Biologically Important Molecules: Carbohydrates, Proteins, Lipids and Nucleic Acids - Lab Report Example

Principles of Biology GSB-101 CM Biologically Important Molecules: Carbohydrates, Proteins, Lipids and Nucleic Acids Exercise 6: Lab Assignment: Week 4 Date Abstract The study of biologically important molecules in the understanding of the body functioning and how metabolism takes place. In this practical we shall focus on carbohydrates, proteins, lipids and nucleic acids. Carbohydrates are divided into three; - monosaccharide, disaccharides and polysaccharides. Examples include glucose, maltose and starch respectively. Chains of monosaccharide yield disaccharides and polysaccharides. A monosaccharide will have a general formula (CnH2nOn) (Helen 8-29). Proteins are made up of long chains of amino acids. Amino acids have an amino group (-NH2), carboxyl group (-COOH) and a variable side chain –R which determines the type of amino acid. Lipids include a variety of molecules that dissolve in non polar solvents such as ether, acetone, methanol or ethanol. RNA and DNA are nucleic acids. They are made up of nucleotide subunits. DNA contains a deoxyribose sugar while RNA contains a ribose sugar. The difference in the sugars is the principle used in differentiating between the two. DNA is also double stranded whereas RNA is single stranded. DNA can be identified chemically using the Dische Diphenylamine test. The acidic conditions it is exposed to convert deoxyribose to a molecule that then binds with diphenyl amine to form a blue complex. In the following set ups, we want to use different methods that are used to detect the presence of various important bio-molecules. At the end of the session I should be able to detect the presence of carbohydrates, proteins, lipids and nucleic acids. I should be able to explain the importance of a positive and a negative control in biochemical tests. Lastly I should be able to identify an unknown compound using the methods learnt. Introduction Carbohydrates Carbohydrates are categorized as monosaccharide, disaccharide or polysaccharide. They contain three elements; - Carbon, Hydrogen and Oxygen in the ratio of 1:2:1. Carbohydrates are mainly broken down in the Krebs cycle to produce energy in the body in form of ATP. Testing for carbohydrates is important, in testing for diabetes abnormal levels of glucose in urine could mean that there is something wrong with the body. Starch is tested to confirm its presence in various food samples. Proteins Proteins are polymers of amino acids covalently linked through peptide bonds to form a long chain. Proteins are important in the body in many functions which include structural roles, catalysts, transportation and hormones (Thomas 4-23). Amino acids contain hydrogen, a carboxyl group, an amino group and an R group which determines the properties and type of amino acid. The biuret test is used to test for the presence of proteins. It is important in identifying presence of proteins in a sample. Nucleic Acids DNA and RNA are nucleic acids. Nucleic acids are responsible for the transfer of genetic information from the parents to the offspring. DNA is double stranded whereas RNA is single stranded (Watson 18-45). The Dische Diphenylamine test is used to test the presence of DNA. The acidic condition leads to converts the deoxyribose sugar to a molecule that binds with diphenylamine to form a blue complex. Lipids Lipids include a variety of molecules that dissolve in non-polar solvents eg. Ether, acetone, methanol or ethanol (Halldor 5-15). An example of a lipid is shown below;- The test for lipids are based on a lipids ability to selectively absorb pigments in fat soluble dyes e.g. Sudan III dye. Purpose of the Experiment Test Bio-molecule Tested Benedict’s Test Reducing Sugars (Monosaccharide) Iodine Starch (Polysaccharide) Sudan III Dye Lipids Biuret Test Proteins Dische Diphenylamine test DNA By performing the above tests it is possible to determine which bio-molecule is present in a sample. However a false positive or a false negative should be expected as there are other materials that also give similar results. The general hypothesis is created by considering the presence or absence of the tested bio-molecule. We can however not be 100 % confident of our results since our results have a small probability of showing a false positive or negative. If a protein is present in the biuret test, a violet color will be seen. If a translucent spot is seen in the Grease spot test, then there is a lipid present. If lipids were present when using Sudan III Dye, red stains will be seen. If starch is present in the Iodine test then a bluish black color will be seen. If DNA is present in the Dische Diphenylamine test a blue complex is formed. Materials and Methods The Biuret test The Biuret test aims at identifying the peptide bonds between amino acids. A violet color shows a positive feedback showing presence of amino acids. Biuret reagent is a 1% copper sulfate (CuSO4). The color intensity relates to the number of peptide bonds that react with Cu++. Procedure 1. Obtain five test tubes and number them 1-5. 2. Add 2 ml of egg albumin into one, 2 ml of honey into the next, 2 ml amino acid solution into the third, and 2 ml distilled water into the fourth and 2 ml protein solution into the fifth one. 3. Add 2 ml of 2.5% sodium hydroxide (NaOH) to each tube. 4. Add three drops of Biuret reagent to each tube and mix. 5. Record the color of the tubes’ contents. Results Tube Solution Color Conclusion 1. 2 ml egg albumin Purple Protein Present 2. 2 ml honey Yellow honey color No proteins present 3. 2 ml amino acid solution blue No peptide bonds broken 4. 2 ml distilled water blue No proteins present 5. 2 ml protein solution purple Proteins present Discussion The formation of purple color confirms the presence of peptides bonds. In or practical, 2 ml egg albumin and 2 ml protein solution formed the purple color which confirmed the presence of peptide bonds. There is no purple color when amino acid solution is used since there are no peptide bonds joining the amino acids. Amino acids are joined together using peptide bonds to form primary, secondary, tertiary and quaternary class of proteins. Honey and water do not contain amino acids and so the latter is used as a negative control to show absence of peptide bonds. They act as the alternative hypothesis. The null hypothesis is the presence of the peptide bonds. Conclusion The knowledge of the important bio-molecules is important and the knowledge of the tests used to decipher their presence ids important in Biology. The procedures discussed above have helped to understand in detail the procedures used and the results expected in each of them. The use of positive and negative controls is also important to generate hypotheses that aide in coming to conclusions. The knowledge of the presence of the bio-molecules is followed by inner details of their structure and their interactions which are used to ensure the organism lives in sound health and coordination. Works Cited Osborn, M, I, Helen. Carbohydrates. Reading: Academic press, 2003. Print. Thorrmar, Halldor. Lipids and Essential Oils as Antimicrobial Agents. Chicester: John Wiley and Sons, 2011. Print. Creighton, E, Thomas. Proteins: Structures and Molecular Properties. Stanford: W. H. Freeman, 1993. Print. Watson, D. James. The Double Helix: A Personal Account of the Discovery of the Structure of DNA. Lexile: Touchstone, 2011. Reprint. Read More