Unformatted text preview: Chapter 3 Proteins 3 THE SHAPE AND STRUCTURE OF PROTEINS In This Chapter DEFINITIONS THE SHAPE AND
PROTEINS A45 PROTEIN FUNCTION A53 3–1 Quaternary structure 3–2 a helix 3–3 Primary structure 3–4 Binding site 3–5 Tertiary structure 3–6 Polypeptide backbone 3–7 b sheet 3–8 Protein domain 3–9 Secondary structure TRUE/FALSE
3–10 True. In both states—stretched like a string and properly folded—a protein
has a highly ordered arrangement of its atoms. A folded protein is stable at a
near entropy minimum because the entropic cost is more than balanced by
the contributions of weak bonds. A stretched out protein, however, is not
stable at this entropy minimum and will assume a more disordered state;
that is, it will maximize its entropy. 3–11 True. In a b sheet the amino acid side chains in each strand are alternately
positioned above and below the sheet. This relationship can be seen in Figure 3–38 (see Answer 3–19), which shows that the carbonyl oxygens alternate
from one side of the strand to the other. Thus, each strand in a b sheet can
be viewed as a helix in which each successive amino acid is rotated 180°. 3–12 True. Chemical groups on such protruding loops can often surround a
molecule, allowing the protein to bind to it with many weak bonds. THOUGHT PROBLEMS
3–13 Free amino acids have an amino group and a carboxylate group, both of
which are charged at neutral pH. In proteins these groups are involved in
peptide bonds, which are uncharged. Thus, the hydrophobicity/hydrophilicity of a free amino acid is not the same as that of its side chain in a protein.
To measure the hydrophobicity/hydrophilicity of the side chains, it is
common to assess the properties of side-chain analogs. Thus, for alanine
one would use methane, for threonine, ethanol, for aspartic acid, acetic A45 A46 Chapter 3: Proteins
hydrophobicity T S
D 10 kcal/mole kcal/
mole increasing hydrophilicity amino
acid increasing hydrophobicity hydrophobicity
(dark bars) L I V F M W A C G H Q K N E Y –10 –20 hydrophilicity Figure 3–34 Measurements of
hydrophilicity and hydrophobicity of
side-chain analogs (Answer 3–13).
Partition coefficients are listed for each
amino acid side-chain analog. A partition
coefficient is the equilibrium constant for
the solubility of a solute in two different
phases, and is expressed as kcal/mole.
The rank order for hydrophobicity is
listed in descending order, while that for
hydrophilicity is listed in ascending order.
If there were a perfect negative
correspondence between hydrophobicity
and hydrophilicity, the amino acids
would appear in the same order in the
two lists. The data are plotted with the
amino acids arranged in order of
decreasing hydrophobicity (dark bars)
from left to right. Hydrophilicity is
indicated by the overlaid gray bars. acid, and so on. To assess hydrophilicity, one can measure the solubility of
the side-chain analogs in water. In general, this is done by determining how
the side-chain analog partitions between a vapor of the analog and water.
Hydrophobicity can be measured in an analogous way by assessing how a
side-chain analog partitions between water and a nonpolar solvent such as
cyclohexane. One might imagine that the rank order for hydrophilicity
would be the reverse of that for hydrophobicity. From the rank order lists in
Figure 3–34, that is mostly true, but there are differences, most notably tyrosine (Y) and tryptophan (W). O8 Reference: Creighton TE (1993) Proteins, 2nd ed, pp 153–155. New York: WH
3–14 O– O O7 Hydrogen bonds, electrostatic attractions, van der Waals attractions, and the
hydrophobic force. N9 N8
A. Heating egg-white proteins denatures them, allowing them to interact with
one another in ways that were not possible at the lower temperature of the
hen’s oviduct. This process forms a tangled meshwork of polypeptide chains.
In addition to these interactions, interchain disulfide bonds also form, so
that hard-boiled egg white becomes one giant macromolecule.
B. Dissolving hard-boiled egg white requires a strong detergent to overcome
the noncovalent interchain interactions and mercaptoethanol to break the
covalent disulfide bonds. Together, but not separately, the two reagents
eliminate the bonds that hold the tangled protein chains in place. Try it for
3–16 3–17 In an a helix, the carbonyl oxygen of the first amino acid hydrogen bonds to
the amide nitrogen of amino acid 5 (Figure 3–35). Thus, there can be no a
helix shorter than five amino acids. The single hydrogen bond that would be
formed with five amino acids gives too little stability to the structure for any
helicity to be detected. Only with six amino acids—two hydrogen bonds—
do you begin to detect some. Helicity becomes increasingly apparent as
more amino acids, hence more hydrogen bonds, are added.
The ends of a helices, like polar amino acids, are almost always found at the
surface of a protein where they can interact with polar water molecules. In
addition to their partial charge, the backbones of the four amino acids at
either end of the helix carry hydrogen-bonding groups that are unsatisfied
by hydrogen-bonding within the helix (Figure 3–36). These groups also add
to the polarity of the termini of a helices. O5
N7 N6 O4 N5
O3 O2 N4 N3 O1 N2 N1 H3+ Figure 3–35 Schematic of an a helix,
showing the pattern of hydrogen bonds
between carbonyl oxygens and amide
nitrogens within the helix (Answer 3–16). THE SHAPE AND STRUCTURE OF PROTEINS A47
PEPTIDE 2 PEPTIDE 1 O
O d- O K R
15 E S 8 1 K E
12 4 S 5 D 11 I 9
I 10 17 V F
12 V 4 5
16 M S 18 9 L R 7 2
14 W 6 8 R 11 13 F
N R 16 T L 18 R
15 3 V V
L 13 T F I I 10
F 17 L 6
L L N
N d+ N Figure 3–36 Representation of an
a helix showing dipole and
groups at its ends (Answer 3–17).
Non-hydrogen-bonded Os and Ns
are labeled. Figure 3–37 Arrangement of amino
acids of the three peptide
sequences around helix wheels
(Answer 3–18). PEPTIDE 3
A 15 R 8 1 V
12 4 S
16 11 9 I 18 3–18
3–19 The first two peptide sequences, but not the third, would give amphiphilic
helices, as shown in Figure 3–37.
As illustrated in Figure 3–38, the first three strands of the sheet are antiparallel to their neighbors, whereas the fourth strand is parallel to the third. 2 V 7
R 14 13
D. Because the side chains of the amino acids alternately project above and
below the sheet, a sequence that could form a strand in an amphiphilic b
sheet should have alternating hydrophobic and hydrophilic amino acids.
Only choice D satisfies this condition.
3–21 None of these folds would give a knot when stretched out. This is a general
principle: proteins fold without forming knots. One might imagine that it
would be difficult to thread the end of a protein through an interior loop to
form a knot. A folding pathway to such a knotted form might not be achieveable through random motions in a reasonable time. 3–22 Antiparallel strands are commonly formed by a polypeptide chain that folds
back on itself. Thus, only a few amino acids are required to allow the
polypeptide chain to make the turn. By contrast, parallel strands must be
connected by a polypeptide chain that is at least as long as the strands in the N N
N N N N
N N N N
N N N N
N N N N
N Figure 3–38 A segment of b sheet
showing the polarity (N to C) of the
individual strands (Answer 3–19). L
L I A48 Chapter 3: Proteins
sheet. For a long peptide, a common solution for satisfying backbone hydrogen-bonding requirements is an a helix (Figure 3–39). 3–23 Proteins obviously can’t search all possible conformations on their way to
finding the correct one. Thus, there must be defined pathways to simplify
the search. It is now thought that weak interactions rapidly cause the protein
to collapse into a molten globule, in which bonding interactions are transient and chains maintain fluidity. Within the molten globule, very weak secondary structures form and disappear, as do tertiary interactions. The formation of small elements of correct secondary structure, stabilized by
appropriate tertiary interactions, then appears to nucleate formation of the
final structure. This general folding pathway represents a fight between the
maximization of entropy, which tends to keep the protein as random as possible, and the minimization of enthalpy through formation of weak bonds.
Increasing numbers of weak interactions pull the structure through a succession of increasingly well defined states to the final conformation. This
conceptualization of folding has been likened to a funnel, and is commonly
referred to as the folding funnel, with multiple routes of progress down the
funnel accompanied by an increase in nativelike structures.
Reference: Fersht A (1999) Structure and Mechanism in Protein Science, pp
575–600. New York: WH Freeman. 3–24 from lysozyme
PARALLEL STRANDS from alcohol dehydrogenase Figure 3–39 Connections between
antiparallel and parallel strands of a
b sheet (Answer 3–22). Many different strings of amino acids can give rise to identical protein folds.
The many amino acid differences between the homeodomain proteins from
yeast and Drosophila are among the many possible ones that do not alter folding and function. This question could have been framed in another way;
namely, how many amino acid changes are required to convert, say, an a helix
into a b sheet? The answer is: surprisingly few. These two answers underscore
the difficulty in predicting protein structures from amino acid sequences. 3–25 ANTIPARALLEL STRANDS The statement is correct. The length of evolutionary separation does not
depend on how long yeast and Drosophila have been around, but rather on
when their last common ancestor existed. Evolutionary separation is calculated as twice the time from when the last common ancestor existed; that is,
counting backwards from one kind of organism to the common ancestor and
then forward to the other kind of organism. 3–26
A. The protein in Figure 3–5 is composed of two domains. The protein can be
cleaved in the exposed peptide segment that links the two domains (Figure
3–40). Fragments that correspond to individual domains are likely to fold
properly. It is common experience that isolated domains are easier to crystallize than the entire protein.
B. The ability to form a crystal depends on the surface characteristics of the
protein because it must be able to interact with itself in a repeating pattern
to form a crystal. Homologous proteins from different species, which fold
the same (like the homeodomain proteins in Problem 3–24), differ subtly in
their surface characteristics. As a result, the protein from one species may
crystallize readily, while that from another species may not crystallize at all.
A single amino acid change sometimes makes all the difference.
3–27 Perhaps this is so. Nevertheless, it seems likely that new, and useful, protein
folds have been invented during evolution by the chance fusion of genes. A
distribution of protein folds within the tree of life would be informative. Protein folds that are distributed in all divisions of the tree—archaea, eubacteria,
and eucaryotes—were very likely present in their last common ancestor.
More recently invented protein folds would likely be confined to a single division, or a few branches. Even if all protein folds were present in the last common ancestor, it seems unlikely that there would be a one-to-one correspondence between folds and genes. Surely, the evolution that led up to the last
common ancestor would already have exploited some of the benefits of gene
duplication and refinement of function that lead to families of related genes. Figure 3–40 Catabolite activator protein
(Answer 3–26). The arrow shows the site
of cleavage in the exposed peptide
segment linking the two domains. THE SHAPE AND STRUCTURE OF PROTEINS
The limited number of protein folds raises a more fundamental question
about the total number of protein folds that are possible. It may be that evolutionary processes have already exploited most of the stable folds that are
possible. Alternatively, it may be that the number of possible stable folds
greatly exceeds those currently used on Earth. The simple cataloguing of
natural protein folds will not address this more basic question.
3–28 Generally speaking, an identity of at least 30% is needed to be certain that a
match has been found. Matches of 20% to 30% are problematical and difficult to distinguish from background ‘noise.’ Searching for distant relatives
with the whole sequence usually drops the overall identity below 30%
because the less conserved portions of the sequence dominate the comparison. Thus, searching with shorter, conserved portions of the sequence gives
the best chance for finding distant relatives. 3–29 As shown in Figure 3–41, the three protein monomers have distinctly different assembly properties because of the three-dimensional arrangement of
their complementary binding surfaces. Monomer A would assemble into a
sheet; monomer B would assemble into a long chain; monomer C would
assemble into a ring composed of four subunits. 3–31 (A) (B) (C) The close juxtaposition of the N- and C-termini of this kelch domain identifies it as a ‘plug-in’ type domain. ‘In-line’ type domains have their N- and Ctermini on opposite sides of the domain. 3–30 A49 Choice B (Æ¨) is the only arrangement of DNA-binding sites that matches
the arrangement of subunits in the ‘head-to-head’ Cro dimer. The DNA that
corresponds to such an arrangement is known as a palindrome: Figure 3–41 Assembly of protein
monomers (Answer 3–30). ATCG.CGAT
Rotation of this sequence 180° about the central dot gives an identical
sequence, just as does rotation of the arrows (Æ¨). This demonstrates that
the Cro dimer and its recognition sequence have the same symmetry, as
3–32 ‘Head-to-tail’ dimers have unsatisfied binding sites at each end, which
would lead to the formation of chains (see Figure 3–41B). 3–33 Proteins 1, 3, 4, and 5 can form head-to-head dimers, as illustrated for protein 1 in Figure 3–42A. All binding surfaces that allow proteins to interact are
complementary. The binding surfaces that allow two copies of a protein to
form a ‘head-to-head’ dimer must be self-complementary because they
bind to themselves. To be self-complementary, one half of the binding site
must be complementary to the other half. This means that the two halves
can be folded on top of one another, with properly matched binding, across
a line drawn through the center of the binding site, as illustrated for the protein 1 binding surface in Figure 3–42B. There is no line across which proteins
2 and 6 can be folded to make their binding partners match. Inclusion of
protrusions and invaginations would not have altered this general principle:
complementary binding surfaces can be folded so that a protrusion on one
side inserts into an invagination on the other side. 3–34 The coil 1A segment of nuclear lamin C matches the heptad repeat at 9 of 11
positions (Figure 3–43), which is very good. The match need not be perfect hydrophobic amino acids * *
***** * *
* * * *
match with heptad repeat (A) +
d − − + a (B) a d + − d a Figure 3–42 Self-complementarity in
proteins (Answer 3–33). (A) Head-to-head
dimer formation. (B) Self-complementary
Figure 3–43 Heptad repeat motif in the
coil 1A region of nuclear lamin C (Answer
3–34). Hydrophobic amino acids are
marked with an asterisk (*). When a
hydrophobic amino acid occurs at an A or
D in the heptad repeat, it is assigned a +.
The start of the heptad repeat was
positioned to maximize matches. A50 Chapter 3: Proteins
Figure 3–44 Vernier assembly of 10-nm
and 14-nm subunits (Answer 3–35).
70 nm to allow formation of a coiled-coil. The matches to the heptad repeat in the
other two marked segments (coil 1B and coil 2, see Figure 3–9) are not as
good, but they are still acceptable for the formation of a coiled-coil.
Reference: McKeon FD, Kirschner MW & Caput D (1986) Homologies in
both primary and secondary structure between nuclear envelope and intermediate filament proteins. Nature 31, 463–468.
A. The final fiber will be 70 nm in length and will contain 7 of the 10-nm subunits and 5 of the 14-nm subunits (Figure 3–44). Assemblies that are any
shorter will have unoccupied binding sites at one end or the other.
B. It makes no difference how the original pair come together. So long as the
unoccupied binding sites of the initial pair are filled in by other subunits, the
fiber will always grow to the same length. CALCULATIONS
3–36 At equilibrium there would be 1 unfolded protein for every 107 folded proteins. This ratio comes from substituting values for DG° (9.9 kcal/mole), R,
and T into the equation and solving for logK:
logK = (9.9)/[( –2.3) ¥ (1.98 ¥ 10–3) ¥ (310)] = 10–7
Since K = [U]/[F],
logK = log ([U]/[F]) = –7
Taking the log of both sides,
[U]/[F] = 10–7, or [U] = 10–7 [F] 3–37
A. The stability of lysozyme at 37°C is 10 kcal/mole. From your measurements
G°unfolded = 128 – 119 = 9 kcal/mole
G°folded = 75 – 76 = –1 kcal/mole
DG° = G°unfolded – G°folded = 9 + 1 = 10 kcal/mole
B. This problem illustrates the essence of the protein folding problem: the
overall stability of a protein derives from small differences between large
numbers. For lysozyme, for example, not only must the hundreds of interactions in the folded state be evaluated, but so also must the hundreds of
interactions in the unfolded state. In each state the sum of the enthalpic
contributions is large, as is the sum of the entropic contributions. As shown
by the calculation above, these large numbers are subtracted from one
another to give a small number. Thus, the enthalpic and entropic values of
the individual interactions must be known with exquisite precision in order
to predict the stability of the folded protein. Yet individual weak bonds can THE SHAPE AND STRUCTURE OF PROTEINS
vary considerably in strength. For example, individual hydrogen bonds in
proteins vary from a few tenths of a kcal/mole to over 1 kcal/mole. These
considerations make prediction of protein structure from these sorts of calculations virtually impossible.
Reference: Creighton TE (1993) Proteins, 2nd edn, pp 297–300. New York:
3–38 Since there are 20 possible amino acids at each position in a 300 amino acid
long protein, there are 20300 (which is 10390) possible proteins. The mass of
one copy of each possible protein would be
mass = 110 d ¥ 300 aa ¥ 10390 proteins ¥
6 ¥ 1023 d
mass = 5.5 ¥ 10370 g
Thus, the mass of protein would exceed the mass of the observable universe
(1080 g) by a factor of about 10290!! 3–39 The fraction of correctly synthesized proteins will be 0.91 for a 1000 amino
acid protein, 0.37 for a 10,000 amino acid protein, and 0.00005 for a 100,000
amino acid protein. The calculation is shown below for a 1000 amino acid
PC = (fC)n
= 0.91 3–40
A. As calculated in the previous problem, synthesis of one 10,000 amino acid
protein would be expected to occur correctly 37% of the time. By contrast,
synthesis of each 200 amino acid subunit would be expected to occur correctly 98% of the time [PC = (fC)n = (0.9999)200 = 0.98]. Assembly of the mixture of subunits into correct ribosomes follows the same equation [PC = (fC)n
= (0.98)50 = 0.37]. Thus, making a ribosome from subunits gives the same
final fraction of correct ribosomes, 37%, as making them from one long protein.
B. The assumption in part A that correct and incorrect subunits are assembled
into ribosomes with equal likelihood is not true. Any mistake that interferes
with the correct folding of a subunit, or that interferes with the ability of the
subunit to bind to other subunits, would eliminate the subunit from assembly into a ribosome. As a result, the fraction of correctly assembled ribosomes would be higher than calculated in part A. Thus, the value of subunit
synthesis lies not in more accurate synthesis, but rather in permitting quality control mechanisms to reject incorrect subunits efficiently. DATA HANDLING
3–41 If unfolding of the protein simply reflected the titration of a buried histidine,
it should require 2 pH units to go from 9% to 91% completion. The actual
unfolding curve takes only 0.3 pH units to span this range. This sharp transition indicates a highly cooperative process; when the protein starts to
unfold, it completes the process rapidly. For example, it might be that several buried histidines can ionize when the chain starts to unfold, so that
when one goes they all go together. Note also that as soon as a buried histidine (pK of 4 in this example) becomes accessible to solvent its pK will shift
toward its normal value of 6, significantly steepening its titration curve.
Reference: Creighton TE (1993) Proteins, 2nd ed, pp 288–289. New York: WH
Freeman. A51 A52 Chapter 3: Proteins 3–42 As expected, hydrophobic amino acid side chains are most frequently
buried and hydrophilic side chains are least commonly buried. Perhaps the
biggest surprise in this list is the high proportion of cysteine (C) side chains
that are buried. Cysteine is generally grouped with polar amino acids
because of its –SH group, but its hydrophobic/hydrophilic properties (see
Answer 3–13) indicate that it is, at best, weakly polar. Tyrosine (Y) also
deserves comment. Tyrosine is usually grouped with polar amino acids
because of its hydroxyl moiety; however, its measured hydrophobic/
hydrophilic properties are ambiguous (see Answer 3–13), indicating that it is
only weakly polar. By the criterion of ‘buriedness,’ it clearly behaves like
other polar amino acids. 3–43
A. These data are consistent with the hypothesis that the springlike behavior of
titin is due to the sequential unfolding of Ig domains. First, the fragment
contained seven Ig domains and there are seven peaks in the force-versusextension curve. In addition, the peaks themselves are what you might
expect for sequential unfolding. Second, in the presence of a protein denaturant, conditions under which the domains will already be unfolded, the
peaks disappear and the extension per unit force increases. Third, when the
domains are cross-linked, and therefore unable to unfold, the peaks disappear and extension per unit force decreases.
B. The spacing between peaks, about 25 nm, is almost exactly what you would
calculate for the sequential unfolding of Ig domains. The folded domain
occupies 4 nm, but when unfolded, its 89 amino acids would stretch to
about 30 nm (89 ¥ 0.34 nm), a change of 26 nm.
C. The existence of separate, discrete peaks means that each domain unfolds
when a characteristic force is applied, implying that each domain has a defined
stability. The collection of domains unfolds in order from least stable to most
stable. Thus, it takes a little more force each time to unfold the next domain.
D. The sudden collapse of the force at each unfolding event reflects an important principle of protein unfolding; namely, its cooperativity. Proteins tend
to unfold in an all-or-none fashion (see Problem 3–41). A small number of
hydrogen bonds are crucial for holding the folded domain together (Figure
3–45). The breaking of these bonds triggers cooperative unfolding.
Reference: Rief M, Gutel M, Oesterhelt F, Fernandez JM & Gaub HE (1997)
Reversible folding of individual titin immunoglobulin domains by AFM.
Science 276, 1109–1112.
A. None are detected in this experiment. Treating first with radiolabeled NEM
shows that many cytosolic proteins have cysteines that are not linked by
disulfide bonds. Treating first with unlabeled NEM to block these sites, followed by DTT to break disulfide bonds, should expose any –SH groups that
were linked by disulfide bonds. These newly exposed –SH groups should be
labeled by subsequent treatment with radiolabeled NEM. The absence of
labeling indicates that no cysteines were involved in disulfide bonds.
B. BSA and insulin are labeled extensively only after their disulfide bonds have
been broken by treatment with DTT. In the absence of DTT treatment, BSA
is weakly labeled. Since BSA has an odd number of cysteines, at least one
cannot be involved in disulfide bonds. Structural analysis confirms that one
of its 37 cysteines is not involved in a disulfide bond.
C. Because the ER is the site where disulfide bond formation is catalyzed in
preparation for export of proteins, it is expected that lysates from cells that
have internal membranes would have many proteins with disulfide bonds.
N Figure 3–45 Hydrogen bonds that lock
the domain into its folded conformation
(Answer 3–43). The indicated hydrogen
bonds (gray lines), when broken, trigger
unfolding of the domain. If you compare
this topological diagram with the threedimensional structure in Figure 3–12A,
you can pick out the two short b strands
that are involved in forming these
hydrogen bonds. PROTEIN FUNCTION PROTEIN FUNCTION
3–45 Scaffold protein 3–46 Feedback inhibition 3–47 Antibody 3–48 Active site 3–49 Enzyme 3–50 Protein phosphatase 3–51 Linkage 3–52 Protein kinase 3–53 Transition state 3–54 SCF ubiquitin ligase 3–55 Allosteric protein 3–56 Proteomics 3–57 Coenzyme TRUE/FALSE
3–58 False. The pK values of specific side-chain groups depend critically on the
environment. On the surface of a protein, in the absence of surrounding
charged groups, the pK of a carboxyl group is usually close to that of the free
amino acid. In the neighborhood of negatively charged groups, the pK of a
carboxyl group is usually higher; that is, the proton dissociates less readily
since the increase in local density of negative charge is not favored. The
opposite is true in a positively charged environment. In hydrophobic surroundings the dissociation of a proton can be substantially suppressed,
since the presence of a naked charge in such an environment is highly disfavored. It is this ability to alter the reactivities of individual groups that
allows proteins to fine-tune their biological functions. 3–59 False. Assuming that the three-dimensional structure of at least one family
member is known, it would be possible to use evolutionary tracing—fitting
the primary sequence to the structure—to determine where the conserved
amino acids cluster on the surfaces of the proteins. Clusters of conserved
amino acids are likely to correspond to important regions such as those
involved in binding to specific ligands or other proteins. Knowing where
such binding sites reside on the surface does not identify the protein’s function. You would not know whether the protein was an enzyme or a structural
protein, or what it bound to. Some other approach, usually biochemical,
would be required to elucidate the function. 3–60 False. The turnover number is constant since it is Vmax divided by enzyme
concentration. For example, a 2-fold increase in enzyme concentration
would give a 2-fold higher Vmax, but it would give the same turnover number: 2 Vmax/2 [E] = k3. 3–61 True. The term cooperativity embodies the idea that changes in the conformation of one subunit are communicated to the other identical subunits in
any given molecule, so that all of these subunits are in the same conformation. 3–62 True. Each cycle of phosphorylation–dephosphorylation hydrolyzes one
molecule of ATP; however, it is not wasteful in the sense of having no benefit. A53 A54 Chapter 3: Proteins
H3C site 2
CH3 Figure 3–46 The two binding sites for
valine and threonine on valyl-tRNA
synthetase (Answer 3–65). CH CH
3 OH H3C CH O CH
COO− +H N
3 C O O
tRNA Constant cycling allows the regulated protein to switch quickly from one
state to another in response to stimuli that require rapid adjustments of cellular metabolism or function. This is the essence of effective regulation.
3–63 False. Although many of the conformational changes induced by ligand
binding are relatively small, in some instances these local changes are propagated through a molecule to give rise to changes of more than a nanometer. The conformational change triggered by hydrolysis of GTP by EF-Tu, for
example, allows two domains of the protein to separate by 4 nm. THOUGHT PROBLEMS
3–64 Antifreeze proteins function by binding to tiny ice crystals and arresting
their growth, thereby preventing the fish from freezing. Ice crystals that form
in the presence of antifreeze proteins are abnormal in that their surfaces are
curved instead of straight. The various forms of the antifreeze proteins in
these fishes are all composed of repeats of a simple glycotripeptide (ThrAla/Pro-Ala) with a disaccharide attached to each threonine. The genes for
these antifreeze proteins were apparently derived by repeated duplication of
a small segment of a protease gene.
References: Cheng CHC & Chen L (1999) Evolution of an antifreeze protein.
Nature 401, 443–444.
Jia Z & Davies PL (2002) Antifreeze proteins: an unusual receptor–ligand
interaction. Trends Biochem. Sci. 27, 101–106. 3–65 To bind to valine, the valyl-tRNA synthetase uses a binding pocket of the
proper shape that is lined with hydrophobic residues. Such a binding site
permits valine to bind well but does not fully exclude threonine, which has
the same shape and a single polar hydroxyl group (Figure 3–46). The second
binding site is much more specific for threonine because it contains an
appropriately positioned hydrogen-bond acceptor that makes a specific
hydrogen bond with threonine but not with valine. Even though valine can
fit into the site, it cannot bind tightly and is thus a very poor substrate for the
hydrolysis reaction. 3–66 Of all the possible pairs, only proteins 2 and 6 can interact in a way that satisfies all the binding groups on their binding surfaces (Figure 3–47). This sort
of complementary arrangement of binding moieties is characteristic of surface–surface interactions between proteins. 3–67 The problem is that the off rate for the antibody–enzyme complex is too slow.
In order for the peptide to displace the enzyme from the column, the enzyme
must first dissociate from the antibody. The antibody binding sites would
then be quickly bound by the peptide, whose high concentration would prevent the enzyme from reattaching to the antibody (any newly exposed antibody binding site would be bound by peptide). In principle, you could soak
the column with peptide for several days (for several dissociation half-times,
see Problem 3–90), but this usually has adverse consequences for the quality
and activity of the enzyme preparation. In general, high-affinity antibodies
have slow off rates and are unsuitable for affinity chromatography. + d d a – + –
a 2 2 a + a – + d – d
6 Figure 3–47 Binding of protein 2 with
protein 6 (Answer 3–66). 6 PROTEIN FUNCTION
Special procedures have been devised for preparing or identifying antibodies that work in such experiments. Usually, lower-affinity antibodies are
used, or chromatography is carried out under special conditions that reduce
the affinity of the antibody.
Reference: Thompson NE & Burgess RR (1996) Immunoaffinity purification
of RNA polymerase II and transcription factors using polyol-responsive
monoclonal antibodies. Methods Enzymol. 274, 513–526.
3–68 The reaction rate for the altered enzyme would be substantially slower than for
the normal enzyme. The reaction rate is related to the activation energy, which
is the difference in energy between the trough labeled ES in Figure 3–16 and the
transition state: the larger the activation energy, the slower the rate. If the
altered enzyme bound the substrate with higher affinity (a lower ES trough),
then the activation energy would increase and the reaction would slow down. 3–69
D. Because an enzyme has a fixed number of active sites, the rate of the reaction cannot be further increased once the substrate concentration is sufficient to bind to all the sites. It is the saturation of binding sites that leads to
an enzyme’s saturation behavior. The other statements are all true, but none
is relevant to the question of saturation.
A. Since k1 corresponds to the on rate and k –1 corresponds to the off rate,
Kd = [E][S]/[ES] = koff/kon = k –1/k1
B. Km is approximately equal to Kd when kcat is much less than k –1; that is to
say, when the ES complex dissociates much more rapidly than substrate is
converted to product. This is true for many enzymes, but not all.
C. Because kcat is in the numerator of the expression, Km will always be somewhat larger that Kd. Since lower values of Kd indicate higher binding affinity,
Km will always underestimate the binding affinity. When kcat is much less than
k –1, the underestimate will be slight and Km will essentially equal Kd.
3–71 All explanations have at their heart the idea that the quantity of active
enzyme per total protein (the specific activity of the enzyme) is 10-fold less
in bacteria. Such a situation could arise for a number of reasons. 90% of the
enzyme may fold incorrectly in bacteria. An essential cofactor of the
enzyme, which is normally tightly bound, may be limiting in bacteria so that
only 10% of the enzyme molecules acquire it. These explanations, which
propose that there are 10% normally active enzymes among otherwise dead
molecules, account naturally for the observation that the Km is identical (Km
is independent of the concentration of active enzyme) while Vmax is lower
(Vmax is dependent on the concentration of active enzyme).
(One common suggestion is that the enzyme in bacteria folds so that each
molecule has 10% of the normal activity. This possibility can be ruled out
because the lower activity of each molecule would show up as a change in
Km as well as Vmax.) 3–72
A. An enzyme composed entirely of mirror-image amino acids would be
expected to fold stably into a mirror-image conformation; that is, it would
look like the normal enzyme when viewed in a mirror.
B. A mirror-image enzyme would be expected to recognize the mirror image of
its normal substrate. Thus, ‘D’ hexokinase would be expected to add a phosphate to L-glucose and to ignore D-glucose.
This experiment has actually been done for HIV protease. The mirrorimage protease recognizes and cleaves a mirror-image substrate.
Reference: Milton RC, Milton SC & Kent SB (1992) Total chemical synthesis
of a D-enzyme: the enantiomers of HIV-1 protease show reciprocal chiral
substrate specificity. Science 256, 1445–1448. A55 A56 Chapter 3: Proteins 3–73 Phosphoglycolate is a transition-state analog for the triosephosphate isomerase reaction. It has the two characteristics that define a transition-state
analog: it resembles the reaction intermediate and it binds more tightly
(here about 15 times more tightly) than the substrates.
References: Kyte J (1995) Mechanism in Protein Chemistry, pp 207–208. New
York: Garland Publishing.
Pauling L (1948) Chemical achievement and hope for the future. Am. Sci. 36,
A. Amino acid side chains in proteins often have quite different pK values from
those in solution. Glu 35 is uncharged because its local environment is nonpolar, which makes ionization less favorable (raises its pK). The local environment of Asp 52 is more polar, permitting ionization near its solution pK.
B. As the pH drops below 5, Asp 52 picks up a proton and becomes nonionized,
interfering with the mechanism. As the pH rises above 5, Glu 35 begins to
release its proton, also interfering with the mechanism.
3–75 Water from rusty pipes provides iron, which is essential for all forms of life.
Egg white contains a special protein, ovotransferrin, which binds iron very
tightly, analogous to the binding of biotin by avidin. Washing eggs in rusty
water allows iron to enter in sufficient quantities to exceed the binding
capacity of ovotransferrin, thereby making free iron available to the
microorganisms. 3–76 This simple question required decades of research to provide a complete
and satisfying answer. At the simplest level, hemoglobin binds oxygen efficiently in the lungs because the concentration (partial pressure) of oxygen is
highest there. In the tissues the concentration of oxygen is lower because it
is constantly being consumed in metabolism. Thus, hemoglobin will tend to
release (bind less) oxygen in the tissues. This natural tendency—an effect on
the binding equilibrium—is enhanced by allosteric interactions among the
four subunits of the hemoglobin molecule. As a consequence, much more
oxygen is released in the tissues than would be predicted by a simple binding equilibrium. 3–77 Although the rate of diffusion cannot be altered by changes to the enzymes,
the average distance over which a molecule must diffuse can be manipulated. Linking the two enzymes together decreases the average distance for
diffusion of the first product to the second enzyme. A decrease in the distance reduces the time for diffusion and, thus, increases the overall rate of
the reactions catalyzed by the pair of enzymes. 3–78 When [S] >> Km, the enzyme will be virtually saturated with substrate at all
times and capable of operating at maximum rate, independent of small fluctuations in substrate concentration. For many enzymes that use ATP and a
second substrate, as protein kinases do, the Km for ATP is usually very low (a
few mM for most protein kinases) relative to the concentration of ATP in the
cell (1–2 mM). This situation allows the kinases to operate effectively regardless of the typical fluctuations in ATP concentration. Under these conditions
the rate of phosphorylation depends solely on the concentration of the
When [S] ≅ Km, the rate of the enzyme-catalyzed reaction will vary in proportion to the changes in substrate concentration. This is the typical situation for most enzymes involved in metabolic pathways, which allows them
to keep up with the changing flow through the pathway.
When [S] << Km, the enzyme will be mostly unoccupied by substrate and
will be operating much below its maximum rate. This is a strategy that might
be used, for example, if multiple different enzymes draw on a common pool
of substrate. An enzyme with a high Km would use only a small fraction of the
pool unless the concentration increased dramatically. Just such a strategy is
employed in animals for routing glucose for metabolism. Most cells in the PROTEIN FUNCTION A57 R5P 50%
B C D F G AMP H I Figure 3–48 Pattern of inhibition in the
metabolic pathway for purine nucleotide
synthesis (Answer 3–80). GMP E
100% body use hexokinase, which has a low Km, to add phosphate to glucose to initiate its metabolism. By contrast, liver cells use glucokinase, which has a high
Km, to carry out this reaction. Between meals the circulating glucose is routed
mainly to nonliver cells, which use the low Km enzyme hexokinase. After
meals when the circulating concentration of glucose is much higher, the liver
captures a much larger fraction (because glucokinase activity increases much
more with higher substrate concentration than does hexokinase activity).
The liver uses much of that glucose to make glycogen, which serves as a glucose reserve for use between meals.
C. Cells cannot influence rates of diffusion, which are limited by physical
parameters beyond a cell’s control. As discussed in Answer 3–77, cells can
decrease the time it takes for substrates to reach an enzyme by increasing
the concentration of enzyme or by linking related enzymes in multienzyme
3–80 One reasonable proposal would be for excess AMP to feedback inhibit the
enzyme for converting E to F, and excess GMP to feedback inhibit the step
from E to H. Intermediate E, which would then accumulate, would feedback
inhibit the step from R5P to A. Some branched pathways are regulated in
just this way. Purine nucleotide synthesis is regulated somewhat differently,
however (Figure 3–48). AMP and GMP regulate the steps from E to F and
from E to H, as above, but they also regulate the step from R5P to A. Regulation by AMP and GMP at this step might seem problematical since it suggests that a rise in AMP, for example, could shut off the entire pathway even
in the absence of GMP The cell uses a very clever trick to avoid this problem.
Individually, excess AMP or GMP can inhibit the enzyme to about 50% of its
normal activity; together they can completely inhibit it. 3–81 In resting muscle ATP usage is at a minimum; hence, the group of ATP-like
signal metabolites accumulate. Specific members of this group inhibit
glycogen phosphorylase and stimulate glycogen synthase, ensuring that
glycogen reserves are maintained or increased.
In exercising muscle ATP usage is high and AMP-like signal metabolites
increase. Specific AMP-like signal metabolites stimulate glycogen phosphorylase and inhibit glycogen synthase, ensuring a breakdown of glycogen to
provide glucose units for ATP production. 3–82 The substrate, phosphate, and the activator, AMP, both bind to the rarer conformation of glycogen phosphorylase, thereby shifting the conformational
equilibrium in favor of the more active species. This makes good biological
sense because phosphate and AMP concentrations both rise when the cell
increases its rate of ATP hydrolysis, and activation of glycogen phosphorylase is one way to provide metabolic substrates for the synthesis of additional ATP. In both cases the overall activity of the enzyme increases because
the fraction of enzymes in the high-activity conformation is increased. 3–83 The first MWC postulate, which states that the subunits are arranged symmetrically, rules out all arrangements except those shown in the leftmost
and rightmost columns of the diagram. If ligand binds much more tightly to
circles, then the allowed arrangements are those shown in Figure 3–49. If the
ligand binds equally to both subunit conformations, then all the arrangements in the leftmost and rightmost columns are allowed, consistent with
MWC postulate 1. Figure 3–49 Arrangements of subunit
conformations that are consistent with
the MWC postulates (Answer 3–83).
Shaded area indicates those
arrangements that are excluded by the
MWC postulates for a ligand with affinity
for one conformation of subunit (circle). A58 Chapter 3: Proteins
Detailed studies on a few cooperative enzymes such as aspartate transcarbamoylase have found no evidence for intermediate, nonsymmetrical conformations. 3–84 The rate of the metabolic reaction depends on the population of enzyme
molecules, not on an individual enzyme molecule. While an individual
molecule is either on or off—depending on whether it is phosphorylated—
the activity of the population of enzyme molecules depends on the proportion of these molecules that are phosphorylated. As the proportion of phosphorylated molecules increases from 0% to 100%, the activity of the population of enzymes (the rate of the metabolic reaction) will decrease smoothly
from 100% to 0%. The phosphorylation state of a population of enzyme
molecules is controlled by the balance between the opposing activities of
protein kinases, which attach phosphates, and protein phosphatases, which
remove them. 3–85
E. A mutation that decreases the rate of GTP hydrolysis by Ras would prolong
its activated state, leading to excessive stimulation of cell proliferation.
Indeed, many cancers contain just such a mutant form of Ras. The mutant
Ras proteins in all other choices would lead to a decreased ability to transmit the downstream signal, thus decreasing cell proliferation. For example,
a mutation that increased the affinity of Ras for GDP (choice B) would prolong the inactive state of Ras, thereby interfering with the growth signal and
decreasing cell proliferation.
3–86 A nonfunctional GAP (choice A) or a permanently active GEF (choice D)
would allow Ras to remain in the active state (with GTP bound) longer than
normal, and thus might cause excessive cell proliferation. 3–87
A. In the absence of ATP a motor protein would stop moving. The conformational shifts that are required for movement are triggered by ATP binding
and hydrolysis. In the absence of ATP the motor protein would be stuck in
its lowest-energy conformation.
B. If the free-energy change for the hydrolysis of ATP by the motor protein were
zero—conditions under which ATP is as easily made as hydrolyzed—the
motor protein would wander back and forth. With zero free-energy change
there would be no barrier between conformations. CALCULATIONS
A. The equilibrium constant, K, equals 106 M–1.
[A][B] (10–6 M) ¥ (10–6 M) K = 106 M–1
B. The same calculation as above, when each component is present at 10–9 M,
gives an equilibrium constant of 109 M–1.
C. This example illustrates that interacting cellular proteins present at low concentrations need to bind to one another with high affinities if a high proportion of the molecules are to be bound together. A three-order of magnitude decrease in the equilibrium constant corresponds to a free-energy difference of about –4.2 kcal/mole. Thus, effective binding at the lower concentration would require the equivalent of 4–5 extra hydrogen bonds. The
free-energy difference between the two equilibrium constants can be calculated. For an equilibrium constant of 106 M–1,
DG° = –2.3 RT log K PROTEIN FUNCTION
DG° = –1.41 kcal/mole ¥ (6)
DG° = –8.46 kcal/mole
For an equilibrium constant of 109 M–1,
DG° = –1.41 kcal/mole ¥ (9)
DG° = –12.69 kcal/mole
Thus, the higher equilibrium constant corresponds to a free-energy difference that is 4.2 kcal/mole more negative. To supply this amount of binding
energy with hydrogen bonds (about 1 kcal/mole) would require about 4–5
extra hydrogen bonds.
3–89 The antibody binds to the second protein with an equilibrium constant, K,
of 5 ¥ 107 M–1.
A useful shortcut to problems of this sort recognizes that DG° is related to
log K by the factor –2.3 RT, which equals –1.4 kcal/mole at 37° C. Thus, a factor of ten increase in the equilibrium constant (an increase in log K of 1) corresponds to a decrease in DG° of –1.4 kcal/mole. A 100-fold increase in K corresponds to a decrease in DG° of –2.8 kcal/mole, and so on. For each factor
of ten increase in K, DG° decreases by –1.4 kcal/mole; for each factor of ten
decrease in K, DG° increases by 1.4 kcal/mole. This relationship allows a
quick estimate of changes in equilibrium constant from free-energy changes
and vice versa. In this problem you are told that DG° increased by 2.8
kcal/mole (a weaker binding gives a less negative DG°). According to the
relationship developed above, this increase in DG° requires that K decrease
by a factor of 100 (a decrease by 2 in log K); thus, the equilibrium constant
for binding to the second protein is 5 ¥ 107 M–1.
The solution to the problem can be calculated by first determining the
free-energy change represented by the binding to the first protein:
DG° = –2.3 RT log K
Substituting for K,
DG° = –2.3 (1.98 ¥ 10–3 kcal/(K mole)) (310 K) log (5 ¥ 109)
DG° = –1.41 kcal/mole ¥ 9.7
DG° = –13.68 kcal/mole
The free-energy change associated with binding to the second protein is
obtained by adding 2.8 kcal/mole to the free-energy change for binding to
the first protein, giving a value of –10.88 kcal/mole. Thus, the equilibrium
constant for binding to the second protein is
log K = (–10.88 kcal/mole)/(–1.41 kcal/mole)
K = 5 ¥ 107 M–1 3–90
A. The equilibrium constants for the two reactions are the same, 108 M–1.
K = [Ab–Pr]/([Ab][Pr]) = kon/koff
For the first antibody–protein reaction,
K = kon/koff = 105 M–1 sec–1/10–3 sec–1
K = 108 M–1 A59 A60 Chapter 3: Proteins
For the second reaction,
K = kon/koff = 103 M–1 sec–1/10–5 sec–1
K = 108 M–1
B. Since the first reaction has both a faster association rate and a faster dissociation rate, it will come to equilibrium more quickly than the second reaction.
C. The time it takes for half the complex to dissociate can be calculated from
the relationship given in the problem:
2.3 log [Ab–Pr]t/[Ab–Pr]0 = –kofft
Substituting 0.5 for [Ab–Pr]t/[Ab–Pr]0 and rearranging the equation,
t= 2.3 log 0.5
–koff For the first complex, with koff = 10–3 sec–1,
t= 2.3 log 0.5
–koff t = 692 seconds, or 11.5 minutes
For the second complex, with koff = 10–5 sec–1, the calculation gives
t = 6.9 ¥ 104 seconds, or about 19 hours
Thus, the first complex, which falls apart relatively quickly, would be much
more difficult to work with than the second complex, which falls apart more
slowly. Inappropriate reliance on the equilibrium constant, instead of the off
rate constant, can lead an investigator astray in this sort of experiment.
A. At equilibrium, the rates of the forward and reverse reactions are equal. This
is the definition of equilibrium. The overall reaction rate at equilibrium will
B. The equilibrium constant equals 103. At equilibrium, the forward and
reverse reactions are equal. Thus,
kf [A] = kr [B]
kf/kr = [B]/[A] = K
K = 10–4 sec–1/10–7 sec–1 = 103
C. Enzyme catalysis does not alter the equilibrium for a reaction; it only speeds
the attainment of equilibrium. Thus, the equilibrium constant is 103. If the
equilibrium is unchanged and kf is increased by a factor of 109, then kr must
also be increased by a factor of 109.
3–92 At [S] = zero, the rate equals 0/Km and the rate is therefore zero. At [S] = Km,
the ratio of [S]/([S] + Km) equals 1/2 and the rate is 1/2 Vmax. At infinite [S]
the ratio of [S]/([S] + Km) equals 1 and the rate is equal to Vmax. 3–93 If Km increases, then the concentration of substrate necessary to give halfmaximal rate also increases. At a concentration of substrate equal to the Km
of the unphosphorylated enzyme, the phosphorylated enzyme would have a
slower rate; thus, phosphorylation inhibits this enzyme. PROTEIN FUNCTION A61 3–94
D. The substrate concentration must be increased by a factor of 16 to increase
the rate from 20% to 80% Vmax. Substituting a rate of 20% Vmax into the
Michaelis–Menten equation gives
0.2 Vmax = (Vmax) [S]/([S] + Km)
Cancelling Vmax and multiplying both sides by ([S] + Km) gives
0.2 [S] + 0.2 Km = [S]
0.8 [S] = 0.2 Km
[S] = 0.25 Km at 20% Vmax
An analogous calculation shows that
[S] = 4 Km at 80% Vmax
Thus, [S] must increase by a factor of 16 (4 Km/0.25 Km) for the rate to go
from 20% to 80% Vmax.
3–95 The turnover number for carbonic anhydrase is 6.1 ¥ 107/min (or 1.0 ¥
106/sec). For this calculation, it is necessary to express the amount of CO2
hydrated and the amount of the enzyme on the same molecular basis, either
as molecules or moles. For CO2,
0.90 g ¥ 6 ¥ 1023 d ¥ molecule = 1.23 ¥ 1022 molecules/min mL
For carbonic anhydrase,
10 mg ¥ mmole ¥ 6 ¥ 1017 molecules = 2.0 ¥ 1014 molecules/min mL
The turnover number
k3 = Vmax/[E] = 3–96 1.22 ¥ 1022 molecules/min mL = 6.1 ¥ 107/min
2.0 ¥ 1014 molecules/mL The numerical value of the product of the Kd values for the substrates is 3.0 ¥
10–7 [(27 ¥ 10–6) ¥ (11 ¥ 10–3)], which is 9-fold greater than the Kd for PALA (2.7
¥ 10–8), suggesting that PALA might be a transition-state analog. PALA, however, is composed not of aspartate plus carbamoyl phosphate, but of succinate plus a close analog of carbamoyl phosphate. If one substitutes the Kd for
succinate, 0.9 mM, into the calculation, the product of the Kd values is 2.4 ¥
10–8 [(27 ¥ 10–6) ¥ (0.9 ¥ 10–3)], which is very close to the Kd for PALA. Thus,
PALA is likely to be a bisubstrate analog rather than a transition-state analog.
You may have wondered whether it is valid to compare Kd values in this
way. Recall that DG° = – 2.3RT log K. Using this equation, we could have converted Kd values to DG° values and compared the sum of DG° values for
aspartate and carbamoyl phosphate with the DG° for PALA. Since DG° is proportional to log K, this is equivalent to comparing the products of the Kd values for aspartate and carbamoyl phosphate with the Kd for PALA. Do the calculation for DG° values and convince yourself that this is true.
Reference: Fersht A (1999) Structure and Mechanism in Protein Science, pp
360–361. New York: WH Freeman. 3–97
A. An understanding of this phenomenon comes from a consideration of Kd,
which for the binding of PALA to aspartate transcarbamoylase (ATC) is
Kd = [PALA][ATC] = 2.7 ¥ 10–8 M
[PALA–ATC] Since the concentration of enzyme is negligible relative to the concentration A62 Chapter 3: Proteins
of added PALA, we can use 2.7 ¥ 10–6 as the equilibrium concentration of
PALA. Substituting this value into the Kd expression gives
Thus, in the presence of PALA only 1% of the enzyme is free. This is true for
normal cells and resistant cells. For resistant cells, which have 100 times as
much enzyme, 1% free corresponds to the amount that is present in uninhibited normal cells. Thus, they can grow perfectly well in the presence of
B. Mutational resistance to an inhibitor requires a subtle change in the enzyme
that allows it to decrease its binding affinity for the inhibitor, while not significantly altering its binding affinity for substrates. This may not be a feasible response to PALA because it so closely resembles the two substrates. A
change that decreases PALA binding will likely decrease substrate binding as
Reference: Wahl GM, Padgett RA & Stark GR (1979) Gene amplification
causes overproduction of the first three enzymes of UMP synthesis in N(phosphonacetyl)-L-aspartate (PALA)–resistant hamster cells. J. Biol. Chem.
254, 8679–8689. 3–98
A. The relative concentrations of the normal and mutant Src proteins are
inversely proportional to the volumes in which they are distributed. The
mutant Src is distributed throughout the volume of the cell, which is
Vcell = (4/3)pr 3 = (4/3)p (10 ¥ 10–6 m)3 = 4.1888 ¥ 10–15 m3
Normal Src is confined to the 4-nm layer beneath the membrane, which has
a volume equal to the volume of the cell minus the volume of a sphere with
a radius 4 nm less than that of the cell:
Vlayer = Vcell – (4/3)p (r – 4 nm)3
= Vcell – (4/3)p [(10 ¥ 10–6 m) – (4 ¥ 10–9 m)]3
= (4.1888 ¥ 10–15 m3) – (4.1838 ¥ 10–15 m3)
Vlayer = 0.0050 ¥ 10–15 m3
Thus, the volume of the cell is 838 times greater than the volume of a 4-nmthick layer beneath the membrane (4.1888 ¥ 10–15 m3/0.0050 ¥ 10–15 m3).
Even allowing for the interior regions of the cell from which it would be
excluded (nucleus and organelles), the mutant Src would still be a couple of
orders of magnitude less concentrated in the neighborhood of the membrane than the normal Src.
B. Its lower concentration in the region of its target X at the membrane is the
reason why mutant Src does not cause cell proliferation. This notion can be
quantified by a consideration of the binding equilibrium for Src and X:
Src + X Æ Src – X
K = [Src – X]
The lower concentration of the mutant Src in the region of the membrane
will shift the equilibrium toward the free components, reducing the amount
of complex. If the concentration is on the order of 100-fold lower, the
amount of complex will be reduced up to 100-fold. Such a large decrease in
complex formation could readily account for the lack of effect of the mutant
Src on cell proliferation. PROTEIN FUNCTION DATA HANDLING
A. Your results support the idea that the PI 3-kinase interacts with the activated
PDGF receptor through its SH2 domains. The interaction is blocked specifically by the phosphorylated pentapeptides 708 and 719. In their nonphosphorylated forms these same pentapeptides do not block the association.
B. The common features of the seven peptides that can bind to PI 3-kinase are
a phosphotyrosine and a methionine (M) located three positions away in the
C-terminal direction. (Although not shown explicitly here, there seems to be
no requirement for specific amino acids on the N-terminal side of the phosphotyrosine.)
C. Recognition of a couple of amino acids in a short sequence is characteristic
of a surface–string interaction. Indeed, recognition of sequences by SH2
domains is often cited as a prime example of such an interaction.
3–100 The immunoblot shows that antibodies BPA1 and BPA2 react specifically
with a 220-kd protein, which is likely to be Brca1. By contrast, although antibody C20 reacts with the same protein, it seems to react even more strongly
with a second protein of about 180 kd. Thus, a likely explanation for the contradictory cell-localization experiments is that C20 antibodies were identifying the location of the 180-kd protein, whereas BPA1 and BPA2 were showing the location of Brca1. Brca1 is now thought to function in the nucleus.
Additional experiments have identified the epidermal growth factor (EGF)
receptor as the protein with which C20 cross-reacts. The EGF receptor has a
couple of regions of similarity to the peptide that was used to generate the
C20 antibody. Cross-reactivity of antibodies is not an uncommon problem.
For this reason, cell-localization studies are usually performed with antibodies raised against more than one region of a protein. Agreement with different antibodies decreases the likelihood that cross-reactivity is a problem.
References: Jensen RA, Thompson ME, Jetton TL, Szabo CI, van der Meer R,
Helou B, Tronick SR, Page DL, King MC & Holt JT (1996) BRCA1 is secreted
and exhibits properties of a granin. Nat. Genet. 12, 303–308.
Thomas JE, Smith M, Rubinfeld B, Gutowski M, Beckmann RP & Polakis P
(1996) Subcellular localization and analysis of apparent 180-kDa and 220kDa proteins of the breast cancer susceptibility gene, BRCA1. J. Biol. Chem.
271, 28630–28635. 3–101
A. The slopes (–1/Kd) of the lines in Figure 3–26 can be estimated by taking the
difference between two points on the y axis divided by the difference
between the corresponding points on the x axis. Thus, the slope of line A is
–1/Kd = (0.08 – 0.35)/(5 ¥ 10–7 M – 1 ¥ 10–7 M)
= –6.75 ¥ 105 M–1
Kd = 1.5 ¥ 10–6 M
For line B,
–1/Kd = (0.20 – 0.90)/(5 ¥ 10–7 M – 1 ¥ 10–7 M)
Kd = 5.7 ¥ 10–7 M
The precise values are dependent on your estimate of the corresponding values on the x axis.
B. The lower the Kd, the tighter the binding; thus, the tighter IPTG-binding
mutant of the Lac repressor corresponds to line B (Kd = 5.7 ¥ 10–7 M) and the
wild-type Lac repressor corresponds to line A (Kd = 1.5 ¥ 10–6 M). That a
lower value corresponds to tighter binding is apparent from the definition of
Kd in the problem. Tighter binding will give more complex (Pr–L) and fewer A63 A64 Chapter 3: Proteins
free components (Pr + L); thus, the ratio of concentrations, Kd, will be
References: Gilbert W & Muller-Hill B (1966) Isolation of the Lac repressor.
Proc. Natl Acad. Sci. U.S.A. 56, 1891–1898.
Kyte J (1995) Mechanism in Protein Chemistry, pp 175–177. New York: Garland Publishing, 1995. 3–102
A. When the concentrations of free and bound ligand are equal, their ratio
becomes 1 and the concentration of free protein is equal to Kd:
Kd = [Pr][L]
When [L] equals [Pr–L],
Kd = [Pr]
B. Visual inspection of the data in Figure 3–27 shows that the concentrations of
free and bound tmRNA are approximately equal when the concentration of
SmpB is 18.8 nM. Thus, Kd is around 20 nM.
Not all of the added SmpB protein is free, since some is obviously bound
to tmRNA. But because tmRNA was included at a concentration of 0.1 nM,
the bound fraction at an SmpB concentration of 18.8 nM is only
0.05 nM. Thus, the correction for bound SmpB is minuscule (less than 1%)
and can be neglected.
C. It is critical in this kind of experiment that the tmRNA concentration be kept
well below Kd. If the concentration of tmRNA had been 100 nM, for example,
the shift to 50% bound would have occurred at around 50 nM SmpB (and
most of the protein would have been in the bound, not the free, form). If
tmRNA were included at 100 mM, the shift to 50% bound would have
occurred at around 50 mM SmpB. Thus, if tmRNA were included at a concentration above Kd, the point at which 50% was shifted to the bound form
would bear no relationship to Kd.
Reference: Karzai AW, Susskind MM & Sauer RT (1999) SmpB, a unique
RNA-binding protein essential for the peptide-tagging activity of SsrA
(tmRNA). EMBO J. 18, 3793–3799.
3–103 The calculated values of fraction bound versus protein concentration are
shown in Table 3–6. Also shown are rule-of-thumb values, which are easier
to remember (see the answer to Question 2–54).
These relationships are useful not only for thinking about Kd, but also for
enzyme kinetics, which we cover in other problems. The rate of a reaction
expressed as a fraction of the maximum rate is
[S] + Km
which has the same form as the equation for fraction bound. Thus, when the
concentration of substrate, [S], is 10-fold above the Michaelis constant, Km,
the rate is 90% of the maximum, Vmax. When [S] is 100-fold below Km, the
rate is 1% of Vmax.
The relationship also works for the fractional dissociation of an acidic
group, HA, as a function of pH. When the pH is 2 units above pK, 99% of the
acidic group is ionized. When the pH is 1 unit less than pK, 10% is ionized. 3–104
A. Since the concentration of Lac repressor is 105 times the Kd for binding, you
would expect 99.999% of the sites in a bacterial population to be occupied
by the Lac repressor.
B. When inducer is present, the concentration of Lac repressor will be only Table 3–6 Calculated values for
fraction bound versus protein
concentration (Answer 3–103).
10–4 Kd FRACTION
BOUND (%) RULE-OFTHUMB 99.99
0.01 PROTEIN FUNCTION A65 100 times more than the Kd, but you would still expect 99% of the sites to be
C. If 99% of the binding sites were occupied by the repressor even in the presence of the inducer, you would expect that the genes would still be very
effectively turned off. This sort of straightforward calculation, and its nonbiological answer—after all, the genes are known to be turned on by the
inducer—tells you that some critical information is missing.
D. Low-affinity, nonspecific binding of the Lac repressor is the missing information suggested by the calculation in part C. Since there are 4 ¥ 106 nonspecific binding sites in the genome (a number equal to the size of the
genome), there is a competition for repressor between the multitude of lowaffinity sites and the single high-affinity site. This competition reduces the
effective concentration of the repressor. As can be calculated, the competition reduces repressor occupancy at the specific site to about 96% in the
absence of lactose, and to about 3% in the presence of lactose. These numbers account nicely for the genes being turned off in the absence of lactose
and turned on in its presence. 3–106
D. Since NAM cannot occupy site C, that site must normally be occupied by
NAG; and since the cell-wall polysaccharide is an alternating polymer of
NAM and NAG, the NAM monomers must occupy sites B, D, and F. Because
cleavage occurs after NAM monomers, the site of cleavage must be between
sites B and C or between sites D and E. Since tri-NAG occupies sites A–C but
is not cleaved, whereas longer NAG polymers are, the catalytic groups for
cleavage must lie between sites D and E.
A. Binding of aspartate normally shifts the conformation of ATCase from the
low-activity to the high-activity state. At low aspartate concentrations not
all of the ATCase will have been shifted to the high-activity conformation.
The peculiar activating effect of malate occurs because its binding helps Michaelis–Menten plot rate (µmol/min) 2.0
0 5.0 10 S (µM) Lineweaver–Burk plot
1/rate (min/µmol) 3–105
A. It is important that only a small quantity of product is made, because otherwise the rate of the reaction would decrease as the substrate was depleted
and product accumulated. Thus, the measured rates would be lower than
they should be, and the kinetic parameters would be incorrect.
B. The Michaelis–Menten plot, shown in Figure 3–50, is a rectangular hyperbola, as expected if this enzyme obeys Michaelis–Menten kinetics. To determine values for Km and Vmax from this plot by visual inspection, you must
estimate the rate at infinite substrate concentration. From the curve of the
line in the figure you might reasonably estimate Vmax as anywhere from 1.8
to 2.0 mmol/min. (As developed in the answer to Problem 3–103, a useful
rule of thumb is that at a concentration of substrate 10-fold above Km, the
rate is about 90% of Vmax.) If you chose 2.0 mmol/min, then 0.5 Vmax
(1.0 mmol/min) corresponds to a substrate concentration of 1.0 mM, which is
the value of Km. The visual uncertainty in this plot led early researchers to
transform the equation into a straight-line form so a line could be fitted to
the data and the kinetic parameters could be more accurately determined.
C. As indicated in the Lineweaver–Burk plot in Figure 3–50, the y intercept is 0.5
(1/Vmax) and the x intercept is –1.0 (–1/Km). Thus, Vmax equals 2.0 mmol/min
and Km equals 1.0 mM. Although this form of a straight-line plot is commonly
discussed in textbooks, it is rarely used in practice because the data points
that are most reliable are tightly grouped at one end of the line. Consequently, the slope of the line is unduly influenced by the low (and usually
less accurately determined) rates at low substrate concentration. Other
straight-line transformations of the Michaelis–Menten equation such as the
Eadie–Hofstee plot, which is analogous in form to the Scatchard plot shown
in Problem 3–101, are generally preferred. In this era of computers, however,
the data can be fitted perfectly well to the nonlinear Michaelis–Menten
equation, although it is still common to present such data in a linear form. 6
–1.0 5 10 15 1/S (1/µM) Figure 3–50 Michaelis–Menten and
Lineweaver–Burk plots of the data in
Table 3–3 (Answer 3–105). The x and y
intercepts are indicated on the
Lineweaver–Burk plot. A66 Chapter 3: Proteins
complete the shift of ATCase to the high-activity conformation. In the presence of a low concentration of malate the number of active sites in the highactivity conformation increases; thus, enzyme activity increases.
B. This peculiar activating effect of malate is not observed at high aspartate
concentrations because ATCase is already entirely shifted to its high-activity
conformation. Under these conditions each molecule of malate that binds
to an active site will reduce the total number of sites accessible to aspartate
and thus reduce the overall activity of ATCase.
References: Cantor CR & Schimmel PR (1980) Biophysical Chemistry, pp
944–945. New York: WH Freeman.
Gerhart JH & Pardee AB (1963) The effect of the feedback inhibitor, CTP, on
subunit interactions in aspartate transcarbamylase. Cold Spring Harbor
Symp. Quant. Biol. 28, 491–496. 3–108 These changes are exactly what you would expect for an allosteric enzyme
such as ATCase. Because binding at one active site (one of six per ATCase
molecule) is sufficient to shift the conformation of a molecule of ATCase, the
change in global conformation (change in sedimentation) is expected to
‘lead’ the change in occupancy of binding sites (change in spectral measurement).
References: Cantor CR & Schimmel PR (1980) Biophysical Chemistry, pp
954–956. New York: WH Freeman.
Kirschner MW & Schachman HK (1973) Local and gross conformational
changes in aspartate transcarbamylase. Biochemistry 12, 2997–3004. 3–109
A. Both cyclin A and phosphorylation of Cdk2 are required to activate Cdk2 for
efficient phosphorylation of histone H1 (see Figure 3–32, lane 5). Absence of
cyclin A (lane 1) or absence of phosphorylation of Cdk2 (lane 3) results in
much reduced levels of histone H1 phosphorylation.
B. Cyclin A, which binds tightly to both forms of Cdk2 (Kd = 0.05 mM), dramatically improves the binding of both forms to histone H1. In the absence of
cyclin A, P-Cdk2, for example, binds histone H1 with a Kd of 100 mM, whereas
in the presence of cyclin A, it binds histone H1 with a Kd of 0.7 mM, an
increase in the tightness of binding of more than a factor of 100. In addition,
as shown in Figure 3–32, phosphorylation of Cdk2 activates its protein
kinase activity, allowing it to phosphorylate histone H1, when cyclin A is present to increase its ability to bind to histone H1.
C. Given that the intracellular concentrations of ATP and ADP are more than
10-fold higher than the measured dissociation constants, the changes in
affinity for ATP and ADP are unlikely to be critical for the function of Cdk2.
The binding sites for ATP will be nearly saturated regardless of the phosphorylation state of Cdk2. ADP, which binds at the same site as ATP, is unlikely to
interfere significantly with ATP binding, because ADP has a higher Kd and its
cellular concentration is generally lower than that of ATP.
Reference: Brown NR, Noble MEM, Lawrie AM, Morris MC, Tunnah P, Divita
G, Johnson LN & Endicott JA (1999) Effects of phosphorylation of threonine
160 on cyclin-dependent kinase 2 structure and activity. J. Biol. Chem. 274,
A. Mutant 2 is Asp-181 Æ Ala (D181A). It is the best candidate because it has a
Km close to that of the wild-type enzyme, but a very low turnover number
(kcat). With these kinetic parameters it might be expected to bind normally
to its target substrates but not remove phosphate from tyrosine. The next
most likely candidate would be Arg-221Æ Lys, which has a slightly lower Km
than wild-type PTP1B and turns over slowly, although about 20 times faster
than D181A. (Further studies identified the band at 180 kd as the epidermal
growth factor—EGF—receptor.) PROTEIN FUNCTION
B. C215S showed no activity, as expected since the –SH group of cysteine is
required for catalysis. Because C215S was not active, it was not possible to
determine its Km, which might have been similar to that of the wild-type
enzyme. (Measurements of Kd were not made.) Thus, C215S was a reasonable candidate to test. Lack of success with C215S suggests that it binds
phosphotyrosine-containing proteins very poorly.
Reference: Flint AJ, Tiganis T, Barford D & Tonks NK (1997) Development of
substrate-trapping mutants to identify physiological substrates of protein
tyrosine phosphatases. Proc. Natl Acad. Sci. U.S.A. 94, 1680–1685. A67 ...
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This homework help was uploaded on 04/07/2008 for the course BME 50A taught by Professor Botvinick during the Spring '08 term at UC Irvine.
- Spring '08