CAPTER 29 W: AMI ACIDS + PRTEIS For all problems, consult the table of 20 Amino Acids provided in lecture if an amino acid structure is needed; these will be given on exams. Use natural amino acids (L) unless otherwise noted. GEERAL CCEPTS 1. Which amino acid is not chiral? Glycine; side chain = 2. Classify each of the following amino acids as neutral, acidic or basic. A.a. Aspartic Acid Lysine Tyrosine Asparagine Threonine 2 2 (R) Type Acidic Basic Acidic eutral eutral 3. Briefly explain why tryptophan is a neutral amino acid, not a basic one. The lone pair on the nitrogen atom in the sidechain is in a p orbital, and is unavailable to be a base as it is critical to the aromatic ring (loop of p s, 10 p e ). Protonation of the nitrogen in the side chain would destroy some of the aromaticity, so that would be an unfavorable process, making it neutral not basic. 4. Briefly explain why cysteine is an acidic amino acid, yet serine and methionine are neutral. Cysteine Serine Methionine Thiols (RS) are mildly acidic while alcohols (R) are not because the conjugate base of a thiol is RS. RS 2 is a lower energy conjugate base than R because sulfur has a larger orbital than oxygen, S and the charge is more delocalized. S C3 Methionine is not acidic because it doesn t have a on a heteroatom; the s in the sidechain are on carbon, which are generally not acidic (they don t stabilize a negative charge in the conjugate base). Page 1
p-depedet AMI ACID STRUCTURE 5. Draw the three forms of serine, and list the p range where the solution is > 90% in that form. A B C 3 3 pk a of C 2 = 2.2 pk a of 3 + = 9.2 p range 0 1.2 3.2 8.2 10.2 14 Annotate the p scale below to show the p ranges for each form. Some p values will have a mixture of forms. mix of mix of A A +B B B + C C p 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 6. By consulting the previous problem, in what range of p values is serine in its zwitterionic form (>90% of the solution is that form)? p 3.2 8.2 7. Draw the three forms of proline, and list the p range where the solution is > 90% in that form. pk a of C 2 = 2.0 pk a of 3 + = 10.6 p range 0 1.0 3.0 9.6 11.6 14 8. Draw the four forms of glutamic acid, and list the p range where the solution is > 90% in that form. 3 3 3 pk a of C 2 = 2.2 pk a of R = 4.3 pk a of + 3 = 9.7 p range 0 1.2 3.2 3.3 5.3 8.7 10.7 14 Page 2
9. For the amino acid arginine, a. Identify which nitrogen atom in the side chain is basic. Then explain why that nitrogen atom is basic while the other nitrogen atoms are not. Pos. 1 Pos. 3 2 2 Pos. 2 2 2 2 2 The nitrogen lone pairs at positions 1+3 are involved in resonance with the C=. They are therefore not basic, because they re unavailable (protonation would destroy the resonance, as seen in the conjugate acid above). The nitrogen lone pair at position 2 is in an sp 2 orbital, so is not involved in resonance (making it available, and the basic nitrogen atom). When position 2 is protonated, the conjugate acid still has resonance. b. Draw the four forms of arginine, and list the p range where the solution is > 90% in that form. 3 3 2 2 2 2 pk a of C 2 = 2.2 pk a of + 3 = 9.0 pk a of R = 12.5 2 p range 0 1.2 3.2 8.0 10.0 11.5 13.5 14 10. For the amino acid histidine, a. Identify which nitrogen atom in the side chain is basic. Then explain why that nitrogen atom is basic while the other nitrogen atom is not. The top nitrogen lone pair is part of the loop of p s that make up the 6 p e aromatic ring. The LP is therefore unavailable to be a base; protonation would ruin the aromaticity. The bottom nitrogen atom s lone pair is in an sp 2 orbital, so is not involved in the aromatic ring and is available to be a base (aromaticity is maintained in protonation). Bottom is basic. b. Draw the two forms of histidine that would be present in a buffered solution of p 5.9. Then decide which form would be in a slightly greater quantity at this p. 3 A 3 B pk a of R = 6.1 (in transition) At p 5.1 it would be 90% A, 10% B At p 6.1 it would be 50% A, 50% B At p 7.1 it would be 10% A, 90% B At p 5.9 the solution is slightly favoring the A form. Page 3
11. Explain why the pk a of the carboxylic acid on glycine is lower than the pk a of acetic acid. 3 pka 4.8 pka 2.3 3 The + 3 is a strong electron-withdrawing group (EWG), which stabilizes the conjugate base (since withdrawing from an anion is a stabilizing factor). This makes the amino acid C 2 more acidic than acetic acid s C 2, which doesn t have an EWG. 12. Draw the primary form of these amino acids at the indicated p. Glutamine Lysine Tyrosine Aspartic acid pk a 3 + 9.1 pk a C 2 2.2 pk a 3 + 9.0 pk a C 2 2.2 pk a R 10.5 pk a 3 + 9.1 pk a C 2 2.2 pk a R 10.1 pk a 3 + 9.6 pk a C 2 1.9 pk a R 3.7 3 3 3 3 p at p 6.4 at p 4.0 at p 12.0 at p 7.0 13. Explain why the amino acid tyrosine never exists in the form shown below. 3 3 pka 2.2 pka 9.1 3 pka 10.1 These are the forms of tyrosine. The given structure never exists because a carboxylic acid is more acidic than a phenol; it can never happen that a carboxylic acid hasn t reacted (with base) but a phenol has. Page 4
DRAWIG PLYPEPTIDES For the problems in this section, don t worry about any acid-base properties (draw the structures neutral). The structural connectivity is what s important at this point. Draw the backbone of the polypeptide in its natural conformation (zigzag fashion). 14. Draw the dipeptide W E made by connecting the natural amino acids tryptophan and glutamic acid. tryptophan (W) 15. Draw the tripeptide M A D. glutamic acid (E) M A D C 3 SC 3 16. Draw the polypeptide Q R G A I C. Q R G A I C C 3 2 S 2 17. What is the amino acid sequence of each polypeptide? G G S C S Sequence: G-G-S-C T A G 2 2 C Sequence: G--A-T 3 Amino acid sequence is reported from -terminus à C-terminus Page 5
p-depedet PRTEI STRUCTURE 18. Draw the following polypeptides at the indicated p. a. S A K at p 7.4 S A K C 3 3 3 b. M E Q W at p 1.0 pka 4.3 M E Q W 3 pka 2.8 SC 3 2 never protonate the peptide bond (amide!) c. A Y at p 7.6 A Y C 3 3 pka + 6.1 pka 10.1 Page 6
1, 2, 3 AD 4 STRUCTURE F PRTEIS 19. Briefly describe what is referred to by a protein s 1, 2, 3, and 4 structure. Type Description Primary Secondary Tertiary Quaternary Sequence of amino acids Folding to form common shapes like a -helices and b -sheets Further folding of protein by interaction (IMF) between the sidechains (R groups). Aggregation of separate protein strands to form a joint protein 20. ow are alpha helices and beta sheets similar, and how are they different? Similar: both alpha helices and beta sheets are part of a protein s 2 structure; both involve hydrogen bonding between the carbonyl and of the protein s backbone. Different: a -helices are coiled structures and b -sheets are flat pleated sheets; the -bonding partners (carbonyl and of backbone) are near each other in the sequence for an a -helix (~3.5 amino acids away), and further apart in the sequence for a b -sheet. EZYMES 21. The following is a diagram of the active site of an enzyme, with a substrate bound (in bold) and some of the enzyme s side chains shown. Fill in the boxes to label the type of interactions pointed to by each arrow. ionic bond / salt bridge hydrophobic interaction 3 3 C hydrophobic interaction hydrogen bond Page 7
22. Ketone 1 is converted to ketone 2 through an enzyme-mediated reaction. The middle structure shows key interactions between the enzyme (tyrosine-14 and aspartic acid-38) and the substrate. Tyr-14 C 3 C 3 C 3 C 3 C 3 C 3 Ketone 1 Asp-38 Ketone 2 a. ow does the aspartic acid sidechain (Asp-38) help the enzyme function? The deprotonated aspartic acid is the base that forms the enolate, then delivers a hydrogen atom to the end further from the carbonyl. Without the base, there is no low energy mechanism for the alkene to shift (through an enolate), so the ketones cannot be easily converted. C 3 C 3 C 3 C 3 C 3 C 3 Asp-38 Asp-38 Asp-38 b. ow does the tyrosine (Tyr-14) help the enzyme function? The tyrosine stabilizes the enolate intermediate through hydrogen bonding. This lowers the energy of the transition state / intermediate, which is key to lowering the activation barrier and accelerating the reaction. Tyr-14 hydrogen bonding C 3 C 3 c. If a mutation occurs such that the 14 th residue is phenylalanine (Phe-14), what effect would this have on the ability of the enzyme to function? Explain. If the key tyrosine residue were mutated to a phenylalanine, the enzyme could no longer hydrogen bond in this area, and the enzyme would no longer stabilize the transition state. The activation barrier would not be lowered, so the enzyme would no longer catalyze the reaction. Phe-14 no hydrogen bonding C 3 C 3 Page 8
PRTEI DEATURATI 23. Explain how each of these conditions can denature a protein. a. Placing the protein in a solution with a different p than it is active. A change in p affects acidic or basic sidechain R groups, which can change salt bridges into hydrogen bonding sites (or vice versa). This can change the shape of the functional protein (thus denaturing the protein) as salt bridges are not of the same strength as bonds. If the protein is an enzyme, a change in p could change the shape of the active site and make it unable to bind substrates. b. eating the protein. eat provides energy that can temporarily break the IMF holding the protein in a certain shape (the molecular vibrations may be too intense to maintain the weak IMF). When cooled, the IMF may reform in a different shape, which maybe a lower energy conformation, but an inactive one (a denatured protein). c. Placing the protein in an isopropyl alcohol solution. Proteins / enzymes are designed to function in water (biological systems), and isopropyl alcohol is less polar than water. The isopropanol solvent will therefore interact more with the nonpolar sidechain groups that are typically buried in the protein s interior (or the nonpolar sidechains will no longer need to be as buried as they will not cause water to be ordered if they were on the outside of the protein). The isopropanol may cause the protein to loosen, or possibly unravel completely. It may create a different shape than is functional, which is a denatured protein. GEERAL IDEAS 24. Choose whether each statement is True or False. Statement T or F a. All proteins have quaternary structure. False b. A mutation does not always cause a protein to stop functioning. True c. The interior of many globular proteins is mostly nonpolar amino acids. True d. The side chains of glutamine and threonine can interact in a protein through hydrogen bonding. [Glutamine R = amide; Threonine R = alcohol] e. The glutamine-threonine side chain interaction previously described can be altered directly by placing the protein in a more acidic environment. [ot directly; the side chains are neutral and neither are affected by acid] f. The side chains of lysine and aspartic acid interact in a protein more strongly when the protein is in a p 7 solution than when the protein is in a p 1 solution. [Salt bridge at p 7 is stronger than hydrogen bonding at p 1] g. Pepsin is an enzyme in the stomach that degrades proteins into shorter polypeptides. Pepsin should be denatured when placed in a p = 1 solution. [Pepsin is designed to function in p 1 (stomach) so is in the correct shape at that p] True False True False Page 9