BICH 410 1st Edition Exam 2 Study Guide I II Format A 50 multiple choice questions i Mostly conceptual factual information ii Learned from last test lots of detail iii Reading the textbook would probably be a good idea to get the little details down because 100 pages in the textbook cannot be placed in this review B Chapters 5 7 i Proteins Protein structure and folding and carbohydrates ii Memorize important aldoses and ketoses highlighted in blue in the powerpoints C Tips for taking the test i Do NOT overthink things She does not put trick questions on the exam the simplest answer is probably the best correct one ii Read the question carefully Note words like all except none that give you clues on whether you are looking for the answer that is correct or incorrect iii Work efficiently You only have 50 minutes so be fast but don t rush so much that you don t read the question through iv FACTS are important She likes the details Proteins A Protein structure i Primary 1 sequence Secondary 2 local structures H bonds Tertiary 3 overall 3 dimensional shape Quaternary 4 subunit organization ii Solubility is influenced by pH iii Know the purification steps in this figure iv Purification processes to know 1 2 3 4 Dialysis concentration Ion exchange chromatography charge Gel filtration chromatography size SDS PAGE molecular weight 5 Affinity chromatography B Sequencing i Sanger sequenced the two chains of insulin ii Established that all molecules of a given protein have the same sequence iii Proteins can be sequenced by direct amino acid sequencing or sequencing the corresponding DNA in the gene C Six step process for determining sequence memorize this i Separation of chains if any 1 Achieved with extreme pH urea guanidine HCl or high salt conc ii Cleavage of disulfide bridges 1 Sulfhydryl reducing agents used like betaMe or DTT iii Identify N and C terminal residues 1 N terminal use Edman reagent 2 C terminal is enzymatic Use carboxypeptidases Carboxypeptidase A cleaves any residue except Pro Arg and Lys B only works on Arg and Lys iv Fragmentation of the chains steps 4 and 5 1 Enzymatic Trypsin cleavage on the C side of Lys Arg Chymotrypsin C side of Phe Tyr Trp Clostripain like trypsin but attacks Arg much more than Lys Staphylococcal protease C side of Glu Asp in phosphate buffer pH 7 Specific for Glu in acetate pH 5 or bicarbonate buffer pH 10 2 Chemical fragmentation CNBr acts only on methionine residues CNBr is useful because proteins usually have only a few Met residues 3 KNOW the reaction of CNBr with a peptide v Reconstructing the Sequence 1 Use two or more fragmentation agents in separate fragmentation experiments Sequence all the peptides produced usually by Edman degradation Compare and align overlapping peptide sequences to learn the sequence of the original polypeptide chain vi Mass spec can also be used to sequence the chain and separate particles on a basis of mass to charge ratio D Know that leucine alanine and serine are the most abundant AAs Also know that histidine cysteine and tryptophan are the least common i Even though sequence and structure of two proteins can be very similar their function will most likely be different ii III Biological functions of proteins know them Protein Structure and Folding A Overarching principles i Function depends on structure ii Structure depends on sequence and on weak noncovalent forces iii The number of protein folding patterns is large but finite iv Structures of globular proteins are marginally stable v Marginal stability facilitates motion vi Motion enables function B Bonding i Hydrogen bonds are formed wherever possible Hydrophobic interactions drive protein folding Ionic interactions usually occur on the protein surface van der Waals interactions are everywhere electrostatic interactions also important C Amide Plane i Angle about the C N bond is denoted phi ii Angle about the C C bond is denoted psi iii 0 180 is unfavorable iv 180 0 is unfavorable v 0 0 is unfavorable vi G N Ramachandran was the first to demonstrate the convenience of plotting phi psi combinations from known protein structures vii The sterically favorable combinations are the basis for preferred secondary structures C Alpha Helix i First proposed by Linus Pauling and Robert Corey in 1951 Read the box about Pauling on page 143 ii Identified in keratin by Max Perutz Rosalind Franklin s first student iii A ubiquitous component of proteins iv Stabilized by H bonds v Residues per turn 3 6 vi Rise per residue 1 5 Angstroms 0 15 nm vii Rise per turn pitch 3 6 1 5 5 4 Angstroms viii Alpha helix has a substantial net dipole moment ix Four N H groups at the N terminal end of an helix and four C O groups at the C terminal endlack partners for H bond formation Theformation of H bonds with other nearby donorand acceptor groups is referred to as helixcapping Capping may also involve appropriatehydrophobic interactions that accommodatenonpolar side chains at the ends of helicalsegments x Proline is a helix breaker glycine can be under certain conditions D Beta sheet i The pleated sheet is composed of strands ii Also first postulated by Pauling and Corey 1951 iii Strands in a sheet may be parallel or antiparallel iv Rise per residue 3 47 Angstroms for antiparallel strands 3 25 Angstroms for parallel strands E Beta turn i Allows the peptide chain to reverse direction Carbonyl C of one residue is H bonded to the amide proton of residue three residues away Proline and glycine are prevalent in turns There are two principal forms of turns tight turns or beta bends F Coiled Coil i The coiled coil is a bundle of helices wound into a superhelix Thelefthanded twist of the structure reduces the number of resides per turn to 3 5 so that the positions of the side chains repeat every 7 residues G Collagen triple helix i Nearly one residue out of three is Gly ii Proline content is unusually high iii Pro and HyPro together make 30 of residues iv The unusual amino acid composition of collagen is unsuited for alpha helices or beta sheets It is ideally suited for the collagen triple helix three intertwined helical strands v Much more extended than alpha helix with a rise per residue of 2 9 Angstroms vi 3 3 residues per turn vii Long stretches of Gly Pro Pro HyP viii Every third residue faces the crowded center of the helix only Gly fits here ix Pro and HyP suit the constraints of and x Interchain H bonds involving HyP stabilize helix xi Fibrils are further strengthened by intrachain lysine