(a) Minimum M_r 32,000. Remember that the molecular weight of a Trp residue is not the same as the molecular weight of the free amino acid. (b) 2 Trp residues.

(a) At pH 3, +2; at pH 8, 0; at pH 11, −1. (b) pI 8.

Details:

S-4

(a) A1, R5, K20, R28. (b) D2, E4, D27, E29, T30. (c) C7, C23.

The sheets are likely to be antiparallel, because the linkers that connect one β strand to the next are too short to connect parallel strands. The linkers are four residues long and could form β turns.

Alternating R groups are on opposite sides of the β sheet structure. Therefore, in a continuous layer of β sheet, with alternating polar and nonpolar residues in the β strands, the sheet is likely to fold into a β barrel, sequestering hydrophobic residues inside.

Residues 1 and 3 are in β sheets, residue 2 is in a right-handed a helix, and residue 4 is in a left-handed a helix. Residue 5 is an outlier, and the electron density map should be examined to see whether the angles are correctly assigned. If residue 5 were Gly, it might be in an acceptable position in the plot, because Gly residues are more flexible than other residues and can adopt a greater range of acceptable bond angles.

The P3 and P18 residues disfavor helix formation, limiting the region favorable for helix formation to the sequence between residues 4 and 17. The several positively charged residues, 4 through 7, are not likely to initiate a helix, because they would repel one another and interact unfavorably with the helix dipole, which is positively charged at the N-terminus. In contrast, E8 will interact favorably with the helix dipole. Thus, the most likely α-helical region is residues 8 through 17 (boxed). Within this helix, the stabilizing interactions are: N-terminal positive dipole, stabilized by E8; C-terminal negative dipole, stabilized by R17; hydrophobic interactions between two F residues, spaced four residues (one turn) apart; ion-pair interaction between E8 and R12, spaced four residues (one turn) apart.

(a) and (d). Both contain the consensus sequence for an ATP/GTP-binding site: (G/A)XXGXGK(T/S), where X is any amino acid.

As an α helix, 210 Å ([1.5 Å/residue] × 140 residues). As a β strand, 490 Å ([3.5 Å/residue] × 140 residues).

There are many possible answers; any of the following will suffice. NMR uses magnets and radiofrequency irradiation; crystallography uses x rays. NMR is performed on proteins in solution; x-ray crystallography requires a protein crystal. NMR measures a nuclear event; x-ray crystallography measures events in the electron shell. In NMR, irradiated proteins emit radiowaves; in x-ray crystallography, irradiated proteins emit x rays. NMR makes great use of protons; protons are largely ignored in x-ray crystallography. NMR can be applied only to small proteins; x-ray crystallography can solve large proteins and complexes. Both methods irradiate the protein sample with electromagnetic radiation (photons). Both methods yield structures with atomic resolution. Both methods make heavy use of computations.

(a), (b), (c) one domain; (d), (e) two domains. When a protein reaches a size of about 150 to 200 residues (M_r ∼20,000), the polypeptide chain usually folds into two domains.

Completely buried residues are likely to be hydrophobic: L2, F4, I6, V8, V12, L13, L18, and L19 fit this description. Highly polar residues, or at least their polar groups, are likely to be on the surface, exposed to water: D1, K3, T5, S7, T14, R15, E16, Q17, and E20 fit this description.

(a) Destabilizes the positive dipole at the N-terminus of the helix. (b) Stabilizes the positive dipole at the N-terminus of the helix. (c) Destabilizes; the ion pair between R2 and E5, one turn apart, is eliminated. (d) No difference; the ion pair between the residues is maintained. (e) Stabilizes; the hydrophobic interaction is on the same side of the helix, one turn apart. (f) Destabilizes; P (a helix breaker) destabilizes the helix.

(a) The N-terminus is the lower-right end; the C-terminus is the upper-right end; the β turns are the U turns at the lower left and upper right. (b) The more hydrophobic surface is likely to be on the right side of the β sheet.

(a) Bovine serum albumin. (b) Green fluorescent protein.

Details: Much of the structure of bovine serum albumin is in the form of an α helix, while green fluorescent protein is predominantly in the β conformation. The corresponding Ramachandran plots exhibit many residues with torsion angles characteristic of the predominant secondary structures.

The conversion of trans peptide bonds to cis peptide bonds is catalyzed by an enzyme, peptide prolyl cis-trans isomerase.

No correct answer is possible at this stage of the book. However, the answer is yes: virtually all of these complexes, no matter how large, are held together by weak noncovalent interactions.