Testing for Biologically Important Molecules
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1 Testing for Biologically Important Molecules General Principles There are four major classes of organic compounds found in living organisms - arbohydrates, Lipids, Proteins and ucleic Acids. The chemical properties of the different classes depend on the presence of their specific functional groups. In general, the larger molecules in each class are formed by joining one or more building block molecules (monomers) together in a Dehydration Synthesis reaction to create a polymer. This is an energy-requiring process in which a hydroxyl of one unit is removed, and hydrogen is removed from the other creating a molecule of water. The two subunits are bonded covalently. It is also referred to as ondensation. ydrolysis is an energy releasing process which breaks the bond between the subunits and requires the addition of a water molecule. Large polymers are broken down into the smaller monomers by a reaction which is the reverse of dehydration synthesis. The larger molecule is split apart ( Lysed ) by water ( ydro ), with a hydroxyl of water going to one group, and hydrogen to the other. This is also referred to as a atabolic process Glucose (Aldehyde) Fructose (Ketone) Maltose, sucrose and lactose have the same molecular formula, owever, they are isomers since they are have different structures Dehydration Synthesis ydrolysis In this exercise you will learn about the structure, properties, and how to test for the presence of these organic molecules. I. arbohydrates arbohydrates are classified according to the number of sugar molecules they contain. Monosaccharides, such as glucose, fructose, ribose, and galactose, contain only one sugar monomer. Disaccharides, such as sucrose, maltose and lactose, contain two sugar monomers linked together. Polysaccharides, such as starch, glycogen, cellulose and chitin, contain many sugar monomers linked together. Monosaccharides have the formula ( 2 ) n, where n is generally an integer from 3 to 8. Monosaccharides contain many hydroxyl groups and either a ketone or an aldehyde functional group. These polar groups make sugar very soluble in water. Glucose ( ) and contains an aldehyde group. Fructose, has the same formula as glucose, but it is a ketone. This difference in structure (Isomers) gives the two monosaccharides slightly different chemical properties. Glucose + Glucose Maltose (malt sugar) Fructose + Glucose Sucrose (cane sugar) Galactose + Glucose Lactose (milk sugar) Glucose 1 2 Sucrose -(1,2) Linkage Fructose Dehydration Synthesis 2 Polysaccharides are formed by linking many monosaccharides together. They are used as energy storage by both plants and animals. Plants store their glucose as Starch while animals produce Glycogen, which is stored in liver and muscle cells. Polysaccharides are also important as structural compounds. Plant cell walls contain the rigid polymer ellulose. Fungi cell walls and the exoskeletons of arthropods contain a polysaccharide called hitin.
2 Tests for arbohydrates xidation/eduction (edox) eactions: xidation is generally referred to as the loss of electrons, while eduction is the gain of electrons. Benedict s reaction for educing Sugars is shown as: u 2+ + e u + opper (II) ions (u 2+ ) are reduced to copper (I) ions (u + ) by gaining electrons from the reducing sugars. The reducing sugar is oxidized as a result of giving up its electron. It is often difficult to see where electrons are flowing when you can t see an ion charge. owever, since the movement of hydrogen often follows the electrons, we could describe oxidation as the loss of hydrogen and reduction as the gain of hydrogen. The reduction of pyruvate to lactate at the end of glycolysis (fermentation) demonstrates this: 3 Pyruvic Acid + 2 = eduction - 2 = xidation 3 1) Benedict s Test for educing Sugars Lactic Acid Benedict s eagent contains copper (II) ions which cause the reagent to be blue colored. It can detect the presence of reducing sugars and is used to monitor glucose levels in urine. arbohydrates which can be oxidized by u 2+ and are classified as educing Sugars. When a solution containing Benedict s reagent and a reducing sugar are heated, the copper (II) ions in are reduced to copper (I) ions and the solution changes from the initial blue color to green, orange, red-orange, or brick-red (depending upon the quantity of reducing sugar). u 2+ (complex) Blue + () n 2 - u 2 + Brick-ed - () n 2 A brick-red precipitate of copper (I) oxide (u 2 ), may appear in the bottom of the tube. The higher the concentration of reducing sugar present, the more precipitate will form in the tube. Increasing amounts of reducing sugar educing Sugar onreducing Sugar emember that all monosaccharides are reducing sugars and that some disaccharides are reducing sugars, while others are not. If the aldehyde or ketone groups are tied to another compound (this is represented by the in the diagram above), they are not free to open, and cannot reduce the u 2+. Therefore, chains of glucose, such as starch, are not reducing sugars. Again, disaccharides may or not be reducing. It depends on what carbons are being linked together. Sucrose -(1,2) -Linkage 1 2 Acetal onreducing -Glucose 4 -Glucose 1 emi-acetal educing 4 Acetal 1 onreducing Maltose (Disaccharide) Polysaccharides do not test positive for reducing sugars unless they are broken down to form smaller units. -(1,4) Linkage Amylose (Starch) You will not be asked details of which sugars are reducing sugars and which are not. You should know that some are and some aren t and that polysaccarides are not. Green range ed Brown 2
3 2) Iodine Test for Starch Lugol's iodine reagent (I 2 KI) is useful to distinguish starch from other polysaccharides. The -(1,4) linkage between monomers in starch cause the helical structure of the polysaccharide chain. Saturated Fatty Acids: Saturated fatty acids (lard, bacon fat, butter, etc.) contain no double bonds, a maximum of hydrogen atoms, and pack very closely together. Unsaturated Fatty Acids: Double bonds between carbon atoms cause fatty acids chains to bend or kink. The fatty acids of unsaturated fats contain at least one double bond, fewer hydrogen atoms, and the fatty acid chains cannot pack as closely together because at least one of the chains has a kink or bend. Since it takes more energy to break tightly packed saturated fatty acids apart, they are solid at room temperature. The less "orderly" structure of unsaturated fats is responsible for their lower melting point (i.e. liquid at room temperature). Stearic Acid Iodine (I 2 ) can fit into the helices and will cause a dark bluepurple-black color. Monosaccharides and other polysaccharides cause no color change; the solution remains brownish yellow. II. Lipids Glycerol leic Acid arbon 9 Lipids are organic molecules that are insoluble in water and other polar solvents. They are, however, very soluble in nonpolar solvents, such as gasoline, chloroform, benzene, and ether. Dehydration Synthesis Lipids include fats and oils (important as energy storage compounds), phospholipids and glycolipids (part of the structure of cell membranes), waxes (protective surface coatings on many plants and animals), and steroids (found in some cell membranes and many hormones). 2 2 Ester Bonds 3) Sudan IV Test for Lipids Sudan IV (ed) is a lipid soluble dye. When Sudan red is added to a mixture of lipids and water, the dye will move into the lipid layer, coloring it red. When testing a solution for lipids there are two results you should be looking for; 1) Do you get a separation of layers (water & lipid)?, and 2) Does the dye migrate toward one of the layers? If the mixtures are all water soluble, then the Sudan IV will form small micelles and disperse throughout the solution. is A very simple test for lipids is based on their ability to produce translucent grease-marks on unglazed paper. This is a lipid test most of us has performed when eating French Fries out of their paper wrappings or when wiping our hands on paper napkins. Fatty Acids: Long carbon chains containing hydrogen with a carboxyl group ( ) on one end, which makes the molecule an acid. The carboxyl group is involved in bonding each fatty acid to the glycerol molecule through dehydration synthesis. When these fatty acids combine with the three hydroxyl groups on glycerol, they will either form a triglyceride (3 fatty acids) or a phospholipid (2 fatty acids and a phosphate). Making Margarine from Plant ils Many clinical studies have shown a correlation between the melting points of dietary fatty acids and hardening of the arteries. Saturated fatty acids appear to be the major problem; therefore unsaturated vegetable oils tend to be healthier. owever, most people don t want to Pour oil on their morning toast, so vegetable oils can be made solid by adding hydrogen in a process called ydrogenation. Unfortunately, hydrogenation creates its own set of problems. The catalytic process can also cause cis-trans isomerization, and unfortunately trans fatty acids appear to be as bad or possibly worse than saturated fats. 3
4 Several Biologically Important Fatty Acids III. Proteins The building blocks of proteins are the twenty different amino acids all of which similar structure. At the center of the molecule is the alpha carbon which is bonded to four different groups: an amino group ( 2 ), a carboxyl group ( ), a hydrogen atom, and the variable group (also called the side chain). The amino acids have identical structures except for their groups. 2 Amino Acid 1 Amino Acid 2 Dehydration Synthesis 2 Stearic, leic, Linoleic, & Linolenic Acids Stearic Acid MP: 70 Peptide Bond 2 Dipeptide Side hains Lauric Acid MP: 44 leic Acid MP: 13 Polypeptides are formed by joining amino acids together in a long, unbranched chain. The amino acids are linked together by peptide bonds formed when the carboxyl group of one amino acid reacts with the amino group of the next amino acid in a dehydration synthesis reaction. Linoleic Acid (mega-6) MP: -5 Linolenic Acid (mega-3) MP: -11 is Isomers TAS vs. IS Five Amino Acide esidues 4
5 4) Biuret Test for Proteins Biuret eagent is a light blue solution (u 2+ ) which turns purple when mixed with polypeptides containing at least four peptide bonds. The purple color is formed when copper (II) ions in the Biuret reagent react with the lone pair of electrons on the in the peptide bonds to form a complex. u 2+ forms a tetradentate coordination complex through the four nitrogen donor atoms. Biuret - u 2+ omplex u 2+ oordinated w/4 Peptide Amides u 2+ Experiments to Identify ompounds In this experiment we will be conducting tests to identify four types of compounds: Benedict s Test for educing Sugars Iodine Test for Starch Sudan IV Test for Lipids Biuret Test for proteins Each of the tests involves a control and an unknown solution. ontrols are a known solution. We use controls to confirm that our procedure is detecting what we expect it to detect. We compare the unknown s response with the control. There is one positive and one negative control for each of the following tests. Just because a substance reacts to the test does not mean it is a control we are simply testing it to obtain the result. A Positive ontrol contains the variable for which you are testing; it reacts positively and demonstrates the test's ability to detect what you expect. For example, if you are testing for protein in unknown solutions, then an appropriate positive control is a solution known to contain protein. A positive reaction shows that your test reacts correctly; it also shows you what a positive test looks like. u A egative ontrol does not contain the variable for which you are searching. It contains only the solvent (often distilled water with no solute) and does not react in the test. A negative control shows you what a negative result looks like. ontrols are important because they reveal the specificity of a particular test. For example, if water and a reducing sugar solution react similarly in a particular test, the test cannot distinguish water from the sugar. But if the sugar solution reacts differently from distilled water, the test can distinguish water from reducing sugars. In this instance, the distilled water is a negative control for the test, and a known reducing sugar solution is a positive control. 5
6 Procedures EVE Mix Pipettes from one solution to another!!! Benedict s Test for educing Sugars: 1. Set up 8 tubes; label each as shown in the table below. 2. Add 1 ml of each sample to be tested. Make sure you stir the solution before pipeting it into your tube. 3. Add 2 ml of Benedict s reagent to each tube. 4. Place the tubes in a beaker of boiling water for 3 minutes. 5. emove the tubes and allow them to cool. 6. Examine each tube and record the solution color in the table. 7. inse out the tubes; use the same labels for the iodine test. Iodine Test for Starch: 1. Set up 8 tubes as shown in the table below. (You will set up the same 8 tubes twice.) 2. Add 1 ml of each sample to be tested. Make sure you stir the solution before pipeting it into your tube. 3. Add 7-9 drops of iodine (IKI) to each tube. D T heat. 4. Examine each tube and record the color in the table. esults of Benedict s & Iodine/Starch Tests Tube ontents Water Glucose Sucrose educing Sugar Benedict s Test olor esult (+/-) Iodine/Starch Test olor esult (+/-) Sudan IV Test for Lipids: 1. Set up 6 tubes; label each as indicated in the table. 2. Add 3 ml of distilled water to each tube. 3. Add about 1 ml of each sample to be tested. 4. D T add Sudan IV to the first tube (Salad il #1) 5. Add 5 drops of Sudan IV to all tubes, EXEPT the first. 6. Mix each tube very well then let settle for 2 minutes. 7. Examine each tube and record the olors and Separation. esults of Sudan IV Tests Tube ontents Salad il #1 Salad il #2 Water Known Lipid oney Unknown Layer (Y/) Description Biuret Test for Proteins 1. Set up 6 tubes; label each as indicated in the table. 2. Add 2 ml of 2.5% sodium hydroxide (a) to each tube. 3. Add 2 ml of each sample to be tested. 4. Add 3-5 drops of Biuret reagent to each tube, mix well and let settle for 2 minutes. 5. Examine each tube and record the colors. esults of Biuret Tests Starch Potato Juice Tube ontents Water olor esult (+/-) nion Juice Amino Acids Unknown Egg Albumen oney Protein Solution Unknown 6
7 ame: Please hand in this and the last page. Due Date: 1) Look at the table for each experiment and list the positive control, the negative control and the positive results for each biochemical test. educing Sugars : + ontrol: ontrol: Positive esults: Starch: + ontrol: ontrol: Positive esults: Fats: + ontrol: ontrol: Positive esults: Proteins: + ontrol: ontrol: Positive esults: 2.Define in your own words Positive ontrol and egative ontrol. Explain why they are used in experiments. Positive ontrol: egative ontrol: 7
8 3) Give the expected results (i.e., state if the result will be positive or negative) for a Iodine test of (A) starch and (B) starch that has been hydrolyzed. ydrolyzed means that the polymer has undergone hydrolysis. A) Starch: B) ydrolyzed Starch: 4) Give the expected results (i.e., state if the result will be positive or negative) for a Biuret test for a protein (A) before and (B) after hydrolysis. A) Protein: B) ydrolyzed Protein: 5) You did two different tests for different forms of carbohydrates. For what two forms of carbohydrates did you test? 6) Based upon your tests, molecules from which of the four categories was in the unknown solution (there can be more than one). 7) Since the onion rings you are eating pass the translucent bag test you know they have lipids. (ot much of a surprise since they are deep-fat fried in oil.) You ask the chef for the recipe and are told that the batter is made of flour, eggs and water and of course the onions. If you ground up the onion rings, and performed all the tests correctly please state what color you would expect for each of these chemical tests: Benedict s (reducing sugars): Iodine (starch): Biruet (protein): 8
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