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Enzymatic Action in Digestion Procedural Inquiries Preparation 1. What is hydrolysis? 2. Which major molecules must be hydrolyzed in digestion? 3. How do we simulate body conditions in setting up digestive experiments? Experimentation 4. How do we test for starch? 5. What is the principle behind the starch test? 6. How do we test for reducing sugars? 7. What is the principle behind the reducing-sugar test? 8. How do we test for proteins? 9. What is the principle behind the protein test? 10. How do we test for lipids? 11. What is the principle behind the lipid test? 12. 13. 14. Additional Inquiries What is digestion? What is the role of enzymes in the digestive process? What are bile salts? Key Terms Amylase Benedict’s Test Bile Biuret’s Test Carbohydrate Hydrolysis Lipid Lugol’s Iodine Micelles Pancreatin Protein Reducing Sugar Starch Materials Needed Water bath - 37°C Water bath - 100°C Test Tubes Test Tube Rack Test Tube Clamp Pipets Spot Plate Solutions 1% Starch 2% Maltose 2% Glucose Water (pH8) Iodine Benedict’s Solution Albumin 5% Pancreatin (pH8) Biuret’s Solution Cream (or Half and Half) Bile Salts Litmus Solution The foods we eat are composed primarily of carbohydrates, proteins, and lipids. In foods, however, these building blocks are often not in their most usable form and therefore must be digested before they can be transformed into whatever substance the body may need. Digestion is the process of breaking foods down into that usable form. The digestive process is a series of enzyme-catalyzed reactions. Recall that an enzyme is an organic catalyst, a molecule at least part protein whose function is to change the rate of a chemical reaction. Enzymes do not do anything that does not happen naturally. For instance, if you had had a steak for dinner last night, you really would not have needed those digestive enzymes for the digestive processes to have taken place. The only problem is that you would have starved to death waiting for digestion! Without digestive enzymes, what normally occurs in a few hours (digesting the steak) would take about 50 years to happen. Time is the justifying factor for enzyme catalysis. Enzymes, by today’s definition, increase the rate of biochemical reactions without being changed themselves. In this laboratory exercise we will demonstrate the principles of enzymatic action. Preparation I. Introduction A. Hydrolysis Hydrolysis, the basic digestive breakdown mechanism, involves the splitting of a biological molecule by the addition of part of a water molecule. Study Figure 59-1 so that you understand exactly what is happening to the molecules being hydrolyzed. If necessary, go back to your lecture text or to your chemistry book to review the molecular intricacies of hydrolysis. Dehydration synthesis + + H2O (a) During dehydration synthesis two molecules are joined by the removal of a water molecule. Dehydration synthesis + Dehydration synthesis + Hydrolysis H2O (b) Hydrolysis reverses the steps of dehydration synthesis: a complex molecule is broken down by the addition of a water molecule. Figure Figure 59-1b, 59-1a c, d (b) Carbohydrate Hydrolysis Starch Molecule (c) Lipid Hydrolysis (d) Dipeptide Hydrolysis Hydrolysis + H2O 2 Disaccharide (maltose) Two monosaccharides (glucose) (b) Hydrolysis + + H2O Monoglyceride Fatty acid (lauric acid) Glycerol (c) O C O H OH C C C H N NH2 H H C H H O C C N H H H O C C C H N H Hydrolysis O H + H O H H H H H Dipeptide Alanine Glycine Two Amino Acids (d) What are the major molecules that must undergo hydrolysis in the digestive processes? ________________________________________________________ _________________________________________________________________ In digestive hydrolysis, what happens to the water molecule? B. Overview of the Experiments Specific enzymes act on each of the major biochemical molecules so that those molecules can be broken down into usable components. We will be demonstrating this biochemical action. Because each enzyme has its own optimum temperature and pH, we will be performing these experiments at 37°C to simulate the conditions of the body, and at pH 8.0 to simulate the specific conditions in the small intestine where most digestion takes place. For each of the following tests, we will: Demonstrate or identify the substance. Simulate digestion under specific conditions. Demonstrate the substance is no longer present after digestion. Experimentation Be certain you have all your equipment ready for a particular experiment before beginning that part of the exercise. 3 II. Carbohydrates A. Background Starch, a basic storage molecule of plants, is a long chain of glucose molecules that occurs in two forms, amylase and amylopectin, the latter being more highly branched and actually more common in the everyday diet. (As a comparison, glycogen is the animal storage molecule. Glycogen is a highly branched glucose chain.) The same enzyme, amylase, is required to break down either of these starches. Amylase is found in the saliva as salivary amylase and in the small intestine as pancreatic amylase. Pancreatic amylase, which is produced in the pancreas, is virtually identical to salivary amylase. Refer back to Figure 59-1. Note that the principle breakdown products of amylase action are disaccharide maltose and certain other short-chain glucose molecules. B. Iodine Test for Starch Breakdown If you wish to test a substance for starch, add Lugol’s iodine (IKI). If starch is present, the geometric pattern of the starch molecules will interact with the iodine and give a purplish-black color. If no starch is present, no geometric interaction can occur and the substance in question will not become dark. 1 → Obtain a beaker of starch solution. (Technically you could use any starch or starch-like substance, including a hamburger bun or a piece of paper. However, for scientific validity, you will want to use the same starch supply for all your starch experiments. Be careful of your clothes because this experiment even works accidentally on cotton shirts!) 2 → Also obtain: maltose solution, glucose solution, water (pH 8), Lugol’s iodine, and a spot plate. (A clean microscope slide or chemistry watch glass can be substituted for a spot plate.) 3 → On the spot plate put separate drops of starch, maltose, glucose, and water. To each spot add a drop of Lugol’s iodine. What happened? Starch ___Black_________________ Maltose __No Change____________ Glucose ___No Change___________ Water ____No Change___________ Neither water, maltose, nor glucose has the geometric configuration to affect a color change. Only the starch drop should have changed color. 4 C. Benedict’s Test Benedict’s solution tests for the presence of reducing sugars. Glucose, maltose, and fructose are reducing sugars. Sucrose is not a reducing sugar. Neither is starch. Go back to the molecular structures of these carbohydrates to see if you can figure out why certain molecules are reducing sugars while other molecules are not. ________________________________________________________________________ ________________________________________________________________________ Reducing sugars react with Benedict’s solution to form the insoluble red precipitat, cuprous (Copper I) oxide. If only small amounts of a reducing sugar are present, the solution will be green. With more and more reducing sugar, the solution will be yellow, orange, or red. 1 → Obtain four test tubes, labeled #1, #2, #3, #4. 2 → Also obtain: Glucose solution, maltose solution, starch solution, water (pH 8), and Benedict’s solution. 3 → Into test tube #1: Test tube #2: Test tube #3: Test tube #4: 10 drops glucose, 10 drops Benedict’s solution. 10 drops maltose, 10 drops Benedict’s solution. 10 drops starch, 10 drops Benedict’s solution. 10 drops water, 10 drops Benedict’s solution. 4 → Place the tubes in the boiling water bath for 30 seconds. Remove and read your results. #1 __Orange (+)___________ #3 ___Clear__(-)_________ #2 __Yellow_(+)__________ #4 ___Clear___(-)________ Glucose and maltose should have given you a positive Benedict’s test. Starch and water should have yielded a negative test. Why? Glucose and Maltose are reducing sugars, Starch is not a Reducing sugar and water is hydrogen and oxygen only D. Experiment In this experiment we are demonstrating the hydrolysis of starch. As starch is hydrolyzed, more and more maltose and other short-chain glucose molecules should be present. 1→ Obtain two test tubes, labeled #1 and #2. 2→ Also obtain: starch solution, water (pH8), iodine solution, Benedict’s solution, and salivary amylase. 3→ Into test tube #1: 10 drops starch, 10 drops water Test tube #2: 10 drops starch, 10 drops saliva (salivary amylase!) 4→ Place these tubes in the 37° water bath for 30 minutes. Remove from the water bath. 5 5→ Place 1 drop from each tube onto your spot plate. Record your results. #1 Black indicating starch present #2 Yellow indicating starch is being digested Add iodine. 6→ Place 9 drops of Benedict’s solution in each tube and place the tubes in the boiling water for 30 seconds. Remove and read your results. #1 No reaction indicating no reducing sugars present #2 Orange indicating amylase digested the starch leaving reducing sugars E. Explanation If starch was hydrolyzed, you should get no color change upon adding iodine because no starch should be present. If starch was hydrolyzed, you should get a positive Benedict’s test because maltose is a breakdown product of starch and maltose is a reducing sugar. Glucose and other short-chain glucose molecules that may be present in your hydrolyzed starch solution are also reducing sugars. Do your results match the predictions? If they do not, can you figure out what happened? ________________________________________________________________________ ________________________________________________________________________ (Some people do not produce salivary amylase. If you are such a person, you will still have starch present and thus will have a positive iodine test and a negative Benedict’s test. The inability to produce salivary amylase is not usually a problem because most amylase action comes from the pancreatic amylase. Before assuming you did not produce salivary amylase, check to make certain you did not mix up your test tubes.) III. Proteins A. Background Proteins are long chains of amino acids held together by peptide bonds. Pepsin, secreted by the stomach, usually only begins the process of protein digestion by splitting the long amino acid chains into shorter amino acid chains. Refer back to Figure 59-1. Most protein digestion takes place in the small intestine under the influence of the pancreatic enzymes – trypsin, chymotrypsin, and carboxypolypeptidase. Trypsin and chymotrypsin split the amino acid chains into short polypeptides. Carboxypolypeptidase cleaves individual amino acids from the polypeptide. (Several additional enzymes can also be found along the brush border of the epithelium.) 6 B. Biuret’s Test Biuret’s solution is a mixture of sodium hydroxide (NaOH) and copper sulfate (CuSO4). In the presence of protein, the copper sulfate reacts with the peptide bonds of the protein, causing the mixture to turn a deep violet color. If protein was never present, or if protein has been completely digested, no color change will occur. 1. → Obtain two test tubes, labeled #1 and #2. 2. → Also obtain: protein solution (albumin), water (pH 8.0), pancreatin (a mixture of pancreatic enzymes), and Biuret’s solution. 3. → Into test tube #1: 10 drops of protein solution Into test tube #2: 10 drops water (pH 8.0) 4. → Add Biuret’s solution to both tubes, drop by drop, until you observe a definite, deep color change. Did the protein solution in tube #1 turn violet? _Yes____ Tube #2 should have remained light blue. C. Experiment 1. → Obtain two fresh test tubes, labeled #1 and #2. 2. → Also obtain: protein (albumin) solution, water (pH 8.0), Biuret’s solution, and pancreatin. 3. → Into test tube #1: 10 drops protein solution, 10 drops water (pH 8.0). Into test tube #2: 10 drops protein solution, 10 drops pancreatin. 4. → Place these tubes in the 37°C water bath for 30 minutes. Remove from the water bath. 5. → Add about 5 drops of Biuret’s solution and gently shake the tubes. Record the color: #1 Dark Purple indicating protein is present #2 Light Purple indicating pancreatin digested the protein D. Explanation Test tube #1 should have changed color because no protein was digested: the color changed because protein was still present. Test tube #2 may have changed color. If the color change was pronounced, your proteins were not digested. Head back to the drawing board and try to figure out what happened! If you had no color change, your proteins were completely digestested. If you had some color change (say to a light pink or a slight purple), you had some protein digestion but you still have some protein present. 7 IV. Lipids A. Background Triglycerides are the primary lipids in the average diet. If necessary, review the chemistry of the triglycerides so that you have a picture of what we are ding in this experiment. Refer back to Figure 59-1 for lipid breakdown. Although an insignificant amount of lipid is digested in the mouth by lingual lipase and in the stomach by gastric lipase, virtually all fat digestion takes place in the small intestine. The digestive enzymes involved in fat breakdown are water-soluble. Unfortunately, fats themselves are not water-soluble. Therefore, if the fat traversed the intestine in “glob” form, very little of it would be digested before the “glob” reached the ileocecal valve. Enter bile, a secretory product of the liver. Bile does not contain any enzymes so it cannot function as an organic catalyst. Bile is an emulsifier. In other words, bile physically separates parts of the lipid into smaller and smaller pieces. Bile increases the surface area of the lipid, thus enhancing the possibility of lipase action. The polar ionic and carboxyl parts of the bile salt molecule are highly water-soluble, while the sterol portion of the molecule is highly lipid-soluble. The lipid-soluble portion of the bile dissolves the fat, while the polar portion extends into the water-based fluid. This lowers the interfacial tension of the lipid, making the molecules far less cohesive. In other words the lipid becomes more soluble in water. The agitation of the small intestine fragments the lipid, thus facilitating digestion. As stated, the lipases are water-soluble and can only catalyze reactions on the surface of the fat globule. Since bile greatly increases the surface area of the fat, fat digestion is accelerated. Bile salts also form aggregates of 20 to 40 molecules called micelles. In a micelle, all hydrophobic sterol units adhere at a central point while the hydrophilic tails extend out into the fluid. Triglycerides are split into free fatty acids and monoglycerides, which are quickly dissolved into the micelle aggregates. The micelles ferry these products to the brush border of the intestinal epithelium. B. Experiment 1. → Obtain three test tubes, labeled #1, #2, #3. 2. → Also obtain: cream, water (pH 8.0), litmus solution, and bile salts. 3. → Into test tube #1: 2ml cream, 2ml water (pH 8.0), pinch of bile salts, 15 drops litmus solution. Into test tube #2: 2ml cream, 2ml pancreatin(pH 8.0), pinch of bile salts, 15 drops litmus solution. 8 Into test tube #3: 2ml cream, 2ml water (pH 8.0), 15 drops litmus solution. If you are using litmus paper instead of a litmus solution, ignore the directions for adding litmus. When the experiment is complete, dip your litmus paper in the solution and read the color. 4. → If the above solutions are not purple, add more litmus. 5. → Place these tubes in the 37°C water bath for 30 minutes. Remove from the water bath. Record the color changes. #1 light purple color indicating the fat was emulsified #2 Light Purple to pink indicating fat was emulsified and digested by pancreatin #3 Light purple to pink, indicating pancreatin digested small amount of fat. C. Explanation Test tube #1 should have shown no change. Bile salts do not contain digestive enzymes. Although your bile salts emulsified the fats, the fats were not digested. Test tube #2 should have turned a nice shade of pink. The bile salts emulsified the cream, and the enzymes in the pancreatin hydrolyzed the fats. You created an acidic environment in the process. Test tube #3 should have turned a color somewhere between #1 and #2. You put the enzymes in but you did not add the bile salts. Thus, the digestive process was proceeding but at a much slower rate. D. Fatty Acids Fatty acids without carbon-carbon double bonds are classified as saturated, whereas those with at least one carbon-carbon double bond are classified as unsaturated. Saturated fatty acids are solids at room temperature. Grease, lard, and butter are examples of fats containing saturated fatty acids. Unsaturated fatty acids are less dense than saturated fatty acids, have lower melting point, and tend to be liquids at room temperature. Fats composed of unsaturated fatty acids are referred to as oils, such as corn oil. E. Fatty Acid Experiment 1. → Add 10 drops of vegetable oil to a test tube half filled with water. 2. → Shake the test tube and observe that the two liquids do not mix. The oil molecules are hydrophobic; they are insoluble in water. 3. → Add 5 drops of Sudan IV and shake the test tube. Observe in which of the two liquids the dye is soluble. What can you deduce from this observation? Sudan IV is lipid soluable 4. → Prepare another test tube with oil, water, and Sudan IV. To this test tube add several droppers full of diluted detergent water. Shake the test tube. Observe the change that has taken place in the two liquids. Detergent is an emulsifier. It surrounds the oil droplets and allows them to stay suspended in the water. 9 10