Chapter 3

1. (a) Phenyl isothiocyanate; (b) urea; β-mercaptoethanol to reduce disulfides; (c) chymotrypsin; (d) CNBr; (e) trypsin.

2. For each cell within an organism, the genome is a fixed property. However, the proteome is dynamic, reflecting different environmental conditions and external stimuli. Two different cell types will likely express different subsets of proteins encoded within the genome.

3. The S-aminoethylcysteine side chain resembles that of lysine. The only difference is a sulfur atom in place of a methylene group.

4. A 1 mg ml−1 solution of myoglobin (17.8 kDa; Table 3.2) corresponds to 5.62 × 10−5M. The absorbance of a 1-cm path length is 0.84, which corresponds to an I0/I ratio of 6.96. Hence 14.4% of the incident light is transmitted.

5. The sample was diluted 1000-fold. The concentration after dialysis is thus 0.001 M, or 1 mM. You could reduce the salt concentration by dialyzing your sample, now 1 mM, in more buffer free of (NH4)2SO4.

6. If the salt concentration becomes too high, the salt ions interact with the water molecules. Eventually, there will not be enough water molecules to interact with the protein, and the protein will precipitate. If there is lack of salt in a protein solution, the proteins may interact with one another—the positive charges on one protein with the negative charges on another or several others. Such an aggregate becomes too large to be solubilized by water alone. If salt is added, the salt neutralizes the charges on the proteins, preventing protein–protein interactions.

7. Tropomyosin is rod shaped, whereas hemoglobin is approximately spherical.

8. The frictional coefficient, f, and the mass, m, determine s. Specifically, f is proportional to r. Hence, f is proportional to m1/3, and so s is proportional to m2/3. An 80-kDa spherical protein undergoes sedimentation 1.59 times as rapidly as a 40-kDa spherical protein.

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9. The long hydrophobic tail on the SDS molecule disrupts the hydrophobic interactions in the interior of the protein. The protein unfolds, with the hydrophobic R groups now interacting with SDS rather than with one another.

10. 50 kDa

11. The protein may be modified. For instance, asparagine residues in the protein may be modified with carbohydrate units (Section 2.6).

12. A fluorescence-labeled derivative of a bacterial degradation product (e.g., a formylmethionyl peptide) would bind to cells containing the receptor of interest.

13. (a) Trypsin cleaves after arginine (R) and lysine (K), generating AVGWR, VK, and S. Because they differ in size, these products could be separated by molecular exclusion chromatography. (b) Chymotrypsin, which cleaves after large aliphatic or aromatic R groups, generates two peptides of equal size (AVGW) and (RVKS). Separation based on size would not be effective. The peptide RVKS has two positive charges (R and K), whereas the other peptide is neutral. Therefore, the two products could be separated by ion-exchange chromatography.

14. Antibody molecules bound to a solid support can be used for affinity purification of proteins for which a ligand molecule is not known or unavailable.

15. If the product of the enzyme-catalyzed reaction is highly antigenic, it may be possible to obtain antibodies to this particular molecule. These antibodies can be used to detect the presence of product by ELISA, providing an assay format suitable for the purification of this enzyme.

16. An inhibitor of the enzyme being purified might have been present and subsequently removed by a purification step. This removal would lead to an apparent increase in the total amount of enzyme present.

17. Many proteins have similar masses but different sequences and different patterns when digested with trypsin. The set of masses of tryptic peptides forms a detailed “fingerprint” of a protein that is very unlikely to appear at random in other proteins regardless of size. (A conceivable analogy is: “Just as similarly sized fingers will give different individual fingerprints, so also similarly sized proteins will give different digestion patterns with trypsin.”)

18. Isoleucine and leucine are isomers and, hence, have identical masses. Peptide sequencing by mass spectrometry as described in this chapter is incapable of distinguishing these residues. Further analytical techniques are required to differentiate these residues.

19.

Purification procedure

Total protein (mg)

Total activity (units)

Specific activity (units mg−1)

Purification level

Yield (%)

Crude extract

20,000

4,000,000

     200

1

100

(NH4)2SO4 precipitation

  5,000

3,000,000

     600

3

  75

DEAE–cellulose chromatography

  1,500

1,000,000

     667

  3.3

  25

Gel-filtration chromatography

     500

   750,000

  1,500

  7.5

  19

Affinity chromatography

      45

   675,000

15,000

75  

  17

20. (a) Ion exchange chromatography will remove Proteins A and D, which have substantially lower isoelectric point; then gel filtration chromatography will remove Protein C, which has a lower molecular weight. (b) If Protein B carries a His tag, a single affinity chromatography step with an immobilized nickel(II) column may be sufficient to isolate the desired protein from the others.

21. Protein crystal formation requires the ordered arrangement of identically positioned molecules. Proteins with flexible linkers can introduce disorder into this arrangement and prevent the formation of suitable crystals. A ligand or binding partner may induce an ordered conformation to this linker and could be included in the solution to facilitate crystal growth. Alternatively, the individual domains separated by the linker may be expressed by recombinant methods and their crystal structures solved separately.

22. Treatment with urea will disrupt noncovalent bonds. Thus the original 60-kDa protein must be made of two 30-kDa subunits. When these subunits are treated with urea and β-mercaptoethanol, a single 15-kDa species results, suggesting that disulfide bonds link the 30-kDa subunits.

23. (a) Electrostatic repulsion between positively charged ε-amino groups hinders α-helix formation at pH 7. At pH 10, the side chains become deprotonated, allowing α-helix formation.

(b) Poly-l-glutamate is a random coil at pH 7 and becomes α helical below pH 4.5 because the γ-carboxylate groups become protonated.

24. The difference between the predicted and the observed masses for this fragment equals 28.0, exactly the mass shift that would be expected in a formylated peptide. This peptide is likely formylated at its amino terminus, and corresponds to the most N-terminal fragment of the protein.

25. Light was used to direct the synthesis of these peptides. Each amino acid added to the solid support contained a photolabile protecting group instead of a t-Boc protecting group at its α-amino group. Illumination of selected regions of the solid support led to the release of the protecting group, which exposed the amino groups in these sites to make them reactive. The pattern of masks used in these illuminations and the sequence of reactants define the ultimate products and their locations.

26. Mass spectrometry is highly sensitive and capable of detecting the mass difference between a protein and its deuterated counterpart. Fragmentation techniques can be used to identify the amino acids that retained the isotope label. Alternatively, NMR spectroscopy can be used to detect the isotopically labeled atoms because the deuteron and the proton have very different nuclear-spin properties.

27. First amino acid: A

Last amino acid: R (not cleaved by carboxypeptidase).

Sequence of N-terminal tryptic peptide: AVR (tryptic peptide ends in K)

Sequence of N-terminal chymotryptic peptide: AVRY (chymotryptic peptide ends in Y)

Sequence: AVRYSR

28. First amino acid: S

Last amino acid: L

Cyanogen bromide cleavage: M is 10th position

C-terminal residues are: (2S,L,W)

Amino-terminal residues: (G,K,S,Y), tryptic peptide, ends in K

Amino-terminal sequence: SYGK

Chymotryptic peptide order: (S,Y), (G,K,L), (F,I,S), (M,T), (S,W), (S,L)

Sequence: SYGKLSIFTMSWSL

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29. If the protein did not contain any disulfide bonds, then the electrophoretic mobility of the trypsin fragments would be the same before and after performic acid treatment: all the fragments would lie along the diagonal of the paper. If one disulfide bond were present, the disulfide-linked trypsin fragments would run as a single peak in the first direction, then would run as two separate peaks after performic acid treatment. The result would be two peaks appearing off the diagonal:

These fragments could then be isolated from the chromatography paper and analyzed by mass spectrometry to determine their amino acid composition and thus identify the cysteines participating in the disulfide bond.