Essential Cell Biology 5th edition

A:32 Answers
in a membrane. A voltage of 150,000 V would instantly
discharge in an arc across a 1-cm-wide gap (that is, air would
be an insufficient insulator for this strength of field).
ANSWER 12–15
A. Nothing. You require ATP to drive the Na + pump.
B. The ATP becomes hydrolyzed, and Na + is pumped into
the vesicles, generating a concentration gradient of Na +
across the membrane. At the same time, K + is pumped
out of the vesicles, generating a concentration gradient
of K + of opposite polarity. When all the K + is pumped
out of the vesicle or the ATP runs out, the pump would
stop.
C. The pump would initiate a transport cycle and then
cease. Because all reaction steps must occur strictly
sequentially, dephosphorylation and the accompanying
conformational switch cannot occur in the absence
of K + . The Na + pump will therefore become stuck in
the phosphorylated state, waiting indefinitely for a
potassium ion. The number of sodium ions transported
would be minuscule, because each pump molecule
would have functioned only a single time. Similar
experiments, leaving out individual ions and analyzing
the consequences, were used to determine the
sequence of steps by which the Na + pump works.
D. ATP would become hydrolyzed, and Na + and K + would
be pumped across the membrane as described in (B).
However, the pump molecules that sit in the membrane
in the reverse orientation would be completely inactive
(i.e., they would not—as one might have erroneously
assumed—pump ions in the opposite direction)
because ATP would not have access to the site on
these molecules where phosphorylation occurs, which is
normally exposed to the cytosol. ATP is highly charged
and cannot cross membranes without the help of specific
transporters.
E. ATP becomes hydrolyzed, and Na + and K + are pumped
across the membrane, as described in (B). K + , however,
immediately flows back into the vesicles through the
K + leak channels. K + moves down the K + concentration
gradient formed by the action of the Na + pump.
With each K + that moves into the vesicle through a
leak channel, a positive charge is moved across the
membrane, generating a membrane potential that is
positive on the inside of the vesicles. Eventually, K +
will stop flowing through the leak channels when the
membrane potential balances the K + concentration
gradient. The scenario described here is a slight
oversimplification: the Na + pump in mammalian
cells actually moves three sodium ions out of cells
for each two potassium ions that it pumps, thereby
driving an electric current across the membrane and
making a small additional contribution to the resting
membrane potential (which therefore corresponds only
approximately to a state of equilibrium for K + moving via
K + leak channels).
ANSWER 12–16 Ion channels can be ligand-gated, voltagegated,
or mechanically- (stress-) gated.
ANSWER 12–17 The cell has a volume of 10 –12 liters
(= 10 –15 m 3 ) and thus contains 6 × 10 4 calcium ions
(= 6 × 10 23 molecules/mole × 100 × 10 –9 moles/liter
× 10 –12 liters). Therefore, to raise the intracellular Ca 2+
concentration fiftyfold, another 2,940,000 calcium ions have
to enter the cell (note that at 5 μM concentration there are
3 × 10 6 ions in the cell, of which 60,000 are already present
before the channels are opened). Because each of the 1000
channels allows 10 6 ions to pass per second, each channel
has to stay open for only 3 milliseconds.
ANSWER 12–18 Animal cells drive most transport processes
across the plasma membrane with the electrochemical
gradient of Na + . ATP is needed to fuel the Na + pump to
maintain the Na + gradient.
ANSWER 12–19
A. If H + is pumped across the membrane into the
endosomes, an electrochemical gradient of H + results,
composed of both an H + concentration gradient and
a membrane potential, with the interior of the vesicle
positive. Both of these components add to the energy
that is stored in the gradient and that must be supplied
to generate it. The electrochemical gradient will limit
the transfer of more H + . If, however, the membrane also
contains Cl – channels, the negatively charged Cl – in the
cytosol will flow into the endosomes and diminish their
membrane potential. It therefore becomes energetically
less expensive to pump more H + across the membrane,
and the interior of the endosomes can become more
acidic.
B. No. As explained in (A), some acidification would still
occur in their absence.
ANSWER 12–20
A. See Figure A12–20.
B. The transport rates of compound A are proportional
to its concentration, indicating that compound A can
diffuse through membranes on its own. Compound A is
likely to be ethanol, because it is a small and relatively
nonpolar molecule that can diffuse readily through the
lipid bilayer (see Figure 12–2). In contrast, the transport
rates of compound B saturate at high concentrations,
indicating that compound B is transported across the
membrane by some sort of membrane transport protein.
Transport rates cannot increase beyond a maximal rate
at which this protein can function. Compound B is likely
to be acetate, because it is a charged molecule that
could not cross the membrane without the help of a
membrane transport protein.
rate of transport (µmol/min)
200
100
compound B
compound A
0
0 2 4 6 8 10
concentration of solute (mM)
Figure A12–20