Molecular Biology of the Cell by Bruce Alberts, Alexander Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter by by Bruce Alberts, Alexander Johnson, Julian Lewis, David Morg

CHAPTER 3 END-OF-CHAPTER PROBLEMS
171
A. Are the data consistent with your hypothesis that
titin’s springlike behavior is due to the sequential unfolding
of individual Ig domains? Explain your reasoning.
B. Is the extension for each putative domain-unfolding
event the magnitude you would expect? (In an
extended polypeptide chain, amino acids are spaced at
intervals of 0.34 nm.)
C. Why is each successive peak in Figure Q3–2B a little
higher than the one before?
D. Why does the force collapse so abruptly after each
peak?
3–12 Rous sarcoma virus (RSV) carries an oncogene
called Src, which encodes a continuously active protein
tyrosine kinase that leads to unchecked cell proliferation.
Normally, Src carries an attached fatty acid (myristoylate)
group that allows it to bind to the cytoplasmic side of the
plasma membrane. A mutant version of Src that does not
allow attachment of myristoylate does not bind to the
membrane. Infection of cells with RSV encoding either the
normal or the mutant form of Src leads to the same high
level of protein tyrosine kinase activity, but the mutant Src
does not cause cell proliferation.
A. Assuming that the normal Src is all bound to the
plasma membrane and that the mutant Src is distributed
throughout the cytoplasm, calculate their relative concentrations
in the neighborhood of the plasma membrane. For
the purposes of this calculation, assume that the cell is a
sphere with a radius (r) of 10 μm and that the mutant Src
is distributed throughout the cell, whereas the normal Src
is confined to a 4-nm-thick layer immediately beneath the
membrane. [For this problem, assume that the membrane
has no thickness. The volume of a sphere is (4/3)πr 3 .]
B. The target (X) for phosphorylation by Src resides in
the membrane. Explain why the mutant Src does not cause
cell proliferation.
3–13 An antibody binds to another protein with an
equilibrium constant, K, of 5 × 10 9 M –1 . When it binds to
a second, related protein, it forms three fewer hydrogen
bonds, reducing its binding affinity by 11.9 kJ/mole. What
is the K for its binding to the second protein? (Free-energy
change is related to the equilibrium constant by the equation
ΔG° = –2.3 RT log K, where R is 8.3 × 10 –3 kJ/(mole K)
and T is 310 K.)
3–14 The protein SmpB binds to a special species of
tRNA, tmRNA, to eliminate the incomplete proteins made
from truncated mRNAs in bacteria. If the binding of SmpB
to tmRNA is plotted as fraction tmRNA bound versus SmpB
concentration, one obtains a symmetrical S-shaped curve
as shown in Figure Q3–3. This curve is a visual display of
a very useful relationship between K d and concentration,
which has broad applicability. The general expression for
fraction of ligand bound is derived from the equation for
K d (K d = [Pr][L]/[Pr–L]) by substituting ([L] TOT – [L]) for
[Pr–L] and rearranging. Because the total concentration of
ligand ([L] TOT ) is equal to the free ligand ([L]) plus bound
ligand ([Pr–L]),
fraction bound = [Pr–L]/[L] TOT = [Pr]/([Pr] + K d )
fraction bound
1.0
0.75
0.5
0.25
0
10 –11 10 –9 10 –7 10 –5
concentration of SmpB (M)
For SmpB and tmRNA, the fraction bound = [SmpB–
tmRNA]/[tmRNA] TOT = [SmpB]/([SmpB] + K d ). Using
this relationship, Problems calculate p3.28/3.24 the fraction of tmRNA bound
for SmpB concentrations equal to 10 4 K d , 10 3 K d , 10 2 K d ,
10 1 K d , K d , 10 –1 K d , 10 –2 K d , 10 –3 K d , and 10 –4 K d .
3–15 Many enzymes obey simple Michaelis–Menten
kinetics, which are summarized by the equation
rate = V max [S]/([S] + K m )
Figure Q3–3 Fraction
of tmRNA bound versus
SmpB concentration
(Problem 3–14).
where V max = maximum velocity, [S] = concentration of
substrate, and K m = the Michaelis constant.
It is instructive to plug a few values of [S] into the
equation to see how rate is affected. What are the rates for
[S] equal to zero, equal to K m , and equal to infinite concentration?
3–16 The enzyme hexokinase adds a phosphate to
D-glucose but ignores its mirror image, L-glucose. Suppose
that you were able to synthesize hexokinase entirely from
D-amino acids, which are the mirror image of the normal
L-amino acids.
A. Assuming that the “D” enzyme would fold to a stable
conformation, what relationship would you expect it to
bear to the normal “L” enzyme?
B. Do you suppose the “D” enzyme would add a
phosphate to L-glucose, and ignore D-glucose?
3–17 How do you suppose that a molecule of hemoglobin
is able to bind oxygen efficiently in the lungs, and yet
release it efficiently in the tissues?
3–18 Synthesis of the purine nucleotides AMP and
GMP proceeds by a branched pathway starting with ribose
5-phosphate (R5P), as shown schematically in Figure
Q3–4. Using the principles of feedback inhibition, propose
a regulatory strategy for this pathway that ensures an adequate
supply of both AMP and GMP and minimizes the
buildup of the intermediates (A–I) when supplies of AMP
and GMP are adequate.
R5P A B C D
F G AMP
H I GMP
Figure Q3–4 Schematic diagram of the metabolic pathway for
synthesis of AMP and GMP from R5P (Problem 3–18).
E
Problems p3.19/3.18