solutions ch 03

S-26

Amino Acids, Peptides,and Proteins

chapter

31. Absolute Configuration of Citrulline The citrulline isolated from watermelons has the structure

shown below. Is it a D- or L-amino acid? Explain.

Answer Rotating the structural representation 180� in the plane of the page puts the mosthighly oxidized group—the carboxyl (OCOO�) group—at the top, in the same position as theOCHO group of glyceraldehyde in Figure 3–4. In this orientation, the a-amino group is on theleft, and thus the absolute configuration of the citrulline is L.

2. Relationship between the Titration Curve and the Acid-Base Properties of Glycine A 100 mLsolution of 0.1 M glycine at pH 1.72 was titrated with 2 M NaOH solution. The pH was monitored and theresults 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 yourchoice.

Note: before considering statements (a) through (o), refer to Figure 3–10. The three species involvedin the titration of glycine can be considered in terms of a useful physical analogy. Each ionic species canbe viewed as a different floor of a building, each with a different net charge:

12

2

4

6

8

0

11.30

0.5

OH� (equivalents)

pH

1.0 1.5 2.0

(V)

9.60(IV)

(III)

2.34

(I)

(II)

5.97

10

C C

O

)H (CH NH2 NH22 2

P

H C N�

H3

COO�

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Chapter 3 Amino Acids, Peptides, and Proteins S-27

�H3NOCH2OCOOH �1�H3NOCH2OCOO� 0 (zwitterion)

H2NOCH2OCOO� �1

The floors are connected by steep stairways, and each stairway has a landing halfway between thefloors. A titration curve traces the path one would follow between the different floors as the pHchanges in response to added OH�. Recall that the pKa of an acid (on a halfway landing) representsthe pH at which half of the acid is deprotonated. The isoelectric point (pI) is the pH at which the aver-age net charge is zero. Now you are ready to consider statements (a) through (o).(a) Glycine is present predominantly as the species �H3NOCH2OCOOH.

(b) The average net charge of glycine is � .

(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).(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

group (i.e., it is half deprotonated), changing its charge from 0 to � . The average net

charge of glycine is (� ) � 1 � .

(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.

1�2

1�2

1�2

1�2

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(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 buffer-

ing action occurs.

3. How Much Alanine Is Present as the Completely Uncharged Species? At a pH equal to the iso-electric point of alanine, the net charge on alanine is zero. Two structures can be drawn that have anet charge of zero, but the predominant form of alanine at its pI is zwitterionic.

(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 theircharged (ionized) forms.

(b) From Table 3–1, the pI of alanine is 6.01, midway between the two pKa values 2.34 and9.69. From the Henderson-Hasselbalch equation, pH � pKa � log ([A�]/[HA]). For thecarboxyl group:

log � 6.01 � 2.34 � 3.67

� 10�3.67 �

That is, one molecule in 4,680 is still in the form OCOOH. Similarly, at pH � pI, one moleculein 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.

4. 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 be-tween its pKa and the pH of the solution. This relationship is described by the Henderson-Hasselbalchequation.(a) Histidine has three ionizable functional groups. Write the equilibrium equations for its three ion-

izations and assign the proper pKa for each ionization. Draw the structure of histidine in eachionization 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. Notethat 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 to-ward the anode (�) or cathode (�) when placed in an electric field?

1��4.68 � 103

[HA]�[A�]

[A�]�[HA]

S-28 Chapter 3 Amino Acids, Peptides, and Proteins

�C

CH3

H3N

H

CO

O�

Zwitterionic Uncharged

C

CH3

H2N

H

CO

OH



Answer(a)

We start with the most highly protonated species of histidine (structure 1, found at themost acidic pH). The pKa values are from Table 3–1. As base is added, the group withthe 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.)

HNCH2

1

NH3CH

COOHpK1 � 1.82

H�NH

�

HN

NH

�

HN

N

HN

N

�

CH2

2

NH3CH

COO�

pKR � 6.0

H�

�

CH2

3

NH3CH

COO�

pK2 � 9.17

H�

�

CH2

4

NH2CH

COO�

Structure Net charge

1 �H3NOCH(RH�)OCOOH �2

2 �H3NOCH(RH�)OCOO� �1

3 �H3NOCH(R)OCOO� 0

4 H2NOCH(R)OCOO� �1

pH Structure Net charge Migrates toward:

1 1 �2 Cathode

4 2 �1 Cathode

8 3 0 Does not migrate

12 4 �1 Anode

(b) and (c) See the structures in (a).


5. Separation of Amino Acids by Ion-Exchange Chromatography Mixtures of amino acids can beanalyzed 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 (OSO3

�) groupsflow 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 functionalgroups on the amino acids, and (2) hydrophobic interactions between amino acid side chains and thestrongly hydrophobic backbone of the polystyrene resin. For each pair of amino acids listed, determinewhich 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 aminoacid has a net positive charge; at pH pI, it has a net negative charge. For any pair of aminoacids, the more negatively charged one passes through the sulfonated resin faster. For twoneutral amino acids, the less polar one passes through more slowly because of its strongerhydrophobic interactions with the polystyrene.

6. Naming the Stereoisomers of Isoleucine The structure of the amino acid isoleucine is


pI values Net charge (pH 7) Elution order Basis for separation

(a) Asp, Lys 2.77, 9.74 �1, �1 Asp, Lys Charge

(b) Arg, Met 10.76, 5.74 �1, 0 Met, Arg Charge

(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

H C

H3NH

C

COO�

H

CH2

CH3

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).



(c)

7. Comparing the pKa Values of Alanine and Polyalanine The titration curve of alanine shows theionization 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 largeroligopeptides of alanine also shows the ionization of only two functional groups, although the experi-mental pKa values are different. The trend in pKa values is summarized in the table.

(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:

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, sepa-

rated by an intervening sequence with an “insulating” nonpolar structure. The carboxylgroup becomes a weaker acid, as reflected in its higher pKa, because the electrostaticrepulsion between the carboxyl proton and the positive charge on the NH3

� groupdiminishes as the groups become more distant.

(c) The negative charge on the terminal carboxyl group has a stabilizing effect on thepositively charged (protonated) terminal amino group. With increasing numbers ofintervening Ala residues, this stabilizing effect is diminished and the NH3

� group loses its

COO�

�

CH2

H3N H

CH3

C

CH CH3

COO�

�

CH2

NH3H

CH3

C

CH CH3

COO�

�

CH2

NH3H

CH3

C

C HH3CH3C

COO�

�

CH2

H3N H

CH3

C

C H

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 4 3.42 7.94

CH3

pK2 � 8.03 pK1 � 3.39CH3H

�

H3N CH

CH3

C

O

N

H

CH C N CH

O

C O�

O


proton more easily. The lower pK2 indicates that the terminal amino group has become aweaker 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.

8. The Size of Proteins What is the approximate molecular weight of a protein with 682 amino acidresidues in a single polypeptide chain?

Answer Assuming that the average Mr per residue is 110 (corrected for loss of water in for-mation of the peptide bond), a protein containing 682 residues has an Mr of approximately682 � 110 � 75,000.

9. The Number of Tryptophan Residues in Bovine Serum Albumin A quantitative amino acidanalysis 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 cal-culated using the following proportionality, where n is the number of Trp residues in theprotein:

�

�

A minimum molecular weight can be found by assuming only one Trp residue per BSAmolecule (n � 1).

� 32,000

(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 gelfiltration. When subjected to gel electrophoresis in the presence of sodium dodecyl sulfate (SDS), the pro-tein gives three bands with molecular masses of 180, 160, and 60 kDa. When electrophoresis is carried outin the presence of SDS and dithiothreitol, three bands are again formed, this time with molecular massesof 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. Thetwo 90 kDa subunits (possibly identical) are linked by one or more disulfide bonds.

11. 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 andterminal amino and carboxyl groups as given in Table 3–1.)

(b) Estimate the pI for this peptide.

(100 g)(186)(1)��

0.58g

n(186)�Mr BSA

0.58 g�100 g

n(Mr Trp)��

Mr BSAwt Trp�wt BSA




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 ofthe R groups of the Glu (4.25), His (6.00), and Arg (12.48) residues. Note: here we areusing the available pKa values: those for the free amino acids as given in Table 3–1. Asdemonstrated in Problem 7, however, the pKa values of the a-amino and a-carboxylgroups shift when the amino acid is at the amino or carboxyl terminus, respectively, of apeptide, and this would affect the net charge and the pI of the peptide.

(b) Two different methods can be used to estimate pI. Find the two ionizable groups withpKa values that “straddle” the point at which net peptide charge � 0 (here, two groupsthat ionize near pH 8): the amino-terminal a-amino group of Glu and the His imidazolegroup. Thus, we can estimate

pI = � 7.8

Alternatively, plot the calculated net charges as a function of pH, and determine graphi-cally the pH at which the net charge is zero on the vertical axis. More data points areneeded to use this method accurately.

Note: although at any instant an individual amino acid molecule will have anintegral 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 thereis an equal mixture of �H3NOCH2OCOOH and �H3NOCH2OCOO�, the average or netcharge on the population of glycine molecules is 0.5 (see the discussion on pp. 80–81).You can use the Henderson-Hasselbalch equation to calculate the exact ratio of chargedand uncharged species at equilibrium at various pH values.

12. 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 hydrochloricacid, which dissolves the particulate matter in food, allowing pepsin to enzymatically cleave individualprotein molecules. The resulting mixture of food, HCl, and digestive enzymes is known as chyme andhas a pH near 1.5. What pI would you predict for the pepsin proteins? What functional groups must bepresent to confer this pI on pepsin? Which amino acids in the proteins would contribute such groups?

Answer Pepsin proteins have a relatively low pI (near the pH of gastric juice) in order toremain soluble and thus functional in the stomach. (Pepsin—the mixture of enzymes—has apI of ~1.) As pH increases, pepsins acquire a net charge and undergo ionic interactions withoppositely charged molecules (such as dissolved salts), causing the pepsin proteins toprecipitate. Pepsin is active only in the stomach. In the relatively high pH of the intestine,pepsin proteins precipitate and become inactive.

A low pI requires large numbers of negatively charged (low pKa) groups. These arecontributed by the carboxylate groups of Asp and Glu residues.

9.67 � 6.00��

2

pH �H3N Glu His Arg COO� Net charge

3 �1 0 �1 �1 �1 �2

8 �1 �1 0 �1 �1 0

11 0 �1 0 �1 �1 �1


13. The Isoelectric Point of Histones Histones are proteins found in eukaryotic cell nuclei, tightlybound to DNA, which has many phosphate groups. The pI of histones is very high, about 10.8. Whatamino acid residues must be present in relatively large numbers in histones? In what way do theseresidues contribute to the strong binding of histones to DNA?

Answer Large numbers of positively charged (high pKa) groups in a protein give it a high pI.In histones, the positively charged R groups of Lys, Arg, and His residues interact stronglywith the negatively charged phosphate groups of DNA through ionic interactions.

14. Solubility of Polypeptides One method for separating polypeptides makes use of their differentsolubilities. The solubility of large polypeptides in water depends upon the relative polarity of their Rgroups, particularly on the number of ionized groups: the more ionized groups there are, the more solublethe polypeptide. Which of each pair of the polypeptides that follow is more soluble at the indicated pH?(a) (Gly)20 or (Glu)20 at pH 7.0(b) (Lys–Ala)3 or (Phe–Met)3 at pH 7.0(c) (Ala–Ser–Gly)5 or (Asn–Ser–His)5 at pH 6.0(d) (Ala–Asp–Gly)5 or (Asn–Ser–His)5 at pH 3.0

Answer(a) (Glu)20; it is highly negatively charged (polar) at pH 7. (Gly)20 is uncharged except for

the amino- and carboxyl-terminal groups.(b) (Lys–Ala)3; this is highly positively charged (polar) at pH 7. (Phe–Met)3 is much less

polar and hence less soluble.(c) (Asn–Ser–His)5; both polymers have polar Ser side chains, but (Asn–Ser–His)5 also has

the polar Asn side chains and partially protonated His side chains.(d) (Asn–Ser–His)5; at pH 3, the carboxylate groups of Asp residues are partially protonated

and neutral, whereas the imidazole groups of His residues are fully protonated and posi-tively charged.

15. Purification of an Enzyme A biochemist discovers and purifies a new enzyme, generating the purifi-cation table below.

(a) From the information given in the table, calculate the specific activity of the enzyme after eachpurification procedure.

(b) Which of the purification procedures used for this enzyme is most effective (i.e., gives the greatestrelative increase in purity)?

(c) Which of the purification procedures is least effective?(d) Is there any indication based on the results shown in the table that the enzyme after step 6 is

now pure? What else could be done to estimate the purity of the enzyme preparation?


Procedure Total protein (mg) Activity (units)

1. Crude extract 20,000 4,000,000

2. Precipitation (salt) 5,000 3,000,000

3. Precipitation (pH) 4,000 1,000,000

4. Ion-exchange chromatography 200 800,000

5. Affinity chromatography 50 750,000

6. Size-exclusion chromatography 45 675,000



Answer(a) From the percentage recovery of activity (units), we calculate percentage yield and

specific activity (units/mg).

(b) Step 4, ion-exchange chromatography; this gives the greatest increase in specific activity(an index of purity and degree of increase in purification).

(c) Step 3, pH precipitation; two-thirds of the total activity from the previous step was losthere.

(d) Yes. The specific activity did not increase further after step 5. SDS polyacrylamide gelelectrophoresis is an excellent, standard way of checking homogeneity and purity.

16. Dialysis A purified protein is in a Hepes (N-(2-hydroxyethyl)piperazine-N�-(2-ethanesulfonic acid)) bufferat pH 7 with 500 mM NaCl. A sample (1 mL) of the protein solution is placed in a tube made of dialysis mem-brane and dialyzed against 1 L of the same Hepes buffer with 0 mM NaCl. Small molecules and ions (such asNa�, Cl�, and Hepes) can diffuse across the dialysis membrane, but the protein cannot.(a) Once the dialysis has come to equilibrium, what is the concentration of NaCl in the protein sam-

ple? Assume no volume changes occur in the sample during the dialysis.(b) If the original 1 mL sample were dialyzed twice, successively, against 100 mL of the same Hepes

buffer with 0 mM NaCl, what would be the final NaCl concentration in the sample?

Answer(a) [NaCl] � 0.5 mM

(b) [NaCl] � 0.05 mM.

17. Peptide Purification At pH 7.0, in what order would the following three peptides be eluted from a col-umn filled with a cation-exchange polymer? Their amino acid compositions are:Protein A: Ala 10%, Glu 5%, Ser 5%, Leu 10%, Arg 10%, His 5%, Ile 10%, Phe 5%, Tyr 5%, Lys 10%, Gly 10%, Pro 5%, and Trp 10%.Protein B: Ala 5%, Val 5%, Gly 10%, Asp 5%, Leu 5%, Arg 5%, Ile 5%, Phe 5%, Tyr 5%, Lys 5%, Trp 5%, Ser 5%, Thr 5%, Glu 5%, Asn 5%, Pro 10%, Met 5%, and Cys 5%.Protein C: Ala 10%, Glu 10%, Gly 5%, Leu 5%, Asp 10%, Arg 5%, Met 5%, Cys 5%, Tyr 5%, Phe 5%, His 5%, Val 5%, Pro 5%, Thr 5%, Ser 5%, Asn 5%, and Gln 5%.

Answer Protein C has a net negative charge because there are more Glu and Asp residues thanLys, Arg, and His residues. Protein A has a net positive charge. Protein B has no net charge atneutral pH. A cation-exchange column has a negatively charged polymer, so protein C interactsmost weakly with the column and is eluted first, followed by B, then A.

Procedure Protein (mg) Activity (units) % Yield Specific activity (units/mg) Purification factor (overall)

1 20,000 4,000,000 (100) 200 (1.0)

2 5,000 3,000,000 75 600 � 3.0

3 4,000 1,000,000 25 250 � 1.25

4 200 800,000 20 4,000 � 20

5 50 750,000 19 15,000 � 75

6 45 675,000 17 15,000 � 75


18. Sequence Determination of the Brain Peptide Leucine Enkephalin A group of peptides that in-fluence nerve transmission in certain parts of the brain has been isolated from normal brain tissue.These peptides are known as opioids because they bind to specific receptors that also bind opiatedrugs, such as morphine and naloxone. Opioids thus mimic some of the properties of opiates. Some re-searchers consider these peptides to be the brain’s own painkillers. Using the information below, deter-mine the amino acid sequence of the opioid leucine enkephalin. Explain how your structure is consis-tent with each piece of information.(a) Complete hydrolysis by 6 M HCl at 110 �C followed by amino acid analysis indicated the presence

of Gly, Leu, Phe, and Tyr in a 2:1:1:1 molar ratio.(b) Treatment of the peptide with 1-fluoro-2,4-dinitrobenzene followed by complete hydrolysis and

chromatography indicated the presence of the 2,4-dinitrophenyl derivative of tyrosine. No freetyrosine could be found.

(c) Complete digestion of the peptide with chymotrypsin followed by chromatography yielded freetyrosine and leucine, plus a tripeptide containing Phe and Gly in a 1:2 ratio.

Answer(a) The empirical composition is (2 Gly, Leu, Phe, Tyr)n.(b) Tyr is the amino-terminal residue, and there are no other Tyr residues, so n � 1 and the

sequence is Tyr–(2 Gly, Leu, Phe). (c) As shown in Table 3–7, chymotrypsin cleaves on the carboxyl side of aromatic residues

(Phe, Trp, and Tyr). The peptide has only two aromatic residues, Tyr at the aminoterminus and a Phe. Because there are three cleavage products, the Phe residue cannotbe at the carboxyl terminus. Rather, release of free leucine means that Leu must be atthe carboxyl terminus and must be on the carboxyl side of Phe in the peptide. Thus thesequence must be

Tyr–(2 Gly)–Phe–Leu � Tyr–Gly–Gly–Phe–Leu

19. Structure of a Peptide Antibiotic from Bacillus brevis Extracts from the bacterium Bacillus

brevis contain a peptide with antibiotic properties. This peptide forms complexes with metal ions andseems to disrupt ion transport across the cell membranes of other bacterial species, killing them. Thestructure of the peptide has been determined from the following observations.(a) Complete acid hydrolysis of the peptide followed by amino acid analysis yielded equimolar

amounts of Leu, Orn, Phe, Pro, and Val. Orn is ornithine, an amino acid not present in proteinsbut present in some peptides. It has the structure

(b) The molecular weight of the peptide was estimated as about 1,200.(c) The peptide failed to undergo hydrolysis when treated with the enzyme carboxypeptidase. This en-

zyme catalyzes the hydrolysis of the carboxyl-terminal residue of a polypeptide unless theresidue is Pro or, for some reason, does not contain a free carboxyl group.

(d) Treatment of the intact peptide with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysisand chromatography, yielded only free amino acids and the following derivative:

(Hint: note that the 2,4-dinitrophenyl derivative involves the amino group of a side chain ratherthan the a-amino group.)


NO2

CH2 CH2

�NH3

O2N COO�CH2NH C

H

CH2 CH2 CH2 C COO�H3N

H

�NH3

�



(e) Partial hydrolysis of the peptide followed by chromatographic separation and sequence analysisyielded the following di- and tripeptides (the amino-terminal amino acid is always at the left):

Leu–Phe Phe–Pro Orn–Leu Val–Orn

Val–Orn–Leu Phe–Pro–Val Pro–Val–Orn

Given the above information, deduce the amino acid sequence of the peptide antibiotic. Showyour reasoning. When you have arrived at a structure, demonstrate that it is consistent with each

experimental observation.

Answer The information obtained from each experiment is as follows.(a) The simplest empirical formula for the peptide is (Leu, Orn, Phe, Pro, Val)n.(b) Assuming an average residue Mr of 110, the minimum molecular weight for the peptide

is 550. Because 1,200/550 ≈ 2, the empirical formula is (Leu, Orn, Phe, Pro, Val)2.(c) Failure of carboxypeptidase to cleave the peptide could result from Pro at the carboxyl

terminus or the absence of a carboxyl-terminal residue—as in a cyclic peptide.(d) Failure of FDNB to derivatize an a-amino group indicates either the absence of a free

amino-terminal group or that Pro (an imino acid) is at the amino-terminal position. (Thederivative formed is 2,4 dinitrophenyl-�-ornithine.)

(e) The presence of Pro at an internal position in the peptide Phe–Pro–Val indicates that itis not at the amino or carboxyl terminus. The information from these experimentssuggests that the peptide is cyclic. The alignment of overlapping sequences is

Leu–Phe

Phe–Pro

Phe–Pro–Val

Pro–Val–Orn

Val–Orn

Orn–(Leu)

Thus, the peptide is a cyclic dimer of Leu–Phe–Pro–Val–Orn:

where the arrows indicate the –CO → NH–, or C → N, direction of the peptide bonds.This structure is consistent with all the data.

20. Efficiency in Peptide Sequencing A peptide with the primary structure Lys–Arg–Pro–Leu–Ile–Asp–Gly–Ala is sequenced by the Edman procedure. If each Edman cycle is 96% efficient, whatpercentage of the amino acids liberated in the fourth cycle will be leucine? Do the calculation a secondtime, but assume a 99% efficiency for each cycle.

Answer 88%, 97%. The formula for calculating the percentage of correct amino acid liber-ated after sequencing cycle n, given an efficiency x, is xn/x, or xn � 1. If the efficiency is 0.96,the fraction of correct amino acid liberated in the fourth cycle is (0.96)3 � 0.88. If the effi-ciency is 0.99, the fraction is (0.99)3 � 0.97.

LeuOrn Phe

Pro

Val

Val

Pro

OrnPheLeu


The n � 1 term can be explained by considering what happens in each cycle. For exam-ple, in the first cycle, a Lys residue is liberated from 96% of the ends, so that 96% of the ter-mini now have Arg and the remaining 4% still have Lys. However, 100% of the residues actu-ally removed in this cycle are Lys. In the second cycle, Arg is removed from 96% of the endsthat contain it, or from 0.96 � 0.96 � 0.92 � 92% of the ends. Lys is removed from 0.96 �0.04 � 0.04 of the ends. However, the fraction of liberated residues that are Arg is greaterthan 92% because only 96% of the ends had residues removed. Hence, the fraction of residuesliberated as Arg in the second cycle is 0.92/0.96 � 0.96 � 96%—and so forth.

21. Sequence Comparisons Proteins called molecular chaperones (described in Chapter 4) assist in theprocess of protein folding. One class of chaperone found in organisms from bacteria to mammals isheat shock protein 90 (Hsp90). All Hsp90 chaperones contain a 10 amino acid “signature sequence,”which allows for ready identification of these proteins in sequence databases. Two representations ofthis signature sequence are shown below.

(a) In this sequence, which amino acid residues are invariant (conserved across all species)?(b) At which position(s) are amino acids limited to those with positively charged side chains? For

each position, which amino acid is more commonly found?(c) At which positions are substitutions restricted to amino acids with negatively charged side

chains? For each position, which amino acid predominates?(d) There is one position that can be any amino acid, although one amino acid appears much more often

than any other. What position is this, and which amino acid appears most often?

Answer(a) Y (1), F (7), and R (9) (b) Positions 4 and 9; K (Lys) is more common at 4, R (Arg) is invariant at 9 (c) Positions 5 and 10; E (Glu) is more common at both positions (d) Position 2; S (Ser)

22. Biochemistry Protocols: Your First Protein Purification As the newest and least experiencedstudent in a biochemistry research lab, your first few weeks are spent washing glassware and labelingtest tubes. You then graduate to making buffers and stock solutions for use in various laboratory proce-dures. Finally, you are given responsibility for purifying a protein. It is citrate synthase (an enzyme of thecitric acid cycle, to be discussed in Chapter 16), which is located in the mitochondrial matrix. Followinga protocol for the purification, you proceed through the steps below. As you work, a more experiencedstudent questions you about the rationale for each procedure. Supply the answers. (Hint: see Chapter 2for information about osmolarity; see p. 7 for information on separation of organelles from cells.)(a) You pick up 20 kg of beef hearts from a nearby slaughterhouse (muscle cells are rich in mito-

chondria, which supply energy for muscle contraction). You transport the hearts on ice, and perform each step of the purification on ice or in a walk-in cold room. You homoge-nize the beef heart tissue in a high-speed blender in a medium containing 0.2 M sucrose, buffered to a pH of 7.2. Why do you use beef heart tissue, and in such large

quantity? What is the purpose of keeping the tissue cold and suspending it

in 0.2 M sucrose, at pH 7.2? What happens to the tissue when it is homogenized?


Y-x-[NQHD]-[KHR]-[DE]-[IVA]-F-[LM]-R-[ED].

4 5 6 7 8 9 1021N C

3

43

10

2Bit

s



(b) You subject the resulting heart homogenate, which is dense and opaque, to a series of differentialcentrifugation steps. What does this accomplish?

(c) You proceed with the purification using the supernatant fraction that contains mostly intact mito-chondria. Next, you osmotically lyse the mitochondria. The lysate, which is less dense than thehomogenate, but still opaque, consists primarily of mitochondrial membranes and internal mito-chondrial contents. To this lysate you add ammonium sulfate, a highly soluble salt, to a specificconcentration. You centrifuge the solution, decant the supernatant, and discard the pellet. To thesupernatant, which is clearer than the lysate, you add more ammonium sulfate. Once again, youcentrifuge the sample, but this time you save the pellet because it contains the citrate synthase.What is the rationale for the two-step addition of the salt?

(d) You solubilize the ammonium sulfate pellet containing the mitochondrial proteins and dialyze itovernight against large volumes of buffered (pH 7.2) solution. Why isn’t ammonium sulfate in-

cluded in the dialysis buffer? Why do you use the buffer solution instead of water?

(e) You run the dialyzed solution over a size-exclusion chromatographic column. Following the pro-tocol, you collect the first protein fraction that exits the column and discard the fractions thatelute from the column later. You detect the protein by measuring UV absorbance (at 280 nm) inthe fractions. What does the instruction to collect the first fraction tell you about the pro-

tein? Why is UV absorbance at 280 nm a good way to monitor for the presence of protein

in the eluted fractions?

(f) You place the fraction collected in (e) on a cation-exchange chromatographic column. After dis-carding the initial solution that exits the column (the flowthrough), you add a washing solutionof higher pH to the column and collect the protein fraction that immediately elutes. Explain

what you are doing.(g) You run a small sample of your fraction, now very reduced in volume and quite clear (though

tinged pink), on an isoelectric focusing gel. When stained, the gel shows three sharp bands. Ac-cording to the protocol, the citrate synthase is the protein with a pI of 5.6, but you decide to doone more assay of the protein’s purity. You cut out the pI 5.6 band and subject it to SDS poly-acrylamide gel electrophoresis. The protein resolves as a single band. Why were you uncon-

vinced of the purity of the “single” protein band on your isoelectric focusing gel? What did

the results of the SDS gel tell you? Why is it important to do the SDS gel electrophoresis af-ter the isoelectric focusing?

Answer(a) Why do you use beef heart tissue, and why do you need so much of it? The protein

you are to isolate and purify (citrate synthase [CS]) is found in mitochondria, which areabundant in cells with high metabolic activity such as heart muscle cells. Beef hearts arerelatively cheap and easy to get at the local slaughterhouse. You begin with a largeamount of tissue because cells contain thousands of different proteins, and no singleprotein is present in high concentration—that is, the specific activity of CS is low. Topurify a significant quantity, you must start with a large excess of tissue.

Why do you need to keep the tissues cold? The cold temperatures inhibit theaction of lysosomal enzymes that would destroy the sample.

Why is the tissue suspended in 0.2 M sucrose, at pH 7.2? Sucrose is used in thehomogenization buffer to create a medium that is isotonic with the organelles. Thisprevents diffusion of water into the organelles, causing them to swell, burst, and spilltheir contents. A pH of 7.2 helps to decrease the activity of lysosomal enzymes andmaintain the native structure of proteins in the sample.

What happens to the tissue when it is homogenized? Homogenization breaksopen the heart muscle cells, releasing the organelles and cytosol.


(b) What does differential centrifugation accomplish? Organelles differ in size andtherefore sediment at different rates during centrifugation. Larger organelles and piecesof cell debris sediment first, and progressively smaller cellular components can beisolated in a series of centrifugation steps at increasing speed. The contents of eachfraction can be determined microscopically or by enzyme assay.

(c) What is the rationale for the two-step addition of ammonium sulfate? Proteins havecharacteristic solubilities at different salt concentrations, depending on the functionalgroups in the protein. In a concentration of ammonium sulfate just below the precipitationpoint of CS, some unwanted proteins can be precipitated (salted out). The ammoniumsulfate concentration is then increased so that CS is salted out. It can then be recoveredby centrifugation.

(d) Why is a buffer solution without ammonium sulfate used for the dialysis step?

Osmolarity (as well as pH and temperature) affects the conformation and stability ofproteins. To solubilize and renature the protein, the ammonium sulfate must be removed.In dialysis against a buffered solution containing no ammonium sulfate, the ammoniumsulfate in the sample moves into the buffer until its concentration is equal in bothsolutions. By dialyzing against large volumes of buffer that are changed frequently, theconcentration of ammonium sulfate in the sample can be reduced to almost zero. Thisprocedure usually takes a long time (typically overnight). The dialysate must be bufferedto keep the pH (and ionic strength) of the sample in a range that promotes the nativeconformation of the protein.

(e) What does the instruction to collect the first fraction tell you about the protein? TheCS molecule is larger than the pore size of the chromatographic gel. Size-exclusioncolumns retard the flow of smaller molecules, which enter the pores of the columnmatrix material. Larger molecules flow around the matrix, taking a direct route throughthe column.

Why is UV absorbance at 280 nm a good way to monitor for the presence of

protein in the eluted fractions? The aromatic side chains of Tyr and Trp residuesstrongly absorb at 280 nm.

(f) Explain the procedure on the cation-exchange chromatography column. CS has apositive charge (at the pH of the separation) and binds to the negatively charged beadsof the cation-exchange column, while negatively charged and neutral proteins passthrough. CS is displaced from the column by raising the pH of the mobile phase and thusaltering the charge on the CS molecules.

(g) Why were you unconvinced of the purity of the “single” protein band on your

isoelectric focusing gel? Several different proteins, all with the same pI, could befocused in the “single” band. SDS polyacrylamide gel electrophoresis separates on thebasis of mass and therefore would separate any polypeptides in the pI 5.6 band.

Why is it important to do the SDS gel electrophoresis after the isoelectric focusing?

SDS is a highly negatively charged detergent that binds tightly and uniformly along thelength of a polypeptide. Removing SDS from a protein is difficult, and a protein with onlytraces of SDS no longer has its native acid-base properties, including its native pI.

Data Analysis Problem

23. Determining the Amino Acid Sequence of Insulin Figure 3–24 shows the amino acid sequence ofthe hormone insulin. This structure was determined by Frederick Sanger and his coworkers. Most of thiswork is described in a series of articles published in the Biochemical Journal from 1945 to 1955.

When Sanger and colleagues began their work in 1945, it was known that insulin was a small proteinconsisting of two or four polypeptide chains linked by disulfide bonds. Sanger and his coworkers haddeveloped a few simple methods for studying protein sequences.




Treatment with FDNB. FDNB (1-fluoro-2,4-dinitrobenzene) reacted with free amino (but notamido or guanidino) groups in proteins to produce dinitrophenyl (DNP) derivatives of amino acids:

Acid Hydrolysis. Boiling a protein with 10% HCl for several hours hydrolyzed all of its peptide andamide bonds. Short treatments produced short polypeptides; the longer the treatment, the more com-plete the breakdown of the protein into its amino acids.

Oxidation of Cysteines. Treatment of a protein with performic acid cleaved all the disulfide bondsand converted all Cys residues to cysteic acid residues (Fig. 3–26).

Paper Chromatography. This more primitive version of thin-layer chromatography (see Fig. 10–24)separated compounds based on their chemical properties, allowing identification of single amino acidsand, in some cases, dipeptides. Thin-layer chromatography also separates larger peptides.

As reported in his first paper (1945), Sanger reacted insulin with FDNB and hydrolyzed the result-ing protein. He found many free amino acids, but only three DNP–amino acids: �-DNP-glycine (DNPgroup attached to the �-amino group); �-DNP-phenylalanine; and �-DNP-lysine (DNP attached to the �-amino group). Sanger interpreted these results as showing that insulin had two protein chains: onewith Gly at its amino terminus and one with Phe at its amino terminus. One of the two chains alsocontained a Lys residue, not at the amino terminus. He named the chain beginning with a Gly residue“A” and the chain beginning with Phe “B.”(a) Explain how Sanger’s results support his conclusions.(b) Are the results consistent with the known structure of insulin (Fig. 3–24)?

In a later paper (1949), Sanger described how he used these techniques to determine the first fewamino acids (amino-terminal end) of each insulin chain. To analyze the B chain, for example, he carriedout the following steps:

1. Oxidized insulin to separate the A and B chains.2. Prepared a sample of pure B chain with paper chromatography.3. Reacted the B chain with FDNB.4. Gently acid-hydrolyzed the protein so that some small peptides would be produced.5. Separated the DNP-peptides from the peptides that did not contain DNP groups.6. Isolated four of the DNP-peptides, which were named B1 through B4.7. Strongly hydrolyzed each DNP-peptide to give free amino acids.8. Identified the amino acids in each peptide with paper chromatography.

The results were as follows:

B1: �-DNP-phenylalanine onlyB2: �-DNP-phenylalanine; valineB3: aspartic acid; �-DNP-phenylalanine; valineB4: aspartic acid; glutamic acid; �-DNP-phenylalanine; valine

(c) Based on these data, what are the first four (amino-terminal) amino acids of the B chain? Explain your reasoning.

(d) Does this result match the known sequence of insulin (Fig. 3–24)? Explain any discrepancies.

Sanger and colleagues used these and related methods to determine the entire sequence of the Aand B chains. Their sequence for the A chain was as follows (amino terminus on left):

� HF�

O2N

NO2N

H

RNH2R

O2N

NO2F

Amine FDNB DNP-amine

Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–1051 2015

Cys–Ser–Leu–Tyr–Glx–Leu–Glx–Asx–Tyr–Cys–Asx

2608T_ch03sm_S26-S43 2/1/08 1:38PM Page 41 ntt 102:WHQY028:Solutions Manual:Ch-03:

Because acid hydrolysis had converted all Asn to Asp and all Gln to Glu, these residues had to be desig-nated Asx and Glx, respectively (exact identity in the peptide unknown). Sanger solved this problem byusing protease enzymes that cleave peptide bonds, but not the amide bonds in Asn and Gln residues, toprepare short peptides. He then determined the number of amide groups present in each peptide bymeasuring the NH4

� released when the peptide was acid-hydrolyzed. Some of the results for the A chainare shown below. The peptides may not have been completely pure, so the numbers were approximate—but good enough for Sanger’s purposes.

(e) Based on these data, determine the amino acid sequence of the A chain. Explain how youreached your answer. Compare it with Figure 3–24.

Answer(a) Any linear polypeptide chain has only two kinds of free amino groups: a single �-amino

group at the amino terminus, and an �-amino group on each Lys residue present. Theseamino groups react with FDNB to form a DNP–amino acid derivative. Insulin gave twodifferent �-amino-DNP derivatives, suggesting that it has two amino termini and thustwo polypeptide chains—one with an amino-terminal Gly and the other with an amino-terminal Phe. Because the DNP-lysine product is �-DNP-lysine, the Lys is not at anamino terminus.

(b) Yes. The A chain has amino-terminal Gly; the B chain has amino-terminal Phe; and (non-terminal) residue 29 in the B chain is Lys.

(c) Phe–Val–Asp–Glu–. Peptide B1 shows that the amino-terminal residue is Phe. Peptide B2also includes Val, but since no DNP-Val is formed, Val is not at the amino terminus; itmust be on the carboxyl side of Phe. Thus the sequence of B2 is DNP-Phe–Val. Similarly,the sequence of B3 must be DNP-Phe–Val–Asp, and the sequence of the A chain mustbegin Phe–Val–Asp–Glu–.

(d) No. The known amino-terminal sequence of the A chain is Phe–Val–Asn–Gln–. The Asnand Gln appear in Sanger’s analysis as Asp and Glu because the vigorous hydrolysis instep 7 hydrolyzed the amide bonds in Asn and Gln (as well as the peptide bonds),forming Asp and Glu. Sanger et al. could not distinguish Asp from Asn or Glu from Gln atthis stage in their analysis.

(e) The sequence exactly matches that in Figure 3–24. Each peptide in the table gives spe-cific information about which Asx residues are Asn or Asp and which Glx residues areGlu or Gln.Ac1: residues 20–21. This is the only Cys–Asx sequence in the A chain; there is ~1amido group in this peptide, so it must be Cys–Asn:


Peptide name Peptide sequence Number of amidegroups in peptide

Ac1 Cys–Asx 0.7Ap15 Tyr–Glx–Leu 0.98Ap14 Tyr–Glx–Leu–Glx 1.06Ap3 Asx–Tyr–Cys–Asx 2.10Ap1 Glx–Asx–Tyr–Cys–Asx 1.94Ap5pa1 Gly–Ile–Val–Glx 0.15Ap5 Gly–Ile–Val–Glx–Glx–Cys–Cys–

Ala–Ser–Val–Cys–Ser–Leu 1.16

N–Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Glx–Leu–Glx–Asx–Tyr–Cys–Asn–C1 5 10 15 20



Ap15: residues 14–15–16. This is the only Tyr–Glx–Leu in the A chain; there is ~1 amidogroup, so the peptide must be Tyr–Gln–Leu:

Ap14: residues 14–15–16–17. It has ~1 amido group, and we already know that residue15 is Gln, so residue 17 must be Glu:

Ap3: residues 18–19–20–21. It has ~2 amido groups, and we know that residue 21 is Asn,so residue 18 must be Asn:

Ap1: residues 17–18–19–20–21, which is consistent with residues 18 and 21 being Asn.Ap5pa1: residues 1–2–3–4. It has ~0 amido group, so residue 4 must be Glu:

Ap5: residues 1 through 13. It has ~1 amido group, and we know that residue 14 is Glu,so residue 5 must be Gln:

N–Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Gln–Leu–Glx–Asx–Tyr–Cys–Asn–C1 5 10 15 20

N–Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Gln–Leu–Glu–Asx–Tyr–Cys–Asn–C1 5 10 15 20

N–Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Gln–Leu–Glu–Asn–Tyr–Cys–Asn–C1 5 10 15 20

N–Gly–Ile–Val–Glu–Gln–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Gln–Leu–Glu–Asn–Tyr–Cys–Asn–C1 5 10 15 20

ReferencesSanger, F. (1945) The free amino groups of insulin. Biochem. J. 39, 507–515.

Sanger, F. (1949) The terminal peptides of insulin. Biochem. J. 45, 563–574.

N–Gly–Ile–Val–Glu–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Gln–Leu–Glu–Asn–Tyr–Cys–Asn–C1 5 10 15 20


solutions ch 03

Documents

ph pka

titration of glycine

average net charge of

lowest ph

ph c nh3coo2608t

carboxyl group

different net charge

protonated amino group