Chap. 11. Problem 1.
The shape of the curve
indicates that at low
compressive forces, the lipids
in the surface monolayer
behave as if they are in a
two-dimensional gas phase
wherein they are not packed
closely with one another. As
force is increased further,
they form a 2-D liquid and
the surface area each
molecule occupies decreases
further. To determine the
average area per lipid
molecule in the monolayer,
one would divide the total
surface area of the water
when the lateral pressure
rises rapidly as the lipids are
compressed to a solid, by the
number of lipids present. The
number of molecules on the
surface is calculated by
multiplying the number of
moles applied by Avogadro’s
number.
Chap. 11. Problem 3.
The molecular weight of an SDS molecule is 288. Thus, if the
average molecular weight of a micelle is 18,000, then there are
18,000/288 = 63 SDS molecules per micelle.
Chap. 11. Problem 7.
To solve this problem, ignoring
the transmembrane electrical
potential, the equation
∆Gt = 2.303 RT log (C2/C1)
is used.
Given that the pH of blood plasma is pH = 7.4 and the pH of
gastric acid is pH = 1.5, the H+ concentrations in the two
compartments can be calculated using [H+] = 10-pH, which gives
Blood plasma [H+] = 10-7.4 = 4.0 x 10-8 M
Gastric acid [H+] = 10-1.5 = 3.2 x 10-2 M.
Then substituting into the upper equation gives
∆Gt = 2.303 RT log (3.2 x 10-2/4.0 x 10-8) = 35 kJ/mol.
Thus the amount of ATP needed for H+ transport into 1 L of
gastric juice is
35 kJ/58 kJ/mol = 0.6 mol
Chap. 11. Problem 8.
The transport described is against the electrochemical potential
across the membrane. Because an ion is being transported, the
proper equation for this calculation is
∆Gt = 2.303 RT log (C2/C1) + ZF∆
Because Na+ is being transported to the more positively charged
side of the membrane, a positive value of +0.07 V is
substituted into the right term so that ∆G > 0 for this term.
∆Gt = 2.303 RT log (145/12) + (1)(96,480 J/Vmol)(0.07 V)
∆Gt = 6.4 kJ/mol + 6.8 kJ/mol = 13 kJ/mol
Chap. 11. Problem 9.
The kidney is a highly aerobic tissue that utilizes mitochondrial
oxidative phosphorylation to make ATP. In this process, electrons
removed in the oxidation of fuel substrates such as glucose or
fatty acids are passed to O2 releasing energy that is used to
make ATP by the F1Fo ATPase (ATP synthase). If ouabain
reduces O2 consumption by 66%, then ATP hydrolyzed by the
Na+K+ ATPase pump must account for 66% of the total energy
consumption of the kidney.
Chap. 11. Problem 11.
Membrane protein topology is commonly assayed by proteolysis.
If a membrane protein cannot be cleaved by a protease added
from the outside of the intact membrane containing it, then it is
likely located on the other side of the bilayer. Because protein
X is not cleaved by proteases unless the red blood cell
membrane has first been disrupted, the data indicate that the
protein is located inside the cell. Because protein X can be
removed from the membrane by salt treatment, the combined
data indicate that it is a peripheral membrane protein that
attaches to the inner leaflet of the RBC plasma membrane.
Chap. 11. Problem 13.
The temperature of body extremities is slightly less than the
temperature of core tissues. Reindeer, which live in cold
climates around the world, maintain the fluidity of the
membranes in their extremities by increasing their contents of
unsaturated fatty acids in these membrane structural lipids.
This compositional change ensures proper membrane dynamics
and function in the extremities.
Chap. 11. Problem 19.
The data indicate that the
leucine transporter has nearly
equal affinities (Kts) for Lleucine and L-valine.
Therefore, it probably has a
hydrophobic binding pocket
that can accommodate the
side chains of both amino
acids, which are similar in
structure. The data also
indicate that the transporter
prefers the L-stereoisomer of leucine, and presumably, valine.
Based on the much lower Vmax for transport in the absence of
Na+, it is likely that the transporter takes up the amino acids
and Na+ via a symport process. Because the Na+K+ ATPase
maintains the Na+ gradient across intestinal epithelial cells,
transport of L-leucine should be inhibited by ouabain, an inhibitor
of the ATPase.
Chap. 11. Problem 22.
This 18 residue helix is clearly
amphipathic with nonpolar (N)
residues on one face and polar
residues (P) on the other. When
clustered together with other
similarly amphipathic helices, the
helix could orient with its
nonpolar side facing out towards
membrane lipids, and its polar
side facing other helices in the
interior of the protein.