Determining the Amino Acid Sequence of Insulin Figure 3–24 shows the amino acid sequence of the hormone insulin This structure was 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.

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

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Chapter 3 Amino Acids, Peptides, and Proteins S-41

Treatment with FDNB. FDNB (1-fluoro-2,4-dinitrobenzene) reacted with free amino (but not

amido 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 and

amide 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 bonds

and 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 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 result- ing protein. He found many free amino acids, but only three DNP–amino acids: -DNP-glycine (DNP group 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: 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. 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 few amino acids (amino-terminal end) of each insulin chain. To analyze the B chain, for example, he carried out 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 only B2: -DNP-phenylalanine; valine

B3: aspartic acid; -DNP-phenylalanine; valine

B4: 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 A and B chains. Their sequence for the A chain was as follows (amino terminus on left):

⫹ HF ⫹ O2N NO2 N H R NH2 R O2N NO2 F Amine FDNB DNP-amine Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–1 5 10 Cys–Ser–Leu–Tyr–Glx–Leu–Glx–Asx–Tyr–Cys–Asx15 20

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 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 NH4 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.

(e) Based on these data, determine the amino acid sequence of the A chain. Explain how you reached 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. These amino groups react with FDNB to form a DNP–amino acid derivative. Insulin gave two different -amino-DNP derivatives, suggesting that it has two amino termini and thus two 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 an amino 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 B2 also includes Val, but since no DNP-Val is formed, Val is not at the amino terminus; it must 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 must begin Phe–Val–Asp–Glu–.

(d) No. The known amino-terminal sequence of the A chain is Phe–Val–Asn–Gln–. The Asn and Gln appear in Sanger’s analysis as Asp and Glu because the vigorous hydrolysis in step 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 at this 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 are Glu or Gln.

Ac1: residues 20–21. This is the only Cys–Asx sequence in the A chain; there is ~1

amido group in this peptide, so it must be Cys–Asn: S-42 Chapter 3 Amino Acids, Peptides, and Proteins

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 N–Gly–Ile–Val–Glx–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Glx–Leu–Glx–Asx–Tyr–Cys–Asn–C 1 5 10 15 20

Chapter 3 Amino Acids, Peptides, and Proteins S-43

Ap15: residues 14–15–16. This is the only Tyr–Glx–Leu in the A chain; there is ~1 amido

group, so the peptide must be Tyr–Gln–Leu:

Ap14: residues 14–15–16–17. It has ~1 amido group, and we already know that residue

15 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–C 1 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–C 1 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–C 1 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–C 1 5 10 15 20 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.

N–Gly–Ile–Val–Glu–Glx–Cys–Cys–Ala–Ser–Val–Cys–Ser–Leu–Tyr–Gln–Leu–Glu–Asn–Tyr–Cys–Asn–C

1 5 10 15 20

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The Three-Dimensional