phenol-oxidase to complete their action. ○ They also require H2O2 as the source of oxygen, and upon which the phenol-oxidases act. Phenol oxidase: also found abundant in potatoes. Procedure C. OXIDASES FROM POTATO 1. Prepare 3 test tubes and place 5 drops each of the following substances: 3 Oxidases a. 1% Phenol b. 1% Catechol 1. Monophenol Oxidase (Tyrosinase): responsible for oxidizing c. Pyrogallol phenol to catechol, to o-quinone and finally forms 2. Add 1 mL of potato extract (peroxidase) to each. condensation brown compounds of unknown composition. 3. Mix well, then add 3 drops of 3% hydrogen peroxide (H2O2) to 2. Polyphenol Oxidase (Catechol Oxidase): acts on catechol each. forming o-quinone, then the unknown brown compounds; it 4. Observe and record the color changes produced and acts on pyrogallol to form purpurogallin. compare them with the results observed on the action of 3. Cytochrome Oxidase: acts in conjunction with cytochrome, oxidase with the same reagent. oxidizing phenylenediamine, which in the presence of alpha-naphthol forms indophenol. Results and Observations Procedure Q What happened to pyrogallol? 1. Prepare 3 test tubes and place 5 drops each of the following substances: A Peroxidase catalyzes more pyrogallol to purpurogallin. a. 1% Phenol b. 1% Catechol SUBSTANCES INITIAL OBSERVATION AFTER 1 HR/END OF LAB c. Pyrogallol 2. Add 1 mL of potato extract (oxidase) to each test tube and 1% Phenol Light Brown Brownish Red shake. 3. Allow them to stand until the end of the laboratory period, 1% Cathecol Brownish Red Darker Brownish Red observing and recording the initial color change, then the change at the end of the period. Pyrogallol Red Darker Red Results and Observations SUBSTANCES INITIAL OBSERVATION AFTER 1 HR/END OF LAB 1% Phenol Light Brown Brownish Red 1% Cathecol Brownish Red Darker Brownish Red Pyrogallol Red Red 2. LIVER 1. Place 1 gram of liver, 4 mL of water and a little sand in a mortar and grind. 2. Add 2 mL of 3% hydrogen peroxide. 3. Test the gas evolved using a glowing splint. Results and Observations Q How to know the gas liberated is oxygen? The glowing splint flame continues to glow due to the oxygen A emitted by the liver, making it uneasy to put off. Oxygen supports combustion. 2  EXPERIMENT 9: TEST FOR NUTRIENTS IN FOOD Q Purpose of Catalase Catalase from the liver catalyzes the decomposition of hydrogen I. PURPOSE peroxide to water and oxygen gas at a rate of 40 million molecules of hydrogen peroxide per second. To identify and understand the nutritional value with specific presence of A The liver filters the body using enzymes as catalysts. It makes organic molecules in foods. harmful substances less toxic like hydrogen peroxide to water and oxygen, and alcohol to aldehyde. Lack of catalase causes II. APPARATUS the graying of hair in humans. Test Tube, Bunsen Burner, Wire Gauze, Evaporating Dish, Watch Glass, Piece of White Paper Chemical Equation III. MATERIALS REACTANTS ENZYME PRODUCTS Hydrogen Peroxide Water + Oxygen Ripe Banana Catalase from Liver Green Banana H2O2 H2O + O2 Carrots Egg White Egg Yolk Peanut Butter Potato Cheese Fruit Juice Dilis Uncooked Peanuts Cooked Rice Iodine Solution: I2 Benedict’s Solution: CuSO4 + Na2CO3 + Sodium Citrate Ether: (C2H5)2O Biuret Reagent: 10% NaOH + 0.5% CusO4 Chloroform: CHCl3 Cold Saturated Antimony Trichloride: SbCl3 Acetic Anhydride: (CH3CO)2O Conc. Sulphuric Acid: H2SO4 IV. PROCEDURES A. STARCH Boil a small amount of the food sample in a test tube with 7 mL of water. Cool by holding it under running cold water. Add a drop of I2 solution. A deep blue color indicates starch. Results and Observations Positive Result: Deep Blue Color Positive Food Samples: ○ Ripe Banana ○ Green Banana ○ Peanut Butter ○ Potato ○ Cheese ○ Peanuts (Uncooked) ○ Rice (Cooked) B. GLUCOSE Add a few drops of a fruit juice to 1 mL of water in a test tube. Then add Benedict’s solution sufficient to give a very light blue solution. Boil. A yellow to red precipitate indicates the presence of a simple sugar such as glucose or fructose. 3 Results and Observations F. CARR-PRICE TEST FOR VITAMIN A Positive Result: Yellow to Red Precipitate To 1 mL chloroform, add 2 drops or a pinch of the food sample. Cool in Positive Food Samples: an ice bath. Add 2 mL of a cold saturated solution of antimony ○ Ripe Banana trichloride. Observe the changes in color to blue. Perform this test on ○ Green Banana margarine/butter and carrots only. ○ Carrots ○ Peanut Butter Results and Observations ○ Potato ○ Cheese Positive Result: Blue Solution ○ Fruit Juice Positive Food Samples: Carrots C. FATS Q How long did the color persist? To examine foods for oil or fat, shake a small portion of the food with 3 A 10–15 minutes mL of ether in a test tube for several minutes, then decant to an evaporating dish or watch glass. Allow the ether to evaporate G. CHOLESTEROL spontaneously (without heating). Place the residue on a piece of white paper. Warm. When held to the light, it will be translucent (refraction of Preparation of Cholesterol light) as an indication of a fat. Evaporate the acetone-ether filtrate from no. 1 to about one-fifth its Results and Observations original volume until crystals are formed. If it evaporates to dryness, redissolve the solid in a very small amount of acetone and then allow to Positive Result: Translucent Spot cool. Use the solid for the following tests: Positive Food Samples: Microscopical Examination: examine the crystals under a ○ Egg Yolk (Cooked) microscope. Sketch the crystals ○ Peanut Butter Acetic Anhydride–H2SO4 Test or Lieberman-Burchard Test ○ Cheese Sulphuric Acid Test or Salkowski’s Test ○ Dilis ○ Peanuts (Uncooked) 1. ACETIC ANHYDRIDE–H2SO4 TEST / LIEBERMANN-BURCHARD TEST D. PROTEINS Acetic Anhydride–H2SO4 Test or Liebermann-Burchard Test: basis for the quantitative determination of cholesterol Use 2 mL of the aqueous food solutions. Perform the Biuret Test or Piotrowski’s Test (NaOH + CuSO4). A violet solution indicates the Dissolve a few crystals of cholesterol in 2 mL of chloroform in a dry test presence of proteins. tube. Add 10 drops of acetic anhydride and 2 drops of conc. sulphuric acid and mix. Allow to stand and note the production of colors. Results and Observations Results and Observations Positive Result: Violet Solution Positive Food Samples: Positive Result: Red Sol’n → Blue Sol’n → Bluish-Green Sol’n ○ Egg White (Cooked) Positive Food Samples: ○ Egg Yolk (Cooked) ○ Egg Yolk (Cooked) ○ Peanut Butter ○ Cheese ○ Cheese ○ Dilis 2. SULPHURIC ACID TEST OR SALKOWSKI’S TEST ○ Peanuts (Uncooked) Dissolve a small portion of cholesterol in a little amount of chloroform E. MINERAL MATTER and add an equal volume of conc. sulphuric acid layer. Note the color produced both in the chloroform and sulphuric acid layer. Burn a very small amount of cheese on an evaporating dish until all of the black carbon is oxidized. The formation of any white powdery Note: 1 mL of chloroform and sulfuric acid will be sufficient. residue indicates the presence of mineral matter. Results and Observations Results and Observations Positive Result: Positive Result: White Powdery Residue ○ CHCl3 Layer: Bluish-Red to Cherry Red and Purple Positive Food Samples: ○ H2SO4 Layer: Marked Green Fluorescence ○ Cheese Positive Food Samples: ○ Dilis ○ Egg Yolk (Cooked) ○ Cheese 4  V. IMPORTANCE OF VITAMIN A EXPERIMENT 11: MILK Vitamin A (Retinol): the immediate precursor to 2 important I. PURPOSE active metabolites. Retinol: plays a critical role in vision To understand the reactions of milk, prepare casein, and the presence of Retinoic Acid: serves as an intracellular messenger that organic materials in milk. affects the transcription of a number of genes Vitamins A does not occur in plants, but many plants contain II. APPARATUS carrotinoid (such as beta carotene) that can be converted to vitamin A within the intestine and other tissues. Test Tube, Test Tube Rack, Beaker, Funnel, Filter Paper, Bunsen Burner, Clay Flame Shield, Wire Gauze, Evaporating Dish, Vials, Litmus Papers (Blue & Red), Densimeter III. MATERIALS Fresh Milk Diluted Canned Milk Formula of Vitamin A (C20H30O) Skimmed Milk 3,7-dimethyl-9-(2,6,6-trimethylcyclohex-1-yl) nona-2,4,6,8-tetraen-1-ol Congo Red: C32H22N6Na2O6S2 Phenolphthalein Solution: C20H14O4 VI. SUMMARY Acetic Acid: CH3COOH 6 M & Dilute Sodium Hydroxide: NaOH General Compositions and the Amount of Nutrients Ethyl Alcohol: CH3CH2OH depends on the: Ether: (C2H5)2O ○ Variety Dilute Hydrochloric Acid: HCl ○ Source 10% Sodium Chloride: NaCl ○ Chemical Reaction it underwent before the Millon’s Reagent: 1 Hg : 2 HNO3 Identification Test. Biuret Reagent: 10% NaOH + 0.5% CuSO4 Most of the samples contain 3 or more of these food Conc. Nitric Acid: HNO3 nutrients but vary in its concentration. Ammonium Oxalate: (NH4)2C2O4 The observable results will depend on the trace amount of Ammonium Molybdate: (NH4)2MoO4 these nutrients that can be detected by the general test Benedict’s Reagent quantitatively. IV. PROCEDURES Mineral Starch Glucose Protein Fats Cholesterol Vitamin A Matter Milk Deep Yellow Translu- White Violet Bluish-Green Blue to Brick- cent Powdery Blue Sol’n Sol’n Red ppt. Sol’n Spot Residue Sol’n Most complete food that contains ○ Proteins Ripe Banana + + – – – – – ○ Fats Green Banana + + – – – – – ○ Carbohydrates ○ Inorganic Salts Carrots – + – – – – + ○ Vitamins Egg White – – + – – – – Fresh Unboiled Milk: contains (Cooked) ○ Enzymes Egg Yolk – – + + – + – ○ Protease (Cooked) ○ Lactase Peanut Butter + + + + – – – ○ Lipase ○ Phosphatase Potato + + – – – – – ○ Catalase Cheese + + + + + + – ○ Peroxidase Fruit Juice – + – – – – – Proteins in Milk Dilis – – + + + – – Peanuts Casein: Chief Protein of Milk (Uncooked) + – + + – – – ○ Can be precipitated by acid, carrying with it the Rice (Cooked) + – – – – – – milk fat. ○ The fat can be subsequently extracted by organic solvents. ○ The filtrate from the casein and fat contains soluble constituents like lactose and inorganic salts. Proteins of Milk ○ Derived from amino acids of the blood ○ Synthesis: occurring in the mammary glands 5  Milk Fat 2. DETERMINATION OF SPECIFIC GRAVITY ○ Believed to have its origin in the phospholipids of the blood By the use of a densimeter determine the specific gravity of both fresh Milk Lactose milk and skimmed milk. ○ Derived from the glucose of the blood Inorganic Salts Results and Observations ○ Calcium, Magnesium, Sodium, Phosphates, Citrates and Chlorides, and Other Inorganic Constituents Q Which has a greater specific gravity? Why? ○ Likewise derived from the blood, some by simple process of filtration A Skimmed Milk because it does not contain milk fat. ○ Bone: acts as a reserve supply of calcium Human Milk INDIRECTLY PROPORTIONAL RELATIONSHIP Milk Fat & Specific Gravity Differs from cow’s milk in having: ○ LESS: casein and ash MORE milk fat LIGHTER specific gravity ○ MORE: albumin and lactose LESS milk fat HEAVIER specific gravity 1. REACTION OF MILK 3. FILM FORMATION Test 1 mL of milk to both red and blue litmus paper, congo red, and phenolphthalein solution. Place 5 mL of diluted canned milk in a small beaker and boil for a few minutes. Note the formation of a film. Results and Observations Results and Observations FRESH MILK SOUR MILK Q What is the composition of the film? Slightly Acidic Acidic pH pH: 6.7 pH: 4–5 When milk is boiled, soluble milk proteins are denatured and then A coagulate with milk fat and form a sticky film across the top of Red & Blue the liquid, which then dries by evaporation. No Change in Color Turn Red Litmus Paper Red/Pink Solution Dark Brown Solution Remove the film and heat again. Congo Red due to redox reaction due to redox reaction Results and Observations Yellowish/Colorless Yellowish/Colorless Phenolphthalein Q Does the film form again? Phenolphthalein: commonly used indicator for titration A Fresh Milk: YES, it forms film Repeat the experiment using sour milk. Results and Observations Q Is there any film formation? Why? Sour Milk: NO film because the proteins are already denatured by A acetic acid. Q What factors facilitate the formation of surface film? Temperature Presence/Absence of Milk Fat A Stirring/Whisking Type of Milk 6  4. COAGULATION TEST Press the precipitate as dry as possible between filter papers. Open the papers and allow the ether to evaporate spontaneously (without heat). Place 1 mL of fresh milk in a test tube and acidify with acetic acid. Heat Grind the precipitate to powder in a mortar and make the following test: to boiling. Results and Observations Q Is there any coagulation? Why? Coagulation: due to the presence of whey protein which is A mainly composed of beta-lactoglobulin. A. SOLUBILITY Test the solubility of casein in water, dilute HCl, dilute NaOH and 10% NaCl. Results and Observations SOLUBILITY OF CASEIN 5. MOORE’S TEST (ACTION OF HOT ALKALI) Water Mix 1 mL of milk with drops of 6 M NaOH. Heat and describe the changes Dilute HCl Insoluble observed. Dilute NaOH Results and Observations 10% NaCl Slightly Soluble Q What is the cause of these changes? Presence of caramel due to the reducing sugar, lactose, with its A reducing end, glucose. What test in the study of carbohydrates involves the same Q principles? Moore’s Test: the polymerization of aldehyde groups of sugars A (lactose). B. PERFORM MILLON’S TEST ON CASEIN Perform Millon’s test (1 Hg : 2 HNO3) on casein. Results and Observations Q Result? Flocculent Red precipitate formation (+) due to tyrosine, an A amino acid, has phenolic rings that confirms the test. 6. PREPARATION OF CASEIN Q To what class of protein does casein belong? Take 10 mL of milk and dilute it with an equal volume of water. Add Casein: a heterogeneous group of four phosphoproteins and dilute acetic (1%) drop by drop until a flocculent precipitate forms. A phosphoglycoproteins Avoid any excess as dissolution may occur. Allow the precipitate to settle, decant the supernatant fluid (whey) and reserve it for experiment 7. Filter off the precipitate. Remove excess moisture by washing the precipitate with a few mL of ethyl alcohol. Transfer the casein to a dry test tube, cover it with ether and heat on a water bath set at 50°C without flame for 5–10 minutes, shaking continuously. Filter and save the filtrate for no. 8. 7  B. TEST FOR PRESENCE OF PHOSPHORUS, CALCIUM, AND REDUCING SUGAR Boil the second portion of the filtrate and remove every trace of coagulable protein. Filter. Test the filtrate for the presence of P–3, Ca+2, and reducing sugar (lactose). Reagents Phosphorus: conc. HNO3 + (NH4)2MoO4 Calcium: (NH4)2C2O4 Reducing Sugar: Benedict’s Reagent 7. WHEY Results and Observations Take two 3 mL portions of the whey (filtrate from no. 6) and make the following tests: TEST POSITIVE RESULT Results and Observations Yellow ppt. Test for Phosphorus (P–3) (NH4)3PO4 12MoO3 Q Draw your conclusions. White ppt. Whey from milk contains all other substances found in milk, apart Test for Calcium (Ca+2) CaC2O4 from the separated casein, such as: Lactalbumin & Lactoglobin proteins: violet solution (Biuret) Benedict’s Test (Reducing Sugar: Brick-Red ppt. A Phosphorus: yellow ppt. of ammonium phosphomolybdate Lactose/Glucose) Cu2O Calcium: white ppt. of calcium oxalate Lactose: orange ppt. of cuprous oxide (Benedict’s) Chemical Equation A. COAGULATION BY HEATING REACTANTS PRODUCTS TEST FOR PHOSPHORUS Coagulate by heating. Note the formation of a coagulum. This consists of lactalbumin and lactoglobin. Make a biuret test (NaOH + CuSO4) on Phosphate Ion + Ammonium Ammonium Phosphomolybdate the coagulum. Molybdate + Nitric Acid + Water 3NH4+ + 12MoO4–2 + PO4–3 + 24H+ (NH4)3PO4 12MoO3 + 12H2O Results and Observations TEST FOR CALCIUM Calcium Ion + Ammonium Oxalate Calcium Oxalate + Ammonium ion Q Observation? Ca2+ + (NH4)2C2O4 CaC2O4 + 2NH4+ Formation of Coagulation: Coagulum that consists of BENEDICT’S TEST A lactalbumin and lactoglobin Biuret Test: Violet Solution (+) Reducing Sugar Copper (I) Oxide + (Lactose/Glucose) + Gluconic Acid + Water Copper (II) Hydroxide C12H22O11 + 2Cu(OH)2 Cu2O + C6H12O7 + 2H2O BIURET TEST: orange due to oxidation 8. MILK FAT Transfer the filtrate (obtained in no. 6) into a dry evaporating dish and place it on a boiling water bath to evaporate the ether (turn out all flames), leaving a small amount or residue. WARNING: Ether could cause severe headaches if smelled in large quantities. Only the assigned groups will be given ether. Tests on casein could be done if addition of ether is omitted for those groups not given. 8 Results and Observations EXPERIMENT 12: SALIVARY DIGESTION I. PURPOSE Q What is the residue? A Milk Fat To understand the reactions of the saliva and stimulate salivary digestion Touch the residue with a piece of paper. II. APPARATUS Results and Observations Test Tube, Bunsen Burner, Wire Gauze, Tripod, Beaker, Dropper Q What do you observe? III. MATERIALS A Formation of translucent spot. Saliva Phenolphthalein Litmus Paper Congo Red Paraffin Dilute Acetic Acid: CH3COOH Hydrochloric Acid: HCl Sodium Hydroxide: NaOH Silver Nitrate: AgNO3 Ammonium Molybdate: (NH4)2MoO4 Barium Chloride: BaCl2 Ammonium Oxalate: (NH4)2C2O4 Nitric Acid: HNO3 1% Starch Paste Iodine: I2 Benedict’s Reagent IV. PROCEDURES Saliva: secreted by 3 pairs of glands and hundreds of small buccal glands ○ Parotid Gland ○ Submaxillary Gland ○ Sublingual Gland Flow: stimulated by psychic, chemical, and mechanical factors Saliva contains ○ Protein: Mucin ○ Enzyme: Ptyalin (starch hydrolyzing enzyme) ○ Inorganic Salts Salivary Digestion ○ Hydrolysis of starch by salivary amylase (ptyalin) ○ Takes place in the buccal cavity and to a certain extent in the fundic end of the stomach 1. REACTION Place a few drops of resting saliva in three test tubes. Test the reaction with phenolphthalein, litmus, and congo red. From the color produced, estimate the approximate pH of resting saliva. Repeat the experiment using stimulated saliva. Stimulate the flow of the saliva by chewing paraffin vigorously for at least 5 minutes. Do not use chewing gum for this purpose. Results and Observations What is the difference between the pH of the resting saliva Q and the stimulated saliva? Resting Saliva: Slightly Basic (pH = 6.2–7.6) A Stimulated Saliva: More Basic (pH = 7.0–8.0) 9  Is there any relation between the pH of the saliva and the SOLUTION PRECIPITATE PRODUCT Q susceptibility of dental caries? Chloride Turbid White AgCl A Acidic pH: essential in enamel dissolution to produce dental caries Phosphate Clear, Colorless Yellow (NH4)3PO4 12MoO3 Sulfate Slightly Turbid – BaSO4 RESTING SALIVA STIMULATED SALIVA Calcium Clear, Colorless – CaC2O4 Slightly Basic More Basic pH pH: 6.2–7.6 pH: 7.0–8.0 Chemical Equation Red & Blue Turn Blue Turn Blue Litmus Paper REACTANTS PRODUCTS Congo Red Red Solution Darker Red Solution TEST FOR CHLORIDE Chloride Ion + Nitric Acid + Silver Chloride + Phenolphthalein Colorless Colorless Silver Nitrate Nitrate Ion Cl– + AgNO3 + HNO3 AgCl + NO3– TEST FOR PHOSPHATE Phosphate Ion + Nitric Acid + Ammonium Phosphomolybdate Ammonium Molybdate + Water 3NH4+ + 12MoO4–2 + PO4–3 + 24H+ (NH4)3PO4 12MoO3 + 12H2O TEST FOR SULFATE Barium Chloride + Barium Sulfate + Sulfuric Acid Hydrochloric Acid BaCl2 + H2SO4 BaSO4 + 2HCl TEST FOR CALCIUM 2. TEST FOR MUCIN Calcium Ion + Ammonium Oxalate Calcium Oxalate + Ammonium ion To 3 mL of saliva in a test tube add 1-2 drops of dilute acetic acid. Ca2+ + (NH4)2C2O4 CaC2O4 + 2NH4+ Results and Observations 4. DIGESTION OF STARCH PASTE Q What is the precipitate formed? Place 10 mL of 1% starch paste in a small beaker. Add 5 drops of saliva and stir thoroughly. Add another 5 drops of saliva if blue color still forms A Mucin: cloudy solution with white precipitate with iodine after 5 minutes. Gradually, the opalescence of the starch solution disappears due to the formation of soluble starch. In making Q What is its function? the test with iodine (confirm digestion of starch), remove a drop of the Heavily O-glycosylated linear glycoproteins secreted by higher starch-saliva solution and transfer to a spot plate or white tile at A organisms to protect and lubricate epithelial cell surfaces. intervals of one minute. Stir for sometime and test again with iodine. A symptom to any abnormalities in the body, such as cancer. When no more blue color is produced with iodine test, a portion with Benedict’s test (test for reducing sugar) and note the degree of reduction. 3. INORGANIC MATTER Results and Observations Acidify 10 mL of saliva with a drop or two of acetic acid, heat to boiling and filter to remove the protein. Test filtrate for chlorides, phosphates, sulfates, and calcium. What is the reaction with iodine when Benedict's test becomes Q positive? Reagents At achromatic point: Iodine: gives colorless solution Chlorides: dil. HNO3 + 0.1 M AgNO3 A Benedict’s: shows a slightly positive result Phosphate: conc. HNO3 + (NH4)2MoO4 It is an indication of the conversion of starch to achrodextrin. Sulfate: dil. HCl + 0.1 N BaCl2 Calcium: (NH4)2C2O4 Q What is responsible for reducing the action? Results and Observations Product of Starch Digestion: Maltose and Glucose A Product of COMPLETE Starch Digestion: Glucose Q Which of the inorganic salt is found abundant in the saliva? How long did it take for a complete transformation of the Q starch into reducing sugar? A Chloride Salts & Phosphate Salts A 10-20 minutes for it to completely transform into reducing sugars 10  Results and Observations Outline the different stages of starch digestion by ptyalin. How Q does this differ from starch acid hydrolysis? Q In which tube did you obtain the greatest digestion? 1. Boiled Starch + Ptyalin = Soluble Starch 2. Soluble Starch + Ptyalin = Erythrodextrin + Maltose A 2% NaOH 3. Erythrodextrin + Maltose + Ptyalin = Achrodextrin + Maltose 4. Achrodextrin + Maltose + Ptyalin = Isomaltose + Maltose Q What is the relationship? General: ↑ pH = ↑ digestion A For Alkali: ↑ concentration/alkalinity = ↑ digestion Q Which has a greater inhibiting power, acid or alkali? Acid: because it disrupts the basicity of the saliva which is A necessary for buffering capacity. General: ↓ pH = ↑ inhibiting power A What is the effect of gastric juice on the digestive section of Q saliva? Gastric Juice: starts protein digestion A Saliva: starts starch digestion Q How long does salivary digestion continue in the stomach? Starch Digestion by Ptyalin: an enzymatic, specific, and It is inhibited more by acids, therefore ptyalin is slowly controlled process occurring under physiological conditions. inactivated by the acidic gastric juice in the stomach. Starch Acid Hydrolysis: a nonspecific chemical reaction Salivary digestion continues in the stomach for 10-20 minutes requiring harsh conditions to achieve complete breakdown of A due to the slow penetration of acidic gastric juice into the starch into glucose. bolus, a mass of chewed food. After the 10-20 minute period, salivary digestion stops due to 5. INFLUENCE OF ACID the total inactivation of ptyalin. Place 2 mL of one of the following strengths of acids in each of the 5 test tubes: 0.25%, 0.1%, 0.05%, 0.025%, and 0.0075% HCl. To each add 1 mL of 1% starch paste and 1 mL saliva. Shake well. Place in a water bath at 40oC for 20 minutes. At intervals of 5 minutes, test a small portion with iodine and Benedict’s test, until a positive result with Benedict’s and a negative result with iodine is obtained. Results and Observations Q Results: Same with Influence of Alkali Benedict’s Test: Yellow / Orange / Green Precipitate (+) A Iodine Test: Yellowish-Brown / Reagent Color (–) Q In which tube did you obtain the greatest digestion? A 0.0075% HCl Q What is the relationship? General: ↑ pH = ↑ digestion A For Acid: ↓ concentration/acidity = ↑ digestion 6. INFLUENCE OF ALKALI Repeat the above experiment using NaOH, instead of HCl, with the following concentrations: 2%, 1%, 0.5%, 0.125%, and 0.065%. Neutralize the alkalinity using acetic acid before performing the iodine test. 11  EXPERIMENT 14: BILE 2. INORGANIC CONSTITUENTS I. PURPOSE Evaporate 10 mL of bile to dryness. Fuse the residue with the fusion mixture (2 Na2CO3 : 1 KNO3). Cool and extract with 10 mL of water. To understand the reactions of bile, detect inorganic and organic Acidify with HNO3 until slightly acidic. Filter and test the filtrate for nutrients, and study its properties. chloride, sulfate, and phosphate. II. APPARATUS Results and Observations Blue and Red Litmus Paper, Test Tubes, Test Tube Rack, Bunsen Burner, Q What inorganic constituents are found in the bile? Clay Flame Shield, Beaker, Filter Paper, Funnel, Erlenmeyer Flask, Evaporating Dish, Test Tube Holder, Mortar and Pestle A Chloride, Sulfate, and Phosphate III. MATERIALS Q Purpose of Fusion Bile Procedure that changes the compounds into ions because ions Phenolphthalein are more readily active when tested. Congo Red A Chlorine → Chloride Fusion Mixture: 2 Na2CO3 : 1 KNO3 Sulfur → Sulfate Conc. Nitric Acid: HNO3 Phosphorus → Phosphate Hydrochloric Acid: HCl Silver Nitrate: AgNO3 PRECIPITATE PRODUCT Barium Chloride: BaCl2 Ammonium Molybdate: (NH4)2MoO4 Chloride White AgCl 5% Sucrose Solution Conc. Sulfuric Acid: H2SO4 Sulfate White BaSO4 Dilute Aqueous Solution of Furfural (1:1000) Finely Powdered Sulfur: S–2 Phosphate Yellow (NH4)3PO4 12MoO3 Ether: (C2H5)2O Dry Chloroform: CHCl3 Chemical Equation Acetic Anhydride: (CH3CO)2O REACTANTS PRODUCTS IV. PROCEDURES TEST FOR CHLORIDE Bile Chloride Ion + Nitric Acid + Silver Chloride + Silver Nitrate Nitrate Ion Secretion of the Liver Cl– + AgNO3 + HNO3 AgCl + NO3– It is viscid and has an alkaline reaction. TEST FOR SULFATE Color: greenish brown Important Constituents: Barium Chloride + Barium Sulfate + ○ bile acids Sulfuric Acid Hydrochloric Acid ○ bile pigments BaCl2 + H2SO4 BaSO4 + 2HCl ○ inorganic salts TEST FOR PHOSPHATE ○ cholesterol Phosphate Ion + Nitric Acid + Ammonium Phosphomolybdate Ammonium Molybdate + Water 1. REACTION 3NH4+ + 12MoO4–2 + PO4–3 + 24H+ (NH4)3PO4 12MoO3 + 12H2O Test the reaction of bile to litmus paper, phenolphthalein, and congo red solutions. Results and Observations BILE Basic pH pH: 7–8.4 Red & Blue Litmus Paper Turn Blue Congo Red Red Solution Phenolphthalein No Change in Color 12  3. TEST FOR BILE PIGMENTS 4. TEST FOR BILE ACIDS AND BILE SALTS Bile Pigments A. PETTENKOFER’S TEST OR SUCROSE-SULPHURIC ACID TEST Bilirubin: gives the red color in the tests Place 1 mL of dilute bile solution in a test tube. Add 1 drop of 5% sucrose ○ Breakdown product of hemoglobin solution. Slowly run about 0.5 mL of conc. sulphuric acid down the side Biliverdin: gives the green color in the test of the tube. Observe the production of a red ring at the point of Bilicyanin: gives the blue color in tests contact. Shake and the whole solution assumes a red color. Place the tube under running water to prevent the rise of temperature beyond A. GMELIN'S TEST 70ºC. Place 1 mL of conc. nitric acid in a test tube. Superimpose 1 mL of dilute The test is not reliable in the presence of proteins and other bile, care being taken not to mix the two fluids. Note the production of chromogenic substances which give color with sulphuric acid and different colors at the point of contact: green, blue, violet, red and therefore, interfere with the reading of the results. reddish-yellow. Results and Observations Results and Observations Q Results Q Results A The solution turns to cherry red and changes to purple. The color follows a sequence of familiar solar spectrum. A Q Interpret the results. A rainbow-like display of colors appears. Presence of a red ring at the point of contact indicates the A Q Order of Colors presence of bile salts. A Green → Blue → Violet → Red → Reddish-Yellow B. FOAM TEST B. ROSENBACH’S MODIFICATION OF GMELIN’S TEST Place 1 mL of dilute bile solution in a test tube and add 1 drop of dilute aqueous solutions of furfural (1:1000). Shake and observe the Filter some very dilute bile (1:50) through a small filter paper. When all production of a thick foam. Place a drop of conc. H2SO4 on the foam. has been drained through and the paper is still moist throughout, place Note the production of a dark pink color. a drop of conc. nitric acid into the top of the cone of the paper. Note the concentric succession of colors radiating from the tip of the cone. Results and Observations Results and Observations Q Results Q Results The formation of dark pink color is observed due to the A condensations of furfural and the bile acids and salts in dilute Different colors like green, blue, violet, red, and reddish yellow are bile. A observed in the paper. Q Explain the results. Q Order of Colors The presence of bile salts in the test solution reduces the surface A Green → Blue → Violet → Red → Reddish-Yellow A tension of the liquid, allowing the stable foam to form when shaken. 13  C. HAY’S TEST OF SURFACE TENSION TEST EXPERIMENT 15: URINE Hay’s Test of Surface Tension Test: based upon the principle I. PURPOSE that bile acids or bile salts have the property of reducing the surface tension of the fluid in which they are contained. To understand the components, properties, and reactions of urine Cool about 5 mL of diluted bile in a test tube to 17ºC or lower and II. APPARATUS sprinkle a little finely powdered sulphur on the surface of the fluid. The presence of bile salts is indicated by the sinking of sulphur, the Beaker, Test Tube, Test Tube Rack, Bunsen Burner, Clay Flame Shield, rapidity depending on the quantity of bile acids present. Wire Gauze, Filter Paper, Erlenmeyer Flask, Funnel Repeat the experiment using water instead of bile. Compare your III. MATERIALS results. Urine Results and Observations Sodium Nitroprusside: Na2[Fe(CN)5NO] Picric Acid: (O2N)3C6H2OH Ammonium Hydroxide: NH4OH Dilute Bile + Sulfur Water + Sulfur Zinc Chloride: ZnCl2 Sulfur sinks at the bottom Sulfur powder floats at the Acetic Acid: CH3COOH because bile reduces surface surface of water due to surface Barium Chloride: BaCl2 tension between polar water and tension between nonpolar sulfur Nitric Acid: HNO3 nonpolar sulfur. and polar water. Silver Nitrate: AgNO3 Magnesia Mixture/ Magnesium Hydroxide: Mg(OH)2 Benedict’s Reagent Saturated Solution of Benzidine in Glacial Acetic Acid: C12​H12​N2 in CH3COOH Hydrogen Peroxide: H2O2 Sodium Hydroxide: NaOH IV. PROCEDURES Urine Filtrate from the blood Serves a medium for excretion of 5. CHOLESTEROL IN BILE ○ water ○ salts Evaporate to dryness 5 mL of undiluted bile. Extract with a small ○ acids amount of ether at a time. Evaporate the ether extract to dryness in a ○ basses hot water bath (no flame). Dissolve the residue in 3 mL of dry ○ waste products of metabolism chloroform. Perform the Liebermann-Burchard test (acetic anhydride ○ other toxic materials + H2SO4) on the chloroform solution. Helps in the maintenance of water balance and acid-base equilibrium Results and Observations Serves as an important factor in the detoxification of the body Q Results Collection and Preservation of Urine Sample A Red Sol’n → Blue Sol’n → Bluish-Green Sol’n For the study of both the qualitative and quantitative What is the danger of excessive amounts of cholesterol in the composition of urine, it is better to examine the 24-hour Q bile? Explain. specimen. The bladder is emptied at 8:00 a.m. and the urine is discarded. Formation of Gallstones: when bile is oversaturated with excess All the urine from this time up to 8:00 a.m. the next day is A cholesterol, the excess crystallizes and forms gallstones. taken as sample. To Preserve the Urine: a thick layer of toluene is over-layed on its surface. 14  1. GENERAL CHARACTERISTICS 2. DETECTION OF CREATININE Examine the specimen as to color, odor, transparency, and reaction. COLOR PRODUCT Results and Observations Nitroprusside Test (Weyl) Ruby Red → Yellow Red Tautomer Creatinine Picrate Q What is the volume of the 24-hour urine? Picric Acid Reaction (Jaffe) Yellow when Acidified A Volume: 1000–1500 mL What substances are responsible for the normal color of Q urine? Urochrome/Urobilin: yellow A Uroerythrin: red/pink What happens when the urine is allowed to stand for some Q time, exposed to air? Bacteria: causes urea to break down into ammonia and A therefore, increases pH Q What is the normal reaction of urine when freshly voided? A. NITROPRUSSIDE TEST (WEYL) A Acidic: pH 6 To 5 mL of the urine add 3 drops of sodium nitroprusside. Alkalinize with Q To what substances is this reaction due to? NaOH. A ruby red color is produced which turns yellow. This is indicative of the presence of creatinine. A Presence of acid phosphates or organic acids. B. PICRIC ACID REACTION (JAFFE) What happens when allowed to stand without a preservative? Q Why? Place 5 mL urine in a test tube, add an aqueous solution of picric acid and render the solution alkaline with NaOH solution. A red color is Urine develops foul odor and color becomes darker because of: produced due to the formation of a red tautomer of creatinine picrate. the conversion of urea to ammonia A This turns yellow when the solution is acidified with conc. HNO3. Glucose the decompositions or oxidations of the organic substance gives a similar red color, only upon heating. by bacteria in urine What constituents of the urine tend to precipitate when the 3. DETECTION OF PIGMENTS Q reaction is acidic? AMMONIACAL ZINC CHLORIDE TEST Acidic Urine: Wastes A Sodium Urate: C5H3N4NaO3 Place 2 mL of urine in a test tube and add 1 mL of NH4OH, stand for a Uric Acid: C5H4N4O3 while and filter. To the filtrate add 2 drops of zinc chloride solution. The production of greenish fluorescence indicates the presence of urobilin. Q What substances precipitate in alkaline urine? Results and Observations Alkaline Urine: A Calcium Phosphates: Ca3(PO4)2 Calcium Oxalates: CaC2O4 Q Results Q What is the specific gravity range of normal urine? A Production of greenish fluorescence due to urobilin. A Specific Gravity: 1.015–1.025 Q What are the other pigments normally present in the urine? General: ↑ urobilin = ↑ darker yellow a A Urochrome/Urobilin: yellow Uroerythrin: red/pink 15  INORGANIC PHYSIOLOGICAL CONSTITUENTS 6. DETECTION OF PHOSPHATES 4. SULPHATES: DETECTION IN INORGANIC SULPHURIC ACID Make 10 mL of urine alkaline with ammonium hydroxide and warm. The earthy phosphates or phosphates of Ca and Mg separate. Place 5 mL of urine with 5 drops of acetic acid. Add barium chloride solution. A precipitate of barium sulphate is formed. Results and Observations Results and Observations Q Result Q What is produced? It precipitates with the remaining solution containing alkali A phosphates. A Barium Sulfate: BaSO4 (white ppt.) Filter off the earthy phosphates and small amounts of magnesia 5. DETECTION OF CHLORIDE mixture to the filtrate. Warm the solution. The alkali phosphates or the phosphates of Na and K separate. Place 5 mL of urine in a test tube and acidify with 2 drops of HNO3. Add 2 drops of AgNO3. Results and Observations Results and Observations Q Result Q What is produced? (Struvite) Magnesium Ammonium Phosphate: MgNH4PO4 6H2O A (white gelatinous ppt.) A Silver Chloride: AgCl (white ppt.) Determine which form of phosphate is present in a larger Q amount. Add excess of NH4OH. A Alkali Phosphates Results and Observations PATHOLOGIC CONSTITUENTS Q What happened to the precipitate? Healthy Individuals: should have lower concentration of The precipitate dissolved and reacted with excess NH4OH forming Albumin A ○ soluble diamine silver (I) chloride (diamine silver-1-chloride). ○ Glucose ○ Heller’s Ring What is the normal amount of chlorides eliminated in 24 Q hours? 7. ALBUMIN 110–250 mL per day A A. COAGULATION TEST 3.89–8.85 g per day Q In what forms are chlorides in urine eliminated? Heat 5 mL of urine to boiling in a test tube (filter if urine is not clear). If the heated portion becomes cloudy, the turbidity may be due to Sodium Chloride: NaCl phosphates. Add 3–4 drops of very dilute acetic acid and warm, the Potassium Chloride: KCl phosphates will dissolve. If a more flocculent precipitate will be A Magnesium Chloride: MgCl2 produced if only albumin is present. Ammonium Chloride: NH4Cl RESULT PRESENT SUBSTANCES Turbid Solution Phosphates + Albumin Flocculent Precipitate Albumin Only No Flocculent Precipitate No Albumin Present B. HELLER’S RING TEST Heller’s Ring Test: used clinically to detect the presence of albumin in urine. Place 5 mL of conc. nitric acid in a test tube, slant the tube, and very carefully allow an equal amount of urine to slowly run down the side of the tube. The urine will float on the nitric acid, and a white ring (precipitated protein) will appear at the junction of the two liquids. Sometimes the white zone does not appear until allowed to stand for a few minutes. 16 Results and Observations Q Is bilirubin normally present in the urine? Q Result NO: bilirubin is converted to urobilinogen in the intestines and A absorbed in the bloodstream, and the absorbed pigment is A No Change in Color: yellow ring (–) oxidized to urochrome or urobilin. Q What does its presence indicate? 8. GLUCOSE Possible jaundice, liver cirrhosis, hepatitis, and other liver BENEDICT’S TEST A diseases due to obstruction of the flow of bile from the gallbladder. Benedict’s Test: qualitative test for glucose in urine. 11. ACETONE BODIES / KETONE BODIES Add 8 drops of urine to 5 mL of Benedict’s reagent and boil vigorously for 2 minutes. Set aside to cool. The amount of precipitate and its color There is no satisfactory, simple direct fest for β-hydroxyl (brick-red, yellow or green) depend on the quantity of glucose butyric acid in the urine. present in the urine. Aceto-acetic acid, on the other hand, decomposes so rapidly with the formation of acetone, that the usual tests for acetone Results and Observations are also given for aceto-acetic acid. The test for ketonuria therefore, are tests for either acetone Q Result or aceto-acetic acid or both—true of the nitroprusside test. A No Change in Color: blue solution (–) NITROPRUSSIDE TEST (LEGAL’S) 9. BLOOD (DEMONSTRATION) Mix 2 mL of urine and 3 drops of 6% freshly prepared aqueous solution of sodium nitroprusside. Alkalinize with NaOH. Note the color BENZIDINE TEST produced. A ruby red color indicates acetone. Add 0.5 mL of acetic acid and make further observations. Benzidine Test: relies on the reaction between benzidine and the heme group in hemoglobin, producing a blue-green color If the test is made directly on urine, a red color is given by creatinine in the presence of blood. which disappears on the addition of acetic acid. Heat 3 mL of urine to boiling, cool and treat with an equal volume of a Results and Observations saturated solution of benzidine in glacial acetic acid (C12​H12​N2 in CH3COOH). Add 1 mL of 3% H2O2. The development of a blue or green Q Result color indicates the presence of blood. A No Change in Color: yellow solution (–) Results and Observations Q Result A No Change in Color: yellow solution (–) 10. BILE GMELIN’S TEST Place 1 mL of conc. nitric acid in a test tube and upon it superimpose 1 mL of urine. Do not mix the 2 fluids. At the point of contact various colored rings are noted: green, blue, violet, red and reddish-yellow, in the presence of bile pigments. Results and Observations Q Result A No Change in Color: yellow solution (–) Q What is bilirubin? A A pigment that is a breakdown product of hemoglobin. 17  EXPERIMENT 17: PAPER CHROMATOGRAPHY 3. Remove the paper from the beaker and open-up. Mark the position of the solvent front before it dries up completely. I. PURPOSE 4. Spray the paper lightly with 0.2% ninhydrin solution and dry it in the oven at 110°C. Heat if necessary for the color forming reaction. To separate amino acids on the basis of the difference in solubility of The color should be readily visible after 20-30 minutes. amino acids between 2 immiscible solvents. D. CALCULATION OF THE Rf VALUE II. APPARATUS 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑 𝑖𝑛 𝑚𝑚 Rf = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑚𝑚 Pipette, Beaker, Aluminum Foil, Whatman Filter Paper no. 1, Pencil, Capillary Tube Calculate the Rf value of each amino acid and identify the unknown III. MATERIALS amino acid. Butanol: CH3(CH2)3OH DATA AND ANSWER SHEET Acetic Acid: CH3COOH Distilled Water: H2O Distance Traveled Distance Traveled Rf Value 0.5% Glycine: C2H5NO2 by Amino Acid (mm) by Solvent (mm) 0.5% Lysine: C6H14N2O2 Glycine 24 mm 60 mm 0.4 0.5% Aspartic Acid: C4H7NO4 0.5% Unknown Solution Lysine 22 mm 59 mm 0.37 0.2% Ninhydrin Solution: C9H6O4 Aspartic Acid 19 mm 62 mm 0.306 IV. PROCEDURES Unknown 24 mm 55 mm 0.436 A. PREPARATION OF THE DEVELOPING CHAMBER Identify the Unknown according to the Table: ASPARTIC ACID Pipette 8 mL of the solvent consisting of 4:1:5 (by volume) mixture of Table 1. Data and Answer Sheet butanol, acetic acid, and distilled water respectively, and introduce into a dry 50 mL beaker. Avoid splashing the liquid on the sides of the beaker. V. AMINO ACIDS AND THEIR Rf VALUE Cover with a piece of aluminum foil and let stand for 10 minutes for the atmosphere inside to become saturated with the solvent vapor. AMINO ACID Rf VALUE B. PREPARATION OF THE PAPER CHROMATOGRAM Histidine (His | H) 0.11