Structural hierarchy in proteins

1. Structural hierarchy in proteins

2. Color conventions

3. Protein Geometry CORN LAW amino acid with L configuration

4. Greek alphabet

5. The Polypeptide Chain

6. Chapter 5 Covalent structures of proteins Proteins function as: 1. Enzymes:biological catalysts 2. Regulators of catalysis-hormones 3. Transport and store i.e. O2, metal ions sugars, lipids, etc. 4. Contractile assemblies Muscle fibers Separation of chromosomes etc. 5. Sensory Rhodopsin nerve proteins

7. 6. Cellular defense • immuoglobulins • Antibodies • Killer T cell • Receptors • 7. Structural • Collagen • Silk, etc. • Function is dictated by protein structure!!

8. There are four levels of protein structure • 1. Primary structure • 1 = Amino acid sequence, the linear order of AA’s. • Remember from the N-terminus to the C-terminus • Above all else this dictates the structure and function of the protein.

9. There are four levels of protein structure • 2. Secondary structure • 2 = Local spatial alignment of amino acids without regard to side chains. • Usually repeated structures • Examples: a helix, b sheets, random coil, or b turns

10. 3. Tertiary Structure • 3 = the 3 dimensional structure of an entire peptide. • Great in detail but vague to generalize. Can reveal the detailed chemical mechanisms of an enzyme.

11. 4. Quaternary Structure • 4 two or more peptide chains associated with a protein. • Spatial arrangements of subunits. Chapter 5.3 is how to determine a protein’s primary structure. “Protein Chemistry”

12. Example of each level of protein structure

13. Insulin was the first protein to be sequenced • F. Sanger won the Nobel prize for protein sequencing. • It took 10 years, many people, • and it took 100 g of protein! • Today it takes one person several days to sequence the same insulin. • 1021 AA b- glactosidase 1978

14. Steps towards protein sequencing Above all else, purify it first!! Chapter 5.3 then 5.1 and 5.2 1. Prepare protein for sequencing a. Determine number of chemically different polypeptides. b. Cleave the protein’s disulfide bonds. c. Separate and purify each subunit. d. Determine amino acid composition for each peptide.

15. Bovine insulin: note the intra- and inter- chain disulfide linkages

16. 2. Sequencing the peptide chains: • a. Fragment subunits into smaller peptides  50 AA’s in length. • b. Separate and purify the fragments • c. Determine the sequence of each fragment. • d. Repeat step 2 with different fragmentation system.

17. 3. Organize the completed structure. • a. Span cleavage points between sets of peptides • determined by each peptide sequence. • b. Elucidate disulfide bonds and modified amino acids. • At best, the automated instruments can sequence about 50 amino acids in one run! • Proteins must be cleaved into smaller pieces to obtain a complete sequence.

18. End Group Analysis How many peptides in protein? Bovine insulin should give 2 N-terminii and 2 C-terminii N-terminus 1-Dimethylamino - naphthalene-5-sulfonyl chloride Dansyl chloride Reacts with amines: N-terminus + Lys (K) side chains

20. Disadvantage with the Dansyl-chloride method is that you must use 6M HCl to cleave off the derivatized amino acid, this also cleaves all other amide bonds (residues) as well. Edman degradation with Phenyl isothiocyanate, PITC

21. Edman degradation has been automated as a method to sequence proteins. The PTH-amino acid is soluble in solvents that the protein is not. This fact is used to separate the tagged amino acid from the remaining protein, allowing the cycle of labeling, degradation, and separation to continue. Even with the best chemistry, the reaction is about 98% efficient. After sufficient cycles more than one amino acid is identified, making the sequence determination error-prone at longer reads.

22. Demonstration of Edman degradation Use your CD disk- install it and run chapter 5 Edman degradation.

23. Carboxypeptidase cleavage at the C-terminus Carboxypeptidase A Rn R, K, P Rn-1 P If the Tyr-Ser bond is more resistant to cleavage than the Leu-Tyr, the Ser and the Tyr will appear simultaneously and the C-terminus would still be in doubt.

24. Cleavage of disulfide bonds Permits separation of polypeptide chains Prevents refolding back to native structure • Performic acid oxidation • Changes cystine or cysteine to Cystic acid • Methionine to Methionine sulfone • 2-Mercaptoethanol, dithiothreitol, or dithioerythritol • Keeps the equilibrium towards the reduced form • -S-S- 2SH

25. Amino acid composition The amino acid composition of a peptide chain is determined by its complete hydrolysis followed by the quantitative analysis of the liberated amino acids. Acid hydrolysis (6 N HCl) at 120 oC for 10 to 100 h destroys Trp and partially destroys Ser, Thr, and Tyr. Also Gln and Asn yield Glu and Asp Base hydrolysis 2 to 4 N NaOH at 100 oC for 4 - 8 h. Is problematic, destroys Cys Ser, Thr, Arg but does not harm Trp.

26. Amino acid analyzer In order to quantitate the amino acid residues after hydrolysis, each must be derivatized at about 100% efficiency to a compound that is colored. Pre or post column derivatization can be done. These can be separated using HPLC in an automated setup

27. Amino acid compositions are indicative of protein structures Leu, Ala,Gly, Ser, Val, Glu, and Ile are the most common amino acids His, Met, Cys, and Trp are the least common. Ratios of polar to non-polar amino acids are indicative of globular or membrane proteins. Certain structural proteins are made of repeating peptide structures i.e. collagen.

28. Long peptides have to be broken to shorter ones to be sequenced Endopeptidases cleave proteins at specific sites within the chain. Trypsin Rn-1 = positively charged residues R, K; Rn P Chymotrypsin Rn-1 = bulky hydrophobic residues F, W, T; Rn P Thermolysin Rn= I, M, F, W, T, V; Rn-1  P Endopeptidase V8 Rn-1 = E

29. Specific chemical cleavage reagents Cyanogen Bromide Rn-1 = M Cleave the large protein using i.e trypsin, separate fragments and sequence all of them. (We do not know the order of the fragments!!) Cleave with a different reagent i.e. Cyanogen Bromide, separate the fragments and sequence all of them. Align the fragments with overlapping sequence to get the overall sequence.

30. How to assemble a protein sequence 1. Write a blank line for each amino acid in the sequence starting with the N-terminus. 2. Follow logically each clue and fill in the blanks. 3. Identify overlapping fragments and place in sequence blanks accordingly. 4. Make sure logically all your amino acids fit into the logical design of the experiment. 5. Double check your work.

31. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 H3N-_-_-_-_-_-_-_-_-_-_-_-_-_-_-COO K F - A - M - K K - F - A - M Q - M - K D - I - K - Q - M G - M - D - I - K Y - R - G - M Y - R Trypsin cleaves after K or R (positively charged amino acids) Q - M - K G - M - D - I - K F - A - M - K Y - R Cyanogen Bromide (CN Br) Cleaves after Met i.e M - X D - I - K - Q - M K K - F - A - M Y - R - G - M

32. There are a variety of ways to purify peptides All are based on the physical or chemical properties of the protein. Size Charge Solubility Chemical specificity Hydrophobicity/ Hydrophylicity Reverse Phase High Pressure Liquid Chromatography is used to separate peptide fragments.

33. Peptide mapping: digest protein with an appropriate agent, then separate using two dimensional paper chromatography Digested Peptide from normal (HbA) and Sickle cell anemia (Hbs) hemoglobins HbA V - H - L - T - P - E - E - K HbS V - H - L - T - P - V - E - K  1 2 3 4 5 6 7 8 Beta chain position 6 contains altered amino acid

34. Red blood cells : (a) normal (b) sickle cell Electrophoretic separation of hemoglobins

36. Deoxyhemoglobin aggregates and deforms cell. Primary structure changes dictate quaternary structure. Why did the problem not die out? Homozygotic Heterzyatic Homozygotic normalsickle cell traitsickle cell gets malaria resistant gets sickle cell to malaria diesdies

37. Species variation in homologous proteins The primary structures of a given protein from related species closely resemble one another. If one assumes, according to evolutionary theory, that related species have evolved from a common ancestor, it follows that each of their proteins must have likewise evolved from the corresponding ancestor. A protein that is well adapted to its function, that is, one that is not subject to significant physiological improvement, nevertheless continues to evolve. Neutral drift: changes not effecting function

38. Homologous proteins (evolutionarily related proteins) Compare protein sequences: Conserved residues, i.e invariant residues reflect chemical necessities. Conserved substitutions, substitutions with similar chemical properties Asp for Glu, Lys for Arg, Ile for Val Variable regions, no requirement for chemical reactionsetc.

39. Amino acid difference matrix for 26 species of cytochrome c Man,chimp0 Rh. monkey 1 0 Average differences Horse 12 11 0 Donkey 11 10 1 0 10.0 cow,sheep 10 9 3 2 0 dog 11 10 6 5 3 0 gray whale 10 9 5 4 2 3 0 5.1 rabbit 9 8 6 5 4 5 2 0 kangaroo 10 11 7 8 6 7 6 6 0 Chicken 13 12 11 10 9 10 9 8 12 0 penguin 13 12 12 11 10 10 9 8 10 2 0 9.9 Duck 11 10 10 9 8 8 7 6 10 3 3 0 14.3 Rattlesnake 14 15 22 21 20 21 19 18 21 19 20 17 0 12.6 turtle 15 14 11 10 9 9 8 9 11 8 8 7 22 0 Bullfrog 18 17 14 13 11 12 11 11 13 11 12 11 24 10 0 Tuna fish 21 21 19 18 17 18 17 17 18 17 18 17 26 18 15 0 18.5 worm fly 27 26 22 22 22 21 22 21 24 23 24 22 29 24 22 24 0 silk moth 31 30 29 28 27 25 27 26 28 28 27 27 31 28 29 32 14 0 25.9 Wheat 43 43 46 45 45 44 44 44 47 46 46 46 46 46 48 49 45 45 0 Bread mold 48 47 46 46 46 46 46 46 49 47 48 46 47 49 49 48 41 47 54 0 47.0 Yeast 45 45 46 45 45 45 45 45 46 46 45 46 47 49 47 47 45 47 47 41 0 Candida k. 51 51 51 50 50 49 50 50 51 51 50 51 51 53 51 48 47 47 50 42 27 0 Man,chimp monkey Horse Donkey cow,sheep dog gray whale rabbit kangaroo Chicken, penguin Duck Rattlesnake turtle Bullfrog Tuna fish worm fly silkworm Wheat Bread mold Yeast Candida

40. Phylogenetic tree Indicates the ancestral relationships among the organisms that produced the protein. Each branch point indicates a common ancestor. Relative evolutionary distances between neighboring branch points are expressed as the number of amino acid differences per 100 residues of the protein. PAM units or Percentage of Accepted Mutations

42. PAM values differ for different proteins. Although DNA mutates at an assumed constant rate. Some proteins cannot accept mutations because the mutations kill the function of the protein and thus are not viable.

43. Mutation rates appear constant in time Although insects have shorter generation times than mammals and many more rounds of replication, the number of mutations appear to be independent of the number of generations but dependent upon time Cytochrome c amino acid differences between mammals, insects and plants note the similar distances

44. Evolution through gene duplication • Many proteins within an organism have sequence similarities with other proteins. • These are called gene or protein families. • The relatedness among members of a family can vary greatly. • These families arise by gene duplication. • Once duplicated, individual genes can mutate into separate genes. • Duplicated genes may vary in their chemical properties due to mutations. • These duplicate genes evolve with different properties. • Example the globin family.

45. Hemoglobin: • is an oxygen transport protein • it must bind and release oxygen as the cells require oxygen • Myoglobin: • is an oxygen storage protein • it binds oxygen tightly and releases it when oxygen concentrations are very low

46. The globin family history 1. Primordial globin gene acted as an Oxygen-storage protein. 2. Duplication occurred 1.1 billion years ago. lower oxygen-binding affinity, monomeric protein. 3. Developed a tetrameric structure two a and two b chains increased oxygen transport capabilities. 4. Mammals have fetal hemoglobin with a variant b chain i.e. g (a2g2). 5. Human embryos contain another hemoglobin 2e2. 6. Primates also have a d chain with no known unique function.

48. Protein Evolution is not organismal evolution • Chimpanzee  human are about 99% the same amino acid sequences in proteins! • However: • Rapid divergence with few mutational changes suggest altered control of gene expression. • Controlling the amount, where, and when a protein is made.