Activity 5A Characterization of Lipids I. OBJECTIVES
To enumerate the different physical properties of lipids To discuss the principle behind each test To name the components of lipids
II. SCHEMATIC DIAGRAM
III. RESULTS AND OBSERVATION Solubility Test Experimental Result Insoluble
Solvent Water
Expected Result Insoluble
Dilute HCL
Insoluble
Insoluble
Dilute NaOH
Insoluble
Insoluble
Ethyl alcohol
Insoluble
insoluble
Chloroform Ether
Soluble Soluble
Soluble soluble
Observation Formation of layer, big bubbles Formation of layer, smaller bubbles Formation of layer, oil are solidified Oil is found at the bottom of the tube dissolved dissolved
Oil used Coconut oil Linseed oil
Formation of Translucent spot Presence of Time Duration translucent spot + 16 minutes +
Peanut oil
1 hour and 1 minute 4 minutes
Cod liver oil
+ +
21 minutes
Olive oil
+
39 minutes
Corn oil
+
20 minutes
Red litmus Blue litmus
0 minute Red-red Blue-blue
Oil used Coconut oil
Reaction of fats After 3 hours Red-red Blue-blue
Acrolein Formation Expected result Formation of acrolein; acrid odor
result No reaction No reaction
Experimental result/observation Acrid odor (+)
IV. ANALYSIS AND CONCLUSION Guide 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 that contain them, are largely determined by the length and degree of unsaturation of the hydrocarbon chain. 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 statement indicates 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.
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 oil
Corn Oil
Linseed oil
Olive oil
3. Show chemical reaction of Acrolein test.
The principle behind the acrolein test is a specific chemical reaction. This reaction is utilized to determine the presence of glycerin in a fat. By heating the fat sample in the presence of potassium 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 a dehydrating 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 such as 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 nonpolarity. In the test you can see the formation layer and big bubbles of oil on the top surface indicates 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 the experiment it shows formation of layer but smaller bubbles compared to the water-oil mixture. This indicates that HCL somehow dissolved the oil into smaller particles and oil has a lower specific gravity 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. In the experiment it shows formation of layer and the oil was solidified. The solidification of the oil was 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 oil because 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 molecular weight and number of double bonds affect the rate of evaporation of oil. The longer the fatty acyl chain and the fewer the double bonds, the lower the evaporation rate. In our test the rate of evaporation 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.
Conclusion Lipids show many physical and chemical properties. These properties were clearly shown in the results 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 in polar such as water, alcohol, acid and base. Lipids, specifically oils shows formation of translucent spot, the rate of evaporation and formation of this translucent spot vary because of the difference of chemical structure. Lipids when exposed to open air or environment would undergo rancidity specifically oxidation because it is exposed to oxygen, wherein it will become acidic. Lipids show a positive reaction to Acrolein test since it is composed of glycerol which is the main reactant to form the product acrolein. V. DOCUMENTATION
Solubility of oil (from left: water, dilute HCL, dilute NaOH, ethyl alcohol, chloroform, ether)
Formation of Translucent Spot (from left: linseed oil, peanut oil, cod liver oil, olive oil, corn oil, coconut oil)
Reaction Of fats
Acrolein test
ACTIVITY 5B CHARACTERIZATION OF LIPIDS
I. OBJECTIVES
Discuss the saponification property of lipids To enumerate the properties of emulsifying fats & saponifying lard Explain the principle in each test employed for lipids
II. SCHEMATIC DIAGRAM A. 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)
Transfer solution in evaporating dish (with 2ml. H20) Acidify with (5 drops diluted HCl) Remove fatty acid
Neutralize sol’n with (10% NaCO3)
Evaporate to dryness
Extract residue with (1ml. ethyl alcohol)
Remove alcohol (by evaporation on water bath)
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
miscible
Saponification Results and Observation: The product was cream to brownish color substance with paste like consistency. After it is removed from heat it hardened into a powdery like residue.
IV. ANALYSIS AND CONCLUSION GUIDE 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 implications for cooking and for digestion. To emulsify fats, you need substances such as bile salts or other compounds that help connect the fat and water. Droplets of the dispersed component rapidly coalesce 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 and the salt of a carboxylic acid. Saponification is a process that produces soap, usually from fats and lye. 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, such traditional saponification processes produces glycerol. "Saponifiable substances" are those that can be converted into soap. In saponification process, we used alcoholic potash (KOH dissolved in ethanol & neutralized fatty acids in the lard) & NaCO3 to produce hard soap; then a metallic salt of fatty acid is formed. Fats (triglycerides) upon alkaline hydrolysis (either with KOH or NaOH ) yield glycerol and potassium or sodium salts of fatty acids (soap)
or
Analysis A. Emulsification In the experiment, only in test tube 3 that the two liquids, coconut oil and albumin solution became miscible. Since albumin contains an emulsifying agent lecithin. The remaining 2 test tubes didn’t mix due to absence of an emulsifier. B. Saponification In our experiment, the outcome is powdered due to high application of heat, water is then evaporated. Nevertheless, we had a positive result on saponifying lard because of alkaline hydrolysis 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 the aqueous solutions that exist in living things. But, with emulsification, little droplets of fat are formed 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
ACTIVITY 5C QUALITATIVE TEST FOR CHOLESTEROL
I.
OBJECTIVES
II.
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.
SCHEMATIC DIAGRAM Preparation:
Salkowski test:
Leibermann-Burchard test:
III. RESULTS AND OBSERVATION TEST
Salkowski test
THEORETICAL RESULT
Chloroform layer- red to blue
EXPERIMENTAL RESULT RESULT
OBSERVATION
-
Chloroform layeryellow Acid layer- orange
Acid layer- green LeibermannBurchard test
Blue-green complex
-
Upper layer- brown Lower layer- dull orange
IV. 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
is the principle behind Salkowski test. On the other hand, the acetic anhydride in the Leibermann-Burchard test can react with the C3 hydroxyl group of cholesterol and related steroid in the presence of a strong acid such as concentrated sulphuric acid that we used to form 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 LeibermannBurchard 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 future V. DOCUMENTATION
Salkowski Test
Leibermann-Burchard Test
ACTIVITY 6A Carbohydrates Qualitative Tests for Different Types of Carbohydrates
I. OBJECTIVES Become familiar with common carbohydrates Learn significant differences in the chemical properties of carbohydrates Identify the carbohydrates present in common food products
II. SCHEMATIC DIAGRAM A. 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
Add 5 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 liquids
B. 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)
C. Phenylhydrazine Test (Osazone Formation) In each of 8 test tube place 1ml. of 5% sol’n of (#1-glucose, #2maltose, #3-fructose, #4-galactose, #5-lactose, #6-sucrose, #7glycogen, #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 OBSERVATION
A. Molisch Test Solution
Molisch’s test
Anthrone Test
Phenylhydrazine Test
Expected result
Purple ring
Green
Yellow-pale orange
distilled water (control)
(-) Clear liquid
N/A
N/A
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
IV. ANALYSIS AND CONCLUSION GUIDE QUESTION
1. 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 compounds containing carbohydrates in a combined form, based on the dehydration of the carbohydrate by sulfuric acid to produce an aldehyde either (furfural or a derivative) which then condenses with the phenolic structure resulting in a purple-colored compound. Carbohydrates are dehydrated by concentrated sulfuric acid to form hydroxymethylfurfural will react with alpha-naphthol (Molisch reagent) to yield a purple condensation product.
B. Anthrone Test In the anthrone assay, carbohydrates are dehydrated by using conc. H2SO4 to form furfural, which in turn condenses with anthrone (10-keto-9,10 dihydroanthracene) to form a bluishgreen complex. Sugars react with the anthrone reagent under acidic conditions to yield a bluegreen color. Concentrated H2SO4 is a powerful dehydrating agent involved in dehydrating sugars leading 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 is formed. Phenylhydrazine (NH2-NH-C6H5) reacts with carbonyl compounds in neutral or slightly acidic conditions to give phenylhydrazone, which is highly soluble. When hydrazones react further with the phenylhydrazine molecule, the condensation products formed are insoluble osazones, which precipitate out as crystals. Reducing sugars form characteristic osazone crystals when heated with an excess phenylhydrazine. This property is attributed to the presence of aldehyde or ketone group in their molecules. -OH group immediately adjacent to the keto group is oxidized to a keto group and adds to phenylhydrazineto form the yellow to pale orange osazones.
Analysis In the Molisch Test, from the experiment all sugar tested solutions yield positive result since they are all carbohydrates. While the control tube which contains water give a negative result which is a clear solution. Since it’s not a carbohydrate. In the Anthrone Test, only glycogen & starch give positive results but, supposedly all sugar solutions must give positive results in the Anthrone Test since they are all carbohydrates but due to 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. But some sugar solutions, when they are viewed under the microscope they didn’t give expected results (crystals formation). This is maybe because by the time given, they are about to form crystals other sugar tested solutions takes longer than the time given time to form their crystals that’s why some didn’t yield a positive crystals formation (glucose & sucrose), while other yield positive results (galactose, fructose, lactose, maltose, glycogen & starch).
Conclusion In conclusion, for the Activity 6A Qualitative Tests for Different Carbohydrates: Using Molisch, Anthrone & Osazone qualitative test we became familiar with common principles of carbohydrates 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 the process of testing them would affect their reactions & results.
V. DOCUMENTATION
Molisch’s Test
Anthrone Test
Phenylhydrazine Test
Phenylhydrazine Test #1: Glycogen, # 2 Lactose, #3 Galactose, #4 Glucose, #5 Sucrose, #6 Starch, #7 Maltose, #8 Fructose
Activity 6B Qualitative Test For Carbohydrates I. OBJECTIVES Enumerate the qualitative tests for various types of carbohydrates Discuss the principles in each test Explain the characteristic of different type carbohydrate
II. SCHEMATIC DIAGRAM
IV. ANALYSIS AND CONCLUSIONS GUIDE QUESTION: 1. Give the principle for each test for carbohydrates. 1. Moore's Test - When a solution of reducing sugar is heated with an alkali (NaOH), it turns yellow to orange and finally dark brown, liberating the odor of caramel. This is due to the liberation of 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. Aldehydes are detected by the reduction of the deep blue solution of Copper (II) to a red precipitate of insoluble Copper oxide. The Fehling’s test is usually used to detect for the presence of reducing sugars but it is not specific for aldehydes. 3. Benedict's Test - Aqueous glucose is mixed with Benedict's reagent, a solution of copper sulfate, sodium hydroxide, and tartaric acid. The mixture is heated. Carbohydrates which react with Benedict's 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 having potentially free aldehyde or keto group reduce cupric hydroxide in alkaline solution to red coloured cuprous 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 reducing sugars (specifically glucose) in urine. But, basically, the Nylander’s test is still a test for reducing sugars. Nylander’s test depends upon the reduction of bismuth subnitrate in a weakly alkaline solution to metallic bismuth. The Nylander’s reagent is a solution composed of a solution of Rochelle salt (potassium sodium tartrate), potassium or sodium hydroxide, and bismuth subnitrate in water. The sodium hydroxide or potassium hydroxide in the Nylander’s reagent gives the alkaline environment in order for the reaction to occur. The potassium tartrate serves as the solvent for the bismuth subnitrate. 5. Barfoed's Test - Barfoed's reagent, a mixture of ethanoic (acetic) acid and copper(II) acetate, is combined with the test solution and boiled. A red copper(II) oxide precipitate is formed will indicates the presence of reducing sugar. The reaction will be negative in the presence of disaccharide sugars because they are weaker reducing agents. This test is specific for monosaccharide. Due to the weakly acidic nature of Barfoed's reagent, it is reduced only by monosaccharides. 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 used was the picric acid (yellow in color). In order for a reaction between the reducing sugar and Picric acid to take place, the solution must be alkaline; this is the purpose of adding the 10% Sodium Carbonate in the experiment. When Picric acid reacts with the aldehyde or keto group of the reducing 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. Seliwanoff's Test - It is a color reaction specific for ketoses. When concentrated HCl is added ketoses undergo dehydration to yield furfural derivatives more rapidly than aldoses. These derivatives form complexes with resorcinol to yield deep red color. The test reagent causes the dehydration of ketohexoses to form 5-hydroxymethylfurfural. 5-hydroxymethylfurfural reacts with resorcinol present in the test reagent to produce a deep red product . It is a timed colour reaction specific for ketoses.
8. Iodine Test - This test is used for the detection of starch in the solution. The blue black colour is due to the formation of starch-iodine complex. Starch contain polymer of α-amylose and amylopectin 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. When monosaccharides (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 the oxidation of galactose is insoluble and separates as crystals. This acid derivative is known as mucic acid. This test allows for the differentiation of galactose to other monosaccharides as other monosaccharides that undergo this test yield soluble dicarboxylic acid whereas galactose produces insoluble Mucic acid. Lactose is a dimer of glucose and galactose. The nitric acid will hydrolyze the dimer and so lactose will also give a positive result. Analysis 1. 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 by a clear solution. Moore’s test reaction is due to the presence of an aldehyde which in turn liberated to form caramel and it is also specific for reducing sugars. Glucose, galactose, maltose, fructose and lactose are reducing sugars while sucrose, glycogen and starch are not, therefore our experimental results are correct. Sucrose did not give a positive result in the test because it is not a reducing sugar. The non-reducing property of sucrose is due to it’s anomeric carbon which is not "free" since this carbon links fructose and glucose together. This anomeric carbon can't open up the ring structure and react with the reagent (which is Concentrated NaOH). Starch and glycogen did not react in the Moore’s test. This is because both of them are is a polysaccharide. All polysaccharides are non-reducing sugars therefore; its anomeric carbon is “full” and cannot react to another molecule. 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 by a 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 are not. 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 by a 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. The Fehling’s test needs a strong alkaline condition in order for a reaction to take place, unlike in the Benedict’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 are not. 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 was indicated by a clear solution. As mentioned previously, this test is specific for reducing sugars thus follow 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. Theoretically glucose, galactose, maltose, fructose and lactose will be positive since it is a reducing sugar while sucrose, glycogen and starch will be negative since it is a non-reducing sugar. The wrong result of the test is maybe due to the heating process done in the experiment. During the heating of the solution we put it in a water bath and place it in a beaker. Maybe that factor hinders the reaction of heat 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 red color when in fact all reducing sugars including glucose and lactose should have given a positive
result excluding sucrose, glycogen, and starch because they are non-reducing sugars. We might have an error doing the procedure during the experiment thus failing to have the actual results. 7. Seliwanoff's Test -In this experiment we gained a positive result for fructose and sucrose which are 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 a positive blue black color while the glycogen will not. Upon performing the procedure, we had gained a starch which is blue black in color while the glycogen being orange. In our documentation we had two clear solutions because it was the end product of the test. They had lost their color upon adding drops of sodium thiosulfate. 9. Mucic acid Test – Performing this experiment galactose, fructose, and lactose gave a positive insoluble mucic acid. Basing on the principle galactose and lactose must have a positive result and we gained those but having fructose being positive because it is water soluble thus unable to crystallize. Conclusion There are many qualitative tests for carbohydrates. This test is designed to identify the specific carbohydrates, whether it is a monosaccharide, disaccharide, polysaccharide, reducing or nonreducing sugar. Moore’s test, Fehling’s test, Benedict’s test, Nylander’s test and Picric acid test are tests for reducing sugars. Barfoeds test is also for reducing sugar but only specific for monosaccharaides. Seliwanoff’s test is specific for ketoses, Iodine test for starch and Mucic acid test 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.
Benedict’s Test From left to right: Glucose, Galactose, Fructose, Maltose, Lactose, Sucrose, Starch, Glycogen
Nylander’s test From left to right:Fructose, Maltose, Lactose, Glucose, Starch, Galactose, Glycogen.
Barfoed’s Test From left to right: Glucose, Galactose, Fructose, Maltose, Lactose, Sucrose, Starch, Glycogen
Picric Acid Test From left to right:Galactose, Maltose. Fructose, Lactose, Glucose, Starch
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 Microscope From left to right: Fructose, Lactose, Galactose.
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:
Congo Red
Litmus Paper
Phenolphthalein
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
Add 1 drop of
Add 1 drop of
Add 1 drop of
Silver Nitrate
Ammonium
Barium Chloride
solution
Molybdate
solution
Calcium
Add 1 drop of Potassium Oxalate
C. Influence Of Temperature On Ptyalin Activity Prepare 1 ml of 1% starch solution in 4 test tubes
Test Tube 4 Boiling Temperature
Test Tube 1
Test Tube 2
Test Tube 3
Ice Water Mixture
Room Temperature
38◦ C
Add 10 drops of
Add 10 drops
Add 10 drops of
Saliva and shake
Saliva and shake
Saliva and shake
well
well
well
Add 10 drops of Saliva and shake 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 each
Test Tube 1
Test Tube 2
Test Tube 3
Test Tube 4
Test Tube 5
Add 1 ml
Add 1 ml of Mixture From Test tube 1
Add 1ml of Mixture From Test tube 2
Add 1ml of Mixture From Test tube 3
Add 1 ml of Mixture From Test tube 4
Of saliva
Shake well
Place at water bath at temperature about 40◦c for 20 minutes Test with Iodine test and Benedict's reagent
Test Tube 6 Add 1ml Of mixture From Test tube 5
III. RESULTS AND OBSERVATION A. Indicators
Results and observations
Phenolphthalein
Colorless
Blue Litmus Paper
Blue to Red
Red Litmus Paper
Red to Red
Congo Red
Red
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)
B
C. Varying Temperatures
Benedict's 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)
Test Tube No.
Iodine Test
Benedict's 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 (Negative
D.
IV. ANALYSIS AND CONCLUSION GUIDE QUESTION 1. 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 it goes to the stomach, the next site of digestion. Our saliva contains an enzyme that starts breaking down long carbohydrate molecules as soon as we put food in our mouth. Salivary amylase is the enzyme 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.
Digestion is a chemical reaction where enzymes cleaves the complex molecules into simple sugar or fatty 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 our body maintains constant temperature but sometimes temperature changes on change in physical conditions or by malfunction of the body. By change in temperature either decrease or increase, the rate of reaction decreases. If saliva is put into a test tube with starch it will convert this into sugar. At low temperatures this process goes on slowly, the velocity increasing as the temperature increases until it reaches its maximum at about 37° C. Above this temperature the velocity again decreases, the enzyme being destroyed at about 70° C. Analysis A. 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 pH of our saliva (resting) . Upon using phenolphthalein as our indicator we obtained a colorless solution that indicates the acidity of our saliva solution. The Blue Litmus Paper turned to red and the Red Litmus paper remained as it is which again shows the acidity of our saliva solution. When we used the congo red indicator, a red solution was formed which then indicates that the pH of our saliva solution is above 5.2 which is almost neural. We then concluded and estimated that the average pH of our resting saliva ranges from acidic to neutral. The experiment that we conducted supported the fact that the optimal pH of our resting saliva is between 6.4 to 6.9. A reading lower than 6.4 is indicative of insufficient alkaline 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 not assimilate food very well. To deviate from ideal salivary pH for an extended time invites illness. B. Inorganic Matter
In this experiment, we tested for the presence of chlorides, phosphates, sulfates and calcium in our saliva solution. To test for presence of chloride we added 1 drop of silver nitrate solution and white precipitate was formed which indicates a positive result for presence of chloride. In the second test tube we tested for phosphate by adding 1 drop of ammonium molybdate, the solution turned into a pale yellow which again indicates a positive result for presence of phosphate. The third test tube was used to test for sulfate by 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 the last test tube we tested for calcium by adding 1 drop of potassium oxalate, a cloudy solution 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, the optimum 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 both benedict’s and iodine test. In the iodine test the solution that give the darkest color reaction was test tube 6, I think this is somehow confusing because it is the most dilute and as expected it should be the lightest. This factor is maybe because the heating of the solution, the test tubes was not evenly exposed to heat.
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 the esophagus. Saliva contains the enzyme amylase (also called ptyalin), and is thus capable of breaking down starch into simpler sugars that can be later absorbed or further broken down in the small intestine. Salivary glands also secrete salivary lipase (a more potent form of lipase) to begin fat digestion. Salivary lipase plays a large role in fat digestion in newborn infants as their pancreatic lipase still needs some time to develop. It also has a protective function, helping to prevent bacterial 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, chlorides
Influence Of Dilution of Ptyalin Activity
Influence Of Temperature in Ptyalin Activity Iodine Test
Benedict's Test
*From left to right: solution 1,2,3,4
ACTIVITY 12 URINE
I. 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 DIAGRAM Creatinine (Nitroprusside test)
G. Pathological constituents: Glucose Benedict’s test
Bile pigments (Gmelin’s test)
Albumin (Heller’s ring test)
Bile acids (Pettenkoffer’s test)
Acetone test (Nitroprusside test)
III. RESULTS AND OBSERVATION TEST
THEORETICAL RESULT
EXPERIMENTAL RESULT RESULT
Gen. characteristics
Color: pale yellow to deep + amber Odor: slightly nutty
OBSERVATION Color: golden yellow Smell: ammoniacal Clarity: clear
Clarity: clear
Reaction: acidic
Reaction: acidic Creatinine (nitroprusside test)
Red to yellow
+
Light yellow, cloudy
Glucose Benedict’s test
Green
+
Green w/ yellow precipitate
Albumin (Heller’s test)
White ring at the junction of + ring the two liquids
Upper layer: cloudy yellow WHITE RING IS PRESENT Lower layer: clear liquid
Bile pigments Green or violet (Gmelin’s test)
-
Bile acids Red (Pettenkofer’s test)
-
Acetone test Orange (nitroprusside)
-
Upper layer: yellow Lower layer: pale yellow Upper layer: brown Lower layer: dark brown, almost black Clear yellow
IV. ANALYSIS AND CONCLUSION GUIDE 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
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 waste product 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. DOCUMENTATION
Urine (untested
Glucose Benedict’s test
Creatinine (nitroprusside test)
Albumin (Heller’s ring test)
Bile pigments (Gmelin’s test)
Bile acids (Pettenkofer’s test)
Acetone test (nitroprusside)
ACTIVITY 13 BLOOD
I.
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 CONCLUSION A. 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.
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.
