Organic Chemistry

988 CHAPTER 23 Amino Acids, Peptides, and Proteins

The last step in determining the primary structure of a protein is to figure out the lo-

cation of any disulfide bonds. This is done by hydrolyzing a sample of the protein that

has intact disulfide bonds. From a determination of the amino acids in the cysteine-

containing fragments, the locations of the disulfide bonds in the protein can be estab-

lished (Problem 47).

PROBLEM 30

Why won’t cyanogen bromide cleave at cysteine residues?

PROBLEM 31

In determining the primary structure of insulin, what would lead you to conclude that it

had more than one polypeptide chain?

PROBLEM 32 SOLVED

Determine the amino acid sequence of a polypeptide from the following results:

Acid hydrolysis gives Ala, Arg, His, 2 Lys, Leu, 2 Met, Pro, 2 Ser, Thr, Val.

Carboxypeptidase A releases Val.

Edman’s reagent releases PTH-Leu.

Cleavage with cyanogen bromide gives three peptides with the following amino acid

compositions:

1. His, Lys, Met, Pro, Ser 3. Ala, Arg, Leu, Lys, Met, Ser


2. Thr, Val
   Trypsin-catalyzed hydrolysis gives three peptides and a single amino acid:


3. Arg, Leu, Ser 3. Lys


4. Met, Pro, Ser, Thr, Val 4. Ala, His, Lys, Met

SOLUTION Acid hydrolysis shows that the polypeptide has 13 amino acids. The N-ter-

minal amino acid is Leu (Edman’s reagent), and the C-terminal amino acid is Val (car-

boxypeptidase A).

Because cyanogen bromide cleaves on the C-side of Met, any peptide containing Met

must have Met as its C-terminal amino acid. The peptide that does not contain Met must be

the C-terminal peptide. We know that peptide 3 is the N-terminal peptide because it con-

tains Leu. Since it is a hexapeptide, we know that the 6th amino acid in the 13-amino acid

peptide is Met. We also know that the eleventh amino acid is Met because cyanogen bro-

mide cleavage gave the dipeptide Thr, Val.

Because trypsin cleaves on the C-side of Arg and Lys, any peptide containing Arg or Lys

must have that amino acid as its C-terminal amino acid. Therefore, Arg is the C-terminal

amino acid of peptide 1, so we now know that the first three amino acids are Leu-Ser-Arg.

We also know that the next two are Lys-Ala because if they were Ala-Lys, trypsin cleav-

age would give an Ala, Lys dipeptide. The trypsin data also identify the positions of His

and Lys.

Finally, because trypsin successfully cleaves on the C-side of Lys, Pro cannot be adja-

cent to Lys. Thus, the amino acid sequence of the given polypeptide is

Leu Ser Arg Lys Ala Met His Lys Ser Pro Met Thr Val

Leu Ser Arg Lys Ala Met His Lys

Pro, Ser

Met Thr Val

Leu Met

Ala, Arg, Lys, Ser His, Lys, Pro, Ser

Met Thr Val

Leu Val

BRUI23-959-998r2 29-03-2003 1:36 PM Page 988