Essential Cell Biology 5th edition

Answers A:11
on its surface; however, these loops are contributed
by both the folded light and heavy chains (see Figure
4−33).
E. False. The possible linear arrangements of amino acids
that lead to a stably folded protein domain are so few
that most new proteins evolve by alteration of old ones.
F. True. Allosteric enzymes generally bind one or more
molecules that function as regulators at sites that are
distinct from the active site.
G. False. Although single noncovalent bonds are weak,
many such bonds acting together are major contributors
to the three-dimensional structure of macromolecules.
H. False. Affinity chromatography separates specific
macromolecules because of their interactions with
specific ligands, not because of their charge.
I. False. The larger an organelle is, the more centrifugal
force it experiences and the faster it sediments, despite
an increased frictional resistance from the fluid through
which it moves.
ANSWER 4–11 In an α helix and in the central strands of a
β sheet, all of the N –H and C=O groups in the polypeptide
backbone are engaged in hydrogen bonds. This gives
considerable stability to these secondary structural
elements, and it allows them to form in many different
proteins.
ANSWER 4–12 No. It would not have the same or even a
similar structure, because the peptide bond has a polarity.
Looking at two sequential amino acids in a polypeptide
chain, the amino acid that is closer to the N-terminal end
contributes the carboxyl group and the other amino acid
contributes the amino group to the peptide bond that
links the two amino acids. Changing their order would
put the side chains into different positions with respect to
the peptide backbone and therefore change the way the
polypeptide folds.
ANSWER 4–13 As it takes 3.6 amino acids to complete a
turn of an α helix, this sequence of 14 amino acids would
make close to 4 full turns. It is remarkable because its
polar and hydrophobic amino acids are spaced so that all
the polar ones are on one side of the α helix and all the
hydrophobic ones are on the other. It is therefore likely
that such an amphipathic α helix is exposed on the protein
surface with its hydrophobic side facing the protein’s
interior. In addition, two such helices might wrap around
each other as shown in Figure 4−16.
ANSWER 4–14
A. ES represents the enzyme–substrate complex.
B. Enzyme and substrate are in equilibrium between their
free and bound states; once bound to the enzyme, a
substrate molecule may either dissociate again (hence
the bidirectional arrows) or be converted to product.
As the substrate is converted to product (with the
concomitant release of free energy), however, a reaction
usually proceeds strongly in the forward direction, as
indicated by the unidirectional arrow.
C. The enzyme is a catalyst and is therefore liberated in an
unchanged form after the reaction; thus, E appears at
both ends of the equation.
D. Often, the product of a reaction resembles the substrate
sufficiently that it can also bind to the enzyme. Any
enzyme molecules that are bound to the product
(i.e., are part of an EP complex) are unavailable for
catalysis; excess P therefore can inhibit the reaction by
lowering the concentration of free E.
E. Compound X would act as an inhibitor of the reaction
and work similarly by forming an EX complex. However,
since P has to be made before it can inhibit the reaction,
it takes longer to act than X, which is present from the
beginning of the reaction.
ANSWER 4–15 The polar amino acids Ser, Ser-P, Lys, Gln,
His, and Glu are more likely to be found on a protein’s
surface, and the hydrophobic amino acids Leu, Phe, Val,
Ile, and Met are more likely to be found in its interior. The
oxidation of two cysteine side chains to form a disulfide
bond eliminates their potential to form hydrogen bonds
and therefore makes them even more hydrophobic; thus
disulfide bonds are usually found in the interior of proteins.
Irrespective of the nature of their side chains, the most
N-terminal amino acid and the most C-terminal amino acid
each contain a charged group (the amino and carboxyl
groups, respectively, that mark the ends of the polypeptide
chain) and hence are usually found on the protein’s surface.
ANSWER 4–16 Many secondary structural elements are
not stable in isolation but are stabilized by other parts of
the polypeptide chain. Hydrophobic regions of fragments,
which would normally be hidden in the inside of a folded
protein, would be exposed to water molecules in an
aqueous solution; such fragments would tend to aggregate
nonspecifically, and not have a defined structure, and they
would be inactive for ligand binding, even if they contained
all of the amino acids that would normally contribute to
the ligand-binding site. A protein domain, in contrast, is
considered a folding unit, and fragments of a polypeptide
chain that correspond to intact domains are often able to
fold correctly. Thus, separated protein domains often retain
their activities, such as ligand binding, if the binding site is
contained entirely within the domain. Thus the most likely
place in which the polypeptide chain of the protein in Figure
4−20 could be severed to give rise to stable fragments
is at the boundary between the two domains (i.e., at the
loop between the two α helices at the bottom right of the
structure shown).
ANSWER 4–17 Because of the lack of secondary structure,
the C-terminal region of neurofilament proteins undergoes
continual Brownian motion. The high density of negatively
charged phosphate groups means that the C-terminals
also experience repulsive interactions, which cause
them to stand out from the surface of the neurofilament
like the bristles of a brush. In electron micrographs of
a cross section of an axon, the region occupied by the
extended C-terminals appears as a clear zone around
each neurofilament, from which organelles and other
neurofilaments are excluded.
ANSWER 4–18 The heat-inactivation of the enzyme
suggests that the mutation causes the enzyme to have a
less stable structure. For example, a hydrogen bond that
is normally formed between two amino acid side chains
might no longer be formed because the mutation replaces
one of these amino acids with a different one that cannot
participate in the bond. Lacking such a bond that normally
helps to keep the polypeptide chain folded properly, the
protein partially or completely unfolds at a temperature at