MCB 450 1st Edition Lecture 6 Outline of Last Lecture I Tissue Cell Fractionation a Centrifugation II Protein Purification a Chromatography b Electrophoresis III Antibodies IV Peptide fragmentation sequencing a Edman Degradation V Analysis of Protein Structure by X ray Crystallography Outline of Current Lecture I Monosaccharides II Glycosidic bonds III Polysaccharides Current Lecture Why do we need to use X ray diffraction Any resolution that we get is comparable to wavelength The problem we run into when viewing a protein is the wavelength of visible light is 4 000 7 000 To look at them this way we need a protein crystal Growing protein crystals We consider crystals large when they are about 0 5mm 3 4 of the crystal volume is actually filled with water molecules o This environment is very similar to the environment it will be in the cell If a protein is well ordered then they will form parallel planes The reflection mirror like will show density etc and information we are wanting to find out about the protein Analogy between X ray diffraction and light microscopy Similar to light microscopy Generating an electron density map From the diffraction pattern can create a structural model X ray structures determined to a certain A resolution As Angstroms decrease resolution increases o 5 0A is a lower resolution than 1 5A More structures are visible at 1 5A These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute Carbohydrates Carbs are most abundant biomolecules on the planet Play many roles Classes of saccharides Oligosaccharides 2 20 monosaccharides joined together by glycosidic bonds Polysaccharides chains of 20 1000 monosaccharides Can get very different structures depending on how they are connected Monosaccharides 1 Simplest monosaccharides have 3 carbons o Carbons continue growing by adding H C OH groups Monosaccharides have asymmetric centers All monosaccharides except dihydroxylacetone have one or more asymmetric carbon chiral centers D and L configurations of monosaccharides In determining D and L configurations look at the chiral carbon that is furthest from the carbonyl D and L configurations exam questions Carbonyl carbon is labeled carbon 1 Aldopentose 8 possible o 4 D and 4 L Epimers Epimer 2 sugars that differ only in the configuration around one carbon Formation of 2 cyclic forms of glucose Haworth perspective Pyranose form Slide 6 22 popular exercises for homework and exams Interconversion between anomers Mutarotation when alpha and beta anomers introconvert D glucose linear form has the chemical properties of an aldehyde Conformations of pyranoses Subsituents on the ring have two orientations o Axial up or down o Equatorial left or right Chair form relatively rare unless a bulky structure forces it into that form Definitions 1 2 Stereoisomers vs enantiomers Know definitions Monosaccharide derivatives In biology there are sugar derivatives o Beta 2 deoxy D ribose o Fucose in seaweed o Glucosamine o Glucouronate o Gluconate o Sorbitol Know shorthand of derivatives Formation of glycosides Methanol is a simple example Glycogen heavily hydrated Change from alpha to beta and get cellulose Cellulose has very little water o Humans can t digest it