Answers to Problems
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ANSWERS TO PROBLEMS Chapter 1 8. There are 20 choices for each of the 50 amino acids: 5020, or 21 1. The hydrogen-bond donors are the NH and NH2 groups. 5 3 10 . The hydrogen-bond acceptors are the carbonyl oxygen atoms and 9. those ring nitrogen atoms that are not bonded to hydrogen or to O O deoxyribose. + 2. Interchange the positions of the single and double bonds in the H3N CH C N CH C O CH3 six-membered ring. 3. (a) Electrostatic interactions; (b) van der Waals interactions. CH2 H CH2 4. Processes a and b C O 2 2 2 2 5. DS 5 2661 J mol 1 K 1 (2158 kcal mol 1 K 1) system O 21 21 21 21 DSsurroundings 5 1842 J mol K (1201 cal mol K ) 6. (a) 1.0; (b) 13.0; (c) 1.3; (d) 12.7 7. 2.88 Aspartame at pH 7 8. 1.96 10. The (nitrogen–a carbon–carbonyl carbon) repeating unit. 9. 11.83 11. Side chain is the functional group attached to the a-carbon 10. 447; 0.00050 atom of an amino acid. 11. 0.00066 M 12. Amino acid composition refers simply to the amino acids 12. 6.0 that make up the protein. The order is not specified. Amino acid 13. 5.53 sequence is the same as the primary structure—the sequence of 14. 6.48 amino acids from the amino terminal to the carboxyl terminal 15. 7.8 of the protein. Different proteins may have the same amino acid 16. 100 composition, but amino acid sequence identifies a unique protein. 17. (a) 1.6; (b) 0.51; (c) 0.16. 13. (a) Each strand is 35 kd and hence has about 318 residues (the 18. 0.1 M sodium acetate solution: 6.34; 6.03; 5.70; 4.75. mean residue mass is 110 daltons). Because the rise per residue in an 0.01 M sodium acetate solution: 5.90; 4.75; 3.38; 1.40. a helix is 1.5 Å, the length is 477 Å. More precisely, for an a-helical 19. 90 mM acetic acid; 160 mM sodium acetate, 0.18 moles acetic coiled coil, the rise per residue is 1.46 Å; so the length is 464 Å. acid; 0.32 moles sodium acetate; 10.81 g acetic acid; 26.25 g sodium (b) Eighteen residues in each strand (40 minus 4 divided by 2) are acetate. in a b-sheet conformation. Because the rise per residue is 3.5 Å, 20. 0.50 moles of acetic acid; 0.32 moles of NaOH; 30.03 g of acetic the length is 63 Å. acid; 12.80 g of NaOH. 14. The methyl group attached to the b-carbon atom of isoleucine 21. 250 mM; yes; no, it will also contain 90 mM NaCl. sterically interferes with a-helix formation. In leucine, this methyl g 22. 8.63 g Na2HPO4; 4.71 g NaH2PO4 group is attached to the -carbon atom, which is farther from the 23. 7.0; this buffer will not be very useful, because the pH value is main chain and hence does not interfere. far from the pKa value. 15. The first mutation destroys activity because valine occupies 2 2 2 2 24. 1.45 kJ mol 1 (0.35 kcal mol 1); 57.9 kJ mol 1 (13.8 kcal mol 1) more space than alanine does, and so the protein must take a dif- 25. There will be approximately 15 million differences. ferent shape, assuming that this residue lies in the closely packed interior. The second mutation restores activity because of a com- Chapter 2 pensatory reduction of volume; glycine is smaller than isoleucine. 1. (A) Proline, Pro, P; (B) tyrosine, Tyr, Y; (C) leucine, Leu, L; 16. The native conformation of insulin is not the thermodynami- (D) lysine, Lys, K. cally most stable form, because it contains two separate chains 2. (a) C, B, A; (b) D; (c) D, B; (d) B, D; (e) B. linked by disulfide bonds. Insulin is formed from proinsulin, a 3. (a) 6; (b) 2; (c) 3; (d) 1; (e) 4; (f ) 5. single-chain precursor, that is cleaved to form insulin, a 51-residue 4. (a) Ala; (b) Tyr; (c) Ser; (d) His. molecule, after the disulfide bonds have formed. 5. Ser, Glu, Tyr, Thr 17. A segment of the main chain of the protease could hydrogen 6. (a) Alanine-glycine-serine; (b) Alanine; (c and d): bond to the main chain of the substrate to form an extended paral- b Peptide bonds lel or antiparallel pair of strands. 18. Glycine has the smallest side chain of any amino acid. Its size CH3 HHO CH2OH is often critical in allowing polypeptide chains to make tight turns or to approach one another closely. N C C NCCNC C 19. Glutamate, aspartate, and the terminal carboxylate can form salt bridges with the guanidinium group of arginine. In addition, H H O H H H O this group can be a hydrogen-bond donor to the side chains of Alpha carbons glutamine, asparagine, serine, threonine, aspartate, tyrosine, and glutamate and to the main-chain carbonyl group. Histidine can 7. OO OOform hydrogen bonds with arginine at pH7. + – +– H3N CH C N CH C O H3N CH C N CH C O 20. Disulfide bonds in hair are broken by adding a thiol-containing HHCH HHCH reagent and applying gentle heat. The hair is curled, and an oxidiz- 2 2 ing agent is added to re-form disulfide bonds to stabilize the desired +HN N shape. 21. Some proteins that span biological membranes are “the excep- NH NH tions that prove the rule” because they have the reverse distribution At pH 5.5, the net charge is 11 At pH 7.5, the net charge is 0 of hydrophobic and hydrophilic amino acids. For example, consider A1 A2 ANSWERS TO PROBLEMS porins, proteins found in the outer membranes of many bacteria. 5. The sample was diluted 1000-fold. The concentration after Membranes are built largely of hydrophobic chains. Thus, porins dialysis is thus 0.001 M, or 1 mM. You could reduce the salt are covered on the outside largely with hydrophobic residues that concentration by dialyzing your sample, now 1 mM, in more interact with the neighboring hydrophobic chains. In contrast, the buffer free of (NH4)2SO4. center of the protein contains many charged and polar amino acids 6. If the salt concentration becomes too high, the salt ions interact that surround a water-filled channel running through the middle of with the water molecules. Eventually, there will not be enough the protein. Thus, because porins function in hydrophobic environ- water molecules to interact with the protein, and the protein will ments, they are “inside out” relative to proteins that function in precipitate. If there is lack of salt in a protein solution, the proteins aqueous solution. may interact with one another—the positive charges on one protein 22. The amino acids would be hydrophobic in nature. An a helix with the negative charges on another or several others. Such an is especially suited to crossing a membrane because all of the amide aggregate becomes too large to be solublized by water alone. If salt hydrogen atoms and carbonyl oxygen atoms of the peptide back- is added, the salt neutralizes the charges on the proteins, preventing bone take part in intrachain hydrogen bonds, thus stabilizing these protein–protein interactions. polar atoms in a hydrophobic environment. 7. Tropomyosin is rod shaped, whereas hemoglobin is approxi- 23. This example demonstrates that the pKa values are affected mately spherical. by the environment. A given amino acid can have a variety of pKa 8. The frictional coefficient, f, and the mass, m, determine s. values, depending on the chemical environment inside the protein. Specifically, f is proportional to r (see equation 2 on p. 71). Hence, 24. A possible explanation is that the severity of the symptoms f is proportional to m1/3, and so s is proportional to m2/3 (see the corresponds to the degree of structural disruption. Hence, substi- equation on p. 76). An 80-kd spherical protein undergoes sedi- tution of alanine for glycine might result in mild symptoms, but mentation 1.59 times as rapidly as a 40-kd spherical protein. substitution of the much larger tryptophan might prevent little or 9. The long hydrophobic tail on the SDS molecule (see p. 72) dis- no collagen triple-helix formation. rupts the hydrophobic interactions in the interior of the protein. 25. The energy barrier that must be crossed to go from the poly- The protein unfolds, with the hydrophobic R groups now interact- merized state to the hydrolyzed state is large even though the ing with SDS rather than with one another. reaction is thermodynamically favorable. 10. 50 kd. 26. Using the Henderson–Hasselbalch equation, we find the ratio 11. The protein may be modified. For instance, serine, threonine, 2 2 of alanine-COOH to alanine-COO at pH 7 to be 10 4. The ratio and tyrosine may have phosphoryl groups attached. 1 of alanine-NH2 to alanine-NH3 , determined in the same fashion, 12. A fluorescence-labeled derivative of a bacterial degradation 2 is 10 1. Thus, the ratio of neutral alanine to the zwitterionic species product (e.g., a formylmethionyl peptide) would bind to cells 2 2 2 is 10 4 3 10 1 5 10 5. containing the receptor of interest. 27. The assignment of absolute configuration requires the assign- 13.