698
Appendix A
4-3
The net transport will be out of the cell in the
direction from the higher-affi nity site on the
intracellular surface to the lower-affi nity site on
the extracellular surface. More molecules will be
bound to the transporter on the higher-affi nity side
of the membrane, and thus more will move out of
the cell than into it, until the concentration in the
extracellular fl uid becomes large enough that the
number of molecules bound to transporters at the
extracellular surface is equal to the number bound at
the intracellular surface.
4-4
Although ATP is not used directly in secondary
active transport, it is necessary for the primary
active transport of sodium out of cells. Because it
is the sodium concentration gradient across the
plasma membrane that provides the energy for most
secondary active transport systems, a decrease in
ATP production will decrease primary active sodium
transport, leading to a decrease in the sodium
concentration gradient and thus to a decrease in
secondary active transport.
4-5
The solution with the greatest osmolarity will have
the lowest water concentration. Recall that NaCl
forms two ions in solution and CaCl
2
forms three.
Thus, the osmolarities are:
A.
20 + 30 + (2
×
150) + (3
×
10) = 380 mOsm
B.
10 + 100 + (2
×
20) + (3
×
50) = 300 mOsm
C.
100 + 200 + (2
×
10) + (3
×
20) = 380 mOsm
D.
30 + 10 + (2
×
60) + (3
×
100) = 460 mOsm
Solution D has the lowest water concentration.
Solution B is isoosmotic since it has the same
osmolarity as intracellular fl
uid.
4-6
Initially the osmolarity of compartment 1 is (2
×
200)
+ 100 = 500 mOsm and that of 2 is (2 × 100) +
300 = 500 mOsm. The two solutions thus have
the same osmolarity, and there is no difference in
water concentration across the membrane. Because
the membrane is permeable to urea, this substance
will undergo net diffusion until it reaches the same
concentration (200 mM) on the two sides of the
membrane. In other words, in the steady state it
will not affect the volumes of the compartments. In
contrast, the higher initial NaCl concentration in
compartment 1 than in compartment 2 will cause, by
osmosis, the movement of water from compartment 2
to compartment 1 until the concentration of NaCl in
both is 150 mM. Note that the same volume change
would have occurred if there were no urea present
in either compartment. It is only the concentration
of nonpenetrating solutes (NaCl in this case) that
determines the volume change, regardless of the
concentration of any penetrating solutes that are
present.
4-7
The osmolarities and nonpenetrating-solute
concentrations are:
Solution
Osmolarity,
mOsm
Nonpenetrating
solute concentration,
mOsm
A
(2 × 150) + 100 = 400
2 × 150 = 300
B
(2 × 100) + 150 = 350
2 × 100 = 200
C
(2 × 200) + 100 = 500
2 × 200 = 400
D
(2 × 100) +
50 = 250
2 × 100 = 200
Only the concentration of nonpenetrating
solutes (NaCl in this case) will determine the change
in cell volume. The intracellular concentration of
nonpenetrating solute is 300 mOsm, so solution A
will produce no change in cell volume. Solutions B
and D will cause cells to swell because they have a
lower concentration of nonpenetrating solute (higher
water concentration) than the intracellular fl
uid.
Solution C will cause cells to shrink because it has a
higher concentration of nonpenetrating solute than
the intracellular fl
uid.
4-8
Solution A is isotonic because it has the same
concentration of nonpenetrating solutes as
intracellular fl uid (300 mOsm). Solution A is also
hyperosmotic because its total osmolarity is greater
than 300 mOsm, as is also true for solutions B and
C. Solution B is hypotonic because its concentration
of nonpenetrating solutes is less than 300 mOsm.
Solution C is hypertonic because its concentration of
nonpenetrating solutes is greater than 300 mOsm.
Solution D is hypotonic (less than 300 mOsm of
nonpenetrating solutes) and also hypoosmotic (having
a total osmolarity of less than 300 mOsm).
4-9
Exocytosis is triggered by an increase in cytosolic
calcium concentration. Calcium ions are actively
transported out of cells, in part by secondary
countertransport coupled to the downhill entry of
sodium ions on the same transporter. If the intracellular
concentration of sodium ions were increased, the
sodium concentration gradient across the membrane
would be decreased, and this would decrease the
secondary active transport of calcium out of the cell.
This would lead to an increase in cytosolic calcium
concentration, which would trigger increased exocytosis.
Chapter 5
Test Questions
5-1 b
5-2 a
5-3 e
5-4 a
Calmodulin is a calcium-binding protein that is
inactive in the absence of Ca
2+
.
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