BICH 410 1st Edition Exam #2 Study GuideI. FormatA. 50 multiple choice questionsi. Mostly conceptual/ factual informationii. 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 reviewB. Chapters 5-7i. Proteins, Protein structure and folding, and carbohydratesii. Memorize important aldoses and ketoses highlighted in blue in the powerpointsC. Tips for taking the testi. 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.II. ProteinsA. Protein structurei. Primary (1°) – sequenceSecondary (2°) - local structures - H-bonds Tertiary (3°) - overall 3-dimensional shape Quaternary (4°) - subunit organization ii. Solubility is influenced by pHiii. Know the purification steps in this figureiv. Purification processes to know1. Dialysis- concentration2. Ion exchange chromatography-charge3. Gel filtration chromatography-size\4. SDS-PAGE-molecular weight5. Affinity chromatographyB. Sequencingi. Sanger sequenced the two chains of insulinii. Established that all molecules of a given protein have the same sequenceiii. Proteins can be sequenced by direct amino acid sequencing or sequencing the corresponding DNA in the geneC. 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 concii. Cleavage of disulfide bridges1. Sulfhydryl reducing agents used like betaMe or DTTiii. Identify N and C terminal residues1. N-terminal- use Edman reagent2. C-terminal is enzymatic. Use carboxypeptidases. Carboxypeptidase A cleavesany residue except Pro, Arg, and Lys. B only works on Arg and Lysiv. Fragmentation of the chains- steps 4 and 51. Enzymatic- Trypsin - cleavage on the C-side of Lys, ArgChymotrypsin - C-side of Phe, Tyr, TrpClostripain - like trypsin, but attacks Arg much more than LysStaphylococcal proteaseC-side of Glu, Asp in phosphate buffer (pH 7) Specific for Glu in acetate (pH 5) or bicarbonate buffer (pH 10)2. Chemical fragmentationCNBr acts only on methionine residuesCNBr is useful because proteins usually have only a few Metresidues3. KNOW the reaction of CNBr with a peptidev. Reconstructing the Sequence1. Use two or more fragmentation agents in separate fragmentation experimentsSequence all the peptides produced (usually by Edman degradation)Compare and align overlapping peptide sequences to learn the sequence of the original polypeptide chainvi. Mass spec can also be used to sequence the chain and separate particles on a basis of mass to charge ratioD. Know that leucine, alanine and serine are the most abundant AAs. Also know that histidine, cysteine and tryptophan are the least commoni. Even though sequence and structure of two proteins can be very similar, their function will most likely be differentii. Biological functions of proteins—know themIII. Protein Structure and FoldingA. Overarching principlesi. Function depends on structureii. Structure depends on sequence and on weak, noncovalent forcesiii. The number of protein folding patterns is large but finiteiv. Structures of globular proteins are marginally stablev. Marginal stability facilitates motionvi. Motion enables functionB. Bondingi.Hydrogen bonds are formed wherever possibleHydrophobic interactions drive protein foldingIonic interactions usually occur on the protein surfacevan der Waals interactions are everywhere--electrostatic interactions also importantC. Amide Planei. Angle about the Cα-N bond is denoted φ (phi)ii. Angle about the Cα-C bond is denoted ψ (psi)iii. φ = 0°, ψ = 180° is unfavorableiv. φ = 180°, ψ = 0° is unfavorablev. φ = 0°, ψ = 0° is unfavorablevi. G. N. Ramachandran was the first to demonstrate the convenience of plotting phi, psi combinations from known protein structuresvii. The sterically favorable combinations are the basis for preferred secondary structuresC. Alpha Helixi. 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 proteinsiv. Stabilized by H bondsv. Residues per turn: 3.6vi. Rise per residue: 1.5 Angstroms (0.15 nm)vii. Rise per turn (pitch): 3.6 × 1.5Å = 5.4 Angstromsviii. Alpha helix has a substantial net dipole momentix. Four N-H groups at the N-terminal end of an α-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 conditionsD. Beta sheeti. The β-pleated sheet is composed of β-strandsii. Also first postulated by Pauling and Corey, 1951iii. Strands in a β-sheet may be parallel or antiparalleliv. Rise per residue:• 3.47 Angstroms for antiparallel strands• 3.25 Angstroms for parallel strandsE. Beta turni. 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 β-turns• There are two principal forms of β-turns- tight turns or beta bendsF. Coiled Coili. The coiled coil is a bundle of α-helices wound into a superhelix.Thelefthandedtwist 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 helixi. Nearly one residue out of three is Glyii. Proline content is unusually highiii. Pro and HyPro together make 30% of residuesiv. 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 strandsv. Much more extended than alpha helix, with a rise per residue of 2.9 Angstromsvi. 3.3 residues per turnvii. Long stretches of Gly-Pro-Pro/HyPviii. Every