Chapter 3. Exploring Proteins and Proteomes

Question 1

Question 3.1

Valuable reagents. The following reagents are often used in protein chemistry:

• CNBr
• Trypsin
• Urea
• Performic acid
• Mercaptoethanol
• 6 N HCl
• Chymotrypsin
• Phenyl isothiocyanate

Which one is the best suited for accomplishing each of the following tasks?

1. Determination of the amino acid sequence of a small peptide.
2. Reversible denaturation of a protein devoid of disulfide bonds. Which additional reagent would you need if disulfide bonds were present?
3. Hydrolysis of peptide bonds on the carboxyl side of aromatic residues.
4. Cleavage of peptide bonds on the carboxyl side of methionines.
5. Hydrolysis of peptide bonds on the carboxyl side of lysine and arginine residues.

Answer 1

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

Question 2

Question 3.2

The only constant is change. Explain how two different cell types from the same organism will have identical genomes but may have vastly divergent proteomes.

Answer 2

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.

Question 3

Question 3.3

Crafting a new breakpoint. Ethyleneimine reacts with cysteine side chains in proteins to form S-aminoethyl derivatives. The peptide bonds on the carboxyl side of these modified cysteine residues are susceptible to hydrolysis by trypsin. Why?

Answer 3

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

Question 4

Question 3.5

It’s in the bag. Suppose that you precipitate a protein with 1 M(NH4)2SO4 and that you wish to reduce the concentration of the (NH4)2SO4. You take 1 ml of your sample and dialyze it in 1000 ml of buffer. At the end of dialysis, what is the concentration of (NH4)2SO4 in your sample? How could you further lower the (NH4)2SO4 concentration?

Answer 4

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.

Question 5

Question 3.6

Too much or not enough. Why do proteins precipitate at high salt concentrations? Although many proteins precipitate at high salt concentrations, some proteins require salt to dissolve in water. Explain why some proteins require salt to dissolve.

Answer 5

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.

Question 6

Question 3.8

Sedimenting spheres. What is the dependence of the sedimentation coefficient s of a spherical protein on its mass? How much more rapidly does an 80-kDa protein sediment than does a 40-kDa protein?

Answer 6

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.

Question 7

Question 3.9

Frequently used in shampoos. The detergent sodium dodecyl sulfate (SDS) denatures proteins. Suggest how SDS destroys protein structure.

Answer 7

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.

Question 8

Question 3.11

Unexpected migration. Some proteins migrate anomalously in SDS-PAGE gels. For instance, the molecular weight determined from an SDS-PAGE gel is sometimes very different from the molecular weight determined from the amino acid sequence. Suggest an explanation for this discrepancy.

Answer 8

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

Question 9

Question 3.12

Sorting cells. Fluorescence-activated cell sorting (FACS) is a powerful technique for separating cells according to their content of particular molecules. For example, a fluorescence-labeled antibody specific for a cell-surface protein can be used to detect cells containing such a molecule. Suppose that you want to isolate cells that possess a receptor enabling them to detect bacterial degradation products. However, you do not yet have an antibody directed against this receptor. Which fluorescence-labeled molecule would you prepare to identify such cells?

Answer 9

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

Question 10

Question 3.13

Column choice. (a) The octapeptide AVGWRVKS was digested with the enzyme trypsin. Which method would be most appropriate for separating the products: ion-exchange or gel-filtration chromatography? Explain. (b) Suppose that the peptide was digested with chymotrypsin. What would be the optimal separation technique? Explain.

Answer 10

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.

Question 11

Question 3.14

Power(ful) tools. Monoclonal antibodies can be conjugated to an insoluble support by chemical methods. Explain how these antibody-bound beads can be exploited for protein purification.

Answer 11

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.

Question 12

Question 3.15

Assay development. You wish to isolate an enzyme from its native source and need a method for measuring its activity throughout the purification. However, neither the substrate nor the product of the enzyme-catalyzed reaction can be detected by spectroscopy. You discover that the product of the reaction is highly antigenic when injected into mice. Propose a strategy to develop a suitable assay for this enzyme.

Answer 12

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.

Question 13

Question 3.16

Making more enzyme? In the course of purifying an enzyme, a researcher performs a purification step that results in an increase in the total activity to a value greater than that present in the original crude extract. Explain how the amount of total activity might increase.

Answer 13

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.

Question 14

Question 3.17

Divide and conquer. The determination of the mass of a protein by mass spectrometry often does not allow its unique identification among possible proteins within a complete proteome, but determination of the masses of all fragments produced by digestion with trypsin almost always allows unique identification. Explain.

Answer 14

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.”)

Question 15

Question 3.18

Know your limits. Which two amino acids are indistinguishable in peptide sequencing by the tandem mass spectrometry method described in this chapter and why?

Answer 15

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.

Question 16

Answer 16

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.

Question 17

Question 3.21

The challenge of flexibility. Structures of proteins comprising domains separated by flexible linker regions can be quite difficult to solve by x-ray crystallographic methods. Why might this be the case? What are possible experimental approaches to circumvent this barrier?

Answer 17

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.

Question 18

Question 3.22

Quaternary structure. A protein was purified to homogeneity. Determination of the mass by gel-filtration chromatography yields 60 kDa. Chromatography in the presence of 6 M urea yields a 30-kDa species. When the chromatography is repeated in the presence of 6 M urea and 10 mM β-mercaptoethanol, a single molecular species of 15 kDa results. Describe the structure of the molecule.

Answer 18

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.

Question 19

Question 3.23

Helix–coil transitions. (a) NMR measurements have shown that poly-l-lysine is a random coil at pH 7 but becomes a helix as the pH is raised above 10. Account for this pH-dependent conformational transition. (b) Predict the pH dependence of the helix–coil transition of poly-l-glutamate.

Answer 19

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.

Question 20

Question 3.24

Peptide mass determination. You have isolated a protein from the bacterium E. coli and seek to confirm its identity by trypsin digestion and mass spectrometry. Determination of the masses of several peptide fragments has enabled you to deduce the identity of the protein. However, there is a discrepancy with one of the peptide fragments, which you believe should have the sequence MLNSFK and an (M+H)+ value of 739.38. In your experiments, you repeatedly obtain an (M+H)+ value of 767.38. What is the cause of this discrepancy and what does it tell you about the region of the protein from which this peptide is derived?

Answer 20

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.

Question 21

Question 3.25

Peptides on a chip. Large numbers of different peptides can be synthesized in a small area on a solid support. This high-density array can then be probed with a fluorescence-labeled protein to find out which peptides are recognized. The binding of an antibody to an array of 1024 different peptides occupying a total area the size of a thumbnail is shown in the adjoining illustration. How would you synthesize such a peptide array? (Hint: Use light instead of acid to deprotect the terminal amino group in each round of synthesis.)

Answer 21

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.

Question 22

Question 3.26

Exchange rate. The amide hydrogen atoms of peptide bonds within proteins can exchange with protons in the solvent. In general, amide hydrogen atoms in buried regions of proteins and protein complexes exchange more slowly than those on the solvent-accessible surface do. Determination of these rates can be used to explore the protein-folding reaction, probe the tertiary structure of proteins, and identify the regions of protein–protein interfaces. These exchange reactions can be followed by studying the behavior of the protein in solvent that has been labeled with deuterium (2H), a stable isotope of hydrogen. What two methods described in this chapter could be readily applied to the study of hydrogen–deuterium exchange rates in proteins?

Answer 22

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.

Question 23

Question 3.27

Protein sequencing 1. Determine the sequence of hexapeptide on the basis of the following data. Note: When the sequence is not known, a comma separates the amino acids (Table 3.3).

Amino acid composition: (2R,A,S,V,Y)

N-terminal analysis of the hexapeptide: A

Trypsin digestion: (R,A,V) and (R,S,Y)

Carboxypeptidase digestion: No digestion.

Chymotrypsin digestion: (A,R,V,Y) and (R,S)

Answer 23

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

Question 24

Question 3.28

Protein sequencing 2. Determine the sequence of a peptide consisting of 14 amino acids on the basis of the following data.

Amino acid composition: (4S,2L,F,G,I,K,M,T,W,Y)

N-terminal analysis: S

Carboxypeptidase digestion: L

Trypsin digestion: (3S,2L,F,I,M,T,W) (G,K,S,Y)

Chymotrypsin digestion: (F,I,S) (G,K,L) (L,S) (M,T) (S,W) (S,Y)

N-terminal analysis of (F,I,S) peptide: S

Cyanogen bromide treatment: (2S,F,G,I,K,L,M*,T,Y) (2S,L,W)

M*, methionine detected as homoserine

Answer 24

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

Question 25

Question 3.29

Applications of two-dimensional electrophoresis. Performic acid cleaves the disulfide linkage of cystine and converts the sulfhydryl groups into cysteic acid residues, which are then no longer capable of disulfide-bond formation.

Consider the following experiment: You suspect that a protein containing three cysteine residues has a single disulfide bond. You digest the protein with trypsin and subject the mixture to electrophoresis along one end of a sheet of paper. After treating the paper with performic acid, you subject the sheet to electrophoresis in the perpendicular direction and stain it with a reagent that detects proteins. How would the paper appear if the protein did not contain any disulfide bonds? If the protein contained a single disulfide bond? Propose an experiment to identify which cysteine residues form the disulfide bond.

Answer 25

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.