Chapter Review in Chapter 3 Amino Acids, Peptides, and Proteins

Chapter Review

KEY TERMS

Terms in bold are defined in the glossary.

Problems

1. Amino Acid Constituents of Glutathione Glutathione is an important peptide antioxidant found in cells from bacteria to humans.

The figure has a central carbon on the left bonded to N H 3 plus above, C O O minus on the left, H and C H 2 on the right that is further bonded to C H 2 that is further bonded to C that is double bonded to O above and bonded to N on the right that is bonded to H below and to C H on the right that is further bonded to C H 2 bonded to S H below and to C on the right that is double bonded to O above and further bonded to N on the right that is bonded to H below and to C H 2 on the right that is further bonded to C O O minus.

Identify the three amino acid constituents of glutathione. What is unusual about glutathione’s structure?
2. Absolute Configuration of Ornithine Ornithine is an amino acid that is not a building block of proteins. Instead, ornithine is an important intermediate in the urea cycle, the metabolic process that facilitates the excretion of ammonia waste products in animals.

What is the absolute configuration of the ornithine molecule shown here?
3. Relationship between the Titration Curve and the Acid-Base Properties of Glycine A researcher titrated a 100 mL solution of 0.1 m glycine at pH 1.72 with 2 m NaOH solution. She then monitored the pH and plotted the results in the graph shown. The key points in the titration are designated I to V. For each of the following statements, identify the appropriate key point in the titration. Note that statement (k) applies to more than one key point in the titration.
1. The pH is equal to the $mathml alt text242$ $p upper K Subscript a Baseline$ of the carboxyl group.
2. The pH is equal to the $mathml alt text243$ $p upper K Subscript a Baseline$ of the protonated amino group.
3. The predominant glycine species is $mathml alt text244$ $Superscript plus Baseline upper H Subscript 3 Baseline upper N em-dash CH Subscript 2 Baseline em-dash COOH$ .
4. The predominant glycine species is $mathml alt text245$ $Superscript plus Baseline upper H Subscript 3 Baseline upper N em-dash CH Subscript 2 Baseline em-dash COO Superscript minus$ .
5. Glycine exists as a 50:50 mixture of $mathml alt text246$ $Superscript plus Baseline upper H Subscript 3 Baseline upper N em-dash CH Subscript 2 Baseline em-dash COOH$ and $mathml alt text247$ $Superscript plus Baseline upper H Subscript 3 Baseline upper N em-dash CH Subscript 2 Baseline em-dash COO Superscript minus$ .
6. The average net charge of glycine is $mathml alt text248$ $plus one-half$ .
7. Half of the amino groups are ionized.
8. The average net charge of glycine is 0.
9. The average net charge of glycine is −1.
10. This is the isoelectric point for glycine.
11. Glycine has its maximum buffering capacity at these regions.
  
  The graph shows a wavy line that begins at point Roman numeral 1 at (0, 0). It moves upward rapidly, then levels off to an almost diagonal line at about (0.10, 1). A dotted vertical line and a dotted horizontal line meet at point Roman numeral 2 at (0.5, 2.34) beneath text reading, 2.34. The line continues to move upward almost diagonally to (0.8, 3), then begins to move upward more rapidly and begins to run almost vertically as it passes through point Roman numeral 3 at (1.0, 5.97). A dotted vertical line and a dotted horizontal line meet at this point and text on the horizontal line reads, 5.97. The line begins to curve upward more slowly at (1, 8.5), then becomes almost diagonal again. A dotted vertical line and a dotted horizontal line meet at point Roman numeral 4 at (1.5, 9.60) and text on the horizontal line reads, 9.60. The line curves upward more rapidly beginning at about (1.8, 10.0), then passes through point Roman numeral 5 at (2.0, 11.30). A dotted vertical line and a dotted horizontal line meet at this point and text on the horizontal line reads, 11.30. The line ends at (2.1, 12). All data are approximate.
4. Charge States of Alanine at Its pI At a pH equal to the isoelectric point (pI) 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.

The structure labeled, zwitterionic has a central carbon bonded to C H 3 above, C on the right that is further double bonded to O and bonded to O minus, H below, and N H 3 plus on the left. The structure labeled uncharged is similar but the O minus is replaced by O H and the N H 3 plus is replaced by N H 2.
1. Why is alanine predominantly zwitterionic at its pI?
2. What fraction of alanine is in the completely uncharged form at its pI?
5. 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 $mathml alt text250$ $p upper K Subscript a Baseline$ and the pH of the solution. This relationship is described by the Henderson-Hasselbalch equation.
1. Histidine has three ionizable functional groups. Write the equilibrium equations for its three ionizations, and assign the proper $mathml alt text251$ $p upper K Subscript a Baseline$ 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?
2. Draw the structures of the predominant ionization state of histidine at pH 1, 4, 8, and 12. Note that you can approximate the ionization state by treating each ionizable group independently.
3. What is the net charge of histidine at pH 1, 4, 8, and 12? For each pH, will histidine migrate toward the anode (+) or toward the cathode (−) when placed in an electric field?
6. Separation of Amino Acids by Ion-Exchange Chromatography We can analyze mixtures of amino acids 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 $mathml alt text254$ $left-parenthesis minus SO Subscript 3 Superscript minus Baseline right-parenthesis$ 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) aggregation of nonpolar amino acid side chains with the hydrophobic backbone of the polystyrene resin. Note that the ionic attraction is more significant than hydrophobicity for this column media. For each pair of amino acids listed, determine which will be eluted first from the cation-exchange column by a pH 7.0 buffer.
1. Glutamate and lysine
2. Arginine and methionine
3. Aspartate and valine
4. Glycine and leucine
5. Serine and alanine
7. Naming the Stereoisomers of Isoleucine Consider the structure of the amino acid isoleucine.
1. How many chiral centers does isoleucine have?
2. How many optical isomers does isoleucine have?
3. Draw perspective formulas for all the optical isomers of isoleucine.
8. Comparing the $mathml alt text256$ $bold p bold-italic upper K Subscript bold a$ Values of Alanine and Polyalanine The titration curve of alanine shows the ionization of two functional groups with $mathml alt text257$ $p upper K Subscript a Baseline$ 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 $mathml alt text258$ $p upper K Subscript a Baseline$ values are different. The table summarizes the trend in $mathml alt text259$ $p upper K Subscript a Baseline$ values.

Amino acid or peptide $mathml alt text260$ $bold p bold-italic upper K bold 1$ $mathml alt text261$ $bold p bold-italic upper K bold 2$

Ala

2.34

9.69

Ala–Ala

3.12

8.30

Ala–Ala–Ala

3.39

8.03

$mathml alt text262$ $Ala en-dash left-parenthesis Ala right-parenthesis Subscript n Baseline en-dash Ala comma$ n ≥ 4

3.42

7.94
1. Draw the structure of Ala–Ala–Ala. Identify the functional groups associated with $mathml alt text264$ $p upper K 1$ and $mathml alt text265$ $p upper K 2$ .
2. Why does the value of $mathml alt text266$ $p upper K 1$ increase with each additional Ala residue in the oligopeptide?
3. Why does the value of $mathml alt text267$ $p upper K 2$ decrease with each additional Ala residue in the oligopeptide?
9. Bonds Form by Condensation The peptide bond is an amide, generated by eliminating the elements of water from the two amino acids so joined. From which groups are the three atoms of water eliminated?
10. The Size of Proteins Calculate the approximate molecular weight of a protein composed of 682 amino acid residues in a single polypeptide chain.
11. Relationship between the Number of Amino Acid Residues and Protein Mass Experimental results describing a protein’s amino acid composition are useful to estimate the molecular weight of the entire protein. A quantitative amino acid analysis reveals that bovine cytochrome c contains 2% cysteine $mathml alt text268$ $left-parenthesis upper M Subscript r Baseline 121 right-parenthesis$ by weight.
1. Calculate the approximate molecular weight in daltons of bovine cytochrome c if the number of cysteine residues is 2.
  - Bovine chymotrypsinogen has a molecular mass of 25.6 kDa. Amino acid analysis shows that this enzyme is 4.7% Gly $mathml alt text269$ $left-parenthesis upper M Subscript r Baseline 75.1 right-parenthesis$ .
2. Calculate how many glycine residues are present in a molecule of bovine chymotrypsinogen.
12. Subunit Composition of a Protein A protein has a molecular mass of 400 kDa when measured by size-exclusion chromatography. 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 again form, this time with molecular masses of 160, 90, and 60 kDa. How many subunits does the protein have, and what is the molecular mass of each?
13. Net Electric Charge of Peptides A peptide has the sequence Glu–His–Trp–Ser–Gly–Leu–Arg–Pro–Gly.
1. Calculate the net charge of the molecule at pH 3, 8, and 11. (Incorporation into a peptide can alter $mathml alt text271$ $p upper K Subscript a Baseline$ values somewhat, but for this exercise, use $mathml alt text272$ $p upper K Subscript a Baseline$ values for side chains and terminal amino and carboxyl groups as given in Table 3-1.)
2. Estimate the pI for this peptide.
14. Isoelectric Point of Histones Histones are proteins found in eukaryotic cell nuclei, tightly bound to DNA, which has many phosphate groups. The pI of histones is very high, about 10.8. What amino acid residues must be present in relatively large numbers in histones? In what way do these residues contribute to the strong binding of histones to DNA?
15. Solubility of Polypeptides One method for separating polypeptides makes use of their different solubilities. The solubility of large polypeptides in water depends on the relative polarity of their R groups, particularly on the number of ionized groups: the more ionized groups there are, the more soluble the polypeptides are. Which of each pair of polypeptides is more soluble at the indicated pH?
1. $mathml alt text273$ $left-parenthesis Gly right-parenthesis Subscript 20$ or $mathml alt text274$ $left-parenthesis Glu right-parenthesis Subscript 20$ at pH 7.0
2. $mathml alt text276$ $left-parenthesis Lys en-dash Val right-parenthesis Subscript 3$ or $mathml alt text277$ $left-parenthesis Phe en-dash Cys right-parenthesis Subscript 3$ at pH 7.0
3. $mathml alt text279$ $left-parenthesis Ala en-dash Ser en-dash Gly right-parenthesis Subscript 5$ or $mathml alt text280$ $left-parenthesis Asn en-dash Ser en-dash His right-parenthesis Subscript 5$ at pH 6.0
4. $mathml alt text282$ $left-parenthesis Ala en-dash Asp en-dash Phe right-parenthesis Subscript 5$ or $mathml alt text283$ $left-parenthesis Asn en-dash Ser en-dash His right-parenthesis Subscript 5$ at pH 3.0

Amino acid or peptide	$mathml alt text260$ $bold p bold-italic upper K bold 1$	$mathml alt text261$ $bold p bold-italic upper K bold 2$
Ala	2.34	9.69
Ala–Ala	3.12	8.30
Ala–Ala–Ala	3.39	8.03
$mathml alt text262$ $Ala en-dash left-parenthesis Ala right-parenthesis Subscript n Baseline en-dash Ala comma$ n ≥ 4	3.42	7.94

16. Purification of an Enzyme A biochemist discovers and purifies a new enzyme, generating the purification table shown.

Procedure	Total protein (mg)	Activity (units)
1. Crude extract	10,000	68,000
2. Precipitation (salt)	5,000	65,000
3. Precipitation (pH)	4,000	56,000
4. Ion-exchange chromatography	70	49,000
5. Affinity chromatography	12	42,000
6. Size-exclusion chromatography	8	40,000

From the information given in the table, calculate the specific activity of the enzyme after each purification procedure.
Which of the purification procedures used for this enzyme is most effective (i.e., gives the greatest relative increase in purity)?
Which of the purification procedures is least effective?
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?

17. De-salting a Protein by Dialysis A purified protein is in a Hepes (N-(2-hydroxyethyl)piperazine- $mathml alt text292a$ $upper N prime$ -(2-ethanesulfonic acid)) buffer at pH 7 with 500 mm NaCl. A dialysis membrane tube holds a 1 mL sample of the protein solution. The sample in the dialysis membrane floats in a beaker containing 1 L of the same Hepes buffer, but with 0 mm NaCl, for dialysis. Small molecules and ions (such as $mathml alt text292$ $Na Superscript plus Baseline comma Cl Superscript minus Baseline comma$ and Hepes) can diffuse across the dialysis membrane, but the protein cannot.
1. Calculate the concentration of NaCl in the protein sample, once the dialysis has come to equilibrium. Assume that no volume changes occur in the sample during the dialysis.
2. Calculate the final NaCl concentration in the protein sample after dialysis in 250 mL of the same Hepes buffer, with 0 mm NaCl, twice in succession.
18. Predicting Cation Exchange Elution Order Suppose a column is filled with a cation-exchange resin at pH 7.0. In what order would the given peptides elute from the column if each has the same number of residues?

Peptide A: Ala 30%, Asp 10%, Lys 10%, Ser 15%, Pro 25%, Cys 10%

Peptide B: Ile 25%, Asp 20%, Arg 5%, Tyr 15%, His 5%, Thr 30%

Peptide C: Ala 40%, Glu 5%, Arg 20%, Ser 5%, His 5%, Trp 25%
19. Protein Analysis by Gel Electrophoresis Chymotrypsin is a protease with a molecular mass of 25.6 kDa. The figure shows a stained SDS polyacrylamide gel with a single band in lane 1 and three bands of lower molecular weight in lane 2. Lane 1 contains a preparation of chymotrypsin, and lane 2 contains chymotrypsin pretreated with performic acid. Why does performic acid treatment of chymotrypsin generate three bands in lane 2?
20. Sequence Determination of Leucine Enkephalin Suppose a researcher isolates a peptide from brain tissue that binds to the same receptor as opiate drugs. This peptide is an opioid leucine enkephalin, a class of endogenous peptides that act within the brain to lower pain sensation. The researcher performs a series of procedures to determine the peptide’s sequence. First, she completely hydrolyzes the peptide by boiling it in 18% w/w HCl solution. Analysis of the hydrolysis products indicates the presence of Gly, Leu, Phe, and Tyr, in a 2:1:1:1 molar ratio. Second, she treats the peptide with 1-dimethylaminoaphthalene-5-sulfonyl chloride (dansyl chloride) before subjecting it to complete hydrolysis. Chromatography indicates the presence of the dansylamino acid derivative of tyrosine. No free tyrosine is present.
1. Given the empirical composition and the results from the dansyl chloride reaction, where is Tyr located in the peptide?
  Finally, the researcher incubates the peptide with chymotrypsin for two hours at $mathml alt text301$ $10 Superscript degree Baseline upper C$ and analyzes the products using chromatography. Complete digestion of the peptide with chymotrypsin followed by chromatography yields free tyrosine and leucine, plus a tripeptide containing Phe and Gly in a 1:2 ratio.
2. Give the final sequence for the peptide based on the results of the acid hydrolysis, dansyl chloride reaction, and the chymotryptic digestion.
21. Analysis of a Protein by Mass Spectrometry Investigators purify a protein produced by yeast grown under standard growth conditions. They incubate the protein with trypsin and sequence the peptides produced using mass spectrometry. One of the detected peptides, called peptide X, has the sequence Ala–Ser–Ala–Gly–Lys–Glu–Leu–Ile–Phe–Gln. The investigators then isolate the same protein, but this time from yeast grown under the stress of ultraviolet irradiation. When the sample is analyzed, a peptide with the mass of peptide X is no longer found. Instead, detection reveals a new peptide with the same sequence, except for an amino acid that replaces Ser and has a molecular mass of 167 Da. The investigators conclude that the protein has been altered in response to stress, and that the serine residue in the analyzed peptide has been modified. An unmodified series residue has a molecular mass of 87 Da. What modification might account for the change in the peptide’s mass?
22. 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 and seems to disrupt ion transport across the cell membranes of other bacterial species, leading to bacterial death. The structure of the peptide has been determined from a series of observations.
1. 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 proteins but present in some peptides. Ornithine has the structure
2. The molecular weight of the peptide is approximately 1,200 Da.
3. The peptide failed to undergo hydrolysis when treated with the enzyme carboxypeptidase. This enzyme catalyzes the hydrolysis of the carboxyl-terminal residue of a polypeptide unless the residue is Pro or, for some reason, does not contain a free carboxyl group.
4. Treatment of the intact peptide with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis and chromatography, yielded only free amino acids and the derivative shown here.
  
  (Hint: The 2,4-dinitrophenyl derivative involves the amino group of a side chain rather than the α-amino group.)
5. Partial hydrolysis of the peptide followed by chromatographic separation and sequence analysis yielded these di- and tripeptides (the amino-terminal amino acid is always the first amino acid):
  
  Leu–Phe Phe–Pro Orn–Leu Val–Orn
  
  Val–Orn–Leu Phe–Pro–Val Pro–Val–Orn
  
  Given this information, deduce the amino acid sequence of the peptide antibiotic. Show your reasoning. When you have arrived at a structure, demonstrate that it is consistent with each experimental observation.
23. Efficiency in Peptide Synthesis A peptide with the primary structure Lys–Arg–Pro–Leu–Ile–Asp–Gly–Ala must be synthesized by the methods developed by Merrifield. Calculate the percentage of the peptides synthesized that will be full length and have the correct sequence if the addition of each amino acid residue is 96% efficient. Do the calculation a second time but assume a 99% efficiency for each cycle.
24. Sequence Comparisons Proteins called molecular chaperones (described in Chapter 4) assist in the process of protein folding. One class of chaperones, found in organisms from bacteria to mammals, is heat shock protein 90 (Hsp90). All Hsp90 chaperones contain a 10 amino acid “signature sequence” that allows ready identification of these proteins in sequence databases. Two representations of this signature sequence are shown here.

There is text across the top that reads, Y-x-[N Q H D]-[K H R]-[D E]-[I V A]-F-[L M]-R-[E D]. Beneath, a horizontal graph has numbers 1 through 10 across the horizontal axis, with N under 1 and C under 10, plotted against bits on the vertical axis ranging from 0 to 4, labeled in increments of 1. All data are approximate. 1: green Y extending to 4.1; 2: blue K extending from the horizontal axis to 0.25 beneath green T extending from 0.25 to 0.35 beneath green S extending from 0.5 to 3.25; 3: indistinct bars beneath blue H extending from 0.2 to 0.35 beneath red D extending from 0.35-0.50 beneath purple K extending from 0.50 to 3.5; 4: indistinct bars beneath blue R extending from 0.30 to 0.40 beneath blue K extending from 0.40 to 3.75; 5: red D extending from the horizontal axis to 0.50 beneath red E extending from 0.50 go 3.75; 6: black V extending from the horizontal axis to 0.30 beneath black I extending from 0.30 to 3.75; 7: black F extending from the horizontal axis to 4.2; 8: black M extending from the horizontal axis to 0.35 beneath black L extending from 0.35 to 4.0; 9: red R extending from the horizontal axis to 4.1; 10: red E extending from the horizontal axis to 4.1.
1. In this sequence, which amino acid residues are invariant (conserved across all species)?
2. 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?
3. At which positions are substitutions restricted to amino acids with negatively charged side chains? For each position, which amino acid predominates?
4. 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?
25. Chromatographic Methods Three polypeptides, the sequences of which are represented using the one-letter code for their amino acids, are present in a mixture:
1. ATKNRASCLVPKHGALMFWRHKQLVSDPILQKRQHILVCRNAAG
2. GPYFGDEPLDVHDEPEEG
3. PHLLSAWKGMEGVGKSQSFAALIVILA
1. Of the three, which one would migrate most slowly during the following:
  - Anion-exchange chromatography?
  - Cation-exchange chromatography?
  - Size-exclusion (gel-filtration) chromatography?
2. Which peptide contains the ATP-binding motif shown in the sequence logo?
  
  A horizontal graph has numbers 1 through 8 across the horizontal axis, with N under 1 and C under 8, plotted against bits on the vertical axis ranging from 0 to 4, labeled in increments of 2. All data are approximate. 1: black A extending from the horizontal axis to 0.9 beneath green G extending from 0.9 to 3.4; 2 and 3: indistinct bars extending to 0.5; 4: indistinct bars extending to 0.4, black A extending from 0.4 to 0.8 beneath green G extending from 0.8 to 1.0; 5: indistinct bars to 0.7 beneath black A extending from 07 to 0.8; 6: green G extending from the horizontal axis to 4.0; 7: blue K extending from the horizontal axis to 4.0; 8: green S extending from the horizontal axis to 1.4 beneath green T extending from 1.4 to 3.2.

DATA ANALYSIS PROBLEM

26. Determining the Amino Acid Sequence of Insulin Figure 3-24 shows the amino acid sequence of bovine insulin, determined by Frederick Sanger and his coworkers. Most of this work is described in a series of articles published in the Biochemical Journal from 1945 to 1955.

In 1945, researchers knew that insulin was a small protein consisting of two or four polypeptide chains linked by disulfide bonds. Sanger’s team had developed a few simple methods for studying protein sequences.

Treatment with FDNB. FDNB (1-fluoro-2,4-dinitrobenzene) reacted with free amino (but not amide or guanidinium) groups in proteins to produce dinitrophenyl (DNP) derivatives of amino acids (see Fig. 3-25).

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

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

Paper Chromatography. This more primitive version of thin-layer chromatography (see Fig. 10-25) separated compounds based on their chemical properties, allowing identification of single amino acids and, 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 resulting protein. He found many free amino acids, but only three DNP–amino acids: α-DNP-glycine (DNP group attached to the α-amino group), α-DNP-phenylalanine, and ε-DNP-lysine (DNP attached to the ε-amino group). Sanger interpreted these results as showing that insulin had two protein chains: one with Gly at its amino terminus and one with Phe at its amino terminus. One of the two chains also contained a Lys residue, not at the amino terminus. Sanger named the chain beginning with a Gly residue “A” and the chain beginning with Phe “B.”

Explain how Sanger’s results support his conclusions.
Are the results consistent with the known structure of bovine insulin (see Fig. 3-24)?

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

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

The results were as follows:

B1: α-DNP-phenylalanine only
B2: α-DNP-phenylalanine; valine
B3: aspartic acid; α-DNP-phenylalanine; valine
B4: aspartic acid; glutamic acid;
α-DNP-phenylalanine; valine

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

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

Sanger and colleagues used these and related methods to determine the entire sequence of the A and B chains. Their sequence for the A chain was as follows:

A sequence of 21 amino acids is given: G l y (1) - I l e - V a l - G l x - G l x (5)- C y s - C y s- A l a - S e r - V a l (10) - C y s - S e r - L e u - T y r - G l x (15) - L e u - G l x - A s x - T y r - C y s (20) - A s x.

Because acid hydrolysis had converted all Asn to Asp and all Gln to Glu, these residues had to be designated Asx and Glx, respectively (exact identity in the peptide unknown). Sanger solved this problem by using protease enzymes that cleave peptide bonds, but not the amide bonds in Asn and Gln residues, to prepare short peptides. He then determined the number of amide groups present in each peptide by measuring the $mathml alt text312$ $NH Subscript 4 Superscript plus$ released when the peptide was acid-hydrolyzed. Some of the results for the A chain are shown below. The peptides may not have been completely pure, so the numbers were approximate — but good enough for Sanger’s purposes.

Peptide name	Peptide sequence	Number of amide groups in peptide
Ac1	Cys–Asx	0.7
Ap15	Tyr–Glx–Leu	0.98
Ap14	Tyr–Glx–Leu–Glx	1.06
Ap3	Asx–Tyr–Cys–Asx	2.10
Ap1	Glx–Asx–Tyr–Cys–Asx	1.94
Ap5pa1	Gly–Ile–Val–Glx	0.15
Ap5	Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu	1.16

Based on these data, determine the amino acid sequence of the A chain. Explain how you reached your answer. Compare your answer with Figure 3-24.

References

Sanger, 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.