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Lecture

1A03 Lab 1 Theory and Procedures Spr2012.pdf

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Department
Biology
Course
BIOLOGY 1A03
Professor
kudjera
Semester
Winter

Description
Page 1 of 15 EQUIPMENT USED IN LAB # 1 Page 2 of 15 LAB 1 THEORY: BIOLOGICAL MOLECULES Objectives: Upon completion of this lab you should be able to: 1. Understand the structure and properties of amino acids, polypeptides and proteins. 2. Describe how one is able to use paper chromatography to separate and identify individual amino acids from a mixture. 3. Prepare a protein standard curve and ascertain the concentration of protein in an unknown solution. 4. Understand and identify dependent and independent variables. HOMEWORK AND PRE-LAB PREPARATION: ▯ Read through this entire lab, including the Theory. ▯ Read BioSkills 1 in the Freeman Biological Science (B-1, p. 505) ▯ Prepare any tables that might be required in your lab notebook. ▯ Complete the pre-lab quiz on AVENUE. CLASSES OF ORGANIC MOLECULES Organic compounds are based upon the element carbon (valence 4) which is usually found bonded to hydrogen, oxygen, nitrogen or more carbon. Compounds containing carbon and hydrogen are called hydrocarbons . These carbon to hydrogen bonds are covalent nonpolar. Adjacent carbon atoms may form single, double or triple bonds, making such compounds very stable. The arrangement of the four valence electrons in the outer orbital of the carbon atom allows a total of four single bonds to be formed. These bonds point to the corners of an imaginary tetrahedron. If the four bonds are to four different atoms or groups the carbon atom is asymmetrical – two different, non-superimposable, molecules can be made. These molecules are called enantiomers or stereoisomers. Enantiomers are mirror images of each other, like your left and right hands. Fig. 1-1: Optical isomers of alanine. These different forms are chemically identical but differ in how their aqueous solutions rotate plane polarized light. Asymmetric carbon atoms make molecules “optically active”. Page 3 of 15 Biological systems are able to differentiate the optically different forms of carbon molecules (stereoisomers or enantiomers) because their shapes are different and their functional groups are in different places. Biochemical processes in living organisms require enzyme catalysts which function on the basis of a very precise “fit” between enzyme and reactant molecules. The same is true for signal molecules and receptors in and on the surface of cells. This “fit” is disrupted by the wrong stereoisomer and it will not be recognized by the enzyme or receptor. There are FOUR main classes of organic molecules: 1. Carbohydrates are compounds composed of carbon hydrogen and oxygen, usually characterized as (CH 2) on H-C-OH. They constitute 3% of the organic matter in the body. The sugars making up carbohydrates form hydrogen bonds with water and are therefore soluble. 2. Lipids are compounds composed of carbon, hydrogen and oxygen but may contain such elements as phosphorus and nitrogen as well. Lipids account for about 40% of the organic matter in the body. Lipid molecules, because their atoms are linked by nonpolar covalent bonds are insoluble in water, but soluble in nonpolar solvents. 3. Nucleic Acids are long polymers of building-block units called nucleotides. Each nucleotide is composed of a 5- carbon sugar with a phosphate group and a nitrogenous base covalently bonded to it. Nucleic acids are the largest of the organic compounds. They make up all of the genetic material of cells which is stored, replicated, and passed on from parent to offspring and from cell to cell during growth. There are two main types of nucleic acids, deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), differing from each other in the structure of the 5-carbon sugar component. 4. Proteinsare more complex than carbohydrates, nucleic acids or lipids. They account for about 50% of the organic material in the human body being components of most of the body's structures, and are therefore involved in almost all chemical interactions. Most proteins are very large molecules formed from linkages of many small subunits called amino acids. Proteins are polymers of the 20 amino acids, and contain carbon, hydrogen, oxygen, nitrogen and sulfur. Amino acids are characterized by the following formula, and polymerize by condensation reactions as shown: Page 4 of 15 Fig 1-2: Formation of peptide bonds. The ▯ carbon atom has both a carboxyl group and an amino group attached to it, in addition to the side chain R which gives it its unique composition. The ▯ carbon atom is asymmetric since it has four single bonds to four different groups. This means that amino acids exist in D and L forms (see Fig 1-1). Sugars also exist as both D and L forms but in contrast to the sugars, almost all naturally occurring amino acids found in proteins belong to the L series. D-amino acids do exist in nature, but only in certain rare instances, for example, in some types of bacterial cell walls and certain types of antibiotics. Levels of organization in polypeptides and proteins: o Primary structure (1 ) of a protein is the sequence of amino acids in the polypeptide chain formed by the condensation reaction between the COOH group of one amino acid and the NH group of a2other to form peptide bonds during protein synthesis. o Secondary structure (2 ) of a protein is the ▯-helical and ▯-pleated sheet configuration which results from the formation of hydrogen bonds between the carboxyl oxygen of one amino acid and amino group hydrogen of another amino acid brought into close proximity due to bending of the polypeptide chain. o Tertiary structure (3 ) of a protein is the 3-dimensional shape formed as a result of interaction between amino acid R-groups or between R-groups and the peptide backbone. For example the R-group on the amino acid cysteine contains a sulfhydral group (SH) which can form a strong disulphide bond with that of another cysteine. Primary, secondary and tertiary structures are intrinsic properties of polypeptide chains. A functional protein often consists of several polypeptide chains of either the same type or subunit (e.g. Cro protein Page 5 of 15 from bacteriophage ▯) or of two or more different subunits (e.g. hemoglobin) this is the quaternary structure. o Quaternary structure (4 ) of a protein is the structure that forms the functional unit of the protein, be it a structural protein or an enzyme and involves two or more polypeptide chains being held together by bonds or interactions between R-groups or sections of polypeptide backbone. The potential number of proteins formed from the chains of amino acids and side groups is nearly infinite, and proteins, when they function as enzymes, make all the reactions of the living cell possible. Additional information can be found in Chapter 3 of Biological Science by Scott Freeman. Some Biologically Important Functional Groups: GROUP NAME BIOLOGICAL SIGNIFICANCE OH Hydroxyl Polar, thus water-soluble Forms hydrogen bonds O C Carboxyl Weak acid (hydrogen donor) OH When it loses 1 hydrogen ion (H+), it becomes negatively charged H N Amino Weak base (hydrogen acceptor) H When it accepts 1 hydrogen ion (H+), it becomes positively charged O C Aldehyde Polar, thus water-soluble H Characterizes some sugars O Ketone Polar, thus water-soluble C Characterizes other sugars H C H Methyl Hydrophobic (insoluble in water) H O Phosphate Acid (hydrogen donor) P OH OH In solution, usually negatively charged, losing 2 H+ ions Page 6 of 15 The Spectrophotometer When a person looks at a blue book it appears that colour because only light from the blue part of the spectrum (~500 nm) is being reflected from its surface; all other parts of the spectrum have been absorbed by the surface of the book. Similarly, to look at a glass of water, almost all light is passing through i.e. being transmitted, so the water appears clear. A bottle of black ink absorbs almost all light so none is transmitted. The spectrophotometer is an instrument which measures quantitatively how much light is absorbed by a substance and how much light is being transmitted at any given time. Readings are always taken by comparison with a standard or "blank" which, depending on the situation, may be a tube containing all the ingredients of the test solution except the material being tested or varied or it may be water as in this case. From inside the machine, light shines through a filter which may be adjusted to control the colour of the ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ sample, the light strikes a light-sensitive phototube, producing an electric current proportional to the light energy received. Readings may be taken either on the absorbance or the transmission scale. Absorbance is proportional to the intensity of the colour measured and therefore to the concentration of the coloured molecule in solution. This is known as Beer's Law . Page 7 of 15 LAB 1 PROCEDURES: BIOLOGICAL MOLECULES PART A: PAPER CHROMATOGRAPHY Overview The mixture of amino acids to be analyzed has been dissolved in propanol. Several drops are placed near the bottom edge of the filter paper. The filter paper is inserted into a jar containing a small amount of a solvent (the spots must be above the solvent level). The solvent, (the flowing phase), migrates up the filter paper, (the stationary phase) by capillary action, carrying the mixture with it. Those parts of the mixture having a higher affinity for the solvent travel freely up the paper with the solvent. Those parts of the mixture having a higher affinity for the filter paper will quickly transfer from the solvent to the paper. At the end of a measured time interval, the amino acids of the original mixture will be at different places along the paper. Thus, each will have its own characteristic rate of migration; relative to the rate at which the solvent moves (which is always the same for any given substance in a particular solvent system). We call this ratif R - ratio of the fronts. R f= distance travelled by solute (cm) distance travelled by solvent (cm) N.B.: THE FILTER PAPER MUST BE HANDLED WITH FORCEPS TO PREVENT CONTAMINATION. Procedure - Work in pairs at the back bench. 1. Using forceps, take two pieces of chromatography paper. When the exercise is completed, you may call them chromatograms. 2. With a pencil, draw a line one centimeter from the bottom and put your initials in each upper corner. Use your dissecting needle to make a hole in the top of the paper half a centimeter down from the top edge, exactly in the middle. 3. Also with pencil, mark 3 spots along each line which should be labelled as illustrated in figure 1-1. Alanine (ala), histidine (his), valine (val) and leucine (leu) are amino acids whose identity is already known. Hence they are being used as "standards". Your TA will tell you which vial to use for your Unknown #1 and #2 samples. The unknowns may be single amino acids or mixtures which must be identified by comparing their activity with the standards. 4. Using a separate tip for each, deposit 2µl of each solution on the paper at the appropriate spot using the pipettors on the demonstrator’s bench. Be sure you have the pipettor set correctly; you want 2 µl, not 20 µl and not 0.2 µl. 5. Hang the chromatography papers from the hooks on the underside of the rubber stopper. Dry fit the assembled stopper and papers in the jar to be sure that the papers will hang freely. If the papers are too long or too wide, trim them to fit properly. They should not touch the sides of the jar or each other. When you’re satisfied with the fit, remove the stopper and chromatography papers. 5. Pour approximately 15 ml of solvent into the glass jar. Then replace the stopper. Be careful to ensure that the bottom edges of paper are only 2-3 mm below the solvent surface and the spots are not submerged. Page 8 of 15 Fig. 1-1: Labeling your chromatogram 6. Watch the solvent move up the paper by capillary action carrying the amino acids along with it. 7. When the solvent gets to within a few mm of the top of the paper (~20 min) remove the chromatograms from the jar. With pencil, mark the position of the solvent front and hang the chromatograms in the fumehood to dry. Note: Don’t stop the chromatogram too soon or you may not see good separation, but on the other hand, your chromatograph won’t have any usable information if the solvent runs off the top edge! 8. Spray the chromatograms in the fume hood with ninhydrin solution evenly over the front surface. Allow a few minutes for them to dry. o 9. Place the chromatograms in the 85 C oven for 5 min for colour development to take place. 10. With a pencil, draw a line across the top of the chromatogram using the mark indicating the location of the solvent front. Also, carefully outline ea
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