Customized Lab Manual FINALS

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  • 1. Activity 5A Characterization of LipidsI. OBJECTIVES To enumerate the different physical properties of lipids To discuss the principle behind each test To name the components of lipidsII. SCHEMATIC DIAGRAMIII. RESULTS AND OBSERVATION Solubility TestSolvent Expected Result Experimental Result ObservationWater Insoluble Insoluble Formation of layer, big bubblesDilute HCL Insoluble Insoluble Formation of layer, smaller bubblesDilute NaOH Insoluble Insoluble Formation of layer, oil are solidifiedEthyl alcohol Insoluble insoluble Oil is found at the bottom of the tubeChloroform Soluble Soluble dissolvedEther Soluble soluble dissolved
  • 2. Formation of Translucent spot Oil used Presence of Time Duration translucent spot Coconut oil + 16 minutes Linseed oil + 1 hour and 1 minute Peanut oil 4 minutes + Cod liver oil + 21 minutes Olive oil + 39 minutes Corn oil + 20 minutes Reaction of fats 0 minute After 3 hours resultRed litmus Red-red Red-red No reactionBlue litmus Blue-blue Blue-blue No reaction Acrolein FormationOil used Expected result Experimental result/observationCoconut oil Formation of acrolein; acrid Acrid odor (+) odorIV. ANALYSIS AND CONCLUSIONGuide questions:1.) Explain the solubility of the different kinds of lipids in each solvent used. Lipids are non-polar organic. The physical properties of the fatty acids, and of compounds thatcontain them, are largely determined by the length and degree of unsaturation of the hydrocarbonchain. The nonpolar hydrocarbon chain accounts for the poor solubility of fatty acids in water.Solubility of a substance depends on a simple rule of a thumb that "like dissolves this statementindicates that a solute will dissolve best in a solvent that has a similar chemical structure to itself.The overall solvation capacity of a solvent depends primarily on its polarity. Water-Coconut oil is insoluble in water even at room temperature since water is a polar solvent it will not dissolved coconut oil which is non-polar. Dilute HCl- coconut oil is insoluble in dilute acid (dilute HCl) since HCL is a nonpolar inorganic solvent. Coconut oil has higher molecular weight and when dissolved in dilute acid tends to float. Dilute NaOH-coconut oil is insoluble in dilute alkali (NaOH). In the solution the coconut oil it tends to form as a white colloidal matter after the addition of dilute alkali. Ethyl Alcohol- Like water ethyl alcohol is a polar solvent and will not dissolve most nonpolar solutes such as oil. Chloroform - Coconut oil is soluble in chloroform. Oil is an polar organic compound same is the position with the chloroform Ether- coconut oil is soluble in ether since ether is an organic non- polar solvent.
  • 3. 2. Explain why other oils evaporate faster, while other show slow evaporation. Give the structural formula of the different types of oil. Each lipid is different with each other, especially with the number and the branching of their carbon chain, the longer the chain and the more it is branching the longer the time it evaporates. There is also a factor in the degree of unsaturation of the lipid. Saturated fatty acids or saturated fats tend to evaporate a lot longer than unsaturated fat. Saturated fats are composed of many single carbon bonds that make the substance less volatile. Unsaturated fats are composed of one or more double bonds. The double bonds make the fat more volatile. In other words there are differences in the evaporation of different kind of oil because their evaporation rate mainly depends upon their structural formula. The difference in molecular weight and number of double bonds therefore affect the rate of evaporation of oil.Coconut oil Peanut oilCorn OilLinseed oil Olive oil
  • 4. 3. Show chemical reaction of Acrolein test. The principle behind the acrolein test is a specific chemical reaction. This reaction is utilizedto determine the presence of glycerin in a fat. By heating the fat sample in the presence ofpotassium bisulfate (KHSO4), which acts as a dehydrating agent, acrolein (C3H4O, or CH2=CH-CHO)is formed and can easily be detected by its odor. Whenever fat is heated in the presence of adehydrating agent, the fat molecule will shed its glycerol in the form of the unsaturated aldehyde -acrolein.Analysis Solubility Lipids are insoluble in non-polar solvents like chloroform and ether and insoluble polar suchas water, alcohol, acid and base. Thus, it follows the rule of a thumb like dissolves like. Water even called a universal solvent will not dissolved the coconut oil because of its non-polarity. In the test you can see the formation layer and big bubbles of oil on the top surfaceindicates that oil have lower specific gravity thus it float on water. In the 2nd tube that contain dilute HCl which is an inorganic acid and a polar solvent. In theexperiment it shows formation of layer but smaller bubbles compared to the water-oil mixture. Thisindicates that HCL somehow dissolved the oil into smaller particles and oil has a lower specificgravity that dil. HCL, thus float on the mixture. In the 3rd tube which contain dilute NaOH which is an inorganic base and a polar solvent. Inthe experiment it shows formation of layer and the oil was solidified. The solidification of the oilwas due to the chemical property of lipid which is saponification, which turns oil into solid. In the 4th and 5th tube which contains chloroform and ether respectively dissolved the oilbecause it is a non-polar organic solvent. Translucent spot Based on our test results it shows that peanut oil evaporated first, followed by coconut oil,corn oil, cod liver oil, olive oil and lastly the linseed oil. It is stated that the difference in molecularweight and number of double bonds affect the rate of evaporation of oil. The longer the fatty acylchain and the fewer the double bonds, the lower the evaporation rate. In our test the rate ofevaporation was also affected by the quantity of oil dropped in the paper. Reaction of fats In the test there was no change in color of the litmus paper in both blue and red, theoretically the oil should turn into acidic since it was exposed in an open environment and thus undergo rancidity. The result was maybe due to several factors like the oil used was not fresh. Acrolein test In the experiment the oil ghive a positive result since coconut oil is a lipid and is composed of glycerol.
  • 5. ConclusionLipids show many physical and chemical properties. These properties were clearly shown in theresults of the test being performed like the solubility test, translucency test and reaction of fats,acrolein test. Lipids are soluble in non-polar solvents like chloroform and ether and insoluble inpolar such as water, alcohol, acid and base. Lipids, specifically oils shows formation of translucentspot, the rate of evaporation and formation of this translucent spot vary because of the differenceof chemical structure. Lipids when exposed to open air or environment would undergo rancidityspecifically oxidation because it is exposed to oxygen, wherein it will become acidic. Lipids show apositive reaction to Acrolein test since it is composed of glycerol which is the main reactant to formthe product acrolein.V. DOCUMENTATIONSolubility of oil (from left: water, dilute HCL, diluteNaOH, ethyl alcohol, chloroform, ether)Formation of Translucent Spot (from left: linseed oil, peanut oil, cod liver oil, olive oil, corn oil,coconut oil)
  • 6. Reaction Of fatsAcrolein test
  • 7. ACTIVITY 5B CHARACTERIZATION OF LIPIDSI. OBJECTIVES Discuss the saponification property of lipids To enumerate the properties of emulsifying fats &saponifying lard Explain the principle in each test employed for lipidsII. SCHEMATIC DIAGRAMA. Emulsification of Fats Shake a drop of cocunut oil &Stand for sometime Place 1ml. H2O & add 1 drop 0.5% NaCO3 & a drop of coconut oil. Shake Shake a drop of cocunut oil with (1 ml. dilute albumin sol’n)B. Saponification of Lard Place 5g lard in (erlenmeyer flask) Add 25ml. alcoholic potash Warm (in water bath) Transfersolution in evaporating dish (with 2ml. H20) Acidify with (5 drops diluted HCl) Removefatty acid Neutralize sol’n with (10% NaCO3) Evaporate to dryness Extract residue with (1ml. ethyl alcohol) Remove alcohol (by evaporation on water bath)
  • 8. III. RESULTS AND OBSERVATION Test tube Results and Observation Coconut Oil in Distilled water immiscible 0.5% Sodium Carbonate and coconut oil in immiscible Distilled water Coconut oil in Albumin Solution miscibleSaponificationResults and Observation: The product was cream to brownish color substance with paste likeconsistency. After it is removed from heat it hardened into a powdery like residue.IV. ANALYSIS AND CONCLUSIONGUIDE QUESTIONS:1. Explain the principle involved in both saponification and emulsification.2. Write chemical equation.A. Emulsification of Fats Emulsification is a process by which you mix two liquids that are ordinarily immiscible.Emulsification of fats to mix them with water-based substances, which has important implicationsfor cooking and for digestion. To emulsify fats, you need substances such as bile salts or othercompounds that help connect the fat and water. Droplets of the dispersed component rapidlycoalesce to form a separate layer. Emulsifying agent must be present to stabilize the emulsion.Lecithin in the egg will serve as emulsifier.B. Saponification of Lard Saponification is the hydrolysis of an ester under basic conditions to form an alcohol andthe salt of a carboxylic acid. Saponification is a process that produces soap, usually from fats andlye. Saponification involves metallic alkali base (usually NaOH & KOH) hydrolysis of triglycerides,which are esters of fatty acids, to form the sodium salt of a carboxylate. In addition to soap, suchtraditional saponification processes produces glycerol. "Saponifiable substances" are those that canbe converted into soap. In saponification process, we used alcoholic potash (KOH dissolved inethanol & neutralized fatty acids in the lard) &NaCO3 to produce hard soap; then a metallic salt offatty acid is formed. Fats (triglycerides) upon alkaline hydrolysis (either with KOH or NaOH ) yield glycerol andpotassium or sodium salts of fatty acids (soap) or
  • 9. Analysis A. Emulsification In the experiment, only in test tube 3 that the two liquids, coconut oil and albumin solutionbecame miscible. Since albumin contains an emulsifying agent lecithin. The remaining 2 test tubesdidn’t mix due to absence of an emulsifier. B. Saponification In our experiment, the outcome is powdered due to high application of heat, wateris then evaporated. Nevertheless, we had a positive result on saponifying lard because of alkalinehydrolysis yielding glycerol and potassium or sodium salts of fatty acids.Conclusion In conclusion, fats are insoluble in water, so they will not mix homogeneously in theaqueous solutions that exist in living things. But, with emulsification, little droplets of fat areformed which DO mix in water, evenly enough at least, without clumping or floating in a mass. So,they can be efficiently moved around the body and used in reactions. While in saponification of lard, It’s an exothermic chemical reaction that occurs when a animal fat is mixed with a strong alkali (Alocoholic Potash). The products of the reaction are two: soap and glycerin. Water is also present, but it does not enter into the chemical reaction. The water is only a vehicle for the alkali, which is otherwise a dry powder. Therefore, only saponifiable substances are those that can be converted into soap.V. DOCUMENTATION Coconut oil with water. Coconut oil, water, and ,5% sodium carbonate. Saponification product
  • 10. ACTIVITY 5C QUALITATIVE TEST FOR CHOLESTEROLI. OBJECTIVES To discuss the principle behind the qualitative tests for cholesterol. To explain the effects of the different reagents used in the activities. To discuss the properties of cholesterol that makes it responsive to Salkowski and Leibermann-Burchard test.II. SCHEMATIC DIAGRAM Preparation:
  • 11. Salkowski test: Leibermann-Burchard test:III. RESULTS AND OBSERVATION TEST THEORETICAL RESULT EXPERIMENTAL RESULT RESULT OBSERVATION Salkowski test Chloroform layer- red to - Chloroform layer- blue yellow Acid layer- orange Acid layer- green Leibermann- Blue-green complex - Upper layer- brown Burchard test Lower layer- dull orangeIV. ANALYSIS AND CONCLUSION GUIDE QUESTION Give the principle involved for Salkowski and Leibermann-Burchard tests. When concentrated sulphuric acid is added to a chloroform solution of cholesterol, the chloroform layer shows a red to blue color and the acid layer shows a green fluorescence. This
  • 12. is the principle behind Salkowski test. On the other hand, the acetic anhydride in theLeibermann-Burchard test can react with the C3 hydroxyl group of cholesterol and relatedsteroid in the presence of a strong acid such as concentrated sulphuric acid that we used toform a blue-green complex.Analysis The solution and the precipitate yielded after the preparation was the separation of cholesterol (precipitate) and triglycerides or neutral fats (solution). Salkowski Test is a test for cholesterol. It was said that the reaction principle of this test indicates the presence of double bond in one cholesterol ring is responsible for its ability to form colored products in the presence of concentrated inorganic acids and it should yield a layering of colors. When concentrated sulfuric acid is added to a chloroform solution of cholesterol, the chloroform layer shows a red to blue color and the acid layer shows a green fluorescence. The concentrated sulfuric acid reacts with the cholesterol leading to reddish color. But in our experiment, it did not end that way. Ours yielded an orange-brown color. Well, it was expected since we did not follow the right procedure, instead of heating it in the water bath, we heated it directly in the hot plate. Leibermann-Burchard Test is still a test for cholesterols. The color that is supposed to be yielded in this test is deep green color. This is due to the hydroxyl group of cholesterol which reacts with the reagent which is acetic anhydride and sulfuric acid and also due to the increasing conjugation or the transformation of a substance to a hydrophilic state. But in our experiment, it ended badly. We didn’t even saw a sign of greenish color. The reason again for this outcome is the incapability of us students to carefully read and follow what the manual said, but what’s done is done we just need to be careful next time.Conclusion After performing the experiment, our group concluded that knowledge, accuracy and carefulness are necessary in conducting an experiment. In our experiment we tried separating neutral fats and triglycerides, and we did it easily with the help of some reagents. But it went bad in performing the 2 tests which are Salkowski and Leibermann- Burchard tests. Instead of yielding the right colors it yielded a very strong color because of the burnt precipitate we produced in its preparation. Our experiment may be unsuccessful but at least we learned some things that could be useful in the futureV. DOCUMENTATION Salkowski Test Leibermann-Burchard Test
  • 13. ACTIVITY 6A Carbohydrates Qualitative Tests for Different Types of CarbohydratesI. OBJECTIVES Become familiar with common carbohydrates Learn significant differences in the chemical properties of carbohydrates Identify the carbohydrates present in common food productsII. SCHEMATIC DIAGRAMA. Molisch’s Test Place 1ml of distilled H2O (control), 1 ml of 5% (glucose, galactose, maltose, fructose, sucrose, lactose, glycogen & starch sol’n in each test tube Add5 drops of Molisch Reagent Mix content by gently shaking each tube Incline tube while adding 1ml. conc. Sulfuric acid Note the color of the ring formed @the junction of 2 liquidsB. Anthrone Test Mix 1ml. of glucose sol’n & iml. Anthrone reagent Stand (sometime) &note color after 1 hr. Dilute with glacial acetic acid or 50% H2SO4 Do the same with 5% (galactose, maltose, fructose, sucrose, lactose, glycogen & starch)
  • 14. C. Phenylhydrazine Test (Osazone Formation) In each of 8 test tube place 1ml. of 5% sol’n of (#1-glucose, #2- maltose, #3-fructose, #4-galactose, #5-lactose, #6-sucrose, #7- glycogen, #8-starch) Add a pinch of phenylhydrazine /test tube. Mix &stopper the test tubes (using cotton) Place them in boiling H2O for 30 min. Cool slowly @ room temperatue Examine crystals formed under a microscope.III. RESULTS AND OBSERVATIONA. Molisch Test Solution Molisch’s test Anthrone Test Phenylhydrazine Test Expected result Purple ring Green Yellow-pale orange distilled water (-) Clear liquid N/A N/A (control) 5% glucose (+) Purple ring (-) Yellow-black (+)Yellow-black galactose (+)Purple ring (-) Brown-black (+)Yellow-black maltose (+)Purple ring (-) black (+)Orange-black fructose (+)Purple ring (-) black (+)brown sucrose (+)Purple ring (-) black (+)Brown lactose (+)Purple ring (-) Yellow-black (+)Orange-black glycogen (+)Purple ring (+)Green-black (+)Pale yellow starch (+)Purple ring (+)Green-black (+)yellow
  • 15. IV. ANALYSIS AND CONCLUSIONGUIDE QUESTION1. Give the principle involved in each test for carbohydrates.A. Molisch’s Test Molisch Test is a sensitive chemical test for all carbohydrates and some compoundscontaining carbohydrates in a combined form, based on the dehydration of the carbohydrate bysulfuric acid to produce an aldehyde either (furfural or a derivative) which then condenses with thephenolic structure resulting in a purple-colored compound. Carbohydrates are dehydrated byconcentrated sulfuric acid to form hydroxymethylfurfural will react with alpha-naphthol (Molischreagent) to yield a purple condensation product.B. Anthrone Test In the anthrone assay, carbohydrates are dehydrated by using conc. H2SO4 to formfurfural, which in turn condenses with anthrone (10-keto-9,10 dihydroanthracene) to form a bluish-green complex. Sugars react with the anthrone reagent under acidic conditions to yield a blue-green color. Concentrated H2SO4 is a powerful dehydrating agent involved in dehydrating sugarsleading to formation of the furfural, which condenses the anthrone to give a colored a product.C. Phenylhydrazine Test (Osazone Formation) When phenylhydrazine react with the reducing sugars @ boiling temperature, osazone isformed. Phenylhydrazine (NH2-NH-C6H5) reacts with carbonyl compounds in neutral or slightlyacidic conditions to give phenylhydrazone, which is highly soluble. When hydrazones react furtherwith the phenylhydrazine molecule, the condensation products formed are insoluble osazones,which precipitate out as crystals. Reducing sugars form characteristic osazone crystals when heatedwith an excess phenylhydrazine. This property is attributed to the presence of aldehyde or ketonegroup in their molecules. -OH group immediately adjacent to the keto group is oxidized to a ketogroup and adds to phenylhydrazineto form the yellow to pale orange osazones.
  • 16. Analysis In the Molisch Test, from the experiment all sugar tested solutions yield positive resultsince they are all carbohydrates. While the control tube which contains water give a negative resultwhich is a clear solution. Since it’s not a carbohydrate. In the Anthrone Test, only glycogen & starch give positive results but, supposedly all sugarsolutions must give positive results in the Anthrone Test since they are all carbohydrates but dueto some factors in the process of testing them that’s why some didn’t gave positive results. In the Osazone Test, all tested sugar solutions gave a yellow-pale orange coloration. Butsome sugar solutions, when they are viewed under the microscope they didn’t give expectedresults (crystals formation). This is maybe because by the time given, they are about to formcrystals other sugar tested solutions takes longer than the time given time to form their crystalsthat’s why some didn’t yield a positive crystals formation (glucose & sucrose), while other yieldpositive results (galactose, fructose, lactose, maltose, glycogen & starch).Conclusion In conclusion, for the Activity 6A Qualitative Tests for Different Carbohydrates: UsingMolisch, Anthrone&Osazone qualitative test we became familiar with common principles ofcarbohydrates in different qualitative tests. We learned also their significant similarities &differences in their reactions in different qualitative tests. We knew also that external factors in theprocess of testing them would affect their reactions & results.
  • 17. V. DOCUMENTATION Molisch’s Test Anthrone Test Phenylhydrazine TestPhenylhydrazine Test #1: Glycogen, # 2 Lactose, #3 Galactose, #4 Glucose, #5 Sucrose, #6 Starch,#7 Maltose, #8 Fructose
  • 18. Activity 6B Qualitative Test For CarbohydratesI. OBJECTIVES Enumerate the qualitative tests for various types of carbohydrates Discuss the principles in each test Explain the characteristic of different type carbohydrateII. SCHEMATIC DIAGRAM
  • 19. IV. ANALYSIS AND CONCLUSIONSGUIDE QUESTION:1. Give the principle for each test for carbohydrates. 1. Moores Test - When a solution of reducing sugar is heated with an alkali (NaOH), it turnsyellow to orange and finally dark brown, liberating the odor of caramel. This is due to the liberationof aldehyde which subsequent polymerizes to form a resinous substance, caramel. 2.Fehling’s test- a test for the presence of aldehydes but not ketones (still reducing sugars).It is an alternative test of the Benedict’s test as to Benedict’s test being more sensitive. Aldehydesare detected by the reduction of the deep blue solution of Copper (II) to a red precipitate ofinsoluble Copper oxide. The Fehling’s test is usually used to detect for the presence of reducingsugars but it is not specific for aldehydes. 3. Benedicts Test - Aqueous glucose is mixed with Benedicts reagent, a solution of coppersulfate, sodium hydroxide, and tartaric acid. The mixture is heated. Carbohydrates which react withBenedicts reagent to reduce the blue copper (II) ion to form a brick red precipitate of copper (I)oxide are classified as reducing sugars. As in Fehling’s test, the reducing sugars because of havingpotentially free aldehyde or keto group reduce cupric hydroxide in alkaline solution to red colouredcuprous oxide. Depending on the sugar concentration yellow to green colour is developed. 4. Nylander Test - Currently, the Nylander’s test is used to test for the presence of reducingsugars (specifically glucose) in urine. But, basically, the Nylander’s test is still a test for reducingsugars.Nylander’s test depends upon the reduction of bismuth subnitrate in a weakly alkalinesolution to metallic bismuth. The Nylander’s reagent is a solution composed of a solution ofRochelle salt (potassium sodium tartrate), potassium or sodium hydroxide, and bismuth subnitratein water. The sodium hydroxide or potassium hydroxide in the Nylander’s reagent gives the alkalineenvironment in order for the reaction to occur. The potassium tartrate serves as the solvent for thebismuth subnitrate. 5. Barfoeds Test - Barfoeds 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 formedwill indicates the presence of reducing sugar. The reaction will be negative in the presence ofdisaccharide sugars because they are weaker reducing agents. This test is specific formonosaccharide. Due to the weakly acidic nature of Barfoeds reagent, it is reduced only bymonosaccharides. It is based on the reduction of copper(II) acetate to copper(I) oxide (Cu2O),which forms a brick-red precipitate. 6. Picric Acid test is generally a test for the presence of reducing sugars. The reagent usedwas the picric acid (yellow in color). In order for a reaction between the reducing sugar and Picricacid to take place, the solution must be alkaline; this is the purpose of adding the 10% SodiumCarbonate in the experiment. When Picric acid reacts with the aldehyde or keto group of thereducing sugar, it turns into a Picramic acid which gives of the positive color of Mahogany red.Heating was done because in higher temperatures, reaction between substances takes place faster. 7. Seliwanoffs Test - It is a color reaction specific for ketoses. When concentrated HCl isadded ketoses undergo dehydration to yield furfural derivatives more rapidly than aldoses. Thesederivatives form complexes with resorcinol to yield deep red color. The test reagent causes thedehydration of ketohexoses to form 5-hydroxymethylfurfural. 5-hydroxymethylfurfural reacts withresorcinol present in the test reagent to produce a deep red product . It is a timed colour reactionspecific for ketoses.
  • 20. 8. Iodine Test - This test is used for the detection of starch in the solution. The blue blackcolour is due to the formation of starch-iodine complex. Starch contain polymer of α-amylose andamylopectin which forms a complex with iodine to give the blue to puple black colour. 9. Mucic Acid test is a test specific for the monosaccharide galactose. Whenmonosaccharides (galactose) are treated with strong oxidizing agents such as nitric acid or HNO3(was utilized in the experiment), it yields a saccharic acid. The saccharic acid obtained after theoxidation of galactose is insoluble and separates as crystals. This acid derivative is known as mucicacid. This test allows for the differentiation of galactose to other monosaccharides as othermonosaccharides that undergo this test yield soluble dicarboxylic acid whereas galactose producesinsoluble Mucic acid. Lactose is a dimer of glucose and galactose. The nitric acid will hydrolyze thedimer and so lactose will also give a positive result.Analysis1. Moore’s test-In our experiment the solution that yield a positive result was glucose, galactose,maltose, fructose and lactose while sucrose, glycogen and starch was negative and was indicated bya clear solution. Moore’s test reaction is due to the presence of an aldehyde which in turn liberatedto form caramel and it is also specific for reducing sugars. Glucose, galactose, maltose, fructose andlactose are reducing sugars while sucrose, glycogen and starch are not, therefore our experimentalresults are correct. Sucrose did not give a positive result in the test because it is not a reducingsugar. The non-reducing property of sucrose is due to it’sanomeric carbon which is not "free" sincethis carbon links fructose and glucose together. This anomeric carbon cant open up the ringstructure and react with the reagent (which is Concentrated NaOH). Starch and glycogen did notreact in the Moore’s test. This is because both of them are is a polysaccharide. All polysaccharidesare non-reducing sugars therefore; its anomeric carbon is “full” and cannot react to anothermolecule.2. Fehling’s Test- In our experiment the solution that yield a positive result was glucose, galactose,maltose, fructose and lactose while sucrose, glycogen and starch was negative and was indicated bya clear bue solution with no precipitate. Fehling’s test is a test specific for reducing sugars. Glucose,galactose, maltose, fructose and lactose are reducing sugars while sucrose, glycogen and starch arenot. This reaction have the same explanation with the moore’s test.3. Benedict’s Test- In our experiment the solution that yield a positive result was glucose, galactose,maltose, fructose and lactose while sucrose, glycogen and starch was negative and was indicated bya clear bue solution with no precipitate. Benedict like Fehling’s is a test specific for reducing sugars.The main difference between Benedict’s with Fehling’s is Benedict’s test is more sensitive. TheFehling’s test needs a strong alkaline condition in order for a reaction to take place, unlike in theBenedict’s test where even in a weak alkaline environment, a reaction could still occur.Glucose,galactose, maltose, fructose and lactose are reducing sugars while sucrose, glycogen and starch arenot.4. Nylander’s Test- In our experiment the solution that yield a positive result was glucose,galactose, maltose, fructose and lactose while sucrose, glycogen and starch was negative and wasindicated by a clear solution. As mentioned previously, this test is specific for reducing sugars thusfollow the same explanation with Moore’s test about the reducing sugar.5. Barfoed’s Test- - In our experiment all the solution yielded a negative result. Theoreticallyglucose, galactose, maltose, fructose and lactose will be positive since it is a reducing sugar whilesucrose, glycogen and starch will be negative since it is a non-reducing sugar. The wrong result ofthe test is maybe due to the heating process done in the experiment. During the heating of thesolution we put it in a water bath and place it in a beaker. Maybe that factorhinders the reaction ofheat and thus the red precipitate was not seen in the reaction.6. Picric Acid Test - In our experiment only galactose, maltose, and fructose gave the mahogany redcolor when in fact all reducing sugars including glucose and lactose should have given a positive
  • 21. result excluding sucrose, glycogen, and starch because they are non-reducing sugars. We mighthave an error doing the procedure during the experiment thus failing to have the actual results.7. Seliwanoffs Test -In this experiment we gained a positive result for fructose and sucrose whichare in fact ketoses in nature. They turned pink in color as they are heated passing Seliwanoff’s test.8. Iodine Test -Before performing the test, we had a pre-conclude answer that the starch will give apositive blue black color while the glycogen will not. Upon performing the procedure, we hadgained a starch which is blue black in color while the glycogen being orange. In our documentationwe had two clear solutions because it was the end product of the test. They had lost their colorupon adding drops of sodium thiosulfate.9. Mucic acid Test – Performing this experiment galactose, fructose, and lactose gave a positiveinsoluble mucic acid. Basing on the principle galactose and lactose must have a positive result andwe gained those but having fructose being positive because it is water soluble thus unable tocrystallize.ConclusionThere are many qualitative tests for carbohydrates. This test is designed to identify the specificcarbohydrates, whether it is a monosaccharide, disaccharide, polysaccharide, reducing or non-reducing sugar. Moore’s test, Fehling’s test, Benedict’s test, Nylander’s testand Picric acid test aretests for reducing sugars. Barfoeds test is also for reducing sugar but only specific formonosaccharaides. Seliwanoff’s test is specific for ketoses, Iodine test for starch and Mucic acidtest for galactose.V. DOCUMENTATION Moore’s Test From left to right: Maltose, Starch, Sucrose, Glycogen, Lactose, Fructose, Glucose, Galactose. Fehling’s Test From left to right: Glucose, Galactose, Fructose, Lactose, Sucrose, Maltose, Starch, Glycogen.
  • 22. Benedict’s Test From left to right: Glucose, Galactose, Fructose, Maltose, Lactose, Sucrose, Starch, GlycogenNylander’s testFrom left to right:Fructose, Maltose,Lactose, Glucose, Starch, Galactose,Glycogen.Barfoed’s Test From left to right:Glucose, Galactose, Fructose, Maltose,Lactose, Sucrose, Starch, GlycogenPicric Acid TestFrom left to right:Galactose, Maltose. Fructose,Lactose, Glucose, Starch
  • 23. Seliwanoff’s Test From left to right:Galactose, Lactose, Glucose, Fructose, Sucrose, Maltose Iodine Test From left to right: Glycogen, Starch Mucic Acid Test From left to right:Galactose, Glucose, Lactose, Fructose, Sucrose.Mucic Acid Test MicroscopeFrom left to right: Fructose, Lactose, Galactose.
  • 24. Activity 7 Salivary Digestion I. OBJECTIVES To determine the normal pH of saliva To discuss the effect of temperature on the digestion of starch in saliva To discuss the different factors influencing ptyalin activity II. SCHEMATIC DIAGRAM A. Reaction of Saliva Prepare 3 test tubes with a few drops of a resting saliva Test the reaction with: Phenolphthalein Litmus Paper Congo Red B. Inorganic Matter Prepare 5 ml of saliva in a test tube Add 5 drops of Nitric Acid to acidify Heat to boiling (to remove proteins) Divide into 4 test tubes; use the filtrate Chlorides Phosphates Sulfates Calcium Add 1 drop ofAdd 1 drop of Add 1 drop of Add 1 drop of PotassiumSilver Nitrate Ammonium Barium Chloride Oxalate solution Molybdate solution
  • 25. C. Influence Of Temperature On Ptyalin Activity Prepare 1 ml of 1% starch solution in 4 test tubes Test Tube 1 Test Tube 2 Test Tube 3 Test Tube 4 Boiling Ice Water Mixture Room Temperature 38◦ C Temperature Add 10 drops of Add 10 drops Add 10 drops of Add 10 drops of Saliva and shake Saliva and shake Saliva and shake Saliva and shake well well well well Every 15 minutes intervals get a sample from Each test tube and test for Iodine and Benedicts solutions Allow to stand for an hour D. Influence Of Dilution Of Ptyalin Activity Prepare 6 test tubes containing 9 ml Of water eachTest Tube 1 Test Tube 2 Test Tube 3 Test Tube 4 Test Tube 5 Test Tube 6 Add 1 ml Add 1 ml of Add 1ml of Add 1ml of Add 1 ml of Add 1ml Mixture Mixture Mixture Mixture Of mixture Of saliva From From From From From Test tube 1 Test tube 2 Test tube 3 Test tube 4 Test tube 5 Shake well Place at water bath at temperature about 40◦ c for 20 minutes Test with Iodine test and Benedicts reagent
  • 26. III. RESULTS AND OBSERVATIONA. Indicators Results and observations Phenolphthalein Colorless Blue Litmus Paper Blue to Red Red Litmus Paper Red to Red Congo Red RedB Filtrate to test for: Results and Observations Chlorides Cloudy with white precipitate (Positive) Phosphates Pale Yellow with a little white precipitate ( Positive) Sulfates Yellow with white precipitate (Positive) Calcium Cloudy with white precipitate (Positive)C. Varying Temperatures Benedicts Test Iodine Test Ice Water Light Blue (Negative) Yellow (Negative) Room Temperature Light Blue (Negative) Yellow with Dark Black Spots (Positive) 38◦ c Light Blue (Negative) Blackish Red (Positive) Boiling Temperature Light Blue (Negative) Yellow ( Negative)D. Test Tube No. Iodine Test Benedicts Test Test tube #1 Yellow ( Negative) Light Blue ( Negative) Test tube #2 Dark Yellow (Negative) Light Blue ( Negative) Test tube #3 Green ppt(Negative) Lightest Blue (Negative) Test tube #4 Yellow (Negative) Deepest Blue among all (Negative) Test tube #5 Dark Yellow (Negative) Light Blue (Negative) Test Tube #6 Darkest Yellow(Negative) Light Blue (NegativeIV. ANALYSIS AND CONCLUSIONGUIDE QUESTION1. How did you know that digestion has taken place in the mouth?The salivary glands in the mouth secrete enzymes which break down food in the mouth before itgoes to the stomach, the next site of digestion. Our saliva contains an enzyme that starts breakingdown long carbohydrate molecules as soon as we put food in our mouth. Salivary amylase is theenzyme that begins splitting apart the bonds between glucose molecules in a long chain of starch.Like in our experiment if saliva is put into a test tube with starch it will convert this into sugar.2. Explain how temperature influence the process of digestion.
  • 27. Digestion is a chemical reaction where enzymes cleaves the complex molecules into simple sugar orfatty acid molecules. Like every chemical reaction the digestion also needs catalyst (enzyme),temperature, and pH. If the optimum temperature changes the rate of reaction decreases, so ourbody maintains constant temperature but sometimes temperature changes on change in physicalconditions or by malfunction of the body. By change in temperature either decrease or increase, therate of reaction decreases. If saliva is put into a test tube with starch it will convert this into sugar. Atlow temperatures this process goes on slowly, the velocity increasing as the temperature increasesuntil it reaches its maximum at about 37° C. Above this temperature the velocity again decreases,the enzyme being destroyed at about 70° C.AnalysisA. reaction of saliva In the reaction of saliva experiment, we used different pH indicator(phenolphthalein, litmus paper and congo red) to be able to estimate the approximate pHof our saliva (resting) . Upon using phenolphthalein as our indicator we obtained a colorlesssolution that indicates the acidity of our saliva solution. The Blue Litmus Paper turned tored and the Red Litmus paper remained as it is which again shows the acidity of our salivasolution. When we used the congo red indicator, a red solution was formed which thenindicates that the pH of our saliva solution is above 5.2 which is almost neural. We thenconcluded and estimated that the average pH of our resting saliva ranges from acidic toneutral. The experiment that we conducted supported the fact that the optimal pH of ourresting saliva is between 6.4 to 6.9. A reading lower than 6.4 is indicative of insufficientalkaline reserves. After eating, the saliva pH should rise to 7.8 or higher. Unless this occurs,the body has alkaline mineral deficiencies ( mainly Calcium and Magnesium ) and will notassimilate food very well. To deviate from ideal salivary pH for an extended time invitesillness.B. Inorganic Matter In this experiment, we tested for the presence of chlorides, phosphates, sulfatesand calcium in our saliva solution. To test for presence of chloride we added 1 drop of silvernitrate solution and white precipitate was formed which indicates a positive result forpresence of chloride. In the second test tube we tested for phosphate by adding 1 drop ofammonium molybdate, the solution turned into a pale yellow which again indicates apositive result for presence of phosphate. The third test tube was used to test for sulfateby adding 1 drop of barium chloride, a yellow solution with white precipitate was formed.The white precipitate in there indicates that sulfate is present in our saliva solution. In thelast test tube we tested for calcium by adding 1 drop of potassium oxalate, a cloudysolution with white precipitate was formed,which indicates positive presence of calcium.We therefore conclude that our saliva has inorganic components like chlorides, phosphates,sulfates and calcium.C. Influence Of Temperature On Ptyalin Activity In our experiment, we gradually gained negative results on both iodine and benedict’s tests.Since the test mainly focused of the temperature and knowing that saliva contains enzymes, theoptimum level of temperature was surpassed that’s why the test didn’t gave the expected outcome.The saliva might have not contained starch after all since the test didn’t have any positive result.D. Influence Of Dilution Of Ptyalin Activity In our experiment, it can be noted that the solution yielded a negative result in bothbenedict’s and iodine test. In the iodine test the solution that give the darkest colorreaction was test tube 6, I think this is somehow confusing because it is the most dilute andas expected it should be the lightest. This factor is maybe because the heating of thesolution, the test tubes was not evenly exposed to heat.
  • 28. Conclusion The digestive functions of saliva include moistening food and helping to create a food bolus.This lubricative function of saliva allows the food bolus to be passed easily from the mouth into theesophagus. Saliva contains the enzyme amylase (also called ptyalin), and is thus capable of breakingdown starch into simpler sugars that can be later absorbed or further broken down in the smallintestine. Salivary glands also secrete salivary lipase (a more potent form of lipase) to begin fatdigestion. Salivary lipase plays a large role in fat digestion in newborn infants as their pancreaticlipase still needs some time to develop. It also has a protective function, helping to preventbacterial build-up on the teeth and washing away adhered food particles.V. DOCUMENTATION Reaction of saliva: from left to right: phenolphthalein, congo red, litmus paper Inorganic Matter: from left to right: sulfate, calcium, phospahtes, chloridesInfluence Of Dilution of Ptyalin Activity
  • 29. Influence Of Temperature in Ptyalin Activity Iodine Test Benedicts Test*From left to right: solution 1,2,3,4
  • 30. ACTIVITY 12 URINEI. OBJECTIVES To discuss the principle behind every test for urine. To discuss the results and its implications. To name the different diseases associated with every test.II. SCHEMATIC DIAGRAMCreatinine (Nitroprusside test)G. Pathological constituents: Glucose Benedict’s test Albumin (Heller’s ring test)Bile pigments (Gmelin’s test)Bile acids (Pettenkoffer’s test)
  • 31. Acetone test (Nitroprusside test)III. RESULTS AND OBSERVATION TEST THEORETICAL RESULT EXPERIMENTAL RESULT RESULT OBSERVATION Gen. Color: pale yellow to deep + Color: golden yellow characteristics amber Smell: ammoniacal Odor: slightly nutty Clarity: clear Clarity: clear Reaction: acidic Reaction: acidic Creatinine Red to yellow + Light yellow, cloudy (nitroprusside test) Glucose Green + Green w/ yellow precipitate Benedict’s test Albumin White ring at the junction of + Upper layer: cloudy yellow (Heller’s ring the two liquids WHITE RING IS PRESENT test) Lower layer: clear liquid Bile pigments Green or violet - Upper layer: yellow (Gmelin’s test) Lower layer: pale yellow Bile acids Red - Upper layer: brown (Pettenkofer’s test) Lower layer: dark brown, almost black Acetone test Orange - Clear yellow (nitroprusside)
  • 32. IV. ANALYSIS AND CONCLUSIONGUIDE QUESTION 1. How did you identify the pathologic constituents of urine? I identified the pathologic constituents of urine by performing different tests such as Benedict’s test for the presence of glucose, Heller’s ring test for albumin, Gmelin’s test for bile pigments, Pettenkofer’s test for bile acids, and acetone or nitroprusside test. ANALYSIS A. General characteristics Urine’s general characteristics can tell a lot about a person’s health. Many things affect urine colour, including fluid balance, diet, medicines, and diseases. How dark or light the colour is tells you how much water is in it. Urine is normally clear. Bacteria, blood, sperm, crystals, or mucus can make urine look cloudy. Urine does not smell very strong, but has a slightly "nutty" odour. Some diseases cause a change in the odour of urine. For example, an infection with E. coli bacteria can cause a bad odour, while diabetes or starvation can cause a sweet, fruity odour. As for reaction, urine is acidic. Sometimes, the pH of urine is affected by certain treatments. F. Creatinine Creatinine is normal in urine. It is a wasteproduct that is excreted through urination. Creatinine reflects the amount and activity of muscles. H. 1. Glucose Benedict’s test Glucose is present in our sample in very small amounts. Normally, there is very little or no glucose in urine. When blood sugar is very high, as in diabetes, the sugar spills over into the urine. Glucose can also be found in urine when the kidneys are damaged or diseased. 2. Albumin (Heller’s ring test) Protein is not found in urine. Fever, hard exercise, pregnancy, and some diseases (especially kidney diseases) may cause protein to be in urine. 3. Bile pigments (Gmelin’s test) Bile pigments in urine can be associated with jaundice, liver diseases, and hepatitis. In our specimen, bile pigments are not present. 4. Bile acids (Pettenkofer’s test) Bile acids in urine indicate liver diseases. Bile is synthesized in the liver. If bile acids are present in urine, the liver may be functioning improperly and some of the bile leaks into the urine. 5. Acetone (nitroprusside test) Acetone, a ketone, is not present in urine. When acetone is present, it indicates ketonuria, a medical condition in which ketone bodies are present in urine. This happens when the body produces excess ketones as an alternative source of energy
  • 33. because of the shortage of glucose. It is seen during starvation or in type 1 diabetes mellitus. Conclusion Therefore, I conclude that urine is a very essential wasteproduct for it provides vital information about a person’s internal functioning. Also, a good knowledge about the different tests’ procedures and principles would help in determining and understanding the condition of the individual.V. DOCUMENTATIONUrine (untested Creatinine (nitroprusside test)Glucose Benedict’s testAlbumin (Heller’s ring test)
  • 34. Bile pigments (Gmelin’s test)Bile acids (Pettenkofer’s test) Acetone test (nitroprusside)
  • 35. ACTIVITY 13 BLOODI. OBJECTIVES To discuss the physical properties of blood and its reactions. To discuss the principle behind every test for blood. To identify the different constituents of blood.II. ANALYSIS AND CONCLUSIONA. Physical properties of blood Blood is red in colour. The pH of blood is between 7.35-7.45; hence, it is slightly basic. When tested with litmus paper, it turned from red to blue and from blue to blue.B. Determination of bleeding time Bleeding time is dependent upon the efficiency of tissue fluid in accelerating the coagulation process on capillary function and the number of platelets present and their ability to form a platelet plug.C. Hemolysis or laking Hemolysis, or the breaking down of the RBC’s membrane, causing the release of haemoglobin and other internal components into the surrounding fluid, may be due to pathological conditions (autoimmune haemolytic anaemia or transfusion reaction) or improper specimen collection, processing, or transport. It is visually detected by showing a pink to red tinge in the serum or plasma.D. Crenation It is a phenomenon that happens when cells are exposed to a hypertonic solution. Through osmosis, water flows from areas of low solute concentrations to areas of high solute concentrations to stabilize the two solutions. As water leaves the cell, it shrivels or becomes crenated.E. A. Benedict’s test Aqueous solution is mixed with Benedict’s reagent and heated. Carbohydrates which react with the reagent to reduce the blue copper (II) ions to form a brick red precipitate of copper (I) oxide are reducing sugars. B. Chloride A test for the presence of chloride, an electrolyte, that works with other electrolytes to keep the proper balance of body fluids and maintain the body’s acid-base balance. C. Phosphate It is performed to see how much phosphorus is in the blood. The amount of phosphate affects the level of calcium. As blood calcium levels rise, phosphate levels fall. But this relation may be disrupted by some diseases or infections. D. Iron It plays a principal role in the synthesis of RBCs. It is necessary for the proliferation of RBCs and haemoglobin function.
  • 36. F. Fibrin test This test measures the amount of functional fibrin present in the plasma. Fibrin is involved in blood coagulation and platelet activation.G. Solubility test Fibrin is insoluble in water because if it is not so, it will not perform its function as an anticoagulant. Remember, blood is part water.H. Protein colour test Fibrin would be positive in all of the tests (Millon’s, Hopkins-Cole, biuret). These tests are tests for proteins, and fibrin is naturally a protein.