S-44
The Three-Dimensional
Structure of Proteins
chapter
4
1. Properties of the Peptide Bond In x-ray studies of crystalline peptides, Linus Pauling and Robert
Corey found that the CON bond in the peptide link is intermediate in length (1.32 Å) between a typical
CON single bond (1.49 Å) and a CPN double bond (1.27 Å). They also found that the peptide bond is
planar (all four atoms attached to the C—N group are located in the same plane) and that the two
a-carbon atoms attached to the CON are always trans to each other (on opposite sides of the peptide
bond).
(a) What does the length of the CON bond in the peptide linkage indicate about its strength and its
bond order (i.e., whether it is single, double, or triple)?
(b) What do the observations of Pauling and Corey tell us about the ease of rotation about the CON
peptide bond?
Answer
(a) The higher the bond order (double or triple vs. single), the shorter and stronger are
the bonds. Thus, bond length is an indication of bond order. For example, the CPN
bond is shorter (1.27 Å) and has a higher order (n � 2.0) than a typical CON bond
(length � 1.49 Å, n � 1.0). The length of the CON bond of the peptide link (1.32 Å)
indicates that it is intermediate in strength and bond order between a single and
double bond.
(b) Rotation about a double bond is generally impossible at physiological temperatures, and
the steric relationship of the groups attached to the two atoms involved in the double
bond is spatially “fixed.” Since the peptide bond has considerable double-bond character,
there is essentially no rotation, and the OCPO and ONOH groups are fixed in the
trans configuration.
2. Structural and Functional Relationships in Fibrous Proteins William Astbury discovered that
the x-ray diffraction pattern of wool shows a repeating structural unit spaced about 5.2 Å along the
length of the wool fiber. When he steamed and stretched the wool, the x-ray pattern showed a new
repeating structural unit at a spacing of 7.0 Å. Steaming and stretching the wool and then letting it
shrink gave an x-ray pattern consistent with the original spacing of about 5.2 Å. Although these
observations provided important clues to the molecular structure of wool, Astbury was unable to
interpret them at the time.
(a) Given our current understanding of the structure of wool, interpret Astbury’s observations.
(b) When wool sweaters or socks are washed in hot water or heated in a dryer, they shrink. Silk, on
the other hand, does not shrink under the same conditions. Explain.
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Chapter 4 The Three-Dimensional Structure of Proteins S-45
Answer
(a) The principal structural units in the wool fiber polypeptide, a-keratin, are successive
turns of the a helix, which are spaced at 5.4 Å intervals; two a-keratin strands
twisted into a coiled coil produce the 5.2 Å spacing. The intrinsic stability of the helix
(and thus the fiber) results from intra chain hydrogen bonds (see Fig. 4–4a).
Steaming and stretching the fiber yields an extended polypeptide chain with the
b conformation, in which the distance between adjacent R groups is about 7.0 Å.
Upon resteaming, the polypeptide chains again assume the less-extended a-helix
conformation.
(b) Freshly sheared wool is primarily in its a-keratin (a-helical coiled coil) form (see
Fig. 4–10). Because raw wool is crimped or curly, it is combed and stretched to
straighten it before being spun into fibers for clothing. This processing converts the
wool from its native a-helical conformation to a more extended b form. Moist heat
triggers a conformational change back to the native a-helical structure, which shrinks
both the fiber and the clothing. Under conditions of mechanical tension and moist
heat, wool can be stretched back to a fully extended form. In silk, by contrast, the
polypeptide chains have a very stable b-pleated sheet structure, fully extended along
the axis of the fiber (see Fig. 4–6), and have small, closely packed amino acid side
chains (see Fig. 4–13). These characteristics make silk resistant to stretching and
shrinking.
3. Rate of Synthesis of Hair a-Keratin Hair grows at a rate of 15 to 20 cm/yr. All this growth is
concentrated at the base of the hair fiber, where a-keratin filaments are synthesized inside living
epidermal cells and assembled into ropelike structures (see Fig. 4–10). The fundamental structural
element of a-keratin is the a helix, which has 3.6 amino acid residues per turn and a rise of 5.4 Å
per turn (see Fig. 4–4a). Assuming that the biosynthesis of a-helical keratin chains is the
rate-limiting factor in the growth of hair, calculate the rate at which peptide bonds of a-keratin
chains must be synthesized (peptide bonds per second) to account for the observed yearly growth
of hair.
Answer Because there are 3.6 amino acids (AAs) per turn and the rise is 5.4 Å/turn, the
length per AA of the a helix is
� 1.5 Å/AA � 1.5 � 10–10 m/AA
A growth rate of 20 cm/yr is equivalent to
� 6.3 � 10�7 cm/s � 6.3 � 10�9 m/s
Thus, the rate at which amino acids are added is
� 42 AA/s � 42 peptide bonds per second
4. Effect of pH on the Conformation of a-Helical Secondary Structures The unfolding of the
a helix of a polypeptide to a randomly coiled conformation is accompanied by a large decrease in a
property called specific rotation, a measure of a solution’s capacity to rotate plane-polarized light.
Polyglutamate, a polypeptide made up of only L-Glu residues, has the a-helical conformation at pH 3.
6.3 � 10�9 m/s
���
1.5 � 10�10 m/AA
20 cm/year
�����
(365 days/yr)(24 h/day)(60 min/h)(60 s/min)
5.4 Å/turn
��
3.6 AA/turn
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S-46 Chapter 4 The Three-Dimensional Structure of Proteins
When the pH is raised to 7, there is a large decrease in the specific rotation of the solution. Similarly,
polylysine (L-Lys residues) is an a helix at pH 10, but when the pH is lowered to 7 the specific rotation
also decreases, as shown by the following graph.
0
Poly(Glu)
Random conformation
Poly(Lys)
pH
S
pe
ci
fi
c
ro
ta
ti
on
2 4 6 8 10 12 14
� Helix
Random
conformation
� Helix
What is the explanation for the effect of the pH changes on the conformations of poly(Glu) and
poly(Lys)? Why does the transition occur over such a narrow range of pH?
Answer At pH values above 6, deprotonation of the carboxylate side chains of poly(Glu)
leads to repulsion between adjacent negatively charged groups, which destabilizes the a helix
and results in unfolding. Similarly, at pH 7 protonation of the amino-group side chains of
poly(Lys) causes repulsion between positively charged groups, which leads to unfolding.
5. Disulfide Bonds Determine the Properties of Many Proteins Some natural proteins are rich in
disulfide bonds, and their mechanical properties (tensile strength, viscosity, hardness, etc.) are corre-
lated with the degree of disulfide bonding.
(a) Glutenin, a wheat protein rich in disulfide bonds, is responsible for the cohesive and elastic char-
acter of dough made from wheat flour. Similarly, the hard, tough nature of tortoise shell is due to
the extensive disulfide bonding in its a-keratin. What is the molecular basis for the correlation
between disulfide-bond content and mechanical properties of the protein?
(b) Most globular proteins are denatured and lose their activity when briefly heated to 65 �C.
However, globular proteins that contain multiple disulfide bonds often must be heated longer at
higher temperatures to denature them. One such protein is bovine pancreatic trypsin inhibitor
(BPTI), which has 58 amino acid residues in a single chain and contains three disulfide bonds.
On cooling a solution of denatured BPTI, the activity of the protein is restored. What is the
molecular basis for this property?
Answer
(a) Disulfide bonds are covalent bonds, which are much stronger than the noncovalent
interactions (hydrogen bonds, hydrophobic interactions, van der Waals interactions) that
stabilize the three-dimensional structure of most proteins. Disulfide bonds serve to
cross-link protein chains, increasing stiffness, hardness, and mechanical strength.
(b) As the temperature is raised, the increased thermal motion of the polypeptide chains
and the vibrational motions of hydrogen bonds ultimately lead to thermal denaturation
(unfolding) of a protein. Cystine residues (disulfide bonds) can, depending on their
location in the protein structure, prevent or restrict the movement of folded protein
domains, block access of solvent water to the interior of the protein, and prevent the
complete unfolding of the protein. Refolding to the native structure from a random
conformation is seldom spontaneous, owing to the very large number of conformations
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Chapter 4 The Three-Dimensional Structure of Proteins S-47
possible. Disulfide bonds limit the number of conformations by allowing only a few
minimally unfolded structures, and hence the protein returns to its native conformation
more easily upon cooling.
6. Amino Acid Sequence and Protein Structure Our growing understanding of how proteins fold
allows researchers to make predictions about protein structure based on primary amino acid sequence
data. Consider the following amino acid sequence.
(a) Where might bends or b turns occur?
(b) Where might intrachain disulfide cross-linkages be formed?
(c) Assuming that this sequence is part of a larger globular protein, indicate the probable location
(the external surface or interior of the protein) of the following amino acid residues: Asp, Ile,
Thr, Ala, Gln, Lys. Explain your reasoning. (Hint: See the hydropathy index in Table 3–1.)
Answer
(a) Bends or turns are most likely to occur at residues 7 and 19 because Pro residues are
often (but not always) found at bends in globular folded proteins. A bend may also occur
at the Thr residue (residue 4) and, assuming that this is a portion of a larger polypep-
tide, at the Ile residue (residue 1).
(b) Intrachain disulfide cross-linkages can form only between residues 13 and 24 (Cys
residues).
(c) Amino acids with ionic (charged) or strongly polar neutral groups (e.g., Asp, Gln, and
Lys in this protein) are located on the external surface, where they interact optimally
with solvent water. Residues with nonpolar side chains (such as Ala and Ile) are situ-
ated in the interior, where they escape the polar environment. Thr is of intermediate
polarity and could be found either in the interior or on the exterior surface (see
Table 3–1).
7. Bacteriorhodopsin in Purple Membrane Proteins Under the proper environmental conditions,
the salt-loving bacterium Halobacterium halobium synthesizes a membrane protein (Mr 26,000)
known as bacteriorhodopsin, which is purple because it contains retinal (see Fig. 10–21). Molecules of
this protein aggregate into “purple patches” in the cell membrane. Bacteriorhodopsin acts as a light-
activated proton pump that provides energy for cell functions. X-ray analysis of this protein reveals
that it consists of seven parallel a-helical segments, each of which traverses the bacterial cell mem-
brane (thickness 45 Å). Calculate the minimum number of amino acids necessary for one segment of
a helix to traverse the membrane completely. Estimate the fraction of the bacteriorhodopsin protein
that is involved in membrane-spanning helices. (Use an average amino acid residue weight of 110.)
Answer Using the parameters from Problem 3 (3.6 AA/turn, 5.4 Å/turn), we can calculate
that there are 0.67 AA/Å along the axis of a helix. Thus, a helix of length 45 Å (sufficient to
span the membrane) requires a minimum of (45 Å)(0.67 AA/Å) � 30 amino acid residues.
The membrane protein has an Mr of 26,000 and average AA Mr of 110. Thus the protein
contains 26,000/110 � 240 AA. Of these, (30 AA/helix)(7 helices) � 210 AA are involved in
membrane-spanning helices, which is 210/240 � 0.87, or 87%, of the protein.
1 2 3 4 5 6 7 8 9 10
Ile Ala His Thr Tyr Gly Pro Phe Glu Ala
11 12 13 14 15 16 17 18 19 20
Ala Met Cys Lys Trp Glu Ala Gln Pro Asp
21 22 23 24 25 26 27 28
Gly Met Glu Cys Ala Phe His Arg
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S-48 Chapter 4 The Three-Dimensional Structure of Proteins
8. Protein Structure Terminology Is myoglobin a motif, a domain, or a complete three-dimensional
structure?
Answer Myoglobin is all three. The folded structure, the “globin fold,” is a motif found in all
globins. The polypeptide folds into a single domain, which for this protein represents the
entire three-dimensional structure.
9. Pathogenic Action of Bacteria That Cause Gas Gangrene The highly pathogenic anaerobic bac-
terium Clostridium perfringens is responsible for gas gangrene, a condition in which animal tissue
structure is destroyed. This bacterium secretes an enzyme that efficiently catalyzes the hydrolysis of
the peptide bond indicated by an asterisk:
X Gly Pro Y
H2O X COO H3N Gly Pro Y
�*
��
O2N
NO2
NH C
C
CH3 CH3
H
H
COOH
where X and Y are any of the 20 common amino acids. How does the secretion of this enzyme
contribute to the invasiveness of this bacterium in human tissues? Why does this enzyme not affect
the bacterium itself?
Answer Collagen is distinctive in its amino acid composition, having a very high proportion of
Gly (35%) and Pro residues. The enzyme secreted by the bacterium is a collagenase, which
breaks down collagen at the X–Gly bonds and damages the connective-tissue barrier (skin,
hide, etc.) of the host; this allows the bacterium to invade the host tissues. Bacteria do not
contain collagen and thus are unaffected by collagenase.
10. Number of Polypeptide Chains in a Multisubunit Protein A sample (660 mg) of an oligomeric
protein of Mr 132,000 was treated with an excess of 1-fluoro-2,4-dinitrobenzene (Sanger’s reagent)
under slightly alkaline conditions until the chemical reaction was complete. The peptide bonds of the
protein were then completely hydrolyzed by heating it with concentrated HCl. The hydrolysate was
found to contain 5.5 mg of the following compound:
2,4-Dinitrophenyl derivatives of the a-amino groups of other amino acids could not be found.
(a) Explain how this information can be used to determine the number of polypeptide chains in an
oligomeric protein.
(b) Calculate the number of polypeptide chains in this protein.
(c) What other protein analysis technique could you employ to determine whether the polypeptide
chains in this protein are similar or different?
Answer
(a) Because only a single 2,4-dinitrophenyl (DNP) amino acid derivative is found, there is
only one kind of amino acid at the amino terminus (i.e., all the polypeptide chains have
the same amino-terminal residue). Comparing the number of moles of this derivative to
the number of moles of protein gives the number of polypeptide chains.
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Chapter 4 The Three-Dimensional Structure of Proteins S-49
(b) The amount of protein � (0.66 g)/(132,000 g/mol) � 5 � 10�6 mol.
Because Mr for DNP-Val (C11H13O6N3) � 283, the amount of DNP-Val � (0.0055 g)/
(283 g/mol) � 1.9 � 10�5 mol.
The ratio of moles of DNP-Val to moles of protein gives the number of amino-
terminal residues and thus the number of chains per oligomer:
� 4 polypeptide chains
An alternative approach to the problem is through the proportionality (n � number of
polypeptide chains):
�
n � � 3.9 ≈ 4
(c) Polyacrylamide gel electrophoresis in the presence of a detergent (such as sodium
dodecylsulfate [SDS]) and an agent that prevents the formation of disulfide bonds (such
as b-mercaptoethanol) would provide information on subunit structure of a protein. In
the example here, an oligomeric protein of Mr 132,000 that had four identical subunits
would produce a single band on the electrophoretic gel, with apparent Mr ~33,000
(132,000/33,000 � 4). If the protein were made up of different polypeptide subunits,
they would likely appear as multiple discrete bands on the gel.
11. Predicting Secondary Structure Which of the following peptides is more likely to take up an
�-helical structure, and why?
(a) LKAENDEAARAMSEA
(b) CRAGGFPWDQPGTSN
Answer By cursory inspection, peptide (a) has five Ala residues (most likely to take up an
�-helical conformation), and peptide (b) has five Pro and Gly residues (least often found in an
� helix). This suggests that (a) is more likely than (b) to form an � helix. Referring to Table 4–1,
(a) has 15 residues with a total ��Gº of 13 kJ/mol, and (b) has 15 residues with a total ��Gº
of 41 kJ/mol. Given that a lower ��Gº indicates a greater tendency to take up an �-helical
structure, this confirms that peptide (a) is much more likely to form an � helix.
12. Amyloid Fibers in Disease Several small aromatic molecules, such as phenol red (used as a non-
toxic drug model), have been shown to inhibit the formation of amyloid in laboratory model systems. A
goal of the research on these small aromatic compounds is to find a drug that would efficiently inhibit
the formation of amyloid in the brain in people with incipient Alzheimer’s disease.
(a) Suggest why molecules with aromatic substituents would disrupt the formation of amyloid.
(b) Some researchers have suggested that a drug used to treat Alzheimer’s disease may also be ef-
fective in treating type 2 (adult onset) diabetes mellitus. Why might a single drug be effective in
treating these two different conditions?
Answer
(a) Aromatic residues seem to play an important role in stabilizing amyloid fibrils. Thus,
molecules with aromatic substituents may inhibit amyloid formation by interfering with
the stacking or association of the aromatic side chains.
(b) Amyloid is formed in the pancreas in association with type 2 diabetes, as it is in the brain in
Alzheimer’s disease. Although the amyloid fibrils in the two diseases involve different pro-
teins, the fundamental structure of the amyloid is similar and similarly stabilized in both,
and thus they are potential targets for similar drugs designed to disrupt this structure.
(5.5 mg)(132,000 g/mol)
���
(660 mg)(283 g/mol)
5.5 mg
�
660 mg
n(283 g/mol)
��
132,000 g/mol
1.9 � 10–5 mol DNP-Val
���
5 � 10–6 mol protein
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S-50 Chapter 4 The Three-Dimensional Structure of Proteins
Biochemistry on the Internet
13. Protein Modeling on the Internet A group of patients with Crohn’s disease (an inflammatory
bowel disease) underwent biopsies of their intestinal mucosa in an attempt to identify the causative
agent. Researchers identified a protein that was present at higher levels in patients with Crohn’s
disease than in patients with an unrelated inflammatory bowel disease or in unaffected controls. The
protein was isolated, and the following partial amino acid sequence was obtained (reads left to
right):
EAELCPDRCI HSFQNLGIQC VKKRDLEQAI SQRIQTNNNP FQVPIEEQRG
DYDLNAVRLC FQVTVRDPSG RPLRLPPVLP HPIFDNRAPN TAELKICRVN
RNSGSCLGGD EIFLLCDKVQ KEDIEVYFTG PGWEARGSFS QADVHRQVAI
VFRTPPYADP SLQAPVRVSM QLRRPSDREL SEPMEFQYLP DTDDRHRIEE
KRKRTYETFK SIMKKSPFSG PTDPRPPPRR IAVPSRSSAS VPKPAPQPYP
(a) You can identify this protein using a protein database on the Internet. Some good places to start
include Protein Information Resource (PIR; http://pir.georgetown.edu), Structural Classification
of Proteins (SCOP; http://scop.mrc-lmb.cam.ac.uk/scop), and Prosite (http://expasy.org/prosite).
At your selected database site, follow links to the sequence comparison engine. Enter about
30 residues from the protein sequence in the appropriate search field and submit it for analysis.
What does this analysis tell you about the identity of t
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