Essential Cell Biology 5th edition

A:30 Answers
side of the membrane (i.e., although it is still the same
molecule, it is now located in a different environment),
and T is the transporter.
B. This equation is useful because it describes a binding
step followed by a delivery step. The mathematical
treatment of this equation would be very similar to
that described for enzymes (see Figure 4–35); thus
transporters are characterized by a K m value that
describes their affinity for a solute and a V max value that
describes their maximal rate of transfer.
To be more accurate, one could include the
conformational change of the transporter in the reaction
scheme:
equation 2: T + S ↔ TS ↔ T*S* → T* + S*
equation 3: T ↔ T*
where T* is the transporter after the conformational
change that exposes its solute-binding site on the other
side of the membrane. This account requires a second
equation (3) that allows the transporter to return to its
starting conformation.
C. The equations do not describe the behavior of channels
because solutes passing through channels do not bind to
them in the way that a substrate binds to an enzyme.
ANSWER 12–2 If the Na + pump is not working at full
capacity because it is partially inhibited by ouabain or
digitalis, the electrochemical gradient of Na + that the
pump generates is less steep than that in untreated cells.
Consequently, the Ca 2+ –Na + antiport works less efficiently,
and Ca 2+ is removed from the cell more slowly. When
the next cycle of muscle contraction begins, there is still
an elevated level of Ca 2+ left in the cytosol. The entry
of the same number of Ca 2+ ions into the cell therefore
leads to a higher Ca 2+ concentration than in untreated
cells, which in turn leads to a stronger and longer-lasting
muscle contraction. Because the Na + pump fulfills essential
functions in all animal cells, both to maintain osmotic
balance and to generate the Na + gradient used to power
many transporters, the drugs are deadly poisons if too much
is taken.
ANSWER 12–3
A. Each of the rectangular peaks corresponds to the
opening of a single channel that allows a small current
to pass. You note from the recording that the channels
present in the patch of membrane open and close
frequently. Each channel remains open for a very short,
somewhat variable time, averaging about 5 milliseconds.
When open, the channels allow a small current with a
unique amplitude (4 pA; one picoampere = 10 –12 A) to
pass. In one instance, the current doubles, indicating
that two channels in the same membrane patch opened
simultaneously.
B. If acetylcholine is omitted or is added to the solution
outside the pipette, you would measure only the
baseline current. Acetylcholine must bind to the
extracellular portion of the acetylcholine receptor in
the membrane patch to allow the channel to open
frequently enough to detect changes in the currents;
in the membrane patch shown in Figure 12–25B, only
the cytoplasmic side of the receptor is exposed to the
solution outside the microelectrode.
ANSWER 12–4 The equilibrium potential of K + is –90 mV [=
62 mV log 10 (5 mM/140 mM)], and that of Na + is
+72 mV [= 62 mV log 10 (145 mM/10 mM)]. The K + leak
channels are the main ion channels open in the plasma
membrane of a resting cell, and they allow K + to come to
equilibrium; the membrane potential of the cell is therefore
close to –90 mV. When Na + channels open, Na + rushes in,
and, as a result, the membrane potential reverses its polarity
to a value nearer to +72 mV, the equilibrium value for Na + .
Upon closure of the Na + channels, the K + leak channels
allow K + , now no longer at equilibrium, to exit from the cell
until the membrane potential is restored to the equilibrium
value for K + , about –90 mV.
ANSWER 12–5 When the resting membrane potential of
an axon (inside negative) rises to a threshold value, voltagegated
Na + channels in the immediate neighborhood open
and allow an influx of Na + . This depolarizes the membrane
further, causing more voltage-gated Na + channels to open,
including those in the adjacent plasma membrane. This
creates a wave of depolarization that spreads rapidly along
the axon, called the action potential. Because Na + channels
become inactivated soon after they open, the outward
flow of K + through voltage-gated K + channels and K +
leak channels is rapidly able to restore the original resting
membrane potential. (96 words)
ANSWER 12–6 If the number of functional acetylcholine
receptors is reduced by the antibodies, the neurotransmitter
(acetylcholine) that is released from the nerve terminals
cannot (or can only weakly) stimulate the muscle to contract.
ANSWER 12–7 Although the concentration of Cl – outside
cells is much higher than inside, when transmitter-gated Cl –
channels open in the plasma membrane of a postsynaptic
neuron in response to an inhibitory neurotransmitter, very
little Cl – enters the cell. This is because the driving force
for the influx of Cl – across the membrane is close to zero at
the resting membrane potential, which opposes the influx.
If, however, the excitatory neurotransmitter opens Na +
channels in the postsynaptic membrane at the same time
that an inhibitory neurotransmitter opens Cl – channels, the
resulting depolarization caused by the Na + influx will cause
Cl – to move into the cell through the open Cl – channels,
neutralizing the effect of the Na + influx. In this way,
inhibitory neurotransmitters suppress the production of an
action potential by making the target cell membrane much
harder to depolarize.
ANSWER 12–8 By analogy to the Na + pump shown
in Figure 12–12, ATP might be hydrolyzed and donate
a phosphate group to the transporter when—and only
when—it has the solute bound on the cytosolic face of the
membrane (step 1 → 2). The attachment of the phosphate
would trigger an immediate conformational change
(step 2 → 3), thereby capturing the solute and exposing
it to the other side of the membrane. The phosphate
would be removed from the protein when—and only
when—the solute had dissociated, and the now empty,
nonphosphorylated transporter would switch back to the
starting conformation (step 3 → 4) (Figure A12–8).
ANSWER 12–9
A. False. The plasma membrane contains transport proteins
that confer selective permeability to many but not all
charged molecules. In contrast, a pure lipid bilayer
lacking proteins is highly impermeable to all charged
molecules.