Essential Cell Biology 5th edition

A:6 Answers
ANSWER 2–19 Noncovalent bonds form between the
subunits of macromolecules—e.g., the side chains of amino
acids in a polypeptide chain—and cause the polymer chain
to assume a unique shape. These interactions include
hydrogen bonds, ionic bonds, van der Waals attractions,
and hydrophobic forces. Because these interactions are
weak, they can be broken with relative ease; thus, most
macromolecules can be unfolded by heating, which
increases thermal motion.
ANSWER 2–20 Amphipathic molecules have both a
hydrophilic and a hydrophobic end. Their hydrophilic ends
can hydrogen-bond to water, but their hydrophobic ends
are repelled from water because they interfere with the
water structure. Consequently, the hydrophobic ends of
amphipathic molecules tend to be exposed to air at air–
water interfaces, or will always cluster together to minimize
their contact with water molecules—both at this interface
and in the interior of an aqueous solution. (See Figure
A2–20.)
Figure A2–20
hydrophilic
hydrophobic
AIR
water
molecules
hydrogen bond
ANSWER 2–21
A, B. (A) and (B) are both correct formulas of the amino acid
phenylalanine. In formula (B), phenylalanine is shown in
EA2.20/2.20
the ionized form that exists in solution in water, where
the basic amino group is protonated and the acidic
carboxylic group is deprotonated.
C. Incorrect. This structure of a peptide bond is missing a
hydrogen atom bound to the nitrogen.
D. Incorrect. This formula of an adenine base features one
double bond too many, creating a five-valent carbon
atom and a four-valent nitrogen atom.
E. Incorrect. In this formula of a nucleoside triphosphate,
there should be two additional oxygen atoms, one
between each of the phosphorus atoms.
F. This is the correct formula of ethanol.
G. Incorrect. Water does not hydrogen-bond to hydrogens
bonded to carbon. The lack of the capacity to hydrogenbond
makes hydrocarbon chains hydrophobic; that is,
water-hating.
H. Incorrect. Na and Cl form an ionic bond, Na + Cl – , but a
covalent bond is drawn.
I. Incorrect. The oxygen atom attracts electrons more than
the carbon atom; the polarity of the two bonds should
therefore be reversed.
J. This structure of glucose is correct.
K. Almost correct. It is more accurate to show that only
one hydrogen is lost from the –NH 2 group and the –OH
group is lost from the –COOH group.
Chapter 3
ANSWER 3–1 The equation represents the “bottom
line” of photosynthesis, which occurs as a large set of
individual reactions that are catalyzed by many individual
enzymes. Because sugars are more complicated molecules
than CO 2 and H 2 O, the reaction generates a more
ordered state inside the cell. As demanded by the second
law of thermodynamics, this increase in order must be
accompanied by a greater increase in disorder, which
occurs because heat is generated at many steps on the
long pathway leading to the products summarized in this
equation.
ANSWER 3–2 Oxidation is defined as removal of
electrons, and reduction represents a gain of electrons.
Therefore, (A) is an oxidation and (B) is a reduction. The
red carbon atom in (C) remains largely unchanged; the
neighboring carbon atom, however, loses a hydrogen atom
(i.e., an electron and a proton) and hence becomes oxidized.
The red carbon atom in (D) becomes oxidized because it
loses a hydrogen atom, whereas the red carbon atom in (E)
becomes reduced because it gains a hydrogen atom.
ANSWER 3–3
A. Both states of the coin, H and T, have an equal
probability. There is therefore no driving force—that
is, no energy difference—that would favor H turning
to T, or vice versa. Therefore, ΔG° = 0 for this reaction.
However, a reaction proceeds if H and T coins are not
present in the box in equal numbers. In this case, the
concentration difference between H and T creates a
driving force and ΔG ≠ 0; when the reaction reaches
equilibrium—that is, when there are equal numbers of H
and T—ΔG = 0.
B. The amount of shaking corresponds to the temperature,
as it results in the “thermal” motion of the coins. The
activation energy of the reaction is the energy that
needs to be expended to flip the coin, that is, to stand it
on its rim, from where it can fall back facing either side
up. Jigglase would speed up the flipping by lowering
the energy required for this; it could, for example, be
a magnet that is suspended above the box and helps
lift the coins. Jigglase would not affect where the
equilibrium lies (at an equal number of H and T), but it
would speed up the process of reaching the equilibrium,
because in the presence of jigglase more coins would flip
back and forth.
ANSWER 3–4 See Figure A3–4. Note that ΔG° X → Y is
positive, whereas ΔG° Y → Z and ΔG° X → Z are negative. The
graph also shows that ΔG° X → Z = ΔG° X → Y + ΔG° Y → Z .