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• Stryer Biochemistry
• Lehninger

Sep 22, 2017 - Lehninger principles of biochemistry 5th edition solutions manual.

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2608Tch03smS26-S43 2/1/08 11:45AM Page 26 ntt 102:WHQY028:Solutions Manual:Ch-03: chapter 3 Amino Acids, Peptides, and Proteins 1. Absolute Configuration of Citrulline The citrulline isolated from watermelons has the structure shown below. Is it a D- or L-amino acid? ⫹ C H NH3 C P CH2 (CH 2) 2 NH NH2 O COO⫺ Answer Rotating the structural representation 180⬚ in the plane of the page puts the most highly oxidized group—the carboxyl (OCOO⫺) group—at the top, in the same position as the OCHO group of glyceraldehyde in Figure 3–4. In this orientation, the a-amino group is on the left, and thus the absolute configuration of the citrulline is L.

Relationship between the Titration Curve and the Acid-Base Properties of Glycine A 100 mL solution of 0.1 M glycine at pH 1.72 was titrated with 2 M NaOH solution. The pH was monitored and the results were plotted as shown in the following graph. The key points in the titration are designated I to V.

For each of the statements (a) to (o), identify the appropriate key point in the titration and justify your choice. 12 11.30 10 (V) 9.60 (IV) 8 pH 5.97 6 (III) 4 2.34 (II) 2 0 (I) 0.5 1.0 1.5 2.0 ⫺ OH (equivalents) Note: before considering statements (a) through (o), refer to Figure 3–10. The three species involved in the titration of glycine can be considered in terms of a useful physical analogy. Each ionic species can be viewed as a different floor of a building, each with a different net charge: S-26 2608Tch03smS26-S43 2/1/08 11:45AM Page 27 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins ⫹ ⫹ H3NOCH2OCOOH H3NOCH2OCOO⫺ H2NOCH2OCOO⫺ S-27 ⫹1 0 (zwitterion) ⫺1 The floors are connected by steep stairways, and each stairway has a landing halfway between the floors.

A titration curve traces the path one would follow between the different floors as the pH changes in response to added OH⫺. Recall that the pKa of an acid (on a halfway landing) represents the pH at which half of the acid is deprotonated. The isoelectric point (pI) is the pH at which the average net charge is zero. Now you are ready to consider statements (a) through (o). (a) Glycine is present predominantly as the species ⫹H3NOCH2OCOOH. 1 (b) The average net charge of glycine is ⫹ ᎏ. 2 (c) Half of the amino groups are ionized.

(d) The pH is equal to the pKa of the carboxyl group. (e) The pH is equal to the pKa of the protonated amino group. (f) Glycine has its maximum buffering capacity. (g) The average net charge of glycine is zero. (h) The carboxyl group has been completely titrated (first equivalence point). (i) Glycine is completely titrated (second equivalence point).

Stryer Biochemistry

(j) The predominant species is ⫹H3NOCH2OCOO⫺. (k) The average net charge of glycine is ⫺1.

(l) Glycine is present predominantly as a 50:50 mixture of ⫹H3NOCH2OCOOH and ⫹ H3NOCH2OCOO⫺. (m) This is the isoelectric point. (n) This is the end of the titration.

(o) These are the worst pH regions for buffering power. Answer (a) I; maximum protonation occurs at the lowest pH (the highest H⫹). (b) II; at the first pKa, or pK1 (2.34), half of the protons are removed from the a-carboxyl 1 group (i.e., it is half deprotonated), changing its charge from 0 to ⫺ ᎏ. The average net 2 1 1 charge of glycine is (⫺ ᎏ ) ⫹ 1 ⫽ ᎏ. 2 2 (c) IV; the a-amino group is half-deprotonated at its pKa, or pK2 (9.60). (d) II; from the Henderson-Hasselbalch equation, pH = pKa + log (A⫺/HA). If A⫺/HA ⫽ 1, or A⫺ ⫽ HA, then pH ⫽ pKa.

(Recall that log 1 ⫽ 0.) (e) IV; see answers (c) and (d). (f) II and IV; in the pKa regions, acid donates protons to or base abstracts protons from glycine, with minimal pH changes.

(g) III; this occurs at the isoelectric point; pI ⫽ (pK1 ⫹ pK2)/2 ⫽ (2.34 ⫹ 9.60)/2 ⫽ 5.97. (h) III; the pH at which 1.0 equivalent of OH⫺ has been added, pH 5.97 (3.6 pH units away from either pKa). (i) V; pH 11.3 (1.7 pH units above pK2). (j) III; at pI (5.97) the carboxyl group is fully negatively charged (deprotonated) and the amino group is fully positively charged (protonated). (k) V; both groups are fully deprotonated, with a neutral amino group and a negatively charged carboxyl group (net charge ⫽ ⫺1). (l) II; the carboxyl group is half ionized at pH ⫽ pK1.

2608Tch03smS26-S43 S-28 2/1/08 11:45AM Page 28 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins (m) III; see answers (g) and (j). (n) V; glycine is fully titrated after 2.0 equivalents of OH⫺ have been added.

(o) I, III, and V; each is several pH units removed from either pKa, where the best pH buffering action occurs. How Much Alanine Is Present as the Completely Uncharged Species? At a pH equal to the isoelectric point of alanine, the net charge on alanine is zero. Two structures can be drawn that have a net charge of zero, but the predominant form of alanine at its pI is zwitterionic. CH3 ⫹ H3N C H CH3 O H2N C O⫺ Zwitterionic C H O C OH Uncharged (a) Why is alanine predominantly zwitterionic rather than completely uncharged at its pI? (b) What fraction of alanine is in the completely uncharged form at its pI?

Justify your assumptions. Answer (a) The pI of alanine is well above the pKa of the a-carboxyl group and well below the pKa of the a-amino group. Hence, at pH ⫽ pI, both groups are present predominantly in their charged (ionized) forms. (b) From Table 3–1, the pI of alanine is 6.01, midway between the two pKa values 2.34 and 9.69. From the Henderson-Hasselbalch equation, pH ⫺ pKa ⫽ log (A⫺/HA). For the carboxyl group: A⫺ log ᎏ ⫽ 6.01 ⫺ 2.34 ⫽ 3.67 HA HA 1 ᎏ ⫽ 10⫺3.67 ⫽ ᎏᎏ A⫺ 4.68 ⫻ 103 That is, one molecule in 4,680 is still in the form OCOOH. Similarly, at pH ⫽ pI, one molecule in 4,680 is in the form ONH2.

Thus, the fraction of molecules with both groups uncharged (OCOOH and ONH2) is 1 in 4,680 ⫻ 4,680, or 1 in 2.19 ⫻ 107. Ionization State of Histidine Each ionizable group of an amino acid can exist in one of two states, charged or neutral. The electric charge on the functional group is determined by the relationship between its pKa and the pH of the solution.

Lehninger

This relationship is described by the Henderson-Hasselbalch equation. (a) Histidine has three ionizable functional groups. Write the equilibrium equations for its three ionizations and assign the proper pKa for each ionization.

Draw the structure of histidine in each ionization state. What is the net charge on the histidine molecule in each ionization state?

(b) Draw the structures of the predominant ionization state of histidine at pH 1, 4, 8, and 12. Note that the ionization state can be approximated by treating each ionizable group independently. (c) What is the net charge of histidine at pH 1, 4, 8, and 12? For each pH, will histidine migrate toward the anode (⫹) or cathode (⫺) when placed in an electric field?

2608Tch03smS26-S43 2/2/08 7:25AM Page 29 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins Answer (a) COOH HN ⫹ CH2 ⫹ CH NH3 pK1 ⫽ 1.82 N H H⫹ 1 COO⫺ HN CH2 ⫹ ⫹ CH NH3 pKR ⫽ 6.0 N H H⫺ 2 COO⫺ HN CH2 ⫹ CH NH3 pK2 ⫽ 9.17 N H⫹ 3 COO⫺ HN CH2 CH NH2 N 4 We start with the most highly protonated species of histidine (structure 1, found at the most acidic pH). The pKa values are from Table 3–1. As base is added, the group with the lowest pKa loses its proton first, followed by the group with the next lowest pKa, then the group with the highest pKa. (In the following table, R ⫽ OCH2–imidazole.) Structure Net charge 1 ⫹H3NOCH(RH⫹)OCOOH ⫹2 2 ⫹H3NOCH(RH⫹)OCOO⫺ ⫹1 3 ⫹H3NOCH(R)OCOO⫺ 0 4 H2NOCH(R)OCOO⫺ ⫺1 (b) and (c) See the structures in (a). PH Structure Net charge Migrates toward: 1 1 ⫹2 Cathode 4 2 ⫹1 Cathode 8 3 0 12 4 ⫺1 Does not migrate Anode S-29 2608Tch03smS26-S43 S-30 2/1/08 11:45AM Page 30 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins 5. Separation of Amino Acids by Ion-Exchange Chromatography Mixtures of amino acids can be analyzed by first separating the mixture into its components through ion-exchange chromatography.

Amino acids placed on a cation-exchange resin (see Fig. 3–17a) containing sulfonate (OSO⫺ 3 ) groups flow down the column at different rates because of two factors that influence their movement: (1) ionic attraction between the sulfonate residues on the column and positively charged functional groups on the amino acids, and (2) hydrophobic interactions between amino acid side chains and the strongly hydrophobic backbone of the polystyrene resin. For each pair of amino acids listed, determine which will be eluted first from an ion-exchange column by a pH 7.0 buffer. (a) Asp and Lys (b) Arg and Met (c) Glu and Val (d) Gly and Leu (e) Ser and Ala Answer See Table 3–1 for pKa values for the amino acid side chains. At pH ⬍ pI, an amino acid has a net positive charge; at pH ⬎ pI, it has a net negative charge.

For any pair of amino acids, the more negatively charged one passes through the sulfonated resin faster. For two neutral amino acids, the less polar one passes through more slowly because of its stronger hydrophobic interactions with the polystyrene.

PI values Net charge (pH 7) Elution order Basis for separation ⫺1, ⫹1 Asp, Lys Charge 0 Met, Arg Charge (a) Asp, Lys 2.77, 9.74 (b) Arg, Met 10.76, 5.74 ⫹1, (c) Glu, Val 3.22, 5.97 ⫺1, 0 Glu, Val Charge (d) Gly, Leu 5.97, 5.98 0, 0 Gly, Leu Polarity (e) Ser, Ala 5.68, 6.01 0, 0 Ser, Ala Polarity 6. Naming the Stereoisomers of Isoleucine The structure of the amino acid isoleucine is COO⫺ H H3N C H H C CH3 CH2 CH3 (a) How many chiral centers does it have? (b) How many optical isomers? (c) Draw perspective formulas for all the optical isomers of isoleucine.

Answer (a) Two; at C-2 and C-3 (the a and b carbons). (b) Four; the two chiral centers permit four possible diastereoisomers: (S,S), (S,R), (R,R), and (R,S). 2608Tch03smS26-S43 2/1/08 11:45AM Page 31 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins (c) COO⫺ COO⫺ ⫹ COO⫺ ⫹ COO⫺ ⫹ ⫹ H3N C H H3N C H H C NH3 H C NH3 H C CH3 H 3C C H H C CH3 H 3C C H CH2 CH2 CH2 CH2 CH3 CH3 CH3 CH3 7. Comparing the pKa Values of Alanine and Polyalanine The titration curve of alanine shows the ionization of two functional groups with pKa values of 2.34 and 9.69, corresponding to the ionization of the carboxyl and the protonated amino groups, respectively. The titration of di-, tri-, and larger oligopeptides of alanine also shows the ionization of only two functional groups, although the experimental pKa values are different. The trend in pKa values is summarized in the table. Amino acid or peptide pK1 pK2 Ala 2.34 9.69 Ala–Ala 3.12 8.30 Ala–Ala–Ala 3.39 8.03 Ala–(Ala)n –Ala, n ⱖ 4 3.42 7.94 (a) Draw the structure of Ala–Ala–Ala.

Identify the functional groups associated with pK1 and pK2. (b) Why does the value of pK1 increase with each additional Ala residue in the Ala oligopeptide? (c) Why does the value of pK2 decrease with each additional Ala residue in the Ala oligopeptide? Answer (a) The structure at pH 7 is: O O O ⫹ H3N CH CH3 pK2 ⫽ 8.03 C N CH H CH3 C N CH C H CH3 O⫺ pK1 ⫽ 3.39 Note that only the amino- and carboxyl-terminal groups ionize.

(b) As the length of poly(Ala) increases, the two terminal groups move farther apart, separated by an intervening sequence with an “insulating” nonpolar structure. The carboxyl group becomes a weaker acid, as reflected in its higher pKa, because the electrostatic repulsion between the carboxyl proton and the positive charge on the NH ⫹ 3 group diminishes as the groups become more distant. (c) The negative charge on the terminal carboxyl group has a stabilizing effect on the positively charged (protonated) terminal amino group.

With increasing numbers of intervening Ala residues, this stabilizing effect is diminished and the NH ⫹ 3 group loses its S-31 2608Tch03smS26-S43 S-32 2/1/08 11:45AM Page 32 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins proton more easily. The lower pK2 indicates that the terminal amino group has become a weaker base (stronger acid). The intramolecular effects of the amide (peptide bond) linkages keep pKa values lower than they would be for an alkyl-substituted amine. The Size of Proteins What is the approximate molecular weight of a protein with 682 amino acid residues in a single polypeptide chain? Answer Assuming that the average Mr per residue is 110 (corrected for loss of water in formation of the peptide bond), a protein containing 682 residues has an Mr of approximately 682 ⫻ 110 ⫽ 75,000.

The Number of Tryptophan Residues in Bovine Serum Albumin A quantitative amino acid analysis reveals that bovine serum albumin (BSA) contains 0.58% tryptophan (Mr 204) by weight. (a) Calculate the minimum molecular weight of BSA (i.e., assuming there is only one tryptophan residue per protein molecule).

(b) Gel filtration of BSA gives a molecular weight estimate of 70,000. How many tryptophan residues are present in a molecule of serum albumin? Answer (a) The Mr of a Trp residue must be adjusted to account for the removal of water during peptide bond formation: Mr ⫽ 204 ⫺ 18 ⫽ 186. The molecular weight of BSA can be calculated using the following proportionality, where n is the number of Trp residues in the protein: wt Trp n(Mr Trp) ᎏ ⫽ ᎏᎏ wt BSA Mr BSA 0.58 g n(186) ᎏ ⫽ ᎏ 100 g Mr BSA A minimum molecular weight can be found by assuming only one Trp residue per BSA molecule (n ⫽ 1).

(100 g)(186)(1) ᎏᎏ ⫽ 32,000 0.58g (b) Given that the Mr of BSA is approximately 70,000, BSA has 70,000/32,000 ⫽ 2.2, or 2 Trp residues per molecule. (The remainder from this division suggests that the estimate of Mr 70,000 for BSA is somewhat high.) 10. Subunit Composition of a Protein A protein has a molecular mass of 400 kDa when measured by gel filtration. When subjected to gel electrophoresis in the presence of sodium dodecyl sulfate (SDS), the protein gives three bands with molecular masses of 180, 160, and 60 kDa. When electrophoresis is carried out in the presence of SDS and dithiothreitol, three bands are again formed, this time with molecular masses of 160, 90, and 60 kDa.

Determine the subunit composition of the protein. Answer The protein has four subunits, with molecular masses of 160, 90, 90, and 60 kDa.

The two 90 kDa subunits (possibly identical) are linked by one or more disulfide bonds. Net Electric Charge of Peptides A peptide has the sequence Glu–His–Trp–Ser–Gly–Leu–Arg–Pro–Gly (a) What is the net charge of the molecule at pH 3, 8, and 11? (Use pKa values for side chains and terminal amino and carboxyl groups as given in Table 3–1.) (b) Estimate the pI for this peptide.

2608Tch03smS26-S43 2/1/08 11:45AM Page 33 ntt 102:WHQY028:Solutions Manual:Ch-03: Chapter 3 Amino Acids, Peptides, and Proteins S-33 Answer (a) When pH ⬎ pKa, ionizing groups lose their protons. The pKa values of importance here are those of the amino-terminal (9.67) and carboxyl-terminal (2.34) groups and those of the R groups of the Glu (4.25), His (6.00), and Arg (12.48) residues.

Note: here we are using the available pKa values: those for the free amino acids as given in Table 3–1. As demonstrated in Problem 7, however, the pKa values of the a-amino and a-carboxyl groups shift when the amino acid is at the amino or carboxyl terminus, respectively, of a peptide, and this would affect the net charge and the pI of the peptide. PH ⴙ H3N Glu His Arg COOⴚ Net charge 3 ⫹1 0 ⫹1 ⫹1 ⫺1 ⫹2 8 ⫹1 ⫺1 0 ⫹1 ⫺1 0 11 0 ⫺1 0 ⫹1 ⫺1 ⫺1 (b) Two different methods can be used to estimate pI. Find the two ionizable groups with pKa values that “straddle” the point at which net peptide charge ⫽ 0 (here, two groups that ionize near pH 8): the amino-terminal a-amino group of Glu and the His imidazole group. Thus, we can estimate 9.67 ⫹ 6.00 pI = ᎏᎏ ⫽ 7.8 2 Alternatively, plot the calculated net charges as a function of pH, and determine graphically the pH at which the net charge is zero on the vertical axis.

More data points are needed to use this method accurately. Note: although at any instant an individual amino acid molecule will have an integral charge, it is possible for a population of amino acid molecules in solution to have a fractional charge. For example, at pH 1.0 glycine exists entirely as the form ⫹ H3NOCH2OCOOH with a net positive charge of 1.0. However, at pH 2.34, where there is an equal mixture of ⫹H3NOCH2OCOOH and ⫹H3NOCH2OCOO⫺, the average or net charge on the population of glycine molecules is 0.5 (see the discussion on pp. You can use the Henderson-Hasselbalch equation to calculate the exact ratio of charged and uncharged species at equilibrium at various pH values.

Isoelectric Point of Pepsin Pepsin is the name given to a mix of several digestive enzymes secreted (as larger precursor proteins) by glands that line the stomach. These glands also secrete hydrochloric acid, which dissolves the particulate matter in food, allowing pepsin to enzymatically cleave individual prote.