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BT631-5-primary_secondary_structures_proteins

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  • 1. Different Levels of Protein Structure
  • 2. ACDEFGHIKLMNPQRSTVWY Primary structure Secondary structure Tertiary structure Quaternary structure
  • 3. Primary structure of proteins The primary structure is the linear order of amino acid residues along the polypeptide chain. Every protein is defined by a unique sequence of residues and all subsequent levels of organization (secondary, super secondary, tertiary and quaternary) rely on this primary level of structure. Some proteins are related to one another leading to varying degrees of similarity in primary sequences.
  • 4. How do you determine the primary structure of a protein? Asn-Gly-Phe-Glu-Gln-Ala-Arg-Asp-Cys-Leu-Ile-Trp-Pro-Tyr-Ser-Met-Lys-Val-His-Thr N- -C 1. Determining amino acid composition of a protein I. Hydrolysis (heat at 100-110 C in 6M HCl for 24 hrs or longer) II. Separation (chromatography techniques) III. Quantitative analysis (color producing reagents e.g. ninhydrin) 3. C-terminal amino acid analysis (Carboxypetidases) 2. N-terminal amino acid analysis I. React the peptide with a reagent which selectively label the terminal amino acid (e.g. DFNB) II. Hydrolyze the protein III. Determine the amino acid by chromatography and compare with standards.
  • 5. Enzymatic Analysis Enzymatic C-terminal amino acid cleavage by one of several carboxypeptidase enzymes is a fast and convenient method of analysis. A peptide having a C-terminal sequence: ~Gly-Ser- Leu is subjected to carboxypeptidase cleavage and the free amino acids cleaved in this reaction are analyzed at increasing time intervals. C-Terminal Group Analysis
  • 6. Selective Peptide Cleavage Name Type Specificity Cyanogen Bromide Chemical Carboxyl Side of Methionine Trypsin Enzymatic Carboxyl Side of Basic Amino Acids e.g. Lys & Arg Chymotrypsin Enzymatic Carboxyl Side of Aryl Amino Acids e.g. Phe, Tyr & Trp
  • 7. 1. DNA or mRNA 2. Edman degradation 3. Mass spectrometry 4. Others Protein sequence determination methods Edman Degradation A free amine function, usually in equilibrium with zwitterion species, is necessary for the initial bonding to the phenyl isothiocyanate reagent. The products of the Edman degradation are a thiohydantoin heterocycle incorporating the N-terminal amino acid together with a shortened peptide chain.
  • 8. Amine functions on a side-chain as in lysine may also react with the isothiocyanate reagent. A major advantage of the Edman procedure is that the remaining peptide chain is not further degraded by the reaction. This means that the N-terminal analysis may be repeated several times, thus providing the sequence of the first three to five amino acids in the chain. A disadvantage of the procedure is that is peptides larger than 30 to 40 units do not give reliable results. It however does not give thiohydantoin products.
  • 9. Secondary structure of proteins The local conformation of the polypeptide chain or the spatial relationship of amino acid residues that are close together in the primary sequence. Why do proteins form secondary structures? We know that proteins have hydrophobic cores. To bring the side chains into the core , the main chain must also fold into the interior. The main chain is however highly polar and therefore hydrophilic, with one hydrogen bond donor, N-H, and one hydrogen bond acceptor C=O, for each peptide unit. In a hydrophobic environment, these main chain polar groups must be neutralized by the formation of hydrogen bonds. This problem is solved by the formation of secondary structures. Proteins adapt secondary structures to stabilize their structures to protect themselves from other proteins which can digest them. In proteins, the formation of secondary structures appears to result from the combination of both the entropic effect of compaction and local energetic effects.
  • 10. In globular proteins, the three basic units of secondary structure are the α helix, β strand and turns. Why do proteins form only these kinds of secondary structures? This is because of local interactions. It is the specific hydrogen bonding patterns in protein which favor the formation of α helices and β strands. In secondary structure the main chain amides and carbonyls participate in H-bonds to each other. This neutralizes the polar nature of the peptide bond and enables the main chain to fold into the hydrophobic core.
  • 11. The a helix The hydrogen bonds occur between the backbone carbonyl oxygen (C=O acceptor) of one residue (i) and the amide hydrogen (N-H donor) of residue (i+4) ahead in the polypeptide chain (i+4  i). In 1954 Pauling was awarded his first Nobel Prize "for his research into the nature of the chemical bond and its application to the elucidation of the structure of complex substances" As the first four N-H groups and the last four C=O groups are normally not involved in the hydrogen bonds, the ends of α helices are polar and are almost always at the surface of protein molecules.
  • 12. The regular α helix has 3.6 residues per turn with each residue offset from the preceding residue by 0.15 nm (translation per residue distance) along the helix axis. Thus, the pitch (the vertical distance between one consecutive turn of the helix) of the α helix is 3.6 x 0.15 = 0.54 nm. The width of the helix is 5 Å. The hydrogen bonds are 0.286 nm long from oxygen to nitrogen atoms, linear and lie parallel to the helical axis. What about turn angle of each amino acid along the helix?
  • 13. Each amino acid participates in 2 H-bonds. Thus all the main chain C=O and N-H participate in H-bonds. In globular proteins, α helices vary in length, ranging from four to over forty amino acids. The average length is around ten residues. What will be the average length of an α helix? Proline does not form helical structure for the obvious reason that the absence of an amide proton (NH) precludes hydrogen bonding whilst the side chain covalently bonded to the N atom restricts backbone rotation.
  • 14. Helical conformations of peptide chains may also be described by a two number term, nm, where n is the number of amino acid units per turn and m is the number of atoms in the smallest ring defined by the hydrogen bond. Thus, a helix is 3.613-helix denoting a hydrogen bond between every carbonyl oxygen and the alpha- amino nitrogen of the fourth residue toward the C- terminus, and 13 atoms being involved in the ring formed by the hydrogen bond. Assignment No. 2: Show that left-handed helices are not permissible?
  • 15. The 310 helix The designation 310 refers to the number of backbone atoms located between the donor and acceptor atoms (10) and the fact that there are three residues per turn. The hydrogen bonds in 310 helix are formed between residues (i, i+3) in contrast to (i, i+4) bonds in regular α helix. The angles are: Φ = -49 and Ψ = -26 . The rise for one residue is 2.0 Å. What about turn angle of each amino acid and the pitch along the helix? The 310 helices do occur, but are not very long, they are sometimes found at the end of an α helix.
  • 16. The π helix Whilst 310 helix is a narrower structure than the α helix, a third possibility is a more loosely coiled helix with hydrogen bonds formed between the C=O and N-H groups separated by five residues (i, i+5). There are 4.4 residues per turn and 16 atoms in the H-bonded ring. The angles are: Φ = -57 and Ψ = -70 . The π helix is more compact, more compressed than the α helix. The H-bonds in the π helix are not straight and side chains interfere. The larger radius of the π helix means that backbone atoms do not make van der Waals contact across the helix axis leading to the formation of a hole down the middle of the helix that is too small for solvent occupation. If the pitch of the helix is 5.06 Å per turn, what will be the rise for one residue?
  • 17. Dipole moment has directionality The magnitude of the dipole moment is about 0.5-0.7 unit charge at each end of the helix. These charges attract ligands of opposite charge such as phosphate ions. Why does C-terminus generally not attract positively charged ligands?
  • 18. Bromodomain Class Number of folds Number of super families Number of families All α proteins 284 507 871 Globin domain
  • 19. 1. Combined pattern of pitch and hydrogen bonding. 2. In terms of repeating φ and ψ torsion angles. How do we find the segments of a given protein structure that belong to the α helix?
  • 20. The β strand β strand is a helical arrangement although an extremely elongated form with two residues per turn. The side chains are oriented alternating up and down. This leads to a pitch or repeat distance of ~0.7 nm in a regular β strand. If the pitch of an anti-parallel and parallel β strands are (i.e. 6.84 Å per turn) and (i.e. 6.4 Å per turn), what will be the length between two adjacent Cα atoms?
  • 21. β strands are stable in the sheet form where adjacent strands can align in parallel or anti- parallel arrangements with the orientation established by determining the direction of the polypeptide chain from the N- to the C-terminal.
  • 22. β strands are quite extended but normally don't reach the 180 for the angles completely, thus are not flat, but pleated. Average values for the angles are: Φ = -139 and Ψ = 135 in anti- parallel β sheets and Φ = -119 and Ψ = 113 in parallel β sheets.
  • 23. Class Number of folds Number of super families Number of families All β proteins 174 354 742
  • 24. The turns More commonly found in protein structures are four residues turns (β turns). Loops which connect 2 adjacent anti-parallel β strands are called hairpin loops. A γ turn contains three residues and frequently links adjacent strands of anti-parallel β sheet. Analysis of the amino acid composition of turns reveals that bulky or branched side chains occur rarely. Instead residues with small side chain such as Gly, Asp, Asn, Ser, Cys and Pro are frequently found.
  • 25. In some proteins, the proportion of residues found in turns can exceed 30 percent and in view of this high value it is unlikely that turns represent random structures (Intrinsically Disordered Proteins). Loop regions are found at the surface of the protein molecules mostly because the main chain groups of these loops do not form hydrogen bonds to each other and hence are exposed to the solvent to form hydrogen bonds to water molecules.
  • 26. Some amino acids prefer to be in α helices. However, amino acids are also dependent on the position of the α helix in the protein or the position of the α helix depends on the amino acids it contains. Examples: • An α helix buried in the hydrophobic core (of citrate synthase) contains only uncharged and mostly nonpolar amino acids. • A partially exposed α helix (of alcohol dehydrogenase) contains polar or charged residues at the exposed side and non-polar ones at the other side. • A fully exposed α helix (of troponin C) contains a lot of charged residues.
  • 27. α helix: Glu, Ala, Leu, Met, Gln, Lys, Arg, His β strand: Val, Ile, Tyr, Cys, Trp, Phe, Thr Reverse turn: Gly, Asn, Pro, Ser, Asp Conformational Preferences of Amino Acids Propensity = (# of a particular a.a. in a particular secondary structure / # of a particular a.a. in the whole protein) / (# of all a.a. in the particular structure / # of all a.a. in whole protein). For example, in a protein if there are 30% of all amino acids in α helices and 50% of Glu in α helices, then in this protein the propensity value for Glu being in α helix is 50/30 = 1.66.) How do you calculate the propensity of an a.a. being in a secondary structure?
  • 28. The helical propensity of amino acid residues substituted into Alanine polymers Residue Helix propensity, ΔG (kJ mol-1) Residue Helix propensity, ΔG (kJ mol-1) Ala 0 Ile 0.41 Arg 0.21 Leu 0.21 Asn 0.65 Lys 0.26 Asp0 0.43 Met 0.24 Asp- 0.69 Phe 0.54 Cys 0.68 Pro 3.16 Gln 0.39 Ser 0.50 Glu0 0.16 Thr 0.66 Glu- 0.40 Tyr 0.53 Gly 1.00 Trp 0.49 His0 0.56 Val 0.61 His+ 0.66 0: uncharged All residues form helices with less propensity than poly-Ala hence the positive values for ΔG. In globular proteins, over 30% of all residues are found in helices.
  • 29. The role of a particular amino acid in the conformational change of a protein is studied by considering homopolymers (e.g. poly-Ala) and mutating single amino acids into another and measuring the stability, solubility or secondary structure properties of the mutant compared to the wild type. How do you find the preference of an amino acid to be in a particular secondary structure using ab-initio methods?
  • 30. The Ramachandran Plot Dihedral angles, translation distances and number of residues per turn for regular secondary structure conformations. In poly(Pro) I ω is 0 whilst in poly(Pro) II, ω is 180 Secondary structure element Dihedral angle ( ) Residues/ turn Translation distance per residue (nm) φ ψ α helix -57 -47 3.6 0.150 310 helix -49 -26 3.0 0.200 π helix -57 -70 4.4 0.115 Anti-parallel β strand -119 +113 2.0 0.320 Parallel β strand -139 +135 2.0 0.340 Poly(Pro) I -83 +158 3.3 0.190 Poly(Pro) II -78 +149 3.0 0.312 Poly(Gly) II -80 +150 - -
  • 31. Assignment No. 3: Submit the following details about the protein on which you are working or might work. 1. Protein name 2. Protein’s primary structure 3. Protein’s primary structure a.a. composition 4. Percentage of the residues found in the secondary structure 5. Secondary structure composition 6. Secondary structure propensity of all 20 amino acids for your protein.