2) Here is tyrosine with its pka values noted:

Transcription

1 1. a) At ph 1.0, the following amino acid side chains have (+) charge: Lys, Arg, His b) At ph 12.0, only the side chain of Arg has (+1) charge. 2) Here is tyrosine with its pka values noted: remember that if a solution is more basic than the pka value, the functional group will lose protons to the solution. So pka < ph, the functional group with deprotonate. the opposite also holds. If a solution if more acidic than the pka, the functional group will gain a proton (pka >ph, the functional group protonates) The two possible states of each of the groups above are as follows: NH 2 NH 3 + COOH COO - a) at ph 1.0: the N-terminus will be protonated (+1) the C-terminus will be protonated (0 chg) the side chain will be protonated (0 chg) so the overall charge will be +1. b) at ph 12: the N-terminus will be deprotonated (0) the C-terminus will be deprotonated (-1) the side chain will be deprotonated (-1) so the overall charge will be -2 Here is histidine with its pka values noted:

2 a) at ph 1.0: the N-terminus will be protonated (+1) the C-terminus will be protonated (0 chg) the side chain will be protonated (+1) so the overall charge will be +2. b) at ph 12: the N-terminus will be deprotonated (0) the C-terminus will be protonated (0) the side chain will be deprotonated (0) so the overall charge will be Draw the structure of and determine the charge at ph 6.0 of the following peptide: ACFDR. First decide which groups can ionize: These will be the N-terminus of Ala (pka = 9.9) and the C-terminus of Arg (pka = 1.8). The other amino and carboxyl groups have become part of peptide bonds so no longer can ionize. The ionizable side chains are Cysteine (pka = 8.4), aspartic acid (pka = 3.9), and Arg (pka = 12.5). So for charges compare pka's to ph: pka 9.9 > ph 6.0, so the amino terminus gets a proton = +1 charge pka 1.8 < ph 6.0, so the C- terminus loses a proton = -1 charge pka 8.4 > ph 6.0, so the cysteine gets a proton = 0 charge pka 3.9 < ph 6.0, so the aspartic acid loses a proton = -1 charge pka 12.5 > ph 6.0, so the arginine gets a proton = +1 charge so, the net charge is 0 4. If ph of solution is >pi of peptide, the peptide will have a net negative charge b/c overall, groups will be losing protons to solution. This means amino groups will be neutral and carboxyl groups will be negative. Overall negative.

3 5. What is meant by the "primary structure" of a protein. Basically, the amino acid sequence of the protein. This phrase includes the covalent bonds in a polypeptide, and so it can be meant to include disulfide bonds. 6. What are the differences between parallel and antiparallel Β-sheets? Parallel beta sheets have strands that run in the same direction (both N to C). Antiparallel, they run in opposite directions (one is N to C, one C to N). Parallel sheets have evenly spaced hydrogen bonds. The H-bonds in antiparallel sheets alternately are narrow and widely spaced. The distortion in the H-bonds of parallel sheets make them less stable. This distoration does not occur in antiparallel sheets. 7. Phi and Psi angles. These are the bonds that can rotate to form secondary structures. 8. Can a single polypeptide chain have a quartenary structure? No. Quaternary structure refers to the way in which multiple chains came together. 9. What is the role of chaperone proteins? We didn't actually cover this yet - Briefly, to aid in protein folding, by preventing misfolding. They therefore speed up the overall protein folding process. 10. Explain the driving force behind protein folding. Proteins collapse into a tertiary structure as a result of the hydrophobic effect. Nonpolar amino acids are pushed together in the core of the protein due to entropic concerns. Overview of protein folding amino acids are attached through covalent bonds called peptide bonds into polypeptide units. These are equivalent to proteins. Proteins contain hydrophobic and hydrophilic amino acids. The surrounding water molecules are polar, and therefore would have to use energy to arrange themselves around nonpolar molecules. Because less energy will be used if the waters only have to arrange themselves once around one big molecule, proteins fold into a 3 dimensional shape with the nonpolar amino acids clustered inside the structure. The attainment of this energetically favorable condition is what drives protein folding. This process is called the hydrophobic effect. So, proteins fold into a tertiary structure. The finer organization of the protein structure is referred to as the secondary structure. Alpha helices and beta sheets are the two most common secondary structures; they occur because only certain angles of rotation around the bonds surrounding the alpha carbon are allowed (due to steric constraints). These structures are stabilized by the maximal use of hydrogen bonding networks. Important to keep in mind that structures are dynamic.

4 11. I would say buried inside because it is primarily hydrophobic. 12. Draw the following structures: (a) a hydrogen bond between the side chains of histidine and glutamic acid (b) a salt bridge between the side chains of aspartic acid and arginine. (a) from the H on NH of histidine to the carbonyl oxygen on the side chain of Glu. (b) from the ionized (O - ) on the side chain of Asp to the NH + on Arg chains = dimer; 4 chains = tetramer 14. A mutation that changes an alanine in the interior of a protein to a valine is found to lead to a loss of activity for that protein. However, activity is regained when a second mutation at a different position changes an isoleucine to a glycine. Draw all the amino acids involved in this series of changes and formulate a hypothesis about how the second mutation (Ile to Gly) might lead to a restoration of activity. First mutation is Ala to Val. This causes a loss of function. The second mutation, at a different amino acid, is a change of Ile to Gly. When the Val and the Gly are both present, activity is restored. So, start by thinking about the properties of the amino acids involved. They are all nonpolar, but have different sizes. Valine is bigger than alanine, so the loss in activity must be caused by this change in size. Glycine is smaller than isoleucine and this change, in conjuction with the first, helps restore activity. Thinking about protein structure, we know that amino acid side chains are often packed tightly together. These data suggest that the alanine was packed into the protein somewhere near the isoleucine. The valine is too big and causes disruptions in the protein structure that lead to the loss in activity. When the isoleucine is changed to glycine, the change in size to a smaller amino acid must compensate for the earlier change and make room for the valine to fit in the protein. So, overall, there is now space for all of the amino acids in the protein once again, and activity is restored. 15. Polyhistidine is insoluble in water at ph 7.8, but is soluble at ph 5.5. Explain this observation in as much detail as possible. It's a question about ph so it must be related to pka's. Polyhistidine means there are a lot of histidines connected by peptide bonds, so in terms of pka values, we are only going to be concerned with the side chain pka, which we know to be 6.0. At ph 7.8, this side chain will deprotonate, and be neutral. At ph 5.5, the side chain will be protonated and have a +1 charge.

5 So, we can suggest that polyhistidine is soluble at ph 5.5, but not at ph 7.8 because it will be more soluble in water, a polar solvent, when it is charged! 16. heat, urea, ph changes, SDS 17. Explain why a larger percentage of polar residues is found in the bends and loops in myoglobin than is found in its helical regions. The bends and loops are more in contact with the polar solvent/less packed together with other parts of the protein. In contact with solvent is more likely to be polar, because the nonpolars get pushed away from solvent to minimize the rearrangement of water molecules surrounding the nonpolar amino acids (hydrophobic effect). 18. Discuss how the noncovalent binding of a molecule to a protein can change the protein's conformation. We previously discussed primary structure defining tertiary structure. Can the primary structure of a protein be changed by such noncovalent binding? discuss the idea that when a molecule binds it can cause structural shifts in a protein because of all of the other types of interactions/bonds that are noncovalent. You could use hemoglobin binding of oxygen as an example here. primary structure cannot be changed because it involves only covalent bonds. 19. The existence of so many prolines prevents the formation of alpha helices. 20) A hair fiber can be drawn to twice its original length, but a collagen fiber has little, if any, stretch. Explain these facts on the basis of the molecular structures of hair and collagen. Keratin is formed of a left-handed supercoiled twist of alpha helical polypeptides. Collagen is formed of a right-handed supercoiled twist of left-handed helices. These lefthanded helices are more stretched out than an alpha helix (look at the picture). As a result, the coils that form from them are better packed. Combine these two structural qualities, and the collagen molecule is less stretchable. 21) Give a molecular explanation for why steak from an old steer tends to be tougher than steak from a young steer. Over time, more cross-links form between the hydroxylproline residues that exist in collagen. These are covalent bonds that lock the structure of the protein into place and prevent flexibility.